# Blackbody Radiation - Entropy and Internal Energy

1. Feb 25, 2013

### chris_avfc

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

Expression for the entropy and internal energy of black body radiation.

Using the below relations:

2. Relevant equations

Total free energy for black body:
$$F = (k_b TV/\pi^2) \int k^2 ln[1-exp(-\hbar ck/k_b T)]dk$$
Relationship between partition function and internal energy:
$$E = -\partial ln(z)/ \partial \beta$$

Where $\beta$ is the inverse temperature given by:
$$\beta = (1/k_b T)$$
Relationship between the free energy, internal energy and entropy:
$$F = E - TS$$

3. The attempt at a solution

If I use $F = E - TS$ rearranged to

$$S = (E-F)/T$$

Then substitute the relations in and calculate.

I make a little progress until I hit the $F$ part, the integral gives me some problems as I am having trouble calculating it, I tried using Wolfram Alpha as a guide but it won't actually give me an answer which suggested to me that I'm going about it the wrong way.

Last edited: Feb 25, 2013
2. Feb 25, 2013

### kreil

Use a u-substitution, like

$$u = \frac{\hbar c k}{k_B T}$$

then you get

$$dk = \frac{k_B T}{\hbar c} du$$

and the integral becomes

$$F = \left( \frac{V}{\pi^2}\right) \left(\frac{(k_B T)^4}{(\hbar c)^3} \right) \int u^2 \ln{\left(1-e^{-u}\right)}du$$

which you can easily use Wolfram Alpha to solve

3. Feb 25, 2013

### dextercioby

Don't forget the limits of integration. The variables are independent of photon's (angular) frequency.

4. Feb 25, 2013

### chris_avfc

Oh man, I really should have seen that...

Yeah I have the limits wrote down, I just didn't know how to show them in the post.

5. Feb 25, 2013

### kreil

Code (Text):
\int_{-\infty}^{\infty} x^2 dx
$$=\int_{-\infty}^{\infty} x^2 dx$$

You can also right click any TeX equation to see the code that produced it.