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I don't understand hawking radiation. People say it would eventually evaporate the black hole, but wouldn't the particle falling in increase its mass, whilst th

by sullyj3 11y ago

I don't understand hawking radiation. People say it would eventually evaporate the black hole, but wouldn't the particle falling in increase its mass, whilst the emitted anti-particle would have no effect?

Retric 11y ago
First it only applies to rotating black holes. Second it's particle and virtual particle not anti particle.
- danbruc 11y ago
  I don't think that this is true, non-rotating black holes can also emit Hawking radiation. Besides that, there are probably no non-rotating black holes, having an angular momentum of (exactly) zero is a pretty improbable situation and a single particle falling not exactly radially into a non-rotating black hole would turn it into a rotating black hole albeit with a really tiny angular momentum.
  - Retric 11y ago
    I don't see why it would not work on stationary black holes, but I keep seeing it in reference to rotating black holes. Though, as you say all black holes rotate.
    - danbruc 11y ago
      According to Wikipedia Hawking started his work after another physicist pointed out that rotating black holes should emit radiation. Maybe that stuck somehow. Hawking's work followed his visit to Moscow in 1973 where the Soviet scientists Yakov Zeldovich and Alexei Starobinsky showed him that, according to the quantum mechanical uncertainty principle, rotating black holes should create and emit particles.
      - Retric 11y ago
        Ahh, ok thanks that's the connection I was missing.
Natanael_L 11y ago
They're virtual with no real energy, and would normally cease to exist. The fact that the escaping particle is prevented from rejoining the captured one forces it to become real, and for that to be possible it takes energy/mass from the black hole.
raattgift 11y ago
That's OK, Hawking Radiation (HR) is non-trivial to understand, and you hit on one problem that arises in a pop description of HR. The key to it is that in General Relativity different observers will see different numbers of particles; in particular, an accelerated observer will see more particles than an unaccelerated observer. You can think of this in terms of a blueshift where particle count increases, rather than just individual particle energy; this is particularly the case in the very low energy limit, where there are tiny numbers of particles to start with, or alternatively, where the average particle wavelength is so long as to be undetectable. A blueshift in the latter case causes particles to become detectable because of wavelength-shortening. A gravitational collapse causes an acceleration between freely-falling observers in the past and observers in the future, mainly because of the spacetime curvature exposed just as the event horizon forms. The "empty" space outside the horizon actually has extremely long wavelength radiation in it; it's a very low temperature thermal distribution, related to the CMB and the temperature of the vacuum. The accelerated future observers see a shortening of the wavelengths of this radiation trapped near the event horizon, which leads to particles appearing where a past observer saw none. Because these particles are all outside the event horizon, they can escape; if and when they do they take mass-energy away from the region near the black hole, with the result that the black hole contracts. That in turn reveals more "fossil" curvature, which causes a greater acceleration for future observers, who see more and hotter particles, and so on until some limit (e.g. the black hole completely evaporates or a stable remnant is left). (It's important to note that the curvature is "fossilized", in that it is there just as the black hole is forming, and is strongest near the gravitational singularity, and weakens at increasing radius from there. The larger the mass of the black hole, the weaker the curvature is just outside the horizon; for ultramassive black holes the curvature immediately outside the horizon is almost arbitrarily flat. Conversely, smaller black holes have stronger curvature just outside the horizon.) There are a variety of very different descriptions of the mechanism of Hawking radiation, including the most famous popularizaiton by Hawking himself, which involves virtual particle pairs, where one half of each pair falls into the black hole and the other escapes to infinity. Unfortunately most of them not only hide the actual nature of Hawking radiation, but outright conflict with mathematical tracing of emitted radiation back to the black hole (in particular the stress-energy at the horizon one gets when backtracking this way ends up being obviously wrong). The mathematically stable solution -- and the one that Hawking writes down in scientific papers rather than popular books -- has any pair separation happening at a significant distance from the event horizon. Sabine Hossenfelder has a more technical and detailed write-up of this line of explanation at http://backreaction.blogspot.co.uk/2015/12/hawking-radiation-is-not-produced-at.html http://backreaction.blogspot.co.uk/2015/12/hawking-radiation... and links from there to a few scientific papers.

by sullyj3 11y ago

Retric 11y ago

First it only applies to rotating black holes. Second it's particle and virtual particle not anti particle.

danbruc 11y ago
I don't think that this is true, non-rotating black holes can also emit Hawking radiation. Besides that, there are probably no non-rotating black holes, having an angular momentum of (exactly) zero is a pretty improbable situation and a single particle falling not exactly radially into a non-rotating black hole would turn it into a rotating black hole albeit with a really tiny angular momentum.
- Retric 11y ago
  I don't see why it would not work on stationary black holes, but I keep seeing it in reference to rotating black holes. Though, as you say all black holes rotate.
  - danbruc 11y ago
    According to Wikipedia Hawking started his work after another physicist pointed out that rotating black holes should emit radiation. Maybe that stuck somehow. Hawking's work followed his visit to Moscow in 1973 where the Soviet scientists Yakov Zeldovich and Alexei Starobinsky showed him that, according to the quantum mechanical uncertainty principle, rotating black holes should create and emit particles.
    - Retric 11y ago
      Ahh, ok thanks that's the connection I was missing.

Natanael_L 11y ago

They're virtual with no real energy, and would normally cease to exist. The fact that the escaping particle is prevented from rejoining the captured one forces it to become real, and for that to be possible it takes energy/mass from the black hole.

raattgift 11y ago

That's OK, Hawking Radiation (HR) is non-trivial to understand, and you hit on one problem that arises in a pop description of HR. The key to it is that in General Relativity different observers will see different numbers of particles; in particular, an accelerated observer will see more particles than an unaccelerated observer. You can think of this in terms of a blueshift where particle count increases, rather than just individual particle energy; this is particularly the case in the very low energy limit, where there are tiny numbers of particles to start with, or alternatively, where the average particle wavelength is so long as to be undetectable. A blueshift in the latter case causes particles to become detectable because of wavelength-shortening. A gravitational collapse causes an acceleration between freely-falling observers in the past and observers in the future, mainly because of the spacetime curvature exposed just as the event horizon forms. The "empty" space outside the horizon actually has extremely long wavelength radiation in it; it's a very low temperature thermal distribution, related to the CMB and the temperature of the vacuum. The accelerated future observers see a shortening of the wavelengths of this radiation trapped near the event horizon, which leads to particles appearing where a past observer saw none. Because these particles are all outside the event horizon, they can escape; if and when they do they take mass-energy away from the region near the black hole, with the result that the black hole contracts. That in turn reveals more "fossil" curvature, which causes a greater acceleration for future observers, who see more and hotter particles, and so on until some limit (e.g. the black hole completely evaporates or a stable remnant is left). (It's important to note that the curvature is "fossilized", in that it is there just as the black hole is forming, and is strongest near the gravitational singularity, and weakens at increasing radius from there. The larger the mass of the black hole, the weaker the curvature is just outside the horizon; for ultramassive black holes the curvature immediately outside the horizon is almost arbitrarily flat. Conversely, smaller black holes have stronger curvature just outside the horizon.) There are a variety of very different descriptions of the mechanism of Hawking radiation, including the most famous popularizaiton by Hawking himself, which involves virtual particle pairs, where one half of each pair falls into the black hole and the other escapes to infinity. Unfortunately most of them not only hide the actual nature of Hawking radiation, but outright conflict with mathematical tracing of emitted radiation back to the black hole (in particular the stress-energy at the horizon one gets when backtracking this way ends up being obviously wrong). The mathematically stable solution -- and the one that Hawking writes down in scientific papers rather than popular books -- has any pair separation happening at a significant distance from the event horizon. Sabine Hossenfelder has a more technical and detailed write-up of this line of explanation at http://backreaction.blogspot.co.uk/2015/12/hawking-radiation-is-not-produced-at.html http://backreaction.blogspot.co.uk/2015/12/hawking-radiation... and links from there to a few scientific papers.