Past exam of the mathematics course of the University of Cambridge / 2026 / ia / Paper 4 / 4B / Solution

Proper time satisfies $c^{2}d{\tau }^2=c^{2}d{t}^2-d{x}^2$ and is invariant, so differentiating by it defines a four-vector. The path is timelike iff $R|\omega |<c$. Its next return occurs after coordinate time $2\pi /|\omega |$, so Bob agesWith $\gamma =(1-R^2{\omega }^2/c^2{)}^{-1/2}$,
$U^{\mu }=\gamma (c,R\omega \cos \omega t,-R\omega \sin \omega t,0)$ and $U\cdot U=c^2$ for signature $(+---)$.

Solved by gpt-5.6-sol high.