For a static observer, $dr=d\theta=d\phi=0$, so
$$
d\tau=\sqrt{1-\frac{2M}{r}}\,dt.
$$
The 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 consequently
$$
\frac{\Delta\tau_B}{\Delta\tau_A}
=\sqrt{\frac{1-2M/r_B}{1-2M/r_A}}.
$$
For $r_B\gg M$ the numerator is approximately one, while
$$
1-\frac{2M}{r_A}
=1-\frac{2M}{2M+\varepsilon}
=\frac{\varepsilon}{2M+\varepsilon}
\simeq\frac{\varepsilon}{2M}.
$$
Hence
$$
\boxed{\frac{\Delta\tau_B}{\Delta\tau_A}
\simeq\sqrt{\frac{2M}{\varepsilon}}}.
$$
The 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.

Solution

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

 Solution

ID: past-exam-of-the-mathematics-course-of-the-university-of-cambridge/2025/ii/paper-3/37e/e/solution