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/Einstein–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.

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

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