[Physics] Gibbons-Hawking-York (GHY) boundary term for Schwarzschild metric

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What is the simplest way to calculate Gibbons-Hawking-York boundary term for Schwarzschild metric?

\begin{align}
\int Kd\Sigma&=-32\pi^2m\left(1-2Mr^{-1}\right)^{1/2}\times\frac{d}{dr}\left[ir^2\left(1-2Mr^{-1}\right)^{1/2}\right]\\
&=-32\pi^2iM\left(2r-3M\right).
\end{align}

Best Answer

The Gibbons-Hawking boundary term for a spacetime manifold is explicitly,

$$S_{GH}=\frac{1}{8\pi G}\int_{\partial M} d^3x \, \sqrt{|h|} \, K$$

where $\partial M$ is the boundary of $M$, $K$ the extrinsic curvature, and $h$ the determinant of the metric on the boundary. Let us Wick rotate the Schwarzschild metric to,

$$ds^2 = \left( 1-\frac{2GM}{r}\right)d\tau^2 + \left( 1-\frac{2GM}{r}\right)^{-1} dr^2 + r^2 d\Omega^2$$

We must impose a radial cutoff $R > GM$. The normal vector on the boundary is given by,

$$n=-\sqrt{1-\frac{2GM}{r}}\frac{\partial}{\partial r}$$

with a minus sign since we require the outward pointing normal, which points into the bulk. The metric on the boundary is then given by,

$$ds^2=\left( 1-\frac{2GM}{R}\right)d\tau^2 + R^2d\Omega^2$$

The extrinsic curvature is simply the divergence of the normal:

$$K=\nabla_a n^a = \frac{1}{r^2}\partial_r (r^2 n^r) \biggr\rvert_{r=R}= -\frac{2}{R}\sqrt{1-\frac{2GM}{R}} - \frac{GM}{R^2} \frac{1}{\sqrt{1-\frac{2GM}{R}}}$$ Can you take the calculation from here?

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General Relativity – Explicit Variation of Gibbons-Hawking-York Boundary Term

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.

[Physics] 2nd order variation of Hilbert-Einstein action + Gibbons-Hawking-York boundary term

A long time ago, when I was in graduate school, a friend and I went to Bryce DeWitt with a similar question. He thought about it for a while, and then said to us:

 "Do you know what your problem is? Too much book learning."

He explained that the answer could almost certainly be found in a book or paper, if we looked hard enough, but we would learn much more by working out the answer ourselves. This may be the best advice I ever received as a graduate student.

Why am I telling you this? Because the answer to your question about the first variation of the action (and the notes I linked to) provides you with everything you need to solve this problem. Give it a try! If you get stuck on a specific step in the derivation, come back here and ask about it -- I'll be happy to work through the details with you. But this is a great chance for you to apply what you learned from your last question.

Best Answer

Related Solutions

General Relativity – Explicit Variation of Gibbons-Hawking-York Boundary Term

[Physics] 2nd order variation of Hilbert-Einstein action + Gibbons-Hawking-York boundary term

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