Blackbody Radiation - Entropy and Internal Energy

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The discussion focuses on deriving expressions for the entropy and internal energy of black body radiation using relevant equations. The total free energy is expressed as an integral involving temperature and other constants, while the relationship between free energy, internal energy, and entropy is highlighted. A u-substitution is suggested to simplify the integral for free energy, which can be computed using Wolfram Alpha. Participants emphasize the importance of correctly identifying the limits of integration and the independence of variables from photon frequency. The conversation illustrates the challenges faced in calculating these thermodynamic properties.
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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.
 
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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
 
Don't forget the limits of integration. The variables are independent of photon's (angular) frequency.
 
kreil said:
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...

dextercioby said:
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.
 
Code:
\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.
 

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