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

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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.

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:

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
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Where exactly is the event horizon?

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Arjang
@Arjang The event horizon is where the distance from the Black hole equals the Schwarzschild radius.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Dilaton
If the photon were emitted outside the black hole, how does the black hole lose mass (energy would have to be transferred across the event horizon and out to the photon). I suspect this implies the photon must be created on the horizon. This is related to my unanswered question here: physics.stackexchange.com/questions/21961/…

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user user1247
@Arjang: The Schwarzschild radius and therefore the location of the event horizon is precisely defined as $\frac{2MG}{c^2}$ (except for the uncertainty principle). Stuff inside the event horizon can never escape again but finally reaches the future (true) singularity.

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

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 (80 points)
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Welcome to Physics.SE!

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Manishearth
Thank you for your answer. I wonder if you could expand it to comment on whether photons emitted in this way interact with the falling matter that an outside observer sees as "frozen" just outside the event horizon?

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
@Nathaniel: The answer to that question is a little more tricky--the Hawking radiation arises because the reference frame of the observer near the horizon is different han that of the observer far from the horizon, and this leads to the two observers having a different notion of what a particle is, which leads to the distant observer saying that the infalling observer's vacuum state actually contains particles. So, the frozen'' observer wouldn't actually see any hawking radiation as they fall in--they are freely falling, and see a nice vacuum state, locally.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Jerry Schirmer
@JerrySchirmer I can understand that (see this blog post in which I come to a similar conclusion: jellymatter.com/2011/02/26/falling-into-a-black-hole-part-1 - although of course from the infalling observer's point of view there is still a horizon ahead, which should still be radiating) but would you say that, from my point of view as the external observer, the photons would appear to have interacted with the infalling matter? Would they contain any information about it, in other words?

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

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:

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 (1,070 points)
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Thanks, this looks promising. It would seem to imply that from the distant observer's point of view, the photons would appear to come from the practically-infinitely-hot extended horizon, and thus should appear to have interacted very strongly with the infalling matter.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
There's also an interesting issue about the conservation of energy. In your diagram, all of the matter-energy that ever falls in escapes in the form of radiation before the moment of collapse. This seems as if it should be correct. But that seems to imply that none of that matter-energy can ever pass the horizon. I used to think nothing could pass the horizon, but after learning about Penrose diagrams I thought I had been mistaken - now I'm unsure again.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
@Nathaniel: The infinite temperature inferred by a distant observer is not because of any interaction with infalling matter. Hawking radiation isn't even calculated for a black hole that is forming by astrophysical collapse, because that would be too complicated to model. Hawking radiation happens in a purely vacuum (Schwarzschild) spacetime.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ben Crowell
@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
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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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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 (7,720 points)
The first part of your answer is interesting, thank you. So is it correct to say that in Hawking's original formulation, my option 2 was the correct one: in travelling outward from its instantaneous origin point, the radiation would eventually interact with all the matter that ever fell into the black hole and carry information about it out again? (And hence the information paradox should never have been a serious concern in the first place?)

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
(The second part of your answer is doubtless helpful as well, but I'm an expert in thermodynamics rather than cosmology, so considerations of holography and string theory are a little beyond me for the moment.)

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
@Nathaniel: the problem is that the "interaction" naively can be made arbitrarily small--- you can look at Hawking radiation of neutrinos. There is no way that the light can be "interacting" with infalling stuff electromagnetically to get the information out. The correct idea is that the interaction is gravitational, and the mechanism is that the horizon has deformation degrees of freedom that encode the infalling information. This is exactly the path by which t'Hooft deduced the holographic principle, and there is no more way to understand this than to understand thermo without entropy.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ron Maimon
@RonMaimon ...but the information is finally imprinted into the outgoing Hawking radiation as Lumo says, or not? And if so, how exactly does it get there from the deformation degrees of freedom at the horizon? (Sorry for this interruptive comment but this interests me too :-P ...)

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Dilaton
@RonMaimon that's a helpful example, thanks. I'm now thinking about what would happen if a cloud of neutrinos somehow collapsed to form a black hole. First they'd all appear frozen at the horizon. Then the hole would evaporate, emitting photons that don't interact with the infalling neutrinos. But once the hole has evaporated the neutrinos are free, and still no information is lost. But this doesn't quite work, because the mass of the black hole is actually the mass of the neutrinos, so really it should have been the neutrinos that were converted to EM radiation. So what could cause that?

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
@Nathaniel: This is a start, but neutrinos at high energies actually interact about as strongly as electrons, so it's not a good example. The apparent loss of information during settling to the no-hair state means that you can throw dark-matter in and get photons out. Or throw neutrons in and get electrons/positrons out. This was known well before t'Hooft. The Hawking radiation is paradoxical seeming, because the output is independent of the input semiclassically, and there is no obvious fix by introducing interactions, because the end result is independent of the strength of these.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ron Maimon
@Dilaton: (and Nathaniel too) The way that the information is imprinted in the horizon is crazy--- its not that the photon is interacting with other stuff frozen on the horizon, its that the horizon itself is jiggling thermally, and the jiggles themselves are the infalling photon and the infalling matter. They contain the entire physics of the infalling matter. This is the holographic idea. When the infalling photon gets close, it is described more and more by BH jiggles (rather than cosmological jiggles), and these jiggles obey an information preserving field theory, at least when extremal.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ron Maimon
So the process of black hole excitation and decay can be understood only if the black hole itself reconstructs the spacetime near it and around it. This is the near-horizon region in AdS/CFT, where the whole spacetime is reconstructed from the brane theory, with an insane nonlocal mapping that reproduces a gravitational theory in a different number of dimensions. An infalling photon carries a gravitational field, which is determined by the deformation of the black hole horizon ( a boundary condition), and is completely determined when the photon is close to the horizon. This is t'Hooft's stuff

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ron Maimon
The infalling matter is holographically represented by the deformation lumps on the horizon itself. This deformed horizon also reconstructs the interior spacetime, as well as the nearby exterior, and it constructs different dual interiors in different extensions, which should be related by unitary transformations of the quantum degress of freedom to an approximate emergent space-time description. This transformation has not been worked out in full even in AdS/CFT, where the map is completely precise, and known to work with certainty. The problem is showing locality emerges.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ron Maimon
@RonMaimon Thanks a lot for these additional interesting and for my intuitive understanding of things very helpful comments :-). Maybe you could include them into your answer such that they do not get lost and I can always reread them if I want to ? Darn, how I would like to understand these things a tiny bit technically too, that is what I always need to feel completely satisfied ... And you and other physicists obviously have still very interesting things to do too ... ;-)

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Dilaton
@Dilaton: I'll try... of course, the technical understanding is the string world-sheet and world-volume formalism, at least for the near-extremal charged black holes of string theory. AdS/CFT explains that this is the AdS space that is being described by these theories in the right limit. But the intuition for normal black holes is from Susskind's identification of highly excited ordinary strings with black holes in the 90s. These papers are harder to make precise, because highly excited states are more difficult to trace in time using our usual formalisms, but the qualitative idea is right.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Ron Maimon
@RonMaimon after thinking about this some more, I've realised I'm still confused. I've added some diagrams to the question, which should help to show why I can't come up with a satisfying mental picture from your answer.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
@Ron while I greatly appreciate your edit, it doesn't give me much to go on. Penrose diagram or not, if I observe a photon coming from a black hole and trace the ray back to try and determine its origin, it intersects the surface of a collapsing star. Tracing back further, I reach a point before the formation of the horizon. The rest of the question follows from that - the Penrose diagrams were just a way to illustrate it.

This post imported from StackExchange Physics at 2014-04-08 05:13 (UCT), posted by SE-user Nathaniel
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
@Nathaniel: the radiation is from a transplanckian point, you are missing that all the far-future radiation is essentially hugging the 45-degree line in your diagram, because in the Penrose diagram it just can't get away from it. The star is irrelevant, the photon is transplanckian and hugging the horizon, and the star-matter is like the most diffuse of wispy gas. The way to understand this is to understand the analytic derivation of Hawking radiation, the equivalence principle interpretation, and then to follow Hawking's calculation and see where the mode originates--- it's at that spot.

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