Why do Black Holes evaporate due to Hawking radiation?

In summary, virtual particle-antiparticle pairs are created near a black hole, and one of the pairs is usually captured by the black hole while the other one is freed to move away. This causes the black hole to lose mass and thus evaporate.
  • #1
lightoflife
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A book I read says that when virtual particle-antiparticle pairs are created near a black hole then sometimes one of the particle pairs will be captured by the black hole while the other one will be freed to move away as a real particle - then this causes the black hole to lose mass and thus evaporate.

How does it cause the black hole to lose mass if the particle captured by the black hole is a 'normal' virtual particle instead of the antiparticle of the pair?

I would think that the virtual antiparticle -once captured- might interact with all the normal matter inside the black hole to somehow reduce mass.

But if the captured virtual particle is a 'normal' or in other words not a virtual antiparticle then should not this cause the mass of the black hole to increase?
 
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  • #2
lightoflife said:
A book I read says that when virtual particle-antiparticle pairs are created near a black hole then sometimes one of the particle pairs will be captured by the black hole while the other one will be freed to move away as a real particle - then this causes the black hole to lose mass and thus evaporate.

How does it cause the black hole to lose mass if the particle captured by the black hole is a 'normal' virtual particle instead of the antiparticle of the pair?

I would think that the virtual antiparticle -once captured- might interact with all the normal matter inside the black hole to somehow reduce mass.

But if the captured virtual particle is a 'normal' or in other words not a virtual antiparticle then should not this cause the mass of the black hole to increase?
This whole "virtual particle pair" thing is NOT what actually happens. That was a description given by Hawking for the advanced-math-challenged folks (most of us) who only speak English, as it was the best he could do to translate the math into something that roughly made sense in English. It is a flawed analogy and needs to just be accepted AS an analogy.
 
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  • #3
As I understand it the emitted particles also carry away energy. Since energy is equivalent to mass then from the outside it must look like the BH is loosing mass.

If I've got that wrong or it's also only an analogy perhaps someone could let me know! I'm not an expert on QM by any means.
 
  • #4
CWatters said:
As I understand it the emitted particles also carry away energy. Since energy is equivalent to mass then from the outside it must look like the BH is loosing mass.

If I've got that wrong or it's also only an analogy perhaps someone could let me know! I'm not an expert on QM by any means.
The whole IDEA of it being cause by particle pairs is, according to Hawking, just an analogy so it doesn't matter if you are talking about mass or energy or both.

I wish I understood the math so I could try to explain where the analogy goes wrong, but if Hawking can't do it I figure I wouldn't have much chance :smile:
 
  • #5
Virtual particles can have negative energy with respect to spatial infinity, hence why the black hole loses mass. As an aside, antiparticles don't have negative mass, which seems to be part of your misconception about the mechanism behind the radiation.
 
  • #6
One description I saw (can't give a reference) describes the boundary of a black hole as fuzzy, so the virtual pair scenario can be though of as taking place in this fuzzy zone.
 
  • #7
lightoflife said:
A book I read says that when virtual particle-antiparticle pairs are created near a black hole then sometimes one of the particle pairs will be captured by the black hole while the other one will be freed to move away as a real particle - then this causes the black hole to lose mass and thus evaporate.

How does it cause the black hole to lose mass if the particle captured by the black hole is a 'normal' virtual particle instead of the antiparticle of the pair?

I would think that the virtual antiparticle -once captured- might interact with all the normal matter inside the black hole to somehow reduce mass.

But if the captured virtual particle is a 'normal' or in other words not a virtual antiparticle then should not this cause the mass of the black hole to increase?

In an abstract sense, the key word here isn't anti-matter/particle, it's virtual particle. In accordance with the uncertainty principle, virtual particle pairs can only exist for a certain amount of time before they mutually annihilate (in accordance with the first law of thermodynamics)-

[tex]\Delta E \Delta t=\frac{\hbar}{2}[/tex]

regardless of which particle is taken into the BH, the other particle will become real, if it is the antimatter particle, then it will most likely almost instantly interact with regular matter and annihilate into photons, if the regular particle escapes, then it will simply interact with regular matter. Due to the first law of thermodynamics, the particle that falls into the BH, regardless of whether it is the anti-particle or the regular particle, will become negative (which is a mechanism not fully understood but is result of the uncertainty principle and first law) and as a result, the BH loses mass.

Below is a link to an old article on the subject-

'The Quantum Mechanics of Black Holes' by Stephen Hawking
http://www.phys.uwosh.edu/rioux/thermo/pdf/Black Holes -- Hawking.pdf
 

1. What causes Black Holes to evaporate?

Black Holes evaporate due to a process called Hawking radiation, which was first proposed by physicist Stephen Hawking in 1974. This radiation is caused by quantum effects near the event horizon of the Black Hole, causing particles to be emitted from the Black Hole's surface.

2. How does Hawking radiation lead to Black Hole evaporation?

The process of Hawking radiation involves virtual particle-antiparticle pairs being created near the event horizon of the Black Hole. In this process, one of the particles is pulled into the Black Hole while the other escapes, carrying away energy. Over time, this energy loss causes the Black Hole to lose mass and eventually evaporate.

3. What is the relationship between the size of a Black Hole and its rate of evaporation?

The rate of evaporation for a Black Hole is inversely proportional to its mass. This means that smaller Black Holes will evaporate faster than larger ones. As a Black Hole loses mass, its rate of evaporation also increases, leading to a runaway process that ultimately results in the complete evaporation of the Black Hole.

4. Can Black Holes completely evaporate?

According to Hawking's theory, Black Holes can indeed completely evaporate over a long period of time. However, this process is incredibly slow for large Black Holes and would take trillions of years. Smaller Black Holes, on the other hand, could theoretically evaporate in a matter of seconds or minutes.

5. Does the evaporation of a Black Hole violate the laws of thermodynamics?

At first glance, the evaporation of a Black Hole seems to violate the laws of thermodynamics, which state that energy cannot be created or destroyed. However, Hawking radiation is a result of quantum effects, and the energy of the escaping particles is balanced by a reduction in the mass of the Black Hole. Therefore, the total energy and mass of the universe remains constant, and the laws of thermodynamics are not violated.

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