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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 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). $\endgroup$ – Chiral Anomaly May 8 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 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 at 18:31
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This is a supplement to rob's answer. It highlights a paradox, and then attempts to resolve it, but the reasoning is flawed. It's based on how Hawking radiation works for a free quantum field, and then it tries to use a handwavy argument to account for the infalling sheet, but a proper derivation would need to use an interacting quantum field theory.

I edited the answer to flag the dubious part of the argument, using the word DUBIOUS in all caps, so you can spot it easily. (It's near the end of the post.) I'm leaving the answer un-deleted for now, because by reviewing the free-field case, it might at least help inspire thought about what might be different in the interacting-field case. It also suggests that if an infalling thin sheet could block Hawking radiation, then the star that collapsed to form the black hole should also block Hawking radiation, forever, which seems unlikely... and therefore maybe we can learn something about the seemingly subtle infalling-sheet problem by thinking about the not-so-subtle infalling-star problem.

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, which could be blocked by a sufficiently durable thin sheet. This seems to contradict rob's conclusion... and that's a paradox, because I think rob's conclusion is correct.

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... why can't an infalling sheet block the outgoing Hawking radiation before its wavelength gets redshifted to $\sim R$?

Names for various observers

To derive his famous result, Hawking considered the formation of a black hole by the collapse of a star. To make the rest of this answer easier to write, 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.

Some basic concepts from QFT

The resolution of the paradox refers to some basic concepts from QFT. I'll review those concepts here to make this answer more widely accessible.

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. Therefore, the vacuum state is observer-dependent, and a given observer's vacuum state typically doesn't have any natural unique extension to the whole spacetime. These statements apply in any spacetime, flat or curved. I wrote more about this theme in another answer and another older answer and another even older answer.

Those connections have a useful consequence: they allow us to 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. This is the most natural picture to use in relativistic QFT, especially in curved spacetime, because then time- and space-coordinates are both used only to parameterize the quantum fields, just like they're used only to parameterize the fields in classical field theory. Consider a free massless scalar field $\phi(\vec x, t)$, like Hawking did.

The Heisenberg equation of motion for $\phi(\vec x,t)$ is just the usual wave equation. Therefore, we can write $$ \phi(\vec x, t) = \sum_{f\in\Omega} f(\vec x,t)\phi(f) \tag{1} $$ where $\Omega$ is a complete set of mutually orthogonal solutions of the wave equation (ordinary complex-valued functions) and where the operators $\phi(f)$ are "indexed" by $f$. The functions $f$ are called mode functions, and $\phi(f)$ is called a mode operator.

Here's the key: We can choose $\Omega$ to be the union of two subsets, $\Omega_+$ and $\Omega_-$, which contain only positive- and negative-frequency solutions, respectively. The fact that the energy operator is the generator of time translations has these implications:

  • The mode operator $\phi(f)$ corresponding to a positive-frequency solution $f\in\Omega_+$ is an operator that reduces the energy of any state on which it acts. This is called an annihilation operator.

  • The operator $\phi(f)$ corresponding to a negative-frequency solution $f\in\Omega_-$ is an operator that increases the energy of any state on which it acts. This is called an creation operator.

We can use these operators to characterize the (local and observer-dependent) vacuum state: 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 key point here is that 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.

Technical note: We might worry that the definition of positive- and negative-frequency requires using a Fourier transform over all time, which would be problematic when we're trying to apply the condition locally — I mean, when we're trying to distinguish between the experiences of observers in the distant past and observers in the distant future. As long as we keep our FAPP$^*$ physics hats on, this isn't a problem. A Fourier transform over a finite time interval is sufficient, as long as the duration is long compared to the inverse frequencies of interest.

$^*$ FAPP = For All Practical Purposes.

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 $f\in\Omega$ 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. According to the QFT concepts reviewed above, this means that 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's geometric optics method

Since the paradox involves high-frequency outgoing modes before they are redshifted to wavelengths $\sim R$, and since Hawking's derivation used this feature, I'll review the context where he used it.

Consider a series of radial null geodesics that go in toward the center of the collapsing star, pass through the center, and come back out the other side just in time to outrun the growing event horizon that started to form right after it passed through the center. Since we're considering a sequence of such geodesics, some of them will pass closer to the horizon-initiation event than others. By thinking of these geodesics as proxies for successive wavecrests of a radial wave, we can use this picture to infer what happens to an initially positive-frequency mode by the time it gets out to the L-observers.

I like the way Parker and Toms said it in their book Quantum Field Theory in Curved Spacetime, on page 154 in the first edition:

...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.

Remember the scene from the climax of Return of the Jedi when the Millenium Falcon just barely outpaced the leading edge of the explosion that devoured the Death Star from the inside out? That's the picture.

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. According to the QFT concepts reviewed above, this implies that the L-observers experience a state that is not their vacuum state. That's Hawking radiation.

Resolution of the paradox

After all that review, we're finally prepared to see how the paradox can be resolved.

As explained above, whether or not a given state is empty (vacuum) depends on the observer. It depends on the observer's motion, and also on where/when the observer is present.

The dependence on the observer's motion is illustrated by the Unruh effect in flat spacetime. If the state is the Minkowski vacuum state (the state that all inertial observers agree is completely empty), then the same state is not empty (not vacuum) according to an accelerating observer. The more intense the acceleration, the higher the frequency of the excited modes.

For an observer near the horizon of a sufficiently large black hole, spacetime is essentially flat. Therefore, if the state is empty according to the F-observer (infalling observer), then the H-observer (hovering observer) will experience high-frequency excited modes because of the Unruh effect. That's only an if-then statement, so it doesn't resolve the paradox. Just knowing about the Unruh effect doesn't tell us what the F-observer experiences. But it does illustrate the key concepts of QFT that I reviewed above: vacuumness (emptiness) depends on the observer's motion. That's important because the redshifting argument that led to the paradox didn't specify which, if any, near-horizon observer should actually experience the high-frequency modes.

By the way, it's worth emphasizing that the Unruh effect is a local phenomenon. It compares the experiences of two differently-moving observers in the same part of spacetime.$^\dagger$ If the accelerating observer stops accelerating, the effect disappears immediately. In contrast, the Hawking effect compares the experiences of the E-observers and the L-observers, which are separated from each other by a large time gap.

$^\dagger$ (The local nature of the Unruh effect is sometimes obscured by describing it with Rindler coordinates, which amounts to filling spacetime with uniformly accelerating observers and considering all of their experiences at once. 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.)

To determine what the F-observer (infalling thin sheet) encounters, we need to consder large-scale effects: 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. That requires propagating the solutions through the the black hole formation event, which is difficult, even if we try using the geometric-optics argument that Hawking used. I haven't done that calculation myself, but 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.

By the way, that same reflection effect also leads to the well-known greybody factors — the fact that Hawking radiation deviates from the ideal blackbody spectrum. This is explained briefly in Section 4 of hep-th/9412138 and in detail in Escobedo's thesis (link to pdf).

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.

Altogether, the resolution of the paradox seems to be this: The mode functions corresponding to the Hawking radiation seen by the L-observers have indeed been very redshifted as a result of climbing out of the gravitational well, but that's just a mathematical feature of the functions $f\in \Omega$. Whether or not those modes are actually populated depends on the observer, and according to the preceding excerpts, they're not populated for the F-observer. Hawking radiation is a large-scale effect. (And here's the DUBIOUS part of the argument...) A 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. (This argument is DUBIOUS because it used free field theory to derive the properties of the Hawking radiation, but a free field is incapable of interacting with the sheet under any circumstances, so the handwavy argument is unjustified.)

Those modes are populated (as high-frequency modes, not yet redshifted) according to the hovering H-observer, but that doesn't matter here. For a related question about the Unruh effect, see Can Unruh particles be blocked on their way from the horizon?

Information loss and firewalls

The information loss paradox and the firewall paradox were mentioned in the question, so I'll say one thing about them here to clarify why they're different.

To derive Hawking radiation, and to answer the present question, we can use a simplified model in which the spacetime metric is prescribed — still time-dependent, but not influenced by the quantum field. In this simplified model, black holes do not evaporate (page 5 in arXiv:2010.06602). That's the model that Hawking used and that this answer used.

The information loss paradox and the firewall paradox came from thinking about what would happen in a more realistic model, in which the spacetime metric is influenced by the quantum field so that black holes do tend to evaporate. That's more difficult, because quantum fields cannot influence classical fields (like the classical metric field) in a mathematically self-consistent way, and quantum gravity is hard.

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    $\begingroup$ A powerhouse answer. It’ll take me a while to digest. Many thanks! $\endgroup$ – rob May 10 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 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 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. $\endgroup$ – Chiral Anomaly May 10 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 at 18:18

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