<p>There is quite a nice explanation on this <a href="http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html" rel="nofollow noreferrer">web page</a>. A key passage is this:</p>

<blockquote>
  <p>in curved spacetime there aren't these "best" co-ordinate systems, the inertial ones.  So even very reasonable different choices of co-ordinates can give disagreements about particles vs antiparticles, or what's the vacuum.  These disagreements don't mean that "everything is relative", because there are nice formulas for how to translate between the descriptions in different co-ordinate systems.  These are Bogoliubov transformations.</p>
</blockquote>

<p>In particular, he goes on to say</p>

<blockquote>
  <p>on the one hand we can split solutions of Maxwell's equations into positive frequency in the most blitheringly obvious way that someone far from the black hole and far in the future would do it...</p>
  
  <p>and on the other hand we can split solutions of Maxwell's equations into positive frequency in the most blitheringly obvious way that someone far in the past, before the collapse into a black hole has happened would do it.</p>
</blockquote>

<p>Thus what the observer in the far past thought was genuinely empty space with no (non virtual) particles or antiparticles, an observer in the far future might see as space with perfectly good particles (and antiparticles) in it. Those particles are Hawking radiation.</p>

CC BY-SA 4.0
2019-12-08T01:46:19.063Z
2019-12-08T01:46:19.063Z
Steve Linton
<p>Andy Gould proposed a classical derivation of Hawking radiation in a somewhat <a href="http://inspirehep.net/record/231344/files/slac-pub-4061.pdf" rel="noreferrer">obscure paper from 1987</a>.  The essential argument is that a black hole must have a finite, non-zero entropy (otherwise you could violate the second law of thermodynamics with a black hole).  Moreover, the entropy of the black hole must depend only on its area (otherwise you could change the area of a black hole via the Penrose process and lower its entropy and make a perpetual motion machine).  If a black hole has an entropy and a mass, then it has a temperature.  If it has a temperature, then it must radiate thermally (otherwise you could again violate the second law of thermodynamics).</p>

<p>Of course, if you look at the Hawking radiation temperature, there's a Planck's constant in there, so it has to know something about quantum mechanics, right?  But it turns out that it's actually thermodynamics in general that knows about quantum mechanics, not general relativity --- Planck's constant is only needed to keep entropies finite (and therefore temperatures non-zero).  This is true of black holes and blackbodies alike.</p>

CC BY-SA 3.0
2017-03-23T13:12:00.513Z
It was an interesting read, but I noted this on page 5:  _”One may now consider doing an experiment first proposed by Geroch$^{[8]}$. One adiabatically lowers a perfectly reflecting box filled with electromagnetic radiation at a temperature T >> T$_{BH}$ to a Schwarzschild radius r, close to the event horizon. One then exchanges radiation with the hole…”_ Surely there’s no exchange because of the infinite gravitational time dilation? Geroch’s gedankenexperiment from the 1971 Princeton colloquium seems to be widely referenced but unpublished. An interesting lead, thanks again.
John Duffield
CC BY-SA 3.0
2017-03-23T17:02:33.387Z
You don't lower the box exactly down to the event horizon, only close to the event horizon.  So there is time dilation, but it is not infinite and radiation can be exchanged.
J. O'Brien Antognini
CC BY-SA 3.0
2017-03-23T20:19:03.127Z
I'm missing something here. If you lower the gedanken box to some place near the event horizon then _exchange radiation with the hole_  then when you pull the box up there's no radiation in it. Assuming the black hole swallowed the radiation (or at least some of it) the black hole mass increases. I'll see if I can find another explanation of Geroch's scenario.
John Duffield
CC BY-SA 3.0
2017-03-23T20:55:13.113Z
I found [this](http://faculty.poly.edu/~jbain/heat/Lectures/17.BlackHoleTD.Part2.pdf), see page 2, but it's wrong. When you lower the box and do work, at the event horizon the box has half the energy it started with. And ouch, I found this too: https://arxiv.org/abs/physics/0501056.
John Duffield
CC BY-SA 3.0
2017-03-23T23:06:07.850Z
I wouldn't trust the Arxiv paper you linked --- it's about 12 years old but was never published in a peer reviewed journal and has no citations.  It looks kooky to me.  And in the first (more trustworthy) reference, I'm not sure where you're getting that the box has half the energy that it started with.
J. O'Brien Antognini
CC BY-SA 3.0
2017-03-25T18:02:59.127Z
Joe: published or not, the paper makes some interesting points. Re the box, draw a square box then rotate it 45 degrees and arrange one electromagnetic wave such that it goes round a square path. The horizontals bend downward, but the verticals don't. You cannot extract that component of the wave energy.
John Duffield
CC BY-SA 3.0
2017-03-22T19:19:05.243Z
2017-03-22T19:19:05.243Z
J. O'Brien Antognini
<p>I'm writing a piece on Hawking radiation, and find I have something of a problem. The "given" explanation which I find on <a href="https://en.wikipedia.org/wiki/Hawking_radiation#Overview" rel="noreferrer">Wikipedia</a> and elsewhere is unsatisfactory: </p>

<blockquote>
  <p>"Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles<sup>[10]</sup>. As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole<sup>[11]</sup>. An alternative view of the process is that 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..."</p>
</blockquote>

<p>It relies on virtual particles and an negative-energy particles. However vacuum fluctuations are not the same thing as virtual particles, which <a href="https://physics.stackexchange.com/a/185139/76162">only exist in the mathematics of the model</a>, and we know of know negative-energy particles. So I'm looking for a better explanation. The Wikpedia article also says this:</p>

<blockquote>
  <p>"In another model, the process is a quantum tunnelling effect, whereby particle–antiparticle pairs will form from the vacuum, and one will tunnel outside the event horizon<sup>[10]</sup>."</p>
</blockquote>

<p>However that suggests pair production is occurring inside the event horizon, which seems to disregard the infinite gravitational time dilation, and that one of them a) appears outside of the event horizon and b) escapes as Hawking radiation when pair production typically involves the creation of an electron and a positron. Again it's unsatisfactory. So:</p>

<p><strong>Is there a better explanation of Hawking radiation?</strong></p>

CC BY-SA 3.0
2017-03-22T10:03:41.137Z
The particle falling in doesn't require negative energy.  All that matters is that some photons escape to infinity, which means that some of the energy that was "borrowed" from the gravitational field is lost (in the form of those photons).  So the gravitational field weakens, which reduces the apparent mass/energy.  But "apparent" is just what we see as distant observers.  What happens inside the event horizon is...in the range of conjectural to nothing.  That said, I don't think there's a majority opinion on how the radiation arises, or if it even exists...
zibadawa timmy
CC BY-SA 3.0
2017-03-22T10:54:52.903Z
You might find more on Physics SE given the fairly esoteric nature of this material.
StephenG - Help Ukraine
CC BY-SA 3.0
2017-03-22T12:36:42.707Z
Noted Stephen.  @zibadawa timmy : but how does you "borrow" energy from a gravitational field? And if you do, how does energy then leak out of the event horizon for more of the same until you end up with no black hole at all?
John Duffield
CC BY-SA 3.0
2017-03-22T12:57:45.840Z
John, from your questions it sounds like you don't understand the concepts of potential energy or energy stored in fields (gravitational, electric, etc).  I'd start by reading about those concepts.
Carl Witthoft
CC BY-SA 3.0
2017-03-22T16:03:05.833Z
@JohnDuffield Try standing on a chair and jumping off. You've just "borrowed" energy from the Earth's gravitational field. How else did your speed downwards increase if not by taking energy from somewhere?
zephyr
CC BY-SA 3.0
2017-03-22T17:50:31.427Z
@Carl Witthoft : I understand such concepts. When I lift a brick I do work on it. I add energy to it, increasing its mass. When I then drop the brick gravity converts potential energy in the brick into kinetic energy. When the brick hits the ground the kinetic energy is dissipated, and you're left with a [mass deficit](https://en.wikipedia.org/wiki/Binding_energy#Mass-energy_relation). The potential energy was not stored in the gravitational field, it was stored in the brick. If you lift a brick so much that you give it escape velocity, it takes all the potential energy away with it.
John Duffield
CC BY-SA 3.0
2017-03-22T18:19:31.773Z
I think this is probably better for Physics.  It's a little too quantum theory for Astronomy.    But I like your though experiment approach and I like your question.   Small point to add, the time dilation might only be infinite at the exact horizon, not inside.  That might allow for some tunneling relating to your 2nd quote.  But I'm not an expert.
userLTK
CC BY-SA 3.0
2017-03-22T18:35:51.263Z
1. All these verbal explanations are just metaphors. The real deal is doing the Hawking calculations - that's the real explanation. 2. Here's another metaphor: The black hole is nothing but a tremendous spacetime curvature tied into itself - and the name we have for spacetime curvature is "gravity". The black hole is nothing but gravity, intense enough to persist itself. The p / anti-p pairs are brought into existence the same way that any extremely intense field can generate particles: when you have plenty of energy, particles can pop out of it. E.g. electromagnetic radiation could do it too.
Florin Andrei
CC BY-SA 3.0
2017-03-23T08:28:30.627Z
@Florin Andrei : the real deal is understanding. And I'm afraid spacetime curvature is not "gravity". Spacetime curvature is "the tidal force". And _particles pop out of it_ is not an explanation at all. Because if vacuum energy near the black hole is somehow converted into say electrons and positrons, the chances are they'd both fall into the black hole. The black hole would grow, not evaporate.
John Duffield
CC BY-SA 3.0
2017-03-23T18:34:45.553Z
Due to conservation of momentum, the particles in a pair must move in opposite directions. If it happens that one is pointing away from EH and has enough kinetic energy, it will escape. Also, the BH cannot "grow" out of vacuum fluctuations - it would violate conservation of energy.
Florin Andrei
CC BY-SA 3.0
2017-03-23T23:54:24.393Z
@JohnDuffield No, because the energy would just return to the field that it came from, so no net change.  And massive particles are extremely unlikely to spawn with sufficient momentum to escape the black hole.  It's the photons they produce that have a real chance to actually escape to infinity.  The energy of those photons came, ultimately, from the gravitational field.  Those that don't escape just return their energy.  Either way, there's no net change in the total energy, just some of it has a chance to become (theoretically) observable photons, reducing the apparent gravitational field.
zibadawa timmy
CC BY-SA 3.0
2017-03-24T14:07:41.433Z
Try 'A Brief History Of Time'.  Seriously, what better source than a general public-level book by the man himself.
Jonathan Twite
<black-hole><gravity><general-relativity><hawking-radiation><quantum-mechanics>
Is there a better explanation of Hawking radiation?
John Duffield
# Is there a better explanation of Hawking radiation?

(Asked by: username_0 on 2017-03-22)

I'm writing a piece on Hawking radiation, and find I have something of a problem. The "given" explanation which I find on [Wikipedia](https://en.wikipedia.org/wiki/Hawking_radiation#Overview) and elsewhere is unsatisfactory:

> "Physical insight into the process may be gained by imagining that particle–antiparticle radiation is emitted from just beyond the event horizon. This radiation does not come directly from the black hole itself, but rather is a result of virtual particles being "boosted" by the black hole's gravitation into becoming real particles[10]. As the particle–antiparticle pair was produced by the black hole's gravitational energy, the escape of one of the particles lowers the mass of the black hole[11]. An alternative view of the process is that 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..."

It relies on virtual particles and an negative-energy particles. However vacuum fluctuations are not the same thing as virtual particles, which [only exist in the mathematics of the model](https://physics.stackexchange.com/a/185139/76162), and we know of know negative-energy particles. So I'm looking for a better explanation. The Wikpedia article also says this:

> "In another model, the process is a quantum tunnelling effect, whereby particle–antiparticle pairs will form from the vacuum, and one will tunnel outside the event horizon[10]."

However that suggests pair production is occurring inside the event horizon, which seems to disregard the infinite gravitational time dilation, and that one of them a) appears outside of the event horizon and b) escapes as Hawking radiation when pair production typically involves the creation of an electron and a positron. Again it's unsatisfactory. So:

**Is there a better explanation of Hawking radiation?**

**Tags:** black-hole, gravity, general-relativity, hawking-radiation, quantum-mechanics
> **Comments:**
> **username_1**: The particle falling in doesn't require negative energy.  All that matters is that some photons escape to infinity, which means that some of the energy that was "borrowed" from the gravitational field is lost (in the form of those photons).  So the gravitational field weakens, which reduces the apparent mass/energy.  But "apparent" is just what we see as distant observers.  What happens inside the event horizon is...in the range of conjectural to nothing.  That said, I don't think there's a majority opinion on how the radiation arises, or if it even exists...  
> **username_2**: You might find more on Physics SE given the fairly esoteric nature of this material.  
> **username_0**: Noted Stephen.  @zibadawa timmy : but how does you "borrow" energy from a gravitational field? And if you do, how does energy then leak out of the event horizon for more of the same until you end up with no black hole at all?  
> **username_3**: John, from your questions it sounds like you don't understand the concepts of potential energy or energy stored in fields (gravitational, electric, etc).  I'd start by reading about those concepts.  
> **username_4**: @JohnDuffield Try standing on a chair and jumping off. You've just "borrowed" energy from the Earth's gravitational field. How else did your speed downwards increase if not by taking energy from somewhere?  
> **username_0**: @Carl Witthoft : I understand such concepts. When I lift a brick I do work on it. I add energy to it, increasing its mass. When I then drop the brick gravity converts potential energy in the brick into kinetic energy. When the brick hits the ground the kinetic energy is dissipated, and you're left with a [mass deficit](https://en.wikipedia.org/wiki/Binding_energy#Mass-energy_relation). The potential energy was not stored in the gravitational field, it was stored in the brick. If you lift a brick so much that you give it escape velocity, it takes all the potential energy away with it.  
> **username_5**: I think this is probably better for Physics.  It's a little too quantum theory for Astronomy.    But I like your though experiment approach and I like your question.   Small point to add, the time dilation might only be infinite at the exact horizon, not inside.  That might allow for some tunneling relating to your 2nd quote.  But I'm not an expert.  
> **username_6**: 1. All these verbal explanations are just metaphors. The real deal is doing the Hawking calculations - that's the real explanation. 2. Here's another metaphor: The black hole is nothing but a tremendous spacetime curvature tied into itself - and the name we have for spacetime curvature is "gravity". The black hole is nothing but gravity, intense enough to persist itself. The p / anti-p pairs are brought into existence the same way that any extremely intense field can generate particles: when you have plenty of energy, particles can pop out of it. E.g. electromagnetic radiation could do it too.  
> **username_0**: @Florin Andrei : the real deal is understanding. And I'm afraid spacetime curvature is not "gravity". Spacetime curvature is "the tidal force". And _particles pop out of it_ is not an explanation at all. Because if vacuum energy near the black hole is somehow converted into say electrons and positrons, the chances are they'd both fall into the black hole. The black hole would grow, not evaporate.  
> **username_6**: Due to conservation of momentum, the particles in a pair must move in opposite directions. If it happens that one is pointing away from EH and has enough kinetic energy, it will escape. Also, the BH cannot "grow" out of vacuum fluctuations - it would violate conservation of energy.  
> **username_1**: @JohnDuffield No, because the energy would just return to the field that it came from, so no net change.  And massive particles are extremely unlikely to spawn with sufficient momentum to escape the black hole.  It's the photons they produce that have a real chance to actually escape to infinity.  The energy of those photons came, ultimately, from the gravitational field.  Those that don't escape just return their energy.  Either way, there's no net change in the total energy, just some of it has a chance to become (theoretically) observable photons, reducing the apparent gravitational field.  
> **username_7**: Try 'A Brief History Of Time'.  Seriously, what better source than a general public-level book by the man himself.  

---

## 2 answers

### Answer by username_9 on 2017-03-22. Score: 8 | Accepted answer

Andy Gould proposed a classical derivation of Hawking radiation in a somewhat [obscure paper from 1987](http://inspirehep.net/record/231344/files/slac-pub-4061.pdf). The essential argument is that a black hole must have a finite, non-zero entropy (otherwise you could violate the second law of thermodynamics with a black hole). Moreover, the entropy of the black hole must depend only on its area (otherwise you could change the area of a black hole via the Penrose process and lower its entropy and make a perpetual motion machine). If a black hole has an entropy and a mass, then it has a temperature. If it has a temperature, then it must radiate thermally (otherwise you could again violate the second law of thermodynamics).

Of course, if you look at the Hawking radiation temperature, there's a Planck's constant in there, so it has to know something about quantum mechanics, right? But it turns out that it's actually thermodynamics in general that knows about quantum mechanics, not general relativity --- Planck's constant is only needed to keep entropies finite (and therefore temperatures non-zero). This is true of black holes and blackbodies alike.

> **Comments:**
> **username_0**: It was an interesting read, but I noted this on page 5:  _”One may now consider doing an experiment first proposed by Geroch$^{[8]}$. One adiabatically lowers a perfectly reflecting box filled with electromagnetic radiation at a temperature T >> T$_{BH}$ to a Schwarzschild radius r, close to the event horizon. One then exchanges radiation with the hole…”_ Surely there’s no exchange because of the infinite gravitational time dilation? Geroch’s gedankenexperiment from the 1971 Princeton colloquium seems to be widely referenced but unpublished. An interesting lead, thanks again.  
> **username_9**: You don't lower the box exactly down to the event horizon, only close to the event horizon.  So there is time dilation, but it is not infinite and radiation can be exchanged.  
> **username_0**: I'm missing something here. If you lower the gedanken box to some place near the event horizon then _exchange radiation with the hole_  then when you pull the box up there's no radiation in it. Assuming the black hole swallowed the radiation (or at least some of it) the black hole mass increases. I'll see if I can find another explanation of Geroch's scenario.  
> **username_0**: I found [this](http://faculty.poly.edu/~jbain/heat/Lectures/17.BlackHoleTD.Part2.pdf), see page 2, but it's wrong. When you lower the box and do work, at the event horizon the box has half the energy it started with. And ouch, I found this too: https://arxiv.org/abs/physics/0501056.  
> **username_9**: I wouldn't trust the Arxiv paper you linked --- it's about 12 years old but was never published in a peer reviewed journal and has no citations.  It looks kooky to me.  And in the first (more trustworthy) reference, I'm not sure where you're getting that the box has half the energy that it started with.  
> **username_0**: Joe: published or not, the paper makes some interesting points. Re the box, draw a square box then rotate it 45 degrees and arrange one electromagnetic wave such that it goes round a square path. The horizontals bend downward, but the verticals don't. You cannot extract that component of the wave energy.  

---

### Answer by username_8 on 2019-12-08. Score: 2

There is quite a nice explanation on this [web page](http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html). A key passage is this:

> in curved spacetime there aren't these "best" co-ordinate systems, the inertial ones. So even very reasonable different choices of co-ordinates can give disagreements about particles vs antiparticles, or what's the vacuum. These disagreements don't mean that "everything is relative", because there are nice formulas for how to translate between the descriptions in different co-ordinate systems. These are Bogoliubov transformations.

In particular, he goes on to say

> on the one hand we can split solutions of Maxwell's equations into positive frequency in the most blitheringly obvious way that someone far from the black hole and far in the future would do it...
>
> and on the other hand we can split solutions of Maxwell's equations into positive frequency in the most blitheringly obvious way that someone far in the past, before the collapse into a black hole has happened would do it.

Thus what the observer in the far past thought was genuinely empty space with no (non virtual) particles or antiparticles, an observer in the far future might see as space with perfectly good particles (and antiparticles) in it. Those particles are Hawking radiation.


---

