A rod AB oriented along the x axis of the reference frame K moves in the positive direction of the x axis with a constant velocity v. The point A is the forward end of the rod, and the point B its near end. Find:
(a) the proper length of the rod, if at the moment `t_A` the coorniate of the point A is equal to `x_A`, and at the moment `t_B` the coordinate of the point B is equal to `x_B`,
(b) what time interval should separate the markings of coordinates of the rod's ends in the frame K for the difference of coordinates to become equal to the proper length of the rod.

Explanation

(a) In the reference frame K with respect to which the rod is moving with velocity v, the coordinates of A and B are
`A: t, x_A+v(t-t_A),0,0`
`B: t, x_B+v(t-t_B), 0,0`
Thus `l=x_A-x_B-v(t_A-t_B)=l_0sqrt(1-beta^2)`
So `l_0=(x_A-x_B-v(t_A-t_B))/(sqrt(1-v^2//c^2))`
(b) `+-l_0-v(t_A-t_B)=l=l_0sqrt(1-v^2//c^2)`
(since `x_A-x_B` can be either `+l_0` or `-l_0`.)
Thus `v(t_A-t_B)=(+-1-sqrt(1-v^2//c_2))l_o`
i.e. `t_A-t_B=l_0/v(1-sqrt(1-v^2/c^2))`
or `t_B-t_A=l_0/v(1+sqrt(1-v^2//c^2))`