Interpretation of temperature in General Relativity I am reading the historical article by Hawking and Page. Their exposition of the Hawking-Page transition differentiates the notion of temperature in both AdS and SAdS spacetimes, as both of these spacetimes have a different $\beta\equiv 1/T$ that have to be reconciled at the boundary.
Then, the authors give the difference of the on-shell action
$$\Delta I = \dfrac{\pi r_h^2}{G_N(L^2+3r_h^2)} \left( L^2 - r_h^2\right),$$
where $r_h$ is the horizon radius, and $L$ the curvature radius of AdS. Then, from a semi-classical approximation to the quantum gravity path integral, the difference in free energy becomes
$$\Delta F = \beta^{-1} \Delta I= T\left(\dfrac{\pi r_h^2}{G_N(L^2+3r_h^2)} \left( L^2 - r_h^2\right)\right).$$
What does $T$ mean here? The previous steps seem to indicate that the meaning of temperature is different for both spacetimes, but then this last formula implies that the meaning of $T$ applies to both spacetimes and can simply multiply the two on-shell actions. What am I missing about $T$ here?
 A: If this is referring to the comment after equation 2.7, the difference is actually coming from the difference between the time coordinate $\tau$ and the local proper time. There is one version of energy which is the conserved charged associated with the time translation symmetry of the two geometries, and another which is the amount of energy measured in local coordinates at a given point.
Since one unit of coordinate time corresponds to many ($V^{1/2}$) units of proper time far from the origin, one unit of conserved energy corresponds to very few units of energy measured in the local coordinate frame. This means that, in a system in thermodynamic equilibrium, at constant $T$, the locally measured temperature is $V^{-1/2} T$ and drops away from the origin.
This happens in ordinary AdS too: see the comment about the density dropping for large $r$ after equation 2.5 of the paper. A heuristic reason to see why this should happen is that, since particles like to oscillate around the center of the AdS geometry, there is an energy penalty to being at large $r$ and thus the density should be higher in the center when in thermodynamic equilibrium.
