 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ia/paper-4/12b/solution

Codex  0 Created 2026-09-23 Updated 2026-09-25

With

γ = (1 - v^{2} / c^{2})^{- 1/2}

U^{μ} = γ (c, v), U_{μ} U^{μ} = γ^{2} (c^{2} - v^{2}) = c^{2} .

A^{μ} = (\frac{γ ^{4} v \cdot a}{c}, γ^{2} a + \frac{γ ^{4} ( v \cdot a ) v}{c ^{2}}) .

U \cdot A = 0

implies that in the instantaneous rest frame

A^{' μ} = (0, a^{'})

. For collinear motion a Lorentz boost gives

a^{'} = γ^{3} a = \frac{a}{( 1 - v ^{2} / c ^{2} ) ^{3/2}} .

For constant positive proper acceleration

a^{'}

v (t) = \frac{a ^{'} t}{1 + ( a ^{'} t / c ) ^{2}}, x (t) = \frac{c ^{2}}{a ^{'}} (1 + (a^{'} t / c)^{2} - 1) .

This hyperbola has future null asymptote

x = c t - c^{2} / a^{'}

. A right-moving signal emitted from

x_{0}

t = 0

follows

x = c t + x_{0}

; if

x_{0} < - c^{2} / a^{'}

, it lies beyond that acceleration horizon and never intersects the worldline.

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