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}


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?

| cite | improve this answer | |
  • $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.