Help in understanding the derivation of Einstein equations

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The discussion centers on understanding the derivation of the Einstein equations, specifically regarding the variation of the Riemann tensor and the integral of a divergence term. The first part addresses the condition that leads to the equality involving the Christoffel symbols, emphasizing their symmetry in the lower indices. The second part explains why the integral of the divergence term vanishes, citing the divergence theorem which allows the total derivative to be evaluated over the boundary of spacetime. The participants clarify that the integral is zero due to its nature as a total derivative, confirming the application of the divergence theorem. Overall, the conversation highlights key mathematical properties essential for deriving the Einstein equations.
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Homework Statement
I am working through the derivation of the Einstein field equations by varying the Einstein-Hilbert action. I need some help in understanding certain steps
Relevant Equations
Given below
There are two parts to my question.

The first is concerns the variation of the Reimann tensor. I am trying to show

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

In order to show the above, it is necessary that ##\Gamma^{\lambda}_{\phantom{\lambda}\nu\mu}\delta\Gamma^{\rho}_{\phantom{\rho}\lambda\sigma}-\Gamma^{\lambda}_{\phantom{\lambda}\mu\nu}\delta\Gamma^{\rho}_{\phantom{\rho}\lambda\sigma}=0##. Why is this true?

The second part concerns the term ##\int\nabla_{\rho}A^{\rho}\sqrt{-g}\,\mathrm{d}^4x=0## where ##A^{\rho}=g^{\sigma\nu}\delta\Gamma^{\rho}_{\phantom{\rho}\nu\sigma}-g^{\sigma\rho}\delta\Gamma^{\mu}_{\phantom{\mu}\mu\sigma}##. Why is the integral zero?
 
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Thomas1 said:
In order to show the above, it is necessary that ##\Gamma^{\lambda}_{\phantom{\lambda}\nu\mu}\delta\Gamma^{\rho}_{\phantom{\rho}\lambda\sigma}-\Gamma^{\lambda}_{\phantom{\lambda}\mu\nu}\delta\Gamma^{\rho}_{\phantom{\rho}\lambda\sigma}=0##. Why is this true?
It's been a long time since I took GR, so I may be misremembering. But aren't Christoffel symbols symmetric in the bottom two indices?
 
Thomas1 said:
The second part concerns the term ##\int\nabla_{\rho}A^{\rho}\sqrt{-g}\,\mathrm{d}^4x=0## where ##A^{\rho}=g^{\sigma\nu}\delta\Gamma^{\rho}_{\phantom{\rho}\nu\sigma}-g^{\sigma\rho}\delta\Gamma^{\mu}_{\phantom{\mu}\mu\sigma}##. Why is the integral zero?
It's a total derivative so vanishes by the divergence theorem upon integration over the whole spacetime. First show that\begin{align*}
\int d^4 x \sqrt{-g} \nabla_{\rho} A^{\rho} = \int d^4 x \partial_{\rho} (\sqrt{-g} A^{\rho})
\end{align*}Then use the divergence theorem of ordinary calculus\begin{align*}
\int d^4 x \partial_{\rho} (\sqrt{-g} A^{\rho}) = \oint d^3 x \sqrt{-\gamma} n_{\rho} A^{\rho}
\end{align*}where the integral on the rhs is taken over the boundary 3-surface of induced metric ##\gamma_{ab}## and normal ##n_{\rho}##
 
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