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On the Electrodynamics of Moving BodiesShow the German source

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by A. EinsteinShow the German source

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I. Kinematical Part.Show the German source

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

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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

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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

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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

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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

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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

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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

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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

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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

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II. Electrodynamical Part.Show the German source

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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

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Bern, June 1905.Show the German source

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(Received 30 June 1905.)Show the German source