Problem 14.4.1

Solution

First situation (system S): A uniform electric field $\mathbf{E}$ is at rest. An electron enters with speed$v = \beta c$ parallel to $\mathbf{E}$, remains for a time$\tau$ and exits.

Second situation (system S'): The electron is initially at rest and it is the field that moves toward it with speed $v = \beta c$. The electron interacts with the field for a time T (to be determined).

Method a) Time dilation

In the second situation, we place ourselves in the reference frame that moves with the field (S). There the electron enters with speed $\beta c$ and takes $\tau$ to exit

In system S' (electron initially at rest), the moving field passes over it between two events (arrival of the leading edge and departure of the trailing edge of the field). The time interval in S' between these two events is

$T = \gamma \tau, \qquad \gamma = \frac{1}{\sqrt{1-\beta^2}}$

$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}}}$

Method b) Lorentz force and relativistic Newton's second law

First situation: Electron moving at $\beta c$ inside the field at rest.
Force on the electron $F = eE$
The equation of motion in relativity

$\frac{dp}{dt} = F \quad\Rightarrow\quad \Delta p = eE\,\tau$

where $p = \gamma m_e v = \dfrac{m_e v}{\sqrt{1-\beta^2}}$

The electron reverses its speed (enters with $+v$ and exits with$-v$). The total change in momentum is

$\Delta p = \gamma m_e v - (-\gamma m_e v) = 2\gamma m_e v = 2\gamma m_e c\beta$

Therefore:

$\tau = \frac{2\gamma m_e c\beta}{eE}$

Second situation: The field "hits" the electron at rest with speed $\beta c.$In the electron's reference frame, it starts from rest and is accelerated by the moving field. At the end, the electron acquires a speed $\beta_1 c$ given by the relativistic velocity addition between the initial speed of the field ($\beta c$) and the final speed in the field's frame ($-\beta c$)

$\beta_1 c = \frac{\beta c + \beta c}{1 + \beta^2} = \frac{2\beta c}{1+\beta^2}$

The change in momentum of the electron (from rest to $\beta_1 c$) is

$\Delta p = \gamma_1 m_e \beta_1 c, \quad \gamma_1 = \frac{1}{\sqrt{1-\beta_1^2}}$

The force on the electron is the same $eE$
The time T that the interaction lasts satisfies

$T = \frac{\Delta p}{eE} = \frac{\gamma_1 m_e \beta_1 c}{eE}$

Using the relations between$\gamma_1, \beta_1$ and $\gamma, \beta$

it can be shown that

$\gamma_1 = \gamma(1+\beta^2), \quad \beta_1 = \frac{2\beta}{1+\beta^2}$

$\gamma_1 \beta_1 = 2\gamma \beta$

Therefore

$T = \frac{2\gamma m_e c\beta}{eE} = \gamma \tau$

$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}$

Both methods yield the same result:

$\boxed{T = \frac{\tau}{\sqrt{1-\beta^2}} = \gamma \tau}$.