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This question is based on page 68 of Thomas Hartman's notes on Quantum Gravity and Black Holes.

To evaluate a path integral in ordinary quantum field theory, we integrate over fields defined on a fixed spacetime manifold.

In quantum gravity, however, we integrate over both the (non-gravitational) fields and the geometry. The (Euclidean) gravitational path integral is therefore

$$\int \mathcal{D}g\mathcal{D}\phi\ e^{-S_{E}[g,\phi]},$$

with the boundary conditions

$$t_{E} \sim t_{E} + \beta, \qquad g_{tt} \to 1\ \text{as}\ r \to \infty.$$

How would you explain these boundary conditions without alluding to finite-temperature quantum field theory?

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Also on page 68 of the same notes, Tom Hartman explains that this is how we choose the boundary conditions for a particular path integral, i.e. the one that should compute the thermal partition function. Therefore we have to allude to finite-temperature quantum field theory, as that is exactly what we'd like to be doing.

The $g_{tt} \to 1$ as $r \to \infty$ condition just says that we want our space to be asymptotically flat. For example we can compute the thermal partition function in AdS space where we have different boundary conditions.

I don't know of a way to understand Euclidean QFT with periodic time other than as being at a finite temperature.

If you'd like a purely gravitational motivation for that boundary condition we can work backwards. We'd like the Euclidean Schwarzschild metric to be a saddle point for that path integral, and therefore we need time to have periodicity $\beta$. See this question for a purely graviational explanation of that condition.

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  • $\begingroup$ See version 1 of my edit to the post. $\endgroup$
    – nightmarish
    Jun 15, 2017 at 22:23
  • $\begingroup$ I don't think your edits came through but hopefully I edited my answer to address your concern. $\endgroup$
    – jswien
    Jun 15, 2017 at 22:35
  • $\begingroup$ I see. But surely, even if you start with the Euclidean Schwarszchild metric, once you find a periodicity $\beta$ in your Euclidean time, this would mean that your gravitational theory is a finite-temperature QFT. Am I correct? $\endgroup$
    – nightmarish
    Jun 15, 2017 at 22:48
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    $\begingroup$ Yes, this is the first step in realizing that black holes are thermodynamic objects that obey laws directly analogous to the laws of thermodynamics. I don't think I would say that it is a "finite-temperature QFT" except in the case of the AdS/CFT correspondence, just because quantum gravity might not be able to be characterized in the language of QFT. $\endgroup$
    – jswien
    Jun 16, 2017 at 0:40
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    $\begingroup$ Yes, you do need to impose boundary conditions on all components of the metric. I think he just left those implied, as the one that matters is the time direction. If that component didn't go to one then there wouldn't necessarily be a clear definition of temperature. You can see an example for how to define temperature in that case here in the first paragraph of section 2 arxiv.org/abs/1605.02803 $\endgroup$
    – jswien
    Jun 16, 2017 at 6:56

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