if light cannot escape a black hole then how do they emit stuff called hawking radiation.
My understanding is that virtual particles are created on the event horizon boundary: one falls in and one escapes, which we see as hawking radiation. The energy for this creation comes from the gravitational field of the black hole which is linked to its mass, thus overtime as this energy is depleted the black hole 'evaporates'.
If the gravitational tidal force does a work equal to 2mc^2 (mc^2 is the mass-energy of a particle) BH can create two real particles (a particle and an antiparticle) in the vicinity of schwarzschild radius. Since these particles are created outside it's possible that one of these fall into the BH whereas the other escape. In this case BH has lost mc^2 of mass-energy (the escaping particle) (hence temperature (which is proportional to 1/(BH mass)) increases) and BH radiates.
Note that this is a very semplified treatment, in reality things are more complicated.
ok thanks for the answers .
You are wrong for using the word real in the first place. Funky rightly said virtual ( because we haven't of cannot directly observed them). The fact is, they are created near the mouth of the blackhole and one has negative while the other has positive energy(this I believe is principally not to violate the conservation of energy).The one with the posite energy is what we see as the radiation. The one with with negative energy falls in and reduces the blackhole's energy. This in turn reduce the mass of the blackhole since the mass is proportional to the energy of the blackhole. reduction in mass means increase in temperature which further speeds up radiation and the blackhole eventually evaporates.
You are obviously right, I was trying to explain the process in the simplest way (that is wrong, but it helps to understand) :)
there is a great vid on youtube done by BBC. just type in hawking radiation and it should be the first vid on the list.
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