Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 37E / e / Solution

For a static observer, $dr=d\theta =d\varphi =0$, soThe coordinate travel time of a radial signal depends only on the fixed endpoint radii. Successive signals emitted a coordinate-time interval $\Delta t$ apart are therefore received with the same coordinate-time separation. The gravitational redshift between static Schwarzschild observers is consequentlyFor $r_B\gg M$ the numerator is approximately one, whileHenceThe received signals become arbitrarily widely separated and redshifted as Alice approaches the horizon. A static observer exactly at the horizon is impossible because the proper acceleration required to remain static diverges there.

Solved by gpt-5.6-sol high.