I should calculate the variation of the Ricci scalar to the metric ##\delta R/\delta g^{\mu\nu}##. According to ##\delta R=R_{\mu\nu}\delta g^{\mu\nu}+g^{\mu\nu}\delta R_{\mu\nu}##, ##\delta R_{\mu\nu}## should be calculated. I have referred to the wiki page: http://en.wikipedia.org/wiki/Einsteinâ€“Hilbert_action which provided the deduction below(adsbygoogle = window.adsbygoogle || []).push({});

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

and this term can be converted to a total derivative when it's multiplied by ##\sqrt{-g}##:

$$\sqrt{-g}\nabla_{\sigma}(g^{\mu\nu}\delta\Gamma^{\sigma}_{\nu\mu}-g^{\mu\sigma}\Gamma^{\rho}_{\rho\mu})=\partial_{\sigma}[\sqrt{-g}(g^{\mu\nu}\delta\Gamma^{\sigma}_{\nu\mu}-g^{\mu\sigma}\Gamma^{\rho}_{\rho\mu})].$$

When we integrate it over the whole spacetime andassume that the variation of the metric ##\delta g^{\mu\nu}## vanishes at infinity, it will contribute nothing to the variation of the action ##\int dx^{4}\sqrt{-g}f(R)##.

However i'm considering things in the spatially flat Friedmann universe ##ds^{2}=dt^{2}-a^2(t)\delta_{ik}dx^{i}dx^{k}.## The assumption above can't be true any more. To make my question specific, could you compute ##\delta R/\delta g^{\mu\nu}## in the spatially flat Friedmann universe? Thanks a lot.

**Physics Forums | Science Articles, Homework Help, Discussion**

Join Physics Forums Today!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

# Help with the variation of the Ricci tensor to the metric

**Physics Forums | Science Articles, Homework Help, Discussion**