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

Write $\Delta x=h(t_B)-f(t_A)$. Substituting $\Delta t=v\Delta x/c^2$ into ii gives $L^2=\Delta x^2(1-v^2/c^2)$. The positive-separation branch therefore hasThe Born-rigid front trajectory from a rear worldline follows by setting $t_A=t$:Every $v$ and $\gamma$ in this formula is evaluated at the rear event with laboratory time $t$, not at the shifted argument of $h$.

For completeness, a smooth future-directed front trajectory requires this time mapping to be locally invertible. If the signed laboratory acceleration is $a_S=dv/dt$ and the rear proper acceleration is ${\alpha }_F={\gamma }^3{a}_S$, thenWhen $1+L{\alpha }_F/c^2>0$, their ratio gives $dh/d{t}_B=v$, so the two simultaneous points indeed share their instantaneous velocity. If that factor reaches zero, an acceleration-horizon obstruction appears; an arbitrary rear trajectory cannot be extended to an arbitrary rigid length while retaining both rockets future-directed. The positive-acceleration example below satisfies the condition.