# Variation of Ricci tensor - perturbative gravitational waves

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.

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.
• I meant that $\Gamma^{a}_{bc} = \Gamma^{a}_{cb}$. Nov 25, 2021 at 7:19
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.