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Lets assume we have a two reference Frames $O$ and $O'$. $O$ is at rest and $O'$ is moving towards $O$ with velocity $v$ along the $x$-axis. Now we look at an object in which is moving with veloctiy $u$ along the $z$-axis in $O$. I have to find the veloctiy of the object in the frame of reference $O'$. So wat i have tried is following: The object is has a four-velocity $u^{\mu}=(c\gamma(u), 0, 0, u\gamma(u))^T$ in $O$. Now I multiply it with the Lorentz-Transformation Matrix $\Lambda_{\nu}^{\mu}$: $$u^{\prime\mu} = \Lambda_{\nu}^{\mu} u^{\mu} = \begin{pmatrix} \gamma(v) & -\beta_x\gamma(v) & 0 & 0 \\ -\beta_x\gamma(v) & \gamma(v) & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c\gamma\\ 0\\ 0\\ u\gamma \end{pmatrix} = \begin{pmatrix} c\gamma(v)\gamma(u)\\ -\beta_x\gamma(v)c\gamma(u)\\ 0\\ u\gamma \end{pmatrix}= \begin{pmatrix} c\gamma(v)\gamma(u)\\ -v\gamma(v)\gamma(u)\\ 0\\ u\gamma(v) \end{pmatrix} $$ But in the solution (in which they use 3-vectors) they get $\vec{u}'=(-v, 0, u/\gamma)$, which makes more sense and I know how this result is achieved. However I'm trying to understand the whole four-vectors thing. If we ignore the first index of four vector, shouldn't the solution be the same? I know it's a really basic problem and that I'm missing something obvious, but I have been looking around for hours and couln't find anything useful. I'd really appreciate your help.

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    $\begingroup$ If we ignore the first index of four vector, shouldn't the solution be the same? Yes, but the spatial components of the 4-velocity are not the 3-velocity. $\endgroup$
    – G. Smith
    Dec 8, 2020 at 18:46

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There are two issues to address here.

First, you made a minor mistake in your Lorentz transformation. It should be:

$$u^{\prime\mu} = \Lambda_{\nu}^{\mu} u^{\mu} = \begin{pmatrix} \gamma(v) & -\beta_x\gamma(v) & 0 & 0 \\ -\beta_x\gamma(v) & \gamma(v) & 0 & 0\\ 0 & 0 & 1 & 0\\ 0 & 0 & 0 & 1 \end{pmatrix} \begin{pmatrix} c\gamma(u)\\ 0\\ 0\\ u\gamma(u) \end{pmatrix} = \begin{pmatrix} c\gamma(v)\gamma(u)\\ -v\gamma(v)\gamma(u)\\ 0\\ u\gamma(u) \end{pmatrix}. $$

Note the last component of the result is $\gamma(u)$, not $\gamma(v)$.

Second, now that you have the four-velocity you need to obtain the three-velocity. Accounting for the correction above, the four-velocity you found is:

$$ u^{\prime\mu} = \begin{pmatrix} c\gamma(v)\gamma(u)\\ -v\gamma(v)\gamma(u)\\ 0\\ u\gamma(u) \end{pmatrix} $$

and the general relationship between the four-velocity and three-velocity is given by the following formula

$$ u'^{\mu} = \begin{pmatrix}c \, \gamma(u') \\ u_x' \, \gamma(u') \\ u_y' \, \gamma(u') \\ u_z' \, \gamma(u') \end{pmatrix}. $$

Comparing the first component in the two equations we see that

$$ \gamma(u') = \gamma(v) \gamma(u) $$

and so we find the components of the three-velocity $u'$

$$ u_x' = -v \\ u_y' = 0 \\ u_z' = \frac{u}{\gamma(v)}. $$

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  • $\begingroup$ thank you very much, that helps a lot. Never thought that I would get help so quickly here, really appreciate it. $\endgroup$ Dec 8, 2020 at 20:07

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