184.

Let us now calculate the change in energy of the system of oscillators which has taken place in the same time $d t$ . According to [eqn:(219)] (219), this energy at the time $t$ is $E = N h ν \sum_{1}^{\infty} (n - \frac{1}{2}) w_{n},$ where the quantities $w_{n}$ whose total sum is equal to $1$ represent the densities of distribution characteristic of the state. Hence the energy change in the time $d t$ is $d E = N h ν \sum_{1}^{\infty} (n - \frac{1}{2}) d w_{n} = N h ν \sum_{1}^{\infty} n d w_{n} . \Label [e q n] (340) \upshape (340)$

To calculate $d w_{n}$ we consider the $n$ th region element. All of the oscillators which lie in this region at the time $t$ have, after the lapse of time $τ$ , given by [eqn:(323)] (323), left this region; they have either passed into the $(n + 1)$ st region, or they have performed an emission at the boundary of the two regions. In compensation there have entered $(1 - η) N w_{n - 1}$ oscillators during the time $τ$ , that is, all oscillators which, at the time $t$ , were in the $(n - 1)$ st region element, excepting such as have lost their energy by emission. Thus we obtain for the required change in the time $d t$ $N d w_{n} = \frac{d t}{τ} N ((1 - η) w_{n - 1} - w_{n}) . \Label [e q n] (341) \upshape (341)$

A separate discussion is required for the first region element $n = 1$ . For into this region there enter in the time $τ$ all those oscillators which have performed an emission in this time. Their number is $η (w_{1} + w_{2} + w_{3} + \dots) N = η N .$

Hence we have $N d w_{1} = \frac{d t}{τ} N (η - w_{1}) .$ We may include this equation in the general one [eqn:(341)] (341) if we introduce as a new expression $w_{0} = \frac{η}{1 - η} . \Label [e q n] (342) \upshape (342)$ Then [eqn:(341)] (341) gives, substituting $τ$ from [eqn:(323)] (323), $d w_{n} = \frac{𝖨 d t}{4 h ν L} ((1 - η) w_{n - 1} - w_{n}), \Label [e q n] (343) \upshape (343)$ and the energy change [eqn:(340)] (340) of the system of oscillators becomes $d E = \frac{N 𝖨 d t}{4 L} \sum_{1}^{\infty} n ((1 - η) w_{n - 1} - w_{n}) .$ The sum $\sum$ may be simplified by recalling that

$\sum_{1}^{\infty} n w_{n - 1} & = \sum_{1}^{\infty} (n - 1) w_{n - 1} + \sum_{1}^{\infty} w_{n - 1} & = \sum_{1}^{\infty} n w_{n} + w_{0} + 1 = \sum_{1}^{\infty} n w_{n} + \frac{1}{1 - η} .$

Then we have $d E = \frac{N 𝖨 d t}{4 L} (1 - η \sum_{1}^{\infty} n w_{n}) . \Label [e q n] (344) \upshape (344)$ This expression may be obtained more readily by considering that $d E$ is the difference of the total energy absorbed and the total energy emitted. The former is found from [eqn:(250)] (250), the latter from [eqn:(324)] (324), by taking account of [eqn:(265)] (265).

The principle of the conservation of energy demands that the sum of the energy change [eqn:(339)] (339) of the field of radiation and the energy change [eqn:(344)] (344) of the system of oscillators shall be zero, which, in fact, is quite generally the case, as is seen from the relations [eqn:(320)] (320) and [eqn:(336)] (336).

221