4

If an object falls into a Schwarzschild black hole and a distant observer watches, they see the object fall slower and slower as it approaches the event horizon, until it is "frozen." (To the observer, it never stops accelerating.) The observer never sees the object cross the event horizon in finite time.

So, with this, how is information about the object lost? Shouldn't the observer have the ability to observe the object indefinitely? To them, shouldn't the object appear "on this side" of the universe indefinitely? Shouldn't the object's light, mass, and charge remain observable "on this side?"

Question: Where does information loss in this setup occur, i.e., which aspects about the object are lost; and—if it makes sense to ask such a thing—when does it occur, i.e., when (with respect to the observer) during the object's approach does the observer lose the ability to measure all of the information about the object?

(If the two questions are not related enough to be asked at the same time, please let me know so I can break this up into multiple questions, Thanks.)

BMF
  • 256

1 Answers1

3

In the proper time of the object falling into the black hole, it crosses the event horizon and hits the singularity in finite time. But the observer outside 'sees' the object never reaching the event horizon. So, for the observer, the object never reached the event horizon. Whatever the observer sees, are events that happened before the object crossed the horizon. Hence the information on what happened to the object after it crossed the horizon is lost to the observer.

Edit: In classical black holes there is no information loss. When the black hole evaporates following thermal Hawking radiation there is no way to retrieve the information of the particle that falls into the black hole. That is to say there is no unitary transformation which can trace back to the state of the particle prior to it hitting the singularity.

abhijit975
  • 1,064