Entropy of a Schwarzchild black hole

  • #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.
 
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Answers and Replies

  • #2
It looks like you got your Latex wrong. Change the [\tex] to [/tex].
 
  • #3
dextercioby
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[tex] S_{Beckenstein-Hawking}=\frac{A}{4\hbar} [/tex]

is more likely defined...

Daniel.
 
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  • #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.
 
  • #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]
 
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  • #6
dextercioby
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Yes,[tex] TdS=\frac{1}{8\pi}\kappa dA [/tex]

Daniel.
 
  • #7
dextercioby
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And one more thing,it's Karl Schwarzschild.

Daniel.
 
  • #8
Thanks for your help.
 

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