\subsection*{Velocity - $\beta$}
The muons oscillate vertically in the field of the quadrupoles. The momentum $p_\phi$ in the azimuthal direction is constant, that is,  independent of the pitch angle $\psi$.
The force on the muons is in the vertical direction.
The momentum in the vertical direction depends on the pitch angle according to $p_z = p_\phi \tan\psi$. Since the muon gains and loses energy
to the electric field of the quadrupole, the magnitude of the energy and velocity $|\bmbetahat|$ will, like the momentum,  be time dependent.  
We can write the azimuthal and vertical components of the velocity as
\begin{eqnarray*}
\beta_\phi &=& \frac{p_\phi}{E}\\
\beta_z &=& \frac{p_z}{E}
\end{eqnarray*}
Then
\begin{eqnarray}
\bmbetahat_\phi &=& \frac{\beta_\phi}{[\beta_\phi^2+\beta_z^2]^\onehalf}\nonumber\\
&=& \frac{p_\phi/E}{[p_\phi^2/E^2+p_z^2/E^2]^\onehalf}\nonumber\\
&=& \frac{p_\phi}{p_\phi[1+p_z^2/p_\phi^2]^\onehalf}\nonumber\\
&\sim& (1-\onehalf\frac{p_z^2}{p_\phi^2}) = 1-\onehalf\tan^2\psi \label{eq:betaphi}
\end{eqnarray}
Also
\begin{eqnarray}
\bmbetahat_z &=& \frac{\beta_z}{[\beta_\phi^2+\beta_z^2]^\onehalf}\nonumber\\
&=& \frac{p_z/E}{[p_\phi^2/E^2+p_z^2/E^2]^\onehalf}\nonumber\\
&=& \frac{p_z}{p_\phi[1+p_z^2/p_\phi^2]^\onehalf}\nonumber\\
&\sim& \frac{p_z}{p_\phi}(1-\onehalf\frac{p_z^2}{p_\phi^2}) = \tan\psi(1-\onehalf\tan^2\psi)\label{eq:betaz}
\end{eqnarray}
We should use this more precise form for $\bmbetahat$ in Equation~\ref{eq:integrate} (rather than what is given in Equation~\ref{eq:beta}).
However, we show below that to order $\psi^2$ the result is the same.
Recall that $\bmbetahat$ appears in the integral in the cross product with $\bfB$ and $\bfB = B\hat\bfz$.
\begin{eqnarray*}
\bmbetahat\times\bfB &=& \bmbetahat_\phi B\\
&\sim & B(1-\onehalf\tan^2\psi)\sim B(1-\onehalf\psi^2)
\end{eqnarray*}
Since $$\cos\psi\sim 1-\onehalf\psi^2,$$  
to order $\psi^2$, $|\bmbetahat\times\bfB| \sim\cos\psi$, and consistent with the approximations used throughout.
That is because $\cos\psi \sim 1-\onehalf\tan^2\psi$ for small $\psi$.
   
(Note that if the focusing were magnetic rather than electric the total energy would constant and independent of pitch.)
