95.

According to the second principle of thermodynamics, the total entropy of radiation of quite arbitrary distribution of energy must remain constant on adiabatic reversible compression. We are now able to give a direct proof of this proposition on the basis of equation [eqn:(119)] (119). For such a process, according to equation [eqn:(113)] (113), the relation holds:

$$\begin{gathered} \delta S\&={\int }_0^{\infty }d\nu (V\delta 𝗌+𝗌\delta V) \\ \&={\int }_0^{\infty }d\nu (V\frac{\partial 𝗌}{\partial 𝗎}\delta 𝗎+𝗌\delta V).\mathrm{\backslash Label}[eqn](120)\text{upshape (120)} \end{gathered}$$

Here, as everywhere, $𝗌$ should be regarded as a function of $𝗎$ and $\nu$, and $\delta \nu =0$.

Now for a reversible adiabatic change of state the relation [eqn:(95)] (95) holds. Let us take from the latter the value of $\delta 𝗎$ and substitute. Then we have $\delta S=\delta V{\int }_0^{\infty }d\nu \{\frac{\partial 𝗌}{\partial 𝗎}(\frac{\nu d𝗎}{3d\nu }-𝗎)+𝗌\}.$ In this equation the differential coefficient of $𝗎$ with respect to $\nu$ refers to the spectral distribution of energy originally assigned arbitrarily and is therefore, in contrast to the partial differential coefficients, denoted by the letter $d$.

Now the complete differential is: $\frac{d𝗌}{d\nu }=\frac{\partial 𝗌}{\partial 𝗎}\frac{d𝗎}{d\nu }+\frac{\partial 𝗌}{\partial \nu }.$ Hence by substitution: $\delta S=\delta V{\int }_0^{\infty }d\nu \{\frac{\nu }{3}(\frac{d𝗌}{d\nu }-\frac{\partial 𝗌}{\partial \nu })-𝗎\frac{\partial 𝗌}{\partial 𝗎}+𝗌\}.\mathrm{\backslash Label}[eqn](121)\text{upshape (121)}$ But from equation [eqn:(119)] (119) we obtain by differentiation $\frac{\partial 𝗌}{\partial 𝗎}=\frac{1}{\nu }\dot{F}\left(\frac{c^3𝗎}{{\nu }^3}\right)\text{and}\frac{\partial 𝗌}{\partial \nu }=\frac{2\nu }{c^3}F\left(\frac{c^3𝗎}{{\nu }^3}\right)-\frac{3𝗎}{{\nu }^2}\dot{F}\left(\frac{c^3𝗎}{{\nu }^3}\right).\mathrm{\backslash Label}[eqn](122)\text{upshape (122)}$ Hence $\nu \frac{\partial 𝗌}{\partial \nu }=2𝗌-3𝗎\frac{\partial 𝗌}{\partial 𝗎}.\mathrm{\backslash Label}[eqn](123)\text{upshape (123)}$ On substituting this in [eqn:(121)] (121), we obtain $\delta S=\delta V{\int }_0^{\infty }d\nu (\frac{\nu }{3}\frac{d𝗌}{d\nu }+\frac{1}{3}𝗌)\mathrm{\backslash Label}[eqn](124)\text{upshape (124)}$ or, $\delta S=\frac{\delta V}{3}[\nu 𝗌]{\mathrm{\backslash Strut}}_0^{\infty }=0,$ as it should be. That the product $\nu 𝗌$ vanishes also for $\nu =\infty$ may be shown just as was done in [sect:83.] Sec. 83 for the product $\nu 𝗎$.