Expression for maximum energy transfer to electron by fast moving charge Suppose a charge $q = +Z e$ is moving along positive $+x$ direction with electron below $x-$axis and the charge is moving fast enough to consider electron at rest, then what is the maximum value of energy that is transferred to electron?  
According to my note, 

"it can be showed that $E_{max} = 2m\gamma^2 v^2$ "

where $\gamma$ is Lorentz factor and $v, m$ are velocity and mass of the charge. My note is probably based on J.D. Jackson's Classical Electrodynamics. I haven't read that whole book. But as far as I know, such results are not derived on that book. How do I arrive at above result.
So far, I know, energy transfer can be calculated as 
$$\Delta E = \frac{2 Z^2 e^4}{mb^2v^2}$$
where $b$ is impact parameter. As $b \to 0, \Delta E \to \infty$, the original problem is to find the bounds for $b$.
 A: We start with a collision between two particles. Particle 1 has mass $m_1$, momentum $\mathbf{p}_{\rm lab}$, and total energy $E_{\rm lab}$ while Particle 2 (mass $m_2$) is stationary. Particle 1 then scatters off at an angle $\psi$ while particle 2 scatters off in angle $\zeta$. By using a transformation of coordinates, we can jump into a frame in which it would appear that the two particles are coming towards each other at the same speed (center of mass (c.m.) frame)

(source)
It's often convenient to work in the c.m. frame, so we'll use that. 
It can be shown that the total energy in the c.m. frame is given by
$$
W^2=m_1^2+m_2^2+2m_2E_{\rm lab}\tag{1}
$$
The change in energy for Particle 2 is then given by
$$
\Delta E=\frac{m_2}{W^2}p^2_{\rm lab}\left(1-\cos\theta\right)=\frac{m_2}{m_1^2+m_2^2+2m_2E_{\rm lab}}p^2_{\rm lab}\left(1-\cos\theta\right)\tag{2}
$$
In order to find the maximum energy transfer, we need to do a little hokey business. First is an easy one: 
$$\left(1-\cos\theta\right)_{max}=2\quad({\rm at}\,\,\theta=\pi)$$
Next, we use the relation $p=\gamma\beta m$ and assume that $m_1\gg m_2$ so that $W\simeq m_1$. Equation (2) then becomes
$$
\Delta E_{max}\simeq\frac{m_2}{m_1^2}\cdot\gamma^2\beta^2m_1^2\cdot2
$$
The $m_1^2$ terms cancel leaving the expected relation:
$$
\Delta E_{max}\simeq2m_2\gamma^2\beta^2
$$
