I'm just starting to learn general relativity (GR), and I'm a beginner, but I came out with this situation which is unclear to me: The trajectory of a charged particle in GR is given from the equation:

$$\dot{u}^{\mu} + \Gamma^{\mu}_{\alpha \beta} u^{\alpha} u^{\beta} = \frac{q}{m} F^{\mu}_{\; \nu} \, u^{\nu}$$

So, if I have a neutral particle $q=0$ the equation reduces to the geodesic equation for a free particle, but because of the Einstein-Maxwell equations:

$$R_{\mu \nu} - \frac{1}{2} R g_{\mu \nu} = T^{EM}_{\mu \nu}$$

the EM stress-energy tensor determines the form of the metric, and consequently the Christoffel symbols that appears in the geodesic equation for the neutral particle. So would the trajectory of this neutral particle in an EM field be different from the case of a space-time with a null EM field?


2 Answers 2


The answer to your question is yes, the metric is influenced by the electromagnetic field, and a neutral particle will follow a geodesic of that metric. Thus, this implies that the neutral particle will indeed "feel" the electromagnetic field, but only in a very indirect way (from the geometry of spacetime around the particle).

The best example of this is the Reissner-Nordström metric, which is a generalisation of the Schwarzschild metric, in case a radial electrostatic field is present in vacuum.

Gravity is universal. It affects everything that has energy-momentum, and electromagnetic fields do have energy-momentum. So spacetime is "deformed" (i.e. curved) by the electromagnetic field, and since matter (or any field) moves in that spacetime, the motion is influenced by the content of spacetime.


Yes, this is not only possible but ubiquitous. A portion of the Earth’s mass is due to electromagnetic field energy in the atoms that make it up. You feel the gravity due to this mass constantly. The fact that all mass-energy contributes to gravity is just the (strong) equivalence principle, which is obeyed by GR.


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