7 ms·

It would make more sense to me if black holes had no internals, so the matter exists on the surface of the horizon. No infinities, and the matter can slowly esc

It would make more sense to me if black holes had no internals, so the matter exists on the surface of the horizon. No infinities, and the matter can slowly escape over time. Maybe something in the nature of gravity which prevents more than a certain density to exist in any place and time.

simonh 6y ago
In my mental model the internal structure of the black hole, composed of all the material that fell into it, is always preserved from some frame of reference. As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it. Thats true of every particle or photon it traps. Of course they get twisted and distorted by the extreme gravitational forces, but in principle there is always structure. I suppose from the internal perspective, Hawking Radiation manifests as a rain of negative energy. I'm sure I'm wildly off the mark, but it's the best I can do.
- macspoofing 6y ago
  >As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it. You have it backwards. From the POV of an external observer you never cross the event horizon, because that's a function time dilation, and light that you emit being more and more red-shifted as you get closer to the event horizon. From the falling observer POV, nothing really changes - or rather, what happens close to the event horizon or just past the event horizon depends on what the theory of quantum gravity says.
  - simonh 6y ago
    I didn't address the POV of an external observer but I'm sure you're right. But what you describe for the POV of someone falling in seems consistent with what I suggested. I don't understand what you think is backwards about it.
    - macspoofing 6y ago
      You said that from the perspective of a falling observer, they never see themselves cross the event horizon. That's not right. The falling observer certainly will see themselves cross the event horizon. In their frame of reference, time flows normally. For smaller black holes, they may get spaghettified, but for larger ones, crossing event horizon will be an unremarkable occurrence.
      - simonh 6y ago
        There will always be an event horizon below them beyond which light cannot reach them, and they will always be able to observe the external universe. So how will they see themselves crossing the event horizon?
        smallnamespace 6y ago
        The usual definition of event horizon is from the perspective of the external observer—the falling astronaut can calculate the threshold at which no amount of thrust can ever move them away from the singularity (in fact, this is what is meant that a time and space coordinate switch places—you can now only move 'down' in space, just like you can only move forward in time).
        simonh 6y ago
        I suppose they could calculate the theoretical event horizon from infinity, but they won't observe it, however they will observe one below them.
        pdonis 6y ago
        > they will observe one below them No, they won't. This is simply wrong. See my other responses upthread.
        zrm 6y ago
        > no amount of thrust Are you sure that's right? Thrust implies acceleration, not just velocity. Suppose you're just inside the event horizon, from the perspective of an external observer. You're traveling outward at a velocity of .95c. That's lower than the escape velocity (which is > c), so you're going to fall back in before you actually get to the external observer. But if you were close to the event horizon you might cross back over it before you fall back in, right? At which point the external observer could see you again. And once you're outside the event horizon, e.g. to the place where escape velocity is .9c, you might by then have slowed to .8c, but if you still have the capacity to accelerate back above .9c again, you could still get out, couldn't you?
  - pdonis 6y ago
    > From the POV of an external observer you never cross the event horizon More precisely, the external observer never sees you cross the horizon. But he can still deduce that you did from other data. So the external observer does not say "you never cross the horizon", period. He just can't see it happen.
    - gizmo686 6y ago
      This still isn't quite right. In the coordinate system of an observer at a constant distance from the horizon, the path of a free-falling object takes an infinite amount of coordinate time. In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike. There is a coordinate singularity around this switch, which is where the infinite coordinate time comes from. An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon, which they would find to be finite.
      - pdonis 6y ago
        > In the coordinate system of an observer at a constant distance from the horizon, the path of a free-falling object takes an infinite amount of coordinate time No: the path of the free-falling observer at and beneath the horizon is not covered at all by the coordinate system of the external observer. That coordinate system simply becomes mathematically undefined, which means you can't use it to make any claims about what happens. > In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike. No: there is a separate coordinate patch that covers the region inside the horizon (but not the horizon itself--the horizon is not covered by either patch), in which the "areal radius" coordinate is timelike, and in which the coordinate corresponding to the extra Killing vector field is spacelike. Calling the former coordinate "radial" is at least justifiable, since it still is the "areal radius" coordinate even though it's timelike. But calling the latter coordinate "the time coordinate" is simply wrong; the fact that the letter t is commonly used for it does not mean it's a "time" coordinate in any meaningful sense. > There is a coordinate singularity around this switch Yes. > which is where the infinite coordinate time comes from. No, which is what makes statements like "infinite coordinate time" incorrect; the correct statement is that neither coordinate patch covers the horizon at all. > An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon Yes. The easiest way to do that is to switch to some other coordinate chart which covers the entire path, above, at, and below the horizon. In such a coordinate chart, the "areal radius" coordinate will not be timelike below the horizon. (In at least one chart commonly used for this, Painleve coordinates, all four coordinates are spacelike below the horizon.)
- jakeinspace 6y ago
  I think you might be a bit reversed; an outside observer will never see you fall past the event horizon (because it would be impossible for light from you to reach them), but unless you happen to fall into a black hole with a "fire wall", you won't experience anything special when passing through the event horizon, aside from the already high tidal forces.
  - flubert 6y ago
    >already high tidal forces. Tidal forces are low for supermassive black holes, right? https://en.wikipedia.org/wiki/Supermassive_black_hole#Description https://en.wikipedia.org/wiki/Supermassive_black_hole#Descri...
- NikolaeVarius 6y ago
  https://www.youtube.com/watch?v=JDNZBT_GeqU&t https://www.youtube.com/watch?v=JDNZBT_GeqU&t
- pdonis 6y ago
  > As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it. No, this is not correct. To the extent that the horizon is something that can be "seen" at all, you see it when you cross it. It doesn't shrink away below you. > I suppose from the internal perspective, Hawking Radiation manifests as a rain of negative energy. This is a common heuristic picture, but it is very limited. No real particles fall into the hole as a result of Hawking radiation. An observer already inside the horizon might be able to deduce that Hawking radiation was being emitted from local observations on, say, the tidal forces in his vicinity, but I'm not sure even that would be true: the models I am familiar with for evaporating black holes do not require any causal connection at all between the emission of Hawking radiation and any particular observer inside the horizon (in more technical language, the two remain spacelike separated for the entire trajectory of the observer inside the horizon).
  - kazagistar 6y ago
    Does that mean the finite duration of a black hole to an outside observer prior to its evaporation corresponds to an infinite duration of an observer within the black hole?
    - pdonis 6y ago
      No. An observer who falls into the hole only experiences a finite duration before he hits the singularity and is destroyed.
macspoofing 6y ago
>It would make more sense to me if black holes had no internals, so the matter exists on the surface of the horizon. That may still be the case, but for that we need a theory of quantum gravity.
jepler 6y ago
Right, except (my lay understanding is) that would require something special to happen to infalling matter at the event horizon, though we don't have a plausible theory for what that thing is--the prevailing interpretation is that passing through the event horizon is a non-event for the observer. On the other hand, the holographic principle says that the event horizon surface is big enough to contain all the information, so in a way of speaking we seem to be free to model black hole information as though it is "on the event horizon".
- emerged 6y ago
  The holographic principle combined with "infinity means the theory fails" is what brought me to post my -3 voted comment. Interestingly, none of the commenters have been negative only the casual down-voters. Where's PBS Space Time when you need them?
sp332 6y ago
To an external observer, new material takes an infinite amount of time to cross the event horizon. So your intuition is kind of right there. But there is matter that starts out inside the horizon, and there needs to be something going on in there because otherwise the gravity would keep pulling it tighter and tighter to infinite density in an infinitely small spot (the original meaning of "singularity").
- pdonis 6y ago
  > To an external observer, new material takes an infinite amount of time to cross the event horizon. While this is a common pop science statement, it is not correct. The correct statement is that the external observer never sees the material cross the event horizon, because, heuristically, it would take an infinite time for light emitted exactly on the event horizon to get out to the external observer. However, this does not equate to the external observer being able to assign an "infinite time" to the material crossing the horizon. The notion of "time" the external observer would have to use to do that is mathematically undefined on the horizon. > there is matter that starts out inside the horizon No, there isn't. When an ordinary object like a star collapses to a black hole, there is no horizon at all to start with, and there is no matter starting out inside any horizon.
- emerged 6y ago
  It's that "keep pulling it tighter and tighter" that I'm wondering about. What if there's a separate force, or an aspect of gravity itself, which prevents a certain amount of gravitational density?

simonh 6y ago

In my mental model the internal structure of the black hole, composed of all the material that fell into it, is always preserved from some frame of reference. As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it. Thats true of every particle or photon it traps. Of course they get twisted and distorted by the extreme gravitational forces, but in principle there is always structure. I suppose from the internal perspective, Hawking Radiation manifests as a rain of negative energy. I'm sure I'm wildly off the mark, but it's the best I can do.

macspoofing 6y ago
>As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it. You have it backwards. From the POV of an external observer you never cross the event horizon, because that's a function time dilation, and light that you emit being more and more red-shifted as you get closer to the event horizon. From the falling observer POV, nothing really changes - or rather, what happens close to the event horizon or just past the event horizon depends on what the theory of quantum gravity says.
- simonh 6y ago
  I didn't address the POV of an external observer but I'm sure you're right. But what you describe for the POV of someone falling in seems consistent with what I suggested. I don't understand what you think is backwards about it.
  - macspoofing 6y ago
    You said that from the perspective of a falling observer, they never see themselves cross the event horizon. That's not right. The falling observer certainly will see themselves cross the event horizon. In their frame of reference, time flows normally. For smaller black holes, they may get spaghettified, but for larger ones, crossing event horizon will be an unremarkable occurrence.
    - simonh 6y ago
      There will always be an event horizon below them beyond which light cannot reach them, and they will always be able to observe the external universe. So how will they see themselves crossing the event horizon?
      - smallnamespace 6y ago
        The usual definition of event horizon is from the perspective of the external observer—the falling astronaut can calculate the threshold at which no amount of thrust can ever move them away from the singularity (in fact, this is what is meant that a time and space coordinate switch places—you can now only move 'down' in space, just like you can only move forward in time).
        simonh 6y ago
        I suppose they could calculate the theoretical event horizon from infinity, but they won't observe it, however they will observe one below them.
        pdonis 6y ago
        > they will observe one below them No, they won't. This is simply wrong. See my other responses upthread.
        zrm 6y ago
        > no amount of thrust Are you sure that's right? Thrust implies acceleration, not just velocity. Suppose you're just inside the event horizon, from the perspective of an external observer. You're traveling outward at a velocity of .95c. That's lower than the escape velocity (which is > c), so you're going to fall back in before you actually get to the external observer. But if you were close to the event horizon you might cross back over it before you fall back in, right? At which point the external observer could see you again. And once you're outside the event horizon, e.g. to the place where escape velocity is .9c, you might by then have slowed to .8c, but if you still have the capacity to accelerate back above .9c again, you could still get out, couldn't you?
- pdonis 6y ago
  > From the POV of an external observer you never cross the event horizon More precisely, the external observer never sees you cross the horizon. But he can still deduce that you did from other data. So the external observer does not say "you never cross the horizon", period. He just can't see it happen.
  - gizmo686 6y ago
    This still isn't quite right. In the coordinate system of an observer at a constant distance from the horizon, the path of a free-falling object takes an infinite amount of coordinate time. In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike. There is a coordinate singularity around this switch, which is where the infinite coordinate time comes from. An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon, which they would find to be finite.
    - pdonis 6y ago
      > In the coordinate system of an observer at a constant distance from the horizon, the path of a free-falling object takes an infinite amount of coordinate time No: the path of the free-falling observer at and beneath the horizon is not covered at all by the coordinate system of the external observer. That coordinate system simply becomes mathematically undefined, which means you can't use it to make any claims about what happens. > In this coordinate system, the time coordinate becomes space-like passed the event horizon and the radial coordinate becomes timelike. No: there is a separate coordinate patch that covers the region inside the horizon (but not the horizon itself--the horizon is not covered by either patch), in which the "areal radius" coordinate is timelike, and in which the coordinate corresponding to the extra Killing vector field is spacelike. Calling the former coordinate "radial" is at least justifiable, since it still is the "areal radius" coordinate even though it's timelike. But calling the latter coordinate "the time coordinate" is simply wrong; the fact that the letter t is commonly used for it does not mean it's a "time" coordinate in any meaningful sense. > There is a coordinate singularity around this switch Yes. > which is where the infinite coordinate time comes from. No, which is what makes statements like "infinite coordinate time" incorrect; the correct statement is that neither coordinate patch covers the horizon at all. > An observer fammiliar with relativity could compute the proper time of a path traversing the event horizon Yes. The easiest way to do that is to switch to some other coordinate chart which covers the entire path, above, at, and below the horizon. In such a coordinate chart, the "areal radius" coordinate will not be timelike below the horizon. (In at least one chart commonly used for this, Painleve coordinates, all four coordinates are spacelike below the horizon.)
jakeinspace 6y ago
I think you might be a bit reversed; an outside observer will never see you fall past the event horizon (because it would be impossible for light from you to reach them), but unless you happen to fall into a black hole with a "fire wall", you won't experience anything special when passing through the event horizon, aside from the already high tidal forces.
- flubert 6y ago
  >already high tidal forces. Tidal forces are low for supermassive black holes, right? https://en.wikipedia.org/wiki/Supermassive_black_hole#Description https://en.wikipedia.org/wiki/Supermassive_black_hole#Descri...
NikolaeVarius 6y ago
https://www.youtube.com/watch?v=JDNZBT_GeqU&t https://www.youtube.com/watch?v=JDNZBT_GeqU&t
pdonis 6y ago
> As you fall into a black hole the event horizon shrinks away below you and from your own POV you never cross it. No, this is not correct. To the extent that the horizon is something that can be "seen" at all, you see it when you cross it. It doesn't shrink away below you. > I suppose from the internal perspective, Hawking Radiation manifests as a rain of negative energy. This is a common heuristic picture, but it is very limited. No real particles fall into the hole as a result of Hawking radiation. An observer already inside the horizon might be able to deduce that Hawking radiation was being emitted from local observations on, say, the tidal forces in his vicinity, but I'm not sure even that would be true: the models I am familiar with for evaporating black holes do not require any causal connection at all between the emission of Hawking radiation and any particular observer inside the horizon (in more technical language, the two remain spacelike separated for the entire trajectory of the observer inside the horizon).
- kazagistar 6y ago
  Does that mean the finite duration of a black hole to an outside observer prior to its evaporation corresponds to an infinite duration of an observer within the black hole?
  - pdonis 6y ago
    No. An observer who falls into the hole only experiences a finite duration before he hits the singularity and is destroyed.

macspoofing 6y ago

>It would make more sense to me if black holes had no internals, so the matter exists on the surface of the horizon. That may still be the case, but for that we need a theory of quantum gravity.

jepler 6y ago

Right, except (my lay understanding is) that would require something special to happen to infalling matter at the event horizon, though we don't have a plausible theory for what that thing is--the prevailing interpretation is that passing through the event horizon is a non-event for the observer. On the other hand, the holographic principle says that the event horizon surface is big enough to contain all the information, so in a way of speaking we seem to be free to model black hole information as though it is "on the event horizon".

emerged 6y ago
The holographic principle combined with "infinity means the theory fails" is what brought me to post my -3 voted comment. Interestingly, none of the commenters have been negative only the casual down-voters. Where's PBS Space Time when you need them?

sp332 6y ago

To an external observer, new material takes an infinite amount of time to cross the event horizon. So your intuition is kind of right there. But there is matter that starts out inside the horizon, and there needs to be something going on in there because otherwise the gravity would keep pulling it tighter and tighter to infinite density in an infinitely small spot (the original meaning of "singularity").

pdonis 6y ago
> To an external observer, new material takes an infinite amount of time to cross the event horizon. While this is a common pop science statement, it is not correct. The correct statement is that the external observer never sees the material cross the event horizon, because, heuristically, it would take an infinite time for light emitted exactly on the event horizon to get out to the external observer. However, this does not equate to the external observer being able to assign an "infinite time" to the material crossing the horizon. The notion of "time" the external observer would have to use to do that is mathematically undefined on the horizon. > there is matter that starts out inside the horizon No, there isn't. When an ordinary object like a star collapses to a black hole, there is no horizon at all to start with, and there is no matter starting out inside any horizon.
emerged 6y ago
It's that "keep pulling it tighter and tighter" that I'm wondering about. What if there's a separate force, or an aspect of gravity itself, which prevents a certain amount of gravitational density?