Problem 14.4.12^∗

Solution

We work in the system S' where the electric field is at rest. In S', the electron moves with speed v' and is decelerated by the constant field E. The penetration depth in S' is l', and the work–energy relation gives:

$eE\,l' = (\gamma' - 1)m_e c^2, \qquad \gamma' = \frac{1}{\sqrt{1 - v'^2/c^2}}$

In the laboratory, the electron travels with speed v and the field with speed u (in opposite directions). In S':

$v' = \frac{v + u}{1 + \dfrac{vu}{c^2}}$

The penetration distance measured in the laboratory l is contracted with respect to l':

$l = \frac{l'}{\gamma_u}, \qquad \gamma_u = \frac{1}{\sqrt{1 - u^2/c^2}}$

From the velocity composition law, it follows:

$\gamma' = \gamma\gamma_u\left(1 + \frac{uv}{c^2}\right), \qquad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$

Substituting$l' = \gamma_u l$ and $\gamma'$into the energy equation and dividing by $\gamma_u$

$E = \frac{m_e c^2}{e l} \left[ \gamma\left(1 + \frac{uv}{c^2}\right) - \frac{1}{\gamma_u} \right]$

Replacing $\gamma = 1/\sqrt{1 - v^2/c^2} and 1/\gamma_u = \sqrt{1 - u^2/c^2}$

$\boxed{E = \frac{m_e c^2}{e l} \left( \frac{1 + \dfrac{uv}{c^2}}{\sqrt{1 - \dfrac{v^2}{c^2}}} - \sqrt{1 - \dfrac{u^2}{c^2}} \right)}$