11
$\begingroup$

There is the problem that is bothering me with the black hole evaporation because of Hawking radiation.

According to Hawking theory the black hole will evaporate in finite time because of quantum effects randomly producing particle-antiparticle pairs on the opposite sites of horizon.

However, the gravity causes the time dilatation, which goes to infinity on the horizon itself, so the arbitrarily small amount of time from horizon's perspective is going to be infinitely long from external observer's perspective?

So, from which perspective is the time the black hole will evaporate calculated? If it is calculated from horizon's perspective, it would mean, the black hole will never evaporate from the external observer's perspective?

Improve this question
$\endgroup$
1

2 Answers 2

Reset to default
3
$\begingroup$

If we refer to the evaporating black hole's Penrose diagram

we would see some important differences between the ideal and the evaporating black holes. For the evaporating black hole, there is a moment in spacetime, where the horizon (and the singularity) disappears, mass left at the center of coordinates is zero, and no time dilatation is left. Hense, this event can be seen for every external observer with no problem.

Improve this answer
$\endgroup$
3
  • $\begingroup$ Yes, but this diagram is constructed based on implication that the black hole will evaporate in finite time. However, that time is calculated, but I don't get it correctly which 'time' is used there, the time on horizon or the time for intertial observer. If I get the idea correctly, this 2 times are separated on horizon. $\endgroup$
    – user20499
    Commented May 12, 2013 at 11:06
  • 1
    $\begingroup$ Rate and time of evaporation are calculated in the time of the static observer at the spatial infinity. Rate of evaporation grows with its progress, so the total time is limited from above. Though, these calculations are valid for the totally empty environment, but in our Universe interstellar or intergalactic gas, or even CMB photons, would fall into black hole and add to its mass faster than it would evaporate. $\endgroup$
    – firtree
    Commented May 12, 2013 at 11:53
  • $\begingroup$ The diagonal line represents infinite time. So, your BH will evaporate after infinite time? $\endgroup$
    – Anixx
    Commented Apr 11 at 16:51
1
$\begingroup$

I think Hawking radiation doesnt originate from 'inside' of the Black Hole. But from the vacuum surface of the horizon, where it is possible for the particle to escape, so it would be finite time from an outsider perspective as the BH is 'evaporating' from barely outside.

But honestly i dont know why the 'vacuum' energy is drained from the horizon and not from the ergosphere.

Improve this answer
$\endgroup$

Your Answer

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