Page:Relativity (1931).djvu/62



PLACE a metre-rod in the $$x'$$-axis of $$K'$$ in such a manner that one end (the beginning) coincides with the point $$x' = 0$$, whilst the other end (the end of the rod) coincides with the point $$x' = 1$$. What is the length of the metre-rod relatively to the system $$K$$? In order to learn this, we need only ask where the beginning of the rod and the end of the rod lie with respect to $$K$$ at a particular time $$t$$ of the system $$K$$. By means of the first equation of the Lorentz transformation the values of these two points at the time $$t = 0$$ can be shown to be

$$\begin{align} x_{\mathrm{(beginning\ of\ rod)}} & = 0 \cdot \sqrt{1 - \frac{v^2}{c^2}} \\ x_{\mathrm{(end\ of\ rod)}} & = 1 \cdot \sqrt{1 - \frac{v^2}{c^2}}\mathrm{,} \end{align}$$

the distance between the points being $$\sqrt{1 - \frac{v^2}{c^2}}$$.

But the metre-rod is moving with the velocity $$v$$ relative to $$K$$. It therefore follows that the length of a rigid metre-rod moving in the direction of its length with a velocity $$v$$ is $$\sqrt{1 - \tfrac{v^2}{c^2}}$$ of a metre. The rigid rod is thus shorter when in motion than