As discussed in
class, we can consider a black body as a large box with a small hole
in it. If we treat the large box a metal cube with side length \(L\)
and metal walls, the frequency of each normal mode will be given by:
\begin{align}
\omega_{n_xn_yn_z} &= \frac{\pi c}{L}\sqrt{n_x^2 + n_y^2 + n_z^2}
\end{align} where each of \(n_x\), \(n_y\), and \(n_z\) will have
positive integer values. This simply comes from the fact that a half
wavelength must fit in the box. There is an additional quantum number
for polarization, which has two possible values, but does not affect
the frequency. Note that in this problem I'm using different
boundary conditions from what I use in class. It is worth learning to
work with either set of quantum numbers. Each normal mode is a
harmonic oscillator, with energy eigenstates \(E_n = n\hbar\omega\)
where we will not include the zeropoint energy
\(\frac12\hbar\omega\), since that energy cannot be extracted from the
box. (See the
Casimir effect
for an example where the zero point energy of photon modes does have
an effect.)
 Note

This is a slight approximation, as the boundary conditions for light
are a bit more complicated. However, for large \(n\) values this gives
the correct result.
Show that the free energy is given by \begin{align}
F &= 8\pi \frac{V(kT)^4}{h^3c^3}
\int_0^\infty \ln\left(1e^{\xi}\right)\xi^2d\xi
\\
&= \frac{8\pi^5}{45} \frac{V(kT)^4}{h^3c^3}
\\
&= \frac{\pi^2}{45} \frac{V(kT)^4}{\hbar^3c^3}
\end{align} provided the box is big enough that
\(\frac{\hbar c}{LkT}\ll 1\). Note that you may end up with a
slightly different dimensionless integral that numerically evaluates
to the same result, which would be fine. I also do not expect you to
solve this definite integral analytically, a numerical confirmation
is fine. However, you must manipulate your integral until it
is dimensionless and has all the dimensionful quantities removed
from it!
Show that the entropy of this box full of photons at temperature
\(T\) is \begin{align}
S &= \frac{32\pi^5}{45} k V \left(\frac{kT}{hc}\right)^3
\\
&= \frac{4\pi^2}{45} k V \left(\frac{kT}{\hbar c}\right)^3
\end{align}
Show that the internal energy of this box full of photons at
temperature \(T\) is \begin{align}
\frac{U}{V} &= \frac{8\pi^5}{15}\frac{(kT)^4}{h^3c^3}
\\
&= \frac{\pi^2}{15}\frac{(kT)^4}{\hbar^3c^3}
\end{align}