Help in understanding the derivation of Einstein equations

Click For Summary
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.
user1139
Messages
71
Reaction score
8
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?
 
Physics news on Phys.org
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}##
 
  • Like
Likes vanhees71 and PeroK
At first, I derived that: $$\nabla \frac 1{\mu}=-\frac 1{{\mu}^3}\left((1-\beta^2)+\frac{\dot{\vec\beta}\cdot\vec R}c\right)\vec R$$ (dot means differentiation with respect to ##t'##). I assume this result is true because it gives valid result for magnetic field. To find electric field one should also derive partial derivative of ##\vec A## with respect to ##t##. I've used chain rule, substituted ##\vec A## and used derivative of product formula. $$\frac {\partial \vec A}{\partial t}=\frac...