5 ms·

That's the thing though - Hawking is saying it's not "inescapable."

ck2 13y ago
I think he is saying the gravity is not even enough to have that effect and therefore would be points that are escapable? I always wondered if a black hole could become completely visible again if it collected enough mass to spread its surface area out so that gravity would become thin enough for light to escape again. Kind of like how only the surface of a pond sometimes freezes.
- lmm 13y ago
  No; adding mass doesn't make it less dense (well, it's a singularity, so it's always infinitely dense. But even if it weren't, normally when you add mass to something it becomes more compressed and gets denser, not less dense), just heavier. (Of course, singularities don't make any sense - which is part of what Hawking's talking about. But that's the classical model of a black hole)
  - waps 13y ago
    It's only necessarily infinitely dense from an outside reference frame. Inside you could have anything you'd want according to GR. One certainty is that the space inside a black hole is bigger than the size of the black hole observed from outside of it. So it's entirely possible that you could have a solar system or even an entire universe on the inside. According to GR the space inside a black hole should pretty much behave like the rest of the universe, as long as the black hole is growing. As for "not getting out", well it's impossible to approach the edge of the universe too, despite that border having a physical location. But it's expanding at light speed, so ... no way to get there. So density of a black hole is an entirely different concept from the outside and the inside. Seen from the outside it is infinite. Seen from the inside it is probably not infinite.
    - fennecfoxen 13y ago
      > One certainty is that the space inside a black hole is bigger than the size of the black hole observed from outside of it. So it's entirely possible that you could have a solar system or even an entire universe on the inside. That is not a certainty at all. The entire premise of Hawking's paper is that the object we call a 'black hole' is just a really degenerate dense lump that slowly turns incoming matter/energy into background radiation with as much entropy as possible. And that's it. No fancy Time Lord bigger-on-the-inside science proposed here.
    - pdonis 13y ago
      it's impossible to approach the edge of the universe too, despite that border having a physical location. No, it doesn't. There is no edge to the universe. Our best current model is that the universe is spatially finite, but even if it turns out to be spatially finite, it is unbounded--the spatially finite model is a 3-sphere; like the surface of the Earth but with one more dimension. The surface of the Earth is finite, but it has no edge; a spatially finite universe would be the same way but with one more dimension. Seen from the outside it is infinite. No, it isn't. Seen from the outside, the hole has a finite mass in a finite volume, so its density is finite. (The finite volume an outside observer would assign, based on the surface area of the hole's horizon, is not the "actual" volume of the hole; but it works well enough as an "effective volume", the volume that the hole occupies from the standpoint of the rest of the universe.) The singularity can be viewed as infinitely dense, but the singularity does not occupy the entire black hole; it only occupies the center at r = 0.
- panzi 13y ago
  What surface? A point has no surface.
  - rosser 13y ago
    The event horizon isn't the surface of the black hole. As classically defined, the event horizon is the point at which the black hole's gravity becomes inescapable. That's not at its surface.
    - panzi 13y ago
      Exactly.
    - pdonis 13y ago
      The event horizon isn't the surface of the black hole. As classically defined, the event horizon is the point at which the black hole's gravity becomes inescapable. But the horizon isn't a point; it is in fact a surface (a 3-surface in 4-dimensional spacetime). Whether you want to call that the "surface" of the black hole is a matter of terminology; the hole is not a solid object--it's not really an "object" at all in the ordinary sense of the term, so it doesn't have a "surface" in the ordinary sense of the term.
- Steuard 13y ago
  The "surface area" of a black hole is defined precisely as the area of the boundary beyond which light cannot escape, so I don't this angle will work the way you're thinking it might. (It's fairly straightforward to show that that surface does indeed expand as more mass is added, too.) It's pretty neat, actually: if you had a black hole containing the mass of a whole galaxy (which isn't unthinkable, especially in the distant future), local space-time at the event horizon would seem completely ordinary (at least as far as we understand from classical relativity and semi-classical quantum corrections like Hawking radiation). You wouldn't have the foggiest idea that you had crossed that fateful line... until you caught on to the fact that you weren't actually making any progress when you tried to turn back.
  - saalweachter 13y ago
    This has always bothered me. I am not a point mass. If I can cross the event horizon and not know it, then there is a time when my feet are on one side and my head on the other. If signals cannot cross the event, how would I not notice my missing feet? If signals from my doomed feet can reach my not yet doomed head, then how is it the event horizon?
    - rca 13y ago
      What if the signals emitted from inside the horizon do reach your head, but only once your head have crossed the horizon? Basically, you can change your reference point and consider that the signal is not moving, but your head is still moving towards the signal. This way, you're still seeing your legs. (Note that I have no idea what I'm talking about)
    - Steuard 13y ago
      Well, your feet aren't ever "missing": as far as classical GR is concerned (even with Hawking's semi-classical quantum corrections included), your experience would be perfectly comfortable and unexceptional as you drifted in. The issue is that in that scenario, your head isn't on an escaping trajectory: the nerve signals from your feet might be moving "outward", but your head will be moving inward, and the signal will only get to your brain after your brain has also crossed the horizon. Now, if you were wearing some sort of rocket collar that fired powerfully just before your head crossed the horizon and carried it safely away from the black hole (or even just held it there), you're right: you wouldn't ever receive that signal from your feet. But I'm pretty sure that would be the least of your worries! (And you would notice your missing feet. And torso.) Roughly speaking, the only non-orbital trajectories close to an event horizon that don't fall straight in must be moving pretty close to light speed as seen by infalling observers.

ck2 13y ago

I think he is saying the gravity is not even enough to have that effect and therefore would be points that are escapable? I always wondered if a black hole could become completely visible again if it collected enough mass to spread its surface area out so that gravity would become thin enough for light to escape again. Kind of like how only the surface of a pond sometimes freezes.

lmm 13y ago
No; adding mass doesn't make it less dense (well, it's a singularity, so it's always infinitely dense. But even if it weren't, normally when you add mass to something it becomes more compressed and gets denser, not less dense), just heavier. (Of course, singularities don't make any sense - which is part of what Hawking's talking about. But that's the classical model of a black hole)
- waps 13y ago
  It's only necessarily infinitely dense from an outside reference frame. Inside you could have anything you'd want according to GR. One certainty is that the space inside a black hole is bigger than the size of the black hole observed from outside of it. So it's entirely possible that you could have a solar system or even an entire universe on the inside. According to GR the space inside a black hole should pretty much behave like the rest of the universe, as long as the black hole is growing. As for "not getting out", well it's impossible to approach the edge of the universe too, despite that border having a physical location. But it's expanding at light speed, so ... no way to get there. So density of a black hole is an entirely different concept from the outside and the inside. Seen from the outside it is infinite. Seen from the inside it is probably not infinite.
  - fennecfoxen 13y ago
    > One certainty is that the space inside a black hole is bigger than the size of the black hole observed from outside of it. So it's entirely possible that you could have a solar system or even an entire universe on the inside. That is not a certainty at all. The entire premise of Hawking's paper is that the object we call a 'black hole' is just a really degenerate dense lump that slowly turns incoming matter/energy into background radiation with as much entropy as possible. And that's it. No fancy Time Lord bigger-on-the-inside science proposed here.
  - pdonis 13y ago
    it's impossible to approach the edge of the universe too, despite that border having a physical location. No, it doesn't. There is no edge to the universe. Our best current model is that the universe is spatially finite, but even if it turns out to be spatially finite, it is unbounded--the spatially finite model is a 3-sphere; like the surface of the Earth but with one more dimension. The surface of the Earth is finite, but it has no edge; a spatially finite universe would be the same way but with one more dimension. Seen from the outside it is infinite. No, it isn't. Seen from the outside, the hole has a finite mass in a finite volume, so its density is finite. (The finite volume an outside observer would assign, based on the surface area of the hole's horizon, is not the "actual" volume of the hole; but it works well enough as an "effective volume", the volume that the hole occupies from the standpoint of the rest of the universe.) The singularity can be viewed as infinitely dense, but the singularity does not occupy the entire black hole; it only occupies the center at r = 0.
panzi 13y ago
What surface? A point has no surface.
- rosser 13y ago
  The event horizon isn't the surface of the black hole. As classically defined, the event horizon is the point at which the black hole's gravity becomes inescapable. That's not at its surface.
  - panzi 13y ago
    Exactly.
  - pdonis 13y ago
    The event horizon isn't the surface of the black hole. As classically defined, the event horizon is the point at which the black hole's gravity becomes inescapable. But the horizon isn't a point; it is in fact a surface (a 3-surface in 4-dimensional spacetime). Whether you want to call that the "surface" of the black hole is a matter of terminology; the hole is not a solid object--it's not really an "object" at all in the ordinary sense of the term, so it doesn't have a "surface" in the ordinary sense of the term.
Steuard 13y ago
The "surface area" of a black hole is defined precisely as the area of the boundary beyond which light cannot escape, so I don't this angle will work the way you're thinking it might. (It's fairly straightforward to show that that surface does indeed expand as more mass is added, too.) It's pretty neat, actually: if you had a black hole containing the mass of a whole galaxy (which isn't unthinkable, especially in the distant future), local space-time at the event horizon would seem completely ordinary (at least as far as we understand from classical relativity and semi-classical quantum corrections like Hawking radiation). You wouldn't have the foggiest idea that you had crossed that fateful line... until you caught on to the fact that you weren't actually making any progress when you tried to turn back.
- saalweachter 13y ago
  This has always bothered me. I am not a point mass. If I can cross the event horizon and not know it, then there is a time when my feet are on one side and my head on the other. If signals cannot cross the event, how would I not notice my missing feet? If signals from my doomed feet can reach my not yet doomed head, then how is it the event horizon?
  - rca 13y ago
    What if the signals emitted from inside the horizon do reach your head, but only once your head have crossed the horizon? Basically, you can change your reference point and consider that the signal is not moving, but your head is still moving towards the signal. This way, you're still seeing your legs. (Note that I have no idea what I'm talking about)
  - Steuard 13y ago
    Well, your feet aren't ever "missing": as far as classical GR is concerned (even with Hawking's semi-classical quantum corrections included), your experience would be perfectly comfortable and unexceptional as you drifted in. The issue is that in that scenario, your head isn't on an escaping trajectory: the nerve signals from your feet might be moving "outward", but your head will be moving inward, and the signal will only get to your brain after your brain has also crossed the horizon. Now, if you were wearing some sort of rocket collar that fired powerfully just before your head crossed the horizon and carried it safely away from the black hole (or even just held it there), you're right: you wouldn't ever receive that signal from your feet. But I'm pretty sure that would be the least of your worries! (And you would notice your missing feet. And torso.) Roughly speaking, the only non-orbital trajectories close to an event horizon that don't fall straight in must be moving pretty close to light speed as seen by infalling observers.