Back to de Broglie, II: Polar twist

Recall $$r^2=x^2+y^2,$$

and $$\theta=\tan^{-1}\frac{y}{x}.$$

If

$$\omega=\frac{d\theta}{dt},$$

we have

$$\omega=(\frac{dx}{dt}, \frac{dy}{dt})\cdot\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)=\textbf{v}\cdot \textbf{k},$$

where $$\textbf{k}:=\left(\frac{-y}{x^2+y^2},\frac{x}{x^2+y^2}\right)$$

so

$$\omega=vk\cos\phi,$$

which gives 

$$v=\frac{\omega}{k},$$

for \(\phi=2n\pi\).

As $$\Arrowvert\textbf{k}\Arrowvert=\frac{1}{\sqrt{x^2+y^2}}=\frac{1}{r},$$

this means to each trajectory  \(\textbf{x}(t)=\big(x(t),y(t)\big),\) we can associate a harmonic wave with frequency $$\omega=\frac{\Arrowvert \textbf{v}(t)\Arrowvert}{\Arrowvert \textbf{x}(t)\Arrowvert}.$$