I. Kinematical Part.Show the German source

II. Electrodynamical Part.Show the German source

§ 8. Transformation of the Energy of Light Rays. Theory of the Radiation Pressure Exerted on Perfect Mirrors.Show the German source

Since $A^2/8\pi$ is equal to the light energy per unit volume, we have to regard $A'^2/8\pi$, by the principle of relativity, as the light energy in the moving system.Show the German source Hence $A'^2/A^2$ would be the ratio of the “measured-in-motion” and the “measured-at-rest” energy of a given light complex, if the volume of a light complex were the same measured in $K$ and measured in $k$.Show the German source This, however, is not the case.Show the German source If $a$, $b$, $c$ are the direction cosines of the wave normal of the light in the stationary system, then through the surface elements of the spherical surface moving with the velocity of lightShow the German source $$(x - V a t)^2 + (y - V b t)^2 + (z - V c t)^2 = R^2$$$x$Position in the stationary system$V$Speed of light$a$Direction cosine (stationary system)$t$Time in the stationary system$y$Position in the stationary system$b$Direction cosine (stationary system)$z$Position in the stationary system$c$Direction cosine (stationary system)$R$Light-sphere radiusExplain this equationShow the German source no energy passes; we can therefore say that this surface permanently encloses the same light complex.Show the German source We ask for the quantity of energy which this surface encloses, regarded in the system $k$, that is, for the energy of the light complex relative to the system $k$.Show the German source

The spherical surface is — regarded in the moving system — an ellipsoidal surface, which at the time $\tau = 0$ has the equation:Show the German source $$\left( \beta \xi - a \beta \frac{v}{V} \xi \right)^2 + \left( \eta - b \beta \frac{v}{V} \xi \right)^2 + \left( \zeta - c \beta \frac{v}{V} \xi \right)^2 = R^2 .$$$\beta$Lorentz factor$\xi$Position in the moving system$a$Direction cosine (stationary system)$v$Speed of the moving system$V$Speed of light$\eta$Position in the moving system$b$Direction cosine (stationary system)$\zeta$Position in the moving system$c$Direction cosine (stationary system)$R$Light-sphere radiusExplain this equationShow the German source If $S$ is called the volume of the sphere and $S'$ that of this ellipsoid, then, as a simple calculation shows:Show the German source $$\frac{S'}{S} = \frac{\sqrt{1 - \left( \frac{v}{V} \right)^2}}{1 - \frac{v}{V} \cos \varphi} .$$$S^{\prime }$Volume of the light packet, moving system$S$Volume of the light packet, stationary system$v$Speed of the moving system$V$Speed of light$\phi$Angle to the motion, stationary systemExplain this equationShow the German source If, therefore, we call $E$ the light energy measured in the stationary system and $E'$ the light energy measured in the moving system, enclosed by the surface under consideration, we obtain:Show the German source $$\frac{E'}{E} = \frac{\frac{A'^2}{8\pi} S'}{\frac{A^2}{8\pi} S} = \frac{1 - \frac{v}{V} \cos \varphi}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} ,$$$E^{\prime }$Energy of the light packet, moving system$E$Energy of the light packet, stationary system$A^{\prime }$Light amplitude (moving system)$S^{\prime }$Volume of the light packet, moving system$A$Light amplitude (stationary system)$S$Volume of the light packet, stationary system$v$Speed of the moving system$V$Speed of light$\phi$Angle to the motion, stationary systemExplain this equationShow the German source a formula which for $\varphi = 0$ passes over into the simpler one:Show the German source $$\frac{E'}{E} = \sqrt{\frac{1 - \frac{v}{V}}{1 + \frac{v}{V}}} .$$$E^{\prime }$Energy of the light packet, moving system$E$Energy of the light packet, stationary system$v$Speed of the moving system$V$Speed of lightExplain this equationShow the German source

It is remarkable that the energy and the frequency of a light complex vary with the observer’s state of motion according to the same law.Show the German source

Now let the coordinate plane $\xi = 0$ be a perfectly reflecting surface, at which the plane waves considered in the last section are reflected.Show the German source We ask for the light pressure exerted on the reflecting surface, and for the direction, frequency and intensity of the light after reflection.Show the German source

Let the incident light be defined by the quantities $A$, $\cos \varphi$, $\nu$ (referred to the system $K$).Show the German source Regarded from $k$, the corresponding quantities are:Show the German source $$A' = A \frac{1 - \frac{v}{V} \cos \varphi}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} ,$$$A^{\prime }$Light amplitude (moving system)$A$Light amplitude (stationary system)$v$Speed of the moving system$V$Speed of light$\phi$Angle to the motion, stationary systemExplain this equationShow the German source $$\cos \varphi' = \frac{\cos \varphi - \frac{v}{V}}{1 - \frac{v}{V} \cos \varphi} ,$$${\phi }^{\prime }$Angle to the motion, moving system$\phi$Angle to the motion, stationary system$v$Speed of the moving system$V$Speed of lightExplain this equationShow the German source $$\nu' = \nu \frac{1 - \frac{v}{V} \cos \varphi}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} .$$${\nu }^{\prime }$Frequency, moving system$\nu$Frequency, stationary system$v$Speed of the moving system$V$Speed of light$\phi$Angle to the motion, stationary systemExplain this equationShow the German source For the reflected light we obtain, when we refer the process to the system $k$:Show the German source $$\begin{aligned} A'' &= A' , \\ \cos \varphi'' &= - \cos \varphi' , \\ \nu'' &= \nu' . \end{aligned}$$$A^{\prime \prime }$Reflected light amplitude (moving system)$A^{\prime }$Light amplitude (moving system)${\phi }^{\prime \prime }$Reflected propagation angle (moving system)${\phi }^{\prime }$Angle to the motion, moving system${\nu }^{\prime \prime }$Reflected light frequency (moving system)${\nu }^{\prime }$Frequency, moving systemExplain this equationShow the German source Finally, by transforming back to the stationary system $K$, we obtain for the reflected light:Show the German source $$A''' = A'' \frac{1 + \frac{v}{V} \cos \varphi''}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} = A \frac{1 - 2 \frac{v}{V} \cos \varphi + \left( \frac{v}{V} \right)^2}{1 - \left( \frac{v}{V} \right)^2} ,$$$A^{\prime \prime \prime }$Reflected light amplitude (stationary system)$A^{\prime \prime }$Reflected light amplitude (moving system)$v$Speed of the moving system$V$Speed of light${\phi }^{\prime \prime }$Reflected propagation angle (moving system)$A$Light amplitude (stationary system)$\phi$Angle to the motion, stationary systemExplain this equationShow the German source $$\cos \varphi''' = \frac{\cos \varphi'' + \frac{v}{V}}{1 + \frac{v}{V} \cos \varphi''} = - \frac{\left( 1 + \left( \frac{v}{V} \right)^2 \right) \cos \varphi - 2 \frac{v}{V}}{1 - 2 \frac{v}{V} \cos \varphi + \left( \frac{v}{V} \right)^2} ,$$${\phi }^{\prime \prime \prime }$Reflected propagation angle (stationary system)${\phi }^{\prime \prime }$Reflected propagation angle (moving system)$v$Speed of the moving system$V$Speed of light$\phi$Angle to the motion, stationary systemExplain this equationShow the German source $$\nu''' = \nu'' \frac{1 + \frac{v}{V} \cos \varphi''}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} = \nu \frac{1 - 2 \frac{v}{V} \cos \varphi + \left( \frac{v}{V} \right)^2}{\left( 1 - \frac{v}{V} \right)^2} .$$${\nu }^{\prime \prime \prime }$Reflected frequency${\nu }^{\prime \prime }$Reflected light frequency (moving system)$v$Speed of the moving system$V$Speed of light${\phi }^{\prime \prime }$Reflected propagation angle (moving system)$\nu$Frequency, stationary system$\phi$Angle to the motion, stationary systemExplain this equationShow the German source

The energy (measured in the stationary system) incident on unit area of the mirror per unit time is evidently $A^2/8\pi (V \cos \varphi - v)$.Show the German source The energy leaving unit area of the mirror in unit time is $A'''^2/8\pi (- V \cos \varphi''' + v)$.Show the German source By the energy principle, the difference of these two expressions is the work done by the light pressure in unit time.Show the German source If we set the latter equal to the product $P \cdot v$, where $P$ is the light pressure, we obtain:Show the German source $$P = 2 \frac{A^2}{8\pi} \frac{\left( \cos \varphi - \frac{v}{V} \right)^2}{1 - \left( \frac{v}{V} \right)^2} .$$$P$Radiation pressure on a mirror$A$Light amplitude (stationary system)$\phi$Angle to the motion, stationary system$v$Speed of the moving system$V$Speed of lightExplain this equationShow the German source To a first approximation we obtain, in agreement with experience and with other theoriesShow the German source $$P = 2 \frac{A^2}{8\pi} \cos^2 \varphi .$$$P$Radiation pressure on a mirror$A$Light amplitude (stationary system)$\phi$Angle to the motion, stationary systemExplain this equationShow the German source

By the method employed here, all problems of the optics of moving bodies can be solved.Show the German source The essential point is that the electric and magnetic force of the light which is influenced by a moving body are transformed to a coordinate system at rest relative to the body.Show the German source In this way every problem of the optics of moving bodies is reduced to a series of problems of the optics of bodies at rest.Show the German source

In conclusion I remark that in working on the problem treated here my friend and colleague M. Besso stood loyally by my side, and that I owe him several valuable suggestions.Show the German source

Bern, June 1905.Show the German source

(Received 30 June 1905.)Show the German source