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The Maxwell equations for the electromagnetic Potential $A_\mu$ are (using Lorenz gauge). $$\nabla_\mu \nabla^\mu A^\nu - R^{\nu \mu} A_\mu = j^\nu$$ However the equivalence principle states:
No local experiment can distinguish between a freely falling nonrotating system and a uniformely moving system without gravitational field.
We usually realise this by choosing Riemannian Normal coordinates, which map geodesics on straight lines. But in the above equation the Ricci tensor enters, which cannot be transformed away by coordinate choice. So I think, that the solutions of the above equation should in general differ from the Minkoski case.

My question is: Is electromagnetism consistent with the equivalence principle? To be so, the new solutions for $A_\mu$ are not allowed to enter a physically measurable object. Does anyone know how to get rid of the Ricci-term using gauge transformation?

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There are many ways of stating the equivalence principle that are either exactly equivalent or not quite equivalent to each other. One way of stating it is that locally, spacetime can be treated as flat. In this limit of small distance scales, curvature is negligible, and we can neglect your $R^{\nu\mu}A_\mu$ term.

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  • $\begingroup$ I might be wrong, but this seems fishy. What if $\nabla^2 A^\mu$ is much smaller than the curvature term? Then locally you can't discard the curvature part. $\endgroup$ – Javier Dec 19 '17 at 21:46

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