Near horizon limit and Hawking Temperature of the horizon

[ShayanJ]

One way that people introduce the Hawking temperature of an event horizon, is by taking the near-horizon limit of the BH metric and then do a Wick rotation of the time coordinate. Then, the regularity of the metric requires that the Euclidean time to be periodic. But how can this give us the temperature of the horizon? What's the relation between the periodicity of the Euclidean time and temperature?

Thanks

 

[Haelfix]

See note 3 and the path integral sections in the following course series:
http://www.hartmanhep.net/topics2015/

Alternatively, the full reasoning is given in this lecture (try to understand the Rindler path integral case first)
https://arxiv.org/abs/1409.1231

It is a peculiar but deep fact of gravity that an identification can be made between the thermal density matrix of finite temperature QFT and the reduced density matrix arrived from a computation of the gravitational path integral of gravity in Rindler space when you trace over the Rindler wedges.

 

[Demystifier]

What's the relation between the periodicity of the Euclidean time and temperature?

See e.g.
http://www.imperial.ac.uk/media/imp...issertations/2011/Yuhao-Yang-Dissertation.pdf
especially Eq. (2.17).

 

[Ben Niehoff]

One way that people introduce the Hawking temperature of an event horizon, is by taking the near-horizon limit of the BH metric and then do a Wick rotation of the time coordinate.

By the way, you don't have to take the near-horizon limit first! It's just often easier that way.

 

[ShayanJ]

By the way, you don't have to take the near-horizon limit first! It's just often easier that way.

Could you give a reference where it is done that way at some detail?

 

[Ben Niehoff]

Could you give a reference where it is done that way at some detail?

Try it with Schwarzschild, it shouldn't be hard. Also try looking up "gravitational instantons".

 

[ShayanJ]

Try it with Schwarzschild, it shouldn't be hard. Also try looking up "gravitational instantons".

Looks like the transformation that takes $ds^2=\frac{dr^2}{1-\frac{2m}r}+(1-\frac{2m}r)dt_E^2$ to $ds^2=d\rho^2+\rho^2 dT_E^2$ is:
$\left\{ \begin{array}{c}\rho=r \sqrt{1-\frac{2m}r}-m\ln\left( \frac{\sqrt{1-\frac{2m}r}-1}{\sqrt{1-\frac{2m}r}+1} \right) \\ \rho T_E=\sqrt{1-\frac{2m}r}t_E\end{array}\right.$