%\begin{flushright}
%D. Rubin\phantom{1111} \\
%September 13,  2020
%\end{flushright}
%\begin{center}
%{\bf \Large ${\vec \omega_S-\vec\omega_c$ and pitch}
%\end{center}
\def\bmhr{{\hat{\bm\rho}}}
\def\bmhp{{\hat{\bm\phi}}}
\def\bmbeta{{{\bm\beta}}}
\def\bmbetahat{{\hat{\bm\beta}}}
\def\bfs{{\bf s}}
\def\bfhz{{\hat{\bf z}}}
\section*{${\bm \omega}_{diff}={\bm\omega}_s -{\bm \omega}_c$ and pitch}
Consider the familiar expression for the difference of the precession frequency and the cyclotron frequency. 
\begin{equation}
{\bm \omega}_{diff} = -\frac{e}{m}\left[a_\mu\bfB - a_\mu\left(\frac{\gamma}{\gamma+1}\right)
(\bmbeta\cdot \bfB)\bmbeta -\left(a_\mu - \frac{1}{\gamma^2-1}\right)\frac{\bmbeta\times\bfE}{c}\right]\label{eq:omegadiff}
\end{equation}
In our experiment we measure the projection of the polarization onto the velocity. We measure that part of ${\bm\omega}_{diff}$, ($\omega_a$) 
that is perpendicular to the velocity. 
%(That is the component of ${\bm\omega}_{diff}$ in the vertical plane that is perpendicular to the velocity.)
We are quite insensitive to rotations about the
direction of motion. We measure $|{\bm \omega}_\perp|=|\bmbetahat\times{\bm\omega}_{diff}|$. 
%The unit vector in the $\bfB-\bmbeta$ plane that is perpendicular to the velocity is
%\begin{eqnarray*}
%\hat {\bf u} &=& \frac{(\bmbeta\times\bfB)\times\bmbeta}{|\bmbeta\times(\bmbeta\times \bfB)|}\\
%\end{eqnarray*}

Then in the absence of an electric field,
\begin{eqnarray*}
 \omega_a = |{\bm\omega}_\perp|  &=& -\frac{e}{m} a_\mu|\bmbetahat\times\left[\bfB - \left(\frac{\gamma}{\gamma+1}\right)(\bmbeta\cdot \bfB)\bmbeta\right]|\\
 &=& -\frac{e}{m} a_\mu |\bmbetahat\times\bfB|
\end{eqnarray*}
%\begin{eqnarray*}
%|{\bm\omega}_\perp| = \omega_a = {\bm \omega}_{diff}\cdot\hat {\bf u} &=& -\frac{e}{m} a_\mu\left[\bfB - \left(\frac{\gamma}{\gamma+1}\right)(\bmbeta\cdot \bfB)\bmbeta\right]
%\cdot \left(\frac{(\bmbeta\times\bfB)\times\bmbeta}{|\bmbeta\times(\bmbeta\times\bfB)|}\right)\\
% &=& -\frac{e}{m} a_\mu \bfB\cdot \frac{((\bmbeta\times\bfB)\times\bmbeta)}{|\bmbeta\times(\bmbeta\times \bfB)|}\\
%&=& -\frac{e}{m} a_\mu  \frac{\bfB\cdot((\bmbeta\times\bfB)\times\bmbeta)}{|\bmbeta\times(\bmbeta\times \bfB)|}\\
%&=& -\frac{e}{m} a_\mu  \frac{(\bmbeta\times\bfB)\cdot(\bmbeta\times\bfB)}{|\bmbeta\times(\bmbeta\times \bfB)|}\\
%&=& -\frac{e}{m} a_\mu  \frac{(\bmbeta\times\bfB)\cdot(\bmbeta\times\bfB)}{\beta|(\bmbeta\times \bfB)|}\\
%&=& -\frac{e}{m} a_\mu  |(\bmbetahat\times\bfB)|\\
%\end{eqnarray*}
%where going from line 3 to 4 we use a standard vector identity. The result is consistent with~\ref{eq:j}.
Note that the second term on the right of Equation~\ref{eq:omegadiff} is parallel to the velocity and therefore does not contribute to $\omega_a$.

Then it follows that
$$\phi_a (T)= -\frac{e}{m}a_\mu\int^T |(\bmbetahat\times\bfB)| dt$$
where $\omega_a = \frac{d\phi_a}{dt}$.

\section*{${\bm \omega}_{diff}={\bm\omega}_s -{\bm \omega}_c$ and electric field correction}
Consider next the contribution from the electric field to ${\bm\omega}_{perp}$. We see that the electric field contribution is exclusively perpendicular to the velocity. 
Therefore
\begin{eqnarray*}
\Delta\omega_a = \Delta{\bm\omega}_\perp &=&  -\left(a_\mu - \frac{1}{\gamma^2-1}\right)\frac{\bmbeta\times\bfE}{c}
\end{eqnarray*}

Then $$\Delta \phi_a({\rm Efield}) = -\frac{e}{m}\left(a_\mu-\frac{1}{\gamma^2-1}\right)\int^T\frac{\bmbeta\times\bfE}{c} dt$$
where $\Delta\omega_a = \frac{d\phi_a(Efield)}{dt}$.

We can write equivalently
\begin{eqnarray*}
\Delta \phi_a({\rm Efield}) &=& -\frac{e}{m}\left(a_\mu-\frac{1}{(E/m)^2-1}\right)\int^T\frac{\bmbeta\times\bfE}{c} dt\\
&=& -\frac{e}{m}\left(a_\mu-\frac{m^2c^2}{p^2}\right)\int^T\frac{\bmbeta\times\bfE}{c} dt\\
&=& -\frac{e}{m}\left(a_\mu-\frac{m^2c^2}{p_m^2}(1-2\frac{\Delta p}{p_m})\right) \int^T\frac{\bmbeta\times\bfE}{c} dt\\
&=& -2\frac{e}{m}a_\mu\frac{\Delta p}{p_m}\int^T\frac{\bmbeta\times\bfE}{c} dt\\
\end{eqnarray*}
where at the magic momentum $p_m$, $a_\mu = \frac{m^2c^2}{p_m}^2$ and $\Delta p =p-p_m$.
