Page:SahaElectrodynamics.djvu/27

 For the reflected light we obtain, when the process is referred to the system k :—

By means of a back-transformation to the stationary system, we obtain K, for the reflected light :—

$$\left. \begin{array}{c} A'=A\frac{1+\frac{v}{c}\cos\Phi''}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=A\frac{1-2\frac{v}{c}\cos\Phi+\frac{v^{2}}{c^{2}}}{1-\frac{v^{2}}{c^{2}}},\\ \\\cos\Phi'=\frac{\cos\Phi+\frac{v}{c}}{1+\frac{v}{c}\cos\Phi''}=-\frac{\left(1+\frac{v^{2}}{c^{2}}\right)\cos\Phi-2\frac{v}{c}}{1-2\frac{v}{c}\cos\Phi+\frac{v^{2}}{c^{2}}},\\ \nu'=\nu\frac{1+\frac{v}{c}\cos\Phi''}{\sqrt{1-\frac{v^{2}}{c^{2}}}}=\nu\frac{1-2\frac{v}{c}\cos\Phi+\frac{v^{2}}{c^{2}}}{1-\frac{v^{2}}{c^{2}}},\end{array} \right\}$$

The amount or energy falling upon the unit surface of the mirror per unit of time (measured in the stationary system) is $$\frac{A^{2}}{8\pi(c\ \cos\Phi-v)}$$. The amount of energy going away from unit surface of the mirror per unit of time is $$A'^{2}/8\pi(-c\cos\Phi+v)$$. The difference of these two expressions is, according to the Energy principle, the amount of work exerted, by the pressure of light per unit of time. If we put this equal to $$P \cdot v$$, where $$P=\text{pressure of light}$$, we have