62.

Let us first establish the equation of the first principle for an infinitesimal change of the system in question. That the cavity enclosing the radiation has a certain energy we have already ([sect:22.] Sec. 22) deduced from the fact that the energy radiation is propagated with a finite velocity. We shall denote the energy by $U$. Then we have $U=Vu,\mathrm{\backslash Label}[eqn](70)\text{upshape (70)}$ where $u$ the volume density of radiation depends only on the temperature $T$ of the black body at the bottom.

The work done by the system, when the volume $V$ of the cavity increases by $dV$ against the external forces of pressure (weight of the loaded piston), is $pdV$, where $p$ represents Maxwell's radiation pressure [eqn:(66)] (66). This amount of mechanical energy is therefore gained by the surroundings of the system, since the weight is raised. The error made by using the radiation pressure on a stationary surface, whereas the reflecting surface moves during the volume change, is evidently negligible, since the motion may be thought of as taking place with an arbitrarily small velocity.

If, moreover, $Q$ denotes the infinitesimal quantity of heat in mechanical units, which, owing to increased emission, passes from the black body at the bottom to the cavity containing the radiation, the bottom or the heat reservoir connected to it loses this heat $Q$, and its internal energy is decreased by that amount. Hence, according to the first principle of thermodynamics, since the sum of the energy of radiation and the energy of the material bodies remains constant, we have $dU+pdV-Q=0.\mathrm{\backslash Label}[eqn](71)\text{upshape (71)}$

According to the second principle of thermodynamics the cavity containing the radiation also has a definite entropy. For when the heat $Q$ passes from the heat reservoir into the cavity, the entropy of the reservoir decreases, the change being $-\frac{Q}{T}.$

Therefore, since no changes occur in the other bodies–-inasmuch as the rigid absolutely reflecting piston with the weight on it does not change its internal condition with the motion–-there

must somewhere in nature occur a compensation of entropy having at least the value ${\displaystyle \frac{Q}{T}}$, by which the above diminution is compensated, and this can be nowhere except in the entropy of the cavity containing the radiation. Let the entropy of the latter be denoted by $S$.

Now, since the processes described consist entirely of states of equilibrium, they are perfectly reversible and hence there is no increase in entropy. Then we have $dS-\frac{Q}{T}=0,\mathrm{\backslash Label}[eqn](72)\text{upshape (72)}$ or from [eqn:(71)] (71) $dS=\frac{dU+pdV}{T}.\mathrm{\backslash Label}[eqn](73)\text{upshape (73)}$

In this equation the quantities $U$, $p$, $V$, $S$ represent certain properties of the heat radiation, which are completely defined by the instantaneous state of the radiation. Therefore the quantity $T$ is also a certain property of the state of the radiation, i.e., the black radiation in the cavity has a certain temperature $T$ and this temperature is that of a body which is in heat equilibrium with the radiation.