Lorentz Transformation of four-velocity using the transformation matrix 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.
 A: 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 $u'$
$$
u_x' = -v \\
u_y' = 0 \\
u_z' = \frac{u}{\gamma(v)}.
$$
