Thermal AdS and the Hawking Page phase transition
I have some difficulty understanding the concept of pure thermal radiation, as described in <PERSON> and <PERSON>'s paper on the <PERSON>-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 <PERSON>-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 <PERSON> and <PERSON>'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 <PERSON> 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. 346
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. 669