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I'm trying to calculate $\delta R_{\mu\nu}$ and prove that it is zero in empty space. I have calculated the variation of the Riemann tensor and substituted in the Christoffel symbols and arrived at this equation.

From $\delta R_{\mu\nu}=\delta R^\alpha_{\mu \alpha \nu}$ I have:

$$ \delta R_{\mu\nu} = \frac{1}{2}g^{\alpha \delta}\left(\nabla_\mu\nabla_\nu\delta g_{\delta\alpha}+\nabla_\mu\nabla_\alpha\delta g_{\delta\nu}-\nabla_\mu\nabla_\delta\delta g_{\nu\alpha}-\nabla_\nu\nabla_\mu\delta g_{\delta\alpha}-\nabla_\nu\nabla_\alpha\delta g_{\delta\mu}+\nabla_\nu\nabla_\delta \delta g_{\mu\alpha}\right). $$

The identities I can think of are $R_{\mu\nu}=0$ and $\nabla_\rho g_{\mu\nu}$ = 0 since this is a vacuum solution of the Einstein equations. The second derivative of the metric cannot be set to zero so this must be a manipulation of the indices that I'm not seeing.

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It is zero. The first and the fourth term cancel out(if there is no torsion you can change the covariant derivatives). Then use the metric outside the parenthesis to make $\nabla_{α}$ and $\nabla_{δ}$, $\nabla^{α}$ and $\nabla^{δ}$ respectively and if you make the change $\alpha \rightarrow \delta$ and $\delta \rightarrow \alpha $ in the remaining terms you're there.

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    $\begingroup$ Thanks! I had not realised how close I actually was! $\endgroup$ Jan 4, 2020 at 19:32
  • $\begingroup$ "if there is no torsion you can change the covariant derivatives" --> Did you mean curvature instead of torsion? $\endgroup$
    – Andrew
    Nov 24, 2021 at 19:33
  • $\begingroup$ I meant that $\Gamma^{a}_{bc} = \Gamma^{a}_{cb}$. $\endgroup$
    – Noone
    Nov 25, 2021 at 7:19
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Your expression for $\delta R_{\mu \nu}$ is incorrect, and it is incorrect that $\delta R_{\mu \nu}$ vanishes on a flat background. See https://en.wikipedia.org/wiki/Einstein%E2%80%93Hilbert_action.

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