4 ms·

According to GR, time stops at the event horizon. It would take infinite time to cross the event horizon. Everything past the event horizon is speculative.

magicalhippo 3y ago
Not really. It just appears to take an infinite amount of time for an outside observer. For someone falling into a black hole, it takes a finite amount of proper time[1] to reach the event horizon. For that infalling observer, the horizon is a boundry where, once beyond, the singularity is always in their finite future. No matter what you do inside, you will reach it at some point. [1]: https://en.wikipedia.org/wiki/Proper_time https://en.wikipedia.org/wiki/Proper_time
- antognini 3y ago
  Yes, this is correct. If you look at the Schwarzschild solution to the Einstein field equations you'll find that there are two apparent singularities: one at the center of the black hole, and another at the event horizon. However, the apparent singularity at the event horizon is not a true singularity because you can perform a coordinate transformation in which it disappears. Mathematically, this is doing what you describe: going from the reference frame of an external observer to one who is falling into the black hole. To the outside observer you appear to take an infinite amount of time to fall in, but from the perspective of the person falling in, it happens in a finite amount of time. The other singularity at r = 0 is different, though. It is a true singularity because there is no coordinate transformation you can make in which it disappears.
  - angiosperm 3y ago
    If the escape velocity from the event horizon is (defined to be) light speed, that seems to mean falling in from indefinitely far away, you would get indefinitely close to light speed (as perceived from at rest outside) at the point when you cross it,.
- Zamicol 3y ago
  How long does it take a infalling observer to cross the event horizon of a black hole?
  - EA-3167 3y ago
    From the perspective of the person falling, it happens from one moment to the next. You as the person being accelerated don't notice any change in your own proper time, only distant observers will see that.
  - magicalhippo 3y ago
    If you were to fall radially inwards towards a non-rotating black hole, that is not spiraling around it but "straight in", it would be the same as with plain Newton's law[1]. There's a nice graph in the midde of this[2] page that shows the difference between the proper time and the aparent time observed by the outside observer. At r = 2m you can see the aparent time goes to infinity and the quickly back again. [1]: https://physics.stackexchange.com/questions/718222/proper-time-of-fall-in-schwarschild-metric https://physics.stackexchange.com/questions/718222/proper-ti... [2]: https://www.mathpages.com/rr/s6-04/6-04.htm https://www.mathpages.com/rr/s6-04/6-04.htm
    - Zamicol 3y ago
      Thank you for the response and links. The second link is particularly helpful. I'm surprised that Newton is equivalent here, but the reason appears simple: although the inbound observer's time slows, their velocity does not. A distant observer is unable to perceive them as they approach the event horizon as the inbound actor grows infinitely dim thanks to time dilation.
- GoblinSlayer 3y ago
  The falling observer is burned by the Hawking radiation before it reaches the horizon.
  - codethief 3y ago
    This is not at all known.
- kbelder 3y ago
  But, if I am concerned solely with my own perspective (since there is no privileged viewpoint), is it fair to say that nothing yet has ever been observed entered a black hole?
  - raattgift 3y ago
    The answer is simply that black holes grow when you throw mass at their horizons. The small region you last saw the mass (dimly and red-shifted) just outside the black hole ultimately ends up being on the inside, even from your perspective. Below I'm going to ignore angular momentum; black hole spin changes the details but not the central thrust of my comment. You could think about it this way: Schwarzschild is an eternal vacuum solution (to the Einstein Field Equations of General Relativity (EFEs)). Like other solutions to the EFEs, it tends to be investigated by tracking the behaviour of "test particles". Test particles don't change the EFEs themselves -- they don't have mass, they don't have any non-gravitational interactions at all, they're not physical, they're just a tool used to trace out the geometry of the spacetime. Throwing in a test particle doesn't change the eternal nature of the Schwarzschild black hole -- it always has the same mass, and the Schwarzschild radius is totally determined by that mass (so the horizon is always of constant size). That is, the test particle is not a perturbation of the Schwarzschild black hole. A significant mass would perturb the Schwarzschild metric though. Unlike test particles, that mass enters into the EFEs. The relevance is that the Schwarzschild radius is totally determined by the central mass. When the significant additional mass (the perturbing infaller) is far away from the central mass, the geometry (described by the EFE's metric tensor) still looks a lot like Schwarzchild. However, as the perturbing mass approaches and falls in, things depart from Schwarzschild for a bit, then returns to Schwarzschild but with an increased central mass. The infaller ultimately ends up at the singularity, leaving M_before < M_after, so the horizon must grow in proportion to the mass that fell in because the Schwarzschild r_s = 2GM/c^2, where M is the central mass. Distant viewers can measure the central mass in several ways; it's observable. Since throwing actual mass (rather than test particles) into Schwarzschild changes the size (and, briefly, shape) of the horizon, you could consider it as if the object you see dimming and moving verrrry slowly doesn't just come to a halt at a constant horizon: instead, the horizon "reaches out" and snatches the infaller just outside the M_before horizon. The worry that nothing actually falls in arises if one insists on keeping M constant (M_before == M_after) even as one has mass outside the centre of the black hole. Constant M is easier to work with mathematically, which leads to things like test particles or Hawking negative energy quanta, and so on. Extrapolating from those convenience uses tends to lead to confusions like "nothing initially outside can actually end up inside", which is just wrong. Finally, using perturbation methods, a compact infaller like a neutron star would raise a significant bump on the horizon, distorting it slightly from ~spherical. That distortion vanishes in a short time (even for an outside observer, who can also detect gravitational waves), and the post-infall result is a bigger spherical horizon. We have observed several neutron star-black hole mergers. The result, generically, is a more massive black hole (and a lot of gravitational radiation).

magicalhippo 3y ago

Not really. It just appears to take an infinite amount of time for an outside observer. For someone falling into a black hole, it takes a finite amount of proper time[1] to reach the event horizon. For that infalling observer, the horizon is a boundry where, once beyond, the singularity is always in their finite future. No matter what you do inside, you will reach it at some point. [1]: https://en.wikipedia.org/wiki/Proper_time https://en.wikipedia.org/wiki/Proper_time

antognini 3y ago
Yes, this is correct. If you look at the Schwarzschild solution to the Einstein field equations you'll find that there are two apparent singularities: one at the center of the black hole, and another at the event horizon. However, the apparent singularity at the event horizon is not a true singularity because you can perform a coordinate transformation in which it disappears. Mathematically, this is doing what you describe: going from the reference frame of an external observer to one who is falling into the black hole. To the outside observer you appear to take an infinite amount of time to fall in, but from the perspective of the person falling in, it happens in a finite amount of time. The other singularity at r = 0 is different, though. It is a true singularity because there is no coordinate transformation you can make in which it disappears.
- angiosperm 3y ago
  If the escape velocity from the event horizon is (defined to be) light speed, that seems to mean falling in from indefinitely far away, you would get indefinitely close to light speed (as perceived from at rest outside) at the point when you cross it,.
Zamicol 3y ago
How long does it take a infalling observer to cross the event horizon of a black hole?
- EA-3167 3y ago
  From the perspective of the person falling, it happens from one moment to the next. You as the person being accelerated don't notice any change in your own proper time, only distant observers will see that.
- magicalhippo 3y ago
  If you were to fall radially inwards towards a non-rotating black hole, that is not spiraling around it but "straight in", it would be the same as with plain Newton's law[1]. There's a nice graph in the midde of this[2] page that shows the difference between the proper time and the aparent time observed by the outside observer. At r = 2m you can see the aparent time goes to infinity and the quickly back again. [1]: https://physics.stackexchange.com/questions/718222/proper-time-of-fall-in-schwarschild-metric https://physics.stackexchange.com/questions/718222/proper-ti... [2]: https://www.mathpages.com/rr/s6-04/6-04.htm https://www.mathpages.com/rr/s6-04/6-04.htm
  - Zamicol 3y ago
    Thank you for the response and links. The second link is particularly helpful. I'm surprised that Newton is equivalent here, but the reason appears simple: although the inbound observer's time slows, their velocity does not. A distant observer is unable to perceive them as they approach the event horizon as the inbound actor grows infinitely dim thanks to time dilation.
GoblinSlayer 3y ago
The falling observer is burned by the Hawking radiation before it reaches the horizon.
- codethief 3y ago
  This is not at all known.
kbelder 3y ago
But, if I am concerned solely with my own perspective (since there is no privileged viewpoint), is it fair to say that nothing yet has ever been observed entered a black hole?
- raattgift 3y ago
  The answer is simply that black holes grow when you throw mass at their horizons. The small region you last saw the mass (dimly and red-shifted) just outside the black hole ultimately ends up being on the inside, even from your perspective. Below I'm going to ignore angular momentum; black hole spin changes the details but not the central thrust of my comment. You could think about it this way: Schwarzschild is an eternal vacuum solution (to the Einstein Field Equations of General Relativity (EFEs)). Like other solutions to the EFEs, it tends to be investigated by tracking the behaviour of "test particles". Test particles don't change the EFEs themselves -- they don't have mass, they don't have any non-gravitational interactions at all, they're not physical, they're just a tool used to trace out the geometry of the spacetime. Throwing in a test particle doesn't change the eternal nature of the Schwarzschild black hole -- it always has the same mass, and the Schwarzschild radius is totally determined by that mass (so the horizon is always of constant size). That is, the test particle is not a perturbation of the Schwarzschild black hole. A significant mass would perturb the Schwarzschild metric though. Unlike test particles, that mass enters into the EFEs. The relevance is that the Schwarzschild radius is totally determined by the central mass. When the significant additional mass (the perturbing infaller) is far away from the central mass, the geometry (described by the EFE's metric tensor) still looks a lot like Schwarzchild. However, as the perturbing mass approaches and falls in, things depart from Schwarzschild for a bit, then returns to Schwarzschild but with an increased central mass. The infaller ultimately ends up at the singularity, leaving M_before < M_after, so the horizon must grow in proportion to the mass that fell in because the Schwarzschild r_s = 2GM/c^2, where M is the central mass. Distant viewers can measure the central mass in several ways; it's observable. Since throwing actual mass (rather than test particles) into Schwarzschild changes the size (and, briefly, shape) of the horizon, you could consider it as if the object you see dimming and moving verrrry slowly doesn't just come to a halt at a constant horizon: instead, the horizon "reaches out" and snatches the infaller just outside the M_before horizon. The worry that nothing actually falls in arises if one insists on keeping M constant (M_before == M_after) even as one has mass outside the centre of the black hole. Constant M is easier to work with mathematically, which leads to things like test particles or Hawking negative energy quanta, and so on. Extrapolating from those convenience uses tends to lead to confusions like "nothing initially outside can actually end up inside", which is just wrong. Finally, using perturbation methods, a compact infaller like a neutron star would raise a significant bump on the horizon, distorting it slightly from ~spherical. That distortion vanishes in a short time (even for an outside observer, who can also detect gravitational waves), and the post-infall result is a bigger spherical horizon. We have observed several neutron star-black hole mergers. The result, generically, is a more massive black hole (and a lot of gravitational radiation).