Question about relativistic quantum hawking radiation.

  • #1
So as I understand it Stephen Hawking and one other person I can't remember the name of worked on an equation to described the "temperature" of a black hole, that you can actually get information out of it. I know that there is one way it could happen with those vacuum particles in vacuum energy that for whatever reason appear out the the nothingness of space and then annihilate itself and near a black hole one get's sucked in while the other doesn't. But, I was wondering I guess more about the other type of particle pair it describes, photons.
As I understand it, a lower frequency photon will be more delocalized, so it would make sense that photons of extremely low frequency would have an uncertainty in position that can exceed the boundaries of a black hole in purely 3 dimensional space. But, when this happens, apparently one photon somehow suddenly becomes two photons. A photon fluctuates against the event horizon from the interior of the black hole and somehow get's separated into two photons, similar to a quantum tunneling effect. But, how does a photon actually "reach" the event horizon from the interior of a black hole when there's length contraction past a zero metric and how do photons of that low of energy exist in a black hole when it has huge concentrations of energy compacted into a small dense space and should be very very very hot?
 

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  • #2
phinds
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Although I can't give you a specific citation (not because there isn't one but just because I don't have a link handy), Hawking himself said that the "virtual particle" explanation was just a sort of layman's description to what really happens and it isn't REALLY what happens, it's just that English does not lend itself to a simple description of what the math really says.

SO ... don't get wrapped up in the virtual particle explanation of Hawking Radiation.

Towards the end of your post you talk about a particle reaching the EH from inside the BH. that can't happen. The virtual particle explanation is based on particles OUTSIDE the EH, not inside.
 
  • #3
Although I can't give you a specific citation (not because there isn't one but just because I don't have a link handy), Hawking himself said that the "virtual particle" explanation was just a sort of layman's description to what really happens and it isn't REALLY what happens, it's just that English does not lend itself to a simple description of what the math really says.

SO ... don't get wrapped up in the virtual particle explanation of Hawking Radiation.

Towards the end of your post you talk about a particle reaching the EH from inside the BH. that can't happen. The virtual particle explanation is based on particles OUTSIDE the EH, not inside.
Hmm so how is it that black holes lose energy and evaporate then? When I talked to a different physicist about it they seemed to suggest a tunneling effect from the interior that allows a black hole to lose energy, like when you shoot a particle at a wall and very small probability passes through the other side.
I guess the problem comes from the fact that photons aren't actually "traveling" backwards to get out of black hole, they are merely following the correlations that describe their probability. I don't get how it happens when space is that contracted though, it seems like a photon would have to have a wavelength of hundreds of miles to do it, which I guess would explain why it would take millions of years for a black hole to evaporate.
 
  • #4
phinds
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Hmm so how is it that black holes lose energy and evaporate then? When I talked to a different physicist about it they seemed to suggest a tunneling effect from the interior that allows a black hole to lose energy, like when you shoot a particle at a wall and very small probability passes through the other side.
No, the virtual particle explanation says that a pair forms just outside the BH and one leaves the vicinity and the other falls into the BH and somehow (and this is where I really don't get it) the one that falls in always has negative energy and thus reduces the mass of the BH.
 
  • #5
No, the virtual particle explanation says that a pair forms just outside the BH and one leaves the vicinity and the other falls into the BH and somehow (and this is where I really don't get it) the one that falls in always has negative energy and thus reduces the mass of the BH.
But negative energy as far as I know only hypothetically occurs in rotating black holes within the Ergo-sphere, otherwise there's really no reason to assume "negative" energy exists. Also, where does the energy go? If I combine matter and anti matter, the net matter is 0, but I still have "stuff" left over in the form of photons. Are you suggesting that somehow that negative energy just somehow causes non-existence?
But I also know that Hawking didn't like the notion that you can't extract information from the interior of a black hole and so it would make sense that he came up with a way to do it.
 
  • #6
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But, how does a photon actually "reach" the event horizon from the interior of a black hole when there's length contraction past a zero metric and how do photons of that low of energy exist in a black hole when it has huge concentrations of energy compacted into a small dense space and should be very very very hot?
None of those asserations represent current theory.

Kip Thorne, BLACK HOLES AND TIME WARPS, PGS 435-440
Hawking concluded that a black hole behaves precisely as though its horizon has a finite temperature. There are several different ways to picture black hole evaporation...all acknowledge vacuum fluctuations as the ultimate source of the emitted radiation...the simplist is that....tidal gravity pulls the virtual photons apart and the one that escapes carries away the energy that the hole's tidal gravity gave it.
Wikipedia describes it this way:

Emission process
Hawking radiation is required by the Unruh effect and the equivalence principle applied to black hole horizons. Close to the event horizon of a black hole, a local observer must accelerate to keep from falling in. An accelerating observer sees a thermal bath of particles that pop out of the local acceleration horizon, turn around, and free-fall back in. The condition of local thermal equilibrium implies that the consistent extension of this local thermal bath has a finite temperature at infinity…, which implies that some of these particles emitted by the horizon are not reabsorbed and become outgoing Hawking radiation
Quoting from "Quantum Fields in Curved Space" by Birrell and Davies,

At first sight, black hole radiance seems paradoxical, for nothing can apparently escape from within the event horizon. However, the average wavelength of the emitted quanta is ~ M, i.e comparable with the size of the hole. As it is not possible to localize a quantum to within one wavelength, it is therefore meaningless to trace the origin of the particles to any particular region near the horizon. The particle concept, which is basically global, is only useful near [infinity]. In the vicinity of the hole, the spacetime curvature is comparable with the radiation wavelength in the energy range of interest, and the concept of locally-defined particles breaks down." .... So saying the particles are emitted by the horizon, or asking what particles you see when you get near the horizon - both of these are meaningless.
 
  • #7
None of those asserations represent current theory.

Kip Thorne, BLACK HOLES AND TIME WARPS, PGS 435-440


Wikipedia describes it this way:

Emission process


Quoting from "Quantum Fields in Curved Space" by Birrell and Davies,
But that seems to suggest that the uncertainty of the thermal energy of a black hole exceeds the boundary of the event horizon...which still suggests particles aren't actually "traveling", just "appearing" as it says.
I'm really asking more about the transition from finite length contraction to whatever is on the other side.
 
  • #8
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Here is another view...I was looking for Leonard Susskind's decription but this is virtually identical.

from a prior discussion in these forums, Mitchell Porter posts:

... the idea is that the interior of the black hole has a dual (holographic) description in terms of states on the horizon; a lot like AdS/CFT, with the horizon being the boundary to the interior. So when someone crosses the horizon from outside, there's a description which involves them continuing to fall inwards, until they are torn apart by tidal forces and their degrees of freedom redistributed among the black hole's degrees of freedom, all of which will later leak away via Hawking radiation; but there's another description in which, when you arrive at the horizon, your degrees of freedom get holographically smeared across it, once again mingling with all the black hole's prior degrees of freedom (also located on the horizon), which all eventually leak away as Hawking radiation
edit: Keep in mind these various descriptions are in part coordinate based descriptions based on idealized spacetimes....static spacetimes, views from infinity, etc, etc. There is certainly some ambigutiy about what the math means 'physically'....I have seen Penrose statements say 'there is no mass inside a BH' which makes sense because the simple solution is a vacuum solution of the Einstein equations.....elsewhere, others say nobody knows for sure....Is there pressure, vacuum energy.....it seems the event horizon is 'coordinate independent' and therefore 'physical' but just how idealized is that??

I think everybody agrees so far BH seem like the ultimate roach motel: you can get in but you can't get out.

You can read more under "Holographic Principle' of t'hooft and Susskind......say, in wikipedia...
 
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  • #9
Here is another view...I was looking for Leonard Susskind's decription but this is virtually identical.

from a prior discussion in these forums, Mitchell Porter posts:
Yes I've come cross that so called "halogram" theory, but from my knowledge it doesn't make sense. One is that it seems based on an old view of falling towards a black-hole from Shwarzild's equations, which I've actually read in some books that says "something falling towards a blck hole appears completely stopped at the event horizon". If this was physically true, a black hole would never gain mass. What actually happens is someone later came up with better equations and it was shown that an observer would keep traveling normally from their frame of reference no problem right past the event horizon supposedly without really noticing it, assuming they aren't torn apart, meanwhile the outside world sees you traveling velocity but becoming indefinitely red-shifted, eventually turning black or invisible as you cross the event horizon while your clock comes to a halt. Although one thing I don't known is how to balance tidal forces and metric contraction. Near the event horizon, you should appear squished, but as you approach the black hole, tidal forces rip you apart. So do you actually stretch out or appear indefinitely squished as you cross the event horizon?
And then, the event horizon of a black hole grows. I a black hole gains mass, it's even horizon won't grow uniformally, it's event horizon will sort of "stretch" and form a bulge and gradually form a perfect hairy-sphere again, and so the spontaneous-uniform growth of the event horizon would have to mess that up.
 
  • #10
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But negative energy as far as I know only hypothetically occurs in rotating black holes within the Ergo-sphere, otherwise there's really no reason to assume "negative" energy exists. Also, where does the energy go? If I combine matter and anti matter, the net matter is 0, but I still have "stuff" left over in the form of photons. Are you suggesting that somehow that negative energy just somehow causes non-existence?
But I also know that Hawking didn't like the notion that you can't extract information from the interior of a black hole and so it would make sense that he came up with a way to do it.
A virtual particle energy and momentum is not required to follow the usual energy-momentum relationship: [itex]E^2=(cp)^2+(mc^2)^2.[/itex] That's means that a virtual particle can have negative energy.
 
  • #11
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No, the virtual particle explanation says that a pair forms just outside the BH and one leaves the vicinity and the other falls into the BH and somehow (and this is where I really don't get it) the one that falls in always has negative energy and thus reduces the mass of the BH.
The one that falls must have negative energy because the one that leaves has positive energy and energy is conserved. The one that leaves must have positive energy because it is a real particle and only virtual particles are allowed to have negative energy.
 
  • #12
A virtual particle energy and momentum is not required to follow the usual energy-momentum relationship: [itex]E^2=(cp)^2+(mc^2)^2.[/itex] That's means that a virtual particle can have negative energy.
And you can also have imaginary mass too. I also don't understand where they come from in the first place if not from the uncertainty in thermal energy.

The one that falls must have negative energy because the one that leaves has positive energy and energy is conserved. The one that leaves must have positive energy because it is a real particle and only virtual particles are allowed to have negative energy.
But didn't both particles start out as "virtual"? And why is it only the one with negative energy that leaves? And doesn't "virtual" only describe a particle's limitation in space caused by it's own limitations in uncertainty from relative mass like with strong and weak force bosons?
 
  • #14
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Yes I've come cross that so called "halogram" theory, but from my knowledge it doesn't make sense. One is that it seems based on an old view of falling towards a black-hole from Shwarzild's equations, which I've actually read in some books that says "something falling towards a blck hole appears completely stopped at the event horizon". If this was physically true, a black hole would never gain mass. What actually happens is someone later came up with better equations and it was shown that an observer would keep traveling normally from their frame of reference no problem right past the event horizon supposedly without really noticing it, assuming they aren't torn apart, meanwhile the outside world sees you traveling velocity but becoming indefinitely red-shifted, eventually turning black or invisible as you cross the event horizon while your clock comes to a halt. Although one thing I don't known is how to balance tidal forces and metric contraction. Near the event horizon, you should appear squished, but as you approach the black hole, tidal forces rip you apart. So do you actually stretch out or appear indefinitely squished as you cross the event horizon?
And then, the event horizon of a black hole grows. I a black hole gains mass, it's even horizon won't grow uniformally, it's event horizon will sort of "stretch" and form a bulge and gradually form a perfect hairy-sphere again, and so the spontaneous-uniform growth of the event horizon would have to mess that up.
The spaghettification (That's the technical term for the stretching you're describing) happens near the singularity. If the blackhole is large enough, it is safe to cross the horizon which lies away from the singularity itself.
 
  • #15
phinds
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The one that falls must have negative energy because the one that leaves has positive energy and energy is conserved. The one that leaves must have positive energy because it is a real particle and only virtual particles are allowed to have negative energy.
OK, but if the one that leaves becomes real, why doesn't the one that falls in?
 
  • #16
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And you can also have imaginary mass too. I also don't understand where they come from in the first place if not from the uncertainty in thermal energy.



But didn't both particles start out as "virtual"? And why is it only the one with negative energy that leaves? And doesn't "virtual" only describe a particle's limitation in space caused by it's own limitations in uncertainty from relative mass like with strong and weak force bosons?
A virtual particle is one that doesn't obey the equation I quoted. In empty space away from the B-hole it is impossible to to produce real particles out of the vacuum because there is no way to produce particles that will both obey conservation of energy (even virtual particles do) and the dispersion relation for energy and momentum. The black hole makes it possible.
 
  • #17
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OK, but if the one that leaves becomes real, why doesn't the one that falls in?
Because it has negative energy. a real particle must have positive energy that obeys the dispersion relation equation. virtual particle are free from those requirements but are shortlived.
 
  • #18
A virtual particle is one that doesn't obey the equation I quoted. In empty space away from the B-hole it is impossible to to produce real particles out of the vacuum because there is no way to produce particles that will both obey conservation of energy (even virtual particles do) and the dispersion relation for energy and momentum. The black hole makes it possible.
Ok so I think I better understand the situation with virtual particles, but why do those virtual particles appear in the first place? And why "can't" the situation I described take place anyway? As long as a photon isn't actually "traveling", it should be fine, just as an electron isn't actually "traveling" when it changes orbitals at technically infinite speed.
 
  • #19
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Ok, but why do those virtual particles appear in the first place?
They appear all the time everywhere continuously.
 
  • #20
They appear all the time everywhere continuously.
Yeah and they annihilate themselves upholding the conservation outside of a black hole right? So...why do they appear in the first place? And how exactly do we measure the difference in vacuum "pressure" in the casimir effect if the virtual pairs annihilate themselves?
 
  • #21
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Away from the black hole the virtual particles must annihilate after a short time. They have no choice. The B-hole creates the possibility for some of those particle to become real and appear as radiation.
 
  • #22
Away from the black hole the virtual particles must annihilate after a short time. They have no choice. The B-hole creates the possibility for some of those particle to become real and appear as radiation.
Yeah that's fine the model you presented makes more sense now, but you're still not addressing why they appear or really even why they must uphold some random conservation law that someone wrote down years ago, nor was it addressed how it is impossible for the uncertainty of the interior radiation to exceed the event horizon, which I could have sworn was how it was describe as in a book I read.
 
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  • #23
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Yeah that's fine the model you presented makes more sense now, but you're still not addressing why they appear or really even why they must uphold some random conservation law that someone wrote down years ago, nor was it addressed how it is impossible for the uncertainty of the interior radiation to exceed the event horizon, which I could have sworn was how it was describe as in a book I read.
That's a question of epistemology, not physics. Why should nature follow some rules in a dusty ol' book? I don't know, but it does - and I sure am glad it does.
 
  • #24
That's a question of epistemology, not physics. Why should nature follow some rules in a dusty ol' book? I don't know, but it does - and I sure am glad it does.
Alright so what about the uncertainty of the thermal radiation? Wasn't it postulated that black holes have a temperature?
And also, what's making those particles appear in the first place? I keep asking that.
 
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  • #25
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Alright so what about the uncertainty of the thermal radiation? Wasn't it postulated that black holes have a temperature?
And also, what's making those particles appear in the first place? I keep asking that.
Yes you keep asking that. I guess you didn't like my answers :( .Basically, some of what starts as part of the virtual particle sea that is present everywhere all the time turns into thermal radiation near a B-hole, just because it can.
 

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