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Are there any references that present the explicit variation of the Hilbert-Einstein action plus the Hawking-Gibbons-York boundary term, and demonstrate the cancellation of the normal derivatives of metric variations? I have tried to read the original papers by York and Gibbons&Hawking, but they are not that pedagogical to me... Thanks to anyone who can help me!

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I've never seen a paper where the calculation is performed in a manifestly covariant manner. However, I've posted a set of reference notes on my website (http://jacobi.luc.edu/notes.html) that contains the variations needed to carry out the calculation. Let me summarize the calculation here.

The action for gravity on a compact region $M$ with boundary $\partial M$ is $$I_{EH} + I_{GHY} = \frac{1}{2 \kappa^2} \int_{M}d^{d+1}x \sqrt{-g} R + \frac{1}{\kappa^2} \int_{\partial M} d^{d}x \sqrt{-h} K ~.$$ The metric on $M$ is $g_{\mu\nu}$, and $R = g^{\mu\nu} R_{\mu\nu}$ is the Ricci Scalar. The induced metric on the boundary $\partial M$ is $h_{\mu\nu} = g_{\mu\nu} - n_{\mu} n_{\nu}$, where $n^{\mu}$ is the (spacelike) unit vector normal to $\partial M \subset M$. Now consider a small variation in the metric: $g_{\mu\nu} \to g_{\mu\nu} + \delta g_{\mu\nu}$. The quantities appearing in the Einstein-Hilbert part of the action change in the following manner: $$ \delta \sqrt{-g} = \frac{1}{2} \sqrt{-g} g^{\mu\nu} \delta g_{\mu\nu}$$ $$ \delta R = -R^{\mu\nu} \delta g_{\mu\nu} + \nabla^{\mu}\left(\nabla^{\nu} \delta g_{\mu\nu} - g^{\nu\lambda} \nabla_{\mu} \delta g_{\nu\lambda} \right)$$ Thus, the change in $I_{EH}$ is $$\begin{aligned}\delta I_{EH} = & \frac{1}{2\kappa^{2}}\int_{M} d^{d+1}x \sqrt{-g} \left(\frac{1}{2} g^{\mu\nu} R - R^{\mu\nu} \right)\delta g_{\mu\nu}\\ & + \frac{1}{\kappa^2} \int_{\partial M} d^{d}x \sqrt{-h} \frac{1}{2} n^{\mu} \left(\nabla^{\nu} \delta g_{\mu\nu} - g^{\nu\lambda} \nabla_{\mu} \delta g_{\nu\lambda}\right)~,\end{aligned}$$ with the boundary term coming from the volume integral of the total derivative in $\delta R$. The variations of the quantities in the GHY term are a bit more complicated to work out, but they all basically follow from standard definitions and this result for the variation of the normal vector: $$\delta n_{\mu} = \frac{1}{2} n_{\mu} n^{\nu} n^{\lambda} \delta g_{\nu\lambda} = \frac{1}{2} \delta g_{\mu\nu} n^{\nu} + c_{\mu}~.$$ In the second equality I've introduced a vector $c_{\mu}$ that is orthogonal to $n^{\mu}$; it is given by $$c_{\mu} = - \frac{1}{2} h_{\mu}{}^{\lambda} \delta g_{\nu\lambda} n^{\nu} ~.$$ The reason I've introduced this vector is that the variation in the trace of the extrinsic curvature can be written as $$\delta K= - \frac{1}{2} K^{\mu\nu} \delta g_{\mu\nu} - \frac{1}{2} n^{\mu}\left(\nabla^{\nu} \delta g_{\mu\nu} - g^{\nu\lambda} \nabla_{\mu} \delta g_{\nu\lambda} \right) + D_{\mu} c^{\mu}$$ where $D_{\mu}$ is the covariant derivative along $\partial M$ that is compatible with the induced metric $h_{\mu\nu}$. So, the change in the GHY part of the action is $$\delta I_{GHY} = \frac{1}{\kappa^2} \int_{\partial M} d^{d}x \sqrt{-h}\left(\frac{1}{2}h^{\mu\nu} \delta g_{\mu\nu} K + \delta K \right)~.$$ Combining this with $\delta I_{EH}$ we see that the several terms cancel, leaving $$\begin{aligned} \delta I = & \frac{1}{2\kappa^2}\int_{M} d^{d+1}x \sqrt{-g}\left(\frac{1}{2} g^{\mu\nu} R - R^{\mu\nu} \right)\delta g_{\mu\nu}\\ & + \frac{1}{\kappa^2}\int_{\partial M} d^{d}x \sqrt{-h}\left(\frac{1}{2}(h^{\mu\nu} K - K^{\mu\nu})\delta g_{\mu\nu} + D_{\mu} c^{\mu} \right) ~.\end{aligned}$$ We discard the term $D_{\mu} c^{\mu}$, which is a total boundary derivative.

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Thank you very much for your detailed reply. But still I don't understand why should the normal vector $n_\mu$ transform according to the formula above? I heard people saying that the transformation is designed to preserve the unit length of $n_\mu$, but I think the sufficient condition for this requirement is simply $n^\mu \delta n_\mu=\frac{-1}{2}n^\mu n^\nu \delta g_{\mu\nu}$, which cannot determine $\delta n_\mu$ completely. Thanks again for the help! –  Michael Shaw Jun 1 '11 at 3:27
    
You're right -- simply preserving $n^{\mu}n_{\mu} = 1$ would only determine $\delta n^{\mu}$ up to a vector orthogonal to $n^{\mu}$. Instead, you should consider the definition of $n^{\mu}$. Let $\partial M$ be an isosurface of some coordinate $r$. Then $\alpha_{\mu} = \nabla_{\mu} r$ is orthogonal to the surface. Now normalize that vector to obtain: $$ n_{\mu} = \frac{\alpha_{\mu}}{\sqrt{g^{\nu\lambda}\alpha_{\nu}\alpha_{\lambda}}}$$ Consider how this expression behaves under a small variation of the metric and you obtain the result for $\delta n_{\mu}$. –  Robert McNees Jun 3 '11 at 4:03
    
Aha, I see! Many thanks again! –  Michael Shaw Jun 7 '11 at 8:34
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