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?


1 Answer 1


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.

  • $\begingroup$ See version 1 of my edit to the post. $\endgroup$ 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? $\endgroup$ Jun 15, 2017 at 22:48
  • 1
    $\begingroup$ 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
  • 1
    $\begingroup$ 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

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge you have read our privacy policy.

Not the answer you're looking for? Browse other questions tagged or ask your own question.