Unifying Gravity and EM

[Mentz114]

Hi Doug:

I was too picky in challenging your beliefs, you're entitled to them and I suppose we all have some rigidities built in.

The proposal is gauge invariant in the usual sense of the word for all mass less particles. When one has massive particles, the proposal does involve choice in how things are measured, either as changes in the derivatives of the 4-potential, or changes in the connection which are changes in the metric. I have had no luck in communicating what the preceding sentence means to those skilled in the mathematical physics arts, which is exasperating since it falls out of looking at the definition of a covariant derivative (it is the sum of these two things, so how much comes from one or the other is arbitrary).

You've definitely failed to communicate it to me, although I wouldn't count myself amongst "those skilled in the mathematical physics arts".

What one has to do is make a decision on how things are measured, either as a change in potential or a change in the metric or a combination of both, and proceed from there.

Are the equations of motion invariant under this choice ?

Regarding your vacuum solution ( flat potential, no mass or charge ?) which gave you the exponential (Rosen) metric - did you discard the off-diagonal elements of the metric, because you've only four equations, and the full metric would have 10 independent terms.

 

[sweetser]

Covariant Derivative Symmetries

Hello Lut:

One of the big ideas I am trying to accept myself is that the covariant differential operator, $\nabla^{\mu}$ plays an equal role to the 4-potential, $A^{\nu}$ in the GEM action. This is a message of quantum mechanics, where operators are what gets measured. I used to look at the derivative and think it was just a math widget - we are trained that way. Yet in quantum mechanics, what we are measuring is the average of a derivative.

I need to view the covariant differential operator this way for two reasons. The first is to have a theory that has enough degrees of freedom to described both gravity and EM. The second reason is to have the symmetry required for a geometric theory for gravity. If I were to use an ordinary derivative, $\partial^{\mu} A^{\nu}$, then the only thing that could be included in the measurement of change is the change of the potential. With a covariant derivative, I get a change in potential minus a change in the metric, $\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}$. So how much of the covariant derivative comes from $\partial^{\mu} A^{\nu}$ and how much comes from $- \Gamma_{\sigma}^{\mu \nu} A^{\sigma}$? You get to decide. That is the symmetry needed for gravity. Whereas GR says gravity is exclusively a geometric theory, GEM says that gravity could be a geometric theory, a potential theory, or any combination of the two consistent with the other choices. Any equation that has a covariant derivative in it has this symmetry.

As far as the (possibly electrically charged) Rosen metric, I worked with the simplest case. This is "The Revenge of Schwarzschild II", same assumptions Karl made on the Russian front so many years ago.

doug

 

[sweetser]

Generalizing Current Coupling Spin 1 & Spin 2

Hello:

Steve Carlip had pointed out the problem with the spin of the coupling term, $-J^{\mu} A_{\mu}$. He gave me a wonderful reference to a description of the problem in chapter 3 of "Feynman Lectures on Gravity". Based on that work, I was able to think about current moving along the z axis, and show that it could have both spin 1 and spin 2 symmetry in the phase (reread post 319, 320, 363, 367 if you want more details). It was great to do a direct variation off of Feynman!

After the celebration of getting around a killer technical problem, there is the hangover, the parts that one does not like about the solution. I was uncomfortable with the way direction played a role. I started with moving along the z axis because that is exactly what Feynman did. I like to follow the footsteps of great ones :-) Yet one could imagine an inertial observer where the motion along z was zero. It felt like my spin argument might fall apart for that observer. That gave me a headache.

I generalized this a bit by introducing two vectors, parallel $I_{||}$ and transverse $I_{tr}$. This will work for motion along x, y, z or any direction between, a definite improvement. But does this still have the problem of what happens when one of these goes to zero? I worried about the meaning of these directions: was it the direction of the potential and its real current, or was it about the two currents? I also did not like introducing new letters into the GEM equations. The algebraic problem was solved, but I had headaches.

Recently I have started working with a representation of quaternions that commute (post 415 and 416). To get the right scalar product and phase with both spin 1 and spin 2 requires the product of both the Hamilton and Even quaternions (which I while symbolize by tacking on a "2").

First write out the standard scalar generated from a tensor contraction:

[tex]-J^{\mu} A_{\mu} = -\rho \phi + Jx Ax + Jy Ay + Jz Az[/itex]

This does not have any information about the phase as is done in chapter 3 of Feynman. What Feynman does is take the Fourier transform of the potential to get a current-current interaction, with A->J':

[tex]-J^{\mu} J'_{\mu} = -\rho \rho' + Jx Jx' + Jy Jy' + Jz Jz'[/itex]

Now form the products for the current-current interaction using both the Hamilton and Even quaternion representations:

$$-J J' = -(\rho, Jx, Jy, Jz)(\rho', Jx', Jy', Jz')$$
$$= (-\rho \rho' + Jx Jx' + Jy Jy' + Jz Jz',$$
$$-\rho Jx' - Jx \rho' + Jz Jy' - Jy Jz',$$
$$-\rho Jy' - Jy \rho' - Jz Jx' + Jx Jz',$$
$$-\rho Jz' - Jz \rho' + Jy Jx' - Jx Jy')$$

$$-J2 J2'^* = -(\rho, Jx, Jy, Jz)(\rho', -Jx', -Jy', -Jz')$$
$$=(- \rho \rho' + Jx Jx' + Jy Jy' + Jz Jz',$$
$$\rho Jx' - Jx \rho' + Jz Jy' + Jy Jz',$$
$$\rho Jy' - Jy \rho' + Jz Jx' + Jx Jz',$$
$$\rho Jz' - Jz \rho' + Jy Jx' + Jx Jy')$$

These two products have the same Lorentz invariant scalar, as they must if they hope to sit in for the tensor 4-vector contraction. For the Hamilton representation, it is the source terms $\rho J_i$ that move together, and thus will represent spin 2 symmetry. The curl terms have opposite signs, and will require the normal 2 pi radians to get around as is the case for spin 1 symmetry. For the Even representation, the source terms have spin 1 symmetry, while the symmetric curl terms have spin 2 symmetry.

Sounds like a complete story to me. Headache gone.

Doug

 

[sweetser]

GEM fields written with quaternions

Hello:

In this post I will write out the GEM field strength tensor using only the Hamilton and Even representations of quaternions. The five fields, E, B, g, e, and b (e and b being symmetric counterparts to E and B respectively) will be written succinctly, a good sign.

Here is the short version. The E and B fields arise from the Hamilton representation where the 4-derivative acts on the 4-potential. The fields E and B can be separated by changing the order of the derivative with the potential. The symmetric fields e and b arise from the Even representation where the conjugate acts on the 4-derivative or 4-potential.

In both cases, there is a scalar field g, but the g has opposite sign and cancels out. This may turn out to be a good thing for approximate gauge symmetry (I don't think it will be perfect because like electric charges repel while like mass charges attract, creating a dipole).

Let's start calculating. Take the 4-derivative of a 4-potential in the Hamilton representation, but put the derivative after, which flips the signs of the curl, nothing else:

$$-A \nabla =$$

$$(-\frac{\partial \phi }{\partial t}+\frac{\partial Ax}{\partial x}+\frac{\partial Ay}{\partial y}+\frac{\partial Az}{\partial z},$$
$$-\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},$$
$$-\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},$$
$$-\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$

$$= -g + E + B$$

[technical note: I will be attaching a Mathematica notebook at the end. It took a few emails to tech support and a bit of work on my part, but I got Mathematica to write the 4-derivative of a 4-potential in a more human-readable way. Then I used "cut as Latex" to plug this in. This is why I am confident this large a volume of partial differential equations is accurate.]

To separate E from B, switch the order of the differential, and either have the same sign for E or different signs for B.

$$=\frac{1}{2}(-\nabla A - A \nabla) =$$

$$(-\frac{\partial \phi }{\partial t}+\frac{\partial Ax}{\partial x}+\frac{\partial Ay}{\partial y}+\frac{\partial Az}{\partial z},$$
$$-\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},$$
$$-\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y},$$
$$-\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z})$$

$$= -g + E$$

$$=\frac{1}{2}(\nabla A - A \nabla) =$$

$$(0,$$
$$c \left(\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right),$$
$$c \left(\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right),$$
$$c \left(\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) )$$

$$= B$$

One sign flip on $\nabla A$ is all it takes. Nice.

Now on to the symmetric fields. Here we use the Even representation where all the signs of all products are positive. For this to be a division algebra, events like photons must be excluded. That doesn't sound so bad in this context, since photons make up the E and B fields. The three symmetric fields are generated in a simple expression:

$$\nabla^* A2 =$$

$$(\frac{\partial \phi }{\partial t}-\frac{\partial Ax}{\partial x}-\frac{\partial Ay}{\partial y}-\frac{\partial Az}{\partial z},$$
$$\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x}-c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},$$
$$\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}-c \frac{\partial Ax}{\partial z}-c \frac{\partial Az}{\partial x},$$
$$\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}-c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$

$$= g + e + b$$

It is important to note that the A2 is a different representation of the same functions or values that are in the Hamilton representation of the 4-potential. In other words, it is the same 4-vector. What has changed is what happens when an operator acts on it, or it gets multiplied by another even quaternion.

Let's isolate the symmetric e field using a conjugate:

$$\frac{1}{2}\left(\nabla^*A2-\nabla A2^*\right) =$$

$$(0,$$
$$\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},$$
$$\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y},$$
$$\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z})$$

$$= e$$

The time derivative of A has flipped its sign as happens for the symmetric e.

Isolating the symmetric b involves one more conjugate applied to the previous result:

$$\frac{1}{2}\left(\nabla^*A2-(\nabla A2^*)^*\right) =$$

$$(0,$$
$$-c \left(\frac{\partial \text{Ay}}{\partial z}+\frac{\partial \text{Az}}{\partial y}\right),$$
$$-c \left(\frac{\partial \text{Ax}}{\partial z}+\frac{\partial \text{Az}}{\partial x}\right),$$
$$-c \left(\frac{\partial \text{Ax}}{\partial y}+\frac{\partial \text{Ay}}{\partial x}\right) )$$

$$= b$$

Every partial derivative has a minus sign.

Once again, the fields are easy to state, easy to separate. Nice.

Now to look at the big sum by adding the two field strength tensors together:

$$\frac{1}{2}\left(-A \nabla + \nabla^* A2\right) =$$

$$(0,-c \left(\frac{\partial Az}{\partial y}+\frac{\partial \phi }{\partial x}\right),-c \left(\frac{\partial Ax}{\partial z}+\frac{\partial \phi }{\partial y}\right),-c \left(\frac{\partial \phi }{\partial z}+\frac{\partial Ay}{\partial x}\right) )$$

This has zero gauge, the gradient of the scalar and one part of a curl. I don't know if I have ever seen a post with this many partial differential equations in it, but the way this all falls together cannot be an accident.

Doug

 

[sweetser]

Gauge symmetry

Hello Lut:

The way the g fields cancel which I write the fields using the Hamilton and Even representations of quaternions, does that mean I now have gauge symmetry for GEM? And would you remind me of why gauge symmetry is so important to have?

The morning realization: to be inverted, the Even representation must exclude its own Eigenvectors and values. Not sure of the consequences of that...

Thanks,
Doug

 

[Lawrence B. Crowell]

To Sweetser,

I got a couple of messages from P-forum about this. I have not followed your stuff in some time. But I will illustrate what a gauge theory is.

Say you have a manifold M with a bundle over it which has some vector space or algebraic system of roots, that are eigenvector of the weights. For the manifold of dimension = d and the bundle of dimension = p we can define a section s over the base manifold. The principal bundle group is then a set of transformations between bundle sections

$$s'~=~gs.$$

Now consider the action of a differential operator d on this transformed section

$$ds'~=~d(gs)~=~(dg)s'~+~gds$$

which can be seen in elementary terms with $ds~=~As$ as

$$ds'~=~A's'~=~\Big((dg)g^{-1}~+~gAg^{-1}\Big)s'$$

which clearly gives the transformation of the gauge connection A. This equation is not homogeneous with respect to the transformation. Yet for the gauge fields $F~=~dA~+~A\wedge A$ the fields transform as $F'~=~gFg^{-1}$, which is a homegenous covariant transformation.

Now the field term can be seen as due to the action of the two-form

$$\Omega^2~=~(d~+~A)\wedge(d~+~A)$$

on the bundle section. Now this differential form is antisymmetric in its tensor components for the fields $F_{ab}~=~-F_{ba}$. This can be extended to spacetime curvature as well, where the bundle is a fibre bundle with a hyperbolic group structure and so forth. This antisymmetric structure is grounded in some basic differential geometry and topology, where these systems of forms can define cohomology rings and groups.

Now what you have is a theory which has symmetric field components. As a result you are doing something else. This is not to say that what you are doing is utterly wrong, but honestly it is outside the standard canon of field theory and the underlying system of differential geometry it is formulated around. As such what you are proposing is something other than gauge theory, or is an extension of gauge theory into some other arena.

Cheers,

Lawrence B. Crowell

 

[Mentz114]

Hi Doug,

after a great struggle I realized what you are attempting with the quaternions. You've found a representation that allows you to separate the fields ( symmetric and anti-symmetric curls ) in a way that you find correct, satisfactory. Right now I can't comment on the meaning or validity of this.

I understand that the differential operators correspond to ordinary partial differentiation.

Getting to gauge invariance: the parts of your Lagrangian density that contain the potential are -

$$-A^{\mu}J_{\mu} + \frac{1}{4c}\partial^{\mu}A_{\nu}\partial^{\nu}A_{\mu}$$

One can split the second term into symmetric and antisymmetric parts without changing the energy.

Provided those operators are not covariant derivatives, this part is globally gauge invariant, because adding a constant to A will not affect the energy. Local gauge invariance which applies when we add a value to A which depends on x, is assured by specifying a gauge condition, which ensures that the Lagrangian doesn't change when the potential is altered.
My complaint has always been that when the derivatives are covariant, products of metric terms with the potential itself appear in the symmetric terms. Now if you add a constant to A, the energy changes, and gauge invariance is lost. Unless more constraints are added to make those terms disappear.

Everything I know about this subject comes from the EM field, and I'm not even sure this can be applied to the symmetric fields. For instance, not requiring a fixed value for A ensures energy conservation, as does the gauge condition.

But as Lawrence B. Crowell tells us, you are going outside standard differental geom. definitions ( at least, I think that is part of what he's saying).

As such what you are proposing is something other than gauge theory, or is an extension of gauge theory into some other arena.

This is not bad news. If you keep track of the energy you won't go wrong.

But I don't think your calculations above affect my ( simplistic ?) view about gauge invariance, because there is only a problem when the derivatives are covariant and introduce terms in A itself.

I've got to go now, but I'll be revisiting soon.

 

[Lawrence B. Crowell]

My comments are meant to show that if you have a theory of fields with

$$F_{ab}~=~\{Q_a,~{\bar Q}_b\}$$

then this symmetric term is an anticommutator of grassmannian fields. In the case I write above this defines gauge connections for N > 2 supersymmetry, for $A_i~=~{\sigma_i}^{ab}F_{ab}$. Yet these fields are physically Fermionic. Gauge fields are vector (or chiral) and don't obey this sort of rule.

To have symmetric gauge fields F_{ab} = F_{ba}, or components which obey this, runs to my mind in a host of mathematical questions. This appears to be something other than supermanifold theory. There a gauge field will be contained in a super-multiplet with

$$\Phi~=~A~+~{\bar\xi}\psi~+~\xi{\bar\psi}~+~F$$

for $\psi$ the fermionic super pair (gaugino) of the gauge potential, $\xi$ the Grassmannian spinor super fields. The gaugino will obey anti-commutator rules. Yet the gauge fields are still bosonic.

This GEM theory appears to be something else, and there are to my mind a host of open questions. The biggest question is whether this even defines an appropriate geometry. A crucial fact is "boundary of a boundary is zero," which is from d^2 = 0 why gauge fields are anti-symmetric. In the case of Grassmannians which obey anti-commutators we have $Q^2~=~0$, which is equivalent to the Pauli exclusion principle. the d^2 = 0 principle is also why the Riemannian curvature tensor has anti-symmetric indices. Once you twist this around with symmetric terms then the whole geometric meaning of things is lost to me --- outside of a super-manifold theory.

So without a blizzard of tensor analysis and differential terms and the like it seems important to give some indication of what this means geometrically. Without some sense of what this means geometrically, such as the symmetries of a manifold, and how this connects up to conservation laws (eg d^2 = 0 is what gives Bianchi identities etc) then I have a very difficult time knowing how to evaluate this.

Lawrence B. Crowell

 

[Mentz114]

Lawrence,
thanks for taking the time to post all this. I don't have the math to understand it all, unhappily. I think I can see how energy conservation is linked to the geometry.

The GEM lagrangian has no explicit time dependence, which ought to be enough to ensure energy conservation. I can't remember now how gauge invariance came up.

There's a lot for Doug to think about still.

M

 

[Lawrence B. Crowell]

Gauge theory is a bed rock of the physics of fields and forces. In fact the subject is mathematically rich. If you are familiar with Uhlenbeck, Atiyah, Donaldson, Freed, etc, the structure of gauge theories, in particular their moduli spaces, have proven surprising results on the categorization of four dimensional manifolds. The work I have been doing for the last two years involves these results.

My sense is that GEM is geometrically mysterious. All of the stuff we do with physics is ultimately based on on fundamentals of differential geometry. For instance in ordinary three dimensions a function F under the action of a differential operator d is

$$dF~=~\frac{\partial F}{\partial x^i} dx^i,$$

which is the gradient. For a one form $\omega~=~\omega_idx^i$ the action of d is

$$d\omega~=~\frac{\partial\omega_i}{\partial x_i}dx^j\wedge dx^i,$$

and since $dx^i\wedge dx^j~=~-dx^j\wedge dx^i$ clearly i can't equal j. This in vector language is the curl. You can actually prove all the div, grad & curl stuff this way if you also throw in something called Poincare duality. All the anti-symmetric tensors and the rest in gauge theory are grounded on this, and this has incredible generalizations for manifolds and bundles.

GEM appears to throw all of this to the wind, which means there has to be some sort of other foundation to this. As yet I have not seen it.

Lawrence B. Crowell

 

[sweetser]

The field g and gauge transformations

Hello Lawrence:

Thanks for all your technical responses. I wish to make clear that there is something brand new in the GEM proposal as of posts 424 and 425. It is fair to say I am working on the extension of gauge theory into quaternion representations, but one that does tie in directly to some of what we know of gauge theory.

There are also some clear breaks. Let me point one out. The Bianchi identities will never be relevant to this work. The Bianchi identities come out of the Riemann curvature tensor. While I do work with the connection if and only if it appears as part of a covariant derivative, I never work with the connection outside of a covariant derivative as happens with the Riemann curvature tensor. This makes communicating with someone with your level of training darn near impossible because you find some reason to reintroduce the curvature tensor or its stand ins.

[digression]
Why should the Riemann curvature tensor create problems? After all, it is part of a huge intellectual effort, done by some of the smartest math and physics people ever (Lawrence listed a few of the many authors working on this subject). It is both courageous and daft to claim Riemann curvature tensor is irrelevant to how Nature works. Here is the logic most readers of this thread should be able to follow.

The 4-potential $A^{\nu}$ transforms like a tensor.
The 4-derivative $\partial^{\mu}$ transforms like a tensor.
The 4-derivative of a 4-potential $\partial^{\mu} A^{\nu}$ does not transform like a tensor.
To correct this problem, we need to add in the connection. It turns out there are lots of options with the choice of the connection. I choose to work with the same one used in GR, which is to say the connection is torsion-free and metric compatible, so the connection is the Christoffel symbol of the second kind. The Christoffel symbol has three first derivatives of the metric. Thus we get the definition of the covariant derivative:

$$\nabla^{\mu} A^{\nu} = \partial^{\mu} A^{\nu} - \Gamma_{\sigma}^{\mu \nu} A^{\sigma}$$

Einstein learned this stuff (he didn't invent it) from is math tutors. He focused on the Gamma which contains the first derivatives of the metric. He then asked his math buddies for the math object that transforms like a tensor but has second order derivatives of the metric. He was thinking about good old $F=Ma$, how Nature accomplishes things with second order derivatives. That was his inspired guess, and he needed some direction to implement it. The answer he was given was the Riemann curvature tensor. That was a great answer that has led to lots of great results. The way math wonks sell it, they say it is the only answer without getting into really obscure math.

Yet Einstein missed the obvious way to get second order differential equations of the metric. Take a closer look at what Newton did. Newton was saying $m \frac{dR}{dt}$ is not interesting, but acting again on this with another time derivative operator, $\frac{d}{dt}$, then you get $F=m \frac{d^2 R}{dt^2}$. Follow that program exactly. Einstein is saying $\nabla^{\mu} A^{\nu}$ with one derivative of the metric is not interesting, so take another covariant derivative: $\nabla_{\nu} \nabla^{\mu} A^{\nu}$. There will be the divergence of the Christoffel symbol. Drop the Rosen metric into the Christoffel, take its divergence, and see something even Poisson would recognize. Why I cannot lead a well-trained mathematical horse to do that calculation is beyond me, but that's the way it is.

GEM is about potentials in bed with geometry. The lights are on, the camera is rolling. Someday I'll sell a lot of film, but until then, I will post to the Independent Research forum.
[/digression]

I have an even simpler view of gauge theory, the one that applies directly to EM. I look at it in terms of this specific Lorenz gauge transformation:

$$(\phi, A) \rightarrow (\phi', A')=(\phi, A)+(\frac{\partial f}{\partial t}, - \frac{\partial f}{\partial x}, - \frac{\partial f}{\partial y}, - \frac{\partial f}{\partial z})$$

The function f needs to have a few derivatives. Drop this into the anti-symmetric field strength tensor, form the field equations and all the dependence on f drops out. Drop A' into any symmetric tensor, and any field equations that come out will depend on f. This is a point you made long ago, and looking back, I don't think I addressed it. Why not? Your point is technically correct. That's the way tensors are.

In post 425, I used no tensors. Instead I used two representations of quaternions. The first is the well known Hamilton representation of the quaternion division algebra. It is an asymmetric tensor, with the symmetric part being made of the scalar, and the antisymmetric part being the 3-vector. The other one I am calling the Even representation. This is symmetric and is a division algebra so long as one does not use the Eigenvalues and Eigenvectors of the 4x4 matrix representation. We can point to gulfs in our communication because I am using these tools, and you are skilled with differential geometry.

When I added all 5 of the fields in my proposal together, E, B, e, b, and g, this was the result:

$$\frac{1}{2}\left(-A \nabla + \nabla^* A2\right)$$

$$= -g + E + B + g + e + b$$

$$= E + B + e + b$$

where
$$g = \frac{\partial \phi }{\partial t}-\frac{\partial Ax}{\partial x}-\frac{\partial Ay}{\partial y}-\frac{\partial Az}{\partial z}$$
$$E = (-\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},-\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}, -\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z})$$
$$B = (c \left(\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right), c \left(\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right), c \left(\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) )$$
$$e = (\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}, \frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z})$$
$$b = (-c \left(\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right), -c \left(\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right), -c \left(\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) )$$

Notice how the field g, no matter what it is, drops out. That means you are completely free to work with whatever g field you choose.
Consider a function h such that $\nabla h = g$. Then:

$$A \rightarrow A' = A + h$$

Hit this with a derivative, out comes the field g, and the field g drops. That is a gauge transformation.

GEM appears to throw all of this to the wind, which means there has to be some sort of other foundation to this. As yet I have not seen it.

I noticed no references to the Hamilton or even representation of quaternions in your replies. That is where the foundation is. We do not have to feel bad if we fail to communicate. You have demonstrated an impressive knowledge of both the history and current efforts to do productive research using the power tools of differential geometry. I respect that. New math is for the next generation. I have no expectation that I will ever be comfortable with it all, but that is the way research goes.

Here is my specific plan of action. I have shown how to write the 5 GEM fields in a compact way using the Hamilton and Even representations of quaternions. I have shown how to write the coupling term using the Hamilton and Even representation in a way that shows the coupling represents both spin 1 and spin 2 fields. I have to form the action from these fields. Then I have to generate the field equations from the action. Finally Lut, I will calculate the stress energy tensor to address your energy conservation question which is a good one. All this work must be checked with Mathematica. There is much to do, like prepare for three talks, make new quaternion animations, do outreach, work full time on something else, walk the dogs, and play nice with the wife.

Doug

 

[sweetser]

Correction to post 424, Generalized current coupling

Hello:

I found a clear technical error with post 424 titled "Generalizing Current Coupling Spin 1 & Spin 2". The two current coupling terms look like so:

$$-J J' -J2 J2'^*$$

These both have the same sign. When dropped into the Lagrangian, they will have like charges repel. That is good for EM, but failure for gravity.

Now that the issue is defined, let's create a solution. Post 424 was trying to generalize a specific case for motion along the z axis. There I needed to use the first or second conjugates, which I had defined somewhere in this vast discussion as:

$$(i q i)^* = (-t, x, -y, -z) === q^{*1}$$
$$(j q j)^* = (-t, -x, y, -z) === q^{*2}$$

We can continue this theme and define the third conjugate like so:

$$(k q k)^* = (-t, -x, -y, z) === q^{*3}$$

As an indication of generalization, all three conjugates are needed:

$$+J2^{*1}^{*2} J2'^{*3}=(\rho, -Jx, -Jy, Jz)(-\rho', -Jx', -Jy', Jz')=$$
$$=(- \rho \rho' + Jx Jx' + Jy Jy' + Jz Jz',$$
$$-\rho Jx' + Jx \rho' - Jz Jy' - Jy Jz',$$
$$-\rho Jy' + Jy \rho' - Jz Jx' - Jx Jz',$$
$$-\rho Jz' + Jz \rho' - Jy Jx' - Jx Jy')$$

This current coupling term has three defining characteristics:
1. The scalar is invariant under a Lorentz transformation
2. The phase has both spin 1 and spin 2 symmetry, necessary for a field theory where like charges repel for EM and attract for gravity respectively
3. Is a positive coupling between the two currents, so when put in the action will apply to like charges that attract.

The analysis of the -J J' coupling is unaltered. The generalzized coupling term using the Hamilton and Even representations is:

$$-J J' + J2^{*1}^{*2} J2'^{*3}$$

That looks better. Dodged another bullet tonight.

Doug

 

[sweetser]

Overview of quaternion-based Lagrangians

Hello:

I said I had to "form the action from these fields" using quaternions only, no tensors. This was not an easy exercise. Once again, the simplest term gave me problems, specifically the g field. I spent much time thinking about what I was trying to do with these fields before I could proceed effectively. In this post I will give an overview of the math to follow in later posts.

The five fields defined in post 432 - the standard electromagnetic E and B fields, their symmetric counterparts e and b, and a diagonal g - together transform like a second rank tensor. That tensor gets contracted to make a Lorentz invariant scalar that goes into the action.

I was not able to build an exact analogy with a rank 2 tensor contraction. That does not mean there isn't one, just that I couldn't find it. There are no convenient indexes to work with when using quaternions, so plucking out the right terms did not work.

We need to create a Lorentz invariant scalar. I recalled reading in Jackson's "Classical Electrodynamics" that $E^2 - B^2$ is Lorentz invariant. To use a little high school algebra, we know:

$$E^2 - B^2 = (E + B)(E - B)$$

There are nice, compact quaternion expressions for -g + E + B and -g + E - B. When will g be zero? If the system has only massless particles. Using that assumption, I am able to calculate $E^2 - B^2$ using quaternions.

The next step is to apply the Euler-Lagrange equation to calculate the field equations from the Lagrangian. I suspect the majority of readers have not done this sort of calculation. It looks scary, but the details are quite simple. I am enough decades away from my calculus classes that I can recall only a few derivatives, simple ones like $\frac{d x y}{d x} = y$ and $\frac{1}{2} \frac{d x^2}{d x} = x$. These are the only two derivatives that are required to get to the Maxwell and GEM field equations. What is needed is to have the courage to write everything down, even though it looks complicated. Instead of the simple x, one uses the derivative of a function, $x \rightarrow \frac{\partial \phi}{\partial t}$, along with 15 other partial derivatives. When I do this in subsequent posts, I will point out the simplicity of what is going on in taking the Lagrangian and getting the field equations.

Does anyone know if physicists make much of the invariant $E^2 - B^2$? As far as I can remember, it was mentioned once in Jackson, and he did not say much about it at all. Yet in my quaternion calculation, this invariant leads directly to the Maxwell equations. If true, the difference of the squares of these two fields should be on the center stage of EM theory. That has been a surprise benefit of my recent struggles.

Doug

 

[Mentz114]

Hi Doug:

Do you mean apart from the term $$F^{\mu\nu}F_{\mu\nu} = -2(E^2 - B^2)$$ appearing in the lagrangian of the EM field ? It seems to be center stage.

 

[sweetser]

Good, it has been done before

Hello Lut:

Good to hear you are familiar with that result. Perhaps it is somewhere in Jackson, but I had not seen it before in my physics browsing. I gain confidence when I see other people have done something I tried to do before. It indicates I am barking up the right tree.

I will go through all the details because that is my idea of a fun technical time. It also sets the stage for variations needed to see where GEM is going.
Doug

 

[Lawrence B. Crowell]

Hello Lawrence:

I noticed no references to the Hamilton or even representation of quaternions in your replies. That is where the foundation is. We do not have to feel bad if we fail to communicate. You have demonstrated an impressive knowledge of both the history and current efforts to do productive research using the power tools of differential geometry. I respect that. New math is for the next generation. I have no expectation that I will ever be comfortable with it all, but that is the way research goes.

Doug

I didn't bring up quaternions because they didn't seem relevant.

In what you wrote here it looks a bit like standard Kaluza Klein theory. I am not sure if this differs significantly from rather standard fair. At lest with this the approach appears within the standard construction of gauge theories.

Lawrence B. Crowell

 

[sweetser]

The Maxwell equations, and more

Hello:

In this post, I will show in detail how to generate the Maxwell equations using quaternions exclusively. This is achieved by getting to exactly the same terms that appear in the standard tensor approach for the electromagnetic field strength contraction. After the contraction, every step is the same. I will do them anyway, since it may be instructive to the viewers of this thread.

Simple quaternion expressions generate combinations of three fields, the standard definition of E and B, along with one I call g:

$$g = (\frac{\partial \phi}{\partial t}- \frac{\partial Ax}{\partial x}- \frac{\partial Ay}{\partial y}- \frac{\partial Az}{\partial z})$$

Notice how this one has all the parts of a typical gauge field in standard EM.

There are two simple ways to generate g, B, and E using quaternions:

$$1. \nabla A = (\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})(\phi, Ax, Ay, Az)$$

$$(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},\frac{\partial Ax}{\partial t}+c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},\frac{\partial Ay}{\partial t}+c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},\frac{\partial Az}{\partial t}+c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$

$$= g - E + B$$

$$2. - A \nabla = -(\phi, Ax, Ay, Az)(\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})$$

$$(-\frac{\partial \phi }{\partial t}+c \frac{\partial Ax}{\partial x}+c \frac{\partial Ay}{\partial y}+c \frac{\partial Az}{\partial z},-\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},-\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},-\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$

$$= -g + E + B$$

By changing the order of the derivative with the potential, we only flip the sign of the curl. Let's get rid of that pesky g by subtracting away the conjugate:

$$1. \nabla A - (\nabla A)^* = (0, -E + B)$$

$$2. - A \nabla - (A \nabla)^* = (0, E + B)$$

Take the product of these two:

$$(\nabla A - (\nabla A)^*)(- A \nabla - (A \nabla)^*) = (0, -E + B)(0, E + B) = (E^2 - B^2, 2 ExB)$$

It is the scalar that will be used to get the Maxwell field equations. It is the same one that appears in standard EM field theory as Lut pointed out in post 435 if I can get my signs right. The 3-vector is the Poynting vector which plays a role in energy conservation laws. Let's work only with the scalar, and toss in a factor of minus a half, and include a charge coupling term. Write out all of the components:

$$-\rho \phi + Jx Ax + Jy Ay + Jz Az$$

$$+\frac{1}{2} c^2 \left(\frac{\partial Ax}{\partial z}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial z}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial z}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial Ax}{\partial y}\right)^2$$
$$-c^2 \frac{\partial Ay}{\partial z} \frac{\partial Az}{\partial y}+\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial y}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial y}\right)^2-c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x}$$
$$+\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial x}\right)^2-c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}+\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial x}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial x}\right)^2-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t}$$
$$-\frac{1}{2} \left(\frac{\partial Ax}{\partial t}\right)^2-c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t}-\frac{1}{2} \left(\frac{\partial Ay}{\partial t}\right)^2-c \frac{\partial \phi }{\partial z} \frac{\partial \text{Az}}{\partial t}$$
$$-\frac{1}{2} \left(\frac{\partial \text{Az}}{\partial t}\right)^2$$

From here on out, this is the standard way to derive the Maxwell equations. This is the part that looks scary, but its bark is worse that its bite. The goal is to take the derivative of the above Lagrangian with respect to the four potentials, $\phi, Ax, Jy, Az$, and all 16 derivatives of the four potentials. The first step, taking the derivative with respect to the 4-potentials, gives us back the current, $-\rho, Jx, Jy$, and $Jz$.

The derivatives of the Lagrangian with respect to the changing potential is more complicated, but once the rule is learned, it is simple to apply, over and over again. Here are the first 4 of 16:

$$\frac{\partial }{\partial x_{\mu }}\left(\frac{\partial \mathcal{L}}{\frac{\partial \phi }{\partial x_{\mu }}}\right)$$

The mu brings in the derivative of phi with respect to t, x, y, z. The derivative out in front will be generating second order derivatives. Let's do this for one of these, say, d phi/dx:

$$\frac{\partial }{\partial x}\left(\frac{\partial \mathcal{L}}{\frac{\partial \phi }{\partial x}}\right)=-c \frac{\partial ^2\phi }{\partial x^2}-\frac{\partial ^2 Ax}{\partial t\partial x} = \frac{\partial E}{\partial x}$$

Repeat this for y and z, and you will have Gauss' law, the divergence of E equals rho, $\rho = \nabla . E$. Now we need to take the derivative of the Lagrangian with respect to Ax, running through t, x, y, and z.

$$c^2 \frac{\partial ^2 Ax}{\partial z^2}+c^2 \frac{\partial ^2 Ax}{\partial y^2}-c^2 \frac{\partial ^2 Az}{\partial x\partial z}-c^2 \frac{\partial ^2 Ay}{\partial x\partial y}-c \frac{\partial ^2\phi }{\partial t\partial x}-\frac{\partial ^2 Ax}{\partial t^2}$$

The last two terms are a time derivative of Ex. If you were to calculate the curl of B, then you would recognize the first four terms. The way I spot it, is the two "pure" second order derivatives, which don't have an x, are positive, meaning this is minus the curl of Bx. We also have a +J, so tossing all the terms on the other side generates:

$$J = \nabla X B - \frac{\partial E}{\partial t}$$

This is Ampere's law.

What has been done

In this post I showed how to get to the standard EM Lagrangian using the first term of a particular quaternion product. One interesting subtle issue is that the field g was explicitly removed, and so the proposal is gauge invariant - no matter the choice for the terms in g, it does not change a thing because all terms in the g field were subtracted away. In the standard approach, we observe the field equations are gauge invariant. With quaternions, we see the step where the field disappears. There is no difference in the end result, but it is fun to think about.

What will be done

In the coming posts, I will repeat the exercise done in this post for the symmetric fields, e and b. The resulting equations are gauge invariant in exactly the same sense as the Maxwell equations were gauge invariant in this exercise, because g will be subtracted away.

In the third installment, I will not subtract g from either E+B or e+b. The field g goes in, but I hope to show that the g field does not come out. The field equations are the same no matter what g one chooses. It is quite remarkable that g politely disappears from the stage.

Doug

 

[Lawrence B. Crowell]

Hello:

Simple quaternion expressions generate combinations of three fields, the standard definition of E and B, along with one I call g:

$$g = (\frac{\partial \phi}{\partial t},- \frac{\partial Ax}{\partial x},- \frac{\partial Ay}{\partial y},- \frac{\partial Az}{\partial z})$$

Notice how this one has all the parts of a typical gauge field in standard EM.

There are two simple ways to generate g, B, and E using quaternions:

$$1. \nabla A = (\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})(\phi, Ax, Ay, Az)$$

$$(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},\frac{\partial Ax}{\partial t}+c \frac{\partial \phi }{\partial x}+c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},\frac{\partial Ay}{\partial t}+c \frac{\partial \phi }{\partial y}+c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},\frac{\partial Az}{\partial t}+c \frac{\partial \phi }{\partial z}+c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$

$$= g - E + B$$

Doug

I am a bit unsure about the nature of g. You have a "nabla" operator on the vector potential, or four potential, which gives a vector quantity. Maybe this nabla is supposed to be a "d" operator?

If you want to work in quaternions things should be in a Clifford basis on Cl(3,1) with the generators

$$\gamma_1~=~\sigma_2\otimes\sigma_1,~\gamma_2~=~\sigma_2\otimes\sigma_2,~\gamma_3~=~\sigma_2\otimes\sigma_3,~\gamma_4~=~i\sigma_1\otimes{\bf 1}$$

The electromagnetic potential are then on the Clifford frame if

$${\underline A}~=~(e_a^{\mu})\gamma_\mu A^a$$

for $e_a^{\mu}$ a tetrad or vierbein. The Maxwell equations will then arise from $d\wedge{\underline A}~=~{\underline F}$ and the gauge condition will come from $\nabla_a A^a~=~C$, for C a constant, set to zero in the Lorentz gauge. This can come from setting $(e_a^{\mu})\gamma_\mu~=~D_a$ which is orthogonal to the vector potential A_a on the bundle section with $D_aA^a~=~0$ --- again for the Lorentz gauge.

Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

I wrote out g incorrectly. I have corrected post #438. Nabla is a quaternion, the 4-potential is a quaternion, the Nabla acting on the 4-potential makes a quaternion, and the three fields of g, E, and B together form a quaternion. The g is the first term of the quaternion 4-derivative of the 4-potential.

Doug

 

[Mentz114]

Doug,

Very nice. I look forward to the next installment. Could you number your equations, please ?

Also using ExB for $$E\times B$$ caused me some confusion until I repeated the calculation.

Lut

 

[sweetser]

The Symmetric Field Maxwell Equations for Gravity

Hello:

In this post, I will apply the techniques used to generate the Maxwell equations with quaternions to the symmetric analogs of the E and B fields.

The three symmetric fields involved in this analysis are:

$$g = (\frac{\partial \phi}{\partial t}- \frac{\partial Ax}{\partial x}- \frac{\partial Ay}{\partial y}- \frac{\partial Az}{\partial z}) \quad eq 1$$

$$e = (\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x},\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}, \frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}) \quad eq 2$$

$$b = (c \left(-\frac{\partial Ay}{\partial z}-\frac{\partial Az}{\partial y}\right), c \left(-\frac{\partial Az}{\partial x}-\frac{\partial Ax}{\partial z}\right), c \left(-\frac{\partial Ax}{\partial y}-\frac{\partial Ay}{\partial x}\right) ) \quad eq 3$$

Although I first spotted these fields by peering into a symmetric rank 2 tensor, for this exercise I will use quaternions. There are two reasons to do so. First, we see exactly why the field equations are invariant under a gauge transformation - such a field will be subtracted away. The second point is more subtle. To do quantum field theory, one needs to take the field equations and invert them to make a propagator. If we consistently use a division algebra, then we necessarily will be able to do that step. With gauge theories such as Maxwell and general relativity, an arbitrary gauge must be chosen before the equation can be inverted. I am trying to get a theory that one is free to choose the gauge, yet remains invertible to get the propagator.

The Hamilton representation of quaternion multiplication will not suffice for generating the symmetric b field, where all the terms that go into a curl have a minus sign. For this reason, I have introduced the Even representation of quaternion multiplication which excludes the Eigen vectors and Eigen values in order to be a division algebra. This representation will not be a Clifford algebra, since $e_0^2 = e_1^2 = e_2^2 = e_3^2 = +1$. Two simple expressions can generate g, e, and b:

$$\nabla A2^* = (\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})(\phi, Ax, Ay, Az)$$

$$=(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},$$
$$-\frac{\partial Ax}{\partial t}+c \frac{\partial \phi }{\partial x}-c\frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},$$
$$-\frac{\partial Ay}{\partial t}+c \frac{\partial \phi }{\partial y}-c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},$$
$$-\frac{\partial Az}{\partial t}+c \frac{\partial \phi }{\partial z}-c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$

$$= g - e + b \quad eq 4$$

$$A2 \nabla^* = (\phi, Ax, Ay, Az)(\frac{\partial}{\partial t},\frac{\partial}{\partial x},\frac{\partial}{\partial y},\frac{\partial}{\partial z})$$

$$=(\frac{\partial \phi }{\partial t}-c \frac{\partial Ax}{\partial x}-c \frac{\partial Ay}{\partial y}-c \frac{\partial Az}{\partial z},$$
$$\frac{\partial Ax}{\partial t}-c \frac{\partial \phi }{\partial x}-c \frac{\partial Ay}{\partial z}-c \frac{\partial Az}{\partial y},$$
$$\frac{\partial Ay}{\partial t}-c \frac{\partial \phi }{\partial y}-c \frac{\partial Az}{\partial x}-c \frac{\partial Ax}{\partial z},$$
$$\frac{\partial Az}{\partial t}-c \frac{\partial \phi }{\partial z}-c \frac{\partial Ax}{\partial y}-c \frac{\partial Ay}{\partial x})$$

$$= g + e + b \quad eq 5$$

To demonstrate this is a subtle proposal, I cut and paste this calculation from my derivation of the Maxwell equations, and made the appropriate sign and letter changes. There are not many, and hopefully I did it right. changing who gets conjugated flips the sign of e, but not b.

Let's get rid of that pesky g by subtracting away the conjugate:

$$\nabla A2^* - (\nabla A2^*)^* = (0, -e + b) \quad eq 6$$

$$- A2 \nabla^* - (A2 \nabla^*)^* = (0, e + b) \quad eq 7$$

Take the product of these two:

$$(\nabla A2^* - (\nabla A2^*)^*)(- A2 \nabla^* - (A2 \nabla^*)^*) = (0, -e + b)(0, e + b) = (-e^2 + b^2, -e.X2.e + b.X2.b) \quad eq 8$$

It is the scalar that will be used to get the symmetric field Maxwell equations. No Poynting vector this time, I don't know what that means. Write out all the components, the same ones as before, but with a different collection of signs:

$$-\rho \phi + Jx Ax + Jy Ay + Jz Az$$

$$+\frac{1}{2} c^2 \left(-\frac{\partial Ax}{\partial z}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial z}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial z}\right)^2-\frac{1}{2} c^2 \left(\frac{\partial Ax}{\partial y}\right)^2$$
$$-c^2 \frac{\partial Ay}{\partial z} \frac{\partial Az}{\partial y}-\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial y}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial y}\right)^2-c^2 \frac{\partial Ax}{\partial y} \frac{\partial Ay}{\partial x}$$
$$-\frac{1}{2} c^2 \left(\frac{\partial Ay}{\partial x}\right)^2-c^2 \frac{\partial Ax}{\partial z} \frac{\partial Az}{\partial x}-\frac{1}{2} c^2 \left(\frac{\partial Az}{\partial x}\right)^2+\frac{1}{2} c^2 \left(\frac{\partial \phi }{\partial x}\right)^2-c \frac{\partial \phi }{\partial x} \frac{\partial Ax}{\partial t}$$
$$+\frac{1}{2} \left(\frac{\partial Ax}{\partial t}\right)^2+c \frac{\partial \phi }{\partial y} \frac{\partial Ay}{\partial t}+\frac{1}{2} \left(\frac{\partial Ay}{\partial t}\right)^2-c \frac{\partial \phi }{\partial z} \frac{\partial \text{Az}}{\partial t}$$
$$+\frac{1}{2} \left(\frac{\partial \text{Az}}{\partial t}\right)^2 \quad eq 9$$

Fear not, the derivative of this Lagrangian with respect to the potential again generates the current density. Now take the derivative with respect to the for derivatives of phi:

$$-\rho +\left(c^2 \frac{\partial ^2\phi }{\partial x^2}+c^2 \frac{\partial ^2\phi }{\partial y^2}+c^2 \frac{\partial ^2\phi }{\partial z^2}-c\frac{\partial ^2 Ax}{\partial t\partial x}-c\frac{\partial ^2 Ay}{\partial t\partial y}-c\frac{\partial ^2 Az}{\partial t\partial z}\right)==0 \quad eq 10$$

Plug in the definition of e and rearrange:

$$-\nabla . e = \rho \quad eq 11$$

This looks similar to Gauss' law, but there is an important difference. In the static case, the time derivatives of A make no contribution, and we have:

$$\nabla^2 \phi = \rho \quad eq 12$$

For Gauss' law of EM, under the same static conditions:

$$-\nabla^2 \phi = \rho \quad eq 13$$

The reason like charges repel in EM has to do with the above minus sign, while for gravity to attract, the Laplacian of phi must have the same sign as the current density.

Take the derivative with respect to the 4 derivatives of Ax:

$$Jx-c^2 \frac{\partial ^2 Ax}{\partial z^2}-c^2 \frac{\partial ^2 Ax}{\partial y^2}-c^2 \frac{\partial ^2 Az}{\partial x\partial z}-c^2 \frac{\partial ^2 Ay}{\partial x\partial y}-c \frac{\partial ^2\phi }{\partial t\partial x}+\frac{\partial ^2\text{Ax}}{\partial t^2}==0 \quad eq 14$$

The time derivative of e is the last two terms. The symmetric curl is negative, but two symmetric curls is positive, so this is minus the symmetric curl of b. Plug in and rearrange:

$$\nabla . X2 . b - \frac{\partial e}{\partial t} = J \quad eq 15$$

Looks similar to Ampere's law. The symmetric Gauss and Ampere's law should not be shocking, since the symmetric fields e and b are built from the same terms as E and B. There are important differences, such as the symmetric Gauss' law has like charges that attract as happens for gravity, whereas the EM Gauss' law has like charges repel. The two Ampere's laws have two different sorts of curls (since curl is one of the most confusing concepts I have come across, I will not venture about the meaning of two different curls).

Again we can choose whatever we want for $g = (\frac{\partial \phi}{\partial t}- \frac{\partial Ax}{\partial x}- \frac{\partial Ay}{\partial y}- \frac{\partial Az}{\partial z})$ will not change these equations at all. Some of my readers will say that is a good thing :-)

Doug

 

[Mentz114]

Hi Doug,
very interesting. I'm impressed with your 'theory engineering' approach.

I've gone back looking for the definition of the X2 operator ( even multiplication ?) and also, is conjugation the same in the even algebra ? I need the operators because I'm repeating it with tensors and pseudo-tensors.

I presume the aim is to get the scalar part of this term

$$(\nabla A2^* - (\nabla A2^*)^*)(- A2 \nabla^* - (A2 \nabla^*)^*) = (0, -e + b)(0, e + b) = (-e^2 + b^2, -e.X2.e + b.X2.b) \quad (8)$$

into the lagrangian.

Lut

 

[sweetser]

Even multiplication tools

Hello Lut:

The X2 is the symmetric analog to the cross product.

$$e.X2.e = (2 ey ez, 2 ex ez, 2 ex ey) \quad eq 1$$

Yes, the conjugate for Even representation of quaternions does exactly the same thing as for the Hamilton quaternions. When it came to programming this in Mathematica, I had be taking the conjugate of the real 4x4 matrix representation. That no longer works. Instead, I had to implement it on the 4-vector:

$$(t, x, y, z)^* \rightarrow (t, -x, -y, -z) \quad eq 2$$

This definition of a conjugate works for either 4x4 real matrix representation of a quaternion, which is a good sign.

When I did the previous post, I kept getting eq 8 wrong because I am too familiar with the Hamilton representation of quaternions. If you are doing this all with tensors, see if you can make use of symmetric tensors. The algebra should line up (loads of details left to you).

Doug

 

[sweetser]

The Maxwell equations using quaternion operators

Hello:

I find the hand drawn derivations more satisfying than what LaTeX produces because it has that human element. Doing things by hand is kind of like Sudoku: the rules are simple, there is a right way, and you get in a grove, so repetition acts as confirmation. Anyone trying to do this on their own can use these as crib sheets.

These three URLs are relevant to this thread. I have shrunk each page down to 65k, so it shouldn't be so bad to load. I recommend hitting the magnifier, using the hand cursor to move up and down. Right now there are three low resolution, and three high resolution derivations of field equations.

1. http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations/photo#5171869015477455458

2. http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations/photo#5171869019772422770

3. http://picasaweb.google.com/dougsweetser/MaxwellFieldEquations/photo#5171869024067390082

I have not discussed case 3 yet, but hopefully will get to it this weekend.

Doug

 

[sweetser]

Current coupling detail omitted in post 442

Hello:

For the symmetric field Maxwell equations, in eq 9 of post 442, I wrote out the current coupling terms in its component parts:

$$-\rho \phi +Jx Ax + Jy Ay + Jz Az$$

What I omitted was how to form this contraction. The details here matter, because if one just does the same as for the Maxwell equations, [itex]-J A[/tex], Feynman and others have show that the phase - stuff other than the scalar - has spin 1 symmetry where like charges repell. I should have included this:

$$+\frac{1}{2}(J^{*1} A^{*2 *3} + (J^{*1} A^{*2 *3})^*) = -\rho \phi +Jx Ax + Jy Ay + Jz Az$$

The scalar is the same, but the phase has spin 2 symmetry which is needed for an equation if like charges attract.

Doug

 

[Lawrence B. Crowell]

accelerated charge I

I recently worked out some stuff involving the emission of radiation by accelerated charges, the Unruh effect and the emission of radiation by a charge falling into a gravity field. I figured I'd send these here, maybe they will help with some thing. This is part I of III


The Lamour formula gives a relationship between the acceleration of a charge and the radiation power emitted by the charge

$$P~=~-\frac{2}{3}\frac{q^2}{c^3}a_\mu a^\mu.$$

This has lead to an ideological stance that accelerated charges emit radiation. Yet it is clear that a charge sitting on the surface of a gravitating body will not emit radiation. The work exerted on such a charge $W~=~\int {\vec F}\cdot d{\vec r}$ is zero as there is no displacement of charge. Similarly, we might suppose that with the Einstein equivalence principle that a freely falling charge should also not radiate. If we were to consider a small falling frame comoving with the charge then at least within that frame any radiation response should be zero. It would then indicate the appearance of electromagnetic radiation is coordinate dependent. This might in fact be the case even if in other coordinate systems the infalling charge is found to be emitting radiation.

Maxwell’s electrodynamics indicates that for a four-vector potential $A^\mu$ the associated four-current is

$$\square A^\mu~=~4\pi j^\mu.$$

For a current from a discrete point charge the current is then determined by a delta function

$$j^\mu(x)~=~q\int_{-\infty}^\infty U^\mu(s)\delta(x~-~y(s))ds$$

for $U^\mu~=~dy^\mu/ds$ the four velocity of the charge. The Maxwell result for the four-vector potential demands that

$$A^\mu~=~4\pi\int d^4x^\prime G(x~-~x^\prime)j^\mu(x^\prime)$$

where the propagator obeys the harmonic condition $\square G(x~-~x^\prime)~=~\delta(x~-~x^\prime)$, which under the integration reduces is to an integration of the delta function along the worldline of the particle. A standard form for the Green’s function is the Pauli-Jordan function

$$G(x~-~y)~=~\frac{1}{2\pi}\theta(x^0~-~y^0)\delta\big((x^\mu~-~y^\mu)(x_\mu~-~y_\mu\big)$$

where the step function turn on the potential for $x^0~-~y^0~>~0$ and defines a retarded four-vector potential. Using the identity for Dirac delta functions $\delta(f(x))~=\sum_i\delta(x~-~x_i)/f^\prime(x)$ the four-potential is then

$$A^\mu(x)~=~q\int_{-\infty}^\infty U^\mu\frac{\delta(s~-~\sigma)}{U_\nu(z)(x^\nu~-~y^\nu)}ds,$$

where the particle path is parameterized by $\sigma$ with $U^\nu(\sigma)$ and $y^\nu(\sigma)$. The distance along the null cone is identified as $r~=~U_\nu(x^\nu~-~y^\nu)$, which gives the Lenard-Wiechart potential $A^\mu(x)~=~qU^\mu(x)/r$.

This may now be extended to accelerated frames. For a frame of constant acceleration the coordinates paramterized by $\sigma$ are given by

$$y^0~=~g^{-1}sinh(g\sigma),~y^i~=~g^{-1}cosh(g\sigma),~U_0~=~cosh(g\sigma),~U^i~=~sinh(g\sigma)$$

and the vector $x^\nu~=~(t,~0,~0,~0)$ for a rest frame. The propagators is then

$$G(\sigma)~=~\frac{1}{2\pi}\frac{1}{t~cosh(g\sigma)}$$

The vector potential will then have the components

$$A^0(x)~=~qc/\rho,~A^i~=~(qc/\rho)tanh(gs),$$

where $\rho~=~ct$. What is of particular importance is the evaluation of fields on the frame. We consider the change in the energy of a probe with states $\{E\}$ and we consider the transition of the probe from states E to a set of possible states E’ due to a change in the vacuum states from $|0\rangle$ to $|0^\prime\rangle$ from the inertial to accelerated frames. The probability is then given by

$$\sum_{E^\prime}\langle 0,~E^\prime|0,~E\rangle~=~\sum_{E^\prime}\langle E^\prime|\mu|~E\rangle\int dse^{i(E^\prime~-~E)s}G(s)~=~ \sum_{E^\prime}\langle E^\prime|\mu|~E\rangle g^{-1}\sqrt{\frac{\pi}{2}}sech\Big(\frac{\pi}{2g}(E^\prime~-~E)\Big)$$

since the Fourier transform of the sech is itself. Here the term [itex]\mu[/tex] is a coupling term for the probe which couples with the vector potential of the field measured. This result is given according to the Fourier set of frequencies of photon an accelerated charged particle will interact with. As indicated later the sech term can be seen to be approximately $\simeq~e^{(E^\prime~-~E)/2g}$, which is a Boltzmann distribution of photons.

The condition on the null cone $(x^\mu~-~y^\mu)(x_\mu~-~y_\mu)$, and the gradient $\nabla_\nu$ operator on this condition is still zero with

$$\Big({\delta^\mu}_\nu~-~\frac{dy^\mu}{d\sigma}\nabla_\nu\sigma\Big) (x_\mu~-~y_\mu)~=~0$$

or

$$x_\mu~-~y_\mu~=~r\nabla_\nu\sigma$$

which defines the null momentum vector for a photon as $k_\mu~=~(x_\mu~-~y_\mu)/r$, where for the case of an accelerated charged particle $r~\sim~sech(g\sigma)$. The acceleration $a^\mu~=~dU^\mu/d\sigma$, the spatial position vector $y^\mu~=~\tau a^\mu~-~U^\mu$ and with the null vector $k^\mu$ defines a normal vector $n^\mu~=~k^\mu~-~U^\mu$ so that $1/\tau~=~a_\nu k^\nu$. These conditions are used in the differentiation of the Lenard-Wiechart potential

$$A_{\mu,\nu}(x)~=~\frac{q}{r}k_\nu(a_\mu~-~\tau^{-1}U_\mu)~+~\frac{q}{r^2}n_\nu U_\mu$$

which gives the field tensor components as $F_{\nu\mu}~=~A_{[\mu,\nu]}$. The four potential and the resulting field strength tensor contains $1/r,~1/r^2$ terms which respectively contain the radiation field due to accelerations and velocity changes and the second term which is the field term comoving with a charge at a velocity.

Lawrence B. Crowell

 

[Lawrence B. Crowell]

accelerated charge II

From the four vector potential of the electromagnetic field we have the field tensor

$$F_{\mu\nu}~=~A_{[\mu,\nu]}~=~\frac{q}{4\pi r}k_{[\mu}y_{\nu]}~+~\frac{q}{r^2}n_{[\mu}U_{\nu]}$$

The Lagrangian for the electromagnetic field is ${\cal L}~=~(1/4)F^{\mu\nu}F_{\mu\nu}$ and for the momentum-energy tensor $\partial{\cal L}/\partial g^{mu\nu}~-~g_{\mu\nu}{\cal L}$ and so

$$T_{\mu\nu}~=~-F^{\mu\gamma}{F^\mu}_{\gamma}~-~{1\over 4}g^{\mu\nu}F^{\sigma\gamma}F_{\sigma\gamma}~=~\frac{q^2}{r^2}(y_\sigma y^\gamma)k_\mu k_\nu$$

For the case with an accelerated frame $r~\rightarrow~r cosh(g)$. The four momentum density of the electromagnetic field is $P_\mu~=~\int_V d^3x T_{\mu\nu}n^\nu$, for the volume specified between two spatial surfaces. The continuity condition on the electromagnetic field $\partial_\mu T^{\mu\nu}$ means that the derivative of the energy along the proper time is

[tex]
\frac{dP_\mu}{ds}~=~\int_V d^3x T_{\mu\nu}\frac{n^\nu}{ds}
[/itex]

and for $dn^\mu/ds~=~dk^\mu/ds~-~dU^\mu/ds$ and letting the volume extend to $\pm$null infinity, as $r~\rightarrow~\infty$ it can be shown that

$$\frac{dP_\mu}{ds}~=~-\frac {q^2}{4\pi }(a_\nu a^\nu~+~(1/\tau)^2)k_\mu~=~-q^2(a_\nu a^\nu~+~(a_\nu k^\nu)^2)k_\mu$$

When integrated over $2\pi r^2 dr$ the first term gives the Lamour formula for radiated power. Since the spacetime acceleration is orthogonal to the four velocty and $a_\nu a^\nu~=~d/ds(U_\nu a^\nu)~-~(da^\nu/ds)U^\nu$ the formula is put in the form which involves the derivative of the acceleration. It is for this reason that the Lamour formula is defined for systems which have a non-constant acceleration, such as the cyclical motion of a charge in a magnetic field and Bremmstrahlung radiation. The second term depends upon the “acceleration curvature” $1/\tau$, and it is from this term where the Unruh effect emerges.

Lawrence B. Crowell

 

[Lawrence B. Crowell]

accelerated charge III

It is now time to turn our attention to the case of a charged particle in a curved spacetime. In particular the sphereically symmetric spacetime with metric

$$ds^2~=~(1~-~F(r))dt^2~-~(1~-~F(r))^{-1}dr^2~-~r^2d\Omega^2$$

which is the Reissner-Nordstrom spacetime for $F(r)~=~2M/r~-~Q^2/r^2~+~\Lambda r^2/3$, for $M,~Q,~\Lambda$ a the mass of a spherically symmetric body, the charge and the deSitter parameter or cosmological constant. This metric is asymptotically flat and corresponds to a Minkowski spacetime as $r~\rightarrow~\infty$, where the tetrad $e^a_\mu~=~(T_\mu,~R_\mu,~\Theta_\mu,~\Phi_\mu)$ defines the above line element as $ds^2~=~e^a_{\mu}e^a_\nudx^\mu dx^\nu$ and the tetrad components necessarily defined so

$$T_\mu dx^\mu~=~(1~-~F(r))^{1/2}dt$$
$$R_\mu dx^\mu~=~(1~-~F(r))^{-1/2}dr$$
$$\Theta_\mu dx^\mu~=~rd\theta$$
$$\Phi_\mu dx^\mu~=~rsin(\theta)d\phi,$$

with the appropriate signature change on the spatial parts on contraction of the tetrad. The electromagnetic field tensor is easily found by computing it in the flat region and with the tetrads extending it to the RN spacetime

$$F_{mu\nu}~=~-\frac{Q}{r^2}(T_\mu R_\nu~-~T_\nu R_\mu)$$

and the electromagnetic contribution to the Ricci curvature is the trace free components

$$R_{\mu\nu}~=~-\frac{Q^2}{r^4}(T_\mu T_\nu~-~R_\mu R_\nu~+~\Theta_\mu\Theta_\nu~+~\Phi_\mu\Phi_\nu)$$

The RN metric permits the Killing vector $k^\mu\partial_\mu~=~(1~-~F(r))^{-1/2}$ where an observer falling into the black hole measure the electric field $E_\mu~=~F_{\mu\nu}k^\nu~=~(q/r^2)$ in the radial direction.

For a charge which falls into a neutral Schwarzschild gravity field we set the RN term $F(r)~=~1~-~M/r$. For a body falling radially in the spacetime its proper time $d\tau~=~dt~-~F(r)^{1/2}(1~-~F(r))^{-1}$ defines the transformed line element

$$ds^2~=~(1~-~F(r))d\tau^2~-~2F(r)^{1/2}d\tau dr~-~dr^2~-~r^2d\Omega$$

with the tetrad basis

$$\tau_\mu dx^\mu~=~(1~-~F(r))^{1/2}d\tau~-~\Big(\frac{F(r)}{1~-~F(r)}\Big)^{1/2}dr,$$
$$R_\mu dx^\mu~=~(1~-~F(r))^{-1/2}dr.$$

The dual vector basis which obeys $e^a_\mu dx^\mu(e^{b\nu}\partial_\nu)~=~{\delta^a}_b$ are

$$\tau^\mu \partial_\mu~=~(1~-~F(r))^{-1/2}\partial_\tau$$
$$R^\mu \partial_\mu~=~-(1~-~F(r))^{1/2}\partial_r~-~\Big(\frac{F(r)}{1~-~F(r)}\Big)^{1/2}\partial_\tau$$

The duality between the differential and vector elements give the tangent vectors for the free fall as

$$U_\mu dx^\mu~=~d\tau,~ U^\mu\partial_\mu~=~\partial_\tau~-~F(r)^{1/2}\partial_r$$

The geodesic condition ${U^\mu}_\nu U^\nu~=~0$ results in the radial acceleration

$$\partial^2_\tau r~=~-\frac{M}{r^2}$$

which for a weak field case with $t~\simeq~\tau$ this reduces to the Newtonian case. Since this approach to the RN metric is according to variables which fit into the asymptotic Minkowki spacetime the power formula for the change in the the four momentum vector as

$$\frac{dP_\mu}{d\tau}~=~-\frac {q^2}{4\pi }(\partial^2_\tau r\partial^2_\tau r~+~(1/B)^2)k_\mu,$$

where the acceleration curvature has been relabed with B. So this calculation indicates that a charge falling into a gravity will indeed emit electromagnetic radiation.

So what is going on? We have this idea of the Einstein equivalence principle. A local infalling frame with a charge should be equivalent to a charge in an inertial frame in Minkowski spacetime. Indeed this is the case, and this is the case for a region where the tidal acceleration is $\sim~1/r^3$ extremely small. An observer in a small frame where the Riemann curvature multiplied by a four volume (which for repeated indices is reduces the volume by one dimension) is small $R_{abcd}\delta vol^{abcd}~\sim~0$ will observe no electromagnetic radiation being emitted. A small enough of a local frame is in the near field domain, where the change in the electromagnetic field, or its wavelength, is small enough so the field appears constant or Coulombic. The electromagnetic field only appears in a region far removed from the infalling charge where the tidal acceleration is significant.

There is one last little bit with this part of the problem. If the charge is emitting radiation by falling into the gravitating body, then this radiation is being produced from the gravitational potential energy. This should slow down the infalling charge or equivalently reduce its kinetic energy. The change in the energy will be $dP_\mu~=~Qk_\mu d\tau$,for Q the above acceleration dependent power term. The power $P~=~F\cdot U$ in relativistic terms and the above equation means that there is a counter acceleration term $A~=~q^2/r^3$. If we were to continue the same calculation with the RN term $F(r)~=~M/r~-~q^2/r^2$ we will obtain an identical term. The presence of the charge on the black hole effectively reduces the gravitational mass and thus curvature for the geodesic of a neutral particle. This gives an identical result for the acceleration of a charged particle approaching a neutral black hole. There is a symmetry here. For a neutral particle falling into a black hole the acceleration is reduced by the effective mass reduction of the black hole by the charge. The charged particle falling into a neutral black hole is slowed by an equal amount by radiating an equal amount of mass-energy in a radiation field.

Lawrence B. Crowell

 

[sweetser]

Mass and Electric Charge

Hello Lawrence:

This looks like good, honest technical work. I wish you luck with it. However it is not relevant to the topic at hand. The charged metric for the GEM proposal is:

$$d \tau^2 = exp(2(\sqrt{G} Q - GM)/c^2 R) dt^2 - exp (-2(\sqrt{G} Q - GM)/c^2 R) dR^2/c^2 - R^2d\Omega^2/c^2$$

The Taylor series of this metric does not match the Reissner-Nordstrom metric. The simplest way to see this is M and Q come in on equal footing for GEM. It should be obvious that Nature works this way, not the affect of M is comparable to Q². I did a thought experiment on this subject, all of three slides, here:

http://theworld.com/~sweetser/quaternions/talks/rank1/4221.html

The gist is quite simple. You have an electron. You measure that the electron is attracted to a box. You do not know what is inside the box. If you have one proton, or 421 kg of mass inside the box, you will not be able to tell any difference in the attraction from the effect of electric charge attraction, or the gravitational mass attraction of the 421 kg brick.

Good luck in your studies, but please let's stay focused on the metric, field equations, and the derivation of the field equations which is more than enough to keep us busy.

Doug

 

[sweetser]

Reissner-Nordstrom metric is not physical

Hello:

A basic property of electromagnetism is that like charges repel and different charges attract. That must be a property of any proposal related to EM. The Reissner-Nordstrom can only accept like charges that repel, since it uses Q², which always has the same sign (and opposite the M/R term). That looks like a deadly flaw to me, one that prevents me for doing anything further with the metric.

In the GEM proposal, the sign of the relative electric charge (ie same sign for source and probe means positive) and the mass charge are different. That way like electric charges can repel, and like mass charges can attract. If the relative electric charge is negative, as happens with a positive and negative charge, then the electric charge contribution would be attractive in exactly the same way as gravitational charge. The two add up. There is no negative mass charge in the GEM proposal because the mass field is about symmetric fields. There are two charges for EM because of the antisymmetric fields.

Doug

 

[Lawrence B. Crowell]

Obviously with the 421 kg mass in a box or a unit charge it will be easy to tell the difference. The mass will attract all masses equally, while the charge will attract or repel the same whether that has the same or opposite charge. The proton will have negligable attraction to an uncharged mass.

The NS metric is for a spacetime determined by a sperically symmetric mass with a charge. One could well enough compute the orbit of a charged particle in this spacetime. Yet what I compare is the fall of a neutral mass around a charged black hole and a charge falling into a neutral black hole. What I find is that the "braking" of a charged particle fall by the emission of radiation by purely classical means is equivalent to the geodesic rate of fall for a neutral particle entering a black hole with the equivalent charge. This result illustrates I must be at least on the right track. It is in line with the "no hair" theorem for black holes is that the details of how a black hole reaches a final state by the acquisition of charge and mass by the infall of matter and fields is independent of the details by which that occurred. A black hole will end up with the same final mass $M~=~M_0~-~(q/r)^2$ indepdendent of which masses brought in what charge.

Lawrence B. Crowell

 

[sweetser]

Hello Lawrence:

The thought experiment did not concern itself with the ability to tell the difference between the behavior of electric charge or mass charge. Rather, it was an effort to find a condition where the behavior was indistinguishable.

> One could well enough compute the orbit of a charged particle in this spacetime.

If and only if the charged mass had the same sign as the charge on the test mass would the NS metric work. If th test had the opposite sign, it would fail because the metric must demonstrate attraction.

I appreciate that you looked into the consequences of the NS metric, and that other skilled people have worked with it. But it looks unacceptable to me because it cannot model electrical attraction.

Doug

 

[Lawrence B. Crowell]

Hello:

A basic property of electromagnetism is that like charges repel and different charges attract. That must be a property of any proposal related to EM. The Reissner-Nordstrom can only accept like charges that repel, since it uses Q², which always has the same sign (and opposite the M/R term). That looks like a deadly flaw to me, one that prevents me for doing anything further with the metric.

Doug

The RN metric uses $(Q/r)^2$ as the self-energy of a charged mass. This can be derived by using the electromagnetic field tensor as a source of the gravitational field. From there the RN metric is a solution to the Einstein field equation. In more advanced settings solutions of this type are BPS black holes, where gauge "charges" are the source of the black hole metric. The charge here is the charge of the black hole. One might then consider the motion of a charged mass in this spacetime, where that charge can have either sign.

This is fairly standard stuff.

Lawrence B. Crowell

 

[Lawrence B. Crowell]

Hello Lawrence:



> One could well enough compute the orbit of a charged particle in this spacetime.

If and only if the charged mass had the same sign as the charge on the test mass would the NS metric work. If th test had the opposite sign, it would fail because the metric must demonstrate attraction.

Doug

I am not sure where you got this idea. You can well enough compute the orbit of a charge with any value or sign in an RN metric.

Lawrence B. Crowell

 

[sweetser]

Coupling to a current, not a stress-energy tensor

Hello Lawrence:

I agree, you have represented standard analysis, which is darn good most of the time. If I see a killer flaw, I will point it out.

Even better is when I realize I was wrong about the killer flaw. I "get" the standard way, and can pinpoint where I drive a different direction. That is what happened with regards to the Reissner-Nordstrom metric.

In general relativity, gravity binds to the stress energy tensor. How much energy will the electric field contribute to the metric? [itex]G Q^2/c^4 R^2[/tex]. If the charge is positive, or the charge is negative, it does not matter, the same amount goes in.

For the GEM proposal, the charge coupling term is $-\frac{1}{4 c}(J A + (J A)^* - J^{*1} A^{*2 *3} - (J^{*1} A^{*2 *3})^*)$ With GEM, the effort is to do both gravity and EM. All these conjugates are required to get the phase to have both spin 1 and spin 2 symmetry. The coupling of gravity is not to energy, but instead to the 4-current density. The energy of the electric field makes zero contribution to gravitational effects.

What we get in return for giving up the energy connection is a new equivalence principle for particles that attract each other. The attraction might be caused by electric charge or by a mass charge. Attraction looks the same either way. Sure, you could do other tests to tell which one was at work, but one could have videos of a particle being attracted by gravity or by EM that are absolutely identical. That is the bridge between gravity and EM.

People who work on EM almost never deal with the tools of differential geometry. Calculate the divergence of the Christoffel symbol of the second kind for this metric:

$$d \tau^2 = exp(2(\sqrt{G} Q - GM)/c^2 R) dt^2 - exp (-2(\sqrt{G} Q - GM)/c^2 R) dR^2/c^2 - R^2d\Omega^2/c^2$$

Since this is for a static, spherically symmetric metric, to be logically consistent with EM, the answer had better be both Gauss' and Newton's potential theory for EM and gravity,$\nabla^2 \frac{\sqrt{G} Q - GM)}{R}$, which it is. Lucky? I don't think so.Doug