Section 12

The Behaviour Of Measuring–rods And Clocks In Motion

April 8, 2022

I place a metre-rod in the x’-axis of moving K' in a way that its beginning coincides with the point x' = 0 while its end coincides with the point x' = 1.

What is the length of the metre rod relative to the non-moving K?

This is answered by asking where the beginning and end of the rod lie with respect to non-moving K at a time t of non-moving K. The first equation of the Lorentz transformation shows that the values of these two points at the time t = 0 is=

x (beginning of rod) = 0  (1  (v2/c2) 

x (end of rod) = 1  (1  (v2/c2)

the distance between the points being=


But the metre-rod is moving with the velocity v relative to non-moving K. It therefore follows that the length of a rigid metre-rod moving in the direction of its length with a velocity v is=

1  (v2/c2) 

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=

 (1  (v2/c2)) = 0

For faster speeds, the square-root becomes imaginary. This means 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.

This c follows from the equations of the Lorentz transformation. These become meaningless if v becomes greater than c.

If, on the contrary, we had considered a metre-rod at rest in the x-axis with respect to non-moving K, then the length of the rod as viewed from K' would have been 1 − v 2 c 2. This is in accordance with my principle of relativity.

The magnitudes x, y, z, t, are merely the results of measurements of the measuring-rods and clocks. If we used Galilei transformation, the rod would not contract 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' = 1 are two successive ticks of this clock. The 1st and 4th equations of the Lorentz transformation give for these two ticks= t = 0 and

t = 1 /  (1-(v2/c2))

As judged from non-moving K, the clock is moving with the velocity v as judged from this viewpoint, the time which elapses between two strokes of the clock is not 1 second, but

1 / (1-(v2/c2))

seconds, 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.