81.

In calculating the radiant energy which enters the interval $(ν, d ν)$ in the time $δ t$ on account of reflection from the moving reflector, the rays falling on the reflector at different angles of incidence must be considered separately. Since in the case of a positive $v$ , the frequency is increased by the reflection, the rays which must be considered have, at the time $t$ , the frequency $ν_{1} < ν$ . If we now consider at the time $t$ a monochromatic pencil of frequency $(ν_{1}, d ν_{1})$ , falling on the reflector at an angle of incidence $θ$ , a necessary and sufficient condition for its entrance, by reflection, into the interval $(ν, d ν)$ is $ν = ν_{1} (1 + \frac{2 v \cos θ}{c}) and d ν = d ν_{1} (1 + \frac{2 v \cos θ}{c}) .$ These relations are obtained by substituting $ν_{1}$ and $ν$ respectively in the equations [eqn:(83)] (83) and [eqn:(86)] (86) in place of the frequencies before and after reflection $ν$ and $ν^{'}$ .

The energy which this pencil carries into the interval $(ν_{1}, d ν)$ in the time $δ t$ is obtained from [eqn:(89)] (89), likewise by substituting $ν_{1}$ for $ν$ . It is $2 𝖪_{ν_{1}} d σ \cos θ d Ω d ν_{1} (1 + \frac{2 v \cos θ}{c}) δ t = 2 𝖪_{ν_{1}} d σ \cos θ d Ω d ν δ t .$ Now we have $𝖪_{ν_{1}} = 𝖪_{ν} + (ν_{1} - ν) \frac{\partial 𝖪}{\partial ν} + \dots$ where we shall assume $\frac{\partial 𝖪}{\partial ν}$ to be finite.

Hence, neglecting small quantities of higher order, $𝖪_{ν_{1}} = 𝖪_{ν} - \frac{2 ν v \cos θ}{c} \frac{\partial 𝖪}{\partial ν} .$ Thus the energy required becomes $2 d σ (𝖪_{ν} - \frac{2 ν v \cos θ}{c} \frac{\partial 𝖪}{\partial ν}) \sin θ \cos θ d θ d ϕ d ν δ t,$ and, integrating this expression as above, with respect to $d σ$ , $ϕ$ , and $θ$ , the total radiant energy which enters into the interval $(ν, d ν)$ in the time $δ t$ becomes $2 π F (𝖪_{ν} - \frac{4}{3} \frac{ν v}{c} \frac{\partial 𝖪}{\partial ν}) d ν δ t . \Label [e q n] (93) \upshape (93)$