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\title{Brief Article}
\author{The Author}
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\begin{equation}
\vec\omega_a = -\frac{q}{m}\left[a_\mu\vec B -\left(a_\mu -\frac{1}{\gamma^2-1}\right)\frac{\vec\beta\times\vec E}{c}\right]
\end{equation}
Energy dependence of $\vec\omega_a$
\begin{eqnarray*}
d\vec\omega_a&=&-\frac{q}{m}d(\gamma^2-1)^{-1}\frac{\vec\beta\times\vec E}{c}\\
\frac{d\vec\omega_a}{d\gamma}&=& \frac{q}{m}\frac{2\gamma}{(\gamma^2-1)^2}\frac{\vec\beta\times\vec E}{c}\\
\Delta\omega_a&=& \frac{q}{m}\frac{2\gamma\Delta\gamma}{(\gamma^2-1)^2}\frac{\vec\beta\times\vec E}{c}
\end{eqnarray*}
Since $\gamma\gg 1$
\begin{eqnarray*}
\Delta\omega_a&\approx& \frac{q}{m}\frac{2\Delta\gamma}{\gamma^3}\frac{\vec\beta\times\vec E}{c}
\end{eqnarray*}
The electric field is roughly proportional to $\Delta\gamma$. Let $E=E_0\Delta\gamma$. Note that the electric field
changes sign with change in sign of energy offset. This is assuming that $\Delta\gamma = 0$ corresponds to $E=0$.
\begin{eqnarray*}
\Delta\omega_a&\approx& \frac{q}{mc}\frac{2\Delta\gamma}{\gamma^3}{E_0\Delta\gamma}\\
&\approx& \frac{qE_0}{mc}\frac{2(\Delta\gamma)^2}{\gamma^3}\\
\end{eqnarray*}

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