1
$\begingroup$

This question is an extension of my previous Phys.SE question, but now in $AdS$ spacetime. I am attempting to derive the Temperature of the Schwarzschild solution in this space, which is given by: $$-f(r)dt^2+f(r)dr^2+r^2d\Omega_{3}^{2}$$

where $f(r)=1-\frac{2m}{r^2}+\frac{r^2}{b^2}$. Differentiating with respect to $r$ yields: $f'(r)=\frac{4m}{r^3}+\frac{2r}{b^2}=\frac{4mb^2+2r^4}{r^3b^2}$. We know that the formula for calculating the temperature of a black hole is $$T_{BH}=\frac{f'(r_+)}{4\pi}$$ where $r_+$ is the radial distance of the horizon. Hence based on my calculations, the corresponding temperature is:

$$T_{BH}=\frac{2mb^2+r_{+}^4}{2\pi r_{+}^3b^2}$$

I have compared this to the generalised temperature of black holes in AdS space - found here link Here in equation 2.3 - and found that they conflict. Is there an error in my calculations? I am thinking that maybe the Temperature equation I have stated is not valid in AdS space.

$\endgroup$
  • $\begingroup$ Your result is consistent, just use $f(r_+)=0$ to express $m$ in terms of $r_+$. $\endgroup$ – ScroogeMcDuck Feb 23 '16 at 14:19
  • $\begingroup$ Wow thanks so much, that solves the problem. I've missed such a trivial step! $\endgroup$ – user280325 Feb 24 '16 at 1:39
3
$\begingroup$

So, you clearly see that your result is consistent if you express $m$ in terms of $r_+$ using $f(r_+)=0$. From the derivation of the equation for the temperature, you should also see that it is still valid if the spacetime is, for example, asymptotically AdS. I think the easiest way to derive the temperature is by performing a Wick rotation on the metric to Euclidean time $\tau=it$ and identifying the period of imaginary time with the inverse temperature. There are some subtleties regarding the Wick rotation, but for metrics like these I wouldn't worry about it.

For a metric of the form $$ds^2=-f(r)dt^2+\frac{1}{f(r)}dr^2+r^2 d\Omega_{d-1}^2,$$ the derivation goes as follows. First we Wick rotate and get $$ds^2=f(r)d\tau^2+\frac{1}{f(r)}dr^2+r^2 d\Omega_{d-1}^2.$$

Next, expand around $r_+$: $$ds^2\approx f'(r_+)(r-r_+)d\tau^2+\frac{1}{f'(r_+)(r-r_+)}dr^2+r_+^2 d\Omega_{d-1}^2.$$ where we used $f(r_+)=0$. Now introduce the coordinate $R$ defined by $R^2=\frac{4(r-r_+)}{f'(r_+)}$, in terms of which the metric becomes $$ds^2=\frac{f'(r_+)^2}{4}R^2 d\tau^2+dR^2+r_+^2 d\Omega_{d-1}^2.$$ You can see that the metric (ignoring the sphere part) looks like a conical metric $R^2 d\phi^2+dR^2$ with $\phi=\frac{\tau f'(r_+)}{2}$. Therefore, it has a conical singularity (see this question if you don't understand this), unless $\phi$ is periodic with period $2\pi$, in which case it's just polar coordinates. So requiring there's no conical singularity at the horizon, we see that $\tau$ must be periodic in $4\pi/f'(r_+)$. But the period of imaginary time is just the inverse temperature, which indeed gives you $T_{BH}=f'(r_+)/4\pi$.

Probably calculating the surface gravity is a more thorough and more coordinate-independent derivation, but I think if you can apply the above it's much easier. Anyhow, note there's no assumption regarding whether the spacetime is asymptotically flat or AdS, so there's no reason the derivation should not work in AdS. However, the interpretation of $T_{BH}$ as the temperature measured by an observer at infinity is true for an asymptotically flat spacetime, but not in AdS.

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

Not the answer you're looking for? Browse other questions tagged or ask your own question.