Value of the invariant $R_{\mu \nu}F^{\mu \nu}$

Is there a simple way to find the value of $R_{\mu \nu}F^{\mu \nu}$ (where $R_{\mu \nu}$ is the Ricci tensor and $F^{\mu \nu}$ is the electromagnetic tensor), knowing that it is an invariant?

Inputting the definitions of $R_{\mu \nu}=R^{\lambda}_{\mu \lambda \nu}= \partial_\lambda \Gamma^\lambda_{\mu \nu}-\partial_\nu \Gamma^{\lambda}_{\lambda \mu}+\Gamma^\lambda_{\lambda \sigma}\Gamma^{\sigma}_{\mu \nu}-\Gamma^\lambda_{\nu \sigma} \Gamma^\sigma_{\lambda \mu}$ and $F^{\mu \nu}=\partial^\mu A^\nu - \partial^\nu A^\mu$ would be a straightforward way, but I have a hunch that there is a much more simple solution if I could interpret the expression physically or transform it to a specific frame of reference where it is easy to evaluate.

• It's zero because $R_{\mu \nu}$ is symmetric while $F^{\mu \nu}$ is antisymmetric. – Avantgarde Aug 24 '18 at 11:10
• @Avantgarde Oh my, that was more obvious than I had expected. Thanks a bunch. – sivlomme Aug 24 '18 at 11:21
• @Avantgarde given that your comment is the full answer to this question, it might be worth it to just make it an answer. – enumaris Aug 24 '18 at 23:09
• If you are interested in nonzero scalars composed from EM and gravitational fields look at the invariants with $R F F$ structure: $R F^{\mu\nu}F_{\mu\nu}$, $R_{\mu\nu}F^{\mu\lambda}F^{\nu} {}_\lambda$, $R_{\mu\nu\lambda\rho}F^{\mu\nu}F^{\lambda\rho}$. These scalars appeared, for example, in the context of superluminal propagation of photons in curved spacetimes discovered by Drummond & Hathrell. – A.V.S. Aug 25 '18 at 6:43

$R_{\mu \nu}F^{\mu \nu}$ is simply zero. No computations are needed to see this. Just note that $R_{\mu \nu}$ is a symmetric tensor while $F^{\mu \nu}$ is antisymmetric.
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