Problem 14.1.6^∗

Solution

We have:

· System S: attached to the first station (E1), at rest.

· System S': attached to the second station (E2), which moves with velocity v along the x-axis relative to S.

We consider$v > 0$ if E2 moves away from E1.

· At $t = 0$ (time of S), E1 is at$x = 0$ and E2 is at $x = L$.

· From E1, the following are emitted simultaneously:

· A light pulse (speed c).
· A test probe (constant speed u along x, to be determined).
· The light travels: E1 → E2 (bounce) → E1 (bounce) → E2.

The probe arrives at E2 exactly at the same instant that the light pulse returns to E2 for the third time.

The position of E2 as a function of time:$x_{\text{E2}}(t) = L + v t$.

· First leg (E1 → E2):
The light catches up to E2 when $c t_1 = L + v t_1 \Rightarrow t_1 = \dfrac{L}{c - v}.$

· Second leg (E2 → E1):
At $t_1$, the light reflects and returns to E1. The position of E1 is $x = 0$.
$0 = c t_1 - c(t_2 - t_1) \Rightarrow t_2 = 2t_1 = \dfrac{2L}{c - v}.$

· Third leg (E1 → E2):

The light leaves E1 again at $t_2$and reaches E2 at$t_3$.
$c(t_3 - t_2) = L + v t_2 + v(t_3 - t_2)$.
Solving:$t_3 - t_2 = \dfrac{L + v t_2}{c - v}$.
Substituting $t_2$ and simplifying gives:
$t_3 = \frac{2L}{c - v} + \frac{L(c+v)}{(c - v)^2} = L\,\frac{3c - v}{(c - v)^2}.$

This is the total time the light takes to complete its path.

The probe moves with speed u and must catch E2 at the same instant $t_3$:

$u t_3 = L + v t_3 \quad\Rightarrow\quad t_3 = \frac{L}{u - v} \quad (u > v)$.

Equating with the expression for $t_3$:

$\frac{L}{u - v} = L\,\frac{3c - v}{(c - v)^2} \quad\Rightarrow\quad u - v = \frac{(c - v)^2}{3c - v}$.

Therefore, the speed of the probe as measured from the first station is:

$\boxed{u = v + \frac{(c - v)^2}{3c - v} = \frac{c(c + v)}{3c - v}}$.

For an observer at E2 (system S'), the probe moves with speed u' given by the relativistic velocity addition formula.
Since S' moves at speed v relative to S:

$u' = \frac{u - v}{1 - \dfrac{u v}{c^2}}$.

Substituting u and simplifying:

$u - v = \frac{(c - v)^2}{3c - v}, \qquad 1 - \frac{u v}{c^2} = 1 - \frac{v\,c(c+v)}{c^2(3c - v)} = \frac{(3c - v)c - v(c+v)}{c(3c - v)} = \frac{(3c + v)(c - v)}{c(3c - v)}$.

$u' = \frac{\dfrac{(c - v)^2}{3c - v}}{\dfrac{(3c + v)(c - v)}{c(3c - v)}} = \frac{c(c - v)}{3c + v}$.

$\boxed{u' = \frac{c(c - v)}{3c + v}}$.

This is the speed at which the probe approaches the second station according to E2's instruments.

By reciprocity between inertial systems, if E2 moves with speed v relative to E1, then E1 moves with speed -v relative to E2. Hence:

$\boxed{v_{\text{E1 from E2}} = -v \quad \text{(magnitude } v\text{)}}$.

The probe constitutes another inertial system (S''). For the probe:

· The first station moves with speed -u (magnitude u).
· The second station approaches with speed u' calculated above.
Thus, the probe's team records:

$\boxed{v_{\text{E1}} = u = \frac{c(c+v)}{3c - v},\qquad v_{\text{E2}} = u' = \frac{c(c - v)}{3c + v}}$.