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.

From the perspective of an outside observer it takes infinitely long for the black hole to form.

But if the black hole is no extremal black hole, it emits Hawking radiation. So the outside observer can detect the Hawking radiation of a black hole that hasn't yet formed from his point of view.

How can this be?

share|improve this question
    
Possible duplicates: physics.stackexchange.com/q/34816/2451 and links therein. –  Qmechanic Jan 11 '13 at 17:22

1 Answer 1

The "infinitely long to form" is only if you allow stuff to compress obscenely close to the horizon, and don't ever cut off the space. You need to cut off this ultra-near-horizon region physically somehow, if you don't you would include the entropy that you can store in a quantum field in this region as part of the black hole entropy, and if you do, then the entropy answer is infinity, it isn't Hawking's answer.

The cutoff means that once something has gotten close enough to the black hole, it's absorbed from the exterior point of view. The location of the cutoff doesn't matter very much because the infalling frozen stuff redshifts away and mashes on the horizon in an exponential way. So it is just a mathematical artifact of horizons that you say "the black hole doesn't form", it isn't significant.

There is a legitimate question here, though, the Hawking radiation is traced back to the initial point, and so it has to go through all the piled up layers on the horizon (from the exterior view). This is no problem, because these piled up layers from the point of view of the outgoing photon are essentially transparent, since the photon is essentially infinitely blueshifted at the moment it encounters them (the photon is redshifting infinitely as it goes outward).

These confusions regarding Hawking radiation are resolved in the standard picture, where the horizon is an ordinary traversible patch of space-time which is locally equivalent to Minkowski space. The issue that "black holes never form" is only formally correct, and it is more correct to just say black holes quantumly form when all the stuff is within a Planckian skin of the horizon. The full resolution to the formation/evaporation question is only in string theory, and it isn't completely worked out, but Hawking's result is a beacon of light that one must trust, because it essentially uses only known physics and plausible assumptions to produce a result of universal validity, one of whose deduced results, the entropy/area relation, is confirmed in many regimes in string theory.

share|improve this answer
    
How is event horizon defined in quantum gravity ? The classical definition is teleological but due to the Hawking radiation we can measure in a finite time that a black hole has formed. –  jjcale Oct 21 '12 at 6:54
    
@jjcale: It's defined by a classical limit plus a teleological limit. The point is that it is an approximate notion, and you can't mash arbitrarily cloase, there's a cutoff. –  Ron Maimon Oct 21 '12 at 12:34
1  
@jjcale: "almost black holes" don't emit until they are "almost to the planck scale" at which point they are no longer almost, but true black holes. The quantum picture resolves the intuitive paradoxes, it's holography and complementarity, but this easier frozen star paradox was essentially resolved in the 1960s, when black holes were understood to have consistent interiors. –  Ron Maimon Oct 21 '12 at 15:24
1  
But "almost to the planck scale" is not sharply defined. So what is a true black hole ? That the Hawking radiation is first exactly zero and then suddenly non zero is not very plausible to me. Or is there a phase transition ? –  jjcale Oct 21 '12 at 15:46
1  
@jjcale: The Hawking radiation begins as the stresses on the horizon asymptote to zero. This is exponentially fast. The interior is extrapolation, but the exterior is approaching the empty space stress quickly, even though technically speaking, the infalling stuff is kneaded into thin layers from the external point of view. This is not paradoxical, there is no "turning on", and from a string theory point of view, it's just strings merging with the horizon-string and thermalizing to its temperature and then other strings peeling off the thermalized horizon-string randomly. –  Ron Maimon Oct 21 '12 at 17:38

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.