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In solutions to Einsteins equations one normally look at how a mass curves spacetime and then how that effect the geodesics of a test particle. So one only consider free movement of the particle. I know there is a solution with a black hole that has an electric charge and one can then describe how a neutrally charged testparticle will be attracted and repelled and so on. But what if the testparticle has an electric charge? Are there any solutions which describe how electric charges interact in general relativity or is it too difficult to figure out?

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  • $\begingroup$ Just to be clear as to what you are asking: are you looking for actual solutions, for the force acting between charged point particles in curved spacetime or for the equations of motion of charged particles subject to an arbitrary electromagnetic field in curved spacetime? $\endgroup$ – Giorgio Comitini Jun 30 '18 at 18:31
  • $\begingroup$ I guess: The equations of motion of charged particles subject to an arbitrary electromagnetic field in curved spacetime. I was just wondering if one could handle this in general relativity since I haven´t seen any mention of it in books. They allways consider geodesics and I was wondering if Einsteins equations only can deal with that. I am not an expert in general relativity at all though so it could just be my ignorance. $\endgroup$ – Kasper Falkenberg Andersen Jun 30 '18 at 18:58
  • $\begingroup$ Have a look at the paper "The Motion of Point Particles in Curved Spacetime" by Poisson et al., here. $\endgroup$ – A.V.S. Jun 30 '18 at 19:55
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Well, of course General Relativity can handle charged particles in electromagnetic fields! The equations of motion for such a particle are the generalization to a curved spacetime of the Lorentz equation,

$$ \frac{d^{2}x^{\mu}}{ds^{2}}=-\Gamma^{\mu}_{\nu\sigma}\ \frac{d x^{\nu}}{ds} \frac{d x^{\sigma}}{ds}+\frac{q}{m}\ F^{\mu}_{\ \ \nu}\ \frac{d x^{\nu}}{ds}\qquad\qquad (\star) $$

$(c=1)$, where $s$ is the proper time along the particle's world line, $F_{\mu\nu}$ is the electromagnetic field strength tensor and $F^{\mu}_{\ \ \nu}=g^{\mu\sigma}\ F_{\sigma \nu}$. The equation is just the geodesic equation plus an electromagnetic force term and can be derived from the action

$$ S=-m\int ds\ \bigg\{\sqrt{g_{\mu\nu}\frac{dx^{\mu}}{ds}\frac{d x^{\nu}}{ds}}+\frac{q}{m}\ A_{\mu}\,\frac{d x^{\mu}}{ds}\bigg\} $$

by defining $F_{\mu\nu}=\partial_{\mu}A_{\nu}-\partial_{\nu}A_{\mu}$. Equation $(\star)$ differs from the Lorentz equation in flat spacetime in that gravity enters both through the Christoffel symbols $\Gamma^{\mu}_{\nu\sigma}$ and through the inverse metric used to raise the index of $F^{\mu}_{\ \ \nu}$. Moreover, the Maxwell equations too are modified on a curved spacetime, as follows:

$$ \nabla_{\mu}F^{\mu\nu}=J^{\nu} $$

Here $J^{\mu}$ is the four-current source for the electromagnetic field and $\nabla_{\mu}$ is the covariant derivative associated to the metric. Since gravity enters the Maxwell equations through the covariant derivative and raising of the indices of $F^{\mu\nu}$, its solutions are different from their analogue on a flat spacetime. Therefore, also the $F_{\mu\nu}$ itself in the Lorentz equation is in general different from that on flat spacetime.

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  • $\begingroup$ And the presence of mass, fields, and charges affect the metric via Eistein's field equations T(mu, nu) ~ G(mu, nu). $\endgroup$ – ggcg Jun 30 '18 at 23:22
  • $\begingroup$ Yup, that's right. $\endgroup$ – Giorgio Comitini Jun 30 '18 at 23:22
  • $\begingroup$ Now, I wonder how much research physicists do looking at the complete set of field equations. I think in most practical problems we assume the background curvature is fixed and look at movement on that b.g. Of course that depends on the specific problem at hand. Even in Newtonian Gravity the lowest order approximation is: Planets make ellipses about the Sun. Is it worth solving the many body problem in G.R. on a solar system scale? $\endgroup$ – ggcg Jun 30 '18 at 23:27
  • $\begingroup$ I don't think it is. Just consider how small an effect GR has on the perihelion precession of Mercury. $\endgroup$ – Giorgio Comitini Jun 30 '18 at 23:35

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