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While doing Lorentz transformation for position $$x'=\gamma(x-ut)$$ Here I understand that instead of just length contraction we also use the term $$ut$$ To include the distances between the frames of reference. But while doing Lorentz transformation for time that is $$t'=\gamma\left(t-\frac{ux}{c^2}\right)$$ What is the physical interpretation of the term. $$\frac{ux}{c^2}$$ like that of $$ut$$ in above I know the derivation but it doesn't give a clue about its actual meaning as if just pops out of nowhere.

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  • $\begingroup$ Are you asking how Lorentz transformation equations are derived? Because they are prior any consequences like length contraction. Note that $\gamma > 1$, so your first equation, without the $-ut$, would be length dilation... $\endgroup$ Commented Jan 18, 2020 at 1:51

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Let's consider two S clocks sitting at x1 and x2. These are synchronized to an observer in S and read the same time but they will not be synchronized to an observer in S'. The ux/c2 term has units of time and accounts for the phase difference, as observed in S', between the two clocks. I don't know if that is a sufficient "physical interpretation" and would welcome comments.

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  • $\begingroup$ How does the term $$ux/c^2$$ account for synchronization can you please elaborate $\endgroup$ Commented Jan 16, 2020 at 17:55
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To show how the term in question accounts for synchronization, consider Einstein's famous train thought experiment. The station observer sees the light that strikes the front of the frame moving toward the observer in the middle of the train car with relative speed (c+u) and for the light from the back of the train it's (c-u). Both light waves travel the same distance L, where L is half the length of the train car as determined by the station observer. So to the station observer, the time difference for reaching the middle of the car, which the car observer interprets as the lack of synchronization of the two light flashes, is Δt = L/(c-u) - L/(c+u) = (2Lu/c2) ɣ2. (I'm considering here the station frame of reference to have the unprimed coordinates and the train frame of reference to have the primed coordinates). Now 2L is the length of the car, D, as determined by the station observer, so we have Δt = (Du/c2) ɣ2. It's starting to look familiar. The ɣ2 factor goes away if we take account of length contraction, D=D'/ɣ and Δt=ɣΔt' (the train observer's time difference is a proper time interval as he is in the same location for each time measurement). So finally we get Δt'=D'u/c2, the amount by which the signals are not synchronized, expressed in terms of the train observer's coordinates.

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  • $\begingroup$ Thank you very much :-) $\endgroup$ Commented Jan 20, 2020 at 2:27

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