# Entropy and Black Hole Temperature

1. Feb 10, 2013

### erok81

1. The problem statement, all variables and given/known data

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

$S_{BH}=\frac{8 \pi^2 G M^2 k}{h c}$

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

2. Relevant equations

$T=(\frac{\partial S}{\partial u})^{-1}$

3. 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.

$\frac{\partial S}{\partial U} = \frac{\partial S}{\partial M} \cdot \frac{\partial M}{\partial U}$

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.

2. Feb 11, 2013

### Staff: Mentor

Look up "chain rule." As far as I understand it, it is formally equivalent to direct substitution.