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I have a question regarding the concept of length contraction. Consider a rod of rest length $L_o=x_2-x_1$ in frame $S$. Now if we want to measure the length of the rod in $S^{'}$ i.e $L^{'}_o=x^{'}_2-x^{'}_1$ in this case we use the inverse Lorentz transformation.

My question is why must we take measurements $x^{'}_2$ and $x^{'}_1$ at the same time $t^{'}$ in the $S^{'}$ frame which is in uniform motion w.r.t the $S$ frame? Why can't we take the measurements of the end points $x^{'}_2$ and $x^{'}_1$ at different times in the $S^{'}$ frame?

PM 2Ring
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Nikhil Pathak
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1 Answers1

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Consider a rod of length 1 metre, aligned with and on the X axis, moving at 10 m/s in the +X direction. Clearly, this speed is non-relativistic, so the length contraction is negligible.

At time $t=0 s$, we measure that the back of the rod is at $x_1=0 m$. At $t=10 s$, we measure that the front of the rod is at $x_2=101 m$. But the length of the rod is 1 metre, not $x_2-x_1=101$ metres. So to get the actual length of the rod we must determine the positions of the front and back of the rod at the same instant of time.

If we can't perform those position measurements simultaneously, then we need to take the rod's velocity into account.

If the rod has a high velocity relative to our frame, so that relativistic effects aren't negligible, we need to be very careful that our position sensors record the exact time of their measurements, using clocks that are synchronised in our frame. And if those sensors measure the rod's position from a distance, they also need to compensate for the time it takes light to travel (from the part of the rod that they're measuring) to the sensor.

PM 2Ring
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  • FWIW, the length contraction of a 1 m rod with speed 10 m/s is ~0.556 fm, so smaller than the charge radius of a proton. (I used my calculator linked in this answer) – PM 2Ring Feb 06 '21 at 15:32