How can Black Holes emit radiation?

[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?

 

[mathman]

The physical explanation has its basis in virtual particles.

http://www.obscure.org/physics-faq/Relativity/BlackHoles/hawking.html

 

[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

http://scienceblogs.com/startswithabang/upload/2011/02/can_you_get_something_for_noth/black%20hole%20eva.jpeg

 

[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?

 

[e.bar.goum]

^ 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?

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

 

[DaveC426913]

^ 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?

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!

 

[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.

 

[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.

 

[DaveC426913]

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.

Sorry, what does antimatter have to do with the topic?

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

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.

 

[gamesguru]

Sorry, what does antimatter have to do with the topic?

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.

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.

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

 

[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.

 

[tom.stoer]

... 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.

No, the particle is not ejected with this thermal frequency near the Horizon; Hawking 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.

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

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

 

[jacksonb62]

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

Hawking radiation does not come about because antimatter particles sometimes fall into black holes; it comes about because negative-energy particles (both matter and animatter) sometimes fall into black holes. Some popular-level treatments of black holes obscure this, and even sometime get this completely wrong.

Steve Carlip has written a non-mathematical virtual particle description of Hawking radiation which is more challenging than most non-mathematical descriptions, but which also is more accurate than most non-mathematical descriptions.

http://www.physics.ucdavis.edu/Text/Carlip.html#Hawkrad

What happens, very roughly, is this. Energy is associated with time and spatial momentum is associated with space. When an matter-antimatter pair of virtual particles is created *outside* the event horizon, they can become a little bit separated in the time that the Heisenberg uncertainty principle allows them to live. Tidal forces caused by the curvature of spacetime help them to separate, and, sometimes, the negative-energy particle (which could be either matter or anitimatter) wanders over the event horizon and into the black hole. Inside the event horizon, the roles of time and space coordinates get interchanged. Thus, according to what I wrote above, the roles of energy and spatial momentum get interchanged. What was negative energy becomes a negative spatial component of a local (for an observer inside the horizon) momentum vector. Only a virtual particle can have negative energy, while any particle, real or virtual, can have a negative component of spatial momentum.

Bottom line: the whole process can become a real process. In this real process, an observer outside a black hole "sees" the black hole hole swallow a negative-energy particle while emiitting a positve energy particle (the other member of the matter-antmatter pair). The balck hole radiates.

 

[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.

 

[Cheese Donkey]

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

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?

 

[tom.stoer]

there are no negative-mass particles;

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

 

[DaveC426913]

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.

Sorry. You said antimatter. Obviously you meant antiparticles.

I fail to grasp why the evaporation rate of matter would be consistently greater than the evaporation rate of antimatter by a fixed amount.

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

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?

 

[Naty1]

there are no virtual particles in Hawking's calculation

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.

 

[Naty1]

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

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?

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

 

[Neandethal00]

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.

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 black hole 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.

 

[jacksonb62]

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?

What someone recently told me on this forum is that all virtual particles, either normal matter or antimatter, can be thought of as having negative energy. It is not until one particle is freed from the pair that it become "real" and then has positive energy. Thus, it isn't always the antimatter particle that falls in and the normal matter particle that escapes. the particle that falls in remains a virtual particle with negative energy and the particle that escapes becomes real with positive energy.

 

[jacksonb62]

Actually, let me correct myself. I referenced my handy copy of A Brief History of Time and read the section on virtual particles. Hawking writes

"A real particle close to a massive body has less energy than if it were far away, because it would take energy to lift it far away against the gravitational attraction of the body. Normally, the energy of the particle is still positive, but the gravitational field inside a black hole is so strong that even a real particle can have negative energy there. it is therefore possible, if a black hole is present, for the virtual particle with negative energy to fall into the black hole and become a real particle or antiparticle. In this case it no longer has to annihilate with its partner. Its forsaken partner may fall into the black hole as well. Or, having positive energy, it might also escape from the vicinity of the black hole as a real particle or antiparticle"

Although a rather simplified explanation, I think Hawking helps to clear things up a little. One particle does have positive energy, and one has negative, but the negative one could be either matter or antimatter. It falls into the hole more often than the particle with positive energy because it can't escape as easily due to a lower level of energy

 

[jacksonb62]

antiparticle* sorry not antimatter

 

[tom.stoer]

Neither antiparticle nor antimatter but "virtual particles with negative mass".

Sorry - and with proper respect - Hawking original idea and calculation was ingenious, his popular books aren't! The good thing is that people like them, and he makes a good job in marketing physics; the bad thing is that some 'explanations' are not explanations but 'deceptions' - and that the Caveat is continuously missing!

There are no virtual particles in his calculation, so there is no reason to introduce them in the explanation. The whole thread is about the misleading explanation - not about physics.

 

[Cheese Donkey]

Neitehr antiparticle nor antimatter but "virtual particles with negative mass".

Sorry - and with proper respect - Hawking original idea and calculation was ingenious, his popular books aren't! The good thing is that people like them, and he makes a good job in marketing physics; the bad thing is that some 'explanations' are not explanations but 'deceptions' - and that the Caveat is continuously missing!

There are no virtual particles in his calculation, so there is no reason to introduce them in the explanation. The whole thread is about the misleading explanation - not about physics.

Then can you explain why? All I've gotten out of this thread is that the popular explanation is false. Math is fine - by layman I mean I'm an undergrad who hasn't taken quantum, but taken a full year of Calculus.

 

[tom.stoer]

Let's try.

There are different "vacuum states", i.e. no unique, global vacuum w.r.t. to all observes, but only local definitons of vacuum. Relating the vacuum state as defined near the BH to the vacuum as defined by the asymptotic observer at infinity, one transforms the states in such a way that they appear as real particles asymptotically.

This should be a starting point ...

 

[jacksonb62]

I would have to agree that often Hawking's popular media oversimplifies things, but I am only a junior in high schools without any formal education on the topic so they are helpful introductions. I'm sorry but so far all of you have said how virtual particles aren't part of Hawking's calculations, but no one has explained then, without virtual particles, how Hawking explains radiation. If someone could post a link or a brief mathematical explanation of how Hawking justifies his name-sake radiation I would greatly appreciate it. Thanks!

 

[tom.stoer]

We should look for his original paper online and then discuss it step by step

 

[Cheese Donkey]

Let's try.

There are different "vacuum states", i.e. no unique, global vacuum w.r.t. to all observes, but only local definitons of vacuum. Relating the vacuum state as defined near the BH to the vacuum as defined by the asymptotic observer at infinity, one transforms the states in such a way that they appear as real particles asymptotically.

This should be a starting point ...

So the gravity becomes so strong approaching infinity (the event horizon) that "normal" matter and photons are confined to the smallest uncertainties they can possibly achieve, and vacuum fluctuations become the dominant factor? I hope I am understanding this. And thanks for helping me out as well.

 

[tom.stoer]

No, gravity need not be "strong" at the event horizon; there is no confining effect.

Here's the original paper:
http://prac.us.edu.pl/~ztpce/QM/CMPhawking.pdf
Particle Creation by Black Holes
S. W. Hawking
Commun. math. Phys. 43, 199—220 (1975)