% Bernd Berg Oct 03 2017.
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\centerline{\bf Special and General Relativity (PHZ 4601/5606) Fall 2017 }
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\centerline{\bf Set 6}
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Relative velocity: Assume that $S'$ moves with velocity $\beta\,
\hat{x}$ with respect to a frame $S$. What is the velocity 
$\vec{\beta'_2}$ then in $S$? According to the earlier problem~13 
we have:
\begin{eqnarray} \nonumber
 \beta_2^1 = \frac{\beta'^1_2+\beta}{1+\beta'^1_2\,\beta}~~~{\rm and}
 ~~~\beta_2^2 = \frac{\beta'^2_2}{\gamma\,(1+\beta'^1_2\,\beta)}\,.
\end{eqnarray} 
To have the first particle at rest in $S$ we have to set
$\beta=-\beta'^1_1=-1/2$. As $\beta'^1_2=0$ we get 
$$ \beta^1_2=\beta=-1/2~~~{\rm and}~~~\beta^2_2=
   \frac{\beta'^2_2}{\gamma}=\frac{1/2}{\gamma}\,.$$
Now,
$$\gamma=\frac{1}{\sqrt{1-\beta^2}}=\frac{1}{\sqrt{1-1/4}}=
  \frac{2}{\sqrt{3}}~~\Rightarrow~~\beta^2_2=\frac{\sqrt{3}}{4}\,.$$ 
For the length we have 
$$ \left|\vec{\beta}_2\right| = \sqrt{(\beta_2^1)^2+(\beta_2^2)^2} 
   = \sqrt{\frac{1}{4}+\frac{3}{16}} = \sqrt{\frac{1}{4}+\frac{3}{16}}
   = \sqrt{\frac{4+3}{16}} = \frac{\sqrt{7}}{4}\,.  $$
  
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