4 ms·

As a particle, you continuously accelerate even after the event horizon (which you don't realize). Immediately after you passed the horizon, any photon you can

by juloo 5y ago

As a particle, you continuously accelerate even after the event horizon (which you don't realize). Immediately after you passed the horizon, any photon you can send won't ever leave the black hole. What outside observers will see "frozen" is the instant just before you cross the horizon. A bit like if your image 1ms before you cross the horizon will be seen by the observers years after, your image 1µs before will be seen millions of years after, etc...

colechristensen 5y ago
But... let’s watch that image or at least keep a model of our friends falling into the black hole for 10^100 years. Forever is a long time. If we keep watching that image the black hole eventually stops growing, the “image” never crosses the event horizon and when the universe cools down enough the event horizon starts getting smaller and hotter until we watch our friends getting roasted outside the evaporating black hole which eventually is gone and just normal matter. The timeline of the “image” would seem to reconnect with the real article having never crossed the event horizon. In other words it would seem if we waited long enough the image of our friends outside the event horizon would outlive the black hole and we could go say hello after it evaporated. In other, other words, how do we see the universe outside aging as we fall into a black hole? Do we not get to watch the heat death of the universe as we approach and consequently the black hole very quickly evaporating in front of us as we fall towards it?
- pdonis 5y ago
  > it would seem if we waited long enough the image of our friends outside the event horizon would outlive the black hole and we could go say hello after it evaporated You would see your friend's image, but your friend's image is not your friend. If you tried to go towards where the image appeared to be coming from, you would just find empty space, where the black hole that evaporated away (with your friend inside, having long before hit the singularity) used to be. Your friend would not be there any more. Note, btw, that once you see the final image of your friend (the image of him just crossing the horizon, which you see in a big flash of light when you see the hole finally evaporate), you see nothing more; your friend, and the hole, and all of the other things that fell into the hole (and whose images as they crossed the horizon you also saw in the big flash of light) all vanish after that (no more images are coming).
- jiggawatts 5y ago
  This has been my line of thinking also. Many models of black holes in the past have been way oversimplified, to the point of absurdity. Their temporal existence is bounded in both directions, and neither is an instant cut-off! 1) When a star collapses, the black hole it becomes isn't formed instantly, that would violate the very theory of relativity that predicts their formation. 2) Black holes evaporate, as you've mentioned. But this begs the question: Forget the hypothetical "probe" particle. What about the particles that made up the original star? They're like many probe particles! Imagine the black hole starting at the centre of the collapsing star, expanding outwards, slowing down time for the infalling matter. Does any of the rest of the star fall in to the black hole? Or, like your test particle, does the rest of the star just hang there, frozen in time, a thin layer above the event horizon? Regressing that back to the formation: Does anything ever truly fall into the black hole? I suspect not. My thinking is that a black hole truly is a hole in spacetime, and contains nothing, not even spacetime. It's like poking your finger through a loose-knit wool jumper and making a hole. From the perspective of the threads, you've made a distortion that can be approached, but not entered. Then the evaporation of the black hole is the hole in spacetime closing up, releasing the compressed matter near the horizon as Hawking radiation...
  - saagarjha 5y ago
    You’re correct: for an outside observer watching the formation of a black hole, the matter going into it will never really fall in. It’ll just continuously get dimmer and dimmer.
    - raattgift 5y ago
      Outside observers are more diverse than perhaps you suspect. For instance, what does an observer hovering just above the horizon of a different black hole see? There is more than one astrophysical black hole in our universe, and one should not forget that when reasoning based on the Schwarzschild exact solution. Worse, most of the astrophysical black hole candidates have non-negligible spin. Certainly it would be weird for a stellar black hole to have low spin, as they will have previously have been rotating stars, and so with any sort of trust in the exterior Kerr solution[1], it seems unwise to ignore the observables generated by the ergosphere, in particular the non-stationarity of objects within it. It doesn't really change the thrust of my question two paragraphs up: can an accelerated observer see a blueshifted infall more similar (in terms of counting time by the ticks of the observer's wristwatch) to the left than to the right in https://en.wikipedia.org/wiki/File:Gravitational_time_dilation_around_a_black_hole.gif https://en.wikipedia.org/wiki/File:Gravitational_time_dilati... ? Or, equivalently, are there families of observers who are not (in this case Boyer-Lindquist) coordinate observers ? (The second answer should be, (a) yes, we can find arbitrarily accelerated observers, and (b) we may not be able to find a mapping among the local inertial frames of the infaller, the coordinate observers, and arbitrarily accelerated observers, but we should not ascribe any physical meaning to our failure to do so. cf [2]). Perhaps a more interesting way of thinking about it is that there is essentially no practical difference between a black hole with the infaller collided with the gravitational singularity and a black hole with the infaller at 2GM+epsilon (in units such that c=1 etc) above the gravitational singularity, and if there was a practical difference and it persisted longer than a light-crossing time, we would have disproven the no-hair conjecture. However, LIGO/VIRGO evidence shows a very clear ring-down for BH/BH and BH/NS mergers, which fails to support such a challenge to no-hair. For us weakly accelerated Earthbound observers, neutron stars and black holes at a wide variety of distances from us completely fall into each other in finite (indeed, short in human terms) time. Non-compact infallers don't raise much of a bump on the horizon of the black hole in comparison, but the result is the same: M changes (as does the spin parameter a), and we now "find" the horizon at a new set of points. (Here it's tempting to talk about slicing up spacetime into space and time so we can talk how it is easy to forget what we mean when we talk about e.g. the "before" horizon and the "after" horizon, or alternatively discuss whether, if we had a sensitive gravitational wave detector, we could use that to decide whether a large-mass infalling object was inside or outside the black hole). - -- [1] There are objects like https://carinaemajoris.wordpress.com/2012/07/07/a-very-erratic-black-hole-grs-1915105v1487-aquilae/ https://carinaemajoris.wordpress.com/2012/07/07/a-very-errat... whose observations are much closer matches to Kerr (exterior) solutions with spin parameter ~ 0.9 than to Kerr (exterior) solutions with spin parameter ~ 0 (in particular they're pretty clearly not generating the sorts of timelike geodesics that one would find in a Schwarzschild spacetime). [2] https://physics.stackexchange.com/a/458855 https://physics.stackexchange.com/a/458855

by juloo 5y ago

colechristensen 5y ago

But... let’s watch that image or at least keep a model of our friends falling into the black hole for 10^100 years. Forever is a long time. If we keep watching that image the black hole eventually stops growing, the “image” never crosses the event horizon and when the universe cools down enough the event horizon starts getting smaller and hotter until we watch our friends getting roasted outside the evaporating black hole which eventually is gone and just normal matter. The timeline of the “image” would seem to reconnect with the real article having never crossed the event horizon. In other words it would seem if we waited long enough the image of our friends outside the event horizon would outlive the black hole and we could go say hello after it evaporated. In other, other words, how do we see the universe outside aging as we fall into a black hole? Do we not get to watch the heat death of the universe as we approach and consequently the black hole very quickly evaporating in front of us as we fall towards it?

pdonis 5y ago
> it would seem if we waited long enough the image of our friends outside the event horizon would outlive the black hole and we could go say hello after it evaporated You would see your friend's image, but your friend's image is not your friend. If you tried to go towards where the image appeared to be coming from, you would just find empty space, where the black hole that evaporated away (with your friend inside, having long before hit the singularity) used to be. Your friend would not be there any more. Note, btw, that once you see the final image of your friend (the image of him just crossing the horizon, which you see in a big flash of light when you see the hole finally evaporate), you see nothing more; your friend, and the hole, and all of the other things that fell into the hole (and whose images as they crossed the horizon you also saw in the big flash of light) all vanish after that (no more images are coming).
jiggawatts 5y ago
This has been my line of thinking also. Many models of black holes in the past have been way oversimplified, to the point of absurdity. Their temporal existence is bounded in both directions, and neither is an instant cut-off! 1) When a star collapses, the black hole it becomes isn't formed instantly, that would violate the very theory of relativity that predicts their formation. 2) Black holes evaporate, as you've mentioned. But this begs the question: Forget the hypothetical "probe" particle. What about the particles that made up the original star? They're like many probe particles! Imagine the black hole starting at the centre of the collapsing star, expanding outwards, slowing down time for the infalling matter. Does any of the rest of the star fall in to the black hole? Or, like your test particle, does the rest of the star just hang there, frozen in time, a thin layer above the event horizon? Regressing that back to the formation: Does anything ever truly fall into the black hole? I suspect not. My thinking is that a black hole truly is a hole in spacetime, and contains nothing, not even spacetime. It's like poking your finger through a loose-knit wool jumper and making a hole. From the perspective of the threads, you've made a distortion that can be approached, but not entered. Then the evaporation of the black hole is the hole in spacetime closing up, releasing the compressed matter near the horizon as Hawking radiation...
- saagarjha 5y ago
  You’re correct: for an outside observer watching the formation of a black hole, the matter going into it will never really fall in. It’ll just continuously get dimmer and dimmer.
  - raattgift 5y ago
    Outside observers are more diverse than perhaps you suspect. For instance, what does an observer hovering just above the horizon of a different black hole see? There is more than one astrophysical black hole in our universe, and one should not forget that when reasoning based on the Schwarzschild exact solution. Worse, most of the astrophysical black hole candidates have non-negligible spin. Certainly it would be weird for a stellar black hole to have low spin, as they will have previously have been rotating stars, and so with any sort of trust in the exterior Kerr solution[1], it seems unwise to ignore the observables generated by the ergosphere, in particular the non-stationarity of objects within it. It doesn't really change the thrust of my question two paragraphs up: can an accelerated observer see a blueshifted infall more similar (in terms of counting time by the ticks of the observer's wristwatch) to the left than to the right in https://en.wikipedia.org/wiki/File:Gravitational_time_dilation_around_a_black_hole.gif https://en.wikipedia.org/wiki/File:Gravitational_time_dilati... ? Or, equivalently, are there families of observers who are not (in this case Boyer-Lindquist) coordinate observers ? (The second answer should be, (a) yes, we can find arbitrarily accelerated observers, and (b) we may not be able to find a mapping among the local inertial frames of the infaller, the coordinate observers, and arbitrarily accelerated observers, but we should not ascribe any physical meaning to our failure to do so. cf [2]). Perhaps a more interesting way of thinking about it is that there is essentially no practical difference between a black hole with the infaller collided with the gravitational singularity and a black hole with the infaller at 2GM+epsilon (in units such that c=1 etc) above the gravitational singularity, and if there was a practical difference and it persisted longer than a light-crossing time, we would have disproven the no-hair conjecture. However, LIGO/VIRGO evidence shows a very clear ring-down for BH/BH and BH/NS mergers, which fails to support such a challenge to no-hair. For us weakly accelerated Earthbound observers, neutron stars and black holes at a wide variety of distances from us completely fall into each other in finite (indeed, short in human terms) time. Non-compact infallers don't raise much of a bump on the horizon of the black hole in comparison, but the result is the same: M changes (as does the spin parameter a), and we now "find" the horizon at a new set of points. (Here it's tempting to talk about slicing up spacetime into space and time so we can talk how it is easy to forget what we mean when we talk about e.g. the "before" horizon and the "after" horizon, or alternatively discuss whether, if we had a sensitive gravitational wave detector, we could use that to decide whether a large-mass infalling object was inside or outside the black hole). - -- [1] There are objects like https://carinaemajoris.wordpress.com/2012/07/07/a-very-erratic-black-hole-grs-1915105v1487-aquilae/ https://carinaemajoris.wordpress.com/2012/07/07/a-very-errat... whose observations are much closer matches to Kerr (exterior) solutions with spin parameter ~ 0.9 than to Kerr (exterior) solutions with spin parameter ~ 0 (in particular they're pretty clearly not generating the sorts of timelike geodesics that one would find in a Schwarzschild spacetime). [2] https://physics.stackexchange.com/a/458855 https://physics.stackexchange.com/a/458855