\subsection{Magic Momentum in Electrostatic focusing channel}
\label{ss:magic}
Consider the projection of the polarization onto the direction of motion.

\begin{eqnarray*}
\bfs\cdot\bmbetahat &=& \frac{\beta_x^0\cos(\kappa\gamma_0\beta_z^0\beta_x^0\sin\omega_v t)\cos(\beta_z^0\sin\omega_v t) + \sin(\kappa\gamma_0\beta_z^0\beta_x^0\sin\omega_v t)\sin(\beta_z^0\sin\omega_v t)}
{[{\beta_0^0}^2\cos^2(\beta_z^0\sin\omega_v t)+\sin^2(\beta_z^0\sin\omega_v t)]^{1/2}}\\
&=&\frac{\beta_x^0\cos\phi\cos\theta + \sin\phi\sin\theta}{[{\beta_x^0}^2\cos^2\theta+\sin^2\theta]^{1/2}}
\end{eqnarray*}
Expand to second order in $\theta, \phi$
\begin{eqnarray*}
\bfs\cdot\bmbetahat &\approx& \frac{\beta_x^0(1-\onehalf\phi^2)(1-\onehalf\theta^2) + \phi\theta}{[{\beta_x^0}^2(1-\onehalf\theta^2)^2+\theta^2]^{1/2}}\\
 &\approx& \frac{\beta_x^0(1-\onehalf\phi^2)(1-\onehalf\theta^2) + \phi\theta}{[{\beta_x^0}^2(1-\theta^2)+\theta^2]^{1/2}}\\
&\approx& \frac{\beta_x^0(1-\onehalf\phi^2)(1-\onehalf\theta^2) + \phi\theta}{[{\beta_x^0}^2+\theta^2(1-{\beta_x^0}^2)]^{1/2}}\\
&\approx& \frac{1}{\beta_x^0}\left(\beta_x^0(1-\onehalf\phi^2)(1-\onehalf\theta^2) + \phi\theta\right)\left(1-\frac{\theta^2}{2\gamma_0^2{\beta_x^0}^2}\right)\\
&\approx& \frac{1}{\beta_x^0}\left(\beta_x^0(1-\onehalf(\phi^2+\theta^2)) + \phi\theta\right)\left(1-\frac{\theta^2}{2\gamma_0^2{\beta_x^0}^2}\right)\\
&\approx& \frac{1}{\beta_x^0}\left(\beta_x^0(1-\onehalf(\phi^2+\theta^2)) + \phi\theta -\frac{\beta_x^0\theta^2}{2\gamma_0^2{\beta_x^0}^2}\right)\\
\end{eqnarray*}
Then $$\der{}{t}\bfs\cdot\bmbetahat=0$$ if
\begin{eqnarray*}
0&=&-\frac{\beta_x^0}{2}(\phi^2+\theta^2) + \phi\theta -\frac{\beta_x^0\theta^2}{2\gamma_0^2{\beta_x^0}^2}\\
0&=&-\frac{\beta_x^0}{2}((\kappa\gamma_0\beta_x^0)^2+1) + \kappa\gamma_0\beta_x^0 -\frac{\beta_x^0}{2\gamma_0^2{\beta_x^0}^2}\\
0&=&-\frac{1}{2}((\kappa\gamma_0\beta_x^0)^2+1) + \kappa\gamma_0 -\frac{1}{2\gamma_0^2{\beta_x^0}^2}\\
0&=&((\kappa\gamma_0\beta_x^0)^2+1) - 2\kappa\gamma_0 +\frac{1}{\gamma_0^2{\beta_x^0}^2}\\
\kappa&=& \frac{\left(2\gamma_0\pm[4\gamma_0^2-4({\beta_x^0}^2\gamma_0^2+1)]^{1/2}\right)}{2{\gamma_0^2\beta_x^0}^2}\\
\kappa&=& \frac{1}{\gamma_0{\beta_x^0}^2}
\end{eqnarray*}
Finally
\begin{eqnarray*}
\kappa&=&a_\mu+\frac{1}{\gamma_0+1} = \frac{1}{\gamma_0{\beta_x^0}^2}\\
\rightarrow a_\mu &= &\frac{1}{\gamma_0}\frac{\gamma_0^2}{\gamma_0^2-1} - \frac{1}{\gamma_0+1}\\
 a_\mu &=& \frac{\gamma_0}{\gamma_0^2-1} - \frac{\gamma_0-1}{\gamma_0^2-1}\\
a_\mu &=& \frac{1}{\gamma_0^2-1}
\end{eqnarray*}
