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

  1. Dec 21, 2007 #1
    How can you find out the amount of hawking radiation a black hole exhibits?
    Rube cuber,
    Merry Christmas
     
  2. jcsd
  3. Dec 21, 2007 #2

    George Jones

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    I'm not sure what you want. Numbers?

    In terms of general formulae:

    A black of mass [itex]M[/itex] has temperature

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

    where [itex]k[/itex] is Boltzmann's constant.

    The Stefan-Boltzmann law for radiation by black bodies is

    [tex]P = \sigma A T^4,[/tex]

    where [itex]\sigma[/itex] is Stefan's constant [itex]A[/itex] is the radiating surface area.

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

    [tex]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
  4. Dec 21, 2007 #3

    pervect

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    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 [/tex]

    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
  5. Dec 22, 2007 #4

    pervect

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    Combining the following formulae:

    [tex]P = \sigma A T^4,[/tex]

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

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

    [tex]A = 4 \pi r_s^2[/tex]

    [tex]r_s = \frac{2GM}{c^2}[/tex]

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

    [tex]\frac {h{c}^{6}}{30720 \, {\pi }^{2}{G}^{2}{M}^{2}}[/tex]

    where:
    h is planck's constant
    c is the speed of light
    G is the gravitational constant
    [itex]\hbar = h / 2 \pi[/itex]
    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:

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

    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
  6. Dec 22, 2007 #5

    George Jones

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    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.
     
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