What would happen if you shoot entanglet photons into a black hole, one falling in and another to a detector. Seems, doing so would in principle allow measurements of the horizon and what's inside.

Thinking further, if you create a black hole from entanglet and separated matter, you would be able to measure what's inside, by measuring the not compressed matter of still entangled partner particles outside.


marked as duplicate by Dale, WillO, Jon Custer, Kyle Kanos, ZeroTheHero Dec 20 '18 at 16:32

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  • $\begingroup$ Entanglement cannot be used to send information FTL. The event horizon does not change that. $\endgroup$ – Dale Dec 19 '18 at 12:09
  • $\begingroup$ The question is not about FTL and FTL is not necessary for my setup. Where do you read FTL? $\endgroup$ – dgrat Dec 19 '18 at 13:53
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    $\begingroup$ For information to escape from inside the event horizon it has to travel FTL. $\endgroup$ – PM 2Ring Dec 19 '18 at 14:46
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    $\begingroup$ @dgrat Light cannot escape the horizon so if you are getting information from inside the horizon then it is FTL by definition. You may not have written it, but you are in fact asking for FTL information transfer using entanglement $\endgroup$ – Dale Dec 19 '18 at 14:55

Suppose that entanglement is preserved across the event horizon. Depending on how quantum gravity works, this may not necessarily be true.

If the photon going into the black hole doesn't interact with anything before measurement, then measuring the other photon will allow you to determine what state the first photon is in. But you didn't actually learn anything about the inside of the event horizon by doing this, because you already effectively knew this information when the entanglement was set up. For example, if you don't know how two entangled photons' polarizations are correlated, then measuring one will tell you nothing about the other. If you do know how two entangled photons' polarizations are correlated (e.g. "if this photon is right-handed, then the other photon is left-handed and vice-versa"), then determining the state of the other photon is simply an application of the information you already have.

If, on the other hand, the photon going into the black hole does interact with something before measurement, then that would change the state of that photon, but not in a way that would be accessible just by measuring the state of the photon you have access to. For example, suppose you know that the accessible photon and the photon going into the black hole were set up to always have opposite polarizations. If you measure the accessible photon to be right-handed, then the photon going into the black hole is in whatever state a left-handed photon would be in after it interacted with whatever is in the black hole (which may be a superposition of right-handed and left-handed states), and vice versa. Without knowing the details of the interaction, you can't tell any more information than that.

In non-black-hole circumstances, pairs of entangled photons are used for advanced microscopy, to probe the details of the interaction between one photon and, say, a protein. The trick here is that both photons are measured, which means that information can be extracted from how the correlation between the two entangled photons differed from the correlation assuming no interaction. You can't do this with a black hole, because you can't access the other photon once it crosses the event horizon.

Now suppose that entanglement is not preserved across the event horizon. Then the properties of the accessible photon are not correlated with the properties of the photon in the black hole, so no information can be extracted.

Since the above covers all possible cases, it should be clear that sending one half of an entangled pair of photons into a black hole won't tell you anything you didn't already know.


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