130.

Let us consider the state of thermodynamic equilibrium. According to the second principle of thermodynamics this state is distinguished from all others by the fact that, for a given volume $V$ and a given energy $E$ of the gas, the entropy $S$ is a maximum. Let us then regard the volume $V=\iiint dxdydz\mathrm{\backslash Label}[eqn](182)\text{upshape (182)}$ and the energy $E$ of the gas as given. The condition for equilibrium is $\delta S=0$, or, according to [eqn:(179)] (179), $\sum (\log w_1+1)\delta w_1=0,$ and this holds for any variations of the distribution densities whatever, provided that, according to [eqn:(167)] (167) and [eqn:(180)] (180), they satisfy the conditions $\sum \delta w_1=0\text{and}\sum ({\xi }_1^2+{\eta }_1^2+{\zeta }_1^2)\delta w_1=0.$ This gives us as the necessary and sufficient condition for thermodynamic equilibrium for every separate distribution density $w$: $\log w+\beta ({\xi }^2+{\eta }^2+{\zeta }^2)+\text{const}.=0$ or $w=\alpha e^{-\beta ({\xi }^2+{\eta }^2+{\zeta }^2)},\mathrm{\backslash Label}[eqn](183)\text{upshape (183)}$ where $\alpha$ and $\beta$ are constants. Hence in the state of equilibrium the distribution of the molecules in space is independent of $x$, $y$, $z$, that is, macroscopically uniform, and the distribution of velocities is the well-known one of Maxwell.