% Bernd Berg sep 21 2017.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 
Solutions} \medskip

\centerline{\bf Set 4} 

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\item{\bf 13. Addition theorem for transverse velocity components.}
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\noindent In $K'$ the motion with velocity $\vec{u'}$ is
$$ x'^i\ =\ c^{-1}\,u'^i\,x'^0\,.$$ 
With respect to frame $K$ the origin of $K'$ frame moves with 
speed $v$ along the $x^1$ axis of $K$. For $i=2,3$ the Lorentz 
transformations give then
$$ x^i\ =\ x'^i\ =\ c^{-1}\,u'^i\,\gamma\,(x^0-\beta\,x^1)\,. $$ 
Dividing by $x^0$ gives the velocity components in $K$:
\begin{eqnarray} \nonumber
  c^{-1}\,u^i&=&c^{-1}\,u'^i\,\gamma\,(1-\beta\,c^{-1}\,u^1)\
  =\ c^{-1}\,u'^i\,\gamma\,\left(1-\beta\,c^{-1}\,\frac{u'^1+v}{1+
     u'^1v/c^2}\right)\,,
\end{eqnarray}
where in the second step the already calculated equation for $u^1$
has been inserted. Bringing everything to the common denominator
gives
\begin{eqnarray} \nonumber
 u^i &=& u'^i\,\gamma\,\frac{(1+\beta\,u'^1/c)-\beta\,c^{-1}\,(u'^1+v)}
      {1+u'^1v/c^2} \\ \nonumber &=& u'^i\,\gamma\,\frac{1-\beta^2}
      {1+u'^1v/c^2}\ =\ \frac{u'^i}{\gamma\,(1+u'^1v/c^2)}\,,
\end{eqnarray}
where $1-\beta^2=1/\gamma^2$ was used in the last step.
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