Question

From the perspective of a far-away observer, matter falling into a black hole never crosses the boundary. Why doesn't a basic symmetry argument prove that Hawking radiation is therefore also frozen on the boundary, and therefore not observable? Wouldn't the hawking radiation have to have started its journey before the formation of the black hole? Furthermore, wouldn't the radiation be infinitely red-shifted?

Answer 1

Classically, this is true. Something exiting from a classical static black hole would have had to have started before the universe was created. The frozen-on-the-horizon view of a black hole is the view you get under classical general relativity. When you add quantum mechanics, this view is no longer quite valid, and you get Hawking radiation, which from a far-away observer's viewpoint, interacts with the infalling matter.

Answer 2

Solution: Although, the gravitation of a black hole is so powerful that nothing, not even electromagnetic radiation, can escape from it. Nevertheless, far from the black hole (on or before the onset of event horizon) the gravitational effects can be weak enough for calculations to be reliably performed in the framework of quantum field theory in curved space-time. Hawking showed that quantum effects allow particles near this event horizon to emit exact black body radiation, which is the average thermal radiation emitted by an idealized thermal source known as a black body.

Physically, particle-antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself but vacuum fluctuations cause a particle-antiparticle pair to appear close to the event horizon of a black hole. One of the pair falls into the black hole while the other escapes, in order to preserve total energy. The particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole). By this process, the black hole loses mass to an outside observer; it would appear that the black hole has just emitted a particle. In another model, the process is a quantum tunneling effect, whereby particle-antiparticle pairs will form from the vacuum, and one will tunnel outside the event horizon.

There will be no apparent redshift of Hawking radiation near event horizon according to