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?


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

  • $\begingroup$ did you verify if on the extremal limit that temperature goes to zero? $\endgroup$ – diffeomorphism Jun 10 '15 at 22:40
  • $\begingroup$ Ah, I was not complete enough in my euclidean description but in fact it does. In euclidean coordinates the $ r_- $ is still the value given above but purely imaginary. So when this BH is extremal @ $ Ml=J$, the numerator cancels out. Is this vanishing of the hawking temperature something one always sees when taking a BH to exremal limits? Is there any intuition on why this is the case? $\endgroup$ – 123hoedjevan Jun 11 '15 at 0:00
  • $\begingroup$ no intuition, but it seems to be consistent with other calculations of temperatures for spinning and charged black holes $\endgroup$ – diffeomorphism Jun 11 '15 at 0:05
  • $\begingroup$ actually, there is an intuitive reason: extremality means that a large part of the mass of the black hole can be extracted via the Penrose or the Blandford-Znajek process, so it is in some sense on a ordered state that can be reverted without increasing the entropy. The mass of the black hole when is uncharged, nonspinning represents the entropic energy content that can't be extracted, and seems to be purely thermal $\endgroup$ – diffeomorphism Jun 11 '15 at 0:11
  • $\begingroup$ Hmm thats funny, extremal mass indicates the smallest possible mass, so this would imply a non extremal BH radiates Hawking radiation and shrinks up to its extremal mass and not further (since it does not radiate in the extremal case). Isn't this hard to match with the fact that hawking radiation should lead to the vanishing of BH's instead of bringing them to this non radiating extremal state? $\endgroup$ – 123hoedjevan Jun 11 '15 at 0:20

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