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.