# Gravitational action for BTZ black hole

I am trying to calculate the gravitational action for a BTZ black hole (ie a Schwarzschild-AdS black hole with spacetime dimension D=3). Below I go through my working.

$$ds^2 = l^2fd\tau^2 + l^2f^{-1}dr^2 + l^2r^2d\phi^2$$

$$f(r) = r^2 - 8M \implies r_h^2 = 8M$$, $$l$$ = AdS radius.

The metric is from equation (2.23) of http://www.hartmanhep.net/topics2015/gravity-lectures.pdf.

See equation (6.21) in http://www.hartmanhep.net/topics2015/6-GravityPathIntegral.pdf for the following.

$$n^a = [0, -f^{1/2}/l,0]$$, the inward pointing unit normal to the boundary surface at $$r = r_0$$. A strange observation is that if this was the outward pointing unit normal, then we would find $$S_E = 0$$.

$$K_{\mu\nu} = \frac{1}{2}n^\alpha\partial_\alpha g_{\mu\nu} = -\frac{1}{2}\frac{f^{1/2}}{l^2}\partial_r g_{\mu\nu} = -\frac{f^{1/2}}{2l}diag(2l^2r, \frac{-2rl^2}{f^2}, 2rl^2)$$.

$$\implies K_{ab} = \frac{f_0^{1/2}}{2l}diag(2l^2r_0, 2r_0l^2)$$, where $$r_0$$ will be the cut-off radius.

Note $$a,b$$ are taking values $$\tau$$ and $$\phi$$.

$$h_{ab}dx^adx^b = l^2f_0d\tau^2 + l^2 r_0^2d\phi^2$$, which is the induced metric on the boundary surface of constant $$r = r_0$$.

$$K = h^{ab}K_{ab} = -(\frac{r_0}{lf_0^{1/2}} + \frac{f_0^{1/2}}{r_0l})$$, which is the extrinsic curvature of the boundary surface.

$$\sqrt{h}K = -2lr_0^2+8Ml$$.

$$R-2\Lambda = -4/l^2$$ and $$\sqrt{g} = l^3r$$.

$$S_E = -\frac{1}{16\pi}\int_0^\beta d\tau \int_{r_h}^{r_0} dr \int_0^{2\pi} d\phi \sqrt{g}(R-2\Lambda) -\frac{1}{8\pi}\int_0^\beta d\tau\int_0^{2\pi} d\phi \sqrt{h}K + \frac{a}{8\pi}\int_0^\beta d\tau\int_0^{2\pi} d\phi\sqrt{h}$$, where $$a$$ is chosen to cancel the divergence.

I end up with $$S_E = -4l\beta M$$. Now I can use the general result $$\beta = \frac{4\pi l^2}{f'(r_h)} \implies M = \frac{\pi^2 l^2}{2\beta^2}$$. If I try to calculate the average energy $$E = -\partial_\beta ln(Z) = \partial_\beta S_E$$, I get $$E = 4lM$$. The following link http://www.hartmanhep.net/topics2015/gravity-lectures.pdf above equation (6.21) seems to suggest I should have found $$E = M^2/8$$. Further, I find the entropy to be $$S = (1-\partial_\beta)ln(Z) = -(1-\partial_\beta)S_E = 4\pi l\sqrt{2M}$$, which I believe should have been $$S = 4\pi l \sqrt{\frac{M}{8}}$$.

Have I done anything wrong in this calculation?

Here is the correct calculation for the action of the BTZ black hole.

Set $$l = 1$$, let $$f(r) = r^2 - 8M$$. ($$l$$ is the AdS radius).

$$g = fd\tau^2 + f^{-1}dr^2 + r^2d\phi^2$$. (Initial metric).

$$h = f_0d\tau^2 + r_0^2d\phi^2$$. (Induced metric on boundary surface at $$r = r_0$$).

$$n_\alpha = [0, f_0^{-1/2}, 0]$$. (Outward looking unit normal to boundary surface at $$r = r_0$$, the components are $$\tau, r, \phi$$ in that order).

$$K_{\mu\nu} = \nabla_{(\mu}n_{\nu)} = \frac{1}{2}(\partial_\mu n_\nu + \partial_\nu n_\mu - 2\Gamma^\sigma_{\mu\nu}n_\sigma)$$. (Instrinsic curvature to boundary surface).

The first two terms in $$K_{\mu\nu}$$ are only non-zero when $$\mu = \nu = r$$ and so we can ignore them since we want $$K = h^{\mu\nu}K_{\mu\nu}$$ and $$h^{rr} = 0$$. To see the latter we use $$h_{\mu\nu} = g_{\mu\nu} - n_\mu n_\nu \implies h^{rr} = g^{rr}g^{rr}(g_{rr}-n_rn_r) = 0$$ using $$n_r = f^{-1/2}$$.

$$\Gamma^r_{\tau\tau} = -rf, \Gamma^r_{\phi\phi} = -rf$$. (To calculate Christoffel symbols using python see link in https://takisangelides.wixsite.com/personal/teaching).

$$K = h^{\mu\nu}K_{\mu\nu} = h^{\tau\tau}K_{\tau\tau} + h^{\phi\phi}K_{\phi\phi} = r_0f^{-1/2}_0 + \frac{f_0^{1/2}}{r_0}$$, using $$K_{\tau\tau} = -\Gamma^r_{\tau\tau}n_r$$ and $$K_{\phi\phi} = -\Gamma^r_{\phi\phi}n_r$$.

$$\sqrt{h}K = 2r_0^2-8M$$, $$R-2\Lambda = -4$$, $$\sqrt{g} = r$$.

$$S_E = -\frac{1}{16\pi}\int_0^\beta d\tau \int_{r_h}^{r_0} dr \int_0^{2\pi} d\phi \sqrt{g}(R-2\Lambda) -\frac{1}{8\pi}\int_0^\beta d\tau\int_0^{2\pi} d\phi \sqrt{h}K + \frac{a}{8\pi}\int_0^\beta d\tau\int_0^{2\pi} d\phi\sqrt{h}.$$

$$S_E = -\beta M$$.

For the above I have used $$a = 1$$ and $$r_h = \sqrt{8M}$$.

We now use the general result $$\beta = \frac{4\pi}{f'(r_h)}$$ to get $$\beta = \frac{\pi}{\sqrt{2M}}$$.

Using $$S_E = -ln(Z)$$, $$E = \partial_\beta S_E$$, $$S = (-1+\beta \partial_\beta)S_E$$, we get $$E = M$$ and $$S = \pi\sqrt{2M} = \frac{\pi r_h}{2}$$, which are the expected results.