Activities
These notes from the fourth week of https://paradigms.oregonstate.edu/courses/ph441 cover blackbody radiation and the Planck distribution. They include a number of small group activities.
This small group activity has students reasoning about how the Planck distribution shifts when the temperature is doubled. This leads to a qualitative argument for the Stefan-Boltzmann law.
Problem
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.
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!
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}
Problem
(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
\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 \(j\);
Solve for the relationship between pressure and internal energy.
This lecture is one step in motivating the form of the Planck distribution.
In this activity, students apply the Stefan-Boltzmann equation and the principle of energy balance in steady state to find the steady state temperature of a black object in near-Earth orbit.
This very short lecture introduces Wein's displacement law.
Problem
Calculate the temperature of the surface of the Earth on the assumption that as a black body in thermal equilibrium it reradiates as much thermal radiation as it receives from the Sun. Assume also that the surface of the Earth is a constant temperature over the day-night cycle. Use the sun's surface temperature \(T_{\odot}=5800\text{K}\); and the sun's radius \(R_{\odot}=7\times 10^{10}\text{cm}\); and the Earth-Sun distance of \(1.5\times 10^{13}\text{cm}\).
Problem
A black (nonreflective) sheet of metal at high temperature \(T_h\) is parallel to a cold black sheet of metal at temperature \(T_c\). Each sheet has an area \(A\) which is much greater than the distance between them. The sheets are in vacuum, so energy can only be transferred by radiation.
Solve for the net power transferred between the two sheets.
- A third black metal sheet is inserted between the other two and is allowed to come to a steady state temperature \(T_m\). Find the temperature of the middle sheet, and solve for the new net power transferred between the hot and cold sheets. This is the principle of the heat shield, and is part of how the James Web telescope shield works.
Optional: Find the power through an \(N\)-layer sandwich.
In this small group activity, students work out the steady state temperature of an object absorbing and emitting blackbody radiation.