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I'm trying to interpret a way to solve the following exercise:

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My instinct is to note that the particle is moving to the right of frame $S$, so we use the formula:

$$u_y = \frac{u'_y}{\gamma (1+\frac{u'_x v}{c^2})}$$

Where $u_y$ is the $y$ component of the particle's velocity in $S$, $u'_y$ is the $y$ component of the particle's velocity in $S'$, and $u'_x$ is the particle's $x$ velocity in $S'$.

My question is dealing with resolving whether $u'_x$ is $0$ or $0.8c$.

$u'_x$, in the frame of the nucleus, would say it's equal to $0$ since in its rest frame, it is only moving in the $y'$ direction of $S'$.

I feel like this is definitely the correct logic, although I'm a bit confused as to why $u'_x$ in this case is $0$ while in question $1$ it's $0.8c$. It is perhaps obvious, but, just to be clear, am I correct in saying it's $0.8c$ in (a) because the particle is moving at $0.8c$ relative to $S'$ and $u'_x = 0$ in (b) because the particle is now moving only along $y$ relative to $S'$?

And thus the answer should be $u'_y =0.8/\gamma$.

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  • $\begingroup$ Duplicate : Why isn't the relative y-velocity in this problem .8c? $\endgroup$ – Frobenius Dec 11 '17 at 17:24
  • $\begingroup$ I never understand the term "velocity addition formula" in SR. How many formulas must I memorize ??? So many as the number of configurations of two frames $\:\mathrm S, \mathrm S' \:$??? I always prefer to start with the Lorentz Transformation. $\endgroup$ – Frobenius Dec 11 '17 at 17:33

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