﻿ Solution: Photon Proper Momentum | Primer on Geometric Algebra | David Hestenes

## Solution: Photon Proper Momentum

Exercise: Show that  ħ(ω/k), where  ω2 c2k2  and  ħk, where the energy is given by Planck’s Law (ħω = hf).

Notice for a photon we can't use  P = mcV = mcγ(1 + v/c)  as for a material particle, because m is zero,
while γ is infinity, so that expression is undefined, but the form  P = E/c + p  is still meaningful. Furthermore,
we can use  PP̃ = 0, since the photon's motion should certainly be lightlike. There are no tricky derivations here,
just many expressions for the same thing. Sometimes it's useful to focus on energy, at other times frequency,
at still other times wavelength, etc.

$\Large{\begin{array}{l}P\widetilde P\,\,\, = \,\,\,0\,\,\, = \,\,\,\left( {\frac{E}{c}\,\, + \,\,{\bf{p}}} \right)\,\left( {\frac{E}{c}\,\, - \,\,{\bf{p}}} \right)\,\,\, = \,\,\,\frac{{{E^2}}}{{{c^2}}}\,\, - \,\,{{\bf{p}}^2}\\\\\\{\rm{Planck's}}\,\,{\rm{Law:}}\,\,\,\,\,\,{E_{{\rm{photon}}}}\,\,\, = \,\,\,hf\,\,\, = \,\,\,\hbar \left( {2\pi f} \right)\,\,\, = \,\,\,\hbar \omega \,\,\, = \,\,\,hc/\lambda \\\\\\{\rm{Convenient}}\,\,{\rm{alternative}}\,\,{\rm{symbol:}}\,\,\,\,\,\,{\bf{k}}\,\,\, \equiv \,\,\,{\bf{p}}/\hbar \\\\\\p\,\,\, = \,\,\,\left| {\bf{p}} \right|\,\,\, = \,\,\,\frac{E}{c}\,\,\, = \,\,\,\hbar k\,\,\, = \,\,\,\frac{{\hbar \omega }}{c}\,\,\, = \,\,\,\frac{h}{\lambda }\,,\,\,\,\,\,\,\,\,\,\,\,\,k\,\,\, = \,\,\,\frac{\omega }{c}\,\,\, = \,\,\,\frac{{2\pi }}{\lambda }\\\\\\P\,\,\, = \,\,\,\frac{E}{c}\,\left( {1\,\, + \,\,{\bf{\hat p}}} \right)\,\,\, = \,\,\,\frac{h}{\lambda }\,\left( {1\,\, + \,\,{\bf{\hat p}}} \right)\,\,\, = \,\,\,\hbar k\,\left( {1\,\, + \,\,{\bf{\hat k}}} \right)\end{array}}$