How Do Entropy and Mass Relate in Black Hole Thermodynamics?

In summary, the equations in black hole thermodynamics involve the Schwarzschild radius, entropy, and mass. Using the expression for the Schwarzschild radius and the area equation for a sphere, the entropy of a black hole can be written as S=4πkGM^2/ħc. As mass is lost, the change in entropy will be dS=8πkGmdm/ħc. These equations can be derived from the Bekenstein-Hawking formula and the area equation for a sphere.
  • #1
MrPhysicsGuy
3
1

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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  • #2
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.
 

FAQ: How Do Entropy and Mass Relate in Black Hole Thermodynamics?

What is a black hole?

A black hole is a region of space where the gravitational pull is so strong that nothing, including light, can escape from it. This occurs when a massive star dies and collapses under its own weight, creating a singularity with infinite density and zero volume.

How do black holes have temperature?

According to Hawking's theory, black holes have a temperature due to the quantum effects near the event horizon. This temperature, known as the Hawking temperature, is inversely proportional to the mass of the black hole.

What is the relationship between black hole temperature and entropy?

The Hawking temperature of a black hole is directly proportional to its entropy, which is a measure of the disorder or randomness of a system. As the black hole's temperature increases, its entropy also increases, meaning it becomes more chaotic.

Can black holes emit radiation?

Yes, black holes can emit Hawking radiation, which is a form of thermal radiation caused by quantum fluctuations near the event horizon. This radiation carries away energy and causes the black hole to lose mass, eventually leading to its evaporation.

How does black hole thermodynamics relate to the laws of thermodynamics?

Black hole thermodynamics follows the same laws as traditional thermodynamics, such as the conservation of energy and the second law of thermodynamics. However, the presence of a singularity and the effects of gravity make the calculations more complex and require the use of theories such as quantum mechanics.

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