Take the 2-minute tour ×
Physics Stack Exchange is a question and answer site for active researchers, academics and students of physics. It's 100% free, no registration required.

In other words, how strong does gravity have to be to cause Hawking radiation to occur?

share|improve this question
2  
Hawking radiation has wavelength comparable to the radius of the black hole, so it must have been generated over the entire event horizon. Is this the whole story? No, and I don't understand it completely, so I'll let somebody who knows post an actual answer. –  Peter Shor Oct 6 '13 at 23:44
2  
@John: I thought (I can't swear to it, either) that you needed an actual horizon to get Hawking radiation, and otherwise you just get virtual photons which are annihilated after some time. –  Peter Shor Oct 7 '13 at 19:46
2  
Call me crazy, but the questions in the title and the body seem different to me. Though I don't know the answers, I would expect the answer to the title to be something like "The radiation rate falls off with the distance from the horizon" and the answer to the body to be something like "You need an event horizon". Perhaps I'm misunderstanding the use of the word "deep" in the title? –  Warrick Oct 9 '13 at 5:49
2  
Related (possibly a duplicate): physics.stackexchange.com/q/22498 –  Nathaniel Oct 9 '13 at 10:37
1  
Warrick, what I guess I mean by the sub-question is, if there is enough gravitation to almost create an event horizon, could it still result in Hawking radiation? Like, if the gravitation was strong enough to restrain a pair of virtual particles from re-meeting and cancelling out long enough for one of the pair to be cancelled out by a different particle. –  Lionel Doolan Oct 10 '13 at 0:07
show 3 more comments

1 Answer 1

up vote 1 down vote accepted

Hawking Radiation can occur in conditions influenced by more than just gravity .. temperature can also influence this - (see Bogoliubov Theory of acoustic Hawking radiation in Bose-Einstein Condensates, A. Recati, N. Pavloff, I. Carusotto and Hawking radiation in a two-component Bose-Einstein condensate, P.-É. Larré and N. Pavloff)

This means that gravity is but a single factor to consider when trying to establish the initial conditions for Hawking Radiation. It's not clear this answers your question, so lets assume we're only looking at gravity.

The event horizon typically takes place at the boundary within which the gravitational object's escape velocity is equal to or greater than the speed of light. There are other definitions involving the pathways of lights, but lets go with this one for now. This means that at the event-horizon nothing with mass can escape since nothing with mass can go the speed of light, and particles with 0 rest mass, such as photons, can, at best, remain in orbit, or some other such odd behaviour such as red-shifting itself to match or exceed the circumference of event-horizon itself. In any event, the effectively disappear to an outside observer.

Notwithstanding, suggestions such as light having fractal properties, or debates about the fractal-dimension of event-horizons and hairy black-holes, as you move away from the event-horizon - lets assume, for the sake of illustration, a simple decaying exponential gradient function describes how the black-hole's influence is felt, with it being greatest near the event-horizon, dying off exponentially the further you travel away (meaning as space-time bends less). Let's also assume the black-hole is not spinning (which isn't likely but it simplifies the illustration). The particle pairs responsible for Hawking radiation, are thought to happen anywhere in space, but for this radiation to actually occur (meaning be detectable), the black-hole's influence must be felt on the pair. So the likelihood of this radiation occurring will be diminishing according to some function as you move further away from the event-horizon but absolutely guaranteed on the event-horizon boundary itself.

This means you could, in theory, have Hawking radiation occurring infinitely far away except the likelihood of the black-hole being able to wield influence on the particle pairs is so unlikely at that range that the probability of this occurring or being detectable is effectively 0. Nevertheless, even from some great distance as you approach your black-hole the probability will slowly raise enough that at some point and given sufficient time, you should be able to measure an occurrence.

So, to answer your question, the strength of gravity sufficient for Hawking radiation to occur, given a particular black-hole, is inversely proportional to the degree of patience you posses. The greater your patience, the weaker the gravity needs to be. The less your patience, the stronger it needs to be.

share|improve this answer
add comment

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.