Hawking Temperature of the BTZ Black Hole

Question

The metric of the BTZ Black Hole is given by $$ ds^2 = - N^2 dt^2 + N^{-2} dr^2 +r^2(d\phi + N^\phi dt)^2 $$ with $$ N^2 = -M+ \frac{r^2}{l^2} + \frac{J^2}{4 r^2}, \ \ \ \ \ \ N^\phi = -\frac{J}{2r} $$ The $g_{rr}$ component of the metric is singular at points where $N^2=0$, yielding the horizons $r_\pm$ $$ r_\pm = \sqrt{ \frac{Ml^2}{2}\left( 1 \pm \sqrt{1-\left(\frac{J}{Ml}\right)^2} \right)} $$ Then for these $r_\pm$, the metric component $g_{tt}$ does not vanish but becomes $$ g_{tt} =\frac{J^2}{4r_\pm^2} $$ Now the perscription I learned to find a Hawking Temperature at a horizon (e.g. Schwarzschild BH) is you expand the Wick rotated metric ($t\to i\tau$) around the solution where $g_{\tau\tau}$ vanishes, find the metric is actually flat at this point, and impose $\tau$-periodicity such that there is no conical singularity at the horizon. This period is then the inverse Hawking temperature $T_H^{-1}$ .

I don't see any singularities in the $g_{\tau\tau}$ component right now so no conical singularity will appear, and I don't know how to interpret this. Does this mean there is no restriction on $T_H$ and the $\tau$ periodicity is free? Or is my way to calculate it somehow not applicable to BTZ horizons?

Answer 1

I think I found a coordinate transformation that shines more light on this. Instead of looking for a conical singularity in $g_{tt}$ alone I must look at conical singularities in any part of the metric (apart from $g_{rr}$) First of all the metric can be expressed in terms of $r_\pm$ and Wick rotated $t\to it_E$ such that $$ ds_{E}^2 = \frac{ (r^2-r_+^2)(r^2+r_-^2)}{l^2r}dt_E^2 + \frac{l^2 r^2}{(r^2-r_+^2)(r^2+r_-^2)} dr^2 + r^2(d\phi + \frac{i r_+ r_-}{l r^2} dt_E)^2 $$ which under the coordinate transformation $$ t'_E = r_+ t_E + r_-\phi, \ \ \ \phi' = r_+ \phi - r_-t_E, \ \ \ r'^2= \frac{r^2-r_+^2}{r_+^2+r_-^2} $$ becomes $$ ds_E^2 = \frac{r'^2}{l^2} dt_E'^2 +\frac{l^2}{1+r'^2} dr'^2 + (1+r'^2)d\phi'^2 $$ which for $r\to r_+$ ($r'\to 0$) becomes $$ ds_E^2 = r'^2 dt_E'^2+dr'^2 +d\phi'^2 $$ which represents flat polar coordinates iff $t_E' \sim t_E' +2\pi$. Furthermore $\phi'$ is not periodic. So the periodicity of $t_E'$ is $\Delta t_E'=2\pi$, and of $\phi'$ is $\Delta \phi'=0$. Combining this yields $$ 2\pi = r_+ \Delta t_E + r_-\Delta \phi, \ \ \ \ 0 = r_+\Delta \phi - r_- \Delta t_E $$ So the time periodicity becomes $\beta =\Delta t_E = \frac{2\pi l r_+}{r_+^2+r_-^2}$ whilst also setting a fixed periodicity for $\phi$. Obviously the Hawking temperature is then $$ T_H = \frac{r_+^2+r_-^2}{2\pi l r_+} $$

Answer 2

Another option for finding the Hawking temperature of the BTZ black hole is by the formalism of the surface gravity $\kappa$ which connected to the Hawking temperature via:

$$T_H=\frac{\kappa}{2\pi}$$ According to Wald GR book (section 12.5 page 332) the surface gravity is given by:

$$\kappa =\lim_{r\to r_0}(aV)$$ Where: $$V=\sqrt{-\chi^a\chi_{a}}$$ $$a = \sqrt{\alpha_c\alpha^c},\quad \alpha_c=\frac{\partial_cV^2}{2V^2}$$

$r_0$ is the event horizon (the outer one in a rotating and charged black holes) corresponds to some bifurcate Killing horizon which have a compatible Killing vector $\chi^a$.

In the case of the BTZ black hole, we begin with the metric: $$ ds^2 = -\frac{ (r^2-r_+^2)(r^2-r_-^2)}{l^2r^2}dt^2 + \frac{l^2 r^2}{(r^2-r_+^2)(r^2-r_-^2)} dr^2 + r^2\left(d\theta - \frac{r_+ r_-}{l r^2} dt\right)^2 $$ Which can be written in a compact way: $$ ds^2 = -N^2dt^2 + \frac{1}{N^2} dr^2 + r^2\left(d\theta - N^\theta dt\right)^2 $$ Then we move to Eddington–Finkelstein coordinates: $$\nu = t + \frac{1}{N^2}r,\quad \phi = \theta + \frac{N^\theta}{N^2}r$$ In which the metric becomes: $$ds^2 = -N^2dt^2 + 2d\nu dr + r^2\left(d\phi - N^\theta d\nu\right)^2 $$ In those coordinated $(\nu,r,\phi)$ it is easier to see that $r=r_+$ is an horizon which corresponds to a surface with Killing vector which is given by: $$\chi^a=(1,0,-N^\theta(r_+))$$ or just $$\chi^a=\partial^\nu - N^\theta(r_+)\partial^\phi$$

We will substitute inside $V$ and get: $$V=\sqrt{-g_{\nu\nu}\chi^{\nu}\chi^{\nu}-2g_{\nu\phi}\chi^{\nu}\chi^{\phi}-g_{\phi\phi}\chi^{\phi}\chi^{\phi}}=\sqrt{\frac{\left(r_{+}^{2}-r_{-}^{2}\right)\left(r^{2}-r_{+}^{2}\right)}{l^{2}r_{+}^{2}}}$$ Thus: $$a = \sqrt{g^{rr}\alpha_r^2} =\frac{\partial_{r}V^{2}}{2V^{2}}\sqrt{N^2} = \frac{r}{r^{2}-r_{+}^{2}}\sqrt{\frac{ (r^2-r_+^2)(r^2-r_-^2)}{l^2r^2}}$$ Where we took $g^{rr}$ to be the inverse of the radial part of the original metric. overall we got: $$\kappa = \lim_{r\to r_+}(aV)= \lim_{r\to r_+}\left(\frac{r}{r^{2}-r_{+}^{2}}\sqrt{\frac{ (r^2-r_+^2)(r^2-r_-^2)}{l^2r^2}}\sqrt{\frac{\left(r_{+}^{2}-r_{-}^{2}\right)\left(r^{2}-r_{+}^{2}\right)}{l^{2}r_{+}^{2}}}\right)=\frac{r_{+}^{2}-r_{-}^{2}}{l^{2}r_{+}}$$

Thus the Hawking temperature is given by:

$$T_H=\frac{r_{+}^{2}-r_{-}^{2}}{2\pi l^{2}r_{+}}$$