# Gravitational path integral derivation of black hole temperature and entropy

Using the gravitational path integral we can define the partition function as:

$$Z(\beta) = \int\mathcal{D}g\mathcal{D}\phi e^{-I_E[g,\phi]}$$ with boundary conditions: $$t_E \sim t_E + \beta, \quad g_{tt} \to 1 \text{ at infinity.}$$ Here, $$t_E$$ is the Euclidean time with period $$\beta$$, $$g$$ denotes the gravitational field, $$\phi$$ denotes the matter fields and $$I_E[g,\phi]$$ is the Euclidean action. In the Schwarzschild solution we can approximate the partition function as: $$Z(\beta) \approx e^{-I_G[g_0]}$$ where $$I_G$$ is the Euclidean gravitational action and $$g_0$$ is the classical Schwarzschild solution.

In order to calculate the partition function in this approximation one needs to perform a Wick rotation in the time coordinate and define the Euclidean Schwarzschild metric. Doing so, one finds that the time coordinate must have a period $$\beta = 8\pi M$$. In the context of gravitational path integrals $$\beta$$ is the inverse temperature, so we have the temperature as $$T = \frac{1}{8\pi M}$$. This is usually interpreted as the temperature of the black hole.

Having found the partition function for the Schwarzschild solution one can calculate an entropy, which turns out to be the Bekenstein-Hawking entropy $$S = A/4$$. This result is usually interpreted as the entropy of the black hole.

Why do we interpret these results as the temperature and energy of the black hole? Couldn't some of the entropy have come from the Hawking radiation?

In particular I find it curious that the Euclidean metric being used in the calculations have coordinate limits: $$r \geq 2M, \quad t_E \sim t_E + \beta$$ The gravitational action is then given as an integral where $$r$$ is in the interval $$[2M, \infty]$$. In other words, the partition function is given as an integral over the entire manifold except from the black hole! Why do we then think of the entropy as being located in the black hole?

In the case of temperature, the answer is straightforward: its temperature is Hawking radiation. We are implicitly assuming that the quantum fields that exist and are under the influence of the black hole's gravitational field are in thermal equilibrium. There is a dependence of the Hawking temperature on the distance we have from the black hole due to the gravitational red-shift, but we are assuming that we as are far away from it. The temperature of those fields at the horizon of the black hole as observed by us effectively at infinity is precisely $$T = \frac{1}{8\pi M}$$. Since this thermal spectrum is entirely due the fact that our fields exist on a Schwarzschild background, and the observed temperature is when we calculate it on the horizon, we could then interpret this temperature as that of the black hole.
As for the seeming paradox that our integrals for $$r$$ run from $$2M$$ to infinity, ignoring the interior, my interpretation is this: these are not direct calculations about the black hole itself, but we can deduce some things about it based on the effects it has on things outside the horizon. The effects of gravitation cause a thermal spectrum, so that energy must be taken away from the black hole itself. The black hole loses energy and its horizon has a temperature given by said thermal spectrum, so it must have entropy too. It is no surprise that the two things we have talked about so far are the temperature and the entropy; both depend on just the mass of the black hole which is one of the few things (as per the black hole no-hair theorem) we can know about the black hole from the outside.