# Evidence of Electromagnetic fields causing spacetime curvature?

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1. Apr 29, 2013

### physwizard

Hi All,
Just wanted to know, is there any experimental or observational evidence today, that electromagnetic fields can cause spacetime curvature? Either direct or indirect?

2. Apr 29, 2013

### h_cat

electromagnetic fields originate at electric charges which posses (what coincidence) mass. Now mass does cause spacetime curvature right? So your experimental evidence is all around you.
Now what the problem is, is to know if the gravity is actually caused by the electromagnetic field or by something else. A proton has the same charge but is lot more massive than an electron so it seems that gravity and electromagnetism have no direct connection. Of course the proton could be a component particle (which in fact even in the standard theory it is) out of a lot of positive and negative charges which sum up to one positron (anti electron) charge.
Now from that point their is hope. The biggest goal of all - to unite all fields into one theory.

mo_cat

3. Apr 29, 2013

### WannabeNewton

There is no ambiguity here. An electromagnetic field $F_{ab}$ has an associated stress-energy tensor $T_{ab} = \frac{1}{4\pi}(F_{ac}F_{b}{}{}^{c} - \frac{1}{4}g_{ab}F_{de}F^{de})$. This will generate space-time curvature as per Einstein's equation $G_{ab} = 8\pi T_{ab}$.

4. Apr 29, 2013

### physwizard

thats right, as per the theory it is supposed to contribute to spacetime curvature. but my question is about experiment. is there any experimental or observational evidence, either direct or indirect to support this 'prediction' of the theory? any experimental evidence which suggests that you can club the electromagnetic $T_{ab}$ with the $T_{ab}$ due to matter in Einstein's equation?

5. Apr 29, 2013

### Bill_K

A significant portion of the mass of a nucleus is due to its electrostatic energy. The total Coulomb repulsion increases quadratically with Z, and for heavy nuclei is of the order of 100's of MeV.

If Coulomb energy did not produce a gravitational field the same as everything else, the Cavendish experiment which measures gravitational attraction of two spheres would give apparently different values of G for say, uranium spheres as opposed to aluminum.

6. Apr 29, 2013

### physwizard

Okay let us say that we performed the Cavendish experiment and found that the force of attraction between uranium spheres was the same as the force of attraction for aluminium spheres of the same mass. so how do we conclude from that that the electromagnetic field has gravitational mass?

7. Apr 29, 2013

### Bill_K

For the sake of round numbers, take a nucleus with A = 200 ≈ 200 GeV, and suppose its Coulomb energy is 200 MeV. That is, 1 part in 1000 of its mass is due to electrostatic energy.

If electromagnetism did not produce a gravitational field, the active gravitational mass of the nucleus would be reduced by 1 part in 1000, and the gravitational attraction between the two spheres would also be reduced by that amount. If instead the attraction is the same, it shows this is not the case.

8. Apr 29, 2013

### pervect

Staff Emeritus
See for instance http://arxiv.org/abs/gr-qc/9909014 "Kinetic Energy and the Equivalence Principle" for a detailed discussion.

Note that we know clasically that the electromagnetic field carries momentum. We've known this for a long time. For instance take the wiki entry on the old-fashioned idea of "electromagnetic mass".

The modern idea is that when you have an isolated system, you can integrate T_00 (the energy density) and get the mass (depending on your choice of units, you might have to throw in a factor of c^2). And the electromagnetic field contributes to this integral, so it contributes to the mass.

9. Apr 29, 2013

### WannabeNewton

Just to add on to Pervect, the electromagnetic contributions can affect the Komar mass as well. The Komar mass for the stationary, asymptotically flat space-time of an appropriate isolated body can be written as $M_{\text{Komar}} = -\frac{1}{8\pi}\int _{S}\epsilon_{abcd}\nabla^{c}\xi^{d}$ where $S$ is a topological 2-sphere. If you calculate this for the Reissner–Nordström metric (it is a rather easy calculation) then you will find that $M_{\text{Komar}} = M - \frac{q^{2}}{r}$ where $M,q$ are the mass and charge parameters.

10. Apr 29, 2013