6 ms·

Black holes are relatively "simple". I mean that you can completely define a black hole with 3 properties, one of which isn't really relevant. Classically, yes

by btilly 6y ago

Black holes are relatively "simple". I mean that you can completely define a black hole with 3 properties, one of which isn't really relevant. Classically, yes. But more recently we've discovered not. One of the major conundrums about black holes is how information gets lost in their creation. There is a lot more information in the stuff that creates a black hole than in the black hole itself. Which violates the third law of thermodynamics. This is called the Black Hole Information Paradox. But as https://blogs.scientificamerican.com/observations/have-we-solved-the-black-hole-information-paradox/ https://blogs.scientificamerican.com/observations/have-we-so... explains, the real state of a black hole includes all of the stuff that you can see in the process of falling in. (We never actually see anything hit the event horizon. And in theory something on its way there can still be retrieved a million years after it started falling.) When you track things carefully, a real black hole is a very complex thing indeed. With no information loss.

jihadjihad 6y ago
> Which violates the third law of thermodynamics Wouldn't this be a violation of the second law rather than the third?
- willis936 6y ago
  It's not quite Maxwell's demon because the event horizon is a one-way causal ticket. The information might not be destroyed but it is totally inaccessible on this side of the horizon.
- btilly 6y ago
  I just looked it up, and it is a violation of the fact that the laws of physics should be reversible. So from the wave function at any point of time you should be able to reconstruct the wave at any other. My bad for having stuck thermodynamics in there at all. :-(
colechristensen 6y ago
This brings up a question I have had for a while: what is the life story of a photon shot at a black hole? From an outside observer, presumably you shoot a laser at a black hole and if your photons don't hit anything on the way to the horizon, they never hit it either but just approach asymptotically as time goes to infinity. As, a photon though, you don't "notice" crossing the horizon and in a measurable amount of time you go from being emitted to hitting the singularity. Black holes don't last until infinity though, they evaporate in a large but measurable time. (let's say 10^100 years, it depends on time and how the universe dies and how big the black hole is, but whatever, presuming it happens it is some extraordinarily large number of years) So... as something falling into a black hole an outside observer will watch you infinitely slowly approaching a growing horizon until at some point the growth goes negative and you watch the horizon falling away from your friend the photon until at last the horizon disappears entirely and releases the photon to go about its merry way to hit something sometime around the heat death of the universe. In other words, if you are a conscious very resilient little particle doomed to fall into a black hole... do you really appear from the outside to be falling in until the black hole evaporates? Do you experience going through the event horizon like it's nothing and hit the singularity in a few hours by your own watch... Or is a black hole sort of a time machine to the end of the universe where in the short process of falling in you get to watch the whole history of existence pass you by and you come out at the end having never crossed the horizon in what was for you a very short ride. Or is there some third option where from the outside it takes an infinite amount of time for you to fall in, but that infinity "for math reasons" actually takes place and passes in a Zeno's paradox kind of way at some distinct point in your timeline and you meanwhile pass the horizon and hit the singularity in short order and are no more? I don't really know anything but these are questions and vague thoughts I have had. The central question is reconciling the outside apparent infinite time to watch something cross an event horizon with the finite lifespan of evaporating black holes.
- bpodgursky 6y ago
  Well, the other wrinkle here is that the "singularity" is a mathematical approximation. There's no such thing as hitting it -- from the outside, you just move asymptotically slower as you approach it.
  - btilly 6y ago
    Not so fast. Black holes have a singularity in the middle, and an event horizon around them. There is no singularity at the event horizon, and that is all that we can see..
    - juloo 6y ago
      There might be naked singularities in case of a rapidly rotating black hole. The singularity would have a donut shape and might be outside of the event horizon. https://en.wikipedia.org/wiki/Naked_singularity https://en.wikipedia.org/wiki/Naked_singularity Of course, this is very theoretical and we don't actually expect that to exist.
      - btilly 6y ago
        Huh, I thought that naked singularities were thought to not exist. But apparently in 2018 someone found a case where they could. Thanks!
  - juloo 6y ago
    You move asymptotically slower from the point of view of the particle falling in the black hole. From outside the black hole, the falling particle should fall into the black hole at a speed close to the speed of light. (but we can't observe that from the outside) The image an outsider would see "printed" on the black hole has nothing to do with where the falling particle is.
- btilly 6y ago
  That's officially above my pay grade. :-) Seriously my course in general relativity was about 30 years ago. And work on the information paradox work is about exactly what you're talking about. I honestly don't fully know what happens when you're approaching a growing black hole. Do you cross the horizon then?
cletus 6y ago
Wasn't the issue of black hole information loss the subject of the Hawking-Thorne bet that Hawking eventually conceded (that black holes didn't destroy information)? [1] (This may be a separate issue; I'm genuinely curious). [1]: https://physicsworld.com/a/hawking-loses-black-hole-bet/#:~:text=Stephen%20Hawking%20revealed%20his%20eagerly,of%20Caltech%2C%20about%20black%20holes https://physicsworld.com/a/hawking-loses-black-hole-bet/#:~:....
gus_massa 6y ago
All the stuff around a black hole and in a black hole is under a lot of tidal forces and will be crushed and crumbled. Imagine a blender. According to quantum mechanics, using a blender for an hour is an unitary transformation, so it will not destroy the information, and it's an invertible operation, you can reconstruct the original items. [And even a classic blender is invertible at the molecular level. It's only not invertible if you consider the friction and other average macroscopically properties.] Anyway, after an hour of blending, you will get a horrible homogeneous mix. But (if the current models are correct) a Neutron star is very different in the crust that is a few km near the surface. It has many levels with different properties that are called "nuclear pasta" https://en.wikipedia.org/wiki/Nuclear_pasta https://en.wikipedia.org/wiki/Nuclear_pasta and one with graphics that explain the names https://astrobites.org/2017/10/05/nuclear-pasta-in-neutron-stars/ https://astrobites.org/2017/10/05/nuclear-pasta-in-neutron-s... Going again to the blender example, it's like a bad blended mix, that has foam at the surface, a liquid in the middle and solids at the bottom.
- btilly 6y ago
  It depends on the size of the black hole. One with the mass of the Sun would rip your head from our feet under a tidal force of thousands of gravities. One the size of the monster at the heart of our galaxy has tidal forces so gentle that you wouldn't even notice them until after you were inside the event horizon. If this seems non-intuitive, remember that the Schwarzschild radius varies linearly with mass, and tidal forces scale like mass / radius^3. So tidal forces at the event horizon scale like 1/mass^2. A black hole that is 100 million times heavier will have tidal forces that are 10 quadrillion times smaller at its event horizon.
  - gus_massa 6y ago
    You are correct. But the "no-hair theorem" that says that a black hole can be characterized using only 3 properties does not apply instantly, but only after some time of isolation. For a small black hole the time is very short, but for a huge black hole that will not tear you apart it takes a long time. So even a classic huge black hole has more internal state that should be measurable from outside.
teslaberry 6y ago
the answer is in the black hole jets. there's a lot of research on those jets. invariably much of it is wrong or missing data. many of the assumptions about black hole polar jets is probably bogus. here is why we barely understand polar jets from our own sun.
hnews_account_1 6y ago
It doesn't violate any law of thermodynamics (in as much as it does violate the arrow of time in some sense). The problem is hawking radiation is pure white noise which means if a laplace demon existed (and for our purposes, when quantum mechanics meets real world, we assume an infinite number of laplace demons exist and are created every moment), it cannot reverse a black hole back into its constituents that fell in. This is a HUGE problem. However, there was a solution to this published in 2019. Just look up Black Hole Information Paradox and it should appear as the last section on the wiki page.

by btilly 6y ago

jihadjihad 6y ago

> Which violates the third law of thermodynamics Wouldn't this be a violation of the second law rather than the third?

willis936 6y ago
It's not quite Maxwell's demon because the event horizon is a one-way causal ticket. The information might not be destroyed but it is totally inaccessible on this side of the horizon.
btilly 6y ago
I just looked it up, and it is a violation of the fact that the laws of physics should be reversible. So from the wave function at any point of time you should be able to reconstruct the wave at any other. My bad for having stuck thermodynamics in there at all. :-(

colechristensen 6y ago

This brings up a question I have had for a while: what is the life story of a photon shot at a black hole? From an outside observer, presumably you shoot a laser at a black hole and if your photons don't hit anything on the way to the horizon, they never hit it either but just approach asymptotically as time goes to infinity. As, a photon though, you don't "notice" crossing the horizon and in a measurable amount of time you go from being emitted to hitting the singularity. Black holes don't last until infinity though, they evaporate in a large but measurable time. (let's say 10^100 years, it depends on time and how the universe dies and how big the black hole is, but whatever, presuming it happens it is some extraordinarily large number of years) So... as something falling into a black hole an outside observer will watch you infinitely slowly approaching a growing horizon until at some point the growth goes negative and you watch the horizon falling away from your friend the photon until at last the horizon disappears entirely and releases the photon to go about its merry way to hit something sometime around the heat death of the universe. In other words, if you are a conscious very resilient little particle doomed to fall into a black hole... do you really appear from the outside to be falling in until the black hole evaporates? Do you experience going through the event horizon like it's nothing and hit the singularity in a few hours by your own watch... Or is a black hole sort of a time machine to the end of the universe where in the short process of falling in you get to watch the whole history of existence pass you by and you come out at the end having never crossed the horizon in what was for you a very short ride. Or is there some third option where from the outside it takes an infinite amount of time for you to fall in, but that infinity "for math reasons" actually takes place and passes in a Zeno's paradox kind of way at some distinct point in your timeline and you meanwhile pass the horizon and hit the singularity in short order and are no more? I don't really know anything but these are questions and vague thoughts I have had. The central question is reconciling the outside apparent infinite time to watch something cross an event horizon with the finite lifespan of evaporating black holes.

bpodgursky 6y ago
Well, the other wrinkle here is that the "singularity" is a mathematical approximation. There's no such thing as hitting it -- from the outside, you just move asymptotically slower as you approach it.
- btilly 6y ago
  Not so fast. Black holes have a singularity in the middle, and an event horizon around them. There is no singularity at the event horizon, and that is all that we can see..
  - juloo 6y ago
    There might be naked singularities in case of a rapidly rotating black hole. The singularity would have a donut shape and might be outside of the event horizon. https://en.wikipedia.org/wiki/Naked_singularity https://en.wikipedia.org/wiki/Naked_singularity Of course, this is very theoretical and we don't actually expect that to exist.
    - btilly 6y ago
      Huh, I thought that naked singularities were thought to not exist. But apparently in 2018 someone found a case where they could. Thanks!
- juloo 6y ago
  You move asymptotically slower from the point of view of the particle falling in the black hole. From outside the black hole, the falling particle should fall into the black hole at a speed close to the speed of light. (but we can't observe that from the outside) The image an outsider would see "printed" on the black hole has nothing to do with where the falling particle is.
btilly 6y ago
That's officially above my pay grade. :-) Seriously my course in general relativity was about 30 years ago. And work on the information paradox work is about exactly what you're talking about. I honestly don't fully know what happens when you're approaching a growing black hole. Do you cross the horizon then?

cletus 6y ago

Wasn't the issue of black hole information loss the subject of the Hawking-Thorne bet that Hawking eventually conceded (that black holes didn't destroy information)? [1] (This may be a separate issue; I'm genuinely curious). [1]: https://physicsworld.com/a/hawking-loses-black-hole-bet/#:~:text=Stephen%20Hawking%20revealed%20his%20eagerly,of%20Caltech%2C%20about%20black%20holes https://physicsworld.com/a/hawking-loses-black-hole-bet/#:~:....

gus_massa 6y ago

All the stuff around a black hole and in a black hole is under a lot of tidal forces and will be crushed and crumbled. Imagine a blender. According to quantum mechanics, using a blender for an hour is an unitary transformation, so it will not destroy the information, and it's an invertible operation, you can reconstruct the original items. [And even a classic blender is invertible at the molecular level. It's only not invertible if you consider the friction and other average macroscopically properties.] Anyway, after an hour of blending, you will get a horrible homogeneous mix. But (if the current models are correct) a Neutron star is very different in the crust that is a few km near the surface. It has many levels with different properties that are called "nuclear pasta" https://en.wikipedia.org/wiki/Nuclear_pasta https://en.wikipedia.org/wiki/Nuclear_pasta and one with graphics that explain the names https://astrobites.org/2017/10/05/nuclear-pasta-in-neutron-stars/ https://astrobites.org/2017/10/05/nuclear-pasta-in-neutron-s... Going again to the blender example, it's like a bad blended mix, that has foam at the surface, a liquid in the middle and solids at the bottom.

btilly 6y ago
It depends on the size of the black hole. One with the mass of the Sun would rip your head from our feet under a tidal force of thousands of gravities. One the size of the monster at the heart of our galaxy has tidal forces so gentle that you wouldn't even notice them until after you were inside the event horizon. If this seems non-intuitive, remember that the Schwarzschild radius varies linearly with mass, and tidal forces scale like mass / radius^3. So tidal forces at the event horizon scale like 1/mass^2. A black hole that is 100 million times heavier will have tidal forces that are 10 quadrillion times smaller at its event horizon.
- gus_massa 6y ago
  You are correct. But the "no-hair theorem" that says that a black hole can be characterized using only 3 properties does not apply instantly, but only after some time of isolation. For a small black hole the time is very short, but for a huge black hole that will not tear you apart it takes a long time. So even a classic huge black hole has more internal state that should be measurable from outside.

teslaberry 6y ago

the answer is in the black hole jets. there's a lot of research on those jets. invariably much of it is wrong or missing data. many of the assumptions about black hole polar jets is probably bogus. here is why we barely understand polar jets from our own sun.

hnews_account_1 6y ago

It doesn't violate any law of thermodynamics (in as much as it does violate the arrow of time in some sense). The problem is hawking radiation is pure white noise which means if a laplace demon existed (and for our purposes, when quantum mechanics meets real world, we assume an infinite number of laplace demons exist and are created every moment), it cannot reverse a black hole back into its constituents that fell in. This is a HUGE problem. However, there was a solution to this published in 2019. Just look up Black Hole Information Paradox and it should appear as the last section on the wiki page.