Question about Hawking radiation

  • #1
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So hawking radiation is a phenomenon thwt happens when a pair of particle-anti particle are generated from vacuum according to heisenberg uncertainty principle at the horizon of e event is that right?
And the negative particle "falls" into the black hole while the positive one escapes as hawking radiation

Now here are the questions:

Why doesn't the positive one fall and the negative escape?

Why is hawking radiation a black body radiation so an EM wave?
Can't be the particles generated electrons positrons?
In this case wouldn't escape a particle not an EM wave?
Why does escape an EM wave?
What does really happen? What am i missing?
 

Answers and Replies

  • #2
Nugatory
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So hawking radiation is a phenomenon that happens when a pair of particle-anti particle are generated from vacuum according to heisenberg uncertainty principle at the horizon of e event is that right?
Not right, although that description has been widely repeated in the popular press. There's a pretty decent laymen-friendly explanation here. The real thing is this paper, and if you take a look at page 4 you'll see that what you heard was the "heuristic" explanation for non-specialists.
 
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  • #3
Nugatory
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Can't be the particles generated electrons positrons?
They can be any particles.

However, the effective temperature of a star-sized black hole is quite low (less than the three-odd degrees Kelvin of the cosmic background radiation, so until the universe has cooled more the black hole will be absorbing more energy than it emits) so the Hawking radiation is not very energetic. It takes a lot of energy to generate an electron-positron pair, let alone anything more massive; and that's not something you'll find in the blackbody radiation of something at a temperature of only a few degrees Kelvin.

It gets more exciting as the black hole gets smaller, but that excitement is very far in the future - perhaps ##10^{65}## years for a star-sized black hole. First we have to wait until the universe expands and cools enough that is is colder than the black hole; only then will the black hole start radiating away more energy than it gains from the surrounding space. Then as the black hole shrinks its effective temperature will increases, but at first this process is very slow because the black hole is still very cold and radating very slowly. Only after it has shed a substantial fraction of its mass will it be radiating energetically enough to create particle/anti-particle pairs.
 
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  • #4
Demystifier
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Fundamentally, quantum field theory (QFT) is about fields, not about particles. Particles are only an emergent concept in QFT, not unlike phonons in condensed matter physics. Fields always have energy, but sometimes this energy does not have a form of particles.

So what happens during Hawking radiation? Near the black-hole horizon the field splits into a positive energy part and a negative energy part. The negative energy part falls into the hole, while the positive energy part escapes out of the hole. Far from the hole, where the influence of gravity is negligible, the positive-energy fields condense into positive-energy particles.

Why can't negative-energy field escape from the hole? If that happened, then it would need to condense into negative-energy particles. But it turns out that negative energy particles (when gravity is negligible) have quantum states with negative norms. Negative norms would imply negative probabilities, which doesn't make sense. So essentially, negative-energy particles cannot escape from the hole because it would contradict the probabilistic interpretation of QM.

But what about negative-energy fields which enter the hole? Do they contradict probabilistic interpretation inside the hole? Nobody knows for sure! Inside the hole the standard non-gravitational QFT cannot be applied, and nobody really knows how to do QFT when gravity is strong. It is almost certain that in this regime gravity also needs to be quantized, but nobody really understands quantum gravity in that regime.

One confusing feature of this negative energy falling into the hole is the fact that it carries a positive amount of information. Therefore the black-hole mass decreases while its information content increases, so that at the end we have a very small hole with an arbitrarily large amount of information. How that can be? Is there something wrong in our theory which predicts all this? Something certainly must be wrong, but what exactly? Nobody really knows, but we have a name for this mystery: black-hole information paradox. For some ideas how the paradox could be resolved see
https://en.wikipedia.org/wiki/Black_hole_information_paradox#Postulated_solutions
 
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  • #5
naima
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So what happens during Hawking radiation? Near the black-hole horizon the field splits into a positive energy part and a negative energy part. The negative energy part falls into the hole, while the positive energy part escapes out of the hole. Far from the hole, where the influence of gravity is negligible, the positive-energy fields condense into positive-energy particles.
I read that a free falling observer does not agree with an observer staying near the horizon. For them the cut between positive and negative energy is not at the same place. So one of them does not see evaporation?
We see here that the concept of particle is less fondamental than energy.
 
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  • #6
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I read that a free falling observer does not agree with an observer staying near the horizon. For them the cut between positive and negative energy is not at the same place. So one of them does not see evaporation?
We see here that the concept of particle is less fondamental than energy.
One of them does not see particles. But they all see the flow of energy-momentum. In this sense they all see evaporation.
 
  • #7
naima
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I do not think so. We would have found something which would not br observer dependent. The free falling observer assigns to the BH a temperature proportional to the acceleration she feels. so T = 0 (no evaporation for her).
 
  • #8
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I do not think so. We would have found something which would not br observer dependent. The free falling observer assigns to the BH a temperature proportional to the acceleration she feels. so T = 0 (no evaporation for her).
Well, perhaps the freely falling observer would not call it "evaporation", but she can still see a flux of energy-momentum which takes away the mass from the black hole. Energy-momentum is a covariant tensor, so it cannot disappear by a change of the reference frame.

In fact, the freely falling observer cannot assign any temperature to the system (not even T=0), because temperature can only be assigned to systems in a thermodynamic equilibrium. For the freely falling observer the system changes with time, so it is not in an equilibrium at all.
 
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