How Does Hawking Radiation Affect Black Hole Temperature?

[AbsoluteZer0]

Hi,

I recently came across this formula:

$T = \frac{\hbar c^3}{8 \pi GK_BM}$

As I understand it deals with the radiation that is believed to be emitted by a black hole.
Does it describe the temperature of the radiation?

Thanks

 

[Simon Bridge]

It describes the statistics of the radiation coming from close to the event horizon of a black-hole due to the large gravity gradient there.
"Temperature" is a common way of describing such statistics - the model says that the radiation coming from the black hole follows a blackbody spectrum with a characteristic temperature given by that equation.

 

[AbsoluteZer0]

Thanks.

One more question I have regarding this equation is: how can we quantify the mass of a black hole (M)? Is it the mass of the singularity?

 

[Simon Bridge]

No worries.

In relativity or QM it helps to be carefully pedantic about what things are saying - and that goes squared for when when both of them are used together :)

 

[pmacias]

for your question. Yes, the formula you mentioned does describe the temperature of the radiation emitted by a black hole. It is known as the Hawking temperature, named after physicist Stephen Hawking who first proposed this concept. This formula is a result of combining quantum mechanics and general relativity, and it describes how black holes emit radiation due to quantum effects near the event horizon. This radiation is known as Hawking radiation and is a unique characteristic of black holes. The formula you mentioned is important because it allows us to calculate the temperature of this radiation, which is incredibly high for small black holes and decreases as the black hole size increases. This radiation emission has important implications for the long-term fate of black holes and is an active area of research in astrophysics. I hope this helps answer your question.