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 $T$.
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b.h. | []
interior | falling sheet
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/ 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:
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b.h. |[]
interior | “frozen” falling sheet,
| apparent temperature $T_a$
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/ 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.