By means of a back-transformation to the stationary system, we obtain K,
for the reflected light:—
$$ A''' = A'' \frac{1 + \frac{v}{c}\cos \Phi''}{\sqrt{1 -
\frac{v^2}{c^2}}} = A \frac{1 - 2\frac{v}{c} \cos \Phi +
\frac{v^2}{c^2}}{1 - \frac{v^2}{c^2}} $$ ,
$$ \cos \Phi''' = \frac{\cos \Phi'' + \frac{v}{c}}{1 + \frac{v}{c}\cos
\Phi''} = - \frac{(1 + \frac{v^2}{c^2}) \cos \Phi - 2 \frac{v}{c}} {1 -
2 \frac{v}{c} \cos \Phi + \frac {v^2}{c^2}} $$ ,
$$ \nu''' = \nu'' \frac{1 + \frac{v}{c} \cos \Phi''}{\sqrt{1 -
\frac{v^2}{c^2}}} = \nu \frac{1 - 2 \frac{v}{c} \cos \Phi +
\frac{v^2}{c^2}} {(1 - \frac{v}{c})^2} $$
The amount or energy falling upon the unit surface of the mirror per
unit of time (measured in the stationary system) is A²/(8π (c cos Φ -
_v_)). The amount of energy going away from unit surface of the mirror
per unit of time is A‴²/(8π (-c cos Φ″ + _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._v_, where P = pressure of light, we have
$$ P = 2 \frac{A^2}{8\pi} \frac{(\cos \Phi - \frac{v}{c})^2} {1 -
(\frac{v}{c})^2} $$