How Do Entropy and Mass Relate in Black Hole Thermodynamics?

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MrPhysicsGuy
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Homework Statement


I would very much like getting some help with my problem regarding the equations in some black hole thermodynamics.

"Using the expression for the Schwarzschild radius, the entropy of a black hole of event-horizon area A=πR^2 can be written in terms of its mass using Eq. (1) as S=4πkGM^2/ħc. As mass is lost, the change in entropy will be dS=8πkGmdm/ħc..."

I don't understand how they got S=4πkGM^2/ħc and dS=8πkGmdm/ħc.

Homework Equations


Eq. (1) Entropy
S=kc^3A/4ħG

Eq. (2) Schwarzshild radius
R(s)=2GM/c^2

Eq. (3) A=πR^2

The Attempt at a Solution


?
Thanks for helping and have a wonderful day :)

[/B]
 
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MrPhysicsGuy said:

Homework Statement


I would very much like getting some help with my problem regarding the equations in some black hole thermodynamics.

"Using the expression for the Schwarzschild radius, the entropy of a black hole of event-horizon area A=πR^2 can be written in terms of its mass using Eq. (1) as S=4πkGM^2/ħc. As mass is lost, the change in entropy will be dS=8πkGmdm/ħc..."

I don't understand how they got S=4πkGM^2/ħc and dS=8πkGmdm/ħc.

Homework Equations


Eq. (1) Entropy
S=kc^3A/4ħG

Eq. (2) Schwarzshild radius
R(s)=2GM/c^2

Eq. (3) A=πR^2

The Attempt at a Solution


?
Thanks for helping and have a wonderful day :)
[/B]
First there's a mistake in the problem statement. The equation that was given was [itex]A = \pi R^2[/itex]. That's the area of a circle, not a sphere. You should use the area equation for a sphere,
[tex]A = 4 \pi R^2.[/tex]

Assuming that [itex]S = \frac{4 \pi k G M^2}{\hbar c}[/itex] is correct, you should be able to derive [itex]dS = \frac{8 \pi k G M \ dM}{\hbar c}[/itex] easily enough; it is just a simple derivative.

So are you asking where the [itex]S = \frac{4 \pi k G M^2}{\hbar c}[/itex] comes from? Here's a wiki link on Black Hole thermodynamics that should help:
https://en.wikipedia.org/wiki/Black_hole_thermodynamics

The Schwartzschild radius is typically given by [itex]R = \frac{2 M G}{c^2}[/itex], by the way.

When expressing that in terms of area, make sure you use the area equation for a sphere ([itex]A = 4 \pi r^2[/itex]). Don't use the area equation for a circle.

That, the Bekenstein–Hawking formula given in the above link, and a bit of substitution should get you to your answer.