Space Time Invariant Intervals of a Moving Clock This simple animation shows the trajectory of light rays that propagate transversely relative to a fixed frame observer as wells as the same light rays that move with their source and detector. The separation beween the mirrors is L so in the moving frame the time between mirror intercepts is t0=L/c. Since the speed of light is always c and the trajectory that he sees is slanted because of the motion it will take longer to complete a round trip between the mirrors than if the mirror were not moving. The velocity component along the y direction will be vy=sqrt(c*c-vx*vx) so that the detection time increase factor is t/t0=c/sqrt(c*c-vx*vx) = 1/sqrt(1-vx*vx/(c*c)) which is usually called gamma in relativity convention. The term vx/c is usually called beta. In the animation, the trace with the black dot at its leading end is the one that would be seen by an observer moving at the same velocity as the source/detector. The trace with the red dot at the leading end is what an observer watching from outside the moving source/detector would see. Since we base our time scale on the times between mirror intercepts, the outside observer time dt(vx) time interval between emission and reception is considerably greater than the time dt(0), as expected by the longer pathlengths. For a parallel mirror source/detector like this, it is usually the time for light to go between mirror intercepts that is important. It is clear that the black dot gets ahead of the red dot so these times appear different in this animation. A question arises whether the outside observer will see the moving source/detector being resonant at a different frequency than the stationary one. Since observed time passes more slowly on the moving platform, any oscillator on it (like a laser) would be observed to have a lower frequency and its emissions would then have longer wavelength. The factor by which the wavelength is increased is the same as the factor by which the observed trajectory length is increased so the number of wavelengths between the mirrors would be the same. That means that, if the moving oscillator was resonant in the source/detector before the transverse motion started, it will remain resonant after the motion starts. This must be the case since, if it were not, then Michelson's famous source/detector experiment would have not given a null result. We might want to use this device as a clock where each mirror intercept would correspond to a single time increment or tick. Since we are trying to define elapsed time, we should not use times as starting and stopping events. A reasonable choice of starting event would be departure from the initial source/detector position x=A and then the ending event would be arrival at x=B. How many ticks of each clock should we expect between these two events? It is easy to show that the number of ticks in the source/detector frame is [(B-A)/vx](c/L) while the number in the laboratory frame is [(B-A)/vx](c/L)*sqrt(1-vx*vx/(c*c)) The lab frame observer would conclude that time passes more slowly (less ticks between events) when the source/detector is moving. This agrees with results for time passage (time dilation) of special relativity. It is easy to see that the space time interval dt^2-(dx/c)^2 is the same in the lab frame as in the source/detector frame. The reason for this is. by the Pytharorean theorem, the increase in dt^2 in the lab frame is exactly the value of (dx/c)^2. This simple kinematic diagram proves that space time intervals are independent of the relative speed, vx. The animation, after reception of the light pulse in the lab frame, prints out both intervals and they are the same regardless of the value of vx/c that can be input by the learner.
Click Here for Lorentz Derivation
beta L Single Step