>\begin{flushright}
D. Rubin\phantom{1111111} \\
September 29,  2020
\end{flushright}
\begin{center}
{\bf \Large Pitch and Efield Systematics}
\end{center}
\def\bmhr{{\hat{\bm\rho}}}
\def\bmhp{{\hat{\bm\phi}}}
\def\bmbetahat{{\hat{\bm\beta}}}
\def\bfs{{\bf s}}
\def\bfhz{{\hat{\bf z}}}
\section*{Precession}
Our unperturbed system is a muon precessing in a uniform magnetic field with zero electric field. 
The rate of change of the polarization with respect to the muon velocity, the thing that we measure, is described by Jackson 11.171
\begin{equation}
\der{}{t}(\bmbetahat\cdot\bfs)=-\frac{e}{mc}\bfs_\perp\cdot\left[a_\mu\bmbetahat\times\bfB+\left(a_\mu-\frac{m^2c^2}{p^2}\right)
\beta\bfE\right]\label{eq:j}
\end{equation}
We use a cylindrical coordinate system
where ${\bf B} = B{\bf\hat z}$, $\bmhr$ is the radial direction and $\bmhp$ is the azimuthal direction.
Our trial solution for $\bfs(t)$ in the $\rho,\phi$ plane is 
$\bfs(t) = |\bfs|(\cos\omega_a t \bmhr-\sin\omega_a t \bmhp)$ where in the unperturbed system $\bmbetahat = \bmhp$.
Then
$$\bfs_\perp = \bfs - (\bmbetahat\cdot \bfs)\cdot\bmbetahat = |\bfs|\cos\omega_a t\bmhr$$
and $$\bmbetahat\cdot\bfs = -|\bfs|\sin\omega_a t$$
Substitution into ~\ref{eq:j} gives us
\begin{equation}
\omega_a = \frac{e}{mc}Ba_\mu\label{eq:oa}
\end{equation}

It should be clear from Equation ~\ref{eq:j} that in general,  only the radial component of $\bfs_\perp$ will contribute, since the magnetic field
is in the $\bfhz$ direction and the velocity always in the $\phi-z$ plane so that the cross product is always exclusively in the radial direction.  
With that in mind we will take 
\begin{eqnarray*}
\bfs_\perp &=& |\bfs|\cos\omega_a t\bmhr\\
%\bmbetahat\cdot\bfs &=& -|\bfs|\cos\Delta\psi\sin\omega t
\end{eqnarray*}
as the form for $\bfs_\perp$ as long as the velocity is in the $\phi-z$ plane. That is a convenient starting point as it allows integration of ~\ref{eq:j}.
For example, when $\bmbetahat=\bmhp$ and $\bfE=0$, we can integrate ~\ref{eq:j}
\begin{eqnarray*}
{\bf \hat s}\cdot\bmbetahat = \int -\frac{e}{mc} a_\mu\hat \bfs_\perp\cdot (\bmbetahat\times
\bfB) dt = -\frac{e}{mc}a_\mu B\frac{\sin\omega_m t}{\omega_m} = -\omega_a\frac{\sin\omega_m t}{\omega_m} 
\end{eqnarray*}
Recall
\begin{eqnarray*}
\hat\bfs_\perp& =& \hat\bfs - (\bmbetahat\cdot\hat\bfs)\bmbetahat=\cos\omega_a t\bmhr\\
\hat\bfs_\parallel& =& (\bmbetahat\cdot\hat\bfs)\bmbetahat = -\frac{\omega_a}{\omega_m}\sin\omega_m t\bmbetahat\\
\end{eqnarray*}
Since $\hat \bfs = \hat\bfs_\perp + \hat\bfs_\parallel$, and $|\hat\bfs|=1$, it follows that $\hat\bfs_\perp^2 + \hat\bfs_\parallel^2 = \cos^2\omega_m t+\left(\frac{\omega_a}{\omega_m}\right)^2\sin\omega^2_m t
=1 \rightarrow \omega_m=\omega_a$.
%\begin{eqnarray}    
%\bfs&=& |\bfs|[(\cos\theta\bmhp+\sin\theta \hat{\bf z})]\sin\omega_a t + \bmhr\cos\omega_a t\label{eq:s}\\
%\rightarrow \bfs_\perp &=& |\bfs|\cos\theta\sin\omega_a t \bmhr\nonumber\\
%\rightarrow \bmbetahat\cdot\bfs&=& -|\bfs|\cos\theta\sin\omega_a t\nonumber
%\end{eqnarray}
%Substitution into Equation~\ref{eq:j} yields
%and
%$$\omega_a = \frac{e}{mc}B a_\mu$$
\section*{Pitch}
\subsection*{Time independent pitch}

%where $\Delta\psi = \psi-\theta$. Then
%the time dependence of the component of polarization projected onto the velocity,(the thing that we measure), is
\begin{eqnarray}
{\bf \hat s}\cdot\bmbetahat = \int -\frac{e}{mc} a_\mu B \cos\omega_m t \cos\psi dt\label{eq:integrate}
\end{eqnarray}
If $\psi$ is independent of time then
\begin{eqnarray*}
\bf\hat s\cdot\bmbetahat &=& -\omega_a \frac{\sin\omega_m t}{\omega_m} \cos\psi \\
\end{eqnarray*}
and evidently has sinusoidal time dependence with frequency $\omega_m$.
\begin{eqnarray*}
 -\sin\omega_m t &=& -\omega_a \frac{\sin\omega_m t}{\omega_m} \cos\psi dt\\
\rightarrow \omega_m &=& \omega_a\cos\psi
\end{eqnarray*}
where $\omega_a$ is defined in Equation ~\ref{eq:oa} and $\omega_m$ is what we measure. 

We note that the component of the polarization vector in the $\rho-\phi$ plane rotates about the z-axis. The component perpendicular to that plane is fixed.
And if there happens to be a component of velocity that is parallel to the magnetic field, the polarization vector will continue rotating about the z-axis, confined to the $\rho-\phi$ plane. 

\subsection*{Pitching motion}
Now suppose the muon is executing vertical betatron oscillations in the $\phi-z$ plane. Then
\begin{equation}
\bmbetahat = \cos\psi(t)\bmhp + \sin\psi(t)\hat{\bf z}\label{eq:beta}\\
\end{equation}
where $\psi(t)$ is the angle of the trajectory with respect to the $\rho-\phi$ plane.
If we imagine for a moment that there is no magnetic field, the muon will execute betatron oscillations in the focusing field of the quadrupoles.
At the magic momentum, absent a magnetic field,  the orientation of the polarization of the magic momentum muon with respect to the velocity is frozen.
That is because
$$\der{}{t}(\bmbetahat\cdot\bfs) = 0.$$
%Evidently
%\begin{equation}
%\bfs = |\bfs|\left([\cos(\psi(t)+\Delta\psi)\bmhp + \sin(\psi(t)+\Delta\psi)\hat{\bf z}]\sin\alpha + \bmhr\cos\alpha\right)\label{eq:polv}
%\end{equation}
% where $\Delta\psi$ and $\alpha$ are arbitrary time independent phases.
%Then as required $$\der{}{t}\left(\bmbetahat\cdot\bfs\right) = \der{}{t}\left(|\bfs|\sin\alpha\right)=0$$
%The spin vector is frozen and follows the velocity vector identically.

%In the magnetic field the angular velocity of the spin remains always in the $\phi-z$ plane and perpendicular to $\bmbetahat$.
% Farley asserts the same
%in Equation 6 of his Phys. Let. article. 

Next restore the uniform and time independent magnetic field and assume free betatron oscillations so that $\psi(t) = \psi_0\cos\omega_p t$. 
Let's imagine that at time $t$, $\psi(t)=0$ 
($\bmbetahat\cdot\bfB=0$), and the polarization $\bfs$ is in the $\rho-\phi$ plane. If the E-field dominates ($\omega_p\gg\omega_a$) then as the pitch angle increases, the polarization will evolve a z-component
that follows the z-component of velocity. If the magnetic field dominates ($\omega_a\gg \omega_p$), the polarization will be fixed in the $\rho-\phi$ plane. 
%If $\omega_p \gg\omega_a$, the z-component of polarization will follow the z-component of velocity.
More generally, in the regime where $\omega_p\gg\omega_a$, the effect of the magnetic field is that the z-component of the polarization
lags (or leads) the z-component of the velocity. The plane of the polarization vector will oscillate about the $\bmhr$ axis.
\begin{comment}
We generalize our expression for the spin vector (Equation \ref{eq:polv}) to
$$\bfs = |\bfs|\left([\cos(\psi(t)-\Delta\psi(t))\bmhp + \sin(\psi(t)-\Delta\psi(t))\hat{\bf z}]\sin\alpha(t) + \bmhr\cos\alpha(t)\right)$$
where both $\Delta\psi(t)$ and $\alpha(t)$ depend on time.
Then
$$\bfs\cdot\bmbetahat= |\bfs|(\cos(\psi+\Delta\psi)\cos\psi + \sin(\psi+\Delta\psi)\sin\psi)\sin\alpha = |\bfs|\cos\Delta\psi\sin\alpha,$$
where $\Delta\psi(t)$ represents the phase offset and
\begin{eqnarray*}
\bfs_\perp &=& \bfs-(\bmbetahat\cdot\bfs)\cdot\bmbetahat\\
& =& |\bfs|([\cos(\psi+\Delta\psi)\bmhp+\sin(\psi+\Delta\psi) \hat{\bf z}]\sin\alpha + \bmhr\cos\alpha)- |\bfs|\cos\Delta\psi \sin\alpha(\cos\psi\bmhp+\sin\psi\hat{\bf z})\\
& =& |\bfs|[(-\sin\psi\sin\Delta\psi)\bmhp+(\cos\psi\sin\Delta\psi) \hat{\bf z}]\sin\alpha + \bmhr\cos\alpha)
\end{eqnarray*}


Substitution into Equation \ref{eq:j} gives

\begin{eqnarray}
|\bfs|\der{}{t}(\sin\alpha\cos(\Delta\psi)) &=& -\frac{e}{mc} a_\mu B|\bfs|\cos\alpha \cos\psi\label{eq:g}\\
|\bfs|(\dot\alpha \cos\alpha\cos(\Delta\psi)-\sin\alpha \dot{(\Delta\psi)}\sin\Delta\psi) &=& -\frac{e}{mc} a_\mu B|\bfs|\cos\alpha (\cos\psi)\label{eq:general}
\end{eqnarray}
where $\dot \alpha, \dot\Delta\psi$ indicate derivative with respect to time. 
Equation ~\ref{eq:general} is true in general. Now we introduce approximations. We assume that  now $\alpha = \omega_a t$, $\psi=\psi_0\cos\omega_p t$, $\psi_0$ is small and that $\omega_p\gg\omega_a$, so that $\Delta\psi$ is also small. Then
\begin{eqnarray*}
|\bfs|(\omega_a \cos\omega_a t-\sin\omega_a t \dot{(\Delta\psi)}\sin\Delta\psi) &\sim& -\frac{e}{mc} a_\mu B|\bfs|\cos\omega_a t (1-\onehalf \psi^2) \\
|\bfs|(\omega_a  \cos\omega_a t-\sin\omega_a t \dot{(\Delta\psi)}\Delta\psi) &\sim& -\frac{e}{mc} a_\mu B|\bfs|\cos\omega_a t (1-\onehalf \psi_0^2\cos^2\omega_p t) \\
&\sim& -\frac{e}{mc} a_\mu B|\bfs|\cos\omega_a t (1-\onehalf (\onehalf\psi_0^2(1-\cos 2\omega_p t)) \\
\end{eqnarray*}
The precession plane oscillates about the radial ($\bmhr$) direction many times ($\omega_p/\omega_a$ times) per period of rotation of polarization about ${\bf\hat z}$. We take the average of the rapidly changing contribution (namely zero) leaving 
\begin{eqnarray*}
|\bfs|(\omega_a \cos\omega_a t) &\sim& -\frac{e}{mc} a_\mu B|\bfs|\cos\omega_a t (1-\onehalf (\onehalf\psi_0^2)\\
\rightarrow \omega_a &\sim&  -\frac{e}{mc} a_\mu B(1-\frac{1}{4}\psi_0^2)\\
\end{eqnarray*}
\begin{comment}
Rather than average over the high frequency component we might try to find an explicit solution for $\Delta\psi(t)$.
We have
\begin{eqnarray}
\sin\omega_a t \onehalf\der{(\Delta\psi)^2}{t} &\sim& -\frac{1}{4}\psi_0^2\frac{e}{mc} a_\mu B\cos\omega_a t\cos 2\omega_p t \\
\der{}{t}(\Delta\psi)^2&\sim& -\frac{1}{2}\psi_0\frac{eB}{mc}a_\mu \frac{\cos(\omega_a +2\omega_p)t +\cos(\omega_a - 2\omega_p)t}{\sin\omega_a t}
 \end{eqnarray}

The contribution to $\omega_a$ due to the pitching motion
$$C_p = \frac{\Delta\omega_P}{\omega_a} = -\frac{1}{4}\psi_0^2 = -\frac{1}{2}\vev{\psi^2}$$
In order to get some sense of $\Delta\psi(t)$ let's try integrating Equation~\ref{eq:g}. We assume that $\alpha = \omega_a t$ which will not be true in general, but in any event we measure an average $\omega_a$.
\end{comment}
Returning to Equation ~\ref{eq:integrate}
\begin{eqnarray}
{\bf\hat s}\cdot\bmbetahat &=& \omega_a\int\cos\omega_m t\cos\psi(t) dt \label{eq:sbeta}
\end{eqnarray}
In our experiment the frequency of vertical oscillations is much greater than the precession frequency, and since we measure the average
precession frequency, rather than any component with the time dependence of the vertical oscillations,
we average over that shorter time scale. Then
\begin{eqnarray*}
{\bf\hat s}\cdot\bmbetahat &=& -\omega_a\int\cos\omega_m t\vev{\cos\psi(t)} dt\\
-\sin\omega_m t &=& -\omega_a\frac{\sin\omega_m t}{\omega_m}\vev{\cos\psi(t)} dt\\
\end{eqnarray*}
and where $|\psi|\ll 1$, 
\begin{eqnarray}
\sin\omega_m t &=& \omega_a \frac{\sin\omega_m t}{\omega_m}(1-\onehalf\vev{\psi^2})\nonumber\\
\rightarrow  \omega_m &=& \omega_a (1-\onehalf\vev{\psi^2})\label{eq:om}\\
\end{eqnarray}
By averaging over the pitching time scale, we are ignoring the z-component of polarization that lags or leads the z-component of velocity.
%, and that contributes a time dependence 
%that is not strictly sinusoidal. 
We can determine this component iteratively. For our first iteration we assume that $\bfs_\perp = |\bfs|\cos\omega_m t \bmhr$ as before, so that at $t=0$ the polarization is initially in the $\bmhr$ direction.
%(The choice of initial conditions is for convenience in the comparison with a spin tracking calculation below.)
Substitute the explict form for $\psi(t)$ into Equation ~\ref{eq:sbeta}, expand about small $|\psi|$  and integrate 
\begin{eqnarray}
{\bf\hat s}\cdot\bmbetahat &=& -\omega_a\int\cos\omega_m t (1-\frac{1}{4}\psi_0^2(1+\cos 2\omega_p t))  dt\nonumber\\
&=& -\omega_a\left[\frac{\sin\omega_m t}{\omega_m}(1-\frac{1}{4}\psi_0^2) -\frac{1}{8}\psi_0^2\left(\frac{\sin(\omega_m + 2\omega_p) t}{\omega_m + 2\omega_p}+\frac{\sin(\omega_m - 2\omega_p)t}{\omega_m-2\omega_p}\right)\right]\label{eq:realanswer}
%&=& \left[\cos\omega_m t -\frac{1}{4}\psi_0^2\left(\cos\omega_m t+\frac{\omega_a}{2}\left(\frac{\sin(\omega_m + 2\omega_p) t}{\omega_m + 2\omega_p}+\frac{\sin(\omega_m - 2\omega_p)t}{\omega_m-2\omega_p}\right)\right)\right]
\end{eqnarray}
We have used $\vev{\psi^2} = \vev{(\psi_0\cos\omega_p t)^2} = \onehalf\psi_0^2$ so that $\omega_m = \omega_a(1-\frac{1}{4}\psi_0^2)$ equivalent to Equation ~\ref{eq:om}.
Note that as anticipated, the second term in Equation ~\ref{eq:realanswer} averages to zero on the pitching time scale.

The contribution of the pitching motion to the average precession frequency is
$$\frac{\Delta\omega_{pitch}}{\omega_a}= C_p= -\frac{1}{4}\psi_0^2$$
\subsection*{Measurement}
The vertical betatron pitching motion, assuming linear fields and continuous quadrupoles,  is characterized by  
\begin{eqnarray*}
 y &=&a\sqrt{\beta_z}\sin(\omega_p t)\\ 
 \psi &=&\frac{a}{\sqrt{\beta_z}}\cos(\omega_p t)=\psi_0\cos\omega_p t\\
\end{eqnarray*}
Then  $$\psi^2_0 = 2\vev{y^2}/ \beta^2_z = 2 \frac{n}{R_0^2}\vev{ y^2}$$  
and
$$\vev{C_p} =  -\frac{n}{2}\frac{\vev{y^2}}{R_0^2}$$  
where $\beta_z = R_0/n$
\subsection*{High frequency component}
Equation ~\ref{eq:realanswer} gives ${\bf\hat s}\cdot\bmbetahat = \frac{1}{\beta}(s_z\beta_z + s_\phi \beta_\phi)$. 
%The first term in ~\ref{eq:realanswer} contributes to both terms in the dot product. 
We argue above that at the magic momentum, and when $\omega_p\gg\omega_a$, the component of the spin vector in the z-direction will simply follow the z-component of velocity.
With that approximation, the polarization 
\begin{equation}
\bfs_0 = |\bfs|[\cos\omega_a t\bmhr -\sin\omega_a t(\cos\psi \bmhp + \sin\psi \bfhz)] \label{eq:sz}
\end{equation}
That is, the polarization vector is rotating about a comoving axis ($\bfz^\prime$), that is parallel to $\bmbetahat\times\bmhr$. Then we can write  
\begin{equation}
\hat\beta_z s_z = \beta_z s_{z0} + \Delta[\beta_z s_z]\label{eq:bzsz}
\end{equation}
where
\begin{equation}
\hat\beta_z s_{z0} = \frac{\beta_z}{\beta}s_{z0} = (\psi_0\cos\omega_p t)[(\sin\omega_a t) \sin(\psi_0\cos\omega_p t)] \label{eq:bzsz0}
\end{equation}
using $\beta_z/\beta = \psi_0\cos\omega_p t$, and
 $\Delta[\beta_z s_z]$ is the contribution from that part of $s_z$ that does not follow $\beta_z$, namely the high frequency part of Equation ~\ref{eq:realanswer}.
\begin{equation}
\Delta[\beta_z s_z] = -\frac{1}{4}\psi_0^2\frac{\omega_a}{2}\left(\frac{\sin(\omega_m + 2\omega_p) t}{\omega_m + 2\omega_p}+\frac{\sin(\omega_m - 2\omega_p)t}{\omega_m-2\omega_p}\right)\label{eq:Dbzsz}
\end{equation}
Note that Equation~\ref{eq:bzsz} is not an exact solution since in order to calculate it we assumed that $\bfs_\perp = |\bfs|\cos\omega_m t\bmhr$. 
The exact solution will satisfy $(\bfs_\parallel)^2+(\bfs_\perp)^2 = |\bfs| = 1$. 
We can use this normalization relationship to solve for the next order correction 
to $\bfs_\perp$, that is
\begin{equation}
\bfs_\perp^\prime = \left(1 - (\bmbetahat\cdot\hat\bfs)^2\right)^{1/2}\label{eq:no}
\end{equation}
and then
substitute ${\bfs^\prime}_\perp$ back into Equation~\ref{eq:sbeta} and integrate again.
%The second
%contributes only to the $s_z\beta_z$ term.
%Rearranging ~\ref{eq:realanswer} we identify the pitch dependent term as
%$$ = -\frac{1}{4}\psi_0^2\left[\cos(\omega_m t) + \onehalf\omega_a\left(\frac{\cos(\omega_m+2\omega_p) t}{\omega_m+2\omega_p}+\frac{\cos(\omega_m-2\omega_p)t}{\omega_m-2\omega_p}\right)\right]$$
%The contribution of the pitch to ${\bf\hat s}\cdot\bmbetahat$
%\begin{equation}
%\frac{\beta_z}{\beta} s_z = -\frac{1}{4}\psi_0^2\left[\cos(\omega_m t) + \omega_p\left(\frac{\cos(\omega_m+2\omega_p) t}{\omega_m+2\omega_p}+\frac{\cos(\omega_m-2\omega_p)t}{\omega_m-2\omega_p}\right)\right]
%\label{eq:toplot}
%\end{equation}
We can of course compute the exact time evolution of the polarization by integrating the Thomas-BMT equation. The exact solution 
is compared to Equations~\ref{eq:bzsz0} and~\ref{eq:Dbzsz} in Figure ~\ref{fig:realanswer}. At $t=0$ in the simulation, the magic momentum muon is 
polarized in the $\bmhr$ direction.  The intial velocity is $\beta = (p_m/mc)(\cos\psi\bmhp +\sin\psi\bfhz)$ where $\psi= 2$ mrad.
The initial coordinates are on the magic orbit.
The muon is 
tracked through the fields of the ring. The polarization is computed by integration of the Thomas-BMT equation. 

\begin{figure}[h]
\centering
%\includegraphics{pitch_plot_1.pdf}
%\includegraphics{pitch_plot_2.pdf}
\includegraphics{pitch_plot_3v1.pdf}
\caption{The projection of spin on the z-component of velocity. The black curve with points is from iteration of the Thomas-BMT equation. 
The red line ('$\beta_z s_{z0}$'), Equation ~\ref{eq:bzsz0}, is the projection of polarization onto the z-axis with the approximation that the polarization is given by Equation ~\ref{eq:sz}.
The purple line is the correction term  ($\Delta[\beta_zs_z]$), Equation~\ref{eq:Dbzsz}. 
We see that our approximation is in very good agreement with the exact result. Note also that the contribution of the z-component to the $\bmbetahat\cdot \bfs$
is at the level of a few parts per millon and that the correction $\Delta[\beta_zs_z]$ two orders of magnitude smaller.}
\label{fig:realanswer}
\end{figure}

\subsection*{Next order}
We substitute $\hat\bfs\cdot\bmbetahat$ from~\ref{eq:realanswer} into ~\ref{eq:no} to determine the first order change in magnitude of $\bfs_\perp$.
\begin{eqnarray*}
s^\prime_\perp \sim |\bfs|\left[\cos\omega_m t -\frac{1}{8}\psi_0^2\sin\omega_m t\left({\rm stuff}\right) +\onehalf\left(\frac{\psi_o^2\omega_a}{8}\right)^2\left({\rm stuff}\right)^2\right]
\end{eqnarray*}
Then we can use that in ~\ref{eq:j} to compute the next higher order contribution in $\psi_0^2$
\begin{comment}

      \begin{eqnarray}
       |\bfs|\der{}{t}(\sin\alpha\cos(\Delta\psi)) &=& -\frac{e}{mc} a_\mu B|\bfs|\cos\alpha \cos\psi \nonumber\\
      \sin\omega_a t\cos(\Delta\psi) &\sim& -\frac{e}{mc} a_\mu B\int \cos\omega_a t (1-\frac{1}{4}\psi_0^2(1-\cos 2\omega_p t))dt \nonumber\\
      \sin\omega_a t\cos(\Delta\psi) &\sim& -\frac{e}{mc} a_\mu B\left[\frac{ \sin\omega_a t}{\omega_a} (1-\frac{1}{4}\psi_0^2)+ \frac{\psi_0^2}{4}\int(\cos\omega_at\cos 2\omega_p t)dt\right] \nonumber\\
      &=& -\frac{e}{mc} a_\mu B\left[\frac{ \sin\omega_a t}{\omega_a} (1-\frac{1}{4}\psi_0^2)+ \frac{\psi_0^2}{4}\left(\frac{\sin(\omega_a + 2\omega_p)t}{\omega_a + 2\omega_p} + \frac{\sin(\omega_a-2\omega_p)t}{\omega_a-2\omega_p}\right)\right]\\
       \omega_a\cos(\Delta\psi)&=& -\frac{e}{mc} a_\mu B\left[ (1-\frac{1}{4}\psi_0^2)+ \frac{\psi_0^2}{4}\frac{\omega_a}{\sin\omega_a t}\left(\frac{\sin(\omega_a + 2\omega_p)t}{\omega_a + 2\omega_p} + \frac{\sin(\omega_a-2\omega_p)t}{\omega_a-2\omega_p}\right)\right]\\
     \end{eqnarray}
      Defining $\omega_a = \frac{e}{mc}B(1-\frac{1}{4}\psi_0^2)$ we have that
      \begin{eqnarray*}
        \cos(\Delta\psi) &=& \left[(1-\frac{1}{4}\psi_0^2) + \frac{\psi_0^2}{4}\frac{\omega_a}{\sin\omega_a t}\left(\frac{\sin(\omega_a + 2\omega_p)t}{\omega_a + 2\omega_p} + \frac{\sin(\omega_a-2\omega_p)t}{\omega_a-2\omega_p}\right)\right]\\
        \Delta\psi &=&\cos^{-1} \left[(1-\frac{1}{4}\psi_0^2) + \frac{\psi_0^2}{4}\frac{\omega_a}{\sin\omega_a t}\left(\frac{\sin(\omega_a + 2\omega_p)t}{\omega_a + 2\omega_p} + \frac{\sin(\omega_a-2\omega_p)t}{\omega_a-2\omega_p}\right)\right]\\
        &\sim& \cos^{-1}(1-\frac{1}{4}\psi_0^2) + \frac{1}{\sqrt{[1-(1-\frac{1}{4}\psi_0^2)^2}} \frac{\psi_0^2}{4}\frac{\omega_a}{\sin\omega_a t}\left(\frac{\sin(\omega_a + 2\omega_p)t}{\omega_a + 2\omega_p} + \frac{\sin(\omega_a-2\omega_p)t}{\omega_a-2\omega_p}\right)\\
          &\sim& \cos^{-1}(1-\frac{1}{4}\psi_0^2) + \frac{1}{[\sqrt{\frac{1}{2}}\psi_0]} \frac{\psi_0^2}{4}\frac{\omega_a}{\sin\omega_a t}\left(\frac{\sin(\omega_a + 2\omega_p)t}{\omega_a + 2\omega_p} + \frac{\sin(\omega_a-2\omega_p)t}{\omega_a-2\omega_p}\right)\\
          &\sim& \cos^{-1}(1-\frac{1}{4}\psi_0^2) + \frac{\sqrt{2}\psi_0}{4}\frac{\omega_a}{\sin\omega_a t}\left(\frac{\sin(\omega_a + 2\omega_p)t}{\omega_a + 2\omega_p} + \frac{\sin(\omega_a-2\omega_p)t}{\omega_a-2\omega_p}\right)\\
          &\sim& \sqrt{\frac{1}{2}}\psi_0\left\{1 + \frac{1}{2}\frac{\omega_a}{\sin\omega_a t}\left(\frac{\sin(\omega_a + 2\omega_p)t}{\omega_a + 2\omega_p} + \frac{\sin(\omega_a-2\omega_p)t}{\omega_a-2\omega_p}\right)\right\}\\
      \end{eqnarray*}
      \begin{eqnarray*}
        \sin\omega_a t(1-\onehalf(\Delta\psi)^2) &\sim& -\frac{e}{mc} a_\mu B\left[\frac{ \sin\omega_a t}{\omega_a} (1-\frac{1}{4}\psi_0^2)+ \frac{\psi_0^2}{4}\int(\cos\omega_at\cos 2\omega_p t)dt\right] \nonumber\\
        1-\onehalf(\Delta\psi)^2 &\sim& -\frac{e}{mc} a_\mu B\left[\frac{ 1}{\omega_a} (1-\frac{1}{4}\psi_0^2)+ \frac{\psi_0^2}{4\sin\omega_a t}\int(\cos\omega_at\cos 2\omega_p t)dt\right] \nonumber\\
        \omega_a(1-\onehalf(\Delta\psi)^2) &\sim& -\frac{e}{mc} a_\mu B\left[(1-\frac{1}{4}\psi_0^2)+ \frac{\omega_a\psi_0^2}{4\sin\omega_a t}\frac{\omega_a}{\sin\omega t}\int(\cos\omega_at\cos 2\omega_p t)dt\right] \nonumber\\
      \end{eqnarray*}
      For long times, and as long as $\omega_a \ne 2\omega_p$, the integral (second term in brackets on the right) sums to zero so that $\Delta \psi$  is zero. Near resonance $\omega_a \sim 2\omega_p$ the approximation breaks down. 
      Presumably at resonance
      the plane of polarization is flipped about the $\bmhr$ direction.
\end{comment}




\begin{comment}


  $$F\sim\cos\phi_a\sin\psi_0\sin\phi_p.$$
  If $\bfs$ is parallel to $\bmhr$, $(\phi_a=0)$, and the pitch angle is at a maximum $\phi_p=0$. After all, when $\bfs$ is in the $\rho$ direction, propagation through the E-field generates no component of polarization in the z-direction.
  And when the pitch angle is zero, 
  rotate about the magnetic field, (the z-axis) generating a component 
  We try the simplest possible generalization of our previous zero B-field solution by supposing that $\alpha=(\omega_a+\Delta\omega)t$.
  $$\bfs = |\bfs|\left([\cos\psi(t)\bmhp + \sin\psi(t)\hat{\bf z}]\sin(\omega_a+\Delta\omega)t + \bmhr\cos(\omega_a+\Delta\omega)t\right)$$
  Then 
  $$\bmbetahat\cdot\bfs = |\bfs|\sin(\omega_a +\Delta\omega)t$$
  %suppose in the spirit of perturbation theory that substitution of the unperturbed solution into the differential equation that
  %includes the perturbation will give us the change in $\omega_a$, that is $\omega_a\rightarrow\omega_a+\Delta\omega$. Then 
  and
  $$\bfs_\perp\rightarrow |\bfs|\cos(\omega_a +\Delta\omega)t\bmhr.$$
  %and $$\bmbetahat\cdot\bfs \rightarrow -|\bfs|\sin(\omega_a+\Delta\omega) t$$
  Substitution of the trial solution into Equation ~\ref{eq:j} gives
  \begin{eqnarray*}
    \der{}{t}(-|\bfs|\sin(\omega_a+\Delta\omega) t)&=&-\frac{e}{mc}|\bfs|\cos(\omega_a +\Delta\omega)t\bmhr\cdot
    \left[a_\mu(|\bfB|(\cos\psi \bmhr +\sin\psi \bmhp)\right]\\
    \der{}{t}|\bfs|\sin(\omega_a+\Delta\omega) t)&=&\frac{e}{mc}|\bfs|\cos(\omega_a +\Delta\omega)t\left[a_\mu(|\bfB|\cos\psi)\right]\\
  \end{eqnarray*}
  Rewriting so that the perturbation is explicit
  $$\der{}{t}(|\bfs|\sin(\omega_a+\Delta\omega) t)=\frac{e}{mc}|\bfs|\cos(\omega_a +\Delta\omega)t\left[a_\mu(|\bfB|-|\bfB|(1-\cos\psi)\right]
  $$
  Solve for $\Delta\omega$
  \begin{eqnarray*}
    \omega_a+\Delta\omega&=&\frac{e}{mc}\left[a_\mu(|\bfB|-|\bfB|(1-\cos\psi)\right]\\
    \rightarrow \Delta\omega&=&\frac{e}{mc}\left[a_\mu(-|\bfB|(1-\cos\psi)\right]
  \end{eqnarray*}
  And
  $$\langle \Delta\omega\rangle=\frac{e}{mc}\left[a_\mu\langle -|\bfB|(1-\cos\psi)\rangle\right]
  $$
  Note that $$|\bfB|(1-\cos\psi) = |\bfB|-|\bmbetahat\times \bfB|$$ and
  $$\frac{\Delta \omega_p}{\omega_a} = -\langle 1-|\hat{\bm\beta}\times {\hat\bfB}|\rangle\sim -\langle \onehalf\psi^2\rangle$$

  All very well as long as $\psi$ is independent of time. But of course it is not. Indeed we can assume time dependence
  $$\psi(t)=\psi_0\cos\omega_p t$$ where $\omega_p$ is the vertical betatron frequency. 
\end{comment}

\section*{Efield}
The contribution of the electric field to the precession frequency is more straightforward.
Expand about the magic momentum in Equation ~\ref{eq:j}
\begin{eqnarray*}
\der{}{t}(\bmbetahat\cdot\bfs)&=&-\frac{e}{mc}\bfs_\perp\cdot\left[a_\mu\bmbetahat\times\bfB+\left(a_\mu-\frac{m^2c^2}{p^2}\right)
\beta\bfE\right]\\
&\sim&-\frac{e}{mc}\bfs_\perp\cdot\left[a_\mu\bmbetahat\times\bfB+\left(a_\mu-\frac{m^2c^2}{p_m^2}(1-2\frac{\Delta p}{p_m})\right)
\beta\bfE\right]\\
&\sim&-\frac{e}{mc}\bfs_\perp\cdot\left[a_\mu\bmbetahat\times\bfB -2a_\mu\frac{\Delta p}{p_m}\beta\bfE\right]
\end{eqnarray*}
With the usual substitution of $\bfs_\perp=|\bfs|\cos\omega_m t \bmhr$
 we have
\begin{eqnarray*}
\Delta\omega_E &=& -\frac{e}{mc}a_\mu(-2\frac{\Delta p}{p}\beta E_\rho)\\
\rightarrow \frac{\Delta\omega_E}{\omega_a}=C_e &=&\frac{-2\vev{\frac{\Delta p}{p} \beta E_\rho}}{B}
\end{eqnarray*}
We can write
$\frac{Delta p}{p} = x_e\eta$, where $x_e$ is the momentum dependent displacement of the closed orbit from the magic radius. 
The radial electric field is given by
$$E_\rho = n\left(\frac{\beta B}{R_0}\right) x_e$$
so that
$$\vev{C_e} = -2\beta^2 n(1-n)\frac{\vev{x_e^2}}{R_0^2}$$ 
where we have used $\eta = R_0/(1-n)$.
