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From puppetsock's reply to Is it possible to build a black hole (kugelblitz) gun?, it seems like a black hole of 228 tonnes evaporate after just 1 second. And the lifetime of a black hole in general is proportional to the cube of its mass.


Edit: Clarifying my thought process:

  • Step 1: The object barely surpasses the event horizon
  • Step 2: The event horizon starts shrinking faster than the object is falling; catching up to/surpassing it.
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No, this is unfortunately impossible.

The key here is that Hawking radiation bears no resemblance to the constituent matter that formed the black hole (this turns out to be a bit of a problem - see the black hole information paradox for more details). Black hole evaporation dramatically favors the production of lighter particles. For example, a black hole of temperature $T=100\;\text{GeV}\approx10^{15}\;\text{K}$ emits less than 3% of its emitted energy in the form of protons and antiprotons (see MacGibbon & Webber 1990). Over three quarters ends up as photons and neutrinos, with most of the remainder in the form of electrons and positrons.

As a rule of thumb, for a particle of mass $m$, a black hole needs to have a temperature $T$ such that its thermal energy $E_T\approx k_BT$ is on the order of $mc^2$ for that particle to be emitted significantly. Hence, hotter (read: less massive and closer to death) black holes may emit more massive particles, but the majority of their emission should still take the form of neutrinos and photons.

The upshot of all of this is that the mass of any object that travels inside the event horizon will reemerge in a form completely different from the original - dump a block of cheese into the black hole, and it won't emerge as anything like the original mixture of protons, neutrons and electrons, let alone a block of cheese.

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  • $\begingroup$ Black holes, a.k.a. "matter shredders". $\endgroup$
    – cowlinator
    Aug 6 '20 at 15:39
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A black hole that explodes within the next second is of subatomic size. Anything that falls in must get atomized before it crosses the event horizon.

And anything that approaches will be ripped apart long before it reaches the event horizon due to the Hawking radiation that's coming out of the black hole. This radiation pressure makes feeding such a small black hole virtually impossible. It would even explode if it were embedded in the core of the sun. Once a black hole's radiation reaches a certain level, the only thing you can do is to flee if you don't want to be blown up with it. It's the perfect ticking time bomb, with a strength that dwarves all the nukes that we have built.

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  • $\begingroup$ That makes sense - but what if "the object" is a single photon that you have entangled with another photon? That way you might be able to verify that it is "the same photon" once you get it back out, right? (Or see some kind of statistical deviation from doing it enough times) $\endgroup$ Aug 6 '20 at 10:15
  • $\begingroup$ @TobiasBergkvist Well, if it takes the BH to evaporate from a subatomic size, and the photon moves at light speed, it's evident that the photon cannot be overtaken by the shrinking event horizon. That on top of the fact that the photon won't even hit the black hole, it'll hit one of the electrons/positrons streaming out of the black hole first. Also, there's no single experiment that you can do to check whether two quantum particles are entangled. All you can do is run a measurement on both and check whether the results agree with entanglement, but a positive result can always be coincidence. $\endgroup$ Aug 6 '20 at 10:26
  • $\begingroup$ I guess actually "finding" your photon among all the stuff the black hole spits out while evaporating could be a big challenge in itself though. Maybe this is not really possible. And maybe the event horizon somehow guarantees that the entanglement is broken (which in itself would be interesting) $\endgroup$ Aug 6 '20 at 10:27
  • $\begingroup$ Yes, it could be a coincidence - which is why you would need to do it a lot of times and look for some statistical result. Like they did when looking for the higgs boson with the LHC. Although that was probably a bit different. What if the photon is not moving straight into the black hole - but an angle of 89.999 degrees or something like that. That way the event horizon could catch up more easily, right? $\endgroup$ Aug 6 '20 at 10:30
  • $\begingroup$ I agree that the event horizon won't break the entanglement. Entanglement is a genuinely non-local phenomenon. However, entanglement tends to be carried away: Consider two entangled electrons. If one of them scatters a photon, you most likely have the other electron entangled to that passing photon after the interaction. After that, the two electrons are not entangled anymore. So, unless you manage to track what particles your infaling photon interacts with, you won't be able to measure entanglement. And that's a big problem if the photon is moving into the most violent area in the universe. $\endgroup$ Aug 6 '20 at 11:27
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No

An object doesn't quite 'fall' into a black hole, a more correct explanation would be that it forms an accretion on the hyper-dense mass which forms the black hole. It doesn't matter how little time an object spends in a black hole, once it's there, it's been compacted beyond all recognizable measure to the point where any object smaller than a human wouldn't even be able to be seen in its current state, assuming you could see the mass which makes up the black hole. Not to mention that when a black hole evaporates, it does so emitting radiation, like, a lot of it. Black holes aren't survivable.

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