# How Hawking Radiation can have any meaningful impact on the black hole's life since time is almost stopped right outside the Even Horizon?

Every object experiences time more and more slowly as they approach the event horizon until stops right at it, according to observers outside the event horizon. As far as I know quantum fluxuations which happen in that region are the direct reason why black holes evaporate.

How is it possible for the event horizon to not exist anymore since the insane time dilation there makes it so it doesn't experience time? Consequently any quantum flaxuations there would freeze or happen at incredible slow motion far too slow to have any effect on the black hole's mass at all. No particle would neither fall nor travel very far from the event horizon according to our clocks. But the evaporation of the black hole does happen in our clocks.

There are two aspects to be addressed here.

Firstly it is not true that Hawking radiation is exactly always produced at the Event horizon. Usually for a field theory, we have vacuum fluctuations where there is creation and annihilation of a particle antiparticle pair. Now since the black hole has an event horizon, one particle can fall into inside the event horizon and the other escapes from the horizon, thereby reducing the total energy of the system in question. The thing is, these processes usually happen near the horizon, but not always.

Secondly, since the radius of the horizon keeps shrinking continuously due to Hawking radiation, therefore if a particle was very highly red-shifted initially, if won't be so sometime later. Given a black hole of mass M, you can calculate that it radiates its mass away in $t \approx M^3$, since the emitted Hawking radiation is blackbody radiation

The textbook understanding of Hawking radiation involves Bogoliubov transformations. The idea is that when you quantize lets say the electromagnetic field you take solutions of the classical equations (Maxwell's equations) and write them as a linear combination of positive-frequency and negative-frequency parts. Roughly speaking, one gives you particles and the other gives you antiparticles. Thus the way we choose the vacuum in our theory plays a strong role in the number of particles and the antiparticles. Outside the event horizon, the choice of vacuum is different and as a result we see a Hawking radiation due to particle production.

• Oh I see. So you are sayin that since the fluxation can happen not at the Horizon but some distance from it, the 2 particles can experiance time. And that the loss of the energy of the escape particle is reduced becuase the black hole now has less mass and the time dilation which would redshift the escpape particle now happens to a lesser degree and the energy lost by the black hole is meanigfully greater. – Bill Aug 27 '16 at 13:40
• @Bill the particle creation and annihilation processes occur everywhere. and if one particle falls inside while the other escapes away, then the energy (correspondingly the mass) gets reduced.. also the red shifting decreases due to mass-energy being radiated away... – Bruce Lee Aug 27 '16 at 13:44
• Yes but there is limited amout of time according to Heisenberg's Uncertainty principle (ΔΕ*Δt>h/4π) before the 2 partciles remeet again. One has to fall and the other has to escpape before that happens. – Bill Aug 27 '16 at 13:47
• @Bill so what is your point? – Bruce Lee Aug 27 '16 at 13:49
• If it happens far away from the event horizon they will remeet without having the chance to affect the black hole. So fluxuations which happen lightyears away don't contibute to the hawking radiation – Bill Aug 27 '16 at 13:52