The Einstein-Hilbert action is given by,

$$I = \frac{1}{16\pi G} \int_{M} \mathrm{d}^d x \, \sqrt{-g} \, R \, \, + \, \, \frac{1}{8\pi G}\int_{\partial M} \mathrm{d}^{d-1}x \, \sqrt{-h} \, K$$

including the Gibbons-Hawking-York boundary term. A well-known derivation of the entropy of the Schwarzschild metric requires the evaluation of the boundary term. However, one must introduce a radial regulator $R$, and subtract off a counter-term which is the Gibbons-Hawking action of emtpy space with the same boundary. The final result is finite as $R\to\infty$.

I am attempting to calculate the full action for a solution for which $R\neq0$, hence I need to also compute the pure Einstein-Hilbert piece. However, I must introduce a regulator, and the final action is not finite as I take it to infinity. My question: **is there an analogous procedure to "tame" the infinity of the pure Einstein-Hilbert piece, perhaps similar to the treatment of the Gibbons-Hawking term?**

I actually have to introduce two regulators. For my solution, the Ricci scalar is independent of a particular coordinate, $x_1$, so I get a factor of $x_1$ after integration evaluated at $\pm\infty$, so I introduced regulators, such that $L^{-} < x_1 < L^{+}$. As I take $L^{\pm}\to \pm \infty$, of course it is divergent.

This post imported from StackExchange Physics at 2014-05-04 11:36 (UCT), posted by SE-user user2062542