The Hawking temperature $T_h$ is proportional to the surface gravity $\varkappa$ which, in spherical-symmetric case, one calculates on the Killing horizon. This temperature is very small but where? at infinity? Then if we use the formula $T\sqrt{g_{00}}=const$ then, one obtains infinity temperatures near the event horizon. Help me with this question please. I thought that one measures $T_h$ near the event horizon and the temperature near the horizon is small. Or my opponents are right and this temperature is measured in infinity?
The temperature is measured at infinity. The horizon isn’t a space-like surface, but a light-like surface, so it isn’t possible to go there; furthermore, it’s a coordinate artifact.
Thank You. Is this formula $T\sqrt{g_{00}}=const.$ correct only for static spacetimes? In Vaidya spacetime $g_{00}=0$ is not null surface and, according to Alex B. Nielsen 'The spatial relation between the event horizon and trapping horizon.' the apparent horizon is inside the event horizon and we can't measure the infinity large temperature near the event horizon (although the definition of the surface gravity in dynamical case is still a problem i suppose).
I am confused with the fact that the Hawking temperature grows unboundly near the horizon and at the planck length it is like at the early universe
No, it's not like the early universe (a black hole's event horizon is a future horizon, not a past horizon).
Yes, the formula isn't universally valid, for, among other reasons, because the components of the metric aren't observable quantities-they're gauge fields. The Hawking temperature makes sense as an observable at infinity-assuming that the spacetime is asymptotically flat there.
These lectures (look at the rest of the course, too!): http://www.suvratraju.net/openscholar/suvrat/files/lecture_4_hawking_radiation.pdf might be useful.
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