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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?

James Ko
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1 Answers1

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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 $t$.

DaYu1729
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