How does hawking radiation escape a black hole?

  • #1
SheldonCooper13
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Hello,
Stephen Hawking's book The Future of The Universe says that matter would have to exceed the minimum speed of 5 times the speed of light to exit a black hole. This means that the matter would have to be transferred to Energy using E=MC^2 but all energy is limited by the speed of light which means that the only possible solution would be that matter would be accelerated to five times the speed of light using 5 * Infinity of Energy which would be theoretically impossible discounting the fact that nothing can accelerate past the speed of light. If that was possible(hypothetically) Then the matter or energy would Destroy itself and cease to exist, violating the laws of energy and matter conservation. How then is hawking Radiation able to escape a black hole?
 
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  • #2
SheldonCooper13 said:
How then is hawking Radiation able to escape a black hole?
Not as you describe.

I guess you are referring to The Future of Spacetime?

https://www.amazon.co.uk/Future-Spacetime-Stephen-Hawking/dp/0393020223

That's a popular science book and probably took some liberties in order to explain Hawking radiation. I haven't read it, so I can't comment on it.
 
  • #5
holes weren't as black as they had been painted. The uncertainty principle of quantum mechanics says that particles cannot have both a well-defined position and a well defined speed. The more accurately the position of a particle is defined, the less accurately its speed can be defined, and vice versa.

If a particle is in a black hole, its position is well defined to be within the black hole. This means that its speed cannot be exactly defined. It is therefore possible for the speed of the particle to be greater than the speed of light. This would enable it to escape from the black hole. Particles and radiation will thus slowly leak out of a black hole.

A giant black hole at the center of a galaxy would be millions of miles across. Thus there would be a large uncertainty in the position of a particle inside it. The uncertainty in the particle's speed would therefore be small. This means that it would take a very long time for a particle to escape from the black hole. But it would eventually.

A large black hole at the center of a galaxy could take 1090 years to evaporate away and disappear completely. That is 1 followed by 90 zeros. This is far longer than the present age of the universe, which is a mere 1010. Still, there will be plenty of time if the universe is going to expand forever.(Hawking, 18)

[Post edited for grammar and readability by the Mentors]
 
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  • #6
SheldonCooper13 said:
This means that itwould take a very long time for a particle toescape from the black hole. But it would eventually.
No, it would not. You are mis-applying the HUP.
SheldonCooper13 said:
A large black hole at the center of agalaxy could take 1090 years to evaporate
1090 is not 90 zeros, 10E90 is.

Your reading has, not surprisingly, led you to misunderstand Hawking Radiation. This is a very common misunderstanding, but nothing escapes from INSIDE the BH.

from the original paper on what is now called Hawking Radiation, with emphasis added by me

As the mass of the black hole decreased, the area of the event horizon would have to go down, thus violating the law that, classically, the area cannot decrease [7, 12]. This violation must, presumably, be caused by a flux of negative energy across the event horizon which balances the positive energy flux emitted to infinity. One might picture this negative energy flux in the following way. Just outside the event horizon there will be virtual pairs of particles, one with negative energy and one with positive energy. The negative particle is in a region which is classically forbidden but it can tunnel through the event horizon to the region inside the black hole where the Killing vector which represents time translations is spacelike. In this region the particle can exist as a real particle with a timelike momentum vector even though its energy relative to infinity as measured by the time translation Killing vector is negative. The other particle of the pair, having a positive energy, can escape to infinity where it constitutes a part of the thermal emission described above. The probability of the negative energy particle tunnelling
through the horizon is governed by the surface gravity K since this quantity measures the gradient of the magnitude of the Killing vector or, in other words, how fast the Killing vector is becoming spacelike. Instead of thinking of negative energy particles tunnelling through the horizon in the positive sense of time one could regard them as positive energy particles crossing the horizon on pastdirected world-lines and then being scattered on to future-directed world-lines by the gravitational field. It should be emphasized that these pictures of the mechanism responsible for the thermal emission and area decrease are heuristic only and should not be taken too literally.
ALSO SEE:
https://www.physicsforums.com/threa...ry-a-simple-explanation.1047355/#post-6822501
 
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  • #7
SheldonCooper13 said:
It is therefore possible for the speed of the particle to be greater than the speed of light.
This is false. I'm not aware of any theory that allows faster-than-light particles.
SheldonCooper13 said:
This would enable it to escape from the black hole.
That doesn't make much sense.
SheldonCooper13 said:
Particles and radiation will thus slowly leak out of a black hole. Agiant black hole at the center of a galaxy wouldbe millions of miles across. Thus there would bea large uncertainty in the position of a particleinside it. The uncertainty in the particle's speedwould therefore be small.
This doesn't make much sense either. The moral is that you can't learn any real physics from such sources - no matter what credentials the author has.
 
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  • #8
SheldonCooper13 said:
How then is hawking Radiation able to escape a black hole?
The outgoing particles were never inside the black hole, so there's no problem there.
But as soon as you hear this, you'll be wondering how does the mass/energy inside the hole turn into particles created outside the hole...

There are several things going on here, all of which are pretty much impossible to describe without a fair amount of math. The general idea is that general relativity's black hole solutions do not take quantum mechanics into account. When we do (and this is by no means a fully solved problem) some unexpected and very counterintuitive stuff shows up in the math.
This is as close to a layman-friendly explanation as you will find: http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html
and the real thing from Hawking himself is here: https://projecteuclid.org/download/pdf_1/euclid.cmp/1103899181

You should be very cautious about reading popular explanations, even ones written by bona fide experts like Hawking. The problem, as you can see by comparing the real thing with the popular explanations, is that it the math won't work for a general audience, but without it the explanation has to cut corners and be incomplete in one way or another.
 
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  • #9
SheldonCooper13 said:
It is thereforepossible for the speed of the particle to be greaterthan the speed of light. This would enable it toescape from the black hole.
I'm a little cautious weighing in here - as my understanding is both outdated and likewise pop-sci based - but, as I understand it, the particles that are able to escape a black hole are (the real component of) a virtual particle pair.

Virtual particles are being created and recombined all the time in a vacuum. If this happoens near a black hole's event horizon, one of the pair might fall into the BH. This leaves the other virtual particle with nothing to combine with, and it becomes a real particle, which is outside the BH and may escape it. No FTL needed.

Yes?
 
  • #10
DaveC426913 said:
Yes?
No. The virtual particle pair is a heuristic. See post #6
 
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  • #11
DaveC426913 said:
Yes?
Um,...no. Your first instinct was correct.
 
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  • #12
We have a mess here.

One problem is that the OP expects a popularization to be 100% rigorously correct. This is, of course, not correct, nor is his conclusion that without studying GR and only be reading a popularization, that he he has discovered a paradox in the theory that has eluded those who have for decades. Um.,..no.

Another problem is that his first and second posts seem to be saying different things.

Hawking himself proposed the "one particle of the pair falls in" heuristic and when he does so, he makes a point to tell the reader not to take it too seriously. That point, sadly, was ignored by many, especially popularizers and their ghost writers.

If your real worry is, as @Nugatory explains it should be, on how the energy in the black hole gets out, one way to think of it is as "vacuum sparking". Essentially the energy density of the vacuum due to the gravitational field becomes so large it becomes energetically favorable to produce new particles. And that field was never inside the hole.
 
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FAQ: How does hawking radiation escape a black hole?

How does Hawking radiation escape a black hole if nothing can escape from it?

Hawking radiation doesn't actually escape from within the event horizon of the black hole. Instead, it is generated just outside the event horizon. Quantum fluctuations cause particle-antiparticle pairs to form near the event horizon. One of these particles falls into the black hole while the other escapes into space, making it appear as if radiation is being emitted from the black hole.

What is the role of virtual particles in Hawking radiation?

Virtual particles are temporary pairs of particles and antiparticles that spontaneously form and annihilate in the vacuum of space. Near the event horizon of a black hole, these pairs can be separated by the black hole's gravitational field. One particle falls into the black hole, and the other escapes as Hawking radiation, making the radiation observable.

Does Hawking radiation cause a black hole to lose mass?

Yes, Hawking radiation causes a black hole to lose mass. When one particle of a virtual pair falls into the black hole and the other escapes, the black hole loses a small amount of its mass equivalent to the energy of the escaping particle. Over time, this process can cause the black hole to evaporate completely.

Why is Hawking radiation significant in the study of black holes?

Hawking radiation is significant because it provides a way for black holes to lose mass and energy, which challenges the notion that nothing can escape from a black hole. It also bridges concepts from quantum mechanics and general relativity, offering insights into the behavior of black holes and the nature of quantum gravity.

Can Hawking radiation be observed directly?

As of now, Hawking radiation has not been observed directly. The radiation is incredibly weak and difficult to detect with current technology. However, its theoretical implications are widely accepted in the scientific community, and researchers continue to look for indirect evidence and ways to observe it.

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