A″ = A′, cos Φ″ = -cos Φ″, ν″ = ν′ 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 $$ P = 2 \frac{A^2}{8\pi} \frac{(\cos \Phi - \frac{v}{c})^2} {1 - (\frac{v}{c})^2} $$ As a first approximation, we obtain A² P = 2 -- cos² Φ 8π which is in accordance with facts, and with other theories. All problems of optics of moving bodies can be solved after the method used here. The essential point is, that the electric and magnetic forces of light, which are influenced by a moving body, should be transformed to a system of co-ordinates which is stationary relative to the body. In this way, every problem of the optics of moving bodies would be reduced to a series of problems of the optics of stationary bodies. § 9. Transformation of the Maxwell-Hertz Equations. Let us start from the equations:— $$ \frac{1}{c}(\rho u_{x} + \frac{\partial X}{\partial t}) = \frac{\partial N}{\partial y} - \frac{\partial M}{\partial z} $$ $$ \frac{1}{c}(\rho u_{y} + \frac{\partial Y}{\partial t}) = \frac{\partial L}{\partial z} - \frac{\partial N}{\partial x} $$ $$ \frac{1}{c}(\rho u_{z} + \frac{\partial Z}{\partial t}) = \frac{\partial M}{\partial x} - \frac{\partial L}{\partial y} $$ $$ \frac{1}{c} \frac{\partial L}{\partial t} = \frac{\partial Y}{\partial z} - \frac{\partial Z}{\partial y} $$ $$ \frac{1}{c} \frac{\partial M}{\partial t} = \frac{\partial Z}{\partial x} - \frac{\partial X}{\partial z} $$ $$ \frac{1}{c} \frac{\partial N}{\partial t} = \frac{\partial X}{\partial y} - \frac{\partial Y}{\partial x} $$ where $$ \rho = \frac{\partial X}{\partial x} + \frac{\partial Y}{\partial y} + \frac{\partial Z}{\partial z} $$