# Why do Black Holes evaporate due to Hawking radiation?

1. Aug 7, 2015

### lightoflife

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?

2. Aug 7, 2015

### phinds

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.

3. Aug 8, 2015

### CWatters

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. Aug 8, 2015

### phinds

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

5. Aug 8, 2015

### WannabeNewton

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. Aug 8, 2015

### mathman

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. Aug 9, 2015

### stevebd1

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)-

$$\Delta E \Delta t=\frac{\hbar}{2}$$

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