Unifying Gravity and EM

[sweetser]

Hello Lut:

I appreciate the importance of having an explicit way to show that energy is conserved. One reason I am not worried that we will eventually find it is that there is no "t" in the Lagrangian, so one can vary t without changing the action. A symmetry like that means that there is a conserved quantity, and it turns out to be energy.

There is no t in the action of GR either. Energy is conserved in GR, but it is not localized, one needing a volume of spacetime before one can say energy is conserved. The nonlocalization happens because one varies the metric field.

> Regarding the the current business. You can split the 4 degrees of freedom into 2 orthogonal vectors but you have an electrical and a matter current to describe, and only 2 df each. Or do I misunderstand ?

That sounds right. Both the graviton and photon travel at the speed of light, and thus only need 2 degrees of freedom each to describe a mediating particle where like attract (gravity) and like repel (EM).

doug

 

[sweetser]

Field Theory of Gravitation

Hello:

I was given an optional assignment by Lut to read a paper: "Field Theory of Gravitation: Desire and Reality" by Yurij V. Baryshev, arxiv:gr-qc/9912003. It was a fun read, and I will give you the GEM spin on the article.

Baryshev sees two types of theories out there in the literature: field theories with flat metric backgrounds and spacetime curvature theories. Nearly all the work done with gravity since Einstein has been spacetime curvature. The few papers on field theories bow to the powers that be, and say they are effectively the same thing as a geometric approach. Baryshev points out that while this might be true for weak gravity fields, the treatment of black holes will necessarily be quite different. He argues that there will be no black holes in a field theory approach, even though there is a Disney movie about such hellish places in the Universe. Einstein and Eddington would be pleased, but not the folks who churn out black hole papers, or model the collapse of black holes.

In Baryshev's hands, the argument against black holes is simple: the energy is positive definite, and built to stay that way by energy conservation (see section 5.3 for details, since I still don't completely get it).

With my GEM bias, I enjoyed the clarity with which he said there were only two types of proposals in the literature. I view GEM as being a third approach, exactly between a field theory with a flat metric and a dynamic spacetime connection (notice: I avoided the word curvature and the Riemann tensor on purpose, using the connection which is a derivative of the metric instead).

GEM does work as a pure potential theory with a flat metric. According to Baryshev, that is a great thing because it means energy conservation will work as a local effect, which is good because that is the way energy works for all other field theories we understand well. I have calculated the $T^{00}$ of the GEM proposal before, but want to redo the exercise using only quaternions (my idea of challenging fun).

If GEM only worked with a flat metric, folks like Prof. John Baez would dismiss it with a one liner (he did actually, this very complaint of working only with a fixed background metric like all other field theories to date).

Actually, based on my education in GR, the thought of working with a flat metric did not occur to me (strange, but true). I learned special and general relativity from Edwin F. Taylor, the coauthor with Wheeler of "Spacetime Physics". That book was devoted to special relativity. They also wrote another book together, "Exploring Black Holes: Introduction to General Relativity". This book was all about playing with the Schwarzschild metric of GR. They did not derive the Schwarzschild metric. They avoided any technical discussions about connections and Riemann curvature tensors. Instead, the book looked at all the consequence of the metric.

In August of 1999, I had a field equation with a 4-potential in it. From my training, I had to find a way to a metric. I was not sure I could find such a path, but gave myself the time to look for one. I don't emphasis the first road I found often, fearing the potholes. What I did was to work with a force equation, $F^{\mu} = J_{\nu} d^{\mu} A^{\nu}$. I found a solution to that equation which was a 4-velocity, $c^2 = (k_1 c \frac{dt}{d \tau}, \vec{K_2} \frac{d \vec{R}}{d \tau})$. I squared this result, using the condition that for flat spacetime I should get a flat Minkowski metric to eliminate the constants k1 and K2. This lead to the exponential metric written in the first post of this long thread.

I derived a metric equation in a way no one else has done. Being skeptical of myself means I was not sure such a derivation was right. To the point of being a bore, I repeat that the exponential metric can be tested against the Schwarzschild metric, using the same experiment that was used to rule GR beats Newton: look at light bending around the Sun. In 1919, that was a big technical challenge. The data was not as clear cut as the public was led to believe. In time, we have solidified the GR victory.

The playing field for GR versus GEM involves a million fold improvement in the light bending measurement. We have a ten thousand fold improvement, so a few orders of magnitude refinement are required. Even so, the difference amounts to 12%. At the microarcsecond level, the rotation of the Sun, and the quadrapole moment also have to be accounted for.

If the only path to the testable exponential metric was the funky chicken walk through the force equation, I would not have the confidence that I am "within spitting distance of a unified field theory". I told the story earlier in this thread of finding a differential equation I had to solve using the Christoffel symbol of the exponential metric. For three weeks I avoided doing the calculation because I had never done a calculation with a Christoffel, and was unsure I could figure it out. The exponential metric did solve the field equation.

In all my reading on gravity, I have never seen another approach that got to a metric in the ways I have. So it goes.

doug

 

[Mentz114]

Hi Doug:

With my GEM bias, I enjoyed the clarity with which he said there were only two types of proposals in the literature. I view GEM as being a third approach, exactly between a field theory with a flat metric and a dynamic spacetime connection (notice: I avoided the word curvature and the Riemann tensor on purpose, using the connection which is a derivative of the metric instead).

You could be right. But one can always transform away the first derivatives of the metric. In GEM you'll have to make the action in the connection go to the potential to preserve an energy density.

I came across this today in a PF thread

As a funny aside, it's a theorem in QFT that the ONLY self-consistent theories with spin-1 bosons are gauge theories,
and the analogous theorems get even stronger as you go to higher spin. For example, the ONLY self-consistent theory of
(massless) spin-2 bosons is Einstein's Gravity!

It is a rather wonderful result (proved by Dick Feynman many years ago, if I remember correctly) that the only way to
write down a self-consistent theory of a *massless, spin-2 * object is Einstein's GR. There is simply no other way to
do it! The proof is quite non-trivial, but it's a famous result. Of course, there are some important assumptions, such
as Lorentz invariance, energy conservation, etc. But given some rather weak assumptions like that, then the only theory
that you can write down with massless spin-2 is GR. There is NOTHING at all "ad-hoc" about Einstein's Field Equations
(I'm assuming that's what "EFE" stands for). High-spin theories (s>1) are VERY constraining.

People are out to modify GR all the time, and that's okay. But almost always, the way they do it is to ADD something,
such as new scalar fields, extra dimensions, etc. But they always end up with Einstein PLUS stuff. You can never get
rid of Einstein entirely. Every model I know of that tries always has deep problems!

Is this discussed in your copy of the Feynman's gravity lectures ?

I'm sure you'll say it's wrong in any case.

[aside]
I have been informed that the potential does not transform as a 4-vector and sorted out the boosting of the EM field tensor and the equivalent operation on the potential.
https://www.physicsforums.com/showthread.php?t=197942

Lut

 

[sweetser]

Where the first derivatives of the metric go

Hello Lut:

Understanding this line is central:

> But one can always transform away the first derivatives of the metric.

I have to agree with this, it is true, but I always have my odd slant, promise! When you transform away the first derivative, where does it go? I have a specific technical answer. In GEM, the first derivative of the metric lives in one and only one place - the covariant derivative:

$$\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}$$

All the first derivatives live in the Christoffel symbol of the second kind. In GEM, when you change the first derivative of the metric in any way for any amount, you must make a compensating change in the potential so the derivative of the potential exactly balances it out. The potential and metric are separate, but changes in the potential play with the first derivative of the metric. You are free to play with the first derivative of the metric. You are not free to change the covariant derivative of the potential.

On the aside...
There is a lot of truth in the aside. I've seen it in action, specifically the Rosen bimetric theory that has the exponential metric in it. The problem is that it does add something, a fixed background metric, so that is a place to store energy/momentum, and it predicts dipole moments for gravity waves. We have good data to say the lowest form of gravity wave emission is a quadrapole.

I am not trying to make a theory of EM. I am not trying to make a theory for gravity. I am trying to make a theory of EM and gravity. The two play together in the covariant derivative as discussed about. This is how I avoid the proofs, I claim they do not apply to the "and" situation.

I don't know if Feynman provides a discussion in his book. He is a rank 2 field theory guy, like every other smart person making their money doing research on gravity. I can only follow along before hopping off that train of thought.

Fun discussion you have going on there. The way I read it, both the differential element and the potential do transform like vectors, but each comes with its own technical issue. For the differential, the contravariant tensor, d^u, needs to have minus signs for the 3-vector. For the potential, you need to have the current-coupling term around to keep things honest. So one can either say they do not transform like 4-vectors because they have these extra rules, or they do transform like 4-vectors so long as these rules are applied. People will argue over which is right, even if the net result is identical.

doug

 

[Mentz114]

Hi Doug:

I have to agree with this, it is true, but I always have my odd slant, promise! When you transform away the first derivative, where does it go? I have a specific technical answer. In GEM, the first derivative of the metric lives in one and only one place - the covariant derivative:

$$\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}$$

All the first derivatives live in the Christoffel symbol of the second kind. In GEM, when you change the first derivative of the metric in any way for any amount, you must make a compensating change in the potential so the derivative of the potential exactly balances it out. The potential and metric are separate, but changes in the potential play with the first derivative of the metric. You are free to play with the first derivative of the metric. You are not free to change the covariant derivative of the potential.

Yes, you've made this clear in the past, thank you. I'm asking about the next step. Suppose I make a coordinate transformation X -> X' = f(X) which removes the Christoffel symbols - by what rule or process does the gradient of the potential A(X) change to compensate ?

Another thing that bothers me is the fact that if the connections are non-zero then the Lagrangian ( the energy ) depends on the value of A as well as its derivatives. This is new territory, because GEM is no longer a gauge theory in the normal sense. Obviously adding a local or global field Phi changes the total energy.

I sent you a dumb question by email - it's important for my understanding of GEM to know the answer.

Lut

 

[Mentz114]

Doug:
(This is highly speculative)
following on from above, if the Lagrangian depends on the field itself, like bA^2, then b is the mass of the field. So it seems that because the connections have no source, it's the gravitons themselves that cause the 'curvature'. The amount of action in the connections is related to the mass of the gravitons. Giving gravitons a small mass can be a good thing, because it reduces the range.

Lut

 

[sweetser]

Hello Lut:

Yes, you've made this clear in the past, thank you. I'm asking about the next step. Suppose I make a coordinate transformation X -> X' = f(X) which removes the Christoffel symbols - by what rule or process does the gradient of the potential A(X) change to compensate ?

Good question. I was thinking about this recently. The current coupling term, $J^{\mu} A_{\mu}$ depends on the value of A. The field strength tensor term, $\nabla^{\mu} A^{\nu}$ only depends on how A changes. We need both, but they are independent of each other. In other words, one can have a value of A remain the same, but the way A changes in a local region of spacetime can vary. Do I understand how to do the nuts and bolts of this? No. What have done is two extremes: work with only the potential in totally flat spacetime, and work with only the connection with a static potential. If I was amazingly good, I could work out all the points in between those two extremes.

My hope is that GEM is a gauge theory, but not in the normal, add this field like so, sense. Gauge means to measure. The symmetry is about measuring with the change in the potential or the first derivatives of the metric.

Gravitons must travel at the speed of light and be massless. This happens when the symmetric tensor has zero trace. What happens when the symmetric tensor has a trace? That forms a scalar field. I think this does the work of the Higgs particle. Mass breaks the gauge (in the usual, traceless field strength tensor sense) symmetry. The Higgs mechanism is unnecessary for a unified field theory like GEM. Sorry LHC, I am betting against you folks based on theory.

doug

 

[sweetser]

Correction to current coupling

Hello:

I believe I have made a technical error in how I have represented the current coupling term for gravity, $J_m^{\mu} A_{\mu}$. There are two independent mistakes. First, it postulates two 4-current densities, $J_m^{\mu}$ and $J_q^{\mu}$, with eight degrees of freedom. A 4D wave equation should have a source with only four degrees of freedom.

Second, anyone who understands Feynman's current-current coupling phase analysis could reasonably say $J_m^{\mu} A_{\mu}$ can only represent a spin 1 interaction, where like charges repel.

In any and all documentation on GEM under my control, I will strive to rewrite the coupling term like so:

From:
$$-\frac{1}{c} J_q^{\mu} A_{\mu} +\frac{1}{c} J_m^{\mu} A_{\mu}$$

To:
$$-\frac{1}{2c} J^{\mu}(A_{\mu} + I_{||}A_{\mu}I_{||}) + \frac{1}{2c} J^{\mu}(I_{tr}(A_{\mu} - I_{||}A_{\mu}I_{||})I_{tr})^*$$

Einstein said one should try to be simple, but not too simple. My initial coupling term was too simple. The complication is required because this scalar is formed from a 4-potential that cannot be measured directly, yet can be assigned to point in an arbitrary direction. The potential is connected to a real current, so the part of the potential that is parallel to the current can be separated from that which is transverse.

One good thing about this change is that the scalar formed from the contraction has not changed at all. It is the representation of the scalar that is getting more verbose. The phase of the second current-current interaction is changed by the two rotations around the parallel and transverse axes. The sign of two terms in the phase are the same, so they assist each other, and will require pi radians to return. This indicates that the second term has spin 2 phase symmetry.

Since there is one 4-current density, there are four degrees of freedom. The spin 1 photons are the transverse potential that form the transverse waves of EM. The spin 2 gravitons are the scalar and longitudinal potential that form the scalar and longitudinal modes of GEM (unlike GR where the graviton is transverse).

My primary short-term goal will be to correct the proposal. I dislike wasting anyone's time with faults I understand and can correct with a large volume of grunge work.

doug

 

[Mentz114]

Doug:

The current coupling term, $$J^{\mu} A_{\mu}$$ depends on the value of A.

Yes, and to get rid of it you have to apply the Lorentz gauge. But you can't gauge away the mass terms ...

The field strength tensor term, $$\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}$$ only depends on how A changes.

Do my eyes deceive me ? I can see A in that covariant derivative.

Gravitons must travel at the speed of light and be massless.

Must ? This sounds like a statement of faith. Please see the following reference and the many others on the subject of graviton mass. The experimental evidence is ambiguous, and probably always will be. We cannot experimentally verify with any confidence whether gravitons have any mass. Similarly, we cannot test if gravitational waves travel at c or slower.

Some references -

http://www.iop.org/EJ/article/0264-9381/20/6/101/q306l1.html

Abstract. We show that the graviton has a mass in an anti-de Sitter background given by This is precisely the fine-tuning value required for the perturbed gravitational field tomaintain its two degrees of freedom.


arXiv.org:gr-qc/9705051

Mass for the graviton
Authors: Visser, Matt
Can we give the graviton a mass? Does it even make sense to speak of a massive graviton? In this essay I shall answer these questions in the affirmative. I shall outline an alternative to Einstein Gravity that satisfies the Equivalence Principle and automatically passes all classical weak-field tests (GM/r approx 10-6). It also passes medium-field tests (GM/r approx 1/5), but exhibits radically different strong-field behaviour (GM/r approx 1). Black holes in the usual sense do not exist in this theory, and large-scale cosmology is divorced from the distribution of matter. To do all this we have to sacrifice something: the theory exhibits *prior geometry*, and depends on a non-dynamical background metric.

http://www.journals.uchicago.edu/cgi-bin/resolve?id=doi:10.1086/427773&erFrom=-2247456991476005095Guest

http://digitalphysics.org/Publications/Ostoma-Trushyk/EMQG/

http://www.springerlink.com/content/x52n3827118uk012/


Lut

[edit]re-reading this, it sounds a bit tetchy. I apologise. All done rather quickly in between other things. I'm sure you'll find a way to wriggle out as you always do !

Lut

 

[sweetser]

Covariant derivative

Hello Lut:

Let me separate the questions. This post is only about this question:

Do my eyes deceive me ? I can see A in that covariant derivative.

As I recall learning about this topic, they tell you first that the contravariant 4-derivative has a bunch of minus signs:

IS A TENSOR
$$\partial^{\mu} = (\partial/\partial t, -\partial/\partial x, -\partial/\partial y, -\partial/\partial z)$$

I was like, that is irritating, because I had learned the up index means everything is positive. This is the exception to that rule. Exceptions have consequences. Specifically, the 4-derivative of a 4-potential does not transform like a tensor:

NOT A TENSOR
$$\partial^{\mu} A^{\mu}$$

They usually go through a transform exercise which I rarely can follow. I just accept it, a little easier now with the "exceptions have consequences" idea. The connection has one job, make the derivative transform like a tensor:

IS A TENSOR
$$\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}$$

The connection has 3 indices, the covariant derivative has 2, so it must be contracted with the potential. Then they point out the obvious:

NOT A TENSOR:
$$\Gamma_{\sigma}^{\mu \nu}$$

Although it is not a tensor, the connection gets separated from its home inside a covariant derivative snuggled right against the potential. I consider this very bad practice that is quite common. One never sees potentials in GR papers, unless they wish to point out their failings. The next goal is to form the simplest tensor out of a connection that transforms like a tensor. That quest has one answer, the Riemann curvature tensor. It is announced that this is a great thing, because that has a derivative of a connection inside a tensor, and so a second derivative of a metric. Such pride strikes me as silly. Take two covariant derivatives in a row, and one will also get the derivative of a connection. The difference is that two covariant derivatives in a row leaves you with exactly 1 derivative of a connection. Energy can be local, whether it is in the derivatives of the metric or the potential. The Riemann curvature tensor has the difference of two derivatives of a connection, the source of being "non-local", and even worse, can be transformed away entirely. That cannot happen with the two covariant derivatives in a row approach, because the energy shifts to the potential.

doug

 

[Mentz114]

Doug:
The point is not whether the connection is a tensor, it's that the terms A^2 appear in the Lagrangian. I'll work this out explicitly when I have time.

Lut

PS.

The Riemann curvature tensor has the difference of two derivatives of a connection, the source of being "non-local", and even worse, can be transformed away entirely.

I don't think the Riemann tensor can be transformed away, because it has second derivatives of the metric and represents tidal fields.

 

[sweetser]

Yup, I was wrong, sorry. At any point in spacetime, the connection can be made zero, but you cannot make a bunch of points zero if there is a gravitational source around.

doug

 

[Mentz114]

Hi Doug:

I noticed that because
$$\Gamma ^{\nu\sigma}_{ \mu} = g_{a\mu}g^{b\nu}g^{c\sigma}\Gamma ^{a}_{bc}$$
there will be actual metric terms in the Lagrangian also.

Lut
(it's a sunny day and I'm feeling better...)

 

[sweetser]

Unified force mediating particles

Hello Lut:

I like simplicity and symmetry, and a really tight logical story. I think I get all of those with a massless graviton.

Let's start with the photon and EM. That fellow is massless, and as such cruises at the speed of light, which seams appropriate given that it is light. When folks go to quantize the field, they only need two of four modes of emission.

I am trying to unify gravity and EM by putting the graviton right in the mix. Like there is an antisymmetric E field, there is a symmetric e. Like there is a symmetric B field, there is a symmetric b. These are made up of derivatives of exactly the same things, just different signs.

New Idea

I was driving back from burning a few DVD of the Stand-Up Physicist when I ask myself a new question: what is the exact relationship between gravitons and photons. I am of the opinion that the fundamental currency of the universe is an event that can be resented may ways, the most efficient involving quaternions because the tool remains the same no matter what sort of physics is being described. I view a photon as having 2 degrees of freedom, and a graviton as having two degrees of freedom, and because 2+2=4, together they have four. Now that I have one current 4-vector, the new idea is that I need to work with one force mediating particle that can have more than one kind of spin. Odd, really odd. I am fighting it, but that is a good sign. We don't understand how any fundamental particles really work. We have taken the simplest approach and put each one in a completely separate box. Yet we know particles can transform from one form to another. We have all the math machinery to describe an super-energetic photon turning into an electron-positron pair. How Nature would make gravity so much weaker than EM inside the same force particle is a huge mystery.

Have fun kicking around this unification-at-the-particle level idea.
doug

Sunny in London, it can be done. Cold and rainy here.

 

[Mentz114]

Hi Doug:

Sure it would be tidy to assume a massless graviton. But you may have a problem because the grav field tensor is only traceless in Minkowski space, if
$$\partial_{x}A^{1}+\partial_{y}A^{2}+\partial_{z}A^{3}-\partial_{t}A^{0} = 0$$.

But with a general solution is this still true ? The problem breaks my software, it's just too big to handle. Shifting the indexes up and down multiplies the number of terms by 4 for each index. I'll work out a shorthand or somehing and try to get $$g_{\mu\mu}F^{\mu\mu}$$ worked out in the general case.

[later edit]
I've got a general expression for $$g_{\mu\mu}F^{\mu\mu}$$ and it's not as bad as I thought. Have you ever calculated the trace of F (symmetric) in the general case ?

Your new idea is interesting. Attached is an excerpt from the Baryshev paper. Their potential has spin-2, spin-1 and spin-0 components.

 

[sweetser]

Have it both ways

Hello Lut:

I am trying to have it both ways. For an infinite range, $1/R^2$ force law, I need a massless particle for light, I need one for gravity. They must in in their own separate way be gauge theory in the traceless field strength tensor way.

I am proposing that gauge symmetry is not broken by a false vacuum, but by mass itself (inertial mass being exactly the same thing as gravitational mass). How this gauge-braking fellow interacts with the massless force particles at this time is something I do not understand.

doug

 

[Mentz114]

Hi Doug:

The 1/r^2 force law is a weak field approximation. Otherwise we have infinities. I thought the whole point of quantum gravity is to find the actual 'force' law.

Addressing the issue of the trace of the symmetric field tensor. In Minkowski space the trace is $$\partial_{x}A^{1}+\partial_{y}A^{2}+\partial_{z}A^ {3}-\partial_{t}A^{0}$$ which is only zero if you apply a wave condition on A.

When the metric is general it is easy to calculate the trace exactly in terms of Christoffel symbols like this ( I'm starting with the covariant field tensor because it is much simpler) -

$$F_{mn} = \partial_mA_n+\partial_nA_m - 2\Gamma^{k}_{mn}A_{k}$$

and the trace is
$$F^{m}_{m} = g^{mm}F_{mm} = 2g^{mm}\partial_mA_m - 2g^{mm}\Gamma^{k}_{mm}A_{k}$$

Looking at the second term, we find the coefficient of A_0 is the sum

$$g^{00}\Gamma^{0}_{00}+g^{11}\Gamma^{0}_{11}+g^{22}\Gamma^{0}_{22}+g^{33}\Gamma^{0}_{33}$$
and similar for the other components of A.

I doubt if that sum is zero. I'll try writing it out in full but it doesn't look likely it's zero.

Serieusly, you must demonstrate the tracelessness of the field tensor or admit your graviton has mass. In my opinion that would make it a more realistic theory, but I know you want simpliciry and logic - as if that exists.

You don't seem too bothered by the fact that the contraction of the symmetric field tensor in the Lagrangian contains terms that involve the metric directly, and the potential directly. Not gradients. That changes the character of the Lagrangian a lot.

You'll need to demonstrate that losing gauge invariance does not affect current/energy conservation. That all balances out in flat space, but it's rather difficult to calculate in curved space-time.

Have a good w/e, I'm off so I won't be posting for a couple of days.

Lut

 

[dlgoff]

I've been following your thread since its begining. Wanting to see a new theory in the making.

I can only get an idea of what you are doing from the mathamatics since I never studied much more than diff. equations and vector calc. But somehow I think you are on to something when you say "Now that I have one current 4-vector, the new idea is that I need to work with one force mediating particle that can have more than one kind of spin." Could "graviton part" have a frequency like that of the "photon part"?

Thanks for your efforts

 

[sweetser]

Hello dlgoff:

If you want some help learning the math stuff, spend some time with the videos listed in post 389. I call it "hard TV", not because of excessive violence or sex, but for excessive math. If you do put in the investment to watch the videos, please do so with pencil in hand, and repeat any of the derivations. Preparing the videos helped me big time, particularly the small drawings at the bottom of slides. Nothing like a visual hook for a nasty bit of algebra!

Now to the physics question...

I don't think the gravitational part with have the same frequency EM part. The reason is that the gravitational part comes from this irreducible symmetric tensor:

$$(\nabla_{\mu}A_{\nu}+\nabla_{\nu}A_{\mu})$$

I call this "average Joe", the average amount of change in an 4-potential in spacetime (it covers 10 of the 16 possibilities). A second, independent, irreducible, but antisymmetric tenor is the one for EM:

$$(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu})$$

These are the six deviants, the deviation from the average amount of change in the 4-potential. Because these are really separate, so too can the frequencies and wavelenths.

I also broke the potential cleanly in two. Nature can deal with them separately, me thinks.

What cannot be separate is the speed: they live in the same house. Both travel at the speed of light to mediate the force.

At this point in the development, I know I will not bring this up in a scientific meeting. There are two big problems with it, only one of which I might be able to deal with some day. The first problem is having one collection of events in spacetime that for its transverse modes, has spin 1 symmetry, and for its longitudinal and scalar modes has spin 2 symmetry. I have great hopes for quaternions.sf.net and the animation software to open completely new doors. I have recent data to demonstrate this: a fellow from Mexico wrote to me about gamma matrices - a bit of confusing algebra that appears in relativistic quantum field theory - and a couple days later the animation was up on YouTube, and I get what it is about (taking a quaternions (t, x, y, z) and shuffling however you want, these four, tossing in a negative sign or too, to get things like (-z, -y, t, x)). Until the animation of spin 1 and 2 coexisting is on my hard drive, I accept that the idea of a particle with both spins will not be accepted.

[cool shift just happened: I may want to start thinking about "collections of events" instead of particles, as it would be more consistent with wave/particle duality and my software. It frees the mind from the chain that is the billiard ball->particle association.]

That is a problem in theory. The problem in practice is how could the gravitational field be so absurdly small and be traveling with a photon? If I was drinking beer, I might talk about a few possibilities, but I need something concrete.

Good luck in your studies,
doug

 

[sweetser]

The New Improved Lagrangians

Hello:

The process of updating the GEM Lagrangian as written on my website has begun. I have some happy news to report. The Lagrangian is a scalar. That has not changed in the slightest way. There are two ways to write the Lagrangian. The first is the "unified" approach, like so:

$$\mathcal{L}_{GEM}=-\frac{\rho}{\gamma}-\frac{1}{c} J^{\mu} A_{\mu} -\frac{1}{2 c^{2}}\nabla_{\mu}A_{\nu}\nabla^{\mu}A^{\nu}$$

The tensor $\nabla_{\mu}A_{\nu}$ is called "reducible", and thus cannot be viewed as the source of a fundamental field. I have known for a long time how to split this reducible asymmetric tensor in the irreducible rank 2 symmetric field strength tensor (average Joe for gravity) and the irreducible rank 2 antisymmetric field strength tensor (the deviants for EM).

The issue of spin 1 and spin 2 for the current coupling term, $-J^{\mu} A_{\mu}$, has that same property. It can represent either two like charges that attract, or two that repel. We see that in Nature. Here is how to split up both the current coupling and field strength tensor contraction terms:

$$\mathcal{L}_{GEM}=-\frac{\rho}{\gamma} -\frac{1}{2 c} J^{\mu}(A_{\mu} + I_{||}A_{\mu}I_{||}) + \frac{1}{2 c} J^{\mu}(I_{tr}(A_{\mu} - I_{||}A_{\mu}I_{||})I_{tr})^*$$
$$- \frac{1}{4 c^2}(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}) (\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})$$
$$- \frac{1}{4 c^2}(\nabla_{\mu}A_{\nu}+\nabla_{\nu}A_{\mu}) (\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})$$

A symmetric split, nice.

doug

 

[dlgoff]

I'm ready for some beer

That is a problem in theory. The problem in practice is how could the gravitational field be so absurdly small and be traveling with a photon? If I was drinking beer, I might talk about a few possibilities, but I need something concrete.

I'm ready for some beer. Hope you are.

 

[sweetser]

Mulitple spins

Hello dlgoff:

I always feel more confident that an issue is important if I find out lots of other people have worried before me. Here is the issue as I stated it:

The first problem is having one collection of events in spacetime that for its transverse modes, has spin 1 symmetry, and for its longitudinal and scalar modes has spin 2 symmetry.

The recent news in unified field theory by a respectable theoretical physicists and surfer Garrett Lisi hopes to unify gravity with the standard model using the group E8, which is so big it took a computer to write it all out. A solid technical review folks reading this list should be able to get a few facts from is here:

http://backreaction.blogspot.com/2007/11/theoretically-simple-exception-of.html

Here is the paragraph relevant to my own worry:

On the algebraic level the problem is that fermions are defined through the fundamental representation of the gauge group, whereas the gauge fields transform under the adjoint representation. Now I learned from Garrett that the five exceptional Lie-groups have the remarkable property that the adjoint action of a subgroup is the fundamental subgroup action on other parts of the group. This then offers the possibility to arrange both, the fermions as well as the gauge fields, in the Lie algebra and root diagram of a single group. Thus, Garret has a third way to address the fermionic problem, using the exceptionality of E8.

This problem of getting fermions to play nicely with bosons is part of all this supersymmetry research, including string theory. The common approach, even by Lisi, is to make things bigger.

I may have found a fourth approach this weekend. I call it the platform diving solution. Once every four years, the summer Olympics comes around, and they show all kinds of sports I would not spend a dime to see. One of those sports is platform diving. They do it from platforms of different heights. The commentators are somehow able to spot that the diver does three back flips while also doing one and a half twists. So I wondered, why not do the same for fundamental particles? I generated a bunch of quaternions using this automorphic function:

$$q(t) = (cos t, sin t/2, sin t, sin 2 t)$$

$$t \rightarrow 0 to 4\pi$$

There are 3 different symmetries for the phases: spin 1/2 for x, spin 1 for y, and spin 2 for z. In easier lingo, x is slow, y is medium, and z is fast. I was able to generate gif animations, but the sizes of those files are kind of big, so I'll just include one frame of the animation:

The t-x graph is lower middle, t-y is upper left, and t-z is lower left. In the animation in the upper middle, 4 dots are running along the path shown in the superposition of the upper right. After generating many permutations, I spent a lot of time staring at these, amused, even my lady enjoyed them. Analytical animations may lead to new ideas.

doug

 

[sweetser]

Degrees of Freedom

Hello:

Nudged by Lut, I am starting to think about the issues of degrees of freedom. I know the logic behind the standard approach that says the photon potential A uses all of 2 degrees of freedom. That appears odd. Consider this thought experiment...

You have devices to measure the electric field E in three directions, and the field B in three direction. A man in a long black trench coat wearing mirrored aviator glasses places a black box in front of you. Since he doesn't look like a chatty fellow, you measure the six values:

Ex, Ey, Ez, Bx, By, Bz

That's the way science is, you make measurements quietly. He picks up his black box and leaves. As you wait, you wonder what would happen if the dollar kept going through its free fall. The man returns, you measure:

Ex', Ey, Ez, Bx, By, Bz

Only one number was different, Ex', the other five were the same. Seven times this scenario plays out, and your final data set looks like this:

1. Ex, Ey, Ez, Bx, By, Bz
2. Ex', Ey, Ez, Bx, By, Bz
3. Ex, Ey', Ez, Bx, By, Bz
4. Ex, Ey, Ez', Bx, By, Bz
5. Ex, Ey, Ez, Bx', By, Bz
6. Ex, Ey, Ez, Bx, By', Bz
7. Ex, Ey, Ez, Bx, By', Bz'

This looks like six degrees of freedom to me. The four potential only has four degrees of freedom, not enough, even if we ignored the analysis that says only two modes can be used for EM.

Recently I have argued that a message from quantum mechanics was to view an operator as an equal to what it acted upon. I was thinking about symmetries in the action. The covariant derivative can move changes in the potential to changes in the connection back or forth. This is the symmetry behind conservation of mass.

In the context of this post, let us try and view an operator as having degrees of freedom. I know in standard approaches, one does not view an operator as increasing the degrees of freedom. One of the missions of the GEM programs is seeing a little bit more in simple tools, just enough to get the job done.

When I say the potential $A^{\nu}$ has 4 degrees of freedom, and at a particular point 0, has the values $A^{\nu} = (\phi_0, Ax_0, Ay_0, Az_0)$, how does that constrain the derivative of the 4-potential, $\nabla^{\mu} A^{\nu}$? I argue that the 4 degrees of freedom of the potential tell you precisely nothing about how A changes. Consider a near-by neighbor, $A'^{\nu}$, which has values $A^{\nu} = (\phi_1, Ax_1, Ay_1, Az_1)$. Our four degrees of freedom are already vested in the 0 terms. How can things change going from 0->1? Most of the change for $\phi_1$ probably comes from $\phi_0$, a continuation. Yet some of the change might come from the other three. Same goes for $Ax_1$, $Ay_1$, and $Az_1$. That adds up to 16 ways change can happen going from potential 0 to 1. If say $Ay_0$ helps how $Ax_1$ changes because of a swirling potential, why should that have any link to how $\phi_0$ contributes to $Az_1$?

The sixteen gamma matrices of quantum mechanics appear to do just this sort of mixing of all the components. (I haven't talk about that stuff here yet, quite neat).

This discussion has caused me to broaden my view of a unified field theory. I had previously thought it was all about the four potential. Now I see the operator plays a much deeper roll, it is not just a piece of math machinery. The operator has a symmetry of its own which gives rise to the conservation of mass. The four operator action on a four potential may allow sixteen degrees of freedom, six for the transverse wave of EM, and ten for the dynamic metric of gravity.

doug

 

[Mentz114]

Hi Doug:

I think the logic of your gedanken experiment is flawed. When you measured the E and B fields, you were measuring space and/or time derivatives of the potential. As you point out later on $$\partial^{\mu}A^{\nu}$$ has 16 df. I'm not sure what the point of the argument was ( maybe that is the popint ?)

Some of the degrees of freedom of the potential are not physical, and it requires a choice of gauge to get rid of them.

We could be talking at cross purposes, because I'm talking about GEM as a classical field theory.

You haven't commented on the calculation I did in post #404. If the connections are non-zero, I don't see how your graviton can be massless without a heavy constraint, or maybe not at all.

I've though a lot about how you can get over the spin-1 limitation of the interaction term. Squaring it will do the trick but that changes the theory a lot.

Lut

 

[sweetser]

Hello Lut:

I had been pondering this thread over on sci.physics.research: Maxwell's wave equation and degrees of freedom. There are two degrees of freedom for the polarization of the massless wave equation. When there is a current, mass enters, and that adds another degree of freedom (hope I got that right!). The Maxwell wave equation thus has three degrees of freedom.

In that thread, someone argued that since the B field can be viewed as the result of relativistic motion, it should not be viewed as a degree of freedom. I disagree with a connection between relativity and degrees of freedom.

I am using the word 'graviton' as it is used now - that which is spin-2 and has zero mass so travels at the speed of light, doing the work of gravity. That is when the trace of the symmetric tensor happens to be zero. There will be many situations where as you say, the trace will not be zero. I don't have a name for the non-zero trace situation (I hate making up names, it marks one as a crank). When the trace is non-zero, that is a scalar field that breaks the gauge symmetry. That sounds like what the Higgs mechanism does, but without the false vacuum. Mass breaks gauge symmetry. So maybe gravito-higgs?

Squaring it will do the trick but that changes the theory a lot.

How so? The components are identical, $-\rho \phi + J_x A_x + J_y A_y + J_z A_z$, just the way to write $-J^{\mu} A_{\mu}$ has changed so the phase is right. The phase is not part of the action.

doug

 

[Mentz114]

Hi Doug:

That is when the trace of the symmetric tensor happens to be zero. There will be many situations where as you say, the trace will not be zero.

It's not enough to say this - you have to show mathematically that the trace can be zero at all. My calculations indicate that if the connections are non-zero, it cannot happen. Prove me wrong. I could have made a mistake, or maybe there's some interesting constraint that makes the trace disappear.

The idea of squaring the interaction comes from

$$(A^{\mu}A^{\nu})(j_{\mu}j_{\nu}) \equiv B^{\mu\nu}T_{\mu\nu}$$

which looks like a spin-2 interaction. I don't know if it makes sense.

[good luck Patriots tonight]

 

[sweetser]

A little mass for all

Hello Lut:

We be doing physics, and in physics, there is no situation with an electrical charge that has zero rest mass. All charges have mass, no exceptions. All neutral systems with mass, well, they have mass. Only a system of pure photons and gravitons would have a trace of zero. Since it is my hope that the trace only has to do with the mass charge, and the mass charge compared to the electric charge is stupidly small (16 orders of magnitude smaller), it is reasonable to work with an effective field theory where one works with the trace being effectively zero. Size can matter. Your algebra observation sounds reasonable to me on the physics grounds. So I have to prove energy conservation, I know.

Since I am trying to write the action as a quaternion, I have not made it back out the the stress-energy tensor. What is the B thing?

Let's hope the Pats get a quick lead so I don't have to stay up late like the last game.
doug

 

[Mentz114]

Hi Doug:

Whoops, the thing I wrote is not an equation but a definition. I have edited it.

Surely the trace of the Grav field tensor is related to the mass of the field, not the mass causing the field.

there is no situation with an electrical charge that has zero rest mass. All charges have mass, no exceptions.

This may be true, but there has been a lot of talk about the electron's rest mass being purely EM in origin. Still, we have no idea what mass actually is except a number we have to add into the field equations. Or in this case try to get rid of !

Anyhow, I'm moved to have another look at my 'sums'.

 

[sweetser]

New quaternion representation for symmetric e and b fields

Hello:

I hope everyone had a good holiday season. I got a pair of noise canceling headphones so I can be off in my own little world, which I can do without headphones.

This project to unify gravity and EM was driven by math: how could quaternions be used for basic laws in physics? Maxwell himself predicted someone would figure out how to do this someday in his Treatise. I was the second person (apres Peter Jack) to figure out how to use real quaternions to write out the Maxwell equations.

The way to generate the Maxwell equations was darn complicated, keeping this, tossing away that. These equations govern the simplest stuff in the Universe, photons and electrons, so elegance - simplicity with purpose - will be part of successful approaches. The things I was throwing away were symmetric. That got me thinking my discards could be gravity. That led directly to the field equations, the 4D wave equation at the beginning of this long thread. It has been a long path making precise math statements as a consequence. Even more effort has been required to prune out the four or five technical errors I have made in trying to write out math expressions.

One thing I learned was how to back translate something I figured out with quaternions into tensors. Professional physicists are trained in tensors. They will give you a blank stare if you go on a quaternion riff. This was a great learning experience for me too, since I could tie into the vast knowledge of physics.

What I have been working on lately is math driven by the physics. Although I own the domain quaternions.com, I am not a zealot. It is the phrase, "up to an isomorphism", that gives me balance. I can look at other areas of study that use tensors or Clifford algebras, then, should I succeed in translating to quaternions, can say the two approaches are the same up to an isomorphism.

As an example, it is easy enough to write out the E and B fields using quaternions:

$$E + scalar field = -\frac{1}{2}(\nabla A + A \nabla)$$

$$B = \frac{1}{2}(\nabla A - A \nabla)$$

[I'll discuss the scalar field later, but to keep things simple in comparing the math of E and B, I will include it, otherwise it could be subtracted away]

Mix and match the 4-derivative with the 4-potential, and you get two of the most important fields in physics: the electric an magnetic fields.

One thing that was trivial to write out with tensors was a symmetric tensor, $\nabla^{\mu} A^{\nu} + \nabla^{\nu} A^{\mu}$. This tensor has 3 fields: a symmetric analog to EM's electric field E I call small e, a symmetric analog to EM's magnetic field B I call small b, and the diagonal field g, the place where gravitational and inertial mass call home (the other two, e and b, represent mass on the move). This is the component definition:

$$e = +\frac{\partial A}{\partial t} - \nabla \phi$$

$$b = (-\frac{\partial A_y}{\partial z}-\frac{\partial A_z}{\partial y}, -\frac{\partial A_z}{\partial x}-\frac{\partial A_x}{\partial z}, -\frac{\partial A_x}{\partial y}-\frac{\partial A_y}{\partial x})$$

$$g = (\frac{\partial \phi}{\partial t}, -\frac{\partial Ax}{\partial x}, -\frac{\partial Ay}{\partial y}, -\frac{\partial Az}{\partial z})$$

Look at the symmetric b: I wondered how I could possibly write that using a quaternion. I did find a way. I took the the quaternion differential operator and broke it into individual parts, then took the 4-potential, broke that up too, and put the two together as needed. Although such a micro-slicing approach works, it is not elegant - simplicity with a purpose.

I have been thinking about, how to make a "symmetric curl". OK, most of the time over the break I spent rearranging stuff in my basement, but it was always there to nag me. The basement has improved, but more relevant, I did come up with an idea... The way to make a symmetric curl in a 4x4 real matrix representation of a quaternion is to make all the off diagonal elements have the same sign. That way, when they get together to form a curl, everyone will have the same sign. So this was my first guess:

$$q (t, x, y, z) = \left(\begin{array}{cccc} t & x & y & z\\ x & t & z & y\\ y & z & t & x\\ z & y & x & t \end{array}\right)$$

The problem with this is all the positive signs. I had taken what was antisymmetric and made it symmetric. Now I needed to take what was symmetric, the scalar t, and figure out how to make it antisymmetric. Therefore I dropped in a factor of "I" which commutes with all the real values of t, x, y, and z:

$$q (t, x, y, z) = \left(\begin{array}{cccc} t I & x & y & z\\ x & t I & z & y\\ y & z & t I & x\\ z & y & x & t I \end{array}\right)$$

For this to work as a representation of the quaternions, t, x, y, and z are constrained to evaluate to real numbers. Not only must one exclude t=x=y=z=0, but one needs to exclude certain values where t=0. If these are removed from the set of possible values, then one could use this as a representation of the quaternion division algebra.

The Hamilton 4x4 real matrix representation of quaternions, and what I will call the "I" representation until I get a better name, cannot be mixed together. The I representation does solve the problem at hand, because a little cutting and pasting, with a few edits, leads to this way of writing small e and b:

$$e = \frac{I}{2}(\nabla_I A_I - A_I \nabla_I)$$

$$b + scalar field = \frac{1}{2}(\nabla_I A_I + A_I \nabla_I)$$

The EM E field is a sum, the symmetric e is a difference. The EM B field is a difference, the symmetric b is a sum. The additional scalar field is in EM's E field, and the symmetric b field. The scalar field is made up of the components of the field g. To go from the fields of EM to those of gravity involves switching the way quaternions are represented.

This qualifies as elegant.
doug

 

[sweetser]

Even quaternions

Hello Folks:

I like to make specific technical proposals, even if I am uncomfortable with certain parts. In my previous post, I disliked having the factor of "I". It looked too much like a biquaternion which is not a division algebra. Inside Mathematica, it led to technical difficulties.

The beauty of a specific proposal is that it can be modified. I decided to drop the factor of "I", and see what happened. That was my first guess after all (I'm using a 2 to suggest "evenness"):

$$q2 (t, x, y, z) = \left(\begin{array}{cccc} t & x & y & z\\ x & t & z & y\\ y & z & t & x\\ z & y & x & t \end{array}\right)$$

With this representation, the product of two quaternions commutes. These symmetric matrices will surely work with addition, subtraction, and multiplication. The open question is whether division works for all possible values of t, x, y, and z. We know already that t=x=y=z=0 doesn't work. I asked Mathematica to find the inverse of the above matrix. It returned a fraction. I then asked Mathematica to factor the dominator, to find out where any zeros might live. Here was the answer:

$$denominator = (t-x-y-z)(t+x+y-z)(t+x-y+z)(t-x+y+z)$$

Every possible quaternion will have an inverse except those quaternions where one of the elements happens to equal the sum of the other three. This happens for photons, where c dt = dx + dy + dz. And apparently it applies to permutations of photons, with cdt switching roles with dx, dy, and dz. The gamma matrices in quantum field theory do this sort of shuffling.

Quaternions are a division algebra. The most common representation for quaternions is the one developed by Hamilton were two quaternions do not commute under multiplication. With this representation - so long as we avoid photons - two quaternions can commute.

The standard definition of a derivative will work for the even quaternion, no change needed. Neat.

Any 4x4 matrix that has the same value down the diagonal can be represented by a sum of the Hamilton representation and this even representation. Cool!

doug

 

[Mentz114]

Hi Doug:

That's an interesting matrix. It looks like the multiplication table for the symmetric curl.
It defines a commuting operator ~ so that

x~y=z
x~z=y
y~z=x
and anything~t = anything

I'm impressed with Mathematica's factorizing power.

Lut

 

[sweetser]

Hello Lut:

The q2 matrix is just the standard 4x4 real matrix representation of the Hamilton quaternion, just no minus signs allowed. Perhaps philosophers will discuss that aspect someday, since the always plus guy is used to write the gravity fields.

Mathematica figured out the inverse and the factorization in the time it took to hit the return key. Scary fast stuff. Toss in the factor of "I", and it becomes hard to tell the software t, x, y, and z are real. I concluded that it was not possible to constrain them to the reals without getting rid of the factor of I. Was I needed? No? Great! In fact, so good, I will be taking the misses out to dinner to celebrate. It looks like I am using physics to upend a deeply rooted idea: representations of quaternion multiplication must not commute. The first one did, but there is another option.

doug

 

[sweetser]

+/- Non Associative and +/- Non Commutative Multiplication

Hello:

What is physics is math, what is math is physics.

The core math idea of this proposal is that while we use an antisymmetric tensor which is the 4-derivative of a 4-potential for EM, to be complete, Nature must also use the symmetric 4-derivative of a 4-potential. That is all gravity is.

Earlier in this thread, I was mystified by Lawrence B. Crowell's focus on the antisymmetric tensor and his exclusion of the symmetric one. That's the way technical discussions sometimes go.

What was so neat is that I have now documented myself doing the same thing, but for quaternions. Everything I have ever read on quaternions, the product of two quaternions contains the antisymmetric curl. I never considered that there could a symmetric product. Non-commuting is too basic to what quaternions are, right?

Quaternions are a 4D division algebra. When Hamilton went to implement them, he chose a representation where multiplication was associative and non-commuting. I figured out how to make the multiplication non-associative and non-commutative by defining multiplication as A* B. With the q2 representation, multiplication is associative and commutative. And to complete the set, q2* q2' will be non-associative and commutative.

No wonder Nature is tricky to figure out because she uses all four simultaneously!
doug

 

[Mentz114]

Hi Doug:

What is physics is math, what is math is physics.

Not true. Mathematics is an art of logic and imagination. Physics is the laboratory.
I can write mathematically consistent space-times for instance, whose physical existence would be riddled with impossibilites.

The core math idea of this proposal is that while we use an antisymmetric tensor which is the 4-derivative of a 4-potential for EM, to be complete, Nature must also use the symmetric 4-derivative of a 4-potential. That is all gravity is.

I wish I had your inside line to mother nature. As I said before, this is not science, it's religious faith.

we use an antisymmetric tensor which is the 4-derivative of a 4-potential for EM,

Yes ! WE use it as a matter of convenience because it allows a compact way to ensure Maxwells equations are satisfied. I don't think mother nature is a mathematician.

You'd win more friends in the physics community if you established that your theory is gauge invariant, so you have the freedom to play with your potential the way you propose.

Your field equations may look elegant, but until you sort out what's causing the curvature and the potential ( source) it has no computational power. How do I solve actual problems with it ?

Regards,
Lut

 

[sweetser]

Put it to the test

Hello Lut:

I confess, this is a scientific belief:

What is physics is math, what is math is physics.

It is my opinion that this is also a scientific belief:

Not true. Mathematics is an art of logic and imagination. Physics is the laboratory.
I can write mathematically consistent space-times for instance, whose physical existence would be riddled with impossibilities.

In the radical form of my belief, only once we see the math in Nature are we confident that a math theorem is true, and not bad math. My canonical example is the math wonk who solved the math problem from Hell: develop a polynomial to describe knots, then based on that polynomial, tell if it can be untied. Those polynomials appear in some systems of quantum mechanics. It may be darn hard to find the physical system, but it is my belief that enough searching will find it if the logic is true.

I respect your belief which happens to be different from mine.

The core math idea of this proposal is that while we use an antisymmetric tensor which is the 4-derivative of a 4-potential for EM, to be complete, Nature must also use the symmetric 4-derivative of a 4-potential. That is all gravity is.

I wish I had your inside line to mother nature. As I said before, this is not science, it's religious faith.

My quote is not a belief, it is a hypothesis that can be tested. Measure the bending of light to a tenth of a microarcsecond, and my hypothesis predicts light will bend 0.8 microarcseconds more than GR. I would accept my hypothesis is wrong should the necessary experiment be done and support the bending predicted by GR.

I also don't think the position about completeness is a belief either, it is an observation. Subtraction takes some thing away to make the antisymmetric tensor. A complete story uses everything. That is logic at work.

You'd win more friends in the physics community if you established that your theory is gauge invariant, so you have the freedom to play with your potential the way you propose.

I'd rather be right than popular :-) The proposal is gauge invariant in the usual sense of the word for all mass less particles. When one has massive particles, the proposal does involve choice in how things are measured, either as changes in the derivatives of the 4-potential, or changes in the connection which are changes in the metric. I have had no luck in communicating what the preceding sentence means to those skilled in the mathematical physics arts, which is exasperating since it falls out of looking at the definition of a covariant derivative (it is the sum of these two things, so how much comes from one or the other is arbitrary).

Your field equations may look elegant, but until you sort out what's causing the curvature and the potential ( source) it has no computational power. How do I solve actual problems with it ?

The most important message is that most of physics is about doing almost nothing. To keep going for 13+ billion years, a system has to be extraordinarily passive, not active, get it all done kind of force. The idea is not cause, but more what is the absolute closest thing to absolute nothingness? Absolute nothingness is the Minkowski metric. The closest thing to that is a 4D simple harmonic oscillator.

In the first post I wrote out a physical relevant solution to the field equations. It was for an electrically neutral point charge. Got to get the basics done. Somewhere else in this long thread I solved the problem for an electrically changed point source (I think I did anyway). What one has to do is make a decision on how things are measured, either as a change in potential or a change in the metric or a combination of both, and proceed from there.

doug