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