BehaviourOfMeasuringRodsAndClocksInMotion 2006-03-11 23:59:44 2006-03-11 23:59:44 Topic Relativity: The Special and General Theory % this is the default PlanetMath preamble. as your knowledge % of TeX increases, you will probably want to edit this, but % it should be fine as is for beginners. % almost certainly you want these \usepackage{amssymb} \usepackage{amsmath} \usepackage{amsfonts} % used for TeXing text within eps files %\usepackage{psfrag} % need this for including graphics (\includegraphics) %\usepackage{graphicx} % for neatly defining theorems and propositions %\usepackage{amsthm} % making logically defined graphics %\usepackage{xypic} % there are many more packages, add them here as you need them % define commands here \subsection{The Behaviour of Measuring-Rods and Clocks in Motion} 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 $x'=I$. 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{eqnarray*} x_{\mbox{(begining of rod)}} &=& 0 \overline{\sqrt{I-\frac{v^2}{c^2}}} \\ x_{\mbox{(end of rod)}} &=& 1 \overline{\sqrt{I-\frac{v^2}{c^2}}} \end{eqnarray*} ~ \noindent the distance between the points being $\sqrt{I-v^2/c^2}$. 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 $\sqrt{I-v^2/c^2}$ of a metre. The rigid rod is thus shorter when in motion than when at rest, and the more quickly it is moving, the shorter is the rod. For the velocity $v=c$ we should have $\sqrt{I-v^2/c^2} = 0$, and for stiII greater velocities the square-root becomes imaginary. From this we conclude that in the theory of relativity the velocity $c$ plays the part of a limiting velocity, which can neither be reached nor exceeded by any real body. Of course this feature of the velocity $c$ as a limiting velocity also clearly follows from the equations of the Lorentz transformation, for these became meaningless if we choose values of $v$ greater than $c$. If, on the contrary, we had considered a metre-rod at rest in the $x$-axis with respect to $K$, then we should have found that the length of the rod as judged from $K'$ would have been $\sqrt{I-v^2/c^2}$; this is quite in accordance with the principle of relativity which forms the basis of our considerations. \emph{A Priori} it is quite clear that we must be able to learn something about the physical behaviour of measuring-rods and clocks from the equations of transformation, for the magnitudes $z, y, x, t$, are nothing more nor less than the results of measurements obtainable by means of measuring-rods and clocks. If we had based our considerations on the Galileian transformation we should not have obtained a contraction of the rod as a consequence of its motion. Let us now consider a seconds-clock which is permanently situated at the origin ($x'=0$) of $K'$. $t'=0$ and $t'=I$ are two successive ticks of this clock. The first and fourth equations of the Lorentz transformation give for these two ticks: $$t = 0$$ \noindent and $$t' = \frac{I}{\sqrt{I-\frac{v^2}{c^2}}}$$ ~ As judged from $K$, the clock is moving with the velocity $v$; as judged from this reference-body, the time which elapses between two strokes of the clock is not one second, but $$\frac{I}{\sqrt{I-\frac{v^2}{c^2}}}$$ ~ \noindent seconds, {\it i.e.} a somewhat larger time. As a consequence of its motion the clock goes more slowly than when at rest. Here also the velocity $c$ plays the part of an unattainable limiting velocity.