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Subject: How does a black hole slowly radiate away?

Date: Mon Aug 21 16:29:04 2000
Posted by Philip Chan
Grade level: grad (non-science) School: No school entered.
City: Astoria State/Province: NY Country: USA
Area of science: Astronomy

Message:
I read in your archives that: "The black hole will slowly radiate away?.....Its just that at the edge of the star particle-antiparticle pairs could be created, whereby one of the particles may fall in leaving the other particle free to escape. Since this represents energy leaving the black hole, it will slowly release all of its mass in this fashion." Date: Sat Jun 8 08:29:42 1996; Posted by Malcolm Tobias; Position: Graduate Student, Physics MY QUESTION: How does Hawkings Radiation (as mentioned above) result in a net loss of mass to a black hole (BH). As I understand it the particle-antiparticle pairs leap into existance from nothing and destroy themselves very quickly. If anything the particle that fall into the BH should add to the BH's mass and since the particles creation was independent of the BH (created out of nothing after all), the fact that it "appears" to radiate from the BH should not effect the BH's mass at all? Help I don't understand.

To answer your question, I went right to the source: Stephen Hawking's original paper on this topic titled ``Particle Creation by Black Holes'' (Communications in Mathematical Physics, 1975, volume 43, page 199).

What I got was a headache.

Not surprisingly, this is a complicated topic! There is a part of the paper which I will quote up front; about the idea of creating a particle-antiparticle pair, he says: ``It should be emphasized that these pictures of the mechanism responsible for the thermal emission... are heuristic only and should not be taken too literally.'' In other words, this is just a model and may, in the end, not be the best way to describe what's going on.

I will be the first to admit that the physics of the situation is over my head. However, what I could glean from the paper is that, in the model, the gravitational field itself is responsible for creating the particle pairs, so the event is not decoupled from the black hole. Quantum mechanically, there is an inherent uncertainty in the curvature of spacetime near a black hole, and this uncertainty can be shown (by Hawking) to be linked with the energy density of particles created by the gravitational field.

If a pair of particles is produced which gets its rest mass energy from the gravitational field of the hole, and one particle is lost, then there will be a decrease in the energy of the hole. This can be translated as a loss of mass, which lowers the gravitational field strength. The pair creation rate depends inversely on the strength of the gravitational field, which itself depends on the mass. So, in essence, an emission of a particle corresponds to a loss of mass in the hole. What we see, far from the hole, is what looks like a thermal emission of particles, as if the black hole were radiating like a classical black body (which is a pun of sorts, I suppose). The smaller the mass, the hotter the hole, and the faster it radiates away its mass. The emission increases exponentially, and when the mass is low enough, bang! The hole is gone.

For a detailed (and technical) description of this, take a look at the Relativity FAQ. You might also do a web search on ``Hawking radiation'' and see what that nets you.



©2008 Phil Plait. All Rights Reserved.

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