180.

The total energy emitted in the time element $dt$ by all $N$ oscillators is found from the consideration that a single oscillator, according to [eqn:(249)] (249), takes up an energy element $h\nu$ during the time $\frac{4h\nu L}{𝖨}=\tau ,\mathrm{\backslash Label}[eqn](323)\text{upshape (323)}$

and hence has a chance to emit once, the probability being $\eta$. We shall assume that the intensity $𝖨$ of the exciting vibration does not change appreciably in the time $\tau$. Of the $N{w}_n$ oscillators which at the time $t$ are in the $n$th region element a number $N{w}_n\eta$ will emit during the time $\tau$, the energy emitted by each being $nh\nu$. From [eqn:(323)] (323) we see that the energy emitted by all oscillators during the time element $dt$ is $\sum N{w}_n\eta nh\nu \frac{dt}{\tau }=\frac{N\eta 𝖨dt}{4L}\sum n{w}_n,$ or, according to [eqn:(265)] (265), $\frac{N(1-\eta )dt}{4pL}\sum n{w}_n.\mathrm{\backslash Label}[eqn](324)\text{upshape (324)}$

From this the energy emitted within the conical element $d\Omega$ may be calculated by considering that, in the state of thermodynamic equilibrium, the energy emitted in every conical element is equal to the energy absorbed and that, in the general case, the energy emitted in a certain direction is independent of the energy simultaneously absorbed. For the stationary state we have from [eqn:(160)] (160) and [eqn:(265)] (265) $𝖪={𝖪}^{\prime }=\frac{3c}{32{\pi }^2}𝖨=\frac{3c}{32{\pi }^2}\frac{1-\eta }{p\eta }\mathrm{\backslash Label}[eqn](325)\text{upshape (325)}$ and further from [eqn:(271)] (271) and [eqn:(265)] (265) $w_n=\frac{1}{p𝖨}{\left(\frac{p𝖨}{1+p𝖨}\right)}^n=\eta (1-\eta {)}^{n-1},\mathrm{\backslash Label}[eqn](326)\text{upshape (326)}$ and hence $\sum n{w}_n=\eta \sum n(1-\eta {)}^{n-1}=\frac{1}{\eta }.\mathrm{\backslash Label}[eqn](327)\text{upshape (327)}$ Thus the energy emitted [eqn:(324)] (324) becomes $\frac{N(1-\eta )dt}{4Lp\eta }.\mathrm{\backslash Label}[eqn](328)\text{upshape (328)}$ This is, in fact, equal to the total energy absorbed, as may be found by integrating the expression [eqn:(322)] (322) over all conical elements $d\Omega$ and taking account of [eqn:(325)] (325).

Within the conical element $d\Omega$ the energy emitted or absorbed will then be $\frac{\pi Ndt}{c}{\sin }^2\theta 𝖪d\Omega ,$ or, from [eqn:(325)] (325), [eqn:(327)] (327) and [eqn:(268)] (268), $\frac{\pi h{\nu }^3(1-\eta )N}{c^{3}L}\sum n{w}_n{\sin }^2\theta d\Omega dt,\mathrm{\backslash Label}[eqn](329)\text{upshape (329)}$ and this is the general expression for the energy emitted by the system of oscillators in the time element $dt$ within the conical element $d\Omega$, as is seen by comparison with [eqn:(324)] (324).