(modified from K&K 4.6) We discussed in class that \begin{align} p &= -\left(\frac{\partial F}{\partial V}\right)_T \end{align} Use this relationship to show that

1. \begin{align} p &= -\sum_j \langle n_j\rangle\hbar \left(\frac{d\omega_j}{dV}\right), \end{align} where \(\langle n_j\rangle\) is the number of photons in the mode \(j\);

2. Solve for the relationship between pressure and internal energy.

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 zero-point 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.

1. 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(1-e^{-\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!

2. 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}

3. 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}

1. Solve for the heat capacity of a vacuum, given the above, and assuming that photons represent all the energy present in vacuum.
2. Compare the heat capacity of vacuum at room temperature with the heat capacity of an equal volume of water.

In our week on radiation, we saw that the Helmholtz free energy of a box of radiation at temperature \(T\) is \begin{align} F &= -8\pi \frac{V(kT)^4}{h^3c^3}\frac{\pi^4}{45} \end{align} From this we also found the internal energy and entropy \begin{align} U &= 24\pi \frac{(kT)^4}{h^3c^3}\frac{\pi^4}{45} V \\ S &= 32\pi kV\left(\frac{kT}{hc}\right)^3 \frac{\pi^4}{45} \end{align} Given these results, let us consider a Carnot engine that uses an empty metalic piston (i.e. a photon gas).

1. Given \(T_H\) and \(T_C\), as well as \(V_1\) and \(V_2\) (the two volumes at \(T_H\)), determine \(V_3\) and \(V_4\) (the two volumes at \(T_C\)).

2. What is the heat \(Q_H\) taken up and the work done by the gas during the first isothermal expansion? Are they equal to each other, as for the ideal gas?

3. Does the work done on the two isentropic stages cancel each other, as for the ideal gas?

4. Calculate the total work done by the gas during one cycle. Compare it with the heat taken up at \(T_H\) and show that the energy conversion efficiency is the Carnot efficiency.