Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 311 / 3 / d

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- Solution d

Solution

Let

τ

be proper time along a geodesic. While

c

is finite and positive,

\frac{d}{d τ} (\frac{1}{c}) = - \frac{1}{c ^{2}} \frac{d c}{d τ} \leq - \frac{1}{3} .

Thus

\frac{1}{c ( τ )} \leq \frac{1}{c _{0}} - \frac{τ}{3} .

The right side reaches zero at

τ = 3/ c_{0}

, so the convergence must diverge no later than that proper time.

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