# ADM mass calculation for the BTZ black hole

Considering a non-rotating and non-charged 2+1 dimensional black hole, known as the BTZ black hole which obtained by adding a negative cosmological constant $$\Lambda=-\frac{1}{l^2},l\ne0$$ to the Einstein-Hilbert action, resulting the following metric:

$$ds^2=-\left(\frac{r^2}{l^2}-M\right)dt^2+\left(\frac{r^2}{l^2}-M\right)^{-1}dr^2+r^2d\theta^2,\quad \theta\in[0,2\pi]$$

I want to prove that $$M$$ which appears in the metric is the ADM mass of the BTZ black hole. In order to use the ADM mass formula one should calculate the following quantities: The normal vector to the infinite sphere (circle in our case), given by: $$n^i=(n^r,n^\theta)=\left(\frac{1}{\sqrt{g_{rr}}},0\right)=\left(\sqrt{\frac{r^2}{l^2}-M},0\right)$$ Using that, one can calculate the trace of the extrinsic curvature: $$k=D_in^i=\partial_rn^r+(\Gamma^r_{rr}+\Gamma^\theta_{\theta r})n^r$$ Which is given by: $$k=\frac{1}{r}{\sqrt{\frac{r ^2}{l^2}-M}}$$ $$k_0$$ The extrinsic curvature when there is no source is given by: $$k_0=k|_{M=0}=\frac{1}{l}$$ The ADM mass is given by doing the following surface integral at infinite radius:

$$M_{ADM}=\frac{1}{16\pi}\int^{2\pi}_{0}d\theta\sqrt{g_{\theta\theta}g_{rr}}(k-k_0)\Big{|}_{r\to\infty}$$

Which eventually yields to:

$$M_{ADM}=\frac{ {\sqrt{\frac{r ^2}{l^2}-M}}-\frac{r}{l}}{8 \sqrt{\frac{r ^2}{l^2}-M}}\Big{|}_{r\to{\infty}}=0$$

I don$$`$$t know where it is wrong.