Entropy of a Schwarzchild black hole

In summary, the conversation discusses the surface area and entropy of a Schwarzschild black hole containing quantized matter. The entropy is shown to be equal to a numerical constant multiplied by the surface area. The first law of thermodynamics for a black hole is also mentioned, with work represented by \Omega_{H}dJ. The possibility of equating dM and dE is discussed, with TdS and \frac{1}{8\pi}\kappa dA being equivalent. The correct spelling of Karl Schwarzschild is also mentioned.
  • #1
johnwalton84
16
0
Hi, I'm looking for some help on where to start with this question:

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

is more likely defined...

Daniel.
 
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  • #4
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, that's 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
Yes,[tex] TdS=\frac{1}{8\pi}\kappa dA [/tex]

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

Daniel.
 
  • #8
Thanks for your help.
 

What is the entropy of a Schwarzchild black hole?

The entropy of a Schwarzchild black hole is a measure of the disorder or randomness of its microscopic constituents. It is a physical quantity that describes the number of microstates that a black hole can be in, given its mass, angular momentum, and charge.

How is the entropy of a Schwarzchild black hole calculated?

The entropy of a Schwarzchild black hole is calculated using the famous Bekenstein-Hawking formula, which states that the entropy is equal to one quarter of the black hole's event horizon area in Planck units. This means that the entropy is directly proportional to the size of the black hole's event horizon.

What does the entropy of a Schwarzchild black hole tell us about its properties?

The entropy of a Schwarzchild black hole is closely related to its mass and temperature. As the black hole's mass increases, so does its entropy, indicating an increase in the number of possible microstates. Similarly, as the black hole's temperature decreases, its entropy decreases as well.

What is the significance of the entropy of a Schwarzchild black hole in the study of black hole thermodynamics?

The entropy of a Schwarzchild black hole is a crucial quantity in black hole thermodynamics, as it is directly related to the second law of thermodynamics. It also plays a role in the understanding of black hole evaporation, as the entropy of a black hole decreases as it emits thermal radiation.

Can the entropy of a Schwarzchild black hole ever decrease?

No, according to the second law of thermodynamics, the entropy of a closed system can never decrease. This means that the entropy of a Schwarzchild black hole can only increase or remain constant, but it can never decrease. This is in line with the behavior of black holes, as they are known to only grow in size, never shrink.

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