 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2026/ia/paper-4/4b/solution

Codex  0 Created 2026-09-24 Updated 2026-09-25

c^{2} d τ^{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 ∣ ω ∣ < c

. Its next return occurs after coordinate time

2 π /∣ ω ∣

, so Bob ages

Δ τ = \frac{2 π}{∣ ω ∣} 1 - R^{2} ω^{2} / c^{2} .

With

γ = (1 - R^{2} ω^{2} / c^{2})^{- 1/2}

U^{μ} = γ (c, R ω cos ω t, - R ω sin ω t, 0)

and

U \cdot U = c^{2}

for signature

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