How Can Gravity and Electromagnetism Be Unified Through a Rank 1 Field Theory?

sweetser · Jan 11, 2006

Hello Lawerence:

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

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

Doc Al · Jan 11, 2006

Lawrence:

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

- Doc

sweetser · Jan 11, 2006

Gauge symmetry

Hello Lawrence:

The post deals with gauge symmetry issues.

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

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

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

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

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

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

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

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

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

doug

sweetser · Jan 11, 2006

Localizable theories

Hello Lawrence:

This post is in reply to the nonlocalizable issue.

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

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

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

Which is better? You make the call,
doug

member 11137 · Jan 12, 2006

Lawrence B. Crowell said:

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

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

member 11137 · Jan 12, 2006

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

sweetser · Jan 12, 2006

Blackforest said:

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

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

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

doug

member 11137 · Jan 12, 2006

sweetser said:

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

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

sweetser said:

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

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

sweetser said:

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

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

sweetser said:

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

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

sweetser said:

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

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

Lawrence B. Crowell · Jan 12, 2006

Yang-Mills fields & symmetric field tensor

Doug,

Since you indicated the structure of gauge theory, I figured I would do the same, but on a somewhat more fundamental level. All of this is based on differential forms. The coboundary operator d~=~dx^a\partial_a acts on a section x, or some cut through a principle bundle Pover a manifold \cal M simply as ds. So act upon s with s^\prime~=~gs, where g is a group element of the Lie group of transformations of the theory. Now consider ds^\prime with ds~=~As, where A[/tex] is a gauge connection ds^\prime~=~d(gs) This then leads to ds^\prime~=~(dg)s~+~gAs ds^\prime~=~\big((dg)g^{-1}~+~gAg^{-1}\big)s^\prime. This leads to the transfomation of a gauge connection as A~\rightarrow~(dg)g^{-1}~+~gAg^{-1} For a group element g~=~e^{i\chi} this means that the gauge transformation is A~\rightarrow~A~+~id\chi~+~i[\chi,~A], and for and ableian theory (commutator = 0) and d~=~dx^a\partial_a this leads to A~\rightarrow~A~+~\nabla\chi, where i has been absorbed into \chi. The differential form d satisfies d^2~=~0 because d^2~=~dx^a\wedge dx^b\partial_a\partial_b, and dx^a\wedge dx^b is antisymmetric and \partial_a\partial_b is symmetric. The appropriate differential form is the gauged differential form D~=~d~+~A. The field tensors are then obtained from D^2~=~D\wedge D~=~(d~+~A)\wedge(d~+~A) =~F=~dA~+~A\wedge A, where the A\wedge d is zero acting on unity or 1. If we expand this in coordinates we then have F~=~\big(\partial_aA_b~-~\partial_bA_a~+~[A_a,~A_b]\big)dx^a\wedge dx^b. I have ignored structure constants and the rest here, but those can be included after the fact here. For the spatial variables this leads to the invariance of the magnetic field B~=~\nabla\times A under the gauge transformation. The two-form F~=~F_{ab}dx^a\wedge dx^b contains the antisymmetric field tensor with the electric and magnetic field components which can be easily derived. The point here is that the antisymmetry of gauge theory is seen to emerge from its structure according to differential forms. This is a bit of a problem I see with your symmetric gauge field terms. All that “div-grad-curl” stuff used in physics emerges from the structure of vector fields on space. When it comes to your e,~b,~g fields e=\frac{\partial A}{\partial t}-\nabla \phi-\Gamma_{\sigma}{}^{0u}A^{\sigma} b=(-\frac{\partial A_{z}}{\partial y}-\frac{\partial A_{y}}{\partial z}-\Gamma_{\sigma}{}^{yz}A^{\sigma}, -\frac{\partial A_{x}}{\partial z}-\frac{\partial A_{z}}{\partial x}-\Gamma_{\sigma}{}^{xz}A^{\sigma}, -\frac{\partial A_{y}}{\partial x}-\frac{\partial A_{x}}{\partial y}-\Gamma_{\sigma}{}^{xy}A^{\sigma}) g=(\frac{\partial \phi}{\partial t}-\Gamma_{\sigma}{}^{tt}A^{\sigma}, -\frac{\partial A_{x}}{\partial x}-\Gamma_{\sigma}{}^{xx}A^{\sigma}, -\frac{\partial A_{y}}{\partial y}-\Gamma_{\sigma}{}^{yy}A^{\sigma}, -\frac{\partial A_{z}}{\partial z}-\Gamma_{\sigma}{}^{zz}A^{\sigma}) I would say that the first is simply a covariant form of the electric field, though with a negative sign “issue” on the first term on the right hand side, and where a \Gamma_{0}{}^{i0}\phi should also appear. The second equation for the b field would apply if \Gamma_{\sigma}{}^{xz} is a torsional part of the spacetime connection term, and where nonvanishing part would involve terms \Gamma_{0}{}^{xz}\phi. I am doing this “on the fly” here, but I think I have this right. So there are two issues that I am scratching my head over. The first is that I am uncertain what is meant by a symmetric field tensor. Such things do occur, such as with Fermi-Dirac fields, and under supersymmetry anti-commuting fields do become commuting field theories If we consider the Lorentz operators p_\mu and M_{\mu\nu} as elements of the subalgebra L_0 and operators Q^\alpha_a as elements of the subalgebra L_1, the for \{a,b\}\in L_0 [a,~b]\in L_0, and for a\in L_0,~b\in L_1 [a,~b]\in L_1, and for \{a,~b\}\in L_1 [a,~b]_+\in L_0, where the + means anticommutator. Here the total algebra is the graded L_0\oplus L_1. By this means a field theory can be extended to something involving symmetric field tensors, but they are due to the graded algebraic structure of SUSY, and not the standard Yang-Mills sort of field theory. I would say that a possibility is the Coleman-Mandula theorem, which states that all the symmetries of the S-matrix emerge from the generators p_\mu and M_{\mu\nu}. Lawrence B. Crowell

Lawrence B. Crowell · Jan 12, 2006

Nonlocalizability of energy

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

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

Lawrence B. Crowell

sweetser · 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

How Can Gravity and Electromagnetism Be Unified Through a Rank 1 Field Theory?

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