Past exam of the mathematics course of the University of Cambridge / 2021 / iii / Paper 311 (Black Holes)

A null geodesic congruence is a smooth family of nonintersecting null geodesics filling an open spacetime region. For its affinely parametrized tangent $U^a$,Writing $B_{ab}={\nabla }_b{U}_a$, differentiation of the null norm and the geodesic equation give respectively

Choose at one transverse cross-section a null vector $N^a$ satisfying $N^a{N}_a=0$ and $U^a{N}_a=-1$, then parallel transport it along every generator:Metric compatibility and the affine geodesic equation imply that both $N^2$ and $U\cdot N$ are constant along each generator, so the required normalization persists.

The screen-space projectordefines the optical tensor ${\hat{B}}_{ab}=P_a{}^c{P}_b{}^d{B}_{cd}$. In four spacetime dimensions its irreducible decomposition isThese are the null expansion, null twist, and null shear.

On the null hypersurface, the generator covector is proportional to a normal, locally $U_a=f{\nabla }_{a}u$ for a level-set function $u$. The Frobenius theorem therefore givesContracting once with $N^a$ and projecting the remaining indices with $P^a{}_b$ removes every term containing $U$ or $N$ and leaves $P_b{}^d{P}_c{}^e{\nabla }_{[d}{U}_{e]}=0$. Henceon the hypersurface.

Commuting covariant derivatives and using $U^b{\nabla }_b{U}_a=0$ gives the optical evolution equationTaking its screen trace yieldsThe optical decomposition and the symmetry or antisymmetry of its pieces implyThus the four-dimensional Null Raychaudhuri equation is

The hypersurface generators have zero null twist. Contracting the Einstein field equations twice with the null tangent removes the trace term, and the null energy condition gives $R_{ab}{U}^a{U}^b=8\pi T_{ab}{U}^a{U}^b\ge 0$. Since the squared null shear is nonnegative, the Null Raychaudhuri equation impliesWhile $\theta <0$ this is equivalent toStarting from $\theta (0)={\theta }_0<0$, the right-hand side reaches zero no later than $\lambda =2/|{\theta }_0|$. The reciprocal expansion must therefore vanish andThis is the null focusing theorem.

Suppose the closed trapped surface $T$ were not wholly in the black hole. The stated consequence of strong asymptotic predictability then supplies a point $p\in {\dot{J}}^{+}(T)\cap {\mathcal{I}}^{+}$ lying on a future null generator $\gamma$ orthogonal to $T$, with no point conjugate to $T$ before $p$. Both future null expansions at $T$ are negative. The null focusing theorem therefore forces the expansion along $\gamma$ to diverge after finite affine parameter, producing a conjugate point before the generator reaches future null infinity. A null geodesic with such a point cannot continue to generate the achronal boundary ${\dot{J}}^{+}(T)$, contradicting the property of $p$. Hence no point of $T$ can communicate with future null infinity, and

Every metric coefficient is independent of $t$ and $\varphi$. The corresponding coordinate flows therefore preserve the metric, sosatisfy the Killing equation and are Killing vector fields. They generate stationarity and axial symmetry respectively.

Put $\mu =r_{+}(r_{+}^2+a^2)$. At large $r$, the one-form dual to $K$ hasOnly the $dr\wedge dt$ term contributes to the constant-$t,r$ Komar integral, andWith the stated orientation, the asymptotic Hodge star operator givesThe angular integral is the area of the unit four-sphere,ConsequentlyThis is the Komar mass of the Singly rotating six-dimensional Myers-Perry black hole in the units of the question.

Substituting the two differential coordinate transformations cancels every coefficient singular as ${\Delta }^{-1}$, so the ingoing Kerr-like coordinates are regular at $r=r_{+}$. The inverse metric applied to the normal covector $dr$ hasIts norm is $g^{rr}=\Delta /\Sigma$, which vanishes at $r=r_{+}$. Thus that level set is a null hypersurface. On it the raised normal is proportional toSince the transformed stationary and axial Killing fields are $K={\partial }_v$ and $m={\partial }_{\chi }$, the horizon is a Killing horizon generated by

For this horizon generator, direct evaluation of ${\xi }^b{\nabla }_b{\xi }^a=\kappa {\xi }^a$, or the radial derivative of $\Delta$, givesBecause $\Delta =r^2+a^2-\mu /r$ and $\mu =r_{+}(r_{+}^2+a^2)$,and thereforeThe induced horizon cross-section has volume elementUsing the unit-four-sphere integral from part b gives

On $\theta =\lambda =0$, the radial function $r\Delta =r^3+a^{2}r-\mu$ is strictly increasing and has the single positive root $r_{+}$. The curvature invariant diverges at $r=0$, and $g^{rr}=\Delta /\Sigma <0$ immediately inside the horizon, so this is a spacelike singularity. The maximally extended Penrose diagram therefore has the Schwarzschild form: two asymptotically flat exterior diamonds separated by future and past event horizons, with a spacelike future singularity above the black-hole regions and a spacelike past singularity below the white-hole regions. A collapse spacetime retains one exterior, the future horizon, and the future spacelike singularity.

The Physical-process first law for a rotating black hole states that a small flux of matter through an initially and finally stationary horizon obeysin units $G=1$. Let $k^a$ be an affinely parametrized horizon tangent with affine parameter $\lambda =0$ on a bifurcation surface. Constancy of the surface gravity from the Zeroth law of black-hole mechanics and Gaussian null coordinates give the horizon generatorTo first order about a stationary horizon, the squared expansion and shear in the Null Raychaudhuri equation are second order. The Einstein field equations reduce it toThe final stationary condition $\theta (\infty )=0$ givesSince $d(\delta A)/d\lambda =\int \theta dA$, reversing the order of integration yieldsThe stress-energy current from a Killing vector gives the horizon Killing-energy fluxwhich proves the stated law.

The second law of black-hole mechanics, or Hawking's area theorem, states that the total area of future-event-horizon cross-sections cannot decrease toward the future under the null energy condition and the usual predictability assumptions. If any horizon generator had $\theta <0$, the twist-free Null Raychaudhuri equation and the null focusing theorem would force a conjugate point at finite affine parameter. A generator complete to the future cannot pass through such a point while remaining on the achronal boundary of $J^{-}({\mathcal{I}}^{+})$. Hence $\theta \ge 0$ everywhere on the regular future horizon, and $dA/d\lambda =\theta A\ge 0$ proves the area law.

The two initial Kerr black holes have total areawhereas the final Schwarzschild black hole has $A_f=16\pi M^{\prime 2}$. Hawking's area theorem impliesSince $\eta =1-M^{\prime }/(2M)$, the radiated fraction is bounded byThis upper bound increases with $|J|/M^2$ and reachesfor two initially extremal holes, $|J|=M^2$. For the final hole to be Schwarzschild, their spins must be oppositely directed so the total angular momentum vanishes. The bound assumes an idealized merger saturating the area law.

First, causal futures of black-hole points remain in the black-hole region. Indeed, if $p\in B$ and $q\in J^{+}(p)$ could send a signal to future null infinity, concatenating the causal curves would put $p$ in $J^{-}({\mathcal{I}}^{+})$, a contradiction. HenceGlobal hyperbolicity supplies the following connectedness lemma: if $C$ is connected on one Cauchy hypersurface, then $J^{+}(C)$ intersects any later Cauchy hypersurface in a connected set. To see the relevant mechanism, flow $C$ to the later surface along a continuous future timelike vector field; the image is connected, and every additional causally reachable point is joined to that image by the endpoint deformation of a causal curve inside the globally hyperbolic diamond. Applying the lemma to the connected component $B$ shows that $J^{+}(B)\cap {\Sigma }_2$ is connected. Since it is contained in $B\cap {\Sigma }_2$, it must lie wholly inside one connected component of that set. This is the black-hole non-splitting theorem: later black-hole components may merge, but one earlier connected black hole cannot split into two future components.

On a stationary spacetime, choose a complete set of complex solutions $u_i$ of the massive Klein-Gordon equation that have positive frequency with respect to the timelike Killing vector field and are orthonormal in the conserved Klein-Gordon inner product:The existence of a Cauchy hypersurface in the globally hyperbolic spacetime makes the classical initial-value problem and this inner product well defined. Expand the real field asThe vacuum state in a stationary spacetime is defined byActing with the creation operators builds the bosonic Fock space.

The early and late stationary regions define distinct positive-frequency mode bases and hence an in-vacuum and out-vacuum. Write their Bogoliubov transformation asThe corresponding operators satisfyUsing ${\hat{a}}_j^{\mathrm{in}}|0_{\mathrm{in}}⟩=0$ and the canonical commutator gives the particle number from Bogoliubov coefficients:Time dependence in the sandwich region can make ${\beta }_{ij}\ne 0$, so the in-vacuum contains out-particles.

A gravitational-collapse spacetime is stationary and approximately empty in the remote past and settles to a stationary black-hole exterior in the future, with a dynamical region between them. The same comparison of in- and out-positive-frequency modes therefore applies. Tracing an outgoing late-time mode backward through the collapse produces an exponentially blueshifted mixture of early positive and negative frequencies. Its nonzero Bogoliubov beta coefficient yields the thermal occupation numbers of Hawking radiation, while partner modes pass through the event horizon.

The horizon generator must be null at $r=r_{+}$. Substitution of $\xi ={\partial }_v+{\Omega }_H{\partial }_{\chi }$ in the Kerr black hole metric givesThe surface gravity and Hawking temperature are

For $a=0$, $T_H=(8\pi M{)}^{-1}$, so the Schwarzschild black hole has negative heat capacity. A small energy gain from a reservoir lowers its temperature below the reservoir temperature and causes further absorption; a small energy loss raises its temperature and causes further emission. The equilibrium is therefore unstable in the canonical ensemble.

DefineThe Bekenstein-Hawking entropy and Hawking temperature becomeAt fixed $J$,The fixed-$J$ first law gives $(\partial S_{BH}/\partial M{)}_J=1/T_H$, soTherefore the Kerr black-hole heat capacity at fixed angular momentum isAt $a=0$, this correctly reduces to $C_J=-8\pi M^2$.

Local thermal stability against energy exchange with a fixed-temperature reservoir requires $C_J>0$. Since $s\ge 0$, this meansEquivalently,The lower endpoint is the divergent-heat-capacity transition. The extremal endpoint $|a|=M$ has $T_H=0$ and $C_J=0$ and is obtained only as a limiting case.