Special theory of relativity > The constancy of the light speed. Velocity addition > Problem 14.1.7

Statement

$14.1.7.$ Along the line connecting two stationary stations relative to each other, a spacecraft was moving at a speed of $v$ relative to the stations. ”The stations were at the same distance from our ship when our light signal was reflected on them at the same time, since the light signals were sent simultaneously to the stations and they returned after being reflected from the stations at the same time,” says the observer from the ship. Employees of the station observed that the signals were reflected from the stations at different times. How can these differences be explained? What difference in reflection times was observed by the station staff if the distance between the stations (in their system) is equal to l? At what distances did they fix the ship at the moments of reflections of signals from stations?

Solution

The key idea in this problem is that light takes a finite, non-zero time to travel from one place to another.

First, the signals were sent by the spaceship when the spaceship was at distances $x_1$ and $l-x_1$ from the stations. Henceforth, we will work in the frame where the stations are at rest for this solution. Also, the signal traveling to the left is signal $1$ and is reflected by the first station, and the signal traveling to the right is signal $2$ and is reflected by station $2$.

To calculate $x_1$ we need to use the idea that the signals, after being reflected, reach the ship at the midpoint between the stations; otherwise the signals would not reach the ship at the same time.

The time taken by ray $1$ to reach the spaceship at the midpoint is:

\begin{equation}
t_1 = \frac{x_1 + l/2}{c}
\end{equation}

On the other hand, the spaceship will move a distance $l/2 - x_1 = v t_1$ in that time. Substituting the time into this equation yields:

\begin{equation}
\frac{x_1 + l/2}{c} = \frac{l/2 - x_1}{v} ;\rightarrow; x_1 = \frac{l}{2} \frac{1-v/c}{1+v/c}
\end{equation}

Knowing this, we can find the time difference between signals at one of the stations. The difference $\Delta t$ arises because ray $1$ takes a time
$t_{11} = \frac{x_1}{c} = \frac{l}{2c} \frac{1-v/c}{1+v/c}$ to travel from the emission point to the first station. On
the other hand, ray $2$ takes a time $t_{22} = \frac{l-x_1}{c} = \frac{l}{2c} \frac{1+3v/c}{1+v/c}$ to reach the second station, and a time $t_{21} = \frac{l}{c}$
to travel from the first station to the second station (or vice versa). Thus, the difference between the reflection times as seen from station $1$ is:

\begin{equation}
\Delta t = t_{22} + t_{21} - t_{11} = \frac{l}{c} + \frac{2l (v/c)}{1+v/c} = \frac{l}{c} \frac{1 + 3v/c}{1+v/c}
\end{equation}

Finally, when the spaceship sends the signals, it is at a distance $x_1$ from station $1$. The stations will observe the spaceship's position when the signals reach each station. Note that the signals take time to travel from the emission point to the stations. When signal $1$ reaches station $1$, the spaceship will be at a distance:

\begin{equation}
x'_1 = x_1 + v t_{11} = \frac{l}{2} \left(1-\frac{v}{c}\right)
\end{equation}

And when ray $2$ reaches station $2$, the spaceship will be at:

\begin{equation}
x'_2 = l - x_1 - v t_{22} = (l - x_1)\left(1-\frac{v}{c}\right) = \frac{l}{2} \frac{1+3v/c}{1+v/c} \left(1-\frac{v}{c}\right)
\end{equation}

Answer

\begin{equation}
\Delta t= \frac{l}{c} \frac{1 + 3v/c}{1+v/c}
\end{equation}

\begin{equation}
x'_1 = \frac{l}{2} \left(1-\frac{v}{c}\right)
\end{equation}

\begin{equation}
x'_2 = \frac{l}{2} \frac{1+3v/c}{1+v/c} \left(1-\frac{v}{c}\right)
\end{equation}