Hawking Radiation and the Negative Energy Particle

  • #1
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The following from Wiki re Hawking Radiation:


“... vacuum fluctuations cause a particle–antiparticle pair to appear close to the event horizon of a black hole. One of the pair falls into the black hole while the other escapes. In order to preserve total energy, the particle that fell into the black hole must have had a negative energy (with respect to an observer far away from the black hole).”


The bit I don’t understand is that the particle that falls into the BH must have had a negative energy. Why is this? And is the particle with positive energy simply repelled from the BH while that with negative energy is absorbed into it? Or does the particle that falls in somehow automatically and de facto become that with negative energy?

Also, does this split between elementary virtual particles imply that the event horizon is hyper-fine, a razor-thin surface in space, that it has no depth whatsoever?


IH
 

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  • #2
PeterDonis
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The following from Wiki re Hawking Radiation
Is a heuristic description that is really pop science and doesn't correspond well to the actual physics.

The bit I don’t understand is that the particle that falls into the BH must have had a negative energy. Why is this?
In the terms of the heuristic description, it's because a virtual particle with negative energy, outside the hole's horizon, has to annihilate with its partner that has positive energy immediately; it can't continue to exist. So if the positive energy member of the pair were to fall through the horizon, the negative energy member would have to go with it; it couldn't stay outside because it can't exist there. So the only times we see actual radiation emitted is when the negative energy member of the pair falls in and the positive energy one stays outside.

But, as I said above, this is just a heuristic picture and doesn't correspond well to the actual physics, so you should not expect it to make complete sense.

More here:

http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html
 
  • #3
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Thanx for the feedback...seems very little of what one reads of ‘pop-sci’ ends up being easy to understand at the hard-science, bedrock level...


IH
 
  • #4
kimbyd
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One of the problems with this description in general is that virtual particles aren't necessarily actual things that ever exist.

Virtual particles stem from a particular method for calculating interactions in what is known as scattering theory. Scattering theory involves some number of particles entering from infinity, interacting with one another in some way, and then some number of particles exiting out to infinity. The "scattering amplitude" of a given outcome tells you how likely that outcome is. For example, if you have an electron and a positron collide, there is some probability for them to annihilate into a pair of photons. There is also some probability that they will, if they have enough energy, produce a pair of muons. Or tau particles. Or neutrinos. Given the energies of the incoming particles, you can calculate the probability of each of these using scattering theory.

The problem is: nobody knows how to solve the equations involved in that calculation exactly. We have to use approximations. And the primary approximation that is used is known as perturbation theory. The idea of perturbation theory is that rather than trying to solve the full equation, we break it up into distinct, solvable pieces which can be solved individually and added together. Mathematically, these distinct pieces end up looking a lot like the creation and annihilation of particles: the calculation, once performed, looks like when the electron and positron come together, there are a whole bunch of photons, electrons, positrons, and other more exotic particles which pop into existence for a fraction of a second before annihilating.

But there's a problem: it doesn't look exactly like the above situation. The masses of the temporary (i.e. virtual) particles are all wrong: while a photon has zero mass, the virtual photon which appears in the above calculation has imaginary mass (that is, the square of its mass is negative). It's not even all that hard to prove that virtual photons must have imaginary masses (this fact draws directly from conservation of momentum and energy of the incoming and outgoing particles). The fact that these virtual particles don't have the right masses may be a hint that this perturbation theory approach isn't actually describing anything that actually happens: it may just be a coincidence that the calculation looks sort of like the creation/annihilation of particles.

So, with the black hole case, we know that the outgoing particles necessarily have positive energy (as PeterDonis described). As the theory respects local energy conservation, that energy has to come from the black hole. Thus, if the system is described as virtual particles falling into the black hole, then those virtual particles must have negative energy in order to satisfy the conservation laws that the equations require. Note that this conservation law isn't anything imposed externally: it falls out of the equations. But given that virtual particles may be mathematical figments anyway, it's not worth taking that kind of description too seriously.
 
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  • #5
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Thanx kimbyd...that helps sorting the mathematical instruments from the physical reality...the two may as I understand it coincide...or they may not...


IH
 
  • #6
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Another question re the negative energy particle which falls into the BH.

This particle will fall into the center of the BH and be annihilated by its impact with the positive energy mass which (somehow) lies there. This will decrease the nominal mass lying at the center. But the energy of annihilation which results has no way to escape the BH, so it is trapped forever within it.

In mass-energy terms, the BH has not lost anything, it still has the same mass-energy as before.

Where is the fallacy with my reasoning here?


IH
 
  • #7
PeterDonis
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This particle will fall into the center of the BH and be annihilated by its impact with the positive energy mass which (somehow) lies there.
The mass of the black hole does not lie at the center. The "center" ##r = 0## is not a place in space; it's a moment of time (which is to the future of all events inside the hole's horizon). To the extent that a time can be assigned at all to "when" the hole loses the mass due to Hawking radiation, it's when the (virtual, heuristic) "negative energy particle" falls through the horizon.
 
  • #8
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Thanx for the reply...seems like one cannot really understand Hawking Radiation unless a significant investment is made in its underlying mathematics...the heuristic explanation is perhaps not the optimal...


IH
 
  • #9
kimbyd
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Another question re the negative energy particle which falls into the BH.

This particle will fall into the center of the BH and be annihilated by its impact with the positive energy mass which (somehow) lies there. This will decrease the nominal mass lying at the center. But the energy of annihilation which results has no way to escape the BH, so it is trapped forever within it.

In mass-energy terms, the BH has not lost anything, it still has the same mass-energy as before.

Where is the fallacy with my reasoning here?


IH
I think you're confusing this negative-energy virtual particle for anti-matter.

Most of the particles that leave the black hole in Hawking Radiation are photons. So, most of the particles in this heuristic description would be negative-energy photons. Of the particles that are not their own anti-particles (like photons), half will be matter, half anti-matter.

Once you go past the event horizon, though, the analogy completely breaks down, for the reason that the matter in the interior of a black hole is not described at all by General Relativity (the r=0 point, being a singularity, cannot be located on a manifold, and GR can only describe what happens on manifolds). We would need to have an understanding of quantum gravity to make any statement about how this negative-energy particle interacts with anything on the interior of the black hole.

And if we had an understanding of quantum gravity, there's a pretty good chance that we'd find this heuristic explanation would be complete nonsense (since it already doesn't make sense given what we do know).
 
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the r=0 point, being a singularity, cannot be located on a manifold.
Negative mass always did sound like nonsense to me, even if can be described mathematically.
 
  • #11
PeterDonis
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the matter in the interior of a black hole is not described at all by General Relativity
This is much too strong. The singularity, as you say, is not part of the manifold; but classical GR gives a perfectly good manifold structure for the interior of the hole for all ##r > 0##, and in models such as the Oppenheimer-Snyder model for the spherically symmetric collapse of dust, there is no problem whatever in having a matter region even inside the horizon, whose description works just fine with GR.

It is an open question how much quantum effects change the actual physics from the classical GR description; but that is not the same as saying that classical GR does not give a description at all. It does.
 
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  • #12
Imager
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The best explanation ever!

One of the problems with this description in general is that virtual particles aren't necessarily actual things that ever exist.

Virtual particles stem from a particular method for calculating interactions in what is known as scattering theory. Scattering theory involves some number of particles entering from infinity, interacting with one another in some way, and then some number of particles exiting out to infinity. The "scattering amplitude" of a given outcome tells you how likely that outcome is. For example, if you have an electron and a positron collide, there is some probability for them to annihilate into a pair of photons. There is also some probability that they will, if they have enough energy, produce a pair of muons. Or tau particles. Or neutrinos. Given the energies of the incoming particles, you can calculate the probability of each of these using scattering theory.

The problem is: nobody knows how to solve the equations involved in that calculation exactly. We have to use approximations. And the primary approximation that is used is known as perturbation theory. The idea of perturbation theory is that rather than trying to solve the full equation, we break it up into distinct, solvable pieces which can be solved individually and added together. Mathematically, these distinct pieces end up looking a lot like the creation and annihilation of particles: the calculation, once performed, looks like when the electron and positron come together, there are a whole bunch of photons, electrons, positrons, and other more exotic particles which pop into existence for a fraction of a second before annihilating.

But there's a problem: it doesn't look exactly like the above situation. The masses of the temporary (i.e. virtual) particles are all wrong: while a photon has zero mass, the virtual photon which appears in the above calculation has imaginary mass (that is, the square of its mass is negative). It's not even all that hard to prove that virtual photons must have imaginary masses (this fact draws directly from conservation of momentum and energy of the incoming and outgoing particles). The fact that these virtual particles don't have the right masses may be a hint that this perturbation theory approach isn't actually describing anything that actually happens: it may just be a coincidence that the calculation looks sort of like the creation/annihilation of particles.

So, with the black hole case, we know that the outgoing particles necessarily have positive energy (as PeterDonis described). As the theory respects local energy conservation, that energy has to come from the black hole. Thus, if the system is described as virtual particles falling into the black hole, then those virtual particles must have negative energy in order to satisfy the conservation laws that the equations require. Note that this conservation law isn't anything imposed externally: it falls out of the equations. But given that virtual particles may be mathematical figments anyway, it's not worth taking that kind of description too seriously.
 
  • #13
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How can a black hole evaporate due to Hawking radiation? Aren't they constantly attracting real matter anyway? o_O
 
  • #14
Ibix
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How can a black hole evaporate due to Hawking radiation? Aren't they constantly attracting real matter anyway? o_O
Yes. They both attract matter and evaporate, and stellar mass black holes attract a lot more than they evaporate so they grow. But eventually they'll run out of matter to attract but continue to evaporate, so they'll start to shrink (billions of years in the future).

Microscopic holes can't draw in enough matter to overcome their evaporation rate, so are expected to disappear on very short timescales if we can ever find or create one.
 
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  • #15
kimbyd
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How can a black hole evaporate due to Hawking radiation? Aren't they constantly attracting real matter anyway? o_O
This is a misconception. If the Sun was suddenly replaced by a black hole of the same mass, the other objects in our Solar System would change very little. They'd get a lot colder, naturally, due to the dramatic decrease in radiation. And there would be no solar wind. But they'd continue on their previous orbits, because the gravity they experience won't have changed one iota.

Black holes definitely absorb matter that gets close enough to them: within a certain radius, no stable orbits around a black hole exist. But once they gobble up that matter, they tend to grow very slowly indeed.

Many black holes these days likely are mostly stable in mass. They can't evaporate in net (yet) because their Hawking temperatures are smaller than the CMB temperature, so that they're being fed by the ever-present CMB. Once the CMB temperature drops to below the Hawking temperatures of smaller black holes, they'll start to evaporate.

If a microscopic black hole were ever to form, however, its temperature would be so incredibly high that it would evaporate pretty much instantly. A 1kg black hole would have a temperature of ##10^{23}## Kelvin (that's nearly a trillion trillion degrees), for example, and would have a lifetime of ##10^{-17}## seconds. Microscopic black holes would generally be even smaller, with even higher temperatures. They would rapidly evaporate into high-energy particles.
 
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  • #16
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This particle will fall into the center of the BH and be annihilated by its impact with the positive energy mass which (somehow) lies there. This will decrease the nominal mass lying at the center. But the energy of annihilation which results has no way to escape the BH, so it is trapped forever within it.
The negative energy particles that fall inside the event horizon will totally annihilate positive energy particles inside, and there will be no energy left over from that annihilation. So the black hole will start evaporating from the process of releasing Hawking radiations.
 
  • #17
phinds
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Thanx for the feedback...seems very little of what one reads of ‘pop-sci’ ends up being easy to understand at the hard-science, bedrock level...
Well, a lot of stuff is not terribly hard but does require slightly more explanation that pop science articles have time for and yes, some stuff is complicated enough that pop science gives up on trying for real explanations. One should never think that pop-science teaches science --- it's entertainment, not education.

Thanx for the reply...seems like one cannot really understand Hawking Radiation unless a significant investment is made in its underlying mathematics
Yep
...the heuristic explanation is perhaps not the optimal...
Heuristics and analogies are never "optimal", but without them we are sometime left with no way to discuss some phenomenon at all at the pop science level. Hawking himself is the one who came up with the heuristic for Hawking Radiation since, as he states, it was the only way he could think of to describe in English what really can only be described with the math.
 
  • #18
PeterDonis
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The negative energy particles that fall inside the event horizon will totally annihilate positive energy particles inside
No, it can't, because there aren't any; the black hole is vacuum inside. The negative energy particles simply reduce the mass of the hole. This is one aspect in which the "virtual particle" heuristic description doesn't work well.
 
  • #19
Chronos
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Keep in mind, the expression 'inside' a black hole is used to describe the region enveloped by the event horizon of a black hole. A black hole does not have a surface in any sense that resembles that of a geometric solid. The EH serves as a virtual surface, for the sake of distinguishing regions external to the EH from the region enveloped by the EH. The unique aspect of the EH is it serves as the boundary between the external universe and the causally disconnected region e of space encases by the EH. This is strictly a one way barrier. Events external to the EH may enter but no event originating inside the EH may ever leave. This concept elicits all kinds of interesting questions - like 'how do gravitons escape a black hole?'. The obvious answer being 'gravitons do not originate from inside the EH'.
 
  • #20
Garth
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This is strictly a one way barrier. Events external to the EH may enter but no event originating inside the EH may ever leave. This concept elicits all kinds of interesting questions - like 'how do gravitons escape a black hole?'. The obvious answer being 'gravitons do not originate from inside the EH'.
So the next obvious question is 'where do they (gravitons) originate?'

Garth
 
  • #21
king vitamin
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'how do gravitons escape a black hole?'. The obvious answer being 'gravitons do not originate from inside the EH'.
An even better answer is that gravitons cannot escape a black hole in the first place. The gravitational force which points towards the interior of the black hole does not require any physical gravitons to mediate its field. (There is a description in terms of virtual gravitons, but we've already had some good explanations here about how virtual particles aren't physical.)

Recall also that it is possible for charged black holes to exist, but the existence of a Coulomb field doesn't mean that physical photons are escaping the event horizon.
 
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This video is very useful:
 
  • #23
Ibix
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This video is very useful:
I got to forty seconds (of which twenty two were logos and an intro) and had spotted two errors (point of infinite density, time is frozen) and one claim that needs a gigantic footnote (spacetime cascading inwards). I'm not hopeful that it will accurately describe Hawking radiation.
 
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  • #24
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I got to forty seconds (of which twenty two were logos and an intro) and had spotted two errors (point of infinite density, time is frozen) and one claim that needs a gigantic footnote (spacetime cascading inwards). I'm not hopeful that it will accurately describe Hawking radiation.
The video is not about Hawking radiation it is about gravity field and charge field of black holes. I posted it as an indirect response of the question how gravitons escape BH.
 
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The video is not about Hawking radiation
Still it's not a good source to learn from about anything...
 
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