\subsection{\label{sec:efield} E-field Correction}
\textcolor{orange}{David R.}
\medskip

In the rest frame of the muon the transverse electic field manifests as a magnetic field and alters the instantaneous precession frequency,
as defined by the third term on the right hand size of Eq.~(\ref{eq:betadots}).  
We minimize the contribution of the electric field by operating near the magic momentum, namely where $a_\mu = \frac{1}{\gamma^2-1} = a_\mu - \frac{m^2}{p_m^2}.$ 
The field of the ring magnet is chosen so that the closed orbit of the magic momentum muon corresponds to the center of the good field region, the design orbit.
The fractional contribution of electric field to $\omega_a$ is
\begin{equation}
C_e\sim-2\frac{\Delta p}{p_m}\frac{|\langle\vec\beta\times\vec E\rangle|}{Bc},
\end{equation}
where $p_m$ is the magic momentum, and $\gamma$ and $\beta$ are evaluated at $p_m$ and averaged along the trajectory of the muon.
The electric field depends on transverse displacement in the quadrupoles. Insofar as the field is linear in displacement, 
the average electric field along the trajectory is proportional to the average radial displacement $x_e$, that is the off momentum closed orbit. 
The  average radial electric field is given by 
\begin{equation}
\langle E_r\rangle = n\left(\frac{v_s B}{R_0}\right)x_e,\label{eq:averageE}
\end{equation}
 where $n$ is the focusing index. 
The average radial displacement is related to momentum through the dispersion function $\eta$, $$x_e = \eta\frac{\Delta p}{p}.$$ 
Then the contribution to $\omega_a$ due to electric field is 
\begin{equation}
C_e \sim -2\beta^2n(1-n)\frac{x_e^2}{R_0^2},\label{eq:efieldlinear}
\end{equation}
 where $\eta =\sim\frac{1-n}{R_0}$. 
Determination of the electric-field correction thus depends on measurement of the equilibrium radial distribution. 
%(The magnitude of the E-field correction is reduced by a few percent due to the nonlinearity of the quadrupole field.)


