# How can Black Holes emit radiation?

1. Dec 20, 2011

### BKLounge

I know that black holes are still not entirely understood, but I do know that it's generally accepted that they emit Hawking radiation. But I've also heard that the gravitational pull of a black hole is so strong that nothing can escape, "not even light". So how is it possible that Hawking radiation can pull free from the gravity of a black hole?

2. Dec 20, 2011

### mathman

3. Dec 21, 2011

### jacksonb62

I researched this topic a decent amount I'll do my best to explain. It has to do with virtual particles. I am not sure how much you know about virtual particles so I will offer a brief explanation. Particles in a vacuum appear with a particle-antiparticle pair that come together and vaporize into pure energy. This effect is observed as vacuum energy. At the event horizon of a black hole, particle-antiparticle pairs appear. The way the particles orientate themselves results in the occasional antiparticle falling into the event horizon. The other particle is then freed and is emitted away from the black hole as Hawking radiation. Due to the law conservation of mass, for a particle and antiparticle to spontaneously appear, an antiparticle must have negative mass to counteract the mass of the particle. When an antiparticle falls into the black hole, its negative mass actually lowers the mass of the overall black hole. over time (a LOT of time) black holes will wither away to nothing and explode. this diagram may help

Last edited by a moderator: May 5, 2017
4. Dec 21, 2011

### gamesguru

^ This is just a theory. Why do we need to assume particle-antiparticles come in pairs? Why can't we assume it's just regular blackbody radiation outside the event horizon?

5. Dec 21, 2011

### e.bar.goum

... Because it wouldn't be Hawking radiation then? I fail to see how your description could provide a means for black hole radiation.

6. Dec 21, 2011

### DaveC426913

Blackbody radiation cannot escape from inside the BH's event horizon, therefore it cannot cause evaporation of the BH itself.

It is the slippery nature of the "negative mass antiparticle" that allows it to cause evaporation - nothing is coming out, only falling in - but what is falling in is reducing the mass of the BH. Presto!

7. Dec 22, 2011

### tom.stoer

The arguments using pairs of virtual particles are slightly missleading b/c what one would observe in Hawking radiation are not virtual but real particles. The virtual particles are an interpretation of a rather complex mathematical issue.

The problem is that usually we define "particle" w.r.t. to a certain "vacuum state". In the presence of a horizon (this need not be caused by a gravitational field - see e.g. Unruh effect) different observers will no longer agree on a unique "vacuum state". The attempt to define a vacuum state far away from the horizon transforms the distorted modes of the quantum field in such a way that they appear as 'particles' asymptotically.

Afaik Hawking never used perturbation theory and "virtual particles" in his derivation.

8. Dec 22, 2011

### gamesguru

I fail to grasp why the evaporation rate of matter would be consistently greater than the evaporation rate of antimatter by a fixed amount. In a simplified view, both would evaporate at equal rates since their probabilities would be equal. Obviously this is not the case, and there is an obvious reason against it which I have overlooked.

I still don't see why matter couldn't jump outside of the event horizon due to thermal activity.

9. Dec 22, 2011

### DaveC426913

Sorry, what does antimatter have to do with the topic?
The event horizon is defined at the radius at which the escape velocity exceeds the speed of light. No matter can exceed the speed of light and so no matter can escape. Furthermore, even electromagnetic radiation will be red-shifted to infinity, so it too cannot escape.

10. Dec 22, 2011

### gamesguru

The theory of Hawking Radiation which we have considered above in this topic is that the radiation is due to the emission of particles when particle-antiparticle pairs appear at the boundary of the event horizon and the particle protrudes slightly from the event horizon. According to this theory, a particle gets ejected with a calculable average frequency, and is related to the Hawking Temperature of the black hole (inversely prop to mass of black hole). In order to lose mass, the black hole must, on average, emit more particles than antiparticles.

According to quantum mechanics, there is a possibility that the particle will drift very near to the boundary of the event horizon.

11. Dec 22, 2011

### mathman

Part of the explanation lies in the fact that the event horizon is not a sharp boundary, but is fuzzy due to quantum effects.

12. Dec 22, 2011

### tom.stoer

No, the particle is not ejected with this thermal frequency near the Horizon; Hawkings calculation only makes sense for an asymptotic observer at infinity.

And of course the BH "emitts both particles and antiparticles"; both carry away positive energy.

That doesn't make sense in this context b/c Hawking's calculation is exactly classical for the gravitational field.

Last edited: Dec 22, 2011
13. Dec 22, 2011

### jacksonb62

This might help. It cleared me up on a few things

14. Dec 22, 2011

### tom.stoer

I still think that the 'virtual particle description' is misleading; there are no virtual particles in Hawking's calculation - and you will never achieve a common non-mathematical understanding what virtual particles are.

15. Dec 22, 2011

### Cheese Donkey

If the black hole can swallow a negative-mass particle, can't it just as easily swallow a positive-mass particle? Would that positive-mass absorption out of the pair still cause radiation and the subsequent loss of mass for the black hole? If so, why?

From my layman's point of view, the probability of the black hole swallowing either the positive-mass or negative-mass would be 50/50, so we would see radiation but no loss of mass?

16. Dec 23, 2011

### tom.stoer

there are no negative-mass particles;

one must not take these 'layman's explanations' too literally; the questions you are asking adress problems of this simplified and partially misleading picture, they do scarcely adress physics - and there should always be a big 'CAVEAT'

Last edited: Dec 23, 2011
17. Dec 23, 2011

### DaveC426913

Sorry. You said antimatter. Obviously you meant antiparticles.
But those particles would still need to have a velocity of near c to within many decimal places in order to escape. How would thermal activity generate ordinary matter particles with such velocity?

18. Dec 23, 2011

### Naty1

From what I have read that's entirely correct. Apparently Hawking used the "virtual particle" explanation as an intuitive explanation of what was happening....and it had nothing do
do with the mathematics he used.

19. Dec 23, 2011

### Naty1

Because quantum theory requires the vacuum to produce virtual particles of all energies.

20. Dec 23, 2011

### Neandethal00

You posted it before I did. I had the same argument.
Unless a BH has an affinity to antiparticles, I do not see why more antiparticles will fall into blackhole than particles? If both anti and pro particles cross the event horizon at the same rate, there will be no lowering of BH mass and no eventual evaporation of BH.