5 ms·

> "To maintain consistency with outside observers, this evaporation must occur fast enough that the victim can never reach any surface. Instead, the black hole

> "To maintain consistency with outside observers, this evaporation must occur fast enough that the victim can never reach any surface. Instead, the black hole recedes from them, evaporating faster and faster." If this is radiating a star's mass worth Hawking radiation particles, is it like the Solar Wind, and if it's happening ever faster is there a point where it would start pushing the victim away from the black hole again? (the 'victim' can be a solar sail if that helps)

jiggawatts 1y ago
Yes, infalling victims will have a rather unpleasant time as they discover that black holes are secretly supernovas frozen in time. Outside observers see the victim's own black body radiation become extremely redshifted, asymptotically matching the black hole's black body radiation. If you mathematically "undo" this distortion for both, then what you are really observing from the outside is a star's worth of matter getting converted to pure energy and the infalling victims getting blasted in the face by that. The victims can't make it back out "whole and intact" in the same sense that you're not going to keep your atomic integrity if you're up close and personal to a supernova. Your quantum numbers however... those can be preserved nicely.
- quantadev 1y ago
  How are they getting blasted in the face when such a blast would necessarily have to be moving faster than light?
  - hparadiz 1y ago
    Because time slows down relative the outside observed as you get closer to the event horizon any matter falling inwards starts to compress blocking the matter above it until eventually that matter is compressed to an extreme against the event horizon. The photons that get fired off from that interaction away from the black hole are able to escape and that's why we can see some black holes as being extremely bright. However matter that is spiraling inwards will be blasted by hundreds of years worth of photons from every direction while inside this matter and energy goop and all sorts of other particles in a matter of moments relative to how it is experiencing time.
    - quantadev 1y ago
      Ah, I gotcha, thanks for explaining. Yeah the 'accretion disk' are what this is normally called. Lots of matter is getting smashed right outside the EH, creating heat energy, and like you said it's able to blast out photons.
- Ygg2 1y ago
  > black holes are secretly supernovas frozen in time. I don't think that's true. What kills you isn't radiation of the singularity, but cosmic microwave background (and other infalling radiation) turned to visible light, then x-rays, then gamma rays.
pixl97 1y ago
I don't think the hawking radiation occurs at the edge of the event horizon itself. Arvin Ash just did a video on this https://www.youtube.com/watch?v=UxVssUb0MsA https://www.youtube.com/watch?v=UxVssUb0MsA It appears to occur outside the event horizon in a large area.

jiggawatts 1y ago

Yes, infalling victims will have a rather unpleasant time as they discover that black holes are secretly supernovas frozen in time. Outside observers see the victim's own black body radiation become extremely redshifted, asymptotically matching the black hole's black body radiation. If you mathematically "undo" this distortion for both, then what you are really observing from the outside is a star's worth of matter getting converted to pure energy and the infalling victims getting blasted in the face by that. The victims can't make it back out "whole and intact" in the same sense that you're not going to keep your atomic integrity if you're up close and personal to a supernova. Your quantum numbers however... those can be preserved nicely.

quantadev 1y ago
How are they getting blasted in the face when such a blast would necessarily have to be moving faster than light?
- hparadiz 1y ago
  Because time slows down relative the outside observed as you get closer to the event horizon any matter falling inwards starts to compress blocking the matter above it until eventually that matter is compressed to an extreme against the event horizon. The photons that get fired off from that interaction away from the black hole are able to escape and that's why we can see some black holes as being extremely bright. However matter that is spiraling inwards will be blasted by hundreds of years worth of photons from every direction while inside this matter and energy goop and all sorts of other particles in a matter of moments relative to how it is experiencing time.
  - quantadev 1y ago
    Ah, I gotcha, thanks for explaining. Yeah the 'accretion disk' are what this is normally called. Lots of matter is getting smashed right outside the EH, creating heat energy, and like you said it's able to blast out photons.
Ygg2 1y ago
> black holes are secretly supernovas frozen in time. I don't think that's true. What kills you isn't radiation of the singularity, but cosmic microwave background (and other infalling radiation) turned to visible light, then x-rays, then gamma rays.

pixl97 1y ago

I don't think the hawking radiation occurs at the edge of the event horizon itself. Arvin Ash just did a video on this https://www.youtube.com/watch?v=UxVssUb0MsA https://www.youtube.com/watch?v=UxVssUb0MsA It appears to occur outside the event horizon in a large area.