Unifying Gravity and EM

[sweetser]

Hello:

I will try to meet the terms of the 8 guidelines.

1. The behavior of light is explained with a rank 1 field theory, the
Maxwell equations. Gravity is explained with a rank 2 field theory,
general relativity. The two can be combined in one Lagrange density,
but they are not in any sense unified.

For my unified field proposal, gravity and EM arise from the same
4-potential and form a rank 1 field. Here is the Lagrange density
for my gravity and EM (GEM) unified field proposal:


\mathcal{L}_{GEM}=-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu}
-\frac{1}{2c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu}


where:
J_{q}^{\mu} is the electric charge 4-current density
J_{m}^{\mu} is the mass charge 4-current density, the standard mass 4-density times \sqrt{G}
A_{\mu} is a 4-potential for both gravity and EM
\nabla_{\mu} is a covariant derivative
\nabla_{\mu}A^{\nu} is the reducible unified field strength tensor
which is the sum of a symmetric irreducible tensor (\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu}) for gravity
and an antisymmetric irreducible tensor (\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu}) for EM which uses an exterior derivative

The core variance is how one gets a dynamic metric which solves the
field equations for gravity. With general relativity, one starts with
the Hilbert action, varies the metric field, and generates the second
rank field equations. Here, I work with a symmetry of the Lagrange
density, working directly from the standard definition of a covariant
derivative:


\bigtriangledown_{\mu}A^{\nu}=\partial_{\mu}A^{\nu }+\Gamma_{\sigma\mu}{}^{\nu}A^{\sigma}


Any value contained in the unified field strength tensor could be due
any combination of the change in the potential or due to a change in
the metric. One is free to alter the change in the metric so long as
the change in potential compensates, and likewise the reverse. I
believe this is called a diffeomorphism symmetry (but my training is
spotty). Any symmetry in the Lagrange density must be related to a
conserved charge. For this symmetry, mass is the conserved charge.

The field equations are generated in the standard way, by varying the
action with respect to the potential. One ends up with a 4D wave
equation:


J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}


For the physical situation where the mass density equation is
effectively zero, one gets the Maxwell equations in the Lorentz gauge.
If the equations describe a static, neutral system, then the first
field equation, \rho_{m}=\nabla^{2}\phi, is Newton's
field equation for gravity. If the neutral system is dynamic, then
the equation transforms like a 4-vector under a Lorentz boost.
Because this equation is consistent with special relativity, that
removes a major motivation for general relativity (consistency with
SR).

If the system is neutral, static, and one chooses a gauge such that the
potential is constant, then the first field equation is the divergence
of the Christoffel symbol:


\rho_{m}=2\partial_{\mu}\Gamma_{\nu}{}^{\:0\mu}A^{ \nu}


This contains second order derivatives of the metric, a requirement
for constraining a dynamic metric. The exponential metric solves the
field equation:


g_{\mu\nu}=\left(\begin{array}{cccc}
exp(-2\frac{GM}{c^{2}R}) & 0 & 0 & 0\\
0 & -exp(2\frac{GM}{c^{2}R}) & 0 & 0\\
0 & 0 & -exp(2\frac{GM}{c^{2}R}) & 0\\
0 & 0 & 0 & -exp(2\frac{GM}{c^{2}R})\end{array}\right).


The easiest way to realize this is that for the definition of a
Christoffel of the second kind for a static, diagonal metric will only
involve g_{00} and g^{0}{}_{0}^{u}.
The exponentials will cancel each other, leaving only the divergence
of the derivative of the exponent, or


\rho_{m}=\nabla^{2}(GM/c^2 R)


The 1/R solution should be familiar. This metric gives a point
singular solution to the field equations.

One could have chosen a gauge where the metric was flat. With that
gauge choice, the potential (GM/c^2 R, 0, 0,0) solves the first field
equation with a point singularity, a good check for logical
self-consistency.

2. The exponential metric solution to the GEM field equations for a
static, neutral system is consistent with first-order parameterized
post-Newtonian predictions of weak field theories. The relevant terms
of the Taylor series expansion are:


(\partial\tau)^{2}\cong(1-2GM/c^{2}R+2(GM/c^{2}R)^{2})dt^{2}-(1+2GM/c^{2}R)dR^\{2}/c^{2}


These are identical to those for the Schwarzschild metric of general
relativity. Therefore all the weak field tests of the metric, and all
tests of the equivalence principle will be passed. To second-order
PPN accuracy the metrics are different:

GEM

(\partial\tau)^{2}\cong(1-2GM/c^{2}R+2(GM/c^{2}R)^{2}-4/3(GM/c^{2}R)^{3})dt^{2}


-(1+2GM/c^{2}R+2(GM/c^{2}R)^{2})dR^{2}/c^{2}


GR

(\partial\tau)^{2}\cong(1-2GM/c^{2}R+2(GM/c^{2}R)^{2}-3/2(GM/c^{2}R)^{3})dt^{2}


-(1+2GM/c^{2}R+3/2(GM/c^{2}R)^{2})dR^{2}/c^{2}


This will translate into 0.7 microarcseconds more bending of light
around the Sun according to a paper by Epstein and Shapiro,
Phys. Rev. D, 22:2947, 1980. We currently can measure bending to 100
microarcseconds. Clifford Will responding to a question I posed said
there are _no_ plans in development to get to the 1 microarcsecond
level of accuracy. Darn!

The antisymmetric field strength tensor will be represented by the
spin 1 photon, where like charges repel. These are the transverse
modes of emission. The symmetric field strength tensor will be
represented by the spin 2 graviton, where like charges attract. These
will be the scalar and longitudinal modes of emission. Should we ever
measure a gravity wave, and then determine its polarization, general
relativity and the GEM proposal differ on the polarization. If the
polarization is transverse, GEM is wrong. If the polarization is not
transverse, general relativity is wrong (Will also made this point in
his living review article).

3. Once the Lagrange density is stated, everything else flows from
that. I have discussed this work as it developed and took misteps on
sci.physics.research and my own web site, but that should not be
needed here.

4. To back up the derivations, I have cranked through all this and a
bit more in a Mathematica notebook. It is available here:

http://www.theworld.com/~sweetser/quaternions/gravity/Lagrangian_to_tests/Lagrangian_to_tests.html
http://www.theworld.com/~sweetser/quaternions/ps/Lagrangian_to_tests.nb.pdf
http://www.theworld.com/~sweetser/quaternions/notebooks/Lagrangian_to_tests.nb

[Despite the URL, no quaternions are used in this body of work,
although they continue to be the wizard behind the curtain.]

5. This theory is consistent with strong field tests of gravity, such
as energy loss by binary pulsars. For an isolated mass, the lowest
mode of emission is a quadrapole moment. This proposal does not have
extra fields that can store energy or momentum, which is what is
needed to form a dipole if there is only one sign to the mass charge,
which the proposal claims.

6. I know of no physical experiments that contradict this work. There
are _thought_ experiments that claim that gravity must be non-linear
(there was a primer on GR by Price I recall as an example). These
thought experiments appear to always use electrically neutral sources.
For a unified field theory, one must consider what happens if charge
is included. What Price did was imagine a pair of boxes with 6
particles in each. Then the energy of one of the particles in one box
gets completely converted to kinetic energy of the other 5. Price
argues that the box with 6 particles should not be able to tell the
difference between the two boxes, the one with 6 still particles and
the one with 5 buzzing about. If this is the case, then the field
equations for gravity must be nonlinear. I argue that if the 6
particles were charged, there would be no way to destroy an electric
charge, so the experiment cannot be done in theory. No conclusions
can be drawn. EM puts new constraints on gravity thought experiments.

7. It has been my observation that no one is impressed by the
Mathematica notebook, even people at Wolfram Research. The notebook
is my best unbiased source that no obvious mathematical errors have
been made. Earlier versions of this body of work did have errors that
Mathematica pointed out.

8. I understand how general relativity works well enough to appreciate
that a linear, rank 1 field theory is in fundamental conflict with GR.
That is an observation, nothing more or less. GR works to first order
PPN accuracy. It is an open question if it will work to second order.
My money is riding on the exponential metric, because exponentials
appear to be Nature's favorite function (simple harmonics around the
identity for small exponents).

Sorry to be this l o n g, but the guidelines appeared to require it.

doug sweetser

[vanesch]

The most obvious, although probably naive, "objection" I would have is that we have one and the same field for EM and gravity and that the source is J_tot = J_q-J_m so that two different configurations of J_q and J_m with same J_q-J_m give identical solutions (and hence J_q-J_m should stay identical).
But we know that true EM or gravitational interactions will make J_q and J_m evolve differently, so that J_tot = J_q-J_m will evolve differently according to the exact composition in J_q and J_m.

Or maybe I misunderstood how your system works, and there is somehow a different input for J_q and J_m.

Some simple illustrations:
if I set up an electrostatic field between two plates, then this should influence neutral matter too, no ? And the other way around, doesn't this mean that electic charges of opposite sign would accelerate essentially in opposite ways in a purely gravitational field ? In order to have each time the same solution for J_q-J_m ?

[sweetser]

The most obvious, although probably naive, "objection" I would have is that we have one and the same field for EM and gravity and that the source is J_tot = J_q-J_m so that two different configurations of J_q and J_m with same J_q-J_m give identical solutions (and hence J_q-J_m should stay identical).
But we know that true EM or gravitational interactions will make J_q and J_m evolve differently, so that J_tot = J_q-J_m will evolve differently according to the exact composition in J_q and J_m.


It is natural to think that gravity and EM are too darn close to each other in this theory :-) The total current couples to the 4-potential. If you were to alter the distribution of charges, but not to total charge, a change in the potential would be required.

Here is one way to see this. Imagine a point source that is a single particle. It will be described by a charge/R potential. If the particle is neutral, then it would be be described by \sqrt{G}M/R. Repeat the exercise, but with a negatively charged particle, that would have a potential (q+\sqrt{G}M)/R. Two things to note: all charged particles have mass, so the total effective charge for a negatively charged particle will increase (it is more attractive to a positive charge due to its mass) and the total effective charge for a positively charged particle will decrease (it is less repulsive to another positive electric charge because of gravity). Only because the proposal is linear, one can linearly superimpose solutions. So once you can do one change, you can do an arbitrary number of them, although the actual math gets more complicated. Second, the gravity and EM charges for a proton are more than thirteen orders of magnitude different, but we know electric charge only to ten significant digits, so no electrical engineering needs to change.

Let me attempt to address a deeper issue, the nature of the relationship between gravity and light. If you focus on the current densities, one observation is that the sign is different between the two is consistent with the well-known fact that like charges attract for gravity and repel for EM. The field strength tensor holds another key. The terms symmetric and antisymmetric are too technical for me to feel comfortable with (yes, they are accurate, but cold). The one for gravity, (\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu}), I call "The average amount of change in the 4-potential". For EM, (\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu}) I title it as "The deviation from the average amount of change in the 4-potential." I can see how both are describing the same potential in different ways. It is also easy to see why the deviation would have the possibility to be both positive and negative charges. One could imagine the average value being anything greater than zero, suggestive of one sign for the mass charge. Hope this helps.

[CarlB]

Here is one way to see this. Imagine a point source that is a single particle. It will be described by a charge/R potential. If the particle is neutral, then it would be be described by \sqrt{G}M/R. Repeat the exercise, but with a negatively charged particle, that would have a potential (q+\sqrt{G}M)/R.

The Milliken oil drop experiments showed a balance between the gravitational and electric forces. In other words, he used drops where the number of positive and negative charged particles was exactly balanced, and then added a single negative or positive charge to them. The drops were placed in a combination electric and gravitational field and their velocities were observed.

Your equations would have the net electric charge for a body with balanced positive and negative charges be non zero. The effect would be that certain sizes of oil drops would end up with an electric charge at least equal to e/2. I suspect that this is contrary to the observations of the Milliken experiment.

Do explain further and keep the fresh ideas coming.

Carl

[sweetser]

Hello Carl:

Let's agree to a basic point: if my theory disagrees with Millikan's oil drop experiment, it is wrong. The Millikan experiment established the charge on a single electron.

At first glance I am skeptical Millikan's experiment can in any way effect this proposal, since it is a classical experiment (low speeds) about EM whose forces which vastly exceed gravity. Recall that for our best tests of general relativity, such as light bending around the Sun, the GEM metric gets the identical 1.75 arcseconds of bending around the Sun predicted by general relativity. I have yet to read anyone applying GR to the oil drop experiment, but GR had better be consistent with Millikan :-) If GR at first order PPN accuracy is correct, then GEM is correct. At second order PPN accuracy, the predictions are 10.8 versus 11.5 microarcseconds for GR and GEM. That is the kind of level of accuracy one needs to get at to see any difference. But enough dodging of a direct question.

Your equations would have the net electric charge for a body with balanced positive and negative charges be non zero.

This is governed by a balancing of electrical and gravitational forces:
qV/d=mg
where q is the electric charge, V is the voltage, d is the distance between the plates applying the voltage, m is the mass of the drop, and g is the Newtonian gravitational acceleration. This looks like classical Newtonian gravity, not any of that fancy PPN stuff. GR and GEM get to Newton in two different ways. The first is as a limit of the metric, and with that approach, there is no difference between GR and GEM. With GEM, if one chooses a metric such that the curvature of the metric makes no contribution, and the apparatus can be described accurately by a static field, one gets exactly Newton's field equations. I am not aware of how one can spot Newton's field equations in Einstein's field equations, but there must be a road. Because GEM has the same field equation as Newton, there is no way it could disagree with the results based on Newtonian gravity analysis.

The effect would be that certain sizes of oil drops would end up with an electric charge at least equal to e/2.

I don't see why you think this is so, but will explore the issue anyway. After the oil drops are balanced, the electric field gets shut off, and the oil drop begin to fall. They reach a terminal velocity quickly. Millikan measures the radius and the terminal velocity. The terminal velocity is governed by this equation:
v=2 R^{2}\rho g/9 \eta
where small v is the terminal velocity, R is the radius of the drop, \rho is the density of the oil drop, and \eta is the viscosity of the air. So again classical Newtonian gravity is used, and again, I see no way for my proposal can differ. I have yet to point out that the EM part which regenerates the Maxwell equations must also be the same.

This question brought up a thought experiment that has made my day :-) According to my 1992 Particle Properties Data Booklet, we knew the charge of a single electron is 1.602 177 33\times10^{-19}C. Let's say someone somehow calculated the electric charge of a _massless_ particle to 20 significant digits (it would have to be a calculation, there are not massless charged particles), and they found it to be 1.602 177 331 234 567 890\times10^{-19}C. Let's further imagine that the Millikan oil drop experiment could measure charge to this level of accuracy (not realistic). Add an electron to a drop. This adds one electric charge, and one mass charge, so that would be -q_{e}-\sqrt{G}m_{e} or -1.602 177 331 234 567 964\times10^{-19}C. Repeat for proton, and its charge would be +q_{e}-\sqrt{G}m_{p} or +1.602 177 331 234 431 287\times10^{-19}C. Not the same!

This experiment would prove that the mass of an electron is quantized, it comes in discrete clumps of 7.43\times10^{-36}C whereas protons, no matter their source, always have a mass charge of 1.36\times10^{-32}C. This is different from EM where the quantum electric charge is the same for all charged particles. I have no idea why electric charge is more general that mass charge.

doug
TheStandUpPhysicist.com

[CarlB]

Dear Doug,

Let me try and explain this better.

My thinking is that your theory basically amounts to supposing that the gravitational potential is what is left over after the difference in (combined electric/gravitational) potentials between positive and negative charges are added up and cancel.

Since neutral matter falls, it must be that + and - charges do not exactly cancel. Suppose an object is composed of N atoms and that each atom has as many + as - charges. Then there must be a net electric charge left over. This idea will clearly give the correct gravitational potential, but an object composed of N neutral atoms will not have be exactly neutral electrically. But in a large object like an oil drop we can not count the number of + and - charges literally, so we can neutralize it by adding + or - charges appropriately.

However, since charge comes in units of e, we cannot precisely cancel the net charge exactly. Consequently, our oil drop will end up with an effective electric charge accurate only to the nearest multiple of the charge of the electron. That means that for some sizes of oil drops, the smallest charge we can get on the oil drop (by adding + or - charges) will be (worst case) e/2. For some other masses of oil drops, we could cancel the charge completely.

Carl

[sweetser]

Hello Carl:

Thanks for the further clarification. I'm pretty sure I haven't said something like this:
Dear Doug,
My thinking is that your theory basically amounts to supposing that the gravitational potential is what is left over after the difference in (combined electric/gravitational) potentials between positive and negative charges are added up and cancel.

The 4-potential cannot be observed directly. What is measurable are changes in the potential, the field strength tensor \bigtriangledown_{\mu}A^{\nu}. Fundamental forces are found in irreducible tensors. The GEM tensor is reducible, into the standard one for EM, (\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu}), and the symmetric one I claim is for gravity, (\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu}). I believe this is a standard approach, that the potential cannot be measured directly. I try to avoid thinking about things that cannot be measured. Instead it is the average amount of change in the potential that I argue is about gravity, and the deviation from the average amount of change that is about EM by standard EM theory.

But in a large object like an oil drop we can not count the number of + and - charges literally, so we can neutralize it by adding + or - charges appropriately.

However, since charge comes in units of e, we cannot precisely cancel the net charge exactly. Consequently, our oil drop will end up with an effective electric charge accurate only to the nearest multiple of the charge of the electron. That means that for some sizes of oil drops, the smallest charge we can get on the oil drop (by adding + or - charges) will be (worst case) e/2. For some other masses of oil drops, we could cancel the charge completely.

This sounds to me like you are counting the potential, and not either the average value of change in the potential or the deviation from the average amount of change in the potential. If you are dealing directly with the potential, I will argue that is a misunderstanding of this field theory.

doug

[CarlB]

Doug;

I see now that my comment about your theory being that the gravitational potential (and therefore the gravitational force) is simply what is left over after cancelling the electric potentials between + and - is not right and that my objection, which amounted to saying that your theory doesn't quite allow + and - charges to be the same, is incorrect.

Understanding new theories is very difficult and requires great patience from the person doing the explaining (ask me about it). Let me try and explain your theory back to you and you can correct my misunderstandings. This part got through my skull better: "Instead it is the average amount of change in the potential that I argue is about gravity, and the deviation from the average amount of change that is about EM by standard EM theory."

Let me try and put it another way. First, I would like to restrict myself to the situation where the underlying metric has no curvature, which I believe you say is acceptable. By the way, I am not a gravity expert (my education is in QM and elementary particles), but my favorite gravity theory is also one that can be expressed by a gauge principle on flat space. Here's a review article:

Foundations of Physics, 35: 1-67 (2005)
"Gauge Theory Gravity with Geometric Calculus"
http://modelingnts.la.asu.edu/pdf/GTG.w.GC.FP.pdf

Above was from this page:
http://modelingnts.la.asu.edu/html/GCgravity.html

So my limited understanding now is that basically you're using a sort of freedom in the E&M potential (available when you expand its usual potential from antisymmetric to assymmetric) to supply a gravitational potential and you're using the opposite sort of freedom in the gravitational potential (available when you expand its symmetric potential to an assymmetric one) to supply the E&M potential.

The silly rules on this forum allow only a limited number of posts so I will ask some more stupid questions privately.

Carl

By the way, I looked at the three links in your original post. The first one didn't work for me, but the second was helpful. Joining Newtonian gravitation and E&M is what I will look at, as it is simpler than Misner Thorne and Wheeler.

[sweetser]

Hello Carl:

Wow, real communication, that's tough :-) For parity, let me admit 3 errors I made when working with the Lagrange density for this proposal. First I had the sign of the mass current density be the same as the electric current density. That asserts like mass charges repel. Oops. The second error was trying to talk about trace(\nabla^{\mu}A^{\nu}) as a scalar field. This must be a mixed index tensor to take the trace. Duh. And the third error was to write the field strength tensor as \partial_{\mu}A^{\nu} which does not transform as a tensor. Each of these errors were pointed out with a harsh tone by professionals, but they have been repaired.

This part got through my skull better: "Instead it is the average amount of change in the potential that I argue is about gravity, and the deviation from the average amount of change that is about EM by standard EM theory."
To be honest, although I understood the technical difference between symmetric and antisymmetric tensors, I am far more comfortable with these phrases.

First, I would like to restrict myself to the situation where the underlying metric has no curvature, which I believe you say is acceptable.

I have to be VERY careful about this. In the newsgroup sci.physics.research, my proposal has been dismissed in one line by Prof. John Baez, a well-respected authority in loop quantum gravity in particular and general relativity in, well, general, because he claims my theory fixes the background metric to be flat Minkowski. What you are doing is a gauge choice. Let me be more precise. Here is the gauge you chose for the GEM Lagrangian, and how it may be changed such that the field equations are unaltered:

\partial_{\mu}A^{\nu}\rightarrow(\partial_{\mu}A^{ \nu})'=\partial_{\mu}A^{\nu}+\Gamma_{\sigma\mu}{}^ {\nu}A^{\sigma}

If an expert on GR were to read this, he or she would say, "Oh, that is just Riemann normal coordinates, where the connection happens to be zero at only one point in the spacetime manifold." That is the definition of Riemann normal coordinates, but that is not what is going on here. This is a symmetry of the GEM Lagrange density, not a coordinate choice issue.

Let me make the opposite gauge choice, where I say all of gravity and EM is the connection, nothing is due to the potential. Here then is the gauge transformation:

\Gamma_{\sigma\mu}{}^{\nu}A^{\sigma}\rightarrow(\G amma_{\sigma\mu}{}^{\nu}A^{\sigma})'=\partial_{\mu }A^{\nu}+\Gamma_{\sigma\mu}{}^{\nu}A^{\sigma}

The gauge symmetry here is tough to communicate. It is not between the potential and the metric, it is the change in the potential and the change in metric (via a torsion-free connection which is metric compatible for the GR experts).

Let me try to explain what is going on compared to general relativity. The connection is almost always introduced the same way. An author introduces the idea of a contravariant vector, say A^{\nu}. He then says that the differential operator also transforms like a vector, \partial_{\mu}. Put the two together, \partial_{\mu}A^{\nu}, and the result does NOT transform like a tensor. One needs the connection. The covariant derivative transforms like a tensor, \nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu}+\Gamma_{ \sigma\mu}{}^{\nu}A^{\sigma}.

Now they say, let's divorce the connection from the potential and study it (actually, it is an implicit divorce, not one discussed). The connection does not transform like a tensor. Bummer (OK, they are a bit more formal when the math gets this heavy). One book I recall said you could calculate the divergence of a connection, but there is no reason to because it will not result in a tensor. What is the simplest thing made up of a connection that transforms like a tensor? That is the Riemann curvature tensor, R^{a}{}_{bcd}. From there it is a few steps to Einstein's field equations for gravity only.

Riemann's curvature tensor is often presented as the simplest thing involving the connection that transforms like a tensor. They forget to add the qualifier: Riemann's curvature tensor is the simplest thing involving the connection that transforms like a tensor so long as you completely and utterly ignore the potential. A covariant derivative has the connection, and it transforms like a tensor. That leads to a big picture idea. Maxwell's theory is all about the potential. One has to actually supply the metric as part of the background structure. GR is all about the changes in the metric, the potential being tossed out when forming the Riemann curvature tensor. The reducible, asymmetric field strength tensor \partial_{\mu}A^{\nu}+\Gamma_{\sigma\mu}{}^{\nu}A^ {\sigma} has both changes in the potential and changes in the metric to do both gravity and EM, either as a potential or metric theory or some combination of both.

doug

<Administration>The rule that put limits on the number of posts which will effect me first are probably intended to prevent repetition, folks effectively quoting themselves. If we demonstrate a conversation that moves to different topics (there are plenty here), then we may be able in this situation to get the folks here to change that parameter. I prefer to keep the conversation public as much as possible.

The first link contained a spurious "\", part of cutting and pasting under x windows to the forum form.

I hope to keep this thread focused on rank 1 unified field theory. From my scan of Hestene's paper, he hopes to bring new insights to rank 2 field theories of gravity, a good goal, but off-topic.
</Administration>

[sweetser]

Hello:

I thought I would point out something cool about the Millikan experiment to 20 digits. If a GR expert were to read the proposal and be in a dismissive mood, he would point out that the theory was linear. Based on GR theory, any linear theory is only an approximation of the real thing. Doing the Millikan experiment for 1, 3, 10, 30, 100 charges would give direct evidence about whether gravity is linear. My proposal says the multiples would be exactly 1, 3, 10,... and general relativity would claim that since gravity fields gravitate, there would be greater than 3, greater than 10,... Another way to test the proposal in theory, if not in practice.

Nature above all is consistent. If I do the Millikan experiment looking at the electric charge, we know that will be linear. To be consistent, mass charge would behave exactly the same way in my opinion.

doug

[Chronos]

Hi sweetser! I like your presentation. My confusion is centered on the way you mix rank 1 and 2 tensors. I have a hard enough time separating the two as is. I'm not saying it's wrong, but, I need my hand held to walk through that mine field.

[sweetser]

Hello Chronos:

No problem. It is a simple game of count the Greek letters that do not cancel each other because they are repeats. That is it, honest!

\nabla_{\mu} is rank 1, 1 Greek letter
\nabla_{\mu}A^{\nu} is rank 2, 2 different Greek letters (mu and nu)
\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu} is rank 0, the two sets of same letters cancelling each other out.

Then there are the names that mean the same thing: rank 0 = scalar, rank 1 = vector, rank 2 = matrix, rank 3 = be scared because it requires three fingers.

Tensors have a bad reputation as being difficult to understand. I'm sure you have a bunch of other questions, just like I am sure I have the answers for you on my web site, www.TheStandUpPhysicist.com (seriously!). There are two ways to extract information you seek. Click on slide/mp3 link, and then Lecture 1. Slides 8-16 discuss tensors, with words, and better yet, simple pictures. I found that trying to explain physics with only words left the visual part of my brain in a coma, so I decided to have a policy that every slide had to have a simple graphic that somehow explained a little bit of what was going on.

If you are a person like most people, you will prefer the video. Click on the shows link, show 7. Your computer will have to learn how to play an mp4 file which is 200 MB. By the end, you might have more of a sense of the difference between a regular derivative and a covariant derivative. Now, you will not be a pro, but the fear of this stuff may decrease.

doug

[vanesch]

It is natural to think that gravity and EM are too darn close to each other in this theory :-) The total current couples to the 4-potential. If you were to alter the distribution of charges, but not to total charge, a change in the potential would be required.


Yes, this is what I don't understand. The lagrangian (if I understand well) determines just as well the action of the potential on the charge (total charge rho - M), as it determines the action of the charge on the potential, right ?
I mean, for the same total current density, we should find the same solution to the potential problem, or is this not the case ? Is there some extra dynamics then that is NOT described by the lagrangian ? If not, I don't see how IDENTICAL J_total can give rise to DIFFERENT potentials ? And if IDENTICAL J_total give rise to identical potentials, then J_total will be a certain solution to the entire system. Now, if that's the case, this would mean that for two different combinations of J_e and J_M into the SAME J_total, this J_total should be "conserved" independent of how it is split into J_e and J_M, but we know that this is not true.

Let me be more specific. Again, this might just be a part of your model I simply didn't understand, so please try to find the error in my reasoning.

Let us have a spacelike sheet with a certain "initial potential" A(sheet) and a certain "initial J_total" J_total(sheet). If I understand lagrangian dynamics sufficiently, this implies that the knowledge of J_total(sheet) and A(sheet) COMPLETELY FIX J_total for all events as well as A for all events. So given these initial conditions on the spacelike (t0-like) sheet, J_total and A are now fixed for all (x,y,z,t).
But consider now two different physical situations:

situation 1) J_total (sheet) = J_q1(sheet) - J_M1(sheet)

This means that somehow at any later time, J_total(t) = J_q1(t) - J_M1(t)

situation 2) J_total (sheet) = J_q2(sheet) - J_M2(sheet)

This means that somehow at any later time, J_total(t) = J_q2(t) - J_M2(t)

So for two DIFFERENT charge and mass configurations 1 and 2, we have that if they sum at a certain moment to J_total, then they will always sum to J_total.

Consider now two extreme cases:
1) J_total = J_q1 (so J_M1 = 0) No mass on the sheet

2) J_total = - J_M1 (so J_q1 = 0) No charge on the sheet

3) or even other combinations, such as -J_q1 = J_M1 on the sheet.
They obey exactly the same evolution ?
I don't think that J_q1 - J_M1 is a conserved quantity when you change the amounts of q and M, no ?

Probably I simply misunderstood a part of your proposal.

[sweetser]

The lagrangian (if I understand well) determines just as well the action of the potential on the charge (total charge M + rho), as it determines the action of the charge on the potential, right ?

Sorry, but I don't understand what you wrote above. Let me give you my understanding of a Lagrange density. The key for me was learning the units: it is mass per unit volume. So a Lagrange density is an expression which is suppose to contain any and all ways that energy can interact inside a box. The Lagrange density has several uses. First, it is unsurpassed in pointing out things that are conserved. This is done by integrating the Lagrange density over volume and time. Integrating over the volume give back all the energy, but why also do time? One uses the calculus of variations, an important branch of integral calculus. Instead of trying to get a number, one gets a function. Strange, but true. The goal is to find a parameter to vary in Lagrange density that no matter what arbitrary time it is integrated over, the integral remains the same. That parameter that doesn't change a darn thing is a symmetry of the Lagrange density, and is always associated with a conserved quantity. For the EM and GEM Lagrangians, one can vary a time t without changing the integral, and that leads to energy conservation. For the EM and GEM Lagrange densities, there is no distance R or angle \theta, so linear and angular momentum are conserved quantities. For EM, there is a rank one gauge symmetry, A^{\mu}\rightarrow A'^{\mu}=A^{\mu}+\nabla\phi. This symmetry is why electric charge is conserved. GEM has a rank two symmetry \partial_{\mu}A^{\nu}\rightarrow(\partial_{\mu}A^{ \nu})'=\partial_{\mu}A^{\nu}+\Gamma_{\sigma\mu}{}^ {\nu}A^{\sigma}. This probably has to do with both charge and mass conservation, but I am not professional enough to work through those points. If anyone wants a further intro to Lagrange densities, that would be lecture 2 at www.TheStandUpPhysicist.com, slides 11-17, and the corresponding video.

One can take different derivatives of the Lagrange density to generate the field equations, or the stress-energy tensor, or forces. To get to the field equations, one takes the derivative with respect to the 4-potential and the derivative of the 4-potential. This is called the Euler-Lagrange equation, and can take about an hour and fifteen minutes using 6 blackboards if no shortcuts are taken.

So now I will try to address the question at hand. The Lagrange density has two sorts of ways for energy interactions. One is known at the charge coupling term, it is \frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu} The difference in sign for these two is absolutely essential, so that like charges attract for mass and repel for EM. This term however is not very interesting. Applying the Euler-Lagrange equations, one ends up with just the J's.

The interesting part of the Lagrange density is the field strength tensor contraction, -\frac{1}{2c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu}. One often hears: "consider the field equations for a vacuum." In that situation, J is zero, and the field equations are all about this field strength tensor contraction, nothing else. The div, grad, and curl stuff is all going to be generated from here. This is the heart of the proposal, but from my reading, vanesch focused exclusively on the charge coupling term.

So what is going on with the charge coupling? As usual, not much. All it really says is that the normal electric current, whatever it is, is effectively a little be less due to interia. Realize there is nothing radical in that statement, that is standard stuff. Like charges will repel from each other a little less due to their interia. If you calculate how much less, it will make effectively no difference.


Consider now two extreme cases:
1) J_total = J_q1 (so J_M1 = 0) No mass on the sheet

All charged particles have mass. The amount of mass they have is usually small enough that it can be utterly ignored.


2) J_total = J_M1 (so J_q1 = 0) No charge on the sheet

This can happen.


They obey exactly the same evolution ?
I don't think that J_q1 +[sic] J_M1 is a conserved quantity when you change the amounts of q and M, no ?

This proposal is linear for EM (standard) and linear for gravity (in conflict with GR). We know you can add in arbitrary combinations of positive and negative charges, and there is zero problems handling that situations with Maxwell's field equations written in the Lorentz gauge, J_{q}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}
The field equations for GEM are nearly the same, but there is an additional current density. All that current does is decrease the EM current density by an amount that is difficult to measure unless the system is electrically neutral. The field strength tensor is much more complicated for GEM, since it has 16 terms instead of 6.
I hope this clarified a point or two, but this proposal necessarily is not easy because it is the Maxwell equations generalized in a way to embrace gravity as a metric theory via a rank 2 Lagrange symmetry.

[sweetser]

<The Setup>
I have only talked with one professor one time three years ago in his office about this project to unify gravity and EM (I am quite independent). I had to pull some strings to talk to a string theorist (an infa-red astronomer who taught a Perl class knew an emeritus MIT professor who knew my grandfather who was the highest ranking American diplomat at the League of Nations and who founded the hight school the emeritus professor and string theorist attended). I emailed the string theorist about my hopes for a rank 1 unification model, and here was his reply:
Francis phoned me yesterday. As a rule, I really think there is nothing new to be done at Lagrangian level-there is no disease. However, if we can keep it short, I'll meet you. I assume you do understand the usual Einstein-Maxwell story, so there is some common ground. Tomorrow around 3 PM? If not, maybe next week. Do you know the way here? ...

The tone was dismissive enough that I had to talk myself into going. The string theorist was pleasant on a personal level, chatting about the International School, wondering how a biologist could ever discuss a Lagrangian. When the topic turned to physics, he was the most arrogant man I have had to deal with.
</The Setup>

<The Challenge>
General Relativity is a beautiful theory. I say that based on my experience studying Sean Carroll's lecture notes. I don't understand every nut and bolt, but the structure is elegant. Is it flawless? The challenge here is to find a specific, entirely mathematical flaw in GR at the Lagrangian level.
</The Challenge>

<The Math of Lagrangians>
I suspect that most people reading this forum do not work with Lagrange densities often, so I'll review a bit. A Lagrange density (mass/volume) encompasses every energy interaction that can happen inside a box. There are two things one can do with a Lagrange density:
Task 1: Vary a field. This will generate the field equations, or calculate the energy density, or the momentum density.
Task 2: Look for things that do not vary after integrating the 4-volume and varying something. This will lead to conserved quantities.
The Maxwell equations, and my unified field equations, are generated by varying the potential field, task 1 work.
Why is energy conserved? That is a task 2 result, there being no time t in the EM or GEM Lagrangians. Why is momentum conserved? Again that is a task 2 result. What about mass? We know from special relativity that energy and momentum have the following relationship:
m^{2}c^{4}=E^{2}-P^{2}c^{2}
Mass is not only a conserved quantity, but mass is invariant under a Lorentz transformation. In the standard model, a kind of generalization of the Maxwell equations, the mass of every particle is zero. The Higgs mechanism is a way to add mass without breaking the symmetry of the standard model. For GEM, mass is a scalar field, the trace of the symmetric field strength tensor. One can see mass charge is conserved by a task 2 process, because it has to do with the second rank tensor symmetry.
</The Math of Lagrangians>

<The Problem with the GR Lagrange Density>
In the same month that Einstein published his field equations for general relativity, Hilbert did the physics the right way, starting from a Lagrange density, a deceptively simple looking one at that:
\mathcal{L}=\frac{c^{4}}{16\pi G} R
where R is the Ricci scalar, formed from two contractions of the Riemann curvature tensor.

Simple, isn't it? One might imagine that energy conservation will be easy to find. Unfortunately, there is a serious technical roadblock. The action S is the spacetime integral of the Lagrange density, or
S=\frac{c^{4}}{16\pi G}\int R\sqrt{-g}dx^{4}
Now vary the metric field g_{\mu\nu}. Oops, that effects both the Ricci scalar, and that factor in the square root which is needed in curved spacetime. The problem gets even more tricky. Recall the Riemann normal coordinates, the coordinates where at only one place in spacetime, the connection is zero. One is free to choose whatever coordinates one wants. For any point in spacetime, one could choose the Riemann normal coordinates, and the energy density there would be zero. That is the problem.

In defense of GR, it must be noted that the theory conserves both energy and momentum. This is a more subtle problem: that energy is not defined locally. Some people are so familiar with this old problem that they do not consider it an issue: old wounds become proud scars. To me, it is a clear blemish.
</The Problem with the GR Lagrange Density>

<The Source>
Why would GR have the localization of energy problem, but the GEM model not? It is because GR uses the Riemann curvature tensor, and GEM does not. Recall that for GEM, the changes in potential and metric lay in the same bed together, as they should in my opinion. In GR, the connection (a fancy name for a change in the metric for a torsion-free, metric compatible connection) gets a divorce from the potential to become a pure geometric object. The pure object does not transform like a tensor. So one looks at the difference between two paths that are very close to one another to define the Riemann curvature tensor. Fundamentally, the Riemann curvature tensor cannot be about one place in spacetime. The Riemann curvature tensor is about a comparison between two places in spacetime. It is no wonder that a problem occurs for one place in spacetime.
</The Source>

<The Solution>
I don't think there is a way to solve the localization of energy problem if one depends anywhere on the Riemann curvature tensor, which the Ricci scalar is as a pair of contractions of the Riemann tensor. This is a problem with how to handle gravity at the level of the Lagrange density. The GEM model has a chance because it uses the connection, but not the Riemann curvature tensor.
</The Solution>

doug
<The End/>

[hellfire]

For the physical situation where the mass density equation is effectively zero, one gets the Maxwell equations in the Lorentz gauge.I have a question for my understanding. If one takes J = 0 in the GEM Lagrangian, why does not the GEM Lagrangian be the same as the EM Lagrangian for the free field (\int \frac{1}{4} F^{\mu\nu} F_{\mu\nu})? How do Maxwell's equations follow?

[hellfire]

Another question. I do not understand how the metric arises. In GR it appears as a dynamic variable in the Lagrangian and is therefore the variable for which the equations of motion are solved. In the GEM Lagrangian the dynamic variable is A and the equations of motion depend on A. However, its covariant derivative is computed according to the metric in spacetime. But that metric itself is a function of the dynamic variable A, am I right? If yes, why don’t you write explicitely the GEM Lagrangian only in terms of A (as the covariant derivative operator is also a function of A)? Sorry if my understanding is completely wrong and if this is trivial, but it is not easy to understand (although you did a very good job explaining it)…

[sweetser]

Hello:

There are two questions, so I'll provide two answers in separate posts, this one about Maxwell equations. When J=0, there are no sources, and the Lagrange density applies to a vacuum. For folks reading this, let me clarify the difference between an action S and a Lagrange density \mathcal{L}:
\mathrm{the\ action }S=\int\mathrm{the\ Lagrangian\ }\mathrm{L}\mathrm{\ over\ space\ and\ time}
In this case:
S_{EM}=\int \frac{1}{4} F^{\mu\nu} F_{\mu\nu}
\mathcal{L}_{EM}=\frac{1}{4} F^{\mu\nu} F_{\mu\nu}
You probably know this, and I was just using this as a chance to explain some terminology that can be scarier than it should be. What we need to define is the electromagnetic field strength tensor:
F^{\mu\nu}=(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})
The electromagnetic field strength tensor is what I have been calling the antisymmetric tensor, or the deviation from the average amount of change tensor for the GEM proposal (for the precise reader, they will note I use a mixed tensor in the GEM Lagrangian so I can define a scalar field by taking the trace). This is 6/16ths of the the story, in that there are six terms in the antisymmetric tensor, and there are ten in the symmetric tensor, or (\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu}).
Please note the difference between the partial derivatives and the covariant derivatives. A metric is a symmetric object. The change in a symmetric object - the connection - is symmetric. The antisymmetric tensor can have zero information about the change in the metric, that is why it has the partial derivatives. The antisymmetric tensor will be exactly the same no matter how spacetime is curved, although the contraction of the antisymmetric tensor will be changed because tensor contractions depend on metrics.

Now to your question: How do the Maxwell equations follow? Well, we follow the standard approach: start from the Lagrange density, integrate that over a volume of spacetime to form the action, then vary the action with respect to the 4-potential, and look for a minimum. That is by the way what the Euler-Lagrange equation effectively does. People more skilled with actions than I can look at the action, I think they apply integration by parts, and pull out the field equations:

0=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}

This is one way to write the Maxwell equations in a vacuum. One thing about field equations: different Lagrange densities can get to the same field equations. It would be incorrect to presume that the F^{\mu\nu} is a unique, necessary path.

If you want to see the Maxwell or GEM equations written out in terms of the fields E, B, ..., well, I could do that by making a killer long post, but let me sketch it out instead. F^{\mu\nu}=E+B. One needs to do the same darn thing for the symmetric part. I HATE to make up new terms, that is a crank calling card, but there is a symmetric counterpart to both E and B I write at e and b, along with a field for the diagonal. It is a confusing game to define these and make sure that E, B, e, b, and g together make up the field equation above. It is done explicitly - every darn term written out - in the lecture 3 notes at www.TheStandupPhysicist.com.

[sweetser]

In the GEM Lagrangian the dynamic variable is A and the equations of motion depend on A. However, its covariant derivative is computed according to the metric in spacetime. But that metric itself is a function of the dynamic variable A, am I right? If yes, ...
The answer here is no. To have a viable theory for gravity, you must be able to have an equation whose solution is a dynamic metric consistent with experimental tests of gravity. GR does this is the most straight forward way possible, by taking the Hilbert action and varying the metric field to generate a rank 2 field theory. I cannot do that for the GEM proposal which has rank 1 field equations. Instead, there is a symmetry in the Lagrange density involving the change in the potential and the change in the metric. Here is the differential equation that a metric must solve if the gauge choice is for a constant potential:
\rho_{m}=2\partial_{\mu}\Gamma_{\nu}{}^{\:0\mu}A^{ \nu}
I have to rush off to work, but I encourage people to check for themselves that the exponential metric referenced in the first post solves this equation. The answer turns out to be reasonable: since the exponential metric is both diagonal and symmetric, only the g_{00} term makes a contribution, the derivative of the exponential has two parts, the exponential that cancels with g^{00} and the derivative of a 1/R, and Laplace knew the answer to that one.

[vanesch]

Sorry, but I don't understand what you wrote above. Let me give you my understanding of a Lagrange density.

What I meant was the following. The Lagrangian (density) is a "prescription" for the equations of motion of ALL things in there which are declared "dynamical variable". You can give it a physical interpretation if you want to, but that's not my argument.
I guess your dynamical variables are:
- the metric
- the field A
- the M and Q current densities, J_M and J_Q.

right ?


When you write down the lagrangian, you've now specified ALL equations of motion.
The metric only appears in the tensor contractions, and has no "dynamical" term of itself.
But the observation I'm making is that J_M and J_Q only can occur in a very specific combination: J_M - J_Q. So if I now define a new dynamical variable, J_T = J_M - J_Q, and replace J_M and J_Q by this new variable then everything should remain exactly the same in ALL EQUATIONS OF MOTION. So you have a dynamics which cannot in any way distinguish any combination of J_M and J_Q, as long as they result in the same J_T, because the only quantity that appears in the equations of motion is J_T. This comes down to saying that there is no dynamical distinction between "mass" and "charge".

To give you a simple example of what I mean: let us consider electromagnetism (Maxwell). Now suppose that I define "yellow charge" and" "green charge" respectively described by J_y and J_g. I now say that J_q = J_y + J_g and this is the J_q that goes in the Maxwell equations. Purely based upon the Maxwell equations, only J_q has a dynamical meaning, and for a given J_q (no matter how it is composed of yellow and green charge), I have a certain solution to my dynamical problem (J_q over spacetime, and A over spacetime). So we observe that there is no dynamical distinction between yellow and green charge, that they are INTERCHANGABLE, and that the only physically meaningful quantity is J_q. Yellow charge could be fluctuating in green charge and back at just any rate I wish, it wouldn't change the dynamics. I see exactly the same happening in your theory, where "mass" and "charge" seem to play the same role as yellow and green charge in my toy example. What stops mass from changing into charge and back without your lagrangian "noticing" ?

I hope I made clear now my difficulty I see with your approach. Maybe it is due to the fact that you do something else but "standard lagrangian dynamics" or that I have missed something else.

cheers,
Patrick.

[sweetser]

What I meant was the following. The Lagrangian (density) is a "prescription" for the equations of motion of ALL things in there which are declared "dynamical variable". You can give it a physical interpretation if you want to, but that's not my argument.
I guess your dynamical variables are:
- the metric
- the field A
- the M and Q current densities, J_M and J_Q.

right ?

I am confident that this is two thirds right:
The field A is a dynamic variable.
The current densities J_M and J_Q are dynamic variables.

I am less certain about the metric, but do not think that it is a dynamic variable of the Lagrange density. Let me ask about the EM Lagrange density and its gauge symmetry. That is a choice, a bit of freedom. I might be making an error here, but I don't think the gauge is viewed as a dynamic variable. The action (integral of the Lagrange density) is not varied with respect to the gauge to generate field equations. Instead the field equations are generated by varying the potential, and then one notices that a symmetry of the Lagrange density applies to the field equations also.

Once again, the key symmetry is not about the metric, but about the changes in the metric (the connection) and the changes in the potential. Still a good technical question.

But the observation I'm making is that J_M and J_Q only can occur in a very specific combination: J_M - J_Q. So if I now define a new dynamical variable, J_T = J_M - J_Q, and replace J_M and J_Q by this new variable then everything should remain exactly the same in ALL EQUATIONS OF MOTION. So you have a dynamics which cannot in any way distinguish any combination of J_M and J_Q, as long as they result in the same J_T, because the only quantity that appears in the equations of motion is J_T. This comes down to saying that there is no dynamical distinction between "mass" and "charge".

I am going to completely and utterly agree with this observation (and then see what happens :-) Let us prove experimentally that one cannot distinguish between attractive gravitational and attractive electrical charges. You are in control of the probe, a single charged particle. I get to play with the source which I keep inside a black box.

Test 1: The probe is positive, the probe is repelled by the black box.
We know that the black box must contain a positive source because only for similar electrical charges can repel from each other.

Test 2: The probe is negative, the probe is attracted by the black box.
Being a good thought experimentalist, you are able to measure and see there is exactly one positive charge in the box. I confess that the box has a solitary positron.

Test 3: The probe is negative, the probe is attracted by a new black box.
This time, there is no charge inside the box, it is as neutral as can be. There is a 421 kg mass in the box. Why that size? Because e^{2}=G m_{e} 421, the electrical charge coupling is exactly equal to the gravitational coupling.

If two particles repel, electricity must be involved (with the charge effectively decreased by mass, since there are no massless electrical charges). If two particles attract, then it could be due to either mass alone or the sum of mass and electric charge. The only way to tell which sort of charge is doing the work is by bringing in another charge of electrically opposite sign.

J_T can have a negative sign, in which case we know it is composed of an electric charge made smaller by its mass or J_T can have a positive sign, either as the sum of electric and mass charges or purely mass charges. I hope this is not a problem because it looks like unification of gravity and EM to me.

doug

[vanesch]

I am less certain about the metric, but do not think that it is a dynamic variable of the Lagrange density.


This is a possibility, but then it must be "given" (for instance, a given Minkowski metric). So OR the metric is given (does not FOLLOW from the Lagrangian dynamics), OR the metric is a dynamical variable and then it is determined by the Lagrangian dynamics. My initial suspicion was that you had a SEPARATE DYNAMICAL PRESCRIPTION for the metric (that wasn't derived from the Lagrangian but from something else, only it wasn't clear how) and that you then PLUGGED IN THAT METRIC into the lagrangian.
That's what I meant with "non standard Lagrangian dynamics".

EDIT: so my question to you is: if it is not the lagrangian that determines your metric, where does it come from ?


Let me ask about the EM Lagrange density and its gauge symmetry. That is a choice, a bit of freedom. I might be making an error here, but I don't think the gauge is viewed as a dynamic variable.


This is correct: the gauge is in fact the "part of the potential that has not been fixed by the dynamics". This comes about by a symmetry of the Lagrangian density. This is a very good suggestion you make because it allows me to reformulate my problem:
My "complaint" is in fact exactly similar to a gauge symmetry:
Under the symmetry:

J_m -> J_m + lambda
J_q -> J_q + lambda

your Lagrangian (and hence your dynamics) is invariant. This is what I was always saying about "equal J_T".



Test 1: The probe is positive, the probe is repelled by the black box.
We know that the black box must contain a positive source because only for similar electrical charges can repel from each other.


Hey, here you are using some knowledge outside of your theory. Negative mass density gives you exactly the same effect. There's a priori nothing which forbids negative mass density in your theory.


Test 2: The probe is negative, the probe is attracted by the black box.
Being a good thought experimentalist, you are able to measure and see there is exactly one positive charge in the box. I confess that the box has a solitary positron.


The question is: how do I know that my probe is NEGATIVE, and not just MASSIVE ? You use extra knowledge outside of your theory. In your theory, negative charge can be exchanged for positive mass.


Test 3: The probe is negative, the probe is attracted by a new black box.
This time, there is no charge inside the box, it is as neutral as can be. There is a 421 kg mass in the box. Why that size? Because e^{2}=G m_{e} 421, the electrical charge coupling is exactly equal to the gravitational coupling.


(I guess you meant: the probe is positive)

I would even be inclined to think that in your theory, equal (positive) masses REPULSE (because they are equivalent to equal NEGATIVE charges). Indeed, according to my "symmetry" above this should not change any dynamics.

Except, of course, if masses and charges somehow give rise to OTHER metrics (but this can then not follow from the Lagrangian dynamics).

[sweetser]

EDIT: so my question to you is: if it is not the lagrangian that determines your metric, where does it come from ?

There are two things that one can do with a Lagrangian: vary a field or look for things that do not change the Lagrange density. You appear to define the standard approach to the Lagrangian as only being the first sort: here is my field, I form the action, I vary the field, and thus get my field equations.

The information about the dynamics of the metric comes out of the second process: looking for a symmetry. One cannot tell the difference between a change in a potential or a change in a metric. If I make a choice that I will pretend that there is no change in potential to see, then the metric must solve this differential equation:

\rho_{m}=2\partial_{\mu}\Gamma_{\nu}{}^{\:0\mu}A^{ \nu}

So I have to be careful in my language. If the metric is not a variable, then it is fixed. But that is missing an essential clause: "up to a gauge symmetry". So really, I should be saying "the metric is fixed up to a gauge symmetry". Now for a choice of symmetry where the potential contributes not change to the Lagrange density, the metric must necessarily solve the above differential equation for the GEM Lagrangian. So the metric is determined not by Lagrangian dynamics, but by Lagrangian gauge symmetry. I can live with that.


My "complaint" is in fact exactly similar to a gauge symmetry:
Under the symmetry:

J_m -> J_m + lambda
J_q -> J_q + lambda

your Lagrangian (and hence your dynamics) is invariant. This is what I was always saying about "equal J_T".

The only thing I would alter is how this is written, because I like this->that=this+foo, or J_{m}\rightarrow J'_{m}=J_{m}+\lambda and J_{q}\rightarrow J'_{q}=J_{q}+\lambda. Let's see if this is a problem, or the clearest indication that gravity and EM charge are unified in this proposal.

No need to worry about negative mass. Here is what I call "General Gauss' law", which is just like Gauss' law, only it also applies to gravity:
\rho_{q} - \rho_{m}=\Box^{2}A
For the static case, this simplifies to:
\rho_{q} - \rho_{m}=-\nabla^{2}A
[technical sidebar: I am still using a covariant derivative, so this still has the change in potential/metric symmetry, and it would be incorrect to say the solution was just charge/R without making a gauge choice, specifically a flat Minkowski background, because the exponential metric written in the first post with an electric and mass charge solves this differential equation. Try it yourself!].

General Gauss' law is within the scope of the theory. In EM books they make clearer than I can that like electric charges reply because \rho_{q}=-\nabla^{2}A. Nothing more is needed. Likewise, for the same Gaussian surface kind of reasons, \rho_{m}=+\nabla^{2}A indicates that like charges attract.

Nature is complete: some things attract each other, other things repel each other. One of the most compelling reasons to like this theory is for the sake of completeness. Please try and defend this view of the status quo: a huge part of the global economy is devoted to EM, the stuff of the electromagnetic field strength tensor (\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu}). This is six out of ten possible changes in the potential. Nature - according to standard physics - has no need for the other ten terms of the field strength tensor, which are (\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu}). I don't find that defensible, ignored yes, but I doubt Nature drops those terms.

doug

[straycat]

For my unified field proposal, gravity and EM arise from the same
4-potential and form a rank 1 field.
doug sweetser

Hey Doug,

I was wondering whether you know anything about Myron Evans' work, and whether there is any similarity between his and yours? Evans has published lots of articles, mainly in Foundations of Physics Letters I think, in which he proposes to unify GR and EM through his "Evans wave equation," which is intended iiuc to supercede the (more restrictive) Einstein equation. My understanding of this is scant, but I think it has something to do with including not only symmetric, but also antisymmetric components into the wave equation.

Here's a short description that I got from one of his websites [1]:

"Einstein used Riemann geometry which describes curvature, but not
torsion. Cartan developed a theory of torsion for electromagnetism.
Evans' development using metric vectors and the tetrad is initially
pure geometry, but the equations combine curvature and torsion. The
first step to unification is achieved by development of the Evans
Field Equation that allows both curvature and torsion to be expressed
in the same set of equations. Gravitation and electromagnetism are
derived from the same geometric equation. This completes Einstein's
goal to show that gravitation and electromagnetism are geometric
phenomena."

David

[1]
http://www.atomicprecision.com/
http://www.aias.us
(The secone one is non-functional right now, not sure why ...)

[vanesch]

General Gauss' law is within the scope of the theory. In EM books they make clearer than I can that like electric charges reply because \rho_{q}=-\nabla^{2}A. Nothing more is needed. Likewise, for the same Gaussian surface kind of reasons, \rho_{m}=+\nabla^{2}A indicates that like charges attract.


This is what I don't understand. Let's stay in pure electrodynamics, and imagine that we would have chosen the electron's charge to be "positive" but kept the "electic field" the way we know it. Then the only thing that that does is to change the sign of rho in Maxwell's equations AND in the Lorentz force, which simply means we change the sign in the lagrangian. So I don't see how such an arbitrary choice of sign can influence whether LIKE charges repell or attract ! You simply flipped the direction of the E-field (which would now point from negative to positive charges), and you also made a positive charge accelerate away from the field (along - E).
You would now flip of course the sign in the above equation.
But it still means that like charges repell ! A positive charge would have the E-field point TOWARDS IT (so it would generate a LOW potential, as of your modified equation), and another positive charge would accelerate AWAY from it.
The reason for this is that "the lorentz force" and "the potential" are derived from the same term in the Lagrangian and you cannot change the sign of one without changing the sign of the other.

So that's why I thought that your negative sign before the rho_m didn't make "like charges attract", but just changed the sign convention of what was "positive charge" and what was "negative charge" and how the E-field pointed.


cheers,
Patrick.

[sweetser]

Hey Doug,

I was wondering whether you know anything about Myron Evans' work,...
http://www.aias.us
(The secone one is non-functional right now, not sure why ...)
Yes I have looked through this site, and I do not believe it could pass the criteria for this particular forum. We should not allow a backdoor method for the IR forum.

I want to remain focused on this rank 1 proposal, because my goal is to find a technical reason to retire the idea from my life.

doug

[sweetser]

Hello Patrick:
This is what I don't understand. Let's stay in pure electrodynamics, and imagine that we would have chosen the electron's charge to be "positive" but kept the "electic field" the way we know it. Then the only thing that that does is to change the sign of rho in Maxwell's equations AND in the Lorentz force, which simply means we change the sign in the lagrangian. So I don't see how such an arbitrary choice of sign can influence whether LIKE charges repell or attract!

Good, now the issue is not with my proposal per se, but with a standard property of EM theory which can be tricky to really understand. This is one of those many fact that I know to be true from books, but is not rock solid in my own mind, so let's address it.

I am consulting "Electricity and Magnetism" by Purcell. He makes this observation on page 24:
Gauss' law and Coulomb's law are not two indpenedent physical laws, but the same law expressed in different ways. [in a footnote he comments on the difference that happens for moving charges]
I feel differently about them myself, Gauss' law being about adding up the E field around a change, and Coulomb about force. From Coulomb's law, it is easy to see that like charges repel: the force law is positive, so put in a pair of charges with the same sign, and the force will be positive, a sign that the particles will be running away from each other. Rewrite Coulomb's F=+q q/R^{2} into the charge and electric field, F=+{q E}. Now the definition of E doesn't appear so abitrary: It is a different collection of the terms in Coulomb's law, which clearly has like charges repelling.

Hope this helps,
doug

[vanesch]

I feel differently about them myself, Gauss' law being about adding up the E field around a change, and Coulomb about force. From Coulomb's law, it is easy to see that like charges repel: the force law is positive, so put in a pair of charges with the same sign, and the force will be positive, a sign that the particles will be running away from each other. Rewrite Coulomb's F=+q q/R^{2} into the charge and electric field, F=+{q E}. Now the definition of E doesn't appear so abitrary: It is a different collection of the terms in Coulomb's law, which clearly has like charges repelling.


Consider the definition of q to be with a minus sign (that we would have choosen the "electron" to have positive charge, and not the proton) ; moreover, that we would have choosen the symbol s and not q, to represent charge.

So for an electron:
q = - 1.602 10^(-19) Coulomb and s = + 1.602 10^(-19) Patrick
For a proton:
q = + 1.602 10^(-19) Coulomb and s = -1.602 10^(-19) Patrick.

In that case, we would introduce a zelectric field Z obeying Gauss' law:
div Z = s/epsilon_0

We would also have the Lawrense force: F = s (Z + v x B) (let's forget the B field for the moment).

Coulomb's law is still s1 s2 / R^2 which tells us that like charges repel.
The field Z is of course MINUS the E field and the s-current would be minus the q-current.

But we could also *MIX* both the s/Z and the q/E theory:
we could work with the *E* field but keep the s charge, or work with the Z field, but keep the q charge.
This would of course come down to writing:
div E = MINUS s / epsilon_0

And the lorentz force: F = MINUS s (E + v x B)

Coulomb's force is all the same because it only depends on s1 x s2 = q1 x q2.

And the fun thing is: the theory where:
div E = - s / epsilon_0

F = - s (E + v x B)

is the theory that you get from the classical EM lagrangian... BUT WITH THE SIGN OF THE CHARGE/CURRENT 4-VECTOR CHANGED.

So if you change the sign of the charge-current 4-vector in the lagrangian density, you GET EXACTLY THE SAME EM THEORY, but with the sign of the E-field flipped.
This means that if you flip the sign of the charge/current 4-vector in the lagrangian density, you change the sign of the fields, but you STILL OBTAIN THAT LIKE CHARGES REPEL.
As such, the -j_M in your lagrangian doesn't make masses attract themselves, they repel in exactly the same way as the charges in the +j_Q term.

cheers,
Patrick.

[sweetser]

Hello Patrick:

Now I think I remember an answer this EM question. You can define the E field so that it points this other way. Our current convention is to define the E field by its effect on a positive charge coming in from infinitely (where the E field will be zero) to a field generated by a positive charge. That takes work, so energy=\vec{E}\cdot\vec{d} is greater than zero.

Now define a new E field, you call it Z, for a charge going out to infinity. That releases energy for a +E field on a positive charge. Not the physics is identical, the charges repel, only the definition of E/Z changes. Like charges repel, so with the field generated by a positive charge, a positive charge would release energy running away to infinity. We have: energy=-\vec{Z}\cdot\vec{d}.

To be honest, I find EM confusing, and do everything I can to avoid changing how anything is defined. So I'll stick with the standard definition of an E field, and the standard interpretations which are that Gauss' law indicates that like charges repel. See Purcell or other EM books for different answers to your question.

doug

[vanesch]

Hello Patrick:

Now I think I remember an answer this EM question. You can define the E field so that it points this other way. Our current convention is to define the E field by its effect on a positive charge coming in from infinitely (where the E field will be zero) to a field generated by a positive charge. That takes work, so energy=\vec{E}\cdot\vec{d} is greater than zero.

Now define a new E field, you call it Z, for a charge going out to infinity. That releases energy for a +E field on a positive charge. Not the physics is identical, the charges repel, only the definition of E/Z changes. Like charges repel, so with the field generated by a positive charge, a positive charge would release energy running away to infinity. We have: energy=-\vec{Z}\cdot\vec{d}.



Yes, of course you also have to flip the sign for the energy, as you do. In fact, you'd have simply to replace E by -Z EVERYWHERE. I hope you see that you then have just exactly the same physics as in standard EM, except for the fact that we wrote now the Z-field everywhere, and not the E-field, and that this scatters around some minus signs as compared what we're used to in standard EM (with an E-field).

But I didn't write this as just a small disgression on EM. I wrote it because I think (not because I want to be mean :smile:) that it points to a flaw in your theory. You seem to think that because you flipped the sign of j_M in the lagrangian, that this suddenly makes positive masses attract. But (let us forget for the moment the j_EM) if there's only j_M present, we have exactly the theory I presented: we have a sign change in the E-field by changing it into the Z-field (and hence in the potential A, and hence in the term containing the j_M in the Lagrangian) BUT WE HAVE STILL A THEORY IN WHICH LIKE CHARGES (MASSES) REPEL. Isn't that a problem for you ?

cheers,
Patrick.

[sweetser]

Hello Patrick:

Critics are not mean, they are useful to me. Due to discussions on this forum, I will not try to always say "The metric is fixed up to a gauge transformation." The question is still open as to whether your criticism is on the mark. It sounds too general to me, like there is no way to form a Lagrange density where like charges attract.

Let us not focus on the field E and how it is defined. The reason is that E, along with B is how one characterizes the second rank antisymmetric field strength tensor. We both accept there is a convention involved in mapping E (and B) to the the field strength tensor.

What is not subject to conventions is writing out Coulomb's law in terms of charges and potentials: F=+qQ/R^{2}. That can be derived from the Lagrange density. I have seen it done in Landau and Lif****z - oops, a 4-letter Russian - but did not follow the derivation so well which is why I don't repeat it here. If the exercise is repeated for the gravity term coupled to the potential, for algebraic consistency there must be one more minus sign.

You appear to ignore a lesson I learned from a caustic professor: it is NOT the sign of E that matters, it is NOT the sign of J that matters, it IS the sign of the charge coupling term J^{\mu}A_{mu} relative to the field strength tensor contraction \nabla_{\mu}A^{\nu}\nabla^{\mu}A_{\nu} that matters. If the sign of the charge coupling term (J and A contracted) is the same as the sign of the field strength tensor contraction term, then such a field has like charges repel. If the sign of the charge coupling term is different from the field strength tensor contraction term, then like charges will attract.

Taking a step back, one can see why disagreements happen. You have been talking about the field E and the current J. Those terms don't appear in isolation in the GEM Lagrange density. They are parts of two contractions, and the information of interest is in my technical opinion in the relative sign of two contractions.

doug

[vanesch]

What is not subject to conventions is writing out Coulomb's law in terms of charges and potentials: F=+qQ/R^{2}. That can be derived from the Lagrange density. I have seen it done in Landau and Lif****z - oops, a 4-letter Russian - but did not follow the derivation so well which is why I don't repeat it here. If the exercise is repeated for the gravity term coupled to the potential, for algebraic consistency there must be one more minus sign.

No, that's what I'm trying to point out. You use TWO TIMES the sign of this term in the Lagrangian in order to deduce the Coulomb force law, so the sign in the lagrangian cannot change the sign in the Coulomb force law.
The reason is the following: the coupling term determines on one hand HOW THE FIELD FOLLOWS FROM A SOURCE CHARGE (call it Gauss' law), and that same coupling term DETERMINES HOW A TEST CHARGE RESPONDS TO THE FIELD GENERATED BY THE SOURCE.
So if you flip the sign of that term, you flip 1) the sign of the field by the source charge, but 2) you also flip the sign of the response of the test charge to the field.
Let us write the EM lagrangian (I take it from Jackson p 599, with c=1):
L = -\frac{1}{16 \pi} F_{ab}F^{ab} - S J_a A^a
where I used S to indicate the sign (S = +1 in standard EM).
Now, the Euler-Lagrange equation becomes:
{\partial}^b \frac{1}{4 \pi} F_{ab} = S J_a
which is the covariant form of the inhomogeneous Maxwell equations (of which we will use Gauss' law a = 0), up to a sign which is given by S.
However, our system is not complete. In order for us to have a total dynamical system, we need to include the dynamics of the "matter" part, namely the inertia of the mass particles making up the current, which leads us directly into matter fields, or by adding a lagrangian of a finite number of particles, which leads us into troubles with self-energy.
Without going into all these tricky details (which I don't master myself), however, this comes down to adding a "matter dynamical term" to the Lagrangian, which we call "M":
L = -\frac{1}{16 \pi} F_{ab}F^{ab} - S J_a A^a + M
Now, whatever is in M, it will not depend on the field, but it will be function of another dynamical quantity (positions of particles, or a matter field...) which will determine the current density J. So J is to be a function of this other dynamical quantity.
The Euler-Lagrange equation for this second dynamical quantity (matter field...) will then take on a general form which is:
(stuff regarding only M and which will essentially result in m.a, the left hand side of Newton's equation) + derivative of L towards J_a through the chain rule.
And it should be clear that this second term is the Lorentz force on the dynamical system described by M. Clearly, this second term has the S - sign in front of it:
So we find something of the kind of m.a = S "lorentz force".
I know that the second part is much more handwaving :redface: but it is because in practice it is quite difficult to do because of all kinds of infinities which pop up. But with a bit of good will you can see that the Lorentz force of the field on the matter system ALSO finds its origin in the J_a A^a term.

You appear to ignore a lesson I learned from a caustic professor: it is NOT the sign of E that matters, it is NOT the sign of J that matters, it IS the sign of the charge coupling term J^{\mu}A_{mu} relative to the field strength tensor contraction \nabla_{\mu}A^{\nu}\nabla^{\mu}A_{\nu} that matters. If the sign of the charge coupling term (J and A contracted) is the same as the sign of the field strength tensor contraction term, then such a field has like charges repel. If the sign of the charge coupling term is different from the field strength tensor contraction term, then like charges will attract.

Your caustic professor was wrong :bugeye: Whether like charges attract or repel through an intermediate field is depending on the tensor order of the intermediate field. If it is an even order tensor (scalar, or 2-tensor), then like charges attract ; if it is first order (vector field), like charges repel.
There's an intuitive reason for that given in Zee (Quantum Field Theory in a nutshell). I don't know of any hard proof of the statement.

[sweetser]

Hello Patrick:
I think we are making progress because more equations are appearing in the discussion :-) You make an accurate point: there is no way to get the force equation with the GEM Lagrange density I wrote in the first post. That requires the inertia term. Unfortunately, there is a sign error in the three term Lagrange density for EM that you posted. Here is the correct one based on my reading of L&L's "The Classical Theory of Fields", chapters 3 & 4 (more specifically, eq 16.4 and 28.6):
L = -\frac{1}{16 \pi} F_{ab}F^{ab} - S \rho_{q} U_{a} A^{a} - \rho_{m}/\gamma
where
U_{a}=(\gamma,\gamma\beta_{x},\gamma\beta_{y},\gam ma\beta_{z})
\beta=v/c
gamma=\frac{1}{\sqrt{1-\beta^{2}}}
So there is a minus sign in front of the mass density term, not a positive sign. It took me so long to get the following point, that I will quote the source:
In finding the field equations [meaning Gauss' law, the Maxwell equations] with the aid of the principle of least action we must assume the motion of the charges to be given and vary only the potentials (which serve as the "coordinates" of the system); on the other hand, to find the equations of motion [meaning the Lorentz force equation] we assumed the field to be given and varied the trajectory of the particle.
Looking back at the EM Lagrangian, the first two terms have a potential. That is what gets varied to generate Maxwell's field equations. The last two terms have velocity in them: by varying that, one gets the Lorentz force (the details of those steps are still unclear to me).
I hope we can agree that in my first post, the Lagrange density was incomplete for describing a Lorentz force equation. Here is the addition:

\mathcal{L}_{GEM}=-\rho_{m}/\gamma-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu}
-\frac{1}{2c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu}

Sorry it took me so long to recognize this! Now I must extend on what the caustic professor so briefly said: the field equations will have like charges repel if the coupling term has the same sign as the field strength contraction term, and the force equation will have like charges repel if the inertia and charge coupling terms have the same sign. If the coupling term flips its sign, then both the field and force equations will have like charges attract.
doug

[sweetser]

Hello Patrick:
This sounds like a separate, field theory argument:
Whether like charges attract or repel through an intermediate field is depending on the tensor order of the intermediate field. If it is an even order tensor (scalar, or 2-tensor), then like charges attract ; if it is first order (vector field), like charges repel.
I'd rather quote an expert on the topic, so this is Brian Hatfield in his intro to "Feynman Lectures on Gravitation":
In order to produce a static force and not just scattering, the emission or absorption of a single graviton by either particle must leave both particles in the same internal state. This rules out the possibility that the graviton carries half-integer spin (for example, related to the fact that it takes a rotation of 720 to return a spin-1/2 wavefunction back to itself). Therefore the graviton must have integer spin. Next, to decide which integer spins are possible, we examine the two cases where particle 2 is identical to particle 1 and where particle 2 is the antiparticle of 1, so that when charged, the two particles will carry the same and opposite charges respectively. When the potential is computed in both cases and the appropriate limits are taken, we find that when the exchanged particle carries odd integer spin, like charges repel and opposite charges attract, just as in the example of electrodynamics. On the other hand, when the exchanged particle carries even integer spin, the potential is universally attractive (like charges and opposite charges attract). Hence, the spin of the graviton must be 0, 2, 4, ...
According to Hatfield, the issue has to do with spin. Consistent with this, in my initial proposal I made a point to say the antisymmetric tensor is represented by a spin 1 field, and the symmetric tensor is represented by a spin 2 field. The rank of both tensors is 2.
doug

[sweetser]

Hello:

Since we are more than half way through the 60 post limit, I thought I would summarize the two technical points that I have learned (or relearned) as a result of this discussion.

The first is that my theory does fix the metric <b>up to a gauge symmetry</b>. I have been uncomfortable with the word "gauge", figuring it was a term only really smart people could use properly. When the word is introduced in the beginning of physics books, they always say it has to do with how things get measured. The central symmetry of the GEM unification proposal has to do with how the asymmetric field strength tensor \nabla_{\mu}A^{\nu} gets measured. Each of its 16 parts could come from a change in the potential (the A^{\nu}) or from a change in the metric (part of the \nabla_{\mu}) or any combination of the two. We know that 6 of these are EM, but what about the other 10? Where there is a symmetry, there is a conserved quantity. The conserved quantity must have something to do with a metric that can change. Mass charge is a reasonable candidate.

The second lesson is that the form of a Lagrange density where like charges repel is:
\mathcal{L}_{like repel}=-interia-charge coupling
-field tensor contraction

Both the equations of motion and field equations can be generated, the first by varying the velocity, keeping the potential fixed, and the second by varying the potential, fixing the motion. If one hopes to construct a theory where like charges attract, the form of a Lagrange density must be:
\mathcal{L}_{like attract}=-interia+charge coupling
-field tensor contraction

doug

ps. If anyone is in Berkeley, CA on Sunday, Oct. 30, please check out the West Cost Premier of "The Stand-Up Physicist" in "Why Quantum Mechanics is Weird", winner of "Best in Festival", education category, 2005 Berkeley Video and Film Festival.

[Chronos]

Sweetser, I too am confused by Lifsh*itz notation. I think I see where you are going with this, but, question the assumptions. How do you derive the extra sign change from the gravitational potential?

[sweetser]

Hello Chronos:

Let's break up the problem into smaller parts. Patrick had made a good technical point: a force equation is not a field equation, so there must be two separate ways to generate those equations starting from a Lagrange density. Both the force and field equations must have either like charges repelling as happens in EM, or like charges attracting as happens in gravity. So let's start with just enough terms in the Lagrange density to cover force. That would be those terms that have velocity in them, because one does the calculus of variations with velocity to generate the force or equations of motion. So...
\mathcal{L}_{force, like repel}=-\rho_{m}/\gamma-\rho_{q}U^{\mu}A_{\mu}

\mathcal{L}_{force, like attract}=-\rho_{m}/\gamma+\rho_{m}U^{\mu}A_{\mu}
[tex]\gamma=\frac{1}{\sqrt{1-\beta^{2}}}
U^{\mu}=(\gamma,\gamma\beta_{x},\gamma\beta_{y},\g amma\beta_{z})
To generate the field equations, the calculus of variations is done on the 4-potential, A_{\mu}. The inertia term, -\rho_{m}/\gamma, has no 4-potential, so plays no role in generating the field equations. We need to add the field strength tensor contraction. Now I cannot just tack on any field strength tensor contraction because this is the term that represents the field particles, and it must have the correct symmetry: odd integral spin if like charges repel, even integral spin if particles attract. So...
\mathcal{L}_{field, like repel}=-\rho_{q}U^{\mu}A_{\mu}-(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})
(\partial^{\mu}A_{\nu}-\partial^{\nu}A_{\mu})

\mathcal{L}_{field, like attract}=+\rho_{m}U^{\mu}A_{\mu}-(\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu})
(\nabla^{\mu}A_{\nu}+\nabla^{\nu}A_{\mu})

Let's count the changes between the Lagrange densities where like charges attract or repel. First there is the one in the charge coupling term, -/+\rho. Second and third are the two signs in the field strength tensor contraction. The final difference is between the kind of derivatives used: an exterior derivative for the anti-symmetric tensor (or deviation from the average amount of 4-change in the 4-potential) that is independent of how the symmetric tensor changes, and the covariant derivative (or average amount of 4-change in the 4-potential) that depends on how the metric changes up to a gauge transformation (we get to decide if the change is due to changes in the potential or the metric or both).

I don't think I can claim to "derive" any of the signs. Instead, it is my hypothesis that the the GEM Lagrange density,

\mathcal{L}_{GEM}=-\rho_{m}/\gamma-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu}
-\frac{1}{2c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu}

can do both gravity and EM. You should be able to see all the preceding Lagrange densities are part of this one. At least that is my hope :-)

doug

[sweetser]

In a Yahoo discussion group, a question was raised as to whether the GEM unified field proposal was consistent with the precession of the perihelion of Mercury results. My stock reply is that the coefficients of the GEM metric are identical to those of the Schwarzschild metric only to first order PPN accuracy, the level used in the calculations, ergo the results must be identical. I like a short, solid reply.
I also like a long-winded one, because it shows all the nuts and bolts. I have read in many places about the precession of the perihelion of Mercury, yet didn't get how they actually did the darn calculation. There were always a few steps that I did not follow. While reading through the Sean Carroll's Lecture notes on GR, I decided to try and figure out the details. Here I write it all out. This is not easy or short, but for those willing to work at it, might be a unique information source.
...OK, after I did the calculation, it was clear it took too darn long for this forum. The GEM theory did pass the test. If you want to see the details, here is the documentation:
http://theworld.com/~sweetser/quaternions/gravity/precession/precession.html
http://theworld.com/~sweetser/quaternions/ps/precession.pdf
This was one of those calculations that always scared me, so I am happy that all the little steps along the way to the 6 \pi GM/a(1-\epsilon^{2})c^{2} are clear to me (and any parts that are unclear I can discuss off line with folks).
doug

[hello1]

We put ball A and ball B by distance R, every atom in ball A has some positive charges and some negative charges, same as atoms in ball B.

Now suppose every positive charge in ball A attracts every negative charge and repells every positive charge in ball B, and verse visa.

Now we have all the single force add up, we find it equals to f=gxm1xm2/rxr

I once did some calculation, it sounds looked right, but now I forgot the details.

So, I believe, gravity is the shadow of electrostatic force.

An easier way to see this is to put two atoms apart and calculate the forces between all positive and negative charges.

You may find I was right. If not, please let me know why, be appreciate.

[sweetser]

Hello Hello1:

There are not enough details here to say you are right or wrong. It appears like you have made a basic observations: that Newton's gravitational force law, \vector{F}=G M m/R^{2} \hat{R} looks similar to Coulomb's static force law, \vector{F}=-Q q/R^{2} \hat{R}. It was Joseph "The Frenchman" Priestly who first made this observation after chatting with Ben "All-American" Franklin (yes, the guy on the c-note) about Ben's observation of no electric field inside a conducting cup. These force laws are purely classical. How do I know? The constants. OK, there is only one constant, G, so the force law is classical gravity. The precession of the perihelion of Mercury is relativistic gravity effect because it has both a G and a c. At this point, I have not done a calculation using G, c, and h, relativistic quantum gravity. First I need to confirm or reject the proposal that I am doing relativistic gravity. The current theory is general relativity, a metric theory based on a simple Lagrange density where one varies the metric field to create Einstein's field equations. In contrast, I vary the potential, which fixes the metric up to a gauge symmetry. The two theories agree at the current level tested, but disagree at levels yet to be reached. That is an incredibly rare place for any proposal to be. String theory is not there today because it postulates energy scales far beyond our reach.

To quote Gertrude Stein, "There's not enough there, there" in your suggestion, but that is not unusual, and is a good form of training. It is odd to find a different place that is testable.

doug

[hello1]

doug,

thanks a lot!

You know way better than me, I will try hard to understand your posts, too bad my math is too bad.

Some people in other forums thought I am having a stupid idea.

I really hope that you can proof your theory, unify the two forces.

Joe

[Don J]

In a Yahoo discussion group, a question was raised as to whether the GEM unified field proposal was consistent with the precession of the perihelion of Mercury results. ...
snip
...OK, after I did the calculation, it was clear it took too darn long for this forum. The GEM theory did pass the test. If you want to see the details, here is the documentation:
http://theworld.com/~sweetser/quaternions/gravity/precession/precession.html
http://theworld.com/~sweetser/quaternions/ps/precession.pdf

Have you compare the accuracy obtained by Einstein
Agreement between the observed precession of Mercury's perihelion and that
predicted by the combination of classical gravitational theory and Einstein's ...
GEM = 42.8
General relativity = 43.0 observed = 43.1
http://www.whfreeman.com/modphysics/PDF/2-1bw.pdf

[sweetser]

Hello Don:

The prediction for GR and GEM are identical because the precession for the perihelion equations are the same, \delta \phi =6 \pi GM/a(1-\epsilon^{2})c^{2}. I happened to do the calculations to three significant digits, and so the way I handled round-off errors is the reason for the difference. So there cannot be a measureable difference between GR and GEM for this particular measurement at this level of accurcacy.

I went through all the details in my pdf, and know that I could not do the calculation to the next level of accuracy. I suspect it could only be done numerically. Not that it would matter. The "next level" requires 6 orders of magniture improvement in the precession data, and that is not going to happen. Light bening around the Sun will require 3 orders of magnitude improvement, and no plans are being made to do that.

doug

[sweetser]

Hello:

I think I am obligated to point out possible logical flaws in general relativity. After all, GR has done brilliantly for 90 years, passing many difficult tests. Many people work on the theory today. Its intellectual structure is elegant. However, I do see specific flaws that I will try to point out in this message.

The vector A^{\nu} transforms like a tensor. The vector \partial_{\mu} transforms like a tensor. The 4-derivative of a 4-vector, \partial_{\mu}A^{\nu} does not transform like a tensor. Instead, the covariant derivative does:
\nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu }+\Gamma_{\sigma\mu}{}^{\nu}A^{\sigma}
where
\Gamma_{\sigma\mu}{}^{\nu}A^{\sigma}= 1/2 g_{\beta\sigma}(\frac{\partial g_{\mu\beta}}{\partial q_{\nu}}+\frac{\partial g^{\nu}{}_{\beta}}{\partial q^{\mu}}-\frac{\partial g_{\mu}{}^{\nu}}{\partial q^{\nu}})
is the Christoffel symbol of the second kind, a measure of how a metric g_{\mu\nu} changes, as indicated by the three derivatives of the metric. The Christoffel symbol, complicated as it is, does not transform like a tensor. Instead, it must be teamed up with another non-tensor, \partial_{\mu}A^{\nu} to transform like a tensor.

Up to this point, I am in complete agreement.

The next question is to ask: "What tensor can be formed out the the Christoffel symbol?" The correct answer provided in GR books is the rank 4 Riemann curvature tensor. No argument with that. I object to the question itself. Why not ask: "What tensor can I form with \partial_{\mu}A^{\nu}?" There may be such a rank 4 tensor, but I am not aware of the question being raised.

If one works only with the Chrisoffel symbol and not the potential, then the potential and metric are effectively divorced. That is what I object to. The divorce is an accurate description of our current understanding of GR and EM. The Maxwell equations are a potential theory, and by extension, the standard model. EM requires a metric be provided as part of the background structure, a sure sign of a divorce. The standard model needs mass to be introduced via the Higgs mechanism. GR is exclusively about gravity. All efforts since 1930 have failed to unify GR with the rest of physics, particularly quantum mechanics. This is not a temporary separation. String theory in my opinion cannot bind the metric to the potential. In my GEM proposal, the changes in the potential and the changes in the metric are united at the most logical and elegant way, right in the asymmetric, reducible tensor \nabla_{\mu}A^{\nu}.

What is the Riemann curvature tensor? It is a measure of the amount of curvature at each point in spacetime. On essential looks at the differences between two paths. Here is the definition:
R^{\rho}{}_{\sigma \mu \nu}=\partial_{\mu} \Gamma^{\rho}{}_{\sigma \nu}-\partial_{\nu} \Gamma^{\rho}{}_{\sigma \mu}+\Gamma^{\rho}{}_{\mu \lambda}\Gamma^{\lambda}{}_{\nu \sigma}-\Gamma^{\rho}{}_{\nu \lambda}\Gamma^{\lambda}{}_{\mu \sigma}
This tensor looks too complicated to me to ever understand in detail because each of those Christoffel symbols already has three metric derivatives inside it. None-the-less, the Reimann curvature tensor is the difference of two paths, which creates another problem in my opinion. Einstein's field equations conserve energy, a good thing. But at any point in spacetime, one can choose Riemann normal coordinates where the Christoffel symbol and all its derivatives are zero (but only one point in spacetime, since spacetime has to curve everywhere else). The energy density at that point will be zero. Thus energy density cannot be defined locally like it is for nearly all other field theories. People have gotten used to this difference in how energy is defined in GR, and do not consider it a flaw, just a property of the theory. I beg to differ because Nature i logically consistent. There should be no way to make a choice of coordinate frame such that the energy density is zero. In EM, one can choose difference coordinate frames, and the amounts of energy contributed separately be E and B fields cand shift, but not go to zero. In GEM, one can choose the Reimann normal coordinates, but the energy density would then by in the potential, and not zero. That to me is a good thing.

Since I am so close, I thought I'd sketch the rest of the way to Einstein's field equations for those reading this message and have not seen the path to those equations. Einstein figured Nature would want to use a simpler tensor to describe curvature. So he decided to use the Ricci curvature tensor, which is the Riemann curvature tensor with the first and third indices contracted with each other, R^{\rho}{}_{\sigma \rho \nu}=R_{\sigma \nu}. A problem with the Ricci tensor is that its divergence is not zero, a problem for energy conservation. One needs to subtract the Ricci scalar to get to the zero, leading to Einstein's vacuum field equations:
R_{\sigma \nu}-1/2 g_{\sigma \nu}R=0
Hilbert did it the proper way. He started with a super simple Lagrange density, \mathcal{L}_{GR}=R. Varying the action with respect to the metric field g_{\mu \nu}, one gets the Einstein field equations.

One final clarification. I use the Chistoffel symbol in the gauge symmetric central to proposal, but at no time is the Reimann curvature tensor, or either of its contractions the Ricci tensor or Ricci scalar needed.

doug

[Don J]

Hello:
I think I am obligated to point out possible logical flaws in general relativity.
(snip)
.....
Hilbert did it the proper way. He started with a super simple Lagrange density, \mathcal{L}_{GR}=R. Varying the action with respect to the metric field g_{\mu \nu}, one gets the Einstein field equations.
One final clarification. I use the Chistoffel symbol in the gauge symmetric central to proposal, but at no time is the Reimann curvature tensor, or either of its contractions the Ricci tensor or Ricci scalar needed.
doug
Some of the problems you mention were worked out in the early 60
Einstein Equations in Arnowitt, Deser and Misner (ADM) 3+1 Form
http://www.tat.physik.uni-tuebingen.de/~koellein/bericht-WEB/node19.html

Recent improvements helping for a near solution linking quantum gravity with GR
http://arxiv.org/abs/gr-qc/9807041

a little history (Explanations and Maths)
http://cgpg.gravity.psu.edu/research/articles/final.pdf
See the contribution made by Dirac, Bergmann, Arnowitt, Deser and Misner on page 4


Phenomenological Approach to a Unified Field Theory
R. L. Arnowitt*

Institute for Advanced Study, Princeton, New Jersey

Received 2 October 1956
http://prola.aps.org/abstract/PR/v105/i2/p735_1

[sweetser]

Hello Don:
Thanks for the references. I'll try and give my own brief summary of what is going on, and how it relates to the GEM proposal made here.

Finding solutions that one can use to make calculation of Einstein's field equations is difficult. One approach is called ADM, the initials of the three initial workers in the area, Arnowitt, Deser and Misner. They took a spacetime metric, and split it into two parts, the space constraint (3) and a time evolution equation (1). If you were to go to a gravity meeting, and they were talking about things like "foliations", "slicing", or "Hamitonian constraints", then the person is probably working on a part of this approach to GR. In quantum mechanics, if you have a Hamilton, you have something to work with. One of the hopes of the ADM approach is that the Hamiltonian is part of its structure, so a connection to quantum is built in. Still, there are technical problems to this approach that have not all been resolved.

How does the ADM approach relate to the GEM proposal? Well, if and only if the GEM proposal is correct, then gravity and EM can be described by a rank 1 field theory, with the dynamic metric as a gauge symmetry, not a field variable as in GR. If and only if it is correct, then any work that is constructed from a rank 2 field theory, like ADM, is not relevant to a description of Nature. It is more threatening to say an area of physics is unnecessary than to say it is wrong.

Abhay Ashtekar is a great writer, making some technical points very clear, so that I can clearly disagree with him :-) Let me quote his opening paragraph:
General relativity and quantum theory are among the greatest intellectual achievements of the 20th century. Each of them has profoundly altered the conceptual fabric that underlies out understanding of the physical world. Furthermore, each has been successful in describing the physical phenomena in its own domain to an astonishing degree of accuracy. And yet, they offer us strikingly different pictures of physical reality. Indeed, at first one is surprised that physics could keep progressing blissfully in the face of so deep a conflict.
Brilliant!
The reason of course is the ‘accidental’ fact that the values of fundamental constants in our universe conspire to make the Planck length so small and the Planck energy so high compared to laboratory scales.
If you want a paying job in theoretical physics, this is a good thing to profess. As an utterly independent researcher, I do not think the Planck length has anything to do with the problem, zero, zippo. It is all about math. EM uses a field strength tensor with an exterior derivative, a derivative which tosses out all information about the connection, how a metric changes (presuming the connection is torsion free and metric compatible as is done in GR). General relativity is exclusively about the connection, having gotten its divorce from the potential in the covariant derivative to end up in the Riemann curvature tensor, or its contractions, the Ricci tensor or Ricci scalar. By working with the covariant 4-derivative of a 4-vector in a reducible asymmetric field strength tensor, there are the six parts of the deviation from the average about of change in the potential second rank irreducible antisymmetric tensor to do EM, and the ten parts of the average amount of change in the potential second rank irreducible symmetric tensor to do gravity.

doug

[Don J]

If you were to go to a gravity meeting, and they were talking about things like "foliations", "slicing", or "Hamitonian constraints", then the person is probably working on a part of this approach to GR. In quantum mechanics, if you have a Hamilton, you have something to work with. One of the hopes of the ADM approach is that the Hamiltonian is part of its structure, so a connection to quantum is built in. Still, there are technical problems to this approach that have not all been resolved.

Thats right but your GEM proposal dont even deal with the quantum at all.


How does the ADM approach relate to the GEM proposal? Well, if and only if the GEM proposal is correct, then gravity and EM can be described by a rank 1 field theory, with the dynamic metric as a gauge symmetry, not a field variable as in GR.

Your GEM seem equivalent to gravitomagnetism also called gravitoelectromagnetism if it is the case you have probably rediscovered it via a different approach.
"gravitoelectromagnetism ("GEM") describes effects expected from the motion of "gravitational charges" (i.e. the motion of conventional matter), which are at least partly analogous to electromagnetic effects associated with the motion of electric charges."

http://en.wikipedia.org/wiki/Gravitoelectromagnetism


If and only if it is correct, then any work that is constructed from a rank 2 field theory, like ADM, is not relevant to a description of Nature. It is more threatening to say an area of physics is unnecessary than to say it is wrong.
Abhay Ashtekar is a great writer, making some technical points very clear, so that I can clearly disagree with him :-) Let me quote his opening paragraph:
Brilliant!
If you want a paying job in theoretical physics, this is a good thing to profess.

Agree ! because gravitomagnetism is probably the key.

[CarlB]

None-the-less, the Reimann curvature tensor is the difference of two paths, which creates another problem in my opinion. Einstein's field equations conserve energy, a good thing. But at any point in spacetime, one can choose Riemann normal coordinates where the Christoffel symbol and all its derivatives are zero (but only one point in spacetime, since spacetime has to curve everywhere else). The energy density at that point will be zero. Thus energy density cannot be defined locally like it is for nearly all other field theories.

Doug, there is a good introduction to the effect of global modifications of the energy as seen in quantum mechanics in Sakurai. It compares the effect of changing potentials in E&M with the effect of changing the gravitational potential. Both turn out to be gauge transformations. So I'm not sure that the quantum mechanics would completely agree with what you're writing here. As I've said before, gravitation is out of my bounds.

Carl

[sweetser]

Hello Don:
I believe my proposal does talk about how to quantize the theory. The approach is simple: go to the book shelf and pick up a book on quantum field theory. Go to the index, look up Gupta/Bleuler quantization of the EM field. The answers are almost written right there. For those that don’t have such a book, here’s a sketch.

The classical EM Lagrange density cannot be quantized. Why? Here it the Lagrange density:

\mathcal{L}_{Classical EM}=-\frac{1}{c}J_{q}^{\mu}A_{\mu}
-1/4c^{2}(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})
(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})

Calculate the cannonical momentum 4-density:
\pi^{\mu}=(\frac{\partial\mathcal{L}}{\partial (\partial \phi/\partial t)},
\frac{\partial\mathcal{L}}{\partial (\partial A_{x}/\partial t)},
\frac{\partial\mathcal{L}}{\partial (\partial A_{y}/\partial t)},
\frac{\partial\mathcal{L}}{\partial (\partial A_{z}/\partial t)})

If you do that for the classical EM Lagrangian, the first term (energy density) is zero. The 4-momentum cannot be quantized because there is no way to form a non-zero conjugate operator, \phi \pi_{0}-\pi_{0}\phi=0.

To correct this problem, folks choose a gauge in the Lagrange density. To make the approach appear manifestly covariant, a favorite choice first done by Gupta and independently by Bleuler was the Lorenz gauge:

\mathcal{L}_{G-B}=-\frac{1}{c}J_{q}^{\mu}A_{\mu}
-1/2c^{2}(\partial A_{\mu}A^{\mu})^{2}
-1/4c^{2}(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})
(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})

Now there are terms with \partial \phi/\partial t, so the energy term of the cannonical momentum is not zero. The field equation is calculated in the standard way, using the Euler Lagrange equation [note: if you want to see all the details of that for GEM, it is in the Lecture 4 notes of TheStandUpPhysicist.com]. The field equations are a 4D wave.

J_{q}^{\mu}=(\partial^{2}/\partial t^{2}-\partial^{2}/\partial x^{2}-\partial^{2}/\partial y^{2}-\partial^{2}/\partial z^{2})A^{\mu}

Because like charges repel, the particles must be spin 1, which makes sense looking at the G-B Lagrange density. There are 4 modes of emission with the choice of the Lorenz gauge. Two modes of emission are transverse. They do the work of EM. There is also a longitudinal mode, and a scalar mode. It is the scalar mode of a spin 1 field that causes a technical problem. It allows for negative probability densities. Oops. So a supplementary condition is added so that the scalar and longitudinal modes always cancel each other, making the modes always virtual. It is common to be suspicious of this supplementary condition. It looks like it is there to hide something inconvenient. What is the deep idea driving the need for it? Still you can write out all the standard tools of quantum field theory, from commutators to creation/destruction operators.

The field equations for GEM look darn similar, the only difference being another current density for mass:

J_{q}^{\mu}-J_{m}^{\mu}=(\partial^{2}/\partial t^{2}-\partial^{2}/\partial x^{2}-\partial^{2}/\partial y^{2}-\partial^{2}/\partial z^{2})A^{\mu}

The key technical difference is that one needs a spin 2 field because like J_{m} charges attract. The transverse modes do the work of EM. Now the scalar mode does the work of gravity, and it will not have the negative probability density problem. The commutator and creation/destruction operators should work fine as they are. The reason for the supplimentary condition is that the Lagrange density of EM in the Lorenz gauge is incomplete, missing gravity.

I am too far away from my training in quantum field theory to make a scattering cross section calculation. That would go a long way to prove that this approach can be quantized. The calculation would be very similar to EM scattering of two electrons. The two differences are that electric charge would have to be replaced by \sqrt{G}m_{e} and the spin-1 propagator would have to be replaced by a spin-2 propagator. I went so far as to get Wienberg’s papers in the 60’s which are suppose to give me what a massless spin-2 propagator should be, but was unable to follow the technical discussion.

I have avoided the term “gravitomagnetism” because that work originated in the analysis of rank 2 field theories, whereas GEM is rank 1. Gravitomagnetism is manifestly non-linear for isolated charges in a vacuum (gravity fields gravitate), but Gem is linear (gravity fields do not gravitate, just like electric fields do not contribute to the electric charge). On a technical level, I do not think the proposal represents a different way to present what is known in the literature as gravitomagnetism.

doug

[sweetser]

Hello:

<preamble>
A well-known expert in GR came to give a talk, and I decided to make a one page pitch of the unification idea. Experts in field theory often talk about the action in a vacuum. For whatever reasons, I had always thought about the Lagrangian when there are charges. For a one page pitch, I thought I would adapt to the intended audience. As it turns out, he gave an hour long talk, then was grilled for an hour by an energetic grad student, and only was able to leave the room by saying he was exhausted, so I only handed off the pitch that follows.
</preamble>

Unifying Gravity and EM or GEM by sweetser@alum.mit.edu

Start with the EM action in a (possibly curved) vacuum:
S_{\tmop{EM}} = \int \sqrt{- g} {} d^4 x ( \partial^{\mu} A^{\nu} -
\partial^{\nu} A^{\mu} )
EM symmetries
\delta S_{\tmop{EM}} = \int \sqrt{- g} d^4 x ( \partial^{\mu} A^{\nu} -
\partial^{\nu} A^{\mu} ) \delta \psi
Vary: \delta t : t \rightarrow t' = t + \delta t
Conserve: Energy, m \frac{d t}{d \tau}

Vary: \delta R : R \rightarrow R' = R + \delta R
Conserve: Momentum, m \frac{d R}{d \tau}

Not the complete story of 4-change of a 4-potential

( \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu} ) has 6 parts of 16 part story

GEM action in a vacuum
S_{\tmop{GEM}} = \int \sqrt{- g} d^4 x ( ( \partial_{\mu} A^{\nu} -
\partial_{\nu} A^{\mu} ) + ( \nabla_{\mu} A^{\nu} + \nabla_{\nu} A^{\mu} ) )
GEM Symmetry
\delta S_{\tmop{GEM}} = \int \sqrt{- g} d^4 x\mathfrak{L}_{\tmop{GEM}}
\delta \psi
Vary how 4-change in the 4-potential is measured:
Vary: \delta ( \partial_{\mu} A^{\nu} ) : ( \partial_{\mu} A^{\nu} )
\rightarrow ( \partial_{\mu} A^{\nu} )' = ( \partial_{\mu} A^{\nu} ) +
\delta ( \Gamma_{\sigma \mu}^{\nu} A^{\sigma} )
Conserve: Mass charge \frac{d \tmop{trace} ( \nabla_{\mu} A^{\nu} )}{d \tau}

Vary: \delta ( \Gamma_{\sigma \mu}^{\nu} A^{\sigma} ) : ( \Gamma_{\sigma
\mu}^{\nu} A^{\sigma} ) \rightarrow ( \Gamma_{\sigma \mu}^{\nu} A^{\sigma} )'
= ( \Gamma_{\sigma \mu}^{\nu} A^{\sigma} ) + \delta ( \partial_{\mu} A^{\nu}
)
Conserve: Mass charge \frac{d \tmop{trace} ( \nabla_{\mu} A^{\nu} )}{d \tau}

Field equations in a vacuum, vary A^{\mu}, fix g_{\mu \nu} up to the gauge symmetry transformation
\Box^2 A^{\mu} = 0
Vacuum Solutions
The dynamic potential, flat metric solution:
A^{\mu} = ( \frac{1}{R}, \vec{0} )
and
g_{\mu \nu} =
\left(\begin{array}{cc}
1 & 0\\
0 & - \hat{1}
\end{array}\right) so
\nabla^2 \frac{1}{R} = 0 \checked

The constant potential, dynamic metric solution:
A^{\mu} = constants and
g_{\mu \nu} =
\left(\begin{array}{cc}
\exp ( - 2 G M / c^2 R ) & 0\\
0 & - \hat{1} \exp ( 2 G M / c^2 R )
\end{array}\right) static, diagonal

so 0 = \partial_{\mu} \Gamma_{\sigma 0}^{\mu} A^{\sigma} = \nabla
g_{00} g^{00, \vec{R}} = \nabla^2 \frac{G M}{c^2 R} = 0 \checked

The Rosen exponential metric = Schwarzschild to 1st order PPN
accuracy, not 2nd order PNN, so it is consistent with current first order
tests, and could be confirmed or rejected for higher order tests. Example: GEM
predicts 0.8 \muarcseconds more bending by the Sun than GR.

Quantization

Gupta/Blueler quantization of a 4D wave equation with a twist.

Spin 1 field is 2 transverse modes of EM, like charges repel

Spin 2 field is scalar, longitudinal mode of Gravity, like charges attract.

[Don J]

The Rosen exponential metric = Schwarzschild to 1st order PPN
accuracy, not 2nd order PNN, so it is consistent with current first order
tests, and could be confirmed or rejected for higher order tests. Example: GEM
predicts 0.8 \muarcseconds more bending by the Sun than GR.


Hello sweetser
However obsevationals results confirm GR accuracy "By 1991 the observations of radio waves from stars consistently showed that the ratio of observed deflections to the deflections predicted by general relativity is 1.0001 ± 0.00001."
Do you have observationals results agreeing with your GEM predictions with an equivalent ratio of accuracy?

http://www.mathpages.com/rr/s6-03/6-03.htm

Fortunately, much more accurate measurements can now be made in the radio wavelengths, especially of quasars, since such measurements can be made from observatories with the best equipment and careful preparation (rather than hurriedly in a remote location during a total eclipse). In particular, the use of Very Long Baseline Interferometry (VBLI), combining signals from widely separate observatories, gives a tremendous improvement in resolving power. With these techniques it’s now possible to precisely measure the deflection (due to the Sun’s gravitational field) of electromagnetic waves from stars at great angular distances from the Sun. By 1991 the observations of radio waves from stars consistently showed that the ratio of observed deflections to the deflections predicted by general relativity is 1.0001 ± 0.00001. Thus the dramatic announcement of 1919 has been retro-actively justified.

[sweetser]

Hello Don:

A fair question. Two technical comments on the URL provided. It doesn't really make sense to write 1.0001 ± 0.00001, because one should have the uncertainty on a particular value. Experimentalist use the measure of arcseconds, and according to a living review article by Clifford Will, the resolution is down to 100 \muarcseconds (p. 36 of "The Confrontation between General Relativity and Experiment").

A second issue has to do with the factor in front of the (m/R_0)^2 term. I have a paper by Epstein and Shapiro ("Post-post-Newtonian deflection of light by the Sun", Phys. Rev D, 22:12, p 2947, 1989) which has the 15pi/4 factor, but not the -4. The difference numerically is 11.8 versus 7.8.

So how big is the (G M/c^2 R)^2 in terms of \muarcseconds?

(6.67 x 10^{- 11} \frac{\mathrm{m}^3}{\mathrm{kg } \mathrm{s}^2} 1.99 x 10^{30}
\mathrm{kg} / ( ( 3.00 x 10^8 \frac{\mathrm{m}}{\mathrm{s}})^2 6.97 x 10^8
\mathrm{m}))^{2} \frac{180^{\circ}}{\pi} \frac{3600''}{^{\circ}} \frac{10^6 \mu
\mathrm{arcsec}}{\mathrm{arcsec}}=0.924 \mu \mathrm{arcsec}

By the Epstein and Shapiro paper, that leads to bending of light around the Sun by 10.9 \muarcseconds. GEM predicts 11.6. The difference is 0.7 \muarcseconds. According to Will - and I spent $500 to fly to a meeting and ask him - there is no research effort under discussion to go from where we are today (100 \muarcseconds) down to the level needed to prove or reject my proposal (tens of \muarcseconds). Things like the rotational velocity of the Sun and its quadrapole moment come into play at that level.

Bottom line: yes GEM is consistent with current experiments, and awaits a future test.

doug

[Don J]

Hello Don:

Two technical comments on the URL provided. It doesn't really make sense to write 1.0001 ± 0.00001, because one should have the uncertainty on a particular value.

They make reference about tests using very-long-baseline radio interferometry which produced greatly improved determinations of the deflection of light. These techniques now have the capability of measuring angular separations and changes in angles as small as 100 microarcseconds.
More details
http://relativity.livingreviews.org/open?pubNo=lrr-2001-4&page=node10.html

[sweetser]

Hello Don:

Yup. I went to the 8th Eastern Gravity Meeting specifically to ask Will if plans were in the works to push the sensitivity beyond 100 \muarcseconds. He gave the first talk of the meeting, and I asked the first question. Since he is the leading figure in tests of gravity, he would know. When he said no plans are even in the works, I pressed him for more detail, and he remembered once in one planning session discussing tests to second order PPN accuracy, the level GEM goes head to head with GR. So it will not be happening anytime soon.

[Don J]

So you are in agreement than there is ACTUALLY absolutely no observationals results you can show which are in agreement with the level of accuracy you claim about your GEM prediction and the bendind of light by the sun.

Dont you consider this a problem?

[sweetser]

Hello Don:

This is a "good" problem in the same way it was for GR in 1915. For the data we have today, GEM is consistent with every test to first order PPN accuracy, including the bending of light around the Sun. If and when we get the second order data, GEM or GR will win. I like that kind of clarity! In 1915, it was either GR or Newton, and the one that bent more won. I hope that is the case again.

Although we have to increase the accuracy of light bending by three orders of magnitude, it can be imagined. The physics community would have to really think there was something to the exponential metric before it invested the time and money in trying to detect second order PPN effects. There is nothing conceptually difficult here at all. The details are currently out of reach.

I feel good about the theory because the exponential metric is more elegant than the Schwarzschild metric, which looks like a truncated power series with a really odd choice of coordinates. Schwarzschild is even worse in isotropic coordinates where it looks like a hack job (brought up once in MTW). It is not an accident that exponentials appear in so many fundamental physics equations. The reason is that if the exponent is zero, the identity element appears, and if there is a small displacement, then simple harmonic motion arises.

doug

[Don J]

I feel good about the theory because the exponential metric is more elegant than the Schwarzschild metric, which looks like a truncated power series with a really odd choice of coordinates. Schwarzschild is even worse in isotropic coordinates where it looks like a hack job (brought up once in MTW). It is not an accident that exponentials appear in so many fundamental physics equations. The reason is that if the exponent is zero, the identity element appears, and if there is a small displacement, then simple harmonic motion arises.

doug
Considering the 60 posts limited discussion on this board
If you are interested to try the ultimate test about your theory I invite you to joint another board where you can have all the room needed for discussion with hard Einstein relativistic theorician defenders.They can check in details the maths using by your theory applied to the bending of light by the sun for example.Or the precession of the orbit of Mercury.

You are very confident about your theory than a little challenge must only be "good".


Example of an actual discussion which is related in part about the Schwarzschild metric
"Celestial Mechanic wrote"
http://www.bautforum.com/showthread.php?p=614832#post614832

[Don J]

Hello Don:

This is a "good" problem in the same way it was for GR in 1915. For the data we have today, GEM is consistent with every test to first order PPN accuracy, including the bending of light around the Sun. If and when we get the second order data, GEM or GR will win. I like that kind of clarity! In 1915, it was either GR or Newton, and the one that bent more won. I hope that is the case again.


Well it seem than the time before they set up higher order term is very near based upon this paper.I think you can even take upon the opportunity as a test for your GEM theory. Comments?
A Double-Pulsar System - A Rare Laboratory for Relativistic Gravity and Plasma Physics
http://arxiv.org/abs/astro-ph/0401086

[sweetser]

Hello:

I did a pitch of an earlier post titled "GEM action in a vacuum" to a physics professor friend of mine at BU. I realized two mistakes, one minor, and one that scared me.

The action is the integral over a volume of spacetime of a Lagrange density. A Lagrange density is a scalar, all the ways energy can be exchanged for a system. In the actions I wrote the fields without contracting them against each other. Here is the corrected actions:
S_{\tmop{EM}} = \int \sqrt{- g} {} d^4 x ( \partial^{\mu} A^{\nu} -
\partial^{\nu} A^{\mu} )( \partial_{\mu} A_{\nu} -
\partial_{\nu} A_{\mu} )
S_{\tmop{GEM}} = \int \sqrt{- g} d^4 x ( ( \partial_{\mu} A^{\nu} -
\partial_{\nu} A^{\mu} )( \partial^{\mu} A_{\nu} -
\partial^{\nu} A_{\mu} ) + ( \nabla_{\mu} A^{\nu} + \nabla_{\nu} A^{\mu} ) ( \nabla^{\mu} A_{\nu} + \nabla^{\nu} A_{\mu} ) )
I consider this a minor error, but it does indicate I am not a professional.

When I got to this symmetry, at the core of my proposal, I recognized a problem:
\delta ( \partial_{\mu} A^{\nu} ) : ( \partial_{\mu} A^{\nu} ) \rightarrow ( \partial_{\mu} A^{\nu} )' = ( \partial_{\mu} A^{\nu} ) \delta ( \Gamma_{\sigma \mu}^{\nu} A^{\sigma} )
This will leave the symmetric field strength tensor invariant. It will however alter the antisymmetric field strength tensor of EM. As written, it is plain old wrong, not a symmetry of the action. Yes, this did cause me to stammer and feel bad in the stomach. My previous experience developing this idea has been to accept technical errors straight on, then give the problem time, and answer has to date always shown up. This one was pretty quick, within two hours. The symmetry is written above for the asymmetric tensor. In needs to be recast as a symmetry of the symmetric field strength tensor:
\delta ( \partial_{\mu} A^{\nu}+\partial_{\nu} A^{\mu} ) : ( \partial_{\mu} A^{\nu}+\partial_{\nu} A^{\mu} ) \rightarrow ( \partial_{\mu} A^{\nu}+\partial_{\nu} A^{\mu} )' = ( \partial_{\mu} A^{\nu}+\partial_{\nu} A^{\mu} ) + 2\delta ( \Gamma_{\sigma \mu}^{\nu} A^{\sigma} )
Crisis averted.

doug

[vanesch]

Hi,

First of all, I think that Tom will announce that the 60 post limit has been lifted so we shouldn't refrain from continuing the discussion. I stopped my postings because I didn't want to spoil your 60 posts, but now that this is not a limit anymore, I feel free to shoot :biggrin:

I still have a hard time believing that you do not have troubles having a single and unique interaction for mass and charge, but my initial objection of total symmetry between rho_m and rho_e has been lifted with the presence of a rho_m-pure kinetical term.

The thing that seems to me to "go obviously wrong" is of course that a neutral particle must see just as well an electric field as a charged one ; that's at least what my gut feeling tells me about it. This is, in another way, still the same initial objection of course.

So my question is: does your theory handle well the interaction between a neutral and a charged particle ?

[Careful]

Hi Doug,

Just a few remarks (I do not know wether other people have made them already; this thread is simply too long to read)

**
\mathcal{L}_{GEM}=-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu}
-\frac{1}{2c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu}


A_{\mu} is a 4-potential for both gravity and EM
\nabla_{\mu} is a covariant derivative
\nabla_{\mu}A^{\nu} is the reducible unified field strength tensor
which is the sum of a symmetric irreducible tensor (\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu}) for gravity
and an antisymmetric irreducible tensor (\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu}) for EM which uses an exterior derivative**

(a) the first term in your Lagrangean is a density (if the currents are); the second term however is *not*; so there is the determinant of a vierbein lacking (or otherwise you confine yourself to a fixed background flat coordinate system). Ah, I see you did that in your last messages... although you should make the currents vectors (densities should be determined dynamically (!))
(b) You have no EM gauge invariance (so you introduce *different* physics since the vector potential is now promoted to a physical quantity) !
(c) Assume now that the dynamical variables are g_{\mu \nu}
and A_{\mu} , variation gives both dynamical equations of motion (put c=1 - I assume here that the currents are still densities which should be changed):

(\textrm{difference of current densities - containing G the gravitational constant})_{\nu} = 2 \nabla_{\mu} \nabla^{\mu} A_{\nu} \sqrt(g)
and
\frac{\sqrt{g} g^{\mu \nu}}{2} \nabla_{\alpha} A_{\beta} \nabla^{\alpha} A^{\beta} \Delta g_{\mu \nu} + \sqrt{g} A^{\alpha} ( \partial_{\nu }(\Delta g_{\mu \alpha}) + \partial_{\mu }(\Delta g_{\nu \alpha}) - \partial_{\alpha }(\Delta g_{\mu \nu}) ) \nabla^{\mu} A^{\nu} = 0

Do partial integration on the last line and see what it gives.

(d) you have only two equations of motion (for the metric and for the four potential); but you do not get the Maxwell current conservation law since for that purpose the gradient terms of the vectorpotential can only ``live´´ in the field strength square term (and it also lives in the square of the symmetric part).

Cheers,

Careful

[sweetser]

Hello Careful:

Sorry for the long thread, but we have learned a few things along the way, and that takes give and take. I want to see if I can understand the points you make.

a). The way I figure out that something is a density is by looking at its units. The currents are definitely densities, so we agree on that point. We also know the units for the contraction of a field strength tensor must be correct since one appears in the standard EM Lagrange density. I like units, so here they are:
A^{\nu}\mathrm{ has units of }\sqrt{m/L}
Take a time derivative:
\nabla_{\mu}A^{\nu}\mathrm{ has units of }\sqrt{m/LT^{2}}
and square this, tossing in a pair of c's:
\frac{1}{c^{2}}\nabla_{\mu}A^{\nu}\nabla^{mu}A_{\n u}\mathrm{ has units of }m/L^{3}
I do like to stay grounded in pedestrian details, or I get disoriented. Your critique is more sophisticated than mere units, which is:
there is the determinant of a vierbein lacking (or otherwise you confine yourself to a fixed background flat coordinate system)
This brings up an central lesson I have learned here, so central I will write it out in bold: the metric for the GEM Lagrangian is fixed up to a second rank gauge symmetry transformation. Fixed in absolutely no way means flat. In standard EM, there are no way to determine how the metric changes, so it must be supplied as part of the mathematical structure. Yet it could definitely be a curved metric. In this proposal, the metric is definitely fixed, but there is a symmetry that allows one to change the metric so long as there is a change in the 4-potential.

b). The asymmetric field strength tensor is reducible, so it is not a fundamental field strength tensor. It splits into two irreducible tensors, one that is antisymmetric for EM, the other symmetric for gravity. One can look at gauge symmetries for each separately.
(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})->(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})'=(\partial_{\mu}(A^{\nu}+\n abla ^{\nu}\lambda)-\partial_{\nu}(A^{\mu}+\nabla^{\mu}\lambda)
That looks like EM gauge symmetry to me. Since the potential has been changed, that will change the derivative of the potential, so one will need to make a change in the connection. A new insight: EM has a rank 1 gauge symmetry while gravity has a rank 2 symmetry. Cool!

c). I absolutely cannot treat g_{\mu \nu} as a variable. Gravity is treated as a symmetry of the Lagrange density in this proposal. While the exercise could be done, it is not relevant to this proposal.

d). Conserved quantities come out of symmetries. If I have two symmetries, roughly: A^{\nu}->A^{\nu}'=A^{\nu}+\nabla^{\nu}\lambda \nabla_{\mu}A^{\nu}->\nabla_{\mu}A^{\nu}'=\nabla_{\mu}A^{\nu}+\Gamma_{\ sigma \mu}{}^{\nu}A^{\sigma})
then I have two conserved quantities.

doug

[Careful]

**The way I figure out that something is a density is by looking at its units. **

Nope, that is incorrect. For example det(g) is dimensionless but it still transforms as a scalar density of rank two - because of the Levi civita symbol (this to compensate for dx^1 \and dx^2 ... \and dx^n)

** The currents are definitely densities, so we agree on that point. **

No, we disagree: the currents should be vectors. Your playing with units is not correct: a covariant derivative has dimension of one over length instead of one over time.

**
This brings up an central lesson I have learned here, so central I will write it out in bold: the metric for the GEM Lagrangian is fixed up to a second rank gauge symmetry transformation. **

But that is nonsense: such transformation does not leave the curvature properties and neither the *signature* of the ``metric tensor´´ invariant.


**Fixed in absolutely no way means flat.**

HUH ???

**In standard EM, there are no way to determine how the metric changes, so it must be supplied as part of the mathematical structure. **

But Einstein Maxwell theory does that for you. Moreover if you would like to associate DIRECTLY the metric to the the EM field, then you should do this by means of the field strength (and not by the field potential) - as is clearly evident for Einstein - Maxwell theory.

** Yet it could definitely be a curved metric. In this proposal, the metric is definitely fixed, but there is a symmetry that allows one to change the metric so long as there is a change in the 4-potential. **

Isn't it contradictory to state in (b) that this change does not influence EM, but it does alter the gravitational force ?? Again, this is different ``physics´´.

** b). The asymmetric field strength tensor is reducible, so it is not a fundamental field strength tensor. It splits into two irreducible tensors, one that is antisymmetric for EM, the other symmetric for gravity. One can look at gauge symmetries for each separately.
(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})->(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})'=(\partial_{\mu}(A^{\nu}+\n abla ^{\nu}\lambda)-\partial_{\nu}(A^{\mu}+\nabla^{\mu}\lambda) **

No, you *cannot* separate both terms, the proposed symmetry should be one of the FULL Lagrangian !


**That looks like EM gauge symmetry to me. **

Wrong, again only symmetries of the FULL dynamics count.

** I absolutely cannot treat g_{\mu \nu} as a variable. Gravity is treated as a symmetry of the Lagrange density in this proposal. While the exercise could be done, it is not relevant to this proposal **

What do you mean? Gravity is a symmety? When something is a symmetry there is no physics ! Only local propagating degrees of freedom are relevant (in gravity that means: the two polarization of grav. waves - at least in 4-D).


** Conserved quantities come out of symmetries. If I have two symmetries, roughly: A^{\nu}->A^{\nu}'=A^{\nu}+\nabla^{\nu}\lambda \nabla_{\mu}A^{\nu}->\nabla_{\mu}A^{\nu}'=\nabla_{\mu}A^{\nu}+\Gamma_{\ sigma \mu}{}^{\nu}A^{\sigma})
then I have two conserved quantities. **

Neither of these transformations are symmetries of the FULL Lagrangean. The second transformation does not even make sense mathematically since the quantity added is NOT a tensor (density) - therefore does depend nontrivially upon a choice of coordinate system.

Cheers,

Careful

[sweetser]

Hello Careful:

Good, a spirited debate. We have yet to agree on a thing. Let's try, by at least finding common ground on two points where my theory may be wrong.

a). If the asymmetric field strength \nabla_{\mu}A^{\nu} does not transform like a tensor, then the approach is not worth investigating because all fundamental laws need to be expressed as tensors.

b). If the GEM Lagrange density does not allow for the transformation:
A^{\nu}->A^{\nu}'=A^{\nu}+\nabla^{\nu}\lambda
then it will not be able to describe EM.

It's getting close to 2AM after a night of cursing at Final Cut Pro, so I will spend Saturday crafting a reply. I would like some clarification on a) through a mini quiz. Which of the following objects would you consider to transform like a tensor, that if contracted with itself, would be a valid term in a Lagrange density:
1. \partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}
2. \partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu}
3. \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}
4. \nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu}
5. \nabla^{\mu}A^{\nu}
6. \nabla_{\mu}A^{\nu}
where \partial_{\mu} is a 4-derivative and \nabla_{\mu} is a covariant 4-derivative, defined in the standard way, \nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu}+\Gamma_{ \sigma \mu}{}^{\nu}A^{\sigma}.

doug

[Careful]

Hi Doug,

**a). If the asymmetric field strength \nabla_{\mu}A^{\nu} does not transform like a tensor, then the approach is not worth investigating because all fundamental laws need to be expressed as tensors. **

Of course \nabla_{\mu}A^{\nu} is a tensor, but your \nabla_{\mu}A'^{\nu} in the second transformation law is not.

**b). If the GEM Lagrange density does not allow for the transformation:
A^{\nu}->A^{\nu}'=A^{\nu}+\nabla^{\nu}\lambda
then it will not be able to describe EM.**

Sure...

**

1. \partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}
2. \partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu}
3. \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}
4. \nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu}
5. \nabla^{\mu}A^{\nu}
6. \nabla_{\mu}A^{\nu}
where \partial_{\mu} is a 4-derivative and \nabla_{\mu} is a covariant 4-derivative, defined in the standard way, \nabla_{\mu}A^{\nu}=\partial_{\mu}A^{\nu}+\Gamma_{ \sigma \mu}{}^{\nu}A^{\sigma}.

**

Sigh, let me do your litte quiz:
3, 5 and 6 transform as tensors. 2 and 4 are not even well defined. 1 does not transform as a tensor since \partial_{\mu} does not kill the metric (if both indices would be down, then it would be all right).

Cheers,

Careful

[sweetser]

Hello Careful:

No need to sigh, no one is born knowing all the rules for tensors. At this point, I do not understand your answers. I thought in some ways I was asking a few trick questions.

Let's start with Q1 and Q2. I know I have seen Q1 referred to in text book as the field strength tensor of EM. I have read and tried to follow Sean Carroll's lecture notes on GR, so that is my extent of training in tensor formalism. One can only say that \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} transforms like a tensor if the connection is metric compatible and torsion free. Then the Christoffel symbol of the second kind is symmetric. It will automatically be dropped out of this derivative which goes by the name of an exterior derivative. If Q5 is a tensor, then \nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu} transforms like a tensor because the difference of two tensors transforms like a tensor. Being a guy who hopes for simple things, I thought the definition of the covariant derivative of \nabla^{\mu}A^{\nu} would look identical to \nabla_{\mu}A_{\nu}, except that all the upper indicies become lower indices, and all the lower indices become upper indices. Only if that is the case, then the Christoffel symbol is symmetric and \nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}.

Why do I consider this a trick question? In framing the question, I said: "if contracted with itself". Let's do that for Q1:
contraction1=(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu})(\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})
All the indices are dummy indices, in the technical sense of the word. None of the indices are in the final contraction, which transforms like a scalar (and has units of mass/volume). Let's say I lowered the indices for the partial derivatives as would be the case for Q2:
contraction2=(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})(\partial^{\mu}A_{\nu}-\partial^{\nu}A_{\mu})
Again, all the indices go away, all the indices are dummies. It looks to me like contraction1 must equal contraction2. Only if this line of reasoning is correct must the answers for Q1=Q2, Q3=Q4, and Q5=Q6, whatever the answers are.

If you are convinced on technical grounds that only tensor with the partial derivatives is the one with the lower indices, then you will need to explain why it can be contracted with Q1 in the classical EM Lagrange density. One can only contract a tensor with another tensor, so to me Q1 has to be a tensor if the lowered one is a tensor.

I am confused about the relationship between Q2, Q4, and Q6. Q6 you say transforms like a tensor. From my perspective, I see Q6 as an asymmetric tensor, which can always be represented by the sum of an antisymmetric and symmetric tensor, Q2 and Q4 respectively.

If Q6 is a tensor, then the sum of two tensors should transform like a tensor, which is what Q4 is.

Hopefully I have been clear about my confusion, which is a nearly impossible task,

doug

[Careful]

Hello Doug,

**Hello Careful:
No need to sigh, no one is born knowing all the rules for tensors. **

?? Any good relativity student should know them (this is a minimal prerequisite before you start doing physics)

** One can only say that \partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu} transforms like a tensor if the connection is metric compatible and torsion free. **

??? There is no need of a metric compatible connection here !! This is much more elementary and is called exterior differential calculus (this is a first chapter thing, connections are third chapter stuff)

** Being a guy who hopes for simple things, I thought the definition of the covariant derivative of \nabla^{\mu}A^{\nu} would look identical to \nabla_{\mu}A_{\nu}, except that all the upper indicies become lower indices, and all the lower indices become upper indices. Only if that is the case, then the Christoffel symbol is symmetric and \nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}=\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}. **

This last equation is false. The mistake you do is the following: consider F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} , then you seem to think that F^{\mu \nu} = g^{\mu \alpha} g^{\nu \beta} F_{\alpha \beta} = \nabla^{\mu} A^{\nu} - \nabla^{\nu} A^{\mu} = \partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu} . This last equality is false since \partial does not commute with the metric (calculate this on a sheet of paper).

**Only if this line of reasoning is correct must the answers for Q1=Q2, Q3=Q4, and Q5=Q6, whatever the answers are. **

This is nonsense, contraction 2 is not even well defined as simple distributivity shows you. That is, there is a term like \partial_{\nu} A^{\mu} \partial^{\mu} A_{\nu} which makes no sense.

I advise you to study a decent elementary course on tensor calculus. Can someone find such a book for Doug?


Cheers,


Careful

[sweetser]

Hello Careful:

I went to a pretty technical school in Massachusetts. They had no undergrad classes in general relativity. At the time, general relativity was taught every other year to graduate students. I suspect most people with physics degrees do not learn the subtleties of vector spaces, dual spaces, tangent bundles, and all that jazz. The distinctions are difficult to keep absolutely clear.

Since you have claimed I need remedial education, I will have to quote credible sources. Let's start with the second edition of Jackson's "Classical Electrodynamics", the chapter on the special theory of relativity, page 550:

These equations imply that the electric and magnetic fields, six components in all, are the elements of a second-rank,antisymmetric field-strength tensor,
F^{\alpha \beta}=\partial^{\alpha} A^{\beta}-\partial^{\beta} A^{\alpha}

This is a direct statement that Q1 is a tensor.

Jackson provides a transformation law:

For reference, we record the field-strength tensor with two covariant indices
F_{\alpha \beta}=g_{\alpha \gamma}F^{\gamma \delta} g_{\delta \beta}

As you noted, if the two metric tensors are put on one side of the tensor, that creates a problem since a partial derivative does not commute with the metric. According to Jackson, that is not the correct question to ask. Let's work this one out:

g_{\alpha \gamma}F^{\gamma \delta} g_{\delta \beta} =
g_{\alpha \gamma}(\partial^{\gamma} A^{\delta}-\partial^{\delta} A^{\gamma}) g_{\delta \beta}
= g_{\alpha \gamma}\partial^{\gamma} A^{\delta} g_{\delta \beta}-g_{\alpha \gamma}\partial^{\delta} A^{\gamma} g_{\delta \beta} = \partial_{\alpha} A_{\beta}-\partial_{\beta} A_{\alpha}=F_{\alpha \beta}

At no time did the derivative have commute with the metric, a necessary thing. I believe we agree to the following equality for a metric compatible, torsion-free connection (which may be too highbrow):
\nabla_{\mu} A_{\nu} - \nabla_{\nu} A_{\mu} =
\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}
If we place inverse metrics on both sized of this expression as was done earlier in this reply, then I think it is mathematically proper to write:
\nabla^{\mu} A^{\nu} - \nabla^{\nu} A^{\mu} =
\partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu}
If you disagree, I need to know why.

doug

[sweetser]

Hello Careful:

Great question on the contraction of two symmetric mixed tensors. In some ways, I'd rather stay at home with my ideas warmed by delusion instead of standing out on a box in public subject to cold confrontation. I "get" why your critique looks solid. Tensors can be tricky, which I hope to show applies in this case. I even wonder if an advanced book on tensors has an index heading under symmetric mixed tensors.

Here is the guiding idea: if we start with a symmetric tensor, then after any indexing operation, we must end up with a symmetric tensor.

Here's the quick story: an index operation on a symmetric contravariant tensor must generate a symmetric mix tensor:
g_{\sigma \mu}(A^{\mu} B^{\nu}+A^{\nu}B^{\mu})=(A_{\sigma} B^{\nu}+A_{\nu}B^{\sigma})
If the indexes are swapped, \sigma\leftrightarrow\nu, the tensor is invariant. Done.

OK, not really, because I found that unsettling. I had expected the second B to drop its index. It pointed out a limitations to the tensor notation: there is no visual clue about the relationship of the two tensors. It helped me to write out all the components.

Let's begin with a symmetric, second rank contravariant tensor in two dimensions:
(A^{\mu}B^{\nu}+A^{\nu}B^{\mu})=
\left(\begin{array}{cc}
2 a_0 b_0 & a_0 b_1 + a_1 b_0\\
a_1 b_0 + a_0 b_1 & 2 a_1 b_1
\end{array}\right)
Act on this tensor with metric to lower one of the indices. That will put a minus sign in front of one of the a1's. Since the matrix is symmetric, what happens to one a1 must happen to the second a1:
(A_{\sigma}B^{\nu}+A_{\nu}B^{\sigma})=
\left(\begin{array}{cc}
2 a_0 b_0 & a_0 b_1 - a_1 b_0\\
-a_1 b_0 + a_0 b_1 & 2 a_1 b_1
\end{array}\right)
There is no choice in the matter: if one changes the sign of one a1, then the other a1 has to change also.

What is the limitation in tensor notation? The visual clue that second matrix has a connection to the first is too darn subtle. There is a pair of Greek letters written in reverse order. To make this idea more solid, I call the first tensor "flop" (it flops down in the way I expect it to), and the second tensor "flip" (it does the opposite of what I expect). Contract a flop tensor with a flop tensor, it looks like it should. Same holds true for a flip and flip tensor contraction. Contract a flip and a flop, and, well, it looks wrong. That does not mean it is wrong, rather it is a limitation of the notation and our experiences with that notation.

doug

[vanesch]

I advise you to study a decent elementary course on tensor calculus. Can someone find such a book for Doug?


Please let's keep it nice here. Let's refrain from such remarks because it quickly degenerates in namecalling which would be a pity.

Doug:
It was me who hurdled Careful in here, because he's a general relativity specialist which could give some interesting input.

Careful:
This place is for amateurs showing their creations - so please be indulgent with them ; if the discussion comes to a conclusion, we have two possibilities: our amateur goes to Stockholm, or he has learned stuff (and so do we) :smile:

[Careful]

Hi Doug,

No need to be mad at me. This was not meant in any way as a disrespectful comment, merely as a kind invitation to learn (as we all do every day).

**Since you have claimed I need remedial education, I will have to quote credible sources. **

I think actually most physicists do need to learn again GR (most of them got a diploma without even studying it).

** Let's start with the second edition of Jackson's "Classical Electrodynamics", the chapter on the special theory of relativity, page 550:


This is a direct statement that Q1 is a tensor.**


I thought that you meant this, but it is only true in SR for the following reasons : (a) you refrain yourself to intertial frames - that means that you bother only about the affine Poincare group (b) in all those frames the flat connection symbol *is* zero and therefore \partial equals the covariant derivative (and therefore the last equality *is* valid).
Obviously, in GR (when nonlinear coordinate transformations are involved) this does not hold anymore (buy the way, in your last derivation, you do use that the partial derivative commutes with the metric).

Your second message: in a sum of two tensors, an index like \nu cannot appear as a covariant and a contravariant one (that is like adding apples with peers).

Perhaps a good place to get intuition for these things is the book ``gravitation´´ of Weinberg, he describes tensor calculus at an intuitive level (without formalising too much) without loosing any content (and a lot of nice physics is involved).

Cheers,

Careful

[sweetser]

Hello:

No harm done. We can all learn more. Technical disagreements always cause tension. There is no way to avoid that. I feel FAR more confident about my line of logic when I get to quote from a knowledgeable source. Now back to the technical stuff.

****This is a direct statement that Q1 is a tensor.****

**I thought that you meant this, **

That is how I read Jackson, that the equation is valid no matter what metric one uses. The metric does not even have to satisfy the Einstein field equations so the exponential metric of this GEM proposal in the first post will also do.

** but it is only true in SR for the following reasons : **
(followed by two valid statements about SR) I concur with this trivial case, but since it is trivial, let's move on.

** Obviously, in GR **

Let me make clear where GEM is different from GR, and will lead to more technical tension. The GEM proposal uses the Christoffel symbol of the second kind. It does not use the Riemann curvature tensor or its contractions the Ricci tensor and the Ricci scalar. Since those tensors are central to GR as an area of study, communication will be difficult. People trained in GR will view the approach as too linear to work. I would argue that is a requirement to make the approach quantizable. Still, the belief that gravity must be treated with nonlinear equations is so strong I have my doubts such trained folks will look at the exponential metric which is a solution of a differential equation and makes predictions that can be tested as second order PPN accuracy.

** (buy [sic] the way, in your last derivation, you do use that the partial derivative commutes with the metric). **

I believe this is the term in question:
g_{\alpha \gamma}\partial^{\delta} A^{\gamma} g_{\delta \beta}
I have tried to be careful to use partial derivatives, \partial^{\mu}, where appropriate, distinct from covariant derivatives, \nabla^{\mu}, which are defined as \nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}{}^{\mu \nu}A^{\sigma}. There are many choices one could make for the connection: some folks study connections with torsion, some do not even work with metric. I am using assumptions made in GR, that the connection is metric compatible and torsion-free. The torsion-free part make the connection symmetric. The metric compatible means there is one metric for the connection. These two assumptions are necessary to set up the role played by the Christoffel symbol of the second kind.

I believe that a metric can commute with a partial derivative, but not a covariant derivative. If that belief is wrong, I would appreciate a source citation. Write the EM strength tensor using covariant derivatives. Because covariant derivatives are used, the field strength tensor will work as is no matter what the metric, even those that do not solve Einstein's field equations:

g_{\alpha \gamma}(\nabla^{\gamma} A^{\delta}-\nabla^{\delta} A^{\gamma}) g_{\delta \beta}
Apply the definition of a covariant derivative:
g_{\alpha \gamma}(\partial^{\gamma} A^{\delta}-\Gamma_{\sigma}{}^{\mu \nu}A^{\sigma}-\partial^{\delta}A^{\gamma}+\Gamma_{\sigma}{}^{\nu \mu}A^{\sigma}) g_{\delta \beta}
The Christoffel symbols are symmetric for \mu and \nu, so they drop:
g_{\alpha \gamma}(\partial^{\gamma} A^{\delta}-\partial^{\delta}A^{\gamma}) g_{\delta \beta}
Proceed as before, this time confident there is nothing wrong with commuting the metric with the partial derivative:
= g_{\alpha \gamma}\partial^{\gamma} A^{\delta} g_{\delta \beta}-g_{\alpha \gamma}\partial^{\delta} A^{\gamma} g_{\delta \beta} = \partial^{\alpha} A^{\beta}-\partial^{\beta} A^{\alpha}=F_{\alpha \beta}
I am aware that should one want to take a covariant derivative of the EM field strength tensor, there is an issue of ordering the covariant derivatives. That is not the goal. Here, all I need to be able to do is contract two tensors, \partial^{\gamma} A^{\delta}-\partial^{\delta}A^{\gamma}[/tex] and [itex]\partial_{\gamma} A_{\delta}-\partial_{\delta}A_{\gamma}. Simple can be good if it is well formed.

doug

[sweetser]

Hello Careful:

I understand that must not mix up their covariant and contravariant indices. Let me try and state this as a problem, and see if anyone can find a solution other than the one I wrote.

The asymmetric mixed second rank field strength tensor, \nabla_{\mu}A^{\nu}, like all asymmetric tensors, can be represented by a sum of symmetric tensor and an antisymmetric tensor. I don't know the theorem that says this, bue here is a reason why it can be done. The symmetric tensor is the average amount of change in the the potential, and the antisymmetric tensor is the deviation from the average amount of change tensor. Appropriately chosen averages and deviations can represent an arbitrary asymmetric tensor.

The exercise would be trivial without the word "mixed", like so:
\nabla^{\mu}A^{\nu}=\frac{1}{2}(\nabla^{\mu}A^{\nu }+\nabla^{\nu}A^{\mu})+\frac{1}{2}(\nabla^{\mu}A^{ \nu}-\nabla^{\nu}A^{\mu})
The question can be made concrete: how would you write this tensor using standard indicies? (only showing 2 dimensions for clarity):

\left(\begin{array}{cc}
2 \nabla_0 b_0 & \nabla_0 b_1 - \nabla_1 b_0 \\
-\nabla_1 b_0 + \nabla_0 b_1 & 2 \nabla_1 b_1 \\
\end{array}\right)=?=\nabla{}{}B+\nabla{}{}B

This looks like a reasonable matrix to represent with tensors. I am not sure how to write it in a proper way with indices.

doug

[Careful]

Hi,

**That is how I read Jackson, that the equation is valid no matter what metric one uses. The metric does not even have to satisfy the Einstein field equations so the exponential metric of this GEM proposal in the first post will also do. **

Nope, this is not valid for general metrics and also not for your exponential metric as an easy calculation shows. Note that you require \partial ^{\mu} g^{\nu \alpha} - \partial^{\nu} g^{\mu \alpha} to be zero for all \mu , \nu , \alpha ; you should derive the rest yourself.


**Let me make clear where GEM is different from GR, and will lead to more technical tension. The GEM proposal uses the Christoffel symbol of the second kind. **

What is second kind ? You simply restrict to flat Lorentz indices I assume...

**It does not use the Riemann curvature tensor or its contractions the Ricci tensor and the Ricci scalar. Since those tensors are central to GR as an area of study, communication will be difficult. People trained in GR will view the approach as too linear to work. **

Well what you do is the following: you choose a class of global Lorentz frames, then you write out the linear part (with respect to a particular frame) of the Ricci scalar and notice that this is the Klein Gordon operator applied to the graviton. At that moment you remark that the same operator plays an important role in free Maxwell theory and there you go. All your tensors are LORENTZ tensors and you raise and lower indices w.r.t. to the flat background metric (in this way it is even legitimate to see a connection as a tensor (!) ). This is nothing new: (a) several people have done this before you (check out one Dr. Johan Masreliez) and reject vigorously standard cosmology (b) it is obvious you will never be able to describe something like the big bang or any strong gravitational effect whatsoever. You get out the newtonian limit in the same way as this is done in the graviton approximation (calculated to first order); your exponential metric is not conformally flat and not translation invariant so you have a source of a gravitational wave there.

** I would argue that is a requirement to make the approach quantizable. Still, the belief that gravity must be treated with nonlinear equations is so strong I have my doubts such trained folks will look at the exponential metric which is a solution of a differential equation and makes predictions that can be tested as second order PPN accuracy. **

We all know that the world is non linear (this is even so in Newtonian mechanics). And indeed, in quantum mechanics, nobody knows how to deal with non-linear systems (for example QFT's) rigorously (except in two spacetime dimensions). You should learn from this that QM is not complete and not the other way around.

**
I believe this is the term in question:
g_{\alpha \gamma}\partial^{\delta} A^{\gamma} g_{\delta \beta} **

Indeed, as I mentioned before...

**
I believe that a metric can commute with a partial derivative, but not a covariant derivative. If that belief is wrong, I would appreciate a source citation. **

See any book on gravitation. Indeed, the statement that the connection is metric compatible simply means that the covariant derivative COMMUTES with the metric. :smile:

So, you do first order perturbation theory and notice that the operators involved are the same for EM which is obvious since there is only one Lorentz invariant second order differential operator :-) Moreover, you give up gauge invariance (and that is far worse ).

[sweetser]

Hello:

Physics can be awkward if an error is pointed out by others. In this thread, Careful argued that the expressions labeled earlier as Q2 and Q4 were nonsense. I hoped they were tensors. He is correct, I am wrong.

There were a series of constraining issues that lead to this problem. Let me try and establish the context.

1. Reducible versus irreducible tensors.
All fundamental theories of physics are expressed as irreducible tensors. They cannot be split into smaller parts. The GEM field strength tensor \nabla_{\mu} A^{\nu} is a reducible tensor. It can be split into two parts that are independent of each other. To have a fundamental theory of forces, I must find a proper way to split them.

2. A Lorentz invariant scalar field for mass
I had written in a newsgroup once about the trace of \nabla^{\mu} A^{\nu}, both indices up. I was informed caustically, that such an expression was utter nonsense: I could only take the trace of a mixed tensor. That caused me to change the GEM field strength tensor to the mix tensor form. Why bother? The Lorentz invariant trace of \nabla_{\mu} A^{\nu} might be able to do the work of the Higgs particle, another scalar field that is used to add mass to particles in the standard model via the Higgs mechanism. There would be no need for the Higgs mechanism as the scalar field is part of the GEM field strength tensor.

3. A symmetric field for gravity, and an antisymmetric field for EM
There has been a tension about how gravity and EM are separate or should be viewed as unified. I thought it would be good if the two lived completely separate in irreducible tensors, unified only at the level of the 4-potential.

Now I can answer the problem posed in my previous post. Physics is the most fun for me when I spot my own biases. I thought if I want to find a symmetric tensor, I would need pair of matrices with a plus sign between them. In the world of mixed derivatives, that is not the case. Here is the symmetric mixed tensor:

\left(\begin{array}{cc}
0 & \frac{\partial \phi}{\partial R} + \frac{\partial A_{R}}{\partial t} \\
\frac{\partial A_{R}}{\partial t} + \frac{\partial \phi}{\partial R} & 0 \\
\end{array}\right)=\nabla_{\mu} A^{\nu}-\nabla^{\nu} A_{\mu}

The other irreducible tensor is still asymmetric:
\left(\begin{array}{cc}
2 \frac{\partial \phi}{\partial t} & \frac{\partial \phi}{\partial R} - \frac{\partial A_{R}}{\partial t} \\
\frac{\partial A_{R}}{\partial t} - \frac{\partial \phi}{\partial R} & 2 \frac{\partial A_{R}}{\partial R} \\
\end{array}\right)=\nabla_{\mu} A^{\nu}+\nabla^{\nu} A_{\mu}

If the diagonal happens to be zero, then this matrix is antisymmetric.

I thought it was fun to think about the EM field strength tensor using only partial derivatives. Scanning several different sources on my book shelf, and on the Internet, I never saw this done. I conclude I was wrong. This does not alter the GEM proposal, only positions I had recently debates with Careful.

** So finally tell us: what do you want to do ? **

I wish to study the following action:

S_{GEM}=\int \sqrt{-g} d^4 x (\frac{-\rho_{m}}{\gamma}-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu}
-\frac{1}{2c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu})

This contains the GEM action as was written in the first post of this thread, plus an inertial term to calculate the force equation by varying the 4-velocity There will be people who want to see the irreducible field strength tensors written in the Lagrangian:

S_{GEM}=\int \sqrt{-g} d^4 x (\frac{-\rho_{m}}{\gamma}-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu}
-\frac{1}{4c^{2}}(\nabla_{\mu} A^{\nu}-\nabla^{\nu} A_{\mu})(\nabla^{\mu} A_{\nu}-\nabla_{\nu} A^{\mu}) - \frac{1}{4c^{2}}(\nabla_{\mu} A^{\nu}+\nabla^{\nu} A_{\mu})(\nabla^{\mu} A_{\nu}+\nabla_{\nu} A^{\mu}))

A Careful pointed out, the GEM action is not invariant under a gauge transformation. This statement is too strong. The GEM action is invariant under a gauge transformation for massless particles, where the trace of the irreducible tensor is zero. For any charged particles with mass, it is their mass that breaks the gauge symmetry (the Higgs mechanism is unnecessary). Note that all electrically charged particles do have a mass. The GEM action is effectively gauge invariant, because the size of the mass charge is thirteen orders of magnitude smaller than the electric charge for a proton, and we only know the charge of a proton to ten significant figures.

Only if these are well-formed actions should the discussion continue.

doug

[Careful]

Hi,

** A symmetric field for gravity, and an antisymmetric field for EM
There has been a tension about how gravity and EM are separate or should be viewed as unified. I thought it would be good if the two lived completely separate in irreducible tensors, unified only at the level of the 4-potential. **

But that is not enough ! You must show that your symmetric tensor has the correct signature (I assume that g is the background flat metric, right?)

**I thought it was fun to think about the EM field strength tensor using only partial derivatives. Scanning several different sources on my book shelf, and on the Internet, I never saw this done. I conclude I was wrong. This does not alter the GEM proposal, only positions I had recently debates with Careful. **

It is ok when you consider only global Lorentzindices (you break covariance then).


**
A Careful pointed out, the GEM action is not invariant under a gauge transformation. This statement is too strong. The GEM action is invariant under a gauge transformation for massless particles, where the trace of the irreducible tensor is zero. **


No, that does not seem to be correct (you would expect it to do that, but it soes not happen).


**For any charged particles with mass, it is their mass that breaks the gauge symmetry (the Higgs mechanism is unnecessary). **

Huh ?? We know that the U(1) symmetry of EM is *not* a broken one (the photon is massless). The mass of the particles is put in by hand in your Lagrangian (that is what the currents are for).


** The GEM action is effectively gauge invariant, because the size of the mass charge is thirteen orders of magnitude smaller than the electric charge for a proton, and we only know the charge of a proton to ten significant figures. **

I don't get it, the coupling constant in front of one of the U(1) breaking terms is the same as the one for the field strength squared (that is 1/c^2).

Cheers,

Careful

[sweetser]

Hello Careful:

Things are in a state of flux, so let me take inventory.

The asymmetric action has not changed.

I am still having trouble with the irreducible tensors.

Math theorem: any asymmetric tensor can be represented by a symmetric tensor, and an antisymmetric tensor.
http://mathworld.wolfram.com/AntisymmetricTensor.html

Math theorem: the number of elements in a symmetric rank 2 tensor in 4 dimensions is n+(n-1)(n-2)=10, the diagonal plus off diagonal parts.

Math theorem: the number of elements in an antisymmetric rank 2 tensor in 4 dimensions is (n-1)(n-2)=6, only the off diagonal parts.

Based on the number counting, this is the symmetric tensor:
\left(\begin{array}{cc}
2 \frac{\partial \phi}{\partial t} & \frac{\partial \phi}{\partial R} - \frac{\partial A_{R}}{\partial t} \\
\frac{\partial A_{R}}{\partial t} - \frac{\partial \phi}{\partial R} & 2 \frac{\partial A_{R}}{\partial R} \\
\end{array}\right)=\nabla_{\mu} A^{\nu}+\nabla^{\nu} A_{\mu}

I am confused. I look across the diagonal, and I see signs flip, which usually is the calling card for an antisymmetric tensor. Perhaps writing a mixed tensor as a matrix has been the point that has tripped me up.

Putting all other valid questions aside for a moment, do you think this must be the symmetric tensor?

doug

[Careful]

**
Math theorem: the number of elements in a symmetric rank 2 tensor in 4 dimensions is n+(n-1)(n-2)=10, the diagonal plus off diagonal parts.**

The correct number is n(n+1)/2 = 10


**Math theorem: the number of elements in an antisymmetric rank 2 tensor in 4 dimensions is (n-1)(n-2)=6, only the off diagonal parts.**

The correct formula is n(n-1)/2 = 6


** Based on the number counting, this is the symmetric tensor:
\left(\begin{array}{cc}
2 \frac{\partial \phi}{\partial t} & \frac{\partial \phi}{\partial R} - \frac{\partial A_{R}}{\partial t} \\
\frac{\partial A_{R}}{\partial t} - \frac{\partial \phi}{\partial R} & 2 \frac{\partial A_{R}}{\partial R} \\
\end{array}\right)=\nabla_{\mu} A^{\nu}+\nabla^{\nu} A_{\mu} **

This is *not* a symmetric tensor since both indices transform differently (if you lower \nu then it is ok).


Cheers,

Careful

[sweetser]

Hello Careful:

I can see in Wald, p. 26, that the theorem for symmetric and antisymmetric indices applies to pairs of covariant or pairs of contravariant indices. It does not appear to apply to a mixed tensor.

** This is *not* a symmetric tensor since both indices transform differently (if you lower \nu, then it is ok). **

Agreed in the general.

I could say this was a symmetric tensor for the choice of a flat Minkowski background, correct? Is there any sort of constraint on the metric that would be more general that fixing a Minkowski metric?

doug

[Careful]

**
I could say this was a symmetric tensor for the choice of a flat Minkowski background, correct? Is there any sort of constraint on the metric that would be more general that fixing a Minkowski metric?
**

This can be done in more generality : although you must keep in mind that symmetry of a (1,1) tensor is a statement wrt to an invertible (0,2) tensor (ie a metric). The question then is wether there exists a coordinate system in which the (1,1) tensor can be written as a symmetric matrix in the usual sense (so that we can apply standard spectral theorems) - note this is a *basis* dependent statement. Here, the source of potential trouble is to be found in the signature of the metric (you might want to investigate that).

I would kindly request you to study tensor calculus in more detail (I am sorry, but I have no time to answer all your questions concerning tensor calculus :smile: ).

Cheers,

Careful

[sweetser]

Hello Careful:

** I would kindly request you to study tensor calculus in more detail (I am sorry, but I have no time to answer all your questions concerning tensor calculus). **

The request has been kindly noted. In no way do I expect you to answer all my questions on tensor calculus. I have tried to make clear I was reading background material, and that did change my views.

I know I have tested your patience, but there is method to the madness. Linux Pauling was asked how he came up with so many good ideas, and it was by having so many bad ones. In biology, the things we understand best have the shortest lifespans, so more experiments can be made in a day. I'd rather make a clear but incorrect mathematical statement than a fuzzy claim. By rapid rough approximations, a solution can be converged to quickly.

A casual reader to this thread would realize that you were an expert on general relativity as promised, and had issues with the proposal. It is important to demarcate these issues. Focus on the positives first.

Claim 1. The GEM action as written below is a well-formed, covariant action:


S_{GEM}=\int \sqrt{-g} d^4 x (\frac{-\rho_{m}}{\gamma}-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu}
-\frac{1}{2c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu})


Claim 2. The GEM action as rewritten is also well-formed:


S_{GEM}=\int \sqrt{-g} d^4 x (\frac{-\rho_{m}}{\gamma}-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu}
-\frac{1}{4c^{2}}(\nabla_{\mu} A^{\nu}-\nabla^{\nu} A_{\mu})(\nabla^{\mu} A_{\nu}-\nabla_{\nu} A^{\mu}) - \frac{1}{4c^{2}}(\nabla_{\mu} A^{\nu}+\nabla^{\nu} A_{\mu})(\nabla^{\mu} A_{\nu}+\nabla_{\nu} A^{\mu}))


Biggest problem: Well-formed statements about gauge and other symmetries.

As to what I will do with the GEM action, I see little choice. The field to vary is the 4-potential. Folks that are good with actions can look at the action in claim one and get to the field equations as a one liner,
J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}. In the static case, if one chooses to work with a flat Minkowski metric, the solution is charge/distance. If one chooses to work with the exponential metric,

g_{\mu\nu}=\left(\begin{array}{cccc}
exp(-2\frac{GM}{c^{2}R}) & 0 & 0 & 0\\
0 & -exp(2\frac{GM}{c^{2}R}) & 0 & 0\\
0 & 0 & -exp(2\frac{GM}{c^{2}R}) & 0\\
0 & 0 & 0 & -exp(2\frac{GM}{c^{2}R})\end{array}\right).

the potential is static. It is a fun exercise to show this metric solves the field equations for a static potential.

doug

[Careful]

Hi,

I am not going to repeat here the things I said before (apart from the fact that Pauling probably meant that at least the math of your theory should be clear). You should really take a look at the work of Masreliez and others as well who have roughly the same ideas...

Good Luck,

Careful

[sweetser]

Much on Masreliez work can be found here: www.estfound.org. A Google site search did not reveal any words on "Lagrange" or "Lagrangians", which is the starting place for field theories, and the fundamental way to compare two field theories. At first glance, Maxreliez looks like a variation on GR to deal with problems in cosmology that has an exponential as part of the metric. I suspect the differences are greater than the similarities, but I will look into it further.

doug

[Careful]

Much on Masreliez work can be found here: www.estfound.org. A Google site search did not reveal any words on "Lagrange" or "Lagrangians", which is the starting place for field theories, and the fundamental way to compare two field theories. At first glance, Maxreliez looks like a variation on GR to deal with problems in cosmology that has an exponential as part of the metric. I suspect the differences are greater than the similarities, but I will look into it further.

doug

That's true: Masreliez does not work with a Lagrangian but that is no problem (you can equally well start from the field equations) - almost any theory can be reformulated in terms of a covariant Lagrangian but that is not the issue. I referred you to this work since he does more or less the same to GR than you seem to do (I recall you that the way you get out the metric is not satisfactory because of the problems with EM gauge transformations). His ideas about cosmology and quantum mechanics are not relevant for this thread.

Cheers,

Careful

[sweetser]

Hello Careful:

One can certainly have this perspective:

** Masreliez does not work with a Lagrangian but that is no problem (you can equally well start from the field equations) - almost any theory can be reformulated in terms of a covariant Lagrangian but that is not the issue. **

I will give two examples why I do not adopt it in my own outlook.

Rosen was the first person to work with an exponential metric exactly like the one I use in this thread (equation 67 in GRG, vol. 4, No 6, 1973, p 435). The metric is consistent with all weak field tests of GR done to date, and will be slightly different at the next level of precision for tests of gravity. Why is there not more interest in his approach?

Let's look at the action for GR. Hilbert deserves much more credit than he gets for finding this piece of the GR puzzle - Einstein guessed the field equations. The action is austere in its simplicity:
S_{Hilbert}=\int \sqrt{-g} d^{4} x R
The square root of g is needed to get volumes correctly in curved spacetime and R is the Ricci scalar, a contraction of a contraction of the Riemann curvature tensor. Vary this action with respect to the metric, and one gets the second rank, nonlinear Einstein field equations of GR.

For an isolated source, the only way to generate waves is through what I like to call the water-balloon wobble: sides come in, the top and bottom blob out. The wobble is a quadrupole kind of thing. We have experimental data from binary pulsars that indicates that the rate of gravity wave emission is consistent with a quadrupole momentum, not a dipole emitter. If a binary pulsar could emit as a dipole, we would expect more energy loss from gravity waves than is seen.

The Lagrange density for Rosen's proposal adds in another field. That field is for a flat metric, so the proposal is known as the bi-metric theory of gravitation. The additional term in the action creates a problem for strong field tests of gravity. The other metric could store energy and momentum. This would make dipole gravity wave emission possible. The experimental data for quadrupole emissions of gravity waves is why the Rosen's approach has not attracted much interest. It can be seen by looking at the Lagrange density.

It is quite the challenge to construct a Lagrange density so simple it will not emit dipole gravity waves. Here is one candidate, the Einstein-Maxwell equations, which is just the sum of the two separately:

\mathcal{L}_{Einstein-Maxwell}=\int \sqrt{-g} d^{4} x (R-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))

[note to self: it would be wrong to use \partial^{\mu} instead of \nabla^{\mu} because spacetime here is curved even though this is the contractions of an antisymmetric tensor.] The Einstein-Maxwell equations cannot be quantized with our current techniques. Vary the metric, one gets GR. Vary the 4-potential, Maxwell. There is no unity.

I am skeptical that Masreliez's Lagrange density is so simple. If the action was available, it would be possible to think about in detail.

The second story is a personal one. Back in 2000, I had an audience with one of the most well known physicists in Boston. I said here are my field equations:

J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}

See how wonderful they are?. If the mass current density in the same units as electric charge are thirteen orders of magnitude smaller than electric charge, one has the Maxwell equations in the Lorenz gauge. If the system is electrically neutral and static, one has Newton's law of gravity. If the system is neutral and not static, the field equations transform like a 4-vector, and thus gets along with SR, the motivation for GR being disolved.

He replied that a field theory requires more than field equations. One needs a Lagrange density, one needs to vary the Lagrange density so that it generates the field questions, one needs a solution to the field equations that is consistent with all current data, and one needs a solution to the field equations that makes it different from our current field theories. Then one can claim they have a field theory.

I thanked him and departed. I accepted his assessment. I was frightened. At that time, although I had hear the word Lagrangian, I had never worked with them. I had never varied an action to generate field equations. But I had no choice, I had to figure these things out that I did not understand. I was scared that I would never be able to do so. I reconnect with that fear when messing up on mixed tensor derivatives and being too liberal with partial derivatives instead of covariant derivatives. It took about a year and a half, but I now have a field theory because that list of requirements has been met. The GEM Lagrange density is simpler than Einstein-Maxwell, because I am about to cut and paste Einstein-Maxwell, then delete a few things:
\mathcal{L}_{GEM}=\int \sqrt{-g} d^{4} x (-\frac{1}{2 c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{\nu})
The Ricci scalar was dropped, the source of problems with quantization. The antisymmetric tensor was made into an asymmetric tensor. With certain choices of basis vectors, the asymmetric tensor can be viewed as a symmetric tensor for gravity and an antisymmetric tensor for EM. I am spending time pondering the apparent lack of gauge symmetry for the 4-potential: is this good, bad, or what? I don't know. It is something worth thinking about, which I am. It is an important open question at this time for the GEM proposal. The issue of gauge symmetry arises because I have the Lagrange density worked out.

Like when one admires art, one can see different things from different angles. It is my own personal option that should you write out a field equation, you are obligated to figure out the Lagrange density. I appreciate this is not a common view, but at least my work is consistent with that view.

Happy vacation days,
doug

[sweetser]

Hello all:

I have decided to ditch the mixed field strength tensor \nabla_{\mu}A^{\nu} for \nabla^{\mu}A^{\nu}: mixed tensors confused me and lead to technical errors Careful pointed out. This is a change in representation, meaning that the GEM Lagrange density is unaltered because:
\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{\nu}=\nabla^{\mu }A^{\nu}\nabla_{\mu}A_{\nu}
I am combing though my seb site, making the appropriate changes. The main benefit is that the symmetric and antisymmetric tensors tensors, \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu} and \nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu} are symmetric and antisymmetric no matter what the manifold is. I prefer to think of these as the average amount of change in the potential, and the deviation from the average amount of change respectively because it sounds more physical, less like a math exercise.

I have enjoyed thinking about gauge symmetry over the last week. I'll write up more later, but some things are clear to me.
1. The GEM Lagrange density breaks the gauge symmetry of EM.
2. Because of 1, the potential must be physically measurable. I believe that mass charge may be the measure of A.

Have a good new year,
doug

[Careful]

I'll write up more later, but some things are clear to me.
1. The GEM Lagrange density breaks the gauge symmetry of EM.
2. Because of 1, the potential must be physically measurable. I believe that mass charge may be the measure of A.
Have a good new year,
doug

Good ! As a far as we know the potential is not measurable apart from some topological winding numbers such as in the Bohm Aharonov effect. So, I am afraid your theory is incorrect :smile:

Have a good new year (and keep on learning :approve: )

Cheers,

Careful

[sweetser]

** As a far as we know the potential is not measurable apart from some topological winding numbers such as in the Bohm Aharonov effect.

In the longer writeup I was thinking about, I was going to point this out, so we are in complete agreement. This is the EM 4-potential of Maxwell's theory which is exclusively about EM, the metric must be supplied as part of the background for the theory.

There is no charged particle that does not have a mass. Mass is a measurable property of every particle with an electric charge. My proposal with the potential being responsible for both electric charge and a measurable mass charge still looks like a plausible way to unify gravity and EM, something the Maxwell equations do not try to do. The mass charge for a proton is 13 orders of magnitude smaller than the electric charge of a proton, and we know electric charge to only 10 orders of magnitude. I don't know quite how to say it, but that may make the symmetry breaking by mass charge decouple from EM in a way consistent with our current approach to the EM potential (yeah, I know that sentence was garbled, need to think some more).

Will keep learning. Enjoy the moment.
doug

[sweetser]

Hello:

I have attended regional APS/AAPT meetings, but have yet to go to a big APS meeting. The discussions here have help refine my proposal. Writing an abstract is a game of word choice efficiency since it is limited to 1300 characters with all the other stuff like the title. Here is my current 1295 character draft:

Title: Unifying Gravity and EM: A Riddle You Can Solve

Abstract: Apply three rules to this riddle:
1. Start from standard theory
2. Work with quantum mechanics
3. No new math
Start from the vacuum Hilbert-Maxwell action:
S_{H-M}=\int\sqrt{-g}d^4x(R-\frac{1}{4c^2}(\nabla^{\mu}
A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))
The Hilbert action cannot be quantized, so drop the Ricci scalar. To do more than EM, use an asymmetric tensor:
S_{GEM}=\int\sqrt{-g}d^4x\frac{1}{4c^2}\nabla^{\mu}A^{\nu}\nabla_{\mu }A_{\nu}
The metric is fixed up to a diffeomorphism. With a constant potential, the Rosen metric solves the field equations, is consistent with current tests, but predicts 0.7 muarcseconds more bending around the Sun than GR. Gauge symmetry is broken by the mass charge of particles.

doug

[Lawrence B. Crowell]

The Lagrangian

L~=~ (J^a_q~-~J^a_m)A_a~-~1/2(\nabla_aA^b\nabla^aA_b

involves the combination of a mass current and a charge current. Further the four potential is defined as being associated with both gravity and EM. Electromagnetism is a U(1) gauge theory. Gravity is an SO(3,1)~\sim~SL(2, C) theory. So this theory appears to have some analogy with the standard model of electroweak interactions. Yet in that case the gauge potential is

A_a = A(em) + A(weak),

for the SU(2)\times U(1) theory. For an SL(2,C)\times U(1) theory one might consider a similar construction, which is a twisted bundle. However, this is not apparent from your Lagrangian. I am presuming that the mass current is defined as

J_a~=~T_{ab}e^b,

or by some similar means. However, \nabla^aJ_a, a term which would emerge from the Euler-Lagrange equation, does not transform homogenously as the connection term emerges. This is related to the so called nonlocalizability of mass-energy in general relativity. So this would indicate that the field equations which emerges from the Lagrangian are not gauge covariant. This can only be recovered if there is are Killing vectors in the direction of this mass current. So without some special considerations the theory appears not to be covariant under the transformations of the theory.

This approach might best be extended to consider a theory that is SO(3,1)\times SO(4) with,

SO(4)~\simeq~SU(2)\times SU(2).

One of the SU(2)'s might be split on a singularity in its moduli space to give U(1)\times U(1), where one of these can play the role of the electromagnetic field. The other U(1) would then correspond to some massive field that is irrelevant to physics if the mass is large enough. The other SU(2) is then the weak interactions.

This might be started by considering a tetrad of the form

E_a^b~=~\gamma^be_a,

where \gamma^b is a Dirac matrix in some representation. One then would have

de^a~=~A_be_adx^b,

where A_b is the gauge potential for the Yang-Mills gauge field. Similarly by tr(\gamma_a\gamma_b)~=~4g_{ab}, if the representation of the Dirac matrices is local (changes from chart to chart) the differential on the tetrad gives

\partial_c(E_a^b)~=~\gamma^bA_ae_c~+~{\Gamma^b}_{c d}E_a^d.

From here your general gauge potential, call it {\cal A}, would be

{{\cal A}^b}_{ac}~=~\gamma^bA_ae_c~+~{\Gamma^b}_{cd}E_a^d

The field tensor for this theory would be defined from

\partial_d{{\cal A}^b}_{ac}~-~\partial_c{{\cal A}^b}_{ad},

which if one were to work out the bits should result in the gravity and EM sectors.
Lawrence B. Crowell

[sweetser]

Hello Lawerence:

A small note on how to post LaTeX at physics forums: the $ does not play the expected role. Instead one needs to use square brackets [] and the word tex to start, and /tex to end it. If you ever want to "borrow" an equation, just click on it and a pop up shows the tex needed for this site. To drop an equation into the middle of a sentence, use [] with itex to start, /itex to finish. Best of all, you can edit a post until the equations are correct. I do that a dozen times until all the parts look right.

A good reply will take me a few hours to compose, so I'll save that for this evenings activities. Thanks for your comments.
doug

[Doc Al]

Lawrence:

I took a crack at inserting the appropriate Latex delimiters into your post (See Doug's last post). You may need to tune it up and repost. You can find more Latex info, should you need it, here: http://www.physicsforums.com/showthread.php?t=8997

- Doc

[sweetser]

Hello Lawrence:

The post deals with gauge symmetry issues.

My proposal breaks U(1) gauge symmetry. Let's be clear for readers what that means. This is the transformation we have all seen before:
A^{\mu} \rightarrow (\phi,\vector{A})'=(\phi-\frac{\partial \Lambda}{\partial t},A+\nabla \Lambda)
The antisymmetric field strength tensor \nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu} can be represented by the fields E and B defined as follows:
E=-\frac{\partial A}{\partial t}-\nabla \phi
B=\nabla \times A
Plug in the U(1) gauge transformation into those definitions:
E \rightarrow E' = -\frac{\partial A}{\partial t}-\frac{\partial \nabla \Lambda}{\partial t}-\nabla \phi+\nabla \frac{\partial \Lambda}{\partial t}=E
B \rightarrow B'=\nabla \times A+\nabla \times \nabla \Lambda=B
For the E field, the mixed time/space derivatives cancel. For the B field, the curl of curl of a scalar is zero.

The GEM proposal has exactly these two fields E and B. But there are also fields to represent the symmetric tensor. I call them small e and small b, the symmetric analogues to EM's big E and big B. There is also a field for the four along the diagonal. Here are the definitions for the 5 fields in the GEM field strength tensor:
E=-\frac{\partial A}{\partial t}-\nabla \phi
B=\nabla \times A
e=\frac{\partial A}{\partial t}-\nabla \phi-\Gamma_{\sigma}{}^{0u}A^{\sigma}
b=(-\frac{\partial A_{z}}{\partial y}-\frac{\partial A_{y}}{\partial z}-\Gamma_{\sigma}{}^{yz}A^{\sigma},
-\frac{\partial A_{x}}{\partial z}-\frac{\partial A_{z}}{\partial x}-\Gamma_{\sigma}{}^{xz}A^{\sigma},
-\frac{\partial A_{y}}{\partial x}-\frac{\partial A_{x}}{\partial y}-\Gamma_{\sigma}{}^{xy}A^{\sigma})
g=(\frac{\partial \phi}{\partial t}-\Gamma_{\sigma}{}^{tt}A^{\sigma}, -\frac{\partial A_{x}}{\partial x}-\Gamma_{\sigma}{}^{xx}A^{\sigma}, -\frac{\partial A_{y}}{\partial y}-\Gamma_{\sigma}{}^{yy}A^{\sigma}, -\frac{\partial A_{z}}{\partial z}-\Gamma_{\sigma}{}^{zz}A^{\sigma})
Apply the U(1) gauge symmetry, and it becomes apparent that the E and B fields are fine, but the fields I think deal with gravity, g, e, and b, are not. Gravity and breaking gauge symmetry are linked in the GEM proposal.

Gauge theory is very powerful. Starting from the U(1) symmetry in 4D, people good at this sort of thing can derive the Maxwell equations. That is a reason why if one states their proposal breaks U(1) gauge symmetry, it is reasonable to think the theory cannot regenerate the Maxwell equations. I am trying to do something more, to fundamentally include mass.

Look at one limitation of gauge theories. Let me quote extensively from Michio Kaku's "Quantum Field Theory: A Modern Introduction" p. 106:
Because of gauge invariance, there are also complications when we quantize the theory. A naive quantization of the Maxwell theory fails for a simple reason: the propagator does not exist. To see this let us write down the action in the following form:
\mathcal{L}=1/2 A^{\mu}P_{\mu \nu}\partial^{2}A^{\nu}
where:
P_{\mu \nu}=g_{\mu \nu}-\partial_{\mu}\partial_{\nu}/(\partial)^2
The problem with this operator is that it is not invertible, and hence we cannot construct a propagator for the theory. In fact, this is typical of any gauge theory, not just Maxwell's theory. This also occurs in general relativity and in superstring theory. The origin of the noninvertibility of this operator is because P_{\mu \nu} is a projection operator, that is, its square is equal to itself:
P_{\mu \nu}P^{\nu \lambda}=P_{\mu}^{\lambda}
and it projects out longitudinal states:
\partial^{\mu}P_{\mu \nu}=0
The fact that P_{\mu \nu} is a projection operator, of course goes to the heart of why Maxwell's theory is a gauge theory. This projection operator projects out any states with the form \partial_{\mu}\Lambda, which is just the statement of gauge invariance.
Physicists understand exactly how to deal with this issue: pick a gauge. With the GEM proposal, this choice is not available. That may be a good thing for quantizing the theory.

There is the problem of mass in the Standard Model. The symmetry U(1) \times SU(2)\times SU(3) justifies the number of particles needed for EM (one photon for U(1), the weak force (three W+, W-, and Z for SU(2)), and the strong force (8 gluons for SU(3)). Straight out of the box, the Standard Model works only if all the masses of particles are zero. Something else is needed to break the symmetry. Readers here know the standard answer: the Higgs mechanism uses spontaneous symmetry breaking to introduce mass into the standard model. As far as I know, there is no compelling connection between the Higgs and the graviton.

Let's think on physical grounds about how mass and charge relate to each other. Consider a pair of electrons and a pair of protons, each held 1 cm apart from each other. Release them, and the electrons repel each other, as do the protons. Measure the acceleration. The electrons accelerate more for two distinct reasons. First, there is the difference in inertial mass because an electron weighs 1800x less than a proton, good old F=mA. Second, the gravitational masses will change the total net force, more attraction for the heavier protons, good old F=-Gmm/R^{2}, which would be too subtle to measure directly. One could say that both inertial and gravitational mass break the symmetry of the standard model. In the GEM proposal, the 3 fields (10 total components) of g, e, and b make up the symmetric field strength tensor \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu} that could do the work of the graviton, while the trace of that matrix could do the work of the Higgs. I am no where near good enough to make those connections solid. I am just pointing out what looks like a duck might be a duck.

Lawrence has pointed out several ways to be a good gauge theory proposal, but I think GEM proposal is heading a different direction. There is a need to break gauge symmetry in a way consistent with gravity and quantum field theory.

doug

[sweetser]

Hello Lawrence:

This post is in reply to the nonlocalizable issue.

Let me explain for folks why mass-energy in GR is nonlocalizable. Pick a point in spacetime, any point in spacetime. You are free to choose whatever coordinates you want. Riemann normal coordinates set the connection to zero at that point (they cannot set the connection to zero everywhere). Since gravity in GR depends only on the metric, the energy density of the gravitational field is zero there. This is one way to see that the energy density of the gravitational field is nonlocalizable. The three forces in Nature we know how to quantize using gauge theories, EM, the weak force, and the strong force, are localizable. No coordinate choice can make the fields zero at a point.

With the GEM proposal, go ahead, pick the Riemann normal coordinates. The energy density of the gravity field is not zero because the gravitational field depends on both the connection and the changes in the potential. Riemann normal coordinates may set the connection to zero but the energy density could still be in the change of potential. One could in fact choose to work in entirely flat spacetime background - I am often accused of this - and all would be explained by the potential. There is nothing wrong with doing everything with the potential. But I could also decide to work with a dead dull potential, and do all of gravity with the connection (see the definitions of g, e, and b in the preceding post).

In GR, mass-energy density in the gravity field is nonlocalizable.
In GEM, mass charge - strictly similar to electric charge - is localizable.

Which is better? You make the call,
doug

[Blackforest]

The Lagrangian

L~=~ (J^a_q~-~J^a_m)A_a~-~1/2(\nabla_aA^b\nabla^aA_b

involves the combination of a mass current and a charge current. Further the four potential is defined as being associated with both gravity and EM. Electromagnetism is a U(1) gauge theory. Gravity is an SO(3,1)~\sim~SL(2, C) theory. So this theory appears to have some analogy with the standard model of electroweak interactions. Yet in that case the gauge potential is

A_a = A(em) + A(weak),

for the SU(2)\times U(1) theory. For an SL(2,C)\times U(1) theory one might consider a similar construction, which is a twisted bundle. However, this is not apparent from your Lagrangian. I am presuming that the mass current is defined as

J_a~=~T_{ab}e^b,

or by some similar means. However, \nabla^aJ_a, a term which would emerge from the Euler-Lagrange equation, does not transform homogenously as the connection term emerges. This is related to the so called nonlocalizability of mass-energy in general relativity. So this would indicate that the field equations which emerges from the Lagrangian are not gauge covariant. This can only be recovered if there is are Killing vectors in the direction of this mass current. So without some special considerations the theory appears not to be covariant under the transformations of the theory.
This approach might best be extended to consider a theory that is SO(3,1)\times SO(4) with,

SO(4)~\simeq~SU(2)\times SU(2).

One of the SU(2)'s might be split on a singularity in its moduli space to give U(1)\times U(1), where one of these can play the role of the electromagnetic field. The other U(1) would then correspond to some massive field that is irrelevant to physics if the mass is large enough. The other SU(2) is then the weak interactions.
This might be started by considering a tetrad of the form

E_a^b~=~\gamma^be_a,

where \gamma^b is a Dirac matrix in some representation. One then would have

de^a~=~A_be_adx^b,

where A_b is the gauge potential for the Yang-Mills gauge field. Similarly by tr(\gamma_a\gamma_b)~=~4g_{ab}, if the representation of the Dirac matrices is local (changes from chart to chart) the differential on the tetrad gives

\partial_c(E_a^b)~=~\gamma^bA_ae_c~+~{\Gamma^b}_{c d}E_a^d.

From here your general gauge potential, call it {\cal A}, would be

{{\cal A}^b}_{ac}~=~\gamma^bA_ae_c~+~{\Gamma^b}_{cd}E_a^d

The field tensor for this theory would be defined from

\partial_d{{\cal A}^b}_{ac}~-~\partial_c{{\cal A}^b}_{ad},

which if one were to work out the bits should result in the gravity and EM sectors.
Lawrence B. Crowell
These explainations sound good to me; I have a hope to be on the right road. Thanks

[Blackforest]

The present GEM discussion is one possible way to consider the problematic of the connections between EM and gravitation. The (E) approach (see other proposed discussion on this subforum) is another one. The "general gauge potential" in the approach proposed by Lawrence B. Crowell could perhaps find an equivalence in my approach under the label of what I have called a local "cube" defining the extended vector product. So far I understand this difficult discussion, we are looking for mechanismus able to explain the symmetry breaking. I don't know if my point of view is relevant, but couldn't we see the begining of an explaination in the not necessary coherent behavior of two mathematical operations defined in any frame: the scalar product and of the extended vector product. Can't we relate this eventually incoherence to the so-called Palatini's principle?

[sweetser]

...couldn't we see the beginning of an explanation in the not necessary coherent behavior of two mathematical operations defined in any frame: the scalar product and of the extended vector product. Can't we relate this eventually incoherence to the so-called Palatini's principle?

I am sorry to report, but I don't think so. I have tried to make clear that I am using standard math tools: the Lagrange density, the variation of the action, the connection, and tensors. The reliance on standard methods has allowed me to make corrections to the proposal, specifically that I should include the kinetic energy term (\rho/\gamma) if I want to get to a Lorentz force law, and that I should avoid using a mixed asymmetric tensor because it mixes me up.

Another weakness in my proposal is I do not yet have a way to discuss it precisely in terms of group theory. I apologize for those schooled in the craft, but here is my effort to connect to group theory. The electric charge of an electron is 16 orders of magnitude greater than its mass charge measured in the same units. That means the antisymmetric "deviation from the average change in the potential" field strength tensor, \nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu} will dwarf the symmetric "average amount of change in the potential", \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}, which together compose the asymmetric field strength tensor, \nabla^{\mu}A^{\nu}. The GEM action thus has an approximate U(1) symmetry, the symmetry of the antisymmetric field strength tensor of EM. Here is where the ice gets thin for me. If one can work in a completely flat spacetime, or work on a manifold where spacetime is curved, I believe that may be called a diffeomorphism symmetry. The energy of curvature is information about how a symmetric math tool, the metric, changes (I am presuming both a torsion-free, and metric compatible connection, or this statement would not be valid). Changes in a symmetric metric are symmetric. That information must be stored in the symmetric field strength tenor, the tensor that breaks U(1) symmetry. U(1) symmetry is barely broken due to the incredible flatness of the Universe where we live.

doug

[Blackforest]

I am sorry to report, but I don't think so.
Don't be sorry, it's your right to have a different opinion than mine. It's making the debate interesting.
I have tried to make clear that I am using standard math tools: the Lagrange density, the variation of the action, the connection, and tensors.
In someway me too... but with a small variation and fantasy relatively to the absolutely standard math tools. I get also a Lagrange density but it is a little bit different than yours. The other difference is that I can connect it with some underground stream...
The reliance on standard methods has allowed me to make corrections to the proposal, specifically that I should include the kinetic energy term (\rho/\gamma) if I want to get to a Lorentz force law, and that I should avoid using a mixed asymmetric tensor because it mixes me up.
... and quite naturally get the Lorentz-Einstein force law
Another weakness in my proposal is I do not yet have a way to discuss it precisely in terms of group theory. I apologize for those schooled in the craft, but here is my effort to connect to group theory. The electric charge of an electron is 16 orders of magnitude greater than its mass charge measured in the same units. That means the antisymmetric "deviation from the average change in the potential" field strength tensor, \nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu} will dwarf the symmetric "average amount of change in the potential", \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}, which together compose the asymmetric field strength tensor, \nabla^{\mu}A^{\nu}. The GEM action thus has an approximate U(1) symmetry, the symmetry of the antisymmetric field strength tensor of EM. Here is where the ice gets thin for me.Don't panic, you cannot imagine how I admire the brightness of your knowledge and for me the ice is so thin that I can see the water... If one can work in a completely flat spacetime, or work on a manifold where spacetime is curved, I believe that may be called a diffeomorphism symmetry. The energy of curvature is information about how a symmetric math tool, the metric, changes (I am presuming both a torsion-free, and metric compatible connection, or this statement would not be valid). Changes in a symmetric metric are symmetric. That information must be stored in the symmetric field strength tenor, the tensor that breaks U(1) symmetry. U(1) symmetry is barely broken due to the incredible flatness of the Universe where we live.
doug
The flatness (average) of our universe is a mystery that can perhaps be explained if my approach makes sense... but here I need the help and the critics of professionnals

[Lawrence B. Crowell]

Doug,

Since you indicated the structure of gauge theory, I figured I would do the same, but on a somewhat more fundamental level. All of this is based on differential forms. The coboundary operator d~=~dx^a\partial_a acts on a section x, or some cut through a principle bundle Pover a manifold \cal M simply as ds. So act upon s with s^\prime~=~gs, where g is a group element of the Lie group of transformations of the theory. Now consider ds^\prime with ds~=~As, where A[/tex] is a gauge connection

ds^\prime~=~d(gs)

This then leads to

ds^\prime~=~(dg)s~+~gAs


ds^\prime~=~\big((dg)g^{-1}~+~gAg^{-1}\big)s^\prime.

This leads to the transfomation of a gauge connection as

A~\rightarrow~(dg)g^{-1}~+~gAg^{-1}

For a group element [itex]g~=~e^{i\chi} this means that the gauge transformation is

A~\rightarrow~A~+~id\chi~+~i[\chi,~A],

and for and ableian theory (commutator = 0) and d~=~dx^a\partial_a this leads to A~\rightarrow~A~+~\nabla\chi, where i has been absorbed into \chi.

The differential form d satisfies d^2~=~0 because

d^2~=~dx^a\wedge dx^b\partial_a\partial_b,

and dx^a\wedge dx^b is antisymmetric and \partial_a\partial_b is symmetric. The appropriate differential form is the gauged differential form D~=~d~+~A. The field tensors are then obtained from

D^2~=~D\wedge D~=~(d~+~A)\wedge(d~+~A)


=~F=~dA~+~A\wedge A,

where the A\wedge d is zero acting on unity or 1. If we expand this in coordinates we then have

F~=~\big(\partial_aA_b~-~\partial_bA_a~+~[A_a,~A_b]\big)dx^a\wedge dx^b.

I have ignored structure constants and the rest here, but those can be included after the fact here. For the spacial variables this leads to the invariance of the magnetic field B~=~\nabla\times A under the gauge transformation. The two-form F~=~F_{ab}dx^a\wedge dx^b contains the antisymmetric field tensor with the electric and magnetic field components which can be easily derived.

The point here is that the antisymmetry of gauge theory is seen to emerge from its structure according to differential forms. This is a bit of a problem I see with your symmetric gauge field terms. All that “div-grad-curl” stuff used in physics emerges from the structure of vector fields on space. When it comes to your e,~b,~g fields


e=\frac{\partial A}{\partial t}-\nabla \phi-\Gamma_{\sigma}{}^{0u}A^{\sigma}

b=(-\frac{\partial A_{z}}{\partial y}-\frac{\partial A_{y}}{\partial z}-\Gamma_{\sigma}{}^{yz}A^{\sigma},
-\frac{\partial A_{x}}{\partial z}-\frac{\partial A_{z}}{\partial x}-\Gamma_{\sigma}{}^{xz}A^{\sigma},
-\frac{\partial A_{y}}{\partial x}-\frac{\partial A_{x}}{\partial y}-\Gamma_{\sigma}{}^{xy}A^{\sigma})

g=(\frac{\partial \phi}{\partial t}-\Gamma_{\sigma}{}^{tt}A^{\sigma}, -\frac{\partial A_{x}}{\partial x}-\Gamma_{\sigma}{}^{xx}A^{\sigma}, -\frac{\partial A_{y}}{\partial y}-\Gamma_{\sigma}{}^{yy}A^{\sigma}, -\frac{\partial A_{z}}{\partial z}-\Gamma_{\sigma}{}^{zz}A^{\sigma})
I would say that the first is simply a covariant form of the electric field, though with a negative sign “issue” on the first term on the right hand side, and where a \Gamma_{0}{}^{i0}\phi should also appear. The second equation for the b field would apply if \Gamma_{\sigma}{}^{xz} is a torsional part of the spacetime connection term, and where nonvanishing part would involve terms \Gamma_{0}{}^{xz}\phi. I am doing this “on the fly” here, but I think I have this right.

So there are two issues that I am scratching my head over. The first is that I am uncertain what is meant by a symmetric field tensor. Such things do occur, such as with Fermi-Dirac fields, and under supersymmetry anti-commuting fields do become commuting field theories

If we consider the Lorentz operators p_\mu and M_{\mu\nu} as elements of the subalgebra L_0 and operators Q^\alpha_a as elements of the subalgebra L_1, the for \{a,b\}\in L_0 [a,~b]\in L_0, and for a\in L_0,~b\in L_1 [a,~b]\in L_1, and for \{a,~b\}\in L_1 [a,~b]_+\in L_0, where the + means anticommutator. Here the total algebra is the graded L_0\oplus L_1. By this means a field theory can be extended to something involving symmetric field tensors, but they are due to the graded algebraic structure of SUSY, and not the standard Yang-Mills sort of field theory. I would say that a possibility is the Coleman-Mandula theorem, which states that all the symmetries of the S-matrix emerge from the generators p_\mu and M_{\mu\nu}.

Lawrence B. Crowell

[Lawrence B. Crowell]

I think there is a problem here. The energy is defined by

E = T_{0a}U^a

where the four velocity is is tied to the metric by

ds^2 = g_{ab}dx^adx^b

and so

1 = U^aU_a

as such the nabla_aE will involve connection terms from the covariant derivative of the four-vector. Remember that for "dust" the momentum-energy term is

T_{ab} = \rho U_aU_b + pg_{ab}

and so any potential term is wrapped up in \rho. This means that the potential can also be set to zero at a Riemannian normal coordinate by an appropriate choice of the connection.

When it comes to charge verses mass, these are the roots of the algebra for the two gauge fields. For U(1) the roots are \pm 1, as the two real values on the circle in the argand plane. For SL(2,~C) roots are again \pm 1, but for the negative root there are violations of the Hawking-Penrose energy conditions, which is why gravity has a positive mass-energy.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

What is a gauge theory? (I need to start very basic, and make those basics concrete, so I am really just talking down to myself here :-) A gauge is the way things get measured. In EM, the potential A^{\mu} cannot be measured, only its changes, because we can add in the derivatives of an arbitrary scalar field. For the GEM proposal, I cannot add in such a field, so A^{\mu} now must be a physically measurable thing, which I am hoping to show is related to mass charge.

What measurement symmetry is at the basis of the GEM proposal? It is a simple observation about any covariant derivative. Here is the definition:
\nabla^{\mu}A^{\nu}=\partial^{\mu}A^{\nu}-\Gamma_{\sigma}{}^{\mu \nu}A^{\sigma}
To make things simple and concrete, imagine measuring the component A^{01} and getting a 7. The question becomes how much of that 7 came from the change in the potential, \partial^{\mu}A^{\nu} and how much came from the changes in the metric via the Christoffel symbol, \Gamma_{\sigma}{}^{\mu \nu}A^{\sigma}? If one chose to measure this component in Riemann normal coordinates, then the answer would be 7 came from \partial^{\mu}A^{\nu}, and zero came from \Gamma_{\sigma}{}^{\mu \nu}A^{\sigma}. If one chose to work with a constant potential where all derivatives of the potential are zero, then zero comes from \partial^{\mu}A^{\nu}, and 7 comes from the changes in the metric, \Gamma_{\sigma}{}^{\mu \nu}A^{\sigma}. Between these two are a continuous set that weighs \partial^{\mu}A^{\nu} and \Gamma_{\sigma}{}^{\mu \nu}A^{\sigma} to come up with a 7 for the component A^{01}. I believe that group goes by the name Diff(M). I may be trying to say that for a torsion-free, metric compatible connection, Diff(M) symmetry gently breaks U(1) symmetry.

I have not learned how to cast my proposal into differential forms. That certainly is a great way to do standard EM. If, when people use the phrase "gauge theory", it comes with implications of differential forms and antisymmetric field strength tensors, then I will avoid it.

So there are two issues that I am scratching my head over. The first is that I am uncertain what is meant by a symmetric field tensor.

This is a math thing: any asymmetric rank 2 tensor can be represented by the sum of a symmetric tensor, and an antisymmetric tensor. If the two indices are reversed, nothing changes for the symmetric tensor, and all signs flip for the antisymmetric tensor. It is like a mini su duko puzzle to find a pair of matrices that do this. Here is one concrete example:

\left(\begin{array}{cccc}
1 & 1 & 2 & 3\\
5 & 6 & 9 & 8\\
0 & 7 & 11 & 12\\
13 & 0 & 16 & 0
\end{array}\right) = \left(\begin{array}{cccc}
1 & 3 & 1 & 8\\
3 & 6 & 8 & 4\\
1 & 8 & 11 & 14\\
8 & 4 & 14 & 0
\end{array}\right) + \left(\begin{array}{cccc}
0 & - 2 & 1 & - 5\\
2 & 0 & 1 & 4\\
- 1 & 1 & 0 & - 2\\
5 & - 4 & 2 & 0
\end{array}\right)

The method, once realized, is easy. For the symmetric tensor, take two numbers across the diagonal and average them. This is the "average Joe" matrix. Now find the numbers that must be added in to get back the original matrix. That is the antisymmetric matrix, or "the deviants". Any matrix full of a random collection of numbers can be viewed as "average Joe and the deviants".

When I think about the field strength tensor of EM, instead of using a shorthand like F^{\mu \nu}, I use a longhand of "the deviation from the average amount of 4-change of the 4-potential". Said that way, it begs the question, "What in Nature uses the average amount of 4-change of the 4-potential"? It must be a bit more symmetric than EM, travel at the speed of light, and depend on 10 components. Gravity sounds like the only answer.

I provided definitions for E, B, e, b, and g. Those are just ways to rewrite \nabla^{\mu}A^{\nu}, or more fine-grained, E and B represent \nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu} and g, e, and b, represent \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}. The field equations also look darn simple, a 4D wave equation with two currents, one for E, B, the other for g, e, and b:

J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}

Let's focus only on the first one:

\rho_{q}-\rho_{m}=\frac{1}{c}\partial^{2}\phi/\partial t^{2}-c\nabla^{2} \phi

I call this equation "General Gauss' law", a small joke for math guys who like Chinese food. Here it is rewritten in terms of the five fields:
\rho_q - \rho_m = \frac{\partial^2 \phi}{c \partial t^2} - c
\frac{\partial^2 \phi}{\partial x^2} - c \frac{\partial^2 \phi}{\partial y^2}
- c \frac{\partial^2 \phi}{\partial z^2}

= \frac{\partial^2 \phi}{c \partial t^2} + \frac{1}{2}
\frac{\partial}{\partial x} ( (-\frac{\partial A_x}{\partial
t} - c \frac{\partial \phi}{\partial x} ) + (
\frac{\partial A_x}{\partial t} - c
\frac{\partial \phi}{\partial x} ) )


+ \frac{1}{2} \frac{\partial}{\partial y} ( ( - \frac{\partial
A_y}{\partial t} - c \frac{\partial \phi}{\partial y} ) +
( \frac{\partial A_y}{\partial t} - c
\frac{\partial \phi}{\partial y} ) )


+ \frac{1}{2} \frac{\partial}{\partial z} ( ( - \frac{\partial
A_z}{\partial t} - c \frac{\partial \phi}{\partial z} ) +
( \frac{\partial A_z}{\partial t} - c
\frac{\partial \phi}{\partial z} ) )


= \frac{\partial g_t}{c \partial t} + \frac{1}{2} ( \vec{\nabla} \cdot \vec{E}
+ \vec{\nabla} \cdot \vec{e} )

I could also do General Amp, but I don't want to scare the children. It is a frightening amount of partial derivatives. If someone requests it, I will print it out here.

doug

I will have to address the energy question later...

[Lawrence B. Crowell]

Dear Doug,

You make a right assessment of gauge theory. The problem is that you then go off to make a statement about meauring A^{0a}, when in fact the potentials are never measured.

Even if one has an asymmetric tensor and decompose it into symmetric plus symmetric parts, the fundamental thing that counts is the two-form F = dA + A/\A. The two-form is F_{ab}dx^a/\dx^b, which by necessity has to be antisymmetric. If one adds a symmetric part to it these don't count, for they are projected out by the two form dx^a/\dx^b.

The only way in which there can be a symmetric part is if one takes the coordinate direction x^a and find that

x^a --> x^a + bar-@^a z + @^a bar-z,

where @ is an antisymmetric variable and z is a Grassmann variable.

{@^a, @^b} = g^{ab}.

then

dx^a/\dx^b ---> dx^a/\dx^b + {@^a, bar-@^b}dzd(bar-z),

The matrix F_{ab} will then contain a symmetric part which will involve a gaugino field that is the supersymmetric pair of the gauge theory.

It is important to learn differential forms, for this is a far more fundamental way of looking at this sort of physics. Given that g is the group for the theory there is the elliptic complex of the Atiyah-Singer theorem

/\^1(ad g) --m--> /\^1(ad g)x/\^0(ad g) --d---> /\^2(ad g),

where ad g is the adjoint action of the group and m is a "map" that removes the group actions from the gauge potential, or defines A/g, which is the moduli for the theory. The second cohomology on the right end gives the set of two-forms which are the gauge fields. In the case of general relativity this moduli is M/diff, and a similar definition obtains for Polyakov path integrals with "mod-Weyl transforms."

I have to figure out how to properly activate the TeX stuff here, but for now things are not too intense.

Anyway, this is why outside of supersymmetry there are no symmetric field tensors.

While it might be a bit worrisome or daunting, learning differential geometry and topology from the veiwpoint of differential forms is most advised, for it provides very powerful machinery to work on these matters.

cheers,

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

In EM, A^{\nu} is not measured because of gauge symmetry. In my GEM proposal, A^{\nu} is measurable. I do still call it a potential in this thread because that is what it has been called for such a long time.

I appreciate how useful differential forms are for doing gauge theory. Perhaps there is no better tool. Yet if a tool clearly states its limitations, then it is time for a skeptic to doubt the tool itself.
Anyway, this is why outside of supersymmetry there are no symmetric field tensors.
This says to me that the tools of differential forms are putting unreasonable limitations on physical descriptions of Nature. I very much doubt I will understand what an "elliptic complex of the Atiyah-Singer theorem" means on a physical level. That sort of thing happens all the time, welcome to the world of physics. I do fell rock solid on saying that \nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu} is the deviation from the average amount of change in A^{\nu}. Logic then dictates to me that I must work with the average amount of change in A^{\nu}. If differential forms are not up to this particular task, then for this particular proposal, I cannot use them.

On LaTeX here:
Thanks all for efforts, seeing the equations helps clarify issues immensely. For simple LaTeX, I click on equations which pops up another window for cutting and pasting. For complicate LaTeX, I use LyX or Texmacs on the expression in question, export to LaTeX, and cut and paste from there. The idea is that this site uses straight old LaTeX with different delimiters. Those delimiters are like tags, but with square braces. For the italics, it was square brakets around i and /i. I always edit until all the darn \'s are in the right place.

doug

[Lawrence B. Crowell]

I am redoing this. I think this should be better

Hello Doug,
\\
To be honest I was a bit afraid of this. You state:
\\
In EM, A^\mu is not measured because of gauge symmetry. In my GEM proposal, A^\mu is measurable.
\\
The problem is that potentials are fictions. Even in the most elementary of physics, say V~=~mg(x~-~x_0) the potential can be set to anything by liberally assigning an x_0 to any value. The force, which contributes to dynamics, F~=~mg is independent of how the potential is set.
\\
In gauge theory gauge potentials are really little more than mathematical artifacts, where they enter into gauge conditions or gauge fixing Lagrangians as constriants to provide sufficient information to solve the DEs. Of course things get a bit mysterious, as found with the Ahrahnov-Bohm experiment. Yet even though the electrons pass around a solonoid and do not interact with the magnetic field inside it is by Stokes' law

\oint_{C=\partial{\cal A}} A\cdot dx~=~\int\int_{\cal A}B\cdot dA

the magnetic field which determines the phase shift

\Delta\phi~=~exp(\oint_{C=\partial{\cal A}} A\cdot dx)

\\
The Atiyah-Singer index theorem and the elliptic complex gives a construction for the moduli space. For a gauge connection the moduli is [tex]A/g[/itex], which is a set of gauge equivalent connections: gauge connections moduli group actions. A moduli space is a space of such moduli, which for [tex]SU(2)[/itex] is five dimensional. The theory of moduli spaces has emerged as the 800 pound gorilla in the Yang-Mills theory. It is a big aspect of superstring theory these days. Maybe in a few days I will post a tutorial on this.
\\
The obvious problem that your GEM theory has is that it goes against a lot of methodology used in physics. This is not to say that any particular "canon" in physics should be regarded as utterly beyond question, but I think that any theory which considers potentials as something physically real (measureable) is bound to run into a lot of resistance.
\\
Lawrence B. Crowell

[sweetser]

Hello Lawrence:

I feel good about this situation because we are getting more precise about the relationship of the GEM proposal to our current understanding.

Differential forms are used in gauge theory, so that means they are used to understand EM, the weak force, the strong force, and general relativity. That is all the fundamental forces in Nature. I might prefer to say that potentials are not directly measurable instead of fictions, but the meaning is the same. Fixing the gauge is the easiest approach to getting differential equations to solve, although there are other approaches I don't fully understand.

As I said earlier in this thread, I am struggling to understand the issue of symmetry and gauges in my proposal. I don't yet get it :-) The way I thrash around like a fish out of water is to formulate the clearest statement I can, then look at its consequences. Then I form an opposite but clear statement, and see how that goes.

Recently I said, "In my GEM proposal "A^\mu" is measurable. The basis of that trial balloon was the observation that the U(1) symmetry clearly did not hold, namely A \rightarrow A' = A + \nabla \lambda. That observation is accurate, and first pointed out here by Careful. I noted that the symmetric field strength tensor that breaks U(1) symmetry may be some sixteen orders of magnitude smaller that the antisymmetric tensor.

At the same time, I also knew that a choice must be made before one can solve a differential equation. That is the calling card of a gauge theory. Take the GEM field equations:

J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}

Given current densities, the 4-potential cannot be found. The reason is that no information has been provide about the metric. If the metric is Euclidean and flat, it is easy enough to calculate the potential. Yet the same differential equation holds in curved spacetime. Choose a different metric, and a completely different potential is required. Let me make this point as concrete as possible. Let's solve General Gauss' equation for a static electrically charged point source. Choose to work in flat, Euclidean spacetime. The answer would have been known to Poisson:
\rho_q - \rho_m = -\nabla^{2}\frac{(q_{q} + q_{m})}{R}
Choose to work in the exponential metric at the start of this thread, or to save you from clicking,

g_{\mu\nu}=\left(\begin{array}{cccc}
exp(-2\frac{\sqrt{G}q + GM}{c^{2}R}) & 0 & 0 & 0\\
0 & -exp(2\frac{\sqrt{G}q+GM}{c^{2}R}) & 0 & 0\\
0 & 0 & -exp(2\frac{\sqrt{G}q+GM}{c^{2}R}) & 0\\
0 & 0 & 0 & -exp(2\frac{\sqrt{G}q+GM}{c^{2}R})\end{array}\right ).

Most people have not had to calculate a Christoffel symbol of the second kind. This is one of few examples where that calculation is easy. The metric is diagonal and static, therefore the only term that matters is g_{00}. The derivative of an exponential gives back the exponential times the derivative of the exponent, which you will notice is charge/R. In the definition of the Christoffel, the derivative of g_{00} contracts with g^{00}. The sign of g^{00} flips, so the exponential will drop out, leaving only the derivative of the exponent:
g^{00}g_{00,R} = -\nabla^{2}\frac{(q_{q} + q_{m})}{R}
That should look familiar. A constant potential will solve the fields equations if one chooses to use the exponential metric.

Now my position is the potential is not measurable until a choice has been made about the metric. This may have to do with the symmetry Diff(M), the group of all smooth coordinate transformations. I know that sounds too tricky to understand at a practical level, but remember the Taylor series expansion for an exponential, it is one plus the exponent plus the exponent squared ...(ignoring all signs and cofactors). The GEM field equations were first solved here with a flat Minkowski background. Then we took a smooth step away from that with the exponential metric. Therefore the symmetry of the proposal looks like Diff(M), and the GEM proposal is a gauge theory under that group transformation.

doug

Your posts are definitely looking better (and are getting me to think). There is no need for \\ as returns do the trick here. The edit button at the bottom of a post is my favorite feature of this forum, so there is no need to resubmit a post, but I alwas have a need to edit.

[sweetser]

Hello All:

I am feeling warm and fuzzy about the GEM proposal today. In high school, I got to study Kuhn's "The Structure of Scientific Revolutions". When I began my independent research in physics after the 1988 Christmas gift from both mom and sis of Hawkings' "Brief History of Time", I could have adopted the paradigm shift model. One would focus on the things that do not fit, dark matter, dark energy, and the difficulty of quantizing GR. Since I was trained scientifically as a bench biologist (go in every day and do a half dozen little experiments, day in, day out), I decided to use a model of intellectual evolution. Evolution uses three processes:

1. Mostly be like your parents.
2. You are not exactly like your parents
3. If something works, do it again (like your parents might have, only not exactly).

Never ever stop repeating cycles 1-3. Instead of jumping into advanced graduate school classes, I chose to take a history of physics class at Harvard Extension School where we got to repeat experiments done by Galileo, Newton, and Franklin. I took a class on special relativity three times, once with Edwin Taylor who wanted student input into his book under development, "Spacetime Physics", once at Harvard, once at MIT. What makes me most happy is not doing something way out there. Instead I love to see a specific, solid connection to work of the past. With that in mind, I will revisit the result of yesterday on the Diff(M) symmetry for the GEM proposal.

I was aware of the Einstein-Maxwell action, but had not worked with it. When Careful point it out (post 63), I decided to look at it again, out of respect for work done by past masters. Here is the action:
S_{Einstein-Maxwell}=\int \sqrt{-g} d^{4} x (R-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))
This action is symmetric under a U(1) transformation. That means this theory can deal with light. What is the symmetry for Einstein's approach to gravity? Special relativity worked only for inertial observers, folks gliding along at a constant velocity at all times. Einstein wanted to find invariants for folks that were accelerating in arbitrary ways. The group Diff(M) of continuous transformations of coordinates is the key. To learn more about it, read this page: http://math.ucr.edu/home/baez/symmetries.html. The Ricci scalar R has all the relevant information about how the metric is curved. It is all that is needed for the action to describe curvature since it is the contraction of a contraction of the Riemann curvature tensor.

The next move sounds almost as radical as the queen sacrifice Bobby Fischer played in 1956 (http://www.chessgames.com/perl/chessgame?gid=1008361&kpage=19). Drop the Ricci scalar R. What one is left with is the Maxwell theory on a (possibly)curved manifold:
S_{?-Maxwell}=\int \sqrt{-g} d^{4} x (-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))
The problem with this approach is that a metric must be supplied as part of the background mathematical structure. A good summary of the issue is here: http://math.ucr.edu/home/baez/background.html. Are we really completely free to choose a metric? Let's look at a concrete example of changing the metric. First choose to work with the flat Minkowski metric:

g_{\mu\nu}=\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
0 & 0 & 0 & -1\end{array}\right).

We know how to work with the Maxwell equations with this metric, this one is easy! Now let's choose a different metric that is a baby step away from flat spacetime:

g_{\mu\nu}=\left(\begin{array}{cccc}
1 & 0 & 0 & \delta\\
0 & -1 & 0 & 0\\
0 & 0 & -1 & 0\\
\delta & 0 & 0 & -1\end{array}\right).

I'll call it Minkowski delta z. This metric could be smoothly merged into Minkowski by a limit process on the delta z, so the Minkowski delta z metric that is part of the Diff(M) group. If we choose to use the Minkowski metric, then there will be zero energy stored in the curvature of spacetime. There is no problem accounting for zero. Now we choose to work with the delta z metric. There is energy stored in the curvature of spacetime. Where does the Lagrange density account for the energy of this curvature? If one tried to put it in (\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}), it would be eliminated since the EM tensor is antisymmetric and filters out what would be a symmetric contribution. In order to be able to freely change the symmetric metric, a symmetric tensor is required:
S_{GEM}=\int \sqrt{-g} d^{4} x (-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})(\n abla_{\mu}A_{\nu}+\nabla_{\nu}A_{\mu}))
S_{GEM}=\int \sqrt{-g} d^{4} x (-\frac{1}{2 c^{2}}\nabla^{\mu}A^{\nu}\nabla_{\mu}A_{\nu})
If we decide to use a metric with a delta y, there is a logical place to put its energy contribution: the y slot of the symmetric tensor. The symmetric tensor is required by the energy accountants for smooth changes in coordinates.

I don't know when people understood Diff(M) was the key group involved in understanding general relativity. Folks who are adept at group theory might claim it was the core idea behind the curtain where the Wizard of Einstein worked. That group is also at the core of the GEM proposal which is shining a bit brighter today.

doug

[Lawrence B. Crowell]

There are a number of things which "conspire" to create your Gauss' law result. Your metric is asymptotically flat. In the case of GR this conspires to recover Newtonian gravity for the [itex]\Gamma^0_{00}[\itex] geodesic equation. The specialness of the solution creates the conditions similar to what you have. Also the t-t and r-r entries of your metric look alright, but the angular parts should just have the same terms as the Schwarzschild solution. This seems to be a bit of an oversight.

Again I am unclear on this kluging of the EM and gravity connections. I am also a bit unclear on how one can really have symmetric field tensors.

cheers,

Lawrence B. Crowell

[sweetser]

Hello:

For a general reader, I thought I would provide a sense of what asymptotically flat means. Basically I will be cribbing from this source:
http://relativity.livingreviews.org/open?pubNo=lrr-2004-1&page=articlesu3.html
I cannot talk this technical unless I am looking directly at technical sources.

How should one compare the quality of one metric to another? A spacetime involves both a manifold and a metric. That begs the question, what is a manifold? A manifold is a topological space where locally it looks like R^{n}. An ice cream cone is not a manifold because of the point, but a donut is a manifold. A golf ball is also a manifold, so lets use that. The golf ball has a boundary which is the surface of the ball. A Lorentz spacetime will have a manifold M (the ball) plus a metric with a signature (+---). Out of the sea of all possible spacetimes, many that are not physical, we want to pluck out useful ones. We start with a submanifold of the the Lorentz manifold, and make sure it has a smooth boundary. There is a scalar field that can relate the Lorentz metirc to the metric of the submanifold. The scalar field on the boundary is zero. Every null geodesic has its future and past end points on the boundary. Such a submanifold is called an asymptotically simple spacetime. My imprecise was of thinking about asymptotically simple spacetime is there is a limit process that would merge the exponential metric into the Minkowski metric. A spacetime which is not asymptotically flat are those with non-zero cosmological constants, such as de Sitter and anti-de Sitter spacetimes.

If an asympotically simple spacetime also solves the Einstein vacuum equations, R_{ab}=0, then the spacetime is asympotically flat.

I would agree that the exponential metric is asympotically simple. I believe the exponential metric does not solve the Einstein vacuum equations. Rosen did the first work with this metric, and he said it did not solve the Einstein field equations. Misner worked the metric recently, making the same point.

Now I have to calculate the Ricci tensor and scalar for the exponential metric. Note: I wrote that metric in Euclidean, x, y, z, coordinates, not spherical coordinates. In spherical coordinates, it looks like so:

g_{\mu\nu}=\left(\begin{array}{cccc}
exp(-2\frac{\sqrt{G}q + GM}{c^{2}R}) & 0 & 0 & 0\\
0 & -exp(2\frac{\sqrt{G}q+GM}{c^{2}R}) & 0 & 0\\
0 & 0 & -R^{2} & 0\\
0 & 0 & 0 & -R^{2} sin^{2}(\theta)\end{array}\right ).

This is the form usually used in books on GR, and I was being unconventional, althought the missing R's and sin's should have been a hint. I decided to try and confirm what I had read by calculating the Ricci tensor for the exponential metric. I had to fire up Mathematica for some assistence.

Define the Minkowski metric in spherical coordinates:

gMinkowski=
\begin{array}{cccc}
1 & 0 & 0 & 0 \\
0 & -1 & 0 & 0 \\
0 & 0 & -R^2 & 0 \\
0 & 0 & 0 & -R^2\text{Sin}[\theta ]^2
\end{array}

Define functions that will calculate the Ricci curvature tensor and scalar (see "Gravitation and Spacetime, second edition" by Ohanian and Ruffini for details, p 337):


\noindent\(\text{RicciTensor}[\text{N$\_$},\text{L$\_$}]\text{:=}\text{Module}[\{\text{R00},\text{R11},\text{R22},\text{R33}\},\\


\text{R00}=\text{Exp}[N-L](-\frac{1}{2}D[N,\{R,2\}]+\frac{1}{4}D[L,R]D[N,R]-\frac{1}{4}D[N,R]^2-\frac{1}{R}D[N,R]);\\


\text{R11}=\frac{1}{2}D[N,\{R,2\}]-\frac{1}{4}D[L,R]D[N,R]+\frac{1}{4}D[N,R]^2-\frac{1}{R}D[L,R];\\


\text{R22}=\text{Exp}[-L](1+\frac{1}{2}R(D[N,R]-D[L,R]))-1;\\


\text{R33}=\text{Sin}[\theta ]^2\text{Exp}[-L](1+\frac{1}{2}R(D[N,R]-D[L,R]))-\text{Sin}[\theta ]^2;\\


\text{RT}=(
\begin{array}{cccc}
R00 & 0 & 0 & 0 \\
0 & R11 & 0 & 0 \\
0 & 0 & R22 & 0 \\
0 & 0 & 0 & R33
\end{array}
)];\)


\noindent\(\text{RicciScalar}[\text{N$\_$},\text{L$\_$}]\text{:=}\\
\text{Exp}[-L](-D[N,\{R,2\}]+\frac{1}{2}D[L,R]D[N,R]-\frac{1}{2}D[N,R]^2+\frac{2}{R}(D[L,R]-D[N,R])-\frac{2}{R^2})+\\
\frac{2}{R^2};\)

Test the functions withe the Schwarzschld solution:

\noindent\(\text{MatrixForm}[\text{Simplify}[\text{RicciTensor}[\text{Log}[1-2/R],\text{Log}[1/(1-2/R)]]]]\)


\noindent\((
\begin{array}{llll}
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0 \\
0 & 0 & 0 & 0
\end{array}
)\)


\noindent\(\text{Simplify}[\text{RicciScalar}[\text{Log}[1-2/R],\text{Log}[1/(1-2/R)]]]\)


\noindent\(0\)

Test with exponential metric:

\noindent\(\text{MatrixForm}[\text{Simplify}[\text{RicciTensor}[-2/R,2/R]]]\)


\noindent\((
\begin{array}{llll}
-\frac{2 e^{-4/R}}{R^4} & 0 & 0 & 0 \\
0 & \frac{2}{R^4} & 0 & 0 \\
0 & 0 & -1+\frac{e^{-2/R} (2+R)}{R} & 0 \\
0 & 0 & 0 & (-1+\frac{e^{-2/R} (2+R)}{R}) \text{Sin}[\theta ]^2
\end{array}
)\)


\noindent\(\text{Simplify}[\text{RicciScalar}[-2/R,2/R]]\)


\noindent\(\frac{e^{-2/R} (-4-4 R+2 (-1+e^{2/R}) R^2)}{R^4}\)


\noindent\(\text{MatrixForm}[\text{Simplify}[\text{RicciTensor}[-2/R,2/R]-\frac{1}{2}\text{gMinkowski} * \text{RicciScalar}[-2/R,2/R]]]\)


\noindent\((
\begin{array}{llll}
\frac{e^{-4/R} (-2-e^{4/R} R^2+e^{2/R} (2+2 R+R^2))}{R^4} & 0 & 0 & 0 \\
0 & \frac{e^{-2/R} (-2-2 R-R^2+e^{2/R} (2+R^2))}{R^4} & 0 & 0 \\
0 & 0 & -\frac{2 e^{-2/R}}{R^2} & 0 \\
0 & 0 & 0 & -\frac{2 e^{-2/R} \text{Sin}[\theta ]^2}{R^2}
\end{array}
)\)

This calculation is consistent with the literature: the exponetial metric is not a solution to the Einstein vacuum equations. I am not sure this calculation is valid because the Ricci tensor may be in Schwarzschild coordinates, but at least I tried.

doug

[Lawrence B. Crowell]

Hello Doug,

There are a number of things that still have me concerned. The “kluging” of the EM potential and the gravitational connection covers over a number of things with a broad brush. The structure of these two potentials is markedly different. For pp-waves it can be demonstrated that the gravity wave has a helicity of two, which leads in a quantization scheme to the conclusion that the graviton (which at best can only be said to exist for certain special solutions such as pp-waves) is spin = 2. Classically a gravity wave has two directions of polarization.
A gravity wave is a perturbation on a background metric \eta_{ab} with the total metric

g_{ab}~=~\eta_{ab}~+~h_{ab}.

The flat background metric has zero Ricci curvature so that to first order in the perturbation expansion

R_{ab}~=~\delta R_{ab},

which enters into the Einstein field equation
R_{ab}~-~1/2Rg_{ab}~=~\kappa T_{ab}, where \kappa~=~8\pi G/c^4 is the very small coupling constant between the momentum-energy source and the spacetime configuration or field. The Ricci curvature to first order is then

R_{ab}~=~{1\over
2}\Big(\partial_c\partial_a{h^c}_b~+~\partial_c\pa rtial_b{h^c}_a~-~\partial_a\partial_bh)~-~\partial_c\partial^ch_{ab}\Big).

The harmonic gauge g^{bc}\Gamma^a_{bc}~=~0, to first order as \partial_c{h^c}_a~=~1/2\partial_mu h the Einstein field equation gives

\partial^c\partial_ch_{ab}~-~{1\over 2}\eta_{ab}\partial^c\partial_ch~=~{{16\pi G}\over {c^4}}T_{ab},

which is well defined for the traceless metric term {\bar h}_{ab}~=~h_{ab}~-~{1\over 2}\eta_{ab}h with the simple wave equation

\partial^c\partial_c{\bar h}_{ab}~=~{{16\pi G}\over {c^4}}T_{ab}.

For the wave in vacuum the momentum energy source is set to zero, and the wave equation is \partial^c\partial_c{\bar h}_{ab}~=~0. This is a bi-vector analogue to the simple wave equation for an electromagnetic wave in free space. The wave is a transverse traceless wave {\bar h}_{ab}~=~A^{TT}_{ab}exp(ik_cx^c) with

A^{TT}_{ab}~=~\left(\begin{array{ccc}0 & 0 & 0 & 0\\ 0 & A_{xx} & A_{yx} & 0\\ 0 & A_{xy} & -A_{xx} & 0\\ 0 & 0 & 0 & 0 \end {array}\right).

The term A_{xx} is the ++ polarization term and A_{xy} is the \times\times polarization term, where each represent a polarization direction. The linearized gravity wave then has a helicity of two, which has its quantum analogue in the di-photon state in quantum optics.
Pp-waves have the nice property of being linear, and the above case is a simple example. This gravity wave, which may be found by LIGO in the near future, is one that adds linearly in a way similar to EM waves. Though it must be stressed that gravity fields (spacetime curvatures) do not in general add this way. Yet in this special case the polarization structure of the wave is different than EM waves. I still think that the different gauge connections need to be built up from a tetard formalism E^a_\mu~=~\gamma^ae_\mu

cheers,

Lawrence B. Crowell

[Lawrence B. Crowell]

opps, the latex equation that failed to show up is redone below



A^{TT}_{ab}~=~\left(\begin{array{ccc}0 & 0 & 0 & 0\\ 0 & A_{xx} & A_{yx} & 0\\ 0 & A_{xy} & -A_{xx} & 0\\ 0 & 0 & 0 & 0 \end{array}\right).

[Lawrence B. Crowell]

A^{TT}_{ab}~=~\left(\begin{array{cccc}0 & 0 & 0 & 0\\ 0 & A_{xx} & A_{yx} & 0\\ 0 & A_{xy} & -A_{xx} & 0\\ 0 & 0 & 0 & 0 \end{array}\right).

[Blackforest]

I still think that the different gauge connections need to be built up from a tetard formalism E^a_\mu~=~\gamma^ae_\mu
cheers,
Lawrence B. Crowell
Are the gamma matrices in your proposition the Dirac matrices? In this case, if yes, I think that it is easy to demonstrate that if the Dirac matrices can vary within a GR approach around the average value that they own in the SR approach, then the metric can vary too around the Minkowski one. Best regards

[CarlB]

Let me try:

A^{TT}_{ab}~=~\left(\begin{array}{cccc}0 & 0 & 0 & 0\\ 0 & A_{xx} & A_{yx} & 0\\ 0 & A_{xy} & -A_{xx} & 0\\ 0 & 0 & 0 & 0 \end{array}\right).

Carl

[sweetser]

Hello Lawrence:

Let's see if we can pinpoint the source of this gut feeling:

**
The “kluging” of the EM potential and the gravitational connection covers over a number of things with a broad brush.
**

Reflection on this issue might clarify why GR and the standard model stand separately from each other on the stage of modern physics.

Start with a covariant tensor, A_{\nu}. The partial derivative operator is also a covariant tensor, \partial_{\mu}. Take the partial derivative, \partial_{\mu} A_{\nu}. It is a common exercise to show the partial derivative of a 4-potential does not transform like a tensor. If spacetime is curved, then the partial derivative will not account for all the changes that happen due to that curvature. A covariant derivative is constructed so that the result transforms like a tensor.
\nabla_{\mu}A_{\nu}=\partial_{\mu} A_{\nu}-\Gamma^{\sigma}{}{\mu\nu}
There is considerable choice in what to use for the connection, \Gamma. I choose to work with the one used in GR that is metric compatible and torsion-free. It is known as the Christoffel symbol of the second kind. Like the partial derivative, the Christoffel symbol does not transform like a tensor. Together they do, separately they do not. The GEM approach keeps then together. Modern physics has developed ways to separate the connection from the partial derivative so both transform like tensors on their own.

In GR, one takes the connection and constructs the simplest object one can out of Christoffel symbols that transforms like a tensor. It is know as as Riemann curvature tensor:
R^{\rho}{}_{\sigma \mu \nu}=\partial_{\mu} \Gamma^{\rho}{}_{\sigma \nu}-\partial_{\nu} \Gamma^{\rho}{}_{\sigma \mu}+\Gamma^{\rho}{}_{\mu \lambda}\Gamma^{\lambda}{}_{\nu \sigma}-\Gamma^{\rho}{}_{\nu \lambda}\Gamma^{\lambda}{}_{\mu \sigma}
Roughly what is happening is that two paths are being compared, and this is the net difference. Contract this once for the Ricci tensor, do that again for the Ricci scalar.

An appealing feature of differential forms is that they are true no matter what the connection is. In other words, the connection is irrelevant to differential forms. This is very convenient. The antisymmetry of differential forms guarantees the symmetric, torsion-free connection will not be on stage. Differential forms are a great platform for understanding the standard model.

If one chooses to work with the Riemann curvature tensor and differential forms, one can successfully understand GR and the standard model. To unify gravity and EM may require going back to square one, exploring anew a reducible asymmetric tensor where the connection and the changes in the potential sleep in the same expression.

doug

[sweetser]

Hello Lawrence:

This note is about gravity waves. There are several lines of logic that argue that the graviton must be a spin-2 particle represented by a second rank symmetric tensor. I agree with that analysis.

Let me try and explain a little about this "tetrad" issue for folks not familiar with the topic. Einstein's approach to GR used the connection. There are technical problems with dealing with half integral spin using the connection. That lead people to work with the spin connection. The spin connection is not torsion free. It can be used with spinors. Another great strength is that the spin connection integrates with gauge theories very well. One can write the Riemann curvature tensor in terms of the spin connection.

Your analysis of gravity waves looks right to me. You note that the polarization will be transverse. I recall reading in Clifford Will's review of experimental tests of GR that were a wave detected, and furthermore the mode of polarization determined, and the wave was not transverse, that would be a serious challenge to GR.

For the GEM proposal, the waves have nothing to do with the Ricci tensor, nor with solutions to Einstein's field equations. Instead the 4D wave equation has two different spin fields for the massless particles. The spin 1 field represents the transverse modes of EM. A spin 2 field represents the scalar and longitudinal modes of gravity. The answer to the quantum gravity riddle is already published in most graduate quantum field theory books. Everything in the method to quantize the 4D wave equation with two spin fields is standard, right down to the gamma matrices. The solution is in the section on the Gupta/Bleuler approach to quantizing the EM field. There are 4 modes of emission: two transverse, one scalar, and one longitudinal. The two transverse modes do EM, no problem. The scalar mode for a spin 1 field is a problem because it predicts negative energy densities. To ensure those modes are always virtual, a supplementary condition is imposed so that both scalar and longitudinal modes are always virtual. There is a bit of resistance to this idea - it sounds like a hack is created just to hide otherwise valid modes of emission - but a diligent student will learn that it must be done for a spin 1 field. In the GEM proposal, the 4D wave equation needs a spin 1 field for EM where like charge repel, and a spin 2 field for gravity where like charges attract. The spin 2 field will not have the negative energy density problem. The scalar and longitudinal modes can do the work of gravity. Since the spin 1 field of the Gupta/Bleuler approach to quantizing a 4D wave equation has no problem with spinors and half integral spin particles, it is my hope that the inclusion of the spin 2 field for gravity will not present a technical challenge requiring tetrads. I do not think I will ever be educated enough on the topic to go beyond this "hope" stage.

I asked Clifford Will what he thought were the odds of determining the mode of polarization. He was not optimistic. We have yet to measure one gravity wave. Should we accomplish that difficult task, the wave will have to be measured on six different axes to determine the polarization.

If a gravity wave is transverse, GR is right, GEM is wrong. If a gravity wave is not transverse, GR is wrong, GEM is right. Let the distant future data decide.

doug

[Lawrence B. Crowell]

The gamma is indeed the Dirac matrix. Since [itex]g^{ab}=1/4\gamma^a\gamma^b[\itex] if these matrices have chart dependent representations (eg they are bundle sections) then gravity connections and curvatures can be found from them.

As for gravity waves, anyone familiar with GR knows they are predicted to be transverse. A direct detection of transversality may not be needed. If a source of gravity waves is identified an indirect inference might be made. For instance if the source has an optical signature, where we know telescopically where the source is, then with various LIGO detectors it would be possible to back out the nature of the wave. For instance if a line to the source is normal to a LIGO then this would be a pretty clear signature for travservse waves.

I would prefer to see this GEM proposal, or a related problem, done within the format of tetrads. I am a bit uncertain about separating out the gauge connection from the gravity connection. Further, why limit this to EM? It seems to me that an extension to a general Yang-Mills theory with [itex]GL(2,C)[\itex].

BTW, when it comes to TeX-ing problems, for some reason often the post preview does not show the TeX'ed equations. I just get a box.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

Minor TeX things. the itex box needs a forward slash, not the backward one of TeX. That makes it more like an HTML tag. Post preview also does not work for me. The edit button does. You can come back weeks later and modify posts, although I think folks prefer old posts to stay as is for anyone reading the conversation later.


**
For instance if a line to the source is normal to a LIGO then this would be a pretty clear signature for travservse waves.
**
My sense from chatting with Will is this sort of calculation is of major interest, but may be beyond their reach. If we ever detect a gravity wave, we cannot be sure that the source can be found in the sky.


**
I am a bit uncertain about separating out the gauge connection from the gravity connection.
**
I am uncertain about how to discuss the Diff(M) symmetry, gauge transformations, and at what level the theory is unified. When I get to deal with equations, it is crystal clear: work in a coordinate system where the connection is zero everywhere, and the system can be completely described by the potential, OR work with a constant potential, and the system can be completely described by the Christoffel symbol, OR work with some combination of potential and connection. I am not confident and finding the right phrase for this situation I know how to calculate for charged and uncharged particles.


**
Further, why limit this to EM?
**
That is entirely due to my own limitations. If the proposal works for gravity and EM, it must be extended to the weak and strong forces. I don't have the training to do that.

doug

[Blackforest]

Hello Doug,
There are a number of things that still have me concerned. ...which is well defined for the traceless metric term {\bar h}_{ab}~=~h_{ab}~-~{1\over 2}\eta_{ab}h with the simple wave equation

\partial^c\partial_c{\bar h}_{ab}~=~{{16\pi G}\over {c^4}}T_{ab}.

For the wave in vacuum the momentum energy source is set to zero, and the wave equation is \partial^c\partial_c{\bar h}_{ab}~=~0. This is a bi-vector ...
Lawrence B. Crowell
Once more time I beg your pardon to be interferring in your private discussion, but I feel concerned too. What is playing the role of a bi-vector here? The traceless metric? A metric within the GR approach has to be symmetric. A bi-vestor is, per definition, anti-symmetric... where is the mistake? Thanks for explaination. Best regards

[Lawrence B. Crowell]

To start, I am so used to backslashes in TeX that it is hard to kick the habit with [/tex].
At this stage I would say with the GEM proposal that one of two things need to be done. Either the graviton and photon sectors, the abuse with the term graviton with standing, need to be clearly indicated in some way. This might be done with some tetrad formalism or with the embedding of GR and EM into some larger GL(n,~C) group. Another approach is to somehow show that the spin-2 field of gravity, say in particular in the pp-wave solutions, can be built up from some coupling of photons. A gravity wave is a bivector, and in quantum optics there are phenomena of photon bunching or "bi-photons" which are similar to gravity waves. I say similar, for they still interact with electric charges and so forth. In such a theory two photons with alligned spins would interact to form a graviton. Currently such an interaction for counter spin alligned photons generate the Z_0 particle of weak interactions. In this way at very high energy, probably approaching the Planck energy, two photons would generate a graviton. How this would fit into Doug's theory theory is a bit unclear. Maybe if Randall et al. are right with so called "soft black holes" that occur at the TeV range in energy the other fields that the photon interacts with have some mass matrix so that there are oscillations between gravitons and photons. By this two photons correlated in a Hanbury Brown-Twiss manner will have some probability of being a graviton.
Yet this is pretty speculative. The dust bin of physics is littered with a lot of quantum field theory speculations.
On a related manner, my "hobby horse" is with information physics of quantum gravity. I have worked out how it is that black holes will preserve quantum information. I am not sure how one starts a thread on this list. Yet I would like to start throwing out some trial balloons before I try to publish things. If so this should prove to be interesting, for I have done some preliminary work on how symmetries of quantum gravity involves error correction codes. Things get into Riemann zeta functions and the like.
Cheers,
Lawrence B. Crowell

[Lawrence B. Crowell]

I said that two high energy photons could generate a Z_0 particle. This is wrong, such an interaction will generate pair W^{\pm}.

Lawrence B. Crowell

[Blackforest]

To start, I am so used to backslashes in TeX that it is hard to kick the habit with [/tex].
At this stage I would say with the GEM proposal that one of two things need to be done. Either the graviton and photon sectors, the abuse with the term graviton with standing, need to be clearly indicated in some way. This might be done with some tetrad formalism or with the embedding of GR and EM into some larger GL(n,~C) group. Another approach is to somehow show that the spin-2 field of gravity, say in particular in the pp-wave solutions, can be built up from some coupling of photons. A gravity wave is a bivector, and in quantum optics there are phenomena of photon bunching or "bi-photons" which are similar to gravity waves. I say similar, for they still interact with electric charges and so forth. In such a theory two photons with alligned spins would interact to form a graviton. Currently such an interaction for counter spin alligned photons generate the Z_0 particle of weak interactions. In this way at very high energy, probably approaching the Planck energy, two photons would generate a graviton. How this would fit into Doug's theory theory is a bit unclear. Maybe if Randall et al. are right with so called "soft black holes" that occur at the TeV range in energy the other fields that the photon interacts with have some mass matrix so that there are oscillations between gravitons and photons. By this two photons correlated in a Hanbury Brown-Twiss manner will have some probability of being a graviton.
Yet this is pretty speculative. The dust bin of physics is littered with a lot of quantum field theory speculations.
On a related manner, my "hobby horse" is with information physics of quantum gravity. I have worked out how it is that black holes will preserve quantum information. I am not sure how one starts a thread on this list. Yet I would like to start throwing out some trial balloons before I try to publish things. If so this should prove to be interesting, for I have done some preliminary work on how symmetries of quantum gravity involves error correction codes. Things get into Riemann zeta functions and the like.
Cheers,
Lawrence B. Crowell
I don't know if Doug appreciates this parasite but interesting discussion and that's why I want to thank him and you, Lawrence. I don't know who you are and I guess that you are flying 2 km over my head with your knowledge but if you are interseting in, we could continue this thread on my subforum. I did make a try to investigate situations where the Lorentz Einstein Law is a differential operator and got a unique familly of solutions where the Christoffel's symbols can only be 0 or 1; that is intuitively appearing to be a possible junction with your "horse" and for a vision of the geometry as being a kind of computing machine: don't you think so?
Thank's for the precision concerning the bi-vector.

[Lawrence B. Crowell]

I am not sure how to interpret the word “parasite.” The issue of including EM into gravity is almost as old as general relativity. The first suggestion was made by Kaluza & Klein in the early 1920s. Here the space is extended to 5 dimensions with the usual connection coefficients. Here one has metric terms g_{5\mu}, but one imposes the so called cyclic condition that g_{\mu\nu,5}~=~0. This means that one has a connection term of the form {\Gamma^5}_{\mu\nu}~-~{\Gamma^5}_{\nu\mu}, which is interpreted as B~=~\nabla\times A on the spacial terms. Of course one observation is that the Yang-Mills field, in this case electromagnetism, is on the connection level with GR, and the field potentials for this field are on the metric level in GR.
As for my information approach to quantum gravity, this involves the so called information paradox with black hole evaporation. The entropy of a black hole is only reversible if the black hole is in equilibrium with the environment so that dS~=~\frac{dM}{T}. Yet the effective heat capacity of spacetime is negative, which means that as the black hole decays to lower entropy it heat up, contrary to normal thermodynamics. This means that even a quantum fluctuation will bump the black hole away from equilibrium and so dS~>~\frac{dM}{T}, which is contrary to standard thermodynamics where systems tend towards equilibrium. Quantum fields in curved spacetime are related to those in a flat spacetime by Bogoliubov transformations. The entropy of quantum states is given by the von Neumann equation S~=~-k~\rho~log_2\rho. Because of this transformation I compute a conditional entropy \rho_{A|B}~=~lim_{n\rightarrow\infty}{{\rho_{AB}}^ {1/n}({\bf 1}_A\otimes\rho_B)^{-1/n} for the teleporation of quantum states with this conditional entropy negative. In a setting where quantum information is not included this negative entropy is ignored and so things appear irreversible. However, this negative conditional entropy means that excess information may be had “for free” so the apparent entropy increase of a black hole is accounted for. However, to use the analogy of a lottery, if one does not read the ticket to claim the winnings one plays a losing game. Of course reading that ticket is difficult, for this means an accounting of a vast number of entangled EPR pairs must be tracked, and tracked from the past to future of the black hole. This is in principle feasible, but is from a practical point of view intractible. A stellar mass black hole is then for all practical purposes an irreversible entropy machine.
Cheers,
Lawrence B. Crowell

[sweetser]

This thread is intended to focus on the GEM proposal. As I interpret it, Blackforest was trying to be polite in shifting the discussion to things like bi-vectors which do not play a key role in my proposal. I am working on a (hopefully) clear statement about group theory and the GEM work which will require a few more days of work.

Efforts to unify gravity and EM go back to Priestly's discussions with Franklin about electricity. Franklin told Priestly about his observation that there was no electric field inside a conducting can. That is similar to there being on gravitational field inside a hollow massive shell. Priestly was the one to deduce the inverse square law of EM by this analogies to gravity. I have seen the GEM Lagrangian in the context of a purely EM proposal.

doug

[Lawrence B. Crowell]

Doug,

Actually I brought up the issue of bivectors within the context of gravity waves. It is a real issue, for gravity waves are dyadic, while EM waves are vector. From a particle physics perspective this is why photons (as well as gluons and weak vector bosons) are spin = \hbar and gravitons are spin = 2\hbar. The issue is then how it is that the two sectors are intertwined.

Agreed that this forum is primarily for your GEM proposal. Things have sort of bled over a bit. I am not sure how strictly things are compartmentalized here. I have looked at Blackforest’s proposal, but have yet to dig into the meat there. The pdf files are a bit dense.

Cheers,

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

Let me clarify what is know about gravity waves. From observing binary pulsars, people have been able to infer that the lowest mode of emission is has a quadrupole moment. Rosen was the first person to write about the exponential metric, but his proposal is considered in error because it has a second metric metric field that could allow for a dipole moment in the wave emission. It is difficult to come up with an alternative proposal that adds fields yet has the quadrupole moment as the lowest mode of emission. The GEM proposal does not add fields, and I believe has the quadrupole, or as I like to think of it, wobbly-water balloon mode, as the lowest form of emission. I thought this issue was all about conservation of energy and momentum, not spin.

The reason that a gravity field must be spin 2 was spelled out for me in Brian Hatfield's introduction to "Feynman lectures on gravitation". If one wants like particles to attract, it takes particles of even spin: 0, 2, 4.... He eliminates the spin 0 graviton because gravity bends light. That leaves the spin 2 field as the simplest possibility.

**
The issue is then how it is that the two sectors are intertwined.
**
Unfortunately, I don't understand this issue yet, so I will explain how I see the relationship. There is the 4D wave equation. The gravitational and electric behavior of massless and massive particles can be characterized by this equation according to the GEM proposal. Focus only on the massless particles. These travel at the speed of light, which in a way is a constraint (I have forgotten how to discuss that constraint, something about polarity perhaps). A massless spin-1 field represents the photon, and has two of the four degrees of freedom of the wave equation. A massless spin-2 field represents the graviton, and has the other two degrees of freedom of the wave equation. Together the spin-1 and spin-2 fields completely describe long distance forces, one where like charges attract, the other where like charges repel.

Particles arise from field strength tensors. For the photon, that would be \nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}. For the graviton, the irreducible tensor \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu} is the source.

It might help to list all the things that I think are related in groups"

Gravity
like charged attract
graviton
spin-2
symm. tensor

EM
like charges repel
photon
spin-1
anti-symm. tensor

It still is necessary to explain what is going on on a group theory level. If that is what you mean by intertwine - clarify the group theory story - I remain guilty for a little while longer.

Researchers need the ability to wander, which is why I made an effort to explain Blackforest's unusual word choice, not complain about the digression. I too printed out three of the PDFs in his thread, but have not made a comment because I was not able to grasp why EM which works so great in 4D should be represented by 3D equations. To prevent a digression along the E question lines in this thread, I'll post something over there.

doug

[Blackforest]

Doug,
Actually I brought up the issue of bivectors within the context of gravity waves. It is a real issue, for gravity waves are dyadic, while EM waves are vector. From a particle physics perspective this is why photons (as well as gluons and weak vector bosons) are spin = \hbar and gravitons are spin = 2\hbar. The issue is then how it is that the two sectors are intertwined.
Agreed that this forum is primarily for your GEM proposal. Things have sort of bled over a bit. I am not sure how strictly things are compartmentalized here. I have looked at Blackforest’s proposal, but have yet to dig into the meat there. The pdf files are a bit dense.
Cheers,
Lawrence B. Crowell
Thanks for the patience. Parasite was absolutely not negative but exactly what sweeter has said.
Coming to your point: "The issue is then how it is that the two sectors are intertwined."
In fact, if you read my homepage and not only the work proposed on this subforum, you will discover another alternative to this question and I am myself astonished. If you construct (as I tried) a GR theory with the moving frame method but without neglecting terms of the second order, only trying to incorporate them via the introduction of the ad hoc mathematical terms, then you find very funny things concerning the first partial derivates of the basis vector: they are isotropic vectors, they can generate symplectic forms, they can generate matrices with a formalism analogue to the formalism of the Maxwell EM field tensor 4D... Because I am not owning all appropriate tools to develop rapidly my theory, I stay in front of these results, perceiving roughly a fundamental relation between geometry and EM fields.... For me, in this very special approach, it's like if variations of the geometric structure would generate EM fields in the tangent space of each event...
In principle The GEM is not far from this discussion. The difference is that I try to developp a more general approach (not reduced to the formulation of a Lagrangian). The problem is that I don't have the necessary mathematical tools to do that. That's why I need your helps. If I have the good ticket !
Best Regards

[Lawrence B. Crowell]

The reason that gravity waves are quadrupole is because a dipole gravitational moment is P~=~mx, and so by conservation of momentum dP/d\tau~=~constant. This makes a further point in illustrating the differences between gravitation and EM.

cheers,

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

That statement looks accurate about GR. There are other theories for gravity that predict dipole moments because there are other fields that can contain the momentum, so the field for gravity can trade momentum with the other field, leaving dP/d\tau~=~constant.

Would it be correct to say that the difference between gravitation and EM in this context is equivalent to saying there is one sign for mass, but two for EM? Momentum is conserved for an isolated electric dipole too.

doug

[sweetser]

Hello Lawrence:

This is a valid request:

**
Either the graviton and photon sectors, the abuse with the term graviton with standing, need to be clearly indicated in some way.
**
(To be honest, I am not familiar with what the word "sectors" means in this context, but I'll accept the task).

Before posting here, I did not have any position about how my proposal worked with group theory, although I wish I had one! Now I have a position.

EM arises due to U(1) symmetry of the action. Gravity is accounted for by the Diff(M) symmetry of the Hilbert action. I consider these two groups, U(1) and Diff(M), to be indispensable groups if one wishes to characterize EM and gravity respectively.

The GEM proposal looks not to replace these groups, but rather to find a relationship between them, how the two groups relate to each other. I propose the Diff(M) symmetry exceptionally weakly breaks U(1) symmetry. For an electron, the mass charge associated with Diff(M) symmetry (\sqrt{G}m_{e}) is 16 orders of magnitude smaller than the electric charge associated with U(1) symmetry. At this point I have no explanation why there is such a large difference.

Contrast this to our best effort to unify gravity and EM, the Einstein-Maxwell equation. That action has Diff(M) symmetry for gravity, U(1) symmetry for EM, and the Higgs mechanism to introduce mass into the Lagrangian while preserving U(1) symmetry. The GEM proposal eliminates the need for the Higgs boson. There is a need for a scalar field to couple with all particles with mass, but that is simply the trace of the asymmetric tensor, or tr(g_{\mu \nu}\nabla^{\mu}A^{\nu}). Another big problem with the Higgs mechanism is there is no obvious connection to the graviton. To my eyes, the symmetric second rank tensor \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu} can do the work of the graviton for gravitational mass, while its trace does the work of the scalar boson need for inertial mass.

doug

[Lawrence B. Crowell]

I am not going to comment much on alternate theories of gravitation. Dicke proposed an alternative as a benchmark for testing deviations from expected results by GR.
//
As for group theory, gravitation is the group SO(3,~1)~=~Z_2\times SL(2,~C). The group SL(2,~C) may in turn be written as

SL(2,~C)~=~SU(1,~1)\times SU(2).

The algebra for this is then su(2) given by the standard Pauli matrices \sigma_{\pm},~\sigma_3 and su(1,~1) has the elements \sigma_{\pm},~\tau_3~=~i\sigma_3. The latter change gives the pseudoEuclidean nature to spacetime.
//
Now consider a connection one-form

A~=~A^+\sigma_+~+~A^3\sigma_3

and a gauge transformation determined by the group action of g~\in~{\cal G}, g~=~e^{i\lambda\sigma_3}. The gauge transformed connection is then

A^\prime~=~g^{-1}Ag~+~g^{-1}dg~=~e^{-2\lambda}A^+\sigma_+~+~A^3\sigma_3,

where d\lambda~=~A^3. Thus \lambda is a parameterization of the gauge orbit for this connection. This leads to the observation

\lim_{\lambda\rightarrow\infty}A(\lambda)~\rightar row~A^3\sigma_3,

where A^+\sigma_+~+~A^3\sigma_3 and A^3\sigma_3 have distinct holonomy groups and thus represent distinct points in the moduli space \cal M. However by the last equation we must have

F_\mu(A^+\sigma_+~+~A^3\sigma_3)~=~ F_\mu(A^3\sigma_3),

and similarly for any gauge invariant function. Hence there exist two distinct point in the moduli space that define the same set of gauge invariant functions. Hence there does not exist a measure over these two points that separates them, and \cal M is then nonHausdorff and has Zariski topology.
\\
So the moduli space for gravitation is nonHausdorff, due to the hyperbolic nature of its group structure. This is also tied into why it is that gravity only has one “charge,” for with EM the two charges as roots are related by a simple rotation in the argand plane, but in gravity the two charges are not so connected. Thus for physics only one root is chosen, that for positive mass, while the negative root is not considered in the theory. Besides negative mass-energy creates all types of problems with GR, such as closed timelike curves and require quantum field sources that are not bounded below. Thus such solutions, wormholes and warp drives, are suspected of being pathological.
//
I tried to start a thread here, but it seems to have vanished into informationn to entropy land. I am not sure what I am doing wrong here.

Cheers,

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

It looks like we are talking past each other. Although I have the Diff(M) symmetry group in the title of two of my recent posts, you did not mention the group in your reply. Likewise, I have not really worked with any of the groups you have suggested. I understand why this can happen: technical discussion tend to stay near the technical topics one is most familiar with. Yet I always challenge myself: what is it I am missing or getting wrong? It is fun for me to make information move in my head.

So I decided to get out all my gravity books this weekend, and read up on Diff(M). I've got MTW, Wald, Feynman, Ohanian & Ruffini, and a few books on quantum field theory and groups. I went to the index, and none of them listed Diff(M). This completely takes you off the hook! (OK, you never were really on a hook, you gave me an honest reply, so this is a nonjudgmental observation). Then I got kind of worried: where is my documentation on this group if it is not featured in these books?

It was on the web at a URL I had posted before. This time I will quote the relevant section:

Now I should point out that for each of the symmetry groups I've listed above, people have worked out the "representations" of these groups. A representation of a group is a way to think of its elements as operators, and this is what we need to understand symmetries in quantum physics. The representation theory of the Poincar´ group dominates relativistic physics, while the representation theory of the Galilei group dominates nonrelativistic physics. I encourage everyone to learn the derivation of Schrödinger's equation straight from the representation theory of the Galilei group! It's cool. (I think it appears in the books by Mackey and Jauch listed here.)

In any event, we can ask for still more symmetry than conformal symmetry. We can ask for symmetry under all smooth coordinate transformations! The first to demand this effectively was Einstein, who got his wishes when he devised general relativity as a theory of gravity. So gravity is the most symmetrical of field theories so far - it's "generally covariant", or invariant under the group Diff(M) all diffeomorphisms of a spacetime M! One can also come up with a (classical) theory of gravity coupled to Yang-Mills fields, or whatever fields you like.

http://math.ucr.edu/home/baez/symmetries.html
There are a couple of really neat things about this quote. There are quite a good number of things in your posts that I do not understand. That indicates that I am not well-educated in quantum physics, otherwise I would know what Zariski topology is. A man has to know his limitations, and one of mine is an in-depth knowledge of quantum theory.

My favorite line: "So gravity is the most symmetrical of field theories so far - it's "generally covariant", or invariant under the group Diff(M) all diffeomorphisms of a spacetime M!" This is the kind of property I think I see in the GEM proposal. Starting from the flat Minkowski metric, it is a smooth transformation out to the exponential metric. That metric is NOT a solution to the Einstein field equations. It makes predictions which are consistent with current tests, and could be accepted or rejected on experimental grounds alone.

And the final line is also very relevant: "One can also come up with a (classical) theory of gravity coupled to Yang-Mills fields, or whatever fields you like." You have gone through the effort to list some of the possibilities. I do not have an issue with your proposal. Instead it is the question I don't get along with. It presupposes the structure of the answer: you have gravity, and it couples politely with other stuff. In the GEM proposal, gravity breaks U(1), SU(2), and SU(3). It does not break it much, in fact it is so darn weak those symmetries appear to our ability to measure, perfect. I probably need to devise a more technical way of saying that, but that is the thrust of the non-debate.

doug

[Lawrence B. Crowell]

Hello Doug,

I did mention the group, that group is diff({cal M})~=~SO(3,~1) . This is the analogue of the gauge group for gravitation. Gauge theory rests upon the idea that the particular gauge choice used is unimportant. The elementary example in electrodynamics is with the magnetic field B~=~\nabla A , and so any change in the vector potential A~\rightarrow~A~+~\nabla\chi does not change the magnetic field by the “curl-div = 0” rule. This can be generalized to higher dimensions, and in one my posts on this discussions I do describe gauge theory in spacetime or {\cal M}^4 .
//
This construction leads to the idea of a moduli space. Given a gauge potential its explicit construction has no basis on the fields. Let \Omega^{1}(ad~g) be the set of one-forms on a manifold which are transformed by the action of a group g . The coboundary operator D then determines these elements by its operation on zero-forms (functions) with

\Omega^0(ad g)~^D\rightarrow~\Omega^1(ad~g).

Thus a gauge connection may be determined as a “pure gauge” potential for

A^\prime~=~gAg^{-1}~+~gd(g^{-1})

with A~=~0 , and the group elements define the one-forms of \Omega^0(ad~g) . Now the set of one-forms \Omega^1(ad~g) are “all over the gauge space map.” In other words nothing is specified about their gauge. However, within this space the moduli may be defined. The following set

{\cal S}_D~=~\{A~\in~\Omega^1(ad~g): *D*A~=~0\}
\}

Here * is the Hodge star operator which takes a p-form on an n-dimensional space p~<~n to an n-p-form. This uses forms of the Levi-Civita antisymmetric symbols. Also the condition *D*A~=~0 may be other conditions --- the Coulomb gauge in 3-d or the Feynman gauge or ‘t Hooft-Veltman gauge or ... . This defines the slice {\cal S}_D through the bundle, or equivalently the tangent space over the base manifold.
\\
Now to round this out the fields are given by the two forms in the set \Omega^2(ad~g) under the map

\Omega^1(ad~g)~^{P_\pm D}\rightarrow~\Omega^2(ad~g)

where the P_\pm is projector operator that send the two-forms to self-dual or anti self dual forms. Now the set of gauge equivalent forms {\cal S}_D is given by the pullback on \Omega^2(ad~g) , and since these two-forms are independent of the gauge choice it is then the case that the slice {\cal S}_D is an element of a space {\cal M}_{mod}~=~A/g , or the set of points “modulo group actions.” So this is the meaning of diff({\cal M})
\\
In the case of gravitation the group action is the Lorentz group, on in general the Poincare group with 6 generators --- 3 boosts plus 3 rotations (or angular momenta). Globally these are the transformation of special relativity, which are described by the orthogonal group SO(3,~1) . In a Euclidean format the group is SO(4)~\sim~SU(2)\times SU(2) , and SO(3,~1) reflects the signature of spacetime. I could go on about how SO(4) and SO(3,~1), as the Euclidean representation and pseudoEuclidean representations, but I’ll do that later. This might have some relevancy for the thread here on Euclideanized spacetime, but I have not yet had time to comment on that. However, in general relativity they are defined for local inertial frames, and how these local regions patch together gives the connection structure and curvatures of general relativity. A set of connections under a coordinate condition define the moduli and the curvatures of GR are \Omega^2(ad~g). Since the moduli space for GR has this pseudoEuclidean structure a particular moduli as {\cal S}_D, a point in {\cal M}_{mod}, is not separable from another point. Hence the moduli space is nonHausdorff, in the parlance of point-set topology (compact, paracompact and all that jazz). The Hausdorf condition on a space says that for any two points in that space sufficiently small neighborhoods may be found around these points that do no overlap. This however does not appear to obtain for the moduli space of general relativity. Hence the solution space (a’la the Frobenius condition) is difficult to understand. However, there are some bright lights here, for this means that apparently independent solutions are not that separate. Pp-waves (Petrov type N solutions) have structure which are related to Black holes (type D solutions) and this Zariski topology may illuminate how solutions in GR transform between each other.

Sincerely,

Lawrence B. Crowell

[Lawrence B. Crowell]

Hello Doug,

I did mention the group, that group is diff({cal M})~=~SO(3,~1) . This is the analogue of the gauge group for gravitation. Gauge theory rests upon the idea that the particular gauge choice used is unimportant. The elementary example in electrodynamics is with the magnetic field B~=~\nabla A , and so any change in the vector potential A~\rightarrow~A~+~\nabla\chi does not change the magnetic field by the “curl-div = 0” rule. This can be generalized to higher dimensions, and in one my posts on this discussions I do describe gauge theory in spacetime or {\cal M}^4 .
//
This construction leads to the idea of a moduli space. Given a gauge potential its explicit construction has no basis on the fields. Let \Omega^{1}(ad~g) be the set of one-forms on a manifold which are transformed by the action of a group g . The coboundary operator D then determines these elements by its operation on zero-forms (functions) with

\Omega^0(ad g)~^D\rightarrow~\Omega^1(ad~g).

Thus a gauge connection may be determined as a “pure gauge” potential for
[/tex]
A^\prime~=~gAg^{-1}~+~gd(g^{-1})
[/tex]
with A~=~0 , and the group elements define the one-forms of \Omega^0(ad~g) . Now the set of one-forms \Omega^1(ad~g) are “all over the gauge space map.” In other words nothing is specified about their gauge. However, within this space the moduli may be defined. The following set

{\cal S}_D~=~\{A~\in~\Omega^1(ad~g): *D*A~=~0\}
\}

Here * is the Hodge star operator which takes a p-form on an n-dimensional space p~<~n to an n-p-form. This uses forms of the Levi-Civita antisymmetric symbols. Also the condition *D*A~=~0 may be other conditions --- the Coulomb gauge in 3-d or the Feynman gauge or ‘t Hooft-Veltman gauge or ... . This defines the slice {\cal S}_D through the bundle, or equivalently the tangent space over the base manifold.
\\
Now to round this out the fields are given by the two forms in the set \Omega^2(ad~g) under the map

\Omega^1(ad~g)~^{P_\pm D}\rightarrow~\Omega^2(ad~g)

where the P_\pm is projector operator that send the two-forms to self-dual or anti self dual forms. Now the set of gauge equivalent forms {\cal S}_D is given by the pullback on \Omega^2(ad~g) , and since these two-forms are independent of the gauge choice it is then the case that the slice {\cal S}_D is an element of a space {\cal M}_{mod}~=~A/g , or the set of points “modulo group actions.” So this is the meaning of diff({\cal M})
\\
In the case of gravitation the group action is the Lorentz group, on in general the Poincare group with 6 generators --- 3 boosts plus 3 rotations (or angular momenta). Globally these are the transformation of special relativity, which are described by the orthogonal group SO(3,~1) . In a Euclidean format the group is SO(4)~\sim~SU(2)\times SU(2) , and SO(3,~1) reflects the signature of spacetime. I could go on about how SO(4) and SO(3,~1), as the Euclidean representation and pseudoEuclidean representations, but I’ll do that later. This might have some relevancy for the thread here on Euclideanized spacetime, but I have not yet had time to comment on that. However, in general relativity they are defined for local inertial frames, and how these local regions patch together gives the connection structure and curvatures of general relativity. A set of connections under a coordinate condition define the moduli and the curvatures of GR are \Omega^2(ad~g). Since the moduli space for GR has this pseudoEuclidean structure a particular moduli as {\cal S}_D, a point in {\cal M}_{mod}, is not separable from another point. Hence the moduli space is nonHausdorff, in the parlance of point-set topology (compact, paracompact and all that jazz). The Hausdorf condition on a space says that for any two points in that space sufficiently small neighborhoods may be found around these points that do no overlap. This however does not appear to obtain for the moduli space of general relativity. Hence the solution space (a’la the Frobenius condition) is difficult to understand. However, there are some bright lights here, for this means that apparently independent solutions are not that separate. Pp-waves (Petrov type N solutions) have structure which are related to Black holes (type D solutions) and this Zariski topology may illuminate how solutions in GR transform between each other.

Sincerely,

Lawrence B. Crowell

[Don J]

Considering the 60 posts limited discussion on this board
If you are interested to try the ultimate test about your theory I invite you to joint another board where you can have all the room needed for discussion with hard Einstein relativistic theorician defenders.They can check in details the maths using by your theory applied to the bending of light by the sun for example.Or the precession of the orbit of Mercury.

You are very confident about your theory than a little challenge must only be "good".


Example of an actual discussion which is related in part about the Schwarzschild metric
"Celestial Mechanic wrote"
http://www.bautforum.com

Just a message of WARNING .I feel very sorry to have linked to BAUT forum.
I realise now after verification than that board called Against the Mainstream dont really allow for serious discussion about allternative concepts or researchs.
In fact every posters with new concept are litterally attacked and ridiculised by a bunch of as they called themsefve "debunkers".
Using a very disloyal technique describe here
See section
HOW TO DEBUNK JUST ABOUT ANYTHING
http://www.tcm.phy.cam.ac.uk/~bdj10/scepticism/drasin.html

Dont post to BAUT forum. :surprised :tongue2:

Edit to add example of debating technique used by BAUT forum members describe in the above link .HOW TO DEBUNK JUST ABOUT ANYTHING

"Portray science not as an open-ended process of discovery but as a holy war against unruly hordes of quackery- worshipping infidels. Since in war the ends justify the means, you may fudge, stretch or violate the scientific method, or even omit it entirely, in the name of defending the scientific method."

[sweetser]

Hello:

Another weekend, another nice technical refinement. Discussions with Lawrence had made me confront the issue of group theory for the GEM proposal. I have realized that Diff(M) symmetry breaks U(1) symmetry slightly. A problem with this statement is that it is not precise enough in defining how U(1) symmetry is broken. In this note I will clarify this issue.

Let's start with an example of symmetry breaking. The symmetry of the standard model for massless particles is U(1)\timesSU(2)\timesSU(3). Most particles in the standard model are not massless. The Higgs mechanism is able to introduce mass for fundamental particles such as quarks and electrons through a false vacuum. A scalar field made up of Higgs bosons does the trick. Several billion dollars are being spent by the global physics community to detect the Higgs, which is thought to have a mass between 120 and 200 proton masses (see the July 2005 Scientific American for a gentle introduction to this topic). This is the proverbial Mexican hat: it is easy to visualize how the mass arises from particles going to the low lands of the hat.

There are three players in the standard model:

U(1) The unit complex numbers
SU(2) The unit quaternions
SU(3) A special (ie norm of 1) group with 8 elements

One thing is shared by these three groups: one. Let's form complex-valued vectors that would be elements of each group:

U^{\mu}/|U| \in U ( 1 )
V^{\mu a} / |V| \in SU(2), \mathrm{a goes from 1 - 3}
W^{\mu b} / |W| \in SU(3), \mathrm{b goes from 1 - 8}

Calculate the norms of the three vectors in flat spacetime.

g_{\mu \nu} ( U^{\mu} )^{\ast} U^{\nu} / |U|^2 = 1
g_{\mu \nu} ( V^{\mu a} )^{\ast} V^{a \nu} / |V|^2 = 1
g_{\mu \nu} ( W^{\mu b} )^{\ast} W^{b \nu} / |W|^2 = 1

[NOTE: this is technically in error, as this is the interval, not the norm. Please see the next post, but I will leave this mistake on the public record]

Now we want to break symmetry a bit using the group Diff(M). That effects the metric, and the metric only:

g'_{\mu \nu} ( U^{\mu} )^{\ast} U^{\nu} / |U|^2 = 1 + \Delta
g'_{\mu \nu} ( V^{\mu a} )^{\ast} V^{a \nu} / |V|^2 = 1 + \Delta
g'_{\mu \nu} ( W^{\mu b} )^{\ast} W^{b \nu} / |W|^2 = 1 + \Delta

[This WILL equal whatever it was with g, a basic property of norms]

The vectors for EM, the weak, and the strong forces have not been changed. The metric has changed, and by the equivalence principle, the only thing that can do that is mass. The vectors with this metric are no longer in the corresponding groups because the norm is different from one. They will form a group, one where the norm is [itex]1 + \Delta[/tex]. Essential the unit circle or spheres change there size a little bit. That is how symmetry is broken for the GEM proposal.

On ontological grounds (a fancy way to say "the reason why things work"), I like the visual of the groups that are a bit different from one. It is easy to see circles and spheres getting bigger or smaller. Now that I have an alternative that can be experimentally tested, I can give up the false vacuum of the Higgs mechanism as I have other false gods.

doug

[sweetser]

Oops, my bad, I am doing my notation wrong:
g_{\mu \nu} ( U^{\mu} )^{\ast} U^{\nu} / |U|^2
will calculate the Lorentz invariant interval. The norm is not the interval. Essentially, instead of t^2 - R^2, I want t^2 + R^2. Right now I am confused about what symbols to use to get to the norm, and how calculating a norm is connected to the metric.

doug

[Terry Giblin]

Are you familiar with Theordor Kaluza paper on combining GR and EM?

Does your work confirm or use the same ideas?

Regards

Terry Giblin

[sweetser]

Hello Terry:

I guess you are refering to what is known as the Kaluza-Klein approach. In some ways that is a basis for work on string theory. If that is your question, the GEM proposal is different. Kaluza-Klein is a 5 dimensional theory; GEM is 4 dimensional. As far as I am aware (and I do not know the literature in any depth), the predictions for light bending around the Sun are identical for Kaluza-Klein compared with GR, but the GEM proposal will be different at second rank PPN accuracy (not that anyone is planning on getting to that level of accuracy). I cannot make an accurate statement about how Kaluza-Klein works in terms of group theory, but I doubt it has anything to do with breaking U(1) symmetry by changing the norm.

There appear to me fundamental differences between the approaches.
doug

[Lawrence B. Crowell]

I am not sure about the reply by Don J.
//
What I wrote about moduli space is basically "bioler plate" stuff on this, dating back to the work of Atiyah and Singer back in the early 1980s. This leads to the stunning results of Uhlenbeck and Donaldson. In the first case singularities on the base manifold may be lifted to the moduli space in projective spaces. In the second it turns out that there exists an uncountably infinite number of 4-dim spaces, where only a measure epsilon of them are diffeomorphic, and most are homeomorphic but not diff - able.
//
My pet approach to what lies beyond standard theoretical approaches is with nonassociative observables. QM and GR have noncommutative structures [x, p] = hbar and commutators in curvatures. The extension beyond this is not to my mind likely to lead to symmetric field theoretic structures, but with nonassociative ones where
//
(ab)c - a(bc) =/=0.
//
I advise taking a look at my book "Quantum Fluctuations of Spacetime," World Scientific (2005) to see how this works. Essentially I have found that quantum mechanics and general relativity are equivalent as categorical structures of observables over Galois fields. Then GR and QM are quaternionic pairs that exist in a system of octonions. The Galois code for this is then developed for the 24-cell and ..., well I'll leave it to you to look this up.
//
At any rate, I strongly doubt that field theoretic structures are going to be symmetric, except for those under supersymmetry transformations where commutators "go to" anti-commutators.
//
cheers,
//
Lawrence B. Crowell

[sweetser]

Hello Lawrence:

I formulated my proposal for this forum using standard tensor notation throughout. That said, I am pretty good with quaternions, it was the algebra I used to figure this stuff out in the first place. I own the domain name quaternions.com which features ways for doing physics with quaternions, from classical physics, to special relativity with local transformations instead of the global Lorentz group, EM, quantum mechanics, and my efforts to unify gravity and EM. Rewriting fundamental laws with quaternions is good training for theoretical physics.

Octonions are not necessary if you want a non-associative quaternion product. I have defined what I call the "Euclidean product" as being the standard Hamiltonian product with a conjugate operator tossed in:

Hamilton product:
(a,B)(c,D)=(ac - B.D,aD+Bc+B\times D)
Euclidean product:
(a,B)*(c,D)=(ac + B.D,aD-Bc-B\times D)
John Baez pointed out to me that:
(AB)*C - A*(BC) =/= 0
See, no octonians are needed.
At any rate, I strongly doubt that field theoretic structures are going to be symmetric, except for those under supersymmetry transformations where commutators "go to" anti-commutators.
I can understand the source of the feeling: none are used now, so that pattern should continue. For me, there is no reasonable justification for working the the deviation of the average amount of change, but completely ignoring the average amount of change itself. At least the differences are clear.

doug

[sweetser]

Hello:

Let's start with things researchers agree on:

1. A spin 0 quantum field is necessary to give fundamental particles like electrons and quarks inertial mass.

2. A spin 2 quantum field is required to mediate the effects of gravity.

3. The equivalence principle is a classical doctrine confirmed by experiments that the inertial mass is equal to the gravitational mass.

4. The correspondence principle shows that classical laws arise from the aggregate of quantum events.

In my limited exposure, folks who are searching for the Higgs particle do not bring up gravitons. Likewise, the gravity wave detector crowd appears unconcerned with the Higgs. Yet the equivalence principle must arise from an absolutely unwavering link between the quantum source for rests mass, a scalar field, and the quantum source for gravitational mass, a spin 2 field. The two fields can never "get out of step" with each other, or the equivalence principle will be violated, which it is not.

In the GEM proposal, the link between the scalar field and the spin-2 field is direct and rock solid: the spin 2 field for gravitational mass is the symmetric field strength tensor \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}, and the spin 0 field for inertial mass is the trace of the same symmetric tensor, tr(g_{\mu \nu}(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})). Einstein was the first to argue the elegant case for the classical equivalence of gravitational and inertial mass. I am pleased the GEM proposal sings the same song on the quantum level.

doug

[Lawrence B. Crowell]

The nonassociative product

(e_ie_j)e_k~-~e_i(e_je_k)~=~\upsilon_{ijkl}e_l

can be defined where \upsilon_{ijkl} is defined without reference to octonions. However, this system will not satisfy a closed algebra unless it defines the Cayley four-form. One can of course restrict things to this rather myopic view.
\\
I am not sure what you mean by the Hamilton product. It appears that you are referencing a Jordan product in some form. The Hamilton product to me is

i^2~=~j^2~=~k^2~=~ijk~=~-1.

and a general quaternion is defined by

H~=~a\sigma_x~+~b\sigma_y~+~c\sigma_z~+~d{\bf 1}

In four dimensions the Pauli matrices are generalized to Dirac matrices.
\\
At any rate, nonassociativity without reference to the octonions means that there is no closed multiplication table defined by the structure constant \upsilon{ijkl}, of which there are 480 possible representations. I could go into the cyclic group and PSL(2,~7) Hurzewitz discrete group theory behind this, but that gets too far into maths.
\\
Gauge theory has some sort of algebraic structure, eg. SU(n), and the like. One of course defines gauge connections according to representations of this group. However, for there curvatures on the principle bundle are found by F~=~dA~+~A\wedge A, where the tensor components of the two-form F in the case of an abelian field (EM) is the antisymmetric field strength tensor containing the electric and magnetic field components. The antisymmetry is imposed not by any group structure, but rather from a much more basic theory of chains and cycles on any manifold.
\\
Physics was not built from the ground up, but rather from a domain of experience that existed at the time. Faraday’s description of changing magnetic fields, currents and the like were based on observational experience of the time. Of course now we can easily see that this all stems from F~=~dA, and things appear elementary. Neither Faraday nor Maxwell had any notion of differential forms or integrating on chains in n-dimensional manifolds. Maxwell indeed had things according to spinning ether vortices and the like as his “description.” In part because of this his original theory was actually quaternionic. Now in our modern age we can see that these types of theories have antisymmetric structure because of these basic “facts,” and hold for any gauge theory --- even gravity.
\\
Lawrence B. Crowell

[sweetser]

Hello Lawrence:

This is the definition for a Hamilton quaternion product I am using:

i^2~=~j^2~=~k^2~=~ijk~=~-1.

While this is true:
The nonassociative product.

(e_ie_j)e_k~-~e_i(e_je_k)~=~\upsilon_{ijkl}e_l

can be defined where \upsilon_{ijkl} is defined without reference to octonions.

it is not relevant because I am using a conjugate operator, or more technically an anti-involutive automorphism. The conjugate flips the sign of the 3-vector.

What I call a Euclidean product has a closed multiplication table:
i^{*} i= j^{*} j = k^{*} k= (i^{*} j)^{*} k = 1
The standard definition of a non-associative product you provide does not use a conjugate operator as I do. Because the automorphism keeps the quaternions within the quaternion manifold, the multiplication table is closed: i^{*} i = 1. Let's see this in action, how always taking the conjugate of the first quaternion in the binary quaternion multiplication makes the triple product non-associative.
a (b c) = a^{*} (b^{*} c) = a^{*} b^{*} c
(a b) c = (a^{*} b)^{*} c = a b^{*} c
These are not the same. The norm of these two are same, but they point in different directions.

Our current gauge theories for gravity and EM are both spectacular and inadequate. Both match experimental data extraordinarily well. Yet we cannot quantize EM without choosing a gauge. Even if a gauge is chosen, gravity cannot be quantized.

Let me reestablish how I am working from the "domain of experience". Gravity and EM as currently formulated work in 4 dimensions, as does GEM. It is odd that simple statement is at odds with the majority of work on gravity today. The group Diff(M) was used by Einstein to make his equations covariant, and it plays a role in GEM. EM symmetry must be broken by a scalar field, as it is in GEM by the trace of the field strength tensor. The 4D EM wave equation can be quantized by the Gupta-Bleuler method, but two of the modes of emission must be eliminated by a supplementary condition. GEM uses the same creation and annihilation operators, but it also has a spin 2 field that will not have the same technical issues the spin 1 scalar mode has. And in keeping with the most important of all scientific traditions, should we refine the level of accuracy of measuring light around the Sun, GEM predicts 0.7 microacrsecond more than GR, so unlike all the work in string theory, I am willing to put every chip I own on the line and let the data decide.

How things get written matters. You write the field strength tensor as F = dA. Nothing can be simpler than that. Sure, it has all the differential forms stuff behind it, but since that is the starting point, I can appreciate that there appears nothing simpler. I write it differently, also in an established way, as a F=\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}. In my notation, I can explain even to my girlfriend that the F thingie is not a complete story because of the act of subtraction: there are 4 parts in the triangle, 4 parts in the A, sixteen parts total, but the study of light needs only 6 of the 16. What I cannot explain to her is why the brightest folks on the planet do not share my feelings of obligation to work with the other 10 terms.

doug


Historical note: Maxwell was familiar with quaternions because Hamilton trumpeted their cause (too loudly). Scalar, vector, div, grad, curl were inventions of Hamilton in his quest to understand quaternions. In Maxwell's first edition of his treatise, he used "pure quaternions" which is one with a zero for the scalar, effectively as a stand-in for 3-vectors. He did nothing at all fancy with quaternion algebra. By the third edition, he only hoped that quaternions would play a role in describing EM. Many folks have figured out how to toss in an extra imaginary number and get the Maxwell equations. Peter Jack was the first person to do so with real quaternion (1996?) and I did the same independently a year later. There is a rumor on the Internet that some 200 quaternion equations were deleted from the first to the third edition. From my viewing of an on-line version of the first treatise, that is an inaccurate description: the quaternions were used like any 3-vector, and he did not write the homogeneous or source equations as an exercise in quaternion algebra.

[Lawrence B. Crowell]

Below is a multiplication table of octonions that I devised for fermions. It was derived with respect to black hole physics and Bogoliubov transformations --- but that is another story. These fields are the operators for ingoing and outgoing modes into a black hole. These fields include their conjugations. I hope this shows up right!
\\

\left(\begin({\array}{}* & * & e_1 & e_2 & b_k & e_4 & b_k^\dagger & b_{-k} &
b_{-k}^\dagger \\ *& * & *& * & * & * & * & * & *\\ e_1 & &-1 & e_4 &
b_{-k}^\dagger & -e_2 & b_{-k} & -b_k^\dagger & -b_k \\ e_2 & &
-e_4 & -1 & b_k^\dagger & e_1 & -b_k & b_{-k}^\dagger & b_{-k} \\
b_k & & -b_{-k}^\dagger & -b_k^\dagger & 0 & b_{-k} & e_2 &-e_4 &
e_1 \\ e_4 & & e_2 & -e_1 & -b_{-k} & -1 & b_{-k}^\dagger & b_k &
-b_k^\dagger \\ b_k^\dagger & & -b_{-k} & b_k & -e_2 &
-b_{-k}^\dagger & 0 & e_1 & e_4 \\ b_{-k} & & b_k^\dagger &
-b_{-k}^\dagger & e_4 & -b_k & -e_1 & 0 & e_2\\ b_{-k}^\dagger &
& b_k & b_{-k} & -e_1 & b_k^\dagger & -e_4 & -e_2 &
0}\end{array}\right)


\\
You say that (ab)c and a(bc) have the same norm but differ in their direction. That is standard with octionions. You might notice this if you examine this table.
\\
A couple of other points. A choice of gauge amounts to lifting the base manifold up the fibre. Think of the base manifold as a sheet of paper. Then perpendicular to each point is a line coming up, which for the case of triviality defines the tensor product of the vector space in the line with the manifold. Now to compute the action of this internal vector space one needs to move from fibre to fibre. But how does one do this? One takes the base manifold (the sheet of paper) and lifts a copy of it so it cuts through the fibres. One can lift the copy paper any way one wants, but once done things are computed with that choice. This is what a gauge choice is. One needs to define a bundle section in some way. It does not matter how this is done, but it must be done and used consistently. Well, 't Hoft and Veltmann did a gauge choice change in mid-stream with QCD to get renormalizability, but one has to be very careful about that! At any rate a gauge choice is nothing more than a way of fixing a frame, akin to defining a reference frame in relativity (or Newtonian mechanics for that matter) in order to do calculations.
\\
As for why only 6 of the possible terms end up as real is because the antisymmetry imposes a structure on the field. Anyway, we only have 3 E-fields and 3 B-fields. Imposing a structure on the theory involves the elimination of some of the possible tensor terms. The antisymmetry of gauge field 2-forms does that automatically.
\\
Lawrence B. Crowell

[Creator]

In the GEM proposal, the link between the scalar field and the spin-2 field is direct and rock solid: the spin 2 field for gravitational mass is the symmetric field strength tensor \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}, and the spin 0 field for inertial mass is the trace of the same symmetric tensor, tr(g_{\mu \nu}(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})).
doug

Hello Doug;
In QFT isn't mass obtained, at least some fraction of it, by radiative corrections? How does that relate here ?

[sweetser]

Hello Creator:

The short answer is I do not know how to incorporate radiative corrections into the proposal.

I do know that for compound particles such as protons and neutrons, most of the mass is from the movement of gluons and quarks. Any radiative corrections would be just another source of energy to add into the picture. I am unable to calculate what percentage of the proton mass comes from the Higgs mechanism. I'm sure some folks on the planet know how to do that calculation, but I do not. The ice is very thin where I am skating!

doug

[sweetser]

Hello Lawrence:

Unfortunately, your table for the Octonians did not work out :-( I would love to see it. If you come back to the forum, there should be an edit button at the bottom if you have logged in. The first two lines look something like this:

\left(\begin({\array}{}* & * & e_1 & e_2 & b_k & e_4 & b_k^\dagger & b_{-k} &
b_{-k}^\dagger \\ *& * & *& * & * & * & * & * & *\\

So the "*"'s need to be replaced with what goes in there. A second effort for this multiplication table would be appreciated. Note that after the save button is hit and the page appears, the new graphic does NOT appear. The equation is a graphic which the computer keeps in cache. One must hit the reload button while holding down the control key to see the improvements.

doug, whose LaTeX here is NEVER done perfectly the first time.

[Lawrence B. Crowell]

I have an inordinate difficulty with the LaTeX here, which I do not have otherwise. Anyway, I have taken a dvi page with this table and prduced an image file. I am going to try to attach that to this post.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

Thanks for the dvi page. I am a bit confused about one aspect of this table. Octonions are a non-associative, non-commutative division algebra. The zeros that appear in the table would create a problem for the division algebra. I think those should be -1's. For completeness, the identity should be included. Here is a table from the web without zeroes: http://www.geocities.com/zerodivisor/obasis.html

doug

[Lawrence B. Crowell]

The zeros are meant to convey fermionic content with b^2 = 0. The anticommutators of the Fermionic fields are "treated" as commutators in order to obtain associators

[a, b, c] = (ab)c - a(bc).

I suppose I should have considered the anti-associator

{a, b, c} = (ab)c + a(bc)

instead to avoid this matter. However, the basic import is that the xero reflect a topological content of fermions. Since b^2 = 0 this is equivalent to d^2 = 0, and that the fields are defined in ker(b)/im(b). This topological element is what skirts the problem with division algebras. I think this extends to the sedenions, but that is another issue.

Maybe this was not the best example for this, for in most cases there is none of this topological implication underlying it. I attach a more familiar octonion multiplication table.

[Lawrence B. Crowell]

This might be a bit outside of the GEM theory, but I figured I would try to clarify a couple of things.
\\
Let b_q~\rightarrow~\phi_q b_q for q~=~\pm k and \phi_q a scalar field that obeys

[\phi_q,~\phi_{q^\prime}^\dagger]~=~f(q,~q^\prime),


[\phi_q,~\phi_{q^\prime}]~=~g(q,~q^\prime)

Then the commutator of \phi b_q is

[\phi_q b_q~,\phi_{q^\prime}b_{q^\prime}]~=~[\phi_q,~\phi_{q^\prime}]b_q b_{q^\prime}


=~1/2/{b_q,~b_{q^\prime}/}g(q,~q^\prime)~=~e_i,

for the appropriate i on the table. In this way one can treat fermions as nonassociative
//
The division algebra is one that want e_ie_j = e_k =/=0. If e_k is zero without either e_i or e_j being zero then there is said to be no algebra. At the level of octonions it is said that this is the final algebra.
//
The whole process of construction from reals, complexes, quaterions and octonions involves a pairing of each other. The simplest of course is the complex plane where z~=~(x,~y) and the defined multiplication

c*z~=~(a,~b)*(x,~y)~=~(ax~-~by)~+~i(bx~+~ay)


(a,~b)*(x,~y)~=~(ax~-~by,~bx~+~ay)

The same goes for the quaternions, they are a pairing of complex numbers. Octonions are then in turn a pairing of quaternions. At each level one loses ordering, commutivity and finally associativity. This also reflects the so called Cayley numbers and the multiplication rule with pairing defines what is called the Cayley-Dickson algebra.
\\
The octonions are pairs of quaternions. Consider the octonion O~=~(A,~B) and O^\prime~=~(X,~Y). The multiplication of these two is then

O\cdot O^\prime~=~(AB~+~e^{i\phi} Z^\dagger B,~BX^\dagger~+~AP).

where the argument is usually taken as \phi~=~\pi/2. For a system of quaternions \sigma_i and \bf 1 this defines four additional elements e_i.
\\
Now for quaternionic valued operators as fields which satisfy the BRST quantization condition Q^2~=~0, where \psi~\in~ker(Q)/im(Q) gives the field as purely topological. In other words \psi~\ne~Q\chi. In this way it is possible to have the square of an element in the octonions being zero without it in a strict sense being unalgebraic. Of course in the octonions of operators e_je_j =/=0 for i =/=j.
\\
What is this good for? I might go on about this in greater detail in another post, but the Dirac operator \Gamma^a\partial_a has the same topological information as the field. Further, in general the Dirac matrices (the quaterions) may have a representation which depends upon the chart on the base manifold. Thus a more general Dirac operator is \partial_a\Gamma^a\_. In this way the quaternions are operators.
\\
In the case of sedenions one has e_ie_j~\ne~0 and algebra is lost. There are eight octonions within it that are "islands" of algebra, but "outside" of them appears to be algebraic "chaos." However, I think that by Bott periodicity there is structure there and I think it is involved with some sort of topology. The sedenions define S^7\times S^7\times G_2, which probably constrains this topology.
\\
Lawrence B. Crowell

[sweetser]

Hello:

I spend time wondering why I have trouble communicating this proposal. The action looks like the simplest one that can be made! I have fun finding tension between what we now know and the GEM proposal. In this post I will outline an example.

Why is energy conserved in EM? Take a look at the action:
S_{EM}=\int \sqrt{-g} d^{4} x (-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))
Vary time:
\delta S_{EM}=\int \sqrt{-g} d^{4} x (-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu})) \delta t
There is no time in the action, so it is not going to be changed in the slightest: integrate over a nanosecond or a century, the integral will be the same under a variation in time. That makes the action symmetric under time transformations. Where there is a symmetry, there is a conserved quantity. In this case, that is energy, E_{cons.}=m \frac{\partial t}{\partial \tau}. The definition of energy conservation says that there is no change in the derivative of time with respect to the interval \tau.

Why is linear momentum conserved in EM? Look at the action above. No distance R appears in it, so like time, varying R will keep the variation in the integral at an extremum, and the conserved quantity is momentum, P = m \frac{\partial R}{\partial \tau}. The same story.

We know what mass is from special relativity, the difference between the square of energy and the square of momentum. Since energy and momentum both arise from a symmetry in the action, mass which is calculated from this two must also arise exclusively as a symmetry in the action.

That is not how it works for the Hilbert action of general relativity. Instead of hunting for a symmetry, the metric is treated as a field and varied.

In the GEM proposal, one looks at the definition of a covariant derivative to spot a symmetry. One can vary changes in the potential or changes in the metric (the connection). So long as the changes offset each other, there would be no difference in the integral of the action with those changes. There must be a conserved quantity. Since the metric is changing, it is reasonable to propose mass is conserved. Logical consistency appears to favor the GEM proposal.

Convincing someone who has made the investment in understanding general relativity at the nuts and bolds level may be beyond my reach, but I still like the view from my chair: it's drop dead gorgeous.

doug

[Lawrence B. Crowell]

There is something rather mysterious here. You say there is no time in the action for the Lagrangian [/itex]{\cal L}~=~1/4 F_{ab}F^{ab}. Agreed there is no explicit function of time, but one does have elements F_{0j}~=~-\partial A^j/\partial t~-~\nabla_jA^0 which are the electric field components. A variation with respect to time is going to pick out \partial_0F^{0j} terms of the form \partial E/\partial t, and \partial_0F^{ij} of the form \partial B/\partial t. These are just the time derivative parts of the Maxwell equations. In fact your \delta S_{EM} is wrong, for the variation with time is not going to involve just a multiplication by \delta t[itex], but partial derivatives of this with time.
//
Lawrence B. Crowell

[Lawrence B. Crowell]

There is something rather mysterious here. You say there is no time in the the Lagrangian {\cal L}~=~1/4 F_{ab}F^{ab}. Agreed there is no explicit function of time, but one does have elements F_{0j}~=~-\partial A^j/\partial t~-~\nabla_jA^0 which are the electric field components. A variation with respect to time is going to pick out \partial_0F^{0j} terms of the form \partial E/\partial t, and \partial_0F^{ij} of the form \partial B/\partial t. These are just the time derivative parts of the Maxwell equations. In fact your \delta S_{EM} is wrong, for the variation with time is not going to involve just a multiplication by \delta t, but partial derivatives of this with time.
//
Lawrence B. Crowell

[sweetser]

Hello Lawrence:

There are partial derivatives with respect to time, but nothing that is just time. A delta t is not a t. One can imagine a Lagrangian where the effects dissapate after a certain amount of time. That is not what happens with the classical EM Lagrangian. If one comes back in 10 years, the delta t's will still be the same. That is the source of energy conservation as I see it.

I believe my S_{EM} is standard, although most people write it in flat spacetime.

Looks like your first itex bracket should not have a /.
doug

[Lawrence B. Crowell]

The variation of the action results in the Euler-Lagrange equations, which are the equations of motion. Conservation of energy for fields comes from the momentum-energy tensor

T^{ab}~=~\partial{\cal L}/\partial g_{ab}~-~g^{ab}{\cal L}

with the continuity condition \partial_cT^{ab}~=~0. The variation of the action gives dynamical equations, or F~=~ma stuff, and the momentum-energy tensor gives the conservation laws a'la Noether's theorem.
\\
Lawrence B. Crowell

[sweetser]

Hello Lawrence:

My equation about \delta S_{EM} above is wrong. Partial derivatives with respect to t are required.

I can see that my description of the issue was informal, but not unconventional. Here is a quote from http://en.wikipedia.org/wiki/Noether's_theorem:
The word "symmetry" in the previous paragraph really means the covariance of the form that a physical law takes with respect to a one-dimensional Lie group of transformations which satisfies certain technical criteria. The conservation law of a physical quantity is usually expressed as a continuity equation.

The most important examples of the theorem are the following:

* the energy is conserved iff the physical laws are invariant under time translations (if their form does not depend on time)
* the momentum is conserved iff the physical laws are invariant under spatial translations (if the laws do not depend on the position)
* the angular momentum is conserved iff the physical laws are invariant under rotations (if the laws do not care about the orientation); if only some rotations are allowed, only the corresponding components of the angular momentum vector are conserved

A Noether charge is a physical quantity conserved as an effect of a continuous symmetry of the underlying system.
The form of the GEM and EM Lagrangians do not depend on time, so energy is conserved.

Landau and Lif****z gave me a really great lesson on this topic. I knew that once the Lagrangian is set, then everything can flow from that, be it equations of motion or dynamical equations or stress tensors. But how are these related? Let's start with the GEM Lagrangian:

\mathcal{L}_{GEM}=\frac{-\rho_{m}}{\gamma}-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu}
-\frac{1}{4c^{2}}(\nabla^{\mu} A^{\nu}-\nabla^{\nu} A^{\mu})(\nabla_{\mu} A_{\nu}-\nabla_{\nu} A_{\mu}) - \frac{1}{4c^{2}}(\nabla^{\mu} A^{\nu}+\nabla^{\nu} A^{\mu})(\nabla_{\mu} A_{\nu}+\nabla_{\nu} A_{\mu})

Vary A and its derivative, keeping the velocity field fixed, and that generates the equations of motion, the stuff of the Maxwell equations. The second, third, and fourth terms come into play.

Now start with the same Lagrange density, but keep A and its derivative fixed, and vary the velocity V, which appears in the terms \frac{-\rho_{m}}{\gamma} and in the current densities J_{q}^{\mu}-J_{m}^{\mu}. That will generate the Lorentz force. One can also get to the Lorentz force through the stress tensor. Only the first and second terms come into play.

One important thing to notice is the role played by the second term, the so-called charge coupling term. In EM, the equations of motion indicated like charges repel, which is the same message as arises from the Lorentz force equation. By having the opposite sign for the gravitational charge in the second term, like charges attract for the field and force equations.

doug

[Chronos]

Pardon my dumb question sweetser, but is this a quantum treatment?

[sweetser]

Hello Chronos:

I hope to give a sophisticated answer to your dumb question. Bur first, the short answer: their are concrete technical reasons to hope the GEM field equations can be quantized, but due to my own limitations, I have not done any quantum calculations, such as a scattering cross section of an electron fired at a proton. A well-behaved, finite calculation of a scattering cross section would demonstrate the proposal is consistent with quantum mechanics.

How does one know if a theory can or cannot be quantized? There are fancy answers, but I prefer simple ones (but not too simple). Zeroes for observables of a classical theory are bad. In a classical field theory, an observable such as energy or momentum is a number. To make the same theory a quantum theory, an operator must be found for that observable that acts on the wave function. The operator gets plugged into a commutator that should be some multiple of Planck's constant. If a classical observable is zero, the quantum operator is zero, and the commutator will be zero, not a multiple of Planck's constant. Failure.

That is exactly what happens to the classical EM Lagrange density:
\mathcal{L}_{EM}=-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))
What is the energy/momentum 4-vector? One thing not appreciated widely enough is the answer to this question is automatic: all it takes is a calculation, no thought involved, a no-brainer. One takes the derivative of this with respect to the time derivative of the potential A, and out comes the canonical energy momentum. I don't know if you can buy a book that does this in detail. It is not a hard calculation, but it is painful to do the LaTeX. I like to see all the details of an easy calculation, so here it is. First, this is the EM Lagrangian written out in all its component parts:
\mathcal{L}_{EM}=
- \frac{1}{2} ( ( - ( \frac{\partial \phi}{\partial x} )^2 - (
\frac{\partial \phi}{\partial y} )^2 - ( \frac{\partial \phi}{\partial z} )^2
\\
- ( \frac{\partial A_x}{c \partial t} )^2 + ( \frac{\partial A_x}{\partial y}
)^2 + ( \frac{\partial A_x}{\partial z} )^2
\\
- ( \frac{\partial A_y}{c \partial t} )^2 + ( \frac{\partial A_y}{\partial x}
)^2 + ( \frac{\partial A_y}{\partial z} )^2
\\
- ( \frac{\partial A_z}{c \partial t} )^2 + ( \frac{\partial A_z}{\partial x}
)^2 + ( \frac{\partial A_z}{\partial y} )^2
\\
- 2 \frac{\partial A_x}{c \partial t} \frac{\partial \phi}{\partial x} - 2
\frac{\partial A_y}{c \partial t} \frac{\partial \phi}{\partial y} - 2
\frac{\partial A_z}{c \partial t} \frac{\partial \phi}{\partial z}
\\
- 2 \frac{\partial A_y}{\partial z} \frac{\partial A_z}{\partial y} - 2
\frac{\partial A_z}{\partial x} \frac{\partial A_x}{\partial z} - 2
\frac{\partial A_x}{\partial y} \frac{\partial A_y}{\partial x} )$

Now take the derivative of the Lagrangian with respect to 4 things: \frac{\partial \phi}{\partial t}, \frac{\partial A_x}{\partial t}, \frac{\partial A_y}{\partial t}, \frac{\partial A_z}{\partial t}. Looks painful. Start with the first one, \frac{\partial \phi}{\partial t}, and you will notice there is no such term in the classical EM Lagrangian. Even if one's calculus is rusty, if a variable is not there, the derivative with respect to there's no there there is zero. This classical EM Lagrangian cannot be quantized for that reason alone.

So what do people do? They tack in a term to cover up this problem, and surround the patch job with discussions of gauge theory. The sport is called fixing the gauge. There are all kinds of super sophisticated things to say about this topic. I prefer the simple question: why is there this problem? The answer is also simple: there was a subtraction step in the Lagrangian, so of course you are missing something. All the GEM proposal does is keep all the parts of the 4-derivative of a 4-potential together so nothing is missing. Let's look at the GEM Lagrangian written out explicitly:

\mathcal{L}_{GEM}=- \frac{1}{2} ( ( \frac{\partial \phi}{c \partial t} )^2 - ( \frac{\partial
\phi}{\partial x} )^2 - ( \frac{\partial \phi}{\partial y} )^2 - (
\frac{\partial \phi}{\partial z} )^2
\\
- ( \frac{\partial A_x}{c \partial t} )^2 + ( \frac{\partial A_x}{\partial x}
)^2 + ( \frac{\partial A_x}{\partial y} )^2 + ( \frac{\partial A_x}{\partial
z} )^2
\\
- ( \frac{\partial A_y}{c \partial t} )^2 + ( \frac{\partial A_y}{\partial x}
)^2 + ( \frac{\partial A_y}{\partial y} )^2 + ( \frac{\partial A_y}{\partial
z} )^2
\\
- ( \frac{\partial A_z}{c \partial t} )^2 + ( \frac{\partial A_z}{\partial x}
)^2 + ( \frac{\partial A_z}{\partial y} )^2 + ( \frac{\partial A_z}{\partial
z} )^2 )

That certainly looks complete. Calculate the canonical momentum:

\pi^{\mu} = h \sqrt{G} \frac{\partial \mathfrak{L}}{\partial (
\frac{\partial A^{\mu}}{c \partial t} )} = h \sqrt{G} ( - \frac{\partial
\phi}{c \partial t}, \frac{\partial A_x}{c \partial t}, \frac{\partial A_y}{c
\partial t}, \frac{\partial A_z}{c \partial t} )

Nothing is zero, so quantizing the theory is possible. [For fun, notice the units required to get mass*length^2/time^2.]

Being skeptical of my own skills, the only time I get confident is when I see that others have already done a nearly identical thing. The field equations written in the first post have been quantized, the energy and momentum turned into operators. The problem is that this was done for a massless spin 1 field of EM only. There are 4 modes of emission, but only 2 of them could be described by a massless particle (something about constraints on polarization because the particle is travel ling a the speed of light). The scalar mode is non-sense, allowing for negative probability. So Gupta makes the ad hoc supplementary condition whose entire purpose is to eliminate the scalar mode of emission as well as the longitudinal one.

In the GEM approach, the same operators for energy and momentum used by Gupta and Bleuler are used. Instead of just a spin 1 field doing only EM, GEM has two fields, a spin 2 for gravity, a spin 1 for EM.

The fact that I can open a graduate level quantum field theory book, go to the section on the Gupta/Bleuler quantization method, and point to the operators for energy and momentum, says the theory can be quantized. The next step is to do a calculation, for example the scattering of an electron off of a proton. It is known how to do this for EM. This is where Feynman diagrams come in. There would be two changes to go from EM scattering to gravity scattering: the coupling constants and the propagator. The couple constants are easy: e^2 -> G m_{e} m_{p}. If you plug in numbers, the coupling goes from weenie to extra-super-wimpy: 2.56 \times 10^{-38} C^2 -> 1.01 \times 10^{-67} C^2 . The propagator, the squiggly line between two nodes of a Feynman diagram, must change from a spin 1 to a spin 2 propagator. Weinberg wrote papers on this topic in the early 60's. I Xeroxed them, and knew I would never understand the contents. I gave those papers to a friend with a recent Ph. D. in cold matter physics from MIT over his Christmas break, and he could not decipher the information. For someone actively doing scattering calculations in quantum field theory, this calculation is probably simpler than what they are getting paid to investigate: it is a variation on a calculation they are trained to do, and variations are easier to perform than making up something new. I have no idea how to reach a person who could do that work, and at this time have no authority to compel them to crank through it.

doug

[Chronos]

Thanks for the thoughtful reply, doug. I didn't phrase the question very eloquently, but you went right to the heart of the matter. Quantization is a difficult prospect in the best of circumstances. It's not pretty, much less elegant, and that rubs me wrong. I've always been intrigued, and am curious what you think of this paper:

That strange procedure called quantisation
http://www.arxiv.org/abs/quant-ph/0304202

[sweetser]

Hello Folks:

I have been preparing for a 15 minute talk at the 9th Eastern Gravity Meeting happening this Friday and Saturday at MIT. I am slated for noon, March 25. The title is "Unifying Gravity and EM, a Riddle You Can Solve". The slides are available, and the pdf has additional text approximating the main points I will say. Please feel free to check out the slides and give feedback.

I am also moving into a new home which takes mountains of time and effort, hence my delay in replying to Chronos.

doug

The slides:
http://theworld.com/~sweetser/quaternions/talks/riddle/riddle.html
[note: The page uses css & javascript, use the click, arrows, pageup and pagedown to advance]

The slides + comments:
http://theworld.com/~sweetser/quaternions/ps/riddle.pdf

[sweetser]

Hello:

The 9th Eastern Gravity Meeting at MIT was full of folks who work with Einstein's field equations for their daily bread. Some are working on trying to make gravity wave detectors, others work with computers trying to develop models of spiraling black holes that crash into each other. There was one string theory guy who still does not have any way to test the proposal.

My observation of the meeting in general was there was very little communication between different workers. It was only clusters of folks who happened to be working on related problems that could bring up decent questions. That may just be the nature physics: it's too technical today to be a generalist.

I can say with some confidence that no one trained in physics today works with a vector proposal for gravity anywhere in their education. It is either Newton's scalar theory or Einstein's rank 2 theory. I don't consider the two paragraphs a piece in MTW, or papers by Gupta and Thirring to count (I am embarrassed by the logical weaknesses involved in those short comments).

There was one work on general relativity - I cannot remember which one - that after introducing the covariant derivative, pointed out that one could take the divergence of the connection:
\partial_{\mu}\Gamma_{\sigma}{}^{\mu\nu}A^{\sigma}
The author then quickly pointed out that there is no need to ever do this sort of calculation because the term does not transform like a tensor. If we want second order derivatives of the metric that transform like a tensor, then we MUST use the Riemann curvature tensor. The logic is that simple and straightforward.

What's wrong with this logic? It is an omission. If you put the divergence of a connection in the right place, it transforms like a tensor:
\partial_{\mu}\partial^{\mu}A^{\nu}- \partial_{\mu}\Gamma_{\sigma}{}^{\mu\nu}A^{\sigma}
If the potential happens to be constant, the result is the divergence of the connection, the thing that was dismissed as silly to bother to calculate. This has second derivatives of the metric in a term far simpler that the Riemann curvature tensor.

My sense of the project is this: I will not be able to communicate this proposal until by some chance combination I find someone else who actually does that calculation for the Rosen metric. If you don't go through the exercise, you don't see it. That was the [non]reaction I got. Oh sure, I can be a bit more entertaining, the graphics are easier to read than the average presentation, but there was no effective communication about the core idea of the talk. I don't think there is anything too unusual about that.

doug

[Lawrence B. Crowell]

To be honest the objection appears valid. Consider this term

\partial_a\Gamma_b^{ac}A^b

then the gravity connection transforms by U(1) by

A^b~=~U^{-1}A_b'U~+~U^{-1}\partial_b U

the partial on the gravity connection will transform homogeneously by local Lorentz transformations. However, this whole term does not transform homogeneously under the gauge action

\partial_a\Gamma_b^{ac}A^b~=~\partial_a\Gamma_b^{a c}(U^{-1}A_b'U~+~U^{-1}\partial_b U)

The second order term \partial\partial A (indices suppressed) does not cure this, since this will transform homogenously under the gauge action and thus contains no inhomogenous terms which cancel out the inhomogeneous term U^{-1}\partial_b U. So this is not a tensor. Thus the objection is valid.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

We both agree that \partial_a\Gamma_b^{ac}A^b does not transform like a tensor. That is a non-issue.

I will always start instead with this:
\partial^a A^c-\Gamma_b^{ac}A^b
This is a tensor. This is the definition of a covariant derivative I am starting with. I choose to work with a potential A^b that is constant, so the derivatives of A^b are zero. Then I act on it with a partial derivative, \nabla_a (Note: I used the triangle there because it too has a partial derivative and a connection). The second derivative of the potential does not get stapled in at the end. Instead I start only with valid tensors, and act on those tensors with tensor operators.

U(1) is an approximate symmetry of this proposal, not an exact symmetry. For the special case of massless particles, the U(1) symmetry is exact. Mass breaks U(1) symmetry extremely weakly (1 part in 10^16 for an electron). You are correct that the connection breaks the U(1) symmetry, but that is not a constraint on the proposal. If one only works with tensors throughout, the result must be a valid tensor expression.

Breaking symmetry with in this way is a good thing. It means that the Higgs particle and Higgs mechanism are unnecessary. In the standard model, the mass of an electron has to be included via the Higgs and the spontaneous symmetry breaking of a false vacuum. I use something we already associate with mass - the metric, and the changes in the metric or connection - to break very slightly U(1) symmetry with something physically relevant.

doug

[sweetser]

Hello:

I think I am going to be a bit more assertive in the coming months about what I may have accomplished. Unifying gravity and EM in a way that can be quantized should get people's attention. In practice, delivering that line is not enough. This confuses me, because whenever someone makes that claim to me, being a good skeptic, I investigate, and a little digging usually determines there is no math there (and thus for me, no content).

Prof. Clifford Will is going to MIT to give a talk. He is a fellow who is an expert on experimental tests of general relativity. He wrote a great review article on the topic in living reviews (long paper, can be found via Google). Someone suggested I read it because it was an exhaustive survey of approaches to gravity. I read it carefully, and like everyone else, he did not mention the possibility of a vector field theory for gravity. He did cite vector-scalar and vector-tensor theories, but no plain old vector. See, Newton's field theory for gravity is the simplest scalar theory that can be written. It is still a darn good theory, used in all rockets except those that carry atomic clocks. The simple scalar theory has a technical flaw: it does not respect the speed of light limit.

A better theory is the simplest rank 2 field theory, that is general relativity. It too has a technical flaw: it cannot be quantized.

The only thing between the simplest rank 0 and the simplest rank 2 field theory is the simplest rank 1 field theory. It should be a big, honking red warning light that someone like Will has this sort of sin of omission. So I will be going to his talk and seeing if I can point this out to him. The odds are very bad - it is like pointing out a shade of green to someone who is color blind. He has zero experience with a vector field theory for gravity, Jq^{\mu}-Jm^{\mu}=\nabla^{2}A^{\mu}.

doug

[nrqed]

Hello:

I think I am going to be a bit more assertive in the coming months about what I may have accomplished. Unifying gravity and EM in a way that can be quantized should get people's attention. In practice, delivering that line is not enough. This confuses me, because whenever someone makes that claim to me, being a good skeptic, I investigate, and a little digging usually determines there is no math there (and thus for me, no content).

Prof. Clifford Will is going to MIT to give a talk. He is a fellow who is an expert on experimental tests of general relativity. He wrote a great review article on the topic in living reviews (long paper, can be found via Google). Someone suggested I read it because it was an exhaustive survey of approaches to gravity. I read it carefully, and like everyone else, he did not mention the possibility of a vector field theory for gravity. He did cite vector-scalar and vector-tensor theories, but no plain old vector. See, Newton's field theory for gravity is the simplest scalar theory that can be written. It is still a darn good theory, used in all rockets except those that carry atomic clocks. The simple scalar theory has a technical flaw: it does not respect the speed of light limit.

A better theory is the simplest rank 2 field theory, that is general relativity. It too has a technical flaw: it cannot be quantized.

The only thing between the simplest rank 0 and the simplest rank 2 field theory is the simplest rank 1 field theory. It should be a big, honking red warning light that someone like Will has this sort of sin of omission. So I will be going to his talk and seeing if I can point this out to him. The odds are very bad - it is like pointing out a shade of green to someone who is color blind. He has zero experience with a vector field theory for gravity, Jq^{\mu}-Jm^{\mu}=\nabla^{2}A^{\mu}.

doug
Good luck Doug.

I am just starting to read this long and fascinating thread so I will read more before starting to ask questions that I am sure have already been asked by others. Of course, the first comment is that the force between like charges interacting through a vector field repel so it seems as if a gravitational force based on such a field would lead to gravitational repulsion. But I am sure you have debated that with others so I will read on.

Regards

[sweetser]

Hello:

There are two papers in the literature that also claim that a vector field equation must have like charges attract just like EM. That result comes from copy EM too closely! My replies to the question are in posts 23 and 33.

The key is one perfectly placed minus sign in the Lagrange density, which then appears in the subsequent field equations and in the Lorentz force laws. The charge coupling term for EM has the same sign as the rank 2 field strength tensor contraction. That is what generates like electrical charges that repel in Gauss's law. The same sign for the coupling term and the inertial mass term lead to the Lorentz force law where like electric charges repel.

By changing the sign of the charge coupling term, the couping and field strength tensor terms have opposite signs, and like mass charges attract. The different sign for the coupling term and the mass term lead to a Lorentz force law where like mass charges attract.

The signs are actually required due to the asymmetric field strength tensor. See, that gets split into two, an irreducible symmetric field strength tensor, and an irreducible antisymmetric field strength tensor. The symmetric rank 2 field strength tensor will be represented by spin 2 particles, and those must attract (read that in Hatfield's introduction to the Feynman lectures on gravity, go there if you want a good explanation). The antisymmetric field strength tensor will be represented by spin 1 particles, and like charges will repel.

Sorry for my delay in replying, the site forgot to send me email. This is an issue my proposal must get right, or it is not worth bothering anyone.

doug

[sweetser]

Hello:

In this note, I will describe my interaction with Prof. Clifford Will of the University of Washington, St. Louis, an authority on experimental tests of gravity, that happened during his visit to MIT Thursday, May 18, 2006.

I am conflicted about my own body of work. On one hand, I consider the equations to be as gorgeous as the come. I've taught my mailman the field equations which he remembers to this day with the mnemonic, "Always give 2 Brownies to Jim", the reverse of J=Box^2 A. The exponential metric is prettier that the algebraic fragment of the Schwarzschild metric in the Schwarzschild coordinates, and completely trumps the Schwarzschild metric in isotropic coordinates with its 4th power terms (most folks studying GR probably aren't even familiar with it, but it is the form used in experimental tests). I can see the symmetries in the GEM Lagrangian that lead to energy, momentum, and mass conservation. As long as I keep focused on the equations, I can feel joy in their elegance.

On the other hand, I am familiar with every misstep made along the way, even in this forum. I am aware of my own inadequacies as a messenger, without any degree in physics, just the ad hoc collection of graduate-level courses audited from MIT, Harvard, and BU (Boston may be the best town to audit courses). It feels like a cruel cosmic joke that someone with such limited skills in math and physics should have found this cache of rocket fuel. As I age and see those limited skills shrink, so does my confidence.

For about a week, I thought about the questions I would like to ask Will. That was fun. Not fun was the self-doubt which can derail the task. I decided to wear the Turquoise Einstein t-shirt featured at the top of quaternions.com for two reasons: I like the shirt enough that it boosts my self confidence, and it has the Always Give 2 Brownies to Jim equation which might be useful in a technical discussion.

Will was to be at an MIT physics lounge at 3:45, with the talk in 10-250, a large hall, at 4:15. I showed up right on time. Will looks much like Ted Turner, with white hair and mustache. He was seated on a couch, the room sparsely occupied by small clusters of graduate students talking about grading undergrads. I went straight for the cookies, wondering when I should say hello. Rather than have a long debate, I decided to get it over with sooner rather than later.

Eye contact made, I told Will that he was the center of a small debate on the Internet (sci.physics.research several years ago). I was investigating a simple approach to gravity in 4D. A fellow suggested I read Will's Living Reviews article on tests of gravity because it covered EVERY respectable approach to gravity, bar none. I read it. It was a very good article (that brought a smile). The paper cited vector/scalar theories, and vector/tensor theories, but no simple vector-only theories. Newton's law of gravity is the simplest rank 0 field theory one can construct. It is a darn good theory, still used in the guidance systems of most rockets. If those rockets carry atomic clocks, we realize two technical limitations of Newton's law: that the speed of gravity is wrong (no speeds are infinite) and gravity bends spacetime more than Newton's law predicts. A better theory is the simplest rank 2 field theory that can be constructed to explain how gravity works - Einstein's field equations for general relativity. It really is a darn simple approach, with the Ricci scalar sitting alone in the Hilbert action. The speed of light is respected. All weak field tests are passed. All strong field tests are passed. The rank 2 field theory also has a technical problem: it cannot be quantized. The brightest folks in physics have all tried to no avail.

Between the simplest rank 0 field theory and the simplest rank 2 field theory is the simplest rank 1 field theory. This formal possibility was not discussed in this otherwise exemplary review article. I asked him what he thought about that omission.

He replied that there was no need to disprove a vector theory because we already had proof that a metric theory was required. All the tests of the equivalence principle were effectively tests that gravity must be a metric theory.

I nodded. I was aware that for my own efforts to succeed, I would have to precisely deflate this position. I may do that in a subsequent post here, but that was not my purpose in the discussion with Will. Rather, I had a rare chance to talk technically with an expert, and wanted to find out his perspective on several issues. I had it, so it was time to move on.

I pointed to my t-shirt, saying this was the 4-vector field equation I happened to study. Most physicists if asked would think it was the Maxwell equation in the Lorenz gauge, where like charges repel, and thus not applicable to gravity. If the metric had a +2 signature instead of -2, then like charges would attract. The key to understanding the proposal is that the box^2 is not a D'Alembertian operator (a scalar operator consisting only of the second time derivative minus a Laplacian operator). Instead the box^2 represents two covariant derivatives. A covariant derivative has a normal derivative minus the connection, so the equation reads: normal derivative minus a connection, applied to a normal derivative minus a connection, applied to a 4-potential. One could have a differential equation that was the divergence of the connection of the potential. I claimed that the Rosen metric (the one at the start of this forum) solves that very differential equation.

He shrugged his shoulders. That was too dense to follow, and he didn't. I offered to send him a copy of my paper. He told me he was extremely busy, and there was no way he would have the time to look at it. I took him at his word, and promised him I would not email him. I am not going to beg for people to look at elegant equations.

He asked if the theory was a full one. I told him I had a Lagrange density and the field equations from the action, along with solutions to the field equations that were physically relevant.

An objection I was sure people would raise is that the proposal has linear vacuum field equations. Like EM, once particles start interacting, things become nonlinear. My impression was that physicists believe that the vacuum field equations had to be nonlinear, that non-linearity was a way to decide if a proposal was mature or just a toy.

Will said there was no experimental data on the topic. After a bit of going back and forth, he said essentially that the collection of good candidate theories for gravity happened to all be nonlinear, so it was reasonable given that observation for there to be a sense that nonlinear field theory was the correct approach. The Schwarzschild metric was linear in one coordinate system, and nonlinear in another, so linearity is coordinate dependent, and of limited value.

I asked if there were any plans to do gravity tests to second order PPN accuracy. My proposal is identical at first order PPN accuracy, but about 12% different at seccond order PPN accuracy. He said those experiments are very demanding. There were nothing coming soon. He wondered if I had shown my proposal worked for the precession of the perihelion of Mercury. I claimed that it did, a four page calculation (available at quaternions.com). I also asserted that the lowest mode of gravity wave emission should be a quadrupole because the theory conserves momentum, and there is no other field to store momentum.

He nodded along, but was not engaged, so I thanked him for his time, shook his hand, and went back for another cookie. He eventually got off the couch and chatted with a few professors. I realized that had I been delayed, there would have been no way to have had such a long and detailed private discussion. He left at 4:05 to go to the lecture hall.

The talk itself was good. He covered tests of the equivalence principle, solar tests (weak field), binary pulsars (strong field), and gravity waves. He was one of the folks who developed the PPN system. For general relativity, \gamma=1, \beta=1, all 8 others are equal to zero. At the end of the lecture, I asked if there were any other theories where \gamma=1, \beta=1, all 8 others are equal to zero. [I was interested because this is a property of my own proposal.] He said Rosen, of Einstein, Poldolski, Rosen fame, came up with such a theory. It works for some tests, but for binary pulsars, the theory predicts the frequency should increase instead of the decrease observed. It is vital to check a proposal in all regions.

Someone asked a question on MOND, the Modification of Newtonian Dynamics that is consistent with some data on the velocity profile of disk galaxies (as well as other astronomical data sets). He said that someone had figured out a Lagrange density for MOND, he had seen it, and it was UGLY. I found that funny. The strength of the GEM proposal is that the Lagrangian is fit and trim. Can it match all the data collected in eighty years? Not in a brief private discussion.

I did feel it was a good day for the GEM proposal. I certainly did not convert Will: that would require a formal demonstration that the proposal works for all weak field tests, all strong field tests, all tests of the equivalence principle, and gravity waves as a starting point. On an informal level, it was not dismissed out of hand. It felt like there was stuff worth talking about. I was glad I had attended and asked as many questions as I did.

doug

[sweetser]

Hello:

One of the things that gives me confidence is when my points of contention with standard theory become subtle instead of confrontational. Too many folks brighter than myself have thought about these issues before. It is unreasonable to expect them to be wrong. It is natural for a blind spot to remain unseen.

Clifford Will has argued that gravity must be explained using a metric theory. He is not the first one to state that position. I've seen it espoused in Misner, Thorne, and Wheeler (the beginning of chapter 40 to be more precise). I agree, gravity must be explained with a metric theory.

Here is where the subtle issues start. The question not asked is how many ways are there to implement a metric theory? The standard approach is to work with the Riemann curvature tensor. If you are not familiar with this rank 4 monster, it is essentially the difference between two curved paths. It measures how much the curvature of spacetime is changing. I call it a monster because one time I tried to write it out in terms of metrics and it was a futile exercise, there were just too many terms. [More detail than necessary: the Riemann curvature tensor is the difference between two derivatives of a Christoffel symbol, and the difference of two products of two Christoffel symbols, and a Christoffel symbol itself has three derivatives of a metric that get contracted with another metric, so the details of the Riemann curvature tensor can be overwhelming.]

There is an implied message behind the pitch for a metric theory: that the approach cannot use a potential. I forget who did the calculation (someone between Newton and Einstein), but he showed that gravity will bend light. Einstein also did the calculation in 1911. The answer was exactly half right. It did make exactly the same prediction for bending the time term of the metric as does general relativity, the g_00 term of the Schwarzschild metric, which gets a little smaller than one. The error is that Newton's theory leaves the space part unchanged. Experimental tests show the g_11 term is a little greater than one.

What the data unambiguously proves is that a scalar potential theory cannot explain light bending around the Sun. The coefficient in front of the dt part of the metric is less than one, whereas the coefficient in front of the dR part of the metric is greater than one. A one parameter potential can not be both greater and lesser than one. Proof complete.

Let's shine light on the blind spot. A potential that has more than one parameter could be consistent with experimental data. If one used 100 parameters, it would be trivial to match any data set. The question is what is the smallest number of parameters needed? It has been established that one is too small. The next thing to try after a scalar tensor potential is a vector tensor potential. In spacetime, that would have 4 parameters. There is sufficient freedom with 4 parameters to match the experimental results.

Another subtle issue is that people if pressed would view this discussion as a metric theory versus a 4-potential theory. It cannot be that, since I have already agreed the theory must be a metric theory - up to a diffeomorphism. That is a fancy way to say I am proposing a metric theory and/or a 4-potential theory. Some might say that is a slimy political move, but I believe that is a beautiful symmetry slight of hand. It's magic! Remember, general relativity is magic too, saying gravity is exclusively about spacetime geometry, no force needed. Incorporating diffeomorphisms correctly to a vector theory allows me to say gravity is all about spacetime geometry or potential, or any combination of geometry and potential you choose. The GEM proposal is not only a unification of gravity and EM, but the union of Newton's exclusively potential theory with Einstein exclusively metric theory. That rocks!

This work keeps getting sexier than string theory :-)
doug

[sweetser]

Hello:

More of our brain is devoted to visual information processing than any other, so it is vital to have a visualization of the GEM proposal. It can be found in this link:

http://TheWorld.com/~sweetser/quaternions/gravity/em2gem/figure_1.jpg

The image is a collage of two figures most people are familiar with. The Newtonian potential looks a little like a ladle, and particles hang out at the bottom of the dip. The lines on the graph paper in the background for the x and y coordinates are always as straight as can be drawn.

For Einstein's explanation of gravity, artists use a rubber sheet, with the mass stretching the sheet. Now a "straight line" - one that stays within the warped lines - looks curved to us from afar.

What the GEM theory proposes is that either or a combination of both images is correct. Newton's potential theory for gravity does not work because it can only bend one way, and spacetime needs at least four parameters to describe its bending as is possible with a 4-potential versus a scalar potential theory.

Newton's theory gets the light bending around the Sun half right. Specifically, it gets the bending of the time portion correct. If one used a metric that was flat for the time part (g_{00} = 1), but was bent the appropriate amount in the space part (g_{11} = 1 + 2 GM/c^{2}R), then the combination of a Newtonian potential with a curved-only-for-R metric would be consistent with experimental tests of gravity to first order PPN accuracy. The curved-only-for-R metric would not be a solution to Einstein's field equations which explain gravity as exclusively arising from spacetime curvature. It is unfortunate, but the GEM theory says that exclusive metric theories like general relativity are too restrictive. The potential well/rubber sheet symmetry must be embraced to unify gravity and light.

doug

[sweetser]

Hello:

People have suggested I submit a paper to a physics journal. I have set a specific goal to achieve before I go through that process. I would like to find someone who understands general relativity and the Maxwell equations at the level of the action, have that person read my draft paper, and tell me if s/he thinks it is valid. Based on the ensuing give and take, I would either submit the paper, or post to my web site (and here) why the proposal was flawed. In this post I will show my latest effort to find someone with the skills required to review my work on a technical level.

A few posts ago on this thread, I told how I had a productive discussion with Clifford Will, but he was too busy to read my work. The Internet was developed by physicists at CERN for physicists working together over the globe. It is my sense that in measurable ways, the pace of life for theoretical physicists is more frantic than most other professions. I expect doors to be closed not due to aloofness - Will was both approachable and polite - but because there is too much stuff crammed in the room to open the door.

Well-known physicists at MIT, my alma mater, are frantic. I've seen it up close when I worked at the bench for Prof. Richard Young. It is part of my ethics not to bother a busy scientist unless I meet one of two criteria: either I have a well-formed question or I have a result. Being a skeptic, it is important to me that at the very least Mathematica has checked the algebra contained in the draft (Mathematica did not alway approve of my efforts, but that is a different story). Professors that are less well-known are probably even more frantic, trying to establish a name.

What follows is my last email to a less well-know physicist, which references an email to a well-known physicist. The short story is the less-well known prof. has not written back, and the well-known prof. documented why he cannot read my draft. I am not mad or frustrated because I understand why they behave as they do. I remain persistent.

doug

--------
From: Doug <dougsweetser@gmail.com>
To: slloyd
Date: Jun 8, 2006 7:22 AM
Subject: Fwd: The work of the stand-up physicist

Hello Seth:

Dave Pritchard gave me your name (email exchange at the end). I am an
amateur physicist who has a Lagrangian for a unified field theory.
There have been a few times in the history of science when an amateur
- after a few pitches by mail - has mad a contribution. What is not
part of the lore is the number of appeals of limited value. I hope
the details in my note to Dave and the attached pdf file show it is
possible my work has some of the claimed significance.

I hope you get a chance to review the work. Thinking about the
action, the field equations, or the solutions to the field equations
gives me a quiet sense of joy (unusual for such abstractions).

Have a good day despite the rain,
doug

---------- Forwarded message ----------
From: Dave Pritchard
Date: Jun 7, 2006 12:30 PM
Subject: Re: The work of the stand-up physicist
To: sweetser@alum.mit.edu
Cc: Sarah Smith

Doug,

Right now I have about 6 papers of which I'm a coauthor (often senior
authro)to read through and correct - I can't take on another (I spent
the weekend dusting off 3 overdue referee reports ahd nave a big
proposal to review, and another one to do also). So adding to this list
is impossible.

Your ideas sound like something theorists might have tried.

[I'll reply here since it would have been impolite to reply to Dave.
The action would certainly have been tried back in the 19th century.
This was before the notion of a diffeomorphism was developed, a key to
riddle. Clifford Will wrote a Living Review article on the modern
view, and despite being over a hundred pages and quite thorough, it
never brought up the simplest vector theory, as a resonable proposal
between the simplest scalar theory, Newton, and the simplest rank 2
theory, GR. Will recently visited MIT, and I asked him why his paper
had this omission. He said that there is more than enough data to
indicate that gravity must be a metric theory. A potential theory
will only generate half the bending of light that has been measured.
He, like Pritchard, said he was too busy to view my work. Gravity
must be explained via a metric theory, I agree on that point. An open
but overlooked question is how to implement a metric theory. All
efforts to date have relied on the Riemann curvature tensor in some
way or another. I work from a simpler starting point, the connection
as it appears naturally in a covariant derivative. It is true that a
scalar potential theory of gravity can only get the smaller than one
g_00 term correct (the g_11 term of the Schwarzschild metric is
greater than one, and a single parameter cannot do both). It would be
imprecise to presume that a 4-potential theory would be insufficient
to achieve the results seen in experiment.]

[...back to Dave]

I think you might try to talk to someone in the field who's not so famous (Seth Lloyd at MIT has been working on GR and QM at MIT).

I wish you luck, and am sorry I can't help.

Dave

David E. Pritchard 617/253-6812
Associate Director, Research Laboratory of Electronics
Cecil and Ida Green Professor of Physics
Room 26-241
MIT
Cambridge, MA 02139
617/253-4876 fax
http://www.rle.mit.edu/pritchard
Education Research: http://relate.mit.edu



Doug wrote:
> Hello Dave:
>
> I was having lunch at Mary Chung's with Sarah Smith. I was lamenting
> my plight in theoretical physics, and she suggested I drop you a note.
>
> It would be great if gravity and EM could play nicely with each other.
> That is not the case today, with general relativity standing outside,
> refusing all efforts to be quantized.
>
> Mathematica and I think we have a solution: a Lagrange density, the
> field equations generated by varying the action, physically relevant
> solutions to those field equations, the force law, a dynamic metric,
> an appreciation of the group theory perspective for the proposal, and
> two experimental tests to distinguish the approach from general
> relativity. That's a long list. My problem is that I am an amateur
> physicist, no funding from the government or industry. I know it
> takes an hour and a half to go from the classical EM Lagrangian to the
> Maxwell equations. In my approach, I toss in another current and a
> symmetric field strength tensor to do the work of gravity right next
> to EM. The vacuum field equations for gravity are every bit as linear
> as EM, and so quantization is just like EM, but with a spin 2 field so
> like charges can attract.
>
> Of course I should just publish, but a man has to know his
> limitations. I am an amateur. I do not have a Ph.D., a masters, or
> even an undergraduate degree in physics. In the 90's, I sat in on one
> graduate physics class a semester. I worked at a lab bench, and know
> how to shut up and get more data. Writing a technical paper in
> physics is not a craft I have mastered. I have attached to this email
> a pdf draft of my best effort to date. It is highly probably that I
> have a garbled line or two in the text that I am unable to spot. That
> is all that is needed for a paper to be rightly rejected.
>
> I know you are not an expert in the study of gravity. If an amateur
> and a symbolic math program can figure it out, so can you. It is a
> variation on the Maxwell equations, so it requires just as much work
> as the Maxwell equations to understand. You probably know from
> teaching that most folks can get Newton's scalar law of gravity.
> Electromagnetism is far more difficult to teach. No one bothers to
> teach general relativity to undergraduates (almost no one at least,
> Edwin Taylor is giving it a try). My work requires an amount of
> effort between Maxwell and GR, which is to say it takes considerable
> work.
>
> In some ways, the thesis is all about doing no work. That is what
> life in spacetime is like, flat as a pancake. Only there is a little
> bit of other stuff around, so everyone does the very least they can,
> which would be simple harmonic oscillation. Do a SHO in 4D, and there
> are two transverse modes for light, longitudinal and scalar modes for
> gravity. The rest is details.
>
> There are a lot of details. At APS meetings I get all of 12 minutes
> to explain five hours of differential equations. You want to write a
> novel and put it in a fortune cookie format? I have a web site, but
> who could possibly click through that much math? I am the kind of
> fellow who prefers a creative solution no matter what the cost.
> People will listen to someone telling stories, to someone teaching
> with a passion. So I decided to create a community access TV show,
> "The Stand-Up Physicist", and it has a companion web site,
> www.thestandupphysicist.com. The show serves as an outlet for me.
> These equations for unifying gravity and EM are drop-dead gorgeous.
> It all works in 4D, no ten or eleven dimension BS needed. It makes
> sense physically and mathematically. If experimentalists could
> measure the bending of light three orders of magnitude better than
> done today, they might see the 0.8 microarcseconds more bending my
> theory predicts than general relativity. It is funny and a little
> tragic that someone with my limited skill set has uncovered these gems
> (my one strong suit is creativity, and that can be measured by
> painting, piano, French pastries, swing dancing, and recumbent bike I
> designed and ride).
>
> I hope you can scan the paper and see if it looks logically coherent,
> something that cannot be faked. Like "The Old Man and the Sea", I am
> trying to land a fish that I know is too big for me to handle by
> myself. I am willing to talk about it at anytime at your convenience
> (I work for a software company in Waltham, live in Acton, and like to
> eat at Mary Chung's).
>
> Have a good day, elegance governs the heavens.
> doug

I attached a file which can be viewed either as a pdf or in HTML
http://TheWorld.com/~sweetser/quaternions/ps/em2gem.pdf
http://TheWorld.com/~sweetser/quaternions/gravity/em2gem/em2gem.html

Thanks for reading,
dougd

[CarlB]

Doug,

Could there be any chance that this new result has anything to do with an interaction between gravity and E&M that is consistent with what you are doing?

New Experimental Results on the Lower Limits of Local Lorentz Invariance
Fabio Cardone1, 2, 3, Roberto Mignani4, 5, 6 and Renato Scrimaglio1

(1) Istituto per lo Studio dei Materiali Nanostrutturati (ISNM-CNR), via dei Taurini 19, 00185 Roma, Italy
(2) Istituto di Radiologia, Facoltà di Medicina, Università di Roma “La Sapienza”, Roma, Italy
(3) I.N.D.A.M.—G.N.F.M., Sesto Fiorentino, Italy
(4) Dipartimento di Fisica “E. Amaldi”, Università di Roma “Roma Tre”, Via della Vasca Navale, 84, 00146 Roma, Italy
(5) Sezione di Roma III, INFN, Roma, Italy
(6) Dipartimento di Fisica, Università dell’Aquila, Via Vetoio, 67010 Coppito, L’Aquila, Italy

Received: 7 October 2004 Published online: 9 May 2006

An experiment aimed at detecting a DC voltage across a conductor induced by the steady magnetic field of a coil, carried out in 1998, provided a positive (although preliminary) evidence for such an effect, which might be interpreted as a breakdown of local Lorentz invariance. We repeated in 1999 the same experiment with a different experimental apparatus and a sensitivity improved by two orders of magnitude. The results obtained are discussed here in detail. They confirm the findings of the previous experiment, and show, among the others, that the effect is independent of the direction of the current. A possible interpretation of the results is given in terms of a geometric description of the gravitational and the electromagnetic interactions by means of phenomenological, energy-dependent metrics.

Foundations of Physics, Volume 36, Issue 2, Feb 2006, Pages 263 - 290, DOI 10.1007/s10701-005-9014-z, URL:
http://dx.doi.org/10.1007/s10701-005-9014-z

I still haven't forgotten about putting together a program to compare your gravitation with Einstein and Newton. But I've been too busy to get it written. When it's done, it will be written in Java. As a test to see if you can compile my variety of Java, try downloading the java source code for my Sudoku solver, which also has a link to where you can get the free Java development program from Borland:
http://www.brannenworks.com/SU/

Carl

[sweetser]

Hello Carl:

Thanks for the reference to the paper. Based only on the abstract, I was not able to justify the $30 cost for downloading the pdf, so I cannot comment on the specific content.

Here's a skeptical (in the postive sense of the word) view. In the 1800s, people thought that electric and magnetic phenomena should somehow be linked. When Oelmsted found that link, experimentalist got excited and did a barrage of work. Experimentalists are like that today. On the positive side, there was a frenzy to find ever higher temperature superconductors. On the negative side, there was a squall of work to detect a 5th force and to see the signs of cold fusion.

I have not picked up a buzz concerning a local violation of Lorentz symmetry. The experiment was first done in 1998, then apparently repeated by the same folks in 1999, and published some 5 years later. I believe the journal "Foundations of Physics" has a reputation for publishing work of questionable long term value.

What does the phrase "violate local Lorentz symmetry" mean? I'm not sure, but will discuss this symmetry as best as I understand it, and how it relates to my work.

Empty spacetime is governed by the Minkowski metric. Square the spacetime difference between any two events, and all inertial observers agree on the value of the interval (Lorentz invariance), and the all disagree about time and space measurements in a way all observers understand is related to their relativistic velocities (Lorentz covariance).

A complete answer to an observation always included both the invariant and the covariant quantities. This pairing is often forgotten. For example, people will point out that the speed of light is an invariant, and forget to mention that the frequency and wavelength of light are covariant, changing by a relativistic doppler equation. Many popular sciences sources will talk about rulers and clocks disagreeing (Lorentz covariance), without mentioning the invariant interval.

So I have an ecletic rule: I must always talk about the invariant and the covariant measurements in order to be complete. Once I have a rule, I try not to break it ever, so let's see how this goes...

In spacetime with mass in GEM theory, either one continues to use a flat spacetime metric with a dynamic potential, or one uses a dynmaic metric with a constant potential, or some combination of the two (in GR, it must exclusively be the metric that changes). There are only an infinite number of choices to be made! That's diffeomorphic symmetry for you. Let's choose to work with the boring potential, so everything is due a dynamic metric. Another way to say it is that the metric is different depending on where you are. We know how it changes. So the interval is now a covariant quantity, not an invariant one, because the interval changes in a way we understand.

That raises the question: what is the invariant quantity? I know exactly what it is, but I don't know its name! If you recall the metric that appeared in the first post of this thread, the dt term had a exponential with a negative exponent, while the dx, dy, dz all had exponentials with a positive exponent. It turns out that the products dt dx, dt dy, and dt dz are invariant under the introduction of mass into empty spacetime!

I don't know what to call dt dx, dt dy, or dt dz, so for now I'll make up a name: a 3-rope. Inertial observsers conserve the interval, and folks in a Universe with mass conserve the 3-rope. Is there any connection between an interval and a 3-rope?

This is where the story gets downright unbelievable. They have a simple, direct connection. Write an interval as a quaternion (if you are unfamiliar with quaternions,visit my website devoted to the topic, quaternions.com, to learn about the next division algebra after the real and complex numbers, which as 4 parts that can be added, subtracted, mutliplied or divided). Here is an interval written as a quaternion:
\xi=(dt, dx, dy, dz).[/itex]
Now square the interval quaternion:
[tex]\xi^2=(dt^2 - dx^2 - dy^2 -dz^2, 2 dt dx, 2 dt dy, 2 dt dz).
In special relativity, it is the first term of this square that is invariant, while the 3-rope is covariant because we know exactly how it changes. Now toss in a mass and for the GEM theory, the first term is covariant while the 3-rope is invariant. If one wanted to be more precise, I would have to add a caveat that all the theory could claim is that there exists a choice of coordinates such that the 3-rope is invariant (I think there is also a caveat like this for special relativity, that pathological coordinates are pathological and mess nice statements up).

If anyone reading this knows the offical name for the 3-rope symmetry, it would be a big help to me. If I knew the name of the beast, then I could read up on it.


To get back to Carl's question, the electric and magnetic fields live inside the same irreducible field strength tensor, the antisymmetric tensor \nabla^{\mu}A^{\nu} -\nabla^{\nu}A^{\mu}. The reason there are all those interactions between E and M is because they live inside the same irreducible tensor. The fields for gravity live in a different irreducible tensor, the symmetric \nabla^{\mu}A^{\nu} +\nabla^{\nu}A^{\mu}. I identified three gravitational fields, and those should be able to mix with each other as happens for E with B. The gravity and EM fields cannot mingle so directly according to GEM theory. To be completely honest, I still am unclear about their relationship. Both are caused by the same 4-potential, something that cannot be measured directly. What can be measured is the change in the 4-potential, and that change falls into these two irreducible tensors.

The bottom line at this point looks like gravity and EM should not mix together in an obvious way according to GEM theory, and the 3-rope symmetry in the context of quaternions is a fun way to wonder what is going on.

doug

[CarlB]

Here is an interval written as a quaternion:
\xi=(dt, dx, dy, dz).[/itex]
Now square the interval quaternion:
[tex]\xi^2=(dt^2 - dx^2 - dy^2 -dz^2, 2 dt dx, 2 dt dy, 2 dt dz).
In special relativity, it is the first term of this square that is invariant, while the 3-rope is covariant because we know exactly how it changes.

Since my background is in elementary particles, I'd prefer that you go with the flat space and explain it again from that point of view and would appreciate that.

But your equation with quaternions is interesting from a particle point of view. Let's see how you can map your way of doing bizness into the Pauli algebra. If

\xi = idt + \sigma_xdx +\sigma_ydy +\sigma_zdz

then

\xi^2 = (-dt^2 + dx^2 + dy^2 + dz^2) + 2i\sigma_x(dt\;dx) +2i\sigma_y(dt\;dy) +2i\sigma_z(dt\;dz)

and all the other cross terms cancel because of anticommutation of the Pauli sigma matrices, giving a result very similar to your own. I suppose I should multiply through by i.

Now as it turns out, the above use of the Pauli algebra is what I was pushing before I decided to abandon "Euclidean relativity" in favor of submitting to Einstein's forms. This was not because it was wrong, but because I got tired of being the nail that stuck out and kept getting hammered down. I was using an algebra where the natural differential operator \nabla is defined as:

\nabla = \hat{t}\partial_t + \hat{x}\partial_x + \hat{y}\partial_y + \hat{z}\partial_z + \hat{s}\partial_s

where \hat{t}^2 = -1 and all the other hats square to +1, and the t hat commutes with everything while the x,y,z and s hats anticommute with each other. The "s" coordinate is for proper time in a short cyclic coordinate. If you go to massive particles, the s coordinate goes away, so ignore it.

This was written up in my original paper classifying the fermions. See the comment on page 5, "For those manifolds that do not explicitly include time, an extra commuting operator (...) accounting for momentum versus position must be included and this increases the value of k by one." http://brannenworks.com/a_fer.pdf see

What this boils down to is that if you want to model spacetime and you insist that time NOT be a geometric part of the manifold (as time is considered geometric by Minkowski), then you have to have an extra degree of freedom to distinguish between traveling backwards and forwards in time. That is, vectors in space-time can give directions forwards and backwards in time. Vectors in space can not, and thus need an extra degree of freedom. This you can accomplish either by having double sets of canonical variables, i.e. keeping track of Ps and Qs in the traditional classical mechanics manner, or you can accomplish this extra degree of freedom by adding an extra "notation" coordinate to your geometry. As a notation coordinate, it must commute with everything else.

There are also some simple ways of embedding the quaternions into the Dirac algebra that might be less unsettling. The key is that you have to arrange for the time component to commute with the spatial components, while the spatial components anticommute. But every example of this will be analogous to the above Pauli example. That is, the algebra is completely defined by these commutation rules, along with the rules telling what the squares need to be.

Carl

[Don J]

Since my background is in elementary particles, I'd prefer that you go with the flat space and explain it again from that point of view and would appreciate that.


Hello Carl
I think this paper can complete your research ...because you work in that field access to this rare paper should be granted to you.
Your opinion will be appreciated
http://prola.aps.org/abstract/PR/v105/i2/p735_1

[sweetser]

Hello Carl:

> Since my background is in elementary particles, I'd prefer that you go with the flat space and explain it again from that point of view and would appreciate that.

No problem, so long as it is clear the ability to choose between a metric theory like GR, and a potential theory like Newton's approach is at the core of what is "new" about GEM. The quotes are required, because all I am doing is exploiting a symmetry property of a covariant derivative, which has been around for a while.


Quaternions have been discovered and rediscovered many times. The first to find them was Gauss, because Gauss discovered everything. Only much later did they realize this bit of one of his notebooks was an uncredited invention. Hamilton discovered them while trying to come up with a rule for multiplying triplets (can be done, but division does not work). Rodriguez discovered them while trying to do 3D rotations, their one wide use today.

The Pauli algebra is almost the quaternions. For doing calculations in physics, they make things a little easier, because there is an extra factor of i. It is simple to set up the Lorentz group using Pauli matrices. It is not simple to do with real quaternions. De Leo first did it in 1996 or so. Peter Jack was the first person to write the Maxwell equations using only real quaternions, and I repeated the trick independently a year later.

There is a cost to convenience, and it is very subtle. With the Pauli algebra, one can multiply two non-zero numbers and get zero. How many pairs of these are there? About an infinite amount, plus or minus 42. In some ways, these infinite zeroes don't do anything but complicate statistics. It is my belief, without evidence, that these bogus zeroes could be the reason one has to do regularization and renormalization in quantum field theory. This is 100% speculation. I could prove the point by doing a calculation using quaternions exclusively (no gamma matrices), and demonstrating that regularization and renormalization were not needed to make the calculation behave. I am quite certain I could not do such a calculation on my own, only able to act like a adviser for people skilled in the arts of quantum field theory. I have no expectation that this speculation with be confirmed or rejected in my lifetime. It is a favorite speculation of mine though :-)

I am a big fan of time! The reason is that time lies straight down the diagonal of a quaternion. With all of our incredibly tiny relativistic velocities, it says that we change hardly at all in space, that nearly all of the change we experience is through time.

It is one of those funny "size" issues: most people will say that the Dirac algebra is bigger since quaternions fit inside the Dirac algebra, along with all those bogus zeros. I claim that being a division algebra will allow you to do more, and it is the size of what can be done that matters most.

What is the biggest, most important algebra? The one the user thinks they can do the most in :-) In my case that would be the quaternions, in yours, a variation on the Dirac algebra. I actually have a limitation with quaternions, it is the pea under a mountain of mattresses for me. I have no idea how to handle the connection. It is also clear that far too few folks work with quaternions, so to communicate, I had to translate all the work I did initially in quaternions into tensors.


For anyone interested in seeing good old technical conflicts, I posted something in slashdot.com about my work, and got in a classic nasty discussion (well, he was the one to toss insults, and I did learn something, that a simple model of mine was too simple. Apres the name calling part, he is back to the banal defense of the status quo).

HTTP://science.slashdot.org/comments.pl?sid=189525&threshold=-1&commentsort=3&mode=thread&pid=15604834#15605147

[Don J]

For anyone interested in seeing good old technical conflicts, I posted something in slashdot.com about my work, and got in a classic nasty discussion (well, he was the one to toss insults, and I did learn something, that a simple model of mine was too simple. Apres the name calling part, he is back to the banal defense of the status quo).

HTTP://science.slashdot.org/comments.pl?sid=189525&threshold=-1&commentsort=3&mode=thread&pid=15604834#15605147
A good discussion netheless without too much math formulas...only pur discussion about the subject at hand...glad that you finally try to challenge your theory outside the relatively confortable independant research section of this forum...its hard to play in the major league Isn't it?
Maybe its time to try
http://www.bautforum.com/forumdisplay.php?f=18

Warning: -avoid presenting your theory in the against the mainstream section where the discussion can turn harsh about your ideas -

[sweetser]

I consider slashdot and this forum to be the random league. Someone random person comes in with a comment, which I usually have to get them to repeat a few times, and eventually I will see their point. I've learned three things from the first 12 pages of this thread:
1. the field strength tensor indices should both be up, \partial^{\mu}A^{\nu},
2. to get the Lorentz force from the action I need to include the inertial term in the Lagrangian, -\rho/\gamma,
3. the GEM action breaks U(1) symmetry for massive particles.

In the slashdot discussion, I learned that my simple argument for a transverse mode of emission is too simple. I have to rely on how the field is quantized to justify the gravity waves are longitudinal or scalar modes of emission.

Harsh? I don't care about that. I make a note of it, and focus on the technical content.

Let me list the initials of the major leaguers I have dealt with, and by that I mean professors who are good enough to have their research funded: AG, SD, and CW. There were so busy, I spent 45 minutes with #1, and ten minutes with the other two, which is not time to present the idea. Again, this is an observation. The pitch to Seth Lloyd is like triple A league. He might get to reading the paper.

I am going to outsource the critique. I know someone who was a physicist in Russia, so I'll try and get feedback from the other side of the globe. I also have a friend in India who is works in at an institute, and will try to work that too.

This forum has helped me in measurable ways, so thanks to all who have participated so far.

doug

[CarlB]

Doug,

The factorizations of zero in the Pauli algebra are of the form (1\pm \sigma_u)/2. To get equivalent factorization in the quaternions requires that we add an imaginary unit to it, as in:
(1 + k\sqrt{-1}) \; (1 - k\sqrt{-1}) = 0

So is the Pauli algebra equal to a complexified copy of the quaternions?

Carl

[sweetser]

bingo, bingo.

Here is my ontological problem with that approach. The quaternion (a, b, 0, 0) behaves with other quaternions of this form EXACTLY like a complex number. So it would be fair to call this a+bi. The quaternion (a, 0, c, 0) also behaves EXACTLY like a complex number. Then there is (a, 0, 0, d), which also could be written a+di, and nothing about how it plays with (e, 0, 0, f) would be different from complex numbers. Now when we call a quaternion a+bi+cj+dk, and have 3 distinct imaginary numbers, it looks like cheating to add in a forth i that behaves like the first one used to. One could claim that as long as we make the rules, it is a fine and very productive thing to do for a mathematician. Thing is, I don't care about mathematicians, I care about Nature, and she is more demanding of silent perfection in how thing work (if they don't work, they die, so that is as demanding as it gets). I understand that there are gaggles of humans that are skilled at manipulating complexified quaternions. It is my belief that Nature does not use that algebra. An odd belief, but quaternions.com demonstrates my practice.

doug

[sweetser]

Hello:



In a different forum, someone asked me about black holes and this proposal. For politeness sake, if you have any questions related to GEM theory, please do so right here.



The short answer is that I know the math behind very dense gravitational sources is going to be fundamentally different from what appears in general relativity, but I do not understand the details (and the details matter). The differences will be so significant that current efforts to understand black holes will have to be dismissed should GEM theory succeed both on experimental and theoretical grounds.



I have two ways so far of finding physically relevant applications of the GEM
theory. The first is the exponential metric that solves the field equations for a single spherical source with the choice of a constant potential. The second is a potential that in an approximation has a derivative with a 1/R^2 dependence. The potential gets plugged into a Lorentz force law to get to te same exponential metric expression (odd but true). In these two approaches, there is a point in the derivation where I say that the change in time is tiny compared to the changes in distance, so assume the small time contribution approximation. That is how one gets to the kinds of solutions we see often in Nature: changes in space dominate the near vacuum of a Universe we live in 13.6 billion years apres the big bang, with enough of a nod to changes in time that special relativity is respected.



Now consider the case where the changes in time are as significant as those that happen in space, a condition which may appear for very dense gravitational sources. The potential whose derivative is 1/R^2 via an application of perturbation theory will no longer be applicable. The potential will have a derivative that is 1/R^3, and then have a force that had the same inverse cubic dependence on distance. How odd! I have heard it said that an inverse cubic law is unphysical. That is true for a classical law of gravity. The math may give a different story for dense sources.



I recall from classes on differential equations taken decades ago that to say an equation was singular had a precise technical meaning. Nearly all researchers consider the point singularity that appears in general relativity as something worth working on, not a thin ice warning that could open up and drown a large body of work. There is a singularity for my field equations, but it is not a point singularity. Instead the equation blows up lightlike intervals, when tau^2=0. That may turn out to be a better deal, because we know there are particles like the

photon and graviton that live on the lightlike surface.



Should the GEM hypothesis get a following, the behavior of singular solutions will be a fun area of study.



doug

[A_Wanderer]

Doug Sweetser type-person:

This is not in the nature of a reply, yet now and then a question is worth a thousand words - did Godel and his progeny ever have anything discrete to say about completeness:consistency as related to GUT and so your little GEM?

[sweetser]

Hello Wanderer:

My first introduction to Godel was through the book "Godel, Escher, Bach: An Eternal Golden Braid" by Douglas Hofstadter. That inspired me to buy "Godel's Proof" by Nagel and Newman, a book that got to the point in 102 pages instead of 777. I have reread that book more times than any other on my bookshelf, so I have a feel for the technical detail. There is no technical connection whatsoever between GEM (or as far as fields, G, E, B) and Godel, Escher, Bach, other than a happy accident.

I tried a logical consistency argument on Hofstadter himself. Darn, I wish I could find my notes from his talk he gave at Harvard, but I recall it was an overview of physics history. His point may have been - my memory is fuzzy - about how physics has gone for consistency over being radical. Work that goes against consistency loses.

I got all excited by the thesis, and wrote out a question. We have a logically consistent theory for EM, that would be the Maxwell equations. We have a logically consistent theory for gravity, that would be general relativity. These two theories are in conflict with each other. So we have to choose between the two, either the Maxwell equations or general relativity, and if we choose one, the other will become a historical footnote, and is wrong. Given the connection between Maxwell, quantization, and the standard model, I would say the weight of history is for the consistency of Maxwell equations over general relativity. Granted we don't know which one is correct, but given his sense of the history of physics, which theory does he think may be shown to be most consistent in the future.

A little bit about the setting: it was a packed large lecture hall at Harvard because Hofstadter has some media draw what with the Pulitzer Prize for the book most physicists have read and enjoyed. The big names in physics in Boston were in the house. There were plenty of young people, and a few really old ones who you know probably figured out some very important things in physics or math in their prime. I was excited by my question, because it would press Hofstadter on his own thesis to say something edgy: either Maxwell's field equations were flawed, or Einstein's field equations were flawed. Logic is that tough.

After his talk was done, I got to asked my question first. I recall feeling the question was audacious: I was uncomfortable with saying in public there was a choice, and with that choice, one would say Maxwell remains right, GR is wrong, or GR remains right, and the Maxwell equations are wrong. I was that clear in the confrontation.

Hofstadter played defense. He noted he as a historian of science, and was not a researcher. He was unwilling to speculate on any future direction for physics research. As far as I can tell, no one in the audience found my question of interest. I know someone else in the audience who heard the question did not think the choice between Maxwell and GR made sense (he was a string theorist).


Einstein and Godel both worked at the Institute for Advanced Studies at Princeton. They went on walks together, discussing physics. This was late in Einstein's career, so all he was working on was unified field theory and the logical foundations of quantum mechanics. I know they collaborated on a paper together, one on closed loop solutions to the Einstein field equations. I don't know more of the history than that.

GEM is not grand, it is extra ordinary.

doug

[CarlB]

The short answer is that I know the math behind very dense gravitational sources is going to be fundamentally different from what appears in general relativity, but I do not understand the details (and the details matter). The differences will be so significant that current efforts to understand black holes will have to be dismissed should GEM theory succeed both on experimental and theoretical grounds.


Doug, I've got my Java applet for gravitation simulation almost done. It now produces correct Newtonian orbits and the vast majority of the user interface works perfectly. I'm working on the equations of motion for the general relativity version, which will assume a flat space coordinate system as done by Lasenby, Gull and Doran as in:
http://www.mrao.cam.ac.uk/~clifford/publications/abstracts/grav_gauge.html

And what does this have to do with you? Because the Lasenby Doran and Gull gauge version of gravity works on flat space, it is very natural to compare with the Newtonian version. And I'm guessing that since your theory also works on flat space it will also be easy to simulate. And therefore I've also got a spot in the simulation for the Sweetser version of flat space gravity.

What I need from you, if and when you've got it, is a flat space version of force (i.e. change in velocity per unit time) as a function of position and velocity (i.e. phase space). To get you started, here is the Java code for the Newton theory. (All units are natural.)

private DeltaV Newton(PhaseSpace PS) {
DeltaV DV = new DeltaV();
double R = Math.sqrt(PS.X*PS.X+PS.Y*PS.Y);
R = R*R*R;
DV.DVX = -PS.X/R;
DV.DVY = -PS.Y/R;
return DV;
}

The above routine computes "DV" or change in velocity, from "PS" or position in phase space. The calculation is made in two dimensions (i.e. the z direction is ignored and all orbits stay in the same plane). The change in velocity therefore has two components, (DVX,DVY), and the phase space has four components, (X,Y,Vx,Vy).

For the Newtonian calculation, the force does not depend on the velocity, so there is no use of things like "PS.Vx" or "PS.Vy".

Anyway, the orbit is found by integrating the acceleration numerically. The applet allows you to change the initial condition and see how the particle orbits for the different physics assumptions differ.

I've written it so that the physics is encapsulated in a simple subroutine as shown above so that the Java will be easily modified by someone who doesn't want to learn the ins and outs of object oriented programming of a user interface.

Now when this is done, I'll put the applet, along with its source code, up at my website //http:www.GaugeGravity.com , which is intended to promote the Lasenby, Gull and Doran version of GR. I've put my current (test only) version up on the net here:
http://www.gaugegravity.com/testapplet/SweetGravity.html

The buttons Einstein, Newton and Sweetser don't work, and I need to readjust it so that it runs faster, which you can accomplish (with some loss in numerical precision) by hitting the "faster" button a couple times. The numerical precision isn't adjusted according to how close you are to the gravitating point, so it can get a bit dodgy on near misses. And when you alter the initial parameters (with a carriage return), it stops the simulation and to start it again you have to press "go". I'll make the stop/go button be red / green so it is more obvious what is going on.

By the way, to translate from LGD's version of the Schwartzschild metric back to the usual GR theory, the reason that they work on a flat space (i.e. coordinates are x,y, z and t), is that their ds^2 is not diagonal, but instead includes a drdt term. Thus their solution is not symmetric with respect to time, which I think is a great thing.

Carl

[A_Wanderer]

[QUOTE=sweetser]Hello Wanderer:

My first introduction to Godel was through the book "Godel, Escher, Bach: An Eternal Golden Braid" by Douglas Hofstadter. That inspired me to buy "Godel's Proof" by Nagel and Newman, a book that got to the point in 102 pages instead of 777.[QUOTE=sweetser]


Hi.

Thud. Seven hundred and 77 pages of a POPular derivation of a technically and imaginitively profound mathematics. You have more patience and tolerance than I could muster in a hinayana trance! Thank heavens you found the Nagel/Newman. Better yet would be to first extract Godel from that framework and wrestle with it unencumbered by focal bias.


What I was suggesting was: Does not Godel's grand reductio establish that any GUT which expression is more complete and and so (inversely?) LESS consistant (in comparison to any theory on an opposing gradient) shall be closer to the truest description/expression/model of the measurable universe (all its forces from the plancky, pointy ultrino to the BB and any and all strange, possibly stringy, curved, n-dimensional, & etc. in between or amidst) - given that the impetus for GUT from the start IS unity/completeness? Despite, I know a 2500+ year historical record with a rather strong attraction to the beautiful elegance of consistancy.

It is most counterintuitive that inconsistancy then becomes a marker for fundamental veracity/agreement the more complete the GUT (or for that matter the GEM) expression. Yet this surprising in-your-face inconsistancy, oddly, or evenly, may be a consistant balance of inconsistancies when placed next to wave/particle and the where/when of quantum subatomics.

I briefly apologize for being rather off-topic - Sorry!!!
But sometimes, I can just not resist, and have a weakness for the gravity of my own conceits. And groaners.

I dislike my own speculations. (Clearly NOT theories, nor even hypotheses.) I am very fond of consistancy.


the Big Bang - Everything from nothing! Is this not perfect elegance?


fondly,

Bill Snyder
aka
A_Wanderer

[sweetser]

Carl:

Wow!

Sorry, just a gut reaction. It will take a few days to think about this, but thought I should put on record my initial reation.

Another reflex reaction...

>A flat space version of force (i.e. change in velocity per unit time) as a function of position and velocity (i.e. phase space).

Spacetime can be treated as flat. Force cannot be treated as m dv/dt. Instead, use the chain rule Luke.
F = m dv/dt + v c dm/dR = - G M m/R^2
If you get GR correct, you should end up with an animation that is identical.

Damn, need to get ready for work...

Kudos, kudos,
doug

[CarlB]

Kudos, kudos, doug

Yes, the Cambridge geometric algebra guys are awesome.

I've uploaded the next version of the applet. This one draws multiple test bodies, which makes for a more pretty display. I've set the initial conditions so as to give a demonstration of the conservation of angular momentum in the Newtonian potential:

http://www.gaugegravity.com/testapplet/SweetGravity.html

I guess it's possible that I'll add the GR simulation within the next 24 hours, and if so, I'll update this note accordingly.

Carl

[Norman Albers]

I am running hard to catch up with you and am almost there. Forgive me for not yet having read this entire disucssion; I shall, but need to start communicating ideas with you. We need to let go of the zero current and charge densities. I have shown minimal, clear construction of electrons and photons by allowing infinitesimal divergence in the field. Now I am studying GR (last chapter in Adler, Bazin, Schiffer) to find the lay of the land. We must allow the vacuum to cook up this inhomogeneous stuff! I have found fundamental problems, in that my photon fields can express a Lagrangian with three different orders of time derivatives, not useful.

[sweetser]

Hello Bill:

What I was suggesting was: Does not Godel's grand reductio establish that any GUT which expression is more complete and and so (inversely?) LESS consistant (in comparison to any theory on an opposing gradient) shall be closer to the truest description/expression/model of the measurable universe (all its forces from the plancky, pointy ultrino to the BB and any and all strange, possibly stringy, curved, n-dimensional, & etc. in between or amidst) - given that the impetus for GUT from the start IS unity/completeness?

I am unfamiliar with any way of taking Godel's incompleteness theorem and translating that into either a field equation or an action. As far as I know, the theorem remains in the house of logic.

Did you know that logic can be translated directly into algebra? Scientific American had an article on this a number of years ago. The translation is particularly relevent for fuzzy logic. Here's the cannonical example. There is a card which says on one side: "The statement on the other side of this card is true." OK, call that x. On the other side, the card reads: "The statement on the other side of this card is false." That would then be 1-x. So what is x?
x=1-x
2x=1
x=1/2
Math I can solve :-) The card is full of half truthes. The point of this story is that it may become possible to translate something at the foundations of logic into algebra someday. What would be required is a way to translate numbers into differential equations. I hope that sounds like BS, because I am clueless as to how to do that.

...next to wave/particle and the where/when of quantum subatomics.
I'd have to start a new thread on this forum to treat this right, but my one liner reply is that if you do 4D calculus correctly, the reason causality is different between classical physics and quantum mechanics is clear. Check out the video "Why Quantum Mechanics is Weird" at TheStandUpPhysicist.com.

I dislike my own speculations. (Clearly NOT theories, nor even hypotheses.) I am very fond of consistancy.
My level of discomfort is dictated by how much of it can be translated directly into math. I have wondered about Godel, and am totally uncomfortable with it since zero can be translated into algebra. I am very comfortable saying I have a proposal to achieve unification of gravity and EM because it is so specific, I can feed it directly into Mathematica, a fact that so far has failed to make an impression with professional physicists (a busy lot).

I do have speculations in between. Let me give you the details of one of them. The standard model has the gauge symmetry U(1)xSU(2)xSU(3). There are obvious questions to ask: why three groups, why these in particular? The answer we have is clear: we have no idea. This is the kind of problem that is beyond hard. There is nothing you can do to "work" on it. All you can do is be aware of the clarity of our ignorance about the standard model.

What kind of symmetry characterizes the 4D wave equation at the heart of my unified field proposal? Because my background involved much work with quaternions, one of the things I know is that the group SU(2) are the unit quaternions. How does one make a unit quaterion? Easy, just make sure the diagonal is zero, A-A^*. The unit quaternion has three of the four degrees of freedom available to a quaternion. What should the fourth one do? The group U(1) is usually represented as the complex numbers with a norm of one. The group is Abelian, that is to say it commutes with other members of the group. Quaternions are well know as not being Abelian. Under special circumstances, quaternions can behave like an Abelian group. The prime example is if all the quaternions point in the same darn direction. I realized a normalized quaternion would commute with its SU(2) sidekick, like so:
\frac{A}{|A|} (A-A^{*}) = (A-A^{*}) \frac{A}{|A|}
This is the electroweak symmetry. If I write my GEM field equations like so,
J_q - J_m = \square^* \square \frac{A}{|A|} (A-A^*)
this has U(1)xSU(2) symmetry! Now the unified field theory contains gravity through the diffeomorphism symmetry, EM through the U(1) symmetry which is not perfect due to massive particles, and the SU(2) symmetry of the weak force which is behind nuclear radiation.

Now we come to the part that I do not understand: SU(3) symmetry. I know its Lie algebra has to have 8 parts to it. The multiplication table has to be different that the standard quaternion multiplication table. It is possible that the \square^{*} \square part of the GEM field equations have what it takes. There are 8 parts, that is easy. What I (and only I) call the Euclidean product, a^{*} b, is not assiociative, because (a b)^{*} c \not= a^{*} (b c). The Hamilton product, a b is associative because (a b) c = a (b c). Quaternions under the Euclidean product still are a group: there is an identiy (1, 0, 0, 0), there is always an inverse, and all products are still quaternions. I am not enough of a professional math guy to prove SU(3) connection. The one course I failed, got an outright E, was a Harvard class on group theory and particle physics. As for the E, it is so a person failing at Harvard remains a cut above the rest.

Now both questions about the standard model have the chance of being answered. The symmetry of the GEM unified field equations is Diff(M)xU(1)xSU(2)xSU(3). The four forces of Nature, gravity (Diff(M)), EM (U(1)), the weak force (SU(2)) and the strong force (SU(3)) all fit comfortably in the same home. This is the sort of specific speculation that keeps me crazy.

doug

[sweetser]

Hello Norman:

Sean Carroll's lecture notes on General Relativity are freely available on the web. I printed them out, and for each chapter, transcribed them into my own form. That process took two months, and now I think I have a concrete handle on the math behind GR.

This forum is more a call and response venue: a question gets asked, I reply, and we go back and forth a few times. It is not organized in a logical way :-) For that, I recommend either downloading the half hour shows at TheStandUpPhysicist.com, or ordering the DVD (1 sale in the world so far, to a physicist friend). Although each show is only a half hour, it would take more time if you tried to confirm that the math makes sense.

doug

[CarlB]

Now both questions about the standard model have the chance of being answered. The symmetry of the GEM unified field equations is Diff(M)xU(1)xSU(2)xSU(3). The four forces of Nature, gravity (Diff(M)), EM (U(1)), the weak force (SU(2)) and the strong force (SU(3)) all fit comfortably in the same home. This is the sort of specific speculation that keeps me crazy.

I think that trying to get SU(3) in there is just one symmetry too far.

You do not have an explanation for why there are exactly three generations of particles. These kinds of things, that is, the representations of these symmetries that happen to correspond to the elementary particles, are at least as important as the symmetries themselves.

But if you assume that the leptons and quarks are composites of three particles, then the SU(3) becomes natural at the same time as you get three generations automtically. And the masses of the leptons become understandable by the Koide relation.

So long as your algebra contains the 3x3 complex matrices, it is of course possible to force SU(3) into it. But this does not imply anything interesting. One could fit any of a large number of symmetries into 3x3 complex matrices. To show a real derivation, you would need to not only pick out SU(3) and all that, but you also need to show why the particular representations that are seen in the elementary particles show up.

As an example, consider the electroweak symmetry. It is not enough to show SU(2) x U(1). What you nee to show is that one ends up with an SU(2) doublet and two SU(2) singlets for each flavor.

If you take a look at the algebra, you will find that this form, a doublet and two singlets, arises naturally. But the same cannot be said of SU(3). SU(3) arises much more naturally as the result of assuming that the elementary particles are composite.

But hey, if you want to put SU(3)xSU(2)xU(1) into complex 4x4 matrices (as is more natural for a Clifford algebra or quaternions, I suspect), then you should read this paper:
http://arxiv.org/abs/math.GM/0307165

Carl

[bda]

Hi Doug,

Pardon me for hopping in like this but I thought you could possibly use one little piece of information about the exponential metric. You may or may not know that the exponential metric seems to have been first proposed by Houssein Yilmaz in "H. Yilmaz, New approach to general relativity, Phys. Rev. 111(5), pp. 1417-1426 (1958)". You can do a search on "Yilmaz metric" and a lot of references come up.

In its Euclidean form the exponential metric has been used by Hans Montanus and myself; among other references see for instance: "J.M.C. Montanus, Proper-time formulation of relativistic dynamics, Found. Phys. 31(9), pp. 1357-1400 (2001)" and "J.B. Almeida, Geometric drive of the Universe's expansion, http://www.arxiv.org/abs/physics/0507102 ."

Best regards,

Jose

[bda]

But hey, if you want to put SU(3)xSU(2)xU(1) into complex 4x4 matrices (as is more natural for a Clifford algebra or quaternions, I suspect), then you should read this paper:
http://arxiv.org/abs/math.GM/0307165

Carl

As author of that paper may I suggest also http://www.arxiv.org/abs/quant-ph/0606123

Jose

[sweetser]

Hello Jose:

The exponential metric is elegant, in an almost measurable way. The number of strokes of the pen required to write it down is small. If the exponent goes to zero, the terms run to one. Exponentials appear over and over again in critical physics equations. There is a good reason why. The exponential is a small step away from unity that embodies a simple harmonic oscillator.

I am not the best student of the literature, but I do have a copy of the Yilmaz paper. I was able to find 4 papers that had the exponential metric:

N. Rosen. (note: this appears like the first reference)
A bi-metric theory of gravitation.
General Relativity Gravitation, 4(6):435-447, 1973.

H. Yilmaz.
Physical foundations of the new theory of gravitation.
Annals Phys., 101:413-432, 1976.

S. Kaniel and Y. Itin.
Gravity on parallelizable manifolds.
113 B(3):393-400, 1998.

Keith Watt and Charles W. Misner.
Relativistic scalar gravity: A laboratory for numerical relativity.
1999.

The way these four papers generated that metric were not elegant. Misner says he is doing this just so numerical calculations go faster, an argument of convenience over conviction. Rosen tosses in another metric field, and because that can store energy and momentum, it means that for an isolated source, gravity waves can have a dipole mode of emission, which disagrees with experiment. I don't recall the details of the other two, other than I did not get excited by said details. I believe they were variations off of the rank 2 field theory of GR, whereas I am trying to show a rank 1 field theory is a different way to implement a metric theory for gravity via symmetry.

The field equations I have were called beautiful by Feynman (in reference to the EM equations in the Lorentz gauge, not as a unified field theory). The equation that generates the exponential metric is also elegant in its directness - it is the divergence of the connection:

\rho_{m}=2\partial_{\mu}\Gamma_{\nu}{}^{\:0\mu}A^{ \nu}

I remain stuck in 4D where few people do gravity research. Retro man!

doug

[sweetser]

Hello Carl:

To make a direct connect to the physics literature, we need a different kind of applet. The issue is to generate a velocity profile for a disk galaxy given the mass distribution function as a function of the radius. Let me do this in crude ASCII graphics. We start with a mass/area function that has an exponential decay:

m/area
|.
|.
| .
| ....
-------......
R

From that we calculate the velocity profile:

vel. calculated
| .
| . .
|. .
|.
-------------
R

This has the "Keplarian decline", which is what Newton's theory should generate if the applet is written correctly. Test out GR in the flat spacetime if you like, overlay them, and a difference should not be visable at this resolution. What is seen in Nature is this:

vel. observed
| . . .
| .
|.
|.
-------------
R

When programming, there are so many ways to do things, it helps to be specific. Let's use the mass distribution profile for the galaxy NGC3198. It is a galaxy that is too faint for you or I to ever see with our eyes, no matter what size telescope we used (something like a magnitude of 23). The velocity ramps up to 150,000 m/s and stays there. The total mass is 1x10^{40} kg. The mass per area as a function of the radius is m/area = 37 Exp(-R'/2.23') solar masses/pc^2.

The velocity profile is not the only problem with Newton's law. The solution is unstable, so disk galaxies should collapse, but they don't. That's unreasonable, because galaxies last a long time.

Let's consider the forces: it is gravity versus the centrifugal acceleration:
m \frac{V(R)^2}{R} - \frac{G m M}{R^2} = \frac{d m V}{d t}
To see the proverbial Keplarian decline, the centrifugal forces and gravity are in balance, so solve for V:
V=\sqrt{\frac{G M}{R}}
If the mass drops off as the square root of 1/R, the velocity can stay constant. The observed light curve instead makes it look like the mass drops off exponentially, much faster than the square root of 1/R.

Algebraically, there are three things that can be done. The first is to "Stuff the Mass box", which goes under the name of the dark matter hypothesis. Folks cannot figure out what dark matter is, but they do know we need more of it than the stuff we know huge amounts about. We cannot see the stuff directly yet despite the extraordinary care astronomers use to analyze light. Since we cannot see it, and don't know what it is made of, folks at computer terminals make up a dark matter distribution that can generate the velocity profile and lead to a stable visible mass distribution. I hope this area of study sounds suspicious to you.

The second approach is "Switch the Equation". It goes under the name of MOND, or Modification Of Newtonian Dynamics. It is claimed that when gravity gets super wimpy, then it becomes a 1/R force law:
\frac{V(R)^2}{R} - \frac{G m M}{R^2} = \frac{d m V}{d t} iff \frac{G M}{R^2} > 10^{-10}
\frac{V(R)^2}{R} - m \sqrt{\frac{a_0 G M}{R^2}} = \frac{d m V}{d t} iff \frac{G M}{R^2} < 10^{-10}, a_0 = 10^{-10}
MOND does a good job with real data. That is is strength. The theory is a weakness. Recently someone figured out the Lagrangian required to get this sort of thing to work out, and I heard it apparently is not a pretty site. Suspicious? You should be.

And the third possibility is "The relativistic chain rule for a distributed mass source". That probably is not familiar. Well, the chain rule should be. What does a force do? If you say it is mA, technically you are wrong. Force is a change in momentum, so F = m\frac{d V}{d t}+ V \frac{d m}{d t}. The second term is literally the stuff of rocket science. For a spiral galaxy that doesn't change its mass over eons, that term can safely be assumed to contribute nothing.

Technically, my expression for force was also wrong. The relativistic force is F^{\mu}=\frac{d m V^{\mu}}{d \tau}. The important thing to focus on is the d \tau. That Greek letter is for a change with respect to the spacetime interval, not exclusively the time interval. This means that formally, it could be a change in space that the rocket term describes. The classical force would then be:
\frac{V(R)^2}{R} - \frac{G m M}{R^2} = m \frac{d V}{d t}+ V \frac{d m}{d R/c}
It is like the rocket force term, but applies to space. There is not a label for it, so I'll call it the rocket-space term, a flip of space rocket because this is a flippy idea. What a rocket-space term says can be described. What does a force do? It is a change in momentum. There is the familiar sort of change in momentum, when something changes its velocity. Well, in the outer reaches of a spiral galaxy, there is ZERO change in velocity, even though there still is a gravitational force changing momentum. That's a real puzzle. The force must be changing something else: where mass is in space. That is exactly the kind of curve I wrote earlier - the amount of mass per area drops exponentially in th outer reaches. The change in momentum as one moves away from the center is seen as the change in mass times a constant velocity. In words, it is an exact match.

This is really a new idea. That is rare in physics. It is worth a try, so see if it is consistent with data from a specific galaxy. I should say I have a way to derive this expression, but it takes longer to do so.

****
So this is what I am thinking about programming-wise. Presume a mass/area function of m/area = 37 Exp(-R'/2.23') solar masses/pc^2. Use the Newtonian equation to calculate the velocity for R over a range of say .1 to 30 arcseconds. Plot velocity versus R. I know it is easier to skip the units, but try not to as a check that the max velocity is in fact 150,000 m/s, and the total mass is 10^{40}. Newton's theory gets the peak right. Then include the rocket-space term and see what happens to the curve.

doug

[bda]

Hi Doug

Hello Carl:

And the third possibility is "The relativistic chain rule for a distributed mass source".
......

This is really a new idea. That is rare in physics. It is worth a try, so see if it is consistent with data from a specific galaxy. I should say I have a way to derive this expression, but it takes longer to do so.



Actually I wrote about this a few years ago with someone who has since passed away. We never agreed that the paper was fit for even placing in arxiv, so it remaind in my hard disk untill now.

Note that I had not yet got the exponential metric quite right at that time, so now the paper would have to be revised in that aspect; this should not produce significant alterations to the main conclusions.

Jose

[bda]

Doug,


Note that I had not yet got the exponential metric quite right at that time, so now the paper would have to be revised in that aspect; this should not produce significant alterations to the main conclusions.

Jose

I had a quick look at the paper and I now realize there are a few mistakes, none of them serious and all easy to correct. The main difference after correction will consist on doubling the exponent in Eq. (7), but that will have no consequences for the discussion and conclusions.

Jose

[sweetser]

Hello Jose:

The exponential metric does not lead obviously to the space-rocket term. There is an error of omission in this express:
\frac{V(R)^2}{R} - \frac{G m M}{R^2} = m \frac{d V}{d t}+ V \frac{d m}{d R/c}
Force is a vector equation. One must include them. You'll notice that the space-rocket term points along \vec{V}, not along \vec{R}! The correct vector expression is:
\frac{V(R)^2}{R}\hat{R} - \frac{G m M}{R^2}(\hat{R}+\hat{V}) = m \frac{d V}{d t}\hat{R}+ V \frac{d m}{d R/c}\hat{V}
The rocket-space term suggests gravity works classically in a new direction. This is ONLY relevant for masses that are distributed over a significant amount of space. It is the passive mass small m whose distribution is changed. Of course the sum of the passive small mass is the active mass M. If you feel it is a bit confusing to have the mass in different part of the same equation, that is the way rocket science works!

doug

[CarlB]

V = \sqrt{GM/R}
If the mass drops off as the square root of 1/R, the velocity can stay constant.

Please forgive a naive amateur, but to get the velocity constant, don't you have to increase the mass proportional to R^2, that is the mass per unit area has to be constant? [Edit: Okay, now I see it. Integral of mass has to be proportional to R, but mass is proportional to area. Nevermind.]

Were I asked to do the simulation you're talking about, I would do it with a large number (maybe 1000) sample points. It would run amazingly slowly in Java. Amazingly slowly. But you could eventually get a result out of it. That's a second stage operation. For now, you still need to tell me what the dv/dt equation is for just a test mass orbiting around a point mass. It turns out that extracting this information from flat space gravity theorists is harder than I expected.

I've now added the logic to throw three types of balls in the air, Newton and two others. I believe I have the correct equations for general relativity, and for the Cambridge geometric algebra relativity, and I should upload the applet later tonight. I've also changed the colors, and added bright white test masses so that you can see them retracing their orbits.

Carl

[bda]

Doug

Hello Jose:

The exponential metric does not lead obviously to the space-rocket term.

I agree; no matter which metric one uses the main thing is that velocity is basically determined by

\frac{v^2}{r} = -\frac{\mathrm{d}V}{\mathrm{d} r},

where V is the gravitational potential; the metric introduces a correction which is significant only for very small r and large M.
You then plug in V =G M(r)/r and apply the the derivation rules
this then results in

v^2 = \frac{G }{r}}\left(M - r \frac{\mathrm{d} M}
{\mathrm{d} r} \right);

it is then obvious that one can get constant velocity.

In my paper I apply this to measured velocity profiles of real galaxies and derive the respective mass distributions; I then compare those with observed light intensity and H1 profiles.

Jose

[CarlB]

Doug, I found another author working on combining E&M with gravity, but from a different perspective. I found it quite convincing, and likely compatible with what you're doing, but simpler for a flat space particle guy like me to understand:

For example, see "An Electromagnetic Basis for Gravity" listed on:
http://www.mass-metricgravity.net/

Some papers that you can download from arXiv are:
http://www.arxiv.org/abs/physics/0012059
http://www.arxiv.org/abs/physics/0101033
http://www.arxiv.org/abs/physics/0601013

I just ordered this book which seems to be the most inclusive
reference (the links to the APS preprint server on his website no longer work as APS closed it down so the only free papers I could find on the web are the above three arXiv papers):
http://www.buybooksontheweb.com/peek.asp?ISBN=0-7414-1466-X

The two of you talk quite over my head with respect to gravitation. Anyway, you should send him an email, and see what the differences are between your theory and his. If his theory is the flat space version of yours, then I now have an equation of motion that I can simulate for your version of gravity.

I loved his most recent paper on arXiv, (last one listed above). It explains the apparent acceleration in the Hubble data as being a consequence of an interaction between gravitational potential energy and time. This makes the red shift less than what is apparently observed by just half, which is just enough to eliminate the apparent acceleration and turn the Hubble expansion into a constant velocity again.

Maybe the above paper has something to do with the MOND thingy, but like I said, you can talk with him about it, it's over my head.

Carl

[sweetser]

Hello Carl:

I was told by an MIT professor of some standing that I could not claim to have a modern field theory unless I had a Lagrange density. This was very scary at the time because I had the field equations, and had never worked with a Lagrangian before in my life. It was something I had read Feynman talking about, but always thought it was too difficult for me to ever comprehend. Turns out that it is not too tough. If anyone reading this thread thinks the nuts and bolts of a Lagrange density are too difficult, download lectures 1 and 2 from TheStandUpPhysicist.com where I go through the details. If you rewrite the equations I present there yourself, and give yourself some time, the logic is straightforward - it has to be, this is the basic math of how things exchange energy in the Universe.

It was really scary, I didn't think I could find a relevant Lagrange density. I remember how dumb I felt sitting down trying to find one that would generate my field equations. No joking aside, I suck at math. The only way this could work was if it was a bit simpler than EM, which it turned out to be.

The point of this story is that I hold up other research to the same standards that I set for myself. So I did this google search: site:mass-metricgravity.net Lagrangian
That looks specifically at everything Collins has put up for the word Lagrangian. I checked for Lagrange density, action and Hamiltonian too, but found nothing. One needs the Lagrangian to do quantum. So I'll wait on investing effort into understanding his work until he has one.

Nothing like having a well-defined, tough test. It means I can skip over reading most papers. Virtually all of string theory would fail the same test.

doug

[CarlB]

The point of this story is that I hold up other research to the same standards that I set for myself. So I did this google search: site:mass-metricgravity.net Lagrangian
That looks specifically at everything Collins has put up for the word Lagrangian.

Tisk. Tisk. He probably won't read your theory because you have no published articles, nothing up on arXiv, and no academic position. If you went to his website, you'd discover that it is only a one page stub and searching it is pointless. He put his unpublished research onto the APS preprint server and that preprint server is no longer available. So there is no way to search the web for the information you're asking for. When I tell you that I see similarities in your work that should be enough to interest you.

He gave me the equations of motion for his theory and it's easy enough to turn that into a Lagrangian. I've asked him by email if he has done this. I would guess that he has, but if not, it's easy enough to do it myself. L = T + V, uh... well... L = T \pm V or something like that. and he's quite clear as to what kinetic energy and potential energy is in his theory.

My own point of view is that the equations of motion (or, in QFT, the wave equation) is more fundamental than the Lagrangian. Of course you are aware that QM has been stuck for 30 years. The physicist who gave you that particular piece of advice could very well be a part of the problem rather than the solution.

Oh, and I'm still waiting for you to provide me with equations of motion that work in flat space. I expect that I will have simulations with GR and gauge gravity comparisons with Newton by Monday or Tuesday. The other half of working on all this is that I feel that I am begining to get a better understanding of gravitation. This is stuff I should have done back in grad school.

Carl

[sweetser]

Hello Carl:
He probably won't read your theory because you have no published articles, nothing up on arXiv, and no academic position.
Yup, I know that. And I have a simple standard before I try and publish it: I need someone who understands the Maxwell equations at the Lagrange level, and knows the formalism of general relativity to read my work and see if it is valid. I'll remain in this state until that time. And if I get such a review, and they show me where I have made a mistake, I will point it out here. Note: I have made errors in this thread, but so far, none of them are killers.

If you went to his website, you'd discover that it is only a one page stub and searching it is pointless. He put his unpublished research onto the APS preprint server and that preprint server is no longer available. So there is no way to search the web for the information you're asking for. When I tell you that I see similarities in your work that should be enough to interest you.

If you have a mature proposal, you'd mention the action and the Lagrange density outside more formal presentations.

He gave me the equations of motion for his theory and it's easy enough to turn that into a Lagrangian. I've asked him by email if he has done this. I would guess that he has, but if not, it's easy enough to do it myself. L = T + V and he's quite clear as to what kinetic energy and potential energy is in his theory.

It is in the first paper, equations 65 and 66: T + U = 0. Oops. That's a problem. There is no way to take 0, and do a variation on that to generate the field equations. That is a deadly technical flaw. It also makes quantizing the theory impossible.


My own point of view is that the equations of motion (or, in QFT, the wave equation) is more fundamental than the Lagrangian. Of course you are aware that QM has been stuck for 30 years. The physicist who gave you that particular piece of advice could very well be a part of the problem rather than the solution.

If you have a self-consistent theory, you must be able to go from the Lagrangian to the field equation and to the force equations and to the stress-energy tensor and to the Hamiltonian and through the process of quantization. This is a matching set. It requires a lot of calculus, none of it particularly hard, but the math steps are intimidating. For me, one is not better than another, but having the complete set is necessary before you claim to have a field theory for anything. This is an OLD standard, going back to the nineteenth century. It is cool that the Lagrangian of those times still plays a central role once we got a handle on quantum mechanics in the early part of the twentieth century. The question my approach addresses was first raised in he 1760's when Joseph Priestly realized that electrostatics must be an inverse square law based on a discussion with Ben Franklin. Certainly it was clear to all involved by 1930 that any explanation of gravity had to get along with quantum mechanics, and GR did not. So for me, the time line of these issues is longer.


Oh, and I'm still waiting for you to provide me with equations of motion that work in flat space. I expect that I will have simulations with GR and gauge gravity comparisons with Newton by Monday or Tuesday.

The exponential metric is identical to the Schwarzschild metric to first order PPN accuracy. If programmed correctly, there is no way to tell the difference. Go literally a million-fold more accurate, and the exponential metric has 12% more bending than GR. For a point source, this is not the way to see a difference.

from an earlier post...
Were I asked to do the simulation you're talking about, I would do it with a large number (maybe 1000) sample points. It would run amazingly slowly in Java. Amazingly slowly. But you could eventually get a result out of it. That's a second stage operation.

This slow, repetitive calculation for a thin disk galaxy with an exponentially decaying mass distribution is where it's at. You should have the program generate the velocity profile graph and save it. It turns out one needs elliptical integrals to solve the rotation profile for such a galaxy, and no, I am not good enough to do that (Alar Toomre of MIT was the first to do it in the early 60s, kind of amazing since one would have thought the problem was easy, but it is not).

If one eliminates V hat from this equation:
\frac{V(R)^2}{R}\hat{R} - \frac{G m M}{R^2}(\hat{R}+\hat{V}) = m \frac{d V}{d t}\hat{R}+ V \frac{d m}{d R/c}\hat{V}
what remains is Newton's law. Now I could go about and figure out how my proposal tweaks this a bit away from plain-Jane Newton, but if I did it right, it should be no different from GR at the resolution of the applet. It is for a system with a distributed mass that GEM theory is noticeably different.

doug

[CarlB]

It is in the first paper, equations 65 and 66: T + U = 0. Oops. That's a problem. There is no way to take 0, and do a variation on that to generate the field equations. That is a deadly technical flaw. It also makes quantizing the theory impossible.

I went and looked at the equations 65 and 66. In Newtonian theory, when a particle is at rest at infinity, its potential energy is zero and its kinetic energy is zero. These add to zero. If you examine particles which have dropped in from infinity, the kinetic energy increase and the potential energy decreases so that the sum stays at zero.

This is a well known fact of Newtonian gravitation. This is not the place where you would start with an effort at quantizing a theory. To do that, you need to write the T and U for all possible particle positions and velocity, not the values for a very specfic single trajectory.

The exponential metric is identical to the Schwarzschild metric to first order PPN accuracy. If programmed correctly, there is no way to tell the difference. Go literally a million-fold more accurate, and the exponential metric has 12% more bending than GR. For a point source, this is not the way to see a difference.

What I'm hoping for is that you will write me a force equation for movement of uncharged particles on a flat space metric. The Schwarzschild metric is not flat, so under this requirement, you will not be giving me something that is identical to the Schwarzschild metric.

For example, the Cambridge gauge gravity is built on a flat space metric and sure enough, their equations of motion are different from that of the Schwarzschild. But they give orbits that are exactly identical to that of Schwarzschild. I know this very well from the effort of programming the computer to draw the orbits. The Schwarzschild particles will get stuck on the event horizon while the Cambridge particles go on to the singularity. And for orbits that avoid the event horizon, the two theories will give results that are identical (after correcting for their different assumptions about the relationship between radius and coordinate time). This I will demonstrate by showing particles that collide in one theory, collide also in the other, but with their positions as a function of time being otherwise (i.e. at different radii) different.

Can you give me a link to your flat space metric or anything else I might start with so that I could get an equation of motion? The Collins proposal matches GR to well within the current accuracy of measurements of light bending and Mercury precession, but I know that my simulation will show differences. I'm quite good at this sort of thing. I should have it running in a few days as I've half programmed in the (rather difficult) general relativity and gauge solutions.

In the unlikely event that I am unable to distinguish two different theories with double precision arithmetic, then it is time to talk about some other, more complicated simulation. At first, baby steps only. But I doubt that your theory will differ so little from GR that I cannot distinguish it with the accuracy I'm working at.

Carl

[bda]

Hi Doug,

The exponential metric is identical to the Schwarzschild metric to first order PPN accuracy. If programmed correctly, there is no way to tell the difference. Go literally a million-fold more accurate, and the exponential metric has 12% more bending than GR. For a point source, this is not the way to see a difference.

Well, yes and no. The Schwarzschild metric in isotropic coordinates is


\label{eq:isotropic}
\mathrm{d}\tau^2 =
\left(\frac{\displaystyle 1-\frac{M}{2r}}{\displaystyle 1+\frac{M}{2r}}\right)^2
\mathrm{d}t^2 -
\left(1+ \frac{M}{2r}\right)^4 \left[ \mathrm{d}r^2 - r^2 \left(\mathrm{d}\theta^2
+ \sin^2 \theta \mathrm{d}\varphi^2 \right) \right];


and the exponential metric is


\mathrm{d}\tau^2 = \mathrm{e}^{-2M/r} \mathrm{d}t^2 - \mathrm{e}^{2M/r}
\left[ \mathrm{d}r^2 - r^2 \left(\mathrm{d}\theta^2
+ \sin^2 \theta \mathrm{d}\varphi^2 \right) \right].


It is true that they are PPN equivalent but this criterion applies only far away from a black hole. If Carl makes the simulation near a black hole horizon the orbits originated by the two metrics will be very different.

Jose

[sweetser]

Hello Jose:

For now, my focus is on a huge problem in classical gravitational theory, the rotation profile of thin disk galaxies. There might have a black hole in the center, but the big problem is after the peak velocity is reached. Newtonian theory, as an approximation of GR, gets the peak correct. The exponential metric itself does not correct the problem. It is the rocket-space term that is my hypothesis for a solution to the problem. Gravity still is in play after the peak is reached, but the change in momentum is not due to a change in the velocity of the particles. Instead the change in momentum is due to the change in the amount of mass of the galaxy after the peak velocity is reached. I have to show numerically that both terms are in play.

doug

[sweetser]

Hello Carl:

This might do the trick. I have a potential which solves the 4D wave equation. For Newton, a 1/R potential solve \rho=\nabla^2 \phi. For the 4D wave equation in this classical problem, we want as solution really, really close to 1/R, but not quite. Hence the use of perturbation theory. It gets a bit more complicated in details to have the potential apply only to gravity, and not gravity and EM (such is the plight of unified field theory). In the interest of full disclosure, the weak gravity, electrically neutral potential is here:
http://theworld.com/~sweetser/quaternions/talks/IAP_3/1119.html
The important thing is the derivative of the potential. That is much simpler. It is:
\nabla^{\mu}A^{\nu}=-\frac{\sqrt{G}M}{c^2 \tau^2}\left(\begin{array}{cccc}
1 & 0 & 0 & 0\\
0 & 1 & 0 & 0\\
0 & 0 & 1 & 0\\
0 & 0 & 0 & 1\end{array}\right)
This derivative of the potential has the correct inverse square dependence. You drop this into Lorentz force law,
F^{\mu}=\frac{\partial m U^{\mu}}{\partial \tau}=-\sqrt{G} m U_{\nu}/c \nabla^{\mu}A^{\nu}
Contract this and do some simplifications, and one has an expression for the 4-velocity:
(\frac{\partial U_0}{\partial \tau}, \frac{\partial \vec{U}}{\partial \tau}) = (e^{-G M/c^2 R}, e^{G M/c^2 R})
This is still a relativistic expression. In a low velocity limit, one ends up with:
\frac{d v}{d t} = e^{G M/c^2 R}
That should be simple enough to program, a baby step as it were.

doug

[bda]

Doug



For now, my focus is on a huge problem in classical gravitational theory, the rotation profile of thin disk galaxies. There might have a black hole in the center, but the big problem is after the peak velocity is reached. Newtonian theory, as an approximation of GR, gets the peak correct. The exponential metric itself does not correct the problem. It is the rocket-space term that is my hypothesis for a solution to the problem. Gravity still is in play after the peak is reached, but the change in momentum is not due to a change in the velocity of the particles. Instead the change in momentum is due to the change in the amount of mass of the galaxy after the peak velocity is reached. I have to show numerically that both terms are in play.



Look at my paper attached to post #196; it does just what you want.

Jose

[sweetser]

Hello:

I went to see "Jim's Big Ego" at the Lizard Lounge in Cambridge, MA. The band wrote "Math Prof. Rock Star", which I subsequently have used as the title song for the "Stand-Up Physicist" community access TV series. The concert was good, skilled musicians making sardonic slashes at society. I hung out afterward to give Jim a DVD with twelve of the TV shows. We chatted about physics (OK, it was mostly me doing various physics riffs), but Jim had a question: instead of gravity being an attractive force going forward in time, why couldn't gravity be a repulsive force like EM going backward in time? I tried to answer the question there and then, and in this post provide a more permanent record of my reply.


If we think about gravity using general relativity, the theory put forward by Einstein and Hilbert, the question is poorly formed. In general relativity, gravity is exclusively about geometry. There is no force of gravity to be either attractive going forward in time, nor a repulsive force going backward in time. A force is something that can act at a point in spacetime: you apply a force to my nose with your fist, it breaks. General relativity doesn't work

that way, it is nonlocalizable. This can be seen by looking at the Riemann curvature tensor, which is the difference between two connections (the difference of two derivatives and the difference between two products of these two connections):

R^{\rho}{}_{\sigma \mu \nu}=\partial_{\mu} \Gamma^{\rho}{}_{\sigma \nu}-\partial_{\nu} \Gamma^{\rho}{}_{\sigma \mu}+\Gamma^{\rho}{}_{\mu \lambda}\Gamma^{\lambda}{}_{\nu \sigma}-\Gamma^{\rho}{}_{\nu \lambda}\Gamma^{\lambda}{}_{\mu \sigma}

The field equations of GR uses the difference of two contractions of this tensor:

R_{\mu \nu}-g_{\mu \nu}R = 8 \pi T_{\mu \nu}

The field equations can never apply to a point, because they always reference the difference between two paths. One cannot discuss the force or gravity at a point in spacetime, or the energy of gravity at a point in spacetime which is just force times distance, all for this same reason. If force is nonlocalizable, the question makes no sense.

In GEM theory, the question is well-formed. Gravity can be viewed as pure geometry, as a pure force, or a mix of the two consistent with what is seen. I have even found a path through a maze of math where the effect of a force is a dynamic metric of spacetime! Since gravity can be viewed as a force, energy is localize, a big plus in my opinion. Let me write out the force equation:

\frac{\partial m v^{\mu}}{\partial \tau}=J_{\nu}\partial^{\mu}A^{\nu}

Remember those word problems back in 8th grade? The challenge here is to translate the words of Jim's question directly into the expression above. If time were to go backwards, then the derivative on the left hand side would changes signs. To be a force where like charges repel would require a minus sign on the right hand side, just as goes on with the Lorentz force of electromagnetism. Plug those sign changes in:

-\frac{\partial m v^{\mu}}{\partial \tau}=-J_{\nu}\partial^{\mu}A^{\nu}

This is the first force equation multiplied by -1. Jim's question concerns transforming one force equation into another. Unfortunately, I don't think the transformation between these two force equations gives us any insight into how the force of gravity works. So Jim's question is not "wrong", it is not informative. It was great that the question was precise enough to map to a specific equation.

So now I am going to speculate on the origin of the question, one possible reason it may have been asked in the first place. Gravity dominates our lives, otherwise we might get up and leave the planet. Gravity can kill from falls or through avalanches. Gravity appears like a very active force shaping our world. It is hard to imagine an active attractive force. Such a force would have to find you and convince you to head their way. A force that shoved everyone in the face, no matter their race, color, or creed, that is easier to imagine. A pushy, repulsive force is easy for the mind to envision. That is the way two electric charges with the same sign work. Neutral particles ignore all the shoving.

Ah, neutrality, that might be the key. Nothing is neutral as far as the gravitational force is concerned. That is the real mystery: how come there is absolutely nothing that can ignore gravity? Photons which have neither electric charge or a mass, the ultimate cream puffs, still bend in both time and space to the curvature of spacetime that is gravity. I don't think it is reasonable to imagine anything going out, finding every photon without exception, and either pushing or pulling on that photon. The cause of the unsubtle force of gravity is subtle. Because there is other stuff in the Universe, where that stuff is is a place you cannot be in spacetime. Other stuff subtracts the possibility of where. So as stuff goes through spacetime, it goes where it can. Since there is less where near stuff, the easiest darn path goes through the possible. There is less possible near stuff, so to balance the path, it will look like the path travels closer to the stuff.

How does Nature keep track of all the stuff? How does she balance the books? In standard physics, I don't think people consider this question. For me it is central. The counting system is based on events in spacetime. Events can be counted. There is the event of this electron living over there now. A moment later, it is still there. I would use a quaternion to keep track of where it is in time and space because a quaternion has a slot for time and three for space. Quaternions come with an accounting system: they can be added together, subtracted, multiplied and divided like a real number, but they have these 4 sub-parts to map to our continued existence (time) in 3D space. Armed with a number and +. -. x, and /, the Universe can do all the math necessary.

Nearly all the math necessary is math very close to one, so really, almost nothing is happening, all the parts are staying near the identity. Yet we do move in time, and a wee bit in space. How is that measured, a distance involving both time and space? Recall how distance always seem to involve squares, the Pythagorian theorem? Same thing is true with quaternions. I am in Waltham at three minutes to seven, and hope to be in Jamaica Plain around 8PM to watch a bit of the Tour de France. The time change is about and hour, the spatial distance, about 10 kilometers. The distance in spacetime is
((1 hr)(\frac{60 min}{1 hr})(\frac{60 sec}{1 min}))^2 - ((10 km) (\frac{1000 m}{1 km})(\frac{1 sec}{3x10^8m}))^2=12959999.999999999 sec^2
Why all the darn 9's? That is because the 7 kilometers count for almost nothing (the factor of the speed of light is what squashes it to insignificant). There are two take home messages. First, time is money, or at least is the Swiss franc, of spacetime, dominating everything else. Second, changes in space, although tiny, are not nothing, and we have to understand their role.

If there was nothing else in the Universe, the spacetime distance would be:
d \tau^2 = 1^2 dt^2 - 1^2 dR^2/c^2
There is stuff in the Universe. In GEM theory, I could still use this metric, and then be required to use a 4-potential (a scalar potential theory is good enough to account for changes in time, but not space). Or in GEM theory, the presence of stuff could change those constant 1's, making the first 1 a little smaller, the second 1 a little bigger. The amount of change is stupidly small. Why? Well this Earth is stupidly stable. We are hanging around one.

****

I hope no one found the preceding explanation completely satisfying. That is what it feels like to wonder how things really work. There comes a point where what you know runs out. I was riding my bike this evening, some 13,000 miles away from the center of the Earth. I could tell when my bike had to be a mere 5 feet further from that point because it meant I had to climb a hill (0.001 miles). Even with all my math tools, it appears miraculous that the bike knows the difference between a few feet more and a few feet less to the gravitational center. No matter how much you know, there is always another few feet to climb from your center of knowledge. Every once and a while, enjoy the glide down the hill and accept that you live in a Universe whose wonders will aways be more than you can embrace.

doug

[sweetser]

Hello:

In my research, I focus on equations, not the words to describe said equations. The reason is that I work intermittently in isolation. There is not a small group of folks working full time on rank 1 unified field theories, so there is no one to chat with about the work.

By making videos for the Stand-Up Physicist - the educational arm of my research efforts - I do have to think about word choices. In creating a show about 4-potentials, I had to give J^{\mu}_{m} a name. I think of this as the same darn thing as the electric current density J^{\mu}_{q} except that the electric charge q has been replaced by a mass charge \sqrt{G} m. I have been calling it the mass charge density by reason of analogy.

No one works with a mass charge density, so maybe that is a poor choice of words. The electric charge density is the electric charge in motion. A mass in motion is a 4-momentum. I should call J^{\mu}_{m} the 4-momentum charge density. I include the word "charge" because the units for this tensor are not the same as a 4-momentum. Units matter. The appearance of G indicates the expression is connected to gravity. There is also a c, which implies the expression is relativistic.

There might be a deeper message with the shift to "momentum charge". In the current way of thinking, energy-momentum is the most important characteristic of a system. The link to gravity is indirect. Energy-momentum gets added into the stress-energy tensor that curves spacetime according to general relativity. Treated instead as a charge, 4-momentum charge is directly connected to gravity and inertia. A momentum charge is always going to effect what happens with gravity, which is how it should be.

I am off to buy groceries, but I will be playing with the idea that all 4-momentums should be rewritten as 4-momentum charges. It is easy enough to change the units, but it is the implications that are fun to consider while shopping.

doug
TheStandUpPhysicst.com

[sweetser]

Hello:

I did get the groceries. Here was the thought of the day I had while trying to choose matching potatoes.

Newton's force law, \vec{F}=m \vec{A}, has no need for universal constants, c, G, h. It remains a very useful law today for doing work in mechanical engineering and architecture. The enormous advance in the study of natural philosophy inspired directly from Newton was the use of units to quantify observation of Nature. Energy would never be momentum, because the first had units of M L^{2}/T^{2}, while the second had units of M L/T.

With the advent of special relativity, the line between energy and momentum blurred, because by changing inertial reference frames, the amount of energy and momentum can also change so long as the mass of the system remains invariant. Energy and momentum now have a relationship with each other.

The separation between gravity and EM is clear from the GEM Lagrangian: gravity lives in the irreducible symmetric rank 2 tensor, and EM lives in the irreducible antisymmetric field strength tensor. With GEM, one can calculate directly how much the gravitational attraction takes away from electrical repulsion with J^{\mu}_{q} - J^{\mu}_{m}. Using universal constants to match units is a basic step in unifying classically separate forces of Nature.

doug

[Norman Albers]

I just uncorked a two-page study where I equate the Schwarzschild metric terms to a dielectic permittivity. This says we can look at GR as a dielectric 'thickening' in a Euclidean space! Very simple, but works beautifully! I don't mean to interrupt, but this is so germain to what you are doing, and I need your help.

[sweetser]

Hello Norman:
There is no way this could be germain to the topic at hand: the GEM model makes predictions that are different from GR, and a rank 2 theory is very different from a rank 1 theory. You might want to try and start your own thread at Independent Research Forums, so long as your work meets the criteria.
doug

[Don J]

Hello:

In my research, I focus on equations, not the words to describe said equations. The reason is that I work intermittently in isolation. There is not a small group of folks working full time on rank 1 unified field theories, so there is no one to chat with about the work.


Can you explain for us poor laymans and -relativistic-(Einstein sense of the word)what is the ULTIMATE goal of your theory?
Do you think your theory may ultimately replace GR?
For now your theory seem only a "curiosity".
My post is not intended to be harsh!

[sweetser]

Hello Don:

Newton's theory of gravity is a good theory for how gravity works. It is still in use today. We know it has a technical flaw, that a change in a mas density must change everything instantaneously.

GR is a better theory, one that supersedes Newton's theory. If one starts from Newton's law and sets out to make it consistent with special relativity, then one ends up with Einstein's field equations. The vacuum field equations are non-linear, something that happens with the weak and the strong force, but not EM. All efforts to quantize GR proper have failed.

The GEM field theory tries to be better than GR. Like GR, one can be certain the theory gets along fine with special relativity, or in fancy works, the field equations are manifestly Lorentz covariant (meaning the form of the equations is not going to change by changing inertial observers). The field equations can be quantized because it is done in graduate-level quantum field theory books on EM field theory (two of the modes for a spin 1 field must be made virtual, but I propose those modes are a very real spin 2 field doing the work of gravity).

If the GEM model is correct, it will supersede GR because it can be quantized and it naturally unifies gravity and EM.

Nature it this harsh one, in that is unreasonable to expect in a lifetime to find a theory consistent with current tests, different from more stringent tests, yet is mathematically sound. To me, the GEM model is as curious as one can ever hope to find.

doug

[Rade]

To Sweetser: I have a question from another thread where I received no help. What I look for are the equations that would predict the gravity force between the three quarks (uud) within the proton, then for the anti-proton (u-bar, u-bar, d-bar). Can your GEM model help me here ? I understand this force is not thought to be important--but I care not, I just look for the equations. Plus, do we really know gravity at scale less than 10^-15 meter, perhaps it increases as we move to scale of quark interactions ? I would then use these equations to derive gravity force equations for neutron and anti-neutron, then for higher order isotope interactions, both matter and antimatter (deuteron, helium-3, triton, etc).

[sweetser]

Hello Rade:

The short answer is I do not know how to do any calulations with the strong force.

For classical gravity and EM, the source is the electrical charge density minus the 4-momentum charge density. On the other side of the equation, two covariant derivatives act on a 4-potential. The solution can either involve a dynamic metric, a dynamic potential, or some combination of the two. I have done calculations of the field equations for an electron and a neutron, using either the potential exclusively or the neutron exclusive. Those calculations form my proof that GEM has Diff(M) symmetry because I use the same manifold, but different metrics (one Minkowski, one the exponential metric at the start of this forum).

As far as I know, there is no classical way to express the strong force. This is a warning flag for me due to the correspondence principle, which I think should always work both ways. Warning: this warning flag is not many other people's warning flag :-)

The question looks like it must be addressed using quantum mechanics. I have a clear picture of how to do a scattering calculation for an electron. This is a nonlinear equation where one uses Feynman diagrams to make approximations because EM is weak, and gravity is far weaker still. I don't know what the current record is for Feynman diagrams (8?). That calculation required a supercomputer because there are so many permutations. Adding another vertex becomes even harder.

Cool! I just had an idea. The contributions to a scattering calculation depend on the coupling constants. Pretend we have a beam of electrons being fired at a proton. For the Feynman diagram with one vertex, the coupling constant would be
\alpha = \frac{\mu_0 c e^2}{2 h} = 4 \pi 10^{- 6} NA^{- 2} 3 \times 10^8 ms^{- 1} 1.6 \times 10^{- 19} / (2 6.63 \times 10^{- 34} Js) = 0.00727

(note: I've spent the last hour trying to spot my tex error, but have failed, sorry!)
Now consider the Feynman diagrams with two vertices. The coupling would be
\alpha^{2}. There is a simple pattern here: for the number n vertices, the coupling constant is \alpha^{n}.

What is the gravitational coupling? With EM, there is one size for electrical charge. For gravity, there are many values for the gravitational charge, so many gravitational alphas. For a proton and electron interaction, the coupling would be:
\alpha_{GEM, p - e} = \frac{\mu_0 c G m_p m_e}{2 h} = 4 \pi 10^{- 6}
NA^{- 2} 3 \times 10^8 ms^{- 1} 6.67 \times 10^{- 11} m^3
kg s^{- 2} 1.67 \times 10^{- 27} kg 9.11 \times 10^{- 31}
kg / (2 6.63 \times 10^{- 34} Js) = 2.55 \times 10^{- 32}


Now we need to calculate the number of vertices n, such that this gravitational effect will equal that of EM.
\alpha^n = \alpha_{\tmop{GEM}, p - e}
n \log \alpha = \log \alpha \text{_{GEM,p-e}}
n = \frac{\log \alpha \text{_{GEM,p-e}}}{\log \alpha} = 14.8
According to GEM, quantum gravity effects will come into play for Feynman scattering calculations where one has 15 vertices. Even with all the steady improvement in computers, I doubt we will be able to make the calculations in a hundred years time.

***
Sorry for the digression, but it is rare when I start a morning with a calculation. GEM does not provide a reason why gravity is so weak relative to EM or the strong force, the subject of your comment. The difference is so vast they should be treated separately.

doug

[Rade]

...According to GEM, quantum gravity effects will come into play for Feynman scattering calculations where one has 15 vertices. Even with all the steady improvement in computers, I doubt we will be able to make the calculations in a hundred years time.Thank you for your time with my question. Now, your very interesting comment above about quantum gravity effects emerging when there are 15 interacting vertices leads me to another question--how to get 15 quarks interacting to form 15 vertices via quantum gravity to form a stable system--perhaps with added dimensions not now considered. Thus, suppose a thought experiment where we describe the quarks within nucleons to interact (via quantum gravity) in "bags", that is, the proton has 1 bag with 3 quarks, so the neutron. Now, we define a matter Helium-3 isotope, which is macroscopically [PNP], as microscopically 9 matter quarks in 3 bags interacting. Then we attempt to couple this matter isotope [NPN] with an antimatter deuterium isotope [N-bar, P-bar], which has 6 antiquarks in 2 antibags. As you can see, the interaction of this type of asymmetrical matter-antimatter bags of quarks results in a possible quantum gravity solution in which there are a total of 15 interacting quarks, 9 matter and 6 antimatter (can we call these vertices vis-a-vis Feynman??), which should then allow for a mathematical expression using your GEM model ? Taken to a logical (but perhaps not realistic) conclusion, the above would suggest that interactions of 2 and 3 nucleon isotopes (at the level of quarks), both matter and antimatter, form the "fundamental" (lowest energy) physical system in which quantum gravity effects can be experimentally documented, perhaps by your GEM model. Does any of this make any sense to you ? Thanks for your time.

[Lawrence B. Crowell]

The 't Hoft Veltman running parameter

g^2(q) = g^2( mu)[1 + (g^2(mu)/12\pi)log(-q^2/mu^2)(2n_f - 11N)]

can handle these problems as it is. For high enough transverse momenta one can do QCD perfectly well. I am not sure how GEM fits into this matter.

The real outstanding question is whether this will address the gauge heirarchy problem. The GEM theory, with all its symmetrical terms and the like is likely to have problems within any attempted exact renormalization group scheme. Without calculation I can see it is likely that GEM will produce all sorts of renormalization flows which are not gauge invariant (eg will involve gauge connections), which will make renormalization problematic.

On my part with Information Preservation in Quantum Gravity I will be soon making an installment which involves regularization schemes in Ricci flows and foliations of three spaces in cosmologies.

Lawrence B. Crowell

[sweetser]

Hello Rade:

When I say I understand something in physics, it means I know how to relate it directly to a physics equation or calculation. In my educational efforts at TheStandUpPhysicist.com, I have made videos which have too much math to reach a wide audience, but for me, physics is the equations so they must always be directly referenced. When I claim that gravity and EM are unified, I really am referring to Jq-Jm=Box^2 A, where the Box is a covariant 4-derivative, a standard derivative minus a connection. For me, the link must be that tight.

Let me now detail what I understand about the calculation I made. I took a year of graduate-level quantum mechanics over a decade ago, so key details are no longer with me. When a photon interacts with an electron, the math to characterize it is non-linear. This makes finding a solution darn tricky. Feynman diagrams are a visual aid to making a calculation. For a low energy system, just one or two vertices are needed to get a great approximation. One can add more terms, but due to a growing number of permutations, it becomes necessary to program the process. If one goes out to as many as eight vertices, a supercomputer struggles with the volume of math passing through silicon.

What do these calculations do? Usually it has to do with calculating scattering: one fires an electron at a proton, the electron scatters, and the odds of it scattering say at 14 degrees off the center line is shown by the calculation. The calculation I did was suppose to show that no matter what happens in our lifetime, or our children's lifetime, or their children's lifetime, no one will be able to see how much the EM scattering is changed by gravity scattering. We understand the size of the calculation, and why it is beyond reach.

What goes into the source term of the GEM field equations, Jq-Jm? Anything with an electric current (Jq) or a mass current (Jm). The mass of an electron is equal to the mass of a positron, its anti-particle partner, so the total contribution to the Jm term of an electron and a positron is 2*511 MeV times the 4-velocity. The same goes for quarks, antiquarks, leptons, anti-leptons. All 4-momentum charges go into Jm, no exceptions granted.

The calculation I did was to compare gravity to EM. The number that came out of that process was 14.8, kind of close to 15. At this time, I have no sense - no calculation - which explains why the difference should be as big as it is. It is an observation. It looks to me like you took the 15, and did an exercise in numerology - where combination adds up to 15? The actual number is 14.8, which makes any connection to 15 not relevant. Plus I see zero connection between the calculation I did (relative forces of gravity and EM) and quarks/antiquarks.

doug

[sweetser]

Hello Don:

Here is a big advantage of the GEM approach a layman can appreciate: the ontology (a fancy why of saying the "why" of gravity and EM). Let's first look at the why behind Newton's law of gravity and general relativity.

There is no "why" for Newton's law of gravity. That law was constructed and shown to be relevant by connection to Kepler's three laws of planetary motion. In theory, gravity could have been a 1/R or 1/R^3 law with the correct adjustment of G (L^2/M T^2 or L^4/M T^2 respectively). An inverse square law has some nice mathematical properties, but that is not a reason why the law should be the product of two masses over the square of the distance.

The situation for general relativity is no better. The best know justification for Einstein's field equations is the following from John Archibald Wheeler:
Spacetime tells mass how to move and mass tells spacetime how to curve.
There is some debate about whether this is an example of circular reasoning. I don't see how or why mass and spacetime should have this sort of dialog, so I am not happy with this perspective.

With GEM, start with an empty Universe. The distance between a pair of events in such a barren place would be governed by the utterly flat Minkowski metric. Why that metric? Calculating a distance is an act of multiplication, and multiplication of a 4-vector as a numerical field is governed by quaternions. Most laymen have never heard of these 4D numbers, and in my informal survey, only about half of physicists recognize the name (even fewer have worked with them). They are the 4D version of 2D complex numbers, which is how the minus sign shows up between time squared and space squared.

Now introduce a mass. The world is not very different. The question is, how does one implement the small amount the Universe has changed because of the mass? No, better, how does one implement the smallest possible way the Universe has changed because of the mass? For systems where one thing barely effects another, there is a universal answer: a simple harmonic oscillator. A less fancy way to say it would be to reference a slinky: let a slinky hang down, give it the smallest of nudges, and it will wobble for a good while. The weaker the connection, the smaller the wobble, but it also lasts longer because not much energy is involved.

Spacetime is nearly flat. It is not perfectly flat, so the smallest step away from perfectly flat involves a 4D slinky. The 4D wave equation, J_q-J_m=\Box^2 A can describe a 4D simple harmonic oscillator. Just because it has the word "simple" in the label does not make it easy to understand. Lots of issues still have me mystified. But the point is to find the math that is all about almost doing nothing. That is exactly what the 4D wave equation with its simple harmonic oscillator give you. The Earth is wobbling around the Sun, been doing that for 4 billion years, and it doesn't take a lot of energy, but it does require the presence of an enormous amount of mass (2\times10^{30} kg). Right now, me, and everything around me is trying to wobble around the center of the Earth. Because we get in each other's way, we are in a stalemate. If I ever get a path clear from all the clutter in my way, like at the top of a diving board, I begin my wobble through the center of the Earth to my antipode and back, a trip that would last about an hour and a half. Of course my freedom lasts mere seconds, and back into the traffic jam of mass charges I go.

If you do research, you know the edge of where you do not know. I still wonder how it is I collect the information about the 6\times10^{24}kg that makes up the Earth. It is the implementation details that I don't understand. Yet I feel good that gravity is the minimal response spacetime can have due to another mass.

doug

[sweetser]

Hello Lawrence:

I've never work with the 't Hoft Veltman running parameter, so I cannot comment on the issue. Nor was I successful with renormalization calculation in my two term quantum field theory class. I doubt I will ever get the training required to answer these technical questions.

I do keep my eye out for a quick and dirty way of doing a calculation, letting those folks with far more refined skills to worry about details. We know for darn sure exactly how to do the calculation of a scattering of an electron off of a proton. We know exactly what the coupling constant is, \alpha. We know exactly what the propagator is for a spin 1 field.

This part of the calculation is standard. Nothing about it changes one bit. As stated before, I cannot do the calculation myself at the current time. It is the sort of thing taught to graduate students. The scattering of an electron off a proton is considered "easy", and relatively speaking, it is.

There are only two parts that change: the coupling constant is no longer alpha, but instead one puts in G m_1 m_2 where the square of electrical charge used to be in fine structure constant. Secondly, the propagator is for a spin 2 field. Weinberg wrote three articles on the form of such a propagator. I did not understand those articles, so I cannot jot it down and use it in a calculation. Selector rules may change as a consequence, but I don't recall their role either.

If I was handed 6 pages of calculations showing how to calculate the scattering of an electron due to a proton, I could use that as my basis to figuring out the scatter due to gravity. Changing the coupling constant would be the trivial part. Dropping in the correct propagator is probably beyond my reach, but I could spot where the work would have to be focused.

Without doing the calculation, I am more hopeful because I understand with precision where the difference are between a calculation we know how to do (EM scattering using a spin 1 particle) and one we hope to do (gravitational scattering using a spin 2 particle).

doug

[Lawrence B. Crowell]

Doug,

The fine structure constant e^2/hbar-c ~1/137 has the unique feature of being dimensionless. The term GMm has units [g cm^3/s^2]. While "dimensionless" in naturalized units it still has a scale dependency. If we set the masses as M = n*m_p, for m_p the Planck unit of mass m_p = sqrt(c/G hbar), and n an integer, then

GMm = (nn')c/hbar.

hbar = h/2pi for h = oint p*dq is unitless in naturalized units and c is also set to unity in naturalized units. However, c requires the use of rulers and clocks to measure and hbar involves the measure of wavelengths (rulers) and momenta (rulers & clocks). Thus alpha and GMm as coupling constants are fundamentally different.

The problem as I see it is that the symmetric structure of GEM leads to a problem with the effective action. For S = S_eff + S',
[tex]
S' = {{\delta S}\over{\delta A}}R_{\Lambda}{{\delta S}\over{\delta A}}
[/itex]
this latter term is not going to satisfy gauge invariant requirements for Wilson/Polchinski renormalization "flow." Here R_{\Lambda} is the regularization cut-off function. This is the central problem here. These renormalization group equations are found from the scale structure of regularization cut-offs /\, stemming from Feynman.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

The fine structure constant is dimensionless. I was not clear on the unit system I was using. I was using cgs, where Coulomb's law is:
F=Q Q'/R^2 \hat{R}
In cgs, the units for Q Q' are the same as GMm. If one uses Gaussian units, then Coulomb's laws is:
F=\frac{1}{4 \pi \epsilon_0} Q Q'/R^2 \hat{R}
In Gaussian units, the switch would be:
\frac{1}{4 \pi \epsilon_0} Q Q' \rightarrow G M m
Let's just check that this works.
\alpha = k_c e^2/\hbar c
My claim is that for cgs units, the k_c is 1, so a straight swap of GMm should work:
\alpha_{GEM} = G M m/\hbar c = (L^3 M^{-1} T^{-2}) M M /((M L^2 T^{-1}) (L/T^{-1}))
That is dimensionless. There is no need to use a Planck mass.

Viewed this way, do you agree that the fine structure constant of EM and \alpha_{GEM} could play an identical role? It is too bad my LaTeX was mangled in the previous post (and I cannot edit it now) because I had \alpha_{GEM} defined there.

doug

[sweetser]

Hello Lawrence:

I did some browsing on the web to find out about the Wilson/Polchinski renormalization flow because I was not familiar with it (probably like other readers of this thread). Here is a quote from the start of a 72 page review article (http://www.citebase.org/fulltext?format=application%2Fpdf&identifier=oai%3AarXiv.org%3Ahep-th%2F0002034):

By "exact renormalization group equation (ERGE)", we mean the continuous (i.e. not discrete) realization of the Wilson renormalization group (RG) transformation of the action in which no approximation is made and also no expansion is involved with respect to some small parameter of the action. Its formulation - under a differential form - is known since the early seventies. However, due to its complexity (an integro-differential equating), its study calls for the use of approximation (an/or truncation) methods. For a long time it was natural to use a perturbative approach (based on the existence of a small parameter like the famous \epsilon-expansion for example). But, the standard perturbative field theory turned out to be more efficient and, in addition the defenders of the nonperturbative approach have turned toward the discrete formulation of the RG due to the problem of the "stiff" differential equations...This is why it is only since the middle of the eighties that substantial studies have been carried out via:
* the truncation procedures in the scaling field method
* the explicit consideration of the local potential approximation and of the derivative expansion
* an appealing us, for theoreticians, of ERGE
In the nineties there has been a rapid growth of studies in all directions, accounting for scalar (or vector) fields, spinor, gauge fields, finite temperature, supersymmetry, gravity, etc...
In the classes I took, we only worked with a perturbation approach in EM. It is a good way to go for a small coupling constant like that for EM. This is not a good way to go for the strong force.

For gravity, perturbation would be an awesome way to go. Let's calculate the gravity coupling for an electron and a proton:
\alpha_{GEM}=G m_e m_p /\hbar c = 6.67 \times 10^{-11} m^3 kg^{-1} s^{-2} 9.1 \times 10^{-31} kg 1.67 \times 10^{-27}kg/1.62 \times 10^{-34} kg m^2 s^{-1} 3.00 \times 10^8 m/s = 2.02 \times 10^{-42}
The size of the coupling constant is important. For much work in quantum field theory like that with the strong force, I could see why the Wilson/Polchinski renormalization method would be important. It is hard to believe that an alpha_{GEM}^2 will ever be important in a physical calculation! To be clear, I am not contesting the difficulties of using the Wilson/Polchinski renormalization method that Lawrence believes are there. I am saying this issue is not a "deal breaker" because first order perturbation theory will be sufficient for the calculations because the gravitational coupling constant is so very small.

doug

[vebrown]

Hi Sweetser; great work; hope it works out !! I have pondered this myself, but from a different direction. Working from the postulates in the Lorentz Ether Theory that the seat of the Electromagnetic field is the empty space, and that the final irreducible constituent of all physical reality is the electromagnetic field, it turned out like this. (http://photontheory.com/newuniverse.html#gravity)

Edit: I just visited your web site; very interesting. You have the graviton as a spin two particle; I always thought it was spin one, maybe I'm wrong. But isn't there a problem with spin two particles moving at the speed of light.

[sweetser]

Hello Vebrown:

I clicked through the page cited, but saw neither a field equation nor Lagrange density. For me, that is the "starting height" a proposal must clear before it is worth studying (it is nice to have a clearly defined criteria, so there is nothing personal here, just analytical). I wish you good luck in your studies none-the-less, because it took me several years to go from the GEM field equations to the Lagrange density.

You have the graviton as a spin two particle; I always thought it was spin one, maybe I'm wrong. But isn't there a problem with spin two particles moving at the speed of light.

If a particle has zero mass, then it will travel at the speed of light. As for the spin of the particle needed, let me quote from Brian Hatfield's introduction to "Feynman Lectures on Gravitation":

When the potential is computed in both (even and odd integral spin) cases and the appropriate limits are taken, we find that when the exchanged particle carries odd integer spin, like charges repel and opposite charges attract, just as in the example of electrodynamics. On the other hand, when the exchanged particle carries even integer spin, the potential is universally attractive (like charges and opposite charges attract). Hence, the spin of the graviton must be 0, 2, 4, ...

To be completely truthful, I do not know the details of that calculation. I do know it is widely accepted that the graviton is a massless spin 2 particle that travels at the speed of light. Because we have data that says photons couple to gravity, a spin 0 graviton would not work (I again am paraphrasing Hatfield without completely understanding him). The smallest even integer particle to do the work of gravity is spin 2.

I am not a great student of the particle physics literature, but so far have yet to hear what physicists think is the connection between the graviton which mediates the force of gravity, and the Higgs particle, which adds inertial mass to the standard model. The deep insight of general relativity is that gravitational and inertial masses are precisely the same thing. Therefore the relationship between the graviton and the Higgs must be as tight as can be constructed. Thing is, the Higgs is a spin 0 particle.

How could one make an unbreakable connection between a spin 0 and a spin 2 particle? My thought was the symmetric rank 2 field strength tensor naturally is a source for a spin 2 field. That is completely standard logic, nothing at all radical. What I realized is a rank 2 tensor will always have associated with it the trace of that tensor. The trace would form a rank 0 tensor, and that rank0 tensor could be represented by a spin 0 field. The fact that a rank 2 symmetric tensor always has a rank 0 trace is the reason that gravitational and inertial mass are the same. Cool!

doug

[Don J]

Hello Don:

Here is a big advantage of the GEM approach a layman can appreciate: the ontology (a fancy why of saying the "why" of gravity and EM). Let's first look at the why behind Newton's law of gravity and general relativity.

There is no "why" for Newton's law of gravity. That law was constructed and shown to be relevant by connection to Kepler's three laws of planetary motion. In theory, gravity could have been a 1/R or 1/R^3 law with the correct adjustment of G (L^2/M T^2 or L^4/M T^2 respectively). An inverse square law has some nice mathematical properties, but that is not a reason why the law should be the product of two masses over the square of the distance.

The situation for general relativity is no better.


doug
Hi sweetser
Your theory must doing something -utile (in the mathematically sense)- that GR cannot do.
Do you have some examples (tests)?
Is there something in the Universe who need to be -explained - with a rank one theory?
If i remember You qualified GR and other -relativistic based theory- of rank two theory

That specific point about rank one and rank two theory needs a clear comparison about its application in the real world (The universe)?

Thanks!

[sweetser]

Hello Don:

The tests for this proposal were detailed at the start of this thread (point 2 of that initial post). I try and do new things, but appear to repeat two specific tests: light should bend around the Sun 11.5 microarcsecond versus 10.8 mircoarcseconds predicted by GR when second order PPN measurements are made, and a gravitational wave will not be a transverse wave as predicted by GR.

The vacuum GEM equations are linear, whereas the ones for GR are nonlinear. In theory, and definitely not in practice, one could measure the strength of a test mass, then add another equal test mass, and show either the gravitational field was exactly doubled, or there was a nonlinear increase. Although I have not done the calculation, I believe such a direct measurement will be impossible to do for the foreseeable future because it would require absurdly sensitive measurements.

GR is a rank 2 field theory because it has two Greek letters:
R_{\mu \nu} - g_{\mu \nu} R = 8 \pi T_{\mu \nu}

GEM is a rank 1 field theory because it has only one:
J^{\mu}_q - J^{\mu}_m = \nabla^2 A^{\mu}

In GR, gravity couples to the stress energy tensor, T_{\mu \nu}. In GEM, gravity couples to the 4-momentum current density, J^{\mu}_m. These are tensors of different rank, so we cannot compare them directly. We can look at what goes into each. I believe the only difference is that the energy of a gravity field goes into the stress-energy of GR, but it is excluded in GEM. The effect of gravity is very weak, and the energy of a gravity field is very small, so spotting the contribution of a very small thing to a very weak effect probably will not be possible to measure in the real world. It is very significant for the math guys. In GEM, because the gravity field does not gravitate, it will be much simpler to quantize.

doug

[CarlB]

Doug, could you take a look at this paper by Kris Krogh and tell me how it differs from your theory?

Equations of Motion in a Quantum-mechanical Theory of Gravity
Kris Krogh
An earlier paper [1] presented a gravity theory based on the optics of de Broglie waves rather than curved space-time. While the universe's geometry is flat, it agrees with the standard tests of general relativity. A second paper [2] showed that, unlike general relativity, it agrees with Doppler tracking signals from the Pioneer 10 and 11 space probes. There a gravitational acceleration equation plays an important role, accounting for the relative motions of Earth and the probes. Here it's shown that equation also describes Mercury's orbit.
http://www.arxiv.org/abs/astro-ph/0508290

He comes at the problem from a different direction, but he also talks about a relationship with E&M. So I'm wondering if his equations of motion are the same as yours. I figure that I can simulate his with no great effort as they are described clearly enough that a simpleton like me can figure them out.

I've had some success with the gravity simulation program recently, well I should say success in figuring out the equations of motion for GR, and have now updated it to include this. I'm working on Painleve coordinates now.

Carl

[Don J]

Doug wrote
The effect of gravity is very weak, and the energy of a gravity field is very small, so spotting the contribution of a very small thing to a very weak effect probably will not be possible to measure in the real world. It is very significant for the math guys. In GEM, because the gravity field does not gravitate, it will be much simpler to quantize.

I think we will have an answer in the coming year...
Gravity Probe B mission, testing Einstein's theory of gravity, completes first year in space
http://www.physorg.com/news4062.html

[sweetser]

Hello Carl:

The GEM theory and the one one put forth by Kris Krogh are very different. It all flows from the Lagrangian. We start from exactly the same starting place, the classical EM Lagrangian (OK, he works with forces, not fields, so he has the inertia and current coupling term, but not the field strength contraction term, F^{\mu \nu}). What he then does is slap an exponential factor to the side of it. I have issues with that. The particle involved in the force of gravity has got to be different from the photon, because the photon helps like charges repel, but for gravity, like charges attract. Curve spacetime as you like, but one has to account for the particles that will be carrying out the program.

This is one reason I stress the importance of being able to write out a Lagrangian: since a Lagrangian has all the energy interactions that can happen inside a box, looking at it - after some practice - can tell you how the proposal works. There is only one current that binds to the potential, and that is the EM current. In my GEM proposal, there are two currents, one for EM that is made a little bit smaller due to the inertial mass current. It is a flaw of mine how much I like that idea: like electrical charges repel each other because of EM, but they repel a little less due to their own inertia.

So at the level of the Lagrangian, my work does not get along with Krogh's. Krogh has thought through the equations for orbits. I did work in detail on the issue, to get the precession of the perihelion of Mercury. That is a tough calculation! The prediction of GEM is exactly the same as that for GR at this level of accuracy.

Even though it will make this post long, I thought I would include the entire thing here, because you are unlikely to ever see all the steps for the calculation spelled out.

The Precession of the Perihelion of Mercury Based on a Solution to the GEM Field Equations

1. Start with a solution to the GEM field equations for a spherically symmetric, non-rotating, uncharged mass with the gauge choice of a constant 4-potential. The solution is the following metric written in spherical coordinates.

d \tau^2 = e^{- 2 \frac{G M}{c^2 R}} d t^2 - \frac{1}{c^2} e^{2 \frac{G
M}{c^2 R}} d R^2 - ( \frac{R}{c})^2 (d \theta^2 + \sin^2 \theta d \phi^2)

2. Simplify by working in the plane of rotation, so d \theta = 0 and \sin^2 \theta = 1[/itex]:

[tex]d \tau^2 = e^{- 2 \frac{G M}{c^2 R}} d t^2 - \frac{1}{c^2} e^{2 \frac{G M}{c^2 R}} d R^2 - ( \frac{R}{c})^2 d \phi^2

3. Divide by d \tau^2 and multiply through by e^{- 2 \frac{G M}{c^2 R}}:

e^{- 2 \frac{G M}{c^2 R}} = e^{- 4 \frac{G M}{c^2 R}} ( \frac{d t}{d
\tau})^2 - \frac{1}{c^2} ( \frac{d R}{d \tau})_{}^2 - \text{} e^{- 2 \frac{G
M}{c^2 R}} ( \frac{R}{c})^2 ( \frac{d \phi}{d \tau})^2

4. Since the exponent for our solar system is small, use the Taylor series expansion, e^{- 2 \frac{G M}{c^2 R}} \approx (1 - 2 \frac{G M}{c^2 R}). View the -4 term as a -2 term squared:

(1 - 2 \frac{G M}{c^2 R}) = (1 - 2 \frac{G M}{c^2 R})^2 ( \frac{d t}{d \tau})^2 - \frac{1}{c^2} ( \frac{d R}{d \tau})^2 - \text{} (1 - 2 \frac{G M}{c^2 R}) ( \frac{R}{c})^2 ( \frac{d \phi}{d \tau})^2

Note: from here on out, there is nothing different from GEM compared to GR.

5. Define the Killing vectors for this metric. Notice that the expression in 4 is not a function of either time t or angle \phi. This means there is a conserved quantity associated with a change in time (energy E) and a change in angle (angular momentum L). Come back at a later time, and the expression stays the same. Spin around a few degrees, and the metric stays the same. The general structure of a Killing vector is this:

(conserved thing) = (Killing vector)\cdot(velocity vector)

Here are the two Killing vectors for 4:

\frac{E}{m c^2} = K_t V_t = (1 - 2 \frac{G M}{c^2 R}, 0, 0, 0) \cdot ( \frac{d t}{d \tau}, 0, 0, 0) = (1 - 2 \frac{G M}{c^2 R}) \frac{d t}{d \tau}

\frac{L}{m c} = K_{\phi} V_{\phi} = (0, 0, 0, \frac{R}{c}) \cdot (0, 0, 0,
R \frac{d \phi}{d \tau}) = \frac{R^2}{c} \frac{d \phi}{d \tau}

6. Calculate the squares of the energy and angular momentum:

( \frac{E}{m c^2})^2 = (1 - 2 \frac{G M}{c^2 R})^2 ( \frac{d t}{d \tau})^2

( \frac{L}{m c})^2 = \frac{R^4}{c^2} ( \frac{d \phi}{d \tau})^2

7. Plug 6 into 4:

(1 - 2 \frac{G M}{c^2 R}) = ( \frac{E}{m c^2})^2 - \frac{1}{c^2} ( \frac{d R}{d \tau})^2 - \text{} (1 - 2 \frac{G M}{c^2 R}) ( \frac{L}{m c R})^2

What has happened? We have introduced two constant quantities, the energy E and the angular momentum L. We still have the factor, (1 - 2 \frac{G M}{c^2 R}).

8. Prepare for a change of variable, to U = \frac{1}{R}:

\frac{d R}{d \tau} = \frac{d R}{d \phi} \frac{d \phi}{d \tau} = (- \frac{1}{U^2} \frac{d U}{d \phi}) ( \frac{L U^2}{m}) = - \frac{L}{m} \frac{d U}{d \phi}

9. Plug 8 into 7:

(1 - 2 \frac{G M}{c^2} U) = ( \frac{E}{m c^2})^2 - \frac{L^2}{m^2 c^2} (
\frac{d U}{d \phi})^2 - \text{} (1 - 2 \frac{G M}{c^2} U) ( \frac{L U}{m
c})^2

10. Multiply through by U and bring all the terms to one side:

0 = ( \frac{E}{m c^2})^2 - \frac{L^2}{m^2 c^2} ( \frac{d U}{d \phi})^2 -
\text{} 1 - \frac{L^2 U^2}{m^2 c^2} + 2 \frac{G M}{c^2} U + 2 \frac{G M L^2
U^3}{m^2 c^2}

11. Take the derivative of 10 with respect to \phi:

0 = - 2 \frac{L^2}{m^2 c^2} \frac{d U}{d \phi} \frac{d^2 U}{d \phi^2} - 2
\frac{L^2 U}{m^2 c^2} \frac{d U}{d \phi} + 2 \frac{G M}{c^2} \frac{d U}{d
\phi} + 6 \frac{G M L^2 U^2}{m^2 c^2} \frac{d U}{d \phi}

12 Divide all the terms by - 2 \frac{L^2}{m^2 c^2} \frac{d U}{d \phi}:

0 = \frac{d^2 U}{d \phi^2} + U - \frac{G M m^2}{L^2} - 3 \frac{G M}{c^2}
U^2

The first three terms are classical Newtonian gravitational physics (implied by the lack of a factor of c). The fourth is the correction required by GEM.

13. Write out the Newtonian equation:

0 = \frac{d^2 U}{d \phi^2} + U - \frac{G M m^2}{L^2}

14. Solve the Newtonian equation. This is a slight variation on an equation with the cosine solution. We must account for the eccentricity of the circle and the constant factor. Guess a solution:

U = \frac{G M m^2}{L^2} (1 + \epsilon \cos (\phi - \phi_0))

\frac{d U}{d \phi} = - \frac{G M m^2}{L^2} \epsilon \sin (\phi - \phi_0)

\frac{d^2 U}{d \phi^2} = - \frac{G M m^2}{L^2} \epsilon \cos (\phi - \phi_0)

\frac{d^2 U}{d \phi^2} + U - \frac{G M m^2}{L^2} = - \frac{G M m^2}{L^2}
\epsilon \cos (\phi - \phi_0) + \frac{G M m^2}{L^2} (1 + \epsilon \cos (\phi
- \phi_0)) - \frac{G M m^2}{L^2} = 0
OK!

15. At the perihelion, \cos (\phi - \phi_0) = 1 and R = 1 / U = a (1 - \epsilon). Plug into U found in 14:

\frac{1}{a (1 - \epsilon^2)} = \frac{G M m^2}{L^2}

16. Use the Newtonian solution found in 14 as a start for the correction term, U^2:

U^2 = \frac{G^2 M^2 m^4}{L^4} (1 + 2 \epsilon \cos (\phi - \phi_0) + \epsilon^2 \cos^2 (\phi - \phi_0))

17. Plug this into 12:

0 = \frac{d^2 U}{d \phi^2} + U - \frac{G M m^2}{L^2} - 3 \frac{G M}{c^2}
\frac{G^2 M^2 m^4}{L^4} (1 + 2 \epsilon \cos (\phi - \phi_0) + \epsilon^2
\cos^2 (\phi - \phi_0))

18. Keep only the second U^2 correction term. The factor of \frac{G^3 M^3}{c^2} will make the U^2 correction tiny. Only if a term is ``on resonance'' - in effect pushing the swing at the same time as the main solution for U - can a term eventually make a contribution to the orbit:

0 = \frac{d^2 U}{d \phi^2} + U - \frac{G M m^2}{L^2} - 6 \frac{G M}{c^2}
\frac{G^2 M^2 m^4}{L^4} \epsilon \cos (\phi - \phi_0)

19. Guess a solution. It must be composed of the previous solution in 14, plus a way to drop the additional cosine term:

U = \frac{G M m^2}{L^2} (1 + \epsilon \cos (\phi - \phi_0) + 3 \frac{G^2 M^2 m^2}{c^2 L^2} \phi \sin (\phi - \phi_0))

\frac{d U}{d \phi} = - \frac{G M m^2}{L^2} (\epsilon \sin (\phi - \phi_0) +
3 \frac{G^2 M^2 m^2}{c^2 L^2} \phi \cos (\phi - \phi_0) + 3 \frac{G^2 M^2
m^2}{c^2 L^2} \sin (\phi - \phi_0)_{})

\frac{d^2 U}{d \phi^2} = - \frac{G M m^2}{L^2} (\epsilon \cos (\phi -
\phi_0) - 3 \frac{G^2 M^2 m^2}{c^2 L^2} \sin (\phi - \phi_0) + 6 \frac{G^2
M^2 m^2}{c^2 L^2} \cos (\phi - \phi_0))

The extra terms will drop. OK!

20. Bring the correction term into the cosine. Because \frac{G^2 M^2 m^2}{c^2 L^2} is so small, only the first term of \phi \sin (\phi - \phi_0) will make a contribution to cosine. Rewrite U in 19:

U = \frac{G M m^2}{L^2} (1 + \epsilon \cos (\phi - \phi_0 - 3 \frac{G^2 M^2
m^2}{c^2 L^2} \phi))

21. Calculate the ratio of the advance of \phi in one 2 \pi rotation between the Newtonian solution, and
the GEM correction:

\Delta \phi = 2 \pi \frac{1}{1 - 3 \frac{G^2 M^2 m^2}{c^2 L^2}} \approx 2
\pi (1 + 3 \frac{G^2 M^2 m^2}{c^2 L^2})

22. Plug the relation in 15, \frac{1}{a (1 - \epsilon^2)} = \frac{G M m^2}{L^2}, into 21.

\phi_{\tmop{advance}} = 6 \pi \frac{G M}{a (1 - \epsilon^2) c^2}

23. Collect the relevant numbers:

G = 6.67 \times 10^{- 11} \mathrm{\mathtt{\mathrm{m^3 / \tmop{kg} s^2}}}
M = 1.99 \times 10^{30} \tmop{kg}
c = 3.00 \times 10^8 m / s
a = 5.79 \times 10^{10} m
\epsilon = 0.206
( \frac{1 \tmop{revolution}}{88.0 \tmop{days}}) ( \frac{365
\tmop{days}}{\tmop{year}}) ( \frac{100 \tmop{years}}{\tmop{century}}) = 415
\frac{\tmop{rev} .}{\tmop{century}}
( \frac{180^{\circ}}{\pi \tmop{radians}}) ( \frac{60'}{^{\circ}}) (
\frac{60''}{'}) = 2.0610^5 \frac{''}{\tmop{radians}}

24. Plug values in 23 into 22:

6 \pi \tmop{rad} . \frac{6.67 \times 10^{- 11}
\mathrm{\mathtt{\mathrm{\frac{m^3}{\tmop{kg} s^2}}}} 1.99 \times 10^{30}
\tmop{kg}}{5.79 \times 10^{10} m (1 - 0.206^2) (3.00 \times 10^8
\frac{m}{s})^2} 415 \frac{\tmop{rev} .}{\tmop{century}} 2.06 \times 10^5
\frac{''}{\tmop{rad} .} = 42.9 \frac{''}{\tmop{century}}

[CarlB]

So am I right in thinking that if I do not make the approximation: e^{- 2 \frac{G M}{c^2 R}} \approx (1 - 2 \frac{G M}{c^2 R}), and solve for the equations of motion in the usual manner I will get the correct equations of motion for the GEM theory?

Also Doug, I was beginning to worry about you. I would have called you up but I lost your phone number when they reduced the number of personal messages that you can keep on PF. Please send me your latest so I can pester you when you next take a break.

Carl

[sweetser]

yep! [Note, that "yep!" was going to be my entire reply, but I learned there is a 10 character minimum, so this is just filler.]

doug

[sweetser]

Hello Carl:

Let's see if I can define for you the numerical integral I need to do to test my proposal for solving the rotation profile of thing galaxies.

Background:
At zeroth order, Newton's law of gravity, the Schwarzschild metric, and the exponential metric of GEM are identical. This is the level of accuracy that applies to weak gravitational systems, such as the rotation profile of thin disk galaxies.

At first order, Newtonian gravity is off from measurements by a factor of two. GR and GEM are still identical in their predictions. Note that there effects are subtle, requiring extremely accurate measurements. I've done the calculation for the velocity profile of a thin disk galaxy using Newton's theory or GR, and the results look identical.

At second order, the exponential metric has 12% more bending than Schwarzschild. With all the improvements that have been made in measurements over the years, we are still two orders of magnitude away from detecting anything this subtle. The difference between GEM and GR, real as it is, numerically does not matter for a huge, weak field thing like a galaxy.

When I derived the exponential metric from a relativistic 4-force law, I had this term for the 4-force:
\frac{d m V^{\mu}}{d \tau}
Two important things to notice. First, force as a change in momentum is a product, governed by the product rule:
\frac{d m V^{\mu}}{d \tau}=m \frac{d V^{\mu}}{d \tau} + V^{\mu} \frac{d m}{d \tau}
This is the way calculus works, there is no doubt about it. A change in momentum can be due to a change in velocity, the mA of Newton's second law, or a change of momentum could be due to a rocket effect, changing the mass.

The second observation is the change is with respect to the spacetime interval tau. Due to the relationship between the Greek and Roman alphabet, I expected this transition from relativistic to classical physics:
d \tau \rightarrow d t
I am certain that is what happens to generate the mA term. A change in an interval appears in classical physics as a change in time. There is another possibility: that a change in a spacetime interval appears as a change in space classically:
d \tau \rightarrow d R/c
With a thin spiral galaxy, the amount of matter changes with respect to the radius of the galaxy, a change which can be described only by the term \frac{d m }{d R}. In many thin disk galaxies, the amount of visible matter drops off exponentially with the radius. That is a serious dm/dR function caused by gravity that must be transparent in a calculation.

For Newtonian gravity, there will be three parts: Newton's law of gravity, the centrifugal force, and the acceleration:
- \frac{G M_a m_p}{R^2} = m_p \frac{V^2}{R} + m_p \frac{d V}{d t}
where M_a is the active mass, basically the sum of all gravitational masses, and m_p is the passive mass, the mass that is going around and accelerating. I can imagine - but have not done - calculating the velocity profile of a galaxy using this equation plus one other, the mass per area. What would be done is to divide the radius R into say 100 chunks, and to divide the circle into say 100 angles. Start with a passive mass at the center, where the velocity is zero. Calculate the mass for all the active masses, and plug it into the above expression. That will generate an acceleration, from zero to a new velocity. Repeat the calculation for the new passive mass at the next step out along the radius traveling at this new velocity V. Repeat for all positions of the passive gravitational mass along the radius. Plot the resulting velocities with respect to the radius.

If this calculation is done correctly, there should be a steep acceleration, a peak, and then continual drop in the velocity with respect to the radius.

The proposal:
For the galaxy NGC3198, we do the classical Newtonian gravity velocity profile calculation as described above. The validity of the calculation can be checked against the numbers for that galaxy (a peak velocity around 150,000 m/s, the velocity drops off after that). Then repeat the calculation with one more term:
- \frac{G M_a m_p}{R^2} = m_p \frac{V^2}{R} + m_p \frac{d V}{d t} + V c \frac{d m_p}{d R}
If the velocity profile reaches the peak, and remains flat after that, then we have something of significance. If the velocity profile still drops off, then it means this product rule does not explain the velocity profile of thin disk galaxies, and the hypothesis is wrong.

If you understood the problem description, I'll collect the number for that galaxy, and let us both give an independent try at the calculation. Anyone else reading this thread who likes doing a real numerical integral can also join in.

doug

[jhmar]

sweetser

I hope no one found the preceding explanation completely satisfying. That is what it feels like to wonder how things really work. There comes a point where what you know runs out.

Your mathematics are way beyond me but are they necessary? Fahr and Heyl published in Astrophysical Notes and on:
http:// arXiv:astro-ph/0606448 v1 19 Jun 2006
a paper on gravity, showing the relationship between mass and radius.

I recently had a particle physics paper rejected that made a similar proposal in that it suggested a relationship between mass, radius, and force. After reading Fahr and Heyl I was able to extend my proposal to show that it produced the same constant (for all particles) as that found by Fahr and Heyl (for gravity) and a revised paper is now under review.

It seems to me that the main difference in our approach is that you use actions to formulate a theory while Fahr and Heyl (and myself) use entities; the advantage of the entity approach to the problem is that what you know does not run out.

I would very like your opinion on the Fahr and Heyl paper.

[sweetser]

Hello Jhmar:

If my proposal is correct, then the Schwarzschild radius plays no role in the description of Nature. Oops, that is kind of obnoxious to a large body of work, oh well! In advanced books on GR, they will also explain how that radius is more a consequence of the choice of coordinates than anything else. What is independent of the choice of coordinates is that there is a point singularity. At the point singularity, many terms to describe a system become undefined.

In detail, I do not know how the GEM proposal is going to deal with areas of very high mass density. I do know it will be different. My sense of it is that there will be a singular at tau=0, but that is not a point singularity, it is the surface that all light travels on. That will mean that folks skilled in math can do things, that the description of the physical system will remain defined, unlike the case for a point singularity.

I also believe cosmology is missing a vital classical term, what I have been calling the relativistic rocket effect. Briefly this says that gravity accelerates mass (m A R_hat) and gravity also determines the location in space of moving masses (V c dm/dR V_hat). The presence of the factor of "c" and small spatial scales makes this term insignificant for our solar system. For the rotation profile of galaxies or the big bang, this term may become dominant. That is pure speculation at this time because I haven't figured out how to crunch numbers using this term.

doug

[jhmar]

doug

Thanks for a detailed reply which will take some time to digest. But at the risk of making a fool of myself I will make a few comments.

We are approaching the problem of explaining the universe from two completely different points of view; you take the GR approach, I take the classical particle physics approach.

Take for example "there will be a singular at tau=0". Now I would agree that a vacuum field can be reduced to a zero point but the mass (matter, anti-force, Higgs field? call it what you will) cannot be reduced to a point. That is why Fahr and Heyl and myself (although I add 'myself' here with some trepidation) say the constant is 1. The field can be 'zeroed' the field content cannot, but as the content of a single field is a fixed quantity there is a mass,force and radius formula that produces the 1 constant. This is as true of a gravity field as it is of an electron field. Gravity and the other forces can be 'unified' when each is considered in its particle form. That is to say that all force carrying particles have the same structural formula (graviton, lepton and quark are different states of a single fundamental particle), and a universal gravity field, like all formations where particles are contained in a single bonding field; is just another composite particle.

john

[sweetser]

Hello John:

It sounds like my efforts to communicate the GEM proposal have fallen short. My proposal is in direct, technical conflict with general relativity. GR is a rank 2 field theory that is generated by varying the action (just the Ricci scalar) with respect to the metric. GEM is a rank 1 field theory generated by varying the action (the contraction of a rank 2 asymmetric tensor) with respect to the potential. There are two testable differences cited in the first post (and I'll try not to repeat myself about them again).

The conservative me says one cannot claim to have a field theory until the action is defined. For a long time, I too did not have the action, and was darn uncertain I would ever find it. It was something like a year and a half before I did find it because at the time I had no practical experience with Lagrangians. That is quite common in the human population!

Most of your points leave me confused. Classical field theory is about Lagrangians, actions, field equations, and solutions to those field equations, all of which are part of the GEM proposal. I know it takes a good long time to understand that volume of detail in this thread, but it is there.

I don't feel I understand the case where tau=0. I would just point out that only particles with a mass of zero will live on that surface.

Elsewhere in the thread I've discussed how the proposal breaks U(1) symmetry due to mass, so there may be no need for the Higgs field.

In the GEM proposal, the stuff of gravity is in the symmetric, rank 2, irreducible tensor, while the stuff of EM is in the antisymmetric, rank 2, irreducible tensor. The graviton cannot do the job of a proton, particularly since like charges attract for the former but repel for the latter.

One must be ultra careful in unified field theory saying what is similar and what is different.

doug

ps. Learning physics is difficult stuff, so I don't consider you foolish at all.

[jhmar]

doug

Classical field theory is about Lagrangians, actions, field equations,

My lack of a decent education is the cause of our communication problem as demonstrated your statement. I thought Classical particle physics was about what particles arerather than what particles do.

I thought that only when we know what determines mass, force and radius (physical size), can we understand the cause and outcome of actions. Realising that this was a far simpler problem than the study of actions I set about finding a formula that gives the basic structure of all elementary particles; only to find that there is only one elementary particle that can be transformed into an infinite number of states (all with the same content).

I could only make a crude estimate for the graviton state and my constant is related to force (it is still valid). So I was over the moon to find Fahr and Heyl's paper and quickly realised that by a variation of my formula, my table could be extended to produce the same value (1) constant.

The difference between states is determined by the wave structure. The data given by The Particle Data Group is compared to the theoretical data and falls within the margins of error given by PDG in all but two borderline cases (about 150 are given).

PDG reject many experiments and average the remainder, I make no rejections and do not average. Each experiment is considered to find a specific particle state.

Sorry, have to go, cannot stop to edit reply, hope it is readable

john

[Sariaht]

Hello John:

It sounds like my efforts to communicate the GEM proposal have fallen short. My proposal is in direct, technical conflict with general relativity. GR is a rank 2 field theory that is generated by varying the action (just the Ricci scalar) with respect to the metric. GEM is a rank 1 field theory generated by varying the action (the contraction of a rank 2 asymmetric tensor) with respect to the potential. There are two testable differences cited in the first post (and I'll try not to repeat myself about them again).

The conservative me says one cannot claim to have a field theory until the action is defined. For a long time, I too did not have the action, and was darn uncertain I would ever find it. It was something like a year and a half before I did find it because at the time I had no practical experience with Lagrangians. That is quite common in the human population!

Most of your points leave me confused. Classical field theory is about Lagrangians, actions, field equations, and solutions to those field equations, all of which are part of the GEM proposal. I know it takes a good long time to understand that volume of detail in this thread, but it is there.

I don't feel I understand the case where tau=0. I would just point out that only particles with a mass of zero will live on that surface.

Elsewhere in the thread I've discussed how the proposal breaks U(1) symmetry due to mass, so there may be no need for the Higgs field.

In the GEM proposal, the stuff of gravity is in the symmetric, rank 2, irreducible tensor, while the stuff of EM is in the antisymmetric, rank 2, irreducible tensor. The graviton cannot do the job of a proton, particularly since like charges attract for the former but repel for the latter.

One must be ultra careful in unified field theory saying what is similar and what is different.

doug

ps. Learning physics is difficult stuff, so I don't consider you foolish at all.


Hey, I recognise that face! You've been on tv! Coool! What are you working with?

And if i asked you right now, what momentum per time causes gravity, what would you respond?

[jhmar]

doug

I cause much confusion by using the wrong terms (lack of training). That aside, I have submitted my work and await the moderator's decision due in about three weeks.
Basically I believe I have demostrated a relationship between wave length and mass, but must now wait and see if I have managed to convince those with a greater knowledge of this subject.
Please keep returning to this forum where I will notify you of the outcome when received; a positive outcome will, of course, appear as a separate forum.
regards
John

[sweetser]

Hello Sariaht:

"The Stand-Up Physicist" is award winning TV seen by almost no one! The show represents ultra-narrow casting. The show appeared on BNN-TV in Boston at 11:30 on a Tuesday night, and no doubt got killed by the likes of Jay Leno & David Letterman. The town of Auburn outside of Worcester gets several doses a week because the guy helping with the post-production work tries to fill the airways. The third place it shows in the town of Acton where I live, four times on the weekend, but I have no idea what time (they schedule things poorly here).

There are all kinds of video festivals out there, particularly if the video has gay content (perhaps as an excuse to get together and party). So far, I found only one festival that had an education category. I entered the episode, "Why Quantum Mechanics is Weird" in the Berkeley Video and Film Festival, and it won Best of Festival in the Education Category. Sounds good, but it was second place to the Grand Prize winner, a film on polar bears. I went up against a polar bear and got mauled.

After reading an article on digital video recording on Slashdot and having discussions with a friend who does professional video work, I spent about $4k to get a 3-CCD digital camera, a great Sennheiser boom microphone, 1500 watts of lights, a tripod, background set holder, and a green screen.

Why bother? Here's my situation: Mathematica and I feel confident that we have found exactly what Einstein was looking for: a unified field theory for gravity, and a reason why causality is different for quantum mechanics versus classical mechanics. In my world view, I don't know if I could construct a more unlikely thesis. It is reasonable for people not to believe it is true. I have no idea how to get through that social barrier. It could be that I am just wrong. That in so many ways is the easiest. Sincere nerd is a moron. I take that seriously because I have made technical mistakes along the way, from the wrong sign for the charge coupling term, to misstatements about the field strength tensor. My honest appraisal is that the foundation still feels solid, and more work can be done.

If and only if the unified field theory is correct, then it is my responsibility to communicate it. I will continue to put my money there in diverse ways. I have a web site, there is the self-printed book, I go to conferences three or four times a year (never as an invited guest, and that limits the audience to those who also were not invited). I make deliberate efforts to toss the idea up to leading researchers in physics, but those folks are busy, busy, busy. There are many more people with more time, but less training, that are interested in the possibility of a unified field theory. That is the group the show is target at. The show is too technical for a general audience. I work through most of the mathematical detail, which is unusual.

The show is just me yakking in front of a camera, very minimalist. I make sure the equations I am thinking about at the moment are on the screen, because physics is math. I do the post-production work at Auburn Community Access Television, in Auburn, MA (a fifty minute drive I have done many times). I use Final Cut Pro, version 5, on a Mac. Sixteen shows are complete, another 8-10 to write, shoot, and produce. The half hour shows can be downloaded from the web (no idea about the stats). The DVDs are for sale internationally, and as of today, the global sales are up to 1 DVD to a friend of mine.

That is the TV story.

> "what momentum per time causes gravity"?

I am going to work to translated this... The units for momentum per time are the units of force. In general relativity, one cannot write an equation for the force of gravity like you can for EM. Instead, there are geodesic paths that objects follow, paths of no force.

GEM is modeled on EM. The answer to what is the force equation of EM is the Lorentz force:
\frac{d m V^{\mu}}{d \tau} = q_e V_{\nu} (\nabla^{\mu} A^{\nu} - \nabla^{\nu} A^{\mu}) = \gamma q_e (\vec{\beta} . \vec{E}, \vec{E} + \vec{\beta} \times \vec{B})

This is the relativistic force law. At low speed, \beta\rightarrow 0, \gamma \rightarrow 1, so the law becomes Coulomb's force law, or:
\vec{F} = q_e \vec{E}
For my approach, one key change is the field strength tensor is the symmetric partner to the antisymmetric field strength tensor of EM. This has an important consequence: like charges attract. The charge for gravity is independent from EM and must have an opposite sign. This should be enough information to form the gravitational Lorentz force equation:
\frac{d m V^{\mu}}{d \tau} = -q_m V_{\nu} (\nabla^{\mu} A^{\nu} + \nabla^{\nu} A^{\mu}) = -\gamma q_m (g_0 + \vec{\beta} . \vec{e}, \vec{g} + \vec{e} + \vec{\beta} \times \vec{b})

where small e and small b are the symmetric analogues to the E and B of EM, and g is the four parts along the diagonal of the symmetric field strength tensor. In the classical physics realm, the only term that survives is the small e:
\vec{F} = - q_e \vec{e}
Newton's law of gravity, bingo. This is darn similar to the case for EM, but the sign is different, which is essential.

Another way to look at this is that I have found the right way to pluck out Newton's law form a relativistic force law from gravity. That wasn't how it was done, but it could be viewed that way.

Hope all had a good thanksgiving (if in the US),
doug

[CarlB]

Doug,

Lubos Motl wrote a paper that pretty much states my objections to the requirement that Lagrangian or Hamiltonian formalism be at the foundations of physices here:
http://motls.blogspot.com/2006/12/conventions-vs-physics.html

[sweetser]

Hello Carl:

In this forum, I am working on a rank 1 field theory to explain gravity as a gauge choice between changes in the potential and the connection (call it diffeomorphism invariance, which is also at the heart of GR, but the details of implementation are different). One consequence is that any rank 2 field theory for gravity will be superseded, included GR. If GEM is correct, the huge amounts of work on black holes and the singularities of GR is not relevant to the description of Nature. Ouch, that is not going to be popular! So for this forum, the first half of Prof. Motl's blog can be summarily dismissed.

EM theory is completely integrated with the standard model. There are 2 charges, and one massless force particle. It is reasonable to speculate that gravity, with only 1 charge and one massless force particle, should be a wee bit simpler to understand. The Newtonain law of gravity, and Coulomb's law are clones. Only for GEM, the relativistic forces also look similar. Here is the EM Lorentz force:
F_{EM}^{\mu}=q_e U_{\nu}(\nabla^{\mu} A^{\nu} - \nabla^{\nu} A^{\mu})
A force is a coupling of the moving charge (q_e U_{\nu}) with a field strength tensor (\nabla^{\mu} A^{\nu} - \nabla^{\nu} A^{\mu}). Move a charge around in a field, forces happen. Right answers are simple and direct.

One path to GR is to start from Newton's law of gravity, which is flawed, and try to correct that one flaw, ignoring for example that no one tries to quantize Newtonian gravity. The result of that exercise is the field equations of GR. Gupta, Feynman, and Weinberg have all shown that to be the case.

If you start from a bad place, bad things follow. Instead, construct the GEM Lorentz force as happened in EM, as the charge in motion coupling to the field strength tensor:
F_{G}^{\mu}=-q_m U_{\nu}(\nabla^{\mu} A^{\nu} + \nabla^{\nu} A^{\mu})
I copied the EM equation, changed two signs, and swapped two labels. What could be more simple, and direct?

I see no value in his comments about Lagrangians, Hamiltonians, actions, and the Feynman integral approach because it is an all or nothing deal. If you know the right Lagrangian, you can calculate the Hamiltonian. If you know the right Lagrangian, you know the action. If you know the right Lagrangian, you know the Feynman integral which is the exponential of the action. If you first figure out one of the other three, then you can determine the set. They are a logically consistent set.

Prof. Motl like other researchers in gravity does not know one of these, therefore he does not know any of them. So this group flounders no matter what tool they use. It is kind of tragic really. The logic of physics is unkind.

Here is his best lip service:
Unless you're lucky to guess new physics with the complete equations directly, new physics can only be revealed by identifying new possible principles, constraints, or physical mechanisms.
That's exactly how "lucky" I feel.

Why is this lip service? There was a time when both Lubos and I posted to the newsgroup sci.physics.research. He was a strident believer in the value of string theory. I often found his position embarrassing even for other people doing string theory. I know my own lack of intellectual precision was embarassing to professionals reading SPR. Lubos conveyed the message that if you did not understand how right string theory was, you were foolish, or stupid but probably both. I had been listening to string theory - not studying it, just eavesdropping - and it did not make sense to me. The units for a volume of spacetime are just wrong, and if you get the units wrong, you are wrong. Compatification is a fancy name for bold physics BS. Call the stuff that stinks s--- and move on before the stench causes permanent brain damage.

There was a post in SPR years ago where someone said don't complain unless you have something better to offer. Once I reached the point of my unified field theory research where I had testable hypotheses (it is plural), I wrote Lubos, the poster child of string theory, a simple financial reward. If anyone anywhere in the world in the next ten years develops an explanation for gravity that uses more than 4 dimensions, then I would send $100 to Lubos. This was not a bet, there is no risk to Prof. Motl. I wrote out the check, but did not sign it, in April of 2004. I closed that bank account in 2006, so had to write another check. I sent Lubos the jpegs, along with a draft of my paper. No comments have been returned. So I have data that he is not interested in a complete set of equations in four dimensions that make predictions that challenge GR.

doug

[CarlB]

EM theory is completely integrated with the standard model. There are 2 charges, and one massless force particle. It is reasonable to speculate that gravity, with only 1 charge and one massless force particle, should be a wee bit simpler to understand.

I'm confused here. What are the two charges of EM? I can think of only one, Q.

Carl

[sweetser]

Perhaps my bad lingo, there is one electric charge with two signs. With gravity for GEM, there is one type of mass charge that can only have one sign. GR uses the stress energy tensor, not a notion of mass charge density. This really is the vanilla momentum 4-vector with units of electric charge, acheived by multiplying by the square root of Netwon's gravitational constant.

doug

[CarlB]

If you're going to count the number of different signs, it seems to me that you should also count the number of different magnitudes. Then the elementary particles come in a fairly wide number of charges, -1, -2/3, -1/3, 0, +1/3, 2/3, +1, which is 7 in all. And while gravity comes in just positive charges, the number of different charges is fairly large. I want to say 12 just for the fermions, plus you need the W and Z.

[jhmar]

Then the elementary particles come in a fairly wide number of charges, -1, -2/3, -1/3, 0, +1/3, 2/3, +1, which is 7 in all.

Surely in experiments involving quarks, fractional charges are allocated to make the particles observed comply with the charge conservation law? In TFQHE they are part of a proposed mathematical explanation? In neither case are the charges actually observed; they are purely theoretical. Even the minus sign is questionable, it cannot denote a reality that is less than nothing, but, a measurement below an arbitrary (unknown) base line.

I would like a reference to the positively charged graviton as I have not come accross any such charge.

W and Z charges are the same as the leptons -1, +1, and 0.

[sweetser]

Hello Carl and Jhmar:

There is the fundamental electric charge, and so far, we have only measured integral amounts of that charge. The quark model does have the fractional charges noted by Carl, but those have not been measured in an experiment. There is a theory as to why they cannot be measured, and I do not know the details of quark confinement.

> W and Z charges are the same as the leptons -1, +1, and 0.

These particles play the roll of a force particle, like the photon for EM. Why one photon for EM, and three force particles for the weak force? It has to do with the Lie algebra for the group U(1) needing one number for EM, and SU(2) needing three numbers for the weak force. Because the W and Z have a mass, the weak force is short range. Gravity and EM are long range forces, to the force particles must be massless.

> I would like a reference to the positively charged graviton as I have not come across any such charge.

I don't know that I can provide you with a reference. All the work with gravity has gravity coupling to the stress-energy rank 2 tensor, not the 4-momentum vector (rank 1 tensor). Mass charge is going to be different than electric charge. As noted above, electric charge is an integral multiple of a fundamental charge. Mass charge does not appear to work that way. A neutron and a proton have slightly different mass charges (\sqrt{G} M). At this time, I have no idea why the pattern of mass charges is the way it is.

doug

[jhmar]

doug

Thanks a lot for your detailed reply. The purpose of my submission was to find out if there was any experimental data that would nullify my proposed model, and I am pleased to say your reply indicates there is not.

I have split my interpretation down into small sections, the first two parts have been submitted. The third,and possibly final section, is almost ready for submission; so I hope you will be able to see where I am heading, in the near future.

For me, it is not a question of devising new mathematical theories (strings, brans etc.), which are totally beyond my abilities. But, it is a question of the correct interpretation of past experiments. It is in the interpretation that I dissagree with the Standard model, but that, of course; does not mean that I dissagree with Quantum theory. Interpretation and theory are related but different; a correct interpretation will, I hope; place a limitation on the multitude of possible mathematical solutions thrown up by QT.

[sweetser]

Hello Jhmar:

The standard model is really successful, but there are two "weak" points. The first is why should the three particular groups, U(1), SU(2), and SU(3), be so important? There are lots of other possibilities, yet so far no one can provide a reason for these three.

The GEM proposal as it is written can explain these three groups: Diff(M)xU(1)xSU(2) - or gravity and the electro weak force. Gravity comes from the two covariant derivatives. One is free to choose how much the 4D wave propagation is due to changes in the potential or in changes that happen as you move around the manifold. If one write the 4-potential as a normalized quaternion, then the quaternion potential can be written as a unit quaternion times itself, like so:
q = \frac{q}{|q|} exp(q - q^*)
SU(2) is the unit quaternions, the exponential part of the expression above. Quaternions do commute with themselves, so:
\frac{q}{|q|} exp(q - q^*) = exp(q - q^*)\frac{q}{|q|}
The normalized quaternion, \frac{q}{|q|}, is now behaving just like U(1), a complex number with a norm equal to one. Let's rewrite the 4D wave equation in the very first post to look like it justifies the electroweak part of the standard model:
J_q - J_m = \square^2 \frac{A}{|A|} exp(A - A*)
The box has Diff(M), A/|A| has the U(1) part, and SU(2) is the exponential. This is good, but not good enough because we need to spot SU(3). One thing I could do with this equation is to calculate its norm. That kind of thing happens all the time in quantum mechanics. It is possible that the norm operation would give the
equation SU(3) symmetry. That is speculation I don't have the skills to prove.

This is a benefit of GEM I don't discuss much due to my lack of self-confidence in group theory: the GEM field equation in and of themselves justify the symmetry seen in the standard model.

Another weakness of the standard model has to do with mass. The standard model out of the box makes one simplifying assumption: all particles have zero mass. Of course that is not true. So now the accepted way to introduce mass back into the model is known as the Higgs mechanism. There has to be a scalar Higgs field everywhere in the universe ready to break the symmetry of the vacuum such that all massive particles get their inertial mass. One of the main reasons for a multi-billion dollar bet being made at CERN is to detect the
Higgs.

The GEM hypothesis rejects the Higgs mechanism, and the Higgs boson. I've had two people comment on my Lagrange density that it does not have U(1) symmetry. This is mostly true. The Lagrangian has U(1) symmetry if the particles are massless. When there is a mass, the mass charge breaks the U(1) symmetry. The symmetry breaking is EXTREMELY slight - one part in 10^16 for an electron. We only define electric charge to ten significant digits, so the symmetry breaking is beyond our ability to directly measure.

This is actually very reasonable. Take a pair of electrons, and a pair of protons, put them 1 cm apart, then measure the acceleration, which is the F/m ratio. To ten significant digits, they are the same. All electric charges repel the same amount once the inertial mass is taken into account. Now take the same electrons and protons, but measure F/m to twenty significant digits. The answer is no longer the same. The gravitational mass of the electron is less than that of the proton, and now the difference can be measured. Gravitational mass
breaks the electron charge symmetry.

Particle physicists are concerned with the scalar Higgs which is suppose to bring inertial mass to the standard model. People who work with inertial gravity are concerned about the spin 2 graviton. Yet there must be some unbreakable link between the particle of inertia (the Higgs) and the particle of gravity (the graviton). The rank 2 symmetric tensor in GEM is the graviton, and its trace which is a scalar field, does the work of the Higgs. It is clear you can never
have an inertial gravity field without having a gravitation field.

It was an unexpected gift that the 4D wave equation to give me insight into the standard model.

doug

[CarlB]

Doug,

I am still very convinced that you're on the right track here. I just haven't had time to work on my simulation. Thanks for the continuing explanations.

Carl

[sweetser]

Hello Carl:

Thanks. I find struggling to explain the proposal is fun. Remember, I really want a solid technical reason to drop this hypothesis. Although I feel confident about having the right symmetry for gravity and the electroweak forces, if I didn't have a proposal for the symmetry of strong force, that would be a reason to reject the proposal. Three out of four is not good enough, since this fantastic four does all the work in the Universe.

In the eternal kids game of "Why?", now that the 4D wave equation may justify the symmetry of the standard model, why is the 4D wave equation so central? The Universe has lasted a good long time. How could it be doing all that it does for such an expanse of time? The key is to do almost nothing. Almost nothing is not the same as nothing. The next door neighbor to doing nothing is the simple harmonic oscillator. The 4D wave equation is the equation that describes the fundamental family of simple harmonic oscillators in spacetime. If you are some particle that happens to exist in 4D spacetime, the closest thing to doing nothing is simple harmonic oscillation caused by that other crap in the Universe. It is really amazing that the Maxwell equations are just a partial rewrite of the 4D wave equations. There is also a set of equations for the symmetric expression which is not an exact clone of Maxwell (like the area of study known as gravitoelectrodynamics, which had identity equations that are not part of GEM). Mapping the classical field equations for GEM back to the 4D wave equation requires getting lots of signs correct, but it is an impressive wad of algebra. A detailed description starts here:

http://theworld.com/~sweetser/quaternions/talks/IAP_2/925.html

and goes on for 9 slides. It is hard to generate that many partial differential equations that work together unless there is some truth there. This is the kind of thing I checked with Mathematica because there are so many signs that have to be right, no exceptions.

doug

[jhmar]

There is the fundamental electric charge, and so far, we have only measured integral amounts of that charge. The quark model does have the fractional charges noted by Carl, but those have not been measured in an experiment.

Tsui et al measured two dimensional fractional charge. This is described as thin film indicating that the term two dimensional is being used to described a three dimensional layer with one dimension being to thin to observe, rather than a genuine two dimensional layer which would of course be unobservable.
The Nobel prize committee and Scientific American describe this discovery as being of fractionally charged electrons. Note that Tsui makes no reference to pos. or neg.; I read somewhere that this is because the experiment indicated a positive charge that could not be explained by Tsui et al. But have lost the ref..

see: http://www.ee.princeton.edu/people/Tsui.php

Note that the dashed (all fractions) line is accompanied by the stepped experimental results line with steps of decreasing value. This is the basis of my proposal for particle structure which has been rejected on the grounds that I have used the two dimensional experiment as a base for a three dimensional object (particle). But I would maintain that any theory must take into account the stepped nature of particle structure. It should also being able to explain what mass and charge are and what causes them to exist.

Personally I think the interpretation (but not the mathematics) of Tsui work is wrong and I note he does not use the interpretation (i.e. the term fractionally charged particles) himself.

[CarlB]

Remember, I really want a solid technical reason to drop this hypothesis.

Can't help you there. Hey, you thinking about going to GRG18?
http://www.grg18.com/

I find that I'm getting convinced into going. Plenty of time left to think about it.

[sweetser]

Hello Carl:

Unless I become an invited speaker, it conflicts with a different physics event. I am taking a 4 day class at MIT titled "Relativity, Gravity, and Cosmology". It is part of MIT's Professional Institute, a way to milk alumni cows (actually, any cow willing to fork over $2k for 4 days can step into the machine). For a full description of the class, go here: http://web.mit.edu/mitpep/pi/courses/relativity_gravity.html

I will be going to the April APS meeting in Jacksonville FL. I think I forgot to post my abstract, so here it is:

Title:
Geometry + 4-potentials = Unified Field Theory

Abstract:
Geometry without a potential is like a bed without a lover. The Riemann curvature tensor is only about the change in moving around the manifold, the geometry of the bed sheet. The exterior derivative of the EM field strength tensor is only about changes in the potential, isolated from geometry. In my work, changes in the potential lay on top of changes in the manifold. A covariant 4D wave equation can describe both gravity and light. The metric solution passes weak field tests of gravity and tests of the equivalence principle. At higher resolution, 0.8 microacrseconds more bending of light around the Sun is predicted than GR. Quantize using the Gupta/Bleuler method, but the scalar and longitudinal modes are the spin 2 graviton.

The misses found the first line fun. I expect to accomplish zero, but it is great that provocative text comes straight out of the math.

At the end of May, beginning of June, there should be EGM 10 at Columbia, but they are still working out the details for that meeting.

doug

[sweetser]

Hello:

The standard model of particle physics defines three out of the four fundamental forces of physics. The key to organizing the particle zoo is a set of three symmetries: U(1) - complex numbers with a norm of 1, SU(2) - unitary quaternions, and SU(3). Gravity has yet to be incorporated into the Standard Model. Two riddles are why Nature chose these three particular symmetries and how can gravity come into the picture.

I've been writing software to animate quaternion expressions (URL at the end). An obvious target for animation is SU(2) which can be written as
SU(2)\rightarrow exp(q - q*)
Pick a bunch of random quaternions, plug them in to the exponential, then create an animation. It looks like a slice of a sphere that grows and shrinks. From there it was easy enough to get to U(1)xSU(2), which is known as the electro-weak symmetry that unifies
EM with the stuff of radioactive decay, like so:
U(1)xSU(2)\rightarrow \frac{q}{|q|}exp(q - q*)=exp(q - q*)\frac{q}{|q|}
A quaternion commutes with itself, that is why \frac{q}{|q|} can be written on either side. The Abelian property is necessary if this is a valid representation of U(1). It actual is U(1) in the usual sense if y=z=0, but the way it is written, this represents U(1) for arbitrary quaternions. The animation looks like a more complete expanding sphere, but it has a bias, the points the make up the sphere are not evenly distributed. Two thirds of the way there!

The question was how to get to SU(3). The Lie algebra that can be used to generate this group has eight elements to it. My thought was to multiply conjugates together, q* q' (q != q' so it has 8 components like the Lie algebra su(3)):
U(1)xSU(2)xSU(3)\rightarrow (\frac{q}{|q|}exp(q - q*))* \frac{q'}{|q'|}exp(q' - q'*)
The result is an evenly distributed collection of points in an expanding/contracting sphere. When you do get if q=q'? The result is a single dot at (1, 0, 0, 0)! The standard model may all be about the symmetry of one in quaternion spacetime. Fun to think about.

The final question is where is gravity? It all has to do with 1, and not quite 1. For an arbitrary point in a spacetime manifold, you calculate the norm of q* q', and it comes out to be exactly 1.0. Now you go to a neighboring point, repeat the calculation, and, oops, the norm is a little bit smaller than 1. That's gravity, because gravity is about how measurements change as one moves around the spacetime manifold. The act of multiplication means metrics must be involved.

This can be explicitly connected to the GEM field equation:
U(1)xSU(2)xSU(3)\rightarrow g_{\mu \nu} J*^{\mu} J'^{\nu} = g_{\mu \nu} ((\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})\frac{A^{\mu}}{|A^{\mu}|}exp (A^{\mu}-A^{\mu}*))* (\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})\frac{A'^{\nu}}{|A'^{\nu}|}exp (A'^{\nu}-A'^{\nu}*)

Why do this big complicated expression? We are trying to pack all the known forces of Nature into one expression: EM as U(1) represented by \frac{A^{\mu}}{|A^{\mu}|}, the weak force as SU(2) represented by exp (A^{\mu}-A^{\mu}*), the strong force as SU(3) represented by q* q', and gravity as Diff(M) represented by the (possibly) dynamic metric g_{\mu \nu}. That's a long sentence too! The animation gives a clue: this is how one fills up a volume of spacetime consistently with groups. This equation is the most basic spacetime wave equation. The collision of spacetime waves, group theory, and quaternions explains the origin of the fantastic four forces of Nature. Stunning, if true.

doug

You can see all the images that make the story here:
http://www.theworld.com/~sweetser/quaternions/quantum/standard_model/standard_model.html
Warning: it will take getting use to the 6 graphs. The animation is center top. On the
right are three complex planes, ty, tz, and tx. On the left is the superposition of all states the system can be in for the animation, formed by merging everything in the animation. That was inspired by quantum mechanics.

[sweetser]

Hello:

The newsgroup sci.physics.research was very important to me as a way of learning about physics and physics research. I began reading the posts in 1995, almost as soon as I had my first access to the Internet (using trn, if I recall correctly). As I continued to see just how much I could do with quaternions, the moderators of the newsgroup tired of me. I know one of the key players in SPR, a math guy named John Baez, no longer thought my quest to work extensively was interesting. John was supportive of non-professionals, unless they were weasels like myself (it is an obtuse skill I have, the ability to tick of math guys). It is hugely frustrating to get a post rejected as being "overly speculative", a nice vague definition they can whip out on a whim. For the most part, I have stopped posting there.

I did make an expectation recently. This work with the cause of the 4 forces of Nature is just too cool. I was not confident they would accept it, but it got in. No one is discussing it so far, but that is par for the course. Here it is, both playful and technical. Hope you enjoy...

[Post to sci.physics.research]

Hello:

"Where's Waldo" is a cartoon phenomena whose goal is to spot Waldo somewhere in a densely drawn image. Waldo is there, you just have to work to spot him. A simple game that has meant millions for its author.

In contrast, the standard model of physics is super serious, dictating three out of the four forces of Nature: EM, the weak force, and the strong force. If you have ever seen the Lagrangian, it is densely drawn, with generators of groups, gamma matrices, binding constants, and wave functions. I do not find it enlightening.

A simpler approach to the standard model focuses on the symmetries: U(1)xSU(2)xSU(3), which are related to EM, the weak force, and the strong force respectively. The Lie algebras for these continuous groups have vector spaces with 1, 3, and 8 dimensions, which exactly matches the number of bosons involved in these forces: 1 photon for EM, +/-W, Z for the weak force, and 8 gluons for the strong force. The group U(1) has all the properties of the complex numbers with a norm of 1. The group SU(2) is the unitary quaternions, quaternions with a norm of 1. The continuous group SU(3) is also a special (norm of 1) unitary group. So now the hunt is on for these groups somewhere in EM, the weak force, or the strong force.

Although I have worked with EM, I cannot say the same for any equations involving the weak or the strong force. I have read what they do, but in an equation-free way. I cannot start with the weak or the strong force, but need to keep an eye out for them. Let's write out EM in a way Feynman called "a beautiful set of equations!" (Lectures, II, 18-11), the Maxwell equations in the Lorentz gauge:

J^u = (1/c d^2/dt^2 - c Del^2) A^u

Both J^u and A^u are 4-vectors. Quaternions can also be viewed as 4-vectors. So treat J^u and A^u as quaternions that happen to have indices (it's just a label after all, and in this case it is restricted to run from 0-3). Now that this is a quaternion wave equation, how would we write a unitary quaternion? Take the exponential of a quaternion where the scalar has been dropped:

exp(q - q*) is an element of SU(2)

There is a problem though, because the J and A have four degrees of freedom, but exp(q-q*) only has 3. We need to plug 1 degree of freedom back in. We could just grab the scalar using (q+q*), but recall the purpose of the exercise: lets go for the Abelian group U(1) since it's Lie algebra also has 1 degree of freedom. Rewrite A^u like so:

A^u = A^u/|A| exp(A^u - A*^u) = exp(A^u - A*^u) A^u/|A|

A^u commutes with itself, and with the exponential of itself because it points in the same direction so the cross product is zero (that's the non-commuting part). Together, A^u/|A| exp(A^u - A*^u) has electroweak symmetry, U(1)xSU(2)!

The Lie algebra su(3) has eight dimensions, twice the number we have in this equation. What would be a reasonable way to "double" this equation, doing some standard operation in quantum mechanics? Recall the <bra|ket> notation, which is a* b. Imagine 2 current densities, J^u and J'^v. I'd like to calculate the inner product of the two, but need a metric to do so:

g_uv J*^u J'v =
g_uv (1/c d^2/dt^2 - c Del^2)(A^u/|A| exp(A^u - A*^u))* A'^u/|A'| exp(A'^u - A'*^u)

An initial objection for this expression being a representative of U(1)xSU(2)xSU(3) might go like this. If a quaternion represents U(1)xSU(2) when written like q/|q| exp (q - q*), then the product of two quaternions should be in the same group. That's how groups work!

A detail was missed: we are taking the conjugate of one of these quaternions and multiplying it by the other. As John Baez pointed out to me, that means that multiplication is no longer associative, since:

(a b)* c != a* (b c)

but norm ((a b)* c) = norm (a* (b c))

I happen to call this sort of non-associative multiplication "Euclidean multiplication" because the scalar part of q* q is t^2 + x^2 + y^2 + z^2. The multiplication table will necessarily be different because regular (or what I call Hamiltonian) multiplication is associative.

It is important to remember that quaternion multiplication, even Euclidean multiplication, preserves the norm. Therefore the norm of this q* q' will be one. There is still the same identity, (1, 0, 0, 0), and every quaternion will have an inverse under Euclidean multiplication. With eight numbers to plug into q* q' and a norm of 1, I believe q* q' is a way to represent SU(3).

There is a bonus to viewing this 4-vector equation as a 4-vector with the properties of an indexed quaternion, which not only can be added, subtracted and multiplied by a scalar, but also multiplied and divided with each other. The bonus is the explicit appearance of the metric g_uv. We have placed no constraints on the metric. It can be whatever, and this wave equation does not change its form. If the metric is the flat Minkowski metric, then we have the same multiplication rules set out by Hamilton for the scalar part. Calculating the inner product of two current densities will use the same equation no matter what the manifold is, whether it is static or dynamic. In terms of group theory, the equation has Diff(M) symmetry which is at the heart of general relativity.

This was a fun game of algebra, but Where's Waldo is a _visual_ exercise. It is time to translate this algebraic story into pictures. Quaternions are 4 dimensional, so how do we deal with that? Go to the movies! Imagine generating a thousand quaternions at random: different values of t, x, y, and z. Take all of them, and sort them by time t. Make an animation lasting 10 seconds, at 30 frames per second, or 300 frames total. Figure out the range of time, from earliest to latest, and divide that by 300. Any particular frame will correspond to a range of time values. If a quaternion happens to fall in the range, place it at the appropriate place given the x, y, and z values.

The software for analytic animation using quaternions has been written. Since SPR is a text based newsgroup, I'll describe the images (URL at the end if you want to see the results, but they are not trivial to grasp since we are not accustom to seeing functions of spacetime). The easiest one to deal with is SU(2) because all that needs to be done is to generate thousands of quaternions randomly, then calculate exp(q-q*), and plot the result. None of the events have a time less than zero. The norm always has to be equal to 1, so the first points start out at the extremes of +/-x, +/-y, +/-z. The image then forms a ball that shrinks to a small radius, because by that time, most of t^2+x^2+y^2+z^2=1 is the t^2 part.

When exp(q-q*) is multiplied by itself normalized, q/|q| exp(q-q*), the result is a sphere that grows and shrinks, but has a decided bias. Most of the points now have a negative time.

The animation of (q/|q| exp(q-q*))* (q'/|q'| exp(q'-q'*)) is a completely smooth rendering of spacetime from (-t,0,0,0) to (+t,0,0,0). It doesn't appear like one could devise a smoother way to fill up a volume of spacetime with events.

The take home message is simple: the symmetry of taking the inner product of two indexed quaternion 4D wave equations with a norm of one is Diff(M)xU(1)xSU(2)xSU(3). This may be the reason behind the four forces of Nature: gravity, EM, the weak force, and the strong force fill up spacetime. Way to go Waldo!

doug

http://www.theworld.com/~sweetser/quaternions/quantum/standard_model/standard_model.html
http://quaternions.sf.net/

[CarlB]

I am amazed that they put this up on SPR. The only sociological guess I can come up with is that they're a bit jealous now that sci.physics.foundations is up and running and is moderated with a lighter hand.

You know, the longer I look at this the more difficulty I have understanding it. I liked: "exp(q - q*) is an element of SU(2)". When we get to "Together, A^u/|A| exp(A^u - A*^u) has electroweak symmetry, U(1)xSU(2)!" I start having problems.

Now in the above, I'm assuming that you mean A^u to mean a collection of four quaternionic numbers. There are therefore a total of 16 degrees of freedom running around here.

I can't seem to agree that A^u/|A| is at all a U(1) symmetry. The way I understand it, it's not a normed quaternion, it's a normed vector of quaternions and must be something rather complicated. Even if it were A^u/|A^u| it would be a pretty complicated beast.

When I've seen SU(3) naturally pop out of combinations of things, the things involved showed up in 3s. Three is a very important number for SU(3), my intuition says that it won't show up naturally as a product of U(1)xSU(2) even with a lack of associativity. Maybe you would end up with a representation of the octonions.

My guess is that the way to approach this is by computation.

Also, the way that degrees of freedom and bosons were adding up made sense when you combined U(1) and SU(2) to get 1+3 = 4 degrees of freedom. But when you add SU(3) to the list you should end up with 12 degrees of freedom, not the eight you have. You could fix this by adding in another q, which gets back to those 3s that are so important to SU(3).

[sweetser]

Hello Carl:

Sounds like we have miscommunication going on. Hopefully I will clarify.


Now in the above, I'm assuming that you mean A^u to mean a collection of four quaternionic numbers. There are therefore a total of 16 degrees of freedom running around here.

This is not what I meant. There is 1 quaternion 4-potential, A, which I happen to write as A^{\mu} so when I calculate the scalar part of two quaternions, say A^{\mu} B^{\nu}, it makes sense how to use a metric g_{\mu \nu} to calculate the scalar part of that quaternion product. The usual quaternion product always and exclusively uses the Minkowski metric to calculate the scalar part of the quaternion product. Let me write out in detail what I mean:
g_{\mu \nu}A^{\mu}B^{\nu} = (g_{00} A^0 B^0 + g_{11} A^1 B^1 + g_{22} A^2 B^2 + g_{33} A^3 B^3, A^0 B^1 + A^1 B^0 + A^2 B^3 - A^3 B^2, A^0 B^2 + A^2 B^0 + A^3 B^1 - A^1 B^3, A^0 B^3 + A^3 B^0 + A^1 B^2 - A^2 B^1)
For the Minkowski metric, g_{00}=1, g_{11}=-1, g_{22}=-1, g_{33}=-1, and for this choice of metric, the product will be exactly the same. The ability to use a different metric only comes into play at the last step of the game, where I want to see the Diff(M) symmetry that underlies any metric theory for gravity. For the sake of clarity, I'll drop the mu's in the discussion.

So A the quaternion has 4 degrees of freedom. The way to calculate and write out a unitary quaternion is exp(A-A*). It is the A-A* step that tosses away one of the four degrees of freedom, because it gives back only the 3-vector.

The group U(1) is usually introduced as the unit circle in the complex plane. This is an Abelian group, with one degree of freedom in its Lie algebra. A quaternion A/|A| might have the right norm, but it has 4 degrees of freedom, and is non-Abelian.

Now we consider a particular product: A/|A| exp(A-A*). We agree that the exp part is a fine way to represent SU(2), whose Lie algebra has 3 degrees of freedom. Since we are using the same A throughout, and since a quaternion commutes with itself, then for this product: A/|A| exp(A-A*) = exp(A-A*) A/|A|. A/|A| must point in the same direction as exp(A-A*), so it really only has one degree of freedom.

I am going to stop here, to see if we are on the same page as far as metrics, labels, SU(2) and U(1)xSU(2).

doug

[jhmar]

Gravity has yet to be incorporated into the Standard Model.

The key to incorporation is to recall that Relativity is a classical theory and that gravity is carried by gravitons. Then by relating classical particle structure to Einstein's equation; classical particle physics is incorporated into classical relativity. The formula is:
E(c squared) = m = Linear Force /2radius
As 2r = wavelength substitute λ for 2r
Ec2 = m = Fl/λ

E = energy
c = speed of light
m = mass
Fl = linear force (2.8799296)
r = radius

Check this and you will find that it gives the Classical electron radius, and (given the wide experimental margin of error) the proton radius (3 elementary particles = 3Fl) and the neutron radius (5 elementary particles [3 quarks 1 electron and 1 neutrino] = 5Fl).

This is not to say that QT is wrong, it is simply an overly complicated way of finding part of the answer. The formula I have given matches the three radii found by experiment, the fractional waves found in TFQHE and the mass found by Einstein's equation.
I am waiting for my submission to be approved but whilst waiting I have made a major correction to one of the tables. The correct wave fraction and wavelength for the electron is 1/5, λ = 5.635882. The fractional wave sequence is 1 1/2 1/3 1/4 1/5 1/6 etc and all elementary particle radii are proportional to the electron radius.
I used the particle data from the 2004 PDG list to show that all particles found by experiment match the predicted data; no rejections, no averaging.
All particles are compaction's of a graviton.

[sweetser]

Hello jhmar:

Good luck in starting a thread. I won't comment on your proposal, as that belongs to a different thread. I will say what my proposal says about particles of the standard model.

In standard quantum field theory, a force can have like particle attract if and only if the particles that mediate the force are even spin. A force can have like particles repel if and only if the particles that mediate the force are odd spin. In the GEM unified field theory, the 4D wave equation contains 2 fields, the even spin 2 field for gravitons, and the odd spin 1 field for photons.

The standard standard model works for massless particles. The Higgs mechanism was a way to break the symmetry using a scalar field as a false vacuum. The problem with the Higgs is there is no connection to the graviton at the quantum level, so no apparent means of enforcing the equivalence principle. The symmetric rank 2 tensor that generates the gravitons will always have associated with it a scalar field by taking the trace of the graviton field. I propose the trace serves the same function as the Higgs, but there is no need for the false vacuum field.

Recall that in this thread, while we respect the success general relativity has had, the GEM model is necessarily in conflict with GR. The graviton in GR is predicted to be a transverse wave, while GEM predicts the longitudinal and scalar modes of emission are the only ways gravitational energy waves can be set through space.

doug

[jhmar]

doug

Thanks for your comments. As a crude definition I like to think that QT tells us what particles do, classical theory tells us what particles are.
You seem to deal only with one class of wave, but I think that force and anti-force produce two different waves; one is associated with vacuum force and the other with anti-vacuum force. Each particle has one vacuum wave that by torque actions winds up the anti-vacuum wave into an odd number of waves; the middle anti-vacuum wave lies at 90 degrees to the vacuum wave. This pattern is caused by the requirement that both wave patterns rotate around a central Zero Vacuum Point.
My concern with 0 charged particles is again to describe what they are, not to predict there existence, a function that QT already does quite well. QT does not explain why Charged and 0 charge particles behave differently, that's the work of classical theory.
Interpretation (the function of classical theory) should not contradict QT except where mathematical proof linked to experiment can be shown as in the case of the allocation of fractional charge to quarks, which quite simply is an assumption to far.
Time for me to stop or I'll be here all night,
regards
john

[sweetser]

Hello John:

Classical and quantum mechanics are both so precisely defined, they can start from exactly the same place. The GEM 4D wave equation, J_q - J_m = \square^2 A, without a single modification, can be used for either classical or quantum mechanics. What changes are the implementation details, which can get very confusing. In a big picture way, the classical approach will lead to expressions where the measurements are numbers. In the quantum approach, operators are the observables which leads to measurements that are averages.

What bothered Einstein was why causality for classical physics was different from causality in quantum mechanics. Classical physics is simple: if I punch someone in the nose, you can make a high speed film of the assault and order all the picture frames, one after the other. In quantum mechanics, if a photon punches an electron in the face, we must average all the possible ways a photon could contact an electron to get the right answer, confirmed to every digit of accuracy experimentalists can push the data.

The reason for the difference in my opinion has to do with doing 4D calculus correctly with quaternions. Defining a quaternion derivative the simple way by exactly copying the one in any calculus book causes a problem: dividing the differential element on the left is different from dividing on the right. I steal a play from L'Hospital who stole it from Bernoulli, and use a dual limit process, where the pesky 3-vector goes to zero first, then the commuting real part goes to zero. Technically it is called a directional derivative along the real axis. It works just like a real derivative because after the 3-vector goes to zero, it is a real derivative.

The important idea for mathematical physics is what happens if the order is now reversed, so that the real goes to zero, then the pesky 3-vector. Writing the differential on the left is different from the right, unless you take the norm of the vector. That is well defined, and always the same for the differential on the left or the right. So one absolutely must always work with norms of derivatives. That is what goes on in quantum mechanics: an amplitude can be calculated, but it is the averages that are observed.

The "differential element" sounds like an abstraction, but I think of it as a ratio of changes in space over changes in time (more technically, over changes in the interval). When the change in space is smaller than the change in time, information travels at less than the speed of light, and we have classical causality. When the change in space is larger than the change in time, information travels at less than the speed of light, so all we can measure is average values of what happens.

There is a way to visually understand this issue. Take 10 frames of someone getting punched in the nose. In the classical view, these can be ordered in time, and you get a movie out of the event. In the quantum view, one cannot make a movie. Instead all the frames of the movie can be superimposed, so you see all valid states of someone getting punched in the nose. This contains all the information about the system, the proverbial wave function. Making a measurement is the act of picking out a particular frame out of all the possible ones, the collapse of the wave function.


About the vacuum...
I consider this one of the sadder issues in physics, and no, this is not a dig at John, but at the larger professional community. The pros want the vacuum to do important things. Breaking the vacuum symmetry with the Higgs mechanism is suppose to give particles mass. The false vacuum energy of empty space is suppose to be cranking up the acceleration of the Universe. On the fringe of physics, zero point energy folks want to power everything and its brother with the vacuum. To all these people, my message is clear.

A vacuum is empty and can do absolutely nothing. Ever.

To the folks at CERN betting billions on the Large Hadron Collider to detect the Higgs, I will go on record to say they are going to fail. For the people working on a non-zero cosmological constant, I will go on record to say Einstein's greatest mistake will remain a mistake. Anyone chasing after zero point energy is on a fools errand.

The variation for any measurement, even zero energy is not zero, but that is all about the numbers we must use to make measurements in spacetime, namely quaternions, which are not a completely ordered set. Think of baseball for a moment. We all know they talk about the average number of hits for a baseball player: that statistic is important. For each of those players, there is also a deviation from average, and almost no one knows what those numbers are. It may well be that batters with low variations also have higher averages, but I don't know, it is not a statistic that is discussed much. It is important to feel the difference between an average - the thing that matters most - and the deviation from the average. In measuring energy, the deviation from the average cannot be made arbitrarily small because quaternions (three complex numbers) are being used. Details for that claim are made on quaternions.com, under the uncertainty principle (borrowed directly from a lecture on the impact of complex numbers on quantum mechanics).

My three take home messages are these:
1. Doing quaternion calculus in spacetime explains why classical physics uses a directional derivative along the real time line, while quantum mechanics must necessarily use normed derivatives for average measurements.
2. Doing group theory with quaternions explains why there are 4 known forces of Nature via their symmetries Diff(M)xU(1)xSU(2)xSU(3) which are needed to characterize any possible collection of events that appear in a volume of spacetime.
3. The simplest quaternion wave function is a way to unify gravity and EM. This claim can be tested two ways: by measuring light bending to the next level of accuracy, or by measuring the polarity of a gravitational wave.

Things have gelled nicely this year (point #2 being brand new). Have a good day.
doug

[CarlB]

This is not what I meant. There is 1 quaternion 4-potential, A, which I happen to write as A^{\mu} so when I calculate the scalar part of two quaternions, say A^{\mu} B^{\nu}, it makes sense how to use a metric g_{\mu \nu} to calculate the scalar part of that quaternion product.

Actually, this was what I thought you meant the first time I read it. It was only on the third or fourth read that I started thinking of it literally.

Since we are using the same A throughout, and since a quaternion commutes with itself, then for this product: A/|A| exp(A-A*) = exp(A-A*) A/|A|.

Of course A commutes with A. Does A also commute with A*? I think it does too, but that is from my intuition with SU(2). Yes, I'm sure it does.

A/|A| must point in the same direction as exp(A-A*), so it really only has one degree of freedom.

I am going to stop here, to see if we are on the same page as far as metrics, labels, SU(2) and U(1)xSU(2).

Yes, I am back on track.

Sorry for not replying sooner. I relied on PF to send me an email when you replied but it apparently didn't work.

[Pippo]

What do you think about the sentence that EM is just a perturbation of gravity, and the gravity is just a perturbation of the space ? Supposing it is true, can we say that gravity moves at light speed? that is space moves at light speed?

[sweetser]

Hello Carl:

No problem. I knew I was trying to mix an apple with an orange, and come up with a new drink, as it were. As you know the metric tensor's job is to decrease the rank of a tensor expression by two. I want the metric to be used to calculate the distance, but keep the rest of the quaternion expression in its place. I am certain math nerds will say that is not legal. I kind of would prefer to work with the "square root" of a metric, so that the vector part of the product could be g_0 t g_1 x + ... I'm certain the square root of a metric is not a new idea, but haven't read up on the topic. Here is a stunner: for the GEM metric solution, the eponential metric, if I were to work with square root of the metric, then the vector is invariant because g_0 g1 = g_0 g_2 = g_0 g_3 = 1. Special relativity preserves the scalar of a quaternion square product, gravity preserves the 3-vector of the same product. That's a simple invariance principle.

doug

[sweetser]

Hello Pippo:

In my proposal, the wave for gravity and EM both travel at the same speed, c. You can call it the speed of light, or the speed of gravity, and they are both the same. As a practice, I always try to talk about spacetime. I don't think it makes sense to talk about space or spacetime moving.

I am not clear about the statement on perturbations, so this is what I normally say.

The Universe has survived as long as it has by doing almost nothing to nothing. That is durable! A completely empty Universe would have the Minkowski metric rule the rulers. The problem is there is stuff in the Universe (very little, despite your own experience). So the question becomes how to do almost nothing in a mathematically formal way. Think of a slinky that is held up with one hand. Give it a slight nudge, and it will wobble. Watch closely, and it wobbles for quite some time. This is a simple harmonic oscillator. That's the equation I am using for gravity and light. The first path I used to get to the exponential metric used perturbation theory in a critical step (I'll skip the details, there was a lot of math).

doug

[CarlB]

Doug,

I have great sympathy for the way you're doing this. I think that tensors are icky because they make the assumption that there is only one sort of symmetry in an object. I'd much rather leave things in algebraic form.

Quantum physics is now done entirely by enforcing symmetry relations so tensors make a certain amount of sense, but in doing this, it becomes impossible to predict relations between different symmetries. Hence the neverending search for a unifying symmetry. What I'd prefer is a unifying symmetry that just happens to have the symmetries one needs, and this is exactly what you are doing.

Getting back to exp(A^u - A*^u) A^u/|A|. Can you write down what this if A happens to be infinitesimal? That is, for first order in the components of A, what is the above? I'd write down what I think it is, but since you're nearby, my tendency towards laziness is multiplied by my tendency towards bossiness and delegation, and probably squared by my tendency to not want to appear stupid in public (the small residual which has survived my being a physics crank), all these things conspire to induce me to ask you to compute the first order for the thing.

Oh, what the heck. We write

A = a_1 + a_j j + a_k k + a_l l

Then A-A^* = 2(a_jj+a_kk + a_ll)
and the exponential of this is
1 + 2(a_jj + a_kk + a_ll)

I'll leave the rest for you. Please feel free to change notation to something more readable.

Hmm. Sure looks like when we multiply this by A/|A| we're going to get back the whole algebra. In other words, we will get back four degrees of freedom, as advertised. And the commutation rules seem correct.

Carl

[sweetser]

Hello Carl:

I like this line:
What I'd prefer is a unifying symmetry that just happens to have the symmetries one needs, and this is exactly what you are doing.
This was a happy accident, honest!

Since you want me to comment on notation, I'll modify your A by including a basis for the first component, and shift to ijk which is more common. For the sake of consistency, I will make all the "amplitudes" small a_something, and all the basis vectors e_something, like so:
A=a_0 e_0 + a_i e_i + a_j e_j + a_k e_k
The Taylor's series to first order of the exponent is:
exp(A - A^*) = 1 e_0 + 2 (a_i e_i + a_j e_j + a_k e_k) + O((A-A^*)^2)
Now we are going to multiply this by A/|A|. We know this is going to be "easy" in the sense that we do not have to calculate the cross product, it will always be zero:
((a_j (2 a_k) - a_k (2 a_j)) e_j e_k = 0
It is the cross product that makes a quaternion non-Abelian, so if the cross product is necessarily zero, then this particular quaternion product is Abelian. Lazy man does not have to write out both to prove it.

Multiply out the normalized quaternion times the exponent:
\frac {A}{|A|} exp(A - A^*)
=(a_0 e_0 + a_i e_i + a_j e_j + a_k e_k)(1 e_0 + 2 (a_i e_i + a_j e_j + a_k e_k) + O((A-A*)^2))
=a_0 e_0^2 + 2 a_i^2 e_i^2 + 2 a_j^2 e_j^2 + 2 a_k^2 e_k^2, 4 a_0 a_i e_0 e_i, 4 a_0 a_j e_0 e_j, 4 a_0 a_k e_0 e_k)/|A|
In flat spacetime using Hamilton's rules, it looks like this is a standard quaternion. In spacetime where you choose to account for some effect using curvature, the basis vectors may no longer have a norm of 1. This might be the way for me to avoid using some of the tools of differential geometry (and I will keep that claim vague, because that's its state).

So for me, this shows that the product A/|A| exp(A-A^*) has the electroweak symmetry, U(1)xSU(2). Cool.

Now the road out to SU(3). I am not going to calculate q q'. We know that quaternion multiplication without 0 forms a group, so q q' will be in the same group. We also know that quaternion multiplication is associative, (A B) C = A (B C). A little bit of thought will let you see that the Euclidean product, q* q', is not associative. The different groups end up pointing in different directions, so (A B)^* C != A^* (B C). We can look at that more closely if you choose. The Euclidean product still has an identity, there is always an inverse, but you will need to be more careful with parentheses. The norm of A/|A| exp(A-A^*) is one, the norm of B/|B| exp(B-B^*) is one, so will the norm of A^* B = 1? Of course it will! The conjugate points A in a different direction, but that does not change the norm one bit since you square all the a_n's anyway before adding them together. Since A = A/|A| exp (A-A^*) has 4 degrees of freedom, A* B should have 8. Sounds like a way to represent SU(3) to me.

Have you looked at the animations? Those images build my confidence because I get to use a different part of my brain to "get it", as it were.

doug
http://www.theworld.com/~sweetser/quaternions/quantum/standard_model/standard_model.html

[CarlB]

Actually, I was hoping you'd remind me what the multiplication rules are for the e_\chi. You haven't, so now I need to figure them out for myself. Looking in wikipedia:
http://en.wikipedia.org/wiki/Quaternion
I see that I need:
e_i^2 = e_j^2 = e_k^2 = e_ie_je_k = -1
e_ie_j = +e_k and cyclic.
e_je_i = -e_k and cyclic.

In multiplying out A/|A| exp(A-A*) it becomes clear that we need to keep stuff to second order in A. Hmmm.

Let's write A = b + cB where b and c are real numbers, and B is a normalized quaternion. That is B = xe_i + ye_j + ze_k, where (x,y,z) is a unit vector. And I'm ignoring e_0. Then BB = -1.

Compute (b+B)/|b+B| exp(2cB)
= (b+cB)/|b+cB|(1 + 2cB + 4ccBB/2 + 8cccBBB/6 + ...)
= (b+cB)/|b+cB|(1/0! + (2c)B/1! - (2c)^2/2! - (2c)^3B/3! + (2c)^4/4! ...)
= (b+cB)/|b+cB|(cos(2c) + sin(2c)B)
= ( (b cos(2c) -c sin(2c)) + (b sin(2c) +c cos(2c))B )/|b+cB|.

Now I could easily make a mistake computing |b+cB|, perhaps you will take it from here.

Of course what I'm doing here is thinking in Clifford algebra terms. That is, write everything in terms of scalars (i.e. b) and vector (i.e. cB) terms, and then keep stuff grouped according to blade. This sometimes works. That is, it sometimes works in explaining this to somebody in Clifford algebra terms.

(I couldn't stand your notation where a_i is a real number but e_i is a basis vector. This is too similar for my limitations.)

Also, are you planning on going to any conferences this season? I thought I would pretty much sit the year out, perhaps going to the gravity conference in Australia.

[jhmar]

Classical and quantum mechanics are both so precisely defined, they can start from exactly the same place.

But QT predicts, Classical physics does not predict. My paper shows how classical theory can predict and in doing so, it shows that QT, while correct; predicts only a fraction of all possible particles.

In the quantum approach, operators are the observables which leads to measurements that are averages.

This process of averaging is carried over into experimental work (see PDG 2004 tables). I show that the rejection of some experiments and the averaging of the remainder are incorrect. All the experimental results are compactions of a single elementary particle; probably the graviton.

What bothered Einstein was why causality for classical physics was different from causality in quantum mechanics.

The difference is caused by Einstein’s inclusion of movement. In order to understand what particles are and why they have there particular quantities it is necessary to examine structure not movement.

When the change in space is smaller than the change in time, information travels at less than the speed of light, and we have classical causality. When the change in space is larger than the change in time, information travels at less than the speed of light, so all we can measure is average values of what happens.

Knowledge travels at the speed of the knowledge carrying particle, usually photons. This varies according to the density of the particles that the photons are passing through (gravitons). Each gravity field is its own time zone and this is the base of all those theories about time slowing down near black holes. Again there is no need for averaging and no disagreement with QT; classical theory is simpler.

About the vacuum...
A vacuum is empty and can do absolutely nothing. Ever.

Newton’s graph of a gravity field without a central mass shows that absolute vacuum does not exist; the (vacuum) zero point has zero dimensions. It is where absolute vacuum would be if it existed.
It is vacuum force emanating from the ZP that controls the structure of matter creating particles that are vacuum fields with mass (vacuum force carrier).

To the folks at CERN betting billions on the Large Hadron Collider to detect the Higgs, I will go on record to say they are going to fail.

I would agree that there is no Higgs particle, but there are certain similarities between a Higgs field and vacuum force carrier that lead me to suggest that they are the same entity.

I have used the maximum attachment allowance to send you some tables from my proposal, only a fraction of table 4 can be included, but this should be sufficient to show the concept.
john

[sweetser]

Hello Carl:

Of course what I'm doing here is thinking in Clifford algebra terms.
I'll get back to you on this point next week. I have never studied Clifford algebra, oops. I will read up on them, and see if I can say anything sensible.

As far as conferences, I will be wasting some time and money at the April APS meeting. I'm in session "E12. Alternative Theories of Gravity". One guy will talk about plants and gravity. Two people have gamed the system so they can talk for two time slots. At least it is in Florida.

At the end of May, there should be the Eastern Gravity Meeting in NYC at Columbia. They still haven't made an official announcement, but it is in the works.

I may go to SIGGRAPH to show off the quaternion animations. Those folks use quaternions, so I might find an interested audience. That meeting conflicts with the one down under.

doug

[sweetser]

Hello John:

There are quite a few statements in your note I find fall into that class "not even wrong." Sorry about that, but I can be precise.

>But QT predicts, Classical physics does not predict.

Newtonian mechanics, the Maxwell equations, even special and general relativity are considered to be classical physics, for the reason I gave (experimentalist can measure a number, the measurements are not quantized). The classical theories are only theories because they make predictions.

> All the experimental results are compactions of a single elementary particle; probably the graviton.

Different particles have different intrinsic spins. The graviton is predicted to have a spin of 2, while electrons have spin 1/2. One of the biggest implications of the difference in spin to characterize particles is the Pauli exclusion principle which applies to thos with half integral spin, but not integral spin. I would expect that should you make clear that there is a single elementary particle, that your proposal to the Independent Research Forum would be rejected on that ground alone. In the GEM proposal, the unified field strength tensor has a spin 1 particle, the photon, to do the work of EM, and a spin 2 graviton to do the work of gravity. Both are massless and travel at the speed c. The also differ because a photon is a transverse wave, while the graviton is a scalar or longitudinal mode of emission.

>The difference is caused by Einstein’s inclusion of movement. In order to understand what particles are and why they have there particular quantities it is necessary to examine structure not movement.

Modern field theory focuses on groups. The standard model was developed in the 1970's, so Einstein's post-mortem opinions don't matter on the subject at hand. Carl and I are working out the details of the groups U(1)xSU(2)xSU(3) which is at the heart of the standard model.

>Knowledge travels at the speed of the knowledge carrying particle, usually photons. This varies according to the density of the particles that the photons are passing through (gravitons). Each gravity field is its own time zone and this is the base of all those theories about time slowing down near black holes. Again there is no need for averaging and no disagreement with QT; classical theory is simpler.

I keep things simple, and work with the vacuum, or isolated currents in a vacuum. All my work so far has been for low energy densities. The math for high energy densities will be different for the GEM proposal, but I do not have the details.

>Newton’s graph of a gravity field without a central mass shows that absolute vacuum does not exist; the (vacuum) zero point has zero dimensions. It is where absolute vacuum would be if it existed. It is vacuum force emanating from the ZP that controls the structure of matter creating particles that are vacuum fields with mass (vacuum force carrier).

My proposal is not Newton's, so this point is not relevant. A vacuum is both well defined and observable: it is a place with an average energy density of zero. There is not such place, but most of the Universe is an excellent approximation of a vacuum, with a hydrogen hanging out in a cubic meter. There are no vacuum fields doing anything because they have no energy to do anything. The logic is simple. The deviation of the average is not zero, and that does not depend on the experimenter, but on a basic property of quantum mechanics, namely that complex numbers are needed.

>I have used the maximum attachment allowance to send you some tables from my proposal, only a fraction of table 4 can be included, but this should be sufficient to show the concept.

I will refrain from commenting further until/if your work is accepted here. As my comments above should indicate, I don't think your work does fit the criteria in the few bits I have seen. I appreciate your sincerity, but I cannot invest more time in work where I see fundamental errors are made. We will have to respectfully disagree with each other probably on most points I've tried to make.

doug

[CarlB]

Doug,

What is |A^2=|a_0 e_0 + a_i e_i + a_j e_j + a_k e_k|^2 ?

I'm guessing it has to be a_0^2 \pm (a_i^2 + a_j^2 +a_k^2), but I'm not sure which sign to take. Am I right in supposing that it is a real number? I like to work with the + sign, but I'm guessing you're defining it as the - sign.

I need this to continue the analysis of the U(1)xSU(2) thingy. I'm sure it's obvious, thanks for bothering.

[sweetser]

Hello Carl:

The norm of a quaternion is:
|A|^2=norm(A)=scalar(A^* A)=a_0^2+a_i^2+a_j^2+a_k^2
The norm can equal zero if and only if A=0, otherwise it is positive definite. The rule for the inverse of real, complex, and quaternion numbers can all be written exactly the same way:
A^{-1}= A^*/|A|^2
For the real numbers, the conjugate operation does nothing, but also don't hurt anything. For complex numbers, the imaginary part will flip a sign, then get hit by the norm. Same for quaternions.

So the answer is definitely the + sign.

doug

note added for fun: I have an animation of the norms of a bunch of quaternions. They all sit at the same place in 3D space: 0, 0, 0. What changes is how far in the future they are from now, t=0.

[CarlB]

Thanks. Continuing the calculation,

Compute (b+B)/|b+B| exp(2cB)
= (b+cB)/|b+cB|(1 + 2cB + 4ccBB/2 + 8cccBBB/6 + ...)
= (b+cB)/|b+cB|(1/0! + (2c)B/1! - (2c)^2/2! - (2c)^3B/3! + (2c)^4/4! ...)
= (b+cB)/|b+cB|(cos(2c) + sin(2c)B)
= ( (b cos(2c) - c sin(2c)) + (b sin(2c) + c cos(2c))B )/|b+cB|

= ( (b cos(2c) - c sin(2c)) + (b sin(2c) + c cos(2c))B )/sqrt(bb +cc).


In the above, b is the amplitude of the temporal part of the quaternion, c is the amplitude of the spatial part, and B is the direction of the spatial part.

Okay, the spatial SU(2) information is in B and that comes through unchanged. The vector (b,c) codes the relative strength of the time and space parts of the original quaternion. It's divided by its length so it becomes a unit vector in two dimensions. Then it is rotated by the absolute value of the value of 2c:

\frac{1}{|A|}\left(\begin{array}{c}b\\c\end{array} \right) ->
\left(\begin{array}{cc}\cos(2c)&-\sin(2c)\\\sin(2c)&\cos(2c)\end{array}
\right)\;\frac{1}{|A|}\left(\begin{array}{c}b\\c\e nd{array}\right)

[sweetser]

Cool! I'll see if I can confirm this numerially.

[jhmar]

My proposal is not Newton's, so this point is not relevant. A vacuum is both well defined and observable: it is a place with an average energy density of zero. There is not such place, but most of the Universe is an excellent approximation of a vacuum, with a hydrogen hanging out in a cubic meter. There are no vacuum fields doing anything because they have no energy to do anything. The logic is simple. The deviation of the average is not zero, and that does not depend on the experimenter, but on a basic property of quantum mechanics, namely that complex numbers are needed.

Einstein improved on Newton's work, he did not prove or claim Newton was wrong. (NASA still uses Newton's formula for gravity). It a simple calculation to show that Newton's original formula gives the force arising from the combination of two vacuum fields obeying the Standard Inverse Square Law that applies to vacuum fields with a central point.
Einstein's and subsequent adjustments to Newton's law are necessary to take into account external factors they do not alter the cause of the original structure.
Are you now saying that Newton and Einstein are wrong?

[sweetser]

Hello John:

Newton had a huge volume of work. Newton's theory of gravity is wrong and as an approximate theory is useful. Newton's theory of gravity applies to masses which might be surrounded by a vacuum, but masses have a positive energy density, so there is no longer a vacuum if there is a mass.

General relativity is a rank 2 field theory. My GEM proposal is a rank 1 field theory. If my proposal is correct, then general relativity is wrong and as an approximate theory is useful. My proposal will probably be more useful because the non-interacting field equations are linear.

doug

[jhmar]

but masses have a positive energy density, so there is no longer a vacuum if there is a mass.

The flaw in this reply is your assumption that we know what mass is, you are assuming that mass has a positive energy density; I am assuming that the vacuum force carrier is the something that is the cause of density and use bubble chamber experiments to justify that assumption. But that is a debate that will have to wait a decision on my submission. For the present I am content to understand the points on which we disagree; which is why I joined in the debate. Unless you have anything to add I will leave you and CarlB to enjoy your mathematics, thanks for taking the trouble to persevere with my replies.
regards
jhmar

[sweetser]

Hello:

I submitted a paper to a contest, the 2007 Essay on Gravitation sponsored by the Gravity Research Foundation. It was set up in the middle of the last century by the businessman who founded Babson College (I'll let you guess his last name). He had all kinds of hopes for anti-gravity devices, and wanted to fund fun/fringe work.

The first year of the contest did not go well: it sounded too wackie. His friend George Rideout suggested that the description of the contest should be toned down and award "provocative" works on gravity. Now this contest is part of the establishment. I recognized 3 names among the winners: Ellis, Smoot, and Wald. There were 33 honorable mentions, so this is a real contest. There are five cash awards, going from $500 to $5000.

http://www.gravityresearchfoundation.org/competition.html

So far I have avoided the full-fledged peer-review journal because I need to work with someone who reads and understands those journals first. This contest sounded a little looser, which is more my style.

The winners will be announced on May 15. Sometime after that, I'll post it here. For now, I will leave the abstract.

Geometry + 4-potential = Unified Field Theory
(same as for my APS talk in mid April)

Gravity is the study of geometry. Light is the study of potentials. A
unified field theory would have to show how geometry and potentials could
share the work of describing gravity and light. There is a long list of
criteria that must be satisfied to have a reasonable hypothesis, from
recreating the Maxwell equations, to passing the classical tests of
gravity, to demonstrating consistency with the equivalence principle, and
working well with quantum mechanics. This essay works through many of the
common objections.

I stayed up to 3 AM trying to polish it, and then skipped out of work because I am too old to bounce back that quickly. Great day out here in Massachusetts. I like the essay. The same guy, George Rideout, is still running it, as he sent me an enthusiastic email reply.

doug

[sweetser]

Hello:

A week after the national April APS meeting, I'll be up in Orono Maine early on a Saturday morning showing pictures of the standard model for a regional APS meeting. This work was a direct result of writing software to animate quaternions. Here is the abstract I submitted.

(abstract)
Software is used to visualize unit quaternions SU(2) as a 3D
animation. Random quaternions are run through a quaternion exponential
function. The results are sorted by time and placed in a frame of the
animation corresponding to their 3D coordinates. The resulting
animation shows a sphere with an apparent distain for the past. The
visual representation of electro-weak symmetry looks like a complete
sphere with a bias for the past. The animation for U(1)xSU(2)xSU(3) is
the smoothest image of an expanding/contracting sphere that could be
created. Any pattern of events can be represented by this
group. Spheres of slightly different sizes nearby on the manifold
would belong to the group Diff(M) which is at the heart of gravity.
(/abstract)

If you want to look at pictures, click here.

http://www.theworld.com/~sweetser/quaternions/quantum/standard_model/standard_model.html

Later,
doug

[sweetser]

Hello:

I will be jetting to Jacksonville in a week to give a talk, "Geometry + 4-potentials = Unified Field Theory". I bought a Mac for Keynote, a presentation software program I had heard good reviews. They were true. I was able to create a simple presentation with nice transitions. I don't expect many people in the room since this is an "alternative" gravity theory session other than the presenters and the moderator and at most two stragglers.

Keynote makes it easy to export an HTML page, and I put that up on quaternions.com. It would be much better if audio could be included with the slides. A little investigation lead to a program called "profcast". That program will record the audio, synced to the clicking of the slides. It does not get the transition animations, but that is minor. I captured the slides and the audio at a good quality, and put it up on youtube:

http://www.youtube.com/watch?v=p7HyXBuY2dM

This is 17 minutes long, and I only will have 12 on next Saturday, so I'll have to be more brief. It was great to record, compress, and upload in a few hours.

As always, I'd appreciate any comments on the content.
doug

[rewebster]

Nice presentation--One comment:

at 15:08 or so, you call it a 'hard' sell--I might say something like "it becomes a potentially interesting sell" or something else besides 'hard'----'hard' puts up a barrier for the idea

[sweetser]

Good suggestion. I'll change the phrase in the live presentation, see if I can swap it out of the video.

[sweetser]

Hello:

At the APS meeting, I talked to a guy (Oleksandr Pavlyk, don't ask me to pronounce it) from Mathematica about my efforts to visualize the standard model. He pointed out a clear error. The group U(1)xSU(2)xSU(3) has the the Lie algebras u(1), su(2), and su(3), which have degrees of freedom 1, 3, and 8 respectively. Add that up, and the standard model has 12 degrees of freedom in its Lie algebra. What I worked with was two quaternion, which has 8 degrees of freedom.

The fellow also said that SU(3) has U(1) and SU(2) as subgroups, something I was unaware of. Now that I look back on those animations, they are easy enough to spot. At the algebra level, it is clear they belong as subgroups, since they are used to generate SU(3).

I have heard that people who are very sophisticated with the standard model put some sort of caveats on the usual U(1)xSU(2)xSU(3) description. I do not recall what those caveats were.

I have some confidence that the animation I created is SU(3): the norm is one, and it has 8 degrees of freedom. I don't know how to write in group theory the algebra I have done (the conjugate of U(1)xSU(2) times a different element of U(1)xSU(2)). I could just toss in another quaternion, but that strikes me as bogus. These two quaternions smoothly cover any possible event in spacetime, bar none.

doug

[sweetser]

Hello:

YouTube is now hosting new animations of the symmetries of the standard model. Everyone thinks they know what U(1) looks like (a circle in the complex plane). Even that looks different than I expected once animated. As for SU(2) and SU(3), all I've seen is algebra. I have each animation separately, and a 2' 40" collection of them all.

The groups of the standard model and gravity (the 5 videos below):
http://www.youtube.com/watch?v=ExNPiMcVXww
The group U(1) http://www.youtube.com/watch?v=KZeULLKHE7w
The group SU(2) http://www.youtube.com/watch?v=OMnNyyZruuE
The group U(1)xSU(2) http://www.youtube.com/watch?v=Jbdj3Xd_nmI
The group SU(3) http://www.youtube.com/watch?v=8T_aNL8LvCs
The group Diff(M)xSU(3) http://www.youtube.com/watch?v=pYiEV8yEZYA

As always, comments are welcome. Enjoy, I think this makes the standard model fun to think about.

doug

[rewebster]

how did you come up with the angled position of the U(1) in the xyz? is that specific?

[sweetser]

Hello:

The plane for U(1) was chosen at random. The software is setup so a specific quaternion can be used to start forming the group, and that one quaternion would determine the angle.

doug

[rewebster]

Well, it seemed that the others 'could' be of any orientation due to their symmetry and the U(1) stood out to be chosen or selected to be in that exact orientation for some reason.

[sweetser]

Representing U(1) - by itself - is arbitrary with what is effectively a 3D imaginary unit. Complex numbers in contrast have a 1D imaginary unit. When calculating U(1)xSU(2), the generators of U(1) and SU(2) need to point in exactly the same direction. That direction can be arbitrary, but must be shared if the U(1) is to commute with SU(2) as it must to be a faithful representation of an Abelian group.

doug

[rewebster]

Representing U(1) - by itself - is arbitrary with what is effectively a 3D imaginary unit. Complex numbers in contrast have a 1D imaginary unit. When calculating U(1)xSU(2), the generators of U(1) and SU(2) need to point in exactly the same direction. That direction can be arbitrary, but must be shared if the U(1) is to commute with SU(2) as it must to be a faithful representation of an Abelian group.

doug

If the U(1) is arbitrary (random?) by itself, would showing it to be more random be more correct? --The one thing, and I don't know how important it would be, is that IF at a arbitrary/random setting it COULD end up laying in the y-z plane and your 'time transversing' line along the x axis (x-y plane) would have quite a bit different 'read out'---what significance would that be/have?

[sweetser]

Hello:

Sorry for the delay in replying. I am not getting the "something has been posted" emails.

The way I view physics, it is a game of describing events in spacetime. Spacetime can be viewed either as a 4D real manifold, or a 1D quaternion manifold. If you choose to work in the quaternion manifold, then choose a quaternion with a norm of 1, and q^n will form the group U(1) because it depends only on one quaternion and is Abelian because quaternions commute with themselves.

In physics, U(1) is where the E and B fields live, the transverse mode of emission of light. One can make the circle appear in only 1 complex plane, say the tz plane. I should be free to do other things in the tx and ty planes. I'll have to think about how to implement that. From my experience, it is the time + space planes that matter, not space x space. Until I create the animation, I don't know. Good question.

doug

[CarlB]

Doug,

I've been spending a lot more time messing around with gravity stuff than I usually do, and have come to appreciate something I think you said about your theory -- that it is a vector based theory.

The reason this jars my memory is that my favorite version of GR also seems to be expressed as a vector field, a velocity vector field. This is the gauge gravity. Their stuff matches GR exactly (provided you avoid the weird topological stuff inside of where you can't observe anyway). For a black hole, the gauge gravity stuff says that the natural coordinate system is Painleve.

There is a wonderful paper on the "river models of black holes" that explains how Painleve coordinates describe a Schwarzschild black hole as a velocity vector field:
http://www.arxiv.org/abs/gr-qc/0411060

The velocity vector field is
\vec{v}(r) = -\frac{\vec{r}\sqrt{2GM/r}}{r}

Can you comment on how your theory works as a vector theory? Right now I'm interested in the field produced by a single mass point. I'm still interested in simulating it, and I think I know enough now that I might be able to work out the equations of motion by myself. But I'd like you to check my work.

Carl

[MeJennifer]

There is a wonderful paper on the "river models of black holes" that explains how Painleve coordinates describe a Schwarzschild black hole as a velocity vector field:
http://www.arxiv.org/abs/gr-qc/0411060

The Painlevé chart is an excellent description of the Schwarzschild metric!

Flat background, the gravitational force shows up as Galilean not Lorentzian, using Newton's escape velocity. It models a comoving observer of the gravitational pull from infinity.

Also check the Doran chart that models the Kerr metric (a rotating point mass). Again a flat background!

By the way, it seems to me that since the Painlevé and the Doran chart chart resp. a static and stationary spacetime, there is no reason we could not define the orbital equations on a Euclidean 5 space. That should simplify the orbital equations since we should be able to express the effect of the gravitational field completely in terms of a non relativistic translation and a time invariant deformation of the Minkowski spacetime. No?

[rewebster]

Hello:

Sorry for the delay in replying. I am not getting the "something has been posted" emails.

The way I view physics, it is a game of describing events in spacetime. Spacetime can be viewed either as a 4D real manifold, or a 1D quaternion manifold. If you choose to work in the quaternion manifold, then choose a quaternion with a norm of 1, and q^n will form the group U(1) because it depends only on one quaternion and is Abelian because quaternions commute with themselves.

In physics, U(1) is where the E and B fields live, the transverse mode of emission of light. One can make the circle appear in only 1 complex plane, say the tz plane. I should be free to do other things in the tx and ty planes. I'll have to think about how to implement that. From my experience, it is the time + space planes that matter, not space x space. Until I create the animation, I don't know. Good question.

doug

The way I see your animation/diagram is/could be that the 'circle' is gimbled (moving at a high rate) on all three axes with time transversing on all three planes at the same time ('-' -> '+') showing/plotting for least separation of the intersection(s) along any of the axes (x, y and/or z--or, to be able to show it in animation, a combination of any two (x,y,z) tranversing across at one of any given/every possible random vector). --that should still show the two points separating and rejoining in the double (overlapping '+' and '-') harmonic/sine (?) 1/2 wave looking pattern. Would that work?

(writing out, now, may appear to be easier than the diagram/animation--hmmm?)

[sweetser]

Hello Carl and MeJennifer:

I have trouble getting excited about any work done on black holes by anyone, good, bad, or ugly.

In the first derivation I did to get to the exponential metric, I struggled to find the right way, no steps skipped. At one point, I thought I had it. Alas, Mathematica did not agree. A bit of retooling, and all the parts fell in place.

At one specific place, I have to make an assumption: the perturbation of the time term was trivial compared to the one for space. That apparently is what needs to be done to generate the exponential metric is consistent with weak field tests.

When a mass is in a very small volume, the assumptions used to derive the exponential metric are no longer valid. That would drastically change the kind of equations that govern motion. What I had done was take a perturbation from a 1/R 4-potential solution, and include a wee bit of a time variation. What if the I look for the non-perturbation solution? I haven't explored it more is because I would get lost. This is way different. There were two different warning signs. First is that the units for the expressions involve had a G (like Newton's law) and a c (like metric solutions in GR) and an h (like anything to do with quantum mechanics). The non-perturbation theory has the units of a quantum gravity theory. The second bigger issue is that the non-perturbation solution is a 1/R^2 4-potential, which results in a 1/R^3 force law.

I am posting here in "Independent Research" forum because I am walking the crank line. Say you have a 1/R^3 force law, and doors close immediately. I would not get the chance to say that the gravity we know - Newton's law that respects special relativity - is a 1/R^2 force law, it is only the quantum gravity theory that the force becomes a 1/R^3. There I things I know because I've done the math with paper and pencil that I don't like to talk about. Selling the exponential metric has been trying enough. A quantum gravity solution that implies that no work done on black holes is correct - there is no chance that will be accepted until derived independently by others.

There turns out to be a few great ironies at work here. Folks who take a quick glance find the 1/R^2 potential with 1/R^3 force, and thus dismiss this approach. The do so without working out the units, since they are professionals using natural units where G = c = h = 1. It was really amazing to see those three amigos in one expression!

doug

[sweetser]

The way I see your animation/diagram is/could be that the 'circle' is gimbled (moving at a high rate) on all three axes with time transversing on all three planes at the same time ('-' -> '+') showing/plotting for least separation of the intersection(s) along any of the axes (x, y and/or z--or, to be able to show it in animation, a combination of any two (x,y,z) tranversing across at one of any given/every possible random vector). --that should still show the two points separating and rejoining in the double (overlapping '+' and '-') harmonic/sine (?) 1/2 wave looking pattern. Would that work?

(writing out, now, may appear to be easier than the diagram/animation--hmmm?)

Maybe its too late for me, but I did not understand what your wrote. I see nothing "moving at a high rate". I happened to scale the animation to fill 10 seconds. I could also scale it to fit into a time frame of 10 picoseconds, or a thousand years. Nothing is marked. It is quite the abstraction.

The reason time is in the 3 complex planes is because time is the real axis that is shared by the three imaginary unit vectors (I happen to use i, j, k for convenience, but r, theta, phi or some other combination of spatial unit vectors are possible).

I have see what sine and cosine look like. It was not anything like I expected. I think there is one image up on quaternions.sf.net, but I need to add a few more to the collection.

doug

[sweetser]

Can you comment on how your theory works as a vector theory?
This line reminded me of one of the paths to the exponential metric. There is a Lorentz 4-force in my proposal:
F^{\mu}=-Jm_{\nu}(\partial^{\mu} A^{\nu}+\partial^{\nu}A^{\mu})
I drop in the weak gravity, electrical neutral potential into this, and solve for the 4-velocity. Here is the 4-velocity, weak gravity solution to the 4-force equation:
(\frac{d t}{d \tau}, \frac{d R}{d \tau}) = (exp(-G M/c^2 R), exp(G M/c^2 R))
Sorry, but I do think those \tau's are necessary.

doug

[CarlB]

I don't mean to imply that I think that current representations of what happens in black holes are correct. I always thought the "event horizon" of the Schwarzschild metric was silly, and that's why I wrote the Painleve simulator, to show that one could redo GR and eliminate that stuff. But more than that, even though the Painleve coordinates are an improvement, I doubt that they are correct.

I'm doing black hole simulations for pretty much the same reason Kepler would have done epicycle simulations.

There turns out to be a few great ironies at work here. Folks who take a quick glance find the 1/R^2 potential with 1/R^3 force, and thus dismiss this approach.

I don't see this as a problem at all. Heck, I don't even see what your point is. You do get the apprximate 1/r potential with the exponential thing don't you? This is approximately valid in the far field isn't it? Of course it is, otherwise you'd never have invested this much time in it.

Any gravitation theory should give equations of motion (and I'm interested in yours even if they are valid only for large r), and those equations of motion can always be written as a power series in r. Unless the theory is straight Newtonian gravity, there has to be forces other than 1/r^2.

In fact, when I write the equations of motion of the Painleve metric as a power series in r, it has some pretty crazy terms. Letting the particle velocity \dot{x},\dot{y} be of order 1 (i.e. for simulating light rays), the orders of the forces are:

\begin{array}{rcl|r}\ddot{x} &=&-\sqrt{2}\dot{x}(\dot{x}^2+\dot{y}^2)/r^{1.5}&1.5\\&&+1.5\sqrt{2}\dot{x}(x\dot{x}+y\dot{y})^2/r^{3.5}&1.5\\ \hline&&-x/r^3&2.0\\&&+3x(x\dot{x}+y\dot{y})^2/r^5&2.0\\&&-2\dot{y}(x\dot{y}-y\dot{x})/r^3&2.0\\ \hline&&+3\sqrt{2}\dot{x}/r^{2.5}&2.5\\&&+2\sqrt{2}y(x\dot{y}-y\dot{x})/r^{4.5}\;\;\;\;\;&2.5\\ \hline&&+2x/r^4&3.0\\ \hline\end{array}

As far as diverging from Newton in the far field, one only worries about terms that have powers of 2 or lower, and that are not multiplied by the particle velocity. (If they are multiplied by particle velocity, then they go to zero as particle velocity is small compared to speed of light.) In the above, you will note that there are lots of such terms, but they all have particle velocity involved.

To look at forces for small velocities, ignore all the terms multiplied by velocity. What's left is:

\begin{array}{rcl|r}\ddot{x}
&=&-x/r^3&2.0\\ \hline
&&+2x/r^4&3.0\end{array}

In other words, in Painleve coordinates there is also a 1/r^2 potential leading to a 1/r^3 force (i.e. the +2x/r^4). But the Newtonian contribution dominates at large r. Schwarzschild is similar, but it only looks nice when you write it in terms of powers of r and (r-2). And in any case, the actual orbits for Schwarzschild and Painleve are identical. All that differs is a redefinition of time, depending on the radius.

By the way, this talk has influenced me to go and make the calculation for the deflection of light from my equations of motion. I'll do it over on my "independent research" thread.

I'm really looking forward to your equations of motion for the exponential solution. It should be possible to see the various tests approximately achieved. For example, the ratio of the bending of light in Newton's and Einstein's theories is only 2x in the limit of small bends. This is something that should be obvious but (stupidly) I did not realize it until I simulated light bending and found that Einstein's light bends far more than twice Newton's prediction close to a black hole. It amazes me that this surprised me.

The simulation is already good enough to pick off small far field effects. And I'm soon going to implement R-K numerical methods (4th order) and expect it to get a lot more accurate soon.

Carl

[sweetser]

Hello Carl:

At long last, I may be able to deliver the equations of motion for the GEM proposal for a spherically symmetric, non-spinning, electrically neutral mass. All that needs to be done is to rearrange the 4-velocity in my earlier post.

Here is that 4-velocity again:
(\frac{d t}{d \tau}, \frac{d R}{d \tau}) = (Exp(-G M/c^2 R, Exp(G M/c^2 R)
Let's eliminate the pesky \tau. Multiply both sides by the inverse of \frac{d t}{d \tau} which is \frac{d \tau}{d t} or equivalently, Exp(G M/c^2 R)
(1, \frac{d R}{d t}) = (1, Exp(2 G M/c^2 R)
Looks like an elegant answer to me. Does this satisfy your request?

doug

[rewebster]

If I'm understanding your animation of the U(1), it's (yours is) an arbitrary (random) representation as a circle in x,y,z---doesn't the representative circle go through all possibilities 'gimbaling' through the x,y,z co-ordinates through time/conditions and describes a hollow sphere in x,y,z --but as a circle at any given instant/circumstance?---maybe I'm understanding your animation wrong.

[sweetser]

Hello:

There is no circle in x, y, z. There are 3 ellipses, one in the t-x plane, one in the t-y plane, and a third in the t-z plane. If there is a circle in one of these three complex planes, then there are lines in the other two.

In the quaternion animation, the line is straight through space. It is true that the line could point in any direction in 3D space.

doug

[rewebster]

yes--excuse me,--ellipse(s)

I had stopped the play and didn't get the audio

I guess what I have been looking at is if the animation for U(1) is a fair and/or generalized representation, IF the position of the ellipse(s) is/can be random, or if there was a way to show that it COULD be somehow represented AS arbitrary/random and still maintain an accurate visual depiction through all other possibilities.

[sweetser]

I am quite confident the animation is fair because the numbers that are fed in using the rules that generate U(1). Pick out any event, form the product with another event. The result will be another element of U(1). Now take those first two events, but multiply them in reverse. You get the same result because these quaternions commute. The generator of all these quaternions is one quaternion. That is the way the Lie algebra u(1) works (I used a small u).

There is freedom to point U(1) along any unit vector in 3D. The unit vectors do not have to be Cartesian, just part of a set of three unit vectors that span the 3D space. One could represent the group U(1) along a curved line with the choice of spherical coordinates.

doug

[rewebster]

There is freedom to point U(1) along any unit vector in 3D. The unit vectors do not have to be Cartesian, just part of a set of three unit vectors that span the 3D space. One could represent the group U(1) along a curved line with the choice of spherical coordinates.

doug

This is what I've been alluding to, or trying to get to. --and after listening (this time) to the audio, there's not this variable mentioned--as a open variable/freedom/arbitrary choice. Is there any way to animate this as a 'freedom' and still maintain the 'point' model that is the result of the time line? It would seem a little more 'correct' as a more representational model or mentioned that this freedom is there.

[sweetser]

This is what I've been alluding to, or trying to get to. --and after listening (this time) to the audio, there's not this variable mentioned--as a open variable/freedom/arbitrary choice. Is there any way to animate this as a 'freedom' and still maintain the 'point' model that is the result of the time line? It would seem a little more 'correct' as a more representational model or mentioned that this freedom is there.
I've only got 30 seconds to yak in that audio, so it is not surprising I don't go into all the issues. You were right to feel like U(1) does not fill up a volume of spacetime. Yet it is only one of the four known forces of nature. Have you listened to the YouTube video on the standard model? U(1) is doing the work of EM. SU(2) does the work of the weak force, and fills up much more space. U(1)xSU(2) goes everywhere, but not evenly. The group SU(3) could be covered by all the possible angles in space of the group U(1). This was already known because U(1) is a subgroup of SU(3), so play with enough U(1)'s and SU(3) is done.

[CarlB]

(1, \frac{d R}{d t}) = (1, Exp(2 G M/c^2 R)
Looks like an elegant answer to me.

Eventually I need a 2nd order differential equation like:

\frac{d^2R}{dt^2} = f(R, dR/dt).

For example, the Newtonian equations of motion are:

\frac{d^2 R}{dt^2} = -\frac{2RMG}{|R|^3}

so you should probably approach this for large R and small velocity dR/dt (assuming you don't have a MOND effect).

To get to this from the line element I will have to make some assumption about the orbits. In GR, the assumption is that the orbits are geodesics of the line element. I can make that calculation, (by calculus of variations) though it will take some effort.

What I'm saying is that if your orbits follow the geodesics of the line element that I think you are using, i.e.:

d\tau^2 = \exp(-GM/c^2R)dt^2 - \exp(+GM/c^2R) (dx^2+dy^2+dz^2)

then I can eventually figure it out. This is a simpler line element than GR gets, so it should be simpler to get the orbits out of it. Or did I leave out a factor of two and get some signs wrong in the above?

[rewebster]

Have you listened to the YouTube video on the standard model?


sorry---I couldn't get passed the one of you with your foot behind your head

[sweetser]

Hello Carl:

Small correction to the metric you wrote: there needs to be factors of 2 in both the exponents.

I just got a useful book, "Mathematica for theoretical physics: Electrodynamics, quantum mechanics, general relativity, and fractals". They showed how to get the equations of motion for the Schwarzschild metric. Being a good biologist by training, I was able to clone the answer for GEM:


\frac{4 G M R'^2 e^{\frac{4 G M}{c^2 R}}}{c^2 R^2}-\frac{4 G M t'^2 e^{-\frac{4 GM}{c^2 R}}}{R^2}+R \phi '^2-2 R'' e^{\frac{4 GM}{c^2 R}} ==0


\frac{4 GM R' t'}{R^2}+c^2 t''==0


-2 R R' \phi ' - R^2 \phi ''==0

There is a 1-to-1 correspondence between the terms in the Schwarzschild equations of motion and the GEM equations of motion. Corrections appear in only one of the three equations as expected.

Is this form more helpful? It now looks more official.
doug

[CarlB]

There is a 1-to-1 correspondence between the terms in the Schwarzschild equations of motion and the GEM equations of motion. Corrections appear in only one of the three equations as expected.

Cool. Would you kindly edit in the difference? Of course these equations of motion are differential equations in s or tau instead of t, but I'm sure I can convert them. And to simulate them efficiently, they will have to be put into Cartesian form.

[sweetser]

Hello Carl:

Here are the equations of motion, for the Schwarzschild metric first (see if it looks Kosher to you), in Cartesian coordinates, presuming we are in the z=0 plane of a system that stays in said plane, so all z's can be ignored.


\frac{2 G M(x x'+y y')t'}{\left(x^2+y^2\right)^{3/2}}+c^2\left(1-\frac{2 G M}{c^2\sqrt{x^2+y^2}}\right)t''=0



-\frac{G M x t'^2\left(1-\frac{2 G M}{c^2\sqrt{x^2+y^2}}\right)^2}{\left(x^2+y^2\righ t)^{3/2}}+\frac{G M\left(x x'^2-x y'^2+2y x'y'\right)}{c^2\left(x^2+y^2\right)^{3/2}}-x''\left(1-\frac{2 G M}{c^2\sqrt{x^2+y^2}}\right)=0



-\frac{G M y t'^2\left(1-\frac{2 G M}{c^2\sqrt{x^2+y^2}}\right)^2}{\left(x^2+y^2\righ t)^{3/2}}+\frac{G M\left(y y'^2-y x'^2+2x x'y'\right)}{c^2\left(x^2+y^2\right)^{3/2}}-y\text{''}\left(1-\frac{2 G M}{c^2\sqrt{x^2+y^2}}\right)=0


All the derivatives are with respect to the Lorentz invariant distance s = \sqrt{x^2 + y^2 - c^2 t^2}. Looks kind of scary to me.

Proceed anyway. On to the GEM equations of motion in Cartesian coordinates:


\frac{2 G M(x x'+y y')t'}{\left(x^2+y^2\right)^{3/2}}+c^2t\text{''}=0



-\frac{G M x t'^2\text{Exp}\left(\frac{-4 G M}{c^2\sqrt{x^2+y^2}}\right)}{\left(x^2+y^2\right) ^{3/2}}+\frac{G M\left(x x'^2-x y'^2+2y x'y'\right)}{c^2\left(x^2+y^2\right)^{3/2}}-x\text{''}=0



-\frac{G M y t'^2\text{Exp}\left(\frac{-4 G M}{c^2\sqrt{x^2+y^2}}\right)}{\left(x^2+y^2\right) ^{3/2}}+\frac{G M\left(y y'^2-y x'^2+2x x'y'\right)}{c^2\left(x^2+y^2\right)^{3/2}}-y\text{''}=0


All the t, x, y, t', x', y', x'', y'' terms are in the same place. I was able to collect the exponential into one term (exponential are so convenient).

I can provide the Mathematica notebook if anyone is interested. The result is not this pretty (meaning the result had to be simplified "by hand" and is subject to human error). I did check that the units for all the terms are the same.

doug

[sweetser]

Hello:

For the last 50+ years, there has been a contest for the most "provocative" paper dealing with gravity. I entered the 2007 contest, but alas, did not win, nor did it receive one of a few dozen "honorable mentions". I have been trading emails with the guy who won the top prize of $5k. I though I'd plug in my entry, even though it is far longer than is the norm here, coming in just under 1500 words, per the rules of the contest. I hope you enjoy.


Submission for the 2007 Essays on Gravitation
Sponsored by the Gravity Research Foundation

Title: Geometry + 4-potential = Unified Field Theory
Author:
Douglas B. Sweetser
39 Drummer Road
Acton, MA 01720
sweetser@alum.mit.edu

Abstract:
Gravity is the study of geometry. Light is the study of potentials. A unified field theory would have to show how geometry and potentials could share the work of describing gravity and light. There is a long list of criteria that must be satisfied to have a reasonable hypothesis, from recreating the Maxwell equations, to passing the classical tests of gravity, to demonstrating consistency with the equivalence principle, and working well with quantum mechanics. This essay works through many of the common objections.

Paper:

Geometry without a potential is like a bed without a lover. The Riemann curvature tensor, with its divergence of two connections, is exclusively about geometry and all about the bed sheet. Newton's scalar potential theory was the first math to reach and direct the motion of the stars. It is only about the scalar potential. Unfortunately, it is too small, being inconsistent with special relativity. I will try to construct a unified field theory for gravity and electromagnetism as a compromise between Newton and Einstein, the potential and the metric, in a way that will get along with quantum mechanics. My guiding principle is provided by Goldielocks, who might find a scalar theory too small, a rank 2 theory too large, so perhaps a rank 1 proposal will be just right.

I honestly love Newton's potential theory. It is still in use today by rocket scientists who do not put an atomic clock onboard their ship. It gets half the answer right about light bending around the Sun. When a theory comes up short, we can either discard it or figure out the simplest way to lend a hand. A gravitational theory with a thousand potentials instead of one will be able to match every experimental test of gravity. Use Occam's razor: a 4-potential should be more than adequate to match how gravity makes the measurement of time get a little smaller, while the measurement of 3-space gets a little larger.

I am torn between two lovers, Newton and Einstein, feeling like a fool. Thugs from Ulm will insist that gravity must be a metric theory. They have the experimental tests of the equivalence principle to prove it. They punch home the fact that the way to take the derivative of a connection that transforms like a tensor is through the Riemann curvature tensor. Drop the Ricci scalar into an action, vary it with respect to the metric, and out from the heavens flies Einstein's field equations.

What's wrong with that? A metric theory isn't silly at all. One must be able to express gravity in terms of a metric. Based on my respect for Newton, I wonder if it is possible to find a compromise between a larger 4-potential and a metric theory?

When we were young, we would write a covariant tensor as A_{\nu}. The differential \partial_{\mu} also transforms like a tensor. When we bring these two together, the 4-derivative of a 4-potential, \partial_{\mu} A_{\nu}, the result does not transform like a tensor. The reason is that as we move around a manifold, the manifold - not the potential - might change. A means of accounting for a changing surface must be made. Here is the definition of a covariant derivative all students of gravity learn:


\nabla_{\mu} A_{\nu} = \partial_{\mu} A_{\nu} - \Gamma_{\mu \nu}^{\sigma} A_{\sigma}


Can you spot the symmetry and identify the group implied by this definition? Imagine we make a measure of one of these terms, say \nabla_0 A_0, and it happens to be 1.007. If one worked in flat Euclidean spacetime, the connection would be zero everywhere, and everything would come from the change in the potential, \partial_0 A_0. One could also decide to use a constant potential, so the dynamic metric's connection would account for all the change seen, -{\Gamma}_{00}^{\sigma} A_{\sigma}. One has the ability to continuously change the metric and thus the connection so long as there is a corresponding change in the potential which leaves the resulting covariant derivative invariant. This sounds like the group Diff(M) of all diffeomorphisms of a 4D spacetime with the additional constraint that there are changes in the 4-potential such that the covariant derivative is unaltered.

Born background free, as free as general relativity, one must find a differential equation whose solution will dictate the terms of the dynamic metric. That is what the derivative of the connection in the Riemann curvature tensor does: there are second derivatives of the metric whose solutions under simple circumstances can be found. I have chosen to study the simplest vacuum 4D wave equation:


\square^2 A_{\mu} = 0


It is vital to note that I did not write the D'Alembertian operator, which would have been a box without the 2. Instead this is a covariant derivative acting on a contravariant derivative acting on the 4-potential. The first derivative will bring in a connection, and the second derivative will take the derivative of the connection, resulting in a second order differential equation of the metric, precisely what is needed to be background free. Can we find interesting combinations of metrics and potentials that solve this differential equation and is consistent with all tests of gravity to date?

Say we used a constant potential, where all the second derivatives were zero. Make the problem simple: a static, spherically symmetric, and non-rotating mass. For those skilled in the arts of differential geometry, it should be straightforward to show that the divergence of the connection of the exponential metric (below) is a non-trivial, entirely metric solution to the 4D wave equation. Compare the exponential metric in isotropic coordinates:


d \tau^2 = exp(-2 \frac{G M}{c^2 R}) d t^2
- {\frac {1}{c^2} exp(2 \frac{G M}{c^2 R}) (d x^2 + d y^2 + d z^2)


a nicely matched pair of exponentials, with the Schwarzschild solution in isotropic coordinates:


d \tau^2 = (\frac{1 - \frac{G M}{2 c^2 R}}{1 + \frac{G M}{2 c^2 R}})^2 d t^2 -
\frac{1}{c^2} (1 + \frac{G M}{2 c^2 R})^4 (d x^2 + d y^2 + d z^2)


which is inelegant enough to rarely be seen in books on general relativity. Theorist prefer the Schwarschild coordinates while experimentalists must work with isotropic ones. Beauty may be in the eye of the beholder, but an exponential is the calling card of a deep insight into physics.

Either metric satisfies all tests of the equivalence principle because the solution is written as a metric. Either metric satisfies all tests of the weak field because their Taylor series is the same to the terms tested. Either metric satisfies all strong field tests because it is entirely about a metric, so there is no other field to store energy or momentum. For an isolated system, the lowest mode of emission is the quadrapole moment. The metrics differ in second order effects by twenty percent in how much light is bent around the Sun, so it is a shame no one has been funded to get the data.

The 4D wave equation has been quantized, and written up in most books on quantum field theory, in the section on relativistic quantization of the Maxwell equations. Two of the modes of emission are the transverse spin 1 fields of light. That is no surprise. The scalar and longitudinal modes are banished to a virtual state using a ``supplementary condition'' because the scalar mode would allow negative probabilities, a no-no. That is the way it is for a spin 1 field theory where like charges repel. The field strength reducible asymmetric tensor \nabla_{\mu} A_{\nu} for this proposal can be split in two: an irreducible antisymmetric rank 2 tensor to do the work of electromagnetism with a spin 1 field so like electric charges repel, and an irreducible symmetric rank 2 tensor to do the work of gravity with a spin 2 field so like mass charges attract. Gravity couples to the 4-momentum, not the rank 2 stress-energy tensor. All forms of energy go into both sources, except one: the energy of a gravitational field. To be consistent with electromagnetism, gravity fields do not gravitate. Should a gravity wave ever be detected and measured along six axes, the polarization of that wave will be transverse if general relativity is correct, but not if this unified field proposal is accurate. Such data will be hard to get, but the difference would be unambiguous.

The speed of gravity is the speed of light, and so its field strength tensor must be gauge invariant. The field strength tensor \nabla_{\mu} A_{\nu} is only gauge invariant if its trace happens to be zero. That is where the massless graviton lives. When the trace is not zero, then the scalar field formed from the trace of \nabla_{\mu} A_{\nu} will break the U(1) symmetry of electromagnetism. The Higgs particle is unnecessary. There is a quantum expression of the equivalence principle, a link between the spin 2 particle (\nabla_{\mu} A_{\nu} when tr(\nabla_{\mu} A_{\nu} = 0) that mediates gravity and the scalar field needed to establish inertia (tr(\nabla_{\mu} A_{\nu})\neq 0).

There is an important benefit to splitting the load for describing gravity between the connection and the changes in the potential. By using Riemann normal coordinates, an arbitrary point in spacetime can have a connection equal to zero. For that point, the energy will be zero. That has remained a technical problem for people trying to quantize general relativity. For this proposal, the energy contributed by the connection could be zero, but that contributed by the potential would be non-zero. Localized energy is a good thing.

Einstein had a great respect for Newton's towering body of work. He might have appreciated this compromise between geometry and potentials which allows light to lay down with gravity in the same equation.

fini

[CarlB]

Doug, I'm glad you put this up here. That concept of "splitting the load between the connection and the potential" gives me a better intuitive understanding.

I will get back to the equations of motion soon. Right now there is a lot of furious paddling underneath the surface, so don't think that this duck isn't going anywhere.

[sweetser]

Hello Carl:

That concept of "splitting the load between the connection and the potential" gives me a better intuitive understanding.

Thanks. The jargon word - gauge - I know is going to create issues. This is a German word for "measure" which we first get exposure to with model trains. In field theories, it means there is a field that can be added in that will not change a solution to some field equations, just the numbers that pour out. The ability to add in the field means one can impose a gauge constraint - I wish to make the first term always equal to zero, so I'll add in this gauge field.

My proposal involves an issue of measurement, but of a different character. It comes straight out of the definition of a covariant derivative. You, the constructor of the numerical description, get to decide how much of the ball dropping to the Earth is due to a 4-potential, and how much is due to spacetime curvature. Go from one extreme to the other in a continuous way, you make the call.


You are a courageous duck, and I am very patient. Those equations of motion look nasty to me, particularly since they are all derivatives with respect to the interval. I'm glad there are both the GR and GEM equations, because those results should differ only at the second order PPN level of accuracy, and even then only by 20%. My sense is that one would need to be hyper paranoid about error management before any difference between the two was "real" and not an artifact. I have no idea how one distinguishes between two big, complicated calculations that are different by a wee little bit, and rounding error.

doug

[sweetser]

Hello:

I have been having an email discussion with a physics professor about my proposal. He did not agree that the second rank symmetric tensor would be the place where a spin 2 field would live. Instead he said it might be a place for a rank 0 field. That mystified me, since a quality of a scalar field is the complete lack of indexes, and a rank 2 tensor has 2. He also was talking about currents, something I had never done anywhere, not even in this very long thread! He scolded me, told me to read "Feynman Lectures on Gravitation", chapter 3.

Feynman rocks! I completely understood where the critical professor was coming from. His objection was reasonable. Then, once I really get a complaint, a small twist of an under-used math tool is usually all it takes to pull out the thorn. Here goes...

Feynman's analysis deals directly with 2 currents that interact, a viewpoint I have not used often. The charge coupling term, j'^u A_u has one current. Where is the other? One can take the Fourier transform of the 4-potential A_u and in the momentum space representation rewrite the potential like so:

A_u = - \frac{1}{K^2} j_u

This is how the 2 currents interact:

interaction = - \frac{1}{K^2}j'^u j_u

Make things simpler by having the current move along z:

K_u = (\omega, 0, 0, k)
K^2 = \omega^2 - k^2

Write out the interaction by its components:

- \frac{1}{K^2}j'^u j_u = - \frac{1}{\omega^2 - k^2}(\rho' \rho - j_x' j_x - j_y' j_y - j_z' j_z)

Charge is conserved, so:

K^u j_u = 0 = \omega \rho - k j_z

Use this to eliminate j_x:

- \frac{1}{K^2} j'^u j_u = \frac{1}{k^2}\rho' \rho + \frac{1}{\omega^2 - k^2}(j_y' j_y + j_z' j_z)

If we are in the rest frame of j' or j, then only the charge density matters. Move relative to that reference frame, and the other terms come into play.

Feynman now focuses on the jx and jy terms. This is where physics becomes math magic. These two currents always involve virtual photons. Further, Feynman works with the poles, where \omega->k. These virtual photons are the sum of two independent terms, j_x' j_x and j_y' j_y. A different way to say this is that there are 2 independent polarities for photons.

That was fun, but I wanted to think more precisely about the product of two currents. I'm going to use quaternion algebra, but if you are more comfortable with the Dirac algebra - only a twist of i away - go ahead.

(0, j_x', j_y', 0) (0, j_x, j_y, 0)^* = (j_x' j_x + j_y' j_y, 0, 0, j_x' j_y - j_y' j_x)

The phase term is in the z slot. It will require a 2 pi rotation to get back to go. The current-current interaction is a spin 1 photon, so like charges repel. Good.

It occurred to me that there might be another distinct product of these two currents. Consider the conjugate operator, which flips all the signs except the first one. It is known by mathematicians that there is more than one anti-involutive automorphism. Let's break down that jargon. The automorphism means that the function maps back to the same space. Taking two operations brings the function back home. The final bit is (a b)* = b* a*. Big words, but here is a simple idea: fix a term other than the first one, and flip the signs of all others. This little algebra trick is missing from many professional physicists tool drawer. Let me define the second conjugate like so:

(t, x, y, z)^{*2} === ( (0, 0, 1, 0) (t, x, y, z) (0, 0, 1, 0) )^* = (-t, -x, y, -z)

Put this tool to work for two interacting currents:

(0, j_x', j_y', 0) (0, j_x, j_y, 0)^{*2} = (j_x' j_x - j_y' j_y, 0, 0, j_x' j_y + j_y' j_x)

Take a peek at page 39, and you'll realize this product has the character of spin 2! I think the idea is that the two parts of the phase term can add together, able to race back to their initial spot in pi radians. This product describes with two degrees of freedom a current interaction where like charges attract. Cool. A 4-current has 4 degrees of freedom, 2 for a spin 1 current where like charges repel, 2 for a spin 2 current where like charges attract.

This calculation made my memorial day weekend memorable.

[sweetser]

Hello:

In the game of unified field theory, terms involving gravity must be similar to those for EM, but not exactly the same. In the above post, I made the gravity current-current interaction exactly in EM's image, looking only at the j_x and j_y currents. This indicates that gravity is a transverse wave.

Gravity is not a transverse wave. The transverse modes of emission for a 4D wave is light, the photon. That leaves the scalar and longitudinal modes for gravity. Something has to be different to justify the different modes of emission. I decided to see if I could somehow involve the current j_z, since that would be a real current along the direction of motion. Going back over the calculation, instead of eliminating j_z, I tossed out \rho:
- \frac{1}{K^2} j'^u j_u = \frac{1}{\omega^2}j_z' j_z + \frac{1}{\omega^2 - k^2}(j_x' j_x + j_y' j_y)
Now I can form products with j_x and j_z:
(0, j_x', 0, j_z') (0, j_x, 0, j_z)^{*1} = (j_z' j_z - j_x' j_x, 0, - j_x' j_z - j_z' j_x, 0)
Likewise for j_y and j_z:
(0, 0, j_y', j_z') (0, 0, j_y, j_z)^{*2} = (j_z' j_z - j_y' j_y, j_y' j_z + j_z' j_y, 0, 0)
This still has the spin 2 symmetry Feynman refers to on page 39 of his lectures on gravity. What is revealing is to add up these current products:
(0, j_x', j_y', 0) (0, j_x, j_y, 0)^{*} + (0, j_x', 0, j_z') (0, j_x, 0, j_z)^{*1} + (0, 0, j_y', j_z') (0, 0, j_y, j_z)^{*2}
=(2 j_z' j_z, j_y' j_z + j_z' j_y, - j_x' j_z - j_z' j_x, j_x' j_y - j_y' j_x)
The first term, the current density, is all the real current, while the phase has both spin 1 and spin 2 symmetry. This appears to be an improvement to me.

doug

[sweetser]

Hello:

This is another L O N G post. I went to the 10th Eastern Gravity Meeting expecting to give a 12 minute talk like I had since EGM 7. It didn't happen because there were too many people attending. I wrote up this note, and will be shipping it out on Monday to the organizers. This is life for an "independent researcher". I do try to do polite body slams.


Subject: Notes from the fringe of the physics community

Alan Lightman discussed progress in physics as a result of community. I thought I would share a few observations I made based on my experiences at EGM 10 from the position on the edge of the community. The hope is to improve the situation in the future, which is why I cc'ed the past organizers.

I define a fringe physicist as someone not employed as a physicist currently, or not on a path towards a degree in physics, who feels compelled to make a contribution. The definition can be applied without passing judgment on the value of the claimed contribution. By this definition, I am a fringe physicist, working for a software company with degrees from MIT in biology and chemical engineering, hoping to unify gravity with the other three fundamental forces of nature.

One defining characteristic of any community is how it treats the elements at the edge. The issues are never easy: is it better to let people live hard but free lives in the streets of the Northeast or force them to take shelter from an institution? The way to handle fringe physicists is not physically dire, but many have a great emotional devotion to their perspective projects. My own approach is to reflect Nature, which doesn't care if we get answers right or wrong, but she gives us an unmatched opportunity to be. Analytical indifference pulls against human hope, and at least for this email, hope has a lead.

At the bulletin boards provided by physicsforums.org, they grew tired of repetitive posts from fringe physicists and created a special place for "independent researchers". They have eight criteria for admission. My work met the criteria, and a discussion going on over there has been accessed twenty two thousand times, a high number for that site. That discussion has improved my proposal. I know where I've made mistakes in my Lagrangian. Dialog can make a difference.

The EGM has been small enough in the past to let everyone hear from the few east coast fringe physicists willing to make the journey. I had presented my work three times before, receiving a few questions at the end. My intent this year was to answer some of those questions, such as a demonstration that the exponential metric I work with satisfies the data for the precession of the perihelion of Mercury. For me, that was a difficult problem to solve. Although I had flipped through a good number of written solutions, none of them made sense in detail. For more than three years I had lived with the weight of feeling inadequate to answer the question. I am persistent to a fault. Sean Carroll's notes helped me step into the problem on the ground floor. I did complete the calculation. I was pleased enough with the twenty four step process that when I got married in October, for my groomsmen I made lunch boxes with the derivation on the side panels. A box made it to Ithaca, but stayed in my bags. The lunch box is both a prized possession and embarrassing.

EGM 10 had to take a different approach. There were too many talks to do in two days. We can hope that next year a smaller armada will fly out from the west coast. The wind was strong due to the stars appearing for Saul's symposium. People being creatures of habit, I would not be surprised if the attendance stays at this level for EGM 11.

The decision was made to provide a poster session for the fringe physicists. There is nothing inherently wrong with such an approach. My guess was such a decision was made very late in a chaotic and hectic game. There is almost no need to point out that people whose talks were shifted to the poster session should have been sent an email, but that detail of execution was missed. I checked the web site on Wednesday at work and was able to print out my slides. At Ithaca during the lunch break on Thursday, a $34 trip to the bookstore plus some high paced snipping led to a colorful poster I was pleased with.

At the end of a long day of talks on Thursday, the session was closed without announcing the poster session. There were more fringe physicists with posters than members of the physics community proper. Although Alan Lightman feels guilty about his baby seal clubbing incident with Webber, being utterly ignored can be as bad. Between these two extremes, is there something better?

This is what I did: I practiced my skills as a skeptic. This turned out far better than I expected. It was trivial to spot errors and glaring omissions. It requires discipline to resist clubbing. The bigger challenge was to find the roots of the person's area of study. It is a great exercise to find the pea under the pile of mattresses that is bothering this person. Since the list is not long, I will tell you the positive things I learned from three of my fellow fringe physicists.

1. Ed "negative mass" Miksch was the man who drove up from Pittsburgh with his wife. The organizers certainly know, but other folks may not be aware, how assertive he and his wife were about the need for Ed to be heard by this gathering of physicists. He was the guy who gave the three minute speech at the end of Friday's session. He claims to have shown that we should all be working with negative mass because he's done the calculation, one no physics professor at Reed College in the 50s could find an error in. The reason is that from where his calculation starts, there is no trivial math error.

This issue was understood by none other than James Clark Maxwell, but has not reached the wider physics community. The story was written up well at this URL: http://www.mathpages.com/home/kmath613/kmath613.htm. In GR books, they note the link between Newton's potential theory and the Schwarzschild metric, g_00 = 1 - 2 G M/c^2 R = 1 - 2 phi, and so ignoring the constants, phi = - M/R. What Maxwell understood was that such a potential plugged into a field theory like that for electromagnetism implies that like mass charges repel. The correct answer for a Newtonian potential where like charges attract is phi = 1 - G M/c^2 R!!!. Ed started from the wrong place because he was instructed by everyone that phi = -M/R is OK. Maxwell did see the correct way out - add a HUGE positive constant - he just couldn't justify it. Einstein's metric theory gets the potential theory right. Ed should be proud that he saw a problem realized by Maxwell. It is unfortunate that this potential theory is taught incorrectly to this day, because there will be people in the future that will walk down this wrong bend in the road.

To appreciate the consequences of how Ed's issue is misunderstood at the highest levels of the physics community, I asked Clifford Will why a 4-potential theory was not listed even as a possibility in his Living Review article on GR. He claimed one could make a potential theory get h!!!alf of the light bending around the Sun, but not all the bending measured by experimental tests. Will is correct for a scalar potential theory, the g_00 term. A 4-potential theory could easily match the data, with A = (1 - 2 GM/c^2 R, -1 - 2 GM/c^2 R, -1, -1), so the g_00 and g_11 terms will contract time measurements and expand space measurements as seen in tests.

2. Fred "MM" Pierce had a poster on the Michelson-Morley experiment. I was not hopeful as he went into his story. He talked about relativity, not once distinguishing between special and general relativity, two very different theories. He pointed out that if the interferometer was placed vertically instead of horizontally, the vertical machine would have interference fringes, both here and at the antipode, while the horizontal machine would not. I told him that was consistent with our current understanding. He then claimed this was connected to a misunderstanding about an ether. I told him I saw no need for an ether.

He also focused on a spinning satellite. He pointed out how the satellite would provide an accelerated reference frame. Go a different distance out from the spinning hub, and your weight would change although your mass would stay constant. Again it sounded consistent with our knowledge. He claimed it shows a link between motion and the cause of gravity. Gravity must be motion since it is the same as being on a spinning satellite. I told him I'd think about it, and left it at that.

Both Newton and Einstein spent quality time thinking about the spinning bucket. I don't think that debate has been settled. There is the standard elevator thought experiment. Gravity has tides, but the elevator does not. Some say that tides are the only real effect of gravity. The spinning satellite has tides, therefore it is a more faithful representative of gravity than the rocket. Once going, the satellite can maintain its "fake" gravity via angular momentum inertia, unlike the rocket which requires ever increasing amounts of energy.

Consider this thought experiment. Someone has replaced the Earth with a thin shell made of dense material such that the mass of the Earth is the same. You don't notice the change until you find a hole in the shell. Curious, you climb through, and almost by accident because you didn't hang on, you end up (or down?) flying across the hollow inside, there being no gravity field. After quite some time (the Earth is large), you get to the other side, and spring on back. This time you grab on to the edge. You got to collect physical data on Gauss' law that there is no gravity field inside a hollow sphere, how cool. You hear a noise, some creaking, and then stand up. The shell of an Earth is moving. It is picking up speed, and finally, it feels like you weigh your usual weight due to the spinning of the Earth. You can crawl back out the hole, and feel your weight due to gravity, or go into the hole, and feel your weight due to the spinning. Do an experiment with tides, and you realize one is a continuation of the other, no matter if you are inside or outside the shell.

I have come to the conclusion that any proposal for gravity must make clear the connection between gravity and rotational dynamics.

3. John "Two Timing" Kulick works with the two dimensions of time. I have written command line programs to add, subtract, multiply, divide, take sines, cosines, and apply group theory to events in spacetime. None of these will work if there are two times. John's work made no sense to me. Although I challenged him on traveling faster than the speed of light, that discussion went nowhere. His mantra was geometry, and I didn't get it.

What I did understand was his complaints about cosmology. From the rotation profiles of disk galaxies, to the big bang, basically anything big or old, physics fails. I am too skilled a skeptic to believe in two dimensions of time, dark matter, or dark energy. It is our mathematical description of nature that will have to change.

4. Doug "Rank 1" Sweetser has a unified field theory. He has also worked on leprosy (cloning genes from the mycobacteria that cause the disease) which somehow seams appropriate: no one wants to hang out with someone who has anything to do with leprosy or unified field theory. It turns out that leprosy is near impossible to transmit, but is the most visually frightening disease and thus the most feared because we are predominantly visual. Since Einstein worked more than thirty years on a unified field theory and failed, this is a topic to avoid, a kiss of death for an academic career.

Prof. Steve Carlip said through an email exchange that he thought my action could only involve a spin 1 and a spin 0 field. I told him if that was true, my proposal would be wrong. I had trouble following his logic. My field equations are rank 1, but my field strength tensor is rank 2. The part that does the work of gravity is a rank 2 symmetric tensor, so I couldn't understand how it could describe a spin 0 field that arises from an index-free tensor. Steve held his ground however, giving up on the discussion, telling me to go read chapter 3 of the Feynman lectures on gravity. That's what I did during EGM 10.

There are two separate reasons why one can spot a spin 1 field in the EM Lagrangian. The first is the rank 2 antisymmetric field strength tensor, A_u;v - A_v;u. The second arises from the charge coupling term, J^u A_u, that can be rewritten using a Fourier transform in momentum space as a current-current interaction. Take one current and the conjugate of another current, form the product, and the phase of the resulting term will return after 2 pi rotations, thus is spin 1. See section 3.2 for details.

My relationship with Steve broke because I was utterly unaware of this standard approach to field theory. I spent many fun hours going through the details of chapter 3, picking up nuances, working on my speed of creating the logic flow. I had his critique based on the coupling term down pat. I have been fortunate that once I understand an issue, I can see two others: how to get around the problem and why it has stumped people before.

If you are given a 4D vector space, list the anti-involutive automorphisms. Sounds too mathematical, sorry. A conjugate is an example. My guess is that nearly all well-trained physicists believe there is nothing other than the conjugate. That is true in a 2D space like complex numbers. A conjugate has three properties: you stay in the same place, two operators in a row is like doing no operator, and (a b)* = b* a*. One could imagine an operator that flips the sign for all but the x. I call this the first conjugate because it is the first part of the 3-vector. The second conjugate is defined similarly. Although one could define a third conjugate, it does what the other three could do in combination, so I don't include it. My bet is that all who read this note have never used a first or second conjugate, but it should be a simple idea to absorb.

I redid the Feynman current-current interaction calculation, and when the time was right, used a first conjugate instead of the Plain Jane conjugate. The resulting product is written on page 39 for a spin 2 particle. Getting the details solid on that calculation made the trip.

My unobserved poster had two themes. One board was nothing but the Lagrangian, a partial derivative party. I included the recently worked out details of the current-current spin 2 interaction. The second part tried to answer a difficult question asked of me: what does your theory do that's really different? A 4D Lagrangian is not different.

I decided at this meeting to come out of a mathematical closet, and admit publicly that I use quaternions. My observation is that half of technically trained people are familiar with the word, and few have done any serious calculations with them. The exceptions are rocket scientists and game designers who do 3D calculations without the problem of gimbal lock. There are a few people who play near quaternions: Connes with non-commutative geometry, Penrose with twistor theory, Alder with quaternionic quantum mechanics, and Baez with octonions. Out here on the edge, I learned how to take a quaternion expression and make an animation. I figured out how to make a ten second animation of SU(2), the group sitting in the middle of the standard model. I can say with complete confidence you don't know what it looks like, but my ipod does. Small steps from there have led to animations of U(1), U(1)xSU(2), SU(3), and Diff(M)xSU(3). I think a visual justification of the standard model with gravity qualifies as "really different".


FUTURE LINE OF ACTION

The organizers of EGM 11 will face the same issues you have. Please feel free to pass on any of these observations and suggestions.

We should admit that there is a physics fringe. That fringe needs to interact with physics skeptics. I could see a later presentation session, or a poster session for "alternative approaches". We would need to get a good dozen or more grad students and professors, including the organizer of the meeting. The reward for the professionals would be a chance to work on their skills as skeptics. It is a balance of asking, connecting, sifting through the history of physics, and criticizing. The session should not be promoted as such, rather we find a different way to accommodate a group that will hopefully remain on the same scale of a half dozen. I could serve as a liaison or chair of the session.


Was the conference worth my time and eleven hundred dollars? The funds came from an estate left by my mother back in August. She would have wanted me to go but to be careful someone did not steal my ideas. This unified field theory is the elephant in my life. I am proud of its latest trick with a spin 2 current-current interaction. I want to know if this elephant is real or a technical mirage. My own limitations are painfully glaring to me. This note, long as it is, has brought clarity on my feelings concerning the fringe of physics. Sorry for the fire hose of an email and attachments, but I am from MIT.

Thank you for all the effort you put into this meeting.

doug

[CarlB]

Thanks for the report. About "The exceptions are rocket scientists and game designers who do 3D calculations without the problem of gimbal lock." The situation in geometric algebra is similar. Half the users are computer programmers.

[sweetser]

Hello:

I posted this in the newsgroup sci.physic.research. It was a clear explanation about like charges in EM and gravity, so am adding it to this thread.

>give some examples of your calc that do occur in, or do reflect our real world applications. (mech/dynamics/electr)

>I ask this because the EM force carrier is supposed to be of a spin 1 nature, whereas the gravitational force mediator is supposed to be a spin 2 entity.

You are correct, the photon is spin 1, and the graviton which we won't be detecting any time soon is spin 2. A fundamental property of EM is that like charges repel. That is consistent with the force mediating particle being spin 1. Likewise, in gravity like charges attract, so the force mediating particle must be spin 2. Brian Hatfield gave a good explanation of this in the introduction to "The Feynman Lectures on Gravity". Force mediating particles must have integral spin. To always act one way, the gravity mediating particle must be even spin.

Since light is bent by a gravity field, a spin 0 particle will not work (anyone want to provide the reason, I've seen that written a few times, but am not clear on the logic). Ergo the simplest particle would be spin 2.

In EM where like charges repel, the photon must be odd spin. I don't know that I have ever heard a spin 3 particle discussed, but a photon is spin 1.

A Lagrange density describes all the ways a system can trade energy per unit volume. Integrate a Lagrangian over space and an arbitrary amount of time. If you can find something that can be varied without changing the integral, that is a conserved quantity for the action. From the Lagrangian, one can crank out the force equations by varying the 4-velocity keeping the 4-potential fixed, or the field equations by varying the 4-potential and keeping the 4-velocity fixed.

In the real world of EM, how can we look at the Lagrangian and tell that like charges repel and the force mediating particle is spin 1? Here is the Lagrangian:

\begin{equation}\mathcal{L}_{EM} = - \rho_m/\gamma\end{equation}
\begin{equation*}\tag{2}- J_q^{\mu} A_{\mu}/c\end{equation*}
\begin{equation*}\tag{3}- \frac{1}{4 c^2} (\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu})(\partial^{\mu} A^{\nu} - \partial^{\nu} A^{\mu})\end{equation*}

To generate the Lorentz force equation, only (1) and (2) matter since they are the ones that have a 4-velocity inside. Because they have the same sign, the resulting force law will indicate that like charges repel.

To generate the Maxwell field equations, only (2) and (3) matter since they have the 4-potentials. Here again, because they have the same sign, like charges repel.

There are 2 ways you can spot the spin 1 force mediating particle. The first way focuses on (3). If one switches the order of the indexes on the antisymmetric field strength tensor, the sign of the field strength tensor will flip. That is a signal that the spin is odd.

What I did not understand was how to look at term (2) and pick out a spin 1 field. Section 3.2 does a great job of it. The idea is transform the charge couping term into a charge-charge interaction. Look at the product of a charge-charge interaction, and focus on the phase. That takes 2 \pi to get back to where it started as expect for a spin 1 particle.

So to review, there are 4 reasons while like charges repel in the \mathcal{L}_{EM} Lagrangian: the force equation (1&2), the field equation (2&3), the spin 1 particle in the antisymmetric field strength rank 2 tensor (3), and the spin 1 particle in the current-current interaction (2). Nothing like logical consistency!

If you wanted to make a Lagrangian that could do the work of gravity, all four of these chips must stack up. Here is a Lagrangian I play with:

\begin{equation*}\tag{4}\mathcal{L}_G = - \rho_m/\gamma\end{equation*}
\begin{equation*}\tag{5}+ J_m^{\mu} A_{\mu}/c\end{equation*}
\begin{equation*}\tag{6}- \frac{1}{4 c^2} (\partial_{\mu} A_{\nu} + \partial_{\nu} A_{\mu})(\partial^{\mu} A^{\nu} + \partial^{\nu} A^{\mu})\end{equation*}

The force equation will have like charges attract because the signs of (4) and (5) are different.

The field equations will have like charges attract because the signs of (5) and (6) are different. These two are easy and well known.

Look at (6), and you seen an indication of an even spin field because if the indexes are changed, the sign does not change. Because there are indexes, the second rank field strength tensor cannot be spin 0. The trace of this tensor would make a spin 0 field. Effectively there will always be spin 0 field associated with this higher spin field. Cool. If the trace is zero, then the particle characterized by (6) will travel at the speed of light. If not, the particle will have a non-zero mass.

In Misner, Thorne, & Wheeler, problem 7.2, they consider an antisymmetric tensor for (6) which cannot work because it would indicate an odd spin mediating particle, and thus not self-consistent.

The memorial day weekend calculation was about term (5). One has to be able to spot a spin 2 particle in the current-current interaction. That will have a phase that looks like 2 jx jy, so that in \pi radians it will get back to where it started. I used a different kind of conjugate that many physicists are not aware of, basically one that tosses in a pair of basis vectors.

So to review, there are 4 reasons while like charges attract in the \mathcal{L}_G Lagrangian: the force equation (4&5), the field equation (5&6), the spin 2 particle in the symmetric rank 2 field strength tensor (6), and the spin 2 particle in the current-current interaction (5).

Steve Carlip was correct to complain about my lack of understanding about the current-current/spin 2 issue. Hopefully I have made progress on it.

doug

[Mentz114]

[Edit] I'm up to page 13 and it's looking great.

The attached paper is not peer reviewed, and I am not suggesting its content should be compared with your own work. Only the proposed tests of the metric might be of interest.

[sweetser]

Hello Hai-Long:

I've printed out your paper, and will hopefully figure out a comment which I will send via email, not this forum. I hope that reading through this LONG thread you will see that some progress has been made.

doug

[sweetser]

Hello:

I have decided that until I identify a show stopping problem, I should try and promote this work with established physicists once a week. This is a sales job. Sales is difficult, because the answer is always no, no, no, no, no, no, and no. If I try weekly, that is at least 52 knocks on the door. As an example, no one responded to the the EGM 10 note posted earlier here. Presume no one responds to these notes unless I tell you otherwise. Nothing in this is personal, it is how people respond to pitches, whether it is a door-to-door vacuum cleaner salesman or an internet unified field theory promoter.

There was a show stopping problem recently that Prof. Steve Carlip pointed out, the current-current interaction having to be spin 2 when Feynman had shown it was spin 1. I think that one has been resolved by using a different kind of conjugate.

So the first pitch was to a Harvard professor. If you want to read a great, long article on the search for the Higgs by the Large Hadron Collider, click here:

http://www.newyorker.com/reporting/2007/05/14/070514fa_fact_kolbert

Many good people are in on the effort to detect the Higgs. My objections is only mathematical. I believe that gravitational mass breaks the symmetry of the standard model, so no Higgs mechanism with its false vacuum is needed. I work with a unification group theory that is smaller than the standard model - U(1)xSU(2)xSU(3) - which has 12 elements in its Lie algebra. Instead I work with SU(3)=(U(1)xSU(2))* U(1)xSU(2). One of the cool things about this smaller model is that it has the chance to explain why we don't see quarks: we already see particles for the photons and the weak particles. I hope people have clicked through the youtube visualization here:

http://youtube.com/watch?v=ExNPiMcVXww

So these are the reasons I don't accept the math - and it is only about the math, nothing personal.

Here was the email I sent to the good professor:


Hello Prof. Arkani-Hamed:

Based on a recent New Yorker article, I see you are an enthusiastic
believer in the Higgs particle, that we should be able to detect it at
the LHC. If that happens any time in the next ten years, I will fill
out the scanned check for the number of GeV/c^2 for the Higgs. You
have an incentive to hope for a heavy Higgs :-)

I used to play against someone who went on to win the World Series of
Poker (at that time, we were about even in skill for dealer's choice).
It might appear this is a high risk without reward. There is a small
chance that despite how busy you are, you might click through
some of the attached work I've done which makes me believe that the
Higgs is unnecessary.

We know what U(1) looks like, a circle in the complex plane. Do you
know what the group SU(2) looks like? I bought an Ipod because I
figured out how to do this (a unique reason for the purchase, I still
forget to bring the earphones with the device, visualizing the
standard model being more important than a few mp3's). I also have
electroweak symmetry, SU(3), and Diff(M) in the machine. I happen to
be an independent guy who has been playing games with quaternions, the
kind of number sitting at the center of the standard model. Analytic
animations using quaternions led directly to these result.

I'm a lab technician by training. I report what I have. I have a
rank 1 field theory that unifies gravity and EM. That statement can
be supported by the Mathematica notebook that goes from the Lagrangian
out to a metric that is consistent with first order PPN tests of weak
field gravity, and predicts 0.8 microarcseconds more bending of light
around the Sun than the Schwarzschild metric of GR. The exponential
metric is manifestly more elegant than the Schwarzschild metric
because exponentials appear in fundamental laws of physics.

It is clear that if this proposal is correct, there will be a new
stable constant velocity solution to classical gravity. I have yet
to cow rope the equation to problems like the rotation profile of
spiral galaxies or issues with the standard big bang (I've had trouble
both understanding the numbers being used, and how to do the numerical
integration). I point this area out because it may provide a new way
to solve big problems in cosmology and to show I don't overstate my
case.

Stay busy.
doug

ps. The check was written yesterday, before the details of the work at
Fermilab on Cascade B, not the Higgs, came out. This is all about the
math, nothing about the good people who do both the experimental and
theoretical work.

[rewebster]

Just as an aside--In what area of work are you a 'lab technician' ? Is it sort of like a patent clerk?

[Mentz114]

Doug,

in post #313 you give the EOM for the exponential metric. You're in 3 dimensions and theta is missing so I asume you've made a simplifying assumption like theta = pi/2, or dtheta = 0. Can you please confirm exactly which metric you used so I can compare with my calculations ?

Thanks,
M

[edit]I sorted it out. I get the same equations now.

[sweetser]

Hello Rewebster:

I am a oversensitive to your question. This is the Independent Research Forum, which makes it fringe physics turf with a veneer of respectability, a veneer that is important not to scratch. John Baez created the "Crackpot Index", http://math.ucr.edu/home/baez/crackpot.html. For example, I get points for this one:

30. 30 points for suggesting that Einstein, in his later years, was groping his way towards the ideas you now advocate.

It is clear that Einstein, in his later years, was looking for a way to unify gravity and EM. There is great documentation for this. I would not say he used my approach. He did try all kinds of different ways. For a good discussion of this, read chapter 26 of A. Pais' "Subtle is the Lord...". The difference technically is clear: Einstein used the Riemann curvature tensor which is kind of like a divergence of connections, a connection being a measure of how metrics change. The Riemann curvature tensor is thus a measure of the second order changes in metrics that transforms like a tensor.

That works for gravity. The problem is that there is no place for a 4-potential. Potential theory is essential for EM, the weak, and the strong force. I am trying to find the compromise between changing potentials and changing metrics.

Now to your aside. I was a molecular biologist. I cloned and sequenced DNA from the mycobacteria that causes leprosy. It was a remarkable job at many levels. The most relevant part for this thread was the way I would go in every day, feeling utterly inadequate for the scale of what was ahead, yet trying to do a little bit. I have avoided going for top positions in my area of work so there would be time to work on things I cared about that are utterly irrelevant to how I make a dollar. The same holds true today, although now I work for a software company. I am not trying to climb any ladders here, no 60 hour work weeks for me.

doug

[rewebster]

Please forgive me if there was any 'itch' caused by my comment. There is/was a comparable nature of someone pursuing and doing the all fine work that you ARE doing while still laboring at a job that isn't related to your main passion. AND, if ANYONE was to receive the 30 points, they should be allotted/ascribed/deposited into MY account. I KNOW that Einstein was a physicist FIRST. The day to day way that people make a living should, as even in Einstein's case, should be taken as an 'aside', rather than a defining attribute (patent clerk), to a person's greater goal. Too often he is described as a patent clerk who wrote relativity and, also won a Nobel prize for his photoelectric effect, when he was always a TRAINED PHYSICIST first, working for a while to make a living for his family at a job, I think, just to minimize what he had done as if 'anyone' could do it. If your work does succeed, someone else (besides me) will make a similar comparison. (I was trying to make a small compliment of the intensive work you were doing.)

[sweetser]

Fair enough. I am not a trained physicist. I am a trained scientist, and try to remain skeptical about my own efforts in physics because I admit the odds of success are very low.

This was an interesting phrase: "intensive work" (and thanks for the compliment, I am not so gracious at accepting them, my bad). Due to the structure of my life, I cannot devote big blocks of time to the work, forcing me to be efficient. The amount of partial differential equations makes it LOOK pretty darn scary. Yet the nuts and bolts of it are actually a bit easier - or at least on par - with the Maxwell equations. My former mailman Jim committed the field equations to memory (backwards): Always give 2 Brownies to Jim, or J = Box^2 A.

The entire reason why is also simpler: it is about being a 4D slinky, super tiny oscillations around doing nothing. Although I understand some of the math behind GR, I don't have any sense why mass should tell spacetime how to curve.

doug

[rewebster]

Fair enough. I am not a trained physicist. I am a trained scientist, and try to remain skeptical about my own efforts in physics because I admit the odds of success are very low.

This was an interesting phrase: "intensive work" (and thanks for the compliment, I am not so gracious at accepting them, my bad). Due to the structure of my life, I cannot devote big blocks of time to the work, forcing me to be efficient. The amount of partial differential equations makes it LOOK pretty darn scary. Yet the nuts and bolts of it are actually a bit easier - or at least on par - with the Maxwell equations. My former mailman Jim committed the field equations to memory (backwards): Always give 2 Brownies to Jim, or J = Box^2 A.

The entire reason why is also simpler: it is about being a 4D slinky, super tiny oscillations around doing nothing. Although I understand some of the math behind GR, I don't have any sense why mass should tell spacetime how to curve.
doug

1) you're welcome (again, nothing in the way of a 'bad' remark was intented)

2) agreed, as to 'something' doesn't seem to 'fit' to me either


and it's /(your's) not string theory! ('not that there's anything wrong with that!'--to quote Seinfeld)

[sweetser]

Hello:

Here are two specific objections to Hai-Long Zhao's paper.

In equation 4, he claims
E = m_0 c^2 (1 - exp(GM/c^2 R))
In support, he notes that it gives the gravitational potential energy everyone uses, \phi = -GM/R which solves the Laplace equation, \nabla^2 \phi = 0. While both true mathematically, we recently discussed here the problem Maxwell had with the idea, namely it leads to like charges repelling if you use a field theory like the one for EM. Bad, very bad, but oh so common.

The way around this is to add in a big honking constant because \phi = K - GM/R also solves Laplace's equation but gets the energy situation right. This is what happens in GR. When Zhao does the precession of the perihelion equation, he actually is defining the energy like this:
E = (1 - 2 GM/c^2 R) \frac{d t}{d \tau}
When I do the same calculation, I use:
E = exp(-2 GM/c^2 R \frac{d t}{d \tau}
The Taylor series expansion is the same as the line above to first order in M/R. To me, it looks like Zhao uses two definitions of energy that are not the same. The work does not look logically consistent.

Zhao recreates another common error. Gravity has to involve symmetric rank 2 tensors. Why? Because the metric is a symmetric rank 2 tensor. Do something with an antisymmetric tensor, and it makes not a bit of difference to a metric. Now think of a B field - a totally antisymmetric animal. Change the order of indexes, and the sign flips. Oops.

If you want a fun exercise involve LOTS of partial differential equations, write out all the components for this tensor:
\partial^{\mu} A^{\nu} + \partial^{\nu} A^{\mu}
Full credit is ONLY given if you include all 16 Christoffel symbols. I had to use my eraser a lot to get it, terms everywhere, signs flipping, not flipping. If you like physics, that will be a fun, time consuming puzzle, particularly if you want the result to look pretty.

Even if you are insecure about getting each sign right - I know I was - one thing is very clear: all the terms that go into what I call the little b field have the same sign, like b_x = -\frac{\partial A_y}{\partial z} - \frac{\partial A_z}{\partial y}. The gravitomagnetism B field in Zhao's paper can only be represented by an odd spin field where like charges repel. The small b field I work with can be represented by a spin 2 field, where like charges attract. Bingo, bingo.

doug

[Mentz114]

Doug, thanks for the review. I didn't think his derivations were elegant or convincing.


Are these right ? I got 18 Christoffel symbols to be non-zero.
Metric signature is -+++ , co-ords are t,r,theta, phi
g_{00} = -\exp(-kr^{-1})
g_{11} = \exp(kr^{-1})
g_{22} = r^2
g_{33} = r^2\sin(\theta)^2

Christoffel symbols.
3-23 = \cot(\theta)
3-13 = r^{-1}
2-33 = -\cos(\theta)\sin(\theta)
2-12 = r^{-1}
1-33 = -r\sin(\theta)^2\exp(-kr^{-1})
1-22 = -r\exp(-kr^{-1})
1-11 = -0.5kr^{-2}
1-00 = 0.5kr^{-2}\exp(-2kr^{-1})
0-01 = 0.5kr^{-2}

[sweetser]

Hello Lut:

I have done this calculation in the past, but thought I would repeat it, without looking at your answer, so it is independent. I am using the book "Gravitation and Spacetime" by Ohanian and Rufini, page 329. They use a metric signature of +---. This does matter to me because that signature also shows up if one wants to work with quaternions (there are three imaginary basis vectors, each one squared being -1).

They write the metric tensor like so:


g_{\mu \nu }=\left(
\begin{array}{cccc}
\text{Exp}[N] & 0 & 0 & 0 \\
0 & -\text{Exp}[L] & 0 & 0 \\
0 & 0 & -R^2 & 0 \\
0 & 0 & 0 & -R^2\text{Sin}^2[\theta ]
\end{array}
\right)


Outside a spherically symmetric, non-rotating, electrically neutral mass where one has chosen to work with a constant potential, I claim the exponential metric is the solution to the GEM field equations.

N=-\frac{2G M}{c^2R}

L=\frac{2 G M}{c^2R}

There are 13 non-zero terms:

\Gamma _{\text{ }01}^0=\Gamma _{\text{ }10}^0=\frac{1}{2}\frac{\partial N}{\partial R} = \frac{G M}{c^2 R^2}

\Gamma _{\text{ }00}^1=\frac{1}{2}\frac{\partial N}{\partial R} \text{Exp}[N-L]} = \frac{G M}{c^2 R^2} \text{Exp}[-\frac{4 G M}{c^2 R}]

\Gamma _{\text{ }11}^1=\frac{1}{2}\frac{\partial L}{\partial R} = -\frac{G M}{c^2 R^2}

\Gamma _{\text{ }22}^1=-R \text{Exp}[-L] = -R \text{Exp}[-\frac{2 G M}{c^2 R}]

\Gamma _{\text{ }33}^1=-R \text{Sin}[\theta ]^2\text{Exp}[-L] = -R \text{Sin}^2[\theta ] \text{Exp}[{-\frac{2 G M}{c^2 R}}]

\Gamma _{\text{ }12}^2=\Gamma _{\text{ }21}^2=\frac{1}{R}}

\Gamma _{\text{ }33}^2=-\text{Sin}[\theta ]\text{Cos}[\theta ]}

\Gamma _{\text{ }13}^3=\Gamma _{\text{ }31}^3=\frac{1}{R}

\Gamma _{\text{ }23}^3=\Gamma _{\text{ }32}^3=\text{Cot}[\theta ]}

Now I'll compare with your results...They are identical, even down to the signs! We are looking at how the metric changes, so the signature does not matter, cool. I suppose I should have "expected" that, but I like these kinds of surprises.

doug

Note to the casual reader of this thread: this is a "look it up, plug it in'' sort of calculation. At this time, I don't have a feel for what individual terms here mean. I basically accept that this is the way a math person measures how changes happen from one place in a manifold to another if the metric is dynamic - meaning it depends on R and theta, where your happen to be.

[Mentz114]

Hi Doug,
I got in close with GEM and calculated the field tensors and force equations for the Rosen metric in polar and cartesian coordinates. The cartesian is easy to work with as you've observed and all the b field terms disappear. On inspection I can see that choosing a potential A = ( 1,0,0,0) eliminates the EM fields and leaves force equations (signature is -+++ so x^0 is t )

f^0 = \frac{-4GM}{c^2R^3}q_m(x\beta_x+y\beta_y+z\beta_z)

f^\mu = \frac{-4GM}{c^2R^3}q_mx^\mu\gamma

which in the limit gamma ->1 and beta-> 0 give Newtons gravity. I haven't yet examined what paths the equations will give in relativistic cases. Also on the plus side, if you look at the first equation in polar coords,

f^0 = -\frac{4GMq_m}{c^2R^2}

this could be interpreted as gravitational redshift - not predicted by Newton obviously.

Doing this has raised some conceptual problems for me. It seems to me you are using the geodesic equations of motion implicitly or explicitly in GEM. But they are derived by extremizing an action based on Ricci curvature, wrt the metric, something you abjure. Planetary precessions and deflection of light should be based on the GEM equations of motion.

Which brings me to the question - how does one describe the paths of light (null geodesics?) from the force equations ? Because there is a factor of gamma in the spatial forces, naively putting v=c will not do, because no deflection can ever take place if gamma=0. There's also the question of the mass charge of light.

I'm going to have a play with the force laws later.

Regards,
Lut

[sweetser]

Hello Lut:

I don't think this logic holds true, or at least let me give you my slant:

> but they are derived by extremizing an action based on Ricci curvature, wrt the metric, something you abjure.

My father would have called "abjure" a Word Wealth word. It sounded right, but I checked: "to reject solemnly". Bingo! The Riemann curvature tensor, and its contractions, the Ricci tensor and scalar, have done much good. I am hoping to find something better.

> Planetary precessions and deflection of light should be based on the GEM equations of motion.

While this could be done, it is not what I did do. The precession calculation is not easy. After getting over much fear, I finally did figure out all the steps, available here:

http://theworld.com/~sweetser/quaternions/gravity/precession/precession.html

If you go there, you'll see I based on off of the Rosen/exponential metric. Once I do the Taylor series approximation, the derivation is exactly the same. I understand bending of light based on the exponential metric.

The way I first got to the exponential metric used the force equation. First I needed to find a relevant potential, one that was "classical", and only involved light. As I told you, I did find it, but it is not trivial. The potential is a linear time perturbation of a 1/R solution. If the spring constants are chosen with care, one get a symmetric field strength tensor like so:

\nabla^{\mu}A^{\nu} = \frac{\sqrt{G} M}{R^2}I(4)

where I(4) is the 4x4 identity matrix. I was able to solve the force equation and get a 4-velocity solution. I rearranged that solution, and there was the exponential metric. I know that approach is unorthodox, but it was the first way I did it: force eq->velocity solution->metric. I was presuming the equivalence principle was valid, and dropped the inertial and gravitational masses. The photon does have a zero mass and electric charge. A more careful student of the mathematical arts would be concerned about this, getting it in a limit process. I am more practical. Initially, I only discussed the road using the force equation to the exponential metric since it was the only path. Now that I figured out the divergence of the Christoffel is a solution to the field equations, I talk about that path. This is the ignored path because people presume I am using a D'Alembertian operator, not a covariant followed by a contravariant derivative, which has a divergence of a Christoffel in it. What is really encouraging is that two very different paths - one using the force equation in an unusual way, the other from the field equation - lead to exactly the same metric which is consistent with weak field experimental tests, strong field tests, and tests of the equivalence principle.

Hope that helps the conceptual issues, which never go away entirely.

doug

[sweetser]

The email below is part of my "sales" activities, efforts to contact people of some stature. This process will involve some repetition, but I hope to adjust the location of the argument so there are some surprises for readers of this thread.

Clifford Will is a leading expert on experimental tests of gravity who happens to look like Ted Turner. Here is the email I sent off ten minutes ago...


Hello Prof. Will:

I have criticized your Living Review article "The Confrontation between General Relativity and Experiment" for being incomplete, both in private emails to the Eastern Gravity Meeting organizers, and in a public forum at physics.forums. It is my responsibility to inform you of this entirely technical issue, which you may or may not act upon.

<quote>
[I was discussing the value of work by people working on the fringe of physics, who clearly misunderstood a number of issues, yet might be raising a topic that is worthy of addressing.]

1. Ed "negative mass" Miksch was the man who drove up from Pittsburgh with his wife. The organizers certainly know, but other folks may not be aware, how assertive he and his wife were about the need for Ed to be heard by this gathering of physicists. He was the guy who gave the three minute speech at the end of Friday's session. He claims to have shown that we should all be working with negative mass because he's done the calculation, one no physics professor at Reed College in the 50s could find an error in. The reason is that from where his calculation starts, there is no trivial math error.

This issue was understood by none other than James Clark Maxwell, but has not reached the wider physics community. The story was written up well at this URL: http://www.mathpages.com/home/kmath613/kmath613.htm. In GR books, they note the link between Newton's potential theory and the Schwarzschild metric, g_00 = 1 - 2 G M/c^2 R = 1 - 2 phi, and so ignoring the constants, phi = - M/R. What Maxwell understood was that such a potential plugged into a field theory like that for electromagnetism implies that like mass charges repel. The correct answer for a Newtonian potential where like charges attract is phi = 1 - G M/c^2 R. Ed started from the wrong place because he was instructed by everyone that phi = -M/R is OK. Maxwell did see the correct way out - add a HUGE positive constant - he just couldn't justify it. Einstein's metric theory gets the potential theory right. Ed should be proud that he saw a problem realized by Maxwell. It is unfortunate that this potential theory is taught incorrectly to this day, because there will be people in the future that will walk down this wrong bend in the road.

To appreciate the consequences of how Ed's issue is misunderstood at the highest levels of the physics community, I asked Clifford Will why a 4-potential theory was not listed even as a possibility in his Living Review article on GR. He claimed one could make a potential theory get half of the light bending around the Sun, but not all the bending measured by experimental tests. Will is correct for a scalar potential theory, the g_00 term. A 4-potential theory could easily match the data, with A = (1 - 2 GM/c^2 R, -1 - 2 GM/c^2 R, -1, -1), so the g_00 and g_11 terms will contract time measurements and expand space measurements as seen in tests.
</quote>

It is almost as if on one understands what diffeomorphism means in the context of a potential theory. Write a field equation, Del^2 A = J, and people think they can write down an answer for a simple current. People presume Del^2 is the d'Alembertian, a scalar operator, when it is not (unless the space is flat and Euclidean). It is a covariant derivative applied to a potential, followed by a contravariant derivative. Einstein's summation convention should NOT be applied to two covariant derivatives. It is necessarily the case that one could find 2 solutions to the field equations: a flat background metric and the potential contains all the information, or a dynamic metric describes every bend and kink in spacetime, while the potential does nothing. The divergence of the connection is in the field equations, which means there is a second order differential equation involving the metric. Solve that equation, and the metric is determined by the physics of the problem at hand. The choice of how to measure - a changing potential or changes in the connection - is up to the observer. One can choose to make a 4-potential field equation into a metric theory, and thus consistent with all tests of the equivalence principle (the Rosen metric is the solution, but without the fixed background that creates problems for the strong field tests).

I hope you get a chance to think about your deliberate omission. There is a lot of fun math going on there.

doug

[sweetser]

Hello:

I had a choice this summer: go to the big meeting on GR in Australia,http://www.grg18.com/ or sign up for a 4 day class on GR at MIT's Professional Institute. I voted for the latter, which would have had 6 hours of lectures each of 4 days, with discussions. Nerd out!

Unfortunately, Prof. Joss has had a medical emergency, and the class was canceled. This is quite a bummer for me because I had high hopes of making a personal connection to a professional in the field of GR. I want to know why I can leave this project alone, and the class might have provided a path.

If anyone reading this thread is going to GRG 18, I would appreciate a report back on the section: A4 Alternative Theories of Gravity – Gilles Esposito-Farese. My bet it will be on things like scalar-tensor and vector-tensor stuff, certainly nothing like vector-all-alone like we've been discussing here.

doug

[Mentz114]

But they are derived by extremizing an action based on Ricci curvature, wrt the metric...
Wrong ! I withdraw this obviously incorrect statement.

But I'm still not happy mixing geodesics and forces.

Bad luck with the GR course.

[sweetser]

Hello Lut:

Let's try and form a few correct statements, always a good exercise. I remember my surprise at seeing the simplicity of the Hilbert action of GR:

S_{Hilbert} = \int \sqrt{-G} d^4 x R

Not many symbols, but let me explain the few that are here. The action S is an integral over spacetime of all the energy interactions of a system per unit volume. Integrate over a fixed amount of space, but arbitrary amounts of time, and you will likely get arbitrary integrals. The game is to use the calculus of variations to find what things can be varied such that the integral stays the same, no matter what time interval is used.

The integral is in 4D spacetime because we live in a 4D Universe. In my opinion, this is all that is needed to reject work done with strings because they integrate over ten or eleven dimensions which is not the way the world is. Notice where I put the \sqrt{-G}. It turns out the the spacetime volume element, d^4 x does not transform like a tensor. In curved spacetime, it will have a different value. The square root of the determinant of the metric compensates for this, so that \sqrt{-G} d^4 x does transform like a tensor.

Now we get to the heart of the action, the Ricci scalar R which is a contraction of the Ricci tensor, itself a contraction of the Riemann curvature tensor. If the action is varied with respect to the metric tensor g_{\mu \nu}[/tex], that generates 3 terms (here my knowledge becomes less precise: I know what happens, but have not looked at it closely enough to understand all the details). One of these three is zero, something about Gauss' law and the boundary of a boundary is zero. The other two together make the field equations:

R_{\mu \nu} - 1/2 g_{\mu \nu} R = 0

It should be clear why general relativity is a theory only about gravity, having nothing to do with EM: there was only the Ricci scalar R in the action.

Let's compare that with GEM. If we only want to get the field equations in a vacuum, this action should be sufficient:

S_{GEM} = \int \sqrt{-G} d^4 x \frac{1}{c^2} \partial_{\mu} A_{\nu} \partial^{\mu} A^{\nu} R

A basic question is how can we even hope that this will do both EM - which requires a spin 1 force mediating particle, and gravity - which requires a spin 2 field. I hope I am recalling this correctly and not making it up, but there was a group theory jock who said with great disdain that this idea was silly because the tensor [itex]A_{\mu \nu} is reducible, and thus cannot be use to represent a fundamental force of nature. The reducible tensor can be written as the sum of two irreducible tensors, \partial_{\mu} A_{\nu} = 1/2(\partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu}) + 1/2(\partial_{\mu} A_{\nu} + \partial_{\nu} A_{\mu}). The first of these is straight-out-of-the-book EM field strength tensor: swap the order of the indexes, the sign changes, as one would expect for a spin 1 field strength tensor. The second one is a rank 2 symmetric tensor, the kind that if its trace is equal to zero, could be a home for a graviton.

Vary the GEM action with respect to the 4-potential, and one gets a 4D wave equation. That part is direct. Rewriting the 4D wave equation in terms of fields, that detail gets more complicated. The answer is NOT the Maxwell equations with gravity fields that are just clones of EM (as happens in gravitomagnetism). The easiest way to see this is to consider the gravity field which has terms like b_x: \frac{\partial A_y}{\partial z}, \frac{\partial A_z}{\partial y}. What happens in gravitomagnetism is they cheat, copying EM too directly, claiming there should be a minus sign involved between those two terms. Swap the order, and b_x would flip signs, not good for a representative of a spin 2 field which doesn't have that property. In GEM, the symmetric analog is b_x = - \frac{\partial A_y}{\partial z} - \frac{\partial A_z}{\partial y}. This is not a curl operation. There will be no vector identity laws such as no gravity magnetic monopoles, or Faraday's law - the signs don't flip in ways that would "make it so".

For GEM, the road to an expression involving force is direct. The action involved is different than the one used for fields:

S_{GEM} = \int \sqrt{-G} d^4 x (- \frac{\rho}{\gamma} - \frac{1}{c} Jq_{\mu} A^{\mu} + \frac{1}{c} Jm_{\mu} A^{\mu})

Vary this action with respect to the 4-velocity, and one gets the Lorentz force law. There is a velocity in the gamma and the current terms, but not in the field strength tensor, A_{\mu \nu}, which is why it can be ignored when thinking about the force law. For every point in the spacetime manifold, there is a force, there is an energy.

In GR, the story is different. There is no force law. If there was a simple force law, we could go to a simple expression about the energy of the field. The Riemann curvature tensor is the source of the problem, since it involves the difference of two paths in spacetime. Instead one solves the field equations and gets a metric solution. Looking at the metric, one can pick out conserved quantities, the Killing vectors. I think of a geodesic as the easiest path through spacetime. The "force" is zero.

I do need to spend more time thinking about the GEM force law, how it can be a Lorentz force law with the potential doing all the work, and one where the force is zero, but the curved metric is where the action happens.

doug

[Mentz114]

Doug:
thanks for the exposition ( nobody expects the Spanish exposition !).
To save you typing in the future, you may assume I know GR well enough, including the EH action and variations thereof.

Your actions are new to me. The first one is obviously the GR action with a vector potential and seems a direct route to including the potential. But now you have you extremize wrt variations in the potential and the metric, simultaneously ( rather than assuming a metric). This will mean introducing constraints between the potential and the metric, which could be interesting.

The second action also seems to require a background metric.

...how it can be a Lorentz force law with the potential doing all the work, and one where the force is zero, but the curved metric is where the action happens. Bingo ! You have articulated my 'conceptual' problem. Also the practical problem of expressing forces as 'curvature'.

The river model cited by Carl earlier is the nearest I've seen to anyone doing this.

Regards,
Lut

[sweetser]

Hello Lut:

I can tell you are well schooled in the arts of GR. My long posts are both to test my own self-taught knowledge, and to bring along other readers of this forum with less experience in this area.

> But now you have you extremize wrt variations in the potential and the metric, simultaneously ( rather than assuming a metric).

I don't think this proposal makes sense. The 4-potential is a rank 1 tensor, but the metric is rank 2. I have yet to say one should vary the action with respect to the metric because that would yield rank 2 field and force equations, but the field and force equations in GEM are rank 1. The metric must be fixed.

A fixed metric does not mean a flat metric. All it means is that in the variation of the action, it was not varied. The fixed metric can be dynamic. Let's see how this works in EM. In EM, if you want to generated the Lorentz force equation, you vary the 4-velocity keeping the 4-potential fixed. Now you have the Lorentz force equation, you can use a dynamic potential to solve problems. If you want to generate the Maxwell field equations, the potential is varied, keeping the 4-velocity fixed. Solve a problem using the Maxwell equations, and the velocity of charges is likely to be dynamic.

I see you slipped in a killer word for me, the "background" metric. A fixed metric also does not mean a background metric. In EM, there are no differential equations that can be used to determine what the metric should be. As such, one must add a metric as part of the background mathematical structure in order to solve problems.

In the GEM proposal, there are differential equations that can be solved to figure out what the metric is according to physical properties of the system. This is the take the divergence of the Christoffel of the exponential stuff I cite from time to time. Because there is a second order partial differential equation involving the metric that depends on conditions, the metric is not part of the background math structure, but instead something that can be solved for.

EM is a gauge theory. GR is a gauge theory. In both cases, there is a field that can be added in without changing a solution. "Gauge" means measure, so arbitrary fields can be included in the measure of the fields.

GEM is a gauge theory of a different class. Here a measurement can involve either the changes in the 4-potential, or changes in the connection. One cannot add in a gauge field as happens in EM or GR because that would change the system. Once can instead decide to make the measurement all about the 4-potential or all about the connection, or any combination of the two.

doug

[Mentz114]

Hi Doug:

...well schooled in the arts of GR.
Not as well schooled as I think, obviously. But my point is central to what bothers me - the relationship between the metric and the potential. How do you decide how much of what goes where ?

If I begin with flat space-time and assume a matter current, then try to solve for a potential - I get a lot of EM fields from the solution that were not ordered by anyone. If I start with a charge current and do the same thing, the potential I get will contain gravitational fields I didn't want.

I'm not sure what you mean by "decide to make the measurement all about the 4-potential or all about the connection, or any combination of the two".

I appreciate you taking the time to answer my questions - even though they are starting to sound like whinges and quibbles.

Lut

[sweetser]

Hello Lut:

Let me adjust your question just a little:

> But my point is central to what bothers me - the relationship between the metric and the potential. How do you decide how much of what goes where ?

It is the relationship between the connection - derivatives of the metric - and the derivatives of the potential. The metric and the potential don't have a relationship to each other until a covariant derivative comes into play. The field equations then involve the second order derivatives of the metric, and the second order derivatives of the potential.

The relationship between the derivative of the metric and the derivative of the potential is in the standard definition of a covariant derivative:

\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma^{\mu \nu} _{\space\sigma} A^{\sigma}

This equation shows what I am calling "gauge choice", the ability to choose how much of the covariant derivative is due to \partial^{\mu} A^{\nu}, and how much is due to \Gamma^{\mu \nu} _{\sigma} A^{\sigma}. There is not a new equation, just a new way at looking at, and really using, an old definition.


Again, this is a great question:

> If I begin with flat space-time and assume a matter current, then try to solve for a potential - I get a lot of EM fields from the solution that were not ordered by anyone. If I start with a charge current and do the same thing, the potential I get will contain gravitational fields I didn't want.

One thing I have specifically avoided is "really new physics", because it is far too easy to get lost and confused. The simplest kind of solution to the 4D wave equation that uses a potential with factors of 1/\sigma^2 = 1/(x^2 + y^2 + z^2 - c^2 t^2). That leads to a 1/distance^3 force law. That constitutes "really new", so new no one will listen. Since I am both a skeptic and a fringe physicist, I have volumes of self-doubt. One way I manage this is to always try and get my units right. The smart kids in the class and in tenured positions use natural units so they can skip these details. If you work out the units for that most general relativistic potential solution, it looks like so:

A_0 = \frac{\sqrt{G} h}{c^2} \frac{1}{x^2 + y^2 + z^2 - c^2 t^2}

That has the units of relativistic (c), quantum (h), gravity (G) as an inverse cube force law. It is a clear outcome of the math, but I don't know what to do with it.

Hope you appreciate my fears.
doug

[sweetser]

Note added in proof: The Maxwell equations written in the Lorentz gauge also has this problem. It has inverse distance squared solutions, and thuse inverse cube force laws. Does anyone know what the standard approach is to such solutions? My bet is they claim it is "unphysical", which strikes me as empty.

In a second form of denial, anyone who works with the antisymmetric field strength tensor could also calculate the symmetric field tensor from the same terms, and then have to claim that Nature has a means of never ever using any such information in any way. That does not strike me as reasonable.

doug

[sweetser]

Peter Woit wrote a book recently called "Not Even Wrong" which was a critique of string theory. Most of the book was a darn good description of quantum field theory at a deeper level than one usually gets without plowing through the technical literature.

Lubos Motl is a string theorist. He used to post in the newsgroup sci.physics.research. He is the strongest supporter of string theory I have ever encountered, so strong, I was embarrassed by the guy. He now has a faculty position at Harvard.

These two REALLY don't like each other. That's why I wrote them both.


Hello Peter and Lubos:

I am aware how tense the professional relationship between you two
happens to be. I thought it would be an interesting challenge to find
points of agreement, and perhaps spot some new physics.

I am a fringe physicist, which I define precisely to be one without a
job or on the way to a degree who still has specific hopes of making a
contribution to physics. I have read Lubos' strong support for work
with strings in sci.physics.research. Being a skeptic, I have
promised to deliver a check to Lubos if in any time in a decade
(April, 2014) the physics of gravity in more than four dimensions
becomes broadly accepted as the correct way to deal with gravity,
instead of merely promising. Responding to a plea for a testable
hypothesis, I forwarded some of my initial work to Peter. At this
time, neither of you have responded, busy as you are. Those are my
ethereal connections to you both.

I have taken all of two lines of text from "Not Even Wrong...". I
hope to show that we all agree with the first line, yet all - even the
author - agree that the second sentence is wrong. Given the
passionate disagreement of the value of the book, the two chosen lines
have nothing to do with work on strings!

The first line came in the back of the book, where Peter is searching
to find common themes in successful approaches to doing new math.
Peter wrote (p. 258):

"Traditionally, the two biggest sources of problems that motivate
new mathematics have been the study of numbers and the study of
theoretical physics."

Knowing some history of mathematics makes this statement sound
reasonable. What I happen to do is study quaternions, a type of 4D
number, in order to do standard physics. Quaternions are just
4-vectors that can also be multiplied and divided. Most of my work
has been prosaic: I can generate Newton's law for a centrally directed
force in a plane as a one line quaternion expression (it usually
requires a few pages of text). Yet there are times when I have been
pushed to do new physics. I came up with a definition of a quaternion
derivative that uses a two limit process reminiscent of L'Hospital's
rule that may justify why causality for classical physics (a
directional derivative along the real axis) is different from quantum
mechanics (a normed derivative is all that can be properly defined).
From my perspective, the quaternion derivative is a math issue with
implications for physics.

Now it is time to turn to the incorrect statement. It is not a
trivial error. Here is the line (p. 117):

"The ability to visualize the graph of the function [which depends
on two complex numbers] is now lost, since it would take four
dimensions to draw it."

Don't tell animators that they cannot draw in 4D, three for space, one
for time. I have written the software to animate quaternions, which
are three complex numbers that share the same real number. If three
complex numbers can be visualized, then doing two is easy.

Rene Descarte developed analytic geometry which is still in wide use
today. Our brains devote more hardware to visual analysis than
anything else, which may explain the lasting power of analytic
geometry. I am developing the tools for analytic animations.
Addition of quaternions is the simplest operation out there. The
result from a physics perspective is an inertial observer, at the
heart of special relativity. That is the first animation included in
this email.

Peter's book deals with the standard model. I have therefore included
animations of U(1), SU(2), and SU(3). My goal has become to work with
a smaller standard model, one that has the symmetries of U(1), SU(2),
and SU(3), but does not view the three as tensor products which would
depend on a Lie algebra with 12 independent players. It may be
possible to view SU(3) as composed of two electroweak symmetries. The
advantage of the smaller model is that it could provide a
justification for the confinement that happens only for the strong
force.

The group SU(3) looks like an expanding, then contracting, bumpy
tennis ball in the attached animations. Go to a different place in a
spacetime manifold, and the size of the tennis ball might change. To
continuously change the measure of distance would involve the group
Diff(M), a symmetry at the heart of our understanding of gravity.

A new approach to understanding the symmetries of nature should look
different. I had to write the software to take quaternion expressions
off the command line and generate animations. My work will not fit in
a PDF. I chose the simple GIF animation just to be sure anyone can
see it.

I have set this email up as a "Prisoner's Dilemma" problem. You both
can choose to ignore it, which is the easiest thing to do. There is a
potential cost, since I will be publishing this email publicly, so if
and only if quaternion animations are an important step forward for
physics, this email would document how difficult it is to bring a new
way to look at the world to life. The second possibility would be
that one of you would start playing with what is visually new math. I
bought an Ipod only because I wanted to see SU(2), and be able to
carry it around with me. The person who "discovered" this new branch
of work would forever have bragging rights over the other, quite a
payoff.

The third possibility is that both of you were interested by the
animation. Given the diversity of your world views, that would speak
to the power of the animations.

Good luck in your areas of research. I hope to hear from either or both of you.
doug

65 MB 10 second GIF animations:
1. Addition/inertial observer:
http://quaternions.sourceforge.net/inertial_obs.povray.animation.scan.100.1002.gif

2. 3 ellipses in the complex planes, the group U(1) of EM:
http://quaternions.sourceforge.net/u1.povray.animation.scan.100.1000.gif

3. A unitary quaternion, the group SU(2) of the weak force:
http://quaternions.sourceforge.net/su2.povray.animation.scan.100.1000.gif

4. An evenly expanding/contracting 4-sphere, the group SU(3) of the
strong force:
http://quaternions.sourceforge.net/u1xsu2xsu3.povray.animation.scan.100.1000.gif

5. Two different 4-spheres, the group Diff(M) of gravity:
http://quaternions.sourceforge.net/u1xsu2xsu3_delta_scale.povray.animation.scan.100.1 000.gif

[Mentz114]

Doug:
The relationship between the derivative of the metric and the derivative of the potential is in the standard definition of a covariant derivative:

\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma^{\mu \nu} _{\space\sigma} A^{\sigma}

This equation shows what I am calling "gauge choice", the ability to choose how much of the covariant derivative is due to , and how much is due to . There is not a new equation, just a new way at looking at, and really using, an old definition.
Understood. I'll have think about this. Feels dodgy at the time of writing.

I'm not sure I follow the next bit of your post #345 (!). Again, I'll have to reflect on it. The only 1/R^3 law I can remember is the electric field of a dipole.

I don't think fear should come into this. If your theory proves to be inconsistent mathemaically or unphysical - what's to fear ?

In a second form of denial, anyone who works with the antisymmetric field strength tensor could also calculate the symmetric field tensor from the same terms, and then have to claim that Nature has a means of never ever using any such information in any way. That does not strike me as reasonable.

Tensors don't exist, numbers don't exist. They are abstractions of the human mind. Trying to second guess nature is not on. Every physical theory has defects. Little dark corners of unphysicallity. What is the meaning of negative energy, what are advanced solutions, why do energy conservation laws not transform relativistically ?

This adds up to one message for me - it can't be done. There is no complete physical theory. So I'm not betting - hence no fear.

For an amusing read, see "Why the laws of physics lie" by Nancy Cartwright ( no, not that one, this one was prof. of logic(?) at London U some time ago).

Regards,
Lut

PS I admire your chutzpah, in your challenge to Motl and Woit.

[sweetser]

Hello Lut:

I have had a pleasant day thinking about this comment:

> The only 1/R^3 law I can remember is the electric field of a dipole.

Let me first set the stage, which goes back to Heisenburg. When he first tried to formulate quantum mechanics, he made sure it was completely relativistic. Makes sense, but it did not match the result of Bohr's atom. Working with an equation that was not relativistic, a connection to the data of the day was possible. It took another five years for Dirac to repeat the effort, and figure out new physics, like all the antiparticles.

The GEM equation is fully relativistic. It took me more than a year to figure out how to correctly break relativity's grip, and make a connection to physics we know is true, specifically the exponential metric that is consistent with weak field tests of gravity.

Your questions are focused on the "Dirac"-like aspect of the GEM field equations. I like your read of it: this is a dipole. Here is my speculation of what may be happening.

When I wrote the charge coupling term on a black board for a friend,

-(Jq^{\mu} - Jm^{\mu})A_{\mu}

he wondered why I didn't just redefine a new current density, Ju^{\mu}=Jq^{\mu} - Jm^{\mu}. I said the problem with that approach is that the sign difference is required so the force laws have like electric charges repel and like mass charges attract. The sign difference is also required for the field equations, where like electric charges repel and like mass charges attract.

The two irreducible field strength tensors, the antisymmetric one and the symmetric one, each require their own charge because they are separate fundamental forces. If you want to have some sense of what the field strength tensors mean, the symmetric tensor is the average amount of change in the 4-potential with respect to t, x, y, and z, while the anti symmetric tensor is the deviation from the average amount of change. I call it "Average Joe and the Deviants". It should be clear that the average is a different sort of thing from the deviation of the average.

Although we have two types of charges, one can be up to sixteen orders of magnitude larger than the other, as is the case for an electron. For every charged particle, there are necessarily two types of charge: the electric charge where like repel, and a mass charge where like attract. In other words, every charged particle is a dipole. Further, the particle will behave like a permanent dipole because the charges are so different in size.

A fundamental property of particles like electrons and protons is the charge to mass ratio. It is a late night speculation, but it seams to me that ratio may be linked to the permanent dipole of electric and mass charges. I am going to study up on dipoles...

doug

ps. Fear really wasn't the right emotion. It is more strategic: make connections to known physics, and the odds are higher it will be right and get listened to.

[sweetser]

Hello:

My weekly effort to pitch the proposal involved writing to a forum on Bad Astronomy. This site is a minor Internet phenomenon run by Phil Plait where he shoots down odd claims, as well as discusses the good work going on in astronomy today.

I had to clip my post there to get it under 15,000 words. Here's the URL if you want to read it:

http://www.bautforum.com/against-mainstream/61876-gem-rank-1-unified-field-proposal.html

Here's my summary. Much of the stuff on that board is BAD, hard to follow the limited logic presented. I spent about half the post discussing the EM action, pointing out the 4 reasons why like electric charges repel. I then go through and show the sign changes necessary so like charges attract as happens with gravity. I finish with a discussion of the field equations. I invite the reader to "bust my pinata".

We will see if we get a spirited debate over there.

doug

[sweetser]

Hello:

The Bad Astronomy discussion is going OK. If I write up a note over there that I like, I'll port it over here. The following deserves "double posting" because it is a simple calculation with consequences. I was asked to derive the light bending around the Sun effect which is different between GR and GEM. Here goes...

The calculation comes directly out of a paper: “Post-post-Newtonian deflections of light by the Sun” by Reuben Epstein and Irwin Shapiro, Physical Review D, 22:2947, 1980. They write a generalized Schwarzschild metric like so:

dτ2 = A(R) dt2 – B(R) dR.dR

where

A(R) = 1 – 2 G M/c2 R + 2 β (G M/c2 R)2 + O(3)

and

B(R) = 1 + 2 γ G M/c2 R + 3/2 ε (G M/c2 R)2 + O(3)

There are three Greek letters in play here: β, γ, and ε. This is a research paper, so naturally they write "Calculating the deflection angle in the usual manner, we find for the ppN contribution

ΔθppN = π (2 + 2 γ - β + 3/4 ε) (G M/c2 R)2"

For the Schwarschild metric, all the Greeks are 1, so these constants add up to 3.75. For GEM, the constants add up to 4. Now collect the numbers needed. I fetch mine online, http://physics.nist.gov/cuu/Constants/index.html, http://en.wikipedia.org/wiki/Sun

G = 6.674 28 x 10-11 m3 kg-1 s-2
M = 1.989 x 1030 kg
R = 6.955 x 108 m
c = 299 792 458 m s-1
π = 3.14159
206265 arcseconds/degree

Go to Google, and type it in:
206265 * pi * 3.75 * 6.67428^2 * 10^-22 * 1.989^2 * 10^60 / 299792458^4 / 6.955^2 /10^16
and the answer is 10.96 microarcseconds.
For GEM, the answer is 11.69.
The difference is 0.73 microarcseconds.

That's how its done. GEM predicts more bending that GR, as earlier GR predicted more bending than Newtonian theory. Tradition!

doug

[CarlB]

I noticed that the GPB is indicating frame dragging effects that are centered about 25 or 35% higher than GR would indicate, but with large enough error bars that GR is consistent. Does GEM get higher frame drag than GR?

[sweetser]

Hello Carl:

GEM always predicts a bit more bending since the coefficients on the Taylor series are a bit larger. I have NOT tried to do a frame dragging calculation. Do you have a reference that spells it all out in detail?

doug

[sweetser]

Note: by "a bit more", this applies only to second order PPN effects, not first order PPN where the two are exactly the same. I don't have a metric for a rotating reference frame, so the short answer is I don't know at this time.

[sweetser]

Hello:

People are often interested in what other folks thought about an idea. I wrote this up for the Bad Astronomy and Universe Today forums, but thought I would share it here. The short summary is that I did get 6 professional physicists to look at the proposal at earlier stages of its development, and they all made direct contributions by finding flaws in the action. Fortunately the errors were correctable. The rest of the technical folks were too busy to reply, which doesn't mean anything.


1. Alan Guth, MIT

Alan Guth developed the inflation hypothesis for the big bang which addresses two problems with classical big bang theory, the horizon and flatness problems. I have met with Alan Guth four times over a decade by lurking outside of his office. He is kind to such people. The first three times was to discuss a math widget that applied to experimental tests of gravity I was curious about. He convinced me in our ten minute discussions that it was not a worthwhile project to pursue because it was a widget, not a theory.

On my last visit, I told him I had a unified field equation. I wrote it on a blackboard:

Jqu-Jmu = Box2 Au

He told me that I did have a field equation, but I did not have a modern unified field theory. For that, I would need to know the action, how to vary the action to generate the field equations, what were solutions to the field equations that were consistent with current tests, and what solutions would give different results in tests, and need to write down the stress tensor. I thanked him and left.

It took me more than two years to come up with answers to his list of reasonable, but difficult demands. One reason is that I had never worked with a Lagrange density before. I used to make a joke that the only people smart enough to work with Lagrangians were either Jewish or Russian (Goldstein or Landau and Lif****z), but the real masters had to be Russian Jews (since this is a positive spin joke, I hope folks accept it. I've heard that in the day, Einstein loved a good Jewish joke. Feynman was a master from New York, whose father was from Russia). It was the Landau and Lif****z book that taught me these tools are not as bad as I had feared.

Guth is a very busy guy, and certainly does not work on a unified field theory. His advice was general, and I was able to follow it given enough time. Guth got me to look for a Lagrangian, an important first step.


2. Stanley Dresser, Brandeis

He is a accomplished theoretical physicist who went to the International School in Geneva that my grandfather founded, which was why he agreed to meet me. He was of the old school of physicists, where put downs were how the boys played. I showed him the first equation of my paper which had the Lagrange density, and he rejected it immediately. That was because the sign of the coupling term was the same as the sign of the field strength tensor term, ergo like charges repel as happens in EM, not for gravity. In a gentler world, he could have suggested a sign flip, but instead mocked me. Although a challenge to take emotionally, I was proud that I cut through the personal issues and accept the technical critique. Professor Dresser contributed the plus sign to +Jmu Au which makes both the force and field equations have like charges attract.


3. Urs Schreiber, Universitat Hamburg

He was a moderator for sci.physics.research and a postdoc in theoretical physics. I had been writing only about the GEM Lagrangian, and he wanted to start from an action. I thought the only difference was the integral. After a painful number of exchanges, I determined that I was wrong: one needs the square root of the determinant. Urs contributed that factor to the action.


4. John Baez, UC Riverside

Baez is a math guy turned physics guy who loves to chat about technical issues on the Internet, and presumably in person. He had a one line dismissal of my proposal on sci.physics.research. He claimed that my proposal used a fixed Minkowski metric, and thus was inconsistent with all tests of the equivalence principle. This is another critique that I gave more that a year of time to. It eventually inspired me to really read through Sean Carroll's lecture notes on general relativity. I now understand general covariance. Newton's theory can be written in many different coordinates, but time's relationship to space - completely separate, unable to mix - is fixed. The idea of general covariance is that the same darn physical situation can be described on a different spacetime manifold. With Newton, one cannot change time's relationship to space, so Newton's laws are not generally covariant.

Special relativity mandates inertial observers, and that limitation also ensures that the proposal is not generally covariant.

Is GEM generally covariant? This is how I demonstrate it. I go about trying to describe the field for a static, charged, massive particle. I can do that with a flat Minkowski metric, and a 1/R potential. I can also describe the exact same physical situation with a constant potential, but the exponential metric. QED.

Anyone who listens to my dog and pony show for a short period of time will hear me talk about the exponential metric. I found it hard to believe that Baez thought I always, necessarily used the flat Minkowski metric. I found a reference to an effort to use a 4-potential that did use the fixed Minkowski metric. It was in Clifford Will's book, unpublished work by Kustaanheimo. Prof. Baez found a valid reason to reject a 4-potential theory - a fixed Minkowski metric - that did not apply to the GEM proposal.


5. Steve Carlip, UC Davis

He is a professor on the West Coast who does work with black holes. Over on sci.physics.research, and in a series of private emails, Steve Carlip and I debated what the spin of the particles that are in the GEM action. At first, he thought since I was using Maxwell as a starting point, it would be spin 1. I pointed out the symmetric field strength tensor must have an even spin particle. From there, he claimed that GEM must be using a spin 0 particle, which can be rejected because we know that the path of photons are bent by gravity. I initially thought this critique was not justified because the second rank symmetric tensor has two indexes, so cannot be represented by a spin 0 field.

After a number of tense exchanges, he gave up, but gave me the references I needed to understand his position. I was looking at the field strength tensor, but he was thinking about the charge coupling term. He was the one who suggested I read the Feynman lectures on gravity, specifically chapter 3. What I did on a trip to the 10th Eastern Gravity Meeting was to go over that chapter carefully, making sure I understood all the algebra, and had it down cold, so I could repeat it quickly. I understood why Steve thought the coupling term indicates that only a spin 1 or spin 0 particle could be represented. The one funny step in the derivation is multiplying two 4-vector currents together to get another 4-current. This is not a contraction to a scalar, it is a different math animal. That is where I used the first conjugate, and got a product that maps directly to the algebra of a spin 2 field according to Feynman. This will be hard to pitch to physicists since most only use JJ'*, never J(iJ'i)*.

Prof. Carlip's critique led me to understand a spin 2 field for 4-potential, 4-current coupling.


6. Clifford Will, Washington University

Will is the leading expert on experimental tests of gravity. I went to the 8th Eastern Gravity Meeting specifically to ask Prof. Will about second order PPN experiments. It was the first question asked at the meeting, and like I said, there are no plans. I asked another question, which was would we be able to tell the polarization of a gravity wave? He noted that we have yet to detect one, but if we did, we would have to measure it on six different orientations to determine the polarization. Bummer.

Over in sci.physics.research, after I had gone over some of the detail of my proposal, someone told me to go read Clifford Will's Living Review article, because he was sure a pure vector theory would be covered in the 100 page article with more than 300 references. I thanked him for the link, found the relevant section, but no, nothing was said about a vector theory. The guy doubted me, but I was right on this one. He suggested I drop the good professor an email because he was a good guy.

Although true, Will is also very busy. I decided to wait, several years as it turned out. When he came to MIT to give a talk, I went to the pre-talk coffee and cookies. I was able to chat with him for about ten minutes. Nothing about GEM, all about his turf, tests of gravity. I told him the story about the missing vector theory, and he said it was not there because one could not construct a potential theory that would match data for bending of light. It gets the g00 part right, but that is not enough.

After our discussion, I thought about what he said. It is true, a scalar potential theory cannot match the bending of light data. It gets half the answer exactly right. It is false that a vector potential theory cannot match the data: just have the gRR be a result of the second term in the potential.

A cool result of the discussion is I now have a clear case of a mixed potential/connection solution. Use an exponential potential in the scalar slot, and the three out of four exponential metric:

ds2 = - dt2 + exp(2 G M/c2 R)(dx2 + dy2 + dz2)

The ultimate compromise! Notice how this will match all current weak field tests of gravity, but will not be a solution to Einstein's field equations.

I may have in my possession the complete set of published papers that discuss a vector theory of gravity. One is a paper by Gupta, another by Thirring. In under two paragraphs, they both dismissed a vector theory because they claimed one could not have like charges attract. They did not provide a detailed reason why they had reached this conclusion. In the first post of this thread, I gave all the reasons I understood why like charges repel for EM. Two of the reasons are classical, and two have to do with the spin of the photon. For the gravity part of my proposal, I gave two classical reasons why like mass charges attract, and two reasons why the mediating particle is spin 2.

My interactions with Will indicated that the literature on rank 1 gravity proposals is missing, and the physicists are not concerned with this omission. What the physicists don't know may hurt them.


Summary:
S_{GEM}[Alan Guth]
= \int \sqrt{(-G)} [Urs Schreiber] d^4 x
(- \rho/\gamma
- 1/c Jq^u A_u
+[Stanley Desser] 1/c Jm^u A_u [Steve Carlip]
- \frac{1}{4c^2} (\nabla_u [John Baez] A_v - \nabla_v A_u)(\nabla^u A^v - \nabla^v A^u)
- \frac{1}{4c^2} (\nabla_u A_v + \nabla_v A_u)(\nabla^u A^v + \nabla^v A^u)

Each interaction with a professional physicist has benefited this project. I look at the action, I know exactly where each person's contribution was.


The Misses
Not all efforts to reach out have worked. To be complete, I'll include them. I know these folks are absurdly busy, so try to make unusual pitches.


7. Lubos Motl, Harvard

Lubos is a vocal advocate for the work on strings and membranes. I have written Lubos a check for $100, payable as soon as any work on a theory that works in more than four dimensions is accepted by the entire physics community as the way to explain gravity. I put a 10 year limit on the payout. About once a year I send him a progress report on my work. He has yet to reply.


8. Arkani-Hamed, Harvard

Arkani-Hamed is part of a stable of string theorists who work at Harvard. The good professor is quite confident that the Large Hadron Collider (LHC) in Geneva will be able to detect the Higgs particle. It is my belief that when gravity is unified with the other forces, the Higgs mechanism will be unnecessary. I have written a check for the amount of the Higgs particle should it ever be detected. I also included links to the animations for the symmetries underlying the standard model.


9. Peter Woit, Columbia

Peter has trained in both math and quantum field theory. He wrote the book, "Not Even Wrong...". Most of the book is a tour of the development of quantum field theory, things like the standard model and all the work done at modern colliders. The last few chapters are a critique of the work being done with strings. One of his central complaints is the lack of a testable hypothesis. I made an effort to point out that GEM does have several tests that could be done, none of them easy, but all understandable (light bending and the polarization of gravity waves).


10. Brian Greene, Columbia

Brian has made contributions to work on strings, and to educate the public on this area of research, in books, radio commentary, and TV shows ("The Elegant Universe"). I sent him a draft of my paper enclosed in a GEM lunch box.http://theworld.com/~sweetser/Images/lunchbox.png I use an identical box daily for lunch. It has the field equations along with a quote by Wheeler, figure 1 from the draft paper which indicates the choice we have to attribute change either to the changing potential or the connection, and on the side panels, a complete derivation of the precession of the perihelion of Mercury using the GEM theory. I don't know if he is using the lunch box.


11. Lee Smolin, Parimeter Institute

Lee does research in loop quantum gravity. He has written a critique of the work done with strings. He appeared on Science Friday to argue the merits of work on strings with Brian Greene. He also got a lunch box. When he visited Boston, I found out that he was so busy traveling and dealing with a new baby that he did not go through his mail and see the lunch box. Oops.


12. Sheldon Glashow, BU

A Nobel Prize winner, and skeptic of work done with strings, he played a part in "The Elegant Universe". He did not reply to an email where I claimed to have a testable proposal, his central complaint against work on strings.


13. David Pritchard, MIT

Someone I have lunch with recommended I pitch the idea to him. He quantified how busy he was, having to work on six separate papers he was an author on while finishing off a grant.


14. Seth Lloyd, MIT

David Pritchard recommended I drop him an email. He said he might be able to take a look at the paper, but it did not pan out.


15. Max Tegmark, MIT

A real astronomer, but too busy to read and reply to my email.

doug

[Don J]

Just by pure curiosity I want to know if GEM can describe "spin orbit coupling"

Apparently from Wikipedia,
As the master theory of classical physics general relativity has one known flaw: it cannot describe "spin orbit coupling"
So they have developed
http://en.wikipedia.org/wiki/Einstein–Cartan_theory

[sweetser]

Hello Don:

Great question! If I was a stellar student of GR, I would know about this issue. Now I know something based on the wikipedia article.

All the players in the Einstein field equations are symmetric: the Ricci tensor and metric tensor are both symmetric. That creates a problem. Things like spin and angular momentum need an antisymmetric tensor. This is a fundamental problem for GR since there is no place to put such energy that exists in the real world.

The problem is averted in GR by including torsion in the connection. This is tough stuff to follow for me. One no longer uses the Christoffel symbol of the second kind which again is a symmetric tool. Bring in torsion to the connection, and there is a place to handle angular momentum conservation in an antisymmetric tensor.

With GEM, spin orbit coupling is handled easily. It is in the EM part of GEM. The Lagrangian has the rank 2 antisymmetric field strength tensor. To quote from http://en.wikipedia.org/wiki/Spin_tensor:

Examples of materials with a nonzero spin density are molecular fluids, the electromagnetic field and turbulent fluids.

Spin orbit coupling will not be a problem for GEM. Unification does has some advantages!

doug

[sweetser]

Hello:

Here is another one of my pitches, but this time I included a t-shirt, one that has an oil painting I did called "Turquoise Einstein" along with the GEM field equations. I've invested $16.99, so I'll have to see if the investment pays off.

doug


Hello Peter:

I enjoyed your book, "Cosmology: A Very Short Introduction", which was assigned reading for a class I planned to take at MIT's Professional Institute on "Relativity, Gravity, and Cosmology" taught by Paul Joss. Unfortunately he had a significant medical issue which led to a cancellation of the class. Near the end of your book, you speculated that one might be able to put a central equation for understanding the Universe on a t-shirt. You should have by now, or within a few days, received from cafepress.com my effort toward that elusive goal. It should be popular due to the image of "Turquoise Einstein", even if people ignore the math.

I'll spend a few paragraphs describing my idea for how to unify gravity with the rest of physics. In my own experience, I have found it is usually quite easy to spot if someone is taking a logically coherent approach to a problem. If I can be coherent in my reasoning, perhaps like a laser I can project an idea over the Atlantic and share some of the profound excitement I feel.

Newton's law is still the most useful approach to gravity. It is what engineers use for most of their calculations. The exception are those rocket scientists that bring on board atomic clocks. Then the flaw in Newton's approach - that gravity works instantaneously - becomes apparent.

Over time, I have come to appreciate the sophistication of general relativity. I particularly like the Hilbert action, which only needs the Ricci scalar to do its magic. Yet more than eighty years of trying to quantize the approach have come up short.

Our very best theory today for gravity is a rank 2 field equation. Our most useful theory is rank 0. Goldielocks would wonder, what about a rank 1 field theory for gravity? This turns out to be a subtle theory in 4D.

Surely we have a nice, solid reason why a rank 1 field theory is dead on arrival. Someone told me to look up Clifford Will's Living Review article on tests of GR, that I would find the reason there. In the third section, a pure 4-vector theory was not mentioned. Go to his green book, p. 19, a vector theory by Kustaanheimo, which was not published (p. 360).

I found two papers in the 1950's, one by Gupta, the other by Thirring, that said it was not possible to make a rank 1 field theory where like charges attract. This was an assertion, it was not proved. If you were to copy the approach to the Maxwell equations keeping all the signs the same, then yes, like charges repel. If one flips the sign of the charge coupling term, both the force equation which results from varying the action relative to the velocity, and the field equations which result from varying the action with respect to the potential, have like charges attract. A well-placed sign change can work wonders.

Earlier in the Spring, I found another fun problem concerning the spin of the mediating particle for gravity. The graviton has to be spin 2 to bend light. In my action, I have a symmetric, rank 2 field strength tensor, a natural place for a spin 2 particle to call home. If you read chapter 3 of Feynman's "Lectures on Gravitation", he shows how to spot a spin 1 particle in the current coupling term. At a critical point, he multiplies to different currents together, J' J* (p. 34). Looking at the resulting phase, it will take 2 pi to come back around, as we expect of a spin 1 particle. What I realized was if a slightly different current product was used, J' (iJi)*, then the resulting phase will return home in pi, the signal that a spin 2 particle can live in the current coupling term (p. 39).

The immediate reaction to the equation on the t-shirt is to dismiss it as the Maxwell equations written in the Lorenz gauge (at least I care enough about details to know the difference between the Danish Ludwig Lorenz and the more famous Dutch Hendrik A. Lorentz). If I had meant the d'Alembertian operator, I would only need the box, not a box squared. The box squared is suppose to symbolize two covariant 4-derivatives, one taken after the other. A covariant derivative has a normal derivative and a connection, which is the derivative of a metric. Take two in a row, and there will be a derivative of a connection, which is a second order differential of a metric. The shirt has a second order differential equation that can determine the metric, so the metric does not have to be provided as part of the background mathematical structure.

I have this speculation that no one of your station in life will find this body of work of interest until they do a simple calculation: calculate the Christoffel of the Rosen metric. I prefer to call that metric the exponential metric for obvious reasons: static exponentials live on the diagonal. Only one of the three metric derivatives matter, and the contravariant exponential walks into its covariant cousin leaving a charge/distance that would look familiar to Poisson himself. Math magic like that means something.

I'll stop here, although there is much more. I've done my responsibility. I have no doubts you are overwhelmed with the standard pace of life. This is a low-odds, high rewards game. At least you have a t-shirt out of it. Feel free to ask any questions.

doug

[sweetser]

Hello

As it was presented, GEM looks like the fish none of the other fishermen saw. Since this cast of casters is made up of all the brightest kids from all the brightest classrooms on the planet, that doesn't sound likely.

Work in 11 dimensions - oh my! - that will impress the unknowning crowd. Say it is so tiny no one will ever see the stuff claim is amazing!!! Might as well use three bangs since it will forever be beyond reach.

Let me tell you how I dreamed up this old dream. I was preparing to go to the Second Meeting on Quaternionic Structures in Mathematics and Physics, in Rome, September 1999. As usual, I had to pay my way, but me and my traveling partner, Prof. Guido Sandri from BU added a week on to spend time looking at the art that has been gathering at the birthplace of western civilization. I was making up my transparencies for my talk which was how to write the Maxwell equations using only real-valued quaternions. It turns out to be easy to write the Maxwell equations with complex-valued quaternions, which are not a division algebra, and therefore of no more interest than any other arbitrary Clifford algebra. Making things work with real quaternions, that was a trick. James Clerk Maxwell himself speculated that someday someone would be able to do the trick. Knowing I accomplished something Maxwell himself wanted will always be one of my more ridiculous achievements. I should be noted that a year earlier, Peter Jack, another non-professional figured out all the hoops that have to be passed through.

The homogeneous equations:

(\frac{\partial}{\partial t}, \nabla)((\frac{\partial}{\partial t}, \nabla)(, A) - (\phi, A)(\frac{\partial}{\partial t}, \nabla)) + ((\frac{\partial}{\partial t}, \nabla)(\phi, A) - (\phi, A)(\frac{\partial}{\partial t}, \nabla))(\frac{\partial}{\partial t}, \nabla)
+(\frac{\partial}{\partial t}, \nabla)((\frac{\partial}{\partial t}, \nabla)^*(\phi, A)^* + (\phi, A)^*(\frac{\partial}{\partial t}, \nabla)^*) - ((\frac{\partial}{\partial t}, \nabla)^*(\phi, A)^* + (\phi, A)^*(\frac{\partial}{\partial t}, \nabla)^*)(\frac{\partial}{\partial t}, \nabla) = (0, 0)


The source equations:


(\frac{\partial}{\partial t}, \nabla)((\frac{\partial}{\partial t}, \nabla)(\phi, A) - (\phi, A)(\frac{\partial}{\partial t}, \nabla)) - ((\frac{\partial}{\partial t}, \nabla)(\phi, A) - (\phi, A)(\frac{\partial}{\partial t}, \nabla))(\frac{\partial}{\partial t}, \nabla)
-
(\frac{\partial}{\partial t}, \nabla)((\frac{\partial}{\partial t}, \nabla)^*(\phi, A)^* + (\phi, A)^*(\frac{\partial}{\partial t}, \nabla)^*) + ((\frac{\partial}{\partial t}, \nabla)^*(\phi, A)^* + (\phi, A)^*(\frac{\partial}{\partial t}, \nabla)^*)(\frac{\partial}{\partial t}, \nabla) = 4 \pi (\rho, J)


Impressively ugly. I can justify why it took me six months to find this particular combination of terms, and why Maxwell did not find them.

Now imagine me with a transparency and a sharpie, trying to come up with something I could be proud to travel several thousand miles to present to a half dozen people. This was my one result, and it was so bulky particularly in comparison to the Maxwell equations themselves. The technical struggle was to toss away just the right stuff.

But does Nature toss away anything? I felt the answer had to be a flat "no". At this point, I was focusing on EM. Why bother doing all this work to throw things away? It appeared to me that the EM equations liked things that were antisymmetric, and things that were symmetric were getting disposed of (which now makes sense, understanding the antisymmetric field strength tensor are the heart of EM). I recalled a quote at the start of one of the chapters of Misner, Thorne, and Wheeler dealt with symmetry. Those quotes are often the only part I understand, so I have read a decent fraction of them. I pulled the black phone book off the shelf, and started to hunt for the quote. Here it is, Chapter 17:

The physical world is represented as a four-dimensional continuum. If in this I adopt a Riemannian metric, and look for the simplest laws which such a metric can satisfy, I arrive at the relativistic gravitational theory of empty space. If I adopt in this space a vector field, or the antisymmetrical tensor field derived from it, and if I look for the simplest laws which such a field can satisfy, I arrive at the Maxwell equations for free space...at any given moment, out of all conceivable constructions, a single one has always proved itself absolutely superior to all the rest...


After reading that, I thought it was possible that if I did not toss away information, the field equations might be able to do the work of both gravity and EM. I traveled to Rome and had a grand time, even if no one in the group of six appeared that interested in the talk.

I had my unified field equations:



Jq^u - Jm^u = \square^2 A^u


This is when I had the meeting with Prof. Alan Guth (24, #355). He told me I needed to figure out the action, derive the field equations from the action, find solutions consistent with current tests and different for more refined ones. That several year march took me away from quaternions. There also was a strategic decision. I knew my intended audience is trained and comfortable with tensors. I rewrote it all to use tensors, and did not mention quaternions. I found it amusing that quaternions could play such a subtle role.

My experience is that theoretical physicists are frenetically busy. They are only comfortable getting involved in a discussion close to their area of expertise. Since no one works on rank 1 field theories for gravity as documented by the lack of coverage in the literature, there is no one to target.

The discussion here has convinced me that I have to return this proposal to its quaternion roots. The technical reasons have to do with discoveries made only in this calendar year. The first was to address Steve Carlip's complaint about a spin 2 particle in the current coupling term, which was dealt with in post p 22, #319 and #320. One cannot multiply one 4-tensor by another 4-tensor and get a third 4-tensor unless all the machinery for automorphic multiplication is there, which is set if one works with quaternions.

The second reason has to do with the weak and the strong forces. To this day, I still have not seen anyone write something like: "...and this force equation does a similar thing to Coulomb's law but for the weak force". There are excellent reasons for this, but still, it makes the weak and the strong forces feel unapproachable. My one handle on them is their symmetries. In particular, the weak force has the symmetry of SU(2), known as the unit quaternions (I am not making that up). If I formulate the GEM proposal in terms of quaternions, then where I can place a unit quaternion will be the symmetric house where the weak force can do its work. That is so direct and simple, odds are good that it is true.

The sexy math idea is that the properties of quaternions - when done right - dictate every fundamental aspect of Nature, bar none. There are four forces of Nature because when you consider the symmetries of two quaternions interacting:



(\frac{J}{|J|} exp(J - J^*))^* (\frac{J'}{|J'|} exp(J' - J'^*)) = 1 + \delta


this one expression has the symmetries of U(1), SU(2), SU(3) and Diff(M), EM, weak force, strong force and gravity respectively. The video is on YouTube because the new math cannot be shown in a PDF.

doug

[Mentz114]

Hi Doug, I hope you are going well.

I've been catching up in the arXiv and I wonder if you might be interested in these ( take a break from GEM ! ).

[40] arXiv:0708.1507 [ps, pdf, other]
Title: Quaternionic and Poisson-Lie structures in 3d gravity: the cosmological constant as deformation parameter
Authors: C Meusburger, B J Schroers
Comments: 34 pages
Subjects: General Relativity and Quantum Cosmology (gr-qc)

arXiv:0708.1154 (cross-list from hep-th) [ps, pdf, other]
Title: Antisymmetric-Tensor and Electromagnetic effects in an alpha'-non-perturbative Four-Dimensional String Cosmology
Authors: Jean Alexandre, Nick E. Mavromatos, Dylan Tanner
Subjects: High Energy Physics - Theory (hep-th)

[sweetser]

Hello Mentz:

The body is not doing so well. Looks like I have forgotten how to eat. I have diabetes, and one of the possible complications is to mess up neurons that signal things, such as "hey, let's empty the belly". I am working on living with gastroparesis. Life is always a challenge.

I did download both papers, and honestly understood about nothing. I was pleased to see a Lagrange density early one in one paper, but was unable to parse it.

GEM is in a good state. Personally, I am not aware of a "killer" problem with the proposal as it is today. There have been times when there were killer problems. Initially, the current for gravity had the same sign as the one for EM, and I got slapped for writing that one. Just this Spring I had the spin 2 current coupling scare that was avoided by the use of the first conjugate (iqi)* at just the right time.

doug

[sweetser]

Hello:

In this post I will start from the GEM unified field equation, and see what needs to be done to get to a quantum simple harmonic oscillator. This will establish a link between the GEM field equations and quantum mechanics.

A 4D homogeneous wave equation is the GEM vacuum equation:

(\frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 \phi, \frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 A) = 0

The 4D inhomogeneous equation is the source GEM equation:

(\frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 \phi, \frac{1}{c} \frac{\partial^2 phi}{\partial t^2} - c \nabla^2 A) = (\rho_q - \rho_m, J_q - J_m)

We can see that the scalar part of this equation can deal with Gauss' law of EM and Newton's law of gravity in a way consistent with special relativity since this equation in manifestly covariant under a Lorentz transformation (a fancy way of saying we know exactly how everything changes). The waves travel at the speed c. It is critical to note that this equation uses covariant derivatives, the sort with a connection that can work fine on a curved manifold. It is unfortunate that some skilled people see a d'Alemberian operator on a flat Minkowski background (it is there), but do not also see this expression can have a dynamic metric and work in a curved spacetime. I have shown earlier how the Rosen metric solves this 4D inhomogeneous equation for a static point charge.

The next step is to add in a term to convert the wave equation into a simple harmonic oscillator. One needs to subtract the norm of a factor that depends on R times the 4-potential. I'll write out the expression in its smallest parts, so you can notice the similarities:

(\frac{1}{c} \frac{\partial}{\partial t}, c \nabla)^*(\frac{\partial}{\partial t}, \nabla)(\phi, A) - (\epsilon, R)^*(\epsilon, R)(\phi, A)= (\rho_q - \rho_m, J_q - J_m)

Multiply things out.

(\frac{1}{c} \frac{\partial^2 \phi}{\partial t^2} + c \nabla^2 \phi - (\epsilon^2 + R^2) \phi, \frac{1}{c} \frac{\partial^2 A}{\partial t^2} + c \nabla^2 A - (\epsilon^2 + R^2) A) = (\rho_q - \rho_m, J_q - J_m)

To make this simpler, imagine that the potential in no way depends on time. A few terms drop:

(c \nabla^2 \phi - (\epsilon^2 + R^2) \phi, c \nabla^2 A - (\epsilon^2 + R^2) A) = (\rho_q - \rho_m, J_q - J_m)

What we now have looks almost like a quantum simple harmonic oscillator, as written here: quantum SHO (http://scienceworld.wolfram.com/physics/SimpleHarmonicOscillatorQuantumMechanical.html). The one difference is that the \epsilon is positive in that expression, and it is a squared negative here. Since that term contains energy, it may end up depending how positive and negative energy are defined.

The quantum simple harmonic oscillator is a particular class of the time-independent Schrödinger equation. I'll put up a quaternion derivation of the Schrödinger equation soon.

doug

[sweetser]

Hello:

I gave a talk this morning at an APS meeting titled "Seeing Spin 2 Symmetry in GEM, a Unified Field Theory", and already it is up on YouTube :smile:

http://www.youtube.com/watch?v=aeGgmE2wB5I

There were seven people in the room, and due to the nature of the meeting, they were all folks who work with atoms, molecules, and lasers, not gravity. I am using Keynote '08 for my presentations these days, and they have made it simpler to record and upload to YouTube. I did that at 11am on Friday night, and now three people have viewed the 15 minute talk, one of them giving it 5 stars. I like the global reach of YouTube! The number of folks that flock to a Saturday 8am meeting in Connecticut was limited exclusively to the other people giving talks and the moderator.

The video contains another presentation of the content in posts #319 and #320, but I got some new insights in preparing for the talk. Let me explain again what the problem is. There is a current coupling term in my vector unified field theory, J^{\mu}A_{\mu}. In Feynman's lecture book on gravity, chapter 3, he shows why a vector coupling term like J^{\mu}A_{\mu} has spin 1 symmetry, a great thing for EM because spin 1 particles dictate that like charges repel. That creates a HUGE problem for a unified field theory because for gravity, like charges attract. I think this is the key calculation why bright folks don't play with something like GEM.

The calculation that shows the current coupling term has spin 1 symmetry is correct as far as its assumptions go. The hidden assumption has to do with the orientation of the current and the 4-potential. You are free to work with a 4-potential pointing in any darn direction you want relative to the current J. The freedom to choose a different directions for the orientation of the potential means there might be different symmetries to the current coupling term. Look at it this way: J^{\mu}A_{\mu} evaluates to a scalar, and we cannot detect A_{\mu} directly (only how it changes), so we must consider different possible angles between the currents. In short, Feynman shows that J^* J' has spin 1 symmetry, and I show that (i J i)^* J' has both spin 1 and spin 2 symmetry. It is a great little calculation, and helps me understand why folks as bright as Feynman never stopped to take a serious look at a rank 1 field theory for gravity. The particle physics is wrong, until you recognized the freedom to orient the coupled currents differently to each other.

doug

[Graviman]

I feel humbled corresponding here. Your YouTube video is about the only leading edge physics on the internet that makes any sense! I tried to order your DVDs, but there seems to be something wrong with the PayPal setup.

I have studied Maxwell, Relativity, and Quantum with the Open University. I am intending to teach myself Differential Geometry, and will probably try to understand Dirac too. What else do i need to read to do your GEM principle justice?

Thanks in advance,

Mart

[Mentz114]

Hi Doug,

re-the APS talk. I don't know the details of the Feynman calculation, and I'd like to see how the Fourier transforming of the potential leads to the Grassman (?) product between the currents. From that point on, it looks reasonable. But if there's a degree of freedom to rotate one vector independently without affecting the coupling scalar - is it physical ?

I'd like to read what Feynman has to say on gravity - do you have a fuller reference ?

Regards,
Lut

[sweetser]

Hello Mart:

The humbling thing here is how cool Nature is. I work by making every possible mistake along the way, and learning from those mistakes, putting policies in place so I don't repeat a mistake.

Take for example relativity. People say they have studied relativity, they can solve problems, they "get it". All true, but how do you make sure you always work in a way that will be consistent with special relativity and the effects of gravity? [note to careful readers: I did not say general relativity since GEM is in technical conflict with GR, but both approaches must be able to explain gravity as the result of a dynamic metric]. My answer to the question is to always use quaternions which are just 4-vectors that one can also multiply or divide. Some folks might think that silly for problems in classical physics, but I say solving such problems in 4D is great practice: you accept the results, can believe them in a classical context, so when the same problem appears (same diff. eq.), the same approach will solve the relativistic problem. This comment is relevant to GEM, since the central equation for GEM is a 4D simple harmonic oscillator.

To learn the GEM proposal specifically, download and print these PDFs:

http://theworld.com/~sweetser/quaternions/ps/day1.pdf
http://theworld.com/~sweetser/quaternions/ps/day2.pdf
http://theworld.com/~sweetser/quaternions/ps/day3.pdf

Then you need to push a pencil. I do lots of calculations, always with a specific "start", always having 8 or fewer steps. It is important for you to be able to do those yourself, and repeat it until you can do it quickly (that is a sign that the brain does really get it). Redraw the little graphics that accompany the slides. The little graphics represent some of the information contained in the algebra, but the drawing will work a different part of the brain than the algebra, so you will get a wider grip on what is going on with this obtuse stuff. I toss around the phrases symmetric and antisymmetric tensors a bit, but I also have a small doodle that gives me a visualization on the issue. It has black and red squares, could be art, but is physics.

If you understand all the calculations in those notes from 2003, that is the foundation of this proposal. The YouTube lecture on spin 2 is an important technical issue, but could not be a complete explanation of the proposal, which probably would run on the order of five hours (the Maxwell equations are not simple, and the necessary bridges to gravity takes time to develop).

Understanding the GEM proposal take a real block of time and effort. If at any point, you get stuck, I want to know about it.

doug

[sweetser]

Hello Lut:

Here is the reference:

http://www.amazon.com/Feynman-Lectures-Gravitation-Richard-Phillips/dp/0813340381/ref=sr_1_1/105-6806776-8142048?ie=UTF8&s=books&qid=1193071871&sr=8-1]
Feynman Lectures on Gravitation

I liked Brian Hatfield's introduction. I have not read the entire book, instead focusing on chapter 3, particularly pages 34-38. I used to be able to "look inside" on Amazon at those pages, but it is not working for me now.

You bring up a most excellent point, one I have been thinking about. EM is a gauge invariant theory, GEM is not. With EM, you can do a transformation like so:

A \rightarrow A' = A - \nabla \phi

This new A' points in a different direction. That is a "legal move" in EM, not GEM, an allowable gauge transformation. I am making this a bit more specific, by looking at a rotation of A:

A \rightarrow A' = A - \nabla \phi = (i A i)^*

The exercise is to compare the symmetry of the phase of A (spin 1) to A' (spin 1 and spin 2).

For GEM, I am not free to rotate A. Instead, I am free write A like so:

A = 1/2(A + (i A i)^*) + 1/2(A - (i A i)^*)

The 1/2(A + (i A i)*) has spin 1 symmetry, while 1/2(A - (i A i)*) has spin 2 symmetry in the phase. I have the freedom to represent the 4-potential in this way.

doug

[Mentz114]

Doug,

that looks promising.
The book is not available at the moment. I'm lost on the notation. Is your i from the vector basis i,j,k ? What is the '*' operator doing ?

Lut

[Mentz114]

I've been looking at the Lagrangian and your splitting of the potential. I've alway been worried about the hidden (implied) coupling between EM and grav currents.

If you write

A_g = \frac{1}{2}(A + L(\theta)A)
and
A_{em} = \frac{1}{2}(A - L(\theta)A)

Where L is a 3D spatial rotation, you can split the interaction term
into two
A_g^{\mu}J_{\mu} + A_{em}^{\mu}I_{\mu}, where J and I are the currents.
Now the coupling is explicit, and depends on theta. For me, this takes away one major problem with GEM.

The meaning of theta is a bit obscure, but as theta changes, energy moves between the spin 2 and the spin 1 fields.

[later] I've just realised that this changes the Lagrangian too much. Back to the scratch-pad.

[sweetser]

Hello Lut:

The "i" is from the 4-basis vector (e, i, j, k). The way I think about curved spacetime is none of these has the norm of one, although ei=ej=ek=1 (e gets smaller, i,j, and k get larger).

The * is the conjugate operator, so (e, i, j, k)* = (e, -i, -j, -k). On my flight, I plan on playing with two sorts of splits of the potential:

A_g = \frac{1}{2}(A + i A i) = (0, 0, A_y, A_z)
A_{EM} = \frac{1}{2}(A - i A i) = (\phi, A_x, 0, 0)

One thing I like about this is that it has two degrees of freedom for gravity and two degrees of freedom for EM. This is done for the current coupling term, but not the field strength tensor, because that splits the spin 1 and spin 2 fields based on the symmetry of the second rank tensors.

The second possibility was my initial suggestion:

A_g = \frac{1}{2}(A + (i A i)^*) = (\phi, 0, A_y, A_z)
A_{EM} = \frac{1}{2}(A - (i A i)^*) = (0, A_x, 0, 0)

Gravity has too much of the potential, EM to little, so I am skeptical of this one working out.

I had been ignoring the coupling term because it looked too simple to do anything with. Now there may be a way to separate the coupling of like charges that attract from the coupling for like charges that repel. Need to do some work with pencil and paper... And do my day job, and pack.

doug

[sweetser]

Hello:

I realize there is a logical inconsistency in my proposal, and therefore there is a technical problem that must be addressed or the GEM proposal is wrong.

I have discussed how the 4D wave has two transverse modes of emission for EM and two modes for gravity. Yet I have written many times about 2 separate 4-currents, J_q and J_m. Together, those 4-currents have eight degrees of freedom which is too many and thus is wrong. I must make due with one 4-current density.

There are three constraints on the current coupling term.

1. The coupling term must be spilt into two: one for gravity and another for EM since they have different properties in regards to the behavior of like charges.

2. The two coupling terms must have opposite signs, so for EM like charges repel and for gravity like charges attract.

3. The EM coupling term must have a phase with spin 1 symmetry, while the gravity coupling term has spin 2 symmetry.

Nothing like a fun riddle!

To make things easier, consider motion along the z axis. The 4-potential can be split along the the transverse part of the potential, Ax and Ay, from the scalar and longitudinal parts, phi and Az:

A = \frac{1}{2} (A - k A k) + \frac{1}{2} (A + k A k)

A = (0, A_x, A_y, 0) + (\phi, 0, 0, A_z)

If I wanted to generalize this, I could refer to the parallel versus the transverse parts of the potential, A_{tr} and A_{||}. Constraint one solved.

The end result of these games must be the Lorentz invariant contraction of the current and the potential as it appears in the GEM action:

-\rho \phi + J_x A_x + J_y A_y + J_z A_z

Let's get the EM part first. I will be treating the current and potential 4-vectors as quaternions equipped with the usual ijk kind of algebra, but keep only the scalar part since that is all the action needs. Just multiply the current and potential together for the transverse part:

scalar (J \frac{1}{2} (A - k A k)) = - J_x A_x - J_y A_y

Close, but not quite. One needs to include a minus sign, which is great, since it satisfies half of the second constraint. To be explicit:

scalar (- J \frac{1}{2} (A - k A k)) = J_x A_x + J_y A_y

Where there is a scalar, there is a vector, necessary to look at the symmetry of the phase. Fourier transform the 4-potential to a current:

vector (- J \frac{1}{2} (A - k A k)) = J_x J'_y - J_y J'_y

This satisfies half of constraint 3, not a surprise since that is standard EM analysis.

Now on to the gravity part. If we do exactly the same as before, we get exactly the same result:

scalar (J \frac{1}{2} (A + k A k)) = \rho \phi - J_z A_z

These have the wrong signs, and so we would have to toss in a minus sign and fail to meet constraint number 2. This would be good in the sense that I would finally be able to forget the GEM proposal. Yet I am too crafty for such a defeat! This is one scalar product, but not the only one. Let's take what I call the first conjugate, (i q i)^* = (-t, x, -y, -z). This will flip the sign of phi and Az:

scalar (J \frac{1}{2} (i (A + k A k)) i)^*
=scalar((\rho, J_x, J_y, J_z)(-\phi, 0 , 0, -J_z))
=-\rho \phi + J_z A_z

Now we have the correct signs for the scalar with a positive J A contraction. That completes the requirements for constraint number 2. Now look at the vector after the potential has been Fourier transformed into a current:

vector (J \frac{1}{2} (i (A + k A k)) i)^* =
=vector((\rho, J_x, J_y, J_z)(-\rho', 0, 0, -J'_z))
=(0, 0, 0, -\rho J'_z - J_z \rho')

The phase has spin 2 symmetry! These two will subtract together, so will take only pi radians to return.

The rain in London was not heavy enough to prevent my writing this note while waiting for the tickets to Wicked. I hope the math stands up to anyone's analysis.

doug

ps. I did get front row tickets, nice.

[Mentz114]

Hi Doug,

I look forward to seeing the new Lagangian. I've been looking at current conservation and have found the energy-momentum tensor of the EM fileld is (from Stephani )-

T^{mn} = g_{ak}F^{am}F^{nk}-\frac{1}{4}g^{mn}F^{pq}F^{ba}g_{ap}g_{bq}
The condition for energy conservation is -

(1) T^{mn}_n = -\frac{1}{c}F^{mn}j_n

The thing on the left is the divergence of T, as you'll recognise. Having found a nice new tensor problem I set it up in the calculator and tried to verify equation 1. The interesting thing is that I get expressions like
T^0 = B_x(\partial_zE_y-\partial_yE_z)
+B_y(\partial_xE_z-\partial_zE_x)
+B_z(\partial_yE_x-\partial_xE_y)
+E_z(\partial_xB_y-\partial_yB_x)
+E_x(\partial_yB_z-\partial_zB_y)
+E_y(\partial_zB_x-\partial_xB_z)

which goes to

T^0 = -B_x\dot{B_x}-B_y\dot{B_y}-B_z\dot{B_z}+E_x(\dot{E_x}+j_x)+E_y(\dot{E_y}+j_y) +E_z(\dot{E_z}+j_z)

In the source free case, this is zero because B and E are orthogonal and J = 0. Yippee. I'm not so sure what emerges if there is a current.

The other terms of T^n look like this -

T^1 = B_x\partial^i\dot{B_i}+E_x\partial^i\dot{E_i}-B_y\dot{E_z}-B_z\dot{E_y}+E_y\dot{B_z}+E_z\dot{B_y}

The first two terms look like divs of curls, but I'm not sure about the other four. They could be parts of a curl of a curl and so zero.

The reason I've shown all this detail is because it shows that the energy conservation conditions are actually Maxwell's equations.

This means that this analogy, and the use of the E-M tensor could show coupling of the current and potential as the equations we get by working out the divergence of the E-M tensor of the gravitational force tensor.

Watch this space.

[Mentz114]

Doug:
Some time later : I have realised that the terms I get for T_0 are not zero. Something wrong somewhere, back to the drawing board.

[Later]
For the source free case, with an EM field travelling in the z direction, we can say

T^0 = -B_y\dot{B_y}+E_x\dot{E_x}

which comes to

2kB_0E_0 - \omega (B_0^2+E_0^2) multiplied by a phase,

which may well be 0.

[Mentz114]

Hi Doug,
I've been re-reading your last post (#371) and disregarding my dislike of using the same potential and current for gravity and EM, it looks OK.

Back to the energy conservation. I did some pencil work and discovered that the divergence of the EM stress tensor actually hinges on this F^{\mu\nu}_{,\nu}, which is easily seen if you apply a differentiation operator to T^{\mu\nu} above.

If F^{\mu\nu}_{,\nu} = J^{\mu} then T^{\mu\nu}_{,\nu} goes to zero even if there's a current. This is a good thing in my opinion. So F^{\mu\nu}_{,\nu} is an interesting thing to calculate. For the EM anti-symmetric F the result looks like a 4-current. With A^{\mu} = ( \phi, \vec{A} )

J^0 = -\nabla^2\phi + \partial_{t} \vec{\nabla}.\vec{A}

J^1 = \partial_{t}E_x + \partial_{y}B_z + \partial_{z}B_y

Repeating the above with the symmetric field tensor

F^{\mu\nu} = \partial^{\mu}A^{\nu} + \partial^{\nu}A^{\mu}

I get,

F^{\mu\nu}_{,\nu} = \partial^{\mu}(\partial_{k}A^{k}) + \partial^{k}\partial_{k}A^{\mu} = J^{\mu}

[edit] I found the Tex 'Box' operator so is this -
F^{\mu\nu}_{,\nu} = \partial^{\mu}(\Box A) + \Box^2A = J^{\mu}


which looks like your field equations. Or is that an extra term ?

So, provided the conditions above are met, we have energy conservation in the field tensor. I'm not sure if this only works in Minkowski space where covariant differentiation commutes. The gravity is all in the potential, and the field equations tell us how to get the potential from the current. It looks too linear to be a theory of gravity without the coupling term, which maybe will tell us how the current changes the potential, which restores the non-linearity.

I might have a go at constructing a full stress tensor from the gravity part of the Lagrangian, then check the divergence for non-linearity.

[sweetser]

Hello Lut:

It is quite common to think that non-linearity is a requirement for a full theory of gravity. What is going on in my opinion is one of those linked circle of statements one sees in math, where one thing implies another thing implies yet another. They all hang together tough. One can even change the order of the links, and the logic is consistent. If you are familiar with that idea in general, here is how it applies for GR:

1. The field equations are rank 2
2. The gravity field can be a source of gravity
3. The non-interacting vacuum field equations are non-linear
4. No efforts to quantize a GR type of theory of gravity have succeeded

For GEM, there is a different collection:

1. The field equations are rank 1
2. The gravity field cannot be a source of gravity just like in EM
3. The non-interacting vacuum field equations are linear (interactions make the equations non-linear, and must be approximated using thngs like Feynman diagrams)
4. The completely relativistic approach to quantizing EM (known as Gupta/Bleuler) can be applied to GEM, but the scalar and longitudinal modes of emission are spin 2

I asked Clifford Will once if there was direct experimental data about non-linearity, and I think he said no (but worry that my memory read in what it wanted to hear, or at least filtered out whatever technical issues he raised. Damn brain). I know there are people who claim there is data for the non-linearity. What I recall Will saying is that so far, all of our best, most consistent theories for gravity have to date all been non-linear. Time for a change! It is just the non-interacting, vacuum equations that are linear.

doug

[CarlB]

For GEM, there is a different collection:

1. The field equations are rank 1
2. The gravity field cannot be a source of gravity just like in EM
3. The non-interacting vacuum field equations are linear (interactions make the equations non-linear, and must be approximated using thngs like Feynman diagrams)
4. The completely relativistic approach to quantizing EM (known as Gupta/Bleuler) can be applied to GEM, but the scalar and longitudinal modes of emission are spin 2

I agree completely with the first three. I think that they make complete sense from a parrticle theory point of view. Number 4 is unfortunately over my head.

By the way, have you given the question of the equations of motion any more thought?

Carl

[Mentz114]

Doug:

How did you derive your field equations ? Do you have any remark about the formula in my last post ?

I think that can be written ( now I've found 'Box' )
F^{\mu\nu}_{,\nu} = \partial^{\mu}(\Box A) + \Box^2A^{\mu} = J^{\mu}


Lut

[sweetser]

I'll give a try at answering both of these questions over the weekend.

doug

[sweetser]

Hello Lut:

The field equations were the first things I derived when I wrote the GEM Lagrangian. They were the test that I had the right Lagrangian. What I did was write the Lagrangian in terms of the components:

http://theworld.com/~sweetser/quaternions/talks/IAP_2/Images/L_gEM_no_indices.jpg

I used this definition of the field equations:

http://theworld.com/~sweetser/quaternions/talks/IAP_2/Images/Euler-Lagrange_equation.jpg

which if you write out all the components looks like so:

http://theworld.com/~sweetser/quaternions/talks/IAP_2/Images/Euler-Lagrange_no_indices.jpg

Although there are lots of terms, it is kind of simple, since everything ends up being positive. The field equations are generated by taking the derivative of the Lagrangian with respect to the sixteen possible combinations of taking the 4-derivative of the 4-potential. The result is this:

http://theworld.com/~sweetser/quaternions/talks/IAP_2/Images/Euler-Lagrange_applied.jpg

This looks like

\Box^2A^{\mu} = J^{\mu}

Notice that I am using the complete asymmetric tensor, not the symmetric term in isolation. I'll have to look at what the scalar is from the contraction of the symmetric tensor. That should be a fun calculation! In other words, I have not calculated the field equations just for gravity.

doug

[Mentz114]

Hi Doug,

thanks for posting all that. I can see you've used Euler-Lagrange with the
A^{\mu}'s as dynamical variables. I reckon the result adds up.

I've been finding out some interesting wrinkles about the EM Lagrangian from Itzykson&Zuber ( Warning; they write \Box for our \Box^2 )

Starting with

L = \frac{1}{4}F^2+j\cdot A

they get field equations

\Box^{2}A^{\mu} - \partial^{\mu}(\partial_{\nu}A^{\nu}) = j^{\mu}

and by applying the Lorentz condition \partial_{\nu}A^{\nu} = 0 the field equations become

\Box^{2} A^{\mu} = j^{\mu}

Now, under the Lorentz condition, only gauge transformations of the type

A^{\mu} \rightarrow A^{\mu} + \partial^{\mu}\phi where \Box\phi = 0 are allowed.

The point of this is to show how they lose the extra term from the field equations by restricting allowed gauge transformations.

Back to GEM. We have,

for EM field : \partial_{\nu} (\partial^{\nu}A^{\mu}-\partial^{\mu}A^{\nu}) = j_{EM}^{\mu}

for Grav field : \partial_{\nu} (\partial^{\nu}A^{\mu}+\partial^{\mu}A^{\nu}) = j_{G}^{\mu}

which when summed gives your original field equations. There is no need to apply the Lorentz condition - it has cancelled out.

What can it mean ? I'm off for food, more later.

Lut

[sweetser]

Hello Lut:

I haven't really integrated the difference in notation yet, but this looks cool. Although I started out with 2 4-currents, now there is only one. So a bit of simplification is:

For EM:
\partial_{\nu} (\partial^{\nu}A^{\mu}-\partial^{\mu}A^{\nu}) = j_{tr}^{\mu}
For Gravity:
\partial_{\nu} (\partial^{\nu}A^{\mu}+\partial^{\mu}A^{\nu}) = j_{||}^{\mu}
For GEM:
\partial_{\nu} \partial^{\nu}A^{\mu} = j^{\mu}

So far, I have missed the detail concerning the \partial_{\nu} \partial^{\mu}A^{\nu} factor.

doug

[sweetser]

Hello Lut:

Great question about the G field equations. The Lagrangian needed for this looks like so:

\mathcal{L}_G = \frac{1}{c} J^{\mu} (A_{\mu} - I_{||} A_{\mu} I_{||})
-\frac{1}{4c^{2}}(\nabla_{\mu}A_{\nu}+\nabla_{\nu}A _{\mu})(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})


Make sure you write out the coupling term in a way that splits out the transverse from the parallel parts. The parallel part is mediated by a spin 2 field where like charges attract.

Write this out in its component parts:

\mathcal{L}_G = -\frac{1}{c} \rho A_0 + \frac{1}{c}J_{||} A_{||}

+ \frac{1}{2c^2}( -(\frac{\partial \theta}{\partial t})^2 + (\frac{\partial A_x}{\partial x})^2 + (\frac{\partial A_y}{\partial y})^2 + (\frac{\partial A_z}{\partial z})^2

- (\frac{\partial A_x}{\partial t})^2 + 2 \frac{\partial A_x}{\partial t} \frac{\partial \theta}{\partial x} - (\frac{\partial \theta}{\partial x})^2

- (\frac{\partial A_y}{\partial t})^2 + 2 \frac{\partial A_y}{\partial t} \frac{\partial \theta}{\partial y} - (\frac{\partial \theta}{\partial y})^2

- (\frac{\partial A_z}{\partial t})^2 + 2 \frac{\partial A_z}{\partial t} \frac{\partial \theta}{\partial z} + (\frac{\partial \theta}{\partial z})^2

+ (\frac{\partial A_y}{\partial x})^2 + 2 \frac{\partial A_x}{\partial y} \frac{\partial A_y}{\partial x} + (\frac{\partial A_y}{\partial x})^2

+ (\frac{\partial A_z}{\partial x})^2 + 2 \frac{\partial A_x}{\partial z} \frac{\partial A_z}{\partial x} + (\frac{\partial A_x}{\partial z})^2

+ (\frac{\partial A_y}{\partial z})^2 + 2 \frac{\partial A_y}{\partial z} \frac{\partial A_z}{\partial y} + (\frac{\partial A_y}{\partial z})^2)

Wow, that's a lot of partial derivatives! Notice the simple pattern: either a term is squared with a 1/2 in front of it, or the term is mix, and has a coefficient of 1. When we start taking the derivatives of these things, the derivative of d \frac{1}{2}x^2/dx = x and d (x y)/dx = y. Now it is time to apply the Euler-Lagrange equation. First do the current, which is the derivative of the Lagrangian with respect to the potential:

\frac{\mathcal{L}_G}{\partial A^{\mu}} = -\frac{1}{c} \rho + \frac{1}{c}J_{||}}

The tougher part is taking the derivative with respect to all the fields. This part takes some practice, but it is cool if you can get it to all work out. Do this one guy at a time:

\frac{\partial}{\partial x^{\mu}}\frac{\mathcal{L}_G}{\partial \frac{\theta}{x^{\mu}}} = -\frac{\partial^2 \theta}{\partial t^2} + \frac{\partial^2 A}{\partial t \partial x} - \frac{\partial^2 \theta}{\partial x^2} + \frac{\partial^2 A}{\partial t \partial y} - \frac{\partial^2 \theta}{\partial y^2} + \frac{\partial^2 A}{\partial t \partial z} - \frac{\partial^2 \theta}{\partial z^2}

There are a few things to notice from this, the first of four field equations. The current density rho has the same negative sign as the second derivative terms of theta. That is good, because it is consistent with like charges attracting each other. We can rewrite pairs of terms in this kind of way:

+ \frac{\partial^2 A}{\partial t \partial x} - \frac{\partial^2 \theta}{\partial x^2} = + \frac{\partial}{\partial x}(\frac{\partial A}{\partial t} - \frac{\partial \theta}{\partial x})

Notice the difference in the sign between these two terms. For the E field, they are both minus signs. For gravity, they have different signs. Time to get the next field equation:

\frac{\partial}{\partial x^{\mu}}\frac{\mathcal{L}_G}{\partial \frac{A_x}{x^{\mu}}} = -\frac{\partial^2 A_x}{\partial t^2} + \frac{\partial^2 \theta}{\partial t \partial x} + \frac{\partial^2 A_x}{\partial x^2} - \frac{\partial^2 A_x}{\partial y^2} - \frac{\partial^2 A_y}{\partial x \partial y} - \frac{\partial^2 A_x}{\partial z^2} - \frac{\partial^2 A_z}{\partial x \partial z}

Here again we can rewrite a pair of these terms like so:

- \frac{\partial^2 A_x}{\partial y^2} - \frac{\partial^2 A_y}{\partial x \partial y} = - \frac{\partial}{\partial y}(\frac{\partial A_x}{\partial y} + \frac{\partial A_y}{\partial x})

This is "curl-like" in the way the different derivatives swap places, but it certainly is not the curl since both have the same sign. There is also the term + \frac{\partial^2 A_x}{\partial x^2} which is probably the "Lorentz gauge canceling" term.

I did this all first with pencil and eraser. It might be a good exercise in 19th century math to see if you can spot mistakes (it is hard to proof in LaTeX format). Kind of amazing to work with this many partial differential equations, and have it make sense. It does look consistent with Lut's statements based on tensor analysis.

doug

[Mentz114]

Hi Doug:

on first reading this, I recognise those not curl-like terms. I've calculated currents and field-tensor contractions for potential A^m and A^m+phi^m to get the extra terms, and check if they vanish like divergences. I'm struggling to make sense of the extra terms, but I'll post them later so you can have a go at solving the puzzle, or even checking the calculation ( Mathematica ?).

Later ...

Lut

[Mentz114]

Hi Doug,

I have to say that I don't think a current can be represented with 2 degrees of freedom, so I have some doubts about the one-current proposal. I've been thinking about stress tensors instead of currents.

I'm intrigued by the cancellation of the extra terms from the field equations when EM and gravity are included. This could mean something or just be an mathematical accident.

Another thing is that if the potential is augmented thus -

A^{\mu} \rightarrow A^{\mu} + \phi^{\mu} where \phi^{\mu} = \partial^{\mu}\Phi

the field equations \Box^{2}A^{\mu} = j^{\mu} have an extra current

2\partial_{\nu} \partial^{\nu}(\partial_{\mu}\phi^{\mu})

which means we can add any potential as long as the above is zero.

But the important thing is to check the Lagrangian when the gauge transformation is done. After much effort I've ended up with 32 terms which I've emailed to you. It would be good to reduce them, and even to check the calculation, if Mathematica is up to it !

My software takes about 4 minutes to run the script, but with no simplification. Some of the extra terms will be divergences and will go to zero when integrated. The remainder are extra conditions to ensure energy conservation. If they cannot be met physically, the action is not gauge invariant. In EM this is bad thing, but whether it has any significance in the GEM context, I don't know yet.

Check your email...

Lut

[sweetser]

Hello Lut:

Looks like we have a few technical disagreements based on your last post. Let's look at them seperately.

> I have to say that I don't think a current can be represented with 2 degrees of freedom, so I have some doubts about the one-current proposal.

Let me say it this way: a 4-current has 4 degrees of freedom, and a current-current interaction has two types of symmetry, both spin 1 and spin 2 symmetry. The two sorts of symmetry seen in the contraction of a current with a potential comes from the ability to arbitrarily point the potential relative to the current density it represents. Part of the potential will be parallel to the current it represents, part of it will be transverse. If you split the current into these two parts, that algebraic act by itself changes nothing. The phase of the parallel (||) and transverse (tr) potentials are:

J A_{||}->J FT[A_{||}] = J J' = \rho \vec{J'} - \vec{J} \rho'

J A_{tr}->J FT[A_{tr}] = J J' = \rho \vec{J'} - \vec{J} \rho'

The phase of these two contractions are the same, spin 1 symmetry.

What is needed is another twist of the potential for the transverse current:

J (I_{tr} A_{||} I_{tr})^*->J FT[(I_{tr} A_{||} I_{tr})^*] = -J J' = - \rho \vec{J'} - \vec{J} \rho'

Notice those minus sign! When I did this work initially, I had to work with 2 4-currents, and put in the sign difference by hand. Now I get to work with 1 4-current with 4-degrees of freedom, and by twisting the orientation of the 4-potential, get to find phases with spin 1 and spin 2 symmetry and two signs.

Onward...

This gauge transformation:

A^{\mu} \rightarrow A^{\mu} + \phi^{\mu} where \phi^{\mu} = \partial^{\mu}\Phi

is not allowed in the GEM proposal. The reason is there are two fields, E for EM, and e for gravity, and this transformation will mess up one or the other. The fields are defined like so:

E = - \frac{\partial A}{\partial t} - \nabla \phi

e = + \frac{\partial A}{\partial t} - \nabla \phi

If I were to work only with E, there is a gauge transformation of the sort you refer to. If I were to work only with e, there is also such a gauge transformation. But these two gauge transformations are not the same.

The title here refers to an unusual way to think about gauge symmetry. Gauge means "to measure". You are exploring the usual use of gauge symmetry, the ability to add in an arbitrary derivative of a scalar field. When I look at the GEM action, and see the covariant derivative, I think of gauge symmetry as the ability to measure the action as the result of a potential in flat spacetime, or as the curvature of the spacetime manifold with a dull potential, or any combination in between.

I will work on the gauge symmetry, to make certain the kind of gauge transformation you suggest is not allowed for the full GEM proposal.

doug

[Mentz114]

Hi Doug :

I thought that

e = + \frac{\partial A}{\partial t} + \nabla \phi ?

I assume that A^{\mu} = (\phi, \vec{A}).

I can see your point, I think.

The only reason gauge transformations are of interest is to see if energy is conserved, and I think it is because the gauge current can be eliminated with the field condition.

Regarding the the current business. You can split the 4 degrees of freedom into 2 orthogonal vectors but you have an electrical and a matter current to describe, and only 2 df each. Or do I misunderstand ?

Lut

[sweetser]

Hello Lut:

I appreciate the importance of having an explicit way to show that energy is conserved. One reason I am not worried that we will eventually find it is that there is no "t" in the Lagrangian, so one can vary t without changing the action. A symmetry like that means that there is a conserved quantity, and it turns out to be energy.

There is no t in the action of GR either. Energy is conserved in GR, but it is not localized, one needing a volume of spacetime before one can say energy is conserved. The nonlocalization happens because one varies the metric field.

> Regarding the the current business. You can split the 4 degrees of freedom into 2 orthogonal vectors but you have an electrical and a matter current to describe, and only 2 df each. Or do I misunderstand ?

That sounds right. Both the graviton and photon travel at the speed of light, and thus only need 2 degrees of freedom each to describe a mediating particle where like attract (gravity) and like repel (EM).

doug

[sweetser]

Hello:

I was given an optional assignment by Lut to read a paper: "Field Theory of Gravitation: Desire and Reality" by Yurij V. Baryshev, arxiv:gr-qc/9912003. It was a fun read, and I will give you the GEM spin on the article.

Baryshev sees two types of theories out there in the literature: field theories with flat metric backgrounds and spacetime curvature theories. Nearly all the work done with gravity since Einstein has been spacetime curvature. The few papers on field theories bow to the powers that be, and say they are effectively the same thing as a geometric approach. Baryshev points out that while this might be true for weak gravity fields, the treatment of black holes will necessarily be quite different. He argues that there will be no black holes in a field theory approach, even though there is a Disney movie about such hellish places in the Universe. Einstein and Eddington would be pleased, but not the folks who churn out black hole papers, or model the collapse of black holes.

In Baryshev's hands, the argument against black holes is simple: the energy is positive definite, and built to stay that way by energy conservation (see section 5.3 for details, since I still don't completely get it).

With my GEM bias, I enjoyed the clarity with which he said there were only two types of proposals in the literature. I view GEM as being a third approach, exactly between a field theory with a flat metric and a dynamic spacetime connection (notice: I avoided the word curvature and the Riemann tensor on purpose, using the connection which is a derivative of the metric instead).

GEM does work as a pure potential theory with a flat metric. According to Baryshev, that is a great thing because it means energy conservation will work as a local effect, which is good because that is the way energy works for all other field theories we understand well. I have calculated the T^{00} of the GEM proposal before, but want to redo the exercise using only quaternions (my idea of challenging fun).

If GEM only worked with a flat metric, folks like Prof. John Baez would dismiss it with a one liner (he did actually, this very complaint of working only with a fixed background metric like all other field theories to date).

Actually, based on my education in GR, the thought of working with a flat metric did not occur to me (strange, but true). I learned special and general relativity from Edwin F. Taylor, the coauthor with Wheeler of "Spacetime Physics". That book was devoted to special relativity. They also wrote another book together, "Exploring Black Holes: Introduction to General Relativity". This book was all about playing with the Schwarzschild metric of GR. They did not derive the Schwarzschild metric. They avoided any technical discussions about connections and Riemann curvature tensors. Instead, the book looked at all the consequence of the metric.

In August of 1999, I had a field equation with a 4-potential in it. From my training, I had to find a way to a metric. I was not sure I could find such a path, but gave myself the time to look for one. I don't emphasis the first road I found often, fearing the potholes. What I did was to work with a force equation, F^{\mu} = J_{\nu} d^{\mu} A^{\nu}. I found a solution to that equation which was a 4-velocity, c^2 = (k_1 c \frac{dt}{d \tau}, \vec{K_2} \frac{d \vec{R}}{d \tau}). I squared this result, using the condition that for flat spacetime I should get a flat Minkowski metric to eliminate the constants k1 and K2. This lead to the exponential metric written in the first post of this long thread.

I derived a metric equation in a way no one else has done. Being skeptical of myself means I was not sure such a derivation was right. To the point of being a bore, I repeat that the exponential metric can be tested against the Schwarzschild metric, using the same experiment that was used to rule GR beats Newton: look at light bending around the Sun. In 1919, that was a big technical challenge. The data was not as clear cut as the public was led to believe. In time, we have solidified the GR victory.

The playing field for GR versus GEM involves a million fold improvement in the light bending measurement. We have a ten thousand fold improvement, so a few orders of magnitude refinement are required. Even so, the difference amounts to 12%. At the microarcsecond level, the rotation of the Sun, and the quadrapole moment also have to be accounted for.

If the only path to the testable exponential metric was the funky chicken walk through the force equation, I would not have the confidence that I am "within spitting distance of a unified field theory". I told the story earlier in this thread of finding a differential equation I had to solve using the Christoffel symbol of the exponential metric. For three weeks I avoided doing the calculation because I had never done a calculation with a Christoffel, and was unsure I could figure it out. The exponential metric did solve the field equation.

In all my reading on gravity, I have never seen another approach that got to a metric in the ways I have. So it goes.

doug

[Mentz114]

Hi Doug:
With my GEM bias, I enjoyed the clarity with which he said there were only two types of proposals in the literature. I view GEM as being a third approach, exactly between a field theory with a flat metric and a dynamic spacetime connection (notice: I avoided the word curvature and the Riemann tensor on purpose, using the connection which is a derivative of the metric instead).
You could be right. But one can always transform away the first derivatives of the metric. In GEM you'll have to make the action in the connection go to the potential to preserve an energy density.

I came across this today in a PF thread
As a funny aside, it's a theorem in QFT that the ONLY self-consistent theories with spin-1 bosons are gauge theories,
and the analogous theorems get even stronger as you go to higher spin. For example, the ONLY self-consistent theory of
(massless) spin-2 bosons is Einstein's Gravity!

It is a rather wonderful result (proved by Dick Feynman many years ago, if I remember correctly) that the only way to
write down a self-consistent theory of a *massless, spin-2 * object is Einstein's GR. There is simply no other way to
do it! The proof is quite non-trivial, but it's a famous result. Of course, there are some important assumptions, such
as Lorentz invariance, energy conservation, etc. But given some rather weak assumptions like that, then the only theory
that you can write down with massless spin-2 is GR. There is NOTHING at all "ad-hoc" about Einstein's Field Equations
(I'm assuming that's what "EFE" stands for). High-spin theories (s>1) are VERY constraining.

People are out to modify GR all the time, and that's okay. But almost always, the way they do it is to ADD something,
such as new scalar fields, extra dimensions, etc. But they always end up with Einstein PLUS stuff. You can never get
rid of Einstein entirely. Every model I know of that tries always has deep problems!


Is this discussed in your copy of the Feynman's gravity lectures ?

I'm sure you'll say it's wrong in any case.

[aside]
I have been informed that the potential does not transform as a 4-vector and sorted out the boosting of the EM field tensor and the equivalent operation on the potential.
http://www.physicsforums.com/showthread.php?t=197942

Lut

[sweetser]

Hello Lut:

Understanding this line is central:

> But one can always transform away the first derivatives of the metric.

I have to agree with this, it is true, but I always have my odd slant, promise! When you transform away the first derivative, where does it go? I have a specific technical answer. In GEM, the first derivative of the metric lives in one and only one place - the covariant derivative:

\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}

All the first derivatives live in the Christoffel symbol of the second kind. In GEM, when you change the first derivative of the metric in any way for any amount, you must make a compensating change in the potential so the derivative of the potential exactly balances it out. The potential and metric are separate, but changes in the potential play with the first derivative of the metric. You are free to play with the first derivative of the metric. You are not free to change the covariant derivative of the potential.

On the aside...
There is a lot of truth in the aside. I've seen it in action, specifically the Rosen bimetric theory that has the exponential metric in it. The problem is that it does add something, a fixed background metric, so that is a place to store energy/momentum, and it predicts dipole moments for gravity waves. We have good data to say the lowest form of gravity wave emission is a quadrapole.

I am not trying to make a theory of EM. I am not trying to make a theory for gravity. I am trying to make a theory of EM and gravity. The two play together in the covariant derivative as discussed about. This is how I avoid the proofs, I claim they do not apply to the "and" situation.

I don't know if Feynman provides a discussion in his book. He is a rank 2 field theory guy, like every other smart person making their money doing research on gravity. I can only follow along before hopping off that train of thought.

Fun discussion you have going on there. The way I read it, both the differential element and the potential do transform like vectors, but each comes with its own technical issue. For the differential, the contravariant tensor, d^u, needs to have minus signs for the 3-vector. For the potential, you need to have the current-coupling term around to keep things honest. So one can either say they do not transform like 4-vectors because they have these extra rules, or they do transform like 4-vectors so long as these rules are applied. People will argue over which is right, even if the net result is identical.

doug

[Mentz114]

Hi Doug:

I have to agree with this, it is true, but I always have my odd slant, promise! When you transform away the first derivative, where does it go? I have a specific technical answer. In GEM, the first derivative of the metric lives in one and only one place - the covariant derivative:

\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}

All the first derivatives live in the Christoffel symbol of the second kind. In GEM, when you change the first derivative of the metric in any way for any amount, you must make a compensating change in the potential so the derivative of the potential exactly balances it out. The potential and metric are separate, but changes in the potential play with the first derivative of the metric. You are free to play with the first derivative of the metric. You are not free to change the covariant derivative of the potential.

Yes, you've made this clear in the past, thank you. I'm asking about the next step. Suppose I make a coordinate transformation X -> X' = f(X) which removes the Christoffel symbols - by what rule or process does the gradient of the potential A(X) change to compensate ?

Another thing that bothers me is the fact that if the connections are non-zero then the Lagrangian ( the energy ) depends on the value of A as well as its derivatives. This is new territory, because GEM is no longer a gauge theory in the normal sense. Obviously adding a local or global field Phi changes the total energy.

I sent you a dumb question by email - it's important for my understanding of GEM to know the answer.

Lut

[Mentz114]

Doug:
(This is highly speculative)
following on from above, if the Lagrangian depends on the field itself, like bA^2, then b is the mass of the field. So it seems that because the connections have no source, it's the gravitons themselves that cause the 'curvature'. The amount of action in the connections is related to the mass of the gravitons. Giving gravitons a small mass can be a good thing, because it reduces the range.

Lut

[sweetser]

Hello Lut:

Yes, you've made this clear in the past, thank you. I'm asking about the next step. Suppose I make a coordinate transformation X -> X' = f(X) which removes the Christoffel symbols - by what rule or process does the gradient of the potential A(X) change to compensate ?

Good question. I was thinking about this recently. The current coupling term, J^{\mu} A_{\mu} depends on the value of A. The field strength tensor term, \nabla^{\mu} A^{\nu} only depends on how A changes. We need both, but they are independent of each other. In other words, one can have a value of A remain the same, but the way A changes in a local region of spacetime can vary. Do I understand how to do the nuts and bolts of this? No. What have done is two extremes: work with only the potential in totally flat spacetime, and work with only the connection with a static potential. If I was amazingly good, I could work out all the points in between those two extremes.

My hope is that GEM is a gauge theory, but not in the normal, add this field like so, sense. Gauge means to measure. The symmetry is about measuring with the change in the potential or the first derivatives of the metric.

Gravitons must travel at the speed of light and be massless. This happens when the symmetric tensor has zero trace. What happens when the symmetric tensor has a trace? That forms a scalar field. I think this does the work of the Higgs particle. Mass breaks the gauge (in the usual, traceless field strength tensor sense) symmetry. The Higgs mechanism is unnecessary for a unified field theory like GEM. Sorry LHC, I am betting against you folks based on theory.

doug

[sweetser]

Hello:

I believe I have made a technical error in how I have represented the current coupling term for gravity, J_m^{\mu} A_{\mu}. There are two independent mistakes. First, it postulates two 4-current densities, J_m^{\mu} and J_q^{\mu}, with eight degrees of freedom. A 4D wave equation should have a source with only four degrees of freedom.

Second, anyone who understands Feynman's current-current coupling phase analysis could reasonably say J_m^{\mu} A_{\mu} can only represent a spin 1 interaction, where like charges repel.

In any and all documentation on GEM under my control, I will strive to rewrite the coupling term like so:

From:
-\frac{1}{c} J_q^{\mu} A_{\mu} +\frac{1}{c} J_m^{\mu} A_{\mu}

To:
-\frac{1}{2c} J^{\mu}(A_{\mu} + I_{||}A_{\mu}I_{||}) + \frac{1}{2c} J^{\mu}(I_{tr}(A_{\mu} - I_{||}A_{\mu}I_{||})I_{tr})^*

Einstein said one should try to be simple, but not too simple. My initial coupling term was too simple. The complication is required because this scalar is formed from a 4-potential that cannot be measured directly, yet can be assigned to point in an arbitrary direction. The potential is connected to a real current, so the part of the potential that is parallel to the current can be separated from that which is transverse.

One good thing about this change is that the scalar formed from the contraction has not changed at all. It is the representation of the scalar that is getting more verbose. The phase of the second current-current interaction is changed by the two rotations around the parallel and transverse axes. The sign of two terms in the phase are the same, so they assist each other, and will require pi radians to return. This indicates that the second term has spin 2 phase symmetry.

Since there is one 4-current density, there are four degrees of freedom. The spin 1 photons are the transverse potential that form the transverse waves of EM. The spin 2 gravitons are the scalar and longitudinal potential that form the scalar and longitudinal modes of GEM (unlike GR where the graviton is transverse).

My primary short-term goal will be to correct the proposal. I dislike wasting anyone's time with faults I understand and can correct with a large volume of grunge work.

doug

[Mentz114]

Doug:

The current coupling term, J^{\mu} A_{\mu} depends on the value of A. Yes, and to get rid of it you have to apply the Lorentz gauge. But you can't gauge away the mass terms ...

The field strength tensor term, \nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma} only depends on how A changes.
Do my eyes deceive me ? I can see A in that covariant derivative.

Gravitons must travel at the speed of light and be massless.

Must ? This sounds like a statement of faith. Please see the following reference and the many others on the subject of graviton mass. The experimental evidence is ambiguous, and probably always will be. We cannot experimentally verify with any confidence whether gravitons have any mass. Similarly, we cannot test if gravitational waves travel at c or slower.

Some references -

http://www.iop.org/EJ/article/0264-9381/20/6/101/q306l1.html

Abstract. We show that the graviton has a mass in an anti-de Sitter background given by This is precisely the fine-tuning value required for the perturbed gravitational field tomaintain its two degrees of freedom.


arXiv.org:gr-qc/9705051

Mass for the graviton
Authors: Visser, Matt
Can we give the graviton a mass? Does it even make sense to speak of a massive graviton? In this essay I shall answer these questions in the affirmative. I shall outline an alternative to Einstein Gravity that satisfies the Equivalence Principle and automatically passes all classical weak-field tests (GM/r approx 10-6). It also passes medium-field tests (GM/r approx 1/5), but exhibits radically different strong-field behaviour (GM/r approx 1). Black holes in the usual sense do not exist in this theory, and large-scale cosmology is divorced from the distribution of matter. To do all this we have to sacrifice something: the theory exhibits *prior geometry*, and depends on a non-dynamical background metric.

http://www.journals.uchicago.edu/cgi-bin/resolve?id=doi:10.1086/427773&erFrom=-2247456991476005095Guest

http://digitalphysics.org/Publications/Ostoma-Trushyk/EMQG/

http://www.springerlink.com/content/x52n3827118uk012/


Lut

[edit]re-reading this, it sounds a bit tetchy. I apologise. All done rather quickly in between other things. I'm sure you'll find a way to wriggle out as you always do !

Lut

[sweetser]

Hello Lut:

Let me separate the questions. This post is only about this question:

Do my eyes deceive me ? I can see A in that covariant derivative.

As I recall learning about this topic, they tell you first that the contravariant 4-derivative has a bunch of minus signs:

IS A TENSOR
\partial^{\mu} = (\partial/\partial t, -\partial/\partial x, -\partial/\partial y, -\partial/\partial z)

I was like, that is irritating, because I had learned the up index means everything is positive. This is the exception to that rule. Exceptions have consequences. Specifically, the 4-derivative of a 4-potential does not transform like a tensor:

NOT A TENSOR
\partial^{\mu} A^{\mu}

They usually go through a transform exercise which I rarely can follow. I just accept it, a little easier now with the "exceptions have consequences" idea. The connection has one job, make the derivative transform like a tensor:

IS A TENSOR
\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}

The connection has 3 indices, the covariant derivative has 2, so it must be contracted with the potential. Then they point out the obvious:

NOT A TENSOR:
\Gamma_{\sigma}^{\mu \nu}

Although it is not a tensor, the connection gets separated from its home inside a covariant derivative snuggled right against the potential. I consider this very bad practice that is quite common. One never sees potentials in GR papers, unless they wish to point out their failings. The next goal is to form the simplest tensor out of a connection that transforms like a tensor. That quest has one answer, the Riemann curvature tensor. It is announced that this is a great thing, because that has a derivative of a connection inside a tensor, and so a second derivative of a metric. Such pride strikes me as silly. Take two covariant derivatives in a row, and one will also get the derivative of a connection. The difference is that two covariant derivatives in a row leaves you with exactly 1 derivative of a connection. Energy can be local, whether it is in the derivatives of the metric or the potential. The Riemann curvature tensor has the difference of two derivatives of a connection, the source of being "non-local", and even worse, can be transformed away entirely. That cannot happen with the two covariant derivatives in a row approach, because the energy shifts to the potential.

doug

[Mentz114]

Doug:
The point is not whether the connection is a tensor, it's that the terms A^2 appear in the Lagrangian. I'll work this out explicitly when I have time.

Lut

PS.
The Riemann curvature tensor has the difference of two derivatives of a connection, the source of being "non-local", and even worse, can be transformed away entirely. I don't think the Riemann tensor can be transformed away, because it has second derivatives of the metric and represents tidal fields.

[sweetser]

Yup, I was wrong, sorry. At any point in spacetime, the connection can be made zero, but you cannot make a bunch of points zero if there is a gravitational source around.

doug

[Mentz114]

Hi Doug:

I noticed that because
\Gamma ^{\nu\sigma}_{ \mu} = g_{a\mu}g^{b\nu}g^{c\sigma}\Gamma ^{a}_{bc}
there will be actual metric terms in the Lagrangian also.

Lut
(it's a sunny day and I'm feeling better...)

[sweetser]

Hello Lut:

I like simplicity and symmetry, and a really tight logical story. I think I get all of those with a massless graviton.

Let's start with the photon and EM. That fellow is massless, and as such cruises at the speed of light, which seams appropriate given that it is light. When folks go to quantize the field, they only need two of four modes of emission.

I am trying to unify gravity and EM by putting the graviton right in the mix. Like there is an antisymmetric E field, there is a symmetric e. Like there is a symmetric B field, there is a symmetric b. These are made up of derivatives of exactly the same things, just different signs.

New Idea

I was driving back from burning a few DVD of the Stand-Up Physicist when I ask myself a new question: what is the exact relationship between gravitons and photons. I am of the opinion that the fundamental currency of the universe is an event that can be resented may ways, the most efficient involving quaternions because the tool remains the same no matter what sort of physics is being described. I view a photon as having 2 degrees of freedom, and a graviton as having two degrees of freedom, and because 2+2=4, together they have four. Now that I have one current 4-vector, the new idea is that I need to work with one force mediating particle that can have more than one kind of spin. Odd, really odd. I am fighting it, but that is a good sign. We don't understand how any fundamental particles really work. We have taken the simplest approach and put each one in a completely separate box. Yet we know particles can transform from one form to another. We have all the math machinery to describe an super-energetic photon turning into an electron-positron pair. How Nature would make gravity so much weaker than EM inside the same force particle is a huge mystery.

Have fun kicking around this unification-at-the-particle level idea.
doug

Sunny in London, it can be done. Cold and rainy here.

[Mentz114]

Hi Doug:

Sure it would be tidy to assume a massless graviton. But you may have a problem because the grav field tensor is only traceless in Minkowski space, if
\partial_{x}A^{1}+\partial_{y}A^{2}+\partial_{z}A^ {3}-\partial_{t}A^{0} = 0.

But with a general solution is this still true ? The problem breaks my software, it's just too big to handle. Shifting the indexes up and down multiplies the number of terms by 4 for each index. I'll work out a shorthand or somehing and try to get g_{\mu\mu}F^{\mu\mu} worked out in the general case.

[later edit]
I've got a general expression for g_{\mu\mu}F^{\mu\mu} and it's not as bad as I thought. Have you ever calculated the trace of F (symmetric) in the general case ?

Your new idea is interesting. Attached is an excerpt from the Baryshev paper. Their potential has spin-2, spin-1 and spin-0 components.

[sweetser]

Hello Lut:

I am trying to have it both ways. For an infinite range, 1/R^2 force law, I need a massless particle for light, I need one for gravity. They must in in their own separate way be gauge theory in the traceless field strength tensor way.

I am proposing that gauge symmetry is not broken by a false vacuum, but by mass itself (inertial mass being exactly the same thing as gravitational mass). How this gauge-braking fellow interacts with the massless force particles at this time is something I do not understand.

doug

[Mentz114]

Hi Doug:

The 1/r^2 force law is a weak field approximation. Otherwise we have infinities. I thought the whole point of quantum gravity is to find the actual 'force' law.

Addressing the issue of the trace of the symmetric field tensor. In Minkowski space the trace is \partial_{x}A^{1}+\partial_{y}A^{2}+\partial_{z}A^ {3}-\partial_{t}A^{0} which is only zero if you apply a wave condition on A.

When the metric is general it is easy to calculate the trace exactly in terms of Christoffel symbols like this ( I'm starting with the covariant field tensor because it is much simpler) -

F_{mn} = \partial_mA_n+\partial_nA_m - 2\Gamma^{k}_{mn}A_{k}

and the trace is
F^{m}_{m} = g^{mm}F_{mm} = 2g^{mm}\partial_mA_m - 2g^{mm}\Gamma^{k}_{mm}A_{k}

Looking at the second term, we find the coefficient of A_0 is the sum

g^{00}\Gamma^{0}_{00}+g^{11}\Gamma^{0}_{11}+g^{22} \Gamma^{0}_{22}+g^{33}\Gamma^{0}_{33}
and similar for the other components of A.

I doubt if that sum is zero. I'll try writing it out in full but it doesn't look likely it's zero.

Serieusly, you must demonstrate the tracelessness of the field tensor or admit your graviton has mass. In my opinion that would make it a more realistic theory, but I know you want simpliciry and logic - as if that exists.

You don't seem too bothered by the fact that the contraction of the symmetric field tensor in the Lagrangian contains terms that involve the metric directly, and the potential directly. Not gradients. That changes the character of the Lagrangian a lot.

You'll need to demonstrate that losing gauge invariance does not affect current/energy conservation. That all balances out in flat space, but it's rather difficult to calculate in curved space-time.

Have a good w/e, I'm off so I won't be posting for a couple of days.

Lut

[dlgoff]

I've been following your thread since its begining. Wanting to see a new theory in the making.

I can only get an idea of what you are doing from the mathamatics since I never studied much more than diff. equations and vector calc. But somehow I think you are on to something when you say "Now that I have one current 4-vector, the new idea is that I need to work with one force mediating particle that can have more than one kind of spin." Could "graviton part" have a frequency like that of the "photon part"?

Thanks for your efforts

[sweetser]

Hello dlgoff:

If you want some help learning the math stuff, spend some time with the videos listed in post 389. I call it "hard TV", not because of excessive violence or sex, but for excessive math. If you do put in the investment to watch the videos, please do so with pencil in hand, and repeat any of the derivations. Preparing the videos helped me big time, particularly the small drawings at the bottom of slides. Nothing like a visual hook for a nasty bit of algebra!

Now to the physics question...

I don't think the gravitational part with have the same frequency EM part. The reason is that the gravitational part comes from this irreducible symmetric tensor:

(\nabla_{\mu}A_{\nu}+\nabla_{\nu}A_{\mu})

I call this "average Joe", the average amount of change in an 4-potential in spacetime (it covers 10 of the 16 possibilities). A second, independent, irreducible, but antisymmetric tenor is the one for EM:

(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu})

These are the six deviants, the deviation from the average amount of change in the 4-potential. Because these are really separate, so too can the frequencies and wavelenths.

I also broke the potential cleanly in two. Nature can deal with them separately, me thinks.

What cannot be separate is the speed: they live in the same house. Both travel at the speed of light to mediate the force.

At this point in the development, I know I will not bring this up in a scientific meeting. There are two big problems with it, only one of which I might be able to deal with some day. The first problem is having one collection of events in spacetime that for its transverse modes, has spin 1 symmetry, and for its longitudinal and scalar modes has spin 2 symmetry. I have great hopes for quaternions.sf.net and the animation software to open completely new doors. I have recent data to demonstrate this: a fellow from Mexico wrote to me about gamma matrices - a bit of confusing algebra that appears in relativistic quantum field theory - and a couple days later the animation was up on YouTube, and I get what it is about (taking a quaternions (t, x, y, z) and shuffling however you want, these four, tossing in a negative sign or too, to get things like (-z, -y, t, x)). Until the animation of spin 1 and 2 coexisting is on my hard drive, I accept that the idea of a particle with both spins will not be accepted.

[cool shift just happened: I may want to start thinking about "collections of events" instead of particles, as it would be more consistent with wave/particle duality and my software. It frees the mind from the chain that is the billiard ball->particle association.]

That is a problem in theory. The problem in practice is how could the gravitational field be so absurdly small and be traveling with a photon? If I was drinking beer, I might talk about a few possibilities, but I need something concrete.

Good luck in your studies,
doug

[sweetser]

Hello:

The process of updating the GEM Lagrangian as written on my web site has begun. I have some happy news to report. The Lagrangian is a scalar. That has not changed in the slightest way. There are two ways to write the Lagrangian. The first is the "unified" approach, like so:


\mathcal{L}_{GEM}=-\frac{\rho}{\gamma}-\frac{1}{c} J^{\mu} A_{\mu}
-\frac{1}{2 c^{2}}\nabla_{\mu}A_{\nu}\nabla^{\mu}A^{\nu}


The tensor \nabla_{\mu}A_{\nu} is called "reducible", and thus cannot be viewed as the source of a fundamental field. I have known for a long time how to split this reducible asymmetric tensor in the irreducible rank 2 symmetric field strength tensor (average Joe for gravity) and the irreducible rank 2 antisymmetric field strength tensor (the deviants for EM).

The issue of spin 1 and spin 2 for the current coupling term, -J^{\mu} A_{\mu}, has that same property. It can represent either two like charges that attract, or two that repel. We see that in Nature. Here is how to split up both the current coupling and field strength tensor contraction terms:


\mathcal{L}_{GEM}=-\frac{\rho}{\gamma}
-\frac{1}{2 c} J^{\mu}(A_{\mu} + I_{||}A_{\mu}I_{||})
+ \frac{1}{2 c} J^{\mu}(I_{tr}(A_{\mu} - I_{||}A_{\mu}I_{||})I_{tr})^*


- \frac{1}{4 c^2}(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu})
(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})


- \frac{1}{4 c^2}(\nabla_{\mu}A_{\nu}+\nabla_{\nu}A_{\mu})
(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})


A symmetric split, nice.

doug

[dlgoff]

That is a problem in theory. The problem in practice is how could the gravitational field be so absurdly small and be traveling with a photon? If I was drinking beer, I might talk about a few possibilities, but I need something concrete.
I'm ready for some beer. Hope you are.

[sweetser]

Hello dlgoff:

I always feel more confident that an issue is important if I find out lots of other people have worried before me. Here is the issue as I stated it:

The first problem is having one collection of events in spacetime that for its transverse modes, has spin 1 symmetry, and for its longitudinal and scalar modes has spin 2 symmetry.

The recent news in unified field theory by a respectable theoretical physicists and surfer Garrett Lisi hopes to unify gravity with the standard model using the group E8, which is so big it took a computer to write it all out. A solid technical review folks reading this list should be able to get a few facts from is here:

http://backreaction.blogspot.com/2007/11/theoretically-simple-exception-of.html

Here is the paragraph relevant to my own worry:

On the algebraic level the problem is that fermions are defined through the fundamental representation of the gauge group, whereas the gauge fields transform under the adjoint representation. Now I learned from Garrett that the five exceptional Lie-groups have the remarkable property that the adjoint action of a subgroup is the fundamental subgroup action on other parts of the group. This then offers the possibility to arrange both, the fermions as well as the gauge fields, in the Lie algebra and root diagram of a single group. Thus, Garret has a third way to address the fermionic problem, using the exceptionality of E8.

This problem of getting fermions to play nicely with bosons is part of all this supersymmetry research, including string theory. The common approach, even by Lisi, is to make things bigger.

I may have found a fourth approach this weekend. I call it the platform diving solution. Once every four years, the summer Olympics comes around, and they show all kinds of sports I would not spend a dime to see. One of those sports is platform diving. They do it from platforms of different heights. The commentators are somehow able to spot that the diver does three back flips while also doing one and a half twists. So I wondered, why not do the same for fundamental particles? I generated a bunch of quaternions using this automorphic function:

q(t) = (cos t, sin t/2, sin t, sin 2 t)

t \rightarrow 0 to 4\pi

There are 3 different symmetries for the phases: spin 1/2 for x, spin 1 for y, and spin 2 for z. In easier lingo, x is slow, y is medium, and z is fast. I was able to generate gif animations, but the sizes of those files are kind of big, so I'll just include one frame of the animation:

http://www.theworld.com/~sweetser/Images/spin1-2-4pi.jpg

The t-x graph is lower middle, t-y is upper left, and t-z is lower left. In the animation in the upper middle, 4 dots are running along the path shown in the superposition of the upper right. After generating many permutations, I spent a lot of time staring at these, amused, even my lady enjoyed them. Analytical animations may lead to new ideas.

doug

[sweetser]

Hello:

Nudged by Lut, I am starting to think about the issues of degrees of freedom. I know the logic behind the standard approach that says the photon potential A uses all of 2 degrees of freedom. That appears odd. Consider this thought experiment...

You have devices to measure the electric field E in three directions, and the field B in three direction. A man in a long black trench coat wearing mirrored aviator glasses places a black box in front of you. Since he doesn't look like a chatty fellow, you measure the six values:

Ex, Ey, Ez, Bx, By, Bz

That's the way science is, you make measurements quietly. He picks up his black box and leaves. As you wait, you wonder what would happen if the dollar kept going through its free fall. The man returns, you measure:

Ex', Ey, Ez, Bx, By, Bz

Only one number was different, Ex', the other five were the same. Seven times this scenario plays out, and your final data set looks like this:

1. Ex, Ey, Ez, Bx, By, Bz
2. Ex', Ey, Ez, Bx, By, Bz
3. Ex, Ey', Ez, Bx, By, Bz
4. Ex, Ey, Ez', Bx, By, Bz
5. Ex, Ey, Ez, Bx', By, Bz
6. Ex, Ey, Ez, Bx, By', Bz
7. Ex, Ey, Ez, Bx, By', Bz'

This looks like six degrees of freedom to me. The four potential only has four degrees of freedom, not enough, even if we ignored the analysis that says only two modes can be used for EM.

Recently I have argued that a message from quantum mechanics was to view an operator as an equal to what it acted upon. I was thinking about symmetries in the action. The covariant derivative can move changes in the potential to changes in the connection back or forth. This is the symmetry behind conservation of mass.

In the context of this post, let us try and view an operator as having degrees of freedom. I know in standard approaches, one does not view an operator as increasing the degrees of freedom. One of the missions of the GEM programs is seeing a little bit more in simple tools, just enough to get the job done.

When I say the potential A^{\nu} has 4 degrees of freedom, and at a particular point 0, has the values A^{\nu} = (\phi_0, Ax_0, Ay_0, Az_0), how does that constrain the derivative of the 4-potential, \nabla^{\mu} A^{\nu}? I argue that the 4 degrees of freedom of the potential tell you precisely nothing about how A changes. Consider a near-by neighbor, A'^{\nu}, which has values A^{\nu} = (\phi_1, Ax_1, Ay_1, Az_1). Our four degrees of freedom are already vested in the 0 terms. How can things change going from 0->1? Most of the change for \phi_1 probably comes from \phi_0, a continuation. Yet some of the change might come from the other three. Same goes for Ax_1, Ay_1, and Az_1. That adds up to 16 ways change can happen going from potential 0 to 1. If say Ay_0 helps how Ax_1 changes because of a swirling potential, why should that have any link to how \phi_0 contributes to Az_1?

The sixteen gamma matrices of quantum mechanics appear to do just this sort of mixing of all the components. (I haven't talk about that stuff here yet, quite neat).

This discussion has caused me to broaden my view of a unified field theory. I had previously thought it was all about the four potential. Now I see the operator plays a much deeper roll, it is not just a piece of math machinery. The operator has a symmetry of its own which gives rise to the conservation of mass. The four operator action on a four potential may allow sixteen degrees of freedom, six for the transverse wave of EM, and ten for the dynamic metric of gravity.

doug

[Mentz114]

Hi Doug:

I think the logic of your gedanken experiment is flawed. When you measured the E and B fields, you were measuring space and/or time derivatives of the potential. As you point out later on \partial^{\mu}A^{\nu} has 16 df. I'm not sure what the point of the argument was ( maybe that is the popint ?)

Some of the degrees of freedom of the potential are not physical, and it requires a choice of gauge to get rid of them.

We could be talking at cross purposes, because I'm talking about GEM as a classical field theory.

You haven't commented on the calculation I did in post #404. If the connections are non-zero, I don't see how your graviton can be massless without a heavy constraint, or maybe not at all.

I've though a lot about how you can get over the spin-1 limitation of the interaction term. Squaring it will do the trick but that changes the theory a lot.

Lut

[sweetser]

Hello Lut:

I had been pondering this thread over on sci.physics.research: Maxwell's wave equation and degrees of freedom (http://groups.google.com/group/sci.physics.research/browse_thread/thread/6d42cc8a571e328e/ac4a8e2c9cd04387?lnk=gst&q=Maxwell+freedom#ac4a8e2c9cd04387). There are two degrees of freedom for the polarization of the massless wave equation. When there is a current, mass enters, and that adds another degree of freedom (hope I got that right!). The Maxwell wave equation thus has three degrees of freedom.

In that thread, someone argued that since the B field can be viewed as the result of relativistic motion, it should not be viewed as a degree of freedom. I disagree with a connection between relativity and degrees of freedom.

I am using the word 'graviton' as it is used now - that which is spin-2 and has zero mass so travels at the speed of light, doing the work of gravity. That is when the trace of the symmetric tensor happens to be zero. There will be many situations where as you say, the trace will not be zero. I don't have a name for the non-zero trace situation (I hate making up names, it marks one as a crank). When the trace is non-zero, that is a scalar field that breaks the gauge symmetry. That sounds like what the Higgs mechanism does, but without the false vacuum. Mass breaks gauge symmetry. So maybe gravito-higgs?

Squaring it will do the trick but that changes the theory a lot.

How so? The components are identical, -\rho \phi + J_x A_x + J_y A_y + J_z A_z, just the way to write -J^{\mu} A_{\mu} has changed so the phase is right. The phase is not part of the action.

doug

[Mentz114]

Hi Doug:
That is when the trace of the symmetric tensor happens to be zero. There will be many situations where as you say, the trace will not be zero.
It's not enough to say this - you have to show mathematically that the trace can be zero at all. My calculations indicate that if the connections are non-zero, it cannot happen. Prove me wrong. I could have made a mistake, or maybe there's some interesting constraint that makes the trace disappear.

The idea of squaring the interaction comes from

(A^{\mu}A^{\nu})(j_{\mu}j_{\nu}) \equiv B^{\mu\nu}T_{\mu\nu}

which looks like a spin-2 interaction. I don't know if it makes sense.

[good luck Patriots tonight]

[sweetser]

Hello Lut:

We be doing physics, and in physics, there is no situation with an electrical charge that has zero rest mass. All charges have mass, no exceptions. All neutral systems with mass, well, they have mass. Only a system of pure photons and gravitons would have a trace of zero. Since it is my hope that the trace only has to do with the mass charge, and the mass charge compared to the electric charge is stupidly small (16 orders of magnitude smaller), it is reasonable to work with an effective field theory where one works with the trace being effectively zero. Size can matter. Your algebra observation sounds reasonable to me on the physics grounds. So I have to prove energy conservation, I know.

Since I am trying to write the action as a quaternion, I have not made it back out the the stress-energy tensor. What is the B thing?

Let's hope the Pats get a quick lead so I don't have to stay up late like the last game.
doug

[Mentz114]

Hi Doug:

Whoops, the thing I wrote is not an equation but a definition. I have edited it.

Surely the trace of the Grav field tensor is related to the mass of the field, not the mass causing the field.

there is no situation with an electrical charge that has zero rest mass. All charges have mass, no exceptions.
This may be true, but there has been a lot of talk about the electron's rest mass being purely EM in origin. Still, we have no idea what mass actually is except a number we have to add in to the field equations. Or in this case try to get rid of !

Anyhow, I'm moved to have another look at my 'sums'.

[sweetser]

Hello:

I hope everyone had a good holiday season. I got a pair of noise canceling headphones so I can be off in my own little world, which I can do without headphones.

This project to unify gravity and EM was driven by math: how could quaternions be used for basic laws in physics? Maxwell himself predicted someone would figure out how to do this someday in his Treatise. I was the second person (apres Peter Jack) to figure out how to use real quaternions to write out the Maxwell equations.

The way to generate the Maxwell equations was darn complicated, keeping this, tossing away that. These equations govern the simplest stuff in the Universe, photons and electrons, so elegance - simplicity with purpose - will be part of successful approaches. The things I was throwing away were symmetric. That got me thinking my discards could be gravity. That led directly to the field equations, the 4D wave equation at the beginning of this long thread. It has been a long path making precise math statements as a consequence. Even more effort has been required to prune out the four or five technical errors I have made in trying to write out math expressions.

One thing I learned was how to back translate something I figured out with quaternions into tensors. Professional physicists are trained in tensors. They will give you a blank stare if you go on a quaternion riff. This was a great learning experience for me too, since I could tie in to the vast knowledge of physics.

What I have been working on lately is math driven by the physics. Although I own the domain quaternions.com, I am not a zealot. It is the phrase, "up to an isomorphism", that gives me balance. I can look at other areas of study that use tensors or Clifford algebras, then, should I succeed in translating to quaternions, can say the two approaches are the same up to an isomorphism.

As an example, it is easy enough to write out the E and B fields using quaternions:

E + scalar field = -\frac{1}{2}(\nabla A + A \nabla)

B = \frac{1}{2}(\nabla A - A \nabla)

[I'll discuss the scalar field later, but to keep things simple in comparing the math of E and B, I will include it, otherwise it could be subtracted away]

Mix and match the 4-derivative with the 4-potential, and you get two of the most important fields in physics: the electric an magnetic fields.

One thing that was trivial to write out with tensors was a symmetric tensor, \nabla^{\mu} A^{\nu} + \nabla^{\nu} A^{\mu}. This tensor has 3 fields: a symmetric analog to EM's electric field E I call small e, a symmetric analog to EM's magnetic field B I call small b, and the diagonal field g, the place where gravitational and inertial mass call home (the other two, e and b, represent mass on the move). This is the component definition:

e = +\frac{\partial A}{\partial t} - \nabla \phi

b = (-\frac{\partial A_y}{\partial z}-\frac{\partial A_z}{\partial y}, -\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z}, -\frac{\partial A_x}{\partial y}-\frac{\partial A_y}{\partial x})

g = (\frac{\partial \phi}{\partial t}, -\frac{\partial Ax}{\partial x}, -\frac{\partial Ay}{\partial y}, -\frac{\partial Az}{\partial z})

Look at the symmetric b: I wondered how I could possibly write that using a quaternion. I did find a way. I took the the quaternion differential operator and broke it into individual parts, then took the 4-potential, broke that up too, and put the two together as needed. Although such a micro-slicing approach works, it is not elegant - simplicity with a purpose.

I have been thinking about, how to make a "symmetric curl". OK, most of the time over the break I spent rearranging stuff in my basement, but it was always there to nag me. The basement has improved, but more relevant, I did come up with an idea... The way to make a symmetric curl in a 4x4 real matrix representation of a quaternion is to make all the off diagonal elements have the same sign. That way, when they get together to form a curl, everyone will have the same sign. So this was my first guess:

q (t, x, y, z) = \left(\begin{array}{cccc}
t & x & y & z\\
x & t & z & y\\
y & z & t & x\\
z & y & x & t
\end{array}\right)

The problem with this is all the positive signs. I had taken what was antisymmetric and made it symmetric. Now I needed to take what was symmetric, the scalar t, and figure out how to make it antisymmetric. Therefore I dropped in a factor of "I" which commutes with all the real values of t, x, y, and z:

q (t, x, y, z) = \left(\begin{array}{cccc}
t I & x & y & z\\
x & t I & z & y\\
y & z & t I & x\\
z & y & x & t I
\end{array}\right)

For this to work as a representation of the quaternions, t, x, y, and z are constrained to evaluate to real numbers. Not only must one exclude t=x=y=z=0, but one needs to exclude certain values where t=0. If these are removed from the set of possible values, then one could use this as a representation of the quaternion division algebra.

The Hamilton 4x4 real matrix representation of quaternions, and what I will call the "I" representation until I get a better name, cannot be mixed together. The I representation does solve the problem at hand, because a little cutting and pasting, with a few edits, leads to this way of writing small e and b:

e = \frac{I}{2}(\nabla_I A_I - A_I \nabla_I)

b + scalar field = \frac{1}{2}(\nabla_I A_I + A_I \nabla_I)

The EM E field is a sum, the symmetric e is a difference. The EM B field is a difference, the symmetric b is a sum. The additional scalar field is in EM's E field, and the symmetric b field. The scalar field is made up of the components of the field g. To go from the fields of EM to those of gravity involves switching the way quaternions are represented.

This qualifies as elegant.
doug

[sweetser]

Hello Folks:

I like to make specific technical proposals, even if I am uncomfortable with certain parts. In my previous post, I disliked having the factor of "I". It looked too much like a biquaternion which is not a division algebra. Inside Mathematica, it led to technical difficulties.

The beauty of a specific proposal is that it can be modified. I decided to drop the factor of "I", and see what happened. That was my first guess after all (I'm using a 2 to suggest "evenness"):

q2 (t, x, y, z) = \left(\begin{array}{cccc}
t & x & y & z\\
x & t & z & y\\
y & z & t & x\\
z & y & x & t
\end{array}\right)

With this representation, the product of two quaternions commutes. These symmetric matrices will surely work with addition, subtraction, and multiplication. The open question is whether division works for all possible values of t, x, y, and z. We know already that t=x=y=z=0 doesn't work. I asked Mathematica to find the inverse of the above matrix. It returned a fraction. I then asked Mathematica to factor the dominator, to find out where any zeros might live. Here was the answer:

denominator = (t-x-y-z)(t+x+y-z)(t+x-y+z)(t-x+y+z)

Every possible quaternion will have an inverse except those quaternions where one of the elements happens to equal the sum of the other three. This happens for photons, where c dt = dx + dy + dz. And apparently it applies to permutations of photons, with cdt switching roles with dx, dy, and dz. The gamma matrices in quantum field theory do this sort of shuffling.

Quaternions are a division algebra. The most common representation for quaternions is the one developed by Hamilton were two quaternions do not commute under multiplication. With this representation - so long as we avoid photons - two quaternions can commute.

The standard definition of a derivative will work for the even quaternion, no change needed. Neat.

Any 4x4 matrix that has the same value down the diagonal can be represented by a sum of the Hamilton representation and this even representation. Cool!

doug

[Mentz114]

Hi Doug:

That's an interesting matrix. It looks like the multiplication table for the symmetric curl.
It defines a commuting operator ~ so that

x~y=z
x~z=y
y~z=x
and anything~t = anything

I'm impressed with Mathematica's factorizing power.

Lut

[sweetser]

Hello Lut:

The q2 matrix is just the standard 4x4 real matrix representation of the Hamilton quaternion, just no minus signs allowed. Perhaps philosophers will discuss that aspect someday, since the always plus guy is used to write the gravity fields.

Mathematica figured out the inverse and the factorization in the time it took to hit the return key. Scary fast stuff. Toss in the factor of "I", and it becomes hard to tell the software t, x, y, and z are real. I concluded that it was not possible to constrain them to the reals without getting rid of the factor of I. Was I needed? No? Great! In fact, so good, I will be taking the misses out to dinner to celebrate. It looks like I am using physics to upend a deeply rooted idea: representations of quaternion multiplication must not commute. The first one did, but there is another option.

doug

[sweetser]

Hello:

What is physics is math, what is math is physics.

The core math idea of this proposal is that while we use an antisymmetric tensor which is the 4-derivative of a 4-potential for EM, to be complete, Nature must also use the symmetric 4-derivative of a 4-potential. That is all gravity is.

Earlier in this thread, I was mystified by Lawrence B. Crowell's focus on the antisymmetric tensor and his exclusion of the symmetric one. That's the way technical discussions sometimes go.

What was so neat is that I have now documented myself doing the same thing, but for quaternions. Everything I have ever read on quaternions, the product of two quaternions contains the antisymmetric curl. I never considered that there could a symmetric product. Non-commuting is too basic to what quaternions are, right?

Quaternions are a 4D division algebra. When Hamilton went to implement them, he chose a representation where multiplication was associative and non-commuting. I figured out how to make the multiplication non-associative and non-commutative by defining multiplication as A* B. With the q2 representation, multiplication is associative and commutative. And to complete the set, q2* q2' will be non-associative and commutative.

No wonder Nature is tricky to figure out because she uses all four simultaneously!
doug

[Mentz114]

Hi Doug:

What is physics is math, what is math is physics.
Not true. Mathematics is an art of logic and imagination. Physics is the laboratory.
I can write mathematically consistent space-times for instance, whose physical existence would be riddled with impossibilites.

The core math idea of this proposal is that while we use an antisymmetric tensor which is the 4-derivative of a 4-potential for EM, to be complete, Nature must also use the symmetric 4-derivative of a 4-potential. That is all gravity is.

I wish I had your inside line to mother nature. As I said before, this is not science, it's religious faith.

we use an antisymmetric tensor which is the 4-derivative of a 4-potential for EM,
Yes ! WE use it as a matter of convenience because it allows a compact way to ensure Maxwells equations are satisfied. I don't think mother nature is a mathematician.

You'd win more friends in the physics community if you established that your theory is gauge invariant, so you have the freedom to play with your potential the way you propose.

Your field equations may look elegant, but until you sort out what's causing the curvature and the potential ( source) it has no computational power. How do I solve actual problems with it ?

Regards,
Lut

[sweetser]

Hello Lut:

I confess, this is a scientific belief:

What is physics is math, what is math is physics.

It is my opinion that this is also a scientific belief:

Not true. Mathematics is an art of logic and imagination. Physics is the laboratory.
I can write mathematically consistent space-times for instance, whose physical existence would be riddled with impossibilities.

In the radical form of my belief, only once we see the math in Nature are we confident that a math theorem is true, and not bad math. My canonical example is the math wonk who solved the math problem from Hell: develop a polynomial to describe knots, then based on that polynomial, tell if it can be untied. Those polynomials appear in some systems of quantum mechanics. It may be darn hard to find the physical system, but it is my belief that enough searching will find it if the logic is true.

I respect your belief which happens to be different from mine.

The core math idea of this proposal is that while we use an antisymmetric tensor which is the 4-derivative of a 4-potential for EM, to be complete, Nature must also use the symmetric 4-derivative of a 4-potential. That is all gravity is.
I wish I had your inside line to mother nature. As I said before, this is not science, it's religious faith.

My quote is not a belief, it is a hypothesis that can be tested. Measure the bending of light to a tenth of a microarcsecond, and my hypothesis predicts light will bend 0.8 microarcseconds more than GR. I would accept my hypothesis is wrong should the necessary experiment be done and support the bending predicted by GR.

I also don't think the position about completeness is a belief either, it is an observation. Subtraction takes some thing away to make the antisymmetric tensor. A complete story uses everything. That is logic at work.


You'd win more friends in the physics community if you established that your theory is gauge invariant, so you have the freedom to play with your potential the way you propose.

I'd rather be right than popular :-) The proposal is gauge invariant in the usual sense of the word for all mass less particles. When one has massive particles, the proposal does involve choice in how things are measured, either as changes in the derivatives of the 4-potential, or changes in the connection which are changes in the metric. I have had no luck in communicating what the preceding sentence means to those skilled in the mathematical physics arts, which is exasperating since it falls out of looking at the definition of a covariant derivative (it is the sum of these two things, so how much comes from one or the other is arbitrary).

Your field equations may look elegant, but until you sort out what's causing the curvature and the potential ( source) it has no computational power. How do I solve actual problems with it ?

The most important message is that most of physics is about doing almost nothing. To keep going for 13+ billion years, a system has to be extraordinarily passive, not active, get it all done kind of force. The idea is not cause, but more what is the absolute closest thing to absolute nothingness? Absolute nothingness is the Minkowski metric. The closest thing to that is a 4D simple harmonic oscillator.

In the first post I wrote out a physical relevant solution to the field equations. It was for an electrically neutral point charge. Got to get the basics done. Somewhere else in this long thread I solved the problem for an electrically changed point source (I think I did anyway). What one has to do is make a decision on how things are measured, either as a change in potential or a change in the metric or a combination of both, and proceed from there.

doug

[Mentz114]

Hi Doug:

I was too picky in challenging your beliefs, you're entitled to them and I suppose we all have some rigidities built in.

The proposal is gauge invariant in the usual sense of the word for all mass less particles. When one has massive particles, the proposal does involve choice in how things are measured, either as changes in the derivatives of the 4-potential, or changes in the connection which are changes in the metric. I have had no luck in communicating what the preceding sentence means to those skilled in the mathematical physics arts, which is exasperating since it falls out of looking at the definition of a covariant derivative (it is the sum of these two things, so how much comes from one or the other is arbitrary).
You've definately failed to communicate it to me, although I wouldn't count myself amongst "those skilled in the mathematical physics arts".

What one has to do is make a decision on how things are measured, either as a change in potential or a change in the metric or a combination of both, and proceed from there.
Are the equations of motion invariant under this choice ?

Regarding your vacuum solution ( flat potential, no mass or charge ?) which gave you the exponential (Rosen) metric - did you discard the off-diagonal elements of the metric, because you've only four equations, and the full metric would have 10 independent terms.

[sweetser]

Hello Lut:

One of the big ideas I am trying to accept myself is that the covariant differential operator, \nabla^{\mu} plays an equal role to the 4-potential, A^{\nu} in the GEM action. This is a message of quantum mechanics, where operators are what gets measured. I used to look at the derivative and think it was just a math widget - we are trained that way. Yet in quantum mechanics, what we are measuring is the average of a derivative.

I need to view the covariant differential operator this way for two reasons. The first is to have a theory that has enough degrees of freedom to described both gravity and EM. The second reason is to have the symmetry required for a geometric theory for gravity. If I were to use an ordinary derivative, \partial^{\mu} A^{\nu}, then the only thing that could be included in the measurement of change is the change of the potential. With a covariant derivative, I get a change in potential minus a change in the metric, \nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}. So how much of the covariant derivative comes from \partial^{\mu} A^{\nu} and how much comes from - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}? You get to decide. That is the symmetry needed for gravity. Whereas GR says gravity is exclusively a geometric theory, GEM says that gravity could be a geometric theory, a potential theory, or any combination of the two consistent with the other choices. Any equation that has a covariant derivative in it has this symmetry.

As far as the (possibly electrically charged) Rosen metric, I worked with the simplest case. This is "The Revenge of Schwarzschild II", same assumptions Karl made on the Russian front so many years ago.

doug

[sweetser]

Hello:

Steve Carlip had pointed out the problem with the spin of the coupling term, -J^{\mu} A_{\mu}. He gave me a wonderful reference to a description of the problem in chapter 3 of "Feynman Lectures on Gravity". Based on that work, I was able to think about current moving along the z axis, and show that it could have both spin 1 and spin 2 symmetry in the phase (reread post 319, 320, 363, 367 if you want more details). It was great to do a direct variation off of Feynman!

After the celebration of getting around a killer technical problem, there is the hangover, the parts that one does not like about the solution. I was uncomfortable with the way direction played a role. I started with moving along the z axis because that is exactly what Feynman did. I like to follow the footsteps of great ones :-) Yet one could imagine an inertial observer where the motion along z was zero. It felt like my spin argument might fall apart for that observer. That gave me a headache.

I generalized this a bit by introducing two vectors, parallel I_{||} and transverse I_{tr}. This will work for motion along x, y, z or any direction between, a definite improvement. But does this still have the problem of what happens when one of these goes to zero? I worried about the meaning of these directions: was it the direction of the potential and its real current, or was it about the two currents? I also did not like introducing new letters into the GEM equations. The algebraic problem was solved, but I had headaches.

Recently I have started working with a representation of quaternions that commute (post 415 and 416). To get the right scalar product and phase with both spin 1 and spin 2 requires the product of both the Hamilton and Even quaternions (which I while symbolize by tacking on a "2").

First write out the standard scalar generated from a tensor contraction:

-J^{\mu} A_{\mu} = -\rho \phi + Jx Ax + Jy Ay + Jz Az[/itex]

This does not have any information about the phase as is done in chapter 3 of Feynman. What Feynman does is take the Fourier transform of the potential to get a current-current interaction, with A->J':

[tex]-J^{\mu} J'_{\mu} = -\rho \rho' + Jx Jx' + Jy Jy' + Jz Jz'[/itex]

Now form the products for the current-current interaction using both the Hamilton and Even quaternion representations:

[tex]-J J' = -(\rho, Jx, Jy, Jz)(\rho', Jx', Jy', Jz')
= (-\rho \rho' + Jx Jx' + Jy Jy' + Jz Jz',
-\rho Jx' - Jx \rho' + Jz Jy' - Jy Jz',
-\rho Jy' - Jy \rho' - Jz Jx' + Jx Jz',
-\rho Jz' - Jz \rho' + Jy Jx' - Jx Jy')

-J2 J2'^* = -(\rho, Jx, Jy, Jz)(\rho', -Jx', -Jy', -Jz')
=(- \rho \rho' + Jx Jx' + Jy Jy' + Jz Jz',
\rho Jx' - Jx \rho' + Jz Jy' + Jy Jz',
\rho Jy' - Jy \rho' + Jz Jx' + Jx Jz',
\rho Jz' - Jz \rho' + Jy Jx' + Jx Jy')

These two products have the same Lorentz invariant scalar, as they must if they hope to sit in for the tensor 4-vector contraction. For the Hamilton representation, it is the source terms \rho J_i that move together, and thus will represent spin 2 symmetry. The curl terms have opposite signs, and will require the normal 2 pi radians to get around as is the case for spin 1 symmetry. For the Even representation, the source terms have spin 1 symmetry, while the symmetric curl terms have spin 2 symmetry.

Sounds like a complete story to me. Headache gone.

Doug

[sweetser]

Hello:

In this post I will write out the GEM field strength tensor using only the Hamilton and Even representations of quaternions. The five fields, E, B, g, e, and b (e and b being symmetric counterparts to E and B respectively) will be written succinctly, a good sign.

Here is the short version. The E and B fields arise from the Hamilton representation where the 4-derivative acts on the 4-potential. The fields E and B can be separated by changing the order of the derivative with the potential. The symmetric fields e and b arise from the Even representation where the conjugate acts on the 4-derivative or 4-potential.

In both cases, there is a scalar field g, but the g has opposite sign and cancels out. This may turn out to be a good thing for approximate gauge symmetry (I don't think it will be perfect because like electric charges repel while like mass charges attract, creating a dipole).

Let's start calculating. Take the 4-derivative of a 4-potential in the Hamilton representation, but put the derivative after, which flips the signs of the curl, nothing else:

-A \nabla =

(-\frac{\partial \phi }{\partial t}+\frac{\partial Ax}{\partial x}+\frac{\partial Ay}{\partial y}+\frac{\partial Az}{\partial z},
-\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},
-\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},
-\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})

= -g + E + B

[technical note: I will be attaching a Mathematica notebook at the end. It took a few emails to tech support and a bit of work on my part, but I got Mathematica to write the 4-derivative of a 4-potential in a more human-readable way. Then I used "cut as Latex" to plug this in. This is why I am confident this large a volume of partial differential equations is accurate.]

To separate E from B, switch the order of the differential, and either have the same sign for E or different signs for B.

=\frac{1}{2}(-\nabla A - A \nabla) =


(-\frac{\partial \phi }{\partial t}+\frac{\partial Ax}{\partial x}+\frac{\partial Ay}{\partial y}+\frac{\partial Az}{\partial z},
-\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},
-\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y},
-\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z})

= -g + E

=\frac{1}{2}(\nabla A - A \nabla) =


(0,
c \left(\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right),
c \left(\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right),
c \left(\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) )

= B

One sign flip on \nabla A is all it takes. Nice.

Now on to the symmetric fields. Here we use the Even representation where all the signs of all products are positive. For this to be a division algebra, events like photons must be excluded. That doesn't sound so bad in this context, since photons make up the E and B fields. The three symmetric fields are generated in a simple expression:

\nabla^* A2 =

(\frac{\partial \phi }{\partial t}-\frac{\partial Ax}{\partial x}-\frac{\partial Ay}{\partial y}-\frac{\partial Az}{\partial z},
\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x}-c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},
\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}-c \frac{\partial Ax}{\partial z}-c \frac{\partial Az}{\partial x},
\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}-c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})

= g + e + b

It is important to note that the A2 is a different representation of the same functions or values that are in the Hamilton representation of the 4-potential. In other words, it is the same 4-vector. What has changed is what happens when an operator acts on it, or it gets multiplied by another even quaternion.

Let's isolate the symmetric e field using a conjugate:

\frac{1}{2}\left(\nabla^*A2-\nabla A2^*\right) =

(0,
\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},
\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y},
\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z})

= e

The time derivative of A has flipped its sign as happens for the symmetric e.

Isolating the symmetric b involves one more conjugate applied to the previous result:

\frac{1}{2}\left(\nabla^*A2-(\nabla A2^*)^*\right) =

(0,
-c \left(\frac{\partial \text{Ay}}{\partial z}+\frac{\partial \text{Az}}{\partial y}\right),
-c \left(\frac{\partial \text{Ax}}{\partial z}+\frac{\partial \text{Az}}{\partial x}\right),
-c \left(\frac{\partial \text{Ax}}{\partial y}+\frac{\partial \text{Ay}}{\partial x}\right) )

= b

Every partial derivative has a minus sign.

Once again, the fields are easy to state, easy to separate. Nice.

Now to look at the big sum by adding the two field strength tensors together:

\frac{1}{2}\left(-A \nabla + \nabla^* A2\right) =

(0,-c \left(\frac{\partial Az}{\partial y}+\frac{\partial \phi }{\partial x}\right),-c \left(\frac{\partial Ax}{\partial z}+\frac{\partial \phi }{\partial y}\right),-c \left(\frac{\partial \phi }{\partial z}+\frac{\partial Ay}{\partial x}\right) )

This has zero gauge, the gradient of the scalar and one part of a curl. I don't know if I have ever seen a post with this many partial differential equations in it, but the way this all falls together cannot be an accident.

Doug

[sweetser]

Hello Lut:

The way the g fields cancel which I write the fields using the Hamilton and Even representations of quaternions, does that mean I now have gauge symmetry for GEM? And would you remind me of why gauge symmetry is so important to have?

The morning realization: to be inverted, the Even representation must exclude its own Eigenvectors and values. Not sure of the consequences of that...

Thanks,
Doug

[Lawrence B. Crowell]

To Sweetser,

I got a couple of messages from P-forum about this. I have not followed your stuff in some time. But I will illustrate what a gauge theory is.

Say you have a manifold M with a bundle over it which has some vector space or algebraic system of roots, that are eigenvector of the weights. For the manifold of dimension = d and the bundle of dimension = p we can define a section s over the base manifold. The principal bundle group is then a set of transformations between bundle sections


s'~=~gs.


Now consider the action of a differential operator d on this transformed section


ds'~=~d(gs)~=~(dg)s'~+~gds


which can be seen in elementary terms with ds~=~As as


ds'~=~A's'~=~\Big((dg)g^{-1}~+~gAg^{-1}\Big)s'


which clearly gives the transformation of the gauge connection A. This equation is not homogeneous with respect to the transformation. Yet for the gauge fields F~=~dA~+~A\wedge A the fields transform as F'~=~gFg^{-1}, which is a homegenous covariant transformation.

Now the field term can be seen as due to the action of the two-form


\Omega^2~=~(d~+~A)\wedge(d~+~A)


on the bundle section. Now this differential form is antisymmetric in its tensor components for the fields F_{ab}~=~-F_{ba}. This can be extended to spacetime curvature as well, where the bundle is a fibre bundle with a hyperbolic group structure and so forth. This antisymmetric structure is grounded in some basic differential geometry and topology, where these systems of forms can define cohomology rings and groups.

Now what you have is a theory which has symmetric field components. As a result you are doing something else. This is not to say that what you are doing is utterly wrong, but honestly it is outside the standard canon of field theory and the underlying system of differential geometry it is formulated around. As such what you are proposing is something other than gauge theory, or is an extension of gauge theory into some other arena.

Cheers,

Lawrence B. Crowell

[Mentz114]

Hi Doug,

after a great struggle I realised what you are attempting with the quaternions. You've found a representation that allows you to separate the fields ( symmetric and anti-symmetric curls ) in a way that you find correct, satisfactory. Right now I can't comment on the meaning or validity of this.

I understand that the differential operators correspond to ordinary partial differentiation.

Getting to gauge invariance: the parts of your Lagrangian density that contain the potential are -

-A^{\mu}J_{\mu} + \frac{1}{4c}\partial^{\mu}A_{\nu}\partial^{\nu}A_{ \mu}

One can split the second term into symmetric and antisymmetric parts without changing the energy.

Provided those operators are not covariant derivatives, this part is globally gauge invariant, because adding a constant to A will not affect the energy. Local gauge invariance which applies when we add a value to A which depends on x, is assured by specifying a gauge condition, which ensures that the Lagrangian doesn't change when the potential is altered.
My complaint has always been that when the derivatives are covariant, products of metric terms with the potential itself appear in the symmetric terms. Now if you add a constant to A, the energy changes, and gauge invariance is lost. Unless more constraints are added to make those terms disappear.

Everything I know about this subject comes from the EM field, and I'm not even sure this can be applied to the symmetric fields. For instance, not requiring a fixed value for A ensures energy conservation, as does the gauge condition.

But as Lawrence B. Crowell tells us, you are going outside standard differental geom. definitions ( at least, I think that is part of what he's saying).

As such what you are proposing is something other than gauge theory, or is an extension of gauge theory into some other arena.

This is not bad news. If you keep track of the energy you won't go wrong.

But I don't think your calculations above affect my ( simplistic ?) view about gauge invariance, because there is only a problem when the derivatives are covariant and introduce terms in A itself.

I've got to go now, but I'll be revisiting soon.

[Lawrence B. Crowell]

My comments are meant to show that if you have a theory of fields with


F_{ab}~=~\{Q_a,~{\bar Q}_b\}


then this symmetric term is an anticommutator of grassmannian fields. In the case I write above this defines gauge connections for N > 2 supersymmetry, for A_i~=~{\sigma_i}^{ab}F_{ab}. Yet these fields are physically Fermionic. Gauge fields are vector (or chiral) and don't obey this sort of rule.

To have symmetric gauge fields F_{ab} = F_{ba}, or components which obey this, runs to my mind in a host of mathematical questions. This appears to be something other than supermanifold theory. There a gauge field will be contained in a super-multiplet with


\Phi~=~A~+~{\bar\xi}\psi~+~\xi{\bar\psi}~+~F


for \psi the fermionic super pair (gaugino) of the gauge potential, \xi the Grassmannian spinor super fields. The gaugino will obey anti-commutator rules. Yet the gauge fields are still bosonic.

This GEM theory appears to be something else, and there are to my mind a host of open questions. The biggest question is whether this even defines an appropriate geometry. A crucial fact is "boundary of a boundary is zero," which is from d^2 = 0 why gauge fields are anti-symmetric. In the case of Grassmannians which obey anti-commutators we have Q^2~=~0, which is equivalent to the Pauli exclusion principle. the d^2 = 0 principle is also why the Riemannian curvature tensor has anti-symmetric indices. Once you twist this around with symmetric terms then the whole geometric meaning of things is lost to me --- outside of a super-manifold theory.

So without a blizzard of tensor analysis and differential terms and the like it seems important to give some indication of what this means geometrically. Without some sense of what this means geometrically, such as the symmetries of a manifold, and how this connects up to conservation laws (eg d^2 = 0 is what gives Bianchi identities etc) then I have a very difficult time knowing how to evaluate this.

Lawrence B. Crowell

[Mentz114]

Lawrence,
thanks for taking the time to post all this. I don't have the math to understand it all, unhappily. I think I can see how energy conservation is linked to the geometry.

The GEM lagrangian has no explicit time dependence, which ought to be enough to ensure energy conservation. I can't remember now how gauge invariance came up.

There's a lot for Doug to think about still.

M

[Lawrence B. Crowell]

Gauge theory is a bed rock of the physics of fields and forces. In fact the subject is mathematically rich. If you are familiar with Uhlenbeck, Atiyah, Donaldson, Freed, etc, the structure of gauge theories, in particular their moduli spaces, have proven surprising results on the categorization of four dimensional manifolds. The work I have been doing for the last two years involves these results.

My sense is that GEM is geometrically mysterious. All of the stuff we do with physics is ultimately based on on fundamentals of differential geometry. For instance in ordinary three dimensions a function F under the action of a differential operator d is


dF~=~\frac{\partial F}{\partial x^i} dx^i,


which is the gradient. For a one form \omega~=~\omega_idx^i the action of d is


d\omega~=~\frac{\partial\omega_i}{\partial x_i}dx^j\wedge dx^i,


and since dx^i\wedge dx^j~=~-dx^j\wedge dx^i clearly i can't equal j. This in vector language is the curl. You can actually prove all the div, grad & curl stuff this way if you also throw in something called Poincare duality. All the anti-symmetric tensors and the rest in gauge theory are grounded on this, and this has incredible generalizations for manifolds and bundles.

GEM appears to throw all of this to the wind, which means there has to be some sort of other foundation to this. As yet I have not seen it.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

Thanks for all your technical responses. I wish to make clear that there is something brand new in the GEM proposal as of posts 424 and 425. It is fair to say I am working on the extension of gauge theory into quaternion representations, but one that does tie in directly to some of what we know of gauge theory.

There are also some clear breaks. Let me point one out. The Bianchi identities will never be relevant to this work. The Bianchi identities come out of the Riemann curvature tensor. While I do work with the connection if and only if it appears as part of a covariant derivative, I never work with the connection outside of a covariant derivative as happens with the Riemann curvature tensor. This makes communicating with someone with your level of training darn near impossible because you find some reason to reintroduce the curvature tensor or its stand ins.


Why should the Riemann curvature tensor create problems? After all, it is part of a huge intellectual effort, done by some of the smartest math and physics people ever (Lawrence listed a few of the many authors working on this subject). It is both courageous and daft to claim Riemann curvature tensor is irrelevant to how Nature works. Here is the logic most readers of this thread should be able to follow.

The 4-potential A^{\nu} transforms like a tensor.
The 4-derivative \partial^{\mu} transforms like a tensor.
The 4-derivative of a 4-potential \partial^{\mu} A^{\nu} does not transform like a tensor.
To correct this problem, we need to add in the connection. It turns out there are lots of options with the choice of the connection. I choose to work with the same one used in GR, which is to say the connection is torsion-free and metric compatible, so the connection is the Christoffel symbol of the second kind. The Christoffel symbol has three first derivatives of the metric. Thus we get the definition of the covariant derivative:

\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}

Einstein learned this stuff (he didn't invent it) from is math tutors. He focused on the Gamma which contains the first derivatives of the metric. He then asked his math buddies for the math object that transforms like a tensor but has second order derivatives of the metric. He was thinking about good old F=Ma, how Nature accomplishes things with second order derivatives. That was his inspired guess, and he needed some direction to implement it. The answer he was given was the Riemann curvature tensor. That was a great answer that has led to lots of great results. The way math wonks sell it, they say it is the only answer without getting into really obscure math.

Yet Einstein missed the obvious way to get second order differential equations of the metric. Take a closer look at what Newton did. Newton was saying m \frac{dR}{dt} is not interesting, but acting again on this with another time derivative operator, \frac{d}{dt}, then you get F=m \frac{d^2 R}{dt^2}. Follow that program exactly. Einstein is saying \nabla^{\mu} A^{\nu} with one derivative of the metric is not interesting, so take another covariant derivative: \nabla_{\nu} \nabla^{\mu} A^{\nu}. There will be the divergence of the Christoffel symbol. Drop the Rosen metric into the Christoffel, take its divergence, and see something even Poisson would recognize. Why I cannot lead a well-trained mathematical horse to do that calculation is beyond me, but that's the way it is.

GEM is about potentials in bed with geometry. The lights are on, the camera is rolling. Someday I'll sell a lot of film, but until then, I will post to the Independent Research forum.


I have an even simpler view of gauge theory, the one that applies directly to EM. I look at it in terms of this specific Lorenz gauge transformation:

(\phi, A) \rightarrow (\phi', A')=(\phi, A)+(\frac{\partial f}{\partial t}, - \frac{\partial f}{\partial x}, - \frac{\partial f}{\partial y}, - \frac{\partial f}{\partial z})

The function f needs to have a few derivatives. Drop this into the anti-symmetric field strength tensor, form the field equations and all the dependence on f drops out. Drop A' into any symmetric tensor, and any field equations that come out will depend on f. This is a point you made long ago, and looking back, I don't think I addressed it. Why not? Your point is technically correct. That's the way tensors are.

In post 425, I used no tensors. Instead I used two representations of quaternions. The first is the well known Hamilton representation of the quaternion division algebra. It is an asymmetric tensor, with the symmetric part being made of the scalar, and the antisymmetric part being the 3-vector. The other one I am calling the Even representation. This is symmetric and is a division algebra so long as one does not use the Eigenvalues and Eigenvectors of the 4x4 matrix representation. We can point to gulfs in our communication because I am using these tools, and you are skilled with differential geometry.

When I added all 5 of the fields in my proposal together, E, B, e, b, and g, this was the result:

\frac{1}{2}\left(-A \nabla + \nabla^* A2\right)

= -g + E + B + g + e + b

= E + B + e + b

where
g = \frac{\partial \phi }{\partial t}-\frac{\partial Ax}{\partial x}-\frac{\partial Ay}{\partial y}-\frac{\partial Az}{\partial z}
E = (-\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},-\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y},
-\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z})
B = (c \left(\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right),
c \left(\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right),
c \left(\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) )
e = (\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y},
\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z})
b = (-c \left(\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right),
-c \left(\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right),
-c \left(\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) )

Notice how the field g, no matter what it is, drops out. That means you are completely free to work with whatever g field you choose.
Consider a function h such that \nabla h = g. Then:

A \rightarrow A' = A + h

Hit this with a derivative, out comes the field g, and the field g drops. That is a gauge transformation.


GEM appears to throw all of this to the wind, which means there has to be some sort of other foundation to this. As yet I have not seen it.

I noticed no references to the Hamilton or even representation of quaternions in your replies. That is where the foundation is. We do not have to feel bad if we fail to communicate. You have demonstrated an impressive knowledge of both the history and current efforts to do productive research using the power tools of differential geometry. I respect that. New math is for the next generation. I have no expectation that I will ever be comfortable with it all, but that is the way research goes.

Here is my specific plan of action. I have shown how to write the 5 GEM fields in a compact way using the Hamilton and Even representations of quaternions. I have shown how to write the coupling term using the Hamilton and Even representation in a way that shows the coupling represents both spin 1 and spin 2 fields. I have to form the action from these fields. Then I have to generate the field equations from the action. Finally Lut, I will calculate the stress energy tensor to address your energy conservation question which is a good one. All this work must be checked with Mathematica. There is much to do, like prepare for three talks, make new quaternion animations, do outreach, work full time on something else, walk the dogs, and play nice with the wife.

Doug

[sweetser]

Hello:

I found a clear technical error with post 424 titled "Generalizing Current Coupling Spin 1 & Spin 2". The two current coupling terms look like so:

-J J' -J2 J2'^*

These both have the same sign. When dropped into the Lagrangian, they will have like charges repel. That is good for EM, but failure for gravity.

Now that the issue is defined, let's create a solution. Post 424 was trying to generalize a specific case for motion along the z axis. There I needed to use the first or second conjugates, which I had defined somewhere in this vast discussion as:

(i q i)^* = (-t, x, -y, -z) === q^{*1}
(j q j)^* = (-t, -x, y, -z) === q^{*2}

We can continue this theme and define the third conjugate like so:

(k q k)^* = (-t, -x, -y, z) === q^{*3}

As an indication of generalization, all three conjugates are needed:

+J2^{*1}^{*2} J2'^{*3}=(\rho, -Jx, -Jy, Jz)(-\rho', -Jx', -Jy', Jz')=
=(- \rho \rho' + Jx Jx' + Jy Jy' + Jz Jz',
-\rho Jx' + Jx \rho' - Jz Jy' - Jy Jz',
-\rho Jy' + Jy \rho' - Jz Jx' - Jx Jz',
-\rho Jz' + Jz \rho' - Jy Jx' - Jx Jy')

This current coupling term has three defining characteristics:
1. The scalar is invariant under a Lorentz transformation
2. The phase has both spin 1 and spin 2 symmetry, necessary for a field theory where like charges repel for EM and attract for gravity respectively
3. Is a positive coupling between the two currents, so when put in the action will apply to like charges that attract.

The analysis of the -J J' coupling is unaltered. The generalzized coupling term using the Hamilton and Even representations is:

-J J' + J2^{*1}^{*2} J2'^{*3}

That looks better. Dodged another bullet tonight.

Doug

[sweetser]

Hello:

I said I had to "form the action from these fields" using quaternions only, no tensors. This was not an easy exercise. Once again, the simplest term gave me problems, specifically the g field. I spent much time thinking about what I was trying to do with these fields before I could proceed effectively. In this post I will give an overview of the math to follow in later posts.

The five fields defined in post 432 - the standard electromagnetic E and B fields, their symmetric counterparts e and b, and a diagonal g - together transform like a second rank tensor. That tensor gets contracted to make a Lorentz invariant scalar that goes into the action.

I was not able to build an exact analogy with a rank 2 tensor contraction. That does not mean there isn't one, just that I couldn't find it. There are no convenient indexes to work with when using quaternions, so plucking out the right terms did not work.

We need to create a Lorentz invariant scalar. I recalled reading in Jackson's "Classical Electrodynamics" that E^2 - B^2 is Lorentz invariant. To use a little high school algebra, we know:

E^2 - B^2 = (E + B)(E - B)

There are nice, compact quaternion expressions for -g + E + B and -g + E - B. When will g be zero? If the system has only massless particles. Using that assumption, I am able to calculate E^2 - B^2 using quaternions.

The next step is to apply the Euler-Lagrange equation to calculate the field equations from the Lagrangian. I suspect the majority of readers have not done this sort of calculation. It looks scary, but the details are quite simple. I am enough decades away from my calculus classes that I can recall only a few derivatives, simple ones like \frac{d x y}{d x} = y and \frac{1}{2} \frac{d x^2}{d x} = x. These are the only two derivatives that are required to get to the Maxwell and GEM field equations. What is needed is to have the courage to write everything down, even though it looks complicated. Instead of the simple x, one uses the derivative of a function, x \rightarrow \frac{\partial \phi}{\partial t}, along with 15 other partial derivatives. When I do this in subsequent posts, I will point out the simplicity of what is going on in taking the Lagrangian and getting the field equations.

Does anyone know if physicists make much of the invariant E^2 - B^2? As far as I can remember, it was mentioned once in Jackson, and he did not say much about it at all. Yet in my quaternion calculation, this invariant leads directly to the Maxwell equations. If true, the difference of the squares of these two fields should be on the center stage of EM theory. That has been a surprise benefit of my recent struggles.

Doug

[Mentz114]

Hi Doug:

Do you mean apart from the term F^{\mu\nu}F_{\mu\nu} = -2(E^2 - B^2) appearing in the lagrangian of the EM field ? It seems to be center stage.

[sweetser]

Hello Lut:

Good to hear you are familiar with that result. Perhaps it is somewhere in Jackson, but I had not seen it before in my physics browsing. I gain confidence when I see other people have done something I tried to do before. It indicates I am barking up the right tree.

I will go through all the details because that is my idea of a fun technical time. It also sets the stage for variations needed to see where GEM is going.
Doug

[Lawrence B. Crowell]

Hello Lawrence:

I noticed no references to the Hamilton or even representation of quaternions in your replies. That is where the foundation is. We do not have to feel bad if we fail to communicate. You have demonstrated an impressive knowledge of both the history and current efforts to do productive research using the power tools of differential geometry. I respect that. New math is for the next generation. I have no expectation that I will ever be comfortable with it all, but that is the way research goes.

Doug

I didn't bring up quaternions because they didn't seem relevant.

In what you wrote here it looks a bit like standard Kaluza Klein theory. I am not sure if this differs significantly from rather standard fair. At lest with this the approach appears within the standard construction of gauge theories.

Lawrence B. Crowell

[sweetser]

Hello:

In this post, I will show in detail how to generate the Maxwell equations using quaternions exclusively. This is achieved by getting to exactly the same terms that appear in the standard tensor approach for the electromagnetic field strength contraction. After the contraction, every step is the same. I will do them anyway, since it may be instructive to the viewers of this thread.

Simple quaternion expressions generate combinations of three fields, the standard definition of E and B, along with one I call g:

g = (\frac{\partial \phi}{\partial t}- \frac{\partial Ax}{\partial x}- \frac{\partial Ay}{\partial y}- \frac{\partial Az}{\partial z})

Notice how this one has all the parts of a typical gauge field in standard EM.

There are two simple ways to generate g, B, and E using quaternions:

1. \nabla A = (\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})(\phi, Ax, Ay, Az)

(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},\frac{\partial Ax}{\partial t}+c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},\frac{\partial Ay}{\partial t}+c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},\frac{\partial Az}{\partial t}+c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})

= g - E + B

2. - A \nabla = -(\phi, Ax, Ay, Az)(\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})

(-\frac{\partial \phi }{\partial t}+c \frac{\partial Ax}{\partial x}+c \frac{\partial Ay}{\partial y}+c \frac{\partial Az}{\partial z},-\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},-\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},-\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})

= -g + E + B

By changing the order of the derivative with the potential, we only flip the sign of the curl. Let's get rid of that pesky g by subtracting away the conjugate:

1. \nabla A - (\nabla A)^* = (0, -E + B)

2. - A \nabla - (A \nabla)^* = (0, E + B)

Take the product of these two:

(\nabla A - (\nabla A)^*)(- A \nabla - (A \nabla)^*) = (0, -E + B)(0, E + B) = (E^2 - B^2, 2 ExB)

It is the scalar that will be used to get the Maxwell field equations. It is the same one that appears in standard EM field theory as Lut pointed out in post 435 if I can get my signs right. The 3-vector is the Poynting vector which plays a role in energy conservation laws. Let's work only with the scalar, and toss in a factor of minus a half, and include a charge coupling term. Write out all of the components:

-\rho \phi + Jx Ax + Jy Ay + Jz Az

+\frac{1}{2} c^2 \left(\frac{\partial Ax}{\partial z}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial z}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial z}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial Ax}{\partial y}\right)^2
-c^2 \frac{\partial Ay}{\partial z} \frac{\partial Az}{\partial y}+\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial y}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial y}\right)^2-c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x}
+\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial x}\right)^2-c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}+\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial x}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial x}\right)^2-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t}
-\frac{1}{2} \left(\frac{\partial Ax}{\partial t}\right)^2-c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t}-\frac{1}{2} \left(\frac{\partial Ay}{\partial t}\right)^2-c \frac{\partial \phi }{\partial z} \frac{\partial \text{Az}}{\partial t}
-\frac{1}{2} \left(\frac{\partial \text{Az}}{\partial t}\right)^2

From here on out, this is the standard way to derive the Maxwell equations. This is the part that looks scary, but its bark is worse that its bite. The goal is to take the derivative of the above Lagrangian with respect to the four potentials, \phi, Ax, Jy, Az, and all 16 derivatives of the four potentials. The first step, taking the derivative with respect to the 4-potentials, gives us back the current, -\rho, Jx, Jy, and Jz.

The derivatives of the Lagrangian with respect to the changing potential is more complicated, but once the rule is learned, it is simple to apply, over and over again. Here are the first 4 of 16:

\frac{\partial }{\partial x_{\mu }}\left(\frac{\partial \mathcal{L}}{\frac{\partial \phi }{\partial x_{\mu }}}\right)

The mu brings in the derivative of phi with respect to t, x, y, z. The derivative out in front will be generating second order derivatives. Let's do this for one of these, say, d phi/dx:

\frac{\partial }{\partial x}\left(\frac{\partial \mathcal{L}}{\frac{\partial \phi }{\partial x}}\right)=-c \frac{\partial ^2\phi }{\partial x^2}-\frac{\partial ^2 Ax}{\partial t\partial x} = \frac{\partial E}{\partial x}

Repeat this for y and z, and you will have Gauss' law, the divergence of E equals rho, \rho = \nabla . E. Now we need to take the derivative of the Lagrangian with respect to Ax, running through t, x, y, and z.

c^2 \frac{\partial ^2 Ax}{\partial z^2}+c^2 \frac{\partial ^2 Ax}{\partial y^2}-c^2 \frac{\partial ^2 Az}{\partial x\partial z}-c^2 \frac{\partial ^2 Ay}{\partial x\partial y}-c \frac{\partial ^2\phi }{\partial t\partial x}-\frac{\partial ^2 Ax}{\partial t^2}

The last two terms are a time derivative of Ex. If you were to calculate the curl of B, then you would recognize the first four terms. The way I spot it, is the two "pure" second order derivatives, which don't have an x, are positive, meaning this is minus the curl of Bx. We also have a +J, so tossing all the terms on the other side generates:

J = \nabla X B - \frac{\partial E}{\partial t}

This is Ampere's law.

What has been done

In this post I showed how to get to the standard EM Lagrangian using the first term of a particular quaternion product. One interesting subtle issue is that the field g was explicitly removed, and so the proposal is gauge invariant - no matter the choice for the terms in g, it does not change a thing because all terms in the g field were subtracted away. In the standard approach, we observe the field equations are gauge invariant. With quaternions, we see the step where the field disappears. There is no difference in the end result, but it is fun to think about.

What will be done

In the coming posts, I will repeat the exercise done in this post for the symmetric fields, e and b. The resulting equations are gauge invariant in exactly the same sense as the Maxwell equations were gauge invariant in this exercise, because g will be subtracted away.

In the third installment, I will not subtract g from either E+B or e+b. The field g goes in, but I hope to show that the g field does not come out. The field equations are the same no matter what g one chooses. It is quite remarkable that g politely disappears from the stage.

Doug

[Lawrence B. Crowell]

Hello:

Simple quaternion expressions generate combinations of three fields, the standard definition of E and B, along with one I call g:

g = (\frac{\partial \phi}{\partial t},- \frac{\partial Ax}{\partial x},- \frac{\partial Ay}{\partial y},- \frac{\partial Az}{\partial z})

Notice how this one has all the parts of a typical gauge field in standard EM.

There are two simple ways to generate g, B, and E using quaternions:

1. \nabla A = (\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})(\phi, Ax, Ay, Az)

(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},\frac{\partial Ax}{\partial t}+c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},\frac{\partial Ay}{\partial t}+c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},\frac{\partial Az}{\partial t}+c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})

= g - E + B

Doug

I am a bit unsure about the nature of g. You have a "nabla" operator on the vector potential, or four potential, which gives a vector quantity. Maybe this nabla is supposed to be a "d" operator?

If you want to work in quaternions things should be in a Clifford basis on Cl(3,1) with the generators


\gamma_1~=~\sigma_2\otimes\sigma_1,~\gamma_2~=~\si gma_2\otimes\sigma_2,~\gamma_3~=~\sigma_2\otimes\s igma_3,~\gamma_4~=~i\sigma_1\otimes{\bf 1}


The electromagnetic potential are then on the Clifford frame if


{\underline A}~=~(e_a^{\mu})\gamma_\mu A^a


for e_a^{\mu} a tetrad or vierbein. The Maxwell equations will then arise from d\wedge{\underline A}~=~{\underline F} and the gauge condition will come from \nabla_a A^a~=~C, for C a constant, set to zero in the Lorentz gauge. This can come from setting (e_a^{\mu})\gamma_\mu~=~D_a which is orthogonal to the vector potential A_a on the bundle section with D_aA^a~=~0 --- again for the Lorentz gauge.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

I wrote out g incorrectly. I have corrected post #438. Nabla is a quaternion, the 4-potential is a quaternion, the Nabla acting on the 4-potential makes a quaternion, and the three fields of g, E, and B together form a quaternion. The g is the first term of the quaternion 4-derivative of the 4-potential.

Doug

[Mentz114]

Doug,

Very nice. I look forward to the next installment. Could you number your equations, please ?

Also using ExB for E\times B caused me some confusion until I repeated the calculation.

Lut

[sweetser]

Hello:

In this post, I will apply the techniques used to generate the Maxwell equations with quaternions to the symmetric analogs of the E and B fields.

The three symmetric fields involved in this analysis are:

g = (\frac{\partial \phi}{\partial t}- \frac{\partial Ax}{\partial x}- \frac{\partial Ay}{\partial y}- \frac{\partial Az}{\partial z}) \quad eq 1

e = (\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y},
\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}) \quad eq 2

b = (c \left(-\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right),
c \left(-\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right),
c \left(-\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) ) \quad eq 3

Although I first spotted these fields by peering into a symmetric rank 2 tensor, for this exercise I will use quaternions. There are two reasons to do so. First, we see exactly why the field equations are invariant under a gauge transformation - such a field will be subtracted away. The second point is more subtle. To do quantum field theory, one needs to take the field equations and invert them to make a propagator. If we consistently use a division algebra, then we necessarily will be able to do that step. With gauge theories such as Maxwell and general relativity, an arbitrary gauge must be chosen before the equation can be inverted. I am trying to get a theory that one is free to choose the gauge, yet remains invertible to get the propagator.

The Hamilton representation of quaternion multiplication will not suffice for generating the symmetric b field, where all the terms that go into a curl have a minus sign. For this reason, I have introduced the Even representation of quaternion multiplication which excludes the Eigen vectors and Eigen values in order to be a division algebra. This representation will not be a Clifford algebra, since e_0^2 = e_1^2 = e_2^2 = e_3^2 = +1. Two simple expressions can generate g, e, and b:

\nabla A2^* = (\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})(\phi, Ax, Ay, Az)

=(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},
-\frac{\partial Ax}{\partial t}+c \frac{\partial \phi }{\partial x}-c\frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},
-\frac{\partial Ay}{\partial t}+c \frac{\partial \phi }{\partial y}-c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},
-\frac{\partial Az}{\partial t}+c \frac{\partial \phi }{\partial z}-c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})

= g - e + b \quad eq 4

A2 \nabla^* = (\phi, Ax, Ay, Az)(\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})

=(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},
\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x}-c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},
\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}-c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},
\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}-c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})

= g + e + b \quad eq 5

To demonstrate this is a subtle proposal, I cut and paste this calculation from my derivation of the Maxwell equations, and made the appropriate sign and letter changes. There are not many, and hopefully I did it right. changing who gets conjugated flips the sign of e, but not b.

Let's get rid of that pesky g by subtracting away the conjugate:

\nabla A2^* - (\nabla A2^*)^* = (0, -e + b) \quad eq 6

- A2 \nabla^* - (A2 \nabla^*)^* = (0, e + b) \quad eq 7

Take the product of these two:

(\nabla A2^* - (\nabla A2^*)^*)(- A2 \nabla^* - (A2 \nabla^*)^*) = (0, -e + b)(0, e + b) = (-e^2 + b^2, -e.X2.e + b.X2.b) \quad eq 8

It is the scalar that will be used to get the symmetric field Maxwell equations. No Poynting vector this time, I don't know what that means. Write out all the components, the same ones as before, but with a different collection of signs:

-\rho \phi + Jx Ax + Jy Ay + Jz Az

+\frac{1}{2} c^2 \left(-\frac{\partial Ax}{\partial z}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial z}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial z}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial Ax}{\partial y}\right)^2
-c^2 \frac{\partial Ay}{\partial z} \frac{\partial Az}{\partial y}-\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial y}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial y}\right)^2-c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x}
-\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial x}\right)^2-c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}-\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial x}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial x}\right)^2-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t}
+\frac{1}{2} \left(\frac{\partial Ax}{\partial t}\right)^2+c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t}+\frac{1}{2} \left(\frac{\partial Ay}{\partial t}\right)^2-c \frac{\partial \phi }{\partial z} \frac{\partial \text{Az}}{\partial t}
+\frac{1}{2} \left(\frac{\partial \text{Az}}{\partial t}\right)^2 \quad eq 9

Fear not, the derivative of this Lagrangian with respect to the potential again generates the current density. Now take the derivative with respect to the for derivatives of phi:

-\rho +\left(c^2 \frac{\partial ^2\phi }{\partial x^2}+c^2 \frac{\partial ^2\phi }{\partial y^2}+c^2 \frac{\partial ^2\phi }{\partial z^2}-c\frac{\partial ^2 Ax}{\partial t\partial x}-c\frac{\partial ^2 Ay}{\partial t\partial y}-c\frac{\partial ^2 Az}{\partial t\partial z}\right)==0 \quad eq 10

Plug in the definition of e and rearrange:

-\nabla . e = \rho \quad eq 11

This looks similar to Gauss' law, but there is an important difference. In the static case, the time derivatives of A make no contribution, and we have:

\nabla^2 \phi = \rho \quad eq 12

For Gauss' law of EM, under the same static conditions:

-\nabla^2 \phi = \rho \quad eq 13

The reason like charges repel in EM has to do with the above minus sign, while for gravity to attract, the Laplacian of phi must have the same sign as the current density.

Take the derivative with respect to the 4 derivatives of Ax:

Jx-c^2 \frac{\partial ^2 Ax}{\partial z^2}-c^2 \frac{\partial ^2 Ax}{\partial y^2}-c^2 \frac{\partial ^2 Az}{\partial x\partial z}-c^2 \frac{\partial ^2 Ay}{\partial x\partial y}-c \frac{\partial ^2\phi }{\partial t\partial x}+\frac{\partial ^2\text{Ax}}{\partial t^2}==0 \quad eq 14

The time derivative of e is the last two terms. The symmetric curl is negative, but two symmetric curls is positive, so this is minus the symmetric curl of b. Plug in and rearrange:

\nabla . X2 . b - \frac{\partial e}{\partial t} = J \quad eq 15

Looks similar to Ampere's law. The symmetric Gauss and Ampere's law should not be shocking, since the symmetric fields e and b are built from the same terms as E and B. There are important differences, such as the symmetric Gauss' law has like charges that attract as happens for gravity, whereas the EM Gauss' law has like charges repel. The two Ampere's laws have two different sorts of curls (since curl is one of the most confusing concepts I have come across, I will not venture about the meaning of two different curls).

Again we can choose whatever we want for g = (\frac{\partial \phi}{\partial t}- \frac{\partial Ax}{\partial x}- \frac{\partial Ay}{\partial y}- \frac{\partial Az}{\partial z}) will not change these equations at all. Some of my readers will say that is a good thing :-)

Doug

[Mentz114]

Hi Doug,
very interesting. I'm impressed with your 'theory engineering' approach.

I've gone back looking for the definition of the X2 operator ( even multiplication ?) and also, is conjugation the same in the even algebra ? I need the operators because I'm repeating it with tensors and pseudo-tensors.

I presume the aim is to get the scalar part of this term

(\nabla A2^* - (\nabla A2^*)^*)(- A2 \nabla^* - (A2 \nabla^*)^*) = (0, -e + b)(0, e + b) = (-e^2 + b^2, -e.X2.e + b.X2.b) \quad (8)

into the lagrangian.

Lut

[sweetser]

Hello Lut:

The X2 is the symmetric analog to the cross product.

e.X2.e = (2 ey ez, 2 ex ez, 2 ex ey) \quad eq 1

Yes, the conjugate for Even representation of quaternions does exactly the same thing as for the Hamilton quaternions. When it came to programming this in Mathematica, I had be taking the conjugate of the real 4x4 matrix representation. That no longer works. Instead, I had to implement it on the 4-vector:

(t, x, y, z)^* \rightarrow (t, -x, -y, -z) \quad eq 2

This definition of a conjugate works for either 4x4 real matrix representation of a quaternion, which is a good sign.

When I did the previous post, I kept getting eq 8 wrong because I am too familiar with the Hamilton representation of quaternions. If you are doing this all with tensors, see if you can make use of symmetric tensors. The algebra should line up (loads of details left to you).

Doug

[sweetser]

Hello:

I find the hand drawn derivations more satisfying than what LaTeX produces because it has that human element. Doing things by hand is kind of like Sudoku: the rules are simple, there is a right way, and you get in a grove, so repetition acts as confirmation. Anyone trying to do this on their own can use these as crib sheets.

These three URLs are relevant to this thread. I have shrunk each page down to 65k, so it shouldn't be so bad to load. I recommend hitting the magnifier, using the hand cursor to move up and down. Right now there are three low resolution, and three high resolution derivations of field equations.

1. The Maxwell equations using quaternion operators (http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations/photo#5171869015477455458)

2. The symmetric field Maxwell equations for gravity (http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations/photo#5171869019772422770)

3. The two together - where g magically cancels out (http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations/photo#5171869024067390082)

I have not discussed case 3 yet, but hopefully will get to it this weekend.

Doug

[sweetser]

Hello:

For the symmetric field Maxwell equations, in eq 9 of post 442, I wrote out the current coupling terms in its component parts:

-\rho \phi +Jx Ax + Jy Ay + Jz Az

What I omitted was how to form this contraction. The details here matter, because if one just does the same as for the Maxwell equations, [itex]-J A[/tex], Feynman and others have show that the phase - stuff other than the scalar - has spin 1 symmetry where like charges repell. I should have included this:

+\frac{1}{2}(J^{*1} A^{*2 *3} + (J^{*1} A^{*2 *3})^*) = -\rho \phi +Jx Ax + Jy Ay + Jz Az

The scalar is the same, but the phase has spin 2 symmetry which is needed for an equation if like charges attract.

Doug

[Lawrence B. Crowell]

I recently worked out some stuff involving the emission of radiation by accelerated charges, the Unruh effect and the emission of radiation by a charge falling into a gravity field. I figured I'd send these here, maybe they will help with some thing. This is part I of III


The Lamour formula gives a relationship between the acceleration of a charge and the radiation power emitted by the charge


P~=~-\frac{2}{3}\frac{q^2}{c^3}a_\mu a^\mu.


This has lead to an ideological stance that accelerated charges emit radiation. Yet it is clear that a charge sitting on the surface of a gravitating body will not emit radiation. The work exerted on such a charge W~=~\int {\vec F}\cdot d{\vec r} is zero as there is no displacement of charge. Similarly, we might suppose that with the Einstein equivalence principle that a freely falling charge should also not radiate. If we were to consider a small falling frame comoving with the charge then at least within that frame any radiation response should be zero. It would then indicate the appearance of electromagnetic radiation is coordinate dependent. This might in fact be the case even if in other coordinate systems the infalling charge is found to be emitting radiation.

Maxwell’s electrodynamics indicates that for a four-vector potential A^\mu the associated four-current is


\square A^\mu~=~4\pi j^\mu.


For a current from a discrete point charge the current is then determined by a delta function


j^\mu(x)~=~q\int_{-\infty}^\infty U^\mu(s)\delta(x~-~y(s))ds


for U^\mu~=~dy^\mu/ds the four velocity of the charge. The Maxwell result for the four-vector potential demands that


A^\mu~=~4\pi\int d^4x^\prime G(x~-~x^\prime)j^\mu(x^\prime)


where the propagator obeys the harmonic condition \square G(x~-~x^\prime)~=~\delta(x~-~x^\prime), which under the integration reduces is to an integration of the delta function along the worldline of the particle. A standard form for the Green’s function is the Pauli-Jordan function


G(x~-~y)~=~\frac{1}{2\pi}\theta(x^0~-~y^0)\delta\big((x^\mu~-~y^\mu)(x_\mu~-~y_\mu\big)


where the step function turn on the potential for x^0~-~y^0~>~0 and defines a retarded four-vector potential. Using the identity for Dirac delta functions \delta(f(x))~=\sum_i\delta(x~-~x_i)/f^\prime(x) the four-potential is then


A^\mu(x)~=~q\int_{-\infty}^\infty U^\mu\frac{\delta(s~-~\sigma)}{U_\nu(z)(x^\nu~-~y^\nu)}ds,


where the particle path is parameterized by \sigma with U^\nu(\sigma) and y^\nu(\sigma). The distance along the null cone is identified as r~=~U_\nu(x^\nu~-~y^\nu) , which gives the Lenard-Wiechart potential A^\mu(x)~=~qU^\mu(x)/r.

This may now be extended to accelerated frames. For a frame of constant acceleration the coordinates paramterized by \sigma are given by


y^0~=~g^{-1}sinh(g\sigma),~y^i~=~g^{-1}cosh(g\sigma),~U_0~=~cosh(g\sigma),~U^i~=~sinh(g \sigma)


and the vector x^\nu~=~(t,~0,~0,~0) for a rest frame. The propagators is then


G(\sigma)~=~\frac{1}{2\pi}\frac{1}{t~cosh(g\sigma) }


The vector potential will then have the components


A^0(x)~=~qc/\rho,~A^i~=~(qc/\rho)tanh(gs),


where \rho~=~ct. What is of particular importance is the evaluation of fields on the frame. We consider the change in the energy of a probe with states \{E\} and we consider the transition of the probe from states E to a set of possible states E’ due to a change in the vacuum states from |0\rangle to |0^\prime\rangle from the inertial to accelerated frames. The probability is then given by


\sum_{E^\prime}\langle 0,~E^\prime|0,~E\rangle~=~\sum_{E^\prime}\langle E^\prime|\mu|~E\rangle\int dse^{i(E^\prime~-~E)s}G(s)~=~ \sum_{E^\prime}\langle E^\prime|\mu|~E\rangle g^{-1}\sqrt{\frac{\pi}{2}}sech\Big(\frac{\pi}{2g}(E^\p rime~-~E)\Big)


since the Fourier transform of the sech is itself. Here the term \mu[/tex] is a coupling term for the probe which couples with the vector potential of the field measured. This result is given according to the Fourier set of frequencies of photon an accelerated charged particle will interact with. As indicated later the sech term can be seen to be approximately [itex]\simeq~e^{(E^\prime~-~E)/2g}, which is a Boltzmann distribution of photons.

The condition on the null cone (x^\mu~-~y^\mu)(x_\mu~-~y_\mu), and the gradient \nabla_\nu operator on this condition is still zero with


\Big({\delta^\mu}_\nu~-~\frac{dy^\mu}{d\sigma}\nabla_\nu\sigma\Big) (x_\mu~-~y_\mu)~=~0


or


x_\mu~-~y_\mu~=~r\nabla_\nu\sigma


which defines the null momentum vector for a photon as k_\mu~=~(x_\mu~-~y_\mu)/r, where for the case of an accelerated charged particle r~\sim~sech(g\sigma). The acceleration a^\mu~=~dU^\mu/d\sigma, the spatial position vector y^\mu~=~\tau a^\mu~-~U^\mu and with the null vector k^\mu defines a normal vector n^\mu~=~k^\mu~-~U^\mu so that 1/\tau~=~a_\nu k^\nu. These conditions are used in the differentiation of the Lenard-Wiechart potential


A_{\mu,\nu}(x)~=~\frac{q}{r}k_\nu(a_\mu~-~\tau^{-1}U_\mu)~+~\frac{q}{r^2}n_\nu U_\mu


which gives the field tensor components as F_{\nu\mu}~=~A_{[\mu,\nu]}. The four potential and the resulting field strength tensor contains 1/r,~1/r^2 terms which respectively contain the radiation field due to accelerations and velocity changes and the second term which is the field term comoving with a charge at a velocity.

Lawrence B. Crowell

[Lawrence B. Crowell]

From the four vector potential of the electromagnetic field we have the field tensor


F_{\mu\nu}~=~A_{[\mu,\nu]}~=~\frac{q}{4\pi r}k_{[\mu}y_{\nu]}~+~\frac{q}{r^2}n_{[\mu}U_{\nu]}


The Lagrangian for the electromagnetic field is {\cal L}~=~(1/4)F^{\mu\nu}F_{\mu\nu} and for the momentum-energy tensor \partial{\cal L}/\partial g^{mu\nu}~-~g_{\mu\nu}{\cal L} and so


T_{\mu\nu}~=~-F^{\mu\gamma}{F^\mu}_{\gamma}~-~{1\over 4}g^{\mu\nu}F^{\sigma\gamma}F_{\sigma\gamma}~=~\fr ac{q^2}{r^2}(y_\sigma y^\gamma)k_\mu k_\nu


For the case with an accelerated frame r~\rightarrow~r cosh(g). The four momentum density of the electromagnetic field is P_\mu~=~\int_V d^3x T_{\mu\nu}n^\nu, for the volume specified between two spatial surfaces. The continuity condition on the electromagnetic field \partial_\mu T^{\mu\nu} means that the derivative of the energy along the proper time is


\frac{dP_\mu}{ds}~=~\int_V d^3x T_{\mu\nu}\frac{n^\nu}{ds}
[/itex]

and for dn^\mu/ds~=~dk^\mu/ds~-~dU^\mu/ds and letting the volume extend to \pmnull infinity, as r~\rightarrow~\infty it can be shown that

[tex]
\frac{dP_\mu}{ds}~=~-\frac {q^2}{4\pi }(a_\nu a^\nu~+~(1/\tau)^2)k_\mu~=~-q^2(a_\nu a^\nu~+~(a_\nu k^\nu)^2)k_\mu


When integrated over 2\pi r^2 dr the first term gives the Lamour formula for radiated power. Since the spacetime acceleration is orthogonal to the four velocty and a_\nu a^\nu~=~d/ds(U_\nu a^\nu)~-~(da^\nu/ds)U^\nu the formula is put in the form which involves the derivative of the acceleration. It is for this reason that the Lamour formula is defined for systems which have a non-constant acceleration, such as the cyclical motion of a charge in a magnetic field and Bremmstrahlung radiation. The second term depends upon the “acceleration curvature” 1/\tau, and it is from this term where the Unruh effect emerges.

Lawrence B. Crowell

[Lawrence B. Crowell]

It is now time to turn our attention to the case of a charged particle in a curved spacetime. In particular the sphereically symmetric spacetime with metric


ds^2~=~(1~-~F(r))dt^2~-~(1~-~F(r))^{-1}dr^2~-~r^2d\Omega^2


which is the Reissner-Nordstrom spacetime for F(r)~=~2M/r~-~Q^2/r^2~+~\Lambda r^2/3, for M,~Q,~\Lambda a the mass of a spherically symmetric body, the charge and the deSitter parameter or cosmological constant. This metric is asymptotically flat and corresponds to a Minkowski spacetime as r~\rightarrow~\infty, where the tetrad e^a_\mu~=~(T_\mu,~R_\mu,~\Theta_\mu,~\Phi_\mu) defines the above line element as ds^2~=~e^a_{\mu}e^a_\nudx^\mu dx^\nu and the tetrad components necessarily defined so


T_\mu dx^\mu~=~(1~-~F(r))^{1/2}dt


R_\mu dx^\mu~=~(1~-~F(r))^{-1/2}dr


\Theta_\mu dx^\mu~=~rd\theta


\Phi_\mu dx^\mu~=~rsin(\theta)d\phi,


with the appropriate signature change on the spatial parts on contraction of the tetrad. The electromagnetic field tensor is easily found by computing it in the flat region and with the tetrads extending it to the RN spacetime


F_{mu\nu}~=~-\frac{Q}{r^2}(T_\mu R_\nu~-~T_\nu R_\mu)


and the electromagnetic contribution to the Ricci curvature is the trace free components


R_{\mu\nu}~=~-\frac{Q^2}{r^4}(T_\mu T_\nu~-~R_\mu R_\nu~+~\Theta_\mu\Theta_\nu~+~\Phi_\mu\Phi_\nu)


The RN metric permits the Killing vector k^\mu\partial_\mu~=~(1~-~F(r))^{-1/2} where an observer falling into the black hole measure the electric field E_\mu~=~F_{\mu\nu}k^\nu~=~(q/r^2) in the radial direction.

For a charge which falls into a neutral Schwarzschild gravity field we set the RN term F(r)~=~1~-~M/r. For a body falling radially in the spacetime its proper time d\tau~=~dt~-~F(r)^{1/2}(1~-~F(r))^{-1} defines the transformed line element


ds^2~=~(1~-~F(r))d\tau^2~-~2F(r)^{1/2}d\tau dr~-~dr^2~-~r^2d\Omega


with the tetrad basis


\tau_\mu dx^\mu~=~(1~-~F(r))^{1/2}d\tau~-~\Big(\frac{F(r)}{1~-~F(r)}\Big)^{1/2}dr,


R_\mu dx^\mu~=~(1~-~F(r))^{-1/2}dr.


The dual vector basis which obeys e^a_\mu dx^\mu(e^{b\nu}\partial_\nu)~=~{\delta^a}_b are


\tau^\mu \partial_\mu~=~(1~-~F(r))^{-1/2}\partial_\tau


R^\mu \partial_\mu~=~-(1~-~F(r))^{1/2}\partial_r~-~\Big(\frac{F(r)}{1~-~F(r)}\Big)^{1/2}\partial_\tau


The duality between the differential and vector elements give the tangent vectors for the free fall as


U_\mu dx^\mu~=~d\tau,~ U^\mu\partial_\mu~=~\partial_\tau~-~F(r)^{1/2}\partial_r


The geodesic condition {U^\mu}_\nu U^\nu~=~0 results in the radial acceleration


\partial^2_\tau r~=~-\frac{M}{r^2}


which for a weak field case with t~\simeq~\tau this reduces to the Newtonian case. Since this approach to the RN metric is according to variables which fit into the asymptotic Minkowki spacetime the power formula for the change in the the four momentum vector as


\frac{dP_\mu}{d\tau}~=~-\frac {q^2}{4\pi }(\partial^2_\tau r\partial^2_\tau r~+~(1/B)^2)k_\mu,


where the acceleration curvature has been relabed with B. So this calculation indicates that a charge falling into a gravity will indeed emit electromagnetic radiation.

So what is going on? We have this idea of the Einstein equivalence principle. A local infalling frame with a charge should be equivalent to a charge in an inertial frame in Minkowski spacetime. Indeed this is the case, and this is the case for a region where the tidal acceleration is \sim~1/r^3 extremely small. An observer in a small frame where the Riemann curvature multiplied by a four volume (which for repeated indices is reduces the volume by one dimension) is small R_{abcd}\delta vol^{abcd}~\sim~0 will observe no electromagnetic radiation being emitted. A small enough of a local frame is in the near field domain, where the change in the electromagnetic field, or its wavelength, is small enough so the field appears constant or Coulombic. The electromagnetic field only appears in a region far removed from the infalling charge where the tidal acceleration is significant.

There is one last little bit with this part of the problem. If the charge is emitting radiation by falling into the gravitating body, then this radiation is being produced from the gravitational potential energy. This should slow down the infalling charge or equivalently reduce its kinetic energy. The change in the energy will be dP_\mu~=~Qk_\mu d\tau,for Q the above acceleration dependent power term. The power P~=~F\cdot U in relativistic terms and the above equation means that there is a counter acceleration term A~=~q^2/r^3. If we were to continue the same calculation with the RN term F(r)~=~M/r~-~q^2/r^2 we will obtain an identical term. The presence of the charge on the black hole effectively reduces the gravitational mass and thus curvature for the geodesic of a neutral particle. This gives an identical result for the acceleration of a charged particle approaching a neutral black hole. There is a symmetry here. For a neutral particle falling into a black hole the acceleration is reduced by the effective mass reduction of the black hole by the charge. The charged particle falling into a neutral black hole is slowed by an equal amount by radiating an equal amount of mass-energy in a radiation field.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

This looks like good, honest technical work. I wish you luck with it. However it is not relevant to the topic at hand. The charged metric for the GEM proposal is:

d \tau^2 = exp(2(\sqrt{G} Q - GM)/c^2 R) dt^2 - exp (-2(\sqrt{G} Q - GM)/c^2 R) dR^2/c^2 - R^2d\Omega^2/c^2

The Taylor series of this metric does not match the Reissner-Nordstrom metric. The simplest way to see this is M and Q come in on equal footing for GEM. It should be obvious that Nature works this way, not the affect of M is comparable to Q2. I did a thought experiment on this subject, all of three slides, here:

http://theworld.com/~sweetser/quaternions/talks/rank1/4221.html

The gist is quite simple. You have an electron. You measure that the electron is attracted to a box. You do not know what is inside the box. If you have one proton, or 421 kg of mass inside the box, you will not be able to tell any difference in the attraction from the effect of electric charge attraction, or the gravitational mass attraction of the 421 kg brick.

Good luck in your studies, but please let's stay focused on the metric, field equations, and the derivation of the field equations which is more than enough to keep us busy.

Doug

[sweetser]

Hello:

A basic property of electromagnetism is that like charges repel and different charges attract. That must be a property of any proposal related to EM. The Reissner-Nordstrom can only accept like charges that repel, since it uses Q2, which always has the same sign (and opposite the M/R term). That looks like a deadly flaw to me, one that prevents me for doing anything further with the metric.

In the GEM proposal, the sign of the relative electric charge (ie same sign for source and probe means positive) and the mass charge are different. That way like electric charges can repel, and like mass charges can attract. If the relative electric charge is negative, as happens with a positive and negative charge, then the electric charge contribution would be attractive in exactly the same way as gravitational charge. The two add up. There is no negative mass charge in the GEM proposal because the mass field is about symmetric fields. There are two charges for EM because of the antisymmetric fields.

Doug

[Lawrence B. Crowell]

Obviously with the 421 kg mass in a box or a unit charge it will be easy to tell the difference. The mass will attract all masses equally, while the charge will attract or repel the same whether that has the same or opposite charge. The proton will have negligable attraction to an uncharged mass.

The NS metric is for a spacetime determined by a sperically symmetric mass with a charge. One could well enough compute the orbit of a charged particle in this spacetime. Yet what I compare is the fall of a neutral mass around a charged black hole and a charge falling into a neutral black hole. What I find is that the "braking" of a charged particle fall by the emission of radiation by purely classical means is equivalent to the geodesic rate of fall for a neutral particle entering a black hole with the equivalent charge. This result illustrates I must be at least on the right track. It is in line with the "no hair" theorem for black holes is that the details of how a black hole reaches a final state by the acquisition of charge and mass by the infall of matter and fields is independent of the details by which that occurred. A black hole will end up with the same final mass M~=~M_0~-~(q/r)^2 indepdendent of which masses brought in what charge.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

The thought experiment did not concern itself with the ability to tell the difference between the behavior of electric charge or mass charge. Rather, it was an effort to find a condition where the behavior was indistinguishable.

> One could well enough compute the orbit of a charged particle in this spacetime.

If and only if the charged mass had the same sign as the charge on the test mass would the NS metric work. If th test had the opposite sign, it would fail because the metric must demonstrate attraction.

I appreciate that you looked into the consequences of the NS metric, and that other skilled people have worked with it. But it looks unacceptable to me because it cannot model electrical attraction.

Doug

[Lawrence B. Crowell]

Hello:

A basic property of electromagnetism is that like charges repel and different charges attract. That must be a property of any proposal related to EM. The Reissner-Nordstrom can only accept like charges that repel, since it uses Q2, which always has the same sign (and opposite the M/R term). That looks like a deadly flaw to me, one that prevents me for doing anything further with the metric.

Doug

The RN metric uses (Q/r)^2 as the self-energy of a charged mass. This can be derived by using the electromagnetic field tensor as a source of the gravitational field. From there the RN metric is a solution to the Einstein field equation. In more advanced settings solutions of this type are BPS black holes, where gauge "charges" are the source of the black hole metric. The charge here is the charge of the black hole. One might then consider the motion of a charged mass in this spacetime, where that charge can have either sign.

This is fairly standard stuff.

Lawrence B. Crowell

[Lawrence B. Crowell]

Hello Lawrence:



> One could well enough compute the orbit of a charged particle in this spacetime.

If and only if the charged mass had the same sign as the charge on the test mass would the NS metric work. If th test had the opposite sign, it would fail because the metric must demonstrate attraction.

Doug

I am not sure where you got this idea. You can well enough compute the orbit of a charge with any value or sign in an RN metric.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

I agree, you have represented standard analysis, which is darn good most of the time. If I see a killer flaw, I will point it out.

Even better is when I realize I was wrong about the killer flaw. I "get" the standard way, and can pinpoint where I drive a different direction. That is what happened with regards to the Reissner-Nordstrom metric.

In general relativity, gravity binds to the stress energy tensor. How much energy will the electric field contribute to the metric? G Q^2/c^4 R^2[/tex]. If the charge is positive, or the charge is negative, it does not matter, the same amount goes in.

For the GEM proposal, the charge coupling term is [itex]-\frac{1}{4 c}(J A + (J A)^* - J^{*1} A^{*2 *3} - (J^{*1} A^{*2 *3})^*) With GEM, the effort is to do both gravity and EM. All these conjugates are required to get the phase to have both spin 1 and spin 2 symmetry. The coupling of gravity is not to energy, but instead to the 4-current density. The energy of the electric field makes zero contribution to gravitational effects.

What we get in return for giving up the energy connection is a new equivalence principle for particles that attract each other. The attraction might be caused by electric charge or by a mass charge. Attraction looks the same either way. Sure, you could do other tests to tell which one was at work, but one could have videos of a particle being attracted by gravity or by EM that are absolutely identical. That is the bridge between gravity and EM.

People who work on EM almost never deal with the tools of differential geometry. Calculate the divergence of the Christoffel symbol of the second kind for this metric:

d \tau^2 = exp(2(\sqrt{G} Q - GM)/c^2 R) dt^2 - exp (-2(\sqrt{G} Q - GM)/c^2 R) dR^2/c^2 - R^2d\Omega^2/c^2

Since this is for a static, spherically symmetric metric, to be logically consistent with EM, the answer had better be both Gauss' and Newton's potential theory for EM and gravity,\nabla^2 \frac{\sqrt{G} Q - GM)}{R}, which it is. Lucky? I don't think so.


Doug

[sweetser]

Hello:

In this post, I will apply similar, but not quite identical, approaches to generating the field equations for the fields E, B, e, and b.

If one hopes to model particles that travel at the speed of light, that requires that the field theory be invariant under a gauge transformation. This is one of those constraints on finding a solution I have heard, I have accepted, and I don't understand as well I should.

In deriving the Maxwell field equations using quaternion operators, the gauge invariance was achieved by noting all the derivatives that make up a gauge are in the first term of -A \nabla. The first term was subtracted away using the time honored quaternion trick subtracting the conjugate, q - q* = vector(q). The identical method was applied to the symmetric fields e and b. If were were to just add these two results together, there would be no link between the two sets of equations.

This time we will not use the trick of subtracting the conjugate, yet no terms with g appear in the final field equations because the g2 terms in one cancel the g2 terms of the other in the scalar. There is a g in the 3-vector, but that is not used for generating the field equations.

Start by taking the derivatives of 4-potentials in two ways. For quaternions written in the Hamilton basis, change the order of the differential operator with the potential, which flips the sign of B. For quaternions written in the Even basis, change which term gets conjugated. Calculate:

\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))

=((-\frac{\partial \phi}{\partial t} ~+~ c \nabla . A, -\frac{\partial Ax}{\partial t} ~-~ \nabla \phi ~+~ \nabla X A)(\frac{\partial \phi}{\partial t} ~-~ c \nabla . A , \frac{\partial Ax}{\partial t} ~+~ \nabla \phi ~+~ \nabla X A)
~+~ (\frac{\partial \phi}{\partial t} - c \nabla . A,\frac{\partial Ax}{\partial t} ~-~ \nabla \phi ~-~ \nabla .X2. A)(\frac{\partial \phi}{\partial t} ~-~ c \nabla . A,-\frac{\partial Ax}{\partial t} ~+~ \nabla \phi ~-~ \nabla .X2. A))

= ((-g, E ~+~ B)(g, -E ~+~ B) ~+~ (g, e ~+~ b)(g, -e ~+~ b))

(-g^2 ~+~ E^2 ~-~ B^2 ~+~ g^2 ~-~ e^2 ~+~ b^2, 2 E X B ~-~ e .X2. e ~+~ b .X2. b ~+~ 2 gE ~+~ 2 gb) \quad eq 1

The g field is not in the scalar due to a cancelation, but is in the 3-vector. The field equations are generated from the scalar, not the 3-vector, so any choice for the gauge g will not effect the field equations I am about to derive.

The current coupling term is complicated by the need to have spin 1 and spin 2 symmetry in the phase. This was worked out earlier in the thread, and here is the solution:

-\frac{1}{4}(J A ~+~ (J A)^* ~-~ J^{*1} A^{*2 *3} ~-~ (J^{*1} A^{*2 *3})^*) = ~-\rho \phi ~+~ Jx Ax ~+~ Jy Ay ~+~ Jz Az \quad eq 2

Write out the Lagrangian by its components, including the current coupling terms:

\mathcal{L}_{BEbe} ~=~ -c^2 \frac{\partial \Ay}{\partial z} \frac{\partial Az}{\partial y} ~-~ c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x} ~-~ c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}
-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t} ~-~ c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t} ~-~ c \frac{\partial \phi }{\partial z} \frac{\partial Az}{\partial t}
-\rho \phi ~+~ Jx Ax ~+~ Jy Ay ~+~ Jz Az \quad eq 3

There are fewer terms than in either the Maxwell Lagrangian or the symmetric field Lagrangian because terms between the two cancel. The fields in the field equations will need to do the same. Calculate the first field equation by taking the derivative of \mathcal{L}_{BEbe} with respect to the 4 derivatives of phi.

\frac{\partial}{\partial x^{\mu}}\left( \frac{\partial \mathcal{L}_{BEbe}}{ \left( \frac{\partial \phi}{\partial x^{\mu} \right)}} \right) ~=~ -c \frac{\partial ^2 Az}{\partial t\partial z} ~-~ c\frac{\partial ^2 Ay}{\partial t\partial y} ~-~ c\frac{\partial ^2 Ax}{\partial t\partial x}\right) ~-~ \rho

=~ \frac{1}{2}\nabla . (E ~-~ e) ~-~ \rho ~=~ 0 \quad eq 4

Nice, the E and e terms work together to isolate the A derivatives. And yet, you can spot Gauss' law for EM where like charges repel. There is a Gauss-like law for like charges that attract. Repeat for the derivative with respect to Ax:

\frac{\partial}{\partial x^{\mu}} \left( \frac{\partial \mathcal{L}_{BEbe}}{ \left( \frac{\partial Ax}{\partial x^{\mu} \right)}} \right) ~=~ -c^2 \frac{\partial ^2 Az}{\partial x\partial z} ~-~ c^2 \frac{\partial ^2 Ay}{\partial x\partial y} ~-~ c\frac{\partial ^2\phi }{\partial t\partial x}\right) ~+~ Jx ~=~ \frac{1}{2}(-\nabla X B ~-~ \nabla .X2. b ~+~ \frac{\partial(Ex ~+~ ex)}{\partial t}) ~+~ Jx ~=~ 0 \quad eq 5

You should be able to spot Ampere's law.

What has been done

There is now a formulation of the GEM proposal that uses quaternions exclusively. The standard quaternion algebra inherited from the nineteenth century needed to be extended in two ways. First the idea of a conjugate (or anti-involutive automorphism in fancy jargon, or thingie that flips the sign of all but one part in simple words) had to be expanded to *1, *2, and *3 to get the phase symmetry right for the current coupling term.

The second advance in quaternion algebra needed is the Even representation of quaternion multiplication. Here the eigenvectors of the representation must be excluded to make the representation an algebraic field. I have consistently said I am more comfortable when I found out someone else has done this before. The multiplication table is known as the Klein four-group (http://en.wikipedia.org/wiki/Klein_four-group). I will have to see if others have noticed what happens when the eigenvectors are excluded.

With these two innovations, the field equations for E, B, e and b have been generated. These field equations are gauge invariant because g may be whatever one chooses since it is canceled out in the process. This is vital since the graviton and proton both travel at the speed of light.

What needs to be done

I need to develop a non-gauge invariant set of field equations for massive particles, where the gauge symmetry is broken by the mass charge. I have something technical to keep me off the streets.

Doug

[sweetser]

Hello:

A gauge-free unified field equation has been constructed using quaternion operators (previous post), a very good thing because the particles that mediate gravity and EM travel at the speed of light. The particles that interacted with these mediators of force - massive and possibly electrically charged particles - do not travel at the speed of light, so the gauge symmetry must be broken.

In a standard approach to EM, gauge symmetry is broken with the Higgs mechanism. This is done by postulating that there is a Higgs field everywhere any particle ever goes. The vacuum state is a false vacuum, actually higher than a nearby state that will give particles like quarks mass without breaking the symmetry needed by EM.

False vacuums remind me of false gods, something akin to the aether that had to be everywhere, no place in the Universe could be without. The Universe is a clumpy place, and it would be a magical coincidence if the Higgs field got to every place it needed to in the right density in order to give every particle the same symmetry breaking experience needed so all protons, neutrons, and electrons have the same mass. These are reasonable skeptical objections, but they get no air time today for a good reason: there is no alternative. One choice makes things simple.

A thought experiment may show that mass charge does break electric charge symmetry. You have a pair of electrons sitting 1 cm apart, and you measure how fast they accelerate away from each other, a_e = F/m_e[/tex]. Repeat the experiment, but for a pair of protons, a_p = F/m_p[/tex]. Let's say your experimental system is so good you are able to measure the two accelerations to ten significant digits. The value of the two accelerations is identical. This equivalence indicates that electric charge is universal.

Repeat the experiment without changing anything concerning the setup. There are two electrons over here rushing away from each other, two protons doing the same thing. The one difference is the acceleration is now measured to twenty significant digits. Now the two accelerations will not be the same. The reason is the gravitational mass has a trivial effect that will keep the heavier protons from accelerating as fast as the pair of electrons because the protons attract each other gravitationally more than the electrons.

Unlike the standard model + Higgs which characterize inertial mass and ignores gravity, the GEM proposal is about gravity and EM working together ever so lightly. We saw that in post 457, eq 1, where the g2 term contributed from EM canceled with the g2 tossed in by gravity.

This time, I will combine the method used to generate the Maxwell equations (post 438) with its Even quaternion representation clone (post 442). If you look carefully at the two Lagrangians, you would notice all the cross terms are the same, and they all have a minus sign (but don't look too close or you will notice I got the sign of one term wrong). This means if we subtract one from the other, all the mixed terms drop, leaving only 12 squared terms.

When I was using tensors, I found the contraction of the two asymmetric rank 2 tensors, [itex]\nabla_{\mu} A_{\nu} \nabla^{\mu} A^{\nu}, had these same twelve terms, along with 4 others. The field equations that come out of the asymmetric field strength tensor contraction are drop dead gorgeous (from page 1 of this long thread):

J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}

What a beautiful set of equations! They are beautiful, first, because they are nicely separated - with the charge density, goes [itex]\phi; with the current, goes A. Furthermore, although the [right] side looks a little funny - a Laplacian together with a (\partial/\partial t)^2 - when we unfold it we see...it has a nice symmetry in the x, y, z, t - the [c's are] necessary because, of course, time and space are different; they have different units.

To get the context correct, Feynman was extolling the virtues of writing the Maxwell equations in the Lorenz gauge. Earlier in this thread we went over how Feynman showed the current coupling term J^{\mu} A_{\mu} has spin 1 symmetry, necessary for EM where like charges attract. I know Feynman did not analyze the spin of (i J^{\mu} i)^* (k(j A_{\mu} j)^* k)^*. That has spin 2 symmetry, necessary for gravity where like mass charges attract. That's the strength of this proposal, it keeps getting more subtle.

I think I will stop here tonight and let the goal sink in...

Doug

[Lawrence B. Crowell]

Hello:

The field equations that come out of the asymmetric field strength tensor contraction are drop dead gorgeous (from page 1 of this long thread):

J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}


To get the context correct, Feynman was extolling the virtues of writing the Maxwell equations in the Lorenz gauge.
Doug

The problem with this wave equation which gives the currents is that it assigns a mass current to an electromagnetic potential. I doubt many are going to cotten on to this idea.

The electromagnetic and gravity fields have their sources because of their group structure. Electromagnetism is an abelian gauge field A\wedge A~=~0[/tex] which has the unitary group U(1). The two roots of this group are the real parts of the circle on the Argand plane C of compex numbers. Those are of course the numbers +1 and -1. Gravitation is the Lorentz group which is the hyperbolic group SO(3,1). This group is similar to SO(4)~=~SO(2)\times SO(2) though the hyperbolic group is noncompact. SO(3,1)~=~SL(2,C)\times Z_2 and the special linear group is SU(2)\times SU(1,1). These two parts define the set of three rotations and three boosts which give the six generators of the Lorentz group.

Compact groups have nice properties that a set of transformations of the group generator will converge in a Cauchy sequence. Hyperbolic groups are non-compact and so a sequence is likely to go off to asymptopia and not converge.

If we think of gravity as a gauge-like theory with F~=~dA~+~A\wedge A for nonabelian guage fields the DE's for these on the classical level are nonlinear. Yet we can quantize these, but renormalization is a bit complicated. We can well enough quantize a SO(4) theory obtained in euclideanization. But gravity is a strangely different. Why? The gauge group SU(1,1) is hyperbolic. In the Pauli matrix representation we have that \tau_z~=~i\sigma_z. So we form a gauge connection


A~=~A^{\pm}\sigma_{\pm}~+~iA^3 \sigma_3


and for the group element g~=~exp(ix\tau_3) =~exp(-x\sigma_3) the connection term transforms as

[tex]
A'~=~g^{-1}Ag~+~g^{-1}dg~=~e^{-2x}A^{\pm}\sigma_{\pm}~+~iA^3\sigma_3


and for x~\rightarrow~\infty this gives A~\rightarrow~iA^3\sigma_3. Now A^{\pm}\sigma_{\pm} and A^3\sigma_3 have distinct holonomy groups and are thus distinct points (moduli) in the moduli space. But this limit has a curious implication that the field F~=~dA~+~A\wedge A for these two are the same and the moduli are not separable. In other words the moduli space for gravity is not Hausdorff. This is the most serious problem for quantum gravity.

Because of this it is not possible to construct the connection coefficient for gravitation from an internal gauge connection that is the generator of a compact group. The basic point set topologies for the two are different. The point set topology for gravitation is the Zariski topology, and defines certain algebraic varieties and a sheaf bundle system that is fundamentally different from the more trivial principal bundle system for the unitary U(1) group of electromagnetism. Penrose's twistor theory exploits this property and in some ways is a fairly honest approach to quantum gravity, even if it has not proven to be terribly workable in more recent time.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

The problem with this wave equation which gives the currents is that it assigns a mass current to an electromagnetic potential. I doubt many are going to cotten on to this idea.

Yeah, people will probably quickly jump to the wrong conclusion. I cannot stop that, nor do I care about such a null reaction.

If you want to look at the group theory behind the GEM proposal, you must look at the Lagrange density, not the field equations. The field equations don't say a word about the spin symmetries, or the symmetries of the field strength tensor, for Maxwell or GEM.

You fortunately are smarter than that, and discuse F. That is in the GEM proposal as - A \nabla + (A \nabla)^* = (0, E + B)=F. The way we write these - you preferring differential geometry, I choosing to work with quaternions - does not have an impact on group theory: it is the unitary group U(1) as it must be to characterize electromagnetism.

The problem as you note is gravity. We agree on that. You then discuss what the group for gravity is, which is kind of surprizing since we don't have a quantum gravity theory yet. How we know what the group theory for the theory we don't have is beyond me. You do know the limitations of current efforts to apply group theory to the problem of gravity. That was a fun challenge to read.

It would appear like you missed a comment made in post #457 about group theory. There I noted that I am using Klein four group, Z_2~X~Z_2, also known as the dihedral group, Dih_2. Since the group is finite, it is compact, a good thing as you have pointed out. This avoids the problem you cited about hyperbolic groups. My proposal is a linear theory for gravity because the spin-coupling is between 2 4-vectors, just like in EM. It is far better to quantize a linear theory than a nonlinear one!

If the technically skilled take a brief glance at my proposal and assume I only use U(1), I can wait for the more reflective to think about the advantages and challenges of using Z_2~X~Z_2 for gravity and U(1) for EM.

Doug

[Lawrence B. Crowell]

Hello Lawrence:

It would appear like you missed a comment made in post #457 about group theory. There I noted that I am using Klein four group, Z_2~X~Z_2, also known as the dihedral group, Dih_2. Since the group is finite, it is compact, a good thing as you have pointed out. This avoids the problem you cited about hyperbolic groups. My proposal is a linear theory for gravity because the spin-coupling is between 2 4-vectors, just like in EM. It is far better to quantize a linear theory than a nonlinear one!

If the technically skilled take a brief glance at my proposal and assume I only use U(1), I can wait for the more reflective to think about the advantages and challenges of using Z_2~X~Z_2 for gravity and U(1) for EM.

Doug

A dihedral group, or any polytope is a Coxeter-Weyl A-D-E system define a system of roots. These are the discrete symmetries for the bundle framing of a system. The continuous symmetry are the group generators.

I have not looked at your reference, but the relationship between cyclic groups such as Z_n are


D_n~=~\{(x_1,~\dots,~x_n)~\in~Z_n: \sum_{i=1}^n x_i = 2Z (even)\}


which puts constraints on the D_n group that can be used for Voronoi cell "glue vectors." But at any rate the simple cyclic group Z_2 corresponds to the }D_2~\simeq~SO(4) or the Lorentzian version SO(3,1). Each of the x_i corresponds to the rotations in an SU(2). The product of these would correspond to a tetrad where one group is for local transformations of affine connections on the base manifold and the other would correspond to an internal bundle.

I could go on a whole lot more about this and how it leads to D_4~=~SO(8) and how SO(7,1) and how the spin(6)~\rightarrow~spin(4,2) embedded in SO(7,1) is the group for conformal gravity. This group has the cyclotomic ring defined by the Galois field GF(4). The Galois field is GF(4)~=~(0,~1,~z,~z^2) with z~=~{1\over 2}(i\sqrt{3}~-~1) with z^2~=~z^*. GF(4) is the Dynkin diagram for the Lie Algebra D_4~=~spin(8). The properties of the basis elements that produce a commutator are


z^2~=~z~+~I,~z^3~=~I,~{\bar z}~=~z^2,


and defines the hexcode system C_6

None of this gets away from the problem that the holonomy groups for the Euclideanized versions of groups do not contain those in the Lorentzian or hyperbolic case. The Berger classification of groups according to their holonomy groups for loop variables of affine variables does not cover groups of the sort SO(n)\times SL(2,C). This is in spite of the fact that SL(2,C)~=~SU(2)\times SU(1,1) and both SU(2) and S(1,1) have the same Dynkin root system.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

I am trying to decipher what is good or bad about the specific proposal in this thread. A gentle reader would need a good graduate level class on group theory to appreciate all the relationships you have pointed out. For those readers of this thread further behind than myself, I will point out a few things I have learned so far.

The Klein four-group is abelian, so two elements in the group commute. That is consistent with my definition of the Even representation of quaternion multiplication.

The Klein four-group can be viewed as a subgroup of A_4. It does not have a simple graph. Instead the graph has 4 vertices, where only two of them are connected:

.-.
. .

where the dots are the vertices, the dash a connecting edge.

Since I came to this group from a quaternion, it was interesting to read this on wikipedia:

The Klein four-group is the group of components of the group of units of the topological ring of split-complex numbers.

Quaternions in the usual Hamilton representation are 3 complex numbers that share the same real. In the Even representation, the complex numbers are split. Nice.

A cyclic graph is created by taking powers of the group to the n. The graph for the Dih2 has the identity element in the middle, connected to the three other elements of the group. For the chemists in the audience, it looks like an amino group.

Another element of this puzzle are the finite fields or Galois fields. A way to represent the Klein group is as the set of four elements, {1, 3, 5, 7} modulo 8. So 1*3=3, 3*7 Mod 8=5, 5*7 Mod 8 = 3. The representation is Abelian, since 3*5 = 5*3. The Klein group is isomorphic to GF(4) because there are 4 prime numbers in the representation shown.

So far so good.

One way in which the Even representation of quaternions is different from the Klein group is that the group is modulo the eigen vectors of the 4x4 real matrix representation. This is necessary so that the group is a division algebra, a subject I did not come across in my readings on the topic. The moding out of the eigen vectors might have big implications, because then from one element of the group, there would necessarily be a way to get to any other element. This implies that the graph for Dih2 Mod (eigen vectors) is simply connected.

Now we come to the objection:

None of this gets away from the problem that the holonomy groups for the Euclideanized versions of groups do not contain those in the Lorentzian or hyperbolic case.

The gentle reader might wonder what a holonomy group is, having never read about it in the funny pages. This is a topic that is central to differential geometry. It is about the relationship between the connection and the manifold. If one moves around the manifold, and does not quite get back to where one started, that is the subject of this topic.

I can now appreciate the problem at hand. If the graph for a group is not simply connected, then one cannot represent the Lorentz group - a picture of flat spacetime - or the small deviations from the Lorentz group needed for gravity to be a dynamic metric theory. If the GEM proposal used the group Dih2 for gravity, it is reasonable to say it could not represent gravity as a smooth metric theory. Although not sure how to write this, I am using a different group, Dih2 Mod (eigen vectors). One can travel anywhere on the manifold with the connection, a necessary thing. At this time, I don't know the impact on parallel transport.

Doug

[Lawrence B. Crowell]

Doug,

For some reason it took a long time to get on here, so I don't have much time. I have a Schubertiade to atttend to. The Wiki-p page indicates what I said. This is involved with D_2, or its roots. These structures are useful in deriving group structures, and I am actively involved with work along these lines, though more up the ladded to Leecha lattices, Conway groups and up to the Fischer-Griess "monster." Yet at the end this does not directly address the issue of non-compact group structure and holonomies. There is in my thinking a system of projective varieties over these in the form of quantum codes, which give light cone structure. Penrose's twistors are related to this type of structure.

Anyway, I will try to address more directly what it is that you say above later this weekend.

Lawrence B. Crowell

[sweetser]

Hello:

In this post I will propose what graphs describe the Even and Hamilton representations of quaternions.

A graph has a number of vertices and then edges which are made of pairs of vertices. The Hamilton and Even representations both have the same 4 vertices: (e, i, j, k). To keep this possibly related to physics, I like to think that in flat spacetime, the absolute value of each of these is equal to 1, but in the curved spacetime of GEM theory, the absolute value of e is the inverse of the absolute value of i, j, and k.

For the Even representation of quaternions, all the vertices are connected to each other. The graph looks like a box with an X:

.-.
|X|
._.

vertices: (e, i, j, k)
edges: ((e, i), (e, j), (e, k), (i, j), (i, k), (j, k)
name: K4:6

This is known as a complete graph because every pair of distinct vertices is connected to an edge.

Here was a bit of fun I had mixing my physics in with pure math. I was thinking about the labels for the edges. For the edge (e, i), I thought I would just use i, similarly for j and k. For the edges connect two 3-vectors, it would be the third 3-vector. The problem is that no labels use e. That didn't sound good to me, ignoring e. What if I labeled each edge as i/e, j/e, or k/e? The physics of GEM suggest the absolute value of this will be one in curved or flat spacetime, so the edge is invariant. Nice. If one is at a vertex and wants to get to another vertex, read the label, and form the product, such as e (j/e) = j.

What about the Hamilton representation of quaternions? We still have four vertices. The part of the graph that connects with e is exactly the same. What changes are the edges that connect the 3-vectors. Because the 3-vectors do not commute, the edges must be directional. The edge that connects i to j has a label of -k/e because -i k/e = j. The directional edge from j to i has the label k/e. Here is the graph for the Hamilton representation of quaternions:

.=.
|X()
._.
vertices: (e, i, j, k)
edges: ((e, i), (e, j), (e, k))
directed edges: ((i, j), (j, i), (i, k), (k, i), (j, k), (k, j))
name: K4:9 ?
Note: the \ of the X should be two directed lines, the limitations of ASCII graphs.

I am not certain if the K classification allows for including directed edges. It is significant that the graphs for the Hamilton and Even representations of quaternions are not the same. Gravity is not the same as EM, in fact, gravity is a little bit simpler (one charge, only attracts), and it is gratifying that the graph for the Even representation is a little simpler than that for the Hamilton representation.

This was a lot of fun, I hope you enjoyed. I have never applied graph theory to anything before in my life.

Doug

[Lawrence B. Crowell]

This is in part a test. I am having trouble getting a post sent here.

[Lawrence B. Crowell]

What about the Hamilton representation of quaternions? We still have four vertices. The part of the graph that connects with e is exactly the same. What changes are the edges that connect the 3-vectors. Because the 3-vectors do not commute, the edges must be directional. The edge that connects i to j has a label of -k/e because -i k/e = j. The directional edge from j to i has the label k/e. Here is the graph for the Hamilton representation of quaternions:

.=.
|X()
._.
vertices: (e, i, j, k)
edges: ((e, i), (e, j), (e, k))
directed edges: ((i, j), (j, i), (i, k), (k, i), (j, k), (k, j))
name: K4:9 ?
Note: the \ of the X should be two directed lines, the limitations of ASCII graphs.


Doug[/QUOTE]

It is best to go all the way and consider the 120 icosian of quaterions, and its extension to the 240 cell with the 128 elements which give the 240 roots plus the 8 Cartan center elements.

The basic group system for gauge theory is the heterotic group E_8, which is one of the Heterotic groups. There has been a lot of activity with this, and the Vogan DeCloux group found its system of root representations and possible impacts on elementary particles and gravitation. There is also a business called supersymmetry, which has 2^{N} elements in its representation. For something called N = 8 this has 256 elements. There is then the Clifford algebra, noted as CL(16)~=~CL(8)\times CL(8) which gives a relationship between these 256 elements and the 240 roots of E_8 and its 8 weights (weights are the Cartan centers or eigen-matrices of the roots). The roots are what define the physical states. Now "half" of this CL(8) is defined by


CL(8)~=~1~+~8~+~28~+~56~+~70~+~56~+~28~+~8~+~1



for spins 2, 3/2, 1, 1/2, 0, -1/2, -1, -3/2, -2. The 248-dim Lie algebra E_8 = 120-dim adjoint Spin(16) + 128-dim half-spinor spin(16) is rank 8, and has 240 root vectors that form the vertices of an 8-dim polytope called the Grosset polytope.

Now the Clifford basis decomposition of the 256 there are the following elements


CL(8)~=~1~+~8~+~(24+*4*)~+~(24+4+28)~+~(32+3+3+32) ~+~(28+4+24)~+~ (24+*4*)~+~8~+~1,


The * * eclosed give the 4+4 = 8 that corresponds to the 8 E_8 Cartan subalgebra elements that are not represented by root vectors (they are the eigen-matrices), and the black non-underlined 1+3+3+1 = 8 correspond to the 8 elements of 256-dim Cl(8) that do not directly correspond elements of 248-dim E8. Now for the spins \pm 1/2 there is a 24 = 8 + 8 + 8, which are the three generations of fermions. These are the long roots associated with the 24, which corresponds to the elements of gauge fields which are the SU(3)xSU(2)xU(1) of the standard model. Now the 8 + 8 + 8 are the short roots which complement the roots of the SU(3), and define a triality condition on the fermionic sector.

This triality condition appears to be a very central aspect of group theory and irreducible representations (ir-rep). I have been doing some calculations on a three fold structure with mutually unbiased bases on Jordan algebras (an ir-rep of E_8 elements with octonionic OP^2 structure), so that the heterotic E_8 is embeded in a three fold structure within the Leech lattice and a general system of modular forms.

But to bring things to a more common level, the E_8 is a rich structure on symmetries with the numbers 2, 3, 5, 8. The two is on the helicity match up above, the 3 is on the triality condition of subgroup ir-reps, the 5 is a symmetry on an icosahedral system "dual" to the triality symmetry, and the 8 the whole system. This naturally embeds in a further three fold system called the Steiner system with the [5, 8, 24] structure for a "code." So the reason there are three families of quarks, three families of leptons (8 + 8 + 8) etc is due to this rich algebraic structure.

Why there are four forces is again an aspect of group decomposition. The Cartan center for the E_8 is related to the other heterotic groups G_2,~F_4,~E_6~ and~ E_7. A basic representation is F_4~=~C_{E_8}(G_2) (the Cartan center weighted on G_2), where G_2~=~SU(3)~+~3~+~{\bar 3}~+~1, which gives the nuclear force plus a hypercharge "1". The F_4 is a D_4 group \sim~SO(7,1) which embeds gravity and the weak interactions plus the 8 + 8 + 8.

Yet this structure operates equally for Euclidean and Lorentzian signatures. There is a loss of information, which is being glossed over. This does not solve the problem of ambiguities in holonomic structure for non-compact groups. In quantum field theory it is a common practice to Euclideanize time by letting time go to t~\rightarrow it. This time turns out to be related to temperature by


t~=~\frac{\hbar}{kT}


This time is not exactly the same as the Lorentzian time, what we measure on a clock (or might we say what a clock “produces”), but is really a measure of the time where quantum fluctuations may be observed. Sometimes the term quantum fluctuations causes trouble, so it really is more the distance in a Euclidean 4-dim space where an instanton (a tunnelling state etc) with a certain magnitude can appear. As this temperature becomes very small the fluctuation time becomes large and the strength of the fluctuation, if we qualitatively invoke the Heisenberg uncertainty principle

and consider this instanton time t as this uncertainty in time. As the temperature heats up it also means that the fluctuation is stronger or its coupling is made larger and a phase transition will ensue.

This gets into some fascinating stuff! The Lorentzian time implies that the moduli space is not separable. Two moduli, points in the space of “gauge equivalent connections,” are not separable in a Hausdorff point-set topological definition. The topology is Zariski. I can go write more about this if needed, but this gets us into some rather serious stuff. So there is the Lorentzian time and there is this Euclidean time and there is a Wick rotational map between the two. So distinct fluctuation in the Euclidean case which have moduli that are Hausdorff, separable and “nice” correspond to a set of moduli in the Lorentzian case which are not. Physically this means there is some scale invariant physics (again to go into would require a bit of writing work) of phase transitions associated with the correspondence between fluctuations at various pseudo-time scales (or temperatures) and their Lorentzian versions.

This correspondence and the phase transition is a quantum critical point, which have been observed with High temp superconductors and in the physics of Landau electron fluids in metallic crystals in the actinide plus transition range. This also shares some aspects of physics with the Hagedorn temperature of strings.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

Thanks for all the information on the group E8, the largest exceptional Lie group. Garrett Lisi made international news back in November for a unification proposal that used E8 without resorting to strings.

Unfortunately, I am not able to implement your suggestion. A brief description of icosian (http://en.wikipedia.org/wiki/Icosian_Calculus) make it clear one is dealing with "a non-commutative algebraic structure". That does work of the Hamilton representation of a quaternion. I am using the word quaternion to mean a 4D division algebra. Many people work productively with the assumption that the only implementation of a 4D division algebra is the one Hamilton developed so many years ago (Gauss got there first by the way).

In this thread, I found it necessary to formulate a new representation of a 4D division algebra where the elements commute. To make the Hamilton representation a mathematical field, one only needs to exclude the additive inverse 0 from the quaternions. For the Even representation, there are more quaternions that need to be excluded (0 and the eigen vectors) but it can be done.

The Even representation of quaternions is not a Clifford algebra. Clifford algebras have orthogonal basis vectors, e_i e_j = -e_j e_i. For the Even representation, e_i e_j = e_j e_i. Oops. I am going to go to the 8th International Conference on Clifford Algebras and their Applications to Mathematical Physics at the end of May in Brazil, and claim the generalization of quaternions that Clifford Algebras represent is not enough to unify gravity and light (that will not go over well!).

Let's think about the two graphs again, something I have done all weekend with great amusement. My central thesis is that we have yet to do the math of 4D like Nature practices it. In the graph for the Hamilton and Even representations, both have 4 vertices. Sometimes I think of them as events - a t, x, y, and z - but even better might be a difference between two events - a dt, dx/c, dy/c, dz/c. Recall the labels I used, which with this application would be dx/dt c, or \beta_x.

The Even and Hamilton representations also share three edges, those for (e, i), (e, j), and (e, k). These seven shared elements indicate an overlap between the two representations. So how are they different? When one wants to connect two vertices that are in the 3-vector, there are two choices: use and edge or a directed edge. The Even representation uses an edge, .-., while the Hamilton representation uses two directed edges, .=. (please imagine arrows on both lines in opposing directions). I got so excited by this simple, direct message, that I designed and made a button with the graph for gravity and light last Saturday night, wearing it to a Boston swing dance. Fresh math! If anyone reading these notes would like a similar button, just email me off this thread with your snail mail address, and I will put one in the mail in a few weeks (I should reward those that are doing the work of trying to follow these nerdly riffs).

doug
sweetser@alum.mit.edu

[Lawrence B. Crowell]

Hello Lawrence:

Thanks for all the information on the group E8, the largest exceptional Lie group. Garrett Lisi made international news back in November for a unification proposal that used E8 without resorting to strings.


It is an framing scheme which puts the particles into the F_4 and G_2 exceptional subgroups of the E_8. It is a neat idea in some ways, but there are few ambiguities, in particular how spin(4,2) fits into the scheme. The heterotic string has E_8\times E_8 for two chiralities and in this one can supersymmetrize the theory. One does not need to have string theory explicitely in here, but the Jordan subalgebra naturally contains stringy structure, such as the 26-dimensional bosonic string.



Unfortunately, I am not able to implement your suggestion. A brief description of icosian (http://en.wikipedia.org/wiki/Icosian_Calculus) make it clear one is dealing with "a non-commutative algebraic structure". That does work of the Hamilton representation of a quaternion. I am using the word quaternion to mean a 4D division algebra. Many people work productively with the assumption that the only implementation of a 4D division algebra is the one Hamilton developed so many years ago (Gauss got there first by the way).



The quaternionic division algebra is noncommutative by definition. The wiki-p site just discusses the Hamilton vertex permutation on the icosian. There is a whole lot more structure to this system.



In this thread, I found it necessary to formulate a new representation of a 4D division algebra where the elements commute. To make the Hamilton representation a mathematical field, one only needs to exclude the additive inverse 0 from the quaternions. For the Even representation, there are more quaternions that need to be excluded (0 and the eigen vectors) but it can be done.



This is a contradition in definition. The Cayley numbers 1, 2, 4, 8 lead to the reals, complexes, quaterions and octonins. Quaternions as a division algebra are noncommutative.



The Even representation of quaternions is not a Clifford algebra. Clifford algebras have orthogonal basis vectors, e_i e_j = -e_j e_i. For the Even representation, e_i e_j = e_j e_i. Oops. I am going to go to the 8th International Conference on Clifford Algebras and their Applications to Mathematical Physics at the end of May in Brazil, and claim the generalization of quaternions that Clifford Algebras represent is not enough to unify gravity and light (that will not go over well!).



This appears related to a graded structure. The even quaterionic structure involves elements e_i~=~\xi_a e^a_i, for \xi^a a Grassmannian element. The quaterions then define supermanifold coordinates with


y^i~=~x^i~+~{\bar\theta}\sigma^i\theta.


You will then have \{e^a_i,~e^b_j\}~=~\eta^{ab}g_{ij}, which are fermionic correspondences with quaterionic elements. The quaterionic fields are framed within a Clifford basis, such as a connection form


{\cal A}~=~A~+~\Gamma\cdot\psi


So that in the case of QCD the SU(3) is combined with the 3~+~\bar 3 in G_2. Similarly the fields can be again framed with Grassmannian elements.

This will then extend these elements to a graded algebraic correspondence between bosons and fermions. I have had some communciations with Garrett Lisi and others on just this issue of extending the framing system to a graded Lie algebraic structure.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

The question is whether representation theory (http://en.wikipedia.org/wiki/Group_representation) can play a role in the description of quaternions. As a warmup exercise, let's write out two different representations of what I have been carefully calling the Hamilton representation of quaternions.


q1 (t, x, y, z) = \left(\begin{array}{cccc}
t & -x & -y & -z\\
x & t & -z & y\\
y & z & t & -x\\
z & -y & x & t
\end{array}\right)\quad eq 1



q2 (t, x, y, z) = \left(\begin{array}{cccc}
t & x & y & z\\
-x & t & z & -y\\
-y & -z & t & x\\
-z & y & x & t
\end{array}\right)\quad eq 2


These two 4x4 real matrix representation of quaternions are not identical. Yet if we look at the multiplication table, they turn out to be the same. We would say the real 4x4 matrix representation q1 is isomorphic to the one for q2. These matrices have addition, subtraction, and multiplication. The one step that requires a light amount of work is to get the inverse and thus show the matrix is always invertible. Turns out that is not the case because there is a division by the norm, t2 + x2 + y2 + z2. If one excludes zero, then division works.

Representation theory does apply to the definition of quaternions. Working exclusively with isomorphic representations can get dull quickly. Let's find a representation that is not isomorphic to eq 1 (and thus eq 2). That would be the Even representation of quaternions, here written in as a 4x4 real matrix:


q1 (t, x, y, z) = \left(\begin{array}{cccc}
t & x & y & z\\
x & t & z & y\\
y & z & t & x\\
z & y & x & t
\end{array}\right)\quad eq 3


Again it should be obvious that such a matrix has addition, subtraction, and multiplication well defined for all members. Fire up Mathematica, and you find out, just like eq 1, the matrix is not always invertible. To make it invertible, zero and the eigen vectors of this matrix are omitted.

One of the things that gives me confidence is when I spot errors. This might appear contradictory, but mathematical physics can be so abstract, it can be difficult to spot the right direction. If a question can be asked with enough precision, and shown to be in error, then a correction to the course can be made.

I have said several times that one needs to exclude the Eigen vectors to construct the Even representation of the quaternions. Being self-skeptical, I fired up Mathematica to confirm my statements. Here is the output.

The Hamilton representation:

q1[t_, x_, y_, z_] := \left(
\begin{array}{cccc}
t & -x & -y & -z \\
x & t & -z & y \\
y & z & t & -x \\
z & -y & x & t
\end{array}
\right)\quad eq 4

Simplify[Inverse[q1[t,x,y,z]].\{1,0,0,0\}]
\left\{\frac{t}{t^2+x^2+y^2+z^2},
-\frac{x}{t^2+x^2+y^2+z^2},
-\frac{y}{t^2+x^2+y^2+z^2},
-\frac{z}{t^2+x^2+y^2+z^2}\right\} \quad eq 5

The Even representation:

q3[t_, x_, y_, z_] := \left(
\begin{array}{cccc}
t & x & y & z \\
x & t & z & y \\
y & z & t & x \\
z & y & x & t
\end{array}
\right) \quad eq 6

Calculate the inverse:

Factor[Inverse[q3[t,x,y,z]].\{1,0,0,0\}]
\left\{\frac{t^3-t x^2-t y^2+2 x y z-t z^2}{(t+x-y-z) (t-x+y-z) (t-x-y+z) (t+x+y+z)},
-\frac{t^2 x-x^3+x y^2-2 t y z+x z^2}{(t+x-y-z) (t-x+y-z) (t-x-y+z) (t+x+y+z)},
-\frac{t^2 y+x^2 y-y^3-2 t x z+y z^2}{(t+x-y-z) (t-x+y-z) (t-x-y+z) (t+x+y+z)},
-\frac{-2 t x y+t^2 z+x^2 z+y^2 z-z^3}{(t+x-y-z) (t-x+y-z) (t-x-y+z) (t+x+y+z)}\right\} \quad eq 7

Calculate the Eigenvalues:

Eigenvalues[q3[t,x,y,z]]
\{t+x-y-z,t-x+y-z,t-x-y+z,t+x+y+z\} \quad eq 8


The mistake I made was it is the Eigen values that need to be excluded. In the Hamilton representation, the same value of zero must be excluded 4 times. In the Even representation, zero needs to be excluded, along with three other values. That is not too different, but does indicate the representations are not isomorphic.

I have no problem with the statement that quaternions have traditionally been defined exclusively as a 4D division algebra that is noncommutative. I have read that a thousand times. Yet I do research. Sometimes you need to knock heads. I do so using representation theory in this case. That nice every-element-is-positive matrix is 4D in eq 3, can be added to other elements just like itself, can be subtracted from other elements like itself, can be multiplied to create other elements just like it, and should its Eigen values be excluded, can promise that a inverse necessarily exists.

I might be able to decrease conflict by talking about 4D division algebras, one of which happens to be the noncommuting quaternions, the other being a hypercomplex number. I am not a fan of this approach since it does not pay respect to representation theory. There is nothing "hyper" about the matrix in eq 3: all the elements are positive. No wonder it may turn out to be the math behind the one universally attractive force in Nature, gravity.

Doug

[Lawrence B. Crowell]

Hello Lawrence:

The question is whether representation theory (http://en.wikipedia.org/wiki/Group_representation) can play a role in the description of quaternions. As a warmup exercise, let's write out two different representations of what I have been carefully calling the Hamilton representation of quaternions.

Doug

As this goes on I am beginning to think that you have very different definitions for various things. It is a bit as if you have some alternate mathematics, such as symmetric field tensors and the like.

Of course quaternions or a Clifford basis has a representation. The Dirac matrices are a particular represenation of Cl(3,1) in the (0,~1/2)\oplus (1/2,~0) spinor representation of the Lorentz group. It is not too difficult to impose a local representation theory on the Dirac matrices and define connection terms and curvatures for general relativity in a spinorial form. Then the gravity field terms are equivalent to the Clifford spinors, and their product is just a Clifford bispinor.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

It certainly is possible that I am misusing standard technical terms. I point out one such error in my last post, where I should have used Eigen value instead of Eigen vector.

It would be a concern to me if as a general practice I was using "very different definitions of various things". This sort of problem does happen for people who do Independent Research. This happens because those independent people are so isolated, they do not learn correct definitions.

In post 469 I talked about three different representations for quaternions. For two of them, I said they were isomorphic representations. In the examples section of http://en.wikipedia.org/wiki/Group_representation, they show trivially different matrices that end up having the same multiplication table. If every representation was isomorphic, we would not need the word isomorphic, it would be part of what representation theory means. Representations of groups that are not isomorphic are an important branch of representation theory.

I like this comment:

It is a bit as if you have some alternate mathematics, such as symmetric field tensors and the like.

I am a subversive, but I am a bit sensitive to the phrase "alternate mathematics" since it is close to "math that makes no sense". My math is precise enough for Mathematica to understand it, defined in equation 6 of post 469, it found the inverse in equation 7, and the Eigen values in equation 8. One cannot put a vague or BS math into a symbolic math package and have it spit out sensible results.

I have been so precise, so into the standard lexicon of math, that I think I am trying to use the graph K4:6 as the foundation for gravity, and K4:9 for EM.

What you have done consistently is try an throw a noncommuting curveball, and I am not swinging at it. Yes, the Hamilton representation of quaternions - the only ones anyone ever learns - is the Clifford algebra CL(0, 2). That is part of my proposal. Not being a Ph.D., I practice the rare art of intellectual minimalism. I can get by with few things, used creatively.

What was Clifford trying to do? He was trying to generalize quaternions, focusing on that strange animal, the cross product. In my work, one cross product is enough. I don't need many varieties of a cross product. What I need is something that is genuinely different from a cross product, hence the two K4 graphs. I don't need more than 4 dimensions. Actually, I think if a proposal works in 5 or more dimensions, it is wrong on the grounds of dimensional analysis alone. I recognize that people in academia may find such a dismissive stance of all work done on Kaluza-Klien and strings harsh, oh well. If people choose to work with the Riemann curvature tensor, the Ricci tensor, the Ricci scalar, or Einstein's field equations, they may do so to describe gravity. The GEM proposal is dedicated to the idea that work with gravity and the tools of Riemann curvature will continue to fail to connect with quantum mechanics as they have since the 1930s (about the time when people thought GR and quantum mechanics had to be united somehow). CL(0, 2) is enough, 4D is enough, the Christoffel is needed but the Riemann curvature tensor is not. Lucky I am not trying to get a grant.

If by "alternate mathematics" you mean I cannot connect to the vast amount of work done on gravity in the past, I can accept that as a compliment. It is a good sign for originality in an area of study that needs a serious slap in the face.

Doug

[Mentz114]

Hi Doug:

The competition has been busy. You really should have a look at this

Gravitomagnetism in teleparallel gravity

E. P. Spaniol, V. C. de Andrade‡
Instituto de Fisica, Universidade de Bralsilia
C. P. 04385, 70.919-970
Brasilia DF, Brazil

arXiv : 0802.2697v1

extract -

In the present work, a dfferent approach will be adopted to reexamine
gravitomagnetism. Due to the fundamental character of the geometric structure
underlying gauge theories, the concept of charges and currents and, in particular, the
concept of energy and momentum are much more transparent when considered from
a gauge point of view. Accordingly, we shall consider gravity to be described
as a gauge theory for the translation group, which gives rise to the so-called
teleparallel equivalent of GR. In this scenario we recover all the aspects predicted by
GR and moreover we have all the formal structure of a gauge theory, which is naturally
close to electromagnetism due to its abelian character. Therefore, the concepts of
gravitoelectric and gravitomagnetic fields emerge, as we will see, in the same way as
in the electromagnetic theory, that is, as components of the field strength of the gauge
theory.
The paper is divided as follows: in section 2 we review the fundamentals of
teleparallel gravity; in section 3 the gravitational Maxwell equations are introduced
in their exact form...

You might even meet these guys on your jaunt.

I'd like to hear what Lawrence B. thinks about Teleparallel gravity. It's very appealing that it's a simple gauge theory.

[Lawrence B. Crowell]

Hello Lawrence:

It certainly is possible that I am misusing standard technical terms. I point out one such error in my last post, where I should have used Eigen value instead of Eigen vector.

It would be a concern to me if as a general practice I was using "very different definitions of various things". This sort of problem does happen for people who do Independent Research. This happens because those independent people are so isolated, they do not learn correct definitions.

---------

If by "alternate mathematics" you mean I cannot connect to the vast amount of work done on gravity in the past, I can accept that as a compliment. It is a good sign for originality in an area of study that needs a serious slap in the face.

Doug

The problem that I have is that some of what you say simply makes little sense to me. The reasons for anti-symmetric field tensors, for instance, are due to some basic results in differential geometry. The reason for some of these structures are mathematically determined by some very well grounded theorems in mathematics. I say this as someone familiar with the theorems of Uhlenbech, Freedman and Donaldson on the differential topology of gauge theory on four manifolds, and the celebrated Atiyah-Singer index theorem which determines the structure of moduli spaces. There is nothing in the mathematical literature which points to anything which you allude to.

There is a reason why people work with dimensions larger than four, or five in the case of the EM Kaluza-Klein theorem. It is likely that the structure of elementary particles is intimately associated with quantum gravity and the structure of the universe. The Maldacena result on the dual isomorphism between the Anti-deSitter spacetime and the conformal structure of field theories is a clear indication that quantum gravity necessitates a unification with gauge field theory and their fermionic sources (quarks, leptons, Higgs, dilatons etc). And as much as you might not like it this gets one into all the complexity of supersymmetry, some stuff with string theory, loop quantum gravity, maybe twistor theory and ... .

My point about nonholonomic loops and noncompactness indicates that I think there is a major physical (and equivalently mathematical) element which is missing from all of physics out there. I am working on a number of possiblities to address this question. I also lean a bit on subjects such as solid state physics, the theory of quantum gases (boson condensates etc) and quantum liquids, quantum phase transitions and so forth. I am primarily interested in approaching this from a physical basis, and exploiting the mathematics to make it work where necessary.

Also, I and anyone can solve equations on MATHEMATICA. However, if those equations are arrived at by wrong mathematics then the solutions don't mean a whole lot, even if done by computer.

Lawrence B. Crowell

[Lawrence B. Crowell]

Hi Doug:

The competition has been busy. You really should have a look at this

Gravitomagnetism in teleparallel gravity

E. P. Spaniol, V. C. de Andrade‡
Instituto de Fisica, Universidade de Bralsilia
C. P. 04385, 70.919-970
Brasilia DF, Brazil

arXiv : 0802.2697v1

extract -


You might even meet these guys on your jaunt.

I'd like to hear what Lawrence B. thinks about Teleparallel gravity. It's very appealing that it's a simple gauge theory.

This has some interesting prospects. The Finsler geometry can be used to intertwine all possible frames. So potentially in a more general setting, potentially useful for quantum gravity, accelerated frames can be brought into a form of equivalence with inertial frames. Then accelerated frames are in a more general setting a form of comoving frame with an underlying isometry.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

I think we should be able to part on good terms.
The problem that I have is that some of what you say simply makes little sense to me. The reasons for anti-symmetric field tensors, for instance, are due to some basic results in differential geometry. The reason for some of these structures are mathematically determined by some very well grounded theorems in mathematics. I say this as someone familiar with the theorems of Uhlenbech, Freedman and Donaldson on the differential topology of gauge theory on four manifolds, and the celebrated Atiyah-Singer index theorem which determines the structure of moduli spaces. There is nothing in the mathematical literature which points to anything which you allude to.

All the physicists eggs are in the anti-symmetric field tensor basket for good technical reasons. It takes an impressive investment to understand the groundwork of differential geometry as you have done.

Important breakthroughs are the sport of the young who have yet to learn the foundations so well. Those youngsters need to be part of a cell that solve enough nagging questions that the establishment has to pay attention.

I am not particularly young, and I do not have a "cell". I am using the Internet to try and construct one via my web assets (this thread on physicsforums.com, quaternions.com, quaternions.sourceforge.net, TheStandUpPhysicist.com, and probably most important, YouTube.com). I can quantify the number of folks which continue to read this thread, and have statistics on my web sites and YouTube (18k downloads).
There is a reason why people work with dimensions larger than four, or five in the case of the EM Kaluza-Klein theorem. It is likely that the structure of elementary particles is intimately associated with quantum gravity and the structure of the universe. The Maldacena result on the dual isomorphism between the Anti-deSitter spacetime and the conformal structure of field theories is a clear indication that quantum gravity necessitates a unification with gauge field theory and their fermionic sources (quarks, leptons, Higgs, dilatons etc). And as much as you might not like it this gets one into all the complexity of supersymmetry, some stuff with string theory, loop quantum gravity, maybe twistor theory and ... .

I know researchers are rational. My objection was technical, and it was not addressed. The units for spacetime are wrong. You start off in Mathematica wrong, everything that follows is wrong. You start off wrong in theoretical work, and everything that follows is wrong. Same harsh logic.

My preferences are not going to change the flow of research money or efforts into work on strings or loop quantum gravity. I am not bitter that physics research happens to be going that way. Nature doesn't care if we get the right answer or the wrong one, and neither do I. I have no fear of anyone. I have no problem telling the brightest physicists on the planet that because the units of higher dimensional spacetime are wrong, what they work on is wrong and will not last the test of time. They will go off on a compatification riff which as my mother, God bless her soul, would say is a bunch of crap. They will get funded, go to conferences, talk about the latest new result, but in time, all my chips are bet on the collapse of extra dimensional spacetime work.

People who do work on gravity do not even acknowledge the risk that their work could be wrong. This is purely a technical issue: if gravity gets united with the rank 1 field theory for EM by also being a rank 1 field theory, then every paper that presumes gravity must be rank 2 is wrong.
My point about nonholonomic loops and noncompactness indicates that I think there is a major physical (and equivalently mathematical) element which is missing from all of physics out there. I am working on a number of possiblities to address this question. I also lean a bit on subjects such as solid state physics, the theory of quantum gases (boson condensates etc) and quantum liquids, quantum phase transitions and so forth. I am primarily interested in approaching this from a physical basis, and exploiting the mathematics to make it work where necessary.

I wish everyone luck in their research. Your point about both group theory and noncompactness effected me in a good way. I was trying to keep up with all your comments on group theory, and was not satisfied with what I could say about group theory as it applied to GEM. One of the big accomplishments I have had is to visualize the groups SU(2) and SU(3) with animations of quaternions. That work did not link to the discussions on the Hamilton and Even representations of quaternions. I recently got a book, "Quaternions, Clifford Algebras and Relativistic Physics" based on the title alone :-) The author Patrick Girard works within the standard limitations people bring to the topic, not realizing the Maxwell equations can be written with real quaternions as has been done here. On page 4, he wrote out the multiplication of the Klein four-group, and it looked identical to what I had posted here for the Even representation of quaternions. Nice.

So I read more about this group. That eventually led to the graph of the dihedral group:

._.
. .

Although I had a group, it did not look right. We start from a flat metric:

d \tau^2 = dt^2 - dR^2/c^2

Take the Newtonian step away from this flat spacetime:

d \tau^2 = (1 - 2 \frac{G M}{c^2 R}) dt^2 - dR^2/c^2

This is Newton's law written as a metric theory. The potential that is consistent with field theory is (1 - 2 \frac{G M}{c^2 R}), not -\frac{G M}{c^2 R} as is often written. Take one more step away to get to the first-order parameterized post Newtonian metric:

d \tau^2 = (1 - 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2) dt^2 - (1 + 2 \frac{G M}{c^2 R}) dR^2/c^2

The Schwarzschild metric a solution to the Einstein field equations, or the exponential metric a solution to the GEM field equations, both have the same Taylor series expansion that match these terms. It is the next terms where GR and GEM part ways.

GR:
d \tau^2 = (1 - 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2 - 3/2 (\frac{G M}{c^2 R})^3) dt^2 - (1 + 2 \frac{G M}{c^2 R} + 3/2 (\frac{G M}{c^2 R})^2) dR^2/c^2

GEM:
d \tau^2 = (1 - 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2 - 4/3 (\frac{G M}{c^2 R})^3) dt^2 - (1 + 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2) dt^2) dR^2/c^2

At second order parameterized post-Newtonian accuracy, the GEM proposal is a testable hypothesis. Nothing this precise has come out of the vast amount of work done in loop quantum gravity or strings.

So we can see this shift from a flat metric, to Newtonian, to weak field, to strong field, with all four terms gently changing. I felt if my graph was this:

._.
. .

then my proposal about the Even representation of quaternions was wrong. I put that key part of my proposal on the firing line. I felt that graph would never be able to do a smooth transition needed for a metric solution (there is also a potential solution to the GEM field equations, and a metric/potential solution to the GEM field equations, but this is a possibility people trained in GR cannot entertain because there is no potential in the Rienmann curvature tensor, it is exclusively about the metric.)

I read up on graph theory, and realized that the graph that described the Even representation was this one:

._.
|X|
._.

This is both compact and the vertices are all connected to each other. This sort of finite group is compact in the formal sense of the word. It is part of standard math, the group K4:6, but is not part of the literature devoted to gravity. This makes it a candidate for the missing piece of physics.

Also, I and anyone can solve equations on MATHEMATICA. However, if those equations are arrived at by wrong mathematics then the solutions don't mean a whole lot, even if done by computer.

I was trying to figure out what I didn't like about this comment. It was not an attack on me, because we both know from experience that processing equations through Mathematica is an acceptable check of form, but like all computer programs, what one puts in is the most important aspect of what comes out. The first line of your reply indicates you don't get where I am going. You know your land well. The best reply I can think of was written by a folk singer:
Come mothers and fathers
Throughout the land
And don't criticize
What you can't understand
Your sons and your daughters
Are beyond your command
Your old road is
Rapidly agin'.
Please get out of the new one
If you can't lend your hand
For the times they are a-changin'.

If the GEM proposal is correct - or some similar technical variation of a rank 1 field theory to unify gravity and EM - then the work of Uhlenbech, Freedman, Donaldson, Wheeler, Hawking, Feynman, Kaluza, Klein, all string theorist, all loop quantum gravity people, and even the work of Albert on gravity will collapse. I cannot care what your opinion is on that clear yet radical sentence because I know you haven't calculated the Christoffel symbol of the second kind for the Rosen metric, finding some erudite reason to not bother. I am much happier finding wonderful new gems about GEM like the graph theory for the Hamilton and Even representation of quaternions than bonking heads like I have done in this paragraph.

This Mathematica warning is also not germane to the question at hand: can one formulate a 4 dimensional, commuting division algebra? If the answer is yes, that would be interesting because everyone is instructed that the only 4 dimensional division algebra has the property that it is non-commuting. That is math worth talking about, not a banal caution about symbolic math programs.

Just for fun, I will take a different approach on the value of Mathematica. What passes for physics research today is so vague it cannot be translated into a proposal that can be confirmed by machine. A measure of the value of work is the ability to translate it to symbolic code. The Universe is constructed out of parts that do not think, they do. The math used to describe the Universe should be the same way. Sure, the high priests that believe they are above the mundane nuts and bolts of math will look scornfully at such a stance, but that is fine with me. I prefer nuts and bolts, one can build real things with them.

Doug

[Lawrence B. Crowell]

Hello Lawrence:

The Schwarzschild metric a solution to the Einstein field equations, or the exponential metric a solution to the GEM field equations, both have the same Taylor series expansion that match these terms. It is the next terms where GR and GEM part ways.

GR:
d \tau^2 = (1 - 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2 - 3/2 (\frac{G M}{c^2 R})^3) dt^2 - (1 + 2 \frac{G M}{c^2 R} + 3/2 (\frac{G M}{c^2 R})^2) dR^2/c^2

GEM:
d \tau^2 = (1 - 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2 - 4/3 (\frac{G M}{c^2 R})^3) dt^2 - (1 + 2 \frac{G M}{c^2 R} + 2 (\frac{G M}{c^2 R})^2) dt^2) dR^2/c^2

At second order parameterized post-Newtonian accuracy, the GEM proposal is a testable hypothesis. Nothing this precise has come out of the vast amount of work done in loop quantum gravity or strings.

Doug

To be honest at this point I think I can rely upon experimental evidence. Last year the motion of two neutron stars was measured and compared to Parameterized Post Newtonian parameters up to O(1/c^4), or ppN in the standard line element parameters with ds^2 in units of distance. The motion agreed with Einstein's GR up to one part in 10^7 in this order. If your equations were correct over those of GR the deviation to this order, which appears as the O(1/c^6) in your proper time, would have been blatantly apparent.

I hate to say but I think your GEM theory, or at least the metric you claim above, has been falsified.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

If you have a reference for this work, I would appreciate it. I certainly would cite it in discussions of my work.
If your equations were correct over those of GR the deviation to this order, which appears as the O(1/c6) in your proper time, would have been blatantly apparent.

We cannot measure the mass or angular momentum of neutron star number 1 directly, it must be modeled. We cannot measure the mass or angular momentum of neutron star number 2, it too must be modeled. The most we could hope to say is that we can construct a model using the Schwarzschild metric that is consistent with the data we have. I am sure they have done that, which is good news.

I would need to see that they tried to plug in the Rosen or other alternative metrics at the start, which could change the mass and angular momentum descriptions of both neutron stars, and that such a model could not be adjusted to also be consistent with the data they collected. The coefficients of the two metric have the same sign but are 12% different at second order PPN accuracy. There are 4 critical numbers that have to be generated, and that gives one freedom to match a range of alternative metrics. I hope they are not claiming a system with 4 parameters can be used to eliminate alternative metrics.

You, the great skeptic of the value of the Mathematica, I am surprised you would think a system that inherently depends on the construction of two models for two spinning masses would think this data could distinguish alternative metric solutions for gravity. It is vital people do not over-claim what their data shows. They can demonstrate consistency, but I am skeptical about broader claims.

For light bending around the Sun whose mass we know from other effects, doing a measure at second order PPN accuracy is also dependent on models, such as the quadrapole moment of the Sun. That will have an effect in the range of 1 microarcsecond, which is the size of the difference between the Schwarzschild and GEM metrics (10.8 versus 11.5 respectively). The rotation of the Sun also must be accounted for at this level of detail, and that is darn difficult. I would bet the folks with the neutron star data did not put an effort to have a quadrapole moment to each of the stars, yet it would necessarily change the model at the resolution they were investigating.

The only non-model dependent data for higher order effects I know is about gravity waves. The rate of energy loss is consistent with a quadrapole moment. If a system lost energy as a dipole, then the loss would be far greater than what we have seen. This is why the Rosen metric is not a serious competitor. He proposed a fixed background metric as well as a dynamic background metric. An isolated system could store energy in the fixed metric field, which would allow the system to have a dipole moment while still conserving energy. This is why many theories that add on a new field fail: the new field can store energy and allow for dipole emission of gravity waves.

The GEM proposal is simpler than GR in that it uses only one derivative of a connection, not the difference between two derivatives of connections in the Riemann curvature tensor. The lowest order of gravity wave emission will be the water balloon wobble, the quadrapole moment.

If we were to ever measure a gravity wave, there could be a clean way to distinguish GEM from GR. In GR, they copy EM so closely, it is widely claimed that the waves must be transverse. In GEM which has EM integrated, the transverse waves are EM, leaving the longitudinal and scalar modes for gravity. We would need to detect the same gravity wave along 6 axes to to figure out the polarity, but that would be non-model dependent way to confirm or reject the proposal.

I am looking forward to reading the reference. I would not "hate" such a result because I have a professional scientific attitude. I have worked on other research projects that had to be abandon because of the data. The limitations of data must be skeptically assessed. Falsification is tough.

Doug

[Lawrence B. Crowell]

Hello Lawrence:

If you have a reference for this work, I would appreciate it. I certainly would cite it in discussions of my work.


Doug

Here is an overview of this type of work from the Cliff Will, a war horse on the experimental verification of GR:

http://arxiv.org/PS_cache/arxiv/pdf/0709/0709.2589v1.pdf

I tried to look up in APJ or in Science an article which discussed a specific neutron pair system. I'll have to look further. Yet the paper by cliff illustrates aspects of what I said. This paper also illustrates why your objections are not terribly relevant.

Lawrence B. Crowell

[CarlB]

This might be the place to start looking:
http://en.wikipedia.org/wiki/PSR_J0737-3039

This is probably what you're looking for, or at least should be referenced by it:

Tests of general relativity from timing the double pulsar
http://arxiv.org/abs/astro-ph/0609417

[Lawrence B. Crowell]

This might be the plae to start looking:
http://en.wikipedia.org/wiki/PSR_J0737-3039

The references in the Wiki-p site

http://www.physicsweb.org/articles/world/18/3/6/1
http://www.jb.man.ac.uk/news/doublepulsar/

http://www.atnf.csiro.au/research/highlights/2003/manchester/manchester.html

are of help, in particular the last one. The graph in figure 3 I recall appeared in the journal paper. The paper was in Science or APJ. The case was pretty convinceingly made that this was a verification of GR beyond the Hulst result of the late 1970s.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

The cited paper by Clifford Will is not relevant to your assertion that "at least the metric you claim above, has been falsified". What Will does in that paper is try to connect the post-Newtonian math machinery to what folks do in numerical relativity for the final few cycles of collapse of binary systems. There are so many assumptions that go into those models - big ones like the correct field equations for gravity are the Einstein field equations - that zero of this article has to do with the first seven coefficients of the Taylor series expansion for the metric of gravity. He is playing guessing games with eccentricities and M/R errors.

The paper shows a mathematical consistency between PN approaches and numerical general relativity, one nicer than he expected. A fine result, but not germane.

I have asked Clifford Will questions about the GEM proposal twice. The first time was at an Eastern Gravity Meeting where he gave a keynote address. I asked the first question: when will we get data to second order PPN accuracy? He said he did not know, and did not know of any experiments designed specifically to get that data (that is one reason I was taken aback by you referencing a paper by Will and saying this was relevant to the coefficients). I also asked another question, about the polarization of gravity waves, and in post #477 put his reply into my own words: it may not happen.

Will came to MIT to give a talk. There was a coffee session before, and I was sure to go although I don't drink coffee or tea. He looks like Ted Turner. No one was talking to him, so I went up to him and asked him about his Living Review article on experimental tests of GR. I pointed out that in section three, he went through lots of alternatives to GR. The article was pointed out to me by someone who claimed that of course a rank 1 field theory of gravity would be cited in his exhaustive review, but I said it wasn't there. He confirmed that a rank 1 theory was not there. The reason was that no rank 1 theory had been proposed that would pass all the basic tests, from the equivalence principle, to light bending around the Sun, to the precession of the perihelion of Mercury. He also made clear he would not be able to read my efforts to accomplish these goals.

Because there is a solution to the GEM field equations that can be expressed as a metric, it will pass the tests done of the equivalence principle, where all objects no matter what their composition move according to the metric equation.

The light bending calculation was done for GEM in post #351. Light bends a little more than what is predicted by GR, but the differences is only 9.1% according the the calculation in the post (11.69-10.96/10.96). The fact that light can bend at all in the proposal is why I have discussed the coupling term at length in this thread. One has to show spin 2 symmetry in both the coupling term and the field strength tensor term.

The third test is the precession of the perihelion of Mercury, which I did in post #233. It took me along time to get all the details of that one right, 24 steps in all.

Per Will's request, I have not forwarded my draft paper to him. There is a rank 1 theory involving gravity that meets his basic criteria not referenced in his review.

Doug

[sweetser]

Hello Karl and Lawrence:

Thanks for the references to the double pulsar system. Looks like they expect to get a few more terms. From 2003:
In the next few years, we expect to measure several more relativistic effects, some dependent on higher-order terms in the post-Newtonian expansion. These will provide the tightest constraints yet on theories of gravity in the strong-field regime.

From the paper itself in 2006:
In particular, we have measured omega dot so precisely (i.e., to a relative precision approaching 10-5) that we expect corrections at the 2 PN level to be observationally significant within a few years.

That is the level where GR and GEM differ. The error on the 2 PN level would have to be under 10%. Then assumptions put into the models would matter.

I know the gamma is the same for GR and GEM (if I am thinking about the right gamma). I don't know what GEM predicts for omega dot, s or r. More work to be done.

Perhaps my skimming was too quick, but I don't see how this particular binary pulsar rules out the exponential metric at this time. This is better data and is good news, but there is enough data fitting required to cast doubt on the ability to distinguish subtly different metrics.

Doug

[CarlB]

It says the orbital velocity is about 0.1% of the speed of light, or 1 part in 1000. Second level PN would be 1 part in a million. According to their data, GR restricts the masses of the stars at an accuracy of 1.337+/-0.004 which is fairly inaccurate at around 1 part in 334.

I think your theory's still alive.

[sweetser]

Hello Carl:

Thanks for confirming my reading, despite spelling your name incorrectly. There are 10 parameters in the first order parameterized post-Newtonian system, with \gamma=\beta=1 for GR, and zero for the other 8. Other proposals have non-zero values, and my impression was those sorts of proposals are at risk of being shown to be wrong by data such as this. For the GEM proposal, all ten of these parameters are the same as for GR. The exponential metric is a kissing cousin of the Schwarschild metric.

The folks doing these observations think they should get to second order PPN accuracy sooner rather than never. It is not clear how they can rule out metrics for fully conservative theories whose coefficients are not significantly different. I will need to watch for further developments in this area.

Doug

[CarlB]

Check this out. (I have no idea if it's accurate.)

OJ 287: New Testing Ground for General Relativity and Beyond
C. Sivaram, March 14, 2008

The supermassive short period black hole binary OJ287 is discussed as a new precision testing ground for general relativity and alternate gravity theories. Like in the case of binary pulsars, the relativistic gravity effects are considerably larger than in the solar system. For instance the observed orbital precession is forty degrees per period. The gravitational radiation energy losses are comparable to typical blazar electromagnetic radiation emission and it is about ten orders larger than that of the binary pulsar with significant orbit shrinking already apparent in the light curves. This already tests Einstein gravity to a few percent for objects at cosmological distances. Constraints on alternate gravity theories as well as possible detection of long term effects of dark matter and dark energy on this system are described.

http://arxiv.org/abs/0803.2077

P.S. I'm getting huge numbers of hits on my various websites the last 6 hours due to interest in the flat space gravitation theory of Lasenby, Doran, and Gull, and the implications this has for Painleve coordinates. Their theory of gravity is identical to GR to all orders. I don't think GR is accurate to all orders; my interest in their theory is cause it's based on Dirac's gamma matrices and fits into elementary particles really well.

[sweetser]

Hello Carl:

How far can I go with the exponential metric?
A man has to know his [metric's] limitations.

When I derived the metric, I assumed the source was static, spherically symmetric, and not rotating. Zero out of three will be right for a binary pulsar. Those sources are dynamic! They are spinning faster than Dorothy Hamill on fast forward. Since the source is binary, the spherical symmetry is out too. I must disqualify the exponential metric from the competition.

It took me several years before I found a derivation of the precession of the perihelion of Mercury that was explicit enough for me to understand with much work (Sean Carroll's notes were my guide). Most people deal with tough issues using a 12 step program, but it took me 24 steps to get to the 42.8"/century. At several important steps, one had to assume that the precession of the perihelion was super tiny. If not, the math becomes a nightmare. The standard derivation of the perihelion shift will go out the window. I have no sense of how to deal with a system with a strong perihelion shift. I recall reading the befuddlement applies to professionals too.

One assumption I am sensitive to is the 'static' assumption. Near a body of mass so dense that 2 GM/c^2 R \approx 1, that bit of spacetime will be jumping with things moving fast. Long before we reach a singularity, Nature will need to have a metric that has an hbar or time in it. In other words, the exponential metric will not apply in the region that many people have decided to study with the Schwarzschild metric, the proverbial black hole. Long before the singularity, a radically new metric will be needed for gravity sources. If Nature does work with a significantly different metric (one with t or hbar), then no work on black holes to date will survive that switch. Oops, that will not be popular.

Doug

[Lawrence B. Crowell]

Hello Lawrence:

The cited paper by Clifford Will is not relevant to your assertion that "at least the metric you claim above, has been falsified". What Will does in that paper is try to connect the post-Newtonian math machinery to what folks do in numerical relativity for the final few cycles of collapse of binary systems. There are so many assumptions that go into those models - big ones like the correct field equations for gravity are the Einstein field equations - that zero of this article has to do with the first seven coefficients of the Taylor series expansion for the metric of gravity. He is playing guessing games with eccentricities and M/R errors.
Doug

The analysis is essentially the same at any stage in the inspiral. C Will is doing a bit more than playing guessing games.



The light bending calculation was done for GEM in post #351. Light bends a little more than what is predicted by GR, but the differences is only 9.1% according the the calculation in the post (11.69-10.96/10.96). The fact that light can bend at all in the proposal is why I have discussed the coupling term at length in this thread. One has to show spin 2 symmetry in both the coupling term and the field strength tensor term.

The third test is the precession of the perihelion of Mercury, which I did in post #233. It took me along time to get all the details of that one right, 24 steps in all.

Per Will's request, I have not forwarded my draft paper to him. There is a rank 1 theory involving gravity that meets his basic criteria not referenced in his review.

Doug

C. Will is not interested in testing alternatives to GR. He tends to stick to the basic paradigm of gravity.

It is not possible to argue against on theory on the basis of another theory. Yet your idea of "symmetric" curvature terms or bundles simply does not make much mathematical sense. It is not applicable in Riemannian geometry or any differential geometry. There are graded structures on noncommutative geometries and the like, but this is not what you are advancing.

Lawrence B. Crowell

[Lawrence B. Crowell]

Check this out. (I have no idea if it's accurate.)

OJ 287: New Testing Ground for General Relativity and Beyond
C. Sivaram, March 14, 2008

The supermassive short period black hole binary OJ287 is discussed as a new precision testing ground for general relativity and alternate gravity theories. Like in the case of binary pulsars, the relativistic gravity effects are considerably larger than in the solar system. For instance the observed orbital precession is forty degrees per period. The gravitational radiation energy losses are comparable to typical blazar electromagnetic radiation emission and it is about ten orders larger than that of the binary pulsar with significant orbit shrinking already apparent in the light curves. This already tests Einstein gravity to a few percent for objects at cosmological distances. Constraints on alternate gravity theories as well as possible detection of long term effects of dark matter and dark energy on this system are described.

http://arxiv.org/abs/0803.2077

P.S. I'm getting huge numbers of hits on my various websites the last 6 hours due to interest in the flat space gravitation theory of Lasenby, Doran, and Gull, and the implications this has for Painleve coordinates. Their theory of gravity is identical to GR to all orders. I don't think GR is accurate to all orders; my interest in their theory is cause it's based on Dirac's gamma matrices and fits into elementary particles really well.

I wrote five years ago or time back a Found Phys paper on something similar to this.

I have not reread this literature on this matter, which clearly has advanced since last year. Frankly I don't have all the time in the world to check this out. Save it to say for a number of reasons I strongly suspect the GEM hypothesis is incorrect. It does not make much mathematical (differential geometric) sense, and I suspect that this work with orbiting pairs of compact bodies in ever tighter bounded gravitational wells will provide further tests on GR. To be honest as a classical theory I think Einstein's GR is completely spot on.

The deviations in general relativity is likely at orders when scales approach \sqrt{G\hbar/c^3} or the Planck length. There is I think an sort of equivalent breakdown on the cosmological scale. Our first hint of this with the quantum realm is with the Hawking result on black hole radiation. The other is with some curious results with the cosmological constant and the recent discovery of "eternal inflation."

The Hawking-Unruh effect comes about because in curved spacetime a basis of states in one local inertial frame can't be unitarily defined in another frame. So given a fixed spacetime classical background a basis of states in one region are related to a basis of states in another region by a transformation that is not unitary. This is the Bogoliubov transformation. So in a simple quantum model of states in one inertial frame, directed in and out of an event horizon, become mixed when transformed to another frame. So the lowering operator for the ingoing state in one frame becomes mixed with the raising operator for an outgoing state in another frame. This process is due to nonlocal entanglements across event horizons/

With quantum gravity we no longer have a fixed spacetime background. The path integral is a summation over a range of possible metric configurations. We might think of the rapidity or the hyperbolic functions of the Bogoliubov transformations as being elements summed over in the path integral, where each particular one determines a particular "history." So the Bogoliubov algebra is not just a transformation principle for a quantum field theory in a curved spacetime, but represents a particular history in the path integral and the nonunitary equivalence of vacua. Hence a set of Bogoliubov algebras might be thought of as related to a set of all possible holonomies for spacetime configurations, which are the variables for quantum gravity functional. These holonomies are defined on spacetimes with noncompact group structure and there are big issues with the classification of loop group spaces. In particular the Berger classification of such groups determined by affine connections does not entirely operate here. This is related to the problem that there are not general Cauchy sequences which can be constructed for connections on a space with a noncompact group structure. Such a Cauchy sequence involves differences between connections and hence is associated with curvatures. The inability to define such sequences means that there is a loss of information or a breakdown in an isometry between curvatures and the holonomic structure of the space.

So this gets to the question of dark energy and dark matter. A cosmology with a nonzero cosmological constant, or what is likely a parameter set into a constant (or approximately constant) value by the inflaton or dilaton in a spin(4,2)~\sim~su(2,2) model, is one where in general there is no unitarily equivalency between states in all regions of the cosmology. Even if the spatial surface is flat the accelerated expansion of the cosmology means that there is no such equivalency, and this comes about because there is no Killing vector K which when it acts on the energy KE = const. Without a Killing vector of this sort it means there is no isometry in the spacetime which maintains a constant energy on all paths in the cosmology. So the unitarity inequivalence of vacua in the earliest universe, where a vacua of unitary states is defined on a region >~ L_p in a superposition of other such vacua on about the same scale, is frozen into the classical cosmology after inflation. In a more general setting the Coleman-Mandula theorem is then a local principle. This gives the maximal set of symmetries of the S-matrix as the (0,~1/2)\oplus(1/2,~0) spinorial Lorentz group for external symmetry, an internal symmetry [A_i,~A_j]~=~c_{ijk}A_k, and the discrete CPT symmetry. The "maximal" extension on this is called supersymmetry. A cosmology with a non-zero cosmological constant necessarily means this is a local law, it does not apply globally. This is likely a source for what we call dark energy.


Lawrence B. Crowell

[sweetser]

Hello Lawrence:

In Will's paper, he does not know what the eccentricities are, nor the right M/R, so he is making a sophisticated exploration of the possibilities. Calling it a guessing game was unfair to this work, but was a little verbal flare, nothing more.

Good to see you missed the point. You made the claim that this particular work was relevant to assessing if the exponential metric has been shown experimentally to be in error. Will's work is fine as it stands, but is not relevant to this forum.

You have yet to publicly back off the claim that the exponential metric should be rejected on experimental grounds we have today. Recanting that claim would in no way be a support of the GEM proposal, just a recalibration of what is in the literature. I at least have backed away from a claim that the binary pulsar data is relevant to testing a metric that presume the system is static, spherically symmetric and not rotating.

If someone is confused, it is not possible to unconfuse them. This does not sound like what I am trying to do:
Yet your idea of "symmetric" curvature terms or bundles simply does not make much mathematical sense. It is not applicable in Riemannian geometry or any differential geometry. There are graded structures on noncommutative geometries and the like, but this is not what you are advancing.

I can dump whatever I want into a Lagrange density, takes some derivatives, and end up with field equations. I don't know what math label I should use for my work. I was able to explicitly calculate the Maxwell equations in #438. It would be useful to me to know what was wrong with that derivation, since it looked fine on paper and in Mathematica. After completing that calculation, your suggestion was to do the derivation your way. A find suggestion except that I am an independent researcher. A few posts later (#442), I use the same core tools of physics, varying the Lagrangian to generate the field equations for gravity. Your reply to that was the Accelerated change I, II, and III posts. Those posts represent much work, but on a different topic.
Save it to say for a number of reasons I strongly suspect the GEM hypothesis is incorrect. It does not make much mathematical (differential geometric) sense, and I suspect that this work with orbiting pairs of compact bodies in ever tighter bounded gravitational wells will provide further tests on GR. To be honest as a classical theory I think Einstein's GR is completely spot on.

You have stuck to this position from day one. I have made claims - which I cannot independently verify - that you have not done calculations with my proposal, the only way any of this work could become interesting, using the common genuine complaint that you are too busy to do by hand #438, #442, #351, #233, or the divergence of the Christoffel symbol of the Rosen metric.

Let's look at the banal defense of the status quo:
To be honest as a classical theory I think Einstein's GR is completely spot on.

Classical theories, such as the Maxwell equation of EM which can also be integrated seamlessly with quantum mechanics, allows one to define energy at a point. That utterly fails for Einstein's GR. Experts in differential geometry claim this bug is a feature. People who know bugs are bugs, and to distrust people claiming bugs are features. Any proposal that can for any point use the Riemann normal coordinates to make the energy of the field zero at that point, means that boat has no bottom, it is sunk before being pushed off the dock. Einstein had the courage to doubt Einstein when he saw a flaw (I believe he was aware of this flaw and took it seriously, but I might be wrong on this historical detail, anyone recall?). It was impressive how often he tried to rebuild the boat for gravity with so many different approaches. Classical GR is broken in how it deals with energy. The flaw can be seen in the Riemann curvature tensor as the difference of two derivatives of the connection. That problem is fixed in GEM which has only one derivative of a connection.
The deviations in general relativity is likely at orders when scales approach ... the Planck length.

This is so standard, and so wrong in my opinion. What did GR do different from Newtonian theory? Very little. It explained the wee bit of error in the precession of the perihelion of Mercury. Light doesn't bend much around the Sun. Newton gets it half right, and GR (as well as GEM) says the wee bending in time is matched by an equally wee bend in space. That took a lot of effort to measure. There are the "deep" ideas of the equivalence of inertial and gravitational mass, which the well-schooled can discuss for hours on end (it cannot be explained to the average person since they are both mass so should be the same). GR clears up a clear technical problem, that gravity must obey the speed of light.

Nothing large fits with GR today. Let's start small. How about the rotation profile of a thin disk galaxy? Nope, doesn't work. How about how galaxies move in clusters? Nope. How about the big bang? Nope, doesn't work if you just crank back the Universe in time. Well, at least we know where we stand today, except that we don't get the acceleration.

I am aware of what the herd thinks: dark matter, dark matter, inflation, and dark energy respectively. We have the impressive standard model of particle physics, which with its great diversity of particles at this time contributes, zero, zero, zero, and zero particles to these hypotheses.

The modus operandi of a math problem is that big things need to get done by things with no properties (other than what is needed to solve the problem). We have problems with thin disks, galaxy clusters, the big bang, and the current state of the Universe. It is my belief based on one equation I derived two ways, that there is exactly no dark matter, no inflationary matter, and no dark energy in the Universe. It is wonderfully ironic that such a conservative skeptical position will be dismissed out of hand by serious researchers in these areas of study.

So what is that equation that is the foundation of my belief? All it takes is a little standard math, played with in a different way. In the classical limit of GR, we get the work horse of practical cosmology, Newtons' law, and I have included the term for rotation:

m\frac{(V(R))^2}{R} - \frac{G m M}{R^2} = \frac{d m V}{d t}

This is the one that fails for disk galaxies. Since this doesn't work, people have tried two fixes. One is from the Modification of Newtonian Dynamics, or MOND, that transforms the gravity force term from a R-2 law to R-1 when the gravitational acceleration is super small. That worked for more than a hundred disk galaxies, but was frowned on since it lacked a good theory. It also lost out based on data from a bullet galaxy passing though another galaxy. Now people claim the only other possibility is to stuff the M box with dark matter, in just the right amount in the right place to get both the speed and stability down. How convenient.

How can one stop here, when there is that other side of the equation? It is a change in momentum people, changes in mass and velocity in spacetime. The deep, true rule for gravity is completely relativistic. The classical law will reveal a shadow of this. By the product rule of calculus, the change in momentum is the result of a constant mass times the change in velocity with respect to spacetime plus a constant velocity times the change in mass with respect to spacetime. We know what the "constant mass times the change in velocity with respect to spacetime" becomes, mA. That is the only one that every gets any light. There are two possibilities for the other term, "constant velocity times the change in mass with respect to spacetime". One would be a change in time, V dm/dt. That happens for rocket ships, but not the Universe, which does not change rapidly. We are boiling down a relativistic problem to a classical one, so formally the other effect of the force of gravity could be Vc dm/dR (check the units, dimensional analysis matters unless you pay your mortgage doing work on strings). I have had the audacity to jot this down a few times for professional physicists, and they want to know what it is, what is its name? It is the product rule of calculus, with a relativistic twist. It says that the force of gravity determines the mass distribution in space. It will provide an utterly new constant velocity solution for classical gravity problems. The rotation profile of disk galaxies is a problem that needs a stable constant velocity solution. The big bang problem needs a stable constant velocity solution. Here is the new kid on the block:

m \frac{(V(R))^2}{R} - \frac{G m M}{R^2} = m \frac{d V}{d t}+ V \frac{d m}{d R/c}

Maybe not so new, I discussed this in posts #195 and #236. I am trying to get my courage up to try and tackle a real problem with it (specifically, the rotation profile of NGC3198, I like to focus on problems that concrete). If in the earliest time, the constant velocity solution was most important, the flat nature of the cosmic background radiation might have a new math explanation. The tug of war between these two terms could lead to a number of epochs where the amount of force of acceleration apparent as mA varies. At the current time, the amount seen for mA is decreasing a bit.

Discussions of the Planck scalar and quantum gravity strike me as utterly irrelevant. The problems cited are all large scale and classical. You certainly will have plenty of people to talk to. Fortunately, I like climbing alone but try to leave clear directions on how to do a similar climb.

Doug

[Lawrence B. Crowell]

Hello Lawrence:

In Will's paper, he does not know what the eccentricities are, nor the right M/R, so he is making a sophisticated exploration of the possibilities. Calling it a guessing game was unfair to this work, but was a little verbal flare, nothing more.
Doug

He discusses a range of these situations, which is from an experimental perspective relevant.



Good to see you missed the point. You made the claim that this particular work was relevant to assessing if the exponential metric has been shown experimentally to be in error. Will's work is fine as it stands, but is not relevant to this forum.

You have yet to publicly back off the claim that the exponential metric should be rejected on experimental grounds we have today.
Doug

I said I think GEM has been falsified. To be honest I'd have to dig deeper into this matter, but it was my understanding that current astronomical measurements probed to ppN on the order where you claim there is a change. At this stage I will say things might be uncertain.



I can dump whatever I want into a Lagrange density, takes some derivatives, and end up with field equations.
Doug

In a sense, sort of. But your results may be meaningless. Seriously, as this is about Riemannian geometry and differential geometry there is a body of work on this which lead to things such as Bianchi identities. It is all based around noncommutative bases of bundles or affine constructions. Symmetric structures emerge in an sort of "oblique" way in supermanifold theory.

I have a lot of work to do besides pouring over your work here. One problem is that there is an historical trend of:

In vino veritas

In video veritas

In cyber veritas,

where by extension you Mathematica work is claimed to be true because, ... well it is done by ... . Look the problem is that Mathematica will give an output, based on strict Boolean operations which are programs to do math operations perfectly, but if the input is wrong to start then Mathematica will give wrong output.

It would require an exhausting amount of work to pour through your stuff, complete with equations written out in brutal form with \partial all explicitely represented. There is not much manifestly covariant formalism, nor have I seen any mathematical theorem-proof constructions. Actually to be honest I have not read a one to five short paragraph discription/essay of 1000 words or less just exactly what this is really all about. What is the motivation, is there some central physical principle here that can be stated in some short elegant way? To be honest what I see now is the same as what I saw two years ago when I encountered this --- a lot of complicated equations and a lot of "fixes and patches" you seem to keep having to perform.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:
my understanding that current astronomical measurements probed to ppN on the order where you claim there is a change. At this stage I will say things might be uncertain.

Fair enough.
Seriously, as this is about Riemannian geometry and differential geometry there is a body of work on this which lead to things such as Bianchi identities. It is all based around noncommutative bases of bundles or affine constructions. Symmetric structures emerge in an sort of "oblique" way in supermanifold theory.

And just as seriously, I understand why this holds together so tightly from a logical perspective. I have even learned from you how odd it is to try and tack on something symmetric to this construction, which is not what I am trying to do. Every tight web of logic has an underlying assumption. What underlies this is the assumption that the Riemann curvature tensor is necessary to describe the physical force of gravity. GR does work that way, we have darn great data to say GR is correct. All vetted researchers try to recreate GR in a wider context, or do a technical variation on the rank 2 field theory theme. I hope to show that GR, as good as it is, is not good enough for a unified field theory, it will be a challenge to challenge. I heard no reply to the long standing energy problem which is well known and well ignored today. There is no trivial way around that problem. If the Riemann curvature tensor is not relevant to the way unified fields in Nature work, then the Bianchi identities - a property of the Riemann curvature tensor - are also not important, nor are the bundles built on top of it all.
It would require an exhausting amount of work to pour through your stuff...

I certainly will not apologize for that. The Maxwell equations require a huge amount of work to understand, and most undergraduates never get it. Jackson's red book is full of technical details that take work to understand. My proposal contains the Maxwell equations as a formal subset. I also am trying to do gravity, so that makes things quite a bit more complicated, because I have to make a link to the divergence of a connection. I have to show to those oh-so-demanding people who work with GR that there is a metric solution to my field equations that is compatible with all current tests of gravity so far. I also have data that should I achieve such a goal, they say they are too busy to listen (note, this is an observation, not a complaint).

I may take up the 1000 word challenge, but it is a trap. Keep it short, keep it sweet, and people will say I haven't thought though some issue (it must be spin 2, you must get the precession of the perihelion of Mercury, what about strong field tests, demonstrate energy conservation...). Then there are those bogus complaints, such as "There is not much manifestly covariant formalism". I put in some effort to write out \nabla because that is how the covariant derivative is written out. It is the only one that I ever use, even when I resort to looking at derivatives with respect to x, y, and z, and write them as -c \frac{\partial ^2 Az}{\partial t\partial z}, I mean these are covariant partial derivatives. I use x, y, and z because I need to communicate with people who should be generous enough to know I don't mean the theory only makes sense in Euclidean coordinates.

I rather enjoy fixes and patches. I learn by humbly bumbling. My action still takes up only 2 lines. The field equations fit on t-shirt. The graphs for the Hamilton and Even representation of quaternions fit on a button. There are many details behind these compact statements. Kind of like the Lagrangian of GR (just R), and the Einstein field equations: not much text that one can write a 1300+ page book about them.

Oh, and I am very persistent. You tuned in two years ago, but the project started out in April of 1996 when I posted a question on the Internet to form a brief definition of time. In the Fall of that year I tackled a special relativity class at MIT. By April of '97 I had my first mix of Maxwell and quaternions. The GEM field equations were jotted down in August '99. It took a year and a half to find a connection to the exponential metric via the force. The new constant velocity solution for gravity was found in '01. By '02 I had a tensor expression for my action. I think my big break of '04 was showing the divergence of the Christoffel of the exponential metric was something Laplace would recognize. The developments of '07 were analytical animation used to visualize the symmetries U(1), SU(2), and SU(3), along with figuring out how to spot spin 2 symmetry in a 4-current, 4-potential coupling term. This year I have the even representation of quaternions with the group K4:6 to be used in a completely quaternion action. I understand why you don't want to get on this bus, and you might appreciate that the bus has enough momentum to keep it rolling forward.

Doug

[Lawrence B. Crowell]

Hello Lawrence:

And just as seriously, I understand why this holds together so tightly from a logical perspective. I have even learned from you how odd it is to try and tack on something symmetric to this construction, which is not what I am trying to do. Every tight web of logic has an underlying assumption. What underlies this is the assumption that the Riemann curvature tensor is necessary to describe the physical force of gravity. GR does work that way, we have darn great data to say GR is correct. All vetted researchers try to recreate GR in a wider context, or do a technical variation on the rank 2 field theory theme. I hope to show that GR, as good as it is, is not good enough for a unified field theory, it will be a challenge to challenge. I heard no reply to the long standing energy problem which is well known and well ignored today. There is no trivial way around that problem. If the Riemann curvature tensor is not relevant to the way unified fields in Nature work, then the Bianchi identities - a property of the Riemann curvature tensor - are also not important, nor are the bundles built on top of it all.


You have fallen into the trap! There is no energy problem with general relativity, at least if you think about it correctly. What you see as a problem is in fact an astounding fact of cosmology which is vitally important! Here is what you think the problem is, which I will state in fairly precise terms. The deSitter spacetime, which our cosmology appears to be asymptoting towards, as the metric


ds^2~=~-dt^2~+~e^{\beta t}(dr^2~+~r^2d\Omega^2)


The metric terms are time dependent and it is not possible to find a k_t so that there is a stationary condition {\cal L}_{K_t}g~=~0 , for a Lie derivative {\cal L}_{K_t}~=~\kappa\partial/\partial t. This Lie derivative is defined according to brackets so that


{\cal L}_{k_t}g(X,~Y)~=~g([K_t,~X],~Y)~+~g(X,~[K_t,~Y]),


where the brackets [K_t,~X] are not zero because the vectors X are functions of time. So there is no involutary system which defines a conservation of energy on the entire spacetime. The basis vectors for the deSitter cosmology are ]X_i~=~exp(\sqrt{\Lambda/3}~t/2) and so the above expression gives


{\cal L}_{K_t}g_{\mu\nu}~=~\sqrt{\Lambda/3}g_{\mu\nu}


For the cosmological constant \Lambda~=~3H^2\Omega/c^2 this equation is an eigenvalued equation, and the nonvanishing of the Hubble constant is a measure of how k_t fails to be a proper Killing vector.


{\cal L}_{K_t}g_{\mu\nu}~=~Hg_{\mu\nu}


The Hubble constant defines a velocity-distance rule H~=~{\dot R}/R, for R(t) a scale factor for the cosmology.

This gives a nonconservation of energy! There is no symmetry in general which defines a conservation of energy in a cosmology. This might for some be a horrible problem --- for me it is an astounding miracle! Now there is still the continuity equation \nabla_a T^{ab}~=~0, which is related to Bianchi identities etc, but the energy is the projection of a manifold basis vector on the momentum energy tensor


E^a~=~\Big(\int_{V^3}~-~\int_{V'^3}\Big) e_bT^{ab}


which is defined in a region of four spacetime bounded by three bounding spatial surfaces. The generalized Stokes' law then tells us this is the same as the differential of the integrand evaluated on the enclosing four spacetime


E^a~=~\int_{V^4}d(e_bT^{ab})


which can be expressed according to the covariant derivative. This will give a covariant derivative on the basis vector with de_a~=~{\underline\omega_a}^b e_b, and the differential one form is a connection form. This is coordinate dependent and so the energy can't be localized. In the above case with cosmology, this is a similar result, but here the energy can't be defined globally and conservation of energy is not a global law.

A cosmology with a nonzero cosmological constant, or what is likely a parameter set into a constant (or approximately constant) value by the inflaton or dilaton in a spin(4,2)~\sim~su(2,2) model, is one where in general there is no unitarily equivalency between states in all regions of the cosmology. Even if the spatial surface is flat the accelerated expansion of the cosmology means that there is no such equivalency, and this comes about because there is no Killing vector K which when it acts on the energy KE = const. Without a Killing vector of this sort it means there is no isometry in the spacetime which maintains a constant energy on all paths in the cosmology. So the unitarity inequivalence of vacua in the earliest universe, where a vacua of unitary states is defined on a region \sim L_p in a superposition of other such vacua on about the same scale, is frozen into the classical cosmology after inflation. In a more general setting the Coleman-Mandula theorem is then a local principle. This gives the maximal set of symmetries of the S-matrix as the (0,~1/2)\oplus(1/2,~0) spinorial Lorentz group for external symmetry, an internal symmetry [A_i,~A_j]~=~c_{ijk}A_k, and the discrete CPT symmetry. The "maximal" extension on this is called supersymmetry. A cosmology with a non-zero cosmological constant necessarily means this is a local law, it does not apply globally. This is likely a source for what we call dark energy.

I will leave this at this point. This might sound odd, but this is a tremendous blessing. This is not something physicists should try to bury away, but embrace it. If thought about properly the consequences are astounding.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

It is good to read that my powers of prediction are spot on.
Classical theories, such as the Maxwell equation of EM which can also be integrated seamlessly with quantum mechanics, allows one to define energy at a point. That utterly fails for Einstein's GR. Experts in differential geometry claim this bug is a feature. People who know bugs are bugs, and to distrust people claiming bugs are features.

Actually, I was off on tone:
This might sound odd, but this is a tremendous blessing. This is not something physicists should try to bury away, but embrace it. If thought about properly the consequences are astounding.

I would like to explain this conflict in a friendly way. As usual, I learn a few more things about the standard approach to dealing with technical issues from Lawrence's posts. If one understands GR in a non-trivial way as Lawrence does, then the logic of his argument is spot on. There is no misstep along the way. I was aware of the "take home message":
This is coordinate dependent and so the energy can't be localized. In the above case with cosmology, this is a similar result, but here the energy can't be defined globally and conservation of energy is not a global law.

If this is in fact the way that Nature handles energy for gravitational systems, then Lawrence is correct to say the consequences of nonlocal energy are astounding. There might even be links to inflation or dark energy. The opportunity to make a contribution in these areas justifies the excitement reflected in Lawrence's post.

Anyone getting paid a working wage in gravity today would endorse in various tones the logic presented by Lawrence. This is because the logic is consistent, and GR is the only game that pays a working wage. Back in the early days of GR, there were people who thought this was a significant issue (I am not enough of a GR historian to know who took which sides, or how the debate evolved).

Let me start my minority view with the words for a pop tune:

There's nothing I hate more than nothing
Nothing keeps me up at night
I toss & turn over nothing
Nothing could cause a great BIG fight
Hey -- what's the matter?
Don't tell me nothing.

It is my unwavering belief that the vacuum state, that volume of spacetime that is devoid of all events, can accomplish nothing. From the perspective of logic, there are no events to do anything. This apparently conservative belief is radical. It goes against many research themes viable today: the false vacuum need by the Higgs mechanism, perhaps a key to inflation, perhaps a key to dark energy.

This will not cause a great BIG fight. My ultra-conservative view will be summarily dismissed as indicating I do not understand the issue. I recognize that there are measurable effects created by the variation of the energy of the vacuum. These effects, real as they are, cannot be engine that flings entire galaxies apart at a greater rate. The deviation from the average measured amount of energy is not energy.

The one and only true engine of the Universe must be math done right.

The road between a classical field theory to a quantum theory is well defined, even if not discussed often. I read this first in a quantum field theory book by Kaku. The idea is that you take the field equation, which related the potential to the source, and then invert this equation to get the propagator used in Feynman diagram calculations. How straightforward! There are a few hitches, such as gauge theories like the Maxwell equations and GR cannot be inverted until one picks a gauge. This is one reason I am excited that I have expressed the action both EM and GEM using quaternion operators. Those expressions were gauge free because the gauge was explicitly subtracted away when constructing the action. Yet the field equations are necessarily invertible because that is a property of a division algebra. Have I constructed the propagators from my fields and done a few quantum field theory calculations? No, not a one. I would like to do so, but would need technical help, or a pretty significant time block to try and work it out. I do think it holds promise.

Since the 1930s, people far brighter than myself have tried to figure out how to quantize an approach to gravity. One consistent theme holds: despite the current excitement measurable in the day, all efforts have failed. There are people today excited about super symmetry, others are into loops. I think every last one of these sincere people are wrong for the same reason. If energy cannot be localized, then one can find a place where the energy is zero, and will not be able to invert the field equations to get a propagator because dividing by zero is not allowed by ultra-conservatives like myself.

In an if...then statement, if the clause in the if is not true, nothing in the then clause matters. From my lofty station in the Independent Research forum, I am stating clearly that the "If GR is true" clause is false. I have a an alternative: "If GEM is true, then...". A rank 1, linear set of field equations would be a cake walk for those skilled in the quantum field theory arts. It would be so easy, that ease would be a reason people would dismiss it. Ironic, but true.

Nothing you said sounded odd to me. I am taking aim at the foundations which are in the action of any field theory.
Doug

[Lawrence B. Crowell]

All actions are Lagrange multipiers on the bare action


S_b~=~\int pdq


and give a set of constraints required to solve the DE that emerge from the Euler-Lagragne equations. In ADM GR the constraints are NH~+~N_iH^i~+~\dots, where the rest are Gauss' law results and the rest on the number of sources. The gauge term is required because constraints on gauge fields 1/4 F^{ab}F_{ab} are insufficient. So an additional gauge terms is required in the Lagrangian to constrain four additional variables, say in the case of EM.

As a classical theory of gravitation I think basic GR is correct. This might be a "bias," but I let that be as it is. Some results of this are a bit striking, in particular with respect to cosmology. It also leads to some strange results when you look at quantum fields in curved spacetime. I think these are simply forced upon us, and quantum gravity requires that we generalize or abstract certain canonical aspects of physics. In particular it demands that we quantize gravity, or maybe as I have said to "gravitize the quantum," in ways which manage noncompact group structure for exterior symmetries that exist locally, and to "patch" these together (atlas-chart constructions in diff-geom) in some consistent manner for a global theory.

Now one can say that GR is incorrect, but frankly I think this is in line with some late 19th century prosaic ideas about fixing gravity so it is not a 1/r^2 for but a 1/r^n force for some n = 2 + a small bit. To do new physics right it often requires that certain aspects of current physics be abandoned in order that different aspects of the world are viewed as aspects of a single system. I could go at great lengths about this with general relativity and quantum mechanics as "relationships" between particles or observables. One is geometric and local, the other is nonlocal and does not describe these relationships according to metric geometry. I am not sure it is of value to write about this in great depth here.

As for energy conservation, or the conservation of a component of a momentum energy tensor projected out by a local basis element, it simply turns out that in general this is not conserved. In particular for a cosmology with a time dependent metric it is not possible to define energy conservation within standard approach. Some people say this is a "disaster," but frankly to me it means that the vacuum on a local region is not unitarily equivalent to a vacuum "out there." This requires that we address the problem of what we mean by a vacuum in a QFT, where much of what physicists think of might just be an aspect of local quantization.

At any rate if one adopts your view that "anything goes," as Cole Porter said, with Lagrangians then maybe one can make any theory, even if it has no proper connection to differential geometry. Your symmetric field tensors and the rest have no connection to differential or algebraic geometry (where the latter is required for quantum fields) and so in effect you might be able to blast your way to what you want. Yet at the end of the day you might find, if you have not already, that few people are really paying any attention.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

You certainly are free to believe that GR is correct, and I hope I have shown respect for that practice. As a gambling man, GR is a safest bet on the table. Some of the security in working with strings is that one can pick out GR within its formalism. I prefer to call it a belief to a bias since it gives direction to studies on a topic.

It is my belief that GR is not correct. This is part of what guides my research. This is a minority belief. People who hold minority beliefs get falsely accused of many things. Here is one in your post:
Now one can say that GR is incorrect, but frankly I think this is in line with some late 19th century prosaic ideas about fixing gravity so it is not a 1/r^2 for but a 1/r^n force for some n = 2 + a small bit.

Why is GR consistent with a 1/R2 force law? Because g_{00} = 1 - 2 G M/c^2 R, take the derivative, and out pops the 1/R2 force law. There is absolutely no difference between this line of reasoning and the GEM proposal. If the GEM proposal was an n=2+delta, I would consider that a deal breaker, a solid reason to reject this line of research. It is unfortunate that you thought GEM does not have an n=2 exactly force law.

One difference between Newtonian theory and GR is in a force formulation, there will be a 1/R3 term in GR, whereas there will be no such term in Newton's proposal for gravity. GEM also has exactly the same term. It is the coefficients for the 1/R4 that are about 10% different. The difference between GR and GEM is subtle, and according to my knowledge - and a direct question to Clifford Will - there is no experimental data to second order PPN accuracy for weak gravitational systems, and no experiments are being funded at this time to look to that level of precision of static sources.

The GEM proposal could not be a 19th century idea because new math is necessary. Any well-trained physicist who has read the Feynman lectures on gravity would know that the phase of the current coupling term J^{\mu} A_{\mu} would have spin 1 symmetry for the transverse wave. That must be the case for EM, a transverse wave where like charges repel. If one asked about the other parts of that analysis, the scalar and longitudinal parts, one would see it had spin 2 symmetry, the particles needed by gravity where like charges attract, and the flight of photons is changed by gravity. You may have missed that issue, but it is an essential 21th century line of reasoning.

The newest addition is the Even representation of quaternions as a 4D commuting algebra. Nothing is ever completely new: a similar idea is packaged under the name of hypercomplex numbers. That introduces a new imaginary number, the hypercomplex number, but in practice of the Even representation, no new number is needed, just a new representation, based on a real 4x4 matrix. It was shown how this is a division algebra once the Eigen values are excluded. One of the cool things is that for some quaternions, this will exclude any quaternion living on the light cone. Neat, since the Even representation is related to gravity whose particles do not live on the light cone.

As for energy conservation, or the conservation of a component of a momentum energy tensor projected out by a local basis element, it simply turns out that in general this is not conserved. In particular for a cosmology with a time dependent metric it is not possible to define energy conservation within standard approach. Some people say this is a "disaster," but frankly to me it means that the vacuum on a local region is not unitarily equivalent to a vacuum "out there."
If you believe in GR, this is an important and productive issue to think about. I try and be more cold and analytical, not labeling it a "disaster", just a reason why GR, great as it has been, is flawed and must be rejected.

Energy conservation is looking better these days for the GEM proposal. Recall how Lut said I needed gauge invariance to get conservation of energy. I also know it is needed so that the gravitons travel at the speed of light. That was accomplished in post #442 for gravity alone by subtracting it away, and in post #457 for the GEM field equations by a fortuitous cancellation between gravity and EM.

"Anything goes"? I have gone to great effort to construct a theory consistent with experimental tests of weak field gravity to first order PPN accuracy. It differs slightly at second order PPN accuracy for spherically symmetric, static sources. My action is well defined. I have made the necessary link to spin 2 in the current coupling and field strength tensors. Because of my belief that GR is wrong, it eliminates the need to make what you label a "proper connection to differential geometry". In practice, that is always some road to the Riemann curvature tensor or Bianchi identities or some antisymmetric system. There is some antisymmetry in my proposal, just enough to get the job done for EM, no more than that, a rare case of intellectual minimalism. I do use a dash of differential geometry in calculating the Christoffel symbol of the second kind for the Rosen metric, getting a GM/R potential out of the exercise, showing I work with exactly n=2 for the force law.

The only thing I care about are technical arguments about the GEM proposal. I can number the exchanges I have had with Ph.D. level physicists. These have always had a one sided nature: they tell me what I need to work on, I do so, but what I get done does not make it back, they are too busy. Again, this is an observation. I told Alan Guth I had a unified field equation. He said I needed a field theory, with the Lagrangian defined, the field equations derived by varying the action, find solutions, show those solutions are both consistent with current tests and different at higher order. That took more than two years of work, but I did it. It is impossible to explain something long and complicated to the man, he will fall asleep, I have seen it happen :-) The entire issue about spin was created based on comments from Steve Carlip. In your own indirect way, I found the graph for both the Hamilton and Even representation of quaternions based on discussions here.
Yet at the end of the day you might find, if you have not already, that few people are really paying any attention.

There is nothing technical in this comment, it is all social. People get paid enough money to pay for mortgages studying GR and slight variations of GR. At the current time, there are no international meetings on quaternions. They will be on the stage at the International Conference on Clifford Algebras which happens once every three years. This thread is over 40k views which is considerably more than most threads at Physics Forums.

I am patient on resolving technical issues, and indifferent to the social aspects.

Doug

[Lawrence B. Crowell]

Hello Lawrence:

Why is GR consistent with a 1/R2 force law? Because g_{00} = 1 - 2 G M/c^2 R, take the derivative, and out pops the 1/R2 force law. There is absolutely no difference between this line of reasoning and the GEM proposal. If the GEM proposal was an n=2+delta, I would consider that a deal breaker, a solid reason to reject this line of research. It is unfortunate that you thought GEM does not have an n=2 exactly force law.

Doug

I indicated this by way of comparison. There is a standard theorem that a central force given by F~\propto~r^n determines closed elliptical orbits for n~=~\pm 2. There were proposals to modify Newtonian gravity to account for the precession of Mercury's orbit.

To be honest what you are doing is to impose symmetric terms into a field tensor, which must necessarily be zero, in a way so as to break gauge symmetry. You then figure this is some sort of coup, for now your theory requires no gauge fixing term or gauge condition ---- apparently seen as some auxilliary "baggage" you have eliminated. The "procedures" you go through are comparatively elementary, which would suggest that if these were appropriate that some physicists would have discovered and applied them by now. After all people such as Feynman were quite on the bright side of the intelligence scale. These symmetric field terms you fold into the theory, which are presumed to account for gauge conditions, must either be necessarily zero or if not then you are breaking up gauge symmetry. In the first case this means your theory would in some way reduce to standard YM gauge field theory, or if not then you are saying that field theory is not at all about gauge theory. You did make a statement about principal bundles and the rest being irrelevant. Hermann Weyl was one of the smarter guys to do theoretical physics and proved that EM was a gauge system.

As for quaternions, they are noncommutative and were employed initially by Maxwell in his EM equations to account for the curl-equations. To be honest a lot of what you say about quaternions does not make much sense, or at least what you are calliing quaternions appear to be something else.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:
I indicated this by way of comparison.

Fair enough. I felt it necessary to point out that this particular comparison was not accurate on a technical level. You still have your serious reservations, but this is not one of them.
The "procedures" you go through are comparatively elementary, which would suggest that if these were appropriate that some physicists would have discovered and applied them by now. After all people such as Feynman were quite on the bright side of the intelligence scale.

The logical conclusion based on accepting this analysis is that one should not bother to try and do physics, unless one could demonstrate in a measurable way they were smarter than Feynman. One of Feynman's chief characteristics is to challenge anybody, no matter what their station. It is ironic that you place Feynman on an unreachable pedestal.

I do not put Feynman there. He was amazing, and human. Let's challenge Feynman specifically on the completeness of his analysis. Do you have "Feynman Lectures on Gravitation" in your possession? If not, one can look inside at amazon.com to review the pages in question (pages 31-34). He does an analysis of the current coupling term in EM, J^{\mu} A_{\mu}. He takes the Fourier transform of the potential to get a current-current interaction. He restricts the analysis to a current moving along the z axis. The goal is to figure out the phase of the transverse current, the Jx and Jy terms (he used J1 and J2 in his lectures). The equation on page 34 is the basis of the statement that for EM, the transverse wave will take a 2\pi radian change in phase to get back to where it started, a property of spin 1 particles. Spin 1 particles mediate a force where like charges repel, so photons can do the work of gravity.

His reasoning is both flawless and incomplete. He does not considered the rho-Jz current coupling (or J3-J4 in the lecture). For that term, the J and J' work together, and because they work together, it will require a change of \pi radians to get back to the start. That is a property of particles with spin 2. Feynman's analysis was not complete, which is different from wrong.

During my first three years at MIT, I played poker against a guy named Rob that went on to win the World Series of Poker. We were roughly the same level, although we both knew Dean was better. Rob kept up and intensified his study of the game, although I did not. I respect the conservative bet that it is unlikely that anyone posting to Independent Physics at Physics Forums has not found something new. Aware of these odds, what I actively look for are areas that have not been explored. I know neither Feynman or Einstein worked with quaternions. I know P.A.M. Dirac was asked if he was interested in a formulation of relativistic quantum field theory with quaternions (in other words, the Dirac equation), and after pausing a really long time as was his way, said he would only be interested if they were the real-valued quaternions (the person asking the question was crushed, since he had worked with complex-valued quaternions). This implies that Dirac did not work with 19th century quaternions.

Rob won that year on the flip of the last card, where is opponent got a flush, but he pulled a full house. In this particular instance, I have documented how Feynman's published analysis was not complete. It is nice it ties in so closely with the content of this thread.

I always get skeptical when I see "quotes" around "words". It usually indicates a breakdown in communication. A differential equation written with a division algebra should always be invertible. This would be a great time to cite such a proof for that assertion, or just do the proof myself, but I know my limitations.

The games with gauges are precise. One needs EM theory as well as gravity to be invariant under a gauge transformation for two reasons. First, both the particles that mediate the forces - the photon and the graviton - travel at the speed of light. Second, according to Lut, it is much easier to demonstrate a gauge theory conserves energy.

So what exactly do I mean by gauge theory in the context of the GEM proposal? Consider this scalar field:

g=\frac{\partial \phi}{\partial t} - \frac{\partial Ax}{\partial x} -\frac{\partial Ay}{\partial y} -\frac{\partial Az}{\partial z}

Not a one of these terms ends up in the field equations in GEM field equations derived in most 438, 442, or 457. You are free to let \nabla . A = 0, known as the Coulomb gauge. You could work in the static gauge, setting \frac{\partial \phi}{\partial t} = 0, or the Lorenz gauge, \frac{\partial \phi}{\partial t} + \frac{\partial Ax}{\partial x} + \frac{\partial Ay}{\partial y} + \frac{\partial Az}{\partial z} = 0.

Peter Jack was the first person to write operators with real-valued operators to generate the Maxwell field equations. A year later, I did that trick independently. We has the mark of independent researchers: we did not do it the right way, just a way that worked. Post 438 is significant because the derivation is the first to use a quaternion to generate the Lagrange density. Once one gets E2 - B2, the rest is completely triple grade A standard field theory. And I got the Poynting vector as a freebee. An accident? I wouldn't bet against that one. Elegance is an essential guide in the search for truth.

One of the steps used there is familiar to anyone who decides to play with quaternions, and that is to eliminate the scalar using a conjugate, q - q*. That is what was done on the road to E2 - B2. We know the Maxwell equations is gauge invariant, and when the Lagrangian is formulated with quaternions, the reason is clear: it got subtracted away. Nice. Do the same exercise with the Even representation of quaternion - an idea from February 2008 - and one gets the field equations which are the natural relativistic form of Newton's law of gravity.

There was no way I could have planned it, but to find the unified field theory, do the exact same as EM and G separately, just skip the q - q* business. Then both Maxwell and G toss in the squared gauge, but with opposite signs, so they drop. An accident? See above comment again.

I also need to formulate GEM in a way that is not invariant under a gauge transformation. This will apply to the multitude of particles that do not travel at the speed of light. I haven't done that yet here, I got distracted defending the virtue of this work, and preparing talks, and living life (broken arm managements, yadda, yadda).

It is clear that the word symmetric is of concern to someone steeped in the technical nature of approaches to GR. As often repeated, I am not doing a variation on GR, I am doing a variation on the Maxwell equations. As you may know, one needs to supply the background metric as part of the mathematical structure of the Maxwell equations to put it to use (people usually use a flat metric, but it is a choice). There is no differential equation to solve that constrains what the metric can be.

What I am doing is a variation on the Maxwell equations, just barely enough to provide a differential equation whose solution is a dynamic metric based on the physical conditions.

It is a fine thing to question how a symmetric component could integrate into the math structure of GR to provide a non-zero result. As a variation on GEM, the concern is silly. Here are the undoubtedly non-zero terms found in the fields of Maxwell:

E = -\frac{\partial A}{\partial t} - \nabla \phi
B = \nabla \times A

And from the same soil, here are the two fields I refer to as the symmetric analogues needed for gravity:

e = \frac{\partial A}{\partial t} - \nabla \phi - 2 \Gamma_\nu^{\mu 0}A^{\nu}
b = -\nabla \Join A - 2 \Gamma_\nu^{i j}A^{\nu}

where \Join is defined as the symmetric curl, composed of the same terms as the curl, but all the signs are positive.

The fields of E and B are manifestly free to be non-zero. The fields e and b, no matter what labels we attach to them, are also free to be non-zero.

So how are these four different? If one decides to work with a metric compatible, torsion-free connection, then the way a dynamic metric changes will not change a calculation of the fields E and B, but will change e and b.


There are many claims on the Internet about Maxwell and quaternions. The idea of the curl was due to quaternions, as was the gradient, the dot product, the divergence, scalars and vectors. In the first edition, he used pure quaternions, where the scalar is equal to zero. That is a different way to write a 3-vector. The pure quaternions were removed by the third edition. In the introduction, he predicts that someone will someday figure out how to do all the work with quaternions, a point of pride for me, completing a task Maxwell defined.

Maxwell would not have been concerned about the issue of spin 2 symmetry, it was before his time. He certainly wouldn't be concerned with the divergence of the Christoffel of the Rosen metric. Smart guy, but imperfect at seeing into the future, a common problem.

Doug

[jillz]

ok; lots of great info, but what effect does mass have on height??
If all parameters remain constant except the mass varies from 1kg to 10kg, the maximum height remains the same, so whats the effect of mass on height?

[Mentz114]

Doug & Lawrence,

this is getting repetitive. You're not going to agree for the simple reason that Doug doesn't see any deep truth in the 4D manifold or the theorems of diff geom or gauge theory ( even while claiming that 'elegance' is a guide to truth ). Where does the 'truth' lie ? You are both convinced that your theoretical structures are 'right', but it's experimental facts that will decide not the mathematics you espouse.

Doug, your epic posts are getting too long. All the biographical detail does not add to the physics, or make your arguments more convincing. Didn't someone say 'brevity is the soul of something' ?

It's time to shut up and calculate !

Lut

[sweetser]

Hello Lut:

Scientific conflicts are hard to manage. They quickly degenerate to name calling, which Lawrence and I have so far avoided (the personal stories might be one way of dodging that descent, plus I think people unskilled in these arts might find those side notes entertaining). Yes, the underlying difference in opinion stands, and I have no doubt will remain. The focus of exchanges has shifted, and that is informative, at least to me.

To stay rooted in calculations, have you been able to get to field questions for these three Lagrangians:

Maxwell:

(\nabla A - (\nabla A)^*)(- A \nabla - (A \nabla)^*) = (0, -E ~+~ B)(0, E ~+~ B) = (E^2 ~-~ B^2, 2 E \times B) \quad eq 1

Gravity:

(\nabla A2^* ~-~ (\nabla A2^*)^*)(- A2 \nabla^* ~-~ (A2 \nabla^*)^*) = (0, -e ~+~ b)(0, e ~+~ b) = (-e^2 ~+~ b^2, -e \Join e ~+~ b \Join b) \quad eq 2

And GEM:

\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))
= ((-g, E ~+~ B)(g, -E ~+~ B) ~+~ (g, e ~+~ b)(g, -e ~+~ b))=(-g^2 ~+~ E^2 ~-~ B^2 ~+~ g^2 ~-~ e^2 ~+~ b^2, 2 E \times B ~-~ e \Join e ~+~ b \Join b ~+~ 2 gE ~+~ 2 gb) \quad eq 3

I can do that with pencil and paper and Mathematica. How is your tensor software doing with the challenge?

Doug

[Lawrence B. Crowell]

I think Lut has a point. I do have to say that you are working up equations which have n-chains equated to m-chains for m =/= n. This is in part what your scalar equation stuff does, which leads you to the symmetric curl, as you call it, which does not mathematically exist. Sure, once you spin something like that up from whole cloth you can then do calculations with it. Yet even if those calculations are flawless they are ultimately based on mathematical nonsense.

Yet it is clear I am not going to make you see these points. I indicated something about some recent measurement of the orbits of neutron stars and their agreement with GR in ppN expansion. Ultimately this is where the verdict will lie. At this point it may come down to a choice between a theory well grounded in differential geometry and one with questionable differential geometric content. If you are right then it not only overturns GR, but it means that the foundations of manifold mathematics from Gauss to Riemann and all the way up to Taubs and Atiyah needs to be seriously modified as well.

Lawrence B. Crowell

[CarlB]

Dr. Crowell's objection reminds me of the objections to Lisi's E8 paper; that one should not be allowed to add together spinors and scalars.

Rather than get involved in the particular objections, I'd like to note that the value of a theory is in its ability to predict reality, not its ability to be mathematically clean in some theoretical sense. What we need is the ability to calculate, not the ability to theorize.

For relativity and quantum field theory, we have absolutely no experimental verification of these theories. What we have instead is good experimental verification of GR calculations and excellent experimental verification of QFT calculations. All theory that lies below the level of calculation is junk DNA that was convenient to frame the calculations, but need not be a part of a newer, more general, theory. The successful calculations, on the other hand, must be retained or replaced with equivalent.

[Lawrence B. Crowell]

I have decided to resurrect my little site here I started a year or two ago. I worked up an interesting idea on quantum fields in curved spacetime. This is very simple, only relying upon some basic ideas of geometry in QM and a fibration.

http://www.physicsforums.com/showthr...=115826&page=2

I worked this up in my head as I wrote this, so there might be a boo-boo or two here, but I think the basic idea looks reasonable. In way of boo-boo I posted this notice on my site as well.
Lawrence B. Crowell

[Lawrence B. Crowell]

Dr. Crowell's objection reminds me of the objections to Lisi's E8 paper; that one should not be allowed to add together spinors and scalars.



Lisi's paper does add them, but with the application of elements of the Clifford algebra. How Garrett frames spinors and scalars is similar to the application of Grassmannians in supersymmetry. Lisi is a bit glib on this, but it is not fatal to the basic architecture of his paper.

Lawrence B. Crowell

[K.J.Healey]

You mean (for a working link) : http://www.physicsforums.com/showthread.php?t=115826&page=2

[sweetser]

Hello Lawrence:

To keep things short and snappy per Lut's request, if this exists:

(\frac{\partial Ay}{\partial z} ~-~ \frac{\partial Az}{\partial y},\frac{\partial Az}{\partial x} ~-~ \frac{\partial Ax}{\partial z},\frac{\partial Ax}{\partial y} ~-~ \frac{\partial Ay}{\partial x})

...and this does not mathematically exist:

(\frac{\partial Ay}{\partial z} ~+~ \frac{\partial Az}{\partial y},\frac{\partial Az}{\partial x} ~+~ \frac{\partial Ax}{\partial z},\frac{\partial Ax}{\partial y} ~+~ \frac{\partial Ay}{\partial x})

then I am more than willing to challenge the status quo of manifold mathematics because it is an error of omission, not a mistake per se.

Doug

[Mentz114]

Hello Lut:

To stay rooted in calculations, have you been able to get to field questions for these three Lagrangians:

Maxwell:

(\nabla A - (\nabla A)^*)(- A \nabla - (A \nabla)^*) = (0, -E ~+~ B)(0, E ~+~ B) = (E^2 ~-~ B^2, 2 E \times B) \quad eq 1

Gravity:

(\nabla A2^* ~-~ (\nabla A2^*)^*)(- A2 \nabla^* ~-~ (A2 \nabla^*)^*) = (0, -e ~+~ b)(0, e ~+~ b) = (-e^2 ~+~ b^2, -e \Join e ~+~ b \Join b) \quad eq 2

And GEM:

\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))
= ((-g, E ~+~ B)(g, -E ~+~ B) ~+~ (g, e ~+~ b)(g, -e ~+~ b))=(-g^2 ~+~ E^2 ~-~ B^2 ~+~ g^2 ~-~ e^2 ~+~ b^2, 2 E \times B ~-~ e \Join e ~+~ b \Join b ~+~ 2 gE ~+~ 2 gb) \quad eq 3

I can do that with pencil and paper and Mathematica. How is your tensor software doing with the challenge?

Doug

I've been playing about with the third one,

E^2 - B^2 -e^2 + b^2
which comes down to

(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu})

with the condition that

(\partial^{0}A^{0})^2+(\partial^{1}A^{1})^2+(\part ial^{2}A^{2})^2+(\partial^{3}A^{3})^2=0

to get rid of g^2. This is as far as I've got, but I'll get back to it when my schedule allows.

This thread is interesting, please take the time to have a look at it.

http://www.physicsforums.com/showthread.php?t=192422

[sweetser]

Hello Lut:

It is good to focus on eq 3 which represents the GEM Lagrangian, instead of separately Maxwell and gravity, is a good one to focus on. I don't think it is correct to impose the condition:

(\partial^{0}A^{0})^2+(\partial^{1}A^{1})^2+(\part ial^{2}A^{2})^2
+(\partial^{3}A^{3})^2=0 \quad eq 4

What happens algebraically is that a -g2 cancels a +g2. You need to see exactly the same thing happen with your software.

As a practical programmer type, I appreciate your shortcut to the scalar. The scalar is the same, so should flow through from there the same. The theorist does not agree, because your algebra is working based on a particular gauge constraint - no different from any other gauge theory - while eq 3 has a gauge cancellation. Because eq 3 is based on cancellation, there are infinitely more choices for the gauge that work with eq 3 than for your software that dictates the sum of the squares of the partial derivatives.

Don't toss away this work on

(\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) \quad eq 5

That may well be the Lagrangian used for GEM that applies to massive particles. The Lagrangian for the force mediating particles that travel at the speed of light because they are gauge invariant due to cancellation - eq 3 - is not the same as the one for massive particles which must break gauge symmetry in an elegant way, perhaps exactly like eq 5. I have to look at the details of this to see if I like it :-) Lots of things going on, preparing for a few talks, APS talks in New London, Connecticut and St. Louis, Missouri.

Just read the first post in the thread you suggested. My initial reaction - which will be fun to see if anyone else made the argument - is the spin symmetry of the field strength tensor is not consistent with a spin 1 particle needed for like charges to repel.

Doug

[Mentz114]

Hi Doug:

OK, I'll find a way to deal with g.

As a practical programmer type, I appreciate your shortcut to the scalar. The scalar is the same, so should flow through from there the same. The theorist does not agree, because your algebra is working based on a particular gauge constraint - no different from any other gauge theory - while eq 3 has a gauge cancellation. Because eq 3 is based on cancellation, there are infinitely more choices for the gauge that work with eq 3 than for your software that dictates the sum of the squares of the partial derivatives.
Sorry, I don't understand a word of that.

I assume it's correct that E^2 - B^2 -e^2 + b^2 is (\partial_{\mu} A_{\nu})(\partial^{\mu} A^{\nu}) with the g part removed.

[sweetser]

Hello Lut:

The question is why does g make no contribution in the GEM proposal, specifically eq 3 of post 500. There are 4 A's required to generate the squared fields:

\frac{1}{2}(-(A \nabla)(\nabla A) ~+~ (\nabla^* A2)(\nabla A2^*))

The first two have curls, the last two have symmetric curls. The gauge g is definitely not set to zero. The g2 contributed by the first pair cancels with the second pair. You can tell if your software has faithfully done this because g is not set to zero, but none of the components of g appear in the final result.

In the first pair, it is the order of the differentials that changes. In the second pair, it is which one gets conjugates. The contraction of the asymmetric tensor \nabla_{\mu} A_{\nu} does not do this.

Doug

[Mentz114]

Doug:

The easiest way is just to ignore it, so we start by defining the field tensors,

F_{(g)}^{\mu\nu} = \[ \left[ \begin{array}{cccc}
0 & e_x & e_y & e_z\\\
e_x & 0 & b_z & b_y \\\
e_y & b_z & 0 & b_x \\\
e_z & b_y & b_x & 0 \end{array} \right]\]

F_{(em)}^{\mu\nu} = \[ \left[ \begin{array}{cccc}
0 & -E_x & -E_y & -E_z\\\
E_x & 0 & B_z & -B_y \\\
E_y & -B_z & 0 & B_x \\\
E_z & B_y & -B_x & 0 \end{array} \right]\]

L_{(g)} = F_{(g)}^{\mu\nu}F_{(g)}_{\mu\nu} = b^2 - e^2

L_{(em)} = F_{(em)}^{\mu\nu}F_{(em)}_{\mu\nu} = B^2 - E^2

and with the usual definitions of E,B, e and b -

E^i = \partial^0A^i - \partial^iA^0

e^i = \partial^0A^i + \partial^iA^0

B^i = \partial^jA^k - \partial^kA^j, i <> j,k

b^i = \partial^jA^k + \partial^kA^j, i <> j,k

from which I finally get this
B^2 - E^2 + b^2 - e^2 =
(\partial^{x}A^y)^{2}+(\partial^{x}A^z)^{2}+(\part ial^{y}A^x)^{2}+(\partial^{y}A^z)^{2}+(\partial^{z }A^x)^{2}+(\partial^{z}A^y)^{2}-(\partial^{t}A^x)^{2}-(\partial^{t}A^y)^{2}-(\partial^{t}A^z)^{2}-(\partial^{x}A^t)^{2}-(\partial^{y}A^t)^{2}-(\partial^{z}A^t)^{2}

The next step is to apply Euler-Lagrange, which doesn't seem to lead anywhere. Where did I go wrong ? ( Apart from losing some factors of 4 and butchering the notation !).

[later] I seem to be getting

\Box^2A^{\mu} = 0

Too tired to continue right now. I've got a feeling I've made a meal of something simple.

[sweetser]

Hello Lut:

You were only suppose to get the cross terms :-) I scanned in my hand-drawn derivations of the field equations here:

http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations/photo#5171869024067390082

The step that may be tripping up your program is multiplying the even quaternion representation that is (0, b + e)(0, b - e) should give the scalar b2 - e2, not the negative of this. My bet is B2 - E2 - b2 + e2 would work in your software as is.

Doug

[Mentz114]

Doug:

Yes, I changed a sign and now I get

B^2 - E^2 - b^2 + e^2 =

8\partial^xA^2\partial^yA^1
+8\partial^xA^3\partial^zA^1
+8\partial^yA^3\partial^zA^2
-8\partial^tA^1\partial^xA^0
-8\partial^tA^2\partial^yA^0
-8\partial^tA^3\partial^zA^0
(apologies for mixed notation, but I'm sure you get the drift)
In agreement ( up to a sign) with your doodle. I won't repeat the algebra to get the field equations.

I suppose the only point of this is that one can start with the traceless field tensors and go from there to get the same formulation as your funny quaternions. Much easier for the lay ( non-quat.) person to grasp.

Lut

PS : before I closed down the calculator I did this -

g_{ik}F_{(g)}^{mk}g_{mn}F_{(em)}^{ni}. It comes to zero, zip, nothing.

So the field tensors are orthogonal in a way.

[Mentz114]

Doug:

there is a neat way to lose g. The symmetric field tensor has a dual, in analogy with EM theory, defined thus -

\tilde{F_g}_{mn} = s_{mnij}F_g^{mn}

where s is the totally symmetric pseudo-tensor equal to |e|. The dual has no diagonal terms and

\tilde{F_g}_{mn}\tilde{F_g}^{mn} = -8(b^2 - e^2)

So, putting the square of the dual in the Lagrangian density does the job.

The symbol s also allows the symmetric curl of a vector to be defined as s_{ijk}\partial^{j}E^{k}

Lut

[Lawrence B. Crowell]

Doug:


[later] I seem to be getting

\Box^2A^{\mu} = 0

Too tired to continue right now. I've got a feeling I've made a meal of something simple.

Which is the Lorentz gauge. If you put the g in there and let this be something "dynamical" then this inserts group dependent structure into the theory. The elliptic complex determines the bundle curvatures \Omega^{2}(ad(g)) which give fields "mod-g," for here g means group. The gauge condition is set to constrain the appropriate variables.

Lawrence B. Crowell

[sweetser]

Hello Lut:

I concur with the postscript statement:

g_{ik}F_{(g)}^{mk}g_{mn}F_{(em)}^{ni} It comes to zero, zip, nothing. So the field tensors are orthogonal in a way.

When I thought about GEM in terms of tensors, the symmetric one was the average amount of change in the 4-potential in 4 spatial dimensions, while the EM tensor is the deviation from the average amount of change. My recent shift to quaternions puts a stress on the underlying math tools, the Hamilton versus Even representations. At this time I don't understand all the links between these two perspectives, but it is nice having multiple perspectives. Thanks for adding to my views via this calculation.

Doug

[sweetser]

Hello Lut:

I am not clear on post #514. F_{(g)}^{\mu\nu} has nothing down its diagonal - where the terms of a gauge live - so its dual also has this property. That appears to be starting too far down the page as it were.

Here is what I am doing with quaternions, written as your tensors:

\[ \left[ \begin{array}{cccc}
g_t & e_x & e_y & e_z\\\
e_x & g_x & b_z & b_y \\\
e_y & b_z & g_y & b_x \\\
e_z & b_y & b_x & g_z \end{array} \right]\] -
\[ \left[ \begin{array}{cccc}
g_y & -e_x & -e_y & -e_z\\\
-e_x & g_x & b_z & -b_y \\\
-e_y & -b_z & g_y & -b_x \\\
-e_z & -b_y & -b_x & g_z \end{array} \right]\]

The second tensor is the conjugate of a symmetric tensor. It shows operationally why there is a significant difference between F_{(g)} and F_{(em)}, where both define the conjugate as flipping the signs of the later three parts, but F_{(em)} can do so by taking the transpose of the matrix representation. That does not work for F_{(g)}. Nice.

Doug

[Mentz114]

Hi Doug,

in post #514, I'm defining the dual of F_g using a pseudo-tensor, in analogy with the EM case. The pseudo-tensor s is something I just invented, because we go from EM/anti-symmetric -> gravity/symmetric. This dual is traceless and gives the right energy.

It's marginally less dodgy than inventing a new quaternion algebra ( maybe).

Lut

[sweetser]

Hello Lut:

I thought you might be inventing things. This is a good sign. It dovetails with Lawrence's complaint about my work not fitting into the proud tradition of differential geometry.

I think I referred to the wrong post. Post #514 is an extension of #511. It was #511 where you wrote out F(g), saying one should just ignore the gauge. It would be great if you could write the software to make it so. I don't count setting to zero a good answer, or just ignoring it is good practice. I want to catch the math responsible for gauge symmetry. That is what is most interesting.

Doug

[Mentz114]

Hi Doug:

I understand your unwillingness to ignore, or set to zero the trace of the grav. field tensor. We can play at finding mathematical ways to dispose of it, but this might not throw light on the physical justification. I'm not sure why you call it 'the gauge'.

Anyhow, I'll try and justify my use of the 'symmetric dual' for the field tensor.

In EM theory, one gets the dual using the Levi-Civita antisymmetric pseuso-tensor like this,

\tilde{F}_{pq} = \epsilon_{mnpq}F^{mn}..............(1)

furthermore

F_{mn}F^{mn} = -\tilde{F_{mn}}\tilde{F^{mn}} = -2(E^2 - B^2).....(2)

So, to carry on the analogy between the EM theory, with anti-symmetric F and curl, to the gravitation case with a symmetric F and symmetric curl, we make a dual using a symmetric equivalent of the L-C pseudotensor, which we call 's' ( for 'symmetric').

\tilde{F}_{pq} = s_{mnpq}F^{mn}............(3)

from which it follows

F_{mn}F^{mn} - 2g^2 = -\tilde{F_{mn}}\tilde{F^{mn}} = -2(e^2 - b^2)....(4)

because \tilde{F}_{pq} is traceless.

While I was typing the above, I realised that your field Lagrangian can now be very succinctly written as

(s_{\mu\nu\rho\sigma}\partial^{\mu}A^{\nu})^2 - (\epsilon_{\mu\nu\rho\sigma}\partial^{\mu}A^{\nu}) ^2........................(5)

this has no g terms as required, with the two pseudo-tensors doing the work of the odd and even quaternions.

I still think GEM is wrong but the Lagrangian is tidier !

Lut

PS the software worked faultlessly and did all the donkey work.

[sweetser]

Hello Lut:

Good work! I particularly like equation (5) and will make a mental note of it. Many group theory pros would object, saying the tensor is reducible, and reducible tensors cannot be used to represent fundamental forces. This issue has been raised indirectly in that thread you recommended reading in the first reply:
The "trouble" with gauge theories is that you are including unphysical degrees of freedom (such as the longitudinally polarized photon) and you have to make sure that they don't appear in the final calculations, and this complicates matters when you have to calculate things. However, despite the headaches, this can be done.

What I have argue since post #1 is the longitudinal mode of emission is a spin 2 graviton traveling at the speed of light doing the work of gravity. It is the scalar mode of emission that causes more serious problems since for a spin 1 field it implies negative probabilities.

People familiar with EM quantum field theory know these problems, and impose a constraint on quantizing EM such that the scalar and longitudinal modes are always virtual. Grad students grinding through these calculations often complain: it looks like a hack. Yet the hack needs to be there for a spin 1 field, and eventually they shut up and accept it.

The hack looks like a golden opportunity to put two modes of emission to work for gravity. Those modes will not harm EM theory in any way. A spin 2 scalar field would not have the indefinite metric problem, no negative probability problem.

No one else has picked up on the omission in Feynman's analysis of the current coupling term. I think that represents the greatest barrier to a rank 1 field theory.

Someone who bought a DVD from me thought I was within spitting distance of a unified field theory. I like that characterization slightly crude as it is.

Doug

[sweetser]

Hello:

I thought I'd share a small epiphany I had since it sheds a bit of light on the conflict I have had with Lawrence. The best field theory we have is the Maxwell field equations for EM. The second rank field strength tensor is:

F_{\mu \nu} = \partial_{\mu} A_{\nu} - \partial_{\nu} A_{\mu} \quad eq 1

It gives me confidence to know this is the foundation of my GEM proposal. Yet it also was the basis for the weak, the strong force, and Einstein's work on EM. For EM, the weak, and the strong forces, what changes is what group gets plugged into the machinery, U(1), SU(2), and SU(3) for EM, the weak, and strong forces respectively.

Gravity is a bit different, but not so different. I don't know the details, but I recall reading how Einstein looked to EM for guidance on how to proceed. Take a peak at the Riemann curvature tensor:

R^{\rho}{}_{\sigma \mu \nu}=\partial_{\mu} \Gamma^{\rho}{}_{\sigma \nu}-\partial_{\nu} \Gamma^{\rho}{}_{\sigma \mu}+\Gamma^{\rho}{}_{\mu \lambda}\Gamma^{\lambda}{}_{\nu \sigma}-\Gamma^{\rho}{}_{\nu \lambda}\Gamma^{\lambda}{}_{\mu \sigma} \quad eq 2

We have a subtraction thing going on again. This is a much more complicated animal, but we only need two contractions (the Ricci tensor minus the Ricci scalar), for the Einstein field equations.

Now I can see the resistance to tossing in a "+" for the minus in eq 1. That equation is good enough for EM, the weak, and the strong force, which covers 3/4 known forces of Nature. We have GR which has withstood every subtle test, while failing all large scale ones (velocity profiles of galaxies, the big bang, and our current acceleration). I can be at peace with that resistance. I don't think the objection to a symmetric field strength tensor will stand the test of time, but it is rational to not embrace it.

Doug

[Lawrence B. Crowell]

Gravitation differs from the gauge fields as an interaction determined by an exterior symmetry, rather than an inner symmetry modelled as a vector space on a principal bundle. The bundle structure is then a fibration of the symmetries of the space (spacetime) in an atlas of local tangent spaces. This is all spelled out in the Coleman-Mandula theorem, which gives a mathematical reason for the structure of interactions which can possibly exist. It does not tell us their explicit form, but as a "no-go theorem" of sorts it tells us what is not permitted. What you say above about "gravity being different" is frankly a bit on the silly side of things.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

Thank you for clarifying the difference between an inner symmetry and an exterior symmetry. I do lack the experience to speak the jargon of what is known. I thought I had admitted I was speaking imprecisely.

So I go read up about the "Coleman-Mandula theorem, which gives a mathematical reason for the structure of interactions which can possibly exist [under the assumptions used in the theorem]." I hope you do not object to the clause, which is appears banal. Plucking out a one-line summary:
In other words, every quantum field theory satisfying certain technical assumptions about its S-matrix that has non-trivial interactions can only have a symmetry Lie algebra which is always a direct product of the Poincare group and an internal group if there is a mass gap: no mixing between these two is possible.
A Lie algebra has the property that it anti-commutes. If one makes a strategic decision to walk in a new direction with a field strength tensor \partial_{\mu} A_{\nu} + \partial_{\nu} A_{\mu}[/tex] which is a different combination of exactly the same players in [itex]F_{\mu \nu} (when the symmetric field tensor trace is zero). One of the first observations is that any generating algebra for the group would have to commute, ergo any conclusions of the Coleman-Mandula theorem do not apply.

Sorry Lut, this is the same non-debate Lawrence and I have had before. I had to find out where the imposed domination of anti-commutivity was written.

Doug

Big fuzzy picture: If gravity is the force justifying why everything loves and attracts everything else, its algebra must be commuting.

[Mentz114]

Hi Doug:

You needn't apologise, it is always diverting to read your and LBC's posts, as long as they are short and to the point.

The gauge group of teleparallel gravity, the translation group, has a commuting Lie algebra.

I don't believe all mathematical theorems necessarily apply in physics, except in so far as they prevent inconsistencies. For example, von Neumann's 'proof' that QM cannot be modelled by 'hidden variables' is disproved by counter-example. Mathematical theorems always have lots of tight conditions that are often violated in practice.

Lut

[Lawrence B. Crowell]

Hi Doug:

You needn't apologise, it is always diverting to read your and LBC's posts, as long as they are short and to the point.

The gauge group of teleparallel gravity, the translation group, has a commuting Lie algebra.

I don't believe all mathematical theorems necessarily apply in physics, except in so far as they prevent inconsistencies. For example, von Neumann's 'proof' that QM cannot be modelled by 'hidden variables' is disproved by counter-example. Mathematical theorems always have lots of tight conditions that are often violated in practice.

Lut

I am not trying to say that the Coleman-Mandula theorem is proven in physics, where after all in science nothing is ever proven as such. It is just that the CM theorem is a fairly strong aspect of theoretical physics.

Teleparallel connections are commutative for special cases of nongeodesic motion. Christopher Columbus sailed to the new World on a nearly teleparallel path.

Lawrence B. Crowell

[sweetser]

Hello:

I will be going to the New England APS/AAPT Meeting in New London, Connecticut on Saturday to give the talk "Using Quaternions for Lagrangians in EM and GEM." At just under 15 minutes, you will be able to see the Maxwell equations derived from its Lagrangian. Rarely will you see this much detail about this central part of physics. GEM is a small variation on the theme, where you can see the 12 of 24 terms flip signs, generating a relativistic cousin of Newton's field equations for gravity. Enjoy.

Doug

http://youtube.com/watch?v=P9TUqUXGgpE

[Mentz114]

Teleparallel connections are commutative for special cases of nongeodesic motion. Christopher Columbus sailed to the new World on a nearly teleparallel path.

I can see that the connections are symmetric for geodesics (force-free) motion. But are the generators of the translation group identified with the connections ? The connections are derivatives of the gauge field whose generators commute. I'm busking this, maybe you can set me straight.

Lut

[sweetser]

Hello Lut:

On my latest YouTube talk, I got asked a good question: what would the Maxwell and GEM derivations look like if I included all the connections? I know that because the EM field strength tensor uses an exterior derivative, I could just they that is why they drop out. I wish to see these drop in detail. I will make the same presumptions as happen in GR: the connection is metric compatible and torsion-free, so I can use Christoffel symbols of the second kind. These are not tensors and have 3 indexes. What I hope to learn is how exactly to play with Christoffel symbols in EM, so that they all happen to drop out. Seeing this work for the B field gives us a good place to start:

\nabla \times A = \frac{\partial A^u}{\partial x^v} - \Gamma_{\mu}^{v u} A^{\mu} - \frac{\partial A^v}{\partial x^u} + \Gamma_{\mu}^{u v} A^{\mu}

= \frac{\partial A^u}{\partial x^v} - \frac{\partial A^v}{\partial x^u}\quad eq 1

Because the Christoffel is symmetric in the indexes, the two gammas cancel. I figured a similar thing should happen with the E field, but I must be missing a sign:

E = -\frac{\partial A^u}{\partial t} + \Gamma_{\mu}^{0 u} A^{\mu} -\frac{\partial \phi}{\partial x^u} + \Gamma_{\mu}^{u 0} A^{\mu}\quad eq 2

This time the gammas add. Oops. Where did I misstep?

Thanks,
Doug

[Mentz114]

Hi Doug:
I'll try and repeat the calculations above later. The covariant derivatives cancel out in the field tensor.

F_{\mu\nu} = \nabla_{\mu}A_{\nu} - \nabla_{\nu}A_{\mu}

where

\nabla_{\mu}A_{\nu} = \partial_{\mu}A_{\nu} - \Gamma^{k}_{\mu\nu}A_k

and because of the symmetry in the lower indexes the Gamma factors cancel.

As I've pointed out many times, this does not happen with the symmetric field tensor, so you end up with terms containing A^2 in the lagrangian, as well as functions of the metric.

I just don't see how you can mix up a metric theory and a vector potential. Your field equations are four in number and your unknowns are 10 for the metric ( 40 connections) and 4 for the potential. How do you solve that ?

Lut

[sweetser]

Hello Lut:

It looks like my mistake in Eq. 2 of post #529 was on the first gamma, it should have been negative. For the E field, I need:

E = \nabla_{0}A_{u} - \nabla_{u}A_{0} = -\frac{\partial A_u}{\partial t} - \Gamma^{k}_{0 u}A_k - (\frac{\partial \phi}{\partial x^u} - \Gamma^{k}_{u 0}A_k)


So the minus in the first term happens from the lower script for the Au which does not change the sign of the gamma. The minus on the second term does flip the switch on the second gamma, making it positive. Not difficult, but I need to get rock solid on how signs work in standard EM first, then out to GEM. Gammas are always negative unless one tosses a minus sign in front.

Doug

[Mentz114]

Hi Doug:

with non-zero Christoffel symbols, raising and lowering indices is a nightmare -

A_{\mu} = g_{\mu\nu}A^{\nu} = A_{0}g^{\mu 0}+A_{1}g^{\mu 1}+A_{2}g^{\mu 2}+A_{3}g^{\mu 3}

and this

\Gamma^{ bc}_{a} = g_{ka}g^{bm}g^{cn}\Gamma^{k}_{ mn}

has 64 terms.

Note that most of the GEM calculations I did are with the Minkowski metric.

Lut

[sweetser]

Hello:

I am writing code to handle the Even representation of quaternions on the command line so I can make animations using them. In this context, quaternions are always 4 numbers that get piped from one program to another. It is not possible to know if a quaternion being fed into the next program in a chain of these programs was an Even or Hamilton quaternion.

When I got to the nuts and bolts of the programming, I noticed that this issue only matters for binary operations, and then only for that part of the binary operation that has the 2 3-vectors forming a product. In the Hamilton representation, the dot product gets a minus sign, the even dot product stays positive. In the Hamilton representation, the curl is the curl with half the terms positive, half minus. In the Even representation, all the same terms are all positive.

A few issues are missed by using tensors. One is the issue of knowing there is an inverse for both the Hamilton and Even representations of quaternions. I used to justify this on an issue of quantum field theory which is kind of obscure. A more apparent reason might have to do with the observation of the critical importance of group theory in physics. The definition of a group relies on there being both an identity and an inverse for every member of the group. Those are both there for these two representations of quaternions. Tensors don't have to have inverses.

Doug

[sweetser]

Hello:

I am preparing a talk for an APS meeting in St. Louis. One of the points I intend to make is that Feynman's analysis of the phase of the current coupling term is incomplete. The image really rocks, but I am unsure how to upload images here, so I'll have to get by with LaTeX.

The current coupling term is also known as the source term: -J^{\mu} A_{\mu}. Take the Fourier transform of the potential to get J'/k2. Since the indices are the same, this term evaluates to the following scalar:

-J^{\mu} J'_{\mu}/k^2 = (-\rho \rho' ~+~ Ax Ax' ~+~ Ay Ay'~+~ Az Az')/k^2 \quad eq~1

What Feynman does is consider a specific case of a current moving along the z axis. While it makes the algebra simpler, it made my slide longer, and not as general. Skip all the z-axis specific stuff, and just multiply these two together using the Hamilton representation of quaternions:

-J J'/k^2 = (-\rho \rho' ~+~ Ax Ax' ~+~ Ay Ay'~+~ Az Az',

-\rho Ax' ~-~ Ax \rho' ~+~ Ay Az'~-~ Az Ay',
-\rho Ay' ~-~ Ay \rho' ~+~ Az Ax'~-~ Ax Az',
-\rho Az' ~-~ Az \rho' ~+~ Ax Ay'~-~ Ay Ax')/k^2 \quad eq~2

The phase is the final 3 lines of eq 2. What Feynman calculated was part of the third line, Ax Ay'~-~ Ay Ax'. These two terms do not add together, so it will take 2 pi to get back to the starting position, the sign of spin 1 symmetry. That is excellent, since these terms are the transverse modes of EM that need spin 1 symmetry so like charges repel.

Now look at the first two terms of the third line, [tex]-\rho Az' ~-~ Az \rho'[/itex]. These do work together, and will require only pi rotations to get back to the starting position. This is a property of spin 2 symmetry. Spin 2 particles are needed for systems where like charges attract. A complete analysis of the phase looks good to me.

Doug

[Lawrence B. Crowell]

I can see that the connections are symmetric for geodesics (force-free) motion. But are the generators of the translation group identified with the connections ? The connections are derivatives of the gauge field whose generators commute. I'm busking this, maybe you can set me straight.

Lut

I guess I missed this one. Non-geodesic motion involves some other force on the motion of a particle. Teleparallel connections have a number of meaning as I understand. One is that the geodesic motion is given by a horizontal bundle and the "force" by a vertical bundle. This is Finsler geometry. Another definition, related maybe, is where the motion is parallel or Euclidean-like (latitudinal lines on a globe) and where the force or vertical bundle terms act to sustain this flow. I think this can result in torsional gravity.

Lawrence B. Crowell

[CarlB]

New article out that perhaps reads on this subject:

A Spatially-VSL Gravity Model with 1-PN Limit of GRT
Jan Broekaert
In the static field configuration, a spatially-Variable Speed of Light (VSL) scalar gravity model with Lorentz-Poincaré interpretation was shown to reproduce the phenomenology implied by the Schwarzschild metric. In the present development, we effectively cover configurations with source kinematics due to an induced sweep velocity field w. The scalar-vector model now provides a Hamiltonian description for particles and photons in full accordance with the first Post-Newtonian (1-PN) approximation of General Relativity Theory (GRT). This result requires the validity of Poincaré’s Principle of Relativity, i.e. the unobservability of ‘preferred’ frame movement. Poincaré’s principle fixes the amplitude of the sweep velocity field of the moving source, or equivalently the ‘vector potential’ ξ of GRT (e.g.; S. Weinberg, Gravitation and cosmology, [1972]), and provides the correct 1-PN limit of GRT. The implementation of this principle requires acceleration transformations derived from gravitationally modified Lorentz transformations. A comparison with the acceleration transformation in GRT is done. The present scope of the model is limited to weak-field gravitation without retardation and with gravitating test particles. In onclusion the model’s merits in terms of a simpler space, time and gravitation ontology—in terms of a Lorentz-Poincaré-type interpretation—are explained (e.g. for ‘frame dragging’, ‘harmonic coordinate condition’).
http://www.springerlink.com/content/b3758t680gj383j8/

Back on the first of the year, Foundations of Physics (the above journal) picked up a new editor, Gerardus 't Hooft, who writes:

During my first couple of months in this office, it became clear that fundamental questions in physics and philosophy also attract the interest of many laymen physicists.

We receive numerous submissions from people who venture to attack the most basic premises of theories such as Special Relativity, but instead only succeed in displaying a lack of professional insight in how a physical theory is constructed. I suspect that some of these people may have been working somewhere in an attic, deprived from daylight for decades, determined only to reemerge with a Theory of Everything in their hands. Even though they may be very sincere, we have to disappoint such authors. New insights are gained only by intense interactions with professionals all over the globe, and by solidly familiarizing oneself with their findings, and we must make a selection from only those papers whose authors have a solid understanding of the topics they are discussing.

Fortunately they also submit their work, and their clever inventiveness continues to surprise us.
...
I hope to receive your submissions. Acceptation of a paper may not necessarily mean that all referees agree with everything, but rather that the issues put forward by the author were considered to be of sufficient interest to our readership, and the exposition was clear enough that our readers, whom we assume to be competent enough, can judge for themselves.
...
http://www.springer.com/physics/journal/10701

[sweetser]

Hello Carl:

The paper cost $32 to download, ouch. I know I have a bias against spatially-Variable Speed of Light theories. On my dinning room wall is a large artwork, The Speed of Light According to Rene Magritte (http://theworld.com/~sweetser/PopScience/speed/speed.html). I think about c as time's relationship to space. Photons travel at the speed of light because time is space. It is hard to see how that statement can vary.

Any reasonable proposal for gravity must agree to the first parameterized Post-Newtonian accuracy, as these folks do. I am a bit surprised they didn't look at the second level of accuracy.

The "scalar-vector model" might have passing similarities to GEM, but I am skeptical.

The attitude of the editor parallels some of the motivations of the Independent Research section of Physics Forums. 't Hooft has a site, How to become a good theoretical physicist (http://www.phys.uu.nl/~thooft/theorist.html), I always recommend to people working alone in the attic.

Should I ever get satisfied with my draft paper, I would consider sending it there. A number of issues need to be addressed first. I need to shift the action from tensor notation to quaternions because the issue of gauge symmetry must be clearly addressed. Second, I still have to get the energy expression in order.

I am emailing from the April APS Meeting in Saint Louis. I heard a talk by someone who is trying to take EM up a level to use the tools similar to GR, basically the opposite direction I am traveling. That talk gave me a good overview of the various approaches to EM as a geometry phenomena. I asked the speaker to send me his slides.

The talks about dark matter and dark energy make me sad. They certainly don't sound like solid physics. In a question and answer section, I confirmed with the speaker that she knew of no large scale systems where Newtonian gravity produces the observed results. That is what I expect since I think an unusual effect of gravity, labeled the relativistic rocket effect, Vc dm/dR Vhat, is never put into their numerical integration systems. I will recall these talks when trying to approach my fears of doing the calculations (they look hard, and I know my limitations too well).

Doug

[Mentz114]

Doug:
Jan Broekaert has several papers in the arXiv. Do a search in GR-QC for his name in authors. I'm having a look at one later, for diversion. Wiil report if it's interesting.

Carl:
I enjoyed the quote from t'Hooft.

Lut

[edit : this appears to be the very paper. Save $32 now !]

arXiv:gr-qc/0405015

[Lawrence B. Crowell]

New article out that perhaps reads on this subject:

A Spatially-VSL Gravity Model with 1-PN Limit of GRT
Jan Broekaert
In the static field configuration, a spatially-Variable Speed of Light (VSL) scalar gravity model with Lorentz-Poincaré interpretation was shown to reproduce the phenomenology implied by the Schwarzschild metric.

These types of theories are curious, but as Doug pointed out c is just an invariant measure by which space and time are equivalent. The value of c in its particular comes from other units in nature associated with electromagnetism, e, \epsilon_0,~\mu and so forth. If suspect that a variability in c or some energy dependence might come about near the Planck scale, but then again the Planck scale is defined according to c.

Lawrence B. Crowell

[CarlB]

These types of theories are curious, but as Doug pointed out c is just an invariant measure by which space and time are equivalent.

This is true only under the assumption that there is no preferred coordinate system. In that sense, the equivalence is logically circular. Any theory of gravity that is built on a flat underlying space has to have a variable speed of light. And in such a theory, the speed of light is well defined and variable.

My favorite such theory is the gauge theory of gravity by the Cambridge geometry group. The math it uses is based on the Clifford algebra of Dirac's gamma matrices (which they call "geometric algebra" following David Hestenes). Their theory is identical to GR (to all orders) outside of the event horizons but is built on a flat (well, Minkowski) underlying space or coordinates. I started a website supporting their theory here that includes links to the papers:
http://www.gaugegravity.com/
I'm planning on redoing the website to make it more informational soon. There are a lot of articles that use these ideas to model electrons and black holes, and I need to link those in.

When you have a flat gravity theory (and from my point of view, Sweetser's theory is flat), the first thing to do is to write down what the theory does for a singular mass, a black hole. This defines a preferred coordinate system, and that system defines the speed of light in the theory.

Since gauge gravity is identical to GR, the gravity you get from it is equivalent to GR written in a particular choice of coordinates. The stationary black hole becomes Painleve coordinates for the Schwarzschild solution. I created a java applet simulation that shows the relation of these to the usual Schwarzschild coordinates here:
http://www.gaugegravity.com/

Now the original name of the applet was "sweet" because I wanted to include Sweetser's gravity as well as Newton, and Einstein's in Schwarzschild and Painleve coordinates. However, I haven't figured out how to actually write down his equations, in the form of the acceleration felt by a test body, and so I haven't modeled it yet. I quit asking, but if he comes up with an equation, I'll type it up into java and add it to the simulation (hint hint).

[lba7]

(translation from french observation)

I think a new theory for source gravity. Maybe the origin of gravity is an oscillation of electrostatic wave. The oscillation of each particle is very very small compared an electron charge. Each particle send a wave, and each wave is accumulate to another wave. The charge of a particle is 0 but the wave is something around 0, one moment + and later - etc. So, each particle send + and - wave around it. It's possible to synchronize all wave because there is a difference between forces if you imagine two sinusoide waves face to face, there is two waves very close and two waves far, the force is with 1/d² so the difference can create a synchronism. This can explain deviation of light and redshift. The formula of perihelion precession of Mercury is the same with electrostatic forces integrate in the formula.

I have create a site for explain (in french for the moment) my ideas: 3w eisog dot com

Maybe this idea is not new, tell me ...

Ludovic

[Lawrence B. Crowell]

The breaking of Lorentz symmetry has been proposed by some. I am not sure how I stand on this for certain --- leaning against the idea I suppose. One can work with all sort of pseudotensors and the like with this, though I am not sure if that is what Sweetser is doing.

Gauge gravity, or in a general form with the B-F or Plebanski Lagrangian, recovers Einstein's gravitation at least for weak \alpha^2~\sim~G, though corrections on this for stronger G and higher energy (also I suspect cosmological scales) probably lead to consequences beyond a naive breaking of Lorentz symmetry.

Lawrence B. Crowell

[sweetser]

Hello Lawrence:

The GEM proposal does not appear to break Lorentz symmetry.

This issue of tensors and GEM remains a concern. We know that \nabla_{\mu} A_{\nu} transforms like a tensor because that is why the machinery of a covariant derivative was built. We also know that the field strength tensor of EM transforms like a tensor, \nabla_{\mu} A_{\nu} ~-~ \nabla_{\nu} A_{\mu}. Correct me if I am wrong, but I believe that the difference of two tensors of the same order is also a tensor. If that is the case, then:

\nabla_{\mu} A_{\nu} - \frac{1}{2}(\nabla_{\mu} A_{\nu} ~-~ \nabla_{\nu} A_{\mu})

= \frac{1}{2}(\nabla_{\mu} A_{\nu} ~+~ \nabla_{\nu} A_{\mu})\quad eq ~1

Is it fair to call this object with the plus sign a tensor? Sure would hope so. Yes all three have different properties, but they all look like they belong in the family of tensors, not pseudotensors.

Doug

[CarlB]

Gauge gravity, or in a general form with the B-F or Plebanski Lagrangian, recovers Einstein's gravitation at least for weak \alpha^2~\sim~G, though corrections on this for stronger G and higher energy (also I suspect cosmological scales) probably lead to consequences beyond a naive breaking of Lorentz symmetry.

This is a different gauge gravity than that of the Cambridge geometry group. I guess "gauge" is a cool word to use to describe a theory. The Cambridge version is identical to GR to all orders, at least locally. Where it differs is in that you can't do weird topological things like wormholes. Here's how David Hestenes puts it:

The specialization of GA to Minkowski spacetime is called Spacetime Algebra (STA). As explained below, STA clarifies the geometric significance of the Dirac Algebra and thereby extends its range of application to the whole of physics. In particular, STA provides the essential mathematical framework for the new Gauge Theory Gravity (GTG).

The foundations of GTG are fully expounded in a seminal paper, so we can concentrate
on highlights of its unique features. GTG is a gauge theory on Minkowski spacetime, but [I]locally it is equivalent to General Relativity (GR), so it can be regarded as an alternative formulation of GR.8 However, GTG reformulates (or one might say, replaces) Einstein’s vague principles of equivalence and general relativity with sharp gauge principles that have clear physical consequences (Section IV). These gauge principles are more than mere rephrasing of Einstein’s ideas. They lead to intrinsic mathematical methods that simplify modeling and calculation in GR and clarify physical meaning of terms at every stage. In particular, they provide clean separation between gauge transformations and coordinate transformations, thus resolving a point of longstanding confusion in GR. Moreover, GTG simplifies and clarifies the analysis of singularities, for example, in assignment of time direction to a black hole horizon. Finally, since tensors and spinors are fully integrated in STA, GTG unifies classical GR and relativistic quantum mechanics with a common system of gauge principles. Besides facilitating the application of quantum mechanics to astrophysics, this opens up new possibilities for a grand unification of gravitation and electroweak theories, as explained in Section V.
http://modelingnts.la.asu.edu/pdf/procGTG-RQM.pdf

So far, GR has only been verified to 1-PN. I don't think that it will make 3-PN. What I like about Sweetser's gravity is that it seems natural to do as the consequences of a particle theory. The problem I see in doing this is that to end up with a variable speed of light type gravity theory requires that the force of gravity be carried by superluminal particles.

In other words, just as Maxwell's equations end up being a force mediated by a speed of light particle, a theory which produces a variable speed of light needs to be mediated by something that moves at some higher speed. Comments from Doug?

[sweetser]

Hello Carl:

Your email inspired me to go make another effort to wrap my head around geometric algebra. David Hestenes will be giving a keynote address at the conference in Brazil. Both Clifford algebras and geometric algebra fans view there tools as more general that quaternions, since quaternions are just Cl(0, 2) or a 0,2-multivector.

There is not much difference between these three approaches, but there is some. I think of the basis vectors differently, due to looking at this with my physics glasses on. A quaternion is described as a 0,2 multivector, which means a scalar for the first part, and three bivectors, e2/\e3, e3/\e1, and e2/\e1 for the 3-vector.

There are two issues the physicists in me objects. First, the main message of special relativity is to bring the scalar in with the 3-vector on equal footing. I look at quaternons at being 4 scalars with 4 basis vectors: (a0 e0, a1 e1, a2 e2, a3 e3). The GA folks like the wedge because it explains the minus signs. I find that claim hollow since is merely shift where the minus comes in to the wedge itself. In my view of quaternions, the four parts look like equals.

The second objection comes from my view of gravity. In GEM, gravity can be about the potential, the basis vectors, or a combination of both. When it is about the basis vectors, then I want to be able to shift e0, e1, e2, and e3. For the exponential metric, e0 = exp(-2 GM/c^2 R) and e0 e1 = e0 e2 = e0 e3 = 1. I don't see how I could make similar statements with the GA formalism.

In the cited paper, Hestene explains how to do the Dirac algebra using geometric algebra. The details are different from how I do that with quaternions as triple products (all sixteen possibilities for e_u Q e_v make up the action of the gamma matrices).

My guess is the GA crowd will not be happy with the even representation of quaternions. I will get to find out, since that will lead off my 25 minute talk in Brazil.

The Cambridge geometry group looks like it is trying to recreate general relativity, with some more polite features. The GEM work is in line with a long tradition of directly confronting the dominant idea of the day, saying it is good up to 1-PN, but not 2-PN. One of those binary pulsar guys was at the APS meeting. I asked him if it is correct to take the equations developed for super-weak perihelion precession and apply it to a strong, dynamic system. If he writes back, I'll commmunicate to this thread.

I stay away from "faster than light". Using quaternions for everything is radical enough!

Doug

[Lawrence B. Crowell]

This is a different gauge gravity than that of the Cambridge geometry group. I guess "gauge" is a cool word to use to describe a theory. The Cambridge version is identical to GR to all orders, at least locally. Where it differs is in that you can't do weird topological things like wormholes. Here's how David Hestenes puts it:

The specialization of GA [i.e. Geometric Algebra] to Minkowski spacetime is called Spacetime Algebra (STA). As explained below, STA clarifies the geometric significance of the Dirac Algebra and thereby extends its range of application to the whole of physics. In particular, STA provides the essential mathematical framework for the new Gauge Theory Gravity (GTG).



Spinor algebra can reproduce general relativity in a pretty straightforward manner. The Dirac matrices \gamma_\mu define the metric as


\gamma_\mu\gamma_\nu~=~\frac{1}{4}g_{\mu\nu}


and so if the representation of the spinor basis is local or chart dependent one can pretty easily reproduce GR with the Dirac operator. A gauge invariant form of the Dirac operator would then be


\partial_\mu(\gamma^\mu\psi)~=~\gamma^\mu D_\mu\psi~=~\gamma^\mu(\partial_\mu~+~A_\mu)\psi,


for A_\mu a gauge term for GR.

For higher Clifford basis elements or with vierbiens extended GR theories can be derived. I suspect these are related in some ways to the string theory for GR which obtains for scales larger than the string length, but has deviations for scales approaching the string length. Of course there are some conformal invariance abuses with the string theoretic approach to GR as a bimetric form of theory. Yet string theory as a form of math-method does have some suggestive elements to it.

It is my sense that GR will survive to very high PN orders. I think deviations from GR obtain on two complementary scales: one scale near the Planck scale or say \le~10^3\sqrt{G\hbar/c^3}, and the other is on the cosmological scale where time translation invariance is "deformed" and there are inequivalent vacua states on scales across the cosmological event horizon. The verdict on this will likely come from gravity wave detection, such as LIGO, and the connection between these measurements with astrophysical events.

Lawrence B. Crowell

[CarlB]

Doug, I'm swamped right now with correspondence and I don't have internet connectivity at home so let me postpone writing more about what you've said here until later. But here's a link on another exponential gravitation theory. Can you talk about how this differs from your own?

[Uh, read down to the end of the article]
http://www.insidegnss.com/node/451

Carl

[Lawrence B. Crowell]

(translation from french observation)

This can explain deviation of light and redshift. The formula of perihelion precession of Mercury is the same with electrostatic forces integrate in the formula.

I have create a site for explain (in french for the moment) my ideas: 3w eisog dot com

Maybe this idea is not new, tell me ...

Ludovic

What is possible is that the spin quantum Hall effect might result in deviations from the equivalence principle. A circularly polarized photon as the superposition of two polarization states might be split by the effective index of refraction due to a gravity field. This would mean that gravitation has a Berry phase or a topological index associated with the spin of particles. This would be a possible deviation from classical general relativity.

An Einstein lens might provide the way to detect this. A radio telescope that detects photons from a lensed source will for a very narrow band pass filter measure quantum effect for entangled photons in this cosmic beam splitter. In principle the Wheeler Delayed Choice experiment could be performed this way. Also there might be for the two arms of this cosmic beam splitter a spin dependency in how the photons are split.

Lawrence B. Crowell

[sweetser]

Hello:

I can finally write about the New England American Physical Society Meeting on Saturday, April 5. Lut can skip this post since it involves the personal stories. The three posts that follow in quick succession will be technical calculations.

My hopes were not high for this local meeting at the Coast Guard Academy. It was held at a military facility, with an official entry gate. People were marching around all dressed to the military nines at 8 am. One positive aspect of the military is they get to wear sharp hats to work, hats that indicate one's station.

The usual cast of characters where at this APS meeting. There were a dozen people in the room, half of whom I was familiar with. The session started off with a familiar fringe guy. As far as I can tell, he claims that everything is in motion. As a skeptic, I attempt to see what notions I need to stretch to see some grain of truth in his near incomprehensible riffs. He claimed that everything was in motion. What I would say is everything moves in spacetime, mostly as time, as things continue to persist. He does make fun of himself a little which helps, otherwise his talks are tragic.

Larry Gold gave his "Let's doubt global warming". It reminds me of Fox News in the US which claims to be unbiased, then makes every effort to selectively sample data needed to support its position.

My talks was titled: "Using Quaternions for Lagrangians in EM and GEM Unified Field Theory." For an audience this tiny and quirky, I could not expect them to know about Lagrangians, or how to use them to derive field equations. I had thought about doing all the equations in LaTeX, but decided to use a fountain pen, good old black ink written by hand, to indicated that despite the complications, this sort of calculation can be done by a person. These sheets of paper packed full of partial differential equations are prized possessions. It looks so hard core. Two of these sheets - a derivation of the Maxwell equations from the Lagrangian, and a derivation of the gravity part of GEM - were at the core of the talk (the rest being a setup for the heavy duty math).

The talk had one technical glitch. My connecting wire is flaky so everything was yellow. Otherwise, the speech went according to plan. My talks often do not elicit questions, but this time someone started asking about the symmetric curl. It was like he was channeling Lawrence. He was wondering why I had all these coordinate dependent equations. After the session broke up, I asked him some more questions, trying to see what he found unsettling.

It certainly is my intension to write coordinate-independent equations that are valid in flat or curved spacetime. When I write a quaternion as "q", it could be in any coordinate system. When I write \nabla q, that is a covariant derivative, the same whether spacetime is flat or curved, independent of coordinates.

When I went to write out the derivative by its component parts, I invariably wrote things in terms of Cartesian coordinates, the rectilinear t, x, y, and z. The progression of symbols for derivatives goes something like this:

\frac{d}{dt} \rightarrow \frac{\partial}{\partial t} \rightarrow \nabla_0 \quad eq~1

Only the last symbol is independent of coordinates, valid in flat or curved spacetime. I had never seen the Maxwell equations derived in a coordinate-free manifestly covariant way, so it was natural for me to just use partial derivatives with respect to t, x, y, and z.

Based on this criticism, I will be changing how I write component expressions. I will only use subscripts 0-3, not t-z. This way, equations that I think of as being manifestly covariant will have the notation required to be manifestly covariant.

For the record, I will repeat the quaternion operator derivations for the Maxwell equations, the gravity part of GEM, and the GEM unified field equations using the covariant notation in the next three posts.

Doug

Stills:
http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations

The talk:
http://www.youtube.com/watch?v=Kd4nNb7nGOc

[sweetser]

Hello:

In this post I will derive the Maxwell field equations using quaternion operators in a manifestly covariant notation.

Notice that the quaternion differential operator acting on a 4-potential creates a scalar gauge field, the -E and B fields:

\nabla A = (\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~+~ \nabla \times A) = (g, -E ~+~ B)\quad eq ~1

The starting point for the derivation of the Maxwell equations is the Lagrangian which can be viewed as the difference between the scalars of B squared and E squared. This can be achieved by changing the order of the covariant differential operator with respect to the 4-potential, which flips the sign of the curl (B), but not the time derivative of A or gradient of phi which make up E. The scalar gauge field can be subtracted away:

-\frac{1}{8}(\nabla A ~-~ (\nabla A)^*)(A \nabla ~-~ (A \nabla)^*)

=\frac{1}{8}(0, -\nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \times A)(0, \nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \times A)

=\frac{1}{2}(0, E ~-~ B)(0, -E ~-~ B) = \frac{1}{2}(E^2 ~-~ B^2, -2 E \times B)\quad eq ~2

Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms \nabla_0 \phi, \nabla \cdot A because they have been explicitly subtracted away at this early stage. It is also of interest to know the 3-vector part of this expression is the Poynting vector, a conserved quantity in EM.

Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling, -\frac{1}{2 c}(J A + (J A)^*):


\mathcal{L}_{EB} = \frac{1}{2}(-(\nabla_1 \phi)^2 ~-~(\nabla_2 \phi)^2 ~-~(\nabla_3 \phi)^2 ~-~ (\nabla_0 A_1)^2 ~-~ (\nabla_0 A_2)^2 ~-~ (\nabla_0 A_3)^2
~+~ (\nabla_3 A_2)^2 ~+~ (\nabla_2 A_3)^2 ~+~ (\nabla_1 A_3)^2 ~+~ (\nabla_3 A_1)^2 ~+~ (\nabla_2 A_1)^2 ~+~ (\nabla_1 A_2)^2)
~-~ (\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)
-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3

Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable:

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} \phi)}) = -\nabla_1^2 \phi ~-~ \nabla_2^2 \phi ~-~ \nabla_3^2 \phi ~-~ \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho
= \nabla \cdot E - \rho = 0 \quad eq ~4

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_1)}) = -\nabla_0^2 A_1 ~+~ \nabla_3^2 A_1 ~+~ \nabla_2^2 A_1 ~-~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1
= \nabla_0 E_1 - (\nabla \times B)_1 + J_1 = 0 \quad eq ~5

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_2)}) = -\nabla_0^2 A_2 ~+~ \nabla_3^2 A_2 ~+~ \nabla_1^2 A_2 ~-~ \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2
= \nabla_0 E_2 - (\nabla \times B)_2 + J_2 = 0 \quad eq ~6

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_3)}) = -\nabla_0^2 A_3 ~+~ \nabla_2^2 A_3 ~+~ \nabla_1^2 A_3 ~-~ \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3
= \nabla_0 E_3 - (\nabla \times B)_3 + J_3 = 0 \quad eq ~7

This work can be summarized with the Maxwell source equations:

\nabla \cdot E = \rho \quad eq 8

\nabla \times B ~-~ \nabla_0 E = J

This has been a manifestly covariant derivation of the Maxwell equations. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates. Once you get used to keeping track of all these terms, it is kind of fun, honest!

Doug

Stills:
http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations

The talk:
http://www.youtube.com/watch?v=Kd4nNb7nGOc

[sweetser]

Hello:

In this post I will derive the Maxwell field equations using quaternion operators in a manifestly covariant notation.

Notice that the quaternion differential operator acting on a 4-potential creates a scalar gauge field, the -E and B fields:

\nabla A = (\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~+~ \nabla \times A) = (g, -E ~+~ B)\quad eq ~1

The starting point for the derivation of the Maxwell equations is the Lagrangian which can be viewed as the difference between the scalars of B squared and E squared. This can be achieved by changing the order of the covariant differential operator with respect to the 4-potential, which flips the sign of the curl (B), but not the time derivative of A or gradient of phi which make up E. The scalar gauge field can be subtracted away:

-\frac{1}{8}(\nabla A ~-~ (\nabla A)^*)(A \nabla ~-~ (A \nabla)^*)

=\frac{1}{8}(0, -\nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \times A)(0, \nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \times A)

=\frac{1}{2}(0, E ~-~ B)(0, -E ~-~ B) = \frac{1}{2}(E^2 ~-~ B^2, -2 E \times B)\quad eq ~2

Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms \nabla_0 \phi, \nabla \cdot A because they have been explicitly subtracted away at this early stage. It is also of interest to know the 3-vector part of this expression is the Poynting vector, a conserved quantity in EM.

Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling, -\frac{1}{2 c}(J A + (J A)^*):

\mathcal{L}_{EB} = \frac{1}{2}(-(\nabla_1 \phi)^2 ~-~(\nabla_2 \phi)^2 ~-~(\nabla_3 \phi)^2 ~-~ (\nabla_0 A_1)^2 ~-~ (\nabla_0 A_2)^2 ~-~ (\nabla_0 A_3)^2
~+~ (\nabla_3 A_2)^2 ~+~ (\nabla_2 A_3)^2 ~+~ (\nabla_1 A_3)^2 ~+~ (\nabla_3 A_1)^2 ~+~ (\nabla_2 A_1)^2 ~+~ (\nabla_1 A_2)^2)
~-~ (\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)
-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3

Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable:

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} \phi)}) = -\nabla_1^2 \phi ~-~ \nabla_2^2 \phi ~-~ \nabla_3^2 \phi ~-~ \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho
= \nabla \cdot E - \rho = 0 \quad eq ~4

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_1)}) = -\nabla_0^2 A_1 ~+~ \nabla_3^2 A_1 ~+~ \nabla_2^2 A_1 ~-~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1
= \nabla_0 E_1 - (\nabla \times B)_1 + J_1 = 0 \quad eq ~5

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_2)}) = -\nabla_0^2 A_2 ~+~ \nabla_3^2 A_2 ~+~ \nabla_1^2 A_2 ~-~ \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2
= \nabla_0 E_2 - (\nabla \times B)_2 + J_2 = 0 \quad eq ~6

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EB}}{\partial (\nabla_{\mu} A_3)}) = -\nabla_0^2 A_3 ~+~ \nabla_2^2 A_3 ~+~ \nabla_1^2 A_3 ~-~ \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3
= \nabla_0 E_3 - (\nabla \times B)_3 + J_3 = 0 \quad eq ~7

This work can be summarized with the Maxwell source equations:

\nabla \cdot E = \rho \quad eq ~8

\nabla \times B ~-~ \nabla_0 E = J \quad eq ~9

This has been a manifestly covariant derivation of the Maxwell equations. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates. Once you get used to keeping track of all these terms, it is kind of fun, honest!

Doug

Stills:
http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations

The talk:
http://www.youtube.com/watch?v=Kd4nNb7nGOc

[sweetser]

Hello:

<Preamble>
This post is very similar to the last two. This time I have few of the set of the last two, but unify gravity and EM in the process. Nice.
</Preamble>

In this post I will derive gravity and the Maxweill field equations using quaternion operators in a manifestly covariant notation.

The Hamilton representation will be used for the Maxwell field equations. The even representation of quaternions - where all signs are the same, and the Eivenvalues ore excluded so the quaternion can be inverted - will be used in the first phase of this derivation. What the even representation does is flip the relative signs of the E and B signs, so now the e filed has two signs, and the b field has one, like so:

E = -\nabla_0 A ~-~ \nabla_u \phi
e = \nabla_0 A2 ~-~ \nabla_u \phi

B_w = \nabla_u A_v ~-~\nabla_v A_u = \nabla \times A
b_w = -\nabla_u A2_v ~-~\nabla_v A2_u = \nabla \Join A2 \quad eq ~0

The fifth play is the term that makes up the gauge field:

g = \nabla_0 \phi ~-~ \nabla \cdot A

None of these transform like tensors, but together they do from \nabla A.

Let's generate all 5 fields:

\nabla A = (\nabla_0 \phi ~-~ \nabla \cdot A2, \nabla_0 A2 ~+~ \nabla_u \phi ~+~ \nabla \times A) = (g, -E ~+~ B)
\nabla^* A2 = (\nabla_0 \phi ~-~ \nabla \cdot A2, -\nabla_0 A2 ~+~ \nabla_u \phi ~-~ \nabla \Join A2) = (g, -e ~+~ b)\quad eq ~1

The starting point for the derivation of the unified GEM equations is the Lagrangian which can be viewed as the difference between the scalars of E,b squared and B,e squared. This can be achieved by using the first set of quaternion operators used in the previous posts:

\frac{1}{8}((\nabla A)(A \nabla) ~-~(\nabla^* A2)(\nabla A2^*))

=\frac{1}{8}(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~+~ \nabla \times A)(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \times A)
-(\nabla_0 \phi ~-~ \nabla \cdot A, \nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \Join A)(\nabla_0 \phi ~-~ \nabla \cdot A, -\nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \Join A)

=\frac{1}{2}(g, -E ~+~ B)(g, -E ~-~ B) ~-~ (g, e ~+~ b)(g, -e ~+~ b) = \frac{1}{2}(B^2 ~-~ E^2 ~-~ b^2 ~+~ e^2, 2 E \times B ~-~ b \Join b ~+~ e \Join e)\quad eq ~2

Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms \nabla_0 \phi, \nabla \cdot A because they have miraculously been cancelled!

Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling whose phase has spin 1 and spin 2 symmetry[/itex]:

\mathcal{L}_{EBeb} = -~(\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)
-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3

[sidebar: The assertion that the current coupling term has both spin 1 and spin 2 symmetry for all terms in the phase requires a small calculation. I decided to leave that as a problem. My guess is that until someone starts confirming these by hand, I'll be the only one to see that even these cool details work out.[/sidebar]

Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable:

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} \phi)}) = - \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho
= -\frac{1}{2}\nabla \cdot (E ~-~ e) - \rho = 0 \quad eq ~4

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_1)}) = -~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1
= \frac{1}{2}(\nabla_0 (E_1 ~+~ e_1) - (\nabla \times B)_1 - (\nabla \Join b)_1) + J_1 = 0 \quad eq ~5

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_2)}) = - \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2
= \frac{1}{2}(\nabla_0 (E_2 ~+~ e_2) - (\nabla \times B)_2 - (\nabla \Join b)_2) + J_2 = 0 \quad eq ~6

\nabla_{\mu}(\frac{\partial \mathcal{L}_{EBeb}}{\partial (\nabla_{\mu} A_3)}) = - \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3
= \frac{1}{2}(\nabla_0 (E_3 ~+~ e_3) - (\nabla \times B)_3 - (\nabla \Join b)_3) + J_3 = 0 \quad eq ~7

This work can be summarized with the GEM gravity source equations:

\frac{1}{2} \nabla \cdot (E ~-~ e) = \rho \quad eq ~8

\frac{1}{2}(\nabla \times B ~+~ \nabla \Join b - \nabla_0 (E ~+~ e)) = J \quad eq ~9

This has been a manifestly covariant derivation. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates.

I have spent the day trying get all the signs and factors right. If there are any questions, send me a private note, and I will recheck it. If needed, we can put up a new post.

Doug

Stills:
http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations

The talk:
http://www.youtube.com/watch?v=Kd4nNb7nGOc
(this derivation was not part of the talk)

[sweetser]

Hello:

<Preamble>Looks like two of my posts were identical, sorry. The software here doesn't appear to like this much LaTeX.

This post is very similar to the last. I am starting out the draft by cutting and pasting the text, as all the same terms are the same. used in the same locations of the same equations. What changes are a few signs. The E field is made of two parts, -\nabla A_0 and -\nabla \phi. These happen to have the same sign. In this post, the case where these have opposite signs will be explored. A similar thing will be done for the magnetic field, where the two terms \nabla_u A_v and \nabla_v A_u have the same sign.

Those readers concerned about how these objects transform can rest easy. We know that \nabla A transforms like a rank 2 tensor - it was the justification behind developing the covariant tensor \nabla[/tex] in the first place. We also know that the EM field strength tensor, [itex]\frac{1}{2}(\nabla_u A_v - \nabla_v A_u), transforms like a tensor. The difference between these two tensors which is \frac{1}{2}(\nabla_u A_v + \nabla_v A_u), also transforms like a tensor since the difference of two tensors remains a tensor.
</Preamble>

In this post I will derive the gravity part of the GEM proposal using quaternion operators in a manifestly covariant notation.

The even representation of quaternions - where all signs are the same, and the Eivenvalues ore excluded so the quaternion can be inverted - will be used in the first phase of this derivation. What the even representation does is flip the relative signs of the E and B signs, so now the e filed has two signs, and the b field has one, like so:

E = -\nabla_0 A ~-~ \nabla_u \phi
e = \nabla_0 A2 ~-~ \nabla_u \phi

B_w = \nabla_u A_v ~-~\nabla_v A_u = \nabla \times A
b_w = -\nabla_u A2_v ~-~\nabla_v A2_u = \nabla \Join A2 \quad eq ~0

None of these transform like tensors, but together they do from \nabla A.

Notice that the quaternion differential operator acting on a 4-potential creates a scalar gauge field, and two fields the -e and b fields:

\nabla^* A2 = (\nabla_0 \phi ~-~ \nabla \cdot A2, \nabla_0 A2 ~-~ \nabla_u \phi ~-~ \nabla \Join A2) = (g, e ~+~ b)\quad eq ~1

The starting point for the derivation of the GEM gravity equations is the Lagrangian which can be viewed as the difference between the scalars of b squared and e squared. This can be achieved by changing the order of the conjugation operator with respect to the 4-potential, which flips the sign of the time derivative of A and gradient of phi which make up e, but not the symmetric curl b. The scalar gauge field can be subtracted away:

\frac{1}{8}(\nabla^* A2 ~-~ (\nabla^* A2)^*)(\nabla A2^* ~-~ (\nabla A2^*)^*)

=\frac{1}{8}(0, \nabla_0 A ~-~ \nabla_u \phi ~-~ \nabla \Join A)(0, -\nabla_0 A ~+~ \nabla_u \phi ~-~ \nabla \Join A)

=\frac{1}{2}(0, e ~+~ b)(0, e ~-~ b) = \frac{1}{2}(b^2 ~-~ e^2, b \Join b ~-~ e \Join e)\quad eq ~2

Any expression derived from this one will be invariant under a scalar gauge transformation which uses the terms \nabla_0 \phi, \nabla \cdot A because they have been explicitly subtracted away at this early stage. It is also of interest to think about the properties of the 3-vector, since in the EM case it was the Poynting vector.

Write out the Lagrange density - the scalar part of eq 2 - in terms of its components, including the vector coupling, -\frac{1}{2 c}(J A2* + (J A2^*)^*):

\mathcal{L}_{eb} = \frac{1}{2}((\nabla_1 \phi)^2 ~+~(\nabla_2 \phi)^2 ~+~(\nabla_3 \phi)^2 ~+~ (\nabla_0 A_1)^2 ~+~ (\nabla_0 A_2)^2 ~+~ (\nabla_0 A_3)^2
~-~ (\nabla_3 A_2)^2 ~-~ (\nabla_2 A_3)^2 ~-~ (\nabla_1 A_3)^2 ~-~ (\nabla_3 A_1)^2 ~-~ (\nabla_2 A_1)^2 ~-~ (\nabla_1 A_2)^2)
~-~ (\nabla_3 A_2)(\nabla_2 A_3) ~-~ (\nabla_1 A_3)(\nabla_3 A_1) ~-~ (\nabla_1 A_2)(\nabla_2 A_1) ~-~ (\nabla_1 \phi)(\nabla_0 A_1) ~-~ (\nabla_2 \phi)(\nabla_0 A_2) ~-~ (\nabla_3 \phi)(\nabla_0 A_3)
-\rho \phi ~+~ J_1 A_1 ~+~ J_2 A_2 ~+~ J_3 A_3 \quad eq ~3

Apply the Euler-Lagrange equation to this Lagrangian, treating the 4-potential as the variable:

\nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} \phi)}) = \nabla_1^2 \phi ~+~ \nabla_2^2 \phi ~+~ \nabla_3^2 \phi ~-~ \nabla_0 \nabla_1 A_1 ~-~ \nabla_0 \nabla_2 A_2 ~-~ \nabla_0 \nabla_3 A_3 ~-~ \rho
= -\nabla \cdot e - \rho = 0 \quad eq ~4

\nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} A_1)}) = \nabla_0^2 A_1 ~-~ \nabla_3^2 A_1 ~-~ \nabla_2^2 A_1 ~-~ \nabla_1 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_2 ~-~ \nabla_0 \nabla_1 \phi ~+~ J_1
= \nabla_0 e_1 - (\nabla \Join b)_1 + J_1 = 0 \quad eq ~5

\nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} A_2)}) = \nabla_0^2 A_2 ~-~ \nabla_3^2 A_2 ~-~ \nabla_1^2 A_2 ~-~ \nabla_2 \nabla_3 A_3 ~-~ \nabla_1 \nabla_2 A_1 ~-~ \nabla_0 \nabla_2 \phi ~+~ J_2
= \nabla_0 e_2 - (\nabla \Join b)_2 + J_2 = 0 \quad eq ~6

\nabla_{\mu}(\frac{\partial \mathcal{L}_{eb}}{\partial (\nabla_{\mu} A_3)}) = \nabla_0^2 A_3 ~-~ \nabla_2^2 A_3 ~-~ \nabla_1^2 A_3 ~-~ \nabla_2 \nabla_3 A_2 ~-~ \nabla_1 \nabla_3 A_1 ~-~ \nabla_0 \nabla_3 \phi ~+~ J_3
= \nabla_0 e_3 - (\nabla \Join b)_3 + J_3 = 0 \quad eq ~7

This work can be summarized with the GEM gravity source equations:

-\nabla \cdot e = \rho \quad eq ~8
\nabla \Join b - \nabla_0 e = J \quad eq ~9

This has been a manifestly covariant derivation. This derivation remains the same in flat or curved spacetime, in Cartesian or spherical coordinates. Once you get used to keeping track of all these terms, it is kind of fun, honest!

Doug

Stills:
http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations

The talk:
http://www.youtube.com/watch?v=Kd4nNb7nGOc

[sweetser]

Hello Carl:

Ronald Hatch sent me his paper which has an exponential in the force equation. For me as a member of the ultra-conservative fringe, it was not a happy read. He did not define a Lagrangian, so the proposal felt ad hoc. The way to get the force equation is to vary the Lagrangian with respect to velocity.

He thinks there is an absolute ether which strikes me as wrong.

He wants to give up the equivalence principle. His logic for this struck me as muddled. The discussion of mass was particular confusing. It looked like he had the classic self-taught blindspot vis-a-vis E=mc2, that the real relation is between the invariant m2 c4 and the square of the covariant 4-momentum, E2 - P2 c2. One of his equations looked like a trivial rearrangement of terms, a rearrangement he took seriously.

The precession of the perihelion of Mercury is a tough calculation. He wanted to add a second order effect in as if it was a first order effect.

By a strict application of the rules for the Independent Research area of Physics Forums, I don't think his work would be accepted. I forwarded my concerns to him directly, but he said I needed to think about things more carefully. The paper is published in Physics Essays.

Doug

[sweetser]

Hello:

I exchanged a dozen emails with Steve Carlip over the coupling current J^{\mu} A_{\mu}. The discussion started this way: "I have thought about Feynman's analysis of the spin of the vector current coupling." This post represents the back story, what I got from looking at chapter 3, pages 29-39 for about three to four weeks.

The first thing I did was work on my speed of going through the algebra in section 3.2, "Amplitudes and polarizations in electrodynamics, our typical field theory". You can only motor if you are confident about all the steps, just like in video games. The logic runs like this:

1. Start with the current-potential contraction.
2. Take the Fourier transformation of the 4-potential to get a current.
3. Simplify the current-current interaction along one axis, z.
4. Write out the contraction in terms of its components.
5. Use charge conservation to eliminate one term.
6. One term is the standard Coulomb interaction, the others are the relativistic corrections.

Then Feynman wants to know what that correction term is. This is were it gets a little odd. He talks about plane polarized light, and how looking at that you can see the angular momentum projections. I admit, I never quite saw those. What I did instead was try to strip away all the physics-speak, and just find the kernel of math underlying the operation. Looking back, that is what took the time: reducing the physics to a simple math expression. The Rosetta stone was a line on page 39:
...we know that (x ~\pm ~ i y)(x ~\pm~ i y) are evidently of spin 2 and projection \pm 2; these products are (xx ~-~ yy ~\pm~ 2 i xy), which have the same structure as our terms (3.4.1)

This is the pure math way to spot a system with spin 2: start with the product of two complex numbers, and check that the imaginary part has a 2ixy. This will require a change in x of pi radians to get back to the start point since there is an multiplier of 2. To speak like Feynman, I should talk up the projection operators, but I like to keep the math kernel free of that jargon.

Looking back on an earlier calculation, I was able to transcribe an earlier bit of algebra into a similar set of complex numbers:

(x ~\pm ~ i y)(x' ~\pm~ i y')^* = (xx' ~+~ yy' ~+~ (yx' ~-~ xy') i \quad eq~1

This is a system which has spin 1 projection operators, to use physics-speak. In math terms, the xy' does not help out the yx', there are no factors of 2 or 1/2, so this would take 2 pi radians to get the imaginary part of this back to where it was.

In terms of the math, the difference between a spin 1 system (eq 1) and a spin 2 system (see the quote) is no more complicated than looking at the imaginary part of a complex product.

Doug

[sweetser]

Hello:

I have the exchange of emails between Prof. Steve Carlip and myself up on the screen. It is not easy for me to read. Steve is a professional, I am not. On occasion, I babble. Babbling is a form of exploring, a process used to learn how to speak a language like a native. Studies have shown that deaf children do so learning sign language, and that baby birds do so before they can sing exactly like adults do. Recognizing this process, I have great patience for others that babble physics. Steve probably is that way in the right context, but in this email exchange, I got the book tossed at my head.

The discussion began with a slide from a talk (http://picasaweb.google.com/dougsweetser/SpinAndPhaseOfCurrentCoupling/photo#5196141258835741330). I went by the book, section 3.2 of Feynman's lectures, for three of four steps:

1. Start with the coupling term, J^{\mu} A_{\mu}.

2. Take the Fourier transformation, J^{\mu} A_{\mu} = -\frac{1}{K^2} J^{\mu} J'_{\mu} \quad eq~3.2.2

3. Write out 2 in terms of the components:
J^{\mu} A_{\mu} = -\frac{1}{K^2}(\rho \rho' - J_1 J'_1 - J_2 J'_2 - J_3 J'_3 ) \quad eq~3.2.5

[Note on imprecise notation: in the slide I use x, y, and z which imply a coordinate choice. I should have used numbers for subscripts. I also didn't toss in the minus sign as Feynman does for the Fourier transformation step.]

Up to this point, I have exactly walked down the path Feynman wrote about. When it came to writing up the slides, I initially put in a further step Feynman used: he imagines picking a coordinate system such that all the current goes along one direction (the 3 axis). Everyone is accustom to this step. Yet it bothered me. I would have to rewrite the derivation if someone chose the 2 axis instead. A more general position would involve choosing no axis, yet spotting the symmetries of the spin in the phase anyway. I decided to work with that as a goal.

4. Multiply out the two currents as quaternions:

-J J' = (-rho rho' + J1 J'1 + J2 J'2 + J3 J'3,

-rho J'1 - J1 rho' - J2 J'3 + J3 J'2,

-rho J'2 - J2 rho' - J3 J'1 + J1 J'3,

-rho J'3 - J3 rho' - J1 J'2 + J2 J'1)


The terms in italics are equation 3.2.5, the underlined terms are in an expression about spin 2 symmetry on page 39, and the terms in bold are in 3.2.10 in a discussion of spin 1 symmetry.

Steve had no idea what I was doing, none. Communication was broken by step 4. I was at work, trying to do my job, or appear to be doing my job, and quickly come up with a response to someone with far more intellectual precision. It did not work out so well. The worse thing I did was about equation 3.2.10 concerning circularly polarized light. Feynman writes out the currents for two circularly polarized light whose imaginary parts cancel. I got a sign wrong, so they didn't cancel, and I got to look stupid.

Both Steve and Feynman talked about projection operators, the relevant machinery from quantum field theory. I did not talk about projection operators at all. I am not going to do so now since I would probably just babble about them. I understand why the well-trained would say that if I don't discuss projection operators intelligently, then this has nothing to do with the spin of particles. Nothing. For me, projection operators are a patina on the underlying algebra (patina def.: the sheen on the surface of an old object, caused by age and much handling).

Steve left the discussion convinced I didn't even understand the basics of spin in physics. He won the word game, but the algebra in step 4 still stands consistent with what Feynman wrote. At the end of the day, algebra trumps words.

Doug

[sweetser]

Hello:

Preparing for my talk in Brazil, I had an interesting insight. The thesis behind "Doing Physics with Quaternions" at quaternions.com is that physics describes patterns of events in spacetime using quaternions up to an isomorphism. Most of physics works great without quaternions because the quaternion expression would not provide new information.

My epiphany was this: new quaternion math equals new physics. Here in this thread, the new math is the Even representation for quaternions (a reinvention of Clyde Daven's hypercomplex numbers). The current coupling J2 A2 has the spin 2 symmetry in the phase, and the field strength tensor \nabla A2 contains the symmetric curl needed for the symmetric field b.

I have also mentioned here the work in animating quaternions which lead to an understanding why the groups U(1), SU(2), SU(3) and Diff(M) must be all that makes up the symmetry forces of Nature. With the visual perspective, the forces are more tightly linked algebraically. The standard model is written as U(1)xSU(2)xSU(3), which would have 1+3+8=12 generators for its Lie algebra. One mystery of the standard approach is why the electro and weak forces should team up to form the electroweak force. Another mystery is why should confinement exist for the strong force SU(3)? These phenomena are suggesting something more like [U(1)xSU(2)]xSU(3) than three equal players. That is what happens with the quaternion representation of the symmetry of forces:

A^* B = (\frac{A}{|A|} exp (A - A^*))^* (\frac{B}{|B|} exp (B - B^*))\quad eq~1

which is (U(1) SU(2))*(U(1) SU(2)). The Lie algebra only has 8 generators. This smaller model has a chance to provide a cause for the confinement of quarks.

The third quaternion math innovation I do not talk about much because I have yet to see how it impacts a calculation, although it helps with a big riddle, a "why" question, in physics. Many who work with quaternion derivatives accept the idea of a left handed versus right handed derivative. This comes from the limit definition, putting the differential on the left or right. This definition is ineffective since one cannot show that a function as simple as f(q)=q2 is analytic in q. For me that indicates the definition has no utility. What I did was steal a move from L'Hospital's rule and use a dual limit process. Let the pesky 3-vector with its three imaginary basis vectors go to zero first, leaving only the real number which commutes with all. Effectively this is a directional derivative along the real axis. Things work out great for proofs using this definition (if one is good at doing proofs, which I am not). All events are ordered by the real scalar. If this definition is applied to events in spacetime, the scalar is time, and thus all the events are order in time like a movie.

What happens when the limit processes are reversed, and the pesky 3-vector goes to zero after the scalar gets frozen? All one can do in this case is to determine the norm of the derivative. Although not well known, there is a branch of math that studies norms of derivatives. I think this is the domain of quantum mechanics. We cannot order things in time, but we can tell on average how much change is going to happen after making many measurements.

Now to make the slides...
Doug

[mhill]

but when doing the product of quaternions if Q is a quaternion then where you put Q.Q=Q^{2} it should read Q.Q^{*} , for example for Minkowsky metric

dQ=dt-idx-jdy-kdz then ds^{2}=(dQ).(dQ^{*})=dt^{2}-dx^{2}- dy^{2}-dz^{2}

and as i pointed in other part of the forum, you have the problem of non-commutativity so

ij(dx.dy)=-ji (dy.dx) ,

anyway the idea seems very interesting

[sweetser]

Hello mhill:

In the standard Hamilton representation, for a quaternion dQ = (c dt, dx, dy, dz) we have the product:

dQ dQ = (c^2 dt^2 ~-~ dx^2 ~-~ dy^2 ~-~ dz^2, 2 c ~dt~ dx, 2 c~ dt~ dy, 2 c ~dt~ dz)\quad eq~1

When one uses tensors, the Einstein summation convention ignores the three other terms. That is a mistake in my opinion. One thing I have noticed about the GEM proposal: the 3-vector (2 dt dx, 2 c dt dy, 2 c dt dz) is an invariant in the presence of a gravitational source. That is really cool because special relativity is about the invariant scalar, and for GEM, gravity is about the invariance of the 3-vector.

If we look at the product of two quaternions that are nearby each other, a dQ and dQ', we get a similar result with a cross product:

dQ dQ' = (c^2~ dt~ dt' ~-~ dx~ dx' ~-~ dy~ dy' ~-~ dz ~dz',
c~ dt ~dx' ~+~ c ~dx~ dt' ~+~ dy ~dz' ~-~ dz~ dy',
c ~dt ~dy' ~+~ c ~dy ~dt' ~+~ dz ~dx' ~-~ dx ~dz',
c ~dt ~dz' ~+~ c ~dz ~dt' ~+~ dx ~dy' ~-~ dy ~dx')\quad eq~2

Hamilton could have done that one, so this is not new math. For the Even representation of quaternions, the rules are easy to remember: everything is sunny and positive in California, and so it goes for the Even representation:

i^2 = j^2 = k^2 = ijk = 1\quad eq~3

ij = ji = k \quad eq~4

ik = ki = j \quad eq~5

jk = kj = i \quad eq~6

There is not a minus sign to write. This will be a division algebra if and only if the eigenvalues of the real 4x4 matrix representation are excluded from the set of quaternions.

There will be math wonks who insist that the name "quaternions" is reserved for non-commutative 4D division algebras. Historically, that has been the use. Doing new math can cause conflict, so be it. The non-commutative aspect is a result of an arbitrary choice in how to represent a 4D division algebra. I have chosen a different representation. It still is a division algebra, but one where multiplication commutes. Clyde Daven did this first and called them hypercomplex numbers. That makes them sound like a separate animal, and I don't think it is. This is representation theory in practice, applied to quaternions.

To indicate the Even representation is being used, I toss in a "2" every now and then. Repeat eq. 1 with the even representation:

dQ2^* dQ2 = (c^2 ~dt^2 ~-~ dx^2 ~-~ dy^2 ~-~ dz^2, - 2~ dy ~dz, - 2 ~dx ~dz, - 2 ~dx ~dy)\quad eq~7

The scalar is the same, but the 3-vector has changed. Instead of rescaling the (dx, dy, dz) by a 2 dt factor, this time it is all about the symmetric curl. The dt factor does not make it into the 3-vector.

Repeat the dQ2 dQ2' in the Even representation:

dQ2^* dQ2' = (c^2 ~dt ~dt' ~-~ dx ~dx' ~-~ dy ~dy' ~-~ dz ~dz',
c ~dt ~dx' ~-~ c ~dx ~dt' ~-~ dy ~dz' ~-~ dz ~dy',
c ~dt ~dy' ~-~ c ~dy ~dt' ~-~ dz ~dx' ~-~ dx ~dz',
c ~dt ~dz' ~-~ c ~dz ~dt' ~-~ dx ~dy' ~-~ dy ~dx')\quad eq~8

If dQ2' where to get acted on by the conjugate operator instead of dQ2, then the only terms to flip signs involve dt dR.

What makes the Even representation interesting is it gives a more complete way of viewing a quaternion product. In the Hamilton representation, the rescaling (dt dR) was unaltered by changing the order of multiplication, while the cross product flipped signs. In the Even representation, the symmetric curl is unaltered by changing the order of conjugation, while the rescaling flips signs. The scalar remains steady no matter what. That sounds like a more complete survey of the product of two sets of four numbers.

Doug

[sweetser]

Hello:

I spend time and effort pitching this research project to both the upper elite and the technical masses. I took off a Wednesday from work to see a talk by Nobel Laureate Sheldon Glashow, who got one in the 70s for the electroweak theory. I told him I had an animation for U(1)xSU(2) on my MacBook Pro. He was too busy to look right then, but I could send him an email, which I did.

Later that evening, Michio Kaku was doing book promo work. He agreed to sign his latest book, "Physics of the Impossible", which is doing well on the New York Times Best Seller list. I got in line late, and the book was sold out by the time I got there. I bought a different book, and had him sign "Maxwell is the best!" I also dropped off a version of my proposal where I derive the Maxwell equations first - to established I am much better than your average crank - and with a variation get equations for a metric approach to gravity. He thanked me as he went on to sign another book.

Max Tegmark gave a talk on the Physics of Super Heroes, along with a screening of Superman. He is a big survey astronomer by day, so it was fun to cruise through the known Universe with his software. He knew me from a previous outreach program he did, and had traded a few stories back then about quaternions. I told him of the quaternion animation project, how it could be found on YouTube with a search for "Quaternions Standard Model". I gave him a business card with the search instructions after my 20 second pitch.

Low odds on getting a reply from these busy folks, but I need to try, so I do.

There are many more folks who read a high traffic site such as slashdot.com They had an article on Lectures On the Frontiers of Physics Online (http://news.slashdot.org/article.pl?sid=08/05/13/0350215&tid=146). Buch of big names in physics have videos up there:
Neil Turok's 'What Banged?,' John Ellis with 'The Large Hadron Collider,' Nima Arkani-Hamed with 'Fundamental Physics in 2010,' Paul Steinhardt with 'Impossible Crystals,' Edward Witten with 'The Quest for Supersymmetry,' Seth Lloyd with 'Programming the Universe,' Anton Zeilinger with 'From Einstein to Quantum Information,' Raymond Laflamme with 'Harnessing the Quantum World,'

You might be able to see the site in a day or two here, http://perimeterinstitute.ca/index.html

So I posted a note there which reflects my current thoughts:


Title: Maxwell Trumps General Relativity

General Relativity rocks. It is elegant in its minimialism. All efforts to add a little extra have failed, usually by allowing a dipole gravity wave mode of emission which has been ruled out by binary pulsar data.

The only field theory that is manifestly better than GR is the Maxwell field equations. Every time we have added to it in the name of symmetry, the theory has done more. James did it himself by tacking on the Ampere current. Einstein looked to get rid of a duplicate law, and so special relativity was born. With the huge supply of new particles coming out of atom smashers, the gauge symmetry in EM (U(1)) was expanded to SU(2) for the weak force, and SU(3) for the strong.

None of those smart cats listed in the initial post will be talking about the Maxwell equations. Too bad, the history of physics is clear: expand Maxwell, you win.

Max depends on the field strength tensor d_u A_v - d_v A_u. There is a subtraction in there, a great thing (called an exterior derivative). But in the name of symmetry, we need to work with the rest of it, d_u A_v + d_v A_u. Do that right, and you get a unified field theory that Einstein failed to find by looking for workable extensions of GR. Extend Max, not GR.

If anyone here wants to see the nuts and bolts of deriving the Maxwell equations using the Euler-Lagrange equations, search for "GEM action" on YouTube. A small variation - two minus signs - on the Maxwell equations leads to equations for gravity. Yes, I show that there is a metric solution (the Rosen metric if you are up on your GR jargon, a bunch of exponentials if not). Yes I know there is an issue of spin 1 and spin 2 which can be addressed if you get what the phase of current coupling really is.

YouTube can survive being slashdotted.

Doug

[sweetser]

Hello:

I am about to power down the laptop and get it ready for a trip to a foreign land. I will be giving this Thursday, May 29, in Campinas, Brazil at 5:30 pm. Stop by if you are in town. Since I figure about zero of you can do that, I just uploaded a video of the talk to YouTube (http://www.youtube.com/v/BbN7Z3GVWoM). The video is 41 minutes long, but I only have a 25 minute slot, so I will have to be more efficient. As always, critiques are appreciated.

I really like the central thesis: that only by doing new math can I do new physics. Will report back on if I can get others excited by this work.

Doug

[Mentz114]

Bon voyage, Doug.

Lut

[sunsphere]

I have a brand new theory on the mechanics of gravity I believe the whole science world should review for possible validity.

[sweetser]

Hello Sunsphere:

The rules for starting your own thread are here (http://www.physicsforums.com/showthread.php?t=82301). The rules may appear harsh, but there are many people who make grand claims and the rules are there to filter out ideas that are not precise enough to have a discussion about.

Doug
Brazil is tropical, what can I say?

[sweetser]

Hello:

I am back in the US after the trip to the 8th International Conference on Clifford Algebras and the Applications to Mathematical Physics in Campinas Brazil. The exotic location was irrelevant. If the meeting were held in Indiana, it would have made little difference. I went from the plane to the hotel to one building on the campus, and back to the hotel. Sure the vegetation was odd, but when technical talks go from 8:30 AM to 6:30 PM, there was not time or structure to get any exposure to Brazilian culture. Campinas means "grass fields" in Portuguese, a place not far outside San Paulo. thirty years ago it had 100,000 people. Now there are over a million, attracted by high tech companies. When a farm town blows up quickly, the effort is focused on infastructure and not on art and public works. In the opening remarks, one of the organizers said one of the benefits of the meeting was that there was nothing to see in Campinas so we could focus on work.

There were about 70 participants. This was an international conference, with accents in French, Italian, German, British, Polish, and South American (few people from the Far East). Each day had a dozen talks. The morning and early afternoon plenary talks were an hour long. The last afternoon session had half hour talks for the "not invited" who wanted to present their work. This was a PC crowd (I may have had the only Keynote presentation on a Mac). Much of the meeting was video taped, but how they hope to make that available is not known. My own session was not recorded since I was shuffled to the alternate room of a parallel session.

There were many lessons I learned. One of the basic ones is that geometric algebra and Clifford algebra are one and the same (the similarities were clear, but not the formal identity). The story is that David Hestenes has made a big push to bring Clifford algebras onto the center stage of physics. He did not think that the name "Clifford" had information content, preferring to use the name Clifford himself used, geometric algebra, to discuss this area of work.

David attended the meeting. It was clear he was the "star", but he was a decent and approachable guy. It is also clear that despite his efforts, geometric algebra has remained a minor contributor to physics. I could relate to his position (excluding the star part). Quaternions are an even smaller footnote in physics. David said he discovered things using geometric algebra that he needed to translate into tensor lingo to publish. I also find myself translating quaternions insights back to tensors to communicate with the broader physics community.

During a coffee break, I asked him what he thought about quaternions. He said they are just a part of the bigger Clifford algebra (specifically CL(0, 2)). I expected that one, the typical male thing - mine is bigger. I will write a separate post about my critique about such a perspective. He invited me, like I saw him do with others, to grow up and embrace the real math tool of Nature, geometric algebra.

David's talk was great. He brought up an idea proposed by Louis de Broglie, that each atom has an internal clock with a frequency of B = m c^2/h. The math is simple but has been largely ignored by the physics community. The frequency is in the zeptoseconds range (if they skipped this in your schooling, it is because it is new, representing a factor of 10^{-21}). David proposed a way to detect this absurdly fast clock as a resonance in a crystal. The message is darn deep: TIME IS MASS (an ironic aside: I recall Archimedes Plutonium, an Internet crackpot who has passed away, that would always write this IN CAPS).

David did not give a mechanism for the clock. My own work might provide that. In GEM, one has a relativistic 4D wave equation. That has a 1/R2 potential solution, leading to a 1/R3 force law. Dipoles have this sort of potential and force law. Since charged massive particles will both attract and repel their brothers, they are born as dipoles. I have been wondering what those dipoles do. Now I have a candidate: it could be the basis of the de Broglie internal clock and connected to mass. Neat.

Over the course of the next few weeks, I may discuss other talks that have already influenced my research. As for my own talk, I was scheduled to be the next to last talk on the next to last day. I had pitched the talk to a number of people. Most of the people I gave the personal pitch showed up, about 20 folks in the room. This may have been the biggest crowd with good technical credentials I have presented to (Hestenes was in the group).

The thesis was that quaternions, as invented by Hamilton and Rodrigues, cannot make new contributions to physics, but new math with quaternions can lead to new physics. I went through three examples.

First, I have a new definition of a quaternion derivative that splits into a directional derivative along the real line for classical physics, and a normed derivative for quantum mechanics. Getting the derivative of a quaternion derivative on a spacetime manifold might resolve the Bohr/Einstein debate as to why quantum mechanics is different from classical physics.

Second, the analytic animations represents a logical extension of analytic geometry. We can be precise, exact, with the treatment of time's relationship to space. For this crowd, I developed some new animations about cross products which I will discuss in another post.

Third, the Maxwell equations have a better track record at being extendable and connected to quantum mechanics than general relativity. I have figured out how to write the Maxwell action using quaternions. When I wanted to write a symmetric tensor contraction using quaternions, I faced a wall. The way to tunnel through the wall was to find a 4D division algebra where elements commute so long as the Eigen values are excluded. That leads to the GEM action, a way to unify gravity and EM. I showed the solution to the field equations that was physically relevant. There was a slide for the spin of the current coupling term.

The crowd reaction was standard: no one asked a question (other than one clarification). Most other talks had two or three questions. No one came up to ask about the content of the talks during later coffee breaks or over lunch. People were interested in the fact that this was a hobby of mine.

This trip was well worth the time and money, I am glad I went.

Doug

[Mentz114]

Hi Doug,
thanks for ther report. It's very interesting. The thing is that if we're geometrizing something, then to handle curved spaces you need tensors It's very handy that all the observables are scalar contractions of tensors.

I will have to bone up on geometric algebra, which may be has a different way of doing curved space.

Lut

[sweetser]

Hello Lut:

It would take less effort to "bone up" on how to handle curved spacetime with quaternions. In flat spacetime, an event at:

event = (t e_0, x_1 e_1, x_2 e_2, x_3 e_3) \quad eq~1

where:

|e_0| = |e_1| = |e_2| = |e_3| = 1 \quad eq~2

e_0^2 = +1
e_1^2 = -1
e_2^2 = -1
e_3^2 = -1\quad eq~3-6

In spacetime curved by a static, spherically symmetric mass charge, equation 1 remains the same, while the other ones are altered:

|e_0| = \frac{1}{|e_1|} = \frac{1}{|e_2|} = \frac{1}{|e_3|} = exp(-G M/c^2 R) \quad eq~7

e_0^2 = +exp(-2 G M/c^2 R)
e_1^2 = -exp(+2 G M/c^2 R)
e_2^2 = -exp(+2 G M/c^2 R)
e_3^2 = -exp(+2 G M/c^2 R)\quad eq~8-11

This might not be too hard to implement, now that I have written it down clearly (the key step to programming). q_metric -mass 10e-6 1 2 3 4 would return the event (1, 2, 3, 4) in a space curved by a 10e-6 mass.

The problem with my software as written is handling this very tiny numbers.

Doug

[Mentz114]

Hi Doug,
yes, one way of losing tensors is to use differential geometry ( if I haven't got my nomenclatures wrong). I think that's what you did above, by introducing basis vectors in your your quaternion space. Makes sense.

What technology are you using for computing ? I have floating point down to 10^-300.

Lut

[sweetser]

Hello Lut:

I am using C. I print to the command line which by default prints to 8 digits. I bet I can adjust printf to do more. I am not certain how to set the internal precision accurately of a standard C program.

I have always had the basis vectors, but had not put them to use in a calculation.
Doug

[Mentz114]

Hi Doug,

sounds like you need a better C compiler.

You seem to have all the apparatus for space-time in your algebra. Letting gravity appear in the basis and not the coordinates is a good practice.

When you conjugate this

event = (t e_0, x_1 e_1, x_2 e_2, x_3 e_3) \quad eq~1

do you change sign and invert the basis vectors ? I'm thinking about a conjugate space, and what scalars you can make by multiplying a vector times a conjugate ?

Lut

[sweetser]

Hello Lut:

That is a great question! My initial reaction was the signs just flip. The thought is that conjugation is about the work of mirrors. In flat spacetime, three sign flips is all one needs to do a mirror reflection. If the mirror is curved, I am not sure if flipping the sign is enough. I'll have to think about it some more.

I use the gcc compiler. I am sure there is a way to make the precision high, but at this time I don't know how to do it. What is your programming trick/tools to get such high precision?

Doug

[Mentz114]

Hi Doug,
I was free thinking there. I'll have to ponder it too.

I use Delphi Win32 and I've got the following floating type amongst others

Extended 3.6 x 10^-4951 .. 1.1 x 10^4932 digits 19-20 uses bytes 10

10^-4951 !

Lut

[sweetser]

Hello Lut:

I figured out how to crank up the precision. The default single precision float has about 7 significant digits. At double precision, there are 16 significant digits. All I have to do is set the -DFmode flag and set all programs to print 16 digits. There is the third level, -XFmode which will do 24 digits, but that may just be wasteful of resources, using 12 bytes. There is a Tetra Floating number for those that need 16 bytes.

Thanks for getting me to look this up.
Doug

[sweetser]

Hello:

Quaternions as events in spacetime physics are not the Clifford algebra CL(0, 2). Let me first explain what that bit of jargon means.

Clifford algebras were an attempt to generalize the math started by Hamilton, passing through Grassman. These algebras are independent of coordinates, and can be written in arbitrary dimensions. As algebras, it is not important that they are invertible, although that can happen.

The Clifford algebra CL(0, 2) is a multivector, meaning it has two parts. First there is the scalar, the 0 part, a pure number. The bivector is the second part. They use a wedge product, \wedge, a generalized cross product to form a 3-term bivector.

q(t, x_1, x_2, x_3) = t ~+~ x_1 e_2 \wedge e_3 ~+~ x_2 e_3 \wedge e_1 ~+~ x_3 e_1 \wedge e_2 \quad eq~1

Why use a wedge product? The reason is that wedge products are axial vectors, those that switch handedness in a mirror. The other sort is a polar vector. If one looks at an event in a mirror, this makes physical sense.

It is not worth the time to argue with math types who think they can define whatever they want, however they see fit. I have the perspective of a mathematical physicists, where any math definition can be cow-roped by physical meaning (I am a crude mathematical physics who uses rodeo analogies).

The unspoken assumption behind CL(0, 2) is that one should use a mirror on this event. That is like taking (t, x1, x2, x3) to (t, -x1, -x2, -x3). That sort of transformation does have the handedness needed for an axial vector. Yet it is easy enough to think of counter examples. What about a time reflection, where an event (t, x1, x2, x3) goes to (-t, x1, x2, x3)? This transformation does not change handedness, it is represented by a polar vector. Compare the two functions

mirror reflection: q -> q' = q*
time reflection: q -> q' = -q*

Since these are so close to each other as functions, I don't think one should take precedence over the other, as asserting one must use a bivector for the spatial part of a quaternion does.

The fundamental currency of the Universe is a bare event which stands alone in a vacuum, not in front of a mirror waving a left hand. General relativity has a message about basis vectors. Should we have a toy Universe with two events, the interval between these two events is found by taking the dot product of the difference between the two events:

(c dt, dx_1, dx_2, dx_3).(c dt, dx_1, dx_2, dx_3) = c^2 dt^2 ~+~ dx_1^2 (e_2 \wedge e_3).(e_2 \wedge e_3) ~+~ dx_2^2 (e_3 \wedge e_1).(e_3 \wedge e_1) ~+~ dx_3^2 (e_1 \wedge e_2).(e_1 \wedge e_2)

= c^2 dt^2 ~-~ dx_2^2 ~-~ dx_2^2 ~-~ dx_2^2 \quad eq~2

So far, so good. Now repeat the measurement in a toy Universe where nothing has been altered for the generators of the two events, but one has added a mass. A metric theory of gravity does not alter events, but does change the measure of events. In this setup, the differences are the same. The only thing that could change are the sizes of the basis vectors themselves. The problem here is that there is no basis vector associated with the dt2 term.

The deep message of special relativity is to treat time as we do space. In the geometric algebra approach, time is a scalar, whereas the space parts use two basis vectors as a bivector. Contrast this with the way I define a quaternion:

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

This is not a Clifford algebra because the basis vector e_0 commutes with the others. In the definition of a Clifford algebra, all the basis elements anti-commute. Physicists would call this a 4-vector, because one can add them or multiply it by a scalar, and it transforms like a 4-vector. It is this last phrase, on how it transforms under a Lorentz boost, that holds the magic. The goal of a great definition is to remove the magic and let the math speak for itself. Square the difference between two events using the quaternion definition as written in equation 3:

(c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3)
\quad eq~4

To be consistent with special relativity, we make the following map:

e_0^2 = +1
e_1^2 = -1
e_2^2 = -1
e_3^2 = -1 \quad eq~5-8

To be consistent with general relativity to first order tests of weak gravity fields for non-rotating, spherically symmetric sources, all we need is:

e_0 = exp(-G M/c^2 R)
1/e_0 = e_1 = e_2 = e_3 \quad eq~9-10

so according to the GEM proposal,

e_0^2 = exp(-2 G M/c^2 R)
e_1^2 = -exp(2 G M/c^2 R)
e_2^2 = -exp(2 G M/c^2 R)
e_3^2 = -exp(2 G M/c^2 R) \quad eq~11-15

I forget who said it, but one can only hope to find an important law in physics if one also finds an invariance principle. When there is a gravitational source, the terms in the 3-vector part of the square (2 dt dx e_0 e_1, 2 dt dy e_0 e_2, 2 dt dz e_0 e_3) remain invariant under the presence of such a gravitational field.

In summary, I see two flaws to the assertion that quaternions are the Clifford algebra CL(0, 2). First, while some quaternions are axial vectors, others are polar vectors. Second, the scalar needs to have a basis vector associated with it to work with metric theories of gravity. As far as I can tell, everyone working with geometric algebra in physics works with the assumption that the quaternions are faithfully represented by CL(0, 2).

Doug

[sweetser]

Hello:

Today I will discuss the geometric product, a tool used extensively for people applying Clifford algebras to problems in physics and robotics. Here is the definition:

ab = a.b ~+~ a \wedge b \quad eq~1

There are a number of great features to this definition. It will work for whatever dimension you have in a Clifford algebra. The product is conformal, meaning angles do not change. The dot product is the magnitude of a time b times the cosine of the angle between a and b, while the wedge is the magnitude of a time b times the sine of the angle between a an b. How clean!

The road to Clifford's geometric product is convoluted. It started with Hamilton writing out the 4D product one day, carving it into a bridge, and the very next day working to get rid of one dimension with "pure quaternions" that had a zero scalar. The scalar that looked completely alien to Hamilton for the square of a quaternion, t^2 ~-~ x^2 ~-~ y^2 ~-~ z^2, looks familiar to us as the interval of special relativity. By setting t=0, one gets something the Egyptians would have recognized instead of Einstein, so long as one ignores a pesky minus sign.

Yet that minus sign did bother people. The nice thing about a.b that appears in the geometric product is that all the signs are the same and positive (fears of minus signs have wasted more time historically than almost any other math issue). It was Grassman that focused on the wedge product, the new math to come out of the quaternions. He built up an entire system of algebra using basis vectors and wedge products. Clifford put the dot product together with the wedge product to make the geometric product. All is good...

...unless you are a stickler for technical details. Hamilton developed quaternions that are defined for one scalar dimension, and three spatial vectors. One starts with 4 numbers. Look for the product of those four numbers with another set of 4 numbers, and the complete story necessarily has 16 parts. The geometric algebra only has 10. Oops. What it is missing is a scalar times a 3 vector, done twice. This is the math omission needed to make the product conformal - there is no vector stretching.

The geometric product, written as a quaternion operation, looks like so:

ab = (a^* b ~+~ b^* a)/2 ~+~ (a b ~-~ b a)/2 \quad eq~2

Recall the history. Quaternions were used first to form a product that always had an inverse. Grassman isolates on the wedge, then Clifford glues things back together. The result to my eye does not look pretty. The geometric product makes sense if you fear the unknown Lorentz invariant interval, but that is not where we are today. The conformal nature of the geometric product looks to me like a rigged outcome because the 6 problem terms are dropped.

I understand why people have put a huge investment in the geometric product, yet it does not look like a safe bet to me.

Doug

[Mentz114]

Hi Doug,

most illuminating. What exactly does this

The geometric product makes sense if you fear the unknown Lorentz invariant interval,

say ?

Lut

[sweetser]

Hello Lut:

I was projecting to the minds of three dead men, Hamilton, Grassmann, and Clifford, whose lives where lived within the 1800's, well before the insights of 1905.

Doug

[CarlB]

Doug, glad to see you're looking healthy, happy, and in the middle of things, at least as of Brazil:
http://www.ime.unicamp.br/~icca8/images/GRUPOWEB.jpg

[sweetser]

Hello:

How would you described the motion of a screw? It is a combination of linear and angular motion, with the angular motion perpendicular to the linear. At the meeting in Brazil, I learned the way most people handle this problem: they use dual quaternions. I had never heard about dual quaternions, so let me give you a brief introduction.

As the name suggests, the dual quaternions have two quaternions. One of these also has an epsilon that when squared equals zero. Linear motion goes in the first quaternion, rotation in the dual. They don't mix with each other, staying nice and perpendicular. Dual quaternions are vital for many calculations in robotics.

At a dinner at the meeting, one fellow said that it would not be possible to use quaternions to do the work of dual quaternions. This made no sense to me. After all, the motion of a screw is a collection of events in spacetime. It might take some work to see the connection between quaternions and dual quaternions, but I had confidence it could be done. Another person thought it would be big news to find such a trick. He sent me a reference to a short paper on the topic. I read the paper, and with a bit of effort, was able to spot how to make things work without the epsilon.

Here is the idea. Screw motion is most easily characterized by two quaternion: one for linear motion, the other for angular motion about the axis of the linear motion. These two quaternions can point in arbitrary directions to each other. For a screw, we want to ensure that the angular motion is at a right angle to the linear motion, good old 90 degrees. Consider two arbitrary quaternions. The norm of each is zero if and only if the quaternion is zero, otherwise it is positive definite:

a^* a = (a_0^2 ~+~ a_1^2 ~+~ a_2^2 ~+~ a_3^2, 0, 0, 0)
b^* b = (b_0^2 ~+~ b_1^2 ~+~ b_2^2 ~+~ b_3^2, 0, 0, 0) \quad eq ~1,2

What happens when we form the same product with a and b? Here we could get a zero if a and b between them have 4 zeros, like so:

a = (0, 3, 0, 4)
b = (1, 0, 2, 0)

scalar(a^* b) = 0 \quad eq ~ 3-5

A more general case happens if the two 3-vectors are at 90 degrees to each other, and one of the two scalars happens to be zero:

Let A = (0, a1, a2, a3)
A = (0, b1, b2, b3)
a \angle b = 90, so \quad a.b = |a||b| cos(90) = 0

scalar(a^* b) = 0 \quad eq ~ 6-10

A pair of quaternions where one of the scalars happens to be zero has a conformal Euclidean product (my name for a*). The angle does not change because it must remain fixed at 90 degrees.

The most general case happens when the contribution from the 3-vector dot product happens to cancel with the scalars, when:

scalar(a^* b) = (a0 b0 ~-~ a1 b1 ~-~ a2 b2 ~-~ a3 b3) = 0 \quad eq ~ 11

I have been spending my time writing programs to make visualizations of these screw motion equations. Nothing obscene, honest. It has been easiest to set up the case where there are four zeroes between the two quaternions. Then you let the numbers travel linear along the z axis, while doing sin t along x, cos t along y, and t along t. A parameterized quaternion:

q(t) = (t, sin(t), cos(t), k t) \quad eq ~ 12

That was so easy to do, I started tossing in different coefficients:

q2(t) = (t, sin(t), cos(t), k t) \quad eq ~ 13
q3(t) = (t, sin(t), cos(2 t), k t) \quad eq ~ 14
q4(t) = (t, sin(t), cos(.5 t), k t) \quad eq ~ 15

What do these look like? The answer is here, with a few others (http://picasaweb.google.com/dougsweetser/Screws). This evening's fun was to take equation 12 and put the sin(t) in the t position like so:

q(t) = (sin(t), t, t, cos(t)) \quad eq ~ 16

What does that look like? Nothing is linear because time is doing the trig function thing. The tx and ty complex planes look like standard trig functions since it is just a plot of t versus sin(t). The tz plane ends up being a circle. And the animation? Well, you will have to go to picasaweb.

Odd simple math. Neat.

Doug

Note: reached the 50k views of thread milestone. At least I keep doing new things with quaternions.

[sweetser]

Hello:

There are two groups that understand the odd ways of relativistic quantum field theory.

The first group are the elite people who hang around the physics departments of major research universities. They don't bother to teach quantum field theory to undergraduates. I got my exposure at Harvard University while actually working as a molecular biologist over at the School of Public Health (and probably the only person to ever make such a commute). I recall going to Mitch Goldstein's office hours, where he would emphasize as he did in class that the only way to understand quantum field theory was to recognize that it involved the behavior of many particles. There was all that stuff about creation and annihilation operators, details which I could follow at the time but would require some refresher work. Few people in a few cities on the planet know how to work with quantum field theory.

The second group is every collection of particles anywhere in the universe. That is a far larger group, ridiculously large. Each and every member of this group is stupid beyond conception. They all know how to do quantum field theory.

The riddle is how all this stupid stuff knows how to do things the smart boys barely can grasp.

Then I took another look at the image (http://picasaweb.google.com/dougsweetser/Screws/photo#5215627885126508242) created by equation 16 of the previous post

q(t) = (2 sin(3 t), t, t, 2 cos(3t)) \quad implementation ~ of ~eq ~ 16

[note: on a slow, wireless connection, picasaweb shows a still, so you might have to download it, then open it up in a browser to see the animation]. The images starts out blank. Then at 3 separate locations, a pair of events are created. The events move apart, then pairs annihilate each other. It can help to watch the shadows, since that represents a 2D simplifications of the motion in 3D. I had no intention of doing quantum field theory. Yet I know that sines and cosines are solutions to 4D relativistic wave equations. Toss the trig function into the time slot, and you have 3 pairs of events creating and then annihilating each other. This may be how quantum field theory looks.

Doug

[shmengie]

Watched a norwegian documentary downloaded as a torrent. The plasma theory of the universe would seem to give credence to Mr. Sweetser's theory, IMHO. Please accept my apologies if this post is unwelcome.

http://thepiratebay.org/tor/3501592/No_Big_Bang________Universe_-_The_Cosmology_Quest

This is not the exact same one I watched. The one I watched didn't have subs. It was mostly English but there was some spoken Norwegian (I think), which this one may have translated subs.

[sweetser]

Hello Joe:

I have not been able to download the 2.8G torrent yet, so I will present some of my thoughts about cosmology independent of that video.

Cosmology has major problems, and they know it. I try and be careful in discussing these issues, to call it by the problem with the data and not by the hypothesis that has gathered the most attention. Dark matter is a hypothesis. There is a problem with using Newtonian gravity - as a good approximation to GR in these low mass density, non-relativistic situations - to create a mathematically stable solution for the motion of a thin disk galaxy where the outermost stars have the same velocity. One might be tempted to call this the velocity profile problem. My issue with that is velocity is not a conserved quantity. We should be focusing on momentum and changes in momentum. I call this the galaxy rotation problem, where we do not have a force equation that yields a stable, constant velocity solution. I have posted in this thread a sketch of an idea that uses the product rule which stands a chance of providing a new classic stable constant velocity solution involving gravity. That equation has not been applied to the data at hand, so remains an area of study.

The standard big bang is known to have two problems. They have the labels of the horizon and flatness problems. We have measured the cosmic microwave background radiation, and it is absurdly consistent, the same everywhere you look to one part in a hundred thousand. If we roll this Universe back to within 100,000 years after its creation in the big bang model, we realize that there is no way for different parts of the Universe to be in causal contact with each other to agree on what speed all the particles should be traveling. That is a major problem which cosmologist are aware of and discuss openly.

The flatness problem comes about by looking at the stability of the standard big bang theory. There is a math solution for how a pencil can be on a table and balanced on the point. The solution is unstable, and pencils fall over if you try to confirm this experimentally. The standard big bang model has the same problem. A bunch of matter either wants to collapse quickly or expand so fast nothing else in the neighborhood. Instead, the Universe went through a constant velocity stage, and keep going apart at a reasonable clip.

I do not like the labels for these two problems because they don't reflect the math issue clearly. The early cosmology of the Universe needs a new classic stable constant velocity solution involving gravity (editor's note: I cut and pasted that from the galaxy rotation profile above). A different problem, but I hope I can apply the same math solution because the problem when put in math terms is the same: stable constant velocity gravity solutions. Since momentum is the product of mass and velocity, changes in momentum caused by a force are the sum of a constant mass times changes in velocity (acceleration, mA) plus a constant velocity times a change in mass distribution. I call this the relativistic rocket science term, V c \frac{d m}{d R} \hat{V} which teaches gravity to work in a new direction, along V instead of R. I have even less idea how to apply thi product rule term to cosmology data than I do for a spiral galaxy. A few times I looked into doing the spiral galaxy problem numerically, and got too scared by the complexity to proceed.

The best known hypothesis to explain the stable constant velocity problem of the big bang is known as inflation. The idea is that at the start of the Universe, there was an epoch were the Universe grew at an extraordinary rate. The equations that govern this process do produce a mathematical stable solution that drives all the participating matter to the same speed. The hypothesis predicted that there would be a particular type of power series for the wee bit of variation seen the cosmic microwave background radiation spectrum. One should show respect for inflation because it was predictive of data gathered from surveys of the cosmic background radiation.

Scientists are aware of the significant problem with the inflation hypothesis: there is no cause. One could speculate that there is a type of matter that could lead to the inflation epoch, yet it would have properties that are not credible. The problem of cause was there in the 1980's when the hypothesis was first proposed, and remains to day. It is why I am skeptical about the proposal, one I feel comfortable making since I have a specific alternative.

As I said earlier, I have not see the video. A common complaint at the fridge of physics is the there an alternative way to explain the cosmic microwave background radiation and redshifts in spectrums. I will give you a cosmological experiment you can do in your own home that demonstrates the Universe is expanding.

One must completely embrace the idea of spacetime - avoid thinking of space separately from time! In my dinning room hangs "The Speed of Light According to René Magritte" (http://theworld.com/~sweetser/PopScience/speed/speed.html). At the bottom of the work, it reads: "Il n'y a pas d'espace sans temps, pas de temps sans expace" (translation: "There is not time without space, no space without time"). If the Universe uses quaternions as the fundamental system of accounting, there may be situations where the time or space time could be zero, but never a situation where the quaternion was missing a time or space part (zero being a value).

Take a look at your watch. If the hand or digits move, that is proof that spacetime is expanding. The watch is gathering information about spacetime. It so happens that the watch is not moving relative to the person wearing the watch, so all the change in the Universe appears as a change in time. This is a universal property: if the relative velocity between two parts of the Universe is zero, any change will appear as a change in time. The corillary is that if the relative velocity is large, the expansion of spacetime appears as an expansion in space.

In the entire beyond-comprehension-huge Universe, the only matter that travels through spacetime at the same speed as yourself is a thin crust of the Earth. Everything else has a relative velocity to you: we spin here once a day looking at the rest of the Universe. The Universe is a story of motion: once a year we go around the Sun, once every 250,000 our solar system goes around the Milky Way, our Milky Way travels at nearly the speed of light relative to high redshift galaxies.

I view relative velocity as a numerical measure of shared history. The Earth, the only object we can ever know that has zero relative velocity to us, has been together for a least four billion years. The atoms that make up you, me, and the planet made it through a supernova explosion. The high redshift galaxies, the ones traveling at nearly the speed of light, while we may have shared the big bang moment, that is all we share except for a few photons an hour delivered over space. We have no visitation rights for such galaxies only ephimeral signals that ultrasensitve telescopes can see (the human eye is blind to sources below a magnitude of 16).

We have much to learn about cosmology, and cosmologist are collecting the needed data as we speak. I hope I can provide a critical equation and apply it to real data someday.

Doug

[shmengie]

Perhaps I should have posted a the link to the torrent I originally downloaded:

http://thepiratebay.org/tor/4223510/Universe_The_Cosmology_Quest_-_Big_Bang_Big_Crap_%5BDocumentary%5D

It came in faster with a slightly smaller foot print. I'm inclined to repost the torrent with an even smaller foot print, but have yet to do such a thing.

re: your comments on cosmology are somewhat in line with what is said in the documentary. Basicaly, "The Universe, The Cosmology Quest" (http://www.universe-film.com/) series debunks the big bang theory. Halton Arp's research which does not fit with the "big bang" mainstream (http://www.haltonarp.com/articles). I found Arp's papers intriguing. Arp has, from what I gather, concludes redshift changes with the age of galaxies, and younger galaxies are being ejected from older ones. He makes a strong argument.

The Plasma Theory of the Universe (http://bigbangneverhappened.org/index1.htm), which is why I posted here postulates that electric charge in the original plasma formed the super structures of galaxies into filaments. Which sounds to me like EM & GR combined.

[shmengie]

Another methodological problem highlighted by Donald Scott is that most cosmologists do not understand or correctly apply the physics of plasma—electrically conducting gases that constitute nearly all the matter in the universe. (http://bigbangneverhappened.org/p17.htm)

[sweetser]

Hello:

I am taking the General Relativity course 8.06s at MIT's Professional Institute. It is great fun that I get to think and chat about physics for four days. It has been work the 4 vacation days and $2k for the class.

At lunch, one fellow said his biggest concern was with force. He pressed us to define it. I decided to just play it quiet, so he said it was either energy integrated over space or momentum integrated over time. Most discussion emphasize then energy integrated over time, while spending less effort on the impulse force. There is no reason for why one should be more important than the other except for accidents in the history of teaching physics.

This gentleman wanted to know why those two made up force. I looked at the problem with my quaternion hat stapled onto my head, realizing that force involves two quaternions: momenergy (E, Pc) and differential spacetime (dt/c, dR). Multiply and integrate these quaternions:

F = \int{(dt/c, dR)(E, Pc)} = \int{(E dt/c ~-~ c ~P.dR, E dR ~+~ P dt ~+~ c ~ dR \times P)}

What the gentleman did not mention - but was aware of - was the angular momentum. That is a great property of quaternions: the calculation makes it impossible to forget. An additional bonus is this equation calculates the 4-force at little extra cost. If a force is conservative, the scalar is zero. It is unclear to me how the c's effect the meaning.

Doug

[Ghost_of_PL]

Hello Doug,

The algebra you describe below is not the quaternions, that is to say it is not isomorphic to the quaternions. What you have is not a division algebra. For example: what is 1/(1+e_0) ?
1+e_0 has no inverse.

Hello:

Quaternions as events in spacetime physics are not the Clifford algebra CL(0, 2). Let me first explain what that bit of jargon means.

Clifford algebras were an attempt to generalize the math started by Hamilton, passing through Grassman. These algebras are independent of coordinates, and can be written in arbitrary dimensions. As algebras, it is not important that they are invertible, although that can happen.

The Clifford algebra CL(0, 2) is a multivector, meaning it has two parts. First there is the scalar, the 0 part, a pure number. The bivector is the second part. They use a wedge product, \wedge, a generalized cross product to form a 3-term bivector.

q(t, x_1, x_2, x_3) = t ~+~ x_1 e_2 \wedge e_3 ~+~ x_2 e_3 \wedge e_1 ~+~ x_3 e_1 \wedge e_2 \quad eq~1

Why use a wedge product? The reason is that wedge products are axial vectors, those that switch handedness in a mirror. The other sort is a polar vector. If one looks at an event in a mirror, this makes physical sense.

It is not worth the time to argue with math types who think they can define whatever they want, however they see fit. I have the perspective of a mathematical physicists, where any math definition can be cow-roped by physical meaning (I am a crude mathematical physics who uses rodeo analogies).

The unspoken assumption behind CL(0, 2) is that one should use a mirror on this event. That is like taking (t, x1, x2, x3) to (t, -x1, -x2, -x3). That sort of transformation does have the handedness needed for an axial vector. Yet it is easy enough to think of counter examples. What about a time reflection, where an event (t, x1, x2, x3) goes to (-t, x1, x2, x3)? This transformation does not change handedness, it is represented by a polar vector. Compare the two functions

mirror reflection: q -> q' = q*
time reflection: q -> q' = -q*

Since these are so close to each other as functions, I don't think one should take precedence over the other, as asserting one must use a bivector for the spatial part of a quaternion does.

The fundamental currency of the Universe is a bare event which stands alone in a vacuum, not in front of a mirror waving a left hand. General relativity has a message about basis vectors. Should we have a toy Universe with two events, the interval between these two events is found by taking the dot product of the difference between the two events:

(c dt, dx_1, dx_2, dx_3).(c dt, dx_1, dx_2, dx_3) = c^2 dt^2 ~+~ dx_1^2 (e_2 \wedge e_3).(e_2 \wedge e_3) ~+~ dx_2^2 (e_3 \wedge e_1).(e_3 \wedge e_1) ~+~ dx_3^2 (e_1 \wedge e_2).(e_1 \wedge e_2)

= c^2 dt^2 ~-~ dx_2^2 ~-~ dx_2^2 ~-~ dx_2^2 \quad eq~2

So far, so good. Now repeat the measurement in a toy Universe where nothing has been altered for the generators of the two events, but one has added a mass. A metric theory of gravity does not alter events, but does change the measure of events. In this setup, the differences are the same. The only thing that could change are the sizes of the basis vectors themselves. The problem here is that there is no basis vector associated with the dt2 term.

The deep message of special relativity is to treat time as we do space. In the geometric algebra approach, time is a scalar, whereas the space parts use two basis vectors as a bivector. Contrast this with the way I define a quaternion:

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

This is not a Clifford algebra because the basis vector e_0 commutes with the others. In the definition of a Clifford algebra, all the basis elements anti-commute. Physicists would call this a 4-vector, because one can add them or multiply it by a scalar, and it transforms like a 4-vector. It is this last phrase, on how it transforms under a Lorentz boost, that holds the magic. The goal of a great definition is to remove the magic and let the math speak for itself. Square the difference between two events using the quaternion definition as written in equation 3:

(c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3)
\quad eq~4

To be consistent with special relativity, we make the following map:

e_0^2 = +1
e_1^2 = -1
e_2^2 = -1
e_3^2 = -1 \quad eq~5-8

To be consistent with general relativity to first order tests of weak gravity fields for non-rotating, spherically symmetric sources, all we need is:

e_0 = exp(-G M/c^2 R)
1/e_0 = e_1 = e_2 = e_3 \quad eq~9-10

so according to the GEM proposal,

e_0^2 = exp(-2 G M/c^2 R)
e_1^2 = -exp(2 G M/c^2 R)
e_2^2 = -exp(2 G M/c^2 R)
e_3^2 = -exp(2 G M/c^2 R) \quad eq~11-15

I forget who said it, but one can only hope to find an important law in physics if one also finds an invariance principle. When there is a gravitational source, the terms in the 3-vector part of the square (2 dt dx e_0 e_1, 2 dt dy e_0 e_2, 2 dt dz e_0 e_3) remain invariant under the presence of such a gravitational field.

In summary, I see two flaws to the assertion that quaternions are the Clifford algebra CL(0, 2). First, while some quaternions are axial vectors, others are polar vectors. Second, the scalar needs to have a basis vector associated with it to work with metric theories of gravity. As far as I can tell, everyone working with geometric algebra in physics works with the assumption that the quaternions are faithfully represented by CL(0, 2).

Doug

[sweetser]

Hello Ghost_of_PL:

Great question! I had not looked at all the details, and your post got me to think about them.

There is hope that the algebra represented in equations 11-15 will be a division algebra since as M approaches zero, the algebra approaches the standard quaternion division algebra of equations 5-8.

To solve this problem, I fired up Mathematica and defined the quaternion using a 4x4 real matrix. This definition is identical to the standard one except for factors of Exp(G M/c2 (x2 + y2 + z2)1/2 (a minus factor on the t term).

In the limit of M->0, the inverse matrix is exactly the same as a Hamilton's quaternions. Those are defined for all possible values of t, x, y, and z because the divisor is the norm, t + x2 + y2 + z2, which is positive definite. The inverse is only undefined if t=x=y=0.

In the general case, the inverse looks like so:

\frac{1}{q(t,x,y,z)} = (t exp(\frac{G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -x ~ exp(\frac{G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -y ~exp(\frac{3 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -z ~exp(\frac{3 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm) \quad eq ~1

where: ~norm = t^2 ~+~ (x^2 ~+~ y^2 ~+~ z^2) exp(\frac{4 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}) \quad eq ~1

Again the norm is positive definite. With Hamilton's quaternions, the point set that must be excluded is zero for all four numbers. With the curved by mass quaternions, one must also exclude where x=y=z=0, which is the entire real line. This calls for a refinement of the definition of the algebra in equation 11-15, that should R=0, then the algebra becomes the standard Hamilton case, equations 5-8.

To answer your question specifically, the inverse in spacetime curved by mass of (1, 1, 0, 0) is:

\frac{1}{(1, 1, 0, 0)} = (\frac{e^{\frac{G M}{c^2}}}{1 ~+~ e^{\frac{4 G M}{c^2}}},-\frac{e^{\frac{3 G M}{c^2}}}{1~+~ e^{\frac{4 G M}{c^2}}}, 0, 0) \quad eq ~ 2

Let's do a sanity check at least for the first term. If we hope to get a 1 out of multiplying (1, 1, 0, 0) by its inverse, we have to get out a 1~+~ e^{\frac{4 G M}{c^2}}, which shows up in the denominator of the inverse. For the time terms, there will be a positive and a negative exponent that cancel each other out, yielding a 1. From the x terms, we get a positive exponent plus an exponent times 3 which combine for the exponent to the fourth power. Looks good. If anyone is interested, the notebook is here (http://theworld.com/~sweetser/quaternions/notebooks/curved_q.nb).

By refining the definition to cover the case where R=0, I hope I have shown the curved basis quaternions are a division algebra isomorphic to the flat spacetime quaternions defined by Hamilton.

Thanks for the question,
Doug

[Ghost_of_PL]

Hi Doug,
Mathemaica is a wonderful tool but I sometimes think too much of it can be the crutch that cripples so I’m going to use good old fashioned algebra. You have stated (e_0)^2=1, correct? Well then (1+ e_0)(1- e_0)=0 since we have 1^2 – e_0 + e_0 – (e_0)^2 =0. That is to say 1+e_0 and 1-e_0 are zero divisors and have no inverse. Taking the limit as M->0 won’t get us to the quaternion’s so there’s an inconsistency somewhere.

Hello Ghost_of_PL:

Great question! I had not looked at all the details, and your post got me to think about them.

There is hope that the algebra represented in equations 11-15 will be a division algebra since as M approaches zero, the algebra approaches the standard quaternion division algebra of equations 5-8.

To solve this problem, I fired up Mathematica and defined the quaternion using a 4x4 real matrix. This definition is identical to the standard one except for factors of Exp(G M/c2 (x2 + y2 + z2)1/2 (a minus factor on the t term).

In the limit of M->0, the inverse matrix is exactly the same as a Hamilton's quaternions. Those are defined for all possible values of t, x, y, and z because the divisor is the norm, t + x2 + y2 + z2, which is positive definite. The inverse is only undefined if t=x=y=0.

In the general case, the inverse looks like so:

\frac{1}{q(t,x,y,z)} = (t exp(\frac{G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -x ~ exp(\frac{G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -y ~exp(\frac{3 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm, -z ~exp(\frac{3 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}/norm) \quad eq ~1

where: ~norm = t^2 ~+~ (x^2 ~+~ y^2 ~+~ z^2) exp(\frac{4 G M}{c^2 \sqrt{x^2 + y^2 + z^2}}) \quad eq ~1

Again the norm is positive definite. With Hamilton's quaternions, the point set that must be excluded is zero for all four numbers. With the curved by mass quaternions, one must also exclude where x=y=z=0, which is the entire real line. This calls for a refinement of the definition of the algebra in equation 11-15, that should R=0, then the algebra becomes the standard Hamilton case, equations 5-8.

To answer your question specifically, the inverse in spacetime curved by mass of (1, 1, 0, 0) is:

\frac{1}{(1, 1, 0, 0)} = (\frac{e^{\frac{G M}{c^2}}}{1 ~+~ e^{\frac{4 G M}{c^2}}},-\frac{e^{\frac{3 G M}{c^2}}}{1~+~ e^{\frac{4 G M}{c^2}}}, 0, 0) \quad eq ~ 2

Let's do a sanity check at least for the first term. If we hope to get a 1 out of multiplying (1, 1, 0, 0) by its inverse, we have to get out a 1~+~ e^{\frac{4 G M}{c^2}}, which shows up in the denominator of the inverse. For the time terms, there will be a positive and a negative exponent that cancel each other out, yielding a 1. From the x terms, we get a positive exponent plus an exponent times 3 which combine for the exponent to the fourth power. Looks good. If anyone is interested, the notebook is here (http://theworld.com/~sweetser/quaternions/notebooks/curved_q.nb).

By refining the definition to cover the case where R=0, I hope I have shown the curved basis quaternions are a division algebra isomorphic to the flat spacetime quaternions defined by Hamilton.

Thanks for the question,
Doug

[sweetser]

Hello Ghost_of_PL:

Good old fashioned algebra rocks! The only way I really learn something is with paper and pencil (so I can erase a lot). I did not understand what your issue was, but now I have a better grasp of it. I did define what I meant by a quaternion in this expression:

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

Notices that every term has a basis vector, no exceptions. With equation 3, there is never a pure number like the 1 in 1 + e_0. One can work with 1 e_0 and 1 e_0 + 1 e_0, both of which have inverses. So your question looks poorly defined, a way of saying it does not work nicely with equation 3. If I were to allow a term such as 1 + e_0, then there would be 5 numbers that could be put into a quaternion. Oops.

The reason I needed Mathematica was to answer a different problem because I misunderstood your question, namely finding the inverse of a quaternion with the basis vectors in equations 5-8. That was a fun calculation anyway :-)

Doug

[Ghost_of_PL]

Hello Doug,
The problem is equation 3 can not mitigate the fact that you introduced basis elements that square to plus or minus “1” . (e_0)^2= 1 So there you have it, “one”, in all its naked glory. Either the number one is a legitimate element, or our “quaternions” are not closed under the operation of multiplication and therefore not an algebra. Maybe what you really want is e_0 as an idempotent, something along the lines of: (e_0)^2= e_0 , (e_0)(e_1)= e_1, (e_0)(e_2)= e_2 , (e_1)(e_2)= e_3 and so on…

Hello Ghost_of_PL:

Good old fashioned algebra rocks! The only way I really learn something is with paper and pencil (so I can erase a lot). I did not understand what your issue was, but now I have a better grasp of it. I did define what I meant by a quaternion in this expression:

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

Notices that every term has a basis vector, no exceptions. With equation 3, there is never a pure number like the 1 in 1 + e_0. One can work with 1 e_0 and 1 e_0 + 1 e_0, both of which have inverses. So your question looks poorly defined, a way of saying it does not work nicely with equation 3. If I were to allow a term such as 1 + e_0, then there would be 5 numbers that could be put into a quaternion. Oops.

The reason I needed Mathematica was to answer a different problem because I misunderstood your question, namely finding the inverse of a quaternion with the basis vectors in equations 5-8. That was a fun calculation anyway :-)

Doug

[sweetser]

Hello Ghost_of_PL:

There are only 4 slots in a quaternion, something that is obvious from the 4x4 real matrix representation used in the Mathematica notebook. Looking back, I not only used a variation on a common definition (which would normally have i, j, k for e_1, e_2, e_3, no e_0):

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

Sometimes people have complained about the meaning of the plus sign since this is not like the sum of real numbers. I also wrote out explicitly what is meant by squaring a quaternion:

(c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3)\quad eq~4

I did this so I could work with basis vectors whose length was not equal to unity, which is why I am not working with idempotent such that (e_0)^2= e_0. Equations 5-8 (unitary basis vectors) and 11-15 (non-unitary basis vectors) are applied in the context of equation 4. In curved spacetime, e_0^2 = exp(-2 G M/c^2 R) and it goes in the first of the four slots of a quaternion. A quaternion sum of a^2 ~+~ b \ne 0 would always have an inverse in flat or curved spacetime.

There certainly is a chance that my way of defining an algebra in flat spacetime with equations 3-8, and in curved spacetime with 3, 4, 11-15 does not work with people more formally trained define an algebra. If that happens to be the case, then please use the 4x4 real matrix representation which carries exactly the same information:


\[ \left[ \begin{array}{cccc}
e_0 t & -e_1 x & -e_2 y & -e_3 z\\\
e_1 x & e_0 t & -e_3 z & e_2 y \\\
e_2 y & e_3 z & e_0 t & -e_1 x \\\
e_3 z & -e_2 y & e_1 x & e_0 t \end{array} \right]\]

In flat spacetime, all the e's have a magnitude equal to one. If the additive identity (zero) is excluded, this matrix always has an inverse.

Doug

[Ghost_of_PL]

Hi Doug,
most likely this is my last post on this subject:
Hello Ghost_of_PL:

There are only 4 slots in a quaternion, something that is obvious from the 4x4 real matrix representation used in the Mathematica notebook. Looking back, I not only used a variation on a common definition (which would normally have i, j, k for e_1, e_2, e_3, no e_0):

q(t, x_1, x_2, x_3) = t e_0 + x_1 e_1 + x_2 e_2 + x_3 e_3 \quad eq~3

You have defined at least 5 slots since you defined e_0^2 =1


Sometimes people have complained about the meaning of the plus sign since this is not like the sum of real numbers. I also wrote out explicitly what is meant by squaring a quaternion:

(c dt, dx_1, dx_2, dx_3)^2 = (c^2 dt^2~ e_0^2 ~+~ dx_1^2 ~e_1^2 ~+~ dx_2^2 ~e_2^2 ~+~ dx_3^2 ~e_3^2, 2 c ~dt ~dx_1~ e_0 ~e_1, 2 c ~dt ~dx_2 ~e_0 ~e_2, 2 c ~dt ~dx_3 ~e_0 ~e_3)\quad eq~4

Do you not notice eq. 4 violates:
Notices that every term has a basis vector, no exceptions. With equation 3, there is never a pure number like the 1 in 1 + e_0. One can work with 1 e_0 and 1 e_0 + 1 e_0, both of which have inverses. So your question looks poorly defined, a way of saying it does not work nicely with equation 3. If I were to allow a term such as 1 + e_0, then there would be 5 numbers that could be put into a quaternion. Oops.
The first term of eq.4 is a pure number without a factor of e_0 and violates "that every term has a basis vector, no exceptions" since c^2dt^2e_0^2 +dx1^2e_1^2 +dx2^2e_2^2 +dx3^2 = c^2dt^2- dx1^2- dx2^2 -dx3^2 . The right hand side of this last equation has no factor of e_0. Your matrix representation can fare no better since it also admits stand alone real numbers.

[sweetser]

Hello Ghost_of_PL:

> most likely this is my last post on this subject

This discussion is becoming repetitive, so that is good news for general readers of this thread.

Your weakest statement is this assertion:

> Your matrix representation can fare no better since it also admits stand alone real numbers.

What behaves no different from the real numbers is this diagonal matrix:

\[ \left[ \begin{array}{cccc}
e_0 t & 0 & 0 & 0\\\
0 & e_0 t & 0 & 0 \\\
0 & 0 & e_0 t & 0 \\\
0 & 0 & 0 & e_0 t \end{array} \right]\] \quad eq ~1

All real numbers are formally a subgroup of quaternions, so there is no way a real number can stand alone from the mathematical field of quaternions.

What behaves no different from anything that can ever be done with complex numbers or complex analysis are the following three matrices:

\[ \left[ \begin{array}{cccc}
e_0 t & -e_1 x & 0 & 0\\\
e_1 x & e_0 t & 0 & 0 \\\
0 & 0 & e_0 t & -e_1 x \\\
0 & 0 & e_1 x & e_0 t \end{array} \right]\]

\[ \left[ \begin{array}{cccc}
e_0 t & 0 & -e_2 y & 0\\\
0 & e_0 t & 0 & e_2 y \\\
e_2 y & 0 & e_0 t & 0 \\\
0 & -e_2 y & 0 & e_0 t \end{array} \right]\]

\[ \left[ \begin{array}{cccc}
e_0 t & 0 & 0 & -e_3 z\\\
0 & e_0 t & -e_3 z & 0 \\\
0 & e_3 z & e_0 t & 0 \\\
e_3 z & 0 & 0 & e_0 t \end{array} \right]\] \quad eq ~2-4

Quaternions have as subgroups equations 1-4, and can be viewed as the union of three complex numbers that share the same real value.

I remain open to the idea that as an exercise in algebra, my notation could be wrong. The most likely thing would be I have to consider keeping all the basis vectors, so in flat spacetime e02 = +1 e02, whereas in curved spacetime e02 = +exp(-2 GM/c2 R) e02. Then one would have to figure out an efficient way to say you can add an e02 to an e12 to an e0. The way I tried to avoid this complexity was to say the numbers mapped back to equations 3 and 4 of the original post (#574). Instead, equations 5-8 were removed from their context to create the stand-alone issue. The real numbers are a subgroup of quaternions, so are of the form (a, 0, 0, 0). A real number cannot stand alone from quaternions, ever. The relation of real numbers to quaternions may not be understood or used widely in practice, but ignoring rules is not a defense.

Perhaps I can say it this way: the equations that no longer have an e02 explicitly have it implicitly up to a cofactor of +1 in flat spacetime, or +exp(-2 GM/c2 R) in curved spacetime. It would take a well-trained math wonk to know how to say these things correctly.

Doug

[sweetser]

Hello:

Personal stuff:
I spent the month of July drifting. I was bummed that neither the trip to Brazil nor the MIT GR course created lasting dialogs with the participants. Although I am familiar with not connecting, both events involved interactions over a week long period. I also spent much time thinking about cell phones. I almost got an iPhone, but that would require sending too many dollars to AT&T over time, so bought a Nokia N82. Those Nokia folks sure know how to make a cell phone.

Physics stuff:
For more than 5 years, I have wanted to do a calculation, but never got the courage up to work all the way through it. The calculation involves figuring out the momentum profile of a thin disk galaxy. We know three things about disk galaxies: the distribution of the visible mass, the velocity of such mass, and that Newton's law is not consistent with these two observations. General relativity is not necessary in such a system since the speeds and mass densities are low. The angular momentum must balance the force. To explain this problem, the most popular hypothesis is dark matter. I call that hypothesis "stuffing the matter box". We know an absurd amount of things about particles, yet do not have a reasonable candidate for dark matter. Given that the dark matter needed to fix problems with gravity exceeds all known matter, this is a huge missing gap in our knowledge, or an indication we should seek an alternative, testable hypothesis.

I have such a hypothesis. What a force does is change momentum. By the product rule, that means the effect of a force is the sum of a constant mass times a change in velocity plus a constant velocity times a change in mass. The former term is the famous Newtonian mA and is the only term used by folks modeling the momentum profiles of galaxies. The latter term is used in rocket science, because as the rocket moves in time, there is less mass in the rocket itself. The rocket term accounts for this change in mass with respect to time.

Galaxies do not change rapidly in time, so it is reasonable to ignore the change in mass with respect to time. The mass of a galaxy does change with respect to space. The mass density of thin disk galaxies has two parts. One is spherical, falling off approximately like R1/4. The disk has a mass that often falls off from the center as an exponential decay. The velocity of matter of the disk stays constant, while the mass drops off exponentially with increasing distance from the center. As a word description, that says to me we are dealing with a term of the form V c dm/dR, what I call the relativistic rocket effect.

The relativistic rocket effect, as a way to express a change in momentum, has to my knowledge never been discussed, let alone seen. My guess is that the factor of c needed to get the units right means that it is only seen for systems where the mass is spread over areas of space as big as a galaxy or larger. For such grand systems, most of the force of gravity goes into dictating where mass should be on that grand scale, and not into making mass move faster.

One of the nice or scary things about the relativistic rocket hypothesis is that it has no free parameters. The equation either matches the data for the visible mass distribution profile with the velocity profile, or the hypothesis is wrong. I have no parameters to "tweak", the way good science proceeds.

The standard approach to figuring out the rotation profile of a thin disk galaxy uses elliptical integrals. That is a math topic I do not understand. I do not have peers who could walk me through that bit of mathematics.

I must do something simpler, more direct. I could take a galaxy, chop it up into lots of little bits, figure out the effect of gravity on two little bits - with and without the relativistic rocket effect - and add it all up. If I were to do the calculation correctly, I would see the problem with Newton's law alone, and then how the velocity profile goes dead flat if I take into account the relativistic rocket effect. If the relativistic rocket effect does not change the velocity profile much from Newton's law alone, then the hypothesis is wrong.

I work from a position of zero self-confidence, but do embrace the process of science. It is most important to ask a specific question clearly. I will be looking at the momentum profile of one galaxy, NGC 3198, because astronomers have figured out an approximate exponential function for the visible mass distribution profile.

Here is the outline (http://picasaweb.google.com/dougsweetser/GalaxyDiskMomemtumProfile/photo#5229379345512696162) for the work I hope to accomplish over the next few months:
http://picasaweb.google.com/dougsweetser/GalaxyDiskMomemtumProfile/photo#5229379345512696162

Draw it
Label Everything
Write out all the equations
Write out all the kg, m, s units and conversions
mo_profile - a program in Perl to do the calculation

List options, input, and output
List modules and variables
Write test programs to check area, mass calculations

Experiment: Given the mass distribution profile for NGC 3198

Does the pure Newtonian term get the correct maximum velocity, but have the velocity fall with increasing distance from the center?
Does the Newtonian term plus the relativistic velocity term get the correct maximum velocity and a flat velocity curve with increasing radius?



As I make steps along this master plan, I will post the results here. The plan was drawn up on Thursday, and my N82 was able to upload it to the web. I spent Saturday and Sunday drawing pictures of the galaxies and labeling all the parts, so by mid-week I may be able to clear the next hurdle. At least the goals are clear.

Doug

[sweetser]

Hello:

Here is a picture of the plan I have for calculating the rotation profile of a galaxy:

http://theworld.com/~sweetser/Images/mo_profile_plan-600.jpg

I had included a higher resolution image in the previous post, but apparently it was too large to be accepted here (600x450 pixels, 57k works).

Doug

[Varnick]

Doug,

Your grasp of physics is, frankly, much beyond mine, but I would disagree on one point in you previous "Wordy" post. It is of no real consequence to your calculation, but SUSY particles are considered excellent candidates for dark matter, as far as I am informed. Would you disagree, or was it just something you missed? Either way, this calculation appears both impressive and immersive, I look forward to seeing it finished.

V

[sweetser]

Hello Vanick:

Supersymmetric (SUSY) proposals have been with us for over 30 years. During that time, every time a new more powerful collider has come on line, part of the justification was the search for the SUSY particles. The same is claimed for the Large Hadron Collider: it might find the first evidence for supersymmetric. Based on the previous decades of effort, I will not bet on it, although there is 8 billion dollars on the line.

As a hypothesis, supersymmetry has two failings in my eyes. The first is that one of the motivating factors behind the proposal is to explain aspects of the standard model. One big mystery is why there should be three groups, U(1), SU(2) and SU(3) - and why those groups in particular. In my reading of this article on grand unified theories (http://pdg.lbl.gov/2008/reviews/gutsrpp.pdf), using a group like SO(10) does not answer that question. Oops.

Animating physics with quaternions does give a visual reason why Nature is constrained to the symmetries of U(1), SU(2), SU(3) and Diff(M) for EM, the weak force, the strong force, and gravity respectively. If you go though the math of an expanding and contracting unit sphere with quaternions, these are the groups that come into play. If an observer is at the origin and sees an event, that event must be a member of these groups. It is less that SUSY is wrong than I have a concrete counter proposal. I too have limitations in my math skills, and have no idea how to pitch a model that is smaller than the standard model to particle physicists. The smaller nature of the quaternion proposal might have an advantage: it could justify containment, the observation that there are no free gluons anywhere in the Universe. Formalizing that line of logic is beyond my reach. A search on YouTube for "standard model" should have my blue & yellow colored animation.

That is the backdrop. Now to your specific question about SUSY and dark matter (http://pdg.lbl.gov/2007/reviews/darkmatrpp.pdf). It is not enough for a light SUSY particle to only interact with gravity, to have the properties of a dark matter particle. That particle would have to be distributed in space around disk galaxies with a distribution that would lead to a momentum profile where the velocity was constant, but the visible mass falls off exponentially. How Nature using only gravity could get the distribution of dark matter such that it is consistent with what we see is a core issue that is unexplained. The distribution of dark matter must be different for clusters of galaxies. Getting non-interacting matter except for gravity into the right place looks like a tough problem.

Doug

[sweetser]

Hello:

To use Newton's law of gravity, masses and distances must be known. For a disk, one cannot treat it as an effective point mass. Instead, the galaxy needs to be chopped up into little bits, and add up each contribution. My plan is to do this discretely, but it may be possible to do things continuously. I will have to see.

Here is a picture of how I plan to slice up a galaxy:

http://theworld.com/~sweetser/Images/galaxy-600.jpg

There are quite a few R's:

Rmax - the maximum radial distance
Rpi - the radius to the passive mass, i steps from the origin
Raj - the radius to the active mass, j steps from the origin
Rij - the distance between the active and passive masses.
Rr - the portion of Rij in the direction of the radius
Rv - the portion of Rij perpendicular to the radius, parallel to V

The passive mass is the one that appears on both sides of Newton's force law, and always gets cancelled. With the relativistic rocket effect, that cancelation might not happen (since the mass distribution is an exponential, and we are taking the derivative of an exponential which returns the exponential times the derivative of the exponent, it might drop out in this special case). There are three forces that point in these directions: Fij, Fr, and Fv. For the pure Newtonian calculation, only Fr will be needed. The rocket effect uses Fv.

I have partially revealed how I will be slicing up the galaxy: in i steps with the passive gravitational mass and in j steps for the active gravitational mass. The active mass also needs to revolve around the origin, which will be done in k steps. The differential active mass is dMa, and the differential passive mass is dmp.

The force term will be the sum of these differential terms:

F_g = \sum_{i=1}^n \sum_{j=1}^n \sum_{k=1}^n (dFr + dFv) \quad eq~1

For the Newtonian case, the Fv is not included. Can all these terms be written in terms of i, j, k, and Rmax? It is not simple, but a weekend of doodling created this picture of the terms involved:

http://theworld.com/~sweetser/Images/draw_it-600.jpg

The first three terms are about the progression along i, j, and k:

Rpi = \frac{i}{n} ~ Rmax

Raj = \frac{j}{n} ~ Rmax

B = 2 \pi \frac{k}{n} \quad eq ~ 2-4

Although we don't use it everyday, Rij is calculated using the law of cosines, a variation on the Pythagorean theorem with a cosine to account for a non-right angle triangle:

Rij = \frac{Rmax}{n} \sqrt{i^2 + j^2 - 2 i j ~Cos(2 \pi \frac{k}{n})} \quad eq ~5

Break down Rij into that pointing along Rv - a simple application of the definition of a sine, and Rr, which is the venerable Pythagorean theorem at work:

Rv = Rij ~Sin (2 \pi \frac{k}{n})

Rr = Rij \sqrt{1 - \frac{Raj^2}{Rij^2} ~Sin^2 (2 \pi \frac{k}{n})}\quad eq ~6, ~7

There are corresponding equations for the forces that are proportional to these:

dFv = dFij ~\frac{Rai}{Rij} ~Sin (2 \pi \frac{k}{n})

dFr = dFij \sqrt{1 - \frac{Raj^2}{Rij^2} ~Sin^2 (2 \pi \frac{k}{n})} \quad eq ~8,~9

With this much algebra going on, it is good to think of quality controls. The distance Rij should only equal 2 Rmax for one set of values of i, j, and k. The differential forces dFij, dFr, and dFv should satisfy the Pythagorean theorem.

The mass for disk galaxies is often given in terms of mass per unit area. As such, determine what a differential area is. The area of the complete galaxy is pi Rmax^2. This is being sliced into n pieces, so the area of a sector is pi Rmax^2/n. As we step from i-1 to i, how much area is there?

dAi = \pi ~\frac{Rmax}{n} ~ (\frac{i^2}{n^2} ~-~ \frac{(i ~-~ 1)^2}{n^2}) = \pi ~ \frac{Rmax}{n^3} ~ (2 i ~-~ 1) \quad eq ~10

One problem with this approach is that it samples the galaxy near the core more densely than the outer regions, where i is greater. That may be acceptable since the mass distribution is exponential, so most of the mass comes from the center.

I have skimmed from a paper that the number of solar masses/parsec2 is 37 exp (R'/2.23'). Combine the mass/area with the differential area to get the differential masses:

dmp = 37 \pi ~\frac{Rmax}{n^3} ~ (2 i ~-~ 1) ~ exp (Rpi'/2.23')

dMa = 37 \pi ~\frac{Rmax}{n^3} ~ (2 j ~-~ 1) ~ exp (Raj'/2.23')\quad eq ~11, ~12

These are the players needed to calculate the force: the distance Rij and the two masses, dmp and dMa. Yoda has said, simple it is not, the way of relativistic rocket astrophysics.

Doug

[sweetser]

Hello:

Symmetry is useful to think about because it can pinpoint what can be ignored. Feynman emphasized looking for things that add up, knowing we can ignore things that cancel. This is what I drew on my board this morning:

http://theworld.com/~sweetser/Images/opposing_forces-600.jpg

The two Fv's point in opposite directions. Oops, those will never add up and amount to anything. Therefore, none of my calculations should involve Fv, which appeared in the previous post #598, eq 1 and 8.

So is this galaxy momentum profile calculation over? No, because there is a change in momentum that is being omitted from the standard calculation. Recall that energy is force times distance. One only does work in the direction of motion, or the cosine of the angle between two 3-vectors. In the drawing, that would be Fr dotted to Rr. That is what goes into a classical calculation. Fr dotted with Rv is zero. My objection is that our analysis is not complete. Something which has the same units as work, but is not a scalar like energy, is the sine of a force and a distance vector, or Fij ~ Rv ~Sin(\theta). That will not be zero. It is the 3-momentum times c. This is the torque force term that pairs with the relativistic rocket effect.

Doug

[sweetser]

Hello:

All forces must have a cause and an effect. For a disk galaxy, there is the Newtonian force, Fij Rr Cos(\theta), whose effect is the centripital motion. At the current time, this is all that is used to calculate the momentum profiles of disk galaxies. This is not enough, something is missing. The leading hypothesis is dark matter. I am proposing that the omission is the gravitational force in a different direction, Fij Rr Sin(\theta), causes the new relativistic rocket effect, which helps determine where masses are located in space. This would eliminate the need for dark matter. It is important that I get through all the details of this calculation to test the hypothesis.

The proposal at this early stage has consequences for cosmology. As one moves to larger scales, the effect of gravity becomes more about where mass is distributed than about making things move faster or slower. At the big bang - the farthest distance we can go in spacetime, the cosmic background radiation data indicates that gravity was exclusively about the relativistic rocket effect, where the velocity is constant, at least to five significant digits. No matter was changing its velocity relative to other matter at the time of recombination. As the Univserse has aged, it may be that the balance has shifted a bit towards the mA term from the relativistic rocket effect. If so, then the net effect of gravity would be the same, but the Universe would appear to be accelerating more instead of all travelling uniformily. This would eliminate the need for dark energy.

If I could only quantify such a claim...

Doug

[sweetser]

Hello:

In this post we will think about momentum for a disk bicycle wheel and a disk galaxy. The similarities between the two systems may give support to the relativistic rocket effect discussed here.

A disk has all of its mass in a plane. Both of the disks in question spin. The bicycle wheel is a rigid body, while the galaxy is not. All parts of the wheel rotate at the same rotational velocity, but travel at a different tangential velocity which depends on how far out from the center a point is. For disk galaxies, the stars at the center of a galaxy travel slowly, yet quickly reach a maximum velocity outside the core. From there on, the stars, and then the helium gas, rotate at the same tangential velocity.

If one applies a force to the axis of a spinning bicycle wheel, the wheel will move a distance. This is an easy problem. It is just like applying a force to a brick, which moves, so the energy is the distance times the force.

Now apply exactly the same amount of force to the spinning bicycle disk, but somewhere along the rim. The problem gets much messier! The axle might move a bit - so you can do the force times distance calculation for that energy - but there will also be a wobble. I don't know how to deal mathematically with the energy that goes into wobbling, sorry. I doubt many people are confident about the subject since it involves torques and all that jazz. My big picture view is that if I apply the same amount of energy to moving the bicycle wheel along the axis as to somewhere on the disk, then the disk must distribute that energy between moving the center of mass and the energy of wobbling. Energy must be conserved.

Now think about a disk galaxy. The mass for disk galaxies has two components, a spherical one, and the disk, whose visible mass drops off exponentially with increasing distance from the center. That means the vast majority of mass can be viewed as being central: due to spherical symmetry and the exponential decay of the disk.

Newton's law of gravity has a cause, the inverse square attraction of gravity, and an effect, the centripetal force that keeps things moving in toward the center:

-\frac{G M m}{R^2} ~\hat{R} = m~\frac{V \cdot V}{R} ~\hat{R} \quad eq ~ 1

What I would like to point out is both cause and effect point in exactly the same direction, as they must, along the radius, \hat{R}. This is why modeling of galaxies by Newton's law works near the core, getting the right maximal terminal velocity. For a bicycle wheel, the closer to the axis you apply the force, the less you need to worry about all the messy torque stuff.

On physical grounds, I cannot accept it as reasonable that gravity does not exert a torque force. Gravity would somehow have to favor doing work along the radial direction. One might argue that perhaps all the torques cancel out nicely. Please recall the bicycle wheel. That problem is hard, there is no cancellation going on for the rigid body. There is no reason to expect the disk galaxy to be so clean. Yet in all my readings, I have never seen anyone discuss gravity working in any direction other than \hat{R}. I have a paper from 1963 where Alar Toomre did the calculation of the momentum profile for highly flattened galaxies the right way (using Bessel integrals, too technical for me, but one of the papers that started this area of study). It is clear he was only dealing with force in the radial direction.

Gravity from a system with its mass distributed in a disc must have torque forces that need to be taken into account. That would make the cause term point in a new direction, perpendicular to the radius, or \hat{V}. The effect of gravity must point in this same direction. Dimensional analysis leads to this proposal:

-\frac{G M m}{R^2} ~\hat{V} = c~ V~ \frac{d m}{d R} ~\hat{V} \quad eq ~ 2

Since the torque force points in the V direction, a V can only appear once, and naturally points along \hat{V}. As discussed in earlier posts, this term is a direct result of viewing a force as a change in momentum, and using the product rule, which has a constant V and a changing m.

I call myself a member of the "ultra-conservative fringe". Folks as bright as Alar Toomre frighten me. I don't think it is reasonable to say someone with such skills is wrong. On rare occasions, it may be OK to document something they did not account for, the error by omission. All the complexity of a torque force for gravity of a disc galaxy have been omitted. Based on energy conservation, I think that omission must be corrected.

It is reasonable to ignore this proposal since I have yet to go numerical. It might be the effect is trivial. Yet this problem works in the trivial end of the force spectrum. The accelerations are on the order of 10-10 m/s2! They are measuring how helium II gas moves way out from the core. One needs to be complete. I cannot defend the models used today.

Doug

Note added in proof: Notice in equations 1 and 2, the cause term has in the denominator R2, which is equal to the dot product of R with itself, and thus has no direction. Newton tacked on the radial direction part, as has every one of his students since. There is no "natural" directionality to the universal law of gravitation. The terms on the effect side both get there directions from single vectors, the dividing by R of the directionless V dotted to V for the centripetal force, and the V in the relativistic rocket effect. Nice.

[Mentz114]

Hello:

I am taking the General Relativity course 8.06s at MIT's Professional Institute. It is great fun that I get to think and chat about physics for four days. It has been work the 4 vacation days and $2k for the class.

At lunch, one fellow said his biggest concern was with force. He pressed us to define it. I decided to just play it quiet, so he said it was either energy integrated over space or momentum integrated over time. Most discussion emphasize then energy integrated over time, while spending less effort on the impulse force. There is no reason for why one should be more important than the other except for accidents in the history of teaching physics.


Hi Doug,

I don't reconise the gentleman's definitions. Isn't force dP/dt or dE/dx ?

Four days seems rather short for the average professional to learn GR.

Regards,
Lut

[sweetser]

Hello Lut:

I have used 3 definitions for force. One is the derivative of the 4-momentum with respect to the interval tau. I saw that in a discussion of the Lorentz force of EM in Jackson, the end of chapter 11. This is the dP/dt and dE/dx one. The second definition is based on varying a Lagrange density with respect to velocity. Again that was used to derive the Lorentz force, in Landau and Lifshitz "Classical Theory of Fields". The third definition is the integral definition discussed in post #585, p37. All three give the same answer for force, love that consistency.

The class was a survey of issues in gravity and cosmology, not a professional introduction to GR. We got a bit deeper than Scientific American, using a few equations. This thread is more "equation dense" than the course was. A fun aspect was being able to spend the days thinking of physics. I miss that spirit, and have been adrift since then.

Doug

[Mentz114]

Of course, any expression with the right dimensions can be interpreted as force. My view was a bit telescopic (?).

I found gravity as a coordinate acceleration independent of mass, by letting inertial mass depend on a field. It doesn't quite work but it was fun doing it.

Lut

[dtfroedge]

Hello:

This would eliminate the need for dark energy.




Doug:

Feng and Gallo have been working on the luninocity density relation for a solution to the dark matter issue. You may find their paper useful.

"Galactic Rotation Described with Thin-Disk Gravitational Model"
http://arxiv.org/abs/0803.0556

DT

[Max™]

Hey Sweets, I was reading this and noticed something which seemed familiar from how I was considering the problem.

Would you object to saying your interpretation of Gravity is more akin to a "stretching" of Space in the direction of the body in question, and EM could be considered a "stretch and a twist" in the direction of the charge orientation?

I didn't come into physics from the mathematics side, but from the thinking about explanations for processes which can be converted into mathematics side.

It could just be that I'm reading what you're describing mathematically wrong, but when I model it in my head, it looks like the explanation I've been developing from the other "direction", if you will.

[sweetser]

Hello Max:

I do have a word description about these two forces that is based on looking at the equations and solutions for quite some time. Here goes...

Stuff in Nature is pretty well isolated from each other after a separation lasting 13 billion years. In an ideal isolationist Universe, nothing would interact with anything else. The problem is that there is other stuff in the Universe. So the math wonk question is what is the least amount of interaction stuff can do with each other?

The answer is a 4D wave equation, with very little waving going on. If there was zero waving, the Minkowski metric would rule all. It does not, although it is darn close. By darn close, I mean the variations happen in parts per million for some systems, parts per hundreds of millions for others.

If we think about gravity versus EM, that has to do with the field strength tensor of gravity versus the field strength tensor of EM. The one for gravity is the average amount of change in a potential. The one for EM is the deviation from the average amount of change in a potential. In baseball terms, gravity is about the batting average, while EM is the deviation from the average - is the batter steady or does he have hot and cold streaks. The average is a different measure than the deviation, yet both are measured in terms of hits because they both are about hits.

If the average is zero, there is no deviation. In Nature, there are no charged particles that have a mass of zero.

As far as your specific effort, I would make the same critique I make of professionals: you cannot say something about space without saying something about time. To my ear, this common habit of taking about space sounds awful. With this correction in mind, is gravity a stretching of spacetime in the direction of the body in question? Even said this way, the "direction of the body" sounds like it wants to only deal with space. It is hard to work consistently with spacetime! Our language has a time separate from space bias built in! If I gave you an assignment, it would be to work on spotting those biases and avoid them.

Good luck in your studies, and good luck to the Red Sox,
Doug

[Max™]

Well, I say space, but I mean space-time.

The two are very intertwined in my head these days. Relativistic overdose, if you will.

Stretching of space when I say it, implies acceleration through a three space direction, and deceleration through the time direction.

Both bodies are moving through time, so I have a nasty habit of assuming that to be implied. Been working on it.

[sweetser]

Hello:

Last weekend, I went to the Fall APS/AAPT meeting at the University of Massachusetts, Boston. They have a lovely campus, right next to the JFK library on the water. It was a sunny and warm day thanks to global warming.

The first speaker was also the second, third, and fourth speaker. I hate when people game the system and claim they are working with four different people so they can ramble on about their crap. And it was crap. He claimed that rooms in the Egyptian pyramids were useful for neutrino detection. At one point, he was taking about how he felt he could tell the intent of moose. I almost lost it at that point, but I showed my self-control.

There were two talks that were serious. One concerned the wave function. Should we consider it to be real only if the result of an experiment is obtained? That is the sort of thing one could argue over in a bar, so was worth the time. The second talk solidly refuted the claim of people who imagined a pattern in radioactive decay.

My talk went well, I fit it to the time slot. The longer version (a half hour) is up on YouTube (http://www.youtube.com/watch?v=tNiaQ5Lg4ko). There were two clarification questions, no probing ones.

I decided to give the Midwest Relativity Meeting (http://http://www.nd.edu/~astro/MWRM18/) a try. It will be held next weekend at Notre Dame University, South Bend Indiana. I will be the absolute last speaker of the conference at 2:40-3:00 pm on Saturday for the last session, "New Ideas/Novel Approaches" i.e. fringe work. Since my flight leaves at 6PM from O'Hare which is a two hour drive away, I will have to fly out of the meeting and burn some rubber.

I was feeling kind of down about my role in these meetings. What I am doing to lift my spirits back up is to review the logic presented in these talks. It is rock solid. I may do a few posts based on the talk since it clarified a few points. The logic is getting better, a sign that I am "within spitting distance" of the unified field theory Nature uses.

Doug

[sweetser]

Hello:

Preparing the talk for Midwest Relativity Meeting was quite productive. It forced an interesting shift: I actually need to use quaternions for the action, tensors will not suffice. It is ironic that it has taken so long for me to insist on using quaternion on technical grounds since I have owned the domain quaternions.com since 1997.

Tensor have a few great properties. For example, one does not choose a coordinate system for a tensor expression. The same tensor expression applies if one uses Cartesian, spherical, cylindrical, or some sort of oddball coordinate system that applies to a problem at hand. This property is shared by quaternions.

Tensors can be added to other tensors, or multiplied by another scalar. This property is shared by quaternions.

Tensors can be in arbitrary dimensions. Quaternions are limited to 4D. Folks working with strings will find that utterly inadequate. I have no interest in making folks who get the units of spacetime wrong happy. Every measurement of events in spacetime, every measurement of energy and momentum, ever made in physics has been 4D. That covers an absurd volume of data, so I will stick with it.

One cannot take two tensors and form a product. It is this limitation which has forced me to write my Lagrangian only in terms of quaternions. If one wants to make a Lagrangian that has connections to EM, the weak, and the strong forces, then one needs to build a bridge to group theory. Group theory involves having an identity, multiplication, and inverses. That is an option for quaternions, but not so for tensors. Sure, one can tack on a group symmetry operator, but that feels phoney. I recall seeing the symmetry stapled onto terms of the action. Yukko.

Better for group theory is to start with a division algebra to see if various ways to write it are equivalent to the group. In the case of U(1), the canonical representation is a complex number normalized to itself, forming a unit circle in the complex plane, \frac{z}{|z|}. Take either n roots, or n powers, whichever you prefer, and a unit circle in the complex plane can be formed.

The exact same approach can be used for quaternions, \frac{q}{|q|}. The group U(1) is Abelian, members commute with each other. Although quaternions do not commute in general, they do commute if all the quaternions point in the same direction, which would be the case for n roots or powers of a normalized quaternion. Nice.

The weak force has the symmetry SU(2), also known as the unitary quaternions. Sure enough, there is a way to write quaternions as unit quaternions. The Lie algebra has 3 generators. A common approach is to take the exponential of a quaternion where the scalar has been subtracted away, exp(q-q*). It is natural to combine the U(1) and SU(2) together so as to capture the four degrees of freedom in a quaternion with \frac{q}{|q|} exp(q-q^*). So wherever there was the rank 1 tensor A^{\mu}, drop in its place the quaternion \frac{q}{|q|} exp(q-q^*) so the resulting expression has electroweak symmetry.

There is only one more symmetry in the standard model to get, that for the strong force, SU(3). Its Lie algebra has 8 parts, which is what is found in two quaternions. Multiplying two U(1)xSU(2) symmetries just creates another member of the group. That is a basic property of groups. To make the multiplication table different, one needs to take the conjugate of one quaternion, then multiply it by the other, (\frac{q}{|q|} exp(q-q^*))^*\frac{p}{|p|} exp(p-p^*). Drop this one in for A^{\mu} and you have the symmetries required by three of the four known forces of Nature. This connection to the known forces makes it worthwhile to insist on quaternions.

My sense is that those formally trained will reject moving to quaternions out of hand. I have seen conversations stop because I use the q word, so that has been the source of my reservations. That is too bad, because a quaternion is in fact a tensor whose diminsions happened to be constrained to 4 but has additional abilities, namely multiplication and division, useful to reach out to group theory.

Doug

[Max™]

I quite like the restriction to 4D myself, never found it too satisfying to try and extend into arbitrary additional dimensions.

[sweetser]

Hello Max:

We share a common suspicion about the focus of efforts in the most prestigious centers of study on the planet. Lee Smolin and Peter Woit are the biggest voices in the professional community. My focus is on constructing a viable alternative, so I like to stick to the math. I need to be precise.

So what is a tensor, really? One way to view it is with group theory: a tensor forms a group under the operation of addition. The identity element is zero, the inverse of any tensor is -1 times that tensor. Add the two together, and the result is the identity, zero.

From a math perspective, in order to do calculus, one needs a mathematical field. This means one needs the set of numbers to behave like a group with the addition operator, but the numbers also must behave as a group under multiplication - with some elements excluded, usually just zero. Real and complex numbers fit those requirements for calculus. Quaternions also fit that definition, both the Hamilton representation and the even representation I have discussed here so long as zero and the Eivenvalues are excluded.

The identity element for addition is zero. The identity element for multiplication is one. This is no accident, it is a central observation. Once one has zero and one, the rest of the numbers we know about can be constructed according to work by Peano. If we only have addition, we only have zero built into the group theory of the number system. By insisting that the numbers used in physics must be mathematical fields, then zero and one, and by logical extension, all of number theory, is part of the system.

Addition is easy to generalize for arbitrary dimensions. Multiplication can also be generalized, but division is the hurdle. Frobenius worked out a proof years ago that associative division algebras are limited to one, two, and four dimensions, up to an isomorphism. If my thesis is correct, that we must necessarily shift any expression in physics written as a tensor as a quaternion to get the advantages of mathematical field theory, then we will look back on much effort in the twentieth century as ways to construct isomorphism of quaternions. Physicists are not way off the mark, only slightly off center.

Here is a fun benefit of using a normalized quaternion to get U(1) symmetry: it explains why we cannot build a device to measure the potential directly. Because the potential must always be normalized, its size is always the same, namely equal to one.

Doug

[sweetser]

Hello:

Clifford algebra = Geometric Algebra.

David Hestenes has promoted GA as a label over Clifford algebra because Clifford was one guy who worked on some aspects of this approach, while Geometric Algebra suggests what the math does. I only figured that out at the end of the 8th International Conference on Clifford Algebras and their applications to Mathematical Physics. David was a nice guy, and was fun to talk to so long as you didn't disagree with him on a technical point. As a visionary, he is quick to take out his blade.

According to standard GA,

quaternions = Cl(0,2)

meaning it has a scalar - the 0 - and a bivector - the 2. A bivector is a generalization of a cross product to other dimensions. Look in a mirror, and left hand becomes right hand. It was clear from a brief discussion with David that this bit of dogma was not worth discussing because there were people who were confused on the topic. Anyone who didn't understand that quaternions were Cl(0, 2) needed education.

It is one of these size arguments: GA is bigger than quaternions, ergo study GA. The size of your algebra matters. NOT.

What I think matters most in our efforts to describe Nature is completeness. Quaternions certainly have the property that under a mirror reflection M:

M: M(t, x, y, z) -> (t, -x, -y, -z)\quad eq~1

The problem with this approach is that we need to also consider a time reversal operation (TR = -M):

TR: TR(t, x, y, z) -> (-t, x, y, z)\quad eq~2

The handedness stays the same, which indicates a quaternion should not be viewed as a scalar and a bivector. The artificial constraint that the behavior of quaternions should fit within the GA model doesn't fit.

It was Grassman and Clifford who people credit with the great discovery of the geometric product:

a b = a \dot b + a \wedge b\quad eq~3

Having my groundings with quaternions, this result is incomplete. For quaternions, the dot product will have 4 pairs of number, while the cross product has 6, making 10 out of the possible 16 combinations. Grassman and Clifford came after Hamilton, so they partially recreated quaternions. Arg! Let me write the quaternion a as (a, A) and b as (b, B) where the lowercase letter inside a () is one number, while a capital letter inside is a 3-vector. Compare the products of Hamilton to Clifford:

(a, A)(b, -B) = (ab + A \dot B, Ab - aB - A \times B) \quad eq~4
(a, A)(b, B) = (ab + A \dot B, A \times B) \quad eq~5

Aside from silly sign issues that arose from the "pro-positive sign" position of Grassman, I won't work with geometric products because there is no accounting for Ab and aB. It is completely legal to adjust the scale of a 3 vector. That completely reasonable operation is missing from the geometric product. It is such a glaring omission that I don't put an effort in bridging the gap between the GA school and my own work. Bad accounting is indefensible.

The even representation of quaternions - or hypercomplex numbers - also doesn't work within the Clifford algebra constraint.

So it goes for Independent Research.

Doug

[sweetser]

Hello:

On occasion I pitch my quaternion animations to people who have written something on quaternion. The usual reply is no reply. Sometime we exchange a few emails, then go our separate ways.

This time I found out the fellow was visually impared. My animations are meaningless. What I could do is provide a description. Here is what a 4D cube looks like:

We are all familiar with a 3D cube. Mathematically that is the eight spatial vertices formed from all the combinations of plus or minus one for x, y, and z. In 4D, there are sixteen vertices, the additional ones being plus or minus time. So when time is minus one, we have a 3D cube. At plus one for time we have a 3D cube. In between, when say time is minus one third or plus two sevenths, all we have are the eight vertices, none of the connecting points between the vertices. We see the vertices travel through time from one 3D cube to another.

This approach can be generalized:

For fun, you might want to imagine what a polynomial looks like, or a sine function. Take your mind's image of these, and run a pencil along the image to create the animation. Note you are allowed to hold the pencil at any angle, which can create very different looking animations.

Let us think about the sine function, an endless range of mole hills. If the pencil is vertical, we will see one dot oscillating forever. If the pencil is horizontal, then we see nothing until we meet the bottom of the sine functions at -1. Then we see single dots that split in two, run apart quickly at 0, and slow down and join another, then disappear.

I think the single oscillating dot is how classical physics uses a sine function. We can watch the time-ordered oscillation of one particle for as long as we want. The array of creation, then destruction has the look of quantum field theory. To make it a complete field theory, then we would have to pile one sine atop another, making a graph that looks something like chainmail, so we see the creation and annihilation forever.

Doug