Simple derivation of de Broglie waves from Relativity

The gamma factor of special relativity

$$\gamma=\frac{1}{\sqrt{1-(\frac{v}{c})^2}},$$

implies $$-c\le v \le c.$$

We now consider two separate cases: 

1. Light (photons), or more generally bosons, for which the velocity can become equal to \(c\),
2. Matter (electrons, for example), for which the velocity cannot become equal to \(c\) itself.

For light, \(-c\le v \le c\) implies that there exists \(\varphi(t)\) such that $$
\begin{cases}v_x=c\cos\varphi \\ v_y=c\sin\varphi\end{cases}$$ which suggests that \(\textbf{v}\) might satisfy a wave equation. But we knew that already from Electromagnetism. A new problem arises though: which one of \({\bf v}\) and \({\bf E}\) is photon's wave function?

Assuming $$\varphi(t)=\omega t,$$ since $$\omega =\frac{1}{c\sqrt{1-(\frac{v}{c})^2}}\frac{dv}{dt},$$ this means that to get \(E=\hbar\omega\), the energy of a particle must be

$$E\propto a\gamma,$$ in direct contrast to $$E=mc^2\gamma\approx mc^2 + \frac{1}{2}mv^2.$$

For matter, since \(-c< v <c\) $$\Arrowvert\textbf{v}\Arrowvert=c\tanh w,$$ where \(w\) is the rapidity. Similar to the first case, this implies that there exists \(\psi(t)\) such that $$\begin{cases}v_x=c\tanh w\cos\psi\\ v_y=c\tanh w\sin\psi\end{cases}$$ which, again, suggests that \({\bf v}\) might satisfy a wave equation.