Let $\tau$ be proper time along a geodesic. While $c$ is finite and positive,
$$
\frac d{d\tau}\left(\frac1c\right)
=-\frac1{c^2}\frac{dc}{d\tau}\leq-\frac13.
$$
Thus
$$
\frac1{c(\tau)}\leq\frac1{c_0}-\frac\tau3.
$$
The right side reaches zero at $\tau=3/c_0$, so the convergence must diverge no later than that proper time.


Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2022/iii/paper-311/3/d/solution