redshift and time dilatation of Hawking radiation

timmdeeg

Micro black holes should in principle be observable by emitting Hawking radiation. However, as this takes place extremely close to the event horizon, shouldn't one expect then extreme redshifting (z = 1/(1-Rs/R)^-1/2 -1, Rs = Schwarzschildradius, R = Radius of Emission) and time dilatation?
In other words, shouldn't see the far away observer the Hawking radiation almost 'frozen'? How long does it take to reach him in far away coordinates?

Bill_K

timmdeeg, It's important to remember that Hawking radiation is fundamentally a quantum process, and therefore one shouldn't try to take a view of it which is purely mechanistic. All one can say is that the vacuum "in" state contains near future null infinity a thermal bath of outward-going particles. For example where are the particles created? In the vicinity of the hole. You can't say whether it occurs on the surface or near the surface - it is a global effect. And in fact the predominant wavelength of the created particles is about as large as the hole itself.

timmdeeg

Thank you, Bill, your answer is very helpful. So, if the wavelenght is in the order of the black hole, the redshift is much less significant than I was anticipating it. But there should still be some redshift, as the photon climbs out of the gravity well of the hole, right? Are there any calculations?

You mentioned the thermal bath. Could you kindly explain, whether there is any physical difference between Unruh radiation and Hawking radiation near the horizon?

tom.stoer

the Hawking radiation is already expressed for an asymptotic observer, so there's no additional redshift

timmdeeg

Ok, thanks for clarifying.