Blackbody Radiation - Entropy and Internal Energy

[chris_avfc]

Homework Statement

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

Using the below relations:

Homework 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:
$$ F = E - TS $$

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.

 

[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

 

[dextercioby]

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

 

[chris_avfc]

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

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

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

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

 

[kreil]

\int_{-\infty}^{\infty} x^2 dx

$$=\int_{-\infty}^{\infty} x^2 dx$$

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