I. Kinematical Part.Show the German source

§ 4. Physical Meaning of the Equations Obtained, Concerning Moving Rigid Bodies and Moving Clocks.Show the German source

We consider a rigid sphere^[1)] of radius $R$, which is at rest relative to the moving system $k$ and whose centre lies at the origin of coordinates of $k$.Show the German source The equation of the surface of this sphere, which moves relative to the system $K$ with the velocity $v$, is:Show the German source $$\xi^2 + \eta^2 + \zeta^2 = R^2 .$$$\xi$Position in the moving system$\eta$Position in the moving system$\zeta$Position in the moving system$R$Sphere radiusExplain this equationShow the German source The equation of this surface, expressed in $x$, $y$, $z$ at the time $t = 0$, is:Show the German source $$\frac{x^2}{\left( \sqrt{1 - \left( \frac{v}{V} \right)^2} \right)^2} + y^2 + z^2 = R^2 .$$$x$Position in the stationary system$v$Speed of the moving system$V$Speed of light$y$Position in the stationary system$z$Position in the stationary system$R$Sphere radiusExplain this equationShow the German source A rigid body which, measured in a state of rest, has the shape of a sphere therefore has in a state of motion — viewed from the stationary system — the shape of an ellipsoid of revolution with the axesShow the German source $$R \sqrt{1 - \left( \frac{v}{V} \right)^2} , \quad R , \quad R .$$$R$Sphere radius$v$Speed of the moving system$V$Speed of lightExplain this equationShow the German source

Thus, whereas the $Y$- and $Z$-dimensions of the sphere (and hence also of every rigid body of whatever shape) do not appear modified by the motion, the $X$-dimension appears shortened in the ratio $1 : \sqrt{1 - (v/V)^2}$, that is, the more strongly the greater $v$ is.Show the German source For $v = V$ all moving objects — viewed from the “stationary” system — shrink into plane figures.Show the German source For velocities greater than that of light our considerations become meaningless; we shall find, moreover, in the considerations that follow, that in our theory the velocity of light plays physically the part of infinitely great velocities.Show the German source

It is clear that the same results hold for bodies at rest in the “stationary” system, viewed from a uniformly moving system. —Show the German source

We imagine further one of the clocks which are fitted to indicate the time $t$ when at rest relative to the stationary system and the time $\tau$ when at rest relative to the moving system, placed at the origin of coordinates of $k$ and so set that it indicates the time $\tau$.Show the German source How fast does this clock go, viewed from the stationary system?Show the German source

Between the quantities $x$, $t$ and $\tau$, which refer to the position of this clock, the following equations evidently hold:Show the German source $$\tau = \frac{1}{\sqrt{1 - \left( \frac{v}{V} \right)^2}} \left( t - \frac{v}{V^2} x \right)$$$\tau$Time in the moving system$v$Speed of the moving system$V$Speed of light$t$Time in the stationary system$x$Position in the stationary systemExplain this equationShow the German source andShow the German source $$x = v t .$$$x$Position in the stationary system$v$Speed of the moving system$t$Time in the stationary systemExplain this equationShow the German source We therefore haveShow the German source $$\tau = t \sqrt{1 - \left( \frac{v}{V} \right)^2} = t - \left( 1 - \sqrt{1 - \left( \frac{v}{V} \right)^2} \right) t ,$$$\tau$Time in the moving system$t$Time in the stationary system$v$Speed of the moving system$V$Speed of lightExplain this equationShow the German source from which it follows that the reading of the clock (viewed in the stationary system) lags behind by $(1 - \sqrt{1 - (v/V)^2})$ sec. per second or — to within quantities of the fourth and higher order, by $\frac{1}{2} (v/V)^2$ sec.Show the German source

From this the following peculiar consequence results.Show the German source If at the points $A$ and $B$ of $K$ there are clocks at rest which, viewed in the stationary system, run synchronously, and if the clock at $A$ is moved with the velocity $v$ along the line joining it to $B$, then on the arrival of this clock at $B$ the two clocks no longer run synchronously, but the clock moved from $A$ to $B$ lags behind the clock that has been at $B$ from the beginning by $\frac{1}{2} t v^2 / V^2$ sec. (to within quantities of the fourth and higher order), if $t$ is the time the clock needs to go from $A$ to $B$.Show the German source

One sees at once that this result still holds if the clock moves from $A$ to $B$ along any polygonal line, and that it holds even if the points $A$ and $B$ coincide.Show the German source

If one assumes that the result proved for a polygonal line also holds for a continuously curved curve, one obtains the theorem: If there are two synchronously running clocks at $A$, and one of them is moved along a closed curve with constant velocity until it returns to $A$, which may take $t$ sec., then on its arrival at $A$ the latter clock lags behind the one that has remained unmoved by $\frac{1}{2} t (v/V)^2$ sec.Show the German source From this one concludes that a balance-wheel clock located at the earth’s equator must go more slowly, by a very small amount, than a precisely similar clock, subject to otherwise identical conditions, located at one of the earth’s poles.Show 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