Special theory of relativity > Motion of relativistic particles in electric and magnetic fields > Problem 14.4.6

Statement

$14.4.6.$

A stationary electron is impacted by a longitudinal electric field of intensity
E at the speed of light. How deep will the electron penetrate into this field if
it acts on the electron in the direction of its motion?

Solution

Physical interpretation

The electric field acts with a constant force $ F = eE $ in the direction of motion of the electron.
The penetration depth is defined as the distance the electron travels until its kinetic energy equals its rest energy.

Equations of motion (relativistic dynamics)

The force modifies the linear momentum:

$\frac{dp}{dt} = eE,\qquad p = \gamma m_e v, \quad \gamma = \frac{1}{\sqrt{1 - v^2/c^2}}$

The total energy of the electron is

$\mathcal{E} = \gamma m_e c^2$

and its variation satisfies

$d\mathcal{E} = v\, dp$.

Combining:

$d\mathcal{E} = eE\, v\, dt = eE\, dx$

Integrating from

$ x = 0 (where v = 0, \mathcal{E}_0 = m_e c^2)$

$\mathcal{E} - m_e c^2 = eE x$

$\Rightarrow \quad \gamma m_e c^2 - m_e c^2 = eE x \quad\Rightarrow\quad x = \frac{m_e c^2}{eE}(\gamma - 1)$

For $\gamma = 2 (kinetic energy = m_e c^2)$

$\boxed{x = \frac{m_e c^2}{eE}}$

This is the distance the electron must travel for its kinetic energy to reach a value comparable to its rest energy;

Answer

$\boxed{x = \frac{m_e c^2}{eE}}$