I. Kinematical Part.Show the German source

§ 3. Theory of the Transformation of Coordinates and Times from the Stationary System to a System in Uniform Translational Motion Relative to It.Show the German source

Let there be given in the “stationary” space two coordinate systems, that is, two systems each of three rigid material lines, perpendicular to one another and issuing from one point.Show the German source Let the $X$-axes of the two systems coincide, and their $Y$- and $Z$-axes be respectively parallel.Show the German source Let each system be provided with a rigid measuring rod and a number of clocks, and let the two measuring rods, and all the clocks of the two systems, be exactly alike.Show the German source

Now to the origin of one of the two systems ($k$) let a (constant) velocity $v$ be imparted in the direction of increasing $x$ of the other, stationary system ($K$), a velocity which is to be communicated also to the coordinate axes, the measuring rod in question and the clocks.Show the German source To each time $t$ of the stationary system $K$ there then corresponds a definite position of the axes of the moving system, and for reasons of symmetry we are entitled to assume that the motion of $k$ can be such that the axes of the moving system are parallel to the axes of the stationary system at the time $t$ (by “$t$” a time of the stationary system is always meant).Show the German source

We now imagine space to be measured out both from the stationary system $K$ by means of the stationary measuring rod and from the moving system $k$ by means of the measuring rod moving with it, and the coordinates $x$, $y$, $z$ and $\xi$, $\eta$, $\zeta$ respectively to be thus obtained.Show the German source Further, let the time $t$ of the stationary system be determined, by means of the clocks at rest in the stationary system and by light signals in the manner given in § 1, for all points of the latter at which there are clocks; likewise let the time $\tau$ of the moving system be determined for all points of the moving system at which there are clocks at rest relative to it, by applying the method of light signals given in § 1 between the points at which these clocks are located.Show the German source

To every system of values $x$, $y$, $z$, $t$ which completely determines the place and time of an event in the stationary system there belongs a system of values fixing that event relative to the system $k$, namely $\xi$, $\eta$, $\zeta$, $\tau$, and the task now is to find the system of equations connecting these quantities.Show the German source

In the first place it is clear that the equations must be linear on account of the properties of homogeneity which we attribute to space and time.Show the German source

If we put $x' = x - v t$, it is clear that to a point at rest in the system $k$ there belongs a definite system of values $x'$, $y$, $z$, independent of time.Show the German source We first determine $\tau$ as a function of $x'$, $y$, $z$ and $t$.Show the German source To this end we have to express in equations that $\tau$ is nothing other than the totality of the readings of clocks at rest in the system $k$ which have been made synchronous according to the rule given in § 1.Show the German source

From the origin of the system $k$ let a ray of light be sent at the time $\tau_0$ along the $X$-axis to $x'$, and from there be reflected at the time $\tau_1$ towards the origin of coordinates, where it arrives at the time $\tau_2$; we must then have:Show the German source $$\frac{1}{2}(\tau_0 + \tau_2) = \tau_1$$${\tau }_0$Time in the moving system${\tau }_2$Time in the moving system${\tau }_1$Time in the moving systemExplain this equationShow the German source or, if we add the arguments of the function $\tau$ and apply the principle of the constancy of the velocity of light in the stationary system:Show the German source $$\begin{aligned} &\frac{1}{2}\left[ \tau(0,0,0,t) + \tau\left( 0,0,0,\left\{ t + \frac{x'}{V-v} + \frac{x'}{V+v} \right\} \right) \right] \\ &= \tau\left( x',0,0,t + \frac{x'}{V-v} \right). \end{aligned}$$$\tau$Time in the moving system$t$Time in the stationary system$x^{\prime }$Auxiliary Galilean coordinate: x'$V$Speed of light$v$Speed of the moving systemExplain this equationShow the German source Hence, if $x'$ is chosen infinitely small, it follows that:Show the German source $$\frac{1}{2}\left( \frac{1}{V-v} + \frac{1}{V+v} \right) \frac{\partial \tau}{\partial t} = \frac{\partial \tau}{\partial x'} + \frac{1}{V-v} \frac{\partial \tau}{\partial t} ,$$$V$Speed of light$v$Speed of the moving system$\tau$Time in the moving system$t$Time in the stationary system$x^{\prime }$Auxiliary Galilean coordinate: x'Explain this equationShow the German source orShow the German source $$\frac{\partial \tau}{\partial x'} + \frac{v}{V^2 - v^2} \frac{\partial \tau}{\partial t} = 0 .$$$\tau$Time in the moving system$x^{\prime }$Auxiliary Galilean coordinate: x'$v$Speed of the moving system$V$Speed of light$t$Time in the stationary systemExplain this equationShow the German source

It is to be noted that instead of the origin of coordinates we could have chosen any other point as the starting point of the ray of light, and the equation just obtained therefore holds for all values of $x'$, $y$, $z$.Show the German source

An analogous consideration — applied to the $H$- and $Z$-axes — gives, if it is borne in mind that light always propagates along these axes, viewed from the stationary system, with the velocity $\sqrt{V^2 - v^2}$:Show the German source $$\begin{aligned} \frac{\partial \tau}{\partial y} &= 0 \\ \frac{\partial \tau}{\partial z} &= 0 . \end{aligned}$$$\tau$Time in the moving system$y$Position in the stationary system$z$Position in the stationary systemExplain this equationShow the German source From these equations it follows, since $\tau$ is a linear function, that:Show the German source $$\tau = a\left( t - \frac{v}{V^2 - v^2} x' \right) ,$$$\tau$Time in the moving system$a$Undetermined scale$t$Time in the stationary system$v$Speed of the moving system$V$Speed of light$x^{\prime }$Auxiliary Galilean coordinate: x'Explain this equationShow the German source where $a$ is a function $\varphi(v)$ as yet unknown, and where for brevity it is assumed that at the origin of $k$, for $\tau = 0$, $t = 0$.Show the German source

With the help of this result it is easy to determine the quantities $\xi$, $\eta$, $\zeta$, by expressing in equations that light (as the principle of the constancy of the velocity of light in conjunction with the principle of relativity requires) also propagates with the velocity $V$ when measured in the moving system.Show the German source For a ray of light emitted at the time $\tau = 0$ in the direction of increasing $\xi$ we have:Show the German source $$\xi = V \tau ,$$$\xi$Position in the moving system$V$Speed of light$\tau$Time in the moving systemExplain this equationShow the German source orShow the German source $$\xi = a V \left( t - \frac{v}{V^2 - v^2} x' \right) .$$$\xi$Position in the moving system$a$Undetermined scale$V$Speed of light$t$Time in the stationary system$v$Speed of the moving system$x^{\prime }$Auxiliary Galilean coordinate: x'Explain this equationShow the German source But the ray of light moves relative to the origin of $k$, measured in the stationary system, with the velocity $V-v$, so that we have:Show the German source $$\frac{x'}{V-v} = t .$$$x^{\prime }$Auxiliary Galilean coordinate: x'$V$Speed of light$v$Speed of the moving system$t$Time in the stationary systemExplain this equationShow the German source If we substitute this value of $t$ in the equation for $\xi$, we obtain:Show the German source $$\xi = a \frac{V^2}{V^2 - v^2} x' .$$$\xi$Position in the moving system$a$Undetermined scale$V$Speed of light$v$Speed of the moving system$x^{\prime }$Auxiliary Galilean coordinate: x'Explain this equationShow the German source In an analogous way, by considering rays of light moving along the two other axes, we find:Show the German source $$\eta = V \tau = a V \left( t - \frac{v}{V^2 - v^2} x' \right) ,$$$\eta$Position in the moving system$V$Speed of light$\tau$Time in the moving system$a$Undetermined scale$t$Time in the stationary system$v$Speed of the moving system$x^{\prime }$Auxiliary Galilean coordinate: x'Explain this equationShow the German source whereShow the German source $$\frac{y}{\sqrt{V^2 - v^2}} = t ; \quad x' = 0 ;$$$y$Position in the stationary system$V$Speed of light$v$Speed of the moving system$t$Time in the stationary system$x^{\prime }$Auxiliary Galilean coordinate: x'Explain this equationShow the German source henceShow the German source $$\eta = a \frac{V}{\sqrt{V^2 - v^2}} y$$$\eta$Position in the moving system$a$Undetermined scale$V$Speed of light$v$Speed of the moving system$y$Position in the stationary systemExplain this equationShow the German source andShow the German source $$\zeta = a \frac{V}{\sqrt{V^2 - v^2}} z .$$$\zeta$Position in the moving system$a$Undetermined scale$V$Speed of light$v$Speed of the moving system$z$Position in the stationary systemExplain this equationShow the German source

If we substitute for $x'$ its value, we obtain:Show the German source $$\begin{aligned} \tau &= \varphi(v) \beta \left( t - \frac{v}{V^2} x \right) , \\ \xi &= \varphi(v) \beta (x - v t), \\ \eta &= \varphi(v) y, \\ \zeta &= \varphi(v) z, \end{aligned}$$$\tau$Time in the moving system$\phi$Unknown scale factor: phi(v)$v$Speed of the moving system$\beta$Lorentz factor$t$Time in the stationary system$V$Speed of light$x$Position in the stationary system$\xi$Position in the moving system$\eta$Position in the moving system$y$Position in the stationary system$\zeta$Position in the moving system$z$Position in the stationary systemExplain this equationShow the German source whereShow the German source $$\beta = \frac{1}{\sqrt{1 - \left( \frac{v}{V} \right)^2}}$$$\beta$Lorentz factor$v$Speed of the moving system$V$Speed of lightExplain this equationShow the German source and $\varphi$ is a function of $v$ as yet unknown.Show the German source If no presupposition whatever is made about the initial position of the moving system and about the zero point of $\tau$, an additive constant is to be added to the right-hand side of each of these equations.Show the German source

We now have to prove that every ray of light, measured in the moving system, propagates with the velocity $V$ if, as we have assumed, this is the case in the stationary system; for we have not yet furnished the proof that the principle of the constancy of the velocity of light is compatible with the principle of relativity.Show the German source

At the time $t = \tau = 0$ let a spherical wave be emitted from the origin of coordinates, common at that time to both systems, a wave which propagates in the system $K$ with the velocity $V$.Show the German source If $(x, y, z)$ is a point just reached by this wave, we therefore haveShow the German source $$x^2 + y^2 + z^2 = V^2 t^2 .$$$x$Position in the stationary system$y$Position in the stationary system$z$Position in the stationary system$V$Speed of light$t$Time in the stationary systemExplain this equationShow the German source

We transform this equation with the help of our transformation equations and obtain, after a simple calculation:Show the German source $$\xi^2 + \eta^2 + \zeta^2 = V^2 \tau^2 .$$$\xi$Position in the moving system$\eta$Position in the moving system$\zeta$Position in the moving system$V$Speed of light$\tau$Time in the moving systemExplain this equationShow the German source

The wave under consideration is therefore, viewed in the moving system too, a spherical wave with the velocity of propagation $V$.Show the German source It is thereby shown that our two fundamental principles are compatible with each other.Show the German source

In the transformation equations developed there still occurs an unknown function $\varphi$ of $v$, which we now wish to determine.Show the German source

For this purpose we introduce a third coordinate system $K'$, which relative to the system $k$ is in such a motion of parallel translation, parallel to the $\Xi$-axis, that its origin moves with the velocity $-v$ on the $\Xi$-axis.Show the German source At the time $t = 0$ let all three origins of coordinates coincide, and for $t = x = y = z = 0$ let the time $t'$ of the system $K'$ be zero.Show the German source We call $x'$, $y'$, $z'$ the coordinates measured in the system $K'$, and by twofold application of our transformation equations we obtain:Show the German source $$\begin{aligned} t' &= \varphi(-v) \beta(-v) \left\{ \tau + \frac{v}{V^2} \xi \right\} &&= \varphi(v) \varphi(-v) t , \\ x' &= \varphi(-v) \beta(-v) \{ \xi + v \tau \} &&= \varphi(v) \varphi(-v) x , \\ y' &= \varphi(-v) \eta &&= \varphi(v) \varphi(-v) y , \\ z' &= \varphi(-v) \zeta &&= \varphi(v) \varphi(-v) z . \end{aligned}$$$t^{\prime }$Time in the stationary system$\phi$Unknown scale factor: phi(v)$v$Speed of the moving system$\beta$Lorentz factor$\tau$Time in the moving system$V$Speed of light$\xi$Position in the moving system$t$Time in the stationary system$x^{\prime }$Position in the stationary system$x$Position in the stationary system$y^{\prime }$Position in the stationary system$\eta$Position in the moving system$y$Position in the stationary system$z^{\prime }$Position in the stationary system$\zeta$Position in the moving system$z$Position in the stationary systemExplain this equationShow the German source

Since the relations between $x'$, $y'$, $z'$ and $x$, $y$, $z$ do not contain the time $t$, the systems $K$ and $K'$ are at rest relative to each other, and it is clear that the transformation from $K$ to $K'$ must be the identity transformation.Show the German source We therefore have:Show the German source $$\varphi(v) \varphi(-v) = 1 .$$$\phi$Unknown scale factor: phi(v)$v$Speed of the moving systemExplain this equationShow the German source We now inquire into the meaning of $\varphi(v)$.Show the German source We consider the portion of the $H$-axis of the system $k$ which lies between $\xi = 0$, $\eta = 0$, $\zeta = 0$ and $\xi = 0$, $\eta = l$, $\zeta = 0$.Show the German source This portion of the $H$-axis is a rod moving relative to the system $K$ with the velocity $v$ perpendicular to its axis, whose ends have in $K$ the coordinates:Show the German source $$x_1 = v t, \quad y_1 = \frac{l}{\varphi(v)}, \quad z_1 = 0$$$x_1$Position in the stationary system$v$Speed of the moving system$t$Time in the stationary system$y_1$Position in the stationary system$l$Rest length$\phi$Unknown scale factor: phi(v)$z_1$Position in the stationary systemExplain this equationShow the German source andShow the German source $$x_2 = v t, \quad y_2 = 0, \quad z_2 = 0 .$$$x_2$Position in the stationary system$v$Speed of the moving system$t$Time in the stationary system$y_2$Position in the stationary system$z_2$Position in the stationary systemExplain this equationShow the German source The length of the rod, measured in $K$, is therefore $l/\varphi(v)$; this gives the meaning of the function $\varphi$.Show the German source For reasons of symmetry it is now evident that the length of a given rod moving perpendicular to its axis, measured in the stationary system, can depend only on the velocity, and not on the direction and sense of the motion.Show the German source Thus the length of the moving rod measured in the stationary system does not change if $v$ is interchanged with $-v$.Show the German source From this it follows that:Show the German source $$\frac{l}{\varphi(v)} = \frac{l}{\varphi(-v)} ,$$$l$Rest length$\phi$Unknown scale factor: phi(v)$v$Speed of the moving systemExplain this equationShow the German source orShow the German source $$\varphi(v) = \varphi(-v) .$$$\phi$Unknown scale factor: phi(v)$v$Speed of the moving systemExplain this equationShow the German source From this and the relation found earlier it follows that $\varphi(v) = 1$ must hold, so that the transformation equations found become:Show the German source $$\begin{aligned} \tau &= \beta\left( t - \frac{v}{V^2} x \right), \\ \xi &= \beta(x - v t), \\ \eta &= y, \\ \zeta &= z, \end{aligned}$$$\tau$Time in the moving system$\beta$Lorentz factor$t$Time in the stationary system$v$Speed of the moving system$V$Speed of light$x$Position in the stationary system$\xi$Position in the moving system$\eta$Position in the moving system$y$Position in the stationary system$\zeta$Position in the moving system$z$Position in the stationary systemExplain this equationShow the German source whereShow the German source $$\beta = \frac{1}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} ,$$$\beta$Lorentz factor$v$Speed of the moving system$V$Speed of lightExplain this equationShow the German source

II. Electrodynamical Part.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