Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 311 (Black Holes)

Every metric component depends only on $r$, so translations of $t$ and $\varphi$ leave the metric unchanged. Equivalently,Thus $k={\partial }_t$ and $m={\partial }_{\varphi }$ are Killing vector fields, generating stationarity and axial symmetry respectively.

Define Ingoing BTZ coordinates byThus $dt=dv-dr/f$ and $d\varphi =d\chi -\Omega dr/f$, so the potentially singular terms cancel:andThe metric becomesFor $r_{+}>r_{-}$, antiderivatives may be chosen asandwith the extremal case obtained by taking the limit. The transformed metric contains neither $1/f$ nor any other singular coefficient at $r=r_{+}$; since $f$, $\Omega$, and $r^2$ are analytic there, it gives an analytic extension across the outer horizon.

Let a future-directed causal tangent in the ingoing chart beThe chosen time orientation has $\dot{v}\ge 0$; equality occurs only for the ingoing radial null direction, which has $\dot{r}<0$. Causality requiresFor $r_{-}<r<r_{+}$ one has $f<0$. The first and third terms are then nonnegative, so if $\dot{v}>0$ the inequality forces $\dot{r}<0$. If $\dot{v}=0$, it forces $\dot{\chi }=0$, and future direction again gives $\dot{r}<0$. Thus $r$ strictly decreases along every future causal curve in this region. No such curve can cross outward through $r=r_{+}$ into $r>r_{+}$, which proves the one-way causal disconnection characteristic of an event horizon.

The normal to a surface of constant $r$ is $n_a=(dr{)}_a$, and the inverse ingoing metric givesHence $r=r_{+}$ is a null hypersurface. LetThis constant linear combination of Killing vector fields is Killing. Its squared norm iswhich vanishes at $r_{+}$. More strongly, lowering its index in the ingoing metric gives ${\xi }_a=(dr{)}_a$ on the horizon. It is therefore the null normal and generator, so the surface is a Killing horizon.

Using ${\nabla }_a({\xi }^2)=-2\kappa {\xi }_a$ on a Killing horizon, the squared term contributes no first derivative at $r_{+}$ andDifferentiating $f$ at its simple outer root givesso the surface gravity isIt vanishes in the extremal case $r_{+}=r_{-}$.

When $r_{-}=0$, $\Omega =0$ and $f=(r^2-r_{+}^2)/L^2$. A circle $S$ at fixed $v,r$ has circumference $2\pi r$. Choose its future null normals asFor a one-dimensional transverse surface, each null expansion is the logarithmic derivative of its length:Inside the horizon, $0<r<r_{+}$ implies $f<0$, so both expansions are negative. Every such circle is therefore a trapped surface, specifically a Trapped circle inside a nonrotating BTZ black hole.

The Penrose singularity theorem concludes future null-geodesic incompleteness; it does not require a divergent curvature invariant. This spacetime supplies exactly that distinction. For a null geodesic in the static coordinates, the Killing constantsand the null condition giveAs $r\to 0$, $f\to -r_{+}^2/L^2$. A radial geodesic with $J=0$ reaches $r=0$ linearly in finite affine parameter. If $J\ne 0$, then $|\dot{r}|\sim r_{+}|J|/(Lr)$, so it again reaches zero in finite affine parameter. The anti-de Sitter quotient ends these geodesics at $r=0$ even though local curvature remains finite. Thus the trapped circles lead to the BTZ causal singularity, satisfying the theorem through geodesic incompleteness rather than curvature blowup.

Locally write the null hypersurface as a level set $u=0$ and its normal as $n_a={\nabla }_{a}u$. Since a null normal is also tangent, $n^a$ lies within the hypersurface. The symmetry of the Levi-Civita connection givesThe scalar $n^2$ vanishes on the hypersurface, so its gradient there is normal and hence proportional to $n_a$:This is the nonaffine geodesic equation. A rescaling of $n$ removes $\kappa$, proving that the null hypersurface normal generates null geodesics lying in the hypersurface.

Let $U^a$ be the affine null tangent and choose another null vector $N^a$ with $U\cdot N=-1$. The screen-space projectorprojects onto the $(D-2)$-dimensional transverse space. The optical tensor isIts irreducible decomposition definesthe null expansion,the null shear, andthe null twist, also called rotation.

Affine geodesic evolution and the Ricci identity give the screen-projected optical equationTaking the screen trace yieldsThe optical decomposition and the antisymmetry of $\hat{\omega }$ implyTherefore the $D$-dimensional Null Raychaudhuri equation isand the requested constant is $\boxed{A=D-2}$.

For generators of a null hypersurface, the Frobenius theorem gives ${\hat{\omega }}_{ab}=0$. The null energy condition and the Einstein field equations imply $R_{ab}{U}^a{U}^b\ge 0$, while the shear norm is nonnegative. With $A=D-2$, Raychaudhuri's equation givesAs long as $\theta <0$ remains finite,If $\theta (0)={\theta }_0<0$, integration givesThe right-hand side reaches zero at $\lambda =A/|{\theta }_0|$, which a finite negative $\theta$ cannot cross. Hence the null focusing theorem forcesprovided the generator extends that far.

Suppose, for contradiction, that every future null geodesic orthogonal to the compact trapped surface $T$ extends beyond $L=A/|{\theta }_0|$. Both families begin with expansion at most ${\theta }_0<0$, so part d gives a point conjugate to $T$ on every generator by affine length $L$. A generator of the achronal boundary ${\dot{J}}^{+}(T)$ cannot remain on that boundary beyond its first conjugate point, because afterward it can be deformed to a timelike curve from $T$.

The two bundles of initial null directions over compact $T$, restricted to $0\le \lambda \le L$, form a compact set. Their image under the geodesic exponential map contains all of ${\dot{J}}^{+}(T)$, so this achronal boundary is compact. Project it along a complete timelike flow onto the noncompact Cauchy hypersurface $\Sigma$. The projection is both open and closed in the connected Cauchy surface, hence would be all of $\Sigma$; compactness of the source would then make $\Sigma$ compact, a contradiction.

Therefore at least one orthogonal future null geodesic cannot extend to affine length $L$. Its maximal future development is future-inextendible with total affine lengthwhich is the incompleteness conclusion of the Penrose singularity theorem.

For a small amount of matter falling into an initially and finally stationary rotating black hole, the Physical-process first law for a rotating black hole isLet $q^a$ be an affinely parametrized horizon generator, with affine parameter $\lambda =0$ at the background bifurcation surface. The horizon Killing vector field iswhere constancy of $\kappa$ is the Zeroth law of black-hole mechanics. The flux of the conserved Killing current through the horizon is

On the stationary background, expansion and shear vanish. To first order in the perturbation, the quadratic optical terms in the Null Raychaudhuri equation may be dropped, and the Einstein equation givesImpose the teleological final condition $\theta (\infty )=0$. Integration followed by reversal of integration order givesComparison with the Killing-energy flux proves the stated law.

The second law of black-hole mechanics states that, assuming the null energy condition and the usual predictability hypotheses, the area of cross-sections of a future event horizon never decreases toward the future.

Indeed, the horizon generators form a hypersurface-orthogonal null congruence, so their twist vanishes. If the expansion were negative anywhere, the null focusing theorem would make it diverge to $-\infty$ within finite affine parameter, provided the generator extends that far. This produces a conjugate point, after which the generator cannot remain on an achronal set such as an event horizon. Future completeness rules out the alternative that the generator simply ends first. Therefore $\theta \ge 0$ everywhere, and

Along the affine null geodesic write ${\Phi }^{\prime }=U^a{\nabla }_a\Phi$. Contracting the scalar stress tensor with $U^a{U}^b$ removes every term proportional to $g_{ab}$. Since $G_{ab}=T_{ab}$ andone obtainsPut $D=1-\xi {\Phi }^2$. An integration by parts givesThe stated endpoint condition kills the boundary term. If $\xi <0$, then $D>0$ and $\xi (4\xi -1)>0$, so the remaining integrand is nonnegative. Consequently this nonminimally coupled scalar satisfies the averaged null energy condition:

The result is weaker than the pointwise null energy condition used in the elementary proof of the second law of black-hole mechanics: $T_{ab}{U}^a{U}^b$ may be negative over a finite interval, so the expansion and area can have local behavior that the pointwise argument does not control. Nevertheless, the averaged null energy condition for a nonminimally coupled scalar supplies the integrated positivity required by strengthened global focusing and area theorems when their completeness, endpoint, and genericity hypotheses hold. Thus the calculation supports a suitable global second law for $\xi <0$, but the displayed averaged inequality alone does not prove pointwise area monotonicity without those additional assumptions.

Choose complete orthonormal sets of complex Klein-Gordon solutions $\{u_i^{\mathrm{in}}\}$ and $\{u_i^{\mathrm{out}}\}$ that have positive frequency with respect to the asymptotic timelike Killing fields in the remote past and future. Their normalization uses the conserved Klein-Gordon inner product:The real quantum field has either mode expansion of a free field

Completeness relates the mode bases by the Bogoliubov transformationand hencePreservation of the canonical commutation relations requiresThe past and future notions of positive frequency therefore define the generally different in-vacuum and out-vacuum. Nonzero $\beta$ means that the in-vacuum is a squeezed state containing out-particles.

The in-vacuum satisfies $a_j^{\mathrm{in}}|0_{\mathrm{in}}⟩=0$. For the out-mode number operator $N_i^{\mathrm{out}}=a_i^{\mathrm{out}†}{a}_i^{\mathrm{out}}$, substitute the operator transformation and use the canonical commutation relations. Only the $a_j^{\mathrm{in}}{a}_k^{\mathrm{in}†}$ contraction survives, giving the particle number from Bogoliubov coefficientsFor a continuous mode label, the sum becomes the corresponding integral with Dirac-delta normalization.

In a collapsing spacetime, positive-frequency in-modes are defined at past null infinity and outgoing modes at future null infinity. Tracing a late outgoing wave packet backward toward the forming event horizon produces an exponential blueshift. Its rapidly varying phase has both positive- and negative-frequency parts relative to the past time coordinate, so the Bogoliubov coefficient $\beta$ is nonzero. The resulting bosonic occupation numbers have the Bose-Einstein distribution atwhere $\kappa$ is the surface gravity. The outgoing quanta form Hawking radiation, while their correlated partners cross the horizon.

For a metric $d{s}^2=-f(r)d{v}^2+2dvdr+r^{2}d{\Omega }_2^2$ with a simple Killing horizon, the surface gravity of $k={\partial }_v$ is $\kappa =f^{\prime }(r_{+})/2$. HeresoThe Hawking temperature of the Schwarzschild--anti-de Sitter black hole is therefore

LetThe one-form dual to $k={\partial }_v$ is $k^{♭}=-fdv+dr$. For the $r_{+}\to 0$ anti-de Sitter background, ${\bar{k}}^{♭}=-(1+r^2/L^2)dv+dr$, soWith the stated orientation and metric volume form,The Background-subtracted Komar energy is consequently

The horizon area is $A=4\pi r_{+}^2$, whileThus the First law of black-hole mechanicsholds with the Bekenstein-Hawking entropy $S_{BH}=A/4$.

For equilibrium with an infinite fixed-temperature reservoir, local stability requires positive heat capacity. Since $dM/d{r}_{+}>0$, its sign is the sign ofTherefore the Canonical stability of a Schwarzschild--anti-de Sitter black hole occurs precisely forThe equality is the minimum-temperature, marginal case; smaller black holes have negative heat capacity and unstable thermal equilibrium.