\begin{flushright}
D. Rubin\\
August 4,  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*{Pitch}
Our unperturbed system is a muon precessing in a uniform magnetic field with zero electric field. We use a cylindrical coordinate system
where ${\bf B} = B{\bf\hat z}$.
The rate of change of the polarization with respect to the muon velocity 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}
Our trial solution for time dependence of the polarization in the $\rho,\phi$ plane is 
$\bfs = |\bfs|(\cos\omega_a t \bmhr-\sin\omega_a t \bmhp)$ where in the unperturbed system $\bmbetahat = \bmhp$.
We find that
$$\bfs_\perp = \bfs - (\bmbetahat\cdot \bfs)\cdot\bmbetahat = |\bfs|\cos\omega_a t\bmhr$$
and $$\bmbetahat\cdot\bfs = -|\bfs|\sin\omega_a t$$
Substitution into ~\ref{eq:j} gives us
$$\omega_a = \frac{e}{mc}Ba_\mu$$

More generally, the polarization may have a component out of the plane,  
\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
$$\omega_a = \frac{e}{mc}B a_\mu$$

If there is a component of velocity parallel to the magnetic field, $\bmbetahat = \cos\psi\bmhp + \sin\psi\hat{\bf z}$
\begin{eqnarray*}
\bfs_\perp &=& |\bfs|\cos\Delta\psi\sin\omega_a\bmhr\\
\bmbetahat\cdot\bfs &=& -|\bfs|\cos\Delta\psi\sin\omega_a t
\end{eqnarray*}
where $\Delta\psi = \psi-\theta$. Then
\begin{eqnarray*}
\omega_a &=& \frac{e}{mc}B a_\mu \cos\psi
\end{eqnarray*}
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. 


Now suppose te muon is executing vertical betatron oscillations in the $\phi-z$ plane. Then
$$\bmbetahat = \cos\psi(t)\bmhp + \sin\psi(t)\hat{\bf z}$$
where $\psi(t)$ is the angle of the trajectory with respect to the $\rho-\phi$ plane.
If we suppose 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, the orientation of the polarization of the magic momentum muon with respect to the velocity is frozen.
Namely, 
$$\der{}{t}(\bmbetahat\cdot\bfs) = 0.$$
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 $\psi(t) = \psi_0\cos\omega_p t$. 
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 $\omega_p\gg\omega_a$ limit, the effect of the magnetic field is that the z-component of the polarization
lags (or leads) the z-component of the velocity. If $\omega_p\gg\omega_a$ the plane of the polarization vector will oscillate about the $\bmhr$ direction. 
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}
\end{comment}
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 $\omega_a$.

\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 + w\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 + w\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{\psi_0^2}{4\sin\omega_a t}}\left(\frac{\sin(\omega_a + 2\omega_p)t}{\omega_a + w\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.

\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}

\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*}

Substitution of Equation ~\ref{eq:s} into ~\ref{eq:j} and assuming $\bmbetahat=\bmhr$, 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*}
.
