In flat space-time the electric potential energy between two charges is $\frac{k Q_1 Q_2}{r_{12}}$, where $Q$'s are charges and $r_{12}$ is the distance between them. What would happen if the two charges are placed in a strongly curved space-time, which makes "distance" and "duration" measures different from place to place? Will the two feel the same strength of electric force/potential energy?

  • $\begingroup$ I'm sure they will feel the "same" force by some formalization in some frame of reference. I'm curious about this question myself, but I'd prefer to say "how does GR make that work?" $\endgroup$ – Alan Rominger Jun 19 '11 at 0:06

In a general curved spacetime, the problem is complicated and obviously you can't end up with the "same" formula, especially not if you interpret $r_{12}$ in a random way, e.g. as the proper distance between the two points. Such a formula is surely incorrect.

To solve the problem, you have to solve the field equations such as $$\nabla\cdot \vec D = \rho $$ in a given curved spacetime. Well, you should write these equations using $F_{\mu\nu}$ with the general relativistic treatment of the indices - you should use the full Einstein-Maxwell equations. Note that in general relativity, you should actually also accept that the electric field itself will contribute to the curvature of spacetime, too.

The result will depend on the precise shape of the space. You may determine the total mass/energy (note that $E=mc^2$ in relativity) either from $Q_1 \phi_2$, calculated in one way or another, and taking the proper redshifts into account, or as the ADM energy i.e. from the asymptotic curvature induced by the configuration of the charges.

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