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I try to understand two principles formulated by Leonard Susskind in his book The Black Hole War:

1, To any observer who remains outside a black hole, the stretched horizon appears to be a hot layer of horizon-atoms that absorb, scramble, and eventually emit (in the form of Hawking radiation) every bit of information that falls onto the black hole.

2, To a freely falling observer, the horizon appears to be absolutely empty space. [...]

Now, in reaction to (an unanswered) question by Nathaniel, let us suppose that a distant observer A sees (according to the principle #1) Hawking radiation from a black hole (with Planck spectrum and say measurable temperature). Now suppose there is a freely falling gas somewhere between the event horizon and the observer. According to the principle #2, to the gas (considered an observer B) the horizon is empty space so there is no light coming from it to absorp.

Would the observer A detect absorption lines in the black hole spectrum? How does it change with distance of the gas (or the observer) from the black hole? Will the distant observer see the lines if he is free falling?

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  • $\begingroup$ And note that any answer will have to repeat this question for the case of Unruh radiation. $\endgroup$ – Jerry Schirmer Dec 14 '13 at 21:41
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This is not a full answer, since I don't know the full answer, but it is more than a comment.

My contribution is to compare your question with another, simpler one, and then come back to yours.

1. Observer-dependence of radiation in classical physics

Here is the simpler question (one to which an answer is known). A charged particle accelerating in empty space will emit electromagnetic radiation. A charged particle fixed at some point in a static gravitational field will not emit electromagnetic radiation. Both statements are true---relative to a certain natural choice of reference frame in each case. But, in the first example one could step on board a rocket accelerating with the particle, and no electromagnetic waves would be seen in the rocket frame. In the second example, one could go freely falling past the particle, and in this freely falling frame, electromagnetic radiation will be seen. So what is going on here? Does the charged particle emit radiation or doesn't it?

All of this scenario can be treated with special relativity, and it teaches good lessons to prepare us for general relativity. The main lesson here is that the process of emission and subsequent absorption of radiation is not absolute but relative, when accelerating reference frames are considered, but the state changes associated with absorbing radiation, such as a detector clicking, are absolute. It is the way we interpret what caused a detector to click that can change from one frame to another.

To make the above really connect with your question, note that in my simple scenarios I could imagine a cloud of gas possibly absorbing the radiation in between emitter and receiver, just like in your scenario.

2. To resolve a paradox in observer-dependent physics, first convince yourself about the easiest observer, then seek arguments to explain what the other observer finds

The above principle can be applied to resolve puzzles in relativity such as whether a fast pole can fit into a short barn, or whether a fast rivet can squash a bug in a hole.

3. Unruh effect has two complementary physical interpretations, depending on who is accelerating

Hawking radiation is like Unruh radiation, and therefore it is more subtle than classical radiation. A useful tip from the consideration of Unruh radiation is as follows. Unruh's calculation says a detector accelerating through the vacuum picks up internal energy, equivalent to detecting particles. Now if we look at this detector from an inertial frame, we still conclude it picks up excitation, but we interpret differently: we say the force pushing it provided some energy which got converted into internal energy because the process is not perfectly smooth.

4. Answer

Now I will apply all the above to provide an answer to your question. I admit I am not sure and the following answer is just my guess. I only claim it to be an intelligent guess.

My guess is that the distant observer observes Hawking radiation and absorption lines. I say this because it is a consistent summary of what seems to me to be ordinary physics, assuming that there is Hawking radiation coming up from a horizon.

So the puzzle is to explain this from the point of view of the freely falling cloud. I think an observer falling with the cloud looks up at his distant friend and notices that his friend is accelerating through the vacuum, and consequently experiencing internal excitation owing to fluctuation of the forces accelerating him. To account for the absorption lines, i.e. the absence of excitation at certain frequencies, I guess (and this is the speculative part) that now the calculation from quantum field theory would have to take into account that the rest of spacetime is not empty but has the cloud you mentioned, and this cloud affects the overall action of the quantum fields in this way. Note, it is not a case of action at a distance (in either perspective), but it is a highly surprising prediction so I think your question is a very interesting one.

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  • $\begingroup$ Your “simpler question” obfuscates the issue rather than clarifies it. Radiation of a constantly accelerating charge vs. a static charge in a static gravitational field is a purely classical problem. Excitation by Hawking radiation of an infalling gas is a fully quantum problem ultimately solvable by a calculation of response of atoms in the spirit of Unruh-DeWitt detector model. $\endgroup$ – A.V.S. Jan 12 at 18:10
  • $\begingroup$ The very fact that the simpler problem is purely classical is why it helps here: it illustrates that different accounts by observers in relative acceleration already happens with classical radiation. $\endgroup$ – Andrew Steane Jan 12 at 21:49
  • $\begingroup$ Null infinity is an observer independent entity, so what is radiated in classical physics does not depend on observers unlike the local particle content in quantum theory. “Different accounts” you mention occur by applying formulas outside of their context, in the case of accelerating charge this means ignoring the unphysicality of situation, namely that charge cannot be in constant acceleration eternally, it must start and stop sometime, that should reconcile your different observers. Hence my use of the term “obfuscate”. $\endgroup$ – A.V.S. Jan 13 at 5:11
  • $\begingroup$ Thanks for that; I see what you mean and I admit it is relevant. However: (1) a charged body at fixed location relative to a massive body is not unphysical; (2) neither is an accelerating rocket carrying a charged body for a finite time, but according to inertial observer such a charge is radiating continuously, where 'radiating' means 'putting energy into a disturbance of the field which propagates away, conserving its own energy as it goes'. An inertially moving cloud can absorb such radiation without reference to null infinity, and the rocket observer must account for this.. $\endgroup$ – Andrew Steane Jan 13 at 13:05
  • $\begingroup$ And here we are encountering things that are observer (or rather reference frame) dependent, namely the separation of EM field into “radiation” and "bound” parts in a finite region at a given moment. But in a classical problem, that still does not introduce any ambiguity because atom/detector does not interact with just “radiation”, it interacts with EM field, and EM field is observer independent. $\endgroup$ – A.V.S. Jan 13 at 14:46
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Your point #1 has to do with the fact that some physicists believe that if information did not RETURN, it would violate the Unitary principle. The philosophy of Quantum Mechanics demands that Unitary is immutable. Therefore some people had come up with theories to show that information is not really lost in the blackhole.

I have not read the book you mention, but I happen to be reading The Rode to Reality from Penrose, and there is a section in the book where he actually touches on this same subject. However, I am afraid that Penrose, does not share this view, in effect in relation to your Point #1. To quote him (page 841), "I find it inconceivable that somehow 'at the moment just before the horizon is crossed' some sort of signal is emitted to the outside world conveying outwards the full details of all information contained in the collapsing material...Simply a signal would be by itself not be enough, since the material itself is, in a sense, really the "information" that one is concerned with. Once it has fallen through the horizon, the material is trapped, and is inevitably destroyed in the singularity itself"

Perhaps this is not the answer you were looking for, but I hope that it is of some value to you.

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  • $\begingroup$ Penrose is not denying that Hawking radiation happens. He is merely saying that he doubts that such signals convey the full details of the information going into the hole. $\endgroup$ – Andrew Steane Jan 15 at 13:37
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First of all sorry for answering you not having a physics education.

If the gas does not fall into the black hole (there is at least a minimal distance between the event horizon and the gas molecules) the absorption would behave the same way as if the black hole was a "regular" star that has the same spectrum as the Hawking radiation.

Because of the high speed of the gas and the large masses relativistic effects must be considered.

As far as I understand the Hawking radiation's source is the empty space near the event horizon of the black hole and not the black hole itself. That theory says that in empty space photons and anti-parts with a negative (!) energy (and mass) are formed but normally these pairs will destroy themselves at once. Near a black hole the anti-parts fall into the black hole while the photons depart from the black hole. (If this is true then Nathaniel's theory is wrong - the Hawking radiation is formed when the black hole is already existing.)

This would mean that the gas between the event horizon and the point where the Hawking radiation originates does not influence the Hawking radiation (while the gas that is more distant from the black hole behaves like the gas close to a "regular" star).

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...suppose that a distant observer A sees... Hawking radiation from a black hole... Now suppose there is a freely falling gas somewhere between the event horizon and the observer. ... Would the observer A detect absorption lines in the black hole spectrum?

Here's the picture:

enter image description here

The black hole is assumed to be non-rotating, with Schwarzschild radius $R$, and the observer is far away ($r_A\gg R$). Suppose that the velocity of the freely falling gas is close to zero (with respect to the observer and with respect to the location of the black hole), as though it had only recently been dropped. Suppose that the gas absorbs all radiation with a given wavelength $\lambda_0$. Will the distant observer see this absoprtion line in the spectrum of Hawking radiation? These are my conclusions:

  • If the gas is dropped far away from the event horizon ($r_B\gg R$), then the distant observer will see an absorption line at practically the same wavelength $\lambda_0$. (The freely falling gas does not see empty space when $r_B\gg R$. It sees Hawking radiation just like the distant hovering observer does, because the acceleration of gravity is negligible when $r_B\gg R$, so the difference between hovering and free-fall is negligible.)

  • If the gas is dropped at some intermediate value of $r_B$ where the acceleration of gravity is noticeable but not extreme, then two things happen: (1) the absorption line is redshifted to a noticeably longer wavelength $\lambda > \lambda_0$ as seen by the distant observer, and (2) even if the gas adjusts its absorption wavelength $\lambda_0$ to keep the absorption line at the original wavelength from the distant observer's perspective, the absorption line will be very slightly weaker, because the state seen by the infalling gas is very slightly closer to being empty.

  • If the gas is dropped very close to the event horizon (so that $r_B-R\lesssim R$), then I don't know what happens. If $\lambda_0$ is held fixed, then I think the absorption line disappears (in part because it is redshifted into the extreme tail of the distant observer's blackbody spectrum, where there is no significant Hawking radiation anyway); but if $\lambda_0$ is adjusted to keep the absorption line at a fixed wavelength from the distant observer's perspective, then $\lambda_0$ becomes much less than the Planck length, where none of these calculations can be trusted.

Here's a quick review of the foundation. A standard derivation of Hawking radiation uses quantum field theory in curved spacetime, as introduced in 1 and [2]. In the model that I'm using here (which is standard), the spacetime metric is taken to be one corresponding to the formation of a black hole by a collapsing star. In the simplest version of the model, the black hole is non-rotating, and the quantum field is a massless scalar field with no interactions other than the influence of the prescribed spacetime metric. The state is assumed to be empty (no quanta) according to inertial observers in the distant past, long before the formation of the black hole. Here, "empty" refers to the quantum field; the material that constitutes the collapsing star is not explicitly modeled, except to motivate the choice of spacetime metric.

My conclusions are based on the following key points:

  • Key point #1: According to [3] and [4], that assumption about the nature of the state in the distant past implies that that same state (in the Heisenberg picture) has the following property: According to observers that are in free-fall across the event horizon after the black hole is fully formed, the state is practically empty of any field quanta having wavelengths $\lambda\ll R$. This result is attributed to the adiabatic principle. According to page 7 in [4], "practically empty" means that such quanta are suppressed by a factor $\sim \exp(-O(R/\lambda))$. This is a more-careful version of the statement shown in the OP: "To a freely falling observer, the horizon appears to be absolutely empty space."

  • Key point #2: According to [3] and [4], key point #1 is sufficient for deducing that a distant observer will experience Hawking radiation. In other words, the Hawking radiation experienced by the distant observer and the empty space experienced by an object falling across the horizon are not merely consistent with each other; the latter actually implies the former.

Key point #2 is a non-trivial and non-obvious result, highlighted in seciton 6.2.3 of [3] and on page 8 in [4]. This key point is central to the derivation of Hawking radiation. Because of its important role in thinking about the OP's question, I'll briefly outline how it is derived. Start with the usual Schwarzschild metric $$ ds^2=-A(r)dt^2+\frac{dr^2}{A(r)}+\text{(angular part)} \hskip2cm A(r)=1-\frac{R}{r}. \tag{1} $$ The angular part will be omitted from now on because it's less important here. For $r\gg R$, the coordinate $t$ becomes the same as the distant hovering observer's proper time. Therefore, if the field operator $\phi(t,r)$ is decomposed into parts having negative and positive frequency with repsect to the $t$-coordinate, those parts act as creation and annihilation operators ($b^\dagger$ and $b$, respectively) for any field quanta experienced by that observer. On the other hand, for an object falling freely very close to the event horizon, the object's proper time $\tau$ is given by (page 36 in [3]) $$ \tau\approx \tau_0-\tau_1 \exp\left(-\frac{u}{2R}\right), \tag{2} $$ where $\tau_0$ and $\tau_1$ are constants (with $\tau_1$ depending on exactly where the object was dropped) and where $$ u=t-r-R\log\left(\frac{r}{R}-1\right). \tag{3} $$ If the same field operator $\phi(t,r)$ is decomposed into parts having negative and positive frequency with respect to $\tau$, then those parts act as creation and annihilation operators ($a^\dagger$ and $a$, respectively) for any field quanta experienced by this infalling object, which is the gas in our case. Key point #1 says that very near the event horizon, the state $|U\rangle$ experienced by the infalling gas is practically empty, so $a|U\rangle=0$ for all of the annihilation operators $a$. But positive/negative frequency with respect to $\tau$ is not the same as positive/negative frequency with respect to $t$, so $a$ is a mixture (linear combination) of the operators $b$ and $b^\dagger$ that annihilate and create quanta as experienced by the distant observer. This implies $b|U\rangle\neq 0$, so the state is not empty from the distant observer's perspective. Working out the details using the relationships (2)-(3) shows that the distant observer experiences Hawking radiation with temperature $T\sim 1/R$. This is what leads to key point #2: the fact that the state is practically empty according to an object falling across the event horizon implies that the distant observer experiences Hawking radiation.

This suggests that the absorption line should disappear as the gas comes very close to the event horizon (cf the exponential factor in (2)), even if $\lambda_0$ is adjusted to keep the wavelength fixed from the distant observer's perspective, but I'm not sure about that because compensating for the extreme redshift requires that $\lambda_0$ become much less than the Planck length, where none of this can be trusted anymore.

By the way, to put things in quantitative perspective, recall that the Hawking radiation from even a small stellar-mass black hole is much colder than the cosmic microwave background and has an extremely low luminosity. Quantitatively, using $1$ kilometer for the Schwarzschild radius $R$, the Hawking temperature seen by the asymptotically-distant observer would be only $T\sim \hbar c/k_BR\sim 10^{-6}$ Kelvin and the total luminosity (integrated over the whole celestial sphere) would be only $L\sim\hbar c^2/R^2\sim 10^{-23}$ Watts. Larger black holes are even colder and have even lower luminosity. So, in practice, absorption lines would be very difficult to observe, simply because the Hawking radiation itself would be very difficult to observe. But that's beside the point of the question. The point of the question is to make us think more carefully about how Hawking radiation works.


References:

$[1]$ Birrell and Davies (1982), Quantum Fields in Curved Space, Cambridge University Press

[2] Parker and Toms (2009), Quantum Field Theory in Curved Spacetime, Cambridge University Press

[3] Jacobson (2004), "Introduction to quantum fields in curved spacetime and the Hawking effect," https://arxiv.org/abs/gr-qc/0308048

[4] Section 3 in Polchinski (2016), "The black hole information problem," https://arxiv.org/abs/1609.04036

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The large bounty encourages me to post a placeholder answer which I shall endeavor to improve over the next 24 hours... The Unruh effect has already been mentioned a few times - sometimes said to show that the existence of particles is relative to reference frame. Indeed, there is a reference frame in which an Unruh detector appears to emit a particle.

I suggest that the way to model this question is to treat the cloud of gas, as it crosses the event horizon, similarly to an Unruh detector; and to use something like Raju-Papadodimas formalism for AdS/CFT, to describe the two complementary scenarios (observer outside the event horizon, observer crossing the event horizon) in a unified way. I shall search to see if something like this has already been discussed in the literature on the information paradox.

update: Another way to think about this question... For a large black hole, the falling gas sees no significant difference immediately above and below the horizon. But for the distant observer, there is a radical difference between something that is outside the black hole and absorbing some of the Hawking radiation, and something that is part of the black hole and (in effect) producing some of the Hawking radiation. Yet by black hole complementarity, there should also be no significant difference between these two situations in the case of the falling gas. How can that be?

The answer must be a kind of Unruh-Hawking correspondence. The gas falling through the gently curved space just inside the horizon participates in some form of Unruh effect; and for the observer outside this horizon, this now corresponds to a continuing suppression of Hawking radiation on the wavelengths that the gas absorbs. Black hole complementarity demands that this should be so, and the challenge is to devise an account purely in terms of horizon thermodynamics (e.g. certain excited states of the black hole?) which reproduces this suppression.

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