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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}

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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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  • $\begingroup$ A question: why a wick rotation is need here? $\endgroup$ – Zoe Rowa Nov 1 '14 at 12:47
  • $\begingroup$ The path integral goes to partition function for a thermal quantum field theory $Z=\int D\phi e^{-I[\phi]}$(Where fields that contribute are only those that are periodic in $\tau=it$) instead of $Z=\int D\phi e^{iI[\phi]} $ which can be thought of as wick rotation of time. $I=\int\mathscr L d^4x$ So if you wick rotate the time coordinate,the $i$ that pops out of the action integral gets multiplied with $i$ in the action of normal quantum field theory to give you path integral of thermal quantum field. $\endgroup$ – vinaymmp Nov 7 '14 at 5:10
  • $\begingroup$ If you are asking for much deeper question of validity of euclidean path integral of quantum gravity then physics.stackexchange.com/questions/4932/… $\endgroup$ – vinaymmp Nov 7 '14 at 5:19

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