# Why does imaginary periodic time of the Rindler space of black hole give Hawking temperature?

The metric of a Schwarzschild black hole in Rindler coordinates is

$$ds^2= -\frac{\rho^2}{(4MG)^2} dt^2 +d\rho^2 + d\mathbf x_\perp$$

where $\rho$ gives us the distance to the horizon. If we switch to imaginary time $\tau=it$, the metric becomes

$$ds^2= \frac{\rho^2}{(4MG)^2} d\tau^2 +d\rho^2 + d\mathbf x_\perp.$$

It is said that to prevent a conical singularity, time must be periodic with period $8\pi GM$. If we identify the period with $\beta=1/(k_B T)$, this leads to

$$T=\frac{1}{8\pi G M k_B}$$

which is Hawking temperature.

Is there any intuition of why this works? There are no fields present in the calculation, so it's not like the standard proofs. Is it just a coincidence?

## 1 Answer

It is not a coincidence. It has to work like that. The deficit angle has to be zero.

It's most convenient to see it in the Feynman's path integral approach to quantum mechanics. One works in the Euclideanized spacetime to calculate the temperature $T=1/\beta$ partition sum. Let us consider the full finite-size black hole; the Rindler geometry is a local description of the near-horizon geometry of a very large black hole.

There are contributions to the path integral without any black hole, and those with a black hole. The latter must combine to the smooth "cigar" geometry without any deficit (or excess) angle because Einstein's equations have to hold everywhere, even in the Wick-rotated setup, and there's no source at the horizon. The correct periodicity (absence of deficit angle) imposes the relationship between the temperature and the size of the black hole.

These ideas were clarified by Hawking and Gibbons in 1977, building on insights by Perry and Gibbons, see e.g.

http://motls.blogspot.com/2012/01/gibbons-hawking-and-euclidean-path.html?m=1