Why doesn't hawking radiation prevent a space-like singularity

PeterDonis
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Does the horizon shrink down to nothing? or does it evaporate at a certain radius?

As the BH evaporates, the horizon shrinks; eventually, it shrinks to zero radius, and at that instant, the BH has completely evaporated.

However, all this happens long *after* the infalling particle has crossed the horizon, reached the singularity, and been destroyed.

So, the partial reaches the horizon at the same instant that the BH evaporates?

No. The distant observer *sees* the particle reaching the horizon at the same time that he *sees* the BH finally evaporate. But that's because the light that was emitted outward by the particle as it crossed the horizon ends up on the same outgoing trajectory as the light from the final evaporation of the BH. It is *not* because the BH's final evaporation happens at the same event as the particle crossing the horizon.

It is *really* important to keep clear the distinction between what the distant observer sees, as in what light signals he receives, and what happens down at the horizon, because the horizon is an outgoing null surface, so light that is emitted outward at the horizon stays at the horizon, for as long as the horizon exists. That means that, when the BH finally evaporates, all the light that was emitted at the horizon, for the entire time of the horizon's existence, is now mingled with the light from the BH's final evaporation, and all of that light goes outward together.

What is observed happening to the particle?

The distant observer never sees anything of the particle once it has crossed the horizon, as I said before. The actual light images he receives will show the BH evaporating away along with the particle falling through the horizon; but as noted above, that does *not* mean that is what actually happened. The light from events that happened in distinct portions of spacetime gets mingled together because of the way the spacetime is curved.

I am having trouble seeing how from the particles frame, there is a discontinuity as it hits the singularity

Because the singularity has infinite curvature. But as noted, the distant observer can never see that, or indeed anything inside the horizon. See below.

but from the distant observers frame, there is no discontinuity, but it is never observed falling into the hole?

Because the distant observer can never see anything inside the horizon. That's what the horizon *is*--the boundary of a region of spacetime that can't send signals out to distant observers.

Surly either, there is a discontinuity some where, from the distant observers point of view, or the particle doesn't hit the singularity. Am I wrong?

Yes. You've left out a third possibility: that the region of spacetime inside the horizon, even though it exists, can't send light signals out to the distant observer. The reason it can't is that the curvature of spacetime in that region is great enough to force even outgoing light to fall inward towards the singularity.

The below is from http://en.wikipedia.org/wiki/Apparent_horizon, not that I necessarily trust Wikipedia 100%

A wise policy. In this case, the Wiki page is correct in what it says, but what it says does not contradict anything I've said.

I have not been able to read the actual papers that are referenced in that article, since they're behind paywalls, but AFAIK, whatever spacetime slicings would not show an apparent horizon, the ones that are relevant to our discussion--the spacetime slicing of the distant observer, and the spacetime slicing of the infalling observer--both *do* show apparent horizons. (Others who have commented in this thread may have read the actual papers and be able to correct me if I've misstated something, or add more clarification.)

Also, as I've noted before, the boundary of the region that can't send light signals to the distant observer is the *absolute* horizon, not the apparent horizon; and strictly speaking, what I said above about light emitted outward at the horizon mingling with light from the BH's final evaporation applies to the absolute horizon. (As I've noted before, for the case we're discussing the difference between the two kinds of horizons, for the obsevers we're talking about, is not significant.)

Finally, just to clear up a possible misunderstanding of the term "apparent horizon": the word "apparent" in this term does not refer directly to whether, or for how long, the observer actually sees light from the horizon. It is just a way of distinguishing it from the absolute horizon. What has been said about the horizon "appearing" to be in front of the infalling observer, even after he has actually fallen past it, applies to both kinds of horizons, apparent and absolute.

Anyway, as I have been tought, if you fall into a bh, you will see a bh in front of you, all the way to the singularity.

You will *see* the horizon in front of you, yes, because of the bending of light emitted *inward* from the horizon by the curvature of spacetime inside the hole. That does not mean you are actually outside the horizon.

There will always be a horizon in front of you from which light cannot reach you. Or is that wrong?

Yes. Light from the horizon appears to be coming from in front of you, but the curvature of spacetime inside the horizon is extreme enough that you can't trust your intuitions about what that light is telling you.

If not then you could see the singularity couldn't you?

No. You can never see the singularity, because it's in your future; you can't see it for the same reason you can't see tomorrow.

George:

"As the proton is observed over shorter and shorter time intrevals, the region it is localized must grow."

I agree that black hole complementarity is controversial,

George, shame on you for lacking the 'Susskind faith'.....

I can't vouch for his mathematics, but his physical inetrpretations are facinating...

That's an insight that Susskind uses frequently....I immediately recognized your description from Susskind's THE BLACK HOLE WAR where he describes 'Alice's last
visible propeller'...imagining rotating propellers falling into a black hole...just as a local propeller slows down you can see more and more of it ,redshift resulting from falling into a black hole slows stuff down bringing larger portions of the object into view....

bcrowell
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To which solutions do you refer?
I had in mind the Vaidya metric, not because it's necessarily particularly important but because it's simple and I happen to have heard of it.

For these solutions: 1) does Hawking radiation stop once the event horizon has formed
It's a solution to the classical field equations, so it doesn't describe Hawking radiation. But isn't it pretty clear that in real-life cases, the time-scales for formation and evaporation are wildly mismatched?

2) does an external observer see a free falling particle fall through the event horizon in a finite time?
Interesting question, and I don't know the answer. I could cop out and say that this question is ill-defined, because we only have a well-defined notion of how long a process takes to a distant observer if the spacetime is static, but the Vaidya metric isn't static. We could also worry about whether there are both an event horizon and an apparent horizon (boundary of the union of all null timelike surfaces) in the case of the Vaidya metric.

If we want to make this discussion precise, probably the difficult issue is how to define the notion that according to a distant observer, a certain black hole has "now" formed an event horizon, in a case where the spacetime is nowhere static. By definition, a distant observer can't receive a signal from an event on the event horizon. But of course by any reasonable definition, Sagittarius A* has "now" formed an event horizon.

bcrowell
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At the end of this blog post http://backreaction.blogspot.com/2009/11/causal-diagram-of-black-hole.html there's a Penrose diagram for a black hole that forms by collapse and then later evaporates. I don't know if this is right and noncontroversial, but let's assume it is. The reasoning leading up to it seems OK on casual inspection by a non-expert (me).

Based on this diagram, it seems like we have kind of a cute resolution to this whole question. A distant observer sees a relatively short time-scale (years, maybe) for the horizon to be effectively established, where "effectively" means that in practical terms, there are infalling particles that we no longer have any hope of gathering information about. However, the completion of evaporation doesn't lie in the interior of the future light cone of any point on the horizon, nor does it lie in the interior of the light cone of points near the singularity. It's lightlike in relation to the horizon, so in that sense, there is no time delay between formation of the horizon and the completion of evaporation.

However, there is a delay in the following sense. We have the last infalling light ray that, in the Schwarzschild version, would have been able to escape to infinity. This ray moves through some nonzero (in fact, very large) affine parameter between the time when it joins the horizon and the time when it's liberated at evaporation.

At the end of this post http://backreaction..com/2009/11/causal-diagram-of-black-hole.html [Broken] there's a Penrose diagram for a that forms by collapse and then later evaporates. I don't know if this is right and noncontroversial, but let's assume it is. The reasoning leading up to it seems OK on casual inspection by a non-expert (me).

Based on this diagram, it seems like we have kind of a resolution to this whole question. A distant observer sees a relatively short time-scale (years, maybe) for the to be effectively established, where "effectively" means that in practical terms, there are infalling particles that we no longer have any hope of gathering information about. However, the completion of evaporation doesn't lie in the interior of the cone of any point on the , nor does it lie in the interior of the light cone of points near the . It's lightlike in relation to the , so in that sense, there is no time delay between formation of the horizon and the completion of evaporation.

However, there is a delay in the following sense. We have the last infalling light ray that, in the Schwarzschild version, would have been able to escape to infinity. This ray moves through some nonzero (in fact, very large) affine parameter between the time when it joins the horizon and the time when it's liberated at evaporation.

penrose solution involves a loss of unitarity,
for a non-unitary black hole evaporation you need a nonlinear quantum mechanics.

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PeterDonis
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At the end of this blog post http://backreaction.blogspot.com/2009/11/causal-diagram-of-black-hole.html there's a Penrose diagram for a black hole that forms by collapse and then later evaporates. I don't know if this is right and noncontroversial, but let's assume it is.

I've seen diagrams like this one before; I think one of Penrose's popular books has a similar one (maybe The Road To Reality?). The things I've been saying about a spacetime with an evaporating BH in it are based on this type of causal structure for the spacetime, which AFAIK is the currently accepted picture. (For example, it's easy to see from this diagram that any observer who finds himself inside the horizon as it currently exists when he falls in, regardless of how much the BH has evaporated by that point, is doomed to be destroyed in the singularity.)

I agree with your follow-on comments about why and in what sense there is a long time delay between the BH forming and its final evaporation.

PeterDonis
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penrose solution involves a loss of unitarity

I've just been re-reading Susskind's The Black Hole War, where he discusses this in some detail. He doesn't give an explicit Penrose diagram for his proposed solution (black hole complementarity), but from what he says in the book (which is, admittedly, not a scientific paper), it seems like such a diagram would look like the one in the blog post bcrowell linked to.

BH complementarity, as I understand it, entails that any quantum information that comes within the stretched horizon (basically within a Planck length above the horizon) gets transferred to the quantum field modes that eventually emerge in the Hawking radiation and thereby get back out to infinity, so unitarity is preserved. Susskind's answer to the question of why this doesn't violate the quantum no-cloning theorem--since there is also a copy of the infalling information that continues to fall in past the horizon and into the singularity--is that the copy that goes down the hole can't send any signals back out, so it's OK. Which sounds hand-wavy, but presumably there's a lot more technical detail in the actual papers.

bcrowell
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Here is my rendition of the a Penrose diagram for a collapsing and evaporating black hole, with all the bells and whistles. This is basically the one given here http://backreaction.blogspot.com/2009/11/causal-diagram-of-black-hole.html and here http://www.math.unb.ca/~seahra/resources/notes/black_holes.pdf , plus my idea of what apparent horizon should be like, based on the discussion in the second reference. I don't think it's consistent with the one at http://en.wikipedia.org/wiki/File:PENROSE2.PNG , which shows a three-way intersection of the infalling matter, singularity, and horizon, and has the singularity being initially timelike. The bow represents the infalling matter.

http://www.lightandmatter.com/penrose_diagram_for_black_hole.jpg [Broken]

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bcrowell
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The process of formation of the horizon also takes finite time in the following sense. At sufficiently early times, the spacetime is static and flat to an arbitrarily good approximation, so we can define a family of observers, all at rest relative to the dilute cloud of slowly collapsing matter, and let them synchronize their clocks in the usual way. These observers know which one of them is positioned at the center of the cloud, and can predict the reading on his clock at the event E when he crosses the horizon. Event E is, however, never in the past light cone of any observer who is outside the event horizon.

Yes. You've left out a third possibility: that the region of spacetime inside the horizon, even though it exists, can't send light signals out to the distant observer. The reason it can't is that the curvature of spacetime in that region is great enough to force even outgoing light to fall inward towards the singularity.
No, I don't think that is a third possibility. I am trying to ask about a discontinuity in the particles path. If the outsider can't observe anything once the particle reaches the horizon, the outside observer must observe a discontinuity, of the path of the particle at the time the observer *sees* the particle reach the horizon, no? If there is a part of the particles trip that cannot be observed, it is very hard to understand how there could not be a discontinuity.

Grom doing some further reading, it does sounds like evaporating BHs are not that well understood, and some points are an open question. I am wondering how much of what you are stating is actually an open to question and far from certain.

Yes. Light from the horizon appears to be coming from in front of you, but the curvature of spacetime inside the horizon is extreme enough that you can't trust your intuitions about what that light is telling you.

No. You can never see the singularity, because it's in your future; you can't see it for the same reason you can't see tomorrow.
I think you are thinking of a different coordinate system to me. Perhaps I am thinking of Rindler space... although, I don't really know much about anything.

If you are observing light that originated from behind your current position, it still got deeper into the BH then you at some point right?. Is it true that if you are inside the event horizon of a BH, then there is point between you and the singularity where on one side, light can reach you, and on the other side, light can't reach you. Couldn't this be considered a horizon that you can never cross? Would you expect the Unruh effect to take place here?

bcrowell
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I am trying to ask about a discontinuity in the particles path. If the outsider can't observe anything once the particle reaches the horizon, the outside observer must observe a discontinuity, of the path of the particle at the time the observer *sees* the particle reach the horizon, no? If there is a part of the particles trip that cannot be observed, it is very hard to understand how there could not be a discontinuity.

There is no discontinuity in the particle's path. If the particle is emitting a radio signal, then the outside observer simply sees the signal become infinitely weak and infinitely long in wavelength as the particle reaches the event horizon.

Continuity or discontinuity of particles' world-lines is something that all observers agree on. Since an observer free-falling along with the particle observes no discontinuity, neither does a distant observer.

PeterDonis
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No, I don't think that is a third possibility. I am trying to ask about a discontinuity in the particles path.

There isn't one. The path is continuous. Unless I'm misunderstanding the way you are using the word "discontinuity". See below.

If the outsider can't observe anything once the particle reaches the horizon, the outside observer must observe a discontinuity, of the path of the particle at the time the observer *sees* the particle reach the horizon, no?

Perhaps you are using the word "discontinuity" in a different way than I have been taking it. Consider a much simpler and more mundane example. When an object travels away from you on Earth, and eventually passes beyond your visual horizon, you can no longer see it; light from it can no longer reach you. Would you say that you perceive a discontinuity in the object's motion in this case?

If your answer is "yes", then I've misunderstood how you're using the word "discontinuity". If your answer is "no", however, then my answer to your question above stands, because the cases are in fact analogous; the trajectory of the infalling particle is continuous when it crosses the BH horizon, in the same sense as the path of the object moving away from you on Earth is continuous when it crosses your visual horizon.

If there is a part of the particles trip that cannot be observed, it is very hard to understand how there could not be a discontinuity.

See above.

Grom doing some further reading, it does sounds like evaporating BHs are not that well understood, and some points are an open question. I am wondering how much of what you are stating is actually an open to question and far from certain.

See bcrowell's posts in this thread and my comments on them. It's certainly true that our understanding of evaporating BH's is incomplete. However, the key claims I've been making are, I think, fairly solid.

Furthermore, the questions you're asking about things like discontinuities seen by the outside observer do *not* depend on our understanding of evaporating BH's; they apply just as well to "eternal" BHs that are static forever (i.e., they never gain mass and never evaporate), and those are *very* well understood; so, for example, the things I'm saying in this post are *not* open to question, unless you are questioning General Relativity in general (and that's a whole different discussion).

I think you are thinking of a different coordinate system to me.

The things I'm saying are coordinate-independent.

Perhaps I am thinking of Rindler space... although, I don't really know much about anything.

Perhaps a quick overview of some common coordinate charts for Schwarzschild spacetime (i.e., an "eternal" BH spacetime) and Minkowski spacetime (i.e., the flat spacetime of Special Relativity) will help; there are two key correspondences between coordinate charts on these two spacetimes:

Schwarzschild coordinates <=> Rindler coordinates

Kruskal-Szekeres coordinates <=> Minkowski coordinates

The following links might be of interest:

http://en.wikipedia.org/wiki/Schwarzschild_coordinates

http://en.wikipedia.org/wiki/Rindler_coordinates

http://en.wikipedia.org/wiki/Kruskal–Szekeres_coordinates

If you are observing light that originated from behind your current position, it still got deeper into the BH then you at some point right?

In a sense. Inside the horizon, *everything*, including light, is falling inward towards the singularity; but light, if it's moving in the right direction, can fall inward more slowly than an observer who is falling directly radially inward. So the observer "catches up" to the light since it's falling more slowly than he is, and to the observer, it seems like the light is coming at him from further inside the BH.

Is it true that if you are inside the event horizon of a BH, then there is point between you and the singularity where on one side, light can reach you, and on the other side, light can't reach you.

Light from where? As you get closer and closer to the singularity, there is less and less of your worldline remaining for light to reach; so as you get closer and closer to the singularity, light will have to be emitted from closer and closer to you to reach you before you hit the singularity. But there is no single point where light can't reach you from anywhere, until the instant you actually hit the singularity itself.

Would you expect the Unruh effect to take place here?

This is a very different question, and it may well deserve a separate thread. All the treatments I have seen of the Unruh effect only talk about the region outside the horizon.

Perhaps you are using the word "discontinuity" in a different way than I have been taking it. Consider a much simpler and more mundane example. When an object travels away from you on Earth, and eventually passes beyond your visual horizon, you can no longer see it; light from it can no longer reach you. Would you say that you perceive a discontinuity in the object's motion in this case?
I'm trying understand what happens in each reference frame with out having to refere to the other like you are here. However, I would think of hitting a singularity, or a particle fading to black with an infinite redshift as it hits the horizon at the point the BH evaporates as discontinuous. Maybe the terminology is wrong again. Maybe I should say, ill defined, or something else. However, my idea if continuos would mean that you observe see the particles path into the infinite future, however, it seems that it either ends at the singularity, or at the event horizon, as the BH evaporates, depending on weather you are with the particle or not.

Furthermore, the questions you're asking about things like discontinuities seen by the outside observer do *not* depend on our understanding of evaporating BH's; they apply just as well to "eternal" BHs that are static forever (i.e., they never gain mass and never evaporate), and those are *very* well understood; so, for example, the things I'm saying in this post are *not* open to question, unless you are questioning General Relativity in general (and that's a whole different discussion).
No, with an enternal BH, the particle will appear to slowly approach the horzon for all time, the path would apear continuous for all time, but observing an evaporating BH, something happens when the BH evaporates, apparantly, the particle will not be observable at the point the BH evaporates and it hits the horizon.

The things I'm saying are coordinate-independent.
According to the wikipedia quote I have earlier, which you stated was correct, an apparent horizon depends on how you slice the geometry.

Wouldn't it be possible to consider a space from the reference of the particle which is falling into the BH. Doesn't any coordinate system allow for that? I need to learn about rindler coordinates I think.

As I understand it, as a particle is observed approaching an event horizon, will would never be observed to cross the horizon, and it's red shift would tend towards infinity.

When an object that has some rest mass, is observed approaching an event horizon, at some point the light can't escape the object. The object is seen to be frozen and falling. A falling black spot will be observed, because there is horizon around the object, inside of which light can't escape.

Well, except that the two horizons are overlapping, so it looks like the large horizon stretches towards the approaching object.

When the object falling into a black hole is a black hole, that seems to be a quite straight forward thing:
http://www.black-holes.org/explore2.html

PeterDonis
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I'm trying understand what happens in each reference frame with out having to refere to the other like you are here.

Once again, the statements I'm making are coordinate-independent statements; they don't depend on reference frames. Consider again the analogy with an object moving away from you on Earth: is the object in a "different reference frame" when it passes beyond your visual horizon?

It is true that the absolute horizon partitions the spacetime into two distinct *regions*, one of which (the region inside the horizon) can't send light signals to the other. But that does not depend on reference frames. If you want to understand what's going on, I think it's a much better idea to look at the spacetime as a whole, as a single geometric object, with different observers traveling along different curves within that single geometric object. See further comments below under Rindler coordinates.

However, I would think of hitting a singularity, or a particle fading to black with an infinite redshift as it hits the horizon at the point the BH evaporates as discontinuous.

At the singularity at r= 0, yes, there is a true discontinuity: particle worldlines simply end there, destroyed by infinite curvature, in a finite amount of proper time.

At the horizon, no, there is no discontinuity. The worldlines of infalling particles continue smoothly into the interior region, without a break. Light signals emitted below the horizon can't get back out, but that doesn't make anything discontinuous.

Maybe the terminology is wrong again. Maybe I should say, ill defined, or something else. However, my idea if continuos would mean that you observe see the particles path into the infinite future, however, it seems that it either ends at the singularity, or at the event horizon, as the BH evaporates, depending on weather you are with the particle or not.

What you *see* of the particle's path depends on whether you are with it or not. But as I've said before, you need to keep clear the distinction between the particle's path itself, and what you see of it. The latter depends not just on the path, but on how light rays emitted from the path travel to you. The distant observer can't see the particle's entire path because of how the curvature of spacetime bends light rays. However, the distant observer can still calculate that (1) the particle takes only a finite amount of proper time, by its own clock, to reach the horizon, and (2) the curvature of spacetime is finite at the horizon, so there is nothing physically preventing the particle from continuing to fall inward. In other words, the distant observer can calculate that the particle's path is continuous at the horizon, even though he can't see it.

No, with an enternal BH, the particle will appear to slowly approach the horzon for all time, the path would apear continuous for all time, but observing an evaporating BH, something happens when the BH evaporates, apparantly, the particle will not be observable at the point the BH evaporates and it hits the horizon.

I'm not sure what you mean by this. When the distant observer sees the light from the BH's final evaporation, he also sees the light emitted by the infalling particle as it crossed the horizon. The latter event happened long before the former event, but the light from the two gets mingled together because of the way the spacetime is curved.

In any case, once again you are ignoring the key distinction between the particle's path itself, and what the distant observer *sees* of the particle's path. The particle's path itself is continuous at the horizon, regardless of whether the BH later evaporates or not. The evaporation of the BH only affects what the distant observer later sees; it doesn't affect the infalling particle at all (that particle gets destroyed in the singularity long before the BH evaporates).

According to the wikipedia quote I have earlier, which you stated was correct, an apparent horizon depends on how you slice the geometry.

Did you read my comments on that? The things we're talking about really refer to the absolute horizon, not the apparent horizon. The apparent horizon really doesn't enter into the issues we're discussing at all. (Also please note my clarification on the term "apparent" in apparent horizon.)

Wouldn't it be possible to consider a space from the reference of the particle which is falling into the BH.

Yes, take a look at Painleve coordinates:

http://en.wikipedia.org/wiki/Gullstrand–Painlevé_coordinates

Doesn't any coordinate system allow for that?

Not all coordinate charts cover the region inside the horizon; standard Schwarzschild coordinates do not. However, Painleve, Eddington-Finkelstein, and Kruskal-Szekeres coordinates all do. Here's a link on Eddington-Finkelstein coordinates:

http://en.wikipedia.org/wiki/Eddington–Finkelstein_coordinates

I need to learn about rindler coordinates I think.

That may help to understand how a coordinate chart can fail to cover the entire spacetime; Rindler coordinates only cover a portion of Minkowski spacetime, in a way that is very similar to how Schwarzschild coordinates only cover a portion of a black hole spacetime. As I noted before, Kruskal-Szekeres coordinates, in this analogy, correspond to Minkowski coordinates.

The key idea is to look at the chart that *does* cover the entire spacetime (Minkowski or Kruskal), and look for important causal boundaries--paths of light rays that are the boundaries of important regions of spacetime that can or cannot send light signals to each other. You will find that the other chart (Rindler or Schwarzschild) only covers one such region--usually called Region I--and that the other region--usually called Region II--can't send light signals to Region I. So observers that stay in Region I for all time can never see anything that happens in Region II.

I've just been re-reading Susskind's War, where he discusses this in some detail. He doesn't give an explicit Penrose diagram for his proposed solution ( complementarity), but from what he says in the book (which is, admittedly, not a scientific paper), it seems like such a diagram would look like the one in the blog post bcrowell linked to.

complementarity, as I understand it, entails that any quantum information that comes within the stretched (basically within a Planck length above the ) gets transferred to the quantum field modes that eventually emerge in the radiation and thereby get back out to infinity, so unitarity is preserved. Susskind's answer to the of why this doesn't violate the quantum no-cloning theorem--since there is also a copy of the infalling information that continues to fall in past the horizon and into the --is that the copy that goes down the hole can't send any signals back out, so it's OK. Which hand-wavy, but presumably there's a lot more technical detail in the actual papers.

he withdrew his last version of the article from arxiv, "Complementarity And Firewalls"
ever there is a trade off or you relaxes locality or permit the relax of unitarity.
Harlow (complementarity advocate) withdrew his paper (Complementarity, not Firewalls) too.

Black Holes: Complementarity or Firewalls?

...We argue that the following three statements cannot all be true: (i) Hawking
radiation is in a pure state, (ii) the information carried by the radiation is emitted
from the region near the horizon, with low energy eective eld theory valid beyond
some microscopic distance from the horizon, and (iii) the infalling observer encounters
nothing unusual at the horizon. Perhaps the most conservative resolution is that
the infalling observer burns up at the horizon.
Alternatives would seem to require
novel dynamics that nevertheless cause notable violations of semiclassical physics at
macroscopic distances from the horizon.

...It is widely believed that an external observer sees this information
emitted by complicated dynamics at or very near the horizon, while an observer falling
through the horizon encounters nothing special there. These three properties | purity of
the Hawking radiation, emission of the information from the horizon, and the absence of
drama for the infalling observer | have in particular been incorporated into the axioms of
black hole complementarity (BHC)...

...There would be an inconsistency if one were to consider a large Hilbert space that de-
scribes both observers at once. Such a Hilbert space appears when quantum gravity
is treated as an efective field theory, but it cannot be part of the correct theory of quantum
gravity if BHC holds...

...that it uses the naturally produced Hawking pairs rather than introducing ad-
ditional entangled ingoing bits. This leads us to a rather dierent conclusion, that the
thermalization time does not protect us from an inconsistency of BHC....

----
complementarity is not enough
---
or throw the second statement, "emited information".
this is gaining atention now.

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Did you read my comments on that? The things we're talking about really refer to the absolute horizon, not the apparent horizon. The apparent horizon really doesn't enter into the issues we're discussing at all. (Also please note my clarification on the term "apparent" in apparent horizon.)
I'm trying to learn about apparent horizons here, that's the point, but you are telling me that from all intent and purpose they are close enough to the same thing, but I thought that an apparent horizon is not close enough to the same thing at all, and can be quite different depending on how you frame the observation, so I am trying to work out if I am wrong, or your statement was incorrect.

Actually, the original point I was trying to make was that the from the point view of an observer following a BH, that crossing the absolute horizon has no effect on the path of motion when the crossing happens. It's just that I instead said that the would observe the event horizon receding from them as they approach, which may be what they see visually, but my semantics and terminology was all wrong.

Also, when I ask about if a path looks continuous to an observer, if on the surface of the earth, an observer watches an object pass over the horizon, then the observed path would end in a discontinuous way at the horizon.

Once again, the statements I'm making are coordinate-independent statements; they don't depend on reference frames.
I was trying to say, that I want to know what the distant observer sees of a particles path with respect to weather or not it is continous, with out having to refer to it's path after it passes the horizon, because that can only be observed from a different reference frame.

Anyway... although, I think there has been a lot of confusion of terminolgy, different observers and what I am trying to ask in this thread, my original question has definitely been answered now.

I now understand that if a distant observer sees a particle fall into an evaporating BH, they will observe the particle hitting the horizon at the moment the horizon disappears by evaporation. The final fate of the particle would be hidden from the distant observer, so it would not be observed being destroyed by the process hawking radiation antiparticles. The path of the particle after hitting the horizon would be not observable to the distant observer. However, from an observer at the particle, the particle would have been observed going all the way to the singularity. That observer would remain at the singularity until the BH evaporates.

This does bring up some new interesting questions, such as, how can all the light stuck on the horizon all dissipate at once from a singular point when the horizon evaporates, considering QFM. It makes me understand why there might have been be some debating about weather or not a BH can fully evaporate.

I still have some questions about apparent horizons, but I have given up trying to get them answered in this thread. I will think I might try to learn about that another time.

I also am curious now about how the Unruh effect my play out from the point of view of an observer following a particle towards the singularity. But, I there are a few things I probably need to teach my self first before I can ponder that.

I am done here for now. Thank you everybody for your patience and sorry for being confusing.

PeterDonis
Mentor
2020 Award
I'm trying to learn about apparent horizons here, that's the point, but you are telling me that from all intent and purpose they are close enough to the same thing, but I thought that an apparent horizon is not close enough to the same thing at all, and can be quite different depending on how you frame the observation, so I am trying to work out if I am wrong, or your statement was incorrect.

Perhaps a better way of putting what I was trying to say is that, if we are talking about the question you raised in the OP, the apparent horizon is not really relevant; all of the important points that involve the horizon at all, involve the absolute horizon, not the apparent horizon.

Actually, the original point I was trying to make was that the from the point view of an observer following a BH, that crossing the absolute horizon has no effect on the path of motion when the crossing happens.

If by this you mean that the infalling observer doesn't notice anything special about his motion, or about the spacetime in his immediate vicinity, when he crosses the absolute horizon, this is correct.

It's just that I instead said that the would observe the event horizon receding from them as they approach, which may be what they see visually, but my semantics and terminology was all wrong.

It's true that the term "apparent horizon" doesn't refer to this phenomenon at all, yes.

Also, when I ask about if a path looks continuous to an observer, if on the surface of the earth, an observer watches an object pass over the horizon, then the observed path would end in a discontinuous way at the horizon.

Ok, this clarifies how you are interpreting the term "discontinuity". As long as you draw a clear distinction between the "observed path" and the actual path, I think this usage is fine.

I was trying to say, that I want to know what the distant observer sees of a particles path with respect to weather or not it is continous, with out having to refer to it's path after it passes the horizon, because that can only be observed from a different reference frame.

Just to reiterate a point I made before, I don't think it's good terminology to say that the path inside the horizon "can only be observed from a different reference frame". A better way to say it would be that the path inside the horizon can only be observed from inside the horizon--i.e., from a different region of spacetime than the one the distant observer is in. That statement does not depend on "reference frames" at all, at least not in the normal usage of that term, where a "reference frame" refers to some particular set of coordinates. The fact that the path inside the horizon can only be observed from inside the horizon remains true regardless of which coordinates anyone is using.

Anyway... although, I think there has been a lot of confusion of terminolgy, different observers and what I am trying to ask in this thread, my original question has definitely been answered now.

Ok, good!

I now understand that if a distant observer sees a particle fall into an evaporating BH, they will observe the particle hitting the horizon at the moment the horizon disappears by evaporation. The final fate of the particle would be hidden from the distant observer...

...The path of the particle after hitting the horizon would be not observable to the distant observer. However, from an observer at the particle, the particle would have been observed going all the way to the singularity.

These statements are fine.

...so it would not be observed being destroyed by the process hawking radiation antiparticles.

I'm not sure what you're trying to say here, but probably it deserves a separate thread.

That observer would remain at the singularity until the BH evaporates.

Actually, no; that observer will be destroyed as soon as he hits the singularity. There is no way for that observer to go anywhere else; the singularity is a future endpoint for all worldlines that hit it.

(Strictly speaking, the above is our best current belief, but it is true that when we have a more complete theory of quantum gravity, our best current belief might change.)

I still have some questions about apparent horizons, but I have given up trying to get them answered in this thread. I will think I might try to learn about that another time.

I think the topic of apparent horizons deserves a separate thread, since, as I said above, apparent horizons aren't really relevant to the question you asked in the OP of this thread.

I am done here for now. Thank you everybody for your patience and sorry for being confusing.

You're welcome; I'm glad we were able to provide answers to at least some of the questions that you have posed.