# #2 – Hilbert – Einstein action – Part 2

Now we compute $\frac{\delta R}{\delta g_{\mu\nu}}$. First, the Riemann tensor has the following definition:

${R^{\rho}}_{\sigma\mu\nu}=\partial_{\mu}{\Gamma^{\rho}}_{\nu\sigma}-\partial_{\nu}{\Gamma^{\rho}}_{\mu\sigma}+ {\Gamma^{\rho}}_{\mu\lambda}{\Gamma^{\lambda}}_{\nu\sigma}-{\Gamma^{\rho}}_{\nu\lambda}{\Gamma^{\lambda}}_{\mu\sigma},$

so

$\delta{R^{\rho}}_{\sigma\mu\nu}=\partial_{\mu}\delta{\Gamma^{\rho}}_{\nu\sigma}-\partial_{\nu}\delta{\Gamma^{\rho}}_{\mu\sigma}+\delta{\Gamma^{\rho}}_{\mu\lambda}{\Gamma^{\lambda}}+{\Gamma^{\rho}}_{\mu\lambda}\delta{\Gamma^{\lambda}}_{\nu\sigma}-\delta{\Gamma^{\rho}}_{\nu\lambda}{\Gamma^{\lambda}}_{\mu\sigma}-{\Gamma^{\rho}}_{\nu\lambda}\delta{\Gamma^{\lambda}}_{\mu\sigma}.$

$\Gamma$ is not a tensor, but since $\delta\Gamma$ is the difference of two connections, it is a tensor. We can compute the quantity $\nabla_{\lambda}(\delta{\Gamma^{\rho}}_{\nu\mu})$:

$\nabla_{\lambda}(\delta{\Gamma^{\rho}}_{\nu\mu})=\partial_{\lambda}(\delta{\Gamma^{\rho}}_{\nu\mu})+{\Gamma^{\rho}}_{\sigma\lambda}\delta{\Gamma^{\sigma}}_{\nu\mu}-{\Gamma^{\sigma}}_{\nu\lambda}\delta{\Gamma^{\rho}}_{\sigma\mu}-{\Gamma^{\sigma}}_{\mu\lambda}\delta{\Gamma^{\rho}}_{\nu\sigma}$

We can identify then that $\delta{R^{\rho}}_{\sigma\mu\nu}$ is just

$\delta{R^{\rho}}_{\sigma\mu\nu}=\nabla_{\mu}(\delta{\Gamma^{\rho}}_{\nu\sigma})-\nabla_{\nu}(\delta{\Gamma^{\rho}}_{\mu\sigma}).$

Moreover:

$\delta R_{\mu\nu}=\delta R^{\rho}_{\mu\rho\nu}=\nabla_{\rho}(\delta{\Gamma^{\rho}}_{\nu\mu})-\nabla_{\nu}(\delta{\Gamma^{\rho}}_{\rho\mu}),$

so

$\delta R=R_{\mu\nu}\delta g^{\mu\nu}+g^{\mu\nu}\delta R_{\mu\nu}=R_{\mu\nu}\delta g^{\mu\nu}+g^{\mu\nu}\left(\nabla_{\rho}(\delta{\Gamma^{\rho}}_{\nu\mu})-\nabla_{\nu}(\delta{\Gamma^{\rho}}_{\rho\sigma})\right)=R_{\mu\nu}\delta g^{\mu\nu}+$

$+\nabla_{\sigma}\left(g^{\mu\nu}\delta{\Gamma^{\sigma}}_{\mu\nu}-g^{\sigma\mu}\delta{\Gamma^{\rho}}_{\rho\mu}\right).$

The last term is a boundary term that does not contribute when integrated. We obtain the result

$\delta R=R_{\mu\nu}\delta g^{\mu\nu}.$

Finally, the Einstein equations are:

$T^{\mu\nu}=\frac{1}{\kappa}\left(R^{\mu\nu}-\frac{1}{2}Rg^{\mu\nu}\right)\ \implies\ R^{\mu\nu}-\frac{1}{2}Rg^{\mu\nu}=\kappa T^{\mu\nu}.$