Superphysics
Section 12

# The Behaviour Of Measuring–rods And Clocks In Motion

##### 3 minutes  • 465 words

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=

``````√ (1-(v2/c2))
``````

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.

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