Past exam of the mathematics course of the University of Cambridge / 2025 / ii / Paper 3 / 37E (General Relativity)

WriteFor a radial null ray, the Schwarzschild spacetime line element givesOutside the horizon an ingoing ray has $dr/dt<0$, so it takes the minus sign. Since the Schwarzschild tortoise coordinate obeys $d{r}_{\ast }/dr=f^{-1}$,It follows thatis constant along every ingoing radial light ray.

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Since $v=t+r_{\ast }$ and $d{r}_{\ast }=f^{-1}dr$,Substitution into the metric givesThese are Ingoing Eddington-Finkelstein coordinates. Every coefficient is finite at $r=2M$, and the determinant of the $v,r$ block is $-1$. Thus the apparent singularity there in Schwarzschild coordinates is a coordinate singularity; the metric extends smoothly across the Schwarzschild event horizon.

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For radial light rays, the transformed metric givesThe two families of radial null trajectories in ingoing Eddington-Finkelstein coordinates are thereforefor ingoing rays andfor outgoing rays.

Now set $t_{\ast }=v-r$. Along ingoing rays,Along outgoing rays,and integration givesThus the ingoing rays are straight lines of slope $-1$ in the $r$-$t_{\ast }$ plane. Outgoing rays have positive slope outside $r=2M$, become vertical as they approach the horizon, and have negative slope inside it; the horizon itself is the limiting outgoing null ray.

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For the outgoing family,When $r>2M$, this derivative is positive, while ingoing rays move to smaller $r$. The future light cone therefore has one outward and one inward radial edge, and a future-directed massive particle may move either inward or outward, provided its worldline remains inside that cone.

At $r=2M$, the outgoing edge has $dr/dv=0$ and lies along the horizon. When $0<r<2M$, even this nominally outgoing edge has $dr/dv<0$. Both future null directions point toward decreasing $r$, so every future-directed timelike direction between them does too. A massive particle inside the horizon cannot remain at fixed $r$ or return to $r>2M$; it must continue toward the curvature singularity at $r=0$.

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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.

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