Past exam of the symmetry and relativity course of the University of Oxford / 2015 / 4 / b / i

Let $A=(c{t}_A,f(t_A))$ on the rear rocket and $B=(c{t}_B,h(t_B))$ on the front rocket. The instantaneous inertial rest frame of the rear rocket at $A$ moves at $v=f^{\prime }(t_A)$ relative to S. Transform the time difference of these two events:Setting $\Delta t^{\prime }=0$ givesThe velocity belongs to the rear event $A$. It is not to be evaluated at laboratory time $t_B$, nor does the simultaneity condition mean $t_A=t_B$ in S. At nonzero velocity the front event on the same instantaneous-rest slice is at a different laboratory time.