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Following up on this question "Does a diatomic molecule falling into a black hole dissociate?", if a diatomic molecule is falling into a black hole (BH) with the bond connecting the atoms oriented normal to the event horizon (EH), there will be some time period when to a distant observer the bond straddles the EH - that is, one atom is outside the EH, the other is within.

During that time, it seems in the molecular frame of reference the atoms in the molecule can interact for example they could still undergo harmonic motion, repel/attract each other, exchange electronic excitation energy etc. However a distant observer can still make observations about the atom exterior to the EH and this would seem to violate the transfer of information across the EH, since by extension the distant observer now knows what is happening to the atom within the EH. How does one resolve this apparent paradox?

To push this further one can imagine a polymer chain of N atoms crossing the EH and then observing the motion of the effects of the entire chain on the last atom exterior to the EH. How long could the chain be and still have communication between the atoms at the opposite ends of the chain?

Note this does not have to be a quantum mechanical phenomenon, you could choose heavy atoms, or even a macroscopic system of 2 (or more) devices that communicate with each other falling into the BH with orientation perpendicular to the EH.

If the answer is best/easiest done by referring to technical papers etc., then I'm happy to have references provided and dig into those myself.

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For the sake of simplicity, I will take a very large Schwarzschild black hole (large enough that the tidal forces are negligible at the event horizon and no "spaghettification" occurs) and we know that

  • Due to the local equivalence principle the diatomic molecule's frame behaves as if it is in empty space.
  • No information can be sent to outside from inside of event horizon "classically".

From the diatomic molecule's perspective, it will be undergoing oscillations as it would do normally in empty space. From the diatomic molecule's perspective, it will cross the event horizon in finite time. While crossing the event horizon it will release photons that a faraway observer needs to observe to see that it is vibrating. But not all photons it releases will reach the faraway observer. The photon at the middle of the molecule which is emitted on the event horizon will take $\infty$ time to reach the observer. Those photons emitted by the part of the molecule inside the event horizon will go at the speed $c$ away from the center of the black hole in the diatomic molecule's frame. But they will never come outside of the event horizon and they never reach the faraway observer. They will also be falling towards the center of the black hole, but the diatomic molecule falls faster so it will see them as moving away.

From the perspective of the faraway observer, the diatomic molecule will never cross the event horizon. It cannot reach the event horizon in finite time from a far away observer's point of view. So, he can never observe one atom outside and another inside the event horizon.

Just outside the Event horizon:

From the diatomic molecule's perspective, it will be undergoing oscillations as it would do normally.

From the perspective of the far away observer, the time period of the diatomic molecule will be extremely dilated due to gravitational time dilation when he sees the molecule very near to the event horizon. So the diatomic molecule will look as if it is not vibrating at all.

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  • $\begingroup$ Thank you! Just to make sure I understand, for your first bullet point "Due to the local equivalence principle the diatomic molecule's frame behaves as if it is in empty space.", it is that the diatomic molecule's frame behaves as if it is accelerating in empty space - is that correct? $\endgroup$
    – dllahr
    Oct 13 at 10:05
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    $\begingroup$ @dllahr No. What I meant was that any object which is following a geodesic path will feel as if it is not accelerating in empty space. Similar to how you will not feel any acceleration when you are freely falling because of the earth's gravity. $\endgroup$ Oct 13 at 10:35
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    $\begingroup$ @dllahr if you sit inside a room which is at rest on the ground that is equivalent to sitting in a room that is accelerating in empty space. But if you are in a freely falling room on earth's gravity that is equivalent to you are being inside a room that is not accelerating in empty space. $\endgroup$ Oct 13 at 10:37
  • $\begingroup$ Thank you for the responses. But I don't see how this addresses the main point of the question about when the molecule is in free fall straddling the event horizon. $\endgroup$
    – dllahr
    Oct 14 at 9:02
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To make this easier to understand, let’s scale up the system. Suppose we have a spaceship which is 1000 ft long so that light takes 1000 ns to cross the ship. The captain in the front is communicating by EM signals with the engineer at the rear. The ship is free falling into the black hole. Suppose further that the black hole is very large relative to the ship.

The captain sends three messages, one when he is 1 ft from the horizon, the second when he is at the horizon, and the last when he is 1 ft past the horizon. 1000 ns after the captain sends the first signal, the engineer receives it. The horizon itself is a lightlike surface, so it moves at c and thus keeps pace exactly with the second signal. Thus the engineer receives the second signal at the same time that the engineer crosses the horizon. Finally, the engineer receives the third signal from the captain. This signal was broadcast from within the horizon. The engineer received the signal, but only after crossing the horizon themselves.

The key is to recognize that the horizon is a lightlike surface. It moves at c relative to any local inertial frame. So you can send signals from inside the horizon, but those signals cannot beat the horizon to the recipient.

The same thing happens with molecules. They may remain intact without violating the no information transfer property of black holes.

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  • $\begingroup$ Thank you! As a follow up, if the ship you describe is initially at rest with respect to the EH, 1.5 ft outside the EH, then goes into free fall, it will not be traveling particularly fast as the captain in the front crosses the EH. So when the captain sends the 1st message, it still takes 1000 ns to reach the engineer. The second message reaches the engineer when the engineer crosses the EH - which would be time >> 1000 ns as the ship is accelerating arbitrarily slowly (depending on size of BH). Is that correct? $\endgroup$
    – dllahr
    Oct 13 at 10:01
  • $\begingroup$ My follow up question is: are the above results due to the "size" of the ship - if we scale this down to the molecule/polymer scale would this no longer apply? Stated another way, is there any system small enough (yet not quantum mechanical) that would not have the same behavior described above? $\endgroup$
    – dllahr
    Oct 13 at 10:03

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