Question about Hawking radiation in BH

[jnorman]

i have read about Hawking's theory of BH radiation based on the idea of spontaneous particle pair creation at the EH wherein, on occasion, one of the particles is absorbed by the BH while the other escapes to become real. since the one which is absorbed is always of negative energy (whatever that means), the mass of the BH is reduced, and over time, hawking theorizes that the BH would eventually disappear.

assuming hawking is correct and this mechanism does indeed occur, it seems to me that the effect is rather small. and since every BH in existence is constatnly bombarded with enormous numbers of photons from elsewhere in the universe, which add energy/mass to the BH, it seems that even with the small reduction in mass due to hawking radiation, the BH would always increase in total mass over time.

am i missing something or misunderstanding hawking theory in some way, or did hawking consider this in his theory (ie, the radiation is greater than the amount of energy entering the BH due to absorbption of photons)? or am i correct?

 

[PeterDonis]

did hawking consider this in his theory (ie, the radiation is greater than the amount of energy entering the BH due to absorbption of photons)?

Yes. What Hawking's theories says is that the BH will lose energy, on net, if the amount it radiates due to Hawking's mechanism is greater than the amount it absorbs due to radiation and matter from the rest of the universe falling into the hole. You are correct that, for a black hole of any significant size in our current universe, the energy radiated by Hawking's mechanism is far, far smaller than the amount being absorbed. The easiest way to see this is to look at Hawking's formula for the temperature of a black hole:

[tex]T = \frac{\hbar c^{3}}{8 \pi G M k_{B}}[/tex]

where [itex]\hbar[/itex] is Planck's constant divided by [itex]2 \pi[/itex], [itex]c[/itex] is the speed of light, [itex]G[/itex] is Newton's gravitational constant, [itex]M[/itex] is the mass of the hole, and [itex]k_{B}[/itex] is Boltzmann's constant. By simple thermodynamics, a black hole can only lose energy due to radiation if its temperature is higher than the temperature of its surroundings. If you run the numbers, a black hole with a mass of about 10^23 kg would have a temperature the same as the CMBR (2.7 K). That's roughly the mass of the Moon, which is well under the size of any black hole we expect to find out there; so any black hole we would reasonably expect to see would be colder than its surroundings and would be gaining energy.