# Kaluza–Klein theory, a point charge, A_0, geometry of.

1. Jan 31, 2013

### Spinnor

We know that the spacetime of General Relativity with a single electron in otherwise empty space is hardly curved, basically zero.

In Kaluza–Klein theory with a single electron in otherwise empty space is there a type curvature due to the charge of a single electron?

Is the amount of "curvature" (if that is the right word) of the spacetime of Kaluza–Klein theory with a single electron some how related to the electric potential of a single electron, something like e/R where e is the charge of the electron and R is the distance from the charge? I'm guessing that the electric potential of an electron comes into play in describing the geometry of the space of Kaluza–Klein theory?

How complicated is the geometry of a single electron in Kaluza–Klein spacetime?

I'm not sure this is the right group, please move if it belongs elsewhere.

Thanks for any help!

2. Jan 31, 2013

### Mentz114

You don't need Kaluza–Klein and the extra dimension to work out the curvature due to the electric field of a point charge. There is a class of solutions in GR called 'electro-vacuums' where the curvature is attributed to an electric field. The SET of the electric field takes the form shown here

http://en.wikipedia.org/wiki/Electromagnetic_stress–energy_tensor

An example of a spacetime that has an electric field is the Reissner-Nordstrom charged BH

http://en.wikipedia.org/wiki/Reissner–Nordström_metric

Last edited: Feb 1, 2013
3. Feb 2, 2013

### Spinnor

Thank you. I was particularly interested in Kaluza–Klein spacetime with a single charged particle. I hoped to make connection with the 1/R potential of a charged particle I am familiar with and the metric of Kaluza–Klein spacetime of a single charged particle. Seems the geometry of such a curved manifold must be simple?

4. Feb 2, 2013

### Mentz114

I don't know what you mean by "Seems the geometry of such a curved manifold must be simple?".

I did the calculation of the Einstein tensor some time ago with this metric

$$ds^2=\,\left( \frac{k\,{Q}^{2}}{{x}^{2}}-1\right){dt}^{2} +\frac{2\,dt\,du\,k\,Q}{x}+\,k{du}^{2}+{dz}^{2}+{dy}^{2}+{dx}^{2}$$

with this result

$$G_{\mu\nu}= \left[ \begin{array}{cccc} -\frac{3\,k\,{Q}^{2}\,\left( k\,{Q}^{2}+3\,{x}^{2}\right) }{4\,{x}^{6}} & 0 & 0 & 0 & -\frac{k\,Q\,\left( 3\,k\,{Q}^{2}+4\,{x}^{2}\right) }{4\,{x}^{5}}\\ 0 & \frac{k\,{Q}^{2}}{4\,{x}^{4}} & 0 & 0 & 0\\ 0 & 0 & -\frac{k\,{Q}^{2}}{4\,{x}^{4}} & 0 & 0\\ 0 & 0 & 0 & -\frac{k\,{Q}^{2}}{4\,{x}^{4}} & 0\\ -\frac{k\,Q\,\left( 3\,k\,{Q}^{2}+4\,{x}^{2}\right) }{4\,{x}^{5}} & 0 & 0 & 0 & -\frac{3\,{k}^{2}\,{Q}^{2}}{4\,{x}^{4}} \end{array} \right]$$

The EMT of the point charge in 4D is
$$\left[ \begin{array}{cccc} \frac{{Q}^{2}}{2\,{x}^{4}} & 0 & 0 & 0\\ 0 & \frac{-{Q}^{2}}{2\,{x}^{4}} & 0 & 0\\ 0 & 0 & \frac{{Q}^{2}}{2\,{x}^{4}} & 0\\ 0 & 0 & 0 & \frac{{Q}^{2}}{2\,{x}^{4}} \end{array} \right]$$

Unfortunately I can't remember where I got this ( Wiki ?) or what the parameter k is.

I suppose you'd want it in spherical polar coords.

I don't know whether is 'simple' or not.

Last edited: Feb 2, 2013
5. Feb 2, 2013

### stevendaryl

Staff Emeritus
The point of Kaluza-Klein theory is to interpret electric charge as basically momentum in a 5th spatial dimension. So I thought the original poster was asking about what the 5D theory of a single charged point-mass looks like in Kaluza-Klein theory.

6. Feb 2, 2013

### Mentz114

Yes, the OP has clarified this. But what does 'simple', 'complicated', 'looks like' mean ?

The metric I used above is built from the electric potential Q/x.

7. Feb 2, 2013

### stevendaryl

Staff Emeritus
But as I said, Kaluza Klein theory does not have any charges. Instead, what appears to be charge is interpreted as momentum in the curled-up dimension.

8. Feb 2, 2013

### Mentz114

So what ? In the KK attempt to unify EM and gravity, charge ( or is it potential ) is modelled as quantised momentum in the extra dimension. How does this help the OP, who probably knows this.

Spinnor, there's a huge literature on this field, including lecture notes, available on the internet. Try a search. I'm out of time now.

Last edited: Feb 2, 2013
9. Feb 2, 2013

### stevendaryl

Staff Emeritus
Because he wanted to know what spacetime looked like in KK in the case of a single electron. So it doesn't have any Qs in it, because there are no charges in KK.

10. Feb 3, 2013

### Mentz114

Well, perhaps I'm not answering the OPs question because I don't know what he's asking. But charge can enter the KK equations via the 4-potential. Please have a look at the attached and tell me where I'm going wrong.

#### Attached Files:

• ###### kaluza-Klein-Straub.pdf
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11. Feb 3, 2013

### stevendaryl

Staff Emeritus
You are completely right, I just didn't realize what you were doing. The way that I thought one would go about answering the question is to just solve the 5D Einstein Field Equations for a point-mass, and then the terms corresponding to the electromagnetic field would pop out under the identification of $g^{\mu 4}$ (where $x^4$ is the curled-up dimension) with the electromagnetic potential $A^\mu$. You were doing it the other way around, which was to solve the 4D equations for the electromagnetic field of a point-particle, and then reinterpreting the resulting $A^\mu$ as $g^{\mu 4}$. If KK is exactly equivalent to E&M, then the two approaches should be the same.

12. Feb 3, 2013

### stevendaryl

Staff Emeritus
Another question for Kaluza-Klein theory. KK theory was considered interesting from the point of view of unification, because it allowed electromagnetism to be seen as a manifestation of gravity in 5D spacetime. But (until String Theory renewed interest in it) supposedly it didn't make any new predictions, so people lost interest in it.

I don't understand why anyone would say that KK makes no new predictions. As described here: http://www.weylmann.com/kaluza.pdf
it seems that KK is not exactly equivalent to classical electrodynamics, instead, it predicts an extra force term on a charged particle that is nonlinear in the electromagnetic field:

$-\frac{1}{2} k A_\nu F^\lambda_\mu U^\nu U^\mu$

(equation 6)

Are they just saying that given practical values for the electromagnetic field strength, this departure from the Lorentz force law is undetectable?

13. Feb 3, 2013

### Mentz114

Good point. I'm puzzled that the potential can occur in the Lorentz force. That seems to lose the gauge freedom to add an arbitrary constant to the potential. Needs some thinking about ...

14. Feb 3, 2013

### elfmotat

That paper by Straub seems to be rather ambiguous. For example, when he talks about the 4-D Christoffel symbols, is he talking about them as derived from the ordinary 4D metric $g_{\mu \nu }$ or as derived from the 4D part of the 5D metric $\tilde{g}_{\mu \nu }=g_{\mu \nu }+kA_\mu A_\nu$ ? If it's the latter, there should be terms involving the four-potential in the connection which might "counteract" a gauge transformation in the context of the 5D geodesic equation.

15. Feb 3, 2013

### stevendaryl

Staff Emeritus
I haven't gone through all the math, but I believe that the extra term that I mentioned, $−\frac{1}{2}k A_\nu F^\lambda_\mu U^\nu U^\mu$, is exactly the correction to the 4-D christoffel symbol. So I think he is using $g_{\mu \nu}$ to mean the 4-D metric without the corrections due to the electromagnetic potential.

16. Feb 3, 2013

### elfmotat

When I did the calculation to find the 4D Christoffel symbols, I got the following:

$$2\tilde{\Gamma}^a_{bc}=2\Gamma^a_{bc}+kA_b F_c^{~a}+kA_c F_b^{~a}+kA^a (\partial_b A_c+\partial_c A_b)$$

where $\Gamma^a_{bc}$ are the ordinary 4D symbols formed from the usual 4D metric.

17. Feb 4, 2013

### stevendaryl

Staff Emeritus
I have a really hard time doing these sorts of index-laden calculations without making a mistake, but when I did it, I got:
$2\tilde{\Gamma}^\mu_{\nu \lambda}=2\Gamma^\mu_{\nu \lambda}+kA_\nu F_\lambda^{\mu}+kA_\lambda F_\nu^{\mu}$

In the paper, the correction term is just $kA_\nu F_\lambda^{\mu}$, but if you symmetrize that, you get the same thing as I did.

18. Feb 4, 2013

### Spinnor

I think I have a surface that might be close. Consider the three dimensional surface that "lives" in the space RxRxRxC, x,y,z,w, with w the coordinate of C given by

C = e(x^2 + y^2 + z^2)^-.5 = e/R

?

19. Feb 4, 2013

### elfmotat

Sorry for derailing your thread a bit further Spinnor, but I felt the need to post this. Unless I messed something up, I believe you can get the following from Maxwell's equations:

$$\nabla_\mu \nabla_\nu \left ( A_\lambda \frac{dx^\lambda}{ds} \right )=\rho g_{\mu \nu}$$

where $\rho$ is charge density. Perhaps this is significant.

EDIT: Someone on another site said that they get the following as the Noether conserved momentum about the 5th dimension:

$$g_{A4}\frac{dx^A}{ds}=k\left (\frac{dx^4}{ds}+A_\mu \frac{dx^\mu}{ds} \right )$$

(Perhaps the factor 1/2 in the paper is wrong.) So if we identify this conserved quantity with charge, we get the usual Lorentz force with no additional term.

EDIT 2: It turns out that the reason the term $kA_\nu F^\lambda_{~\mu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}$ isn't gauge-invariant is because the term $kF^\lambda_{~\mu} \frac{dx^\mu}{ds} \frac{dx^4}{ds}$ isn't gauge-invariant either. A gauge transformation in the first term corresponds to a change in the 5th coordinate:

$$kF^\lambda_{~\mu} \frac{dx^\mu}{ds} \left (\frac{dx^4}{ds}+\frac{d\zeta}{ds} \right )$$

and in the second term:

$$k(A_\nu-\partial_\nu \zeta ) F^\lambda_{~\mu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds} =kA_\nu F^\lambda_{~\mu} \frac{dx^\mu}{ds} \frac{dx^\nu}{ds}-kF^\lambda_{~\mu} \frac{dx^\mu}{ds} \frac{d \zeta}{ds}$$

So everything works out fine and dandy.

Last edited: Feb 4, 2013