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Length contraction

Length Contraction Derivation:

Moving objects appear shorter in the dimension parallel to their velocity, again by the g factor introduced previously. To derive the contraction we again consider a light clock as the case of time dilation, only in this case we consider the clock on its side such that the motion of the clock pulse is parallel to the clock's velocity. If the clock has length L₀ in the rest frame of the clock, the time to for light to bounce from one side of the clock and back is:

However to an observer who sees the clock pass by with a velocity v, the light appears to take more time to traverse the length of the clock when the pulse is traveling in the same direction as the clock, and it appears that it takes less time for the return trip.

We know from the time dilation that

Therefore the above three equation can be used to eliminate t and t₀ to obtain the result

Thus moving meter sticks appear shorter along the direction of motion.  

Length Contraction for Particles:

Instead of analyzing the motion of the tau from our frame of reference, we could ask what the tau would see in its reference frame. Its half-life in its reference frame is 3.05 x 10^-13 s. This does not change. The tau goes nowhere in this frame.

How far would an observer, sitting in the tau rest frame, see an observer in our laboratory frame move while the tau lives?

We just calculated that the tau would travel 1.8 mm in our frame of reference. Surely we would expect the observer in the tau frame to see us move the same distance relative to the tau particle. Not so says the tau-frame observer --  you only moved 1.8 mm/gamma = 0.09 mm relative to me. This is length contraction.

How long did the tau particle live according to the observer in the tau frame? We can rearrange d = v x t to read t = d/v. Here we use the same speed, Because the speed of the observer in the lab relative to the tau is just equal to (but in the opposite direction) of the speed of the tau relative to the observer in the lab, so we can use the same speed. So time = 0.09 x 10^-3 m/(3 x 10⁸)m/sec  = 3.0 x 10^-13 sec. This is the half-life of the tau as seen in its rest frame, just as it should be!