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This is from Purcell's E&M book, appendix G.

enter image description here

I take the photo in (a) as a given. In frame F, the letter E is stationary, and in frame F', the letter L is stationary. This assumes that $\beta = \sqrt3/2$, $\gamma = 2$, and that at the start in both frames ($t=0$ and $t'=0$), both letters E and L are lined up with their left sides in the same spot. In other words, the ICs are $x(t=0) = 1$, $x'(t=0)=2$

I wanted to clarify how Purcell went from (a) to determine the coordinates in (b). I assume you can argue there's time dilation, so the origin moves 4 units to the right in (a) and $4/\gamma = 2$ units to the left in (b). Is this right?

Assuming this is right, is there a deeper meaning to how events not defined at the origin have strange behavior like this?

Edit: For instance, if the E and L were both located in their respective frames at x=x’=0, the times would be equal, right?

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In both cases, we are given the position of the origin of one frame in the other frame. So, if we can work out where the origin of one frame was in the other frame at the start, we can work out the time.

In case (a), we know that because of length contraction, the position of the origin on $F'$ in $F$ is $x = 0$, since it is displaced by $-2$ from the left edge of the $L$ in $F'$, which corresponds to a displacement of $-1$ in $F$. Then, because we know the origin of $F'$ ends up at $x = 4$, $t = \frac{4}{\beta} = 4.62$ ns.

In case (b), we know that the position of the origin of $F$ in $F'$ is $x' = 1.5$, since it is displaced by $-1$ from the left edge of the $E$ in $F$, which corresponds to a displacement of $-0.5$ in $F'$. Then, because we know the origin ends up at $x' = -2$, $t' = \frac{2 - (-1.5)}{\beta} = 4.04$ ns.

I don't think there is anything deeper going on here - it's simply keeping track of what coordinates are given in the question. In this case, it's easiest to keep track of the movement of the respective origins.

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  • $\begingroup$ Just to clarify, are the diagrams (a) and (b) unrelated? For instance, the time corresponding t' for diagram (a) would be $t/\gamma$, and not 4.04 ns, right? $\endgroup$
    – Mondo Duke
    Jul 20, 2021 at 21:19
  • $\begingroup$ yes, they represent different events in spacetime. $\endgroup$
    – Lili FN
    Jul 20, 2021 at 22:35

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