# Phase Velocity Proof

Using the relativistic energy equation $E^{2} = p^{2}c^{2}+m^{2}c^{4}$, I want to show that the resulting phase velocity for the de Broglie wave of an electron is greater than the speed of light.

I start by deriving an expression for Energy in terms of the phase velocity, which is the rate at which the phase of the wave propagates in space, and the momentum of the electron. Using the equation $v_{p}=f\lambda$, I can then make two substitutions namely that $f =\frac{E}{h}$ which gives the energy required or released when electrons change their energy levels and $\lambda=\frac{h}{p}$ which is the wavelength associated with a particle as postulated by Louis DeBroglie. After canceling like terms this gives me $E=v_{p}p$. Now I can plug this into my relativistic formula to obtain $v_{p}^{2}p^{2} = p^{2}c^{2} + m^{2}c^{4}$. Next I can make a substitution using the equation $p=mv_{p}$, and cancel out the masses. This leaves me with the bi-quadratic equation $y^{2}-c^{2}y-c^{4} = 0$ where $y=v_{p}^{2}$. This equation has only one real solution which is $\pm \sqrt\frac{c^{2}+\sqrt{5c^{4}}}{2}$. Taking the positive root, I obtain the velocity $3.8*10^{8} m/s$ which is greater than the speed of light.

Advertisements