Unifying Gravity and EM

[Chronos]

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

 

[sweetser]

Hello Chronos:

Let's break up the problem into smaller parts. Patrick had made a good technical point: a force equation is not a field equation, so there must be two separate ways to generate those equations starting from a Lagrange density. Both the force and field equations must have either like charges repelling as happens in EM, or like charges attracting as happens in gravity. So let's start with just enough terms in the Lagrange density to cover force. That would be those terms that have velocity in them, because one does the calculus of variations with velocity to generate the force or equations of motion. So...
$$\mathcal{L}_{force, like repel}=-\rho_{m}/\gamma-\rho_{q}U^{\mu}A_{\mu}$$
[tex]\mathcal{L}_{force, like attract}=-\rho_{m}/\gamma+\rho_{m}U^{\mu}A_{\mu}
$$\gamma=\frac{1}{\sqrt{1-\beta^{2}}}$$
$$U^{\mu}=(\gamma,\gamma\beta_{x},\gamma\beta_{y},\gamma\beta_{z})$$
To generate the field equations, the calculus of variations is done on the 4-potential, $A_{\mu}$. The inertia term, $-\rho_{m}/\gamma$, has no 4-potential, so plays no role in generating the field equations. We need to add the field strength tensor contraction. Now I cannot just tack on any field strength tensor contraction because this is the term that represents the field particles, and it must have the correct symmetry: odd integral spin if like charges repel, even integral spin if particles attract. So...
$$\mathcal{L}_{field, like repel}=-\rho_{q}U^{\mu}A_{\mu}-(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu}) (\partial^{\mu}A_{\nu}-\partial^{\nu}A_{\mu})$$
$$\mathcal{L}_{field, like attract}=+\rho_{m}U^{\mu}A_{\mu}-(\nabla_{\mu}A^{\nu}+\nabla_{\nu}A^{\mu}) (\nabla^{\mu}A_{\nu}+\nabla^{\nu}A_{\mu})$$
Let's count the changes between the Lagrange densities where like charges attract or repel. First there is the one in the charge coupling term, $-/+\rho$. Second and third are the two signs in the field strength tensor contraction. The final difference is between the kind of derivatives used: an exterior derivative for the anti-symmetric tensor (or deviation from the average amount of 4-change in the 4-potential) that is independent of how the symmetric tensor changes, and the covariant derivative (or average amount of 4-change in the 4-potential) that depends on how the metric changes up to a gauge transformation (we get to decide if the change is due to changes in the potential or the metric or both).

I don't think I can claim to "derive" any of the signs. Instead, it is my hypothesis that the the GEM Lagrange density,
$$\mathcal{L}_{GEM}=-\rho_{m}/\gamma-\frac{1}{c}(J_{q}^{\mu}-J_{m}^{\mu})A_{\mu} -\frac{1}{2c^{2}}\nabla_{\mu}A^{\nu}\nabla^{\mu}A_{ \nu}$$
can do both gravity and EM. You should be able to see all the preceding Lagrange densities are part of this one. At least that is my hope :-)

doug

 

[sweetser]

Precession of the Perihelion of a Planet

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

 

[hello1]

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

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

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

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

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

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

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

 

[sweetser]

Hello Hello1:

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

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

doug

 

[hello1]

doug,

thanks a lot!

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

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

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

Joe

 

[Don J]

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

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

 

[sweetser]

Hello Don:

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

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

doug

 

[sweetser]

Technical issues with GR

Hello:

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

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

Up to this point, I am in complete agreement.

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

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

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

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

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

doug

 

[Don J]

Hello:
I think I am obligated to point out possible logical flaws in general relativity.
(snip)
...
Hilbert did it the proper way. He started with a super simple Lagrange density, $\mathcal{L}_{GR}=R$. Varying the action with respect to the metric field $g_{\mu \nu}$, one gets the Einstein field equations.
One final clarification. I use the Chistoffel symbol in the gauge symmetric central to proposal, but at no time is the Reimann curvature tensor, or either of its contractions the Ricci tensor or Ricci scalar needed.
doug

Some of the problems you mention were worked out in the early 60
Einstein Equations in Arnowitt, Deser and Misner (ADM) 3+1 Form
http://www.tat.physik.uni-tuebingen.de/~koellein/bericht-WEB/node19.html

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

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


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

Institute for Advanced Study, Princeton, New Jersey

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

 

[sweetser]

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

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

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

Abhay Ashtekar is a great writer, making some technical points very clear, so that I can clearly disagree with him :-) Let me quote his opening paragraph:

General relativity and quantum theory are among the greatest intellectual achievements of the 20th century. Each of them has profoundly altered the conceptual fabric that underlies out understanding of the physical world. Furthermore, each has been successful in describing the physical phenomena in its own domain to an astonishing degree of accuracy. And yet, they offer us strikingly different pictures of physical reality. Indeed, at first one is surprised that physics could keep progressing blissfully in the face of so deep a conflict.

Brilliant!

The reason of course is the ‘accidental’ fact that the values of fundamental constants in our universe conspire to make the Planck length so small and the Planck energy so high compared to laboratory scales.

If you want a paying job in theoretical physics, this is a good thing to profess. As an utterly independent researcher, I do not think the Planck length has anything to do with the problem, zero, zippo. It is all about math. EM uses a field strength tensor with an exterior derivative, a derivative which tosses out all information about the connection, how a metric changes (presuming the connection is torsion free and metric compatible as is done in GR). General relativity is exclusively about the connection, having gotten its divorce from the potential in the covariant derivative to end up in the Riemann curvature tensor, or its contractions, the Ricci tensor or Ricci scalar. By working with the covariant 4-derivative of a 4-vector in a reducible asymmetric field strength tensor, there are the six parts of the deviation from the average about of change in the potential second rank irreducible antisymmetric tensor to do EM, and the ten parts of the average amount of change in the potential second rank irreducible symmetric tensor to do gravity.

doug

 

[Don J]

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

Thats right but your GEM proposal don't even deal with the quantum at all.

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

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

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

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

Agree ! because gravitomagnetism is probably the key.

 

[CarlB]

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

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

Carl

 

[sweetser]

Quantization and Gravitomagnetism

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

The classical EM Lagrange density cannot be quantized. Why? Here it the Lagrange density:
$$\mathcal{L}_{Classical EM}=-\frac{1}{c}J_{q}^{\mu}A_{\mu} -1/4c^{2}(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}) (\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})$$
Calculate the cannonical momentum 4-density:
$$\pi^{\mu}=(\frac{\partial\mathcal{L}}{\partial (\partial \phi/\partial t)}, \frac{\partial\mathcal{L}}{\partial (\partial A_{x}/\partial t)}, \frac{\partial\mathcal{L}}{\partial (\partial A_{y}/\partial t)}, \frac{\partial\mathcal{L}}{\partial (\partial A_{z}/\partial t)})$$
If you do that for the classical EM Lagrangian, the first term (energy density) is zero. The 4-momentum cannot be quantized because there is no way to form a non-zero conjugate operator, $\phi \pi_{0}-\pi_{0}\phi=0$.

To correct this problem, folks choose a gauge in the Lagrange density. To make the approach appear manifestly covariant, a favorite choice first done by Gupta and independently by Bleuler was the Lorenz gauge:
$$\mathcal{L}_{G-B}=-\frac{1}{c}J_{q}^{\mu}A_{\mu} -1/2c^{2}(\partial A_{\mu}A^{\mu})^{2} -1/4c^{2}(\partial^{\mu}A^{\nu}-\partial^{\nu}A^{\mu}) (\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu})$$
Now there are terms with $\partial \phi/\partial t$, so the energy term of the cannonical momentum is not zero. The field equation is calculated in the standard way, using the Euler Lagrange equation [note: if you want to see all the details of that for GEM, it is in the Lecture 4 notes of TheStandUpPhysicist.com]. The field equations are a 4D wave.
$$J_{q}^{\mu}=(\partial^{2}/\partial t^{2}-\partial^{2}/\partial x^{2}-\partial^{2}/\partial y^{2}-\partial^{2}/\partial z^{2})A^{\mu}$$
Because like charges repel, the particles must be spin 1, which makes sense looking at the G-B Lagrange density. There are 4 modes of emission with the choice of the Lorenz gauge. Two modes of emission are transverse. They do the work of EM. There is also a longitudinal mode, and a scalar mode. It is the scalar mode of a spin 1 field that causes a technical problem. It allows for negative probability densities. Oops. So a supplementary condition is added so that the scalar and longitudinal modes always cancel each other, making the modes always virtual. It is common to be suspicious of this supplementary condition. It looks like it is there to hide something inconvenient. What is the deep idea driving the need for it? Still you can write out all the standard tools of quantum field theory, from commutators to creation/destruction operators.

The field equations for GEM look darn similar, the only difference being another current density for mass:
$$J_{q}^{\mu}-J_{m}^{\mu}=(\partial^{2}/\partial t^{2}-\partial^{2}/\partial x^{2}-\partial^{2}/\partial y^{2}-\partial^{2}/\partial z^{2})A^{\mu}$$
The key technical difference is that one needs a spin 2 field because like $J_{m}$ charges attract. The transverse modes do the work of EM. Now the scalar mode does the work of gravity, and it will not have the negative probability density problem. The commutator and creation/destruction operators should work fine as they are. The reason for the supplimentary condition is that the Lagrange density of EM in the Lorenz gauge is incomplete, missing gravity.

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

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

doug

 

[sweetser]

GEM action in a vacuum

Hello:

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

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

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

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

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

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

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

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

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

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

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

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

Quantization

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

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

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

 

[Don J]

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

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

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

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

 

[sweetser]

Deflection of light measurements

Hello Don:

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

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

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

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

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

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

doug

 

[Don J]

Hello Don:

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

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

 

[sweetser]

Hello Don:

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

 

[Don J]

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

Dont you consider this a problem?

 

[sweetser]

Hello Don:

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

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

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

doug

 

[Don J]

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

doug

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

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


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

 

[Don J]

Hello Don:

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

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

 

[sweetser]

One page pitch errors

Hello:

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

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

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

doug

 

[vanesch]

Hi,

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

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

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

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

 

[Careful]

Hi Doug,

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

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

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

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

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

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

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

Cheers,

Careful

 

[sweetser]

Hello Careful:

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

a). The way I figure out that something is a density is by looking at its units. The currents are definitely densities, so we agree on that point. We also know the units for the contraction of a field strength tensor must be correct since one appears in the standard EM Lagrange density. I like units, so here they are:
$$A^{\nu}\mathrm{ has units of }\sqrt{m/L}$$
Take a time derivative:
$$\nabla_{\mu}A^{\nu}\mathrm{ has units of }\sqrt{m/LT^{2}}$$
and square this, tossing in a pair of c's:
$$\frac{1}{c^{2}}\nabla_{\mu}A^{\nu}\nabla^{mu}A_{\nu}\mathrm{ has units of }m/L^{3}$$
I do like to stay grounded in pedestrian details, or I get disoriented. Your critique is more sophisticated than mere units, which is:

there is the determinant of a vierbein lacking (or otherwise you confine yourself to a fixed background flat coordinate system)

This brings up an central lesson I have learned here, so central I will write it out in bold: the metric for the GEM Lagrangian is fixed up to a second rank gauge symmetry transformation. Fixed in absolutely no way means flat. In standard EM, there are no way to determine how the metric changes, so it must be supplied as part of the mathematical structure. Yet it could definitely be a curved metric. In this proposal, the metric is definitely fixed, but there is a symmetry that allows one to change the metric so long as there is a change in the 4-potential.

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

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

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

doug

 

[Careful]

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

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

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

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

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

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


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

HUH ?

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

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

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

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

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

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


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

Wrong, again only symmetries of the FULL dynamics count.

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

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


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

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

Cheers,

Careful

 

[sweetser]

Hello Careful:

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

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

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

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

doug

 

[Careful]

Hi Doug,

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

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

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

Sure...

**

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

**

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

Cheers,

Careful

 

[sweetser]

Hello Careful:

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

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

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

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

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

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

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

doug

 

[Careful]

Hello Doug,

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

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

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

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

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

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

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

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

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


Cheers,


Careful

 

[sweetser]

A valid contravariant expression

Hello Careful:

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

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

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

This is a direct statement that Q1 is a tensor.

Jackson provides a transformation law:

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

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

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

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

doug

 

[sweetser]

Contracting symmetric mixed rank 2 tensors

Hello Careful:

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

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

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

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

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

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

doug

 

[vanesch]

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

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

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

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