§3 · temperature and entropy

Letters: the paper puts the radiation in a volume v, writes ρ for its density at each frequency ν, and writes φ(ρ, ν) for its entropy per unit volume and unit frequency. This explanation writes $\rho_\nu$ and $s_\nu$ for the same two quantities. The paper takes it as given that the observable properties of the radiation are fixed once ρ is known for all frequencies; a footnote calls this assumption arbitrary, to be kept until experiment forces it to be dropped.

1. Radiation of different frequencies can be separated without doing work and without supplying heat, so entropy and energy are sums over frequencies: $S=v\int_0^\infty\varphi(\rho,\nu)\,d\nu$ and $E=v\int_0^\infty\rho\,d\nu$.
2. Black radiation is the equilibrium state: for the given energy, its entropy is as large as it can be. Take v = 1 and change the spectrum by a small amount $δρ$ at each frequency.
3. The energy must not change: $\int_0^\infty δρ\,d\nu=0$. At a maximum, the entropy must not change to first order either: $\int_0^\infty\frac{\partial\varphi}{\partial\rho}δρ\,d\nu=0$.
4. Both statements must hold for every such change. That is possible only if $\int_0^\infty\left(\frac{\partial\varphi}{\partial\rho}-\lambda\right)δρ\,d\nu=0$ with λ a number that does not depend on ν. So $\partial\varphi/\partial\rho$ has the same value at every frequency.
5. A two-band check of that step: move a small amount of energy q out of band 1 and into band 2, keeping the total fixed. The entropy changes by q times (the derivative in band 2 minus the derivative in band 1). If the two derivatives differed, choosing the sign of q would raise the entropy, so at the maximum they are equal.
6. Now warm black radiation of volume 1 by dT. Its entropy changes by $dS=\int_0^\infty\frac{\partial\varphi}{\partial\rho}d\rho\,d\nu$. Because the derivative is the same at every frequency, it comes outside the integral, and what remains is the change of energy: $dS=\frac{\partial\varphi}{\partial\rho}dE$.
7. dE is the heat supplied, and the warming is reversible, so thermodynamics also gives $dS=\frac{1}{T}dE$.
8. Compare the two expressions for dS. The common derivative is the inverse temperature, which in the explanatory letters is the formula below.

1. Read it in words: at fixed frequency, adding a little energy density raises the entropy density by that amount divided by the temperature.
2. So a known radiation law, which gives T as a function of ρ at each ν, gives $\partial\varphi/\partial\rho$ as a function of ρ. Integrating recovers φ, once one more condition fixes the constant of integration: the paper requires φ to vanish when ρ = 0.
3. The paper credits this reasoning to a celebrated work of W. Wien and includes it for completeness. It uses no picture of how radiation is produced or how it travels.