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I have a question concerning the Wald's book: General Relativity. In the appendix E, he derived the Einstein equation by considering the surface term (GHY). I do not understand what he said after the equation (E.1.38).

Actually he considers that $h^{bc}\nabla_c(\delta g_{ab})=0$, because we fix $\delta g_{ab}=0$ on the surface, but therefore why the other term in (E.1.38) is not null, the term $h^{bc}\nabla_a(\delta g_{bc})$.

They look the same for me, and after some algebra, where we replace the covariant derivative by the one compatible with the metric on the surface we should have a total derivative term on the surface that we can integrate away.

Thanks in advance

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If you could quote the book more extensively it would help get an answer--- I know it is tedious to copy, but you might find a place to copy/paste online. – Ron Maimon Apr 13 '12 at 7:50
Hi anubis, and welcome to Physics Stack Exchange! If you've found the answer to your question, we'd love it if you write it up and post it using the answer text box below. Just editing your question to say that it's been solved isn't very helpful, so I've reverted that edit. – David Z Apr 13 '12 at 9:12

Derivatives along the boundary surface are fixed, since the formalism requires fixing coordinates on your initial time-slice, meaning that both the metric and the 3-connexion are fixed. Derivatives of the surface metric with respect to the normal vector pointing out of the boundary are not fixed, since this is, in fact, the direction of evolution in your Hamiltonian formalism.

Since the first term contains a contraction of the connexion along the intrinsic metric, it is inherently a derivative along the boundary surface, which has zero variation by the argument above. Since the second term has a free index on the derivative, it is not fixed, since it may still be contracted on the normal vector.

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Jerry Schirmer is right, I will give a little bit more details. – anubis Apr 14 '12 at 2:20

If you consider a vector tangential to the surface we have:

$$\nabla_\mu V^\mu=D_\mu V^\mu-V^\mu n^\nu\nabla_\nu n_\mu$$

where $D$ is the connexion compatible to the metric induced on the surface $h_{\mu\nu}$.

Hence we have for the first case

$$n^ah^{bc}\nabla_c \delta g_{ab}=\nabla_c(n^ah^{bc}\delta g_{ab})-\delta g_{ab}\nabla_c (n^a h^{bc})$$

The second term has the nice form, so I will not consider it for the next. Actually in my calculations, I did not fixed $\delta g_{\mu\nu}$ on the surface because I wanted to recover also the Junction conditions.

The first term gives, because the vector $n^ah^{bc}\delta g_{ab}$ is tangential to the surface.

$$\nabla_c(n^ah^{bc}\delta g_{ab})=D_c(n^ah^{bc}\delta g_{ab})-n^ah^{bc}\delta g_{ab} n^d\nabla_d n_c$$

Again the second term has the nice form, so it will contribute to the Junction condition, and the first term is a total derivative term on the surface. Hence we finished with the first term

We can't use the same calculations for the second term $n^a h^{bc}\nabla_a\delta g_{bc}$, because if we do

$$n^a h^{bc}\nabla_a\delta g_{bc}=\nabla_a(n^a h^{bc}\delta g_{bc})-\delta g_{bc}\nabla_a(n^a h^{bc})$$

The second term has the nice form, but we will have problems with the first one. In fact the vector field $n^a h^{bc}\delta g_{bc}$ is perpendicular to the surface, so we have

$$\nabla_a(n^a h^{bc}\delta g_{bc})=D_a(n^a h^{bc}\delta g_{bc})- h^{bc}\delta g_{bc} n^a n^d\nabla_d n_a+n^a\nabla_a(h^{bc}\delta g_{bc})$$

The last term can not be suppressed ... this is why we have to add a Gibbons-Hawking-York term on the surface to cancel this factor.

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