# 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?

## 1 Answer

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.

• Sorry for the late answer. I don't think this really answers my question, or at least I fail to understand how it does. The thing here is that there's this whole "locally-measured temperature" stuff to be able to write the difference of the on-shell actions $\Delta I$. Then, the two solutions are compared via their free energy $\Delta F= \beta^{-1} \Delta I$ to be able to state that there is a phase transition at $r_h=L$. Jun 20 at 11:25
• This step ($\Delta F$) is not performed explictly. However, this implies that the difference in free energy is simply the temperature times the difference of the on-shell actions, even though they make a point that the $\beta$'s, hence the temperatures, are not to be compared lightly. Jun 20 at 11:25
• It should be the case that all of the temperatures used in the paper, outside of the parenthetical remark about local temperature, should be the coordinate time version. These are comparable due to the fact that they are both asymptotically AdS and the clocks are required to match far from the black hole. So the thermodynamic phase transition properties use the coordinate time, while the only usage of the local temperature is in justifying why the local energy density of the thermal gas outside the black hole is not constant, despite being in thermodynamic equilibrium. Jun 20 at 18:23
• Thanks for your answer. Could you explain what you mean by coordinate time temperature? My understanding is that we had two $\beta$'s: one for AdS, one for SAdS, and that both of these were to be interpreted as the inverse of the temperature in geometry and are not the same. Then, locally-measured temperature acts a matching condition to compare the two on-shell actions. Are you referring to $T$ in the $\Delta F$ formula as the time coordinate temperature, and why is the same for both spacetimes? Thanks for your patience and your time, but there is really something I fail to understand here. Jun 20 at 18:57
• I'm sorry, maybe I should have tagged you to notice you of my answer. Jun 20 at 19:10