# Thermal AdS and the Hawking Page phase transition

I have some difficulty understanding the concept of pure thermal radiation, as described in Hawking and Page's paper on the Hawking-Page phase transition.

The four-dimensional thermal AdS solution (with cosmological constant $\Lambda<0$) is given by

$ds^2=f(r)d\tau^2+\frac{1}{f(r)}dr^2+r^2d\Omega^2$,

with $f(r)=1+\frac{r^2}{L^2}$, $L^2\equiv {-3/\Lambda}$ and the imaginary time $\tau$ is periodic in the inverse temperature $\beta$. Apparently this describes thermal radiation.

How should I see this thermal radiation? Does it consist of a gas gravitons, since the Einstein-Hilbert action has no other fields than the metric tensor? Or does it consist of other particles and should I add other fields to the action to describe these?

Then in Hawking and Page's paper, it is stated that: "The dominant contribution to the path integral is expected to come from metrics which are near classical solutions to the Einstein equations. Periodically identified anti-de Sitter space is one of these and we take it to be the zero of action and energy. The path integral over the matter fields and metric fluctuations on the anti-de Sitter background can be regarded as giving the contribution of thermal radiation in anti-de Sitter space to the partition function Z. For a conformally invariant field this will be: $\log Z= \frac{\pi^4}{90}g\frac{L^3}{\beta^3}$." Here $g$ is the effective number of spin states.

Since they state thermal AdS is taken as the zero of the action and energy, does this mean that the $\frac{\pi^4}{90}g\frac{L^3}{\beta^3}$ comes from loop corrections? And where does the expression come from?

From the above expression the free energy $F_{AdS}$ of thermal AdS follows. Later on in the paper, the temperature $T_1$ at which the free energy $F_{BH}$ of the (stable) black hole solution becomes negative is determined. It would seem to me that the phase transition occurs when $F_{AdS}=F_{BH}$. But rather than calculating the temperature at which this occurs, it is stated that for $T\gtrsim T_1$ the black hole solution will have lower free energy, hence will be thermodynamically favorable. So it looks to me that $F_{AdS}$ is neglected.

Why can we neglect $F_{AdS}$? Is there a parameter in the expression for $\log Z$ above which is very small? (Or is it just $\hbar$?)

Any help is appreciated, I'd be happy with even one answer to one of the questions in the blockquotes.

• There is some information on this phase transition here hartmanhep.net/topics2015/16-hawkingpage.pdf. Jul 17, 2021 at 6:44
• I guess the main point is that the metric you wrote has no singularity, i.e., f(r) is nowhere zero. Then, I guess one might be able to obtain such a simple metric by considering some kind of non-interacting photon gas? Apr 6, 2022 at 9:10
• Could you give a reference on how to obtain the partition function of thermal radiation, $\log Z=2\pi^2L^3\int_0^T T^{-2}f(T)dT=\frac{\pi^4}{90}g\frac{L^3}{\beta^3}$? Jun 1, 2022 at 4:08

• Thanks for your reply. Some months ago I think I read that the second question is indeed due to quantum effects, i.e. fluctuations of matter fields around their expectation value which is zero. The details I'd have to look up though. I guess this also answers the third question, since the free energy there is calculated using the on-shell action only, such that classically $F_{AdS}=0$. I have heard several times that the thermal gas can be seen as a thermal gas of gravitons in case there are no other fields in the theory. Sorry for not being able to give more details at the moment. Dec 6, 2014 at 12:54
From the notes mentioned in the comment, the answer appears to be the following. One has the boundary conditions on the metric that as $$r\to\infty$$, the metric must obey $$ds^2 \to \frac{r^2}{L^2} dt^2 + \frac{L^2}{r^2} dr^2 + \text{angular part}$$ From this, one is able to find three solutions. Two correspond to black holes, one with a small horizon and another with a larger horizon, the latter always being thermodynamically favored over the former. However, one also finds the solution which you wrote. If we just compare this to the Schwarzschild case where $$f(r) = 1 + r^2/L^2 - M/r^2.$$ Note that your written solution $$f(r) = 1 + r^2/L^2$$, the "new solution", has no mass, i.e., $$M=0$$. It also has no singularity, i.e., nowhere is $$f(r)=0$$. It still has a temperature, though, which in this case turns out to be a free parameter. Plotting the free energies over temperature, one notes that this solution is thermodynamically favorable, i.e., minimizes the free energy, at low temperatures.