Unifying Gravity and EM

sweetser · Jan 12, 2006

Hello Lawrence:

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

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

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

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

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

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

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

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

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

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

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

doug

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

Lawrence B. Crowell · Jan 12, 2006

Dear Doug,

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

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

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

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

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

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

then

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

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

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

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

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

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

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

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

cheers,

Lawrence B. Crowell

sweetser · Jan 13, 2006

Hello Lawrence:

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

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

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

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

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

doug

Lawrence B. Crowell · Jan 13, 2006

gauge theory & moduli

I am redoing this. I think this should be better

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

sweetser · Jan 14, 2006

Diff(M) as a gauge group

Hello Lawrence:

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

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

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

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

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

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

doug

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

sweetser · Jan 15, 2006

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 · Jan 15, 2006

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 \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 · Jan 17, 2006

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 · Jan 18, 2006

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 · Jan 18, 2006

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 · Jan 18, 2006

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 · Jan 18, 2006

Lawrence B. Crowell said:

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 · Jan 18, 2006

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 · Jan 19, 2006

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 · Jan 19, 2006

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 · Jan 19, 2006

transverse g-waves and Dirac matrices

The gamma is indeed the Dirac matrix. Since 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 GL(2,C)[\itex]. BTW, when it comes to TeX-ing problems, for some reason often the post preview does not show the TeX&#039;ed equations. I just get a box. Lawrence B. Crowell

sweetser · Jan 19, 2006

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 · Jan 20, 2006

Lawrence B. Crowell said:

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 · Jan 20, 2006

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 · Jan 20, 2006

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 · Jan 20, 2006

Lawrence B. Crowell said:

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 · Jan 23, 2006

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 · Jan 23, 2006

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 · Jan 23, 2006

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 · Jan 23, 2006

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 · Jan 24, 2006

Lawrence B. Crowell said:

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 · Jan 24, 2006

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 · Jan 24, 2006

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 · Jan 25, 2006

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 · Jan 26, 2006

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 · Jan 31, 2006

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:

Baez said:

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 · Feb 1, 2006

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 · Feb 1, 2006

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 · Feb 5, 2006

Don J said:

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 · Feb 5, 2006

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 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 · Feb 5, 2006

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 · Feb 6, 2006

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 · Feb 6, 2006

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 · Feb 6, 2006

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 · Feb 6, 2006

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.

Crowell said:

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

sweetser · Feb 7, 2006

The quantum equivalence principle

Hello:

Let's start with things researchers agree on:

1. A spin 0 quantum field is necessary to give fundamental particles like electrons and quarks inertial mass.

2. A spin 2 quantum field is required to mediate the effects of gravity.

3. The equivalence principle is a classical doctrine confirmed by experiments that the inertial mass is equal to the gravitational mass.

4. The correspondence principle shows that classical laws arise from the aggregate of quantum events.

In my limited exposure, folks who are searching for the Higgs particle do not bring up gravitons. Likewise, the gravity wave detector crowd appears unconcerned with the Higgs. Yet the equivalence principle must arise from an absolutely unwavering link between the quantum source for rests mass, a scalar field, and the quantum source for gravitational mass, a spin 2 field. The two fields can never "get out of step" with each other, or the equivalence principle will be violated, which it is not.

In the GEM proposal, the link between the scalar field and the spin-2 field is direct and rock solid: the spin 2 field for gravitational mass is the symmetric field strength tensor \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}, and the spin 0 field for inertial mass is the trace of the same symmetric tensor, tr(g_{\mu \nu}(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})). Einstein was the first to argue the elegant case for the classical equivalence of gravitational and inertial mass. I am pleased the GEM proposal sings the same song on the quantum level.

doug

Lawrence B. Crowell · Feb 8, 2006

nonassociativity, octonions & field theory

The nonassociative product
 (e_ie_j)e_k~-~e_i(e_je_k)~=~\upsilon_{ijkl}e_l 
can be defined where \upsilon_{ijkl} is defined without reference to octonions. However, this system will not satisfy a closed algebra unless it defines the Cayley four-form. One can of course restrict things to this rather myopic view.
\\
I am not sure what you mean by the Hamilton product. It appears that you are referencing a Jordan product in some form. The Hamilton product to me is
 i^2~=~j^2~=~k^2~=~ijk~=~-1. 
and a general quaternion is defined by
 H~=~a\sigma_x~+~b\sigma_y~+~c\sigma_z~+~d{\bf 1} 
In four dimensions the Pauli matrices are generalized to Dirac matrices.
\\
At any rate, nonassociativity without reference to the octonions means that there is no closed multiplication table defined by the structure constant \upsilon{ijkl}, of which there are 480 possible representations. I could go into the cyclic group and PSL(2,~7) Hurzewitz discrete group theory behind this, but that gets too far into maths.
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Gauge theory has some sort of algebraic structure, eg. SU(n), and the like. One of course defines gauge connections according to representations of this group. However, for there curvatures on the principle bundle are found by F~=~dA~+~A\wedge A, where the tensor components of the two-form F in the case of an abelian field (EM) is the antisymmetric field strength tensor containing the electric and magnetic field components. The antisymmetry is imposed not by any group structure, but rather from a much more basic theory of chains and cycles on any manifold.
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Physics was not built from the ground up, but rather from a domain of experience that existed at the time. Faraday’s description of changing magnetic fields, currents and the like were based on observational experience of the time. Of course now we can easily see that this all stems from F~=~dA, and things appear elementary. Neither Faraday nor Maxwell had any notion of differential forms or integrating on chains in n-dimensional manifolds. Maxwell indeed had things according to spinning ether vortices and the like as his “description.” In part because of this his original theory was actually quaternionic. Now in our modern age we can see that these types of theories have antisymmetric structure because of these basic “facts,” and hold for any gauge theory --- even gravity.
\\
Lawrence B. Crowell

sweetser · Feb 8, 2006

Non-associative quaternion multiplication

Hello Lawrence:

This is the definition for a Hamilton quaternion product I am using:
 i^2~=~j^2~=~k^2~=~ijk~=~-1. 
While this is true:

The nonassociative product.
 (e_ie_j)e_k~-~e_i(e_je_k)~=~\upsilon_{ijkl}e_l 
can be defined where \upsilon_{ijkl} is defined without reference to octonions.

it is not relevant because I am using a conjugate operator, or more technically an anti-involutive automorphism. The conjugate flips the sign of the 3-vector.

What I call a Euclidean product has a closed multiplication table:
i^{*} i= j^{*} j = k^{*} k= (i^{*} j)^{*} k = 1
The standard definition of a non-associative product you provide does not use a conjugate operator as I do. Because the automorphism keeps the quaternions within the quaternion manifold, the multiplication table is closed: i^{*} i = 1. Let's see this in action, how always taking the conjugate of the first quaternion in the binary quaternion multiplication makes the triple product non-associative.
a (b c) = a^{*} (b^{*} c) = a^{*} b^{*} c
(a b) c = (a^{*} b)^{*} c = a b^{*} c
These are not the same. The norm of these two are same, but they point in different directions.

Our current gauge theories for gravity and EM are both spectacular and inadequate. Both match experimental data extraordinarily well. Yet we cannot quantize EM without choosing a gauge. Even if a gauge is chosen, gravity cannot be quantized.

Let me reestablish how I am working from the "domain of experience". Gravity and EM as currently formulated work in 4 dimensions, as does GEM. It is odd that simple statement is at odds with the majority of work on gravity today. The group Diff(M) was used by Einstein to make his equations covariant, and it plays a role in GEM. EM symmetry must be broken by a scalar field, as it is in GEM by the trace of the field strength tensor. The 4D EM wave equation can be quantized by the Gupta-Bleuler method, but two of the modes of emission must be eliminated by a supplementary condition. GEM uses the same creation and annihilation operators, but it also has a spin 2 field that will not have the same technical issues the spin 1 scalar mode has. And in keeping with the most important of all scientific traditions, should we refine the level of accuracy of measuring light around the Sun, GEM predicts 0.7 microacrsecond more than GR, so unlike all the work in string theory, I am willing to put every chip I own on the line and let the data decide.

How things get written matters. You write the field strength tensor as F = dA. Nothing can be simpler than that. Sure, it has all the differential forms stuff behind it, but since that is the starting point, I can appreciate that there appears nothing simpler. I write it differently, also in an established way, as a F=\nabla^{\mu}A^{\nu}-\nabla^{\nu}A^{\mu}. In my notation, I can explain even to my girlfriend that the F thingie is not a complete story because of the act of subtraction: there are 4 parts in the triangle, 4 parts in the A, sixteen parts total, but the study of light needs only 6 of the 16. What I cannot explain to her is why the brightest folks on the planet do not share my feelings of obligation to work with the other 10 terms.

dougHistorical note: Maxwell was familiar with quaternions because Hamilton trumpeted their cause (too loudly). Scalar, vector, div, grad, curl were inventions of Hamilton in his quest to understand quaternions. In Maxwell's first edition of his treatise, he used "pure quaternions" which is one with a zero for the scalar, effectively as a stand-in for 3-vectors. He did nothing at all fancy with quaternion algebra. By the third edition, he only hoped that quaternions would play a role in describing EM. Many folks have figured out how to toss in an extra imaginary number and get the Maxwell equations. Peter Jack was the first person to do so with real quaternion (1996?) and I did the same independently a year later. There is a rumor on the Internet that some 200 quaternion equations were deleted from the first to the third edition. From my viewing of an on-line version of the first treatise, that is an inaccurate description: the quaternions were used like any 3-vector, and he did not write the homogeneous or source equations as an exercise in quaternion algebra.

Lawrence B. Crowell · Feb 9, 2006

Below is a multiplication table of octonions that I devised for fermions. It was derived with respect to black hole physics and Bogoliubov transformations --- but that is another story. These fields are the operators for ingoing and outgoing modes into a black hole. These fields include their conjugations. I hope this shows up right!
\\
 \left(\begin({\array}{}* & * & e_1 & e_2 & b_k & e_4 & b_k^\dagger & b_{-k} & b_{-k}^\dagger \\ *& * & *& * & * & * & * & * & *\\ e_1 & &-1 & e_4 & b_{-k}^\dagger & -e_2 & b_{-k} & -b_k^\dagger & -b_k \\ e_2 & & -e_4 & -1 & b_k^\dagger & e_1 & -b_k & b_{-k}^\dagger & b_{-k} \\ b_k & & -b_{-k}^\dagger & -b_k^\dagger & 0 & b_{-k} & e_2 &-e_4 & e_1 \\ e_4 & & e_2 & -e_1 & -b_{-k} & -1 & b_{-k}^\dagger & b_k & -b_k^\dagger \\ b_k^\dagger & & -b_{-k} & b_k & -e_2 & -b_{-k}^\dagger & 0 & e_1 & e_4 \\ b_{-k} & & b_k^\dagger & -b_{-k}^\dagger & e_4 & -b_k & -e_1 & 0 & e_2\\ b_{-k}^\dagger & & b_k & b_{-k} & -e_1 & b_k^\dagger & -e_4 & -e_2 & 0}\end{array}\right) 
\\
You say that (ab)c and a(bc) have the same norm but differ in their direction. That is standard with octionions. You might notice this if you examine this table.
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A couple of other points. A choice of gauge amounts to lifting the base manifold up the fibre. Think of the base manifold as a sheet of paper. Then perpendicular to each point is a line coming up, which for the case of triviality defines the tensor product of the vector space in the line with the manifold. Now to compute the action of this internal vector space one needs to move from fibre to fibre. But how does one do this? One takes the base manifold (the sheet of paper) and lifts a copy of it so it cuts through the fibres. One can lift the copy paper any way one wants, but once done things are computed with that choice. This is what a gauge choice is. One needs to define a bundle section in some way. It does not matter how this is done, but it must be done and used consistently. Well, 't Hoft and Veltmann did a gauge choice change in mid-stream with QCD to get renormalizability, but one has to be very careful about that! At any rate a gauge choice is nothing more than a way of fixing a frame, akin to defining a reference frame in relativity (or Newtonian mechanics for that matter) in order to do calculations.
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As for why only 6 of the possible terms end up as real is because the antisymmetry imposes a structure on the field. Anyway, we only have 3 E-fields and 3 B-fields. Imposing a structure on the theory involves the elimination of some of the possible tensor terms. The antisymmetry of gauge field 2-forms does that automatically.
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Lawrence B. Crowell

Creator · Feb 9, 2006

sweetser said:

In the GEM proposal, the link between the scalar field and the spin-2 field is direct and rock solid: the spin 2 field for gravitational mass is the symmetric field strength tensor \nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu}, and the spin 0 field for inertial mass is the trace of the same symmetric tensor, tr(g_{\mu \nu}(\nabla^{\mu}A^{\nu}+\nabla^{\nu}A^{\mu})).
doug

Hello Doug;
In QFT isn't mass obtained, at least some fraction of it, by radiative corrections? How does that relate here ?

sweetser · Feb 10, 2006

Radiative corrections

Hello Creator:

The short answer is I do not know how to incorporate radiative corrections into the proposal.

I do know that for compound particles such as protons and neutrons, most of the mass is from the movement of gluons and quarks. Any radiative corrections would be just another source of energy to add into the picture. I am unable to calculate what percentage of the proton mass comes from the Higgs mechanism. I'm sure some folks on the planet know how to do that calculation, but I do not. The ice is very thin where I am skating!

doug

sweetser · Feb 10, 2006

Hello Lawrence:

Unfortunately, your table for the Octonians did not work out :-( I would love to see it. If you come back to the forum, there should be an edit button at the bottom if you have logged in. The first two lines look something like this:

\left(\begin({\array}{}* & * & e_1 & e_2 & b_k & e_4 & b_k^\dagger & b_{-k} &
b_{-k}^\dagger \\ *& * & *& * & * & * & * & * & *\\

So the "*"'s need to be replaced with what goes in there. A second effort for this multiplication table would be appreciated. Note that after the save button is hit and the page appears, the new graphic does NOT appear. The equation is a graphic which the computer keeps in cache. One must hit the reload button while holding down the control key to see the improvements.

doug, whose LaTeX here is NEVER done perfectly the first time.

Lawrence B. Crowell · Feb 10, 2006

I have an inordinate difficulty with the LaTeX here, which I do not have otherwise. Anyway, I have taken a dvi page with this table and prduced an image file. I am going to try to attach that to this post.

Lawrence B. Crowell

sweetser · Feb 13, 2006

Hello Lawrence:

Thanks for the dvi page. I am a bit confused about one aspect of this table. Octonions are a non-associative, non-commutative division algebra. The zeros that appear in the table would create a problem for the division algebra. I think those should be -1's. For completeness, the identity should be included. Here is a table from the web without zeroes: http://www.geocities.com/zerodivisor/obasis.html

doug

Lawrence B. Crowell · Feb 13, 2006

octonion table

The zeros are meant to convey fermionic content with b^2 = 0. The anticommutators of the Fermionic fields are "treated" as commutators in order to obtain associators

[a, b, c] = (ab)c - a(bc).

I suppose I should have considered the anti-associator

{a, b, c} = (ab)c + a(bc)

instead to avoid this matter. However, the basic import is that the xero reflect a topological content of fermions. Since b^2 = 0 this is equivalent to d^2 = 0, and that the fields are defined in ker(b)/im(b). This topological element is what skirts the problem with division algebras. I think this extends to the sedenions, but that is another issue.

Maybe this was not the best example for this, for in most cases there is none of this topological implication underlying it. I attach a more familiar octonion multiplication table.

Unifying Gravity and EM

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