<p>The black hole loses mass in order to generate the Hawking radiation. The energy carried by the radiation is exactly proportional to the mass lost by the black hole according to Einstein's formula E = mc^2</p>

<p>No new mass is brought into the Universe this way.</p>

<p><strong>EDIT:</strong></p>

<p>(Keep in mind the explanation below is not rigorous. As Stephen Hawking himself has said, this is more like a mental model, good enough to visualize things.)</p>

<p>Virtual particles are called virtual because they don't really "exist" in the common sense. They have mass and they have energy. Due to Heisenberg's uncertainty principle, it's okay if particles with a total energy E almost "exist" for a very short while, as long as they return to nothingness afterwards.</p>

<p>The production / destruction of virtual particles could happen all the time, anywhere, not just at the edge of a black hole. This is fine, as long as each particle only seems to "exist" a short time. Conservation of energy is not violated.</p>

<p>However, at the edge of a black hole, something different happens. You have a pair of virtual particles popping up into existence. Soon after, one of the particles is captured by the black hole, while the other particle moves away from it. Since the captured particle is virtual, it doesn't really have mass and/or energy. No new mass is added to the black hole.</p>

<p>Now the two particles cannot re-annihilate each other. So you have one lone particle moving away from the black hole. It's not a virtual particle anymore, it's a real particle. But it has energy and perhaps it has mass also. Where does the energy come from?</p>

<p>The energy required to make the new particle real comes from the gravitational field of the black hole. The black hole is losing mass in order to power the creation of real particles moving away from it.</p>

<p>Again, don't get stuck in details. This is not what really happens. This is more like a metaphor.</p>

CC BY-SA 3.0
2018-03-15T18:33:42.717Z
Sorry, I don't see how that answers my questions
zuba
CC BY-SA 3.0
2018-03-15T19:10:50.150Z
@zuba I made an edit with additional details.
Florin Andrei
CC BY-SA 3.0
2018-03-15T19:19:06.570Z
@Andrei I'm quite confused as "Virtual particles are called virtual because they don't really "exist" in the common sense. They have mass and they have energy" doesn't match to "Since the captured particle is virtual, it doesn't really have mass and/or energy"
zuba
CC BY-SA 3.0
2018-03-15T19:39:57.587Z
@zuba This is slowly becoming a quantum mechanics seminar :) and I've used a lot of non-rigorous similes in my explanation. I think if you keep reading about virtual particles it will become clear in the end.
Florin Andrei
CC BY-SA 3.0
2018-03-16T13:05:26.993Z
@zuba I would suggest this page by Matt Strassler (https://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/) as a partial explanation of why "virtual particles" -- better understood as complex disturbances in a quantum field -- aren't really like "real particles", which is why intuitions about how real particles behave can lead you astray.
Peter Erwin
CC BY-SA 3.0
2018-03-16T13:29:27.390Z
@Peter, thank you for the link. I see now, that "Perhaps unfortunately, this type of disturbance, whose details can vary widely, was given the name “virtual particle” for historical reasons, which makes it sound both more mysterious, and more particle-like, than is necessary."
zuba
CC BY-SA 3.0
2018-03-15T18:08:27.533Z
2018-03-15T19:10:34.563Z
2018-03-15T19:10:34.563Z
Florin Andrei
<p>As I understand virtual particles pairs, e.g. electron and positron spontaneously pop into existence for tiny period of time, then they annihilate.</p>

<p>When virtual electron and virtual positron annihilate we should see two photons emitted to opposite directions, right? If so, shouldn't we detect photons appearing from vacuum for every such pair?</p>

<p>Also, when a pair of virtual particles pops near event horizon of a black hole one particle may be captured by the black hole, while the other escapes it. Both particles have mass and energy. Thus caught particle brings the black hole its mass. Then how the black hole <em>loses</em> mass through Hawking radiation? Shouldn't it gain mass instead?</p>

CC BY-SA 3.0
2018-03-15T13:04:47.110Z
I'd refer you to a question on Physics SE [An Explanation of Hawking Radiation](https://physics.stackexchange.com/questions/251385/an-explanation-of-hawking-radiation) and similar questions there.
StephenG - Help Ukraine
CC BY-SA 3.0
2018-03-15T14:40:47.510Z
You seem to be missing the point that the black hole is emitting photons, thus (theoretically) has given up mass to generate those photons.
Carl Witthoft
CC BY-SA 3.0
2018-03-15T18:37:39.047Z
Carl, imho particles just appear near a black hole, not emitted by it
zuba
CC BY-SA 3.0
2018-03-16T13:09:49.533Z
"Photons appearing from the vacuum for every such pair" would violate conservation of energy.
Peter Erwin
CC BY-SA 3.0
2018-03-16T13:25:26.083Z
@Peter Ok, how two virtual particles "annihilate" without emitting photons? Maybe they do not "annihilate", but rather disappear?
zuba
<black-hole><hawking-radiation>
Does Hawking radiation in fact bring mass into the universe?
zuba
# Does Hawking radiation in fact bring mass into the universe?

(Asked by: username_0 on 2018-03-15)

As I understand virtual particles pairs, e.g. electron and positron spontaneously pop into existence for tiny period of time, then they annihilate.

When virtual electron and virtual positron annihilate we should see two photons emitted to opposite directions, right? If so, shouldn't we detect photons appearing from vacuum for every such pair?

Also, when a pair of virtual particles pops near event horizon of a black hole one particle may be captured by the black hole, while the other escapes it. Both particles have mass and energy. Thus caught particle brings the black hole its mass. Then how the black hole *loses* mass through Hawking radiation? Shouldn't it gain mass instead?

**Tags:** black-hole, hawking-radiation
> **Comments:**
> **username_1**: I'd refer you to a question on Physics SE [An Explanation of Hawking Radiation](https://physics.stackexchange.com/questions/251385/an-explanation-of-hawking-radiation) and similar questions there.  
> **username_2**: You seem to be missing the point that the black hole is emitting photons, thus (theoretically) has given up mass to generate those photons.  
> **username_0**: Carl, imho particles just appear near a black hole, not emitted by it  
> **username_3**: "Photons appearing from the vacuum for every such pair" would violate conservation of energy.  
> **username_0**: @Peter Ok, how two virtual particles "annihilate" without emitting photons? Maybe they do not "annihilate", but rather disappear?  

---

## 1 answers

### Answer by username_4 on 2018-03-15. Score: 2

The black hole loses mass in order to generate the Hawking radiation. The energy carried by the radiation is exactly proportional to the mass lost by the black hole according to Einstein's formula E = mc^2

No new mass is brought into the Universe this way.

**EDIT:**

(Keep in mind the explanation below is not rigorous. As Stephen Hawking himself has said, this is more like a mental model, good enough to visualize things.)

Virtual particles are called virtual because they don't really "exist" in the common sense. They have mass and they have energy. Due to Heisenberg's uncertainty principle, it's okay if particles with a total energy E almost "exist" for a very short while, as long as they return to nothingness afterwards.

The production / destruction of virtual particles could happen all the time, anywhere, not just at the edge of a black hole. This is fine, as long as each particle only seems to "exist" a short time. Conservation of energy is not violated.

However, at the edge of a black hole, something different happens. You have a pair of virtual particles popping up into existence. Soon after, one of the particles is captured by the black hole, while the other particle moves away from it. Since the captured particle is virtual, it doesn't really have mass and/or energy. No new mass is added to the black hole.

Now the two particles cannot re-annihilate each other. So you have one lone particle moving away from the black hole. It's not a virtual particle anymore, it's a real particle. But it has energy and perhaps it has mass also. Where does the energy come from?

The energy required to make the new particle real comes from the gravitational field of the black hole. The black hole is losing mass in order to power the creation of real particles moving away from it.

Again, don't get stuck in details. This is not what really happens. This is more like a metaphor.

> **Comments:**
> **username_0**: Sorry, I don't see how that answers my questions  
> **username_4**: @zuba I made an edit with additional details.  
> **username_0**: @Andrei I'm quite confused as "Virtual particles are called virtual because they don't really "exist" in the common sense. They have mass and they have energy" doesn't match to "Since the captured particle is virtual, it doesn't really have mass and/or energy"  
> **username_4**: @zuba This is slowly becoming a quantum mechanics seminar :) and I've used a lot of non-rigorous similes in my explanation. I think if you keep reading about virtual particles it will become clear in the end.  
> **username_3**: @zuba I would suggest this page by Matt Strassler (https://profmattstrassler.com/articles-and-posts/particle-physics-basics/virtual-particles-what-are-they/) as a partial explanation of why "virtual particles" -- better understood as complex disturbances in a quantum field -- aren't really like "real particles", which is why intuitions about how real particles behave can lead you astray.  
> **username_0**: @Peter, thank you for the link. I see now, that "Perhaps unfortunately, this type of disturbance, whose details can vary widely, was given the name “virtual particle” for historical reasons, which makes it sound both more mysterious, and more particle-like, than is necessary."  

---

