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nickyrtr

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- Thread starter nickyrtr
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nickyrtr

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As a simple example, suppose an observer in flat spacetime is accelerating continuously in a rocket. There is a region of spacetime from which information can never propagate to this observer, and the edge of this region is seen by this observer as a horizon.

We don't really know for sure how quantum mechanics interfaces to gravity, but one principle that seems to hold true is that observers see Hawking radiation coming from any horizon. It doesn't have to be a black hole's horizon. So the observer accelerating in the rocket in flat spacetime will see Hawking radiation coming from the horizon that he observes -- even though it's not a horizon that an inertial observer even agrees exists.

An observer who has already fallen through the horizon of a black hole does not perceive the region outside the black hole as lying behind a horizon. Information from outside the horizon can still propagate to him (until he goes splat at the singularity). So I don't think any Hawking radiation is observable by him.

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Bill_K

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Yes, it's often described that way, but not the case. Quoting from "Quantum Fields in Curved Space" by Birrell and Davies,That radiation is described as being emitted by the event horizon

"At first sight, black hole radiance seems paradoxical, for nothing can apparently escape from within the event horizon. However,

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.

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As a simple example, suppose an observer in flat spacetime is accelerating continuously in a rocket. There is a region of spacetime from which information can never propagate to this observer, and the edge of this region is seen by this observer as a horizon.

But is is not true that the EH of a black hole, while not actually a physical construct, has nonetheless a local physical effect that allows split virtual particles to be separated, and thus become Hawking radiation?

We don't really know for sure how quantum mechanics interfaces to gravity, but one principle that seems to hold true is that observers see Hawking radiation coming from any horizon. It doesn't have to be a black hole's horizon.

That really puzzles me, because such a horizon is just an observational effect that has no localized physical ability to do anything the way a BH's EH does so why would not split virtual particles just recombine and NOT become Hawking radiation?

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Bill_K

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And so are particles. The important realization that lies behind Hawking radiation is that the presence or absence of particles is not absolute. One observer may detect particles while another observer detects none. In particular a uniformly accelerating observer in Minkowski space detects a thermal bath of particles associated with his Rindler horizon.That really puzzles me, because such a horizon is just an observational effect

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And so are particles. The important realization that lies behind Hawking radiation is that the presence or absence of particles is not absolute. One observer may detect particles while another observer detects none. In particular a uniformly accelerating observer in Minkowski space detects a thermal bath of particles associated with his Rindler horizon.

OK, but this makes it sound like you are saying that Hawking radiation is just an optical illusion, but if it is REAL, as I thought it to be, it has the physical effect of reducing the mass of the BH. How is that consistent with "one observer may detect particles while another observer detects none" ? It either does or does not reduce the mass of the BH. I'm sure you're not saying it is observer dependant, so are you saying it does not?

- #7

soothsayer

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OK, but this makes it sound like you are saying that Hawking radiation is just an optical illusion, but if it is REAL, as I thought it to be, it has the physical effect of reducing the mass of the BH. How is that consistent with "one observer may detect particles while another observer detects none" ? It either does or does not reduce the mass of the BH. I'm sure you're not saying it is observer dependant, so are you saying it does not?

What Bill K is talking about is Unruh Radiation. I think there is significant debate in the scientific community about how to exactly distinguish Unruh radiation from Hawking radiation. The general consensus I have heard is that they are not the same thing, but I have never heard a satisfying answer as to why (at least, none that I could understand). I have talked to a professor of mine extensively about it and he too is unclear as to the distinction. What he told me was, that if we have an observer at infinity, measuring what he believes to be Hawking radiation coming from a black hole, and we have an observer free falling into the black hole, measuring what he believes to be Unruh radiation due to his acceleration, they will agree on the amount of thermal radiation.

- #8

jartsa

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A free falling observer does not observe Hawking radiation. He does observe the mass of the black hole decreasing, and he observers the mass of a Hawking Radiation Absorber increasing.

Hawking Radiation Absorber = non free falling black piece of cardboard.

An observer that feels an acceleration sees some Unruh radiation. Our Hawking Radiation Absorber does feel acceleration, and it sees radiation that looks just like Unruh radiation caused by that acceleration.

Hawking Radiation Absorber = non free falling black piece of cardboard.

An observer that feels an acceleration sees some Unruh radiation. Our Hawking Radiation Absorber does feel acceleration, and it sees radiation that looks just like Unruh radiation caused by that acceleration.

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I think the answer to this is no, and one way to see that it has to be no is that it would violate the equivalence principle. One way of stating the e.p. is that spacetime is always locally Minkowski. Spacetime doesn't locally have special properties.But is is not true that the EH of a black hole, while not actually a physical construct, has nonetheless a local physical effect that allows split virtual particles to be separated, and thus become Hawking radiation?

I haven't dug deeply into the theory of Hawking radiation, but my understanding is that this picture of splitting pairs of virtual particles is really a popularization and not to be taken completely seriously. Here's a discussion of that by Baez: http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.htmlThat really puzzles me, because such a horizon is just an observational effect that has no localized physical ability to do anything the way a BH's EH does so why would not split virtual particles just recombine and NOT become Hawking radiation?

- #10

soothsayer

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I haven't dug deeply into the theory of Hawking radiation, but my understanding is that this picture of splitting pairs of virtual particles is really a popularization and not to be taken completely seriously.

As someone who has dug somewhat deeply into Hawking Radiation (though my understanding of QFT is qualitative at best, so my understanding of HR is somewhat limited) you are correct, the popular idea of particles being split at the horizon of the black hole is a bit inaccurate. At the very least, there's a more formal way of understanding it, and I think your link gets at it pretty well.

- #11

jartsa

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An observer who has already fallen through the horizon of a black hole does not perceive the region outside the black hole as lying behind a horizon. Information from outside the horizon can still propagate to him (until he goes splat at the singularity). So I don't think any Hawking radiation is observable by him.

And what does the observer see below him? Well let me answer:

Things falling down seem to freeze when aproaching some kind of spherical "horizon", and nothing ever seems to cross said horizon.

And also Hawking radiation seems to be coming from where the "horizon" is.

If this observer creates a small black hole, then surely there is a horizon around that black hole. This proves that inside an event horizon, event horizons can be observed.

- #12

nickyrtr

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I think the answer to this is no, and one way to see that it has to be no is that it would violate the equivalence principle. One way of stating the e.p. is that spacetime is always locally Minkowski. Spacetime doesn't locally have special properties.

What about curvature? That is a local property of spacetime, and I thought it was the property associated with Hawking radiation.

Say you have a particle detector in a curved region of spacetime, such as near a black hole. Particle-antiparticle pairs popping out of the vacuum have a certain probability that one is absorbed by the detector before recombining with the other. Wouldn't this probability increase with the local curvature of spacetime?

- #13

stevebd1

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Probably not what the OP was enquiring about but thought it would be of interest-

'Negative Temperature of Inner Horizon and Planck Absolute Entropy of a Kerr-Newman Black' Hole by Liu Bo Liu & Wen-Biao

http://bit.ly/O1L2qD (paper will download and open)

'Entropy of Kerr-Newman Black Hole Continuously Goes to Zero when the Hole Changes from Nonextreme Case to Extreme Case' by ZHAO Zheng, ZHU Jian-yang & LIU Wen-biao

Abstract-

A new formulation of the Bekenstein-Smarr formula of a Kerr-Newman black hole is given. The re-defined black hole entropy continuously goes to zero as the black hole temperature approaches absolute zero, which satisfies the Nernst theorem. Our new result suggests that the Kerr-Newman black hole should be regarded as a composite thermodynamic system composed of two sub-systems, its outer horizon and its inner horizon. There exists a new quantum thermal effect. “Hawking absorption”, near the inner horizon of the black hole.

http://cpl.iphy.ac.cn/EN/Y1999/V16/I9/698 [Broken]

'A Proposed Absolute Entropy of Near Extremal Kerr-Newman Black Hole' by Hai Lin

http://arxiv.org/abs/gr-qc/0104098v2

'Negative Temperature of Inner Horizon and Planck Absolute Entropy of a Kerr-Newman Black' Hole by Liu Bo Liu & Wen-Biao

http://bit.ly/O1L2qD (paper will download and open)

'Entropy of Kerr-Newman Black Hole Continuously Goes to Zero when the Hole Changes from Nonextreme Case to Extreme Case' by ZHAO Zheng, ZHU Jian-yang & LIU Wen-biao

Abstract-

A new formulation of the Bekenstein-Smarr formula of a Kerr-Newman black hole is given. The re-defined black hole entropy continuously goes to zero as the black hole temperature approaches absolute zero, which satisfies the Nernst theorem. Our new result suggests that the Kerr-Newman black hole should be regarded as a composite thermodynamic system composed of two sub-systems, its outer horizon and its inner horizon. There exists a new quantum thermal effect. “Hawking absorption”, near the inner horizon of the black hole.

http://cpl.iphy.ac.cn/EN/Y1999/V16/I9/698 [Broken]

'A Proposed Absolute Entropy of Near Extremal Kerr-Newman Black Hole' by Hai Lin

http://arxiv.org/abs/gr-qc/0104098v2

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