Black holes and Hawking radiation

**JosephRombousky** · Nov15-04, 07:18 AM

Please correct my question if I am again misguided.

Theoretically not even light is able to escape the gravity of a blackhole. Due to the curvature of space the blackhole prevents anything or any information from coming back through the event horizon.

However Blackholes are supposed to give off Hawking radiation(which i believe was x-rays). Now this Hawking radiation is "half"(you know what i mean) of the virtual particle/antiparticle pairs.

So why should only x-rays be emmited from the blackholes. Would only x-rays be emmited from the blackhole suggest that somehow the gravity off the backhole is influencing the quantum vacuum resulting in only energy jumps which coencide with x-rays.

Thank you all! (this has kept me up all night)

**hellfire** · Nov15-04, 08:03 AM

I belive x-rays are mainly due to thermal emission of matter in the accretion disk, heated due to gravitational collapse. There may be also synchrotron x-ray emission from the relativistic particles moving in the magnetic fields near the black hole.

Hawking radiation is of different nature. Strictly speaking it has nothing to do with pair creation in the vacuum (this seams to be a heuristic explanation), but with a difference in the ground states (vacuum) of the quantum fields inside and outside the event horizon. This energetic difference leads to a faint thermal emission of particles of the corresponding field (lets say, some kind of equilibrium reaction).

**JosephRombousky** · Nov15-04, 10:31 AM

okies, but i destinctily remember the virtual particle pairs being involved. But i am aware of the heating of the matter that falls into the black hole.

any more comment? maybe book suggestions, ways to put me in the right direction?

**da_willem** · Nov15-04, 10:58 AM

Virtual paticles are just another interpretation of the zero point fluctuations of quantum fields, so I think this is just an interpretation thing. I once calculated the energy of a photon of Hawking radiation for a black hole with tha mass of the sun. It's something like 10E-8 eV, so definitely no x-rays!! This is VERY low energetic radiation.

A black hole wich is not 'feeded' by the accretion of mass theoretically 'evaporates'. The energy of the outgoing Hawking radiation is substracted from the black hole's mass. As the luminosity of a black hole is inversely proportional to it's mass squared, this means we only have a fair chance of detecting a black hole that is at the end of it's life, so it has only a small mass and radiates very fast.

When the black hole is nearly gone the radiation will be more and more energetic, so this will be seen as a gamma burst. Note that this is totally theoretic, and can only happen for very small black holes. The has not even with complete certainty a large black hole been found, let alone a small black hole... A black hole with the size of our sun will have a lifetime of ~3E60 years. Compare this with the present age of our universe and come to the conclusion that no such gamma burst will be observed if there is no other mechanism for creating small black holes than teh evaporation from large black holes.

For some nice and easy calculations and examples on Hawking radiation see:

http://library.thinkquest.org/C00757...ance/core4.htm

This does not require quantum field theory, as Hawking used. This has the advantage of being easily accessable. But I wouldn't read too much in his 'proof' of the energy of the outgoing photon. And ofcourse he does not prove that a black hole is a blackbody radiator...

**hellfire** · Nov15-04, 11:34 AM

Originally Posted by da_willem

Virtual paticles are just another interpretation of the zero point fluctuations of quantum fields, so I think this is just an interpretation thing.

I am not sure whether this is a matter of interpretation. If one claims that Hawking radiation is produced due to virtual particles becoming real, one is faced with the problem of explaining the mass loss of the black hole: virtual particles are created in pairs; particle and antiparticle, both with positive mass, but the one which falls into the black hole must have a negative mass in order to decrease the mass of the black hole. This sounds very weird to me. I think one is forced here to leave the notion of particle (or virtual particle) and one must treat the field as such. It seams to me that this can inferred from the problem of defining the correct multiple particle spaces (creation and annihilation operators) on both sides of the event horizon (although I am not able to follow the derivation of the Hawking radiation, I think I can reasonably follow the derivation of the Unruh radiation in a Rindler space, which is similar).

**da_willem** · Nov15-04, 12:01 PM

I'm not sure either. But this wikipedia article ( http://en.wikipedia.org/wiki/Virtual_particles ) strongly suggests that virtual particles are a good model for visualizing vacuum fluctuations.

About the negative mass of the virtual particle going into the black hole. Schutz introduction to general relatvity uses virtual particles to describe the Hawking proces. If I recall this correctly he says the mass of the ingoing virtual photon is only negative as seen by an observer very far away from the black hole. If you would be near the event horizon of the black hole and you would measure the energy of the ingoing particle you would get as it should, a positive energy.

~~Wave's_Hand_Particle~~ · Nov15-04, 04:03 PM

Originally Posted by JosephRombousky

Please correct my question if I am again misguided.

Theoretically not even light is able to escape the gravity of a blackhole. Due to the curvature of space the blackhole prevents anything or any information from coming back through the event horizon.

Theoretically, 'Visible-Light' can be Downconverted by Certain Horizonal conditions, thus the rebounded product becomes 'Invisisble-Light', or a certain type of Energy!

The Virtual products of Particle 'Pairs', CAN become E-M Vacuum Energy, a back-scattering process.

Visible Energy can be Polarized, Sunglass lensses can Darken because of certain structure of the Glass??,,does this help?

**jcsd** · Nov15-04, 04:28 PM

Originally Posted by hellfire

I am not sure whether this is a matter of interpretation. If one claims that Hawking radiation is produced due to virtual particles becoming real, one is faced with the problem of explaining the mass loss of the black hole: virtual particles are created in pairs; particle and antiparticle, both with positive mass, but the one which falls into the black hole must have a negative mass in order to decrease the mass of the black hole. This sounds very weird to me. I think one is forced here to leave the notion of particle (or virtual particle) and one must treat the field as such. It seams to me that this can inferred from the problem of defining the correct multiple particle spaces (creation and annihilation operators) on both sides of the event horizon (although I am not able to follow the derivation of the Hawking radiation, I think I can reasonably follow the derivation of the Unruh radiation in a Rindler space, which is similar).

From what I undretsnad the explanation is that one of the virtual photons has negative energy and always has negtaive energy, this virtual photon can only propagate freely (i.e. cannot be real) in the spacetime inside the event horizon.

**JosephRombousky** · Nov15-04, 05:49 PM

Originally Posted by Wave's_Hand_Particle

Theoretically, 'Visible-Light' can be Downconverted by Certain Horizonal conditions, thus the rebounded product becomes 'Invisisble-Light', or a certain type of Energy!

This actually is answering something of another thread i opened reffering to debrolie wavelegth redshift, and i think the energy you are talking about acutally changes into gravitational potential energy.

Originally Posted by da_willem

A black hole wich is not 'feeded' by the accretion of mass theoretically 'evaporates'. The energy of the outgoing Hawking radiation is substracted from the black hole's mass. As the luminosity of a black hole is inversely proportional to it's mass squared, this means we only have a fair chance of detecting a black hole that is at the end of it's life, so it has only a small mass and radiates very fast.

So this evaporation reduces the mass of the black hole and emmits only photons?

Thank you very much for the link!
Would their be an energy cut off for the gamma radiation however? At what point would hte photons of gamma rays decay into other particles?

Originally Posted by hellfire

Hawking radiation is of different nature. Strictly speaking it has nothing to do with pair creation in the vacuum (this seams to be a heuristic explanation), but with a difference in the ground states (vacuum) of the quantum fields inside and outside the event horizon. This energetic difference leads to a faint thermal emission of particles of the corresponding field (lets say, some kind of equilibrium reaction).

this only serves to confuse me a lil more. but why would the equilibrium reaction lead to loss of overal mass of the blackhole?

THANKS FOR ALL the INFO PEOPLE!!

**jcsd** · Nov15-04, 06:10 PM

Originally Posted by JosephRombousky

This actually is answering something of another thread i opened reffering to debrolie wavelegth redshift, and i think the energy you are talking about acutally changes into gravitational potential energy.

This is something different as I told you in the other thread that is more to do with the fact that energy is not conserved as an absolute rule in GR, only in the Netwonian limit. In this case what is being described is the result of the extreme coniditons created by a black hole, infact the Hawking process is not the only process that allows for negative energy photons to be created within a black hole, for comaprison the Penrose process allows for the cretaion of negative enrgy phtons that are confined to the ergoregion of a rotating black hole..

So this evaporation reduces the mass of the black hole and emmits only photons? Thank you very much for the link!
Would their be an energy cut off for the gamma radiation however? At what point would hte photons of gamma rays decay into other particles?

Apparently the radiation is not just limited to photons, the key thing is that it is the radiation has a blackbody spectrum.

**JosephRombousky** · Nov15-04, 08:22 PM

Black body spectrum? (i'm new here)

but i always had problems with the virtual particle antiparticle pairs.
When matter and antimatter are put together, your supposed to get 2 high energy photons which corespond ot the energy of the matter you put together.(lets say we used an electron and a positron) you would get 2 gamma rays or some such. But it seems to me that Photons have an absolute value of energy, while matter can either be positive or negative. I would think +1 + (-1) gives zero energy. Not 2.

But my head hurts again.

~~Wave's_Hand_Particle~~ · Nov16-04, 03:42 AM

Originally Posted by JosephRombousky

Black body spectrum? (i'm new here)

but i always had problems with the virtual particle antiparticle pairs.
When matter and antimatter are put together, your supposed to get 2 high energy photons which corespond ot the energy of the matter you put together.(lets say we used an electron and a positron) you would get 2 gamma rays or some such. But it seems to me that Photons have an absolute value of energy, while matter can either be positive or negative. I would think +1 + (-1) gives zero energy. Not 2.

But my head hurts again.

As if by magic this has come to my attention this morning:http://uk.arxiv.org/PS_cache/gr-qc/pdf/0411/0411070.pdf

**jcsd** · Nov16-04, 11:29 AM

Originally Posted by JosephRombousky

Black body spectrum? (i'm new here)

but i always had problems with the virtual particle antiparticle pairs.
When matter and antimatter are put together, your supposed to get 2 high energy photons which corespond ot the energy of the matter you put together.(lets say we used an electron and a positron) you would get 2 gamma rays or some such. But it seems to me that Photons have an absolute value of energy, while matter can either be positive or negative. I would think +1 + (-1) gives zero energy. Not 2.

But my head hurts again.

Photons are what are called selfconjugate particles i.e. they are their own antiparticles. matter for just about all intents and puproses has postive energy certainly a real particle must alawys have postive energy.

**AVNguyen** · Nov17-04, 06:08 AM

But we can also have a negative energy, is that also mean that we have negative mass

~~Wave's_Hand_Particle~~ · Nov26-04, 07:26 PM

This out just three days late:http://www.physlink.com/News/111804P...dBlackHole.cfm