I was reading Brian Greene's "Hidden Reality" and came to the part about Hawking Radiation. Quantum jitters that occur near the event horizon of a black hole, which create both positive-energy particles and negative-energy particles, produce the so-called Hawking radiation. However, I do not understand why only the negative-energy particles are being absorbed into the black hole, while the positive-energy particles shoot outward. Shouldn't there be 50/50 chance that each type of particle is being absorbed by the black hole? Also, the book mentions that a negative-energy particle would appear to an observer inside the black hole as positive. Why?
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asked Jun 22, 2012 at 7:53
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$\begingroup$ No answer is given as to why negative energy is on average absorbed by the black hole. The explanations appear as a desperate attempt to justify a wrong conclusion by Hawking. Hawking assumed a radiating body will evaporate, found that Black Hole radiate, and therefore must evaporate. Hawking did not consider the quantum weirdness of virtual particles radiating from a body do not evaporate that body. What is the explanation that refutes this conclusion that also does not conflict with quantum randomness? $\endgroup$– William BarrCommented Apr 16, 2023 at 12:27
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4 Answers
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There are two ways to approach your question. The first is to explain what Brian Greene means, and the second is to point out that the "particles being swallowed" explanation is a metaphor and isn't actually how the calculation is done. I'll attempt both, but I'm outside my comfort zone so if others can expand or correct what follows please jump in!
When a pair of virtual particles are produced there isn't a negative energy particle and a positive energy particle. Instead the pair form an entangled system where it's impossible to distinguish between them. This entangled system can interact with the black hole and split, and the interaction guarantees that the emerging particle will be the positive one. NB "positive" and "negative" doesn't mean "particle" and "anti-particle" (for what it does mean see below), and the black hole will radiate equal numbers of particles and anti-particles.
Now onto the second bit, and I approach this with trepidation. When you quantise a field you get positive frequency and negative frequency parts. You can sort of think of these as representing particles and anti-particles. How the positive and negative frequencies are defined depends on your choice of vacuum, and in quantum field theory the vacuum is unambiguously defined. The problem is that in a curved spacetime, like the region near a black hole, the vacuum changes. That means observers far from the black hole see the vacuum as different from observers near the black hole, and the two observers see different numbers of particles (and antiparticles). A vaccum near the event horizon looks like excess particles to observers far away, and this is the source of the radiation.
See the Wikipedia article on the Bogoliubov transformation for more information, though I must admit I found this article largely incomprehensible.
Exactly the same maths gives the Unruh effect, i.e. the production of particles in an accelerated frame. The fact that the Unruh effect also produces particles shows that a black hole is not necessary for the radiation, so it can't simply be virtual particles being swallowed.
answered Jun 22, 2012 at 9:04
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2$\begingroup$ You're not alone in thinking that. The problem is you really need to understand quantum field theory to get more than a very approximate feel for what's going on. I guess that's why the "anti-particle falling into the black hole" analogy is so widely used. $\endgroup$ Commented Jun 22, 2012 at 9:43
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3$\begingroup$ This is a good answer, except for one thing: In curved spacetime, there isn't an unambiguous choice of vacuum. There's an unambiguous choice only when you have a Hamiltonian (in which case the vacuum is the lowest energy eigenstate), and this only happens when the time translation is an isomorphism of the spacetime. Without this symmetry, different observers may disagree about which state is really the vacuum, depending on which time axis they've chosen. $\endgroup$– user1504Commented Jun 22, 2012 at 14:08
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4$\begingroup$ Since particles are excitations from the vacuum, these observers will then disagree about the particle content of a system. So, for example, an observer falling freely into a black hole will see empty space, while an observer off at infinity will see space filled with blackbody radiation. $\endgroup$– user1504Commented Jun 22, 2012 at 14:10
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1$\begingroup$ Re Danu's comment: the draft appears to be titled Introduction to Quantum Fields in Classical Backgrounds, but a quick comparison with the Google Books copy of Quantum effects in gravity suggests they are the same book. Thanks Danu, another addition to the large pile of books to be read one day. $\endgroup$ Commented Nov 28, 2014 at 8:47
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7$\begingroup$ Could you expand a bit on why "the interaction guarantees that the emerging particle will be the positive one"? $\endgroup$– tparkerCommented Jul 24, 2016 at 1:21
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Also, the book mentions that a negative-energy particle would appear to an observer inside the black hole as positive. Why?
Very roughly speaking and in as simple terms as possible, inside the black hole, gravity is so intense that the time coordinate and one of the spatial coordinates (the radial coordinate) swap "roles". That's one way to see why you can't get "back up" and exit the hole. "Back up" is now in the reverse time direction and you can't go back in time.
Anyhow, since energy is associated with the time coordinate and momentum with the spatial coordinates, the energy and radial momentum of a particle also swap "roles" when crossing the horizon. The negative energy of the particle becomes negative momentum and the positive momentum of the particle becomes positive energy.
answered Jun 22, 2012 at 12:23
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1$\begingroup$ Re the other comment I made (under John Rennie's answer), the link it includes was to a 2019 paper by the author of 2009's "Radial motion into an Einstein-Rosen bridge", which, together with intervening papers by him, identifies the locally-traversible region of time with the direction opposite the propagation of the BH's apparent horizon. $\endgroup$– EdouardCommented Feb 12, 2020 at 3:46
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I just watched PBS Space time on this, which also seems to answer your question.
https://www.youtube.com/watch?v=ztFovwCaOik
While I have no doubt the accepted answer is the most complete. I still want to try to explain it in the simple terms I understood it from the video.
Disclaimer: I am not a physicist and my answer is most likely wrong/incomplete.
a) Virtual particles are first and foremost a mathematical trick for us to be able to calculate the constantly wiggling space-time on a quantum scale. In reality it seems like, when we talk about virtual particles we actually talk about small field fluctuations canceling out each other.
b) On the event horizon however those fields are suddenly cut of, so the fluctuation can't cancel out and produce real, positive energy possessing particles instead. The Hawking radiation.
So this explains the radiation. Question is, why does the black hole loose energy.
The easy way out would be to call for the law of energy conservation. I have no clue, but I picture it like when you hold a sheet with others steady that wiggles due to wind. It costs you energy to hold it steady at the border as you absorb the fluctuations.
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$\begingroup$ "I picture it like when you hold a sheet with others steady that wiggles due to wind. It costs you energy to hold it steady at the border as you absorb the fluctuations." - THIS! Hallelujah. I've been trying to understand this issue for days, and no explanation has done it yet. So we might say the gravity gets sort of "jostled" away (likely as tiny gravity waves) by the buffeting from all the virtual particle bombardment and launching. $\endgroup$– jazammCommented Feb 23 at 6:48
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Re Torge’s query, “Why does the black hole lose energy?” In field-free space quantum fluctuations arise constantly out of the vacuum, as virtual pairs of particles. But, if I understand correctly, such particle pairs ‘dissolve’ right back into the vacuum and these events do not permanently cause increase or decrease of real energy. However, near a black hole’s horizon, two things are present to change the outcome: (1) A powerful gravitational field and (2) a horizon cutting off back and forth interactions. The vacuum is no longer field-free. Thus the local quantum fluctuations that are always happening, now have real local energy to draw on, and also the horizon functions to allow permanent splitting of the virtual pair into its two components, one outside and one inside the horizon. Their usual annihilation together back into the vacuum is prohibited and their ‘marriage’ cannot be consummated. Thus the now real particle outside the horizon now has a non-zero chance of propagating out to infinity; though not a guarantee.
