= Solution
In the station frame, intersecting the signal trajectories with the endpoint <world lines> gives
$$
ut'_A=vt'_A+\frac L{2\gamma},\qquad -ut'_B=vt'_B-\frac L{2\gamma}.
$$
Since $u>v$, both receptions occur at positive times,
$$
t'_A=\frac L{2\gamma(u-v)},\qquad t'_B=\frac L{2\gamma(u+v)}.
$$
The inverse <Lorentz transformation> then gives
$$
t_A=\gamma t'_A(1-uv/c^2)=\frac{L(1-uv/c^2)}{2(u-v)},\qquad t_B=\gamma t'_B(1+uv/c^2)=\frac{L(1+uv/c^2)}{2(u+v)}.
$$
Subtracting and simplifying yields
$$
\boxed{t_B-t_A=-\frac{Lv(1-u^2/c^2)}{u^2-v^2},\qquad t'_B-t'_A=-\frac{Lv}{\gamma(u^2-v^2)}.}
$$
For $u<c$, both differences are negative: the rear receives first in both frames, but by different time intervals.
Back to article page
