Let us now ask ourselves about the composition of these waves, when they are investigated by an observer at rest in a moving medium _k_:—By applying the equations of transformation obtained in §6 for the electric and magnetic forces, and the equations of transformation obtained in § 3 for the co-ordinates, and time, we obtain immediately:— X′ = X₀ sin Φ′ v Y′ = β(Y₀ - --- N₀) sin Φ′ c v Z′ = β(Z₀ - --- M₀) sin Φ′ c L′ = L₀ sin Φ′ v M′ = β(M₀ - --- Z₀) sin Φ′ c v N′ = β(N₀ - --- Y₀) sin Φ′ c l′ξ + m′η + n′ζ Φ′ = ω′(t - --------------- ) c where $$ \omega' = \omega \beta (1 - \frac {lv}{c}) $$ , $$ l' = \frac {l - \frac {v}{c}}{1 - \frac {lv}{c}} $$ , $$ m' = \frac {m}{\beta (1 - \frac {lv}{c})} $$ , $$ n' = \frac {n}{\beta (1 - \frac {lv}{c})} $$ From the equation for ω′ it follows:—If an observer moves with the velocity _v_ relative to an infinitely distant source of light emitting waves of frequency ν, in such a manner that the line joining the source of light and the observer makes an angle of Φ with the velocity of the observer referred to a system of co-ordinates which is stationary with regard to the source, then the frequency ν′ which is perceived by the observer is represented by the formula $$ \nu' = \nu \frac {1 - cos \Phi \frac {v}{c}} {\sqrt {1 - \frac {v^2}{c^2}}} $$ This is Döppler’s principle for any velocity. If Φ = 0, then the equation takes the simple form $$ \nu' = \nu (\frac {1 - \frac {v}{c}}{1 + \frac {v}{c}})^{\frac {1}{2}} $$ We see that—contrary to the usual conception—ν = ∞, for _v_ = -_c_. If Φ′ = angle between the wave-normal (direction of the ray) in the moving system, and the line of motion of the observer, the equation for _l´_ takes the form $$ \cos \Phi' = \frac {\cos \Phi - \frac {v}{c}} {1 - \frac {v}{c} \cos \Phi} $$ This equation expresses the law of observation in its most general form. If Φ = π/2, the equation takes the simple form v cos Φ′ = --- c We have still to investigate the amplitude of the waves, which occur in these equations. If A and A′ be the amplitudes in the stationary and the moving systems (either electrical or magnetic), we have $$ A'^2 = A^2 \frac {(1 - \frac {v}{c} \cos \Phi)^2} {1 - \frac {v^2}{c^2}} $$ If Φ = 0, this reduces to the simple form $$ A'^2 = A^2 \frac {1 - \frac {v}{c}} {1 + \frac {v}{c}} $$