I am confused with this question. Does electric charge affect the space time fabric? If so, why? Also if electric charge does not affect the space time fabric, how can we interpret the origin of the electric field around a charge?


The The electric field around a charge has energy, and so contributes to the gravitational field around the charge. You can see the effect in the exact solution for charged black holes, where the curvature is nonzero everywhere because of the electric charge.

$$ ds^2 = - f(r) dt^2 + {1\over f(r)} dr^2 + r^2 (d\theta^2 + \sin^2\theta d\phi^2) $$

$$ f(r) = 1 - {2m\over r} + {Q^2\over r^2} $$

If you set Q to zero, you get a solution of the vacuum Einstein equations, so that there is no curvature away from the singular region at $r=0$. But if you have a nonzero Q, you get curvature at all points, and this is due to the stress-energy of the electric field.

$$ E = {Q\over r^2} \hat{r} $$

leads to a local energy density which is

$$ {Q^2\over 2 r^4} $$

outside of the black hole, and this energy density makes a contribution to the total mass. It is a negligible fraction of the mass of the black hole, unless the black hole is extremal. This happens when $m=Q$, so that the solution degenerates, and the horizon becomes AdS as opposed to Rindler. In such a limit, the electric repulsion is equal and opposite to the gravitational attraction, and two such stationary black holes just sit next to each other, neither replling nor attracting. In this limit, you can consider the entire mass of the black hole as contained in the field.

But the question is not about the effect of electricity on space, but about the geometric interpretation of electromagnetic fields. What electromagnetism is a change in the "phase space fabric", not the "metric space" fabric. When you take a particle around a loop in space, there is quantum mechanical phase from doing this, and the extra phase $\Delta \Phi$ when the particle is charged is given by:

$$ \Delta\Phi = q\oint A \cdot dx $$

This is the definition of the electromagnetic gauge field, and it is the fundamental geometric description of electromagnetism. It has nothing to do with the direct geometry of space time, it has to do with an internal geometry in the shape of a circle, which is the phase of the quantum wavefunction.

In Kaluza Klein theory, this circle becomes a geometrical circle, an extra dimension in the shape of a small circle, and the interpretation of the vector potential is that it is the amount of translation you get in the little circle when you go around a loop in our ordinary 4 dimensions. The charge is then interpreted as the momentum along this circle, and in quantum mechanics, the phase acquired from translation is the same as the momentum, so this embeds electromagnetism in geometry.

But within modern theories, a gauge field is as fundamental as a metric field, and if you have enough supersymmetry requiring the gauge field to be there, you can consider the gauge field to be just as geometrical. It is not the same geometry as the geometry that explains gravity, but it is geometry nonetheless, it is the geometry of a fibre bundle (a separate space attached to the main space at every point, like the extra circle in Kaluza Klein in the limit that it is infinitesimal).

So in the modern understanding, the notion of geometry is just general enough to make any gauge field geometrical, even if it isn't directly a change in the naive thing you would call the "space-time fabric".

  • $\begingroup$ How this equation will be modified if we get too close to the charge? I mean the $1/r^2$ looks like a coulomb term, so how the quantum mechanical/QED related formula look like? $\endgroup$ – Revo Aug 25 '12 at 1:38
  • $\begingroup$ @Revo: The 1/r^2 is exact in GR, tit is geometrical, since Gauss's law still holds, and r in Schwarzschild coordinates is defined by areas of spheres. This is classical, in QED, you can't really ask the question inside the compton wavelength of the charged particle, because it isn't localized in space at this length scale. But the divergence in self-energy is modified from 1/a to log(a), so the field can be thought of as smeared out not quite uniformly in such a way as to leave a log-divergence. $\endgroup$ – Ron Maimon Aug 25 '12 at 1:55
  • $\begingroup$ Yeah, well, does that smearness show up in the ds equation? $\endgroup$ – Revo Aug 25 '12 at 2:15
  • $\begingroup$ @Revo: You can't use a "ds" equation for quantum particles, it doesn't make sense. The particle is superposed, and so is the gravitational field. You can't use classical equations for quantum systems, you need quantum description. I didn't do any of that, and it doesn't really help understand this thing. $\endgroup$ – Ron Maimon Aug 25 '12 at 2:39
  • $\begingroup$ Oh thanks. I did not know that. So it is only pure GR with no quantum effects. $\endgroup$ – Revo Aug 25 '12 at 3:18

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