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2

There is a really nice derivation of this identity using differential forms, and it completely avoids all the messiness of the Christoffel symbols. The nice thing about differential forms is that the exterior derivative can be computed using any derivative operator, so it allows us to compare the expressions we get using the covariant derivative to the ...

2

This is based on the observation that, given some vector $V^\mu$, $$\nabla_\mu V^\mu=\frac{1}{\sqrt{-g}}\partial_\mu(\sqrt{-g}V^\mu)$$ We can show explicitly that this is true: $$\nabla_\mu V^\mu=\partial_\mu V^\mu +\Gamma^\mu_{\mu\lambda}V^\lambda$$ Let's examine the last term: $$\Gamma^\mu_{\mu\lambda}=\frac{1}{2}g^{\mu\rho}(\partial_\mu ... 2 Given a pseudo-Riemannian manifold (M,g), the Laplace-Beltrami operator acts on scalar functions. The formula for the Laplace-Beltrami operator follows from the formula$$\Gamma^{\nu}_{\mu\nu}=\partial_{\mu}\ln\sqrt{|g|}$$for the Levi-Civita connection. 4 What you said is only true if the hypersurface is space-like or time-like. If a non-null hypersurface is defined by f(x) =  constant, then the normal to the hypersurface is given by$$ n_\alpha \propto \partial_\alpha f $$The fact that the hypersurface is non-null implies$$ g^{\alpha\beta} \partial_\alpha f \partial_\beta f = \varepsilon\neq 0  ...

0

The Einstein equations are some of the most complicated PDE's people study. There is no shortcut for this, you just have to do all the horrific algebra. Start with the trace-reversed Einstein equation $$R_{\mu \nu}=8\pi G(T_{\mu \nu}-\frac{1}{2}Tg_{\mu \nu})$$ Use the equation for Ricci in terms of the Christoffel Connection ...

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I suggest you to take a look at Appendix E of Wald's General relativity book. There he derives all boundary terms which appear in the variation of Hilbert's action. There are only 3 terms coming from this variation. Two of them give Einstein's equation. The surface term comes from the other term, $g^{ab}\delta R_{ab}$, which is a total derivative, as ...

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It is, in a sense, just semantics but I'd say the natural choice is $\mathcal{L}=\sqrt{-g}\times \text{something}$. If you take this definition, the general form of the equations of motion is the same as when doing QFT in Minkowski, with the appropriate generalizations to account for curvature. Furthermore, I think it is standard practice to define the ...

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