The rod `A^'B^'` moves with a constant velocity v relative to the rod AB (figure). Both rods have the same proper length `l_0` and at the ends of each of them clocks are mounted, which are synchronized pariwise: A with B and `A^'` with `B^'`. Suppose the moment when the clock `B^'` gets opposite the clock A is taken for the beginning of the time count in the reference frames fixed to each of the rods. Determine:
(a) the readings of the clocks B and `B^'` at the moment when they are opposite each other,
(b) the same for the clocks A and `A^'`.

Explanation

At the instant the picture is taken the coordinates of `A,B,A^',B^'` in the rest frame of AB are
`A:(0,0,0,0)`
`B:(0,l_0,0,0)`
`B^':(0,0,0,0)`
`A^':(0_1-l_0sqrt(1-v^2//c^2),0,0)`
In this frame the coordinates of `B^'` at other times are `B^': (t, vt, 0, 0)`. So `B^'` is opposite to B at time `t(B)=l_0/v`. In the frame in which `B^'`, `A^'` is at rest the time corresponding this is by Lorentz transformation.
`t^0(B^')=(1)/(sqrt(1-v^2/c^2))(l_0/v-(vl_0)/(c^2))=l_0/vsqrt(1-v^2//c^2)`
Similarly in the rest frame of A, B, te coordinates of A at other times are
`A^':(t, -l_0sqrt(1-v^2/c^2)+vt, 0,0)`
`A^'` is opposite to A at time `t(A)=l_0/vsqrt(1-v^2/c^2)`
The corresponding time in the frame in which `A^'`, `B^'` are at rest is
`t(A^')=gammat(A)=l_0/v`