Unifying Gravity and EM

[sweetser]

Diff(M) symmetry

Hello All:

I am feeling warm and fuzzy about the GEM proposal today. In high school, I got to study Kuhn's "The Structure of Scientific Revolutions". When I began my independent research in physics after the 1988 Christmas gift from both mom and sis of Hawkings' "Brief History of Time", I could have adopted the paradigm shift model. One would focus on the things that do not fit, dark matter, dark energy, and the difficulty of quantizing GR. Since I was trained scientifically as a bench biologist (go in every day and do a half dozen little experiments, day in, day out), I decided to use a model of intellectual evolution. Evolution uses three processes:

1. Mostly be like your parents.
2. You are not exactly like your parents
3. If something works, do it again (like your parents might have, only not exactly).

Never ever stop repeating cycles 1-3. Instead of jumping into advanced graduate school classes, I chose to take a history of physics class at Harvard Extension School where we got to repeat experiments done by Galileo, Newton, and Franklin. I took a class on special relativity three times, once with Edwin Taylor who wanted student input into his book under development, "Spacetime Physics", once at Harvard, once at MIT. What makes me most happy is not doing something way out there. Instead I love to see a specific, solid connection to work of the past. With that in mind, I will revisit the result of yesterday on the Diff(M) symmetry for the GEM proposal.

I was aware of the Einstein-Maxwell action, but had not worked with it. When Careful point it out (post 63), I decided to look at it again, out of respect for work done by past masters. Here is the action:
$$S_{Einstein-Maxwell}=\int \sqrt{-g} d^{4} x (R-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))$$
This action is symmetric under a $U(1)$ transformation. That means this theory can deal with light. What is the symmetry for Einstein's approach to gravity? Special relativity worked only for inertial observers, folks gliding along at a constant velocity at all times. Einstein wanted to find invariants for folks that were accelerating in arbitrary ways. The group Diff(M) of continuous transformations of coordinates is the key. To learn more about it, read this page: http://math.ucr.edu/home/baez/symmetries.html. The Ricci scalar $R$ has all the relevant information about how the metric is curved. It is all that is needed for the action to describe curvature since it is the contraction of a contraction of the Riemann curvature tensor.

The next move sounds almost as radical as the queen sacrifice Bobby Fischer played in 1956 (http://www.chessgames.com/perl/chessgame?gid=1008361&kpage=19). Drop the Ricci scalar $R$. What one is left with is the Maxwell theory on a (possibly)curved manifold:
$$S_{?-Maxwell}=\int \sqrt{-g} d^{4} x (-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))$$
The problem with this approach is that a metric must be supplied as part of the background mathematical structure. A good summary of the issue is here: http://math.ucr.edu/home/baez/background.html. Are we really completely free to choose a metric? Let's look at a concrete example of changing the metric. First choose to work with the flat Minkowski metric:
$$g_{\mu\nu}=\left(\begin{array}{cccc} 1 & 0 & 0 & 0\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ 0 & 0 & 0 & -1\end{array}\right).$$
We know how to work with the Maxwell equations with this metric, this one is easy! Now let's choose a different metric that is a baby step away from flat spacetime:
$$g_{\mu\nu}=\left(\begin{array}{cccc} 1 & 0 & 0 & \delta\\ 0 & -1 & 0 & 0\\ 0 & 0 & -1 & 0\\ \delta & 0 & 0 & -1\end{array}\right).$$
I'll call it Minkowski delta z. This metric could be smoothly merged into Minkowski by a limit process on the delta z, so the Minkowski delta z metric that is part of the Diff(M) group. If we choose to use the Minkowski metric, then there will be zero energy stored in the curvature of spacetime. There is no problem accounting for zero. Now we choose to work with the delta z metric. There is energy stored in the curvature of spacetime. Where does the Lagrange density account for the energy of this curvature? If one tried to put it in $(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})$, it would be eliminated since the EM tensor is antisymmetric and filters out what would be a symmetric contribution. In order to be able to freely change the symmetric metric, a symmetric tensor is required:
$$S_{GEM}=\int \sqrt{-g} d^{4} x (-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}-\nabla_{\nu}A_{\mu}))-\frac{1}{4 c^{2}}(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})(\nabla_{\mu}A_{\nu}+\nabla_{\nu}A_{\mu}))$$
$$S_{GEM}=\int \sqrt{-g} d^{4} x (-\frac{1}{2 c^{2}}\nabla^{\mu}A^{\nu}\nabla_{\mu}A_{\nu})$$
If we decide to use a metric with a delta y, there is a logical place to put its energy contribution: the y slot of the symmetric tensor. The symmetric tensor is required by the energy accountants for smooth changes in coordinates.

I don't know when people understood Diff(M) was the key group involved in understanding general relativity. Folks who are adept at group theory might claim it was the core idea behind the curtain where the Wizard of Einstein worked. That group is also at the core of the GEM proposal which is shining a bit brighter today.

doug

 

[Lawrence B. Crowell]

a couple of notes

There are a number of things which "conspire" to create your Gauss' law result. Your metric is asymptotically flat. In the case of GR this conspires to recover Newtonian gravity for the [itex]\Gamma^0_{00}[\itex] geodesic equation. The specialness of the solution creates the conditions similar to what you have. Also the t-t and r-r entries of your metric look alright, but the angular parts should just have the same terms as the Schwarzschild solution. This seems to be a bit of an oversight.

Again I am unclear on this kluging of the EM and gravity connections. I am also a bit unclear on how one can really have symmetric field tensors.

cheers,

Lawrence B. Crowell

 

[sweetser]

Asympotically simple, not flat

Hello:

For a general reader, I thought I would provide a sense of what asymptotically flat means. Basically I will be cribbing from this source:
http://relativity.livingreviews.org/open?pubNo=lrr-2004-1&page=articlesu3.html
I cannot talk this technical unless I am looking directly at technical sources.

How should one compare the quality of one metric to another? A spacetime involves both a manifold and a metric. That begs the question, what is a manifold? A manifold is a topological space where locally it looks like $R^{n}$. An ice cream cone is not a manifold because of the point, but a donut is a manifold. A golf ball is also a manifold, so let's use that. The golf ball has a boundary which is the surface of the ball. A Lorentz spacetime will have a manifold M (the ball) plus a metric with a signature (+---). Out of the sea of all possible spacetimes, many that are not physical, we want to pluck out useful ones. We start with a submanifold of the the Lorentz manifold, and make sure it has a smooth boundary. There is a scalar field that can relate the Lorentz metirc to the metric of the submanifold. The scalar field on the boundary is zero. Every null geodesic has its future and past end points on the boundary. Such a submanifold is called an asymptotically simple spacetime. My imprecise was of thinking about asymptotically simple spacetime is there is a limit process that would merge the exponential metric into the Minkowski metric. A spacetime which is not asymptotically flat are those with non-zero cosmological constants, such as de Sitter and anti-de Sitter spacetimes.

If an asympotically simple spacetime also solves the Einstein vacuum equations, $R_{ab}=0$, then the spacetime is asympotically flat.

I would agree that the exponential metric is asympotically simple. I believe the exponential metric does not solve the Einstein vacuum equations. Rosen did the first work with this metric, and he said it did not solve the Einstein field equations. Misner worked the metric recently, making the same point.

Now I have to calculate the Ricci tensor and scalar for the exponential metric. Note: I wrote that metric in Euclidean, x, y, z, coordinates, not spherical coordinates. In spherical coordinates, it looks like so:
$$g_{\mu\nu}=\left(\begin{array}{cccc} exp(-2\frac{\sqrt{G}q + GM}{c^{2}R}) & 0 & 0 & 0\\ 0 & -exp(2\frac{\sqrt{G}q+GM}{c^{2}R}) & 0 & 0\\ 0 & 0 & -R^{2} & 0\\ 0 & 0 & 0 & -R^{2} sin^{2}(\theta)\end{array}\right ).$$
This is the form usually used in books on GR, and I was being unconventional, althought the missing R's and sin's should have been a hint. I decided to try and confirm what I had read by calculating the Ricci tensor for the exponential metric. I had to fire up Mathematica for some assistence.

Define the Minkowski metric in spherical coordinates:

$$gMinkowski= \begin{array}{cccc} 1 & 0 & 0 & 0 \\ 0 & -1 & 0 & 0 \\ 0 & 0 & -R^2 & 0 \\ 0 & 0 & 0 & -R^2\text{Sin}[\theta ]^2 \end{array}$$
Define functions that will calculate the Ricci curvature tensor and scalar (see "Gravitation and Spacetime, second edition" by Ohanian and Ruffini for details, p 337):

$$\noindent\(\text{RicciTensor}[\text{N\_},\text{L\_}]\text{:=}\text{Module}[\{\text{R00},\text{R11},\text{R22},\text{R33}\},\\$$
$$\text{R00}=\text{Exp}[N-L](-\frac{1}{2}D[N,\{R,2\}]+\frac{1}{4}D[L,R]D[N,R]-\frac{1}{4}D[N,R]^2-\frac{1}{R}D[N,R]);\\$$
$$\text{R11}=\frac{1}{2}D[N,\{R,2\}]-\frac{1}{4}D[L,R]D[N,R]+\frac{1}{4}D[N,R]^2-\frac{1}{R}D[L,R];\\$$
$$\text{R22}=\text{Exp}[-L](1+\frac{1}{2}R(D[N,R]-D[L,R]))-1;\\$$
$$\text{R33}=\text{Sin}[\theta ]^2\text{Exp}[-L](1+\frac{1}{2}R(D[N,R]-D[L,R]))-\text{Sin}[\theta ]^2;\\$$
$$\text{RT}=( \begin{array}{cccc} R00 & 0 & 0 & 0 \\ 0 & R11 & 0 & 0 \\ 0 & 0 & R22 & 0 \\ 0 & 0 & 0 & R33 \end{array} )];\)$$
$$\noindent\(\text{RicciScalar}[\text{N\_},\text{L\_}]\text{:=}\\ \text{Exp}[-L](-D[N,\{R,2\}]+\frac{1}{2}D[L,R]D[N,R]-\frac{1}{2}D[N,R]^2+\frac{2}{R}(D[L,R]-D[N,R])-\frac{2}{R^2})+\\ \frac{2}{R^2};\)$$
Test the functions withe the Schwarzschld solution:
$$\noindent\(\text{MatrixForm}[\text{Simplify}[\text{RicciTensor}[\text{Log}[1-2/R],\text{Log}[1/(1-2/R)]]]]\)$$
$$\noindent\(( \begin{array}{llll} 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \\ 0 & 0 & 0 & 0 \end{array} )\)$$
$$\noindent\(\text{Simplify}[\text{RicciScalar}[\text{Log}[1-2/R],\text{Log}[1/(1-2/R)]]]\)$$
$$\noindent\(0\)$$
Test with exponential metric:
$$\noindent\(\text{MatrixForm}[\text{Simplify}[\text{RicciTensor}[-2/R,2/R]]]\)$$
$$\noindent\(( \begin{array}{llll} -\frac{2 e^{-4/R}}{R^4} & 0 & 0 & 0 \\ 0 & \frac{2}{R^4} & 0 & 0 \\ 0 & 0 & -1+\frac{e^{-2/R} (2+R)}{R} & 0 \\ 0 & 0 & 0 & (-1+\frac{e^{-2/R} (2+R)}{R}) \text{Sin}[\theta ]^2 \end{array} )\)$$
$$\noindent\(\text{Simplify}[\text{RicciScalar}[-2/R,2/R]]\)$$
$$\noindent\(\frac{e^{-2/R} (-4-4 R+2 (-1+e^{2/R}) R^2)}{R^4}\)$$
$$\noindent\(\text{MatrixForm}[\text{Simplify}[\text{RicciTensor}[-2/R,2/R]-\frac{1}{2}\text{gMinkowski} * \text{RicciScalar}[-2/R,2/R]]]\)$$
$$\noindent\(( \begin{array}{llll} \frac{e^{-4/R} (-2-e^{4/R} R^2+e^{2/R} (2+2 R+R^2))}{R^4} & 0 & 0 & 0 \\ 0 & \frac{e^{-2/R} (-2-2 R-R^2+e^{2/R} (2+R^2))}{R^4} & 0 & 0 \\ 0 & 0 & -\frac{2 e^{-2/R}}{R^2} & 0 \\ 0 & 0 & 0 & -\frac{2 e^{-2/R} \text{Sin}[\theta ]^2}{R^2} \end{array} )\)$$
This calculation is consistent with the literature: the exponetial metric is not a solution to the Einstein vacuum equations. I am not sure this calculation is valid because the Ricci tensor may be in Schwarzschild coordinates, but at least I tried.

doug

 

[Lawrence B. Crowell]

Hello Doug,

There are a number of things that still have me concerned. The “kluging” of the EM potential and the gravitational connection covers over a number of things with a broad brush. The structure of these two potentials is markedly different. For pp-waves it can be demonstrated that the gravity wave has a helicity of two, which leads in a quantization scheme to the conclusion that the graviton (which at best can only be said to exist for certain special solutions such as pp-waves) is spin = 2. Classically a gravity wave has two directions of polarization.
A gravity wave is a perturbation on a background metric $\eta_{ab}$ with the total metric
$$g_{ab}~=~\eta_{ab}~+~h_{ab}.$$
The flat background metric has zero Ricci curvature so that to first order in the perturbation expansion
$$R_{ab}~=~\delta R_{ab},$$
which enters into the Einstein field equation
$R_{ab}~-~1/2Rg_{ab}~=~\kappa T_{ab}$, where $\kappa~=~8\pi G/c^4$ is the very small coupling constant between the momentum-energy source and the spacetime configuration or field. The Ricci curvature to first order is then
$$R_{ab}~=~{1\over 2}\Big(\partial_c\partial_a{h^c}_b~+~\partial_c\partial_b{h^c}_a~-~\partial_a\partial_bh)~-~\partial_c\partial^ch_{ab}\Big).$$
The harmonic gauge $g^{bc}\Gamma^a_{bc}~=~0$, to first order as $\partial_c{h^c}_a~=~1/2\partial_mu h$ the Einstein field equation gives
$$\partial^c\partial_ch_{ab}~-~{1\over 2}\eta_{ab}\partial^c\partial_ch~=~{{16\pi G}\over {c^4}}T_{ab},$$
which is well defined for the traceless metric term ${\bar h}_{ab}~=~h_{ab}~-~{1\over 2}\eta_{ab}h$ with the simple wave equation
$$\partial^c\partial_c{\bar h}_{ab}~=~{{16\pi G}\over {c^4}}T_{ab}.$$
For the wave in vacuum the momentum energy source is set to zero, and the wave equation is $\partial^c\partial_c{\bar h}_{ab}~=~0$. This is a bi-vector analogue to the simple wave equation for an electromagnetic wave in free space. The wave is a transverse traceless wave ${\bar h}_{ab}~=~A^{TT}_{ab}exp(ik_cx^c)$ with
$$A^{TT}_{ab}~=~\left(\begin{array{ccc}0 & 0 & 0 & 0\\ 0 & A_{xx} & A_{yx} & 0\\ 0 & A_{xy} & -A_{xx} & 0\\ 0 & 0 & 0 & 0 \end {array}\right).$$
The term $A_{xx}$ is the $++$ polarization term and $A_{xy}$ is the $\times\times$ polarization term, where each represent a polarization direction. The linearized gravity wave then has a helicity of two, which has its quantum analogue in the di-photon state in quantum optics.
Pp-waves have the nice property of being linear, and the above case is a simple example. This gravity wave, which may be found by LIGO in the near future, is one that adds linearly in a way similar to EM waves. Though it must be stressed that gravity fields (spacetime curvatures) do not in general add this way. Yet in this special case the polarization structure of the wave is different than EM waves. I still think that the different gauge connections need to be built up from a tetard formalism $E^a_\mu~=~\gamma^ae_\mu$

cheers,

Lawrence B. Crowell

 

[Lawrence B. Crowell]

failed latex eqn

opps, the latex equation that failed to show up is redone below


$$A^{TT}_{ab}~=~\left(\begin{array{ccc}0 & 0 & 0 & 0\\ 0 & A_{xx} & A_{yx} & 0\\ 0 & A_{xy} & -A_{xx} & 0\\ 0 & 0 & 0 & 0 \end{array}\right).$$

 

[Lawrence B. Crowell]

try again

$$A^{TT}_{ab}~=~\left(\begin{array{cccc}0 & 0 & 0 & 0\\ 0 & A_{xx} & A_{yx} & 0\\ 0 & A_{xy} & -A_{xx} & 0\\ 0 & 0 & 0 & 0 \end{array}\right).$$

 

[member 11137]

I still think that the different gauge connections need to be built up from a tetard formalism $E^a_\mu~=~\gamma^ae_\mu$
cheers,
Lawrence B. Crowell

Are the gamma matrices in your proposition the Dirac matrices? In this case, if yes, I think that it is easy to demonstrate that if the Dirac matrices can vary within a GR approach around the average value that they own in the SR approach, then the metric can vary too around the Minkowski one. Best regards

 

[CarlB]

Let me try:

$$A^{TT}_{ab}~=~\left(\begin{array}{cccc}0 & 0 & 0 & 0\\ 0 & A_{xx} & A_{yx} & 0\\ 0 & A_{xy} & -A_{xx} & 0\\ 0 & 0 & 0 & 0 \end{array}\right).$$

Carl

 

[sweetser]

The connection and differential forms

Hello Lawrence:

Let's see if we can pinpoint the source of this gut feeling:

**
The “kluging” of the EM potential and the gravitational connection covers over a number of things with a broad brush.
**

Reflection on this issue might clarify why GR and the standard model stand separately from each other on the stage of modern physics.

Start with a covariant tensor, $A_{\nu}$. The partial derivative operator is also a covariant tensor, $\partial_{\mu}$. Take the partial derivative, $\partial_{\mu} A_{\nu}$. It is a common exercise to show the partial derivative of a 4-potential does not transform like a tensor. If spacetime is curved, then the partial derivative will not account for all the changes that happen due to that curvature. A covariant derivative is constructed so that the result transforms like a tensor.
$$\nabla_{\mu}A_{\nu}=\partial_{\mu} A_{\nu}-\Gamma^{\sigma}{}{\mu\nu}$$
There is considerable choice in what to use for the connection, $\Gamma$. I choose to work with the one used in GR that is metric compatible and torsion-free. It is known as the Christoffel symbol of the second kind. Like the partial derivative, the Christoffel symbol does not transform like a tensor. Together they do, separately they do not. The GEM approach keeps then together. Modern physics has developed ways to separate the connection from the partial derivative so both transform like tensors on their own.

In GR, one takes the connection and constructs the simplest object one can out of Christoffel symbols that transforms like a tensor. It is know as as Riemann curvature tensor:
$$R^{\rho}{}_{\sigma \mu \nu}=\partial_{\mu} \Gamma^{\rho}{}_{\sigma \nu}-\partial_{\nu} \Gamma^{\rho}{}_{\sigma \mu}+\Gamma^{\rho}{}_{\mu \lambda}\Gamma^{\lambda}{}_{\nu \sigma}-\Gamma^{\rho}{}_{\nu \lambda}\Gamma^{\lambda}{}_{\mu \sigma}$$
Roughly what is happening is that two paths are being compared, and this is the net difference. Contract this once for the Ricci tensor, do that again for the Ricci scalar.

An appealing feature of differential forms is that they are true no matter what the connection is. In other words, the connection is irrelevant to differential forms. This is very convenient. The antisymmetry of differential forms guarantees the symmetric, torsion-free connection will not be on stage. Differential forms are a great platform for understanding the standard model.

If one chooses to work with the Riemann curvature tensor and differential forms, one can successfully understand GR and the standard model. To unify gravity and EM may require going back to square one, exploring anew a reducible asymmetric tensor where the connection and the changes in the potential sleep in the same expression.

doug

 

[sweetser]

Gravity waves

Hello Lawrence:

This note is about gravity waves. There are several lines of logic that argue that the graviton must be a spin-2 particle represented by a second rank symmetric tensor. I agree with that analysis.

Let me try and explain a little about this "tetrad" issue for folks not familiar with the topic. Einstein's approach to GR used the connection. There are technical problems with dealing with half integral spin using the connection. That lead people to work with the spin connection. The spin connection is not torsion free. It can be used with spinors. Another great strength is that the spin connection integrates with gauge theories very well. One can write the Riemann curvature tensor in terms of the spin connection.

Your analysis of gravity waves looks right to me. You note that the polarization will be transverse. I recall reading in Clifford Will's review of experimental tests of GR that were a wave detected, and furthermore the mode of polarization determined, and the wave was not transverse, that would be a serious challenge to GR.

For the GEM proposal, the waves have nothing to do with the Ricci tensor, nor with solutions to Einstein's field equations. Instead the 4D wave equation has two different spin fields for the massless particles. The spin 1 field represents the transverse modes of EM. A spin 2 field represents the scalar and longitudinal modes of gravity. The answer to the quantum gravity riddle is already published in most graduate quantum field theory books. Everything in the method to quantize the 4D wave equation with two spin fields is standard, right down to the gamma matrices. The solution is in the section on the Gupta/Bleuler approach to quantizing the EM field. There are 4 modes of emission: two transverse, one scalar, and one longitudinal. The two transverse modes do EM, no problem. The scalar mode for a spin 1 field is a problem because it predicts negative energy densities. To ensure those modes are always virtual, a supplementary condition is imposed so that both scalar and longitudinal modes are always virtual. There is a bit of resistance to this idea - it sounds like a hack is created just to hide otherwise valid modes of emission - but a diligent student will learn that it must be done for a spin 1 field. In the GEM proposal, the 4D wave equation needs a spin 1 field for EM where like charge repel, and a spin 2 field for gravity where like charges attract. The spin 2 field will not have the negative energy density problem. The scalar and longitudinal modes can do the work of gravity. Since the spin 1 field of the Gupta/Bleuler approach to quantizing a 4D wave equation has no problem with spinors and half integral spin particles, it is my hope that the inclusion of the spin 2 field for gravity will not present a technical challenge requiring tetrads. I do not think I will ever be educated enough on the topic to go beyond this "hope" stage.

I asked Clifford Will what he thought were the odds of determining the mode of polarization. He was not optimistic. We have yet to measure one gravity wave. Should we accomplish that difficult task, the wave will have to be measured on six different axes to determine the polarization.

If a gravity wave is transverse, GR is right, GEM is wrong. If a gravity wave is not transverse, GR is wrong, GEM is right. Let the distant future data decide.

doug

 

[Lawrence B. Crowell]

transverse g-waves and Dirac matrices

The gamma is indeed the Dirac matrix. Since [itex]g^{ab}=1/4\gamma^a\gamma^b[\itex] if these matrices have chart dependent representations (eg they are bundle sections) then gravity connections and curvatures can be found from them.

As for gravity waves, anyone familiar with GR knows they are predicted to be transverse. A direct detection of transversality may not be needed. If a source of gravity waves is identified an indirect inference might be made. For instance if the source has an optical signature, where we know telescopically where the source is, then with various LIGO detectors it would be possible to back out the nature of the wave. For instance if a line to the source is normal to a LIGO then this would be a pretty clear signature for travservse waves.

I would prefer to see this GEM proposal, or a related problem, done within the format of tetrads. I am a bit uncertain about separating out the gauge connection from the gravity connection. Further, why limit this to EM? It seems to me that an extension to a general Yang-Mills theory with [itex]GL(2,C)[\itex].

BTW, when it comes to TeX-ing problems, for some reason often the post preview does not show the TeX'ed equations. I just get a box.

Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

Minor TeX things. the itex box needs a forward slash, not the backward one of TeX. That makes it more like an HTML tag. Post preview also does not work for me. The edit button does. You can come back weeks later and modify posts, although I think folks prefer old posts to stay as is for anyone reading the conversation later.**
For instance if a line to the source is normal to a LIGO then this would be a pretty clear signature for travservse waves.
**
My sense from chatting with Will is this sort of calculation is of major interest, but may be beyond their reach. If we ever detect a gravity wave, we cannot be sure that the source can be found in the sky.**
I am a bit uncertain about separating out the gauge connection from the gravity connection.
**
I am uncertain about how to discuss the Diff(M) symmetry, gauge transformations, and at what level the theory is unified. When I get to deal with equations, it is crystal clear: work in a coordinate system where the connection is zero everywhere, and the system can be completely described by the potential, OR work with a constant potential, and the system can be completely described by the Christoffel symbol, OR work with some combination of potential and connection. I am not confident and finding the right phrase for this situation I know how to calculate for charged and uncharged particles.**
Further, why limit this to EM?
**
That is entirely due to my own limitations. If the proposal works for gravity and EM, it must be extended to the weak and strong forces. I don't have the training to do that.

doug

 

[member 11137]

Hello Doug,
There are a number of things that still have me concerned. ...which is well defined for the traceless metric term ${\bar h}_{ab}~=~h_{ab}~-~{1\over 2}\eta_{ab}h$ with the simple wave equation
$$\partial^c\partial_c{\bar h}_{ab}~=~{{16\pi G}\over {c^4}}T_{ab}.$$
For the wave in vacuum the momentum energy source is set to zero, and the wave equation is $\partial^c\partial_c{\bar h}_{ab}~=~0$. This is a bi-vector ...
Lawrence B. Crowell

Once more time I beg your pardon to be interferring in your private discussion, but I feel concerned too. What is playing the role of a bi-vector here? The traceless metric? A metric within the GR approach has to be symmetric. A bi-vestor is, per definition, anti-symmetric... where is the mistake? Thanks for explanation. Best regards

 

[Lawrence B. Crowell]

bivectors, gravitons & photons

To start, I am so used to backslashes in TeX that it is hard to kick the habit with [/tex].
At this stage I would say with the GEM proposal that one of two things need to be done. Either the graviton and photon sectors, the abuse with the term graviton with standing, need to be clearly indicated in some way. This might be done with some tetrad formalism or with the embedding of GR and EM into some larger $GL(n,~C)$ group. Another approach is to somehow show that the spin-2 field of gravity, say in particular in the pp-wave solutions, can be built up from some coupling of photons. A gravity wave is a bivector, and in quantum optics there are phenomena of photon bunching or "bi-photons" which are similar to gravity waves. I say similar, for they still interact with electric charges and so forth. In such a theory two photons with alligned spins would interact to form a graviton. Currently such an interaction for counter spin alligned photons generate the $Z_0$ particle of weak interactions. In this way at very high energy, probably approaching the Planck energy, two photons would generate a graviton. How this would fit into Doug's theory theory is a bit unclear. Maybe if Randall et al. are right with so called "soft black holes" that occur at the TeV range in energy the other fields that the photon interacts with have some mass matrix so that there are oscillations between gravitons and photons. By this two photons correlated in a Hanbury Brown-Twiss manner will have some probability of being a graviton.
Yet this is pretty speculative. The dust bin of physics is littered with a lot of quantum field theory speculations.
On a related manner, my "hobby horse" is with information physics of quantum gravity. I have worked out how it is that black holes will preserve quantum information. I am not sure how one starts a thread on this list. Yet I would like to start throwing out some trial balloons before I try to publish things. If so this should prove to be interesting, for I have done some preliminary work on how symmetries of quantum gravity involves error correction codes. Things get into Riemann zeta functions and the like.
Cheers,
Lawrence B. Crowell

 

[Lawrence B. Crowell]

erratum on last post

I said that two high energy photons could generate a $Z_0$ particle. This is wrong, such an interaction will generate pair $W^{\pm}$.

Lawrence B. Crowell

 

[member 11137]

To start, I am so used to backslashes in TeX that it is hard to kick the habit with [/tex].
At this stage I would say with the GEM proposal that one of two things need to be done. Either the graviton and photon sectors, the abuse with the term graviton with standing, need to be clearly indicated in some way. This might be done with some tetrad formalism or with the embedding of GR and EM into some larger $GL(n,~C)$ group. Another approach is to somehow show that the spin-2 field of gravity, say in particular in the pp-wave solutions, can be built up from some coupling of photons. A gravity wave is a bivector, and in quantum optics there are phenomena of photon bunching or "bi-photons" which are similar to gravity waves. I say similar, for they still interact with electric charges and so forth. In such a theory two photons with alligned spins would interact to form a graviton. Currently such an interaction for counter spin alligned photons generate the $Z_0$ particle of weak interactions. In this way at very high energy, probably approaching the Planck energy, two photons would generate a graviton. How this would fit into Doug's theory theory is a bit unclear. Maybe if Randall et al. are right with so called "soft black holes" that occur at the TeV range in energy the other fields that the photon interacts with have some mass matrix so that there are oscillations between gravitons and photons. By this two photons correlated in a Hanbury Brown-Twiss manner will have some probability of being a graviton.
Yet this is pretty speculative. The dust bin of physics is littered with a lot of quantum field theory speculations.
On a related manner, my "hobby horse" is with information physics of quantum gravity. I have worked out how it is that black holes will preserve quantum information. I am not sure how one starts a thread on this list. Yet I would like to start throwing out some trial balloons before I try to publish things. If so this should prove to be interesting, for I have done some preliminary work on how symmetries of quantum gravity involves error correction codes. Things get into Riemann zeta functions and the like.
Cheers,
Lawrence B. Crowell

I don't know if Doug appreciates this parasite but interesting discussion and that's why I want to thank him and you, Lawrence. I don't know who you are and I guess that you are flying 2 km over my head with your knowledge but if you are interseting in, we could continue this thread on my subforum. I did make a try to investigate situations where the Lorentz Einstein Law is a differential operator and got a unique familly of solutions where the Christoffel's symbols can only be 0 or 1; that is intuitively appearing to be a possible junction with your "horse" and for a vision of the geometry as being a kind of computing machine: don't you think so?
Thank's for the precision concerning the bi-vector.

 

[Lawrence B. Crowell]

EM in GR & Information Q-gravity

I am not sure how to interpret the word “parasite.” The issue of including EM into gravity is almost as old as general relativity. The first suggestion was made by Kaluza & Klein in the early 1920s. Here the space is extended to 5 dimensions with the usual connection coefficients. Here one has metric terms $g_{5\mu}$, but one imposes the so called cyclic condition that $g_{\mu\nu,5}~=~0$. This means that one has a connection term of the form ${\Gamma^5}_{\mu\nu}~-~{\Gamma^5}_{\nu\mu}$, which is interpreted as $B~=~\nabla\times A$ on the spatial terms. Of course one observation is that the Yang-Mills field, in this case electromagnetism, is on the connection level with GR, and the field potentials for this field are on the metric level in GR.
As for my information approach to quantum gravity, this involves the so called information paradox with black hole evaporation. The entropy of a black hole is only reversible if the black hole is in equilibrium with the environment so that $dS~=~\frac{dM}{T}$. Yet the effective heat capacity of spacetime is negative, which means that as the black hole decays to lower entropy it heat up, contrary to normal thermodynamics. This means that even a quantum fluctuation will bump the black hole away from equilibrium and so $dS~>~\frac{dM}{T}$, which is contrary to standard thermodynamics where systems tend towards equilibrium. Quantum fields in curved spacetime are related to those in a flat spacetime by Bogoliubov transformations. The entropy of quantum states is given by the von Neumann equation $S~=~-k~\rho~log_2\rho$. Because of this transformation I compute a conditional entropy $\rho_{A|B}~=~lim_{n\rightarrow\infty}{{\rho_{AB}}^{1/n}({\bf 1}_A\otimes\rho_B)^{-1/n}$ for the teleporation of quantum states with this conditional entropy negative. In a setting where quantum information is not included this negative entropy is ignored and so things appear irreversible. However, this negative conditional entropy means that excess information may be had “for free” so the apparent entropy increase of a black hole is accounted for. However, to use the analogy of a lottery, if one does not read the ticket to claim the winnings one plays a losing game. Of course reading that ticket is difficult, for this means an accounting of a vast number of entangled EPR pairs must be tracked, and tracked from the past to future of the black hole. This is in principle feasible, but is from a practical point of view intractible. A stellar mass black hole is then for all practical purposes an irreversible entropy machine.
Cheers,
Lawrence B. Crowell

 

[sweetser]

This thread is intended to focus on the GEM proposal. As I interpret it, Blackforest was trying to be polite in shifting the discussion to things like bi-vectors which do not play a key role in my proposal. I am working on a (hopefully) clear statement about group theory and the GEM work which will require a few more days of work.

Efforts to unify gravity and EM go back to Priestly's discussions with Franklin about electricity. Franklin told Priestly about his observation that there was no electric field inside a conducting can. That is similar to there being on gravitational field inside a hollow massive shell. Priestly was the one to deduce the inverse square law of EM by this analogies to gravity. I have seen the GEM Lagrangian in the context of a purely EM proposal.

doug

 

[Lawrence B. Crowell]

bivectors & GEM

Doug,

Actually I brought up the issue of bivectors within the context of gravity waves. It is a real issue, for gravity waves are dyadic, while EM waves are vector. From a particle physics perspective this is why photons (as well as gluons and weak vector bosons) are spin = $\hbar$ and gravitons are spin = 2$\hbar$. The issue is then how it is that the two sectors are intertwined.

Agreed that this forum is primarily for your GEM proposal. Things have sort of bled over a bit. I am not sure how strictly things are compartmentalized here. I have looked at Blackforest’s proposal, but have yet to dig into the meat there. The pdf files are a bit dense.

Cheers,

Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

Let me clarify what is know about gravity waves. From observing binary pulsars, people have been able to infer that the lowest mode of emission is has a quadrupole moment. Rosen was the first person to write about the exponential metric, but his proposal is considered in error because it has a second metric metric field that could allow for a dipole moment in the wave emission. It is difficult to come up with an alternative proposal that adds fields yet has the quadrupole moment as the lowest mode of emission. The GEM proposal does not add fields, and I believe has the quadrupole, or as I like to think of it, wobbly-water balloon mode, as the lowest form of emission. I thought this issue was all about conservation of energy and momentum, not spin.

The reason that a gravity field must be spin 2 was spelled out for me in Brian Hatfield's introduction to "Feynman lectures on gravitation". If one wants like particles to attract, it takes particles of even spin: 0, 2, 4... He eliminates the spin 0 graviton because gravity bends light. That leaves the spin 2 field as the simplest possibility.

**
The issue is then how it is that the two sectors are intertwined.
**
Unfortunately, I don't understand this issue yet, so I will explain how I see the relationship. There is the 4D wave equation. The gravitational and electric behavior of massless and massive particles can be characterized by this equation according to the GEM proposal. Focus only on the massless particles. These travel at the speed of light, which in a way is a constraint (I have forgotten how to discuss that constraint, something about polarity perhaps). A massless spin-1 field represents the photon, and has two of the four degrees of freedom of the wave equation. A massless spin-2 field represents the graviton, and has the other two degrees of freedom of the wave equation. Together the spin-1 and spin-2 fields completely describe long distance forces, one where like charges attract, the other where like charges repel.

Particles arise from field strength tensors. For the photon, that would be $\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}$. For the graviton, the irreducible tensor $\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}$ is the source.

It might help to list all the things that I think are related in groups"

Gravity
like charged attract
graviton
spin-2
symm. tensor

EM
like charges repel
photon
spin-1
anti-symm. tensor

It still is necessary to explain what is going on on a group theory level. If that is what you mean by intertwine - clarify the group theory story - I remain guilty for a little while longer.

Researchers need the ability to wander, which is why I made an effort to explain Blackforest's unusual word choice, not complain about the digression. I too printed out three of the PDFs in his thread, but have not made a comment because I was not able to grasp why EM which works so great in 4D should be represented by 3D equations. To prevent a digression along the E question lines in this thread, I'll post something over there.

doug

 

[member 11137]

Doug,
Actually I brought up the issue of bivectors within the context of gravity waves. It is a real issue, for gravity waves are dyadic, while EM waves are vector. From a particle physics perspective this is why photons (as well as gluons and weak vector bosons) are spin = $\hbar$ and gravitons are spin = 2$\hbar$. The issue is then how it is that the two sectors are intertwined.
Agreed that this forum is primarily for your GEM proposal. Things have sort of bled over a bit. I am not sure how strictly things are compartmentalized here. I have looked at Blackforest’s proposal, but have yet to dig into the meat there. The pdf files are a bit dense.
Cheers,
Lawrence B. Crowell

Thanks for the patience. Parasite was absolutely not negative but exactly what sweeter has said.
Coming to your point: "The issue is then how it is that the two sectors are intertwined."
In fact, if you read my homepage and not only the work proposed on this subforum, you will discover another alternative to this question and I am myself astonished. If you construct (as I tried) a GR theory with the moving frame method but without neglecting terms of the second order, only trying to incorporate them via the introduction of the ad hoc mathematical terms, then you find very funny things concerning the first partial derivates of the basis vector: they are isotropic vectors, they can generate symplectic forms, they can generate matrices with a formalism analogue to the formalism of the Maxwell EM field tensor 4D... Because I am not owning all appropriate tools to develop rapidly my theory, I stay in front of these results, perceiving roughly a fundamental relation between geometry and EM fields... For me, in this very special approach, it's like if variations of the geometric structure would generate EM fields in the tangent space of each event...
In principle The GEM is not far from this discussion. The difference is that I try to developp a more general approach (not reduced to the formulation of a Lagrangian). The problem is that I don't have the necessary mathematical tools to do that. That's why I need your helps. If I have the good ticket !
Best Regards

 

[Lawrence B. Crowell]

gravity wave quadrupole moment

The reason that gravity waves are quadrupole is because a dipole gravitational moment is $P~=~mx$, and so by conservation of momentum $dP/d\tau~=~constant$. This makes a further point in illustrating the differences between gravitation and EM.

cheers,

Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

That statement looks accurate about GR. There are other theories for gravity that predict dipole moments because there are other fields that can contain the momentum, so the field for gravity can trade momentum with the other field, leaving $dP/d\tau~=~constant$.

Would it be correct to say that the difference between gravitation and EM in this context is equivalent to saying there is one sign for mass, but two for EM? Momentum is conserved for an isolated electric dipole too.

doug

 

[sweetser]

Diff(M) breaks U(1) symmetry

Hello Lawrence:

This is a valid request:

**
Either the graviton and photon sectors, the abuse with the term graviton with standing, need to be clearly indicated in some way.
**
(To be honest, I am not familiar with what the word "sectors" means in this context, but I'll accept the task).

Before posting here, I did not have any position about how my proposal worked with group theory, although I wish I had one! Now I have a position.

EM arises due to U(1) symmetry of the action. Gravity is accounted for by the Diff(M) symmetry of the Hilbert action. I consider these two groups, U(1) and Diff(M), to be indispensable groups if one wishes to characterize EM and gravity respectively.

The GEM proposal looks not to replace these groups, but rather to find a relationship between them, how the two groups relate to each other. I propose the Diff(M) symmetry exceptionally weakly breaks U(1) symmetry. For an electron, the mass charge associated with Diff(M) symmetry ($\sqrt{G}m_{e}$) is 16 orders of magnitude smaller than the electric charge associated with U(1) symmetry. At this point I have no explanation why there is such a large difference.

Contrast this to our best effort to unify gravity and EM, the Einstein-Maxwell equation. That action has Diff(M) symmetry for gravity, U(1) symmetry for EM, and the Higgs mechanism to introduce mass into the Lagrangian while preserving U(1) symmetry. The GEM proposal eliminates the need for the Higgs boson. There is a need for a scalar field to couple with all particles with mass, but that is simply the trace of the asymmetric tensor, or $tr(g_{\mu \nu}\nabla^{\mu}A^{\nu})$. Another big problem with the Higgs mechanism is there is no obvious connection to the graviton. To my eyes, the symmetric second rank tensor $\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}$ can do the work of the graviton for gravitational mass, while its trace does the work of the scalar boson need for inertial mass.

doug

 

[Lawrence B. Crowell]

group structure of gravity

I am not going to comment much on alternate theories of gravitation. Dicke proposed an alternative as a benchmark for testing deviations from expected results by GR.
//
As for group theory, gravitation is the group $SO(3,~1)~=~Z_2\times SL(2,~C)$. The group $SL(2,~C)$ may in turn be written as
$$SL(2,~C)~=~SU(1,~1)\times SU(2).$$
The algebra for this is then $su(2)$ given by the standard Pauli matrices $\sigma_{\pm},~\sigma_3$ and $su(1,~1)$ has the elements $\sigma_{\pm},~\tau_3~=~i\sigma_3$. The latter change gives the pseudoEuclidean nature to spacetime.
//
Now consider a connection one-form
$$A~=~A^+\sigma_+~+~A^3\sigma_3$$
and a gauge transformation determined by the group action of $g~\in~{\cal G}$, $g~=~e^{i\lambda\sigma_3}$. The gauge transformed connection is then
$$A^\prime~=~g^{-1}Ag~+~g^{-1}dg~=~e^{-2\lambda}A^+\sigma_+~+~A^3\sigma_3,$$
where $d\lambda~=~A^3$. Thus $\lambda$ is a parameterization of the gauge orbit for this connection. This leads to the observation
$$\lim_{\lambda\rightarrow\infty}A(\lambda)~\rightarrow~A^3\sigma_3,$$
where $A^+\sigma_+~+~A^3\sigma_3$ and $A^3\sigma_3$ have distinct holonomy groups and thus represent distinct points in the moduli space $\cal M$. However by the last equation we must have
$$F_\mu(A^+\sigma_+~+~A^3\sigma_3)~=~ F_\mu(A^3\sigma_3),$$
and similarly for any gauge invariant function. Hence there exist two distinct point in the moduli space that define the same set of gauge invariant functions. Hence there does not exist a measure over these two points that separates them, and $\cal M$ is then nonHausdorff and has Zariski topology.
\\
So the moduli space for gravitation is nonHausdorff, due to the hyperbolic nature of its group structure. This is also tied into why it is that gravity only has one “charge,” for with EM the two charges as roots are related by a simple rotation in the argand plane, but in gravity the two charges are not so connected. Thus for physics only one root is chosen, that for positive mass, while the negative root is not considered in the theory. Besides negative mass-energy creates all types of problems with GR, such as closed timelike curves and require quantum field sources that are not bounded below. Thus such solutions, wormholes and warp drives, are suspected of being pathological.
//
I tried to start a thread here, but it seems to have vanished into informationn to entropy land. I am not sure what I am doing wrong here.

Cheers,

Lawrence B. Crowell

 

[sweetser]

Two groups pass in the night

Hello Lawrence:

It looks like we are talking past each other. Although I have the Diff(M) symmetry group in the title of two of my recent posts, you did not mention the group in your reply. Likewise, I have not really worked with any of the groups you have suggested. I understand why this can happen: technical discussion tend to stay near the technical topics one is most familiar with. Yet I always challenge myself: what is it I am missing or getting wrong? It is fun for me to make information move in my head.

So I decided to get out all my gravity books this weekend, and read up on Diff(M). I've got MTW, Wald, Feynman, Ohanian & Ruffini, and a few books on quantum field theory and groups. I went to the index, and none of them listed Diff(M). This completely takes you off the hook! (OK, you never were really on a hook, you gave me an honest reply, so this is a nonjudgmental observation). Then I got kind of worried: where is my documentation on this group if it is not featured in these books?

It was on the web at a URL I had posted before. This time I will quote the relevant section:

Now I should point out that for each of the symmetry groups I've listed above, people have worked out the "representations" of these groups. A representation of a group is a way to think of its elements as operators, and this is what we need to understand symmetries in quantum physics. The representation theory of the Poincar´ group dominates relativistic physics, while the representation theory of the Galilei group dominates nonrelativistic physics. I encourage everyone to learn the derivation of Schrödinger's equation straight from the representation theory of the Galilei group! It's cool. (I think it appears in the books by Mackey and Jauch listed here.)

In any event, we can ask for still more symmetry than conformal symmetry. We can ask for symmetry under all smooth coordinate transformations! The first to demand this effectively was Einstein, who got his wishes when he devised general relativity as a theory of gravity. So gravity is the most symmetrical of field theories so far - it's "generally covariant", or invariant under the group Diff(M) all diffeomorphisms of a spacetime M! One can also come up with a (classical) theory of gravity coupled to Yang-Mills fields, or whatever fields you like.

http://math.ucr.edu/home/baez/symmetries.html
There are a couple of really neat things about this quote. There are quite a good number of things in your posts that I do not understand. That indicates that I am not well-educated in quantum physics, otherwise I would know what Zariski topology is. A man has to know his limitations, and one of mine is an in-depth knowledge of quantum theory.

My favorite line: "So gravity is the most symmetrical of field theories so far - it's "generally covariant", or invariant under the group Diff(M) all diffeomorphisms of a spacetime M!" This is the kind of property I think I see in the GEM proposal. Starting from the flat Minkowski metric, it is a smooth transformation out to the exponential metric. That metric is NOT a solution to the Einstein field equations. It makes predictions which are consistent with current tests, and could be accepted or rejected on experimental grounds alone.

And the final line is also very relevant: "One can also come up with a (classical) theory of gravity coupled to Yang-Mills fields, or whatever fields you like." You have gone through the effort to list some of the possibilities. I do not have an issue with your proposal. Instead it is the question I don't get along with. It presupposes the structure of the answer: you have gravity, and it couples politely with other stuff. In the GEM proposal, gravity breaks U(1), SU(2), and SU(3). It does not break it much, in fact it is so darn weak those symmetries appear to our ability to measure, perfect. I probably need to devise a more technical way of saying that, but that is the thrust of the non-debate.

doug

 

[Lawrence B. Crowell]

diff(M), moduli and Hausdorff conditions

Hello Doug,

I did mention the group, that group is $diff({cal M})~=~SO(3,~1)$. This is the analogue of the gauge group for gravitation. Gauge theory rests upon the idea that the particular gauge choice used is unimportant. The elementary example in electrodynamics is with the magnetic field $B~=~\nabla A$, and so any change in the vector potential $A~\rightarrow~A~+~\nabla\chi$ does not change the magnetic field by the “curl-div = 0” rule. This can be generalized to higher dimensions, and in one my posts on this discussions I do describe gauge theory in spacetime or ${\cal M}^4$.
//
This construction leads to the idea of a moduli space. Given a gauge potential its explicit construction has no basis on the fields. Let $\Omega^{1}(ad~g)$ be the set of one-forms on a manifold which are transformed by the action of a group $g$. The coboundary operator $D$ then determines these elements by its operation on zero-forms (functions) with
$$\Omega^0(ad g)~^D\rightarrow~\Omega^1(ad~g).$$
Thus a gauge connection may be determined as a “pure gauge” potential for
$$A^\prime~=~gAg^{-1}~+~gd(g^{-1})$$
with $A~=~0$, and the group elements define the one-forms of $\Omega^0(ad~g)$. Now the set of one-forms $\Omega^1(ad~g)$ are “all over the gauge space map.” In other words nothing is specified about their gauge. However, within this space the moduli may be defined. The following set
$${\cal S}_D~=~\{A~\in~\Omega^1(ad~g): *D*A~=~0\} \}$$
Here $*$ is the Hodge star operator which takes a p-form on an n-dimensional space $p~<~n$ to an n-p-form. This uses forms of the Levi-Civita antisymmetric symbols. Also the condition $*D*A~=~0$ may be other conditions --- the Coulomb gauge in 3-d or the Feynman gauge or ‘t Hooft-Veltman gauge or ... . This defines the slice ${\cal S}_D$ through the bundle, or equivalently the tangent space over the base manifold.
\\
Now to round this out the fields are given by the two forms in the set $\Omega^2(ad~g)$ under the map
$$\Omega^1(ad~g)~^{P_\pm D}\rightarrow~\Omega^2(ad~g)$$
where the $P_\pm$ is projector operator that send the two-forms to self-dual or anti self dual forms. Now the set of gauge equivalent forms ${\cal S}_D$ is given by the pullback on $\Omega^2(ad~g)$, and since these two-forms are independent of the gauge choice it is then the case that the slice ${\cal S}_D$ is an element of a space ${\cal M}_{mod}~=~A/g$, or the set of points “modulo group actions.” So this is the meaning of $diff({\cal M})$
\\
In the case of gravitation the group action is the Lorentz group, on in general the Poincare group with 6 generators --- 3 boosts plus 3 rotations (or angular momenta). Globally these are the transformation of special relativity, which are described by the orthogonal group $SO(3,~1)$. In a Euclidean format the group is $SO(4)~\sim~SU(2)\times SU(2)$, and $SO(3,~1)$ reflects the signature of spacetime. I could go on about how $SO(4)$ and $SO(3,~1)$, as the Euclidean representation and pseudoEuclidean representations, but I’ll do that later. This might have some relevancy for the thread here on Euclideanized spacetime, but I have not yet had time to comment on that. However, in general relativity they are defined for local inertial frames, and how these local regions patch together gives the connection structure and curvatures of general relativity. A set of connections under a coordinate condition define the moduli and the curvatures of GR are $\Omega^2(ad~g)$. Since the moduli space for GR has this pseudoEuclidean structure a particular moduli as ${\cal S}_D$, a point in ${\cal M}_{mod}$, is not separable from another point. Hence the moduli space is nonHausdorff, in the parlance of point-set topology (compact, paracompact and all that jazz). The Hausdorf condition on a space says that for any two points in that space sufficiently small neighborhoods may be found around these points that do no overlap. This however does not appear to obtain for the moduli space of general relativity. Hence the solution space (a’la the Frobenius condition) is difficult to understand. However, there are some bright lights here, for this means that apparently independent solutions are not that separate. Pp-waves (Petrov type N solutions) have structure which are related to Black holes (type D solutions) and this Zariski topology may illuminate how solutions in GR transform between each other.



Lawrence B. Crowell

 

[Lawrence B. Crowell]

diff(M), moduli and Hausdorff conditions

Hello Doug,

I did mention the group, that group is $diff({cal M})~=~SO(3,~1)$. This is the analogue of the gauge group for gravitation. Gauge theory rests upon the idea that the particular gauge choice used is unimportant. The elementary example in electrodynamics is with the magnetic field $B~=~\nabla A$, and so any change in the vector potential $A~\rightarrow~A~+~\nabla\chi$ does not change the magnetic field by the “curl-div = 0” rule. This can be generalized to higher dimensions, and in one my posts on this discussions I do describe gauge theory in spacetime or ${\cal M}^4$.
//
This construction leads to the idea of a moduli space. Given a gauge potential its explicit construction has no basis on the fields. Let $\Omega^{1}(ad~g)$ be the set of one-forms on a manifold which are transformed by the action of a group $g$. The coboundary operator $D$ then determines these elements by its operation on zero-forms (functions) with
$$\Omega^0(ad g)~^D\rightarrow~\Omega^1(ad~g).$$
Thus a gauge connection may be determined as a “pure gauge” potential for
[/tex]
A^\prime~=~gAg^{-1}~+~gd(g^{-1})
[/tex]
with $A~=~0$, and the group elements define the one-forms of $\Omega^0(ad~g)$. Now the set of one-forms $\Omega^1(ad~g)$ are “all over the gauge space map.” In other words nothing is specified about their gauge. However, within this space the moduli may be defined. The following set
$${\cal S}_D~=~\{A~\in~\Omega^1(ad~g): *D*A~=~0\} \}$$
Here $*$ is the Hodge star operator which takes a p-form on an n-dimensional space $p~<~n$ to an n-p-form. This uses forms of the Levi-Civita antisymmetric symbols. Also the condition $*D*A~=~0$ may be other conditions --- the Coulomb gauge in 3-d or the Feynman gauge or ‘t Hooft-Veltman gauge or ... . This defines the slice ${\cal S}_D$ through the bundle, or equivalently the tangent space over the base manifold.
\\
Now to round this out the fields are given by the two forms in the set $\Omega^2(ad~g)$ under the map
$$\Omega^1(ad~g)~^{P_\pm D}\rightarrow~\Omega^2(ad~g)$$
where the $P_\pm$ is projector operator that send the two-forms to self-dual or anti self dual forms. Now the set of gauge equivalent forms ${\cal S}_D$ is given by the pullback on $\Omega^2(ad~g)$, and since these two-forms are independent of the gauge choice it is then the case that the slice ${\cal S}_D$ is an element of a space ${\cal M}_{mod}~=~A/g$, or the set of points “modulo group actions.” So this is the meaning of $diff({\cal M})$
\\
In the case of gravitation the group action is the Lorentz group, on in general the Poincare group with 6 generators --- 3 boosts plus 3 rotations (or angular momenta). Globally these are the transformation of special relativity, which are described by the orthogonal group $SO(3,~1)$. In a Euclidean format the group is $SO(4)~\sim~SU(2)\times SU(2)$, and $SO(3,~1)$ reflects the signature of spacetime. I could go on about how $SO(4)$ and $SO(3,~1)$, as the Euclidean representation and pseudoEuclidean representations, but I’ll do that later. This might have some relevancy for the thread here on Euclideanized spacetime, but I have not yet had time to comment on that. However, in general relativity they are defined for local inertial frames, and how these local regions patch together gives the connection structure and curvatures of general relativity. A set of connections under a coordinate condition define the moduli and the curvatures of GR are $\Omega^2(ad~g)$. Since the moduli space for GR has this pseudoEuclidean structure a particular moduli as ${\cal S}_D$, a point in ${\cal M}_{mod}$, is not separable from another point. Hence the moduli space is nonHausdorff, in the parlance of point-set topology (compact, paracompact and all that jazz). The Hausdorf condition on a space says that for any two points in that space sufficiently small neighborhoods may be found around these points that do no overlap. This however does not appear to obtain for the moduli space of general relativity. Hence the solution space (a’la the Frobenius condition) is difficult to understand. However, there are some bright lights here, for this means that apparently independent solutions are not that separate. Pp-waves (Petrov type N solutions) have structure which are related to Black holes (type D solutions) and this Zariski topology may illuminate how solutions in GR transform between each other.



Lawrence B. Crowell

 

[Don J]

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

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


Example of an actual discussion which is related in part about the Schwarzschild metric
"Celestial Mechanic wrote"
http://www.bautforum.com

Just a message of WARNING .I feel very sorry to have linked to BAUT forum.
I realize now after verification than that board called Against the Mainstream don't really allow for serious discussion about allternative concepts or researchs.
In fact every posters with new concept are litterally attacked and ridiculised by a bunch of as they called themsefve "debunkers".
Using a very disloyal technique describe here
See section
HOW TO DEBUNK JUST ABOUT ANYTHING
http://www.tcm.phy.cam.ac.uk/~bdj10/scepticism/drasin.html

Dont post to BAUT forum.

Edit to add example of debating technique used by BAUT forum members describe in the above link .HOW TO DEBUNK JUST ABOUT ANYTHING

"Portray science not as an open-ended process of discovery but as a holy war against unruly hordes of quackery- worshipping infidels. Since in war the ends justify the means, you may fudge, stretch or violate the scientific method, or even omit it entirely, in the name of defending the scientific method."

 

[sweetser]

Break symmetry by enlarging the circles

Hello:

Another weekend, another nice technical refinement. Discussions with Lawrence had made me confront the issue of group theory for the GEM proposal. I have realized that Diff(M) symmetry breaks U(1) symmetry slightly. A problem with this statement is that it is not precise enough in defining how U(1) symmetry is broken. In this note I will clarify this issue.

Let's start with an example of symmetry breaking. The symmetry of the standard model for massless particles is $U(1)\timesSU(2)\timesSU(3)$. Most particles in the standard model are not massless. The Higgs mechanism is able to introduce mass for fundamental particles such as quarks and electrons through a false vacuum. A scalar field made up of Higgs bosons does the trick. Several billion dollars are being spent by the global physics community to detect the Higgs, which is thought to have a mass between 120 and 200 proton masses (see the July 2005 Scientific American for a gentle introduction to this topic). This is the proverbial Mexican hat: it is easy to visualize how the mass arises from particles going to the low lands of the hat.

There are three players in the standard model:

$U(1)$ The unit complex numbers
$SU(2)$ The unit quaternions
$SU(3)$ A special (ie norm of 1) group with 8 elements

One thing is shared by these three groups: one. Let's form complex-valued vectors that would be elements of each group:

$$U^{\mu}/|U| \in U ( 1 )$$
$$V^{\mu a} / |V| \in SU(2), \mathrm{a goes from 1 - 3}$$
$$W^{\mu b} / |W| \in SU(3), \mathrm{b goes from 1 - 8}$$

Calculate the norms of the three vectors in flat spacetime.

$$g_{\mu \nu} ( U^{\mu} )^{\ast} U^{\nu} / |U|^2 = 1$$
$$g_{\mu \nu} ( V^{\mu a} )^{\ast} V^{a \nu} / |V|^2 = 1$$
$$g_{\mu \nu} ( W^{\mu b} )^{\ast} W^{b \nu} / |W|^2 = 1$$

[NOTE: this is technically in error, as this is the interval, not the norm. Please see the next post, but I will leave this mistake on the public record]

Now we want to break symmetry a bit using the group Diff(M). That effects the metric, and the metric only:

$$g'_{\mu \nu} ( U^{\mu} )^{\ast} U^{\nu} / |U|^2 = 1 + \Delta$$
$$g'_{\mu \nu} ( V^{\mu a} )^{\ast} V^{a \nu} / |V|^2 = 1 + \Delta$$
$$g'_{\mu \nu} ( W^{\mu b} )^{\ast} W^{b \nu} / |W|^2 = 1 + \Delta$$

[This WILL equal whatever it was with g, a basic property of norms]

The vectors for EM, the weak, and the strong forces have not been changed. The metric has changed, and by the equivalence principle, the only thing that can do that is mass. The vectors with this metric are no longer in the corresponding groups because the norm is different from one. They will form a group, one where the norm is [itex]1 + \Delta[/tex]. Essential the unit circle or spheres change there size a little bit. That is how symmetry is broken for the GEM proposal.

On ontological grounds (a fancy way to say "the reason why things work"), I like the visual of the groups that are a bit different from one. It is easy to see circles and spheres getting bigger or smaller. Now that I have an alternative that can be experimentally tested, I can give up the false vacuum of the Higgs mechanism as I have other false gods.

doug

 

[sweetser]

Oops, my bad, I am doing my notation wrong:
$$g_{\mu \nu} ( U^{\mu} )^{\ast} U^{\nu} / |U|^2$$
will calculate the Lorentz invariant interval. The norm is not the interval. Essentially, instead of $t^2 - R^2$, I want $t^2 + R^2$. Right now I am confused about what symbols to use to get to the norm, and how calculating a norm is connected to the metric.

doug

 

[Terry Giblin]

Theordor Kaluza wrote a similar paper

Are you familiar with Theordor Kaluza paper on combining GR and EM?

Does your work confirm or use the same ideas?

Regards

Terry Giblin

 

[sweetser]

Hello Terry:

I guess you are referring to what is known as the Kaluza-Klein approach. In some ways that is a basis for work on string theory. If that is your question, the GEM proposal is different. Kaluza-Klein is a 5 dimensional theory; GEM is 4 dimensional. As far as I am aware (and I do not know the literature in any depth), the predictions for light bending around the Sun are identical for Kaluza-Klein compared with GR, but the GEM proposal will be different at second rank PPN accuracy (not that anyone is planning on getting to that level of accuracy). I cannot make an accurate statement about how Kaluza-Klein works in terms of group theory, but I doubt it has anything to do with breaking U(1) symmetry by changing the norm.

There appear to me fundamental differences between the approaches.
doug

 

[Lawrence B. Crowell]

moduli space overview

I am not sure about the reply by Don J.
//
What I wrote about moduli space is basically "bioler plate" stuff on this, dating back to the work of Atiyah and Singer back in the early 1980s. This leads to the stunning results of Uhlenbeck and Donaldson. In the first case singularities on the base manifold may be lifted to the moduli space in projective spaces. In the second it turns out that there exists an uncountably infinite number of 4-dim spaces, where only a measure epsilon of them are diffeomorphic, and most are homeomorphic but not diff - able.
//
My pet approach to what lies beyond standard theoretical approaches is with nonassociative observables. QM and GR have noncommutative structures [x, p] = hbar and commutators in curvatures. The extension beyond this is not to my mind likely to lead to symmetric field theoretic structures, but with nonassociative ones where
//
(ab)c - a(bc) =/=0.
//
I advise taking a look at my book "Quantum Fluctuations of Spacetime," World Scientific (2005) to see how this works. Essentially I have found that quantum mechanics and general relativity are equivalent as categorical structures of observables over Galois fields. Then GR and QM are quaternionic pairs that exist in a system of octonions. The Galois code for this is then developed for the 24-cell and ..., well I'll leave it to you to look this up.
//
At any rate, I strongly doubt that field theoretic structures are going to be symmetric, except for those under supersymmetry transformations where commutators "go to" anti-commutators.
//
cheers,
//
Lawrence B. Crowell

 

[sweetser]

Non-associative quaternion multiplication

Hello Lawrence:

I formulated my proposal for this forum using standard tensor notation throughout. That said, I am pretty good with quaternions, it was the algebra I used to figure this stuff out in the first place. I own the domain name quaternions.com which features ways for doing physics with quaternions, from classical physics, to special relativity with local transformations instead of the global Lorentz group, EM, quantum mechanics, and my efforts to unify gravity and EM. Rewriting fundamental laws with quaternions is good training for theoretical physics.

Octonions are not necessary if you want a non-associative quaternion product. I have defined what I call the "Euclidean product" as being the standard Hamiltonian product with a conjugate operator tossed in:

Hamilton product:
$$(a,B)(c,D)=(ac - B.D,aD+Bc+B\times D)$$
Euclidean product:
$$(a,B)*(c,D)=(ac + B.D,aD-Bc-B\times D)$$
John Baez pointed out to me that:
$$(AB)*C - A*(BC) =/= 0$$
See, no octonians are needed.

At any rate, I strongly doubt that field theoretic structures are going to be symmetric, except for those under supersymmetry transformations where commutators "go to" anti-commutators.

I can understand the source of the feeling: none are used now, so that pattern should continue. For me, there is no reasonable justification for working the the deviation of the average amount of change, but completely ignoring the average amount of change itself. At least the differences are clear.

doug