3 ms·

> How do these two things mix? A particle experiences asymptotic time dilation when it crosses the event horizon and the rest of the universe ends in that momen

> How do these two things mix? A particle experiences asymptotic time dilation when it crosses the event horizon and the rest of the universe ends in that moment. Then the black hole evaporates and lo and behold the universe is not ended. I don't know if that's necessarily correct. From the point of view of an object falling through the event horizon, nothing special happens and they can look outwards normally; whereas an observer some distance from the black hole (and maintaining that distance) sees the falling object fall slower and slower and becoming more redshifted. The falling object also become shorter in the direction of travel (according to the external observer) and end up 'smeared' on the horizon. Leonard Susskind shows this using a Penrose diagram in lecture 5 or 6 (I think, but they're all worth watching) of the Stanford Topics in String Theory lectures[1] [1] https://www.youtube.com/watch?v=NZ-ElsvYKyo https://www.youtube.com/watch?v=NZ-ElsvYKyo

zwkrt 8y ago
Suskind actually has a whole lecture on Youtube devoted to addressing the paradox of falling into a black hole from the perspective of the person falling in versus an outside observer. The outside observer sees the falling observer catastrophically turned into high-entropy soup smeared across the horizon, while the one falling feels nothing particularly special as they cross the event horizon. https://www.youtube.com/watch?v=2DIl3Hfh9tY https://www.youtube.com/watch?v=2DIl3Hfh9tY
- Pharmakon 8y ago
  Just a note, that’s the case for a very high mass black hole. A “smaller” black hole would shred the astronaut before they even reached the event horizon.
dpark 8y ago
I still don’t see how to reconcile these two things. The observer sees the particle essentially slow to a stop as it crosses the event horizon, right (infinite time dilation)? Then the black hole evaporates. Evaporation happens in finite time, so how is that particle “frozen” indefinitely? Is characterizing the dilation as “infinite” incorrect? I’ve heard that and equivalent statements from people who studied and seem to actually understand this stuff, and it seems correct per the article here.
- speakeron 8y ago
  The particle goes through the event horizon in finite time according to its own clock (and will then hit the singularity in finite term however it tries to travel). Think of the view of it by an external observer as being 'frozen' at the event horizon as kind of optical illusion caused by the extreme warping of spacetime. For a non-evaporating black hole (as described by general relativity), the 'last' photon coming off it will indeed be at infinity (or with infinite redshift). If (when) the black hole evaporates, the external observer will see that and of course there will be no more photons from the infalling particle. These two different views are hard to reconcile from our human perspective of how the world works, but they really do come from the mathematics of general relativity. Here's a really excellent and short video which attempts to explain this in ten minutes. (His explanation of how a Penrose diagram works is fleshed-out in more detail in the Susskind lectures I mentioned above). PBS Spacetime - What happens at the Event Horizon https://www.youtube.com/watch?v=mht-1c4wc0Q https://www.youtube.com/watch?v=mht-1c4wc0Q

zwkrt 8y ago

Suskind actually has a whole lecture on Youtube devoted to addressing the paradox of falling into a black hole from the perspective of the person falling in versus an outside observer. The outside observer sees the falling observer catastrophically turned into high-entropy soup smeared across the horizon, while the one falling feels nothing particularly special as they cross the event horizon. https://www.youtube.com/watch?v=2DIl3Hfh9tY https://www.youtube.com/watch?v=2DIl3Hfh9tY

Pharmakon 8y ago
Just a note, that’s the case for a very high mass black hole. A “smaller” black hole would shred the astronaut before they even reached the event horizon.

dpark 8y ago

I still don’t see how to reconcile these two things. The observer sees the particle essentially slow to a stop as it crosses the event horizon, right (infinite time dilation)? Then the black hole evaporates. Evaporation happens in finite time, so how is that particle “frozen” indefinitely? Is characterizing the dilation as “infinite” incorrect? I’ve heard that and equivalent statements from people who studied and seem to actually understand this stuff, and it seems correct per the article here.

speakeron 8y ago
The particle goes through the event horizon in finite time according to its own clock (and will then hit the singularity in finite term however it tries to travel). Think of the view of it by an external observer as being 'frozen' at the event horizon as kind of optical illusion caused by the extreme warping of spacetime. For a non-evaporating black hole (as described by general relativity), the 'last' photon coming off it will indeed be at infinity (or with infinite redshift). If (when) the black hole evaporates, the external observer will see that and of course there will be no more photons from the infalling particle. These two different views are hard to reconcile from our human perspective of how the world works, but they really do come from the mathematics of general relativity. Here's a really excellent and short video which attempts to explain this in ten minutes. (His explanation of how a Penrose diagram works is fleshed-out in more detail in the Susskind lectures I mentioned above). PBS Spacetime - What happens at the Event Horizon https://www.youtube.com/watch?v=mht-1c4wc0Q https://www.youtube.com/watch?v=mht-1c4wc0Q