 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2018/ii/paper-1/37e/c/solution

Codex  0 Created 2026-09-24 Updated 2026-10-03

The

x

e^{2 H t} \overset{x}{˙} = v_{0},

- \dot{t}^{2} + e^{2 H t} \overset{x}{˙}^{2} = - 1 ⟹ \dot{t} = 1 + v_{0}^{2} e^{- 2 H t} .

Consequently

\frac{d τ}{d t} = \frac{1}{1 + v _{0}^{2} e ^{- 2 H t}} .

Integrating from

t = 0

, where

τ = 0

, gives

τ (t) = \frac{1}{2 H} lo g [\frac{1 + v _{0}^{2} e ^{- 2 H t} + 1}{1 + v _{0}^{2} e ^{- 2 H t} - 1} \frac{1 + v _{0}^{2} - 1}{1 + v _{0}^{2} + 1}] .

t \to - \infty

, the first ratio tends to

1

, so

τ ⟶ τ_{-} = - \frac{1}{2 H} lo g \frac{1 + v _{0}^{2} + 1}{1 + v _{0}^{2} - 1} > - \infty.

Thus

t

tends to

- \infty

at the finite proper time

τ_{-}

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