﻿ Solution: Proper Velocity | Primer on Geometric Algebra | David Hestenes

## Solution: Proper Velocity

Exercise: Prove

$\Large{V\,\,\, = \,\,\,\gamma \,\left( {1\,\, + \,\,\frac{{\bf{v}}}{c}} \right)\,\,,\,\,\,\,\,\,\,\,~~~~~~~~\,\,\,\,\gamma \,\,\, = \,\,\,\frac{{dt}}{{d\tau }}\,\,\, = \,\,\,\sqrt {1\,\, - \,\,\frac{{{{\bf{v}}^2}}}{{{c^2}}}} }$

Solution:

$\Large{\begin{array}{l}V\,\,\, \equiv \,\,\,\frac{{dX}}{{c\,\,d\tau }}\,\,\, = \,\,\,\frac{{d\left( {ct + {\bf{x}}} \right)}}{{c\,\,d\tau }}\,\,\, = \,\,\,\frac{{dt}}{{d\tau }}\,\, + \,\,\frac{1}{c}\,\frac{{d{\bf{x}}}}{{d\tau }}\,\,\, = \,\,\,\frac{{dt}}{{d\tau }}\,\, + \,\,\frac{1}{c}\,\frac{{dt}}{{d\tau }}\,\frac{{d{\bf{x}}}}{{dt}}\,\,\, = \,\,\,\frac{{dt}}{{d\tau }}\,\left( {1\,\, + \,\,\frac{{\bf{v}}}{c}} \right)\\\\\\\\\\V\widetilde V\,\,\, = \,\,\,\frac{{dX}}{{c\,\,d\tau }}\,\frac{{d\widetilde X}}{{c\,\,d\tau }}\,\,\, = \,\,\,\frac{{{{\left| {dX} \right|}^2}}}{{{{\left( {c\,\,d\tau } \right)}^2}}}\,\,\, = \,\,\,1\,\,\, = \,\,\,{\left( {\frac{{dt}}{{d\tau }}} \right)^2}\,\left( {1\,\, + \,\,\frac{{\bf{v}}}{c}} \right)\,\,\left( {1\,\, - \,\,\frac{{\bf{v}}}{c}} \right)\\\\\\\\\\\frac{{dt}}{{d\tau }}\,\,\, = \,\,\,\frac{1}{{\sqrt {1\,\, - \,\,{{\bf{v}}^2}/{c^2}} }}\,\,\, \equiv \,\,\,\gamma \,\,\,\,\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,\,\,\,\,V\,\,\, = \,\,\,\gamma \,\left( {1\,\, + \,\,\frac{{\bf{v}}}{c}} \right)\end{array}}$