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Homework Help: Entropy of a Schwarzchild black hole

  1. May 2, 2005 #1
    Hi, i'm looking for some help on where to start with this question:

    The surface area of a Schwarzchild black hole is [tex]A=16 \pi R^2_c[/tex] where [tex]R_c[/tex] is the distance of the event horizon from the centre of the black hole. Show that for such a hole containing quantized matter, its entropy can be written

    [tex]S = \frac{\xi k c}{4\pi h G}A[/tex]

    where [tex]\xi[/tex] is a numerical constant.



    I know that the enropy of a change is

    [tex]S = \int_{initial}^{final} \frac{Q_{rev}}{T}[/tex]

    and can show that using the de Broglie relation

    [tex]\lambda dB <= 2R_c = \frac{4GM}{c^2}[/tex]

    the energy is

    [tex]\frac{hc^3}{4GM} <= E[/tex]

    But i'm not sure where to go with proving that the entropy is the equation given.
     
    Last edited: May 2, 2005
  2. jcsd
  3. May 2, 2005 #2
    It looks like you got your Latex wrong. Change the [\tex] to [/tex].
     
  4. May 2, 2005 #3

    dextercioby

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    [tex] S_{Beckenstein-Hawking}=\frac{A}{4\hbar} [/tex]

    is more likely defined...

    Daniel.
     
    Last edited: May 2, 2005
  5. May 2, 2005 #4

    dextercioby

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    In the section 12.5 of his book [1],Wald shows that the first law of thermodynamics for a black hole can be written

    [tex] dM=\frac{1}{8\pi}\kappa dA+\Omega_{H}dJ [/tex]

    Daniel.

    ----------------------------------------
    [1]Wald R.M."General Relativity",1984.
     
  6. May 2, 2005 #5
    Ok, thats helpful, thanks. I assume [tex]\Omega_{H}dJ [/tex] represents work done.

    Does that mean the two forms

    [tex] dM = \frac{K dA}{8\pi} + work[/tex]
    [tex]dE = T dS + work[/tex]

    could be equated?

    [tex]dE - T dS = dM - \frac{K dA}{8\pi}[/tex]
     
    Last edited: May 2, 2005
  7. May 2, 2005 #6

    dextercioby

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    Yes,[tex] TdS=\frac{1}{8\pi}\kappa dA [/tex]

    Daniel.
     
  8. May 2, 2005 #7

    dextercioby

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    And one more thing,it's Karl Schwarzschild.

    Daniel.
     
  9. May 2, 2005 #8
    Thanks for your help.
     
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