Angular acceleration of light from Relativity

 $$v^2=v_x^2+v_y^2=c^2$$

So

$$\begin{cases}v_x=c\cos\phi \\ v_y=c\sin\phi\end{cases}$$

Therefore $$\textbf{v}=c(\cos\phi,\sin\phi)=:c\hat{\textbf{v}},$$ yielding $$\boxed{\textbf{a}=c\dot{\phi}\hat{\boldsymbol{\phi}}}$$

Assuming $$\dot{\phi}=\omega,$$ that is, \(\dot{\phi}\) is the same as the frequency of (monochromatic) light, $$\boxed{a=c\omega}$$ Together with \(E=\hbar\omega\) this means $$E\propto a,$$ contrary to $$E\propto v^2.$$