\subsection{{Derivation of $\omega_v$}
\label{ss:omegav}}
Equation~\ref{eq:eom} tells us the contribution of the electric field to the revolution frequency.
Neglecting the last term on the right, (that is the energy changing term), 
\begin{eqnarray*}
\frac{1}{c}\partder{^2z}{t^2}&=& \frac{e}{\gamma mc}\frac{E}{\beta}\\
&=& \frac{e}{\gamma mc}\frac{kz}{\beta}\\
\rightarrow z&=&z_0\cos\omega_v t,\ \ \omega_v^2 =\frac{e}{\gamma m c}\frac{ck}{\beta}\\
\omega_v^2 &=&\frac{eB}{\gamma m c}\frac{ck}{\beta B}\\
\omega_v^2 &=&\omega\frac{ck}{\beta B}\\
\rightarrow ck&=&\frac{\omega_v^2}{\omega}\beta B= \omega Q_v^2\beta B = n\omega\beta B =\frac{n\beta c B}{R_0}\\
\end{eqnarray*}
Evidently $\omega_v$ is the contribution of the electric field to the velocity.
Then
$$\omega = \sqrt{\omega_b^2 + \omega_v^2}$$
where $\omega_b = \frac{eB}{\gamma m c}$ and $\omega_v = \frac{eE}{\beta m c \gamma}$.
\label{ss:pitchanglebetaz}
Now that we know how $\omega_a$ depends on $\beta_z$, the next step is to write $\beta_z$ in terms of the pitch angle.
The pitch angle $$\tan\psi \equiv \frac{p_z}{p_\perp} = \frac{\beta_z\gamma_0 m}{p_\perp}.$$ The transverse momentum $p_\perp$ is invariant with respect to the boost in the z-direction. 

where we identify 3 disting Lorentz factors.
\begin{enumerate}
\item In the lab frame, with momentum transverse and parallel to the magnetic field $p_\perp$ and $p_z$, 
$$\gamma_0 = \frac{E_0}{m}=\frac{\sqrt{p_\perp^2+p_z^2+m^2}}{m}= \frac{\sqrt{p_\perp^2(1+\tan\psi^2)+m^2}}{m}$$
\item The Lorentz factor characterizing the boost by velocity $\beta_z$ to the frame where $p_z=0$, $\gamma = (1-\beta_z^2)^{-1/2}$
\item Finally in the frame where $p_z=0$, $\gamma^\prime = \frac{\sqrt{p^2+m^2}}{m}$.
\end{enumerate} 

We find that
\begin{eqnarray*}
\beta_z^2 &=& \frac{p_\perp^2\tan^2\psi}{p_\perp^2(1+\tan^2\psi)+m^2}\\
 &=& \frac{p_\perp^2\sin^2\psi}{p_\perp^2+m^2\cos^2\psi}\\
\frac{1}{\gamma^2} &=& 1-\beta_z^2= \frac{1+(m^2/p_\perp^2)\cos^2\psi-\sin^2\psi}{1+(m^2/p_\perp^2)\cos^2\psi}\\
&=&  \frac{\cos^2\psi(1+(m^2/p_\perp^2))}{1+(m^2/p_\perp^2)\cos^2\psi}\\
\end{eqnarray*}
There are two small parameters $\psi$ and $(m/p)^2$ so we need to take care in the expansion.
\begin{eqnarray*}
 \frac{\cos^2\psi(1+(m^2/p_\perp^2))}{1+(m^2/p_\perp^2)\cos^2\psi}&=&  \frac{\cos^2\psi(1+(m^2/p_\perp^2))}{1+(m^2/p_\perp^2)(1-\sin^2\psi)}\\
&=&  \frac{\cos^2\psi(1+(m^2/p_\perp^2))}{1+(m^2/p_\perp^2)-(m^2/p_\perp^2)\sin^2\psi }\\
&=&  \frac{\cos^2\psi}{(1-\frac{m^2}{m^2+p_\perp^2}\sin^2\psi}\\
&\sim&  \cos^2\psi\left(1+\frac{m^2}{m^2+p_\perp^2}\sin^2\psi\right)\\
&\sim&  \cos^2\psi
\end{eqnarray*}
Finally the dependence of $\omega_a$ on the pitch angle is
$$\omega_a(\psi) = \gamma\omega_a(\psi=0) \sim \omega_a(\psi=0)\cos\psi$$

