Return to current section: The Photon.
Exercise: Show that P = ħ(ω/c + k), where ω2 = c2k2 and p = ħk, where the energy is given by Planck’s Law (E = 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.
\[\Large{\begin{array}{l}P\widetilde P\,\,\, = \,\,\,0\,\,\, = \,\,\,\left( {\frac{E}{c}\,\, + \,\,{\bf{p}}} \right)\,\left( {\frac{E}{c}\,\, - \,\,{\bf{p}}} \right)\,\,\, = \,\,\,\frac{{{E^2}}}{{{c^2}}}\,\, - \,\,{{\bf{p}}^2}\\\\\\{\rm{Planck's}}\,\,{\rm{Law:}}\,\,\,\,\,\,{E_{{\rm{photon}}}}\,\,\, = \,\,\,hf\,\,\, = \,\,\,\hbar \left( {2\pi f} \right)\,\,\, = \,\,\,\hbar \omega \,\,\, = \,\,\,hc/\lambda \\\\\\{\rm{Convenient}}\,\,{\rm{alternative}}\,\,{\rm{symbol:}}\,\,\,\,\,\,{\bf{k}}\,\,\, \equiv \,\,\,{\bf{p}}/\hbar \\\\\\p\,\,\, = \,\,\,\left| {\bf{p}} \right|\,\,\, = \,\,\,\frac{E}{c}\,\,\, = \,\,\,\hbar k\,\,\, = \,\,\,\frac{{\hbar \omega }}{c}\,\,\, = \,\,\,\frac{h}{\lambda }\,,\,\,\,\,\,\,\,\,\,\,\,\,k\,\,\, = \,\,\,\frac{\omega }{c}\,\,\, = \,\,\,\frac{{2\pi }}{\lambda }\\\\\\P\,\,\, = \,\,\,\frac{E}{c}\,\left( {1\,\, + \,\,{\bf{\hat p}}} \right)\,\,\, = \,\,\,\frac{h}{\lambda }\,\left( {1\,\, + \,\,{\bf{\hat p}}} \right)\,\,\, = \,\,\,\hbar k\,\left( {1\,\, + \,\,{\bf{\hat k}}} \right)\end{array}}\]