4 ms·

> Like the coin trap’s slope, the curvature of space becomes infinite at the event horizon of a black hole. So time literally stops [...] so actually you can’t

by tbabb 11y ago

> Like the coin trap’s slope, the curvature of space becomes infinite at the event horizon of a black hole. So time literally stops [...] so actually you can’t reach the event horizon. I think this is completely wrong. Isn't the curvature at the event horizon large but finite? The curvature is only infinite (in theory) at the singularity.

hsk 11y ago
Correct. A safe observer far from the black hole will observe a singularity at the event horizon, but that is a coordinate singularity, not a true one. There is singularity at the event horizon for a falling observer. A falling observer will pass through the event horizon and reach the center, which is a true singularity that physics currently cannot explain.
- iofj 11y ago
  I think you mean there no singularity for a falling observer. Given that time slows down, it is not a certainty that you'll fall through the event horizon. There are two viewpoints: For an outside observer: the black hole will evaporate (over many billions of years) before you ever cross the event horizon. For the falling observer: There shouldn't be an event horizon at all. That brings the question: what happens. Occam's razor would seem to indicate that most likely you'd just keep falling. There are various "solutions" to this problem being worked out. One is that the surface of the event horizon is actually a universe all by itself, with 3d space you can live in. Over time that space would collapse into nothing, and that would happen quickly, but not instantaneously. This space is visible from the outside of the black hole, and you can interact with it but because time goes so much faster in this space any light escaping from it would appear extremely redshifted and weak.
baghira 11y ago
Yes, that part is definitely wrong. The free falling observer crosses the horizon in a finite time. It's the external obsever that see the infalling object redshifted to infinity. An the curvature scalar is finite everywhere except that at the singularity.

by tbabb 11y ago

hsk 11y ago

Correct. A safe observer far from the black hole will observe a singularity at the event horizon, but that is a coordinate singularity, not a true one. There is singularity at the event horizon for a falling observer. A falling observer will pass through the event horizon and reach the center, which is a true singularity that physics currently cannot explain.

iofj 11y ago
I think you mean there no singularity for a falling observer. Given that time slows down, it is not a certainty that you'll fall through the event horizon. There are two viewpoints: For an outside observer: the black hole will evaporate (over many billions of years) before you ever cross the event horizon. For the falling observer: There shouldn't be an event horizon at all. That brings the question: what happens. Occam's razor would seem to indicate that most likely you'd just keep falling. There are various "solutions" to this problem being worked out. One is that the surface of the event horizon is actually a universe all by itself, with 3d space you can live in. Over time that space would collapse into nothing, and that would happen quickly, but not instantaneously. This space is visible from the outside of the black hole, and you can interact with it but because time goes so much faster in this space any light escaping from it would appear extremely redshifted and weak.

baghira 11y ago

Yes, that part is definitely wrong. The free falling observer crosses the horizon in a finite time. It's the external obsever that see the infalling object redshifted to infinity. An the curvature scalar is finite everywhere except that at the singularity.