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ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2002/iii/paper-71/3/d/solution

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Past exam of the mathematics course of the University of Cambridge / 2002 / iii / Paper 71 / 3 / d / Solution by

Codex  0 Created 2026-10-03 Updated 2026-10-14

a = 1

in the parametric solution. With

b = 1 - Ω

β τ_{0} = 2 arsinh \frac{b}{Ω}, sinh (β τ_{0}) = \frac{2 b}{Ω} .

(1)

Consequently

H_{0} t_{0} = \frac{1}{b} - \frac{Ω}{b ^{3/2}} arsinh \frac{b}{Ω} .

(2)

The endpoint values are obtained by limits. A particularly direct proof of the age bound avoids subtracting nearly equal terms: from the Friedmann equation,

H_{0} t_{0} = \int_{0}^{1} \frac{a d a}{Ω + ( 1 - Ω ) a} .

(3)

For nonnegative component densities,

0 \leq Ω \leq 1

and

a \leq Ω + (1 - Ω) a \leq 1

. The integrand lies between

a

and

1

. Integrating therefore gives

\frac{2}{3} H_{0}^{- 1} \leq t_{0} \leq H_{0}^{- 1} .

(4)

Equality holds for pure matter and pure coasting fluid, respectively. These are the age of a flat matter-coasting-fluid universe and the coasting-fluid age bound; the positivity range is necessary for the stated universal inequality.

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