<p>I cannot show this or prove it but the main idea, in simplest terms, is that a matter-antimatter pair, near the event horizon, will form. A member of this pair will decay back into the event horizon and be absorbed by the black hole. The other particle will decay out into the free space.</p>

CC BY-SA 4.0
2019-12-16T12:38:56.633Z
This is inaccurate - especially “The other particle will decay out into the event horizon and then on into space.”   If it “decays” into the event horizon it disappears into the BH. You also fail to explain why a virtual particle falling into the BH makes it lose mass instead of gaining mass. Nor why the effect isn’t cancelled out by the virtual pairs forming just *inside* the EH. Sometimes simple answers can be **too** simple :-)
Chappo Hasn't Forgotten
CC BY-SA 4.0
2019-12-16T23:38:23.470Z
@ChappoSaysReinstateMonica. I changed my answer somewhat.  If its still wrong I will delete it.  But if you know its wrong I encourage you to answer the question yourself.
Natsfan
CC BY-SA 4.0
2019-12-12T16:35:45.993Z
2019-12-16T23:39:23.513Z
2019-12-16T23:39:23.513Z
Natsfan
<p>I saw that Black Holes can evaporate because of Hawking radiation.
Because pair of particle, antiparticle can spawn at any point of the space, the antiparticle could be absorbed by the black hole and the particle go outside of it, like if the black hole where &quot;lightning&quot;.</p>
<p>But, if one element of the pair can be absorbed by the black hole, isn't there an equal probability of that the particle will be absorbed, too?</p>
<p>I could imaginate that the probability of fluctuation of going into 0 with an equal probability of +1 or -1 mass over time is nonnull, also that's why black hole could evaporate, or something like it?</p>

CC BY-SA 4.0
2019-12-11T14:51:19.527Z
Ok i see it's because anti-particle have positive mass
Raphaël Frantz
CC BY-SA 4.0
2019-12-17T07:29:55.063Z
See https://physics.stackexchange.com/a/252236/123208 and http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html
PM 2Ring
<hawking-radiation>
What's the point with Hawking Radiation?
Raphaël Frantz
# What's the point with Hawking Radiation?

(Asked by: username_0 on 2019-12-11)

I saw that Black Holes can evaporate because of Hawking radiation.
Because pair of particle, antiparticle can spawn at any point of the space, the antiparticle could be absorbed by the black hole and the particle go outside of it, like if the black hole where "lightning".

But, if one element of the pair can be absorbed by the black hole, isn't there an equal probability of that the particle will be absorbed, too?

I could imaginate that the probability of fluctuation of going into 0 with an equal probability of +1 or -1 mass over time is nonnull, also that's why black hole could evaporate, or something like it?

**Tags:** hawking-radiation
> **Comments:**
> **username_0**: Ok i see it's because anti-particle have positive mass  
> **username_1**: See https://physics.stackexchange.com/a/252236/123208 and http://math.ucr.edu/home/baez/physics/Relativity/BlackHoles/hawking.html  

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## 1 answers

### Answer by username_2 on 2019-12-12. Score: 0

I cannot show this or prove it but the main idea, in simplest terms, is that a matter-antimatter pair, near the event horizon, will form. A member of this pair will decay back into the event horizon and be absorbed by the black hole. The other particle will decay out into the free space.

> **Comments:**
> **username_3**: This is inaccurate - especially “The other particle will decay out into the event horizon and then on into space.”   If it “decays” into the event horizon it disappears into the BH. You also fail to explain why a virtual particle falling into the BH makes it lose mass instead of gaining mass. Nor why the effect isn’t cancelled out by the virtual pairs forming just *inside* the EH. Sometimes simple answers can be **too** simple :-)  
> **username_2**: @ChappoSaysReinstateMonica. I changed my answer somewhat.  If its still wrong I will delete it.  But if you know its wrong I encourage you to answer the question yourself.  

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