Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 311 / 2

The Penrose diagram is the maximally extended Schwarzschild diagram modified by an ingoing null shell from the chosen ${\mathcal{I}}^{-}$. To its past the vacuum mass is $M$; to its future the exterior mass is $M+E$. Draw the shell as a $45^{\circ }$ ingoing line from the chosen ${\mathcal{I}}^{-}$, crossing the future event horizon and ending on the future spacelike singularity. In that same exterior, label $i^{-}$ and $i^{+}$ at the lower and upper timelike corners, $i^0$ at the spatial corner, and ${\mathcal{I}}^{-}$ and ${\mathcal{I}}^{+}$ as the past and future null boundaries.

The crucial teleological feature is that the final ${\mathcal{H}}^{+}$, traced backward before the shell, lies outside the old $r=2M$ future horizon. It crosses the old past horizon ${\mathcal{H}}^{-}$ at the bifurcation two-sphere and continues into the white-hole region, ending on the past spacelike singularity. Thus the diagram must not draw the relevant ${\mathcal{H}}^{+}$ as beginning at the old bifurcation sphere.

Solved by gpt-5.6-sol high.

After the shell, spherical symmetry and stationarity put the horizon at $r=2(M+E)$. Before the shell, an outgoing radial null geodesic in the mass-$M$ region has $r$ linear in an affine parameter. The normalizations $r(0)=2M$ and $r(1)=2(M+E)$ therefore giveThe degenerate intrinsic metric of the horizon has no $d{\lambda }^2$ term, soThe two expressions agree at the shell, $\lambda =1$.

Solved by gpt-5.6-sol high.

Tracing the generator backward through the mass-$M$ region gives $r=2(M+E\lambda )$. It reaches the past spacelike singularity when $r=0$, namely atThe singular endpoint is not part of the spacetime, while the final stationary horizon extends indefinitely to the future. HenceIn particular, negative $\lambda$ is physically necessary: that part of the relevant event horizon lies inside the old white-hole region before reaching the old bifurcation sphere at $\lambda =0$.

Solved by gpt-5.6-sol high.

Let $k^a=(d/d\lambda {)}^a$ be an affine horizon generator, normalized so that the horizon Killing field is ${\xi }^a=\kappa \lambda k^a$. To first order in the small total injected energy, expansion and shear-squared terms are second order, and the Null Raychaudhuri equation on the horizon becomesThe flux of Killing energy through all the shells isSubstitute the linearized focusing equation and integrate by parts. Stationarity before and after the process makes the endpoint term $\lambda \theta$ vanish, while $dA/d\lambda =\theta A$ to first order. ThereforeUsing the Hawking temperature $T_H=\kappa /(2\pi )$ and Bekenstein-Hawking entropy $S_{BH}=A/4$ gives the Physical-process first law of black-hole mechanicsThe derivation is linear in the stress tensor, so separated shells simply contribute additively.

Solved by gpt-5.6-sol high.