- #1

"Don't panic!"

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However, I'm unsure how one uses this fact to derive the (twice-contracted) Bianchi identity?

Straumann simply writes: $$\delta S=\int_{M}d^{4}x\sqrt{-g}\left(\frac{1}{\sqrt{-g}}\frac{\delta S_{EH}}{\delta g^{\mu\nu}}\right)\delta g^{\mu\nu}=\int_{M}d^{4}x\sqrt{-g}\,G_{\mu\nu}\delta g^{\mu\nu}=-\int_{M}d^{4}x\sqrt{-g}\,G^{\mu\nu}\delta g_{\mu\nu}$$ and notes that for an infinitesimal diff (generated by some vector field ##X##), ##\delta g_{\mu\nu}=2\nabla_{(\mu}X_{\nu)}##, such that $$\delta_{X} S=-2\int_{M}d^{4}x\sqrt{-g}\,G^{\mu\nu}\nabla_{\mu}X_{\nu}=2\int_{M}d^{4}x\sqrt{-g}\,X_{\nu}\nabla_{\mu}G^{\mu\nu}=0$$ and so, since ##X^{\nu}## is arbitrary, it must be that ##\nabla_{\mu}G^{\mu\nu}=0##.

What confuses me about this, is that one neglects the effect of the Lie derivative on ##d^{4}x## in this case (in the proof that the EH action is diff invariant, it was taken into account). Is the point that an infinitesimal diff is carried out in the *same* coordinate chart, and so ##d^{4}x## doesn't change in this case?