139.

Let us now consider the state of thermodynamic equilibrium of the oscillators. According to the second principle of thermodynamics, the entropy $S$ is in that case a maximum for a given energy $E$. Hence we assume $E$ in [eqn:(219)] (219) as given. Then from [eqn:(179)] (179) we have for the state of equilibrium: $\delta S=0={\sum }_1^{\infty }(\log w_n+1)\delta w_n,$ where according to [eqn:(167)] (167) and [eqn:(219)] (219) ${\sum }_1^{\infty }\delta w_n=0\text{and}{\sum }_1^{\infty }(n-\frac{1}{2})\delta w_n=0.$ From these relations we find: $\log w_n+\beta n+\text{const}.=0$ or $w_n=\alpha {\gamma }^n.\mathrm{\backslash Label}[eqn](220)\text{upshape (220)}$ The values of the constants $\alpha$ and $\gamma$ follow from equations [eqn:(167)] (167) and [eqn:(219)] (219): $\alpha =\frac{2Nh\nu }{2E-Nh\nu }\gamma =\frac{2E-Nh\nu }{2E+Nh\nu }.\mathrm{\backslash Label}[eqn](221)\text{upshape (221)}$ Since $w_n$ is essentially positive it follows that equilibrium is not possible in the system of oscillators considered unless the total energy $E$ has a greater value than ${\displaystyle \frac{Nh\nu }{2}}$, that is unless the mean energy of the oscillators is at least ${\displaystyle \frac{h\nu }{2}}$. This, according to [eqn:(218)] (218), is the mean energy of the oscillators lying in the first region element. In fact, in this extreme case all $N$ oscillators lie in the first region element, the region of smallest energy; within this element they are arranged uniformly.

The entropy $S$ of the system, which is in thermodynamic equilibrium, is found by combining [eqn:(173)] (173) with [eqn:(220)] (220) and [eqn:(221)] (221)

$$\begin{gathered} S=kN\backslash \{(\frac{E}{Nh\nu }+\frac{1}{2})\log (\frac{E}{Nh\nu }+\frac{1}{2}) \\ -(\frac{E}{Nh\nu }-\frac{1}{2})\log (\frac{E}{Nh\nu }-\frac{1}{2})\backslash \}.\mathrm{\backslash Label}[eqn](222)\text{upshape (222)} \end{gathered}$$