Question

Two events A and B occur at places separated by 10${}^6$ km, B occurring 5 s after A. (a) Find the velocity of a frame in which these events occur at the same place. (b) What is the time interval between the events in this frame?

Answer 1

(a) Suppose the events A and B occur at points X and Y at times $t_A$ and $t_B$ where $t_B=t_A+5s$. Consider a small train which is at the point X when the event A occurs.
Suppose, this same train moves towards Y and reaches the point Y when the event B occurs. Thus, the events A and B occur at the same place in the train frame. This frame moves ${10}^6$ km in 5 s as seen from the original frame. Thus, the velocity of the train frame is
$v=\frac{{10}^{6}km}{5s}=2\times {10}^{8}m{s}^{-1}$
(b) As the events A and B occur at the same place in the train frame, the time interval between the events measured in this frame is the proper interval. Thus, this time interval is
$=\left(5s\right)\sqrt{1-v^2/c^2}=\left(5s\right)\sqrt{1-{\left(\frac{2}{3}\right)}^2}$
$=3.7s$