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Suppose I have a reference frame $S$. In this system, the spacetime coordinates of a particle are $(ct, x, y)$ where $x = u_x t$, $y = u_y t$. Suppose I have another frame $S'$, moving in the $x$-direction with velocity $v$ relative to $S$, and I wish to find the coordinates $(ct', x', y')$ of the particle for that system. The Lorentz transformations tell us

\begin{align*} ct' &= \gamma(ct - \beta x)\\ x' &= \gamma(x - \beta ct)\\ y' &= y \end{align*}

Focusing on the second component ($x$), $x' = \gamma(u_x t - \beta ct) = \gamma t (u_x - v) = t' (u_x - v)$. (The last step comes from the time dilation formula.) Rearranging, $\frac {x'} {t'} = u_x' = u_x - v$, which disagrees with the Einstein velocity addition formula. Why is this?

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    $\begingroup$ In my opinion, to use the a posteriori results of time dilation and length contraction in cooperation with Lorentz transformation is dangerous to get false results. I think that you must elaborate your problems with math, that is Lorentz transformation, and after this to use time dilation and length contraction to interpret the results. On the other hand, this question is identical to yours here : Why isn't the relative y-velocity in this problem .8c?... $\endgroup$
    – Voulkos
    Commented Nov 28, 2017 at 6:35
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    $\begingroup$ ...and you have already your answer therein. Moreover, if as Einstein velocity addition formula you mean this $$ w=\dfrac{u\pm\upsilon}{1\pm\dfrac{u\upsilon}{c^{2}}} \tag{01} $$ you must note that this formula is valid only for $\:u\:$ and $\:\upsilon\:$ collinear. $\endgroup$
    – Voulkos
    Commented Nov 28, 2017 at 6:42
  • $\begingroup$ You can't just use the time-dilation formula at your whimsy. You have a clear formula written down for $t'$ and it is NOT $t' = \gamma t$. $\endgroup$
    – Prahar
    Commented Nov 28, 2017 at 13:01
  • $\begingroup$ @Prahar OK. I'm asking this question because the answerer in this problem: physics.stackexchange.com/a/371050/138775 invoked time dilation and said that $t' = \gamma t$. Is that answerer wrong? $\endgroup$
    – James Ko
    Commented Nov 28, 2017 at 13:26
  • $\begingroup$ @James Ko - I haven't look at that answer, but I will tell you that there are very specific conditions under which the time-dilation formula $t' = \gamma t$ applies. You should always confirm that all those conditions hold before you use it. Also strictly speaking the time dilation formula relates time differences in one frame to time differences in the other frame and holds if and only if $\Delta x = 0$ (this is easy to check using your formulae above). Here, you are taking $x = u_s t$ so clearly this is not true. You should use your first formula for $t'$ instead. $\endgroup$
    – Prahar
    Commented Nov 28, 2017 at 13:42

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You're actually really close to the answer. You're one mistake, as pointed out in the comments, is you are using the wrong equation for $t$. You need to use the inverse Lorentz transform equation: $$ t = \gamma(t'+\frac{v}{c^2}x'). $$

Starting from before you went wrong in your derivation: $$ x' = \gamma t(u_x-v) $$ we use $t = \gamma(t'+\frac{v}{c^2}x')$: $$ x' = \gamma^2 (t'+\frac{v}{c^2}x')(u_x-v) $$

Solving for $\frac{x'}{t'}$ we get $$ \frac{x'}{t'}=\gamma^2\frac{u_x-v}{1-\gamma^2\frac{v}{c^2}(u_x-v)} $$ With some algebra we find the denominator to be $$ 1-\gamma^2\frac{v}{c^2}(u_x-v)=\gamma^2(1-\frac{u_xv}{c^2}) $$ so that $$ \frac{x'}{t'}=\gamma^2\frac{u_x-v}{\gamma^2(1-\frac{u_xv}{c^2})}=\frac{u_x-v}{1-\frac{u_xv}{c^2}} $$ which is the relativistic addition of velocities formula for the colinear velocity component.

Again, to elaborate on the comments, you have to be careful when using the simple $t' = \gamma t$ equation when transforming between coordinates. This would only be the case at $x = 0$. With a finite velocity, we know the particle's position will no longer be zero after a finite time $t$.

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