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

1. Dec 21, 2007

rubecuber

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

2. Dec 21, 2007

George Jones

Staff Emeritus
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

$$P = \frac{dE}{dt} = \sigma A \left( \frac{\hbar c^3}{8 \pi k G M} \right)^4.[/itex] Detail and extensions of this calculation are in this thread. Last edited: Dec 22, 2007 3. Dec 21, 2007 pervect Staff Emeritus Would it correct to say that A in this equation is the area of the event horizon, this area being the same for a local observer and for an observer at infinity? If so, we should be able to write [tex]A = 4 \pi r_s^2 = 4 \pi \left( \frac{2 G M}{c^2}\right)^2$$

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

Last edited: Dec 21, 2007
4. Dec 22, 2007

pervect

Staff Emeritus
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 this semi-random webpage

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(?).

Last edited: Dec 22, 2007
5. Dec 22, 2007

George Jones

Staff Emeritus
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.