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  From where (in space-time) does Hawking radiation originate?

+ 9 like - 0 dislike
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According to my understanding of black hole thermodynamics, if I observe a black hole from a safe distance I should observe black body radiation emanating from it, with a temperature determined by its mass. The energy from this radiation comes from the black hole's mass itself.

But where (in space-time) does the process of generating the Hawking radiation happen? It seems like it should be at the event horizon itself. However, here is a Penrose diagram of a black hole that forms from a collapsing star and then evaporates, which I've cribbed from this blog post by Luboš Motl.

enter image description here

On the diagram I've drawn the world-lines of the star's surface (orange) and an observer who remains at a safe distance and eventually escapes to infinity (green). From the diagram I can see how the observer can see photons from the star itself and any other infalling matter (orange light rays). These will become red-shifted into gamma rays. But it seems as if any photons emitted from the horizon itself will only be observed at a single instant in time (blue light ray), which looks like it should be observed as the collapse of the black hole.

So it seems that if I observe photons from a black hole at any time before its eventual evaporation, they must have originated from a time before the horizon actually formed. Is this correct? It seems very much at odds with the way the subject of Hawking radiation is usually summarised. How is it possible for the photons to be emitted before the formation of the horizon? Does the energy-time uncertainty relation play a role here?

One reason I'm interested in this is because I'd like to know whether Hawking radiation interacts with the matter that falls in to the black hole. There seem to be three possibilities:

  1. Hawking radiation is generated in the space-time in between the black hole and the observer, and so doesn't interact (much, or at all) with the infalling matter;
  2. Hawking radiation is generated near to the centre of the black hole, at a time before the horizon forms, and consequently it does interact with the matter.
  3. The Hawking radiation is actually emitted by the infalling matter, which for some reason is heated to a very high temperature as it approaches the event horizon.
  4. (With thanks to pjcamp) you can't think of them as coming from a particular point, because they are quantum particles and never have a well-defined location.

All these possibilities have quite different implications for how one should think of the information content of the radiation that eventually reaches the observer, so I'd like to know which (if any) is correct.

The fourth possibility does sound like the most reasonable, but if it's the case I'd like some more details, because what I'm really trying to understand is whether the Hawking photons can interact with the infalling matter or not. Ordinarily, if I observe a photon I expect it to have been emitted by something. If I observe one coming from a black hole, it doesn't seem unreasonable to try and trace its trajectory back in time and work out when and where it came from, and if I do that it will still appear to have come from a time before the horizon formed, and in fact will appear to be originating from surface of the original collapsing star, just before it passed the horizon. I understand the argument that the infalling matter will not experience any Hawking radiation, but I would like to understand whether, from the perspective of the outside observer, the Hawking radiation appears to interact with the matter falling into the black hole. Clearly it does interact with objects that are sufficiently far from the black hole, even if they're free-falling towards it, so if it doesn't interact with the surface of the collapsing star then where is the cutoff point, and why?

In an answer below, Ron Maimon mentions "a microscopic point right where the black hole first formed," but in this diagram it looks like no radiation from that point will be observed until the hole's collapse. Everything I've read about black holes suggests that Hawking radiation is observed to emanate from the black hole continuously, and not just at the moment of collapse, so I'm still very confused about this.

If the radiation is all emitted from this point in space-time, it seems like it should interact very strongly with the in-falling matter:

enter image description here

In this case, crossing the event horizon would not be an uneventful non-experience after all, since it would involve colliding with a large proportion of the Hawking photons all at once. (Is this related to the idea of a "firewall" that I've heard about?)

Finally, I realise it's possible that I'm just thinking about it in the wrong way. I know that the existence of photons is not observer-independent, so I guess it could just be that the question of where the photons originate is not a meaningful one. But even in this case I'd really like to have a clearer physical picture of the situation. If there is a good reason why "where and when do the photons originate?" is not the right question, I'd really appreciate an answer that explains it. (pjcamp's answer to the original version of the question goes some way towards this, but it doesn't address the time-related aspect of the current version, and it also doesn't give any insight as to whether the Hawking radiation interacts with the infalling matter, from the observer's perspective.)

Editorial note: this question has been changed quite a bit since the version that pjcamp and Ron Maimon answered. The old version was based on a time-symmetry argument, which is correct for a Schwartzchild black hole, but not for a transient one that forms from a collapsing star and then evaporates. I think the exposition in terms of Penrose diagrams is much clearer.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
asked Mar 18, 2012 in Theoretical Physics by Nathaniel (495 points) [ no revision ]
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In the third paragraph under the title Black hole information puzzle of this article, Lumo explains that the information about the initial state (the infalling stuff) is due to subtle nonlocal effects (which are over my head) imprinted into the outgoing Hawking radiation. So I´d still vote for (3) choosing from the possibilities (not sure if it is 100% correct to say this).

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Dilaton
Maybe my confusion (which forced me to retract my answer) has something to do with the correspondence between the desciption of a black hole using information at the boundary or in the bulk (the holographic principle) which I do not understand detailed enough :-/ ...

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Dilaton
Why would someone downvote this? If it's unclear, tell me where so I can clarify; if it's obvious, please post an answer; if I've made a mistake somewhere then please post an answer, as I really want to know this. I can't really think of any other possible reason.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
I don't have the time to parse the whole question right now :P but I can answer a small aspect. When a BH gives off hawking radiation, it's is evaporating and shrinking. So, the horizon is not eactly at 45-degrees on that conformal diagram but at a slightly larger angle instead. That would give observers a finite window over which they can observe Hawking radiation, rather than just an instant. EDIT: Seems like @BenCrowell explains that in his answer.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Siva
@Siva really? If that's true it's very interesting, and bizarre that no-one ever mentions it when explaining the Penrose diagram for an evaporating black hole. (It's different from what Ben Crowell said, AFAICT.)

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
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@Dilaton : I thought the radiation was due to the uncertainty principle.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Arjang
@Arjang: Yeah, that`s the first explanation of the Hawking radiation I heard some tima again too ...

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Dilaton

3 Answers

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There are a number of equivalent ways to think about Hawking radiation. One is pair creation, as endolith mentions, where the infalling particle has negative total energy and so reduces the mass of the black hole. Another way, perhaps more useful here, involves Compton wavelength. If the wavelength of a particle (not just photons, by the way) is greater than the Schwarzchild radius, then the particle cannot be thought of as localized within the black hole. There is a finite probability that it will be found outside. In other words, you can think of it as a tunneling process. In fact, you can derive the correct Hawking temperature from the correct wavelength and the uncertainty principle, without deploying the full machinery of quantum field theory.

So I guess that counts as #4 because it isn't on your list. You can't think of quantum particles coming from a specific point because you can't think of them ever having a specific location.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user pjcamp
answered Mar 27, 2012 by pjcamp (80 points) [ no revision ]
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Yeah, de Broglie, sorry. Late night. Nathaniel thinks that all the infalling matter is piled up on the event horizon but that is only from the point of view of a distant observer. Anything falling in crosses the event horizon in finite proper time. So an infalling observer wouldn't see anything piled up anywhere.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user pjcamp
How can (for example) an electron's de Broglie wavelength ever be larger than a macroscopic black hole's radius?

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user user1247
@pjcamp I've resurrected this rather old question. I do understand that the "piling up" is only from the point of view of a distant observer, and I understand why an infalling observer doesn't observe Hawking radiation. However, the question is about what an outside observer will see. I have substantially updated the question, which should make it clearer.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
@Nathaniel I am not sure if I undestood Nathaniel correctly, but myself I would formulate the problem as follows: Distant observer sees Havking radiation from a black hole (with Planck spectrum and say measurable temperature). Now suppose there is a gas between the event horizon and the observer - would the observer detect absorption line in the black hole spectrum? How does this result change if the gas is free falling? How does it change with distance of the gas (or the observer) from the black hole. Will the distant observer observer the lines if he is free falling?

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Leos Ondra
@LeosOndra yes, that's exactly what I'm asking. (Note that for any real black hole there is a gas between the horizon and the observer.)

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
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+1; Never hear the Compton wave-length argument before! Will probably throw this exercise to students, will see how accurate our $T_H$ estimate will be.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Slaviks
@pjcamp You actually meant de Broglie, not Compton, didn't you?

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Slaviks
+ 3 like - 0 dislike

Nice question! I'll try to put together a few ideas into what might be a valid answer, but this could be wrong.

The event horizon is a lightlike surface. Therefore no proper time passes at the horizon, and any photon emitted exactly at the horizon would remain exactly at the horizon until the moment when evaporation was complete, after which it would fly off -- at the same instant as every other photon that had been trapped there. The Penrose diagram would look like this:

enter image description here

The green world-line represents a distant inertial observer. (Geodesics don't generally look straight on a Penrose diagram.) She receives all of the Hawking radiation (purple) at a single instant. Now we know that this is not really right. Hawking radiation is supposed to be detected at some finite rate by a distant observer, and this rate increases continuously until evaporation is complete.

I think this suggests that we need to consider the concept of a "stretched horizon," which is basically the horizon plus an extra distance on the order of the Planck length. The distant observer applies her knowledge of gravitational time dilation and infers that the horizon is infinitely hot, and therefore that the laws of physics she knows break down there. It thus becomes useless to worry about the exact nature of the degrees of freedom that are present there; they could be at the Planck scale, could be photons, could be strings, could be virtual black holes, could be real black holes that fission off from the main hole by a process analogous to alpha decay. So we just say that there's a sort of "atmosphere" that extends at least one Planck length above the horizon. We then have this picture:

enter image description here

The stretched horizon is the red curve. The purple lines represent photons emitted from it at different times, and detected at different times by the distant observer. I've simply guessed the qualitative shape of the stretched horizon on the Penrose diagram (curved, concave down); maybe someone else could check whether this actually checks out with a specific coordinate transformation.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ben Crowell
answered May 14, 2013 by Ben Crowell (1,070 points) [ no revision ]
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I'm aware of the issue to do with simultaneity. When I said "moment of collapse" I should have said "the top-most point of the line representing the horizon", and when I said "before" I should have said "in the past light cone of".

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
@Nathaniel: I'd suggest not going into the issues about the infalling matter here, because it's not really relevant, and it's a complicated issue. This is related to ideas like black hole complementarity.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ben Crowell
But the issues to do with the in falling matter were the entire point of the question!

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
To clarify, the main reason I'm interested in this is that it seems as if the whole mass of the collapsing star should be in between the distant observer and the region where the Hawking radiation is emitted, so it seems that it should get in the way and prevent us from directly observing it, instead being heated up by it and re-emitting thermal radiation of the same spectrum. But the infalling matter itself is not supposed to experience interaction with the Hawking radiation, so this is a paradox.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
Black hole complementarity can't resolve it (afaics) because we're talking about two different observers who are both outside the event horizon and in principle can meet and compare notes. But people say Hawking radiation doesn't contain information about the infalling matter (modulo weird stuff about non-locality and tunnelling past the horizon), which would suggest they think it doesn't interact with the matter at all. This means it either magically passes through it, or it gets generated well away from the horizon, in between the matter and the distant observer.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
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@Nathaniel: In your reasoning about conservation of energy, I think you need to be more careful about how you think about simultaneity. For example, a distant observer and an infalling observer have different notions of time. When you talk about the "moment of collapse," that isn't a time that all observers will agree on.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ben Crowell
I didn't mean to imply that the radiation is caused by interactions with the infalling matter. I'm just saying that if there is any infalling matter then most or all of it be will further away from the event horizon than the region where the Hawking photons appear, and consequently they must interact with it.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
+ 2 like - 0 dislike

The origin of the radiation in Hawking's original calculation is in a microscopic point right where the black hole first formed. This is where your back-tracing procedure ends. Unfortunately, the light is blue shifted insanely during the back-shifting, so that it is far too blue to be physical (it has wavelength past the Planck length by a lot). This made people worry about Hawking's derivation.

The modern picture of holography generally resolves this issue. The black hole is indistinguishable from and quantum mechanically dual to a white hole, so that you can consider the Hawking radiation as coming from the white hole. This is a consistent picture, and the compression of the radiation at the initial point has an exact analog in what happens to infalling matter in a classical white hole, or in an evaporating black hole. Both issues require a duality between the interior and exterior description, and the realization that highly transplanckian objects near the horizon are really best thought of as stringly spread out over the entire surface, or living in a dual interior region.

(The question changed since I answered this, I am getting downvotes: Penrose diagrams are a wrong picture, they are misleading, they are completely incorrect in the proper quantum gravity, don't use it. The proper picture is the naive one, without twisting the horizon to lie at 45 degrees.

When you make the horizon lie at 45 degrees you must choose if the horizon is a past or future horizon. But the two things are quantum-mechanically dual (although classically separate) and you shouldn't force the horizon to be one and not the other. All black holes which are around long enough can be viewed as eternal, and near the end-state, they are all white holes. This was explained by Hawking, and verified by AdS/CFT. Penrose diagrams should be retired, and downvoters, and those who give answers other than mine, have no idea what they're talking about).

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ron Maimon
answered Mar 28, 2012 by Ron Maimon (7,730 points) [ no revision ]
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@RonMaimon ok, if it comes from a Planck-sized region rather than a single point, that's fine. I understand (qualitatively) how the Penrose diagram smooshes space-time up. But why is the star matter like a diffuse, wispy gas? Stars under normal circumstances are made of opaque plasma, and if there is nothing "special" about the event horizon from a falling observer's point of view, the star should continue to be made of opaque plasma as it falls through it.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
@Nathaniel: Yes, of course a planck size region, single point is meaningless. The star is diffuse wispy gas because you are talking about a super-trans-planckian mode, the scale of variation is wildly off--- this mode has (if it were occupied) 10^(10^10) GeV. This is the reason some people were skeptical of Hawking's calculation, because the modes end up so unphysical.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ron Maimon
When you said "the radiation is from a transplanckian point" (i.e. planck sized region), you mean this region is located at the surface in space time where the light cones tip up enough to form a trapped surface? So it's the whole surface's worth of these Planck regions where the HR begins - you didn't mean a single Planck area - or have I misunderstood? (A (non conformal!) diagram would really help)

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user twistor59
@RonMaimon would I be correct in understanding that as "close to the horizon, the photons are of such a high frequency that they pass through just about anything"?

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
@Nathaniel: Yes, locally, the outgoing photons are of such small wavelength that they pass through anything, it's unphysical.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ron Maimon
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So my questions would be (i) if the radiation is emitted from a single point, why are we predicted to observe it over periods of time much longer than one wave cycle; and (ii) there is a very large opaque object in between that point and a distant observer, namely the collapsing star. Does the radiation get absorbed by the matter, or magically shine through it, or what?

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
@Nathaniel: this is why the Penrose diagram is misleading--- when you are looking at late times, the photon that escapes the black hole is SMOOSHED so close to the 45-degree line of the horizon (as is everything else, including you) that you are completely misled. Write down a normal r-t coordinate description, and see how mushed up it gets at late times--- the photons that peel off the horizon and go to infinity in the normal picture stay close to the horizon forever in the Penrose picture. It's good for conformal structure, nothing else, it's not physical.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ron Maimon

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