# Unifying Gravity and EM

• sweetser
In summary, the conversation discusses a proposal for a unified field theory that combines gravity and electromagnetism into a single rank 1 field. The Lagrange density for this proposal is provided, along with a discussion of how the equations are generated and the physical implications of the theory. The proposal is consistent with both weak and strong field tests of gravity and there are no known physical experiments that contradict it.

#### sweetser

Gold Member
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

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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 ?

vanesch said:
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.

sweetser said:
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

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.

CarlB said:
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

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

Hello Carl:

Thanks for the further clarification. I'm pretty sure I haven't said something like this:
CarlB said:
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

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 [Broken]

http://modelingnts.la.asu.edu/html/GCgravity.html [Broken]

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.

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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:

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

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.

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

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.

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

sweetser said:
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.

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vanesch said:
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.

A known problem with GR's Lagrange density

<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/>

sweetser said:
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?

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)…

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.

hellfire said:
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.

sweetser said:
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.

vanesch said:
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

sweetser said:
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).

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vanesch said:
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

sweetser said:
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 ...)

sweetser said:
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.

straycat said:
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

Hello Patrick:
vanesch said:
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:
Purcell said:
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

sweetser said:
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.

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

sweetser said:
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 ) 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.

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

sweetser said:
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 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 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.

Last edited:
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
$$U_{a}=(\gamma,\gamma\beta_{x},\gamma\beta_{y},\gamma\beta_{z})$$
$$\beta=v/c$$
$$gamma=\frac{1}{\sqrt{1-\beta^{2}}}$$
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:
L&L said:
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

Hello Patrick:
This sounds like a separate, field theory argument:
vanesch said:
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":
Hatfield said:
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

Summary of 2 insights

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.