Unifying Gravity and EM

[Careful]

Hi Doug,

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

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

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

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


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


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

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

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

Cheers,

Careful

 

[sweetser]

A valid contravariant expression (cont.)

Hello:

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

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

**I thought that you meant this, **

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

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

** Obviously, in GR **

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

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

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

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

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

doug

 

[sweetser]

Asymmetric mixed rank 2 tensors

Hello Careful:

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

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

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

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

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

doug

 

[Careful]

Hi,

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

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


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

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

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

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

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

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

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

Indeed, as I mentioned before...

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

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

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

 

[sweetser]

The GEM action

Hello:

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

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

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

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

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

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

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

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

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

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

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

I wish to study the following action:
$$S_{GEM}=\int \sqrt{-g} d^4 x (\frac{-\rho_{m}}{\gamma}-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu} -\frac{1}{2c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu})$$
This contains the GEM action as was written in the first post of this thread, plus an inertial term to calculate the force equation by varying the 4-velocity There will be people who want to see the irreducible field strength tensors written in the Lagrangian:
$$S_{GEM}=\int \sqrt{-g} d^4 x (\frac{-\rho_{m}}{\gamma}-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu} -\frac{1}{4c^{2}}(\nabla_{\mu} A^{\nu}-\nabla^{\nu} A_{\mu})(\nabla^{\mu} A_{\nu}-\nabla_{\nu} A^{\mu}) - \frac{1}{4c^{2}}(\nabla_{\mu} A^{\nu}+\nabla^{\nu} A_{\mu})(\nabla^{\mu} A_{\nu}+\nabla_{\nu} A^{\mu}))$$
A Careful pointed out, the GEM action is not invariant under a gauge transformation. This statement is too strong. The GEM action is invariant under a gauge transformation for massless particles, where the trace of the irreducible tensor is zero. For any charged particles with mass, it is their mass that breaks the gauge symmetry (the Higgs mechanism is unnecessary). Note that all electrically charged particles do have a mass. The GEM action is effectively gauge invariant, because the size of the mass charge is thirteen orders of magnitude smaller than the electric charge for a proton, and we only know the charge of a proton to ten significant figures.

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

doug

 

[Careful]

Hi,

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

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

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

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


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


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


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

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


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

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

Cheers,

Careful

 

[sweetser]

Symmetric versus antisymmetric tensors

Hello Careful:

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

The asymmetric action has not changed.

I am still having trouble with the irreducible tensors.

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

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

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

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

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

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

doug

 

[Careful]

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

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


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

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


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

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


Cheers,

Careful

 

[sweetser]

Hello Careful:

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

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

Agreed in the general.

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

doug

 

[Careful]

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

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

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

Cheers,

Careful

 

[sweetser]

Hello Careful:

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

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

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

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

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

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

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

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

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

As to what I will do with the GEM action, I see little choice. The field to vary is the 4-potential. Folks that are good with actions can look at the action in claim one and get to the field equations as a one liner, $J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}$. In the static case, if one chooses to work with a flat Minkowski metric, the solution is charge/distance. If one chooses to work with the exponential metric,
$$g_{\mu\nu}=\left(\begin{array}{cccc} exp(-2\frac{GM}{c^{2}R}) & 0 & 0 & 0\\ 0 & -exp(2\frac{GM}{c^{2}R}) & 0 & 0\\ 0 & 0 & -exp(2\frac{GM}{c^{2}R}) & 0\\ 0 & 0 & 0 & -exp(2\frac{GM}{c^{2}R})\end{array}\right).$$
the potential is static. It is a fun exercise to show this metric solves the field equations for a static potential.

doug

 

[Careful]

Hi,

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

Good Luck,

Careful

 

[sweetser]

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

doug

 

[Careful]

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

doug

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

Cheers,

Careful

 

[sweetser]

Field equations and Lagrange densities

Hello Careful:

One can certainly have this perspective:

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

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

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

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

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

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

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

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

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

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

The second story is a personal one. Back in 2000, I had an audience with one of the most well known physicists in Boston. I said here are my field equations:
$$J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}$$
See how wonderful they are?. If the mass current density in the same units as electric charge are thirteen orders of magnitude smaller than electric charge, one has the Maxwell equations in the Lorenz gauge. If the system is electrically neutral and static, one has Newton's law of gravity. If the system is neutral and not static, the field equations transform like a 4-vector, and thus gets along with SR, the motivation for GR being disolved.

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

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

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

Happy vacation days,
doug

 

[sweetser]

Good bye mixed tensors in 2006

Hello all:

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

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

Have a good new year,
doug

 

[Careful]

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

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

Have a good new year (and keep on learning )

Cheers,

Careful

 

[sweetser]

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

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

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

Will keep learning. Enjoy the moment.
doug

 

[sweetser]

Abstract for APS meeting in Dallas

Hello:

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

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

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

doug

 

[Lawrence B. Crowell]

GEM theory problem

The Lagrangian
$$L~=~ (J^a_q~-~J^a_m)A_a~-~1/2(\nabla_aA^b\nabla^aA_b$$
involves the combination of a mass current and a charge current. Further the four potential is defined as being associated with both gravity and EM. Electromagnetism is a U(1) gauge theory. Gravity is an $$SO(3,1)~\sim~SL(2, C)$$ theory. So this theory appears to have some analogy with the standard model of electroweak interactions. Yet in that case the gauge potential is
$$A_a = A(em) + A(weak),$$
for the $$SU(2)\times U(1)$$ theory. For an $$SL(2,C)\times U(1)$$ theory one might consider a similar construction, which is a twisted bundle. However, this is not apparent from your Lagrangian. I am presuming that the mass current is defined as
$$J_a~=~T_{ab}e^b,$$
or by some similar means. However, $$\nabla^aJ_a$$, a term which would emerge from the Euler-Lagrange equation, does not transform homogenously as the connection term emerges. This is related to the so called nonlocalizability of mass-energy in general relativity. So this would indicate that the field equations which emerges from the Lagrangian are not gauge covariant. This can only be recovered if there is are Killing vectors in the direction of this mass current. So without some special considerations the theory appears not to be covariant under the transformations of the theory.

This approach might best be extended to consider a theory that is $$SO(3,1)\times SO(4)$$ with,
$$SO(4)~\simeq~SU(2)\times SU(2).$$
One of the $$SU(2)'s$$ might be split on a singularity in its moduli space to give $$U(1)\times U(1)$$, where one of these can play the role of the electromagnetic field. The other $$U(1)$$ would then correspond to some massive field that is irrelevant to physics if the mass is large enough. The other $$SU(2)$$ is then the weak interactions.

This might be started by considering a tetrad of the form
$$E_a^b~=~\gamma^be_a,$$
where $$\gamma^b$$ is a Dirac matrix in some representation. One then would have
$$de^a~=~A_be_adx^b,$$
where $$A_b$$ is the gauge potential for the Yang-Mills gauge field. Similarly by $$tr(\gamma_a\gamma_b)~=~4g_{ab}$$, if the representation of the Dirac matrices is local (changes from chart to chart) the differential on the tetrad gives
$$\partial_c(E_a^b)~=~\gamma^bA_ae_c~+~{\Gamma^b}_{cd}E_a^d.$$
From here your general gauge potential, call it $${\cal A}$$, would be
$${{\cal A}^b}_{ac}~=~\gamma^bA_ae_c~+~{\Gamma^b}_{cd}E_a^d$$
The field tensor for this theory would be defined from
$$\partial_d{{\cal A}^b}_{ac}~-~\partial_c{{\cal A}^b}_{ad},$$
which if one were to work out the bits should result in the gravity and EM sectors.
Lawrence B. Crowell

 

[sweetser]

Hello Lawerence:

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

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

 

[Doc Al]

Lawrence:

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

- Doc

 

[sweetser]

Gauge symmetry

Hello Lawrence:

The post deals with gauge symmetry issues.

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

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

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

Look at one limitation of gauge theories. Let me quote extensively from Michio Kaku's "Quantum Field Theory: A Modern Introduction" p. 106:

Because of gauge invariance, there are also complications when we quantize the theory. A naive quantization of the Maxwell theory fails for a simple reason: the propagator does not exist. To see this let us write down the action in the following form:
$$\mathcal{L}=1/2 A^{\mu}P_{\mu \nu}\partial^{2}A^{\nu}$$
where:
$$P_{\mu \nu}=g_{\mu \nu}-\partial_{\mu}\partial_{\nu}/(\partial)^2$$
The problem with this operator is that it is not invertible, and hence we cannot construct a propagator for the theory. In fact, this is typical of any gauge theory, not just Maxwell's theory. This also occurs in general relativity and in superstring theory. The origin of the noninvertibility of this operator is because $P_{\mu \nu}$ is a projection operator, that is, its square is equal to itself:
$$P_{\mu \nu}P^{\nu \lambda}=P_{\mu}^{\lambda}$$
and it projects out longitudinal states:
$$\partial^{\mu}P_{\mu \nu}=0$$
The fact that $P_{\mu \nu}$ is a projection operator, of course goes to the heart of why Maxwell's theory is a gauge theory. This projection operator projects out any states with the form $\partial_{\mu}\Lambda$, which is just the statement of gauge invariance.

Physicists understand exactly how to deal with this issue: pick a gauge. With the GEM proposal, this choice is not available. That may be a good thing for quantizing the theory.

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

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

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

doug

 

[sweetser]

Localizable theories

Hello Lawrence:

This post is in reply to the nonlocalizable issue.

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

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

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

Which is better? You make the call,
doug

 

[member 11137]

The Lagrangian
$$L~=~ (J^a_q~-~J^a_m)A_a~-~1/2(\nabla_aA^b\nabla^aA_b$$
involves the combination of a mass current and a charge current. Further the four potential is defined as being associated with both gravity and EM. Electromagnetism is a U(1) gauge theory. Gravity is an $$SO(3,1)~\sim~SL(2, C)$$ theory. So this theory appears to have some analogy with the standard model of electroweak interactions. Yet in that case the gauge potential is
$$A_a = A(em) + A(weak),$$
for the $$SU(2)\times U(1)$$ theory. For an $$SL(2,C)\times U(1)$$ theory one might consider a similar construction, which is a twisted bundle. However, this is not apparent from your Lagrangian. I am presuming that the mass current is defined as
$$J_a~=~T_{ab}e^b,$$
or by some similar means. However, $$\nabla^aJ_a$$, a term which would emerge from the Euler-Lagrange equation, does not transform homogenously as the connection term emerges. This is related to the so called nonlocalizability of mass-energy in general relativity. So this would indicate that the field equations which emerges from the Lagrangian are not gauge covariant. This can only be recovered if there is are Killing vectors in the direction of this mass current. So without some special considerations the theory appears not to be covariant under the transformations of the theory.
This approach might best be extended to consider a theory that is $$SO(3,1)\times SO(4)$$ with,
$$SO(4)~\simeq~SU(2)\times SU(2).$$
One of the $$SU(2)'s$$ might be split on a singularity in its moduli space to give $$U(1)\times U(1)$$, where one of these can play the role of the electromagnetic field. The other $$U(1)$$ would then correspond to some massive field that is irrelevant to physics if the mass is large enough. The other $$SU(2)$$ is then the weak interactions.
This might be started by considering a tetrad of the form
$$E_a^b~=~\gamma^be_a,$$
where $$\gamma^b$$ is a Dirac matrix in some representation. One then would have
$$de^a~=~A_be_adx^b,$$
where $$A_b$$ is the gauge potential for the Yang-Mills gauge field. Similarly by $$tr(\gamma_a\gamma_b)~=~4g_{ab}$$, if the representation of the Dirac matrices is local (changes from chart to chart) the differential on the tetrad gives
$$\partial_c(E_a^b)~=~\gamma^bA_ae_c~+~{\Gamma^b}_{cd}E_a^d.$$
From here your general gauge potential, call it $${\cal A}$$, would be
$${{\cal A}^b}_{ac}~=~\gamma^bA_ae_c~+~{\Gamma^b}_{cd}E_a^d$$
The field tensor for this theory would be defined from
$$\partial_d{{\cal A}^b}_{ac}~-~\partial_c{{\cal A}^b}_{ad},$$
which if one were to work out the bits should result in the gravity and EM sectors.
Lawrence B. Crowell

These explanations sound good to me; I have a hope to be on the right road. Thanks

 

[member 11137]

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

 

[sweetser]

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

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

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

doug

 

[member 11137]

I am sorry to report, but I don't think so.

Don't be sorry, it's your right to have a different opinion than mine. It's making the debate interesting.

I have tried to make clear that I am using standard math tools: the Lagrange density, the variation of the action, the connection, and tensors.

In someway me too... but with a small variation and fantasy relatively to the absolutely standard math tools. I get also a Lagrange density but it is a little bit different than yours. The other difference is that I can connect it with some underground stream...

The reliance on standard methods has allowed me to make corrections to the proposal, specifically that I should include the kinetic energy term ($\rho/\gamma$) if I want to get to a Lorentz force law, and that I should avoid using a mixed asymmetric tensor because it mixes me up.

... and quite naturally get the Lorentz-Einstein force law

Another weakness in my proposal is I do not yet have a way to discuss it precisely in terms of group theory. I apologize for those schooled in the craft, but here is my effort to connect to group theory. The electric charge of an electron is 16 orders of magnitude greater than its mass charge measured in the same units. That means the antisymmetric "deviation from the average change in the potential" field strength tensor, $\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}$ will dwarf the symmetric "average amount of change in the potential", $\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}$, which together compose the asymmetric field strength tensor, $\nabla^{\mu}A^{\nu}$. The GEM action thus has an approximate $U(1)$ symmetry, the symmetry of the antisymmetric field strength tensor of EM. Here is where the ice gets thin for me.

Don't panic, you cannot imagine how I admire the brightness of your knowledge and for me the ice is so thin that I can see the water...

If one can work in a completely flat spacetime, or work on a manifold where spacetime is curved, I believe that may be called a diffeomorphism symmetry. The energy of curvature is information about how a symmetric math tool, the metric, changes (I am presuming both a torsion-free, and metric compatible connection, or this statement would not be valid). Changes in a symmetric metric are symmetric. That information must be stored in the symmetric field strength tenor, the tensor that breaks U(1) symmetry. U(1) symmetry is barely broken due to the incredible flatness of the Universe where we live.
doug

The flatness (average) of our universe is a mystery that can perhaps be explained if my approach makes sense... but here I need the help and the critics of professionnals

 

[Lawrence B. Crowell]

Yang-Mills fields & symmetric field tensor

Doug,

Since you indicated the structure of gauge theory, I figured I would do the same, but on a somewhat more fundamental level. All of this is based on differential forms. The coboundary operator $d~=~dx^a\partial_a$ acts on a section $x$, or some cut through a principle bundle $P$over a manifold $\cal M$ simply as $ds$. So act upon $s$ with $s^\prime~=~gs$, where $g$ is a group element of the Lie group of transformations of the theory. Now consider $ds^\prime$ with $ds~=~As$, where [itex]A[/tex] is a gauge connection
$$ds^\prime~=~d(gs)$$
This then leads to
$$ds^\prime~=~(dg)s~+~gAs$$
$$ds^\prime~=~\big((dg)g^{-1}~+~gAg^{-1}\big)s^\prime.$$
This leads to the transfomation of a gauge connection as
$$A~\rightarrow~(dg)g^{-1}~+~gAg^{-1}$$
For a group element $g~=~e^{i\chi}$ this means that the gauge transformation is
$$A~\rightarrow~A~+~id\chi~+~i[\chi,~A],$$
and for and ableian theory (commutator = 0) and $d~=~dx^a\partial_a$ this leads to $A~\rightarrow~A~+~\nabla\chi$, where $i$ has been absorbed into $\chi$.

The differential form $d$ satisfies $d^2~=~0$ because
$$d^2~=~dx^a\wedge dx^b\partial_a\partial_b,$$
and $dx^a\wedge dx^b$ is antisymmetric and $\partial_a\partial_b$ is symmetric. The appropriate differential form is the gauged differential form $D~=~d~+~A$. The field tensors are then obtained from
$$D^2~=~D\wedge D~=~(d~+~A)\wedge(d~+~A)$$
$$=~F=~dA~+~A\wedge A,$$
where the $A\wedge d$ is zero acting on unity or $1$. If we expand this in coordinates we then have
$$F~=~\big(\partial_aA_b~-~\partial_bA_a~+~[A_a,~A_b]\big)dx^a\wedge dx^b.$$
I have ignored structure constants and the rest here, but those can be included after the fact here. For the spatial variables this leads to the invariance of the magnetic field $B~=~\nabla\times A$ under the gauge transformation. The two-form $F~=~F_{ab}dx^a\wedge dx^b$ contains the antisymmetric field tensor with the electric and magnetic field components which can be easily derived.

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


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

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

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

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

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

Lawrence B. Crowell

 

[Lawrence B. Crowell]

Nonlocalizability of energy

I think there is a problem here. The energy is defined by
$$E = T_{0a}U^a$$
where the four velocity is is tied to the metric by
$$ds^2 = g_{ab}dx^adx^b$$
and so
$$1 = U^aU_a$$
as such the $nabla_aE$ will involve connection terms from the covariant derivative of the four-vector. Remember that for "dust" the momentum-energy term is
$$T_{ab} = \rho U_aU_b + pg_{ab}$$
and so any potential term is wrapped up in $\rho$. This means that the potential can also be set to zero at a Riemannian normal coordinate by an appropriate choice of the connection.

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

Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

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

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

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

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

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

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

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

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

I provided definitions for E, B, e, b, and g. Those are just ways to rewrite $\nabla^{\mu}A^{\nu}$, or more fine-grained, E and B represent $\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}$ and g, e, and b, represent $\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}$. The field equations also look darn simple, a 4D wave equation with two currents, one for E, B, the other for g, e, and b:
$$J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}$$
Let's focus only on the first one:
$$\rho_{q}-\rho_{m}=\frac{1}{c}\partial^{2}\phi/\partial t^{2}-c\nabla^{2} \phi$$
I call this equation "General Gauss' law", a small joke for math guys who like Chinese food. Here it is rewritten in terms of the five fields:
$$\rho_q - \rho_m = \frac{\partial^2 \phi}{c \partial t^2} - c \frac{\partial^2 \phi}{\partial x^2} - c \frac{\partial^2 \phi}{\partial y^2} - c \frac{\partial^2 \phi}{\partial z^2}$$

$$= \frac{\partial^2 \phi}{c \partial t^2} + \frac{1}{2} \frac{\partial}{\partial x} ( (-\frac{\partial A_x}{\partial t} - c \frac{\partial \phi}{\partial x} ) + ( \frac{\partial A_x}{\partial t} - c \frac{\partial \phi}{\partial x} ) )$$
$$+ \frac{1}{2} \frac{\partial}{\partial y} ( ( - \frac{\partial A_y}{\partial t} - c \frac{\partial \phi}{\partial y} ) + ( \frac{\partial A_y}{\partial t} - c \frac{\partial \phi}{\partial y} ) )$$
$$+ \frac{1}{2} \frac{\partial}{\partial z} ( ( - \frac{\partial A_z}{\partial t} - c \frac{\partial \phi}{\partial z} ) + ( \frac{\partial A_z}{\partial t} - c \frac{\partial \phi}{\partial z} ) )$$
$$= \frac{\partial g_t}{c \partial t} + \frac{1}{2} ( \vec{\nabla} \cdot \vec{E} + \vec{\nabla} \cdot \vec{e} )$$

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

doug

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

 

[Lawrence B. Crowell]

Dear Doug,

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

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

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

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

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

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

then

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

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

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

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

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

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

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

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

cheers,

Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

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

I appreciate how useful differential forms are for doing gauge theory. Perhaps there is no better tool. Yet if a tool clearly states its limitations, then it is time for a skeptic to doubt the tool itself.

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

This says to me that the tools of differential forms are putting unreasonable limitations on physical descriptions of Nature. I very much doubt I will understand what an "elliptic complex of the Atiyah-Singer theorem" means on a physical level. That sort of thing happens all the time, welcome to the world of physics. I do fell rock solid on saying that $\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}$ is the deviation from the average amount of change in $A^{\nu}$. Logic then dictates to me that I must work with the average amount of change in $A^{\nu}$. If differential forms are not up to this particular task, then for this particular proposal, I cannot use them.

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

doug

 

[Lawrence B. Crowell]

gauge theory & moduli

I am redoing this. I think this should be better

Hello Doug,
\\
To be honest I was a bit afraid of this. You state:
\\
In EM, $A^\mu$ is not measured because of gauge symmetry. In my GEM proposal, $A^\mu$ is measurable.
\\
The problem is that potentials are fictions. Even in the most elementary of physics, say $V~=~mg(x~-~x_0)$ the potential can be set to anything by liberally assigning an $x_0$ to any value. The force, which contributes to dynamics, $F~=~mg$ is independent of how the potential is set.
\\
In gauge theory gauge potentials are really little more than mathematical artifacts, where they enter into gauge conditions or gauge fixing Lagrangians as constriants to provide sufficient information to solve the DEs. Of course things get a bit mysterious, as found with the Ahrahnov-Bohm experiment. Yet even though the electrons pass around a solonoid and do not interact with the magnetic field inside it is by Stokes' law
$$\oint_{C=\partial{\cal A}} A\cdot dx~=~\int\int_{\cal A}B\cdot dA$$
the magnetic field which determines the phase shift
$$\Delta\phi~=~exp(\oint_{C=\partial{\cal A}} A\cdot dx)$$
\\
The Atiyah-Singer index theorem and the elliptic complex gives a construction for the moduli space. For a gauge connection the moduli is [tex]A/g[/itex], which is a set of gauge equivalent connections: gauge connections moduli group actions. A moduli space is a space of such moduli, which for [tex]SU(2)[/itex] is five dimensional. The theory of moduli spaces has emerged as the 800 pound gorilla in the Yang-Mills theory. It is a big aspect of superstring theory these days. Maybe in a few days I will post a tutorial on this.
\\
The obvious problem that your GEM theory has is that it goes against a lot of methodology used in physics. This is not to say that any particular "canon" in physics should be regarded as utterly beyond question, but I think that any theory which considers potentials as something physically real (measureable) is bound to run into a lot of resistance.
\\
Lawrence B. Crowell

 

[sweetser]

Diff(M) as a gauge group

Hello Lawrence:

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

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

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

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

At the same time, I also knew that a choice must be made before one can solve a differential equation. That is the calling card of a gauge theory. Take the GEM field equations:
$$J_{q}^{\mu}-J_{m}^{\mu}=(\frac{1}{c}\partial^{2}/\partial t^{2}-c\nabla^{2})A^{\mu}$$
Given current densities, the 4-potential cannot be found. The reason is that no information has been provide about the metric. If the metric is Euclidean and flat, it is easy enough to calculate the potential. Yet the same differential equation holds in curved spacetime. Choose a different metric, and a completely different potential is required. Let me make this point as concrete as possible. Let's solve General Gauss' equation for a static electrically charged point source. Choose to work in flat, Euclidean spacetime. The answer would have been known to Poisson:
$$\rho_q - \rho_m = -\nabla^{2}\frac{(q_{q} + q_{m})}{R}$$
Choose to work in the exponential metric at the start of this thread, or to save you from clicking,
$$g_{\mu\nu}=\left(\begin{array}{cccc} exp(-2\frac{\sqrt{G}q + GM}{c^{2}R}) & 0 & 0 & 0\\ 0 & -exp(2\frac{\sqrt{G}q+GM}{c^{2}R}) & 0 & 0\\ 0 & 0 & -exp(2\frac{\sqrt{G}q+GM}{c^{2}R}) & 0\\ 0 & 0 & 0 & -exp(2\frac{\sqrt{G}q+GM}{c^{2}R})\end{array}\right).$$
Most people have not had to calculate a Christoffel symbol of the second kind. This is one of few examples where that calculation is easy. The metric is diagonal and static, therefore the only term that matters is $g_{00}$. The derivative of an exponential gives back the exponential times the derivative of the exponent, which you will notice is charge/R. In the definition of the Christoffel, the derivative of $g_{00}$ contracts with $g^{00}$. The sign of $g^{00}$ flips, so the exponential will drop out, leaving only the derivative of the exponent:
$$g^{00}g_{00,R} = -\nabla^{2}\frac{(q_{q} + q_{m})}{R}$$
That should look familiar. A constant potential will solve the fields equations if one chooses to use the exponential metric.

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

doug

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