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Imagine you have a large black hole, onto which you drop a small object. This object could have any shape you like, but later on it will look like a thin sheet, so we could imagine it starts that way, too. The important properties are that it it’s a good opaque absorber of radiation — that is, it’s approximately a blackbody — and that it has some finite temperature $T$.

        \
         \
         |
 b.h.    |               []
interior |                 falling sheet
         |
         /
        / horizon

We can make simplifying assumptions about the black hole: it’s nonrotating and uncharged, and large enough that no tidal spaghettification happens outside the horizon.

From the perspective of the falling sheet, there is nothing special about the horizon: it passes through without incident. During its travel, the sheet emits a finite amount of thermal radiation which depends on its temperature, with rest-frame power $P\propto T^4$.

For an outside observer, the information that the horizon has been crossed never becomes available. An outside observer will see the falling object approach the horizon asymptotically, eventually getting “stuck” an infinitesimal distance above the horizon. The finite amount of thermal energy which was emitted by the sheet on its way down gets stretched into infinite time by an ever-increasing redshift. However, a redshifted thermal spectrum is still a thermal spectrum. An outside observer will see our falling sheet approach the horizon and “freeze” there, both in the sense that it doesn’t move any more and also in the sense that the apparent temperature decreases forever, $T_\text{apparent}\to 0$.

The black hole also has a temperature, which goes like $T_\text{h} \propto 1/M$, and emits blackbody radiation corresponding to this temperature. This is cold (colder than the cosmic microwave background for any black hole which meets our assumptions, so the black hole is a net radiation absorber until the universe has cooled down some more), but it’s a finite cold.

I think this means that our falling object ought to eventually form a sunspot-like shadow in the Hawking radiation emitted by the black hole: a region of the surface where an outside observer sees the cold shadow of the infalling object, rather than the slightly-less-cold radiation from the event horizon:

        \
         \
         |
 b.h.    |[]
interior |  “frozen” falling sheet,
         |  apparent temperature $T_a$
         /
        / horizon, $T_h > T_a$

We have, after all, stipulated that the falling object is a good absorber of radiation, and (from our outside observation point) the falling object remains forever outside of the event horizon. We have already decided that the apparent temperature of the infalling object decreases forever, so the contrast between the event horizon and the “shadow” of our object should increase with time.

However, the idea that an object dropped onto a black hole forms a visibly cold “sunspot” makes an uncomfortable prediction. An outside observer with enough stuff to drop could cover the entire surface of the black hole with the shadows of recently-dropped items. This black hole would not emit thermal radiation at its Hawking temperature, because all of the bits of “bare” horizon will have been occluded by more recent stuff — stuff whose apparent temperature is decreasing without bound. This prediction, that you could “paint” a black hole by dropping things onto it, seems to conflict with the no-hair theorem. But the counter-claim, that you can’t paint the black hole, suggests that stuff you drop onto it eventually disappears, which isn’t consistent with the fact that information about stuff crossing the horizon never reaches an outside observer.

If you complain that both of these temperatures are too cold to observe directly, let’s arrange for the object we drop onto the black hole be its last non-radiation interaction with the universe. Eventually Hawking radiation will start to decrease the black hole’s mass and raise its temperature. But our dropped object should still be “stuck” above the event horizon, its apparent temperature only decreasing, still casting its shadow even as the black hole evaporates.

One way out of the muddle is to suggest that the outsider’s (illusory) perception of the infalling object, that it has “frozen” on the event horizon, breaks down in the redshift limit where $T_\text{apparent} < T_\text{horizon}$. As the apparent temperature goes below the Hawking temperature, the radiation from the black hole somehow overwhelms the radiation from the infalling object. This is probable in a statistical sense, if only because the infalling object will eventually be dimmer than an equivalent area of the black hole by a factor $(T_\text{a}/T_\text{h})^4$. With sufficient delay, we’ll eventually run into shot noise in the blackbody radiation from the infalling object, where the thermal photons just aren’t emitted very often any more.

But simply assuming the thermal radiation from an infalling object is overwhelmed by the brightness of the Hawking radiation doesn’t solve the problem of “painting” a black hole with opaque absorbers. Thus the title question: how does the Hawking radiation get out from behind an absorber dropped onto the black hole? Is Hawking radiation perhaps actually emitted from above the absorber, and therefore becomes visible as the dropped object appears (from outside) to squash sufficiently close to the event horizon? If so, is there actually some feature of spacetime near the event horizon associated with this emission?

(There was a period a few years ago when people started talking about a “firewall” at the event horizon, usually in the context of the black hole information paradox. When I started writing this question I thought it was independent of the information paradox, but now I’m not so sure.)

After I’d written most of this question I found this near-duplicate. However I think that breaking spherical symmetry makes the puzzle clearer, so I’m asking this one rather than bountying that one.

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After I’d written most of the question I found this answer, which says in part

Another way [to think about Hawking radiation], perhaps more useful here, involves de Broglie 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 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.

Without having done the due diligence of a sample calculation myself, this applies to the present question in two different ways.

First, it’s a mechanism for the “frozen” image of the infalling object to dissolve on the event horizon. Not only do we have shot noise, where photons from the infalling object are emitted only infrequently as it appears to cool, but also we have diffraction effects. As the light emitted by the infalling object gets redshifted to wavelengths larger than the object’s actual size, diffraction destroys our ability to trace that light back to a specific part of the object.

Second, from the other direction, this means that the Hawking radiation from the black hole can never be traced back to a specific part of the event horizon. (That sounds like a statement that, if I were a proper relativist, I would be able to relate back to the no-hair theorem.) Diffraction means that the best I can do to pinpoint the origin of a particular quantum of Hawking radiation is to say that it came from a volume of space roughly the size of the black hole. And furthermore this breaks the assumption I made at the very beginning of my question: an object which is smaller than a black hole cannot be opaque to Hawking radiation, because the Hawking radiation is sufficiently long-wavelength to get around it by diffraction.

I was missing a key feature of the outside-observer perspective of an object falling onto a black hole. I knew that it would asymptotically approach the horizon, and there its emitted light would get dimmer and redder. But I didn’t understand until today that diffraction meant the image would eventually also get blurrier.

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    $\begingroup$ According to the Hawking radiation calculator the wavelength of the peak photons of the Hawking blackbody spectrum is $20.1385r_s$. Eg, a 1e6 solar mass SMBH has a Schwarzschild radius $r_s$ of 2.95399E9 m & a temperature of 6.16871E-14 K. The peak photons of its spectrum have a wavelength of 5.94890E10 m. So the region where the radiation appears to be emitted from is quite large. Of course, actually detecting such cold low power radiation isn't easy. ;) $\endgroup$
    – PM 2Ring
    May 8, 2021 at 20:53
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    $\begingroup$ @PM2Ring To clarify, that's the wavelength at the far-away observer. To reach that location, the radiation must climb up out of an intense gravitational well, which means it's been intensely redshifted by the time it gets all the way out there. Naively, this might seem to imply that an observer falling into the black hole would encounter very high-frequency outgoing radiation, which could be blocked by a thin screen. But that conclusion is incorrect, and explaining why it's incorrect could make an interesting auxiliary answer, even though it doesn't change rob's conclusion (+1). $\endgroup$ May 8, 2021 at 21:11
  • $\begingroup$ @ChiralAnomaly Indeed! I didn't mention the redshift, since I figured Rob already knew about that. ;) But it'd be good if you do decide to post an auxiliary answer. $\endgroup$
    – PM 2Ring
    May 8, 2021 at 21:18
  • $\begingroup$ this means that the Hawking radiation from the black hole can never be traced back to a specific part of the event horizon That is not completely true. Planckian spectrum of Hawking radiation has photons of arbitrarily high energies and azimuthal quantum number (though with exponentially suppressed probabilities), so if horizon has truly static features, as OP seems to suggest, patient observer can reconstruct them with detectors sensitive only to short wavelengths photons. $\endgroup$
    – A.V.S.
    May 9, 2021 at 18:31
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This is a supplement to rob's answer. It highlights a paradox that was posed in a comment below that answer, and then attempts to resolve it.

Paradox

Here's the paradox: According to the far-away observer, the temperature $T$ of the Hawking radiation is $k_B T\sim \hbar c/R$, where $R$ is the Schwarzschild radius. Therefore, the radiation should have a typical wavelength $\sim R$. However, in order to reach the far-away observer, the radiation had to climb up out of an intense gravitational well, which means it's been intensely redshifted by the time it gets all the way out to the far-away observer. Therefore, it must have had an extremely high frequency when it left the vicinity of the horizon. Naively, this might seem to imply that an observer falling into the black hole would encounter this very high-frequency outgoing radiation.

To reinforce the paradox, recall that Hawking's original derivation of Hawking radiation used the extremely high frequency (short wavelength) of the outgoing modes to justify using the geometric optics approximation, which is how he turned an otherwise-intractable calculation into something tractable.

So... does an infalling sheet encounter high-frequency outgoing radiation near the horizon? If not, then why not?

Names for various observers

Hawking considered the formation of a black hole by the collapse of a star. I'll use these names for the various observers of interest:

  • The E-observers: These are inertial observers at early (E) times, long before the black hole forms, at rest with respect to the center of the star.

  • The L-observers: These are nearly-inertial observers at late (L) times, long after the black hole forms, at rest with respect to the black hole and far away from the black hole.

  • An H-observer: This is a very non-inertial observer hovering (H) just outside the horizon of the black hole after it is fully formed.

  • An F-observer: This is an inertial observer falling (F) radially inward, somewhere near the event horizon (outside, passing through, or inside) of the fully formed black hole. This observer is falling alongside the thin sheet that was described in the question.

Vacuum vs radiation in QFT

By definition, the vacuum (empty) state is the state of lowest energy. And by definition, the energy operator is the operator that generates time-translations. But time is observer-dependent, and a given observer's proper time typically doesn't have any natural unique extension to the whole spacetime, so the concept of vacuum state inherits these ambiguities. I wrote more about this in another answer and another older answer and another even older answer.

More explicitly, we can relate the (local and observer-dependent) concept of vacuum state to the (local and observer-dependent) concept of positive and negative frequency (gr-qc/0308048). Consider the Heisenberg picture, where all time-dependence is carried by the observables, and consider a free massless scalar field $\phi$, like Hawking did. For each complex-valued function $f$ that satisfies the wave equation (the equation of motion for a free massless scalar field), we can define an operator $\phi(f)$ called a mode operator. These operators satisfy $[\phi(f),\phi^\dagger(g)]=(f,g)$, where $(f,g)$ is a particular Lorentz-invariant scalar product of the two functions $f,g$. In this context, $f$ and $g$ are called mode functions.

Here's the key: if $f$ a positive-frequency solution of the wave equation, then the mode operator $\phi(f)$ is an annihilation operator: it reduces the energy of any state on which it acts. Its adjoint $\phi^\dagger(f)=\phi(f^*)$ is a creation operator, which increases the energy of any state on which it acts. The vacuum state is defined to be the state of lowest energy, so it can equivalently be characterized as the state that is annihilated by the annihilation operators. The separation into positive- and negative-frequency parts (annihilation and creation operators) can be different for each observer, because time can be different for each observer.

The initial state

Since "vacuum" is observer-dependent and typically only defined locally, we need to be clear about how we're choosing the global state. In the far past, long before the star collapses, the spacetime metric is flat. We can assume that the "star" is initially so diffuse that it doesn't cause any significant spacetime curvature. At those early times, we can fill the space with E-observers (defined above) and choose a state that all of them agree is empty, everywhere. The fact that such a global vacuum state exists is a special feature of inertial observers in flat spacetime.

As explained in other answers (this one by Joe Schindler and this one by me), this is why we consider the formation of a black hole instead of considering a black hole that has been around forever, even though the latter would make the math easier.

Hawking radiation

A mode function $f$ is a solution of the wave equation. In the dynamic spacetime of a collapsing star, a solution of the wave function that is initially purely positive-frequency according to the E-observers ends up having both positive- and negative-frequency parts according to the L-observers. So even if the state is initially empty according to the E-observers, it ends up being non-empty according to the L-observers. That's Hawking radiation.

Here's a contrived example to illustrate how this frequency-sign mixing can happen. Consider the time-only version of the wave equation: $$ \ddot f + \omega^2 f = 0, \tag{2} $$ where overhead dots denote time-derivatives. If $\omega$ is a constant, then $f(t)=e^{i\omega t}$ is one solution. This solution has positive frequency forever. But now suppose that $\omega$ is not constant. Suppose that $$ \omega = \begin{cases} \omega_1 & \text{ for }t<0 \\ \omega_2 & \text{ for }t>0. \end{cases} \tag{3} $$ If we require that $f(t)$ and $\dot f(t)$ be continuous, then a solution that is equal to $e^{i\omega t}$ for $t<0$ ends up being equal to $ae^{i\omega t}+be^{-i\omega t}$ for $t>0$, with $b/a=(\omega_2-\omega_1)/(\omega_2+\omega_1)$. In words, as a result of redshifting (reducing the frequency), a solution that initially involved only positive frequency ends up involving both positive and negative frequencies.

Hawking used a ray-tracing argument do deduce how a mode function that is purely positive-frequency for and E-observer can end up having a negative-frequency part for an L-observer. Parker and Toms described the relevant geodesicts like this, on page 154 in the first edition of their book:

...the properties of the massless quanta created by the gravitational field of a black hole are governed at late times by the null geodesics that begin far outside the collapsing body at early times, move inward through the body to become outgoing null geodesics, and escape from the collapsing body just before it collapses within the black hole event horizon. These outgoing null geodesics reach future null infinity... at arbitrarily late times.

The details are nontrivial, but careful calculations (reviewed in section 2 in gr-qc/9707062 and very nicely reviewed in section 7.3 in gr-qc/9707012) show that a wave that had only positive frequencies according to the E-observers ends up involving both positive and negative frequencies according to the L-observers, as a result of its close encounter with the nascent event horizon. This implies that the L-observers experience a state that is not their vacuum state. That's Hawking radiation.

Does an infalling observer encounter high-frequency radiation?

Hawking's calculation compared the experiences of E-observers and L-observers. To address the paradox, we need to compare the experiences of E-observers and F-observers (infalling thin sheet) instead. We need to consider modes that have positive frequency according to the E-observers and determine whether or not they end up with significant negative-frequency parts according to the F-observer. I'll quote a couple of results from the literature.

Unruh's paper, the one that introduced what we now call the Unruh effect, analyzed the infalling-observer problem, but for an eternal black hole instead of one that forms by collapse. This is his conclusion on page 888:

The [infalling] geodesic detector therefore sees no outflow of particles, but does see an influx of particles into the black hole.

His argument uses an eternal black hole, but I don't think that's essential. Interestingly, even though he concludes that the F-observer doesn't see any outgoing radiation, the F-observer allegedly does see ingoing radiation. The key input to his argument seems to be the failure of the geometric-optics approximation for large wavelengths: an outgoing radial wave is partly reflected back toward the black hole. This is a large-wavelength effect, though, so it's consistent with the expectation that the F-observer doesn't experience any high-frequency radiation.

For additional confirmation, the first page in arXiv:1609.06584 says this:

while an infalling observer would not see all the particles which are excited in this space-time, he will definitely observe modes whose wavelengths are comparable or larger than his local curvature radius. Only those modes whose wavelength is shorter than the local curvature radius around him would remain invisible.

These quotes suggest that the outgoing high-frequency modes are not populated for the F-observer (to a good approximation, for sufficiently high frequencies), because no significant mixing of positive- and negative-frequency parts occurs in going from E-observers to F-observers. This addresses the paradox, at least in the weak sense of revealing a loophole in the naive argument that the infalling observer encounters high-frequency radiation.

By the way, the high-frequency modes are populated according to the hovering H-observer (as high-frequency modes, not yet redshifted). That's perfectly consistent with the conclusion that they're not significantly populated according to the F-observer, even of the F- and H-observers are temporarily co-located. This is the Unruh effect, which is a local phenomenon. It compares the experiences of two differently-moving observers in the same part of spacetime. If the accelerating observer stops accelerating, the effect disappears immediately. For more about the distinction between the Unruh and Hawking effects, see tparker's answer to the question Intensity of Hawking radiation for different observers relative to a black hole, the answers to the question Hawking radiation and Unruh effect, and the paper gr-qc/9912119. For one perspective on the similarity between the two effects, see hep-th/9809159.


An earlier version of this answer went on to say this:

Hawking radiation is a large-scale effect. An infalling thin sheet cannot block it, because there is nothing to block: the high-frequency outgoing modes are not even significantly populated (according to the F-observer/sheet) until after they've propagated out beyond the near-horizon region, and by then they're already mostly redshifted.

But as emphasized in a comment by A.V.S., the preceding argument doesn't justify this conclusion.

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    $\begingroup$ A powerhouse answer. It’ll take me a while to digest. Many thanks! $\endgroup$
    – rob
    May 10, 2021 at 3:28
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    $\begingroup$ A infalling thin screen cannot block it, because there is nothing to block: the high-frequency outgoing modes are not even significantly populated And with this phrase you are sweeping the “paradox” under the rug, which makes the preceding wall of text irrelevant at best. This is like saying that modification of potential in classically forbidden region (in the sense of ordinary QM) cannot change the tunneling rate because the probability of finding particle here is very small. Details of interaction of sheet with quantum field do matter here, and handwavy argument is not sufficient. $\endgroup$
    – A.V.S.
    May 10, 2021 at 6:55
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    $\begingroup$ To me this question seems like an aspect of “trans-planckian problem” of Hawking radiation, namely that the answer in principle depends on the UV behavior for coupling of infalling matter (which probably must be treated as a quantum field) and the quantum field being radiated by the BH. $\endgroup$
    – A.V.S.
    May 10, 2021 at 16:04
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    $\begingroup$ @A.V.S. Yeah, I'm having second thoughts. Most of my intuition (and of my "answer") is based on free field theory. Even equation (1). The conclusion seems to have stood the test of time (I don't see anybody today questioning whether the infalling star can block Hawking radiation), but that doesn't justify the reasoning. Will probably delete the answer, but will wait until you see this comment. $\endgroup$ May 10, 2021 at 18:14
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    $\begingroup$ Please don't delete it! At least wait for more people to read it & respond to it. But maybe put a disclaimer at the top. $\endgroup$
    – PM 2Ring
    May 10, 2021 at 18:18

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