John Landry

2022-07-21

Let's say someone who is not moving is seeing a rod which has one edge at $x=0$ and the other at $x=a$ so its length is $a$. An observer moving at constant speed $u$ along the $x$-axis uses lorentz transformation to determine the coordinates of these two points in his frame of refrence.
${x}_{a}^{\prime }=\gamma \left(a-ut\right)$, ${x}_{0}^{\prime }=\gamma \left(0-ut\right)$ so to find the length of the rod we take the difference of these two points and it gives us ${x}_{a}^{\prime }-{x}_{0}^{\prime }=\gamma a$.
I know that $\gamma >1$ so shouldn't the moving reference frame observe the rod to be bigger? Why are we talking about length contraction?

Ali Harper

The rod is moving in the observer's reference frame and he has to measure each end at the same time in his frame. You are using the same time, $t$, in the frame of the first observer who is at rest with respect to the rod. What's simultaneous (same $t$) in the first observer's frame is not simultaneous in the moving observer's frame.
So the difference in ${x}^{\prime }$ coordinates that you calculate is not the length in the moving observer's frame.

Nash Frank

You want to measure the location of both ends at the same time, let us say when ${t}^{\prime }=0$. Given the Lorentz transformations, the point $\left(x,t\right)=\left(0,0\right)$ maps to $\left({x}^{\prime },{t}^{\prime }\right)=\left(0,0\right)$, and $\left(L,0\right)$ to $\left(\gamma L,\gamma vL/{c}^{2}\right)$. But this is the position of the other extreme of the rod at t′≠0. The position at ${t}^{\prime }\ne 0$ is ${L}^{\prime }={x}^{\prime }-v{t}^{\prime }$, replacing we get ${L}^{\prime }=L/\gamma$.