Entropy and Black Hole Temperature

In summary, the conversation discusses finding the temperature of a black hole given its energy and entropy. The solution involves using the partial derivative of entropy with respect to energy and then multiplying it with the partial derivative of energy with respect to mass. This method is equivalent to the chain rule in calculus.
  • #1
erok81
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0

Homework Statement



In a previous problem I had to find the entropy of a black hole where I ended with this:

[itex]S_{BH}=\frac{8 \pi^2 G M^2 k}{h c}[/itex]

Now I am to find the temp, given the energy of a black hole is mc2.

Homework Equations



[itex]T=(\frac{\partial S}{\partial u})^{-1}[/itex]

The Attempt at a Solution



Originally I was stumped on how to start and in the process of google'ing different things, I accidentally ran across the solution. Since I can't undo my mistake, I would like to understand how some of this solution works.

I understand in order to get the temperature, I need to take the partail of S with respect to U. In my original black hole entropy, I can't do this since there is no U term to differentiate. However, I do have the energy is = mc2 or U=mc2.

Now I suppose I could substite that in for M2 and get my U term, but the solution I saw showed something different and that is where I am confused.

They split the partial derivative into two parts.

[itex] \frac{\partial S}{\partial U} = \frac{\partial S}{\partial M} \cdot \frac{\partial M}{\partial U}[/itex]

So to get my T equation, we take the partial of the original equation with respect to M and then the partial of M with respect to U.

The second half using ∂U/∂M (mc2) → c2 then the inverse so ∂M/∂U which I need from above. Which is dumb. I could just solve for M and take ∂M/∂U in the first place...

Then mulitplying together gives me the correct equation for T.

My question lies in splitting up the derivative in order to get the form I need at the end. Is that the correct method to solve this? I looked around in my calculus book and didn't see anything that split up derivatives like this. Or anything that really explained this process. I think it makes more sense to sub in the energy like I did in the beginning, but woud like to understand this method as well.

Hopefully that all makes sense.
 
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  • #2
Look up "chain rule." As far as I understand it, it is formally equivalent to direct substitution.
 

Related to Entropy and Black Hole Temperature

1. What is entropy and how does it relate to black hole temperature?

Entropy is a measure of the disorder or randomness in a system. It is closely related to the number of possible microstates that a system can have. In the context of black holes, entropy is directly proportional to the surface area of the black hole's event horizon. This means that as the black hole's surface area increases, so does its entropy. Black hole temperature is also related to entropy, as the temperature of a black hole is inversely proportional to its entropy.

2. How is black hole temperature calculated?

The temperature of a black hole is calculated using the Hawking radiation formula, which takes into account the black hole's mass, charge, and angular momentum. This radiation is predicted to be emitted from the event horizon of a black hole, causing a gradual loss of mass and energy over time. As the black hole loses energy, its temperature increases.

3. Can black holes have a negative temperature?

No, black holes cannot have a negative temperature. According to the laws of thermodynamics, a negative temperature would mean that a system has more energy in its lower energy states than in its higher energy states. However, black holes have a maximum entropy and cannot have more energy in their lower energy states than in their higher energy states.

4. What is the relationship between entropy and information in black holes?

Entropy and information are closely related in the context of black holes. As black holes have a maximum entropy, it is believed that they also have a maximum amount of information that can be stored on their event horizons. This relationship is still being studied and is a topic of ongoing research in the field of black hole thermodynamics.

5. How does the concept of entropy in black holes relate to the second law of thermodynamics?

The second law of thermodynamics states that the total entropy of a closed system will always increase over time. This means that as black holes emit Hawking radiation and lose mass, their entropy will also increase. This is in accordance with the second law of thermodynamics, as the total entropy of the universe continues to increase with the formation and eventual evaporation of black holes.

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