How can you find out the amount of hawking radiation a black hole exhibits?

[rubecuber]

How can you find out the amount of hawking radiation a black hole exhibits?
Rube cuber,
Merry Christmas

 

[George Jones]

I'm not sure what you want. Numbers?

In terms of general formulae:

A black of mass M has temperature

T = \frac{\hbar c^3}{8 \pi k G M},

where k is Boltzmann's constant.

The Stefan-Boltzmann law for radiation by black bodies is

P = \sigma A T^4,

where \sigma is Stefan's constant A is the radiating surface area.

Putting these together gives that a black hole radiates power (mass-energy per unit time) according to

P = \frac{dE}{dt} = \sigma A \left( \frac{\hbar c^3}{8 \pi k G M} \right)^4.[/itex]<br /> <br /> Detail and extensions of this calculation are in <a href="https://www.physicsforums.com/showthread.php?p=1549655#post1549655&quot;" class="link link--internal">https://www.physicsforums.com/showthread.php?p=1549655#post1549655&quot;</a>.

 

[pervect]

Would it correct to say that A in this equation is the area of the event horizon, this area being the same for a local observer and for an observer at infinity?

If so, we should be able to write

A = 4 \pi r_s^2 = 4 \pi \left( \frac{2 G M}{c^2}\right)^2

eliminating it from the equation. And the radiated power we compute in this manner would be that measured by an observer "at infinity".

 

[pervect]

Combining the following formulae:

P = \sigma A T^4,

\sigma = \frac {2 {\pi }^{5}{k}^{4}}{15 {h}^{3}{c}^{2}}

T = \frac {h{c}^{3}}{16 {\pi }^{2}\,kG\,M}

A = 4 \pi r_s^2

r_s = \frac{2GM}{c^2}

I get the following expression for P, the power radiated at infinity:

\frac {h{c}^{6}}{30720 \, {\pi }^{2}{G}^{2}{M}^{2}}

where:
h is Planck's constant
c is the speed of light
G is the gravitational constant
\hbar = h / 2 \pi
k is Boltzman's constant.

Substituting for hbar in terms of h makes the above answer the same as this current wikipedia article, and also http://library.thinkquest.org/C007571/english/advance/core8.htm

The lifetime of the black hole should be Mc^2 / P, where M is the mass of the black hole (Mc^2 is the "energy at infinity" of the black hole, and P is the "power radiated at infinity", so the ratio should be the lifetime of the black hole).

This gives an expression for the lifetime of the black hole of:

\frac {30720 \,{\pi }^{2}{G}^{2}}{{c}^{4}h} M^3= 2.52\,10^{-16} \frac{s}{kg^3} \,M^3

I get a lifetime of a solar mass black hole of 6.3e67 years, however some of the web sources seem to get a different figure in spite of using the same formula(?).

 

[George Jones]

I get a lifetime of a solar mass black hole of 6.3e67 years, however some of the web sources seem to get a different figure in spite of using the same formula(?).

Doing an accurate calculation is a fairly involved business, so books and webpages that make simplifying assumptions get results that range over several orders of magnitude.

I am going to put the details of the simplified calculation in a new thread.