\section{Interaction with an Electric Dipole Moment}
The Thomas BMT equation extended to include an electric dipole moment is
\begin{equation}
\der{}{t}{\bf s} = \frac{e}{mc}\bfs\times\left[(a_\mu+\frac{1}{\gamma})\bfB -a_\mu\frac{\gamma}{\gamma+1}({\bm \beta}\cdot\bfB){\bm \beta}-\left(\frac{g}{2}-\frac{\gamma}{\gamma+1}\right){\bm\beta}\times\bfE\right.\\                    \left.       +\frac{\eta}{2}\left(\bfE-\frac{\gamma}{\gamma+1}({\bm\beta}\cdot\bfE){\bm\beta}+{\bm\beta}\times\bfB\right)\right]
\end{equation}
The Hamiltonian is
\begin{equation}
{\cal H} = -\frac{e}{mc}\frac{\hbar}{2}{\bm\sigma}\cdot\left[(a_\mu+\frac{1}{\gamma})\bfB -a_\mu\frac{\gamma}{\gamma+1}({\bm \beta}\cdot\bfB){\bm \beta}-\left(\frac{g}{2}-\frac{\gamma}{\gamma+1}\right){\bm\beta}\times\bfE\right.\\                    \left.       +\frac{\eta}{2}\left(\bfE-\frac{\gamma}{\gamma+1}({\bm\beta}\cdot\bfE){\bm\beta}+{\bm\beta}\times\bfB\right)\right]
\end{equation}
Suppose the magnetic field $\bfB=B_z{\bf\hat z}$ and the electric field is in the radial direction. Assume no betatron oscillations so the particle circulates on its closed orbit.
The electric field is
\begin{equation}
\bfE=-E_r(\sin\theta\bfhx+\cos\theta\bfhy)
\end{equation}
The velocity has the form
\begin{equation}
{\bm\beta}=\beta(\cos\omega t\bfhx - \sin\omega t\bfhy)
\end{equation}
where $\omega = \frac{e}{mc\gamma}\left(\frac{E_r}{\beta}-B_z\right)$ and $\theta=\omega t$.
The effective electro-magnetic field 
\begin{equation}
{\cal B} = \left[B_z(a_\mu+\frac{1}{\gamma})-\left(a_\mu-\frac{1}{\gamma+1}\right)\beta E_r\right]\bfhz +\frac{\eta}{2}\left[-E_r(\sin\omega t\bfhx+\cos\omega t\bfhy)+\beta B_z (-\cos\omega t\bfhy-\sin\omega t\bfhx)\right]
\end{equation}
The Hamiltonian takes the form
\begin{eqnarray}
{\cal H}=-\frac{e}{mc}\frac{\hbar}{2}{\bm\sigma}\cdot{\cal B} &=& -\frac{\hbar}{2}\begin{pmatrix}\omega_0 & \frac{\eta}{2}\frac{e}{mc}\left[iE_re^{i\omega t} +i\beta B_ze^{i\omega t}\right]\cr
                                                    \frac{\eta}{2}\frac{e}{mc}\left[-iE_re^{-i\omega t} -i\beta B_ze^{-i\omega t}\right] & -\omega_0\end{pmatrix}\\
&=& -\frac{\hbar}{2}\begin{pmatrix}\omega_0 & i\Pi_0 e^{i\omega t}\cr
                                                    -i\Pi_0 e^{-i\omega t} & -\omega_0\end{pmatrix}
\end{eqnarray}
where 
\begin{eqnarray*}
\omega_0 &=& \frac{e}{mc}\left(B_z(a_\mu+\frac{1}{\gamma})-\left(a_\mu-\frac{1}{\gamma+1}\right)\beta E_r\right)\\
\omega_0-\omega &=& \frac{e}{mc}\left(B_z a_\mu -\left(a_\mu -\frac{1}{\gamma+1}\right)\beta E_r - \frac{E_r}{\beta\gamma}\right)\\
 &=& \frac{e}{mc}\left(B_z a_\mu -\left(a_\mu -\frac{1}{\gamma+1}+\frac{1}{\beta^2\gamma}\right)\beta E_r \right)\\
 &=& \frac{e}{mc}\left(B_z a_\mu -\left(a_\mu -\frac{1}{\gamma+1}+\frac{\gamma}{\gamma^2-1}\right)\beta E_r \right)\\
 &=& \frac{e}{mc}\left(B_z a_\mu -\left(a_\mu -\frac{1}{\gamma^2-1}\right)\beta E_r \right)\\
\Pi_0 &=&\frac{\eta}{2}\frac{e}{mc}[E_r +\beta B_z]\\
&=&-\omega_{edm}(1+\frac{E_r}{\beta B_z})
\end{eqnarray*}
In the rotating frame
\begin{eqnarray*}
{\cal H}_{rot}
&=& -\frac{\hbar}{2}\begin{pmatrix}\omega_0-\omega & i\Pi_0 \cr
                                                    -i\Pi_0  & -\omega_0+\omega\end{pmatrix}
\end{eqnarray*}
The precession frequency in the rotating frame is the difference of the eigenvalues
\begin{eqnarray*}
\omega^\prime&=& [(\omega_0-\omega)^2 +\Pi_0^2]^{1/2}
\end{eqnarray*}
At the magic momentum
\begin{eqnarray*}
\omega^\prime&=& [\omega_a^2 + \omega^2_{edm}(1+\frac{E_r}{\beta B_z})^2]^{1/2}
\end{eqnarray*}
