Past exam of the mathematics course of the University of Cambridge / 2025 / iii / Paper 311 (Black Holes)

Write $f(r)=1-r_{+}^2/r^2$ and let a dot denote differentiation with respect to an affine parameter $\tau$. The time-translation and $z$-translation Killing vector fields, together with rotational symmetry of the unit $S^3$, give the conserved quantitiesDefine $\epsilon =-g_{ab}{\dot{x}}^a{\dot{x}}^b$, so $\epsilon =1,0,-1$ for timelike, null and spacelike geodesics respectively. The normalization equation becomesIt therefore has the effective potential form

For radial null geodesics, $d{\Omega }_3=dz=0$ and $d{s}^2=0$, soIntroduce the tortoise coordinate byup to an additive constant. The plus sign gives $d(t-r_{\star })=0$ on outgoing rays, while the minus sign gives $d(t+r_{\star })=0$ on ingoing rays. Henceare respectively constant on the stated radial null families.

The normal to a surface of constant $r$ has squared normwhich vanishes at $r=r_{+}$. Thus this surface is a null hypersurface. The stationary Killing vector fieldhas $K^2=g_{tt}=-f$, so it becomes null there and generates a Killing horizon. Since $f$ has a simple zero, its surface gravity isA horizon cross-section has topology $S^3\times S^1$, where the circle is the periodic $z$ direction. Including a complete generator, the null hypersurface has topology

Using $v=t+r_{\star }$ puts the black string metric into regular ingoing form,Its inverse metric givesFor $r<r_{+}$, $f<0$, so $\nabla r$ is timelike. It is future-directed by continuity from the future horizon, where it equals the future generator ${\partial }_v$. If $X$ is any future-directed causal tangent, thenThus $r$ decreases strictly along every future-directed causal curve in the interior. No such curve can cross back through $r=r_{+}$ or reach future null infinity. The region $r<r_{+}$ is consequently outside the causal past of future null infinity and is part of the black hole region.

Choose an auxiliary null vector $N^a$ with $U\cdot N=-1$. The screen-space projector isand the optical tensor is ${\hat{B}}_{ab}=P_a{}^c{P}_b{}^d{B}_{cd}$ for $B_{ab}={\nabla }_b{U}_a$. Because the screen is two-dimensional, its irreducible decomposition iswhereThese are respectively the null expansion, null shear, and null twist, also called rotation.

Affine geodesic motion gives $U^c{\nabla }_c{U}_a=0$. Commuting the two covariant derivatives and applying the product rule yieldsThis proves the required transport identity. The curvature term is symmetric after screen projection, so taking the antisymmetric screen part and inserting the optical decomposition gives the null-twist propagation equationIn particular, an initially twist-free congruence remains twist-free.

Choose a spacelike two-dimensional cross-section $S$ of the null hypersurface $\mathcal{N}$ and coordinates $y^i$ on $S$. Extend $y^i$ along the null generators, and choose their parameter $\lambda$ to be affine, so $U={\partial }_{\lambda }$ on $\mathcal{N}$. At every point choose the other null normal $L$ with $g(U,L)=1$, shoot out the affinely parametrized null geodesic with tangent $L$, and call its affine parameter $r$. Transport $(\lambda ,y^i)$ along those transverse geodesics.

This construction gives Gaussian null coordinates. The hypersurface is $r=0$, and the coordinate conditions implyBecause $\lambda$ is affine on the generators, ${\partial }_r{g}_{\lambda \lambda }{|}_{r=0}=0$. Smoothness then factors the remaining components as $g_{\lambda \lambda }=r^{2}F$ and $g_{\lambda i}=r{h}_i$, giving

On $r=0$, the screen metric is $h_{ij}$ and $U={\partial }_{\lambda }$. Therefore the null expansion isThe derivative formula for the determinant givesand henceThus $\theta$ is the fractional rate of change of an infinitesimal transverse area carried along the generators: positive expansion enlarges the beam and negative expansion focuses it.

An asymptotically flat spacetime at future null infinity admits a smooth conformal completion $(\overline{\mathcal{M}},\bar{g})$ with the following properties. The physical spacetime $\mathcal{M}$ is the interior of $\overline{\mathcal{M}}$, its metric obeysand the boundary component ${\mathcal{I}}^{+}$ satisfies $\Omega =0$ and $d\Omega \ne 0$. Every future-directed outgoing null geodesic has an endpoint there, the generators of ${\mathcal{I}}^{+}$ are complete, and in four spacetime dimensions ${\mathcal{I}}^{+}\simeq \mathbb{R}\times S^2$. The physical Einstein field equations are vacuum in a neighborhood of this boundary.

Let $n_a={\bar{\nabla }}_a\Omega$. Multiplying the supplied conformal Ricci relation by ${\Omega }^2$, using $R_{ab}=0$, and taking the limit to ${\mathcal{I}}^{+}$ givesThe boundary normal is therefore null. Since a null normal is also tangent to its hypersurface, $n^a$ generates ${\mathcal{I}}^{+}$.

Smoothness makes $s={\Omega }^{-1}{n}^2$ finite at the boundary. Multiplication of the same vacuum equation by $\Omega$ gives thereTaking the trace yields $s=\frac{1}{2}\bar{\square }\Omega$, and substitution givesThe remaining freedom $\Omega \mapsto \omega \Omega$ can be used to impose $\bar{\square }\Omega =0$ on ${\mathcal{I}}^{+}$. In this conformal gauge, ${\bar{\nabla }}_a{n}_b=0$ there, so the generators are affinely parametrized, expansion-free null geodesics of the unphysical metric.

Choose a generator coordinate $u$, the defining function $\Omega$, and angular coordinates $x^A$ whose leading metric $q_{AB}$ is the round metric on the unit sphere. After the conformal and coordinate choices above, the leading unphysical metric iswith the Minkowski term $-{\Omega }^{2}d{u}^2$ entering at the next relevant order. Setting $r={\Omega }^{-1}$ recovers the physical asymptotic formThe displayed leading metric is Minkowski spacetime in outgoing null coordinates. Smooth conformal extendibility controls the lower-order corrections, while the vacuum equations constrain them to the radiative Bondi--Sachs expansion. This is the precise sense in which the permitted spacetimes approach Minkowski spacetime near ${\mathcal{I}}^{+}$ while still allowing outgoing gravitational radiation.

The second law of black-hole mechanics states that the area of spatial cross-sections of a future event horizon cannot decrease toward the future, provided the null energy condition and the standard global assumptions hold.

Let $k^a$ be an affinely parametrized horizon generator. It is hypersurface-orthogonal, so its null twist vanishes. The Null Raychaudhuri equation in $D$ dimensions and the Einstein equation giveIf $\theta ({\lambda }_0)<0$, integration implies that $\theta$ diverges to $-\infty$ within affine distance at most $(D-2)/|\theta ({\lambda }_0)|$. The resulting focal point would make the generator leave the achronal boundary that defines the event horizon, contradicting its assumed future completeness. Hence $\theta \ge 0$ everywhere. Since null expansion obeysevery horizon area element, and therefore every complete cross-section area, is nondecreasing.

For this metric, $\sqrt{-g}=a^2$, while $g^{\eta \eta }=-a^{-2}$, $g^{zz}=a^{-2}$ and $g^{ij}={\delta }^{ij}$. The massless Klein-Gordon equation is consequentlyInsert the spatial Fourier transform given in the question. Each wavenumber mode then satisfiesThus every field mode is a simple harmonic oscillator with a time-dependent angular frequency.

In the two constant regions defineThe normalized positive-frequency modes areup to the common spatial Fourier normalization. They satisfy the unit Wronskian conditionwhich is the mode form of the Klein-Gordon inner product. Because the mode equation contains no delta function at $\eta =0$, both $\Phi$ and ${\Phi }^{\prime }$ are continuous there.

Continue the incoming positive-frequency mode through $\eta =0$ asContinuity of the mode and its first derivative gives the sudden frequency quench coefficientsThe associated Bogoliubov transformation says that the incoming vacuum has expected outgoing occupation numberIt vanishes when $a_{+}=a_{-}$, as required when there is no change of geometry.