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

[Careful]

Hi Doug,

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

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

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

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

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

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

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

Cheers,

Careful

 

[sweetser]

Hello Careful:

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

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

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

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

b). The asymmetric field strength tensor is reducible, so it is not a fundamental field strength tensor. It splits into two irreducible tensors, one that is antisymmetric for EM, the other symmetric for gravity. One can look at gauge symmetries for each separately.
(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})-&gt;(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})&#039;=(\partial_{\mu}(A^{\nu}+\nabla ^{\nu}\lambda)-\partial_{\nu}(A^{\mu}+\nabla^{\mu}\lambda)
That looks like EM gauge symmetry to me. Since the potential has been changed, that will change the derivative of the potential, so one will need to make a change in the connection. A new insight: EM has a rank 1 gauge symmetry while gravity has a rank 2 symmetry. Cool!

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

d). Conserved quantities come out of symmetries. If I have two symmetries, roughly: A^{\nu}-&gt;A^{\nu}&#039;=A^{\nu}+\nabla^{\nu}\lambda \nabla_{\mu}A^{\nu}-&gt;\nabla_{\mu}A^{\nu}&#039;=\nabla_{\mu}A^{\nu}+\Gamma_{\sigma \mu}{}^{\nu}A^{\sigma})
then I have two conserved quantities.

doug

 

[Careful]

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

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

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

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

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

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


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

HUH ?

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

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

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

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

** b). The asymmetric field strength tensor is reducible, so it is not a fundamental field strength tensor. It splits into two irreducible tensors, one that is antisymmetric for EM, the other symmetric for gravity. One can look at gauge symmetries for each separately.
(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})-&gt;(\partial_{\mu}A^{\nu}-\partial_{\nu}A^{\mu})&#039;=(\partial_{\mu}(A^{\nu}+\nabla ^{\nu}\lambda)-\partial_{\nu}(A^{\mu}+\nabla^{\mu}\lambda) **

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


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

Wrong, again only symmetries of the FULL dynamics count.

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

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


** Conserved quantities come out of symmetries. If I have two symmetries, roughly: A^{\nu}-&gt;A^{\nu}&#039;=A^{\nu}+\nabla^{\nu}\lambda \nabla_{\mu}A^{\nu}-&gt;\nabla_{\mu}A^{\nu}&#039;=\nabla_{\mu}A^{\nu}+\Gamma_{\sigma \mu}{}^{\nu}A^{\sigma})
then I have two conserved quantities. **

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

Cheers,

Careful

 

[sweetser]

Hello Careful:

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

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

b). If the GEM Lagrange density does not allow for the transformation:
A^{\nu}-&gt;A^{\nu}&#039;=A^{\nu}+\nabla^{\nu}\lambda
then it will not be able to describe EM.

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

doug

 

[Careful]

Hi Doug,

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

Of course \nabla_{\mu}A^{\nu} is a tensor, but your \nabla_{\mu}A&#039;^{\nu} in the second transformation law is not.

**b). If the GEM Lagrange density does not allow for the transformation:
A^{\nu}-&gt;A^{\nu}&#039;=A^{\nu}+\nabla^{\nu}\lambda
then it will not be able to describe EM.**

Sure...

**

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

**

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

Cheers,

Careful

 

[sweetser]

Hello Careful:

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

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

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

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

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

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

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

doug

 

[Careful]

Hello Doug,

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

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

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

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

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

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

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

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

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


Cheers,


Careful

 

[sweetser]

A valid contravariant expression

Hello Careful:

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

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

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

This is a direct statement that Q1 is a tensor.

Jackson provides a transformation law:

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

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

g_{\alpha \gamma}F^{\gamma \delta} g_{\delta \beta} = <br /> g_{\alpha \gamma}(\partial^{\gamma} A^{\delta}-\partial^{\delta} A^{\gamma}) g_{\delta \beta}
= g_{\alpha \gamma}\partial^{\gamma} A^{\delta} g_{\delta \beta}-g_{\alpha \gamma}\partial^{\delta} A^{\gamma} g_{\delta \beta} = \partial_{\alpha} A_{\beta}-\partial_{\beta} A_{\alpha}=F_{\alpha \beta}

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

doug

 

[sweetser]

Contracting symmetric mixed rank 2 tensors

Hello Careful:

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

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

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

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

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

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

doug

 

[vanesch]

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

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

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

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

 

[Careful]

Hi Doug,

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

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

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

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


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


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

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

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

Cheers,

Careful

 

[sweetser]

A valid contravariant expression (cont.)

Hello:

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

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

**I thought that you meant this, **

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

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

** Obviously, in GR **

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

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

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

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

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

 

[sweetser]

Asymmetric mixed rank 2 tensors

Hello Careful:

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

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

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

\left(\begin{array}{cc}<br /> 2 \nabla_0 b_0 &amp; \nabla_0 b_1 - \nabla_1 b_0 \\<br /> -\nabla_1 b_0 + \nabla_0 b_1 &amp; 2 \nabla_1 b_1 \\<br /> \end{array}\right)=?=\nabla{}{}B+\nabla{}{}B

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

doug

 

[Careful]

Hi,

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

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


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

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

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

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

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

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

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

Indeed, as I mentioned before...

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

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

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

 

[sweetser]

The GEM action

Hello:

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

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

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

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

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

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

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

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

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

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

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

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

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

doug

 

[Careful]

Hi,

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

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

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

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


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


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


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

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


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

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

Cheers,

Careful

 

[sweetser]

Symmetric versus antisymmetric tensors

Hello Careful:

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

The asymmetric action has not changed.

I am still having trouble with the irreducible tensors.

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

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

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

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

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

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

doug

 

[Careful]

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

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


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

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


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

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


Cheers,

Careful

 

[sweetser]

Hello Careful:

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

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

Agreed in the general.

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

doug

 

[Careful]

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

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

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

Cheers,

Careful

 

[sweetser]

Hello Careful:

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

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

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

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

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

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

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

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

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

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

doug

 

[Careful]

Hi,

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

Good Luck,

Careful

 

[sweetser]

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

doug

 

[Careful]

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

doug

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

Cheers,

Careful

 

[sweetser]

Field equations and Lagrange densities

Hello Careful:

One can certainly have this perspective:

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Happy vacation days,
doug

 

[sweetser]

Good bye mixed tensors in 2006

Hello all:

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

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

Have a good new year,
doug

 

[Careful]

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

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

Have a good new year (and keep on learning )

Cheers,

Careful

 

[sweetser]

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

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

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

Will keep learning. Enjoy the moment.
doug

 

[sweetser]

Abstract for APS meeting in Dallas

Hello:

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

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

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

doug

 

[Lawrence B. Crowell]

GEM theory problem

The Lagrangian
<br /> L~=~ (J^a_q~-~J^a_m)A_a~-~1/2(\nabla_aA^b\nabla^aA_b<br />
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
<br /> A_a = A(em) + A(weak),<br />
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
<br /> J_a~=~T_{ab}e^b,<br />
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,
<br /> SO(4)~\simeq~SU(2)\times SU(2).<br />
One of the SU(2)&#039;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
<br /> E_a^b~=~\gamma^be_a,<br />
where \gamma^b is a Dirac matrix in some representation. One then would have
<br /> de^a~=~A_be_adx^b,<br />
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
<br /> \partial_c(E_a^b)~=~\gamma^bA_ae_c~+~{\Gamma^b}_{cd}E_a^d.<br />
From here your general gauge potential, call it {\cal A}, would be
<br /> {{\cal A}^b}_{ac}~=~\gamma^bA_ae_c~+~{\Gamma^b}_{cd}E_a^d<br />
The field tensor for this theory would be defined from
<br /> \partial_d{{\cal A}^b}_{ac}~-~\partial_c{{\cal A}^b}_{ad},<br />
which if one were to work out the bits should result in the gravity and EM sectors.
Lawrence B. Crowell