Past exam of the mathematics course of the University of Cambridge / 2008 / iii / Paper 66 (Black Holes)

Use geometrized units $G=c=1$. For a stationary asymptotically flat spacetime, normalize the timelike Killing vector field $\xi$ to have squared norm $-1$ at infinity. The Komar mass isHere the oriented binormal is chosen so that an ordinary positive-mass static solution gives a positive integral. Equivalently, on a static slice with lapse $N=\sqrt{-{\xi }^2}$ and outward spatial unit normal $s^i$,This second expression fixes the orientation unambiguously and is convenient for the Majumdar–Papapetrou solution.

The spatial metric is ${\gamma }_{ij}=H^2{\delta }_{ij}$ and $N=H^{-1}$. On a large coordinate sphere, $s^i=H^{-1}{\hat{r}}^i$ and $dA=H^2{r}^{2}d\Omega$. ThusThe original PDF has the Euclidean norm in each source denominator, which is missing in the TeX aid. With that norm, the harmonic function has asymptoticsThereforeIf $M_i$ are retained as length parameters while $G$ is restored, the physical mass is ${\sum }_i{M}_i/G$. This is the charge at infinity. A Komar integral is not generally surface-independent here, since Einstein-Maxwell theory has the electromagnetic stress-energy tensor between the surfaces; in particular it should not be confused with a sum of vacuum horizon Komar charges.

Take the first centre at the origin. If some listed centres coincide, first combine their poles; let $M$ be the total residue at the origin, so $M=M_1$ when the centres are distinct. The contributions from centres away from the origin are real analytic there. Their constant term iswhere $F$ is analytic in $r$ and in local angular charts. This follows by substituting $\mathbf{x}=r\mathbf{n}(\theta ,\varphi )$ in the ordinary Taylor expansions of the regular contributions. The constant $h$ is independent of angle, which makes the hinted coordinate differential integrable.

Put $A=M+hr$, $Q=rH=A+r^{2}F$, and defineinitially for $r>0$. Substitution into the metric givesThe apparently singular radial coefficient is actuallysince $Q-A=r^{2}F$. All coefficients therefore extend analytically to negative and positive $r$ near zero, with $Q(0)=M>0$. At the horizon $g_{vv}=0$, $g_{vr}=1$, $g_{rr}=4MF(0,\theta ,\varphi )$ and the angular metric is $M^{2}d{\Omega }_2^2$. The $v,r$ determinant is $-1$ both before and at the extension, so the metric remains nondegenerate and Lorentzian. Angular coordinate singularities are handled by ordinary sphere charts. This explicitly constructs the analytic extension of a Majumdar–Papapetrou horizon.

Inverting the radial block gives $g^{rr}=r^2/Q^2$, which vanishes at $r=0$. Thus the hypersurface is null. Its normal agrees there with the metric dual of $\xi ={\partial }_v$: ${\xi }_{a}d{x}^a=dr$ on the horizon. The metric is independent of $v$, so $\xi$ is a Killing vector field and the null hypersurface is a Killing horizon.

The surface gravity obeys ${\nabla }_a({\xi }^2)=-2\kappa {\xi }_a$ on a Killing horizon. Here ${\xi }^2=-r^2/Q^2$ has a double zero and its entire differential vanishes at $r=0$, while ${\xi }_a$ is the nonzero normal. ConsequentlyThe horizon is degenerate. Its cross-sectional area is $4\pi M^2$, independent of the centres away from the origin. The essential cancellation would fail with an arbitrary constant in place of the displayed regular part $h$.

Let $x^a(\lambda ,s)$ be a smooth family of affinely parametrized null geodesics. The tangent is $U=\partial x/\partial \lambda$, and the deviation vector is $Z=\partial x/\partial s$, comparing nearby rays at the same affine parameter. The two parameter derivatives commute, soThus the separation vector is Lie-transported along the null geodesic congruence; it is not required to be parallel-transported. Differentiating once more gives the geodesic deviation equation. With curvature convention $[{\nabla }_c,{\nabla }_d]V^a=R^a{}_{bcd}{V}^b$,For optical measurements one uses transverse deviations with $U\cdot Z=0$ and identifies longitudinal additions proportional to $U$. The scalar $U\cdot Z$ is constant along affine null rays by the displayed transport relation, so an initially transverse deviation remains transverse.

Choose an auxiliary null vector $N$ with $U\cdot N=-1$, parallel-transported along each affine ray. The screen-space projector and its induced positive metric areThey annihilate $U,N$, obey $P^2=P$, and have trace two in four spacetime dimensions. Since both $U$ and $N$ are parallel along $U$, ${\nabla }_{U}P=0$. Equivalently, the screen represents the two-dimensional quotient $U^{\perp }/\mathrm{span}(U)$.

Define the optical matrix of a null congruence byFor a transverse deviation, its screen projection $\zeta =PZ$ satisfies ${\nabla }_U{\zeta }^a={\hat{B}}^a{}_b{\zeta }^b$. Decompose the matrix asThese are the null expansion, null shear and null twist. Null expansion gives the fractional change of a small transverse area, $\theta =d\log (dA)/d\lambda$. Null shear is symmetric trace-free distortion; null twist is the antisymmetric rotation part. This null expansion convention uses the full trace, not half the trace. The auxiliary screen fixes representatives, while the transverse optical content is intrinsic to the null geodesic congruence.

First let $B_{ab}={\nabla }_b{U}_a$ without projection. Commute derivatives on this covector and use affine geodesic motion:The last equality relabels the two contracted null-vector indices. Nullness and affine parametrization give $U^a{B}_{ab}=0$ and $B_{ab}{U}^b=0$. Therefore inserting $P$ into the contracted middle index changes no screen-projected quadratic term. Since ${\nabla }_{U}P=0$, screen projection proves the Sachs optical equationsThe optical tidal matrix ${\mathcal{R}}_{ab}$ is symmetric by the Riemann curvature tensor's pair symmetries. Raising one screen index gives the corresponding equation for ${\hat{B}}^a{}_b$.

In terms of the defined optical scalars, the equation is explicitlyThis matrix identity contains the null expansion, null shear and null twist evolution equations as its trace, symmetric trace-free part and antisymmetric part.

Take the screen trace of the optical matrix of a null congruence equation. The cross-contractions of symmetric null shear with antisymmetric null twist vanish, andThe minus sign in the null twist contribution follows from tracing the square of an antisymmetric matrix. The tidal trace is $P^{ef}{R}_{cedf}{U}^c{U}^d=R_{cd}{U}^c{U}^d$: the additional null terms in $P^{ef}$ vanish by curvature antisymmetry. Thus the Null Raychaudhuri equation isIn particular, null shear focuses the null geodesic congruence while null twist enters with the opposite sign. With vanishing null twist and the null energy condition, the Einstein field equations imply $U\cdot \nabla \theta \le -{\theta }^2/2$. These signs and the factor $1/2$ depend on using the full-trace null expansion in a two-dimensional screen.

Choose a parallel-transported orthonormal basis of the screen. In that basis a symmetric trace-free null shear and antisymmetric null twist have formsDirect multiplication givesHenceSince the tidal matrix is symmetric, the antisymmetric part of the Sachs optical equations immediately givesThus an initially null twist-free null geodesic congruence remains null twist-free while its optical description is regular.

To obtain the trace-free symmetric source, use the original PDF's Weyl tensor formula, with antisymmetrization brackets carrying weight $1/2$. The TeX aid has lost indices and brackets. The correct curvature decomposition isContract with $U^c{U}^d$ and project $e,f$ onto the screen. Terms containing $U^2$ or a remaining screen-projected $U$ vanish. The only Ricci term left isThe Weyl term has zero screen trace, by its trace-free property and antisymmetry. Consequently the trace-free symmetric part of the optical equation is the null-shear propagation equationThe two-dimensional matrix identities are essential: in a higher-dimensional screen the null shear square can have a trace-free part and the null shear–null twist anticommutator need not vanish. Choosing the parallel-transported auxiliary null vector makes these tensor transport formulae hold without extra moving-screen terms.

Use geometrized units $G=c=1$. The Physical-process first law of black-hole mechanics concerns a small perturbation of a rotating black hole in an initially stationary spacetime by infalling matter, followed by relaxation to another stationary spacetime. For an uncharged black hole, or neutral accretion with charge held fixed, its first-order statement isHere $\kappa >0$ is the background surface gravity, ${\Omega }_H$ is the background horizon angular velocity, and $\delta A$ is the change in event horizon area. The energy $\delta E$ and angular momentum $\delta J$ are the fluxes carried into the event horizon; when these account for the changes of the black hole, they equal $\delta M$ and its change of angular momentum. The result is a linearized law, not an exact identity for a finite violent accretion event. The perturbation must be small enough that focusing does not produce new caustics or invalidate the expansion about the existing generators.

Let $k^a$ be the stationary Killing vector field normalized at infinity, $m^a$ the axial Killing vector field with $2\pi$-periodic orbits, andOn the background Killing horizon, ${\chi }^a$ is its future-directed null generator and satisfies ${\chi }^b{\nabla }_b{\chi }^a=\kappa {\chi }^a$. Choose a parameter $v$ with $\chi ={\partial }_v$. The null expansion in this parameter is $\theta ={\partial }_v\log dA$. The generator is not affinely parametrized; if $\ell ={\partial }_{\lambda }$ is affine, then $\chi =\kappa \lambda \ell$ after an appropriate choice of affine origin, with $\lambda \propto e^{\kappa v}$. Rescaling the Null Raychaudhuri equation givesThe null twist vanishes because the generators are normal to the event horizon. By the Einstein field equations, $R_{ab}{\chi }^a{\chi }^b=8\pi T_{ab}{\chi }^a{\chi }^b$. Background null expansion and null shear vanish on a Killing horizon of the background stationary spacetime. Thus $\theta$ and $\hat{\sigma }$ are first order, their squares are second order, and the linearized equation isThe final stationary spacetime supplies the boundary condition $\theta \to 0$ in the future. Explicitly,This future boundary condition reflects the global definition of the event horizon. For localized infall, $\theta$ also tends to zero in the remote past. Integrating the linearized equation along the generators, and then over a background cross-section with area element $d{A}_0$, givesUsing the background area element introduces only second-order corrections.

With orientations chosen so that future infalling positive energy has positive flux, the conserved Killing energy and axial angular momentum giveConsequently $\delta E-{\Omega }_H\delta J={\int }_{\mathcal{H}}\delta T_{ab}{\chi }^a{\chi }^{b}dvd{A}_0$, proving the displayed Physical-process first law of black-hole mechanics. Restoring $G$ replaces $8\pi$ by $8\pi G$. The null energy condition makes the integral nonnegative and therefore implies area increase in this regime; that condition is needed for the sign conclusion, not for the linearized flux identity itself. Charged infall would require the additional electrostatic work term, and an extremal background needs separate treatment because the proof uses $\kappa >0$.

Let $p^a$ be the future-directed causal four-momentum of the fragment captured by the Kerr black hole. In signature $(-+++)$ its conserved Killing energy and axial angular momentum areThe future-directed null generator of the Killing horizon is ${\chi }^a=k^a+{\Omega }_H{m}^a$, where ${\Omega }_H$ is the Kerr horizon angular velocity. At a crossing of the future event horizon, the scalar product of a future-directed causal vector with a future-directed null vector is nonpositive. For example, in a local orthonormal frame, $\chi =\alpha (1,\mathbf{n})$ with $|\mathbf{n}|=1$ and $\alpha >0$, while $p^0\ge |\mathbf{p}|$; hence $p\cdot \chi =\alpha (-p^0+\mathbf{p}\cdot \mathbf{n})\le 0$. ThereforeThis is a local causal restriction, so it applies also to an extremal Kerr black hole. A massive fragment crossing a regular future event horizon has a strict inequality; equality is the reversible limiting case, with causal four-momentum approaching the generator direction.

The Penrose process exploits the fact that $k^a$ becomes spacelike in the Kerr ergoregion, allowing future-directed fragments with negative Killing energy. For a positive choice of Kerr horizon angular velocity, a captured negative-energy fragment must have sufficiently negative angular momentum to satisfy the inequality. Conservation of energy in the splitting then gives $E_{\mathrm{escape}}=E_{\mathrm{initial}}-E_{\mathrm{capture}}>E_{\mathrm{initial}}$: the escaping fragment extracts rotational energy and the captured fragment reduces the black hole's mass and angular momentum.

For the independent thermodynamic argument, take a subextremal Kerr black hole and a small capture process covered by the Physical-process first law of black-hole mechanics. To first order, $\delta M=E$, $\delta J=L$, soThe second law of black-hole mechanics, under its classical null energy condition and regularity hypotheses, gives $\delta A\ge 0$; since $\kappa >0$, this reproduces $E\ge {\Omega }_{H}L$. The ideal reversible limit has $\delta A\to 0$. One should not infer a general finite-capture equality for an extremal black hole by simply substituting $\kappa =0$: the linearized nonextremal process proof and its differentiability assumptions must first be controlled. The causal proof already establishes the required inequality there without this issue.

Hawking radiation is a prediction of quantum field theory in curved spacetime, not of classical collapse alone. Consider a free massless Klein-Gordon field in the collapse vacuum for Hawking radiation: initially no incoming particles from past null infinity, with the short-distance state regular for freely falling observers through the forming event horizon. At late times use the approximately fixed exterior Schwarzschild metric; neglect backreaction during the observation interval. First set $G=c=\hslash =k_B=1$.

The exterior metric has $f(r)=1-2M/r$. Introduce the tortoise coordinate and outgoing and ingoing null coordinates,The surface gravity is $\kappa =1/(4M)$. The logarithm in $r_{\ast }$ makes a late escaping ray spend a long exterior coordinate time close to the event horizon. One of the regular outgoing Kruskal–Szekeres coordinates is proportional to $-e^{-\kappa u}$. Smooth propagation backwards through the collapsing body relates this regular coordinate to the incoming coordinate on past null infinity. If $v_H$ labels the last ray able to escape, the late-time Hawking exponential ray map is consequentlyHere the symbol $u(v)$ is the outgoing retarded time evaluated on the ray; the first equation is the leading asymptotic relation, with subleading corrections that vanish at late times. The constant $C$ depends on collapse history, whereas the exponential rate is determined by the final surface gravity. The relation follows from a nondegenerate Killing horizon; a linear ray map would not have the same effect.

An outgoing positive-frequency mode at future null infinity behaves as $p_{\omega }\sim e^{-i\omega u}/\sqrt{4\pi \omega }$, where $\omega >0$ is measured with respect to the asymptotic stationary time. In the near-horizon geometric optics approximation its backward-propagated form on past null infinity isThis is not purely positive frequency in $v$. Its decomposition into incoming positive-frequency modes $f_{{\omega }^{\prime }}\sim e^{-i{\omega }^{\prime }v}/\sqrt{4\pi {\omega }^{\prime }}$ and their complex conjugates is a Bogoliubov transformation:The coefficients follow from the Klein-Gordon inner product, or equivalently from the Fourier transform with the indicated flux normalization. Writing $y=v_H-v$ and $a=\omega /\kappa$, the two Fourier integrals, after a convergence regulator, areThe equality is the defining integral of the Gamma function, continued to these arguments from the right half-plane. Up to phases of unit modulus, $\alpha$ is $\sqrt{{\omega }^{\prime }/\omega }{I}_{+}/(2\pi )$ and $\beta$ is $\sqrt{{\omega }^{\prime }/\omega }{I}_{-}/(2\pi )$. Taking $ϵ\downarrow 0$ on the principal branches, the arguments of $ϵ\pm i{\omega }^{\prime }$ tend to $\pm \pi /2$. Thus $|I_{+}|\propto e^{\pi a/2}$ and $|I_{-}|\propto e^{-\pi a/2}$, giving the thermal ratio of Hawking Bogoliubov coefficientsIn particular, using $|\Gamma (1+ia){|}^2=\pi a/\sinh (\pi a)$,The collapse-dependent phases and $C$ disappear from the ratio. The canonical identities for a bosonic Bogoliubov transformation state that the positive-frequency norm minus the negative-frequency norm is one for a normalized outgoing mode. Combining that identity with the thermal ratio yields the late-time mean occupationThese are the Bose-Einstein distribution and the Hawking temperature. Strictly, monochromatic modes have delta-function normalization, and $\int d{\omega }^{\prime }|{\beta }_{\omega {\omega }^{\prime }}{|}^2$ contains the infinite-duration divergence $\int d{\omega }^{\prime }/{\omega }^{\prime }$. Normalized wave packets with a finite frequency band and a finite retarded-time window remove this artifact. For sufficiently late packets the occupation is the frequency average of the displayed thermal factor, tending to it for a narrow band. Fermionic fields instead give the corresponding denominator $e^{2\pi \omega /\kappa }+1$ because their canonical normalization uses anticommutators.

Four-dimensional propagation adds an important qualification to the word thermal. Separation of the massless Klein-Gordon equation into spherical harmonics gives an exterior radial scattering potentialOnly a fraction ${\Gamma }_{\ell }(\omega )$ of the outgoing mode is transmitted to future null infinity; this is the greybody factor. The scalar particle flux there isThe factor $2\ell +1$ counts the angular modes, and multiplication by $\omega$ gives the spectral energy flux. An observer at infinity therefore receives an outgoing thermal population filtered by curvature scattering, rather than an exact unfiltered blackbody spectrum. In physical units, for mass $M_{\mathrm{phys}}$,There is no incoming thermal bath in the collapse vacuum for Hawking radiation. The outgoing flux has positive Killing energy; its counterpart inside the event horizon can carry negative Killing energy. Including the expectation value of the stress-energy tensor in the gravitational evolution allows the mass to decrease. The fixed-background derivation describes the slowly evolving semiclassical regime and does not determine the endpoint of evaporation. Quantum expectation values need not satisfy the classical null energy condition, so this loss of area is not a contradiction of the classical second law of black-hole mechanics. The essential late-time universality is the combination of a regular initial quantum state and the exponential redshift at the newly formed event horizon.