Minkowski-space/Two observers/Relative velocity/Fact/Proof

Proof

1. we have

   ${}\left\langle v,v_{BC}\right\rangle =\left\langle v,-v-{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle =\left\langle v,-v\right\rangle +\left\langle v,-{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle =1-1=0\,,$

   so these vectors are orthogonal to each other. Therefore, ${}v_{BC}$ belongs to the space component of ${}B$.

2. We have

   ${}{\begin{aligned}\left\langle v_{BC},v_{BC}\right\rangle &=\left\langle -v-{\frac {1}{\left\langle v,w\right\rangle }}w,-v-{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle \\&=\left\langle v+{\frac {1}{\left\langle v,w\right\rangle }}w,v+{\frac {1}{\left\langle v,w\right\rangle }}w\right\rangle \\&=\left\langle v,v\right\rangle +2{\frac {\left\langle v,w\right\rangle }{\left\langle v,w\right\rangle }}+{\frac {1}{\left\langle v,w\right\rangle ^{2}}}\left\langle w,w\right\rangle \\&=1-{\frac {1}{\left\langle v,w\right\rangle ^{2}}}.\end{aligned}}$

   By part (1) (or because of fact  (3)), this expression is non-negative. The square root of this number is the relative velocity ${}\rho$.
3. This follows directly from the definition

   ${}\rho ={\sqrt {1-{\frac {1}{\left\langle v,w\right\rangle ^{2}}}}}\,,$

   using

   ${}\left\langle v,w\right\rangle <0\,.$

4. From

   ${}v_{BC}=-v-{\frac {1}{\left\langle v,w\right\rangle }}w\,$

   and (3), we can infer

   ${}w=-\left\langle v,w\right\rangle v-\left\langle v,w\right\rangle v_{BC}={\frac {v}{\sqrt {1-\rho ^{2}}}}+{\frac {v_{BC}}{\sqrt {1-\rho ^{2}}}}\,.$

   Because of part (1), the vector ${}v_{BC}$ belongs to the space component of ${}B$.

5. From (4) we can see directly that the time cefficient of ${}w$ with respect to ${}B$ equals ${}{\frac {1}{\sqrt {1-\rho ^{2}}}}$.

To fact