<p>I'm going to give you an intuitive answer. Keep in mind, this is not the "actual" answer, as the Hawking radiation is quite a bit more complex than the typical pop-sci explanation with virtual particles. But some intuitive justification is possible nevertheless.</p>

<blockquote>
  <p>I don't see how this event contributes to evaporation of the black
  hole (, since the particles do not originate from the black hole).</p>
</blockquote>

<p>You're missing a key point here.</p>

<p>When the pair was generated, those were virtual particles. After one side of the pair was absorbed by the black hole, and the other side was released, the released part is a real particle. Huge difference there - virtual vs real.</p>

<p>Virtual particles don't really exist the same way that you and me exist. They seem to exist for a very short time; the more energetic they are, the shorter the interval of their virtual "existence", per the Heisenberg equation. In many ways they are just a mathematical trick.</p>

<p>Think of the vacuum, where no real particles exist. Before, it's just vacuum. Right now, a virtual pair flickers briefly, then it's gone. In the future, it's vacuum again.</p>

<p>What was the energy before? Zero. What is the energy in the future? Zero. What's the energy during the flicker? Well, it basically zero, within the limits permitted by Heisenberg's equations. Bottom line is, virtual particles come and go, and they do not contribute to the energy balance of some empty chunk of space.</p>

<p>(I am ignoring here the concept of vacuum energy, for the sake of an intuitive explanation.)</p>

<p>But let's say one of the virtual particles gets trapped by the black hole, so it cannot annihilate with its counterpart. The other particle flies off in the opposite direction and escapes the black hole. What's worse, this is now a real particle - we've exceeded the duration permitted by the Heisenberg equations, so the one that escapes is not virtual anymore.</p>

<p>How did that particle become real?</p>

<p>This is a big issue, because virtual particles don't require an energy budget to briefly exist, while real particles do carry energy forever. Something prevented the virtual pair from annihilating itself, and boosted one of the components to the status of real particle. The virtual pair has zero energy. The real particle that gets away has non-zero energy. That energy has to come from somewhere.</p>

<p>It comes from the black hole. The black hole gives up some of its mass / energy (same thing) to boost one particle from virtual to real. The other particle is captured - but being virtual anyway, it doesn't really matter.</p>

<p>What this intuitive explanation doesn't say is how the boost actually happens. I dunno, magic. Somehow one of the virtual particles gets a chunk of energy from the black hole and becomes real.</p>

<p>Again, this is not the actual process. The actual process <a href="http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html" rel="noreferrer">is more complex</a>. This is just a pop-sci fairy tale.</p>

<hr>

<p>EDIT: To hit closer to home, Hawking radiation is more like a close relative to the <a href="https://en.wikipedia.org/wiki/Unruh_effect" rel="noreferrer">Unruh effect</a>. Say an inertial observer sees empty space here in this chunk of volume. An accelerating observer would not see empty space in the same volume, but instead would see blackbody radiation. That's the Unruh effect.</p>

<p>Well, gravity and acceleration are the same thing, per general relativity. So the strong gravity near a black hole is equivalent to strong acceleration. Something similar to the Unruh effect must happen there. That's the Hawking radiation.</p>

<p><a href="http://backreaction.blogspot.com/2015/12/hawking-radiation-is-not-produced-at.html" rel="noreferrer">http://backreaction.blogspot.com/2015/12/hawking-radiation-is-not-produced-at.html</a></p>

<p>EDIT2: The other answers currently on this page provide useful alternative points, so check them out too.</p>

CC BY-SA 4.0
2019-04-25T11:21:49.033Z
This "pop-sci fairytale", as you called it yourself, is quite a pleasant read, I even laughed. Thanks. But it is this "I dunno magic" I am after: how does the virtual particle get it's real state (besides magic) and how does this contribute to BH evaporation, having that nothing can escape the black hole..
marko-36
CC BY-SA 4.0
2019-04-25T12:49:20.730Z
@Marko Also see http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html
PM 2Ring
CC BY-SA 4.0
2019-04-26T12:11:11.757Z
Do you get particles and anti-particles in equal proportion?
Steven Gubkin
CC BY-SA 4.0
2019-04-26T16:52:32.240Z
I'm struggling with one part of the intuition - why is the particle which escapes converted to "real" while the one which is captured (and thus doesn't annihilate any more than the escaping one) remains "virtual"? I think the answer needs to address what is meant by real vs virtual. My intuition tells me that the escaping particle and the captured particle are just as real as each other, and thus the black hole may have lost mass/energy allowing one to escape, but it has gained the mass/energy of the captured one, resulting in an overall loss of zero.
JBentley
CC BY-SA 4.0
2019-04-26T17:53:21.423Z
@JBentley It's just a story anyway.
Florin Andrei
CC BY-SA 4.0
2019-04-26T18:16:42.867Z
@FlorinAndrei I got that. I just would like the missing piece for it to make sense :) ([obligatory xkcd](https://xkcd.com/895/))
JBentley
CC BY-SA 4.0
2019-07-16T04:35:36.653Z
@JBentley The pair retain the relationship that they combine to have (approximately) zero energy.  But there are no "real" particles in the universe that have non-positive energy.  Therefore any particle which escapes must have positive energy, and to preserve the relation the infalling one (which disappears from the universe and past the horizon) must have negative energy.  So the black hole loses from it's mass-energy in the exchange, every single time.
zibadawa timmy
CC BY-SA 4.0
2020-09-13T02:32:57.507Z
Indeed [the "Heisenberg/virtual-particle" explanation is false](https://www.physicsforums.com/insights/vacuum-fluctuation-myth/). [Uhruh radiation is a better analogy given the Bogoliubov transformations' use.](https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html). But a simplified, **accurate** explanation of why mass decreases is still missing from this site.
alexchandel
CC BY-SA 4.0
2020-11-24T04:19:08.697Z
@StevenGubkin Yes, the proportions should be equal or very nearly equal.  Most of the particles are photons, which are their own anti-particle or neutrinos, which may be their own anti particle, so it's not always relevant, but when relevant, it's equal amounts.    Also, I feel I should point, not to you, but in general that anti matter and virtual particles are entirely different things.  I've seen people get that mixed up.
userLTK
CC BY-SA 4.0
2019-04-25T10:36:22.583Z
2019-04-26T05:40:33.677Z
2019-04-26T05:40:33.677Z
Florin Andrei
<p><a href="http://teacher.pas.rochester.edu/Ast102/LectureNotes/Lecture19/Lecture19.pdf" rel="noreferrer">These lecture notes</a> address the issues to some degree, especially on slides 33-35.</p>

<blockquote>
  <p>Because in the strongly warped spacetime near the
   horizon, virtual particles made from vacuum fluctuations
   turn out to have negative energy density.</p>
  
  <p>Energy density = energy per unit volume.</p>
  
  <p>These particles indeed have positive mass -- look at the
   one that escaped! -- but their mass is distributed very
   strangely over spacetime. (Quantum-mechanically
   speaking, particles have nonzero volume; this is an aspect
   of the wave-particle duality.)</p>
  
  <p>Matter with negative energy density is generally called
  <em>exotic matter</em></p>
</blockquote>

<p>and, a bit later:</p>

<blockquote>
  <p>Quantum mechanical vacuum fluctuations in flat
   spacetime - far from any strong gravitational field -
   always have zero net energy density; they can never be
   exotic.</p>
  
  <p>However, in warped spacetime, vacuum fluctuations are
   in general exotic: their net energy density is negative,
   according to a distant observer measuring the energy
   density by observation of the deflection of light by the
   ensemble of fluctuations. The stronger the curvature, the
   more negative the energy density looks. </p>
</blockquote>

<p>This is the best explanation I have seen so far. </p>

CC BY-SA 4.0
2019-04-25T21:46:37.710Z
This is great !
Florin Andrei
CC BY-SA 4.0
2019-04-25T22:46:48.393Z
@jakub_d. Oops sorry. Would you like me to delete mine?
Steve Linton
CC BY-SA 4.0
2019-04-25T23:11:15.543Z
This is a complex phenomenon. I think the question would benefit from having a diversity of answers.
Florin Andrei
CC BY-SA 4.0
2019-04-26T10:38:30.093Z
Thinking about this. I came up with an alternative interesting way of thinking about it   1 -- just as an electron, in vacuum is surrounded by a cloud of virtual photons and electrons (https://www.newscientist.com/article/mg15320662-300-science-more-to-electrons-than-meets-the-eye/) a BH is shielded by a cloud of virttual particles which make its mass seem a little smaller to a distant observer. Occassioanlly some of this shield is consumed by the event horizon, making its mass actually smaller, and resulting in emission of a particle. Disclaimer, I am not an expert and this is just a story.
Steve Linton
CC BY-SA 4.0
2019-07-16T05:03:40.807Z
The exam satistics at the start of those lecture notes is almost more interesting to me.  It basically seems to say "three people didn't take the test, one person got a zero on both the test and the practice test, and scores on the exam were almost directly correlated to the binary 'did you take the practice exam?'."  That last bit is kind of weird.
zibadawa timmy
CC BY-SA 4.0
2019-04-25T21:04:22.083Z
2019-04-25T21:04:22.083Z
Steve Linton
<p>I don't know if the experts will agree with this description, but here is how I understand it:</p>

<p>Both space and the event horizon are in constant quantum fluctuation. Essentially, the event horizon has tiny ripples. At points where the event horizon ripples up (above the average radius of the black hole), it has an above average amount of local energy. The intense gravity rapidly pulls that local bump back down, the falling bump sends that local energy concentration back across the rest of the event horizon.</p>

<p>Now let's consider possible virtual particle-pairs near the hole. If a stationary virtual particle-pair appears just above the event horizon, it will either recombine and vanish or the entire thing gets pulled into the hole and vanishes in zero. We need a virtual particle-pair that has an apparent motion away from the black hole, at nearly the speed of light. If that that virtual particle-pair is going fast enough to completely escape, they recombine and vanish. Zero net effect. We need a virtual particle-pair that is moving away from the black hole at nearly the speed of light, and we need a ripple in horizon which only catches one virtual particle. I believe the ripple must be under extreme downwards acceleration for it to pull away from the second virtual particle, to avoid catching both. And here's the key part: The energy-debt between the particle pair intensely pulls them towards each other. The trapped particle is being pulled upwards, effectively pulling upwards on the horizon that trapped it. This slows the fall of horizon-ripple, diminishing the energy that the falling ripple returns to the rest of the black hole.</p>

<p>The energy required to pull the two virtual particles apart equals the combined energy of the two non-virtual particles. So the falling ripple loses energy equal to two particles, and the hole eats one particle. Everything balances out with the one escaped particle.</p>

<p>I believe it works the same, regardless of whether the virtual particles are photons or a matter-antimatter pair.</p>

CC BY-SA 4.0
2019-04-27T04:55:15.930Z
2019-04-27T17:15:56.997Z
2019-04-27T17:15:56.997Z
Alsee
<p>Here's an analogy to Quantum Mechanics. A particle in QM can tunnel through an impossible barrier, which is how elements heavier than lead can have some of their neutrons "tunnel" out of the nucleus escaping the bonds of the Strong Force.</p>

<p><a href="https://i.sstatic.net/ymt1c.png" rel="nofollow noreferrer"><img src="https://i.sstatic.net/ymt1c.png" alt="enter image description here"></a></p>

<p>A small Black Hole is like a quantum barrier that a particle can be tunnel through to escape. The smaller the barrier (Event Horizon), the more likely it is to be able to tunnel out. So a micro black hole with a mass of <strong>228 tons</strong> and a Event Horizon of <strong>3.4 x 10^-7 femtometers</strong> (literally less than 1 millionth the size of a proton) won't hold on to it's particles for very long and at all. In fact it will explode in a burst of Hawking Radiation after exactly <strong>1 second</strong>.</p>

<p>A larger <strong>Earth Mass</strong> black hole with a radius of a whole <strong>centimeter</strong>, will last much longer: <strong>8 x 10^50 years</strong> because it's much less likely for a particle to tunnel through a whole centimeter to get to freedom.</p>

<hr>

<p>Source: Quantum tunneling from three-dimensional black holes: <a href="https://arxiv.org/abs/1306.6380" rel="nofollow noreferrer">https://arxiv.org/abs/1306.6380</a></p>

<p>Source: Hawking Radiation modelled as a Quantum Effect: <a href="http://cscanada.net/index.php/ans/article/view/j.ans.1715787020120502.1817" rel="nofollow noreferrer">http://cscanada.net/index.php/ans/article/view/j.ans.1715787020120502.1817</a></p>

CC BY-SA 4.0
2019-07-15T21:23:06.213Z
2019-07-15T21:23:06.213Z
SurpriseDog
<p>From what I understand so far, when one of virtual particles crosses the event horizon and the other does not, they can not annihilate each other. The latter wanders off into the universe (btw, is it still virtual at this point, and what does 'virtual' mean at this point, if so?), while the other gets consumed by the black hole. I don't see how this event contributes to evaporation of the black hole (since the particles do not originate from the black hole). Shouldn't the consumed particle actually add-up to the black hole mass?</p>

<p>The closest question to mine is <a href="https://astronomy.stackexchange.com/questions/25501/does-hawking-radiation-in-fact-bring-mass-into-the-universe">Does Hawking radiation in fact bring mass into the universe?</a>, but I don't find the answers satisfactory.</p>

<p>I.e. "<a href="https://astronomy.stackexchange.com/questions/20496/is-there-a-better-explanation-of-hawking-radiation">the escaped virtual particle is 'boosted' by black hole's gravitational field into becoming a real particle</a>", rather adds to the question then answer it.</p>

<p>EDIT: I am humbled by the knowledge presented in the answers and feel incompetent to mark any as the most suitable. I hope this is okay.</p>

CC BY-SA 4.0
2019-04-25T09:32:15.713Z
Forget about virtual particle pairs, that's more like a metaphor. The thing is, black holes emit radiation, regardless of how they do that. Radiation carries energy, which must come from somewhere, there's no free lunch here. But energy equals mass. It all comes out of the black hole's "bank account" of mass, because that's the only thing nearby.
Florin Andrei
CC BY-SA 4.0
2019-04-25T09:35:49.457Z
Ok, thanks @Florin Andrei, but virtual particles are commonly accepted as a real thing, so how (much) are they metaphorical? Why?
Also, black hole radiation is explicitly explained by them. Should they be a metaphorical concept, what is this black hole radiation, you mentioned, really?
Also assuming that "nothing can escape the black hole, not even EM radiation".
marko-36
CC BY-SA 4.0
2019-04-25T10:04:26.753Z
@Marko36 It's not that virtual particles are a metaphor in general (although in a sense they are, all particles, virtual or otherwise are just a way of viewing some aspects of the underlying fields), but they are not really a very good explanation of Hawking Radiation. This, however, doesn't answer your question. I look forward to seeing an answer.
Steve Linton
CC BY-SA 4.0
2019-04-25T12:53:35.863Z
You might like to check out http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html and https://physics.stackexchange.com/questions/185110/do-virtual-particles-actually-physically-exist and other related questions on Physics about virtual particles.
PM 2Ring
CC BY-SA 4.0
2019-04-25T14:43:34.473Z
This question on physics (and its accepted answer come the closest I have yet found to addressing this question) but they still don't answer it completely. https://physics.stackexchange.com/questions/251385/an-explanation-of-hawking-radiation.
Steve Linton
CC BY-SA 4.0
2019-04-26T17:24:41.817Z
@Marko36 "virtual particles" are internal lines in Feynman diagrams and should not be reified more than that.  QFT never assigns a *state* to a "virtual particle". They are *entirely* metaphorical. See the answer by Arnold Neumaier (https://physics.stackexchange.com/a/252183/) and his links.
Robin Ekman
CC BY-SA 4.0
2021-05-09T08:38:52.707Z
Matt Strassler has an interesting explanation of virtual particles here: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/
PM 2Ring
<black-hole><hawking-radiation><particles>
How exactly does Hawking radiation decrease the mass of black holes?
marko-36
# How exactly does Hawking radiation decrease the mass of black holes?

(Asked by: username_0 on 2019-04-25)

From what I understand so far, when one of virtual particles crosses the event horizon and the other does not, they can not annihilate each other. The latter wanders off into the universe (btw, is it still virtual at this point, and what does 'virtual' mean at this point, if so?), while the other gets consumed by the black hole. I don't see how this event contributes to evaporation of the black hole (since the particles do not originate from the black hole). Shouldn't the consumed particle actually add-up to the black hole mass?

The closest question to mine is [Does Hawking radiation in fact bring mass into the universe?](https://astronomy.stackexchange.com/questions/25501/does-hawking-radiation-in-fact-bring-mass-into-the-universe), but I don't find the answers satisfactory.

I.e. "[the escaped virtual particle is 'boosted' by black hole's gravitational field into becoming a real particle](https://astronomy.stackexchange.com/questions/20496/is-there-a-better-explanation-of-hawking-radiation)", rather adds to the question then answer it.

EDIT: I am humbled by the knowledge presented in the answers and feel incompetent to mark any as the most suitable. I hope this is okay.

**Tags:** black-hole, hawking-radiation, particles

> **Comments:**
> **username_1**: Forget about virtual particle pairs, that's more like a metaphor. The thing is, black holes emit radiation, regardless of how they do that. Radiation carries energy, which must come from somewhere, there's no free lunch here. But energy equals mass. It all comes out of the black hole's "bank account" of mass, because that's the only thing nearby.  
> **username_0**: Ok, thanks @username_1, but virtual particles are commonly accepted as a real thing, so how (much) are they metaphorical? Why? Also, black hole radiation is explicitly explained by them. Should they be a metaphorical concept, what is this black hole radiation, you mentioned, really? Also assuming that "nothing can escape the black hole, not even EM radiation".  
> **username_2**: @username_0 It's not that virtual particles are a metaphor in general (although in a sense they are, all particles, virtual or otherwise are just a way of viewing some aspects of the underlying fields), but they are not really a very good explanation of Hawking Radiation. This, however, doesn't answer your question. I look forward to seeing an answer.  
> **username_3**: You might like to check out http://math.ucr.edu/home/baez/physics/Quantum/virtual_particles.html and https://physics.stackexchange.com/questions/185110/do-virtual-particles-actually-physically-exist and other related questions on Physics about virtual particles.  
> **username_2**: This question on physics (and its accepted answer come the closest I have yet found to addressing this question) but they still don't answer it completely. https://physics.stackexchange.com/questions/251385/an-explanation-of-hawking-radiation.  
> **username_4**: @username_0 "virtual particles" are internal lines in Feynman diagrams and should not be reified more than that.  QFT never assigns a *state* to a "virtual particle". They are *entirely* metaphorical. See the answer by Arnold Neumaier (https://physics.stackexchange.com/a/252183/) and his links.  
> **username_3**: Matt Strassler has an interesting explanation of virtual particles here: https://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/  

---

## 4 answers

### Answer by username_1 on 2019-04-25. Score: 31

I'm going to give you an intuitive answer. Keep in mind, this is not the "actual" answer, as the Hawking radiation is quite a bit more complex than the typical pop-sci explanation with virtual particles. But some intuitive justification is possible nevertheless.

> I don't see how this event contributes to evaporation of the black
> hole (, since the particles do not originate from the black hole).

You're missing a key point here.

When the pair was generated, those were virtual particles. After one side of the pair was absorbed by the black hole, and the other side was released, the released part is a real particle. Huge difference there - virtual vs real.

Virtual particles don't really exist the same way that you and me exist. They seem to exist for a very short time; the more energetic they are, the shorter the interval of their virtual "existence", per the Heisenberg equation. In many ways they are just a mathematical trick.

Think of the vacuum, where no real particles exist. Before, it's just vacuum. Right now, a virtual pair flickers briefly, then it's gone. In the future, it's vacuum again.

What was the energy before? Zero. What is the energy in the future? Zero. What's the energy during the flicker? Well, it basically zero, within the limits permitted by Heisenberg's equations. Bottom line is, virtual particles come and go, and they do not contribute to the energy balance of some empty chunk of space.

(I am ignoring here the concept of vacuum energy, for the sake of an intuitive explanation.)

But let's say one of the virtual particles gets trapped by the black hole, so it cannot annihilate with its counterpart. The other particle flies off in the opposite direction and escapes the black hole. What's worse, this is now a real particle - we've exceeded the duration permitted by the Heisenberg equations, so the one that escapes is not virtual anymore.

How did that particle become real?

This is a big issue, because virtual particles don't require an energy budget to briefly exist, while real particles do carry energy forever. Something prevented the virtual pair from annihilating itself, and boosted one of the components to the status of real particle. The virtual pair has zero energy. The real particle that gets away has non-zero energy. That energy has to come from somewhere.

It comes from the black hole. The black hole gives up some of its mass / energy (same thing) to boost one particle from virtual to real. The other particle is captured - but being virtual anyway, it doesn't really matter.

What this intuitive explanation doesn't say is how the boost actually happens. I dunno, magic. Somehow one of the virtual particles gets a chunk of energy from the black hole and becomes real.

Again, this is not the actual process. The actual process [is more complex](http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html). This is just a pop-sci fairy tale.

---

EDIT: To hit closer to home, Hawking radiation is more like a close relative to the [Unruh effect](https://en.wikipedia.org/wiki/Unruh_effect). Say an inertial observer sees empty space here in this chunk of volume. An accelerating observer would not see empty space in the same volume, but instead would see blackbody radiation. That's the Unruh effect.

Well, gravity and acceleration are the same thing, per general relativity. So the strong gravity near a black hole is equivalent to strong acceleration. Something similar to the Unruh effect must happen there. That's the Hawking radiation.

<http://backreaction.blogspot.com/2015/12/hawking-radiation-is-not-produced-at.html>

EDIT2: The other answers currently on this page provide useful alternative points, so check them out too.

> **Comments:**
> **username_0**: This "pop-sci fairytale", as you called it yourself, is quite a pleasant read, I even laughed. Thanks. But it is this "I dunno magic" I am after: how does the virtual particle get it's real state (besides magic) and how does this contribute to BH evaporation, having that nothing can escape the black hole..  
> **username_3**: @Marko Also see http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html  
> **username_5**: Do you get particles and anti-particles in equal proportion?  
> **username_6**: I'm struggling with one part of the intuition - why is the particle which escapes converted to "real" while the one which is captured (and thus doesn't annihilate any more than the escaping one) remains "virtual"? I think the answer needs to address what is meant by real vs virtual. My intuition tells me that the escaping particle and the captured particle are just as real as each other, and thus the black hole may have lost mass/energy allowing one to escape, but it has gained the mass/energy of the captured one, resulting in an overall loss of zero.  
> **username_1**: @username_6 It's just a story anyway.  
> **username_6**: @username_1 I got that. I just would like the missing piece for it to make sense :) ([obligatory xkcd](https://xkcd.com/895/))  
> **username_7**: @username_6 The pair retain the relationship that they combine to have (approximately) zero energy.  But there are no "real" particles in the universe that have non-positive energy.  Therefore any particle which escapes must have positive energy, and to preserve the relation the infalling one (which disappears from the universe and past the horizon) must have negative energy.  So the black hole loses from it's mass-energy in the exchange, every single time.  
> **username_8**: Indeed [the "Heisenberg/virtual-particle" explanation is false](https://www.physicsforums.com/insights/vacuum-fluctuation-myth/). [Uhruh radiation is a better analogy given the Bogoliubov transformations' use.](https://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html). But a simplified, **accurate** explanation of why mass decreases is still missing from this site.  
> **username_9**: @username_5 Yes, the proportions should be equal or very nearly equal.  Most of the particles are photons, which are their own anti-particle or neutrinos, which may be their own anti particle, so it's not always relevant, but when relevant, it's equal amounts.    Also, I feel I should point, not to you, but in general that anti matter and virtual particles are entirely different things.  I've seen people get that mixed up.  

---

### Answer by username_2 on 2019-04-25. Score: 9

[These lecture notes](http://teacher.pas.rochester.edu/Ast102/LectureNotes/Lecture19/Lecture19.pdf) address the issues to some degree, especially on slides 33-35.

> Because in the strongly warped spacetime near the
> horizon, virtual particles made from vacuum fluctuations
> turn out to have negative energy density.
>
> Energy density = energy per unit volume.
>
> These particles indeed have positive mass -- look at the
> one that escaped! -- but their mass is distributed very
> strangely over spacetime. (Quantum-mechanically
> speaking, particles have nonzero volume; this is an aspect
> of the wave-particle duality.)
>
> Matter with negative energy density is generally called
> *exotic matter*

and, a bit later:

> Quantum mechanical vacuum fluctuations in flat
> spacetime - far from any strong gravitational field -
> always have zero net energy density; they can never be
> exotic.
>
> However, in warped spacetime, vacuum fluctuations are
> in general exotic: their net energy density is negative,
> according to a distant observer measuring the energy
> density by observation of the deflection of light by the
> ensemble of fluctuations. The stronger the curvature, the
> more negative the energy density looks.

This is the best explanation I have seen so far.

> **Comments:**
> **username_1**: This is great !  
> **username_2**: @jakub_d. Oops sorry. Would you like me to delete mine?  
> **username_1**: This is a complex phenomenon. I think the question would benefit from having a diversity of answers.  
> **username_2**: Thinking about this. I came up with an alternative interesting way of thinking about it   1 -- just as an electron, in vacuum is surrounded by a cloud of virtual photons and electrons (https://www.newscientist.com/article/mg15320662-300-science-more-to-electrons-than-meets-the-eye/) a BH is shielded by a cloud of virttual particles which make its mass seem a little smaller to a distant observer. Occassioanlly some of this shield is consumed by the event horizon, making its mass actually smaller, and resulting in emission of a particle. Disclaimer, I am not an expert and this is just a story.  
> **username_7**: The exam satistics at the start of those lecture notes is almost more interesting to me.  It basically seems to say "three people didn't take the test, one person got a zero on both the test and the practice test, and scores on the exam were almost directly correlated to the binary 'did you take the practice exam?'."  That last bit is kind of weird.  

---

### Answer by username_10 on 2019-04-27. Score: 2

I don't know if the experts will agree with this description, but here is how I understand it:

Both space and the event horizon are in constant quantum fluctuation. Essentially, the event horizon has tiny ripples. At points where the event horizon ripples up (above the average radius of the black hole), it has an above average amount of local energy. The intense gravity rapidly pulls that local bump back down, the falling bump sends that local energy concentration back across the rest of the event horizon.

Now let's consider possible virtual particle-pairs near the hole. If a stationary virtual particle-pair appears just above the event horizon, it will either recombine and vanish or the entire thing gets pulled into the hole and vanishes in zero. We need a virtual particle-pair that has an apparent motion away from the black hole, at nearly the speed of light. If that that virtual particle-pair is going fast enough to completely escape, they recombine and vanish. Zero net effect. We need a virtual particle-pair that is moving away from the black hole at nearly the speed of light, and we need a ripple in horizon which only catches one virtual particle. I believe the ripple must be under extreme downwards acceleration for it to pull away from the second virtual particle, to avoid catching both. And here's the key part: The energy-debt between the particle pair intensely pulls them towards each other. The trapped particle is being pulled upwards, effectively pulling upwards on the horizon that trapped it. This slows the fall of horizon-ripple, diminishing the energy that the falling ripple returns to the rest of the black hole.

The energy required to pull the two virtual particles apart equals the combined energy of the two non-virtual particles. So the falling ripple loses energy equal to two particles, and the hole eats one particle. Everything balances out with the one escaped particle.

I believe it works the same, regardless of whether the virtual particles are photons or a matter-antimatter pair.


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### Answer by username_11 on 2019-07-15. Score: 1

Here's an analogy to Quantum Mechanics. A particle in QM can tunnel through an impossible barrier, which is how elements heavier than lead can have some of their neutrons "tunnel" out of the nucleus escaping the bonds of the Strong Force.

[![enter image description here](https://i.sstatic.net/ymt1c.png)](https://i.sstatic.net/ymt1c.png)

A small Black Hole is like a quantum barrier that a particle can be tunnel through to escape. The smaller the barrier (Event Horizon), the more likely it is to be able to tunnel out. So a micro black hole with a mass of **228 tons** and a Event Horizon of **3.4 x 10^-7 femtometers** (literally less than 1 millionth the size of a proton) won't hold on to it's particles for very long and at all. In fact it will explode in a burst of Hawking Radiation after exactly **1 second**.

A larger **Earth Mass** black hole with a radius of a whole **centimeter**, will last much longer: **8 x 10^50 years** because it's much less likely for a particle to tunnel through a whole centimeter to get to freedom.

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Source: Quantum tunneling from three-dimensional black holes: <https://arxiv.org/abs/1306.6380>

Source: Hawking Radiation modelled as a Quantum Effect: <http://cscanada.net/index.php/ans/article/view/j.ans.1715787020120502.1817>


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