How Do Entropy and Mass Relate in Black Hole Thermodynamics?

[MrPhysicsGuy]

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]

 

[collinsmark]

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.