Solutions: Relativistic Energy & Momentum

Return to current sectionThe Proper Momentum.


Exercise: Derive expressions for PP̃, vector momentum p, particle energy E, and its kinetic energy K.


Solutions:

\[\Large{\begin{array}{l}{\rm{Remember:}}\,\,\,\,\,\,\,\,\,\,\,\,\gamma \,\,\, \equiv \,\,\,\frac{{dt}}{{d\tau }}\,\,\, = \,\,\,\frac{1}{{\sqrt {1\,\, - \,\,{{\bf{v}}^2}/{c^2}} }}\,\,\,\,\,\,\,\,\,\,\,\,and\,\,\,\,\,\,\,\,\,\,\,\,V\widetilde V\,\,\, = \,\,\,1\\\\\\\\\\P\widetilde P\,\,\, = \,\,\,{m^2}{c^2}V\widetilde V\,\,\, = \,\,\,{m^2}{c^2}\,\,\, = \,\,\,\left( {\frac{E}{c}\,\, + \,\,{\bf{p}}} \right)\,\left( {\frac{E}{c}\,\, - \,\,{\bf{p}}} \right)\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{Therefore}}\,\,\,\,\,\,\,\,\,\,\,\,{m^2}{c^4}\,\,\, = \,\,\,{E^2}\,\, - \,\,{{\bf{p}}^2}{c^2}\\\\\\\\\\P\,\,\, = \,\,\,mcV\,\,\, = \,\,\,mc\,\gamma \,\left( {1\,\, + \,\,\frac{{\bf{v}}}{c}} \right)\,\,\, = \,\,\,\frac{E}{c}\,\, + \,\,{\bf{p}}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{Therefore}}\,\,\,\,\,\,\,\,\,\,\,\,{\bf{p}}\,\,\, = \,\,\,\gamma m{\bf{v}}\,\,\, = \,\,\,\frac{{m{\bf{v}}}}{{\sqrt {1\,\, - \,\,{{\bf{v}}^2}/{c^2}} }}\,\,\, = \,\,\,\frac{{dt}}{{d\tau }}\,m\,\frac{{d{\bf{x}}}}{{dt}}\,\,\, = \,\,\,m\,\frac{{d{\bf{x}}}}{{d\tau }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,{\rm{and}}\,\,\,\,\,\,\,\,\,\,\,\,E\,\,\, = \,\,\,\gamma m{c^2}\,\,\, = \,\,\,\frac{{m{c^2}}}{{\sqrt {1\,\, - \,\,{{\bf{v}}^2}/{c^2}} }}\\\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,\,E\left( {{\bf{v}}\,\, = \,\,0} \right)\,\,\, = \,\,\,m{c^2}\,\,\,\,\,\,\,\, \Rightarrow \,\,\,\,\,\,\,\,K\,\,\, = \,\,\,E\,\, - \,\,m{c^2}\,\,\, = \,\,\,\left( {\gamma \,\, - \,\,1} \right)\,m{c^2}\\\\\\\\\\K\,\,\, = \,\,\,\left[ {\,{{\left( {1\,\, - \,\,{{\bf{v}}^2}/{c^2}} \right)}^{ - 1/2}}\,\, - \,\,1} \right]\,m{c^2}\,\,\, \approx \,\,\,\left( {\frac{1}{2}\,\frac{{{{\bf{v}}^2}}}{{{c^2}}}\,\, + \,\frac{3}{8}\,\frac{{{{\bf{v}}^4}}}{{{c^4}}}\,\, + \,\,\frac{5}{{16}}\,\frac{{{{\bf{v}}^6}}}{{{c^6}}}\,\, + \,\, \ldots } \right)\,m{c^2}\\\,\,\,\,\,\,~~~\, = \,\,\,\frac{1}{2}\,m{{\bf{v}}^2}\,\, + \,\,\frac{3}{8}\,m{{\bf{v}}^2}\,\left( {\frac{{{{\bf{v}}^2}}}{{{c^2}}}} \right)\,\, + \,\,\frac{5}{{16}}\,m{{\bf{v}}^2}\,{\left( {\frac{{{{\bf{v}}^2}}}{{{c^2}}}} \right)^2}\,\, + \,\, \ldots \end{array}}\]



Return to current sectionThe Proper Momentum.


© David Hestenes 2005, 2014