Entropy of a Schwarzchild black hole

1. May 2, 2005

johnwalton84

Hi, i'm looking for some help on where to start with this question:

The surface area of a Schwarzchild black hole is $$A=16 \pi R^2_c$$ where $$R_c$$ 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

$$S = \frac{\xi k c}{4\pi h G}A$$

where $$\xi$$ is a numerical constant.

I know that the enropy of a change is

$$S = \int_{initial}^{final} \frac{Q_{rev}}{T}$$

and can show that using the de Broglie relation

$$\lambda dB <= 2R_c = \frac{4GM}{c^2}$$

the energy is

$$\frac{hc^3}{4GM} <= E$$

But i'm not sure where to go with proving that the entropy is the equation given.

Last edited: May 2, 2005
2. May 2, 2005

TensorKhan

It looks like you got your Latex wrong. Change the [\tex] to [/tex].

3. May 2, 2005

dextercioby

$$S_{Beckenstein-Hawking}=\frac{A}{4\hbar}$$

is more likely defined...

Daniel.

Last edited: May 2, 2005
4. May 2, 2005

dextercioby

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

$$dM=\frac{1}{8\pi}\kappa dA+\Omega_{H}dJ$$

Daniel.

----------------------------------------
[1]Wald R.M."General Relativity",1984.

5. May 2, 2005

johnwalton84

Ok, thats helpful, thanks. I assume $$\Omega_{H}dJ$$ represents work done.

Does that mean the two forms

$$dM = \frac{K dA}{8\pi} + work$$
$$dE = T dS + work$$

could be equated?

$$dE - T dS = dM - \frac{K dA}{8\pi}$$

Last edited: May 2, 2005
6. May 2, 2005

dextercioby

Yes,$$TdS=\frac{1}{8\pi}\kappa dA$$

Daniel.

7. May 2, 2005

dextercioby

And one more thing,it's Karl Schwarzschild.

Daniel.

8. May 2, 2005