Let's say you have a rod that is 10 meters long. Observer O sees the ends of the rod at (t=0, x=0), and (t=0, x=10). Observer O' moves at speed v = 0.8c relative to O. What is the length of the rod in O's perspective?(adsbygoogle = window.adsbygoogle || []).push({});

Using the length contraction formula L' = γL, we find that O' sees the rod as 6 meters long. Geometrically, the ends of the rod are at (t'=0, x'=0) and (t'=0, x'=6).

However, things seem to be different if you use the Lorentz boosts, t' = γ(t-vx/c^2) and x' = γ(x-vt). The ends of the rod are at (t'=0, x'=0) and (t'=13.3/c, x'=3.3).

Even without the foreknowledge that the length contraction formula is the right way to answer this question, it seems like the Lorentz boosts gives the wrong answer, as it mixes length and time for the second end of the rod. However, I don't see why you *can't* apply it here. All the Lorentz boosts tells you is the coordinates of the ends of the rod. But it appears that in this scenario Lorentz boosts gives you the "incorrect" coordinate of the second end of the rod (while the contraction formula + the geometry gives you the "correct" coordinate). Why?

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# I Why can't Lorentz boosts be used for contraction?

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