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D. Rubin\\
December 27, 2018
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{\bf \Large Pitch Systematic}
\end{center}

The conventional strategy for estimating the correction due to the vertical component of the muon velocity is based on measurement of the distribution in the vertical phse space.
The vertical velocity component is then related to the measured vertical distribution, and that the velocity component is proportional to the average of ${\bf(\vec\beta\cdot \vec B)\vec\beta}$
along the trajectory of the muon. We test the methodology in simulation by comparing the shift in $\omega_a$ based on $\langle{\bf(\vec\beta\cdot \vec B)\vec\beta}\rangle$ with spin tracking. 

\section*{Thomas Equation}

According to the Thomas Equation (Jackson 11.170) time dependence of the spin is related to electro-magnetic fields, and particle momentum, charge and mass, according to
\begin{equation}
\frac{d{\bf s}}{dt} = \frac{e}{mc}{\bf s\times}\left[\left(\frac{g}{2}-1 + \frac{1}{\gamma}\right){\bf B} -
\left(\frac{g}{2}-1\right)\frac{\gamma}{\gamma+1}{(\bf\vec\beta\cdot B)\beta} -\left(\frac{g}{2}-\frac{\gamma}{\gamma+1}\right){\bf\beta\times E}\right] \label{eq:thomas}
\end{equation}
The equation of motion for the muon is (Jackson 11.168)
\begin{equation}
\frac{d\bf\vec\beta}{dt} = \frac{e}{mc\gamma}\left[{\bf E } + {\bf\vec\beta\times B}-({\bf\vec \beta\cdot E}){\bf \vec\beta}\right]\label{eq:eom}
\end{equation}
We measure $\frac{d (\bf \hat\beta\cdot s)}{dt}$.
\begin{equation}
\frac{({d\bf\hat\beta\cdot s})}{dt} = {\bf\hat\beta}\cdot\frac{d{\bf s}}{dt}+\frac{1}{\beta}[{\bf s}-({\bf\hat\beta\cdot s}){\bf\hat\beta}]\cdot\frac{d\bf\beta}{t}
\end{equation}
The above are combined to give (Jackson 11.171)
\begin{equation}
\frac{d({\bf\hat\beta\cdot s})}{dt} = -\frac{e}{mc}{\bf s_\perp}\cdot\left[\left(\frac{g}{2}-1\right)({\bf\hat \beta\times B}) + \left(\frac{g}{2}-1 - \frac{1}{\gamma^2-1}\right){\bf E}\right]
\end{equation}
Let $\hat\beta\cdot{\bf s} =s\cos\theta$, and ${\bf s_\perp} =\hat\beta\times {\bf s} = s\sin\theta {\bf\hat s_\perp}$ so that
\begin{eqnarray}
\frac{d({\bf\hat \beta\cdot s})}{dt} &=& -s\sin\theta\frac{d\theta}{dt} =   -\frac{e}{mc}{\bf s_\perp\cdot}\left[\left(\frac{g}{2}-1\right)({\bf\hat \beta\times B}) + \left(\frac{g}{2}-1 - \frac{1}{\gamma^2-1}\right){\bf E}\right]\label{eq:betadots}\\
&\rightarrow& -\omega = -\frac{e}{mc}{\bf \hat s_\perp}\cdot\left[\left(\frac{g}{2}-1\right)({\bf\hat \beta\times B}) + \left(\frac{g}{2}-1 - \frac{1}{\gamma^2-1}\right){\bf E}\right]\\
&\rightarrow&  -\frac{e}{mc}{\bf \hat s_\perp}\cdot\left[\left(\frac{g}{2}-1\right)({\bf\hat \beta\times B}) + \left(\frac{g}{2}-1 - \frac{1}{\gamma^2-1}\right){\bf \hat\beta\times E}\right]\\
\end{eqnarray}
That last step since ${\bf s_\perp\cdot E} = {\bf \hat\beta\times s\cdot E} = {\bf s\cdot (\hat\beta\times E)}= {\bf s_\perp\cdot(\hat\beta\times E)}.$
If ${\bf\hat\beta} = \beta(\cos\psi \hat\phi +\sin\psi\hat z)$ then
\begin{equation}
\omega\rightarrow \omega \frac{({\bf\hat\beta\times B})}{B}= \omega\cos\psi\sim\omega (1-\frac{1}{2}\psi^2)
\rightarrow \Delta\omega = \frac{1}{2}\psi^2
\end{equation}
Let's try another representation.
Assume 
$${\vec\beta} = \beta({\bf\hat k}\cos(\psi_0\cos\omega_v t) + {\bf\hat j}\sin(\psi_0\sin\omega_v t))$$
where $\omega_v$ is the vertical betatron frequency and $\psi$ is the pitch angle.
Define $\theta$ so that ${\bf\hat\beta\cdot s} = s\cos\theta$ and ${\bf\hat\beta\times s} = s\sin\theta{\bf\hat s_\perp}$.
Then ${\bf\hat s_\perp\cdot B} = B\cos\psi$.  ($B$ in the vertical direction).
Now
\begin{eqnarray*}
\frac{d({\bf\hat\beta\cdot s})}{dt} &=& -\frac{e}{mc}{\bf s_\perp}\cdot\left[\left(\frac{g}{2}-1\right)({\bf\hat \beta\times B}) + \left(\frac{g}{2}-1 - \frac{1}{\gamma^2-1}\right){\bf E}\right]\\
\frac{s d(\cos\theta)}{dt} &=& -\frac{e}{mc}{\bf B}\cdot\left(\frac{g}{2}-1\right)({\bf s\times \hat\beta})\\
-\sin\theta \frac{d\theta}{dt} &=& -\frac{e}{mc}B\left(\frac{g}{2}-1\right)\cos\psi (s\sin\theta)\\
\rightarrow \omega_a &=& -\frac{e}{mc}B\left(\frac{g}{2}-1\right)\cos\psi
\end{eqnarray*}


Then ${\bf\beta\times B} = {\bf \hat i}\psi B\cos\omega_v t$

${\bf\hat\beta} = (\cos\alpha {\bf\hat k}+\sin\alpha{\bf\hat j})$. Note that ${\bf s_\perp \cdot (\hat\beta\times B}) = {\bf s \cdot (\hat\beta\times B})$. Then
\ref{eq:betadots} becomes 
\begin{equation}
\frac{d({\bf\hat\beta\cdot s})}{dt}= \frac{d}{dt}\left(s(\sin\alpha\sin\theta\sin\phi + \cos\alpha\cos\theta)\right) \sim \frac{d}{dt}\left(s(\cos\alpha\cos\theta)\right)
\end{equation}
The pitch angle $\alpha$ oscillates with the vertical betatron frequency which is much greater than the precession frequency with average value of zero.
Then
\begin{eqnarray*}
\frac{d({\bf\hat\beta\cdot s})}{dt} &=& -s(\cos\alpha\sin\theta)\frac{d\theta}{dt} = sB\sin\theta\cos\phi\cos\alpha \left(\frac{g}{2}-1\right)\\
&\rightarrow&\omega = -B\cos\phi\left(\frac{g}{2}-1\right)
\end{eqnarray*}
\subsection{E-field correction}
If ${\bf E} = E_x{\bf\hat i}$ and ${\bf\vec\beta}  = \beta{\bf\hat z}$ then from \ref{eq:betadots} 
\begin{equation}
\frac{d({\bf\hat\beta\cdot s})}{dt}= \frac{d}{dt}s(\cos\theta)
\end{equation}
Then
\begin{eqnarray*}
\frac{d({\bf\hat\beta\cdot s})}{dt} &=& -s(\sin\theta)\frac{d\theta}{dt} = sE\sin\theta\cos\phi \left(\frac{g}{2}-1 - \frac{1}{\gamma^2-1}\right)\\
&\rightarrow&\Delta\omega = -E\cos\phi\left(\frac{g}{2}-1- \frac{1}{\gamma^2-1}\right)
\end{eqnarray*}





\begin{comment}
If we assume that the magnetic field ${\bf B}=B{\bf\hat z}$ is vertical, the spin ${\bf s}$ is in the $\rho-\phi$ plane, ${\bf E}=0$, and $\beta$ is in the $z-\phi$ plane, then Equation
Equation \ref{eq:thomas} becomes
\begin{eqnarray*}
\frac{d{\bf s}}{dt} &=& \frac{e}{mc}{\bf s\times}\left[\left(\frac{g}{2}-1 + \frac{1}{\gamma}\right){\bf B} -
\left(\frac{g}{2}-1\right)\frac{\gamma}{\gamma+1}{(\bf\vec\beta\cdot B)\beta} \right]\\ && \rightarrow \omega_p 
=  \frac{e}{mc}\left[\left(\frac{g}{2}-1 + \frac{1}{\gamma}\right){\bf B} -
\left(\frac{g}{2}-1\right)\frac{\gamma}{\gamma+1}{(\bf\vec\beta\cdot B)\beta}\right] 
\end{eqnarray*}
and \ref{eq:eom} becomes
\begin{equation}
\frac{d\bf\vec\beta}{dt} = \frac{e}{mc\gamma}{\bf\vec\beta\times B} \rightarrow \omega_c = \frac{e}{mc\gamma}{\bf\hat\beta\times B}
\end{equation}
We measure
\begin{equation}
\omega_a = \omega_p - \omega_a
\end{equation}
If ${\bf\vec\beta}$ is in the $\rho-\phi$ plane then
\begin{equation}
\omega_{a0} = 
 \frac{e}{mc}\left[\left(\frac{g}{2}-1 + \frac{1}{\gamma}\right) B -
\left(\frac{g}{2}-1\right)\frac{\gamma}{\gamma+1}{(\bf\vec\beta\cdot B)\beta}\right] - \frac{e}{mc\gamma}B =
\frac{e}{mc}\left(\frac{g}{2}-1 \right) B 
\end{equation}
But suppose that $\beta$ lies in the $z-\phi$ plane. 
Then
\begin{equation}
\omega_a = 
 \frac{e}{mc}\left[\left(\frac{g}{2}-1 + \frac{1}{\gamma}\right) {\bf B} -\left(\frac{g}{2}-1\right) \frac{\gamma}{\gamma+1}{\bf B}\beta_z^2\right] - \frac{e}{mc\gamma}{\bf\hat\beta\times B}
\end{equation}
becomes
\begin{equation}
\omega_a = 
 \frac{e}{mc}\left[\left(\frac{g}{2}-1 \right)B  -
\left(\frac{g}{2}-1\right)\frac{\gamma}{\gamma+1}B\beta_z^2 + \frac{1}{\gamma}( B -  \frac{\beta_\phi}{\beta} B)\right]
\end{equation}
If we substitute $\beta_z = \beta\sin\psi$ where $\psi$ is the pitch,and note that at the magic momentum $\frac{g}{2}-1 = \frac{1}{\gamma^2-1}$ then
\begin{eqnarray*}
\omega_a &=& 
 \frac{e}{mc}\left[\left(\frac{g}{2}-1 \right)B  -
\left(\frac{1}{\gamma^2-1}\right)\frac{\gamma-1}{\gamma}B\sin^2\psi + \frac{1}{\gamma}( B -  \cos\psi B)\right]\\
&\sim &
 \frac{Be}{mc}\left[\frac{g}{2}-1   -
\frac{1}{\gamma(\gamma+1)}\psi^2 + \frac{1}{\gamma} \frac{\psi^2}{2}\right]\\
&\sim &
 \frac{Be}{mc}\left[\frac{g}{2}-1+
\frac{1-\frac{1}{2}(\gamma+1)}{\gamma(\gamma+1)}\psi^2\right]\\
&=& \frac{Be}{mc}\left[\frac{g}{2}-1+
\frac{\frac{1}{2}(1-\gamma)}{\gamma(\gamma+1)}\psi^2\right]\\
&\sim & \frac{Be}{mc}\left[\frac{g}{2}-1-
\frac{1}{2\gamma}\psi^2\right] = \omega_{a0}-\omega_c\frac{\psi^2}{2}\\
&=&\omega_{p}-\omega_{c}(1+\frac{\psi^2}{2}
\end{eqnarray*}
\end{comment}

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    D. Stratakis, `` 400,000 particles at the end of M5'', DocDb 4461-v1
\bibitem{stoughton}
C. Stoughton,``Simulated pulses for Kicker upgrades” C. Stoughton in docdb 11721
\end{thebibliography}
