Past exam of the mathematics course of the University of Cambridge / 2015 / iii / Paper 54 (Black Holes)

In the vacuum exterior, the Tolman–Oppenheimer–Volkoff equation gives $m^{\prime }=0$, so $m(r)=M$. The remaining metric function obeysThus $e^{2\Phi }=C(1-2M/r)$. A constant rescaling of the time coordinate removes $C$; equivalently, asymptotic flatness fixes $C=1$ when time is normalized at infinity. Consequently the exterior is the Schwarzschild metric:The mass $M=m(R)$ includes the full perfect fluid in general relativity interior. No equation of state beyond the surface enters this vacuum result.

A regular central expansion of the TOV equations requires $m(0)=0$. Once ${\rho }_c$ is chosen, the barotropic stellar equation of state fixes $p_c=p({\rho }_c)$. Smoothness at the centre forces the radial first derivatives of $p$ and $\rho$ to vanish. Expanding the Tolman–Oppenheimer–Volkoff equation givesSince $dp/d\rho >0$, the equation of state is locally invertible, and these coefficients determine the corresponding energy density expansion as well. Integrating the regular solution outward gives the surface at the first zero of the pressure, when such a finite surface exists. The ordinary differential equation uniqueness theorem then fixes the interior and its exterior mass from ${\rho }_c$.

The additive constant ${\Phi }_c$ is a time-coordinate normalization, not a second physical parameter. Matching to the Schwarzschild metric with time normalized at infinity fixes it. Therefore the regular stellar solutions, modulo time rescaling, have one physical parameter: ${\rho }_c$. This statement uses a suitably smooth fixed equation of state and refers to central densities for which the solution actually forms a regular star; not every central energy density need give a finite-radius configuration.

The useful high-density core bound from TOV compactness follows without choosing the unknown high-density equation of state. Let $a$ be the radius at which the decreasing energy density first reaches ${\rho }_0$, and put $\mu =m(a)$, $p_0=p({\rho }_0)$. The Tolman–Oppenheimer–Volkoff equation gives $p^{\prime }<0$ away from a nonempty centre; since $dp/d\rho >0$, energy density decreases outward. HenceFor the local stellar compactness, if $z\ge 0$, $1-z+\sqrt{1+z}\le 2$, since its derivative is $-1+(2\sqrt{1+z}{)}^{-1}<0$. Applying the supplied compactness inequality at the core boundary therefore gives $\mu /a<4/9$. Combining the two bounds yieldsThe pressure-dependent form further restricts the allowed core-boundary data:In particular, positivity of its right side gives $6\pi a^2{p}_0<3$, so $a^2<1/(2\pi p_0)$ when $p_0>0$. These restrictions depend only on known quantities, not on the equation of state above ${\rho }_0$.

The intended stellar-mass argument now attaches the known low-density envelope to each admissible pair $(a,\mu )$ with pressure $p_0$. Its Tolman–Oppenheimer–Volkoff equation involves only the known equation of state. For a physical low-density envelope with a uniformly bounded mass on this bounded set of core data, taking the supremum of the envelope-plus-core masses gives an upper mass limit independent of any high-density continuation. Stars with ${\rho }_c\le {\rho }_0$ are also determined entirely by the known equation of state. Unknown dense-core physics cannot evade the displayed core bounds.

A qualification is necessary for a claim about the entire star: the printed monotonicity assumptions alone do not guarantee a finite maximum total mass for an arbitrary known low-density equation of state. An explicit low-density polytropic mass divergence occurs for $p=K{\rho }^{5/4}$ with fixed $K>0$. In the low-density Newtonian limit, this is a stellar polytrope of index $n=4$, whose Lane-Emden equation has a finite first zero. With $\rho ={\rho }_c{\vartheta }^4$ and $r=\ell \xi$, hydrostatic balance fixeswhere ${\mu }_4>0$ is the dimensionless Lane-Emden surface mass constant. Thus $M\to \infty$ as ${\rho }_c\to 0$, although $M/R\propto {\rho }_c^{1/4}\to 0$. The first Lane-Emden zero is simple. The continuous dependence of an ODE solution on parameters therefore preserves this finite surface for sufficiently small relativistic parameter. The relativistic correction is controlled by $p_c/{\rho }_c=K{\rho }_c^{1/4}\to 0$, so the corresponding regular finite-radius Tolman–Oppenheimer–Volkoff equation solutions have the same limiting scalings. They never sample $\rho >{\rho }_0$, and satisfy the local positivity and monotonicity assumptions in their nonvacuum interiors. This shows exactly why a suitable bounded-envelope hypothesis, or an appropriate restriction to physically stable stellar models, must accompany the intended total-mass conclusion. The compactness inequality supplies a bound on the unknown core, not a bound on the mass of every conceivable dilute envelope.

For the constant-density model, $m(r)=4\pi \rho r^3/3=M{r}^3/R^3$. This is the incompressible Interior Schwarzschild metric, replacing the earlier finite-slope barotropic stellar equation of state. The Tolman–Oppenheimer–Volkoff equation separates asOn the left, partial fractions giveOn the right the antiderivative is $\frac{3}{4\rho }\log (1-8\pi \rho r^2/3)$. Applying the surface condition $p(R)=0$, and writing $s(r)=\sqrt{1-2M{r}^2/R^3}$, $s_R=\sqrt{1-2M/R}$, givesSolving for the pressure recovers the Interior Schwarzschild solution:In particular the central pressure of a constant-density relativistic star isThe denominator must be positive for a regular positive-pressure centre, requiring $M/R<4/9$. As that limiting compactness is approached, the central pressure diverges, in agreement with Buchdahl's theorem.

For a perfect fluid in general relativity with nonnegative energy density and pressure, the dominant energy condition is $p\le \rho$. Pressure decreases outward, so it is enough to test the central pressure. Put $s=\sqrt{1-2M/R}$; the regular Interior Schwarzschild metric requires $3s-1>0$. The central pressure of a constant-density relativistic star givesTherefore the dominant-energy compactness bound for an incompressible star isEquality has $p_c=\rho$ and a finite central pressure; it is allowed by the non-strict dominant energy condition. This is stronger than the regular-star Buchdahl's theorem limit $M/R<4/9$.

A null geodesic congruence is a smooth family of null geodesics filling a region without intersections there, with one tangent direction assigned smoothly at every point. Choose an affine parameter $\lambda$ and tangent $U^a$, so $U^a{U}_a=0$ and $U^b{\nabla }_b{U}^a=0$. Choose a second null vector $N^a$ with $U\cdot N=-1$. The screen-space projector isIt annihilates $U,N$ and restricts to a positive-definite metric on the two-dimensional transverse screen. Project the covariant derivative of the tangent to obtain the optical tensorIts decomposition defines the three optical quantities:The null expansion is the fractional rate of change of a transverse area, $\theta =A^{-1}dA/d\lambda$. The null shear changes a beam's shape without changing its area to first order. The null twist is its antisymmetric transverse deformation. The factor $1/2$ uses four spacetime dimensions.

For an affinely parametrized null geodesic congruence, put $B^a{}_b={\nabla }_b{U}^a$. Differentiating along a ray and commuting covariant derivatives givesusing $[{\nabla }_c,{\nabla }_b]V^a=R^a{}_{dcb}{V}^d$. The geodesic acceleration term vanishes. Since $U$ is null and affine, $B^a{}_b{U}^b=0$ and $U_a{B}^a{}_b=0$. Its trace is therefore the screen trace $\theta$, and the trace of its square equals that of the optical tensor squared. Taking the trace, with $R_{db}=R^a{}_{dab}$, yieldsNow decompose the optical tensor into null expansion, null shear and null twist. The symmetric trace-free part is orthogonal to the trace and to the antisymmetric part, while ${\hat{\omega }}_{ab}{\hat{\omega }}^{ba}=-{\hat{\omega }}_{ab}{\hat{\omega }}^{ab}$. HenceSubstitution proves the Null Raychaudhuri equation:The squares use the positive-definite screen metric. Thus null shear contributes to focusing, whereas nonzero null twist has the opposite sign.

In the two null coordinates the inverse metric has $g^{UV}=g^{VU}=-f^{-1}$ and $g^{UU}=g^{VV}=0$. ThereforeBoth tangents are null and future-directed, since $f>0$ and both coordinate directions are future-directed. To prove the affine parameter property, use the null-gradient affine-geodesic identity. The one-form $k_a=-{\nabla }_{a}U$ is closed, so ${\nabla }_a{k}_b={\nabla }_b{k}_a$. ConsequentlyThe same argument applies to ${\ell }_a=-{\nabla }_{a}V$. Thus both vector fields satisfy the geodesic equation without a nonaffinity term, and generate affinely parametrized null geodesic congruences. The coordinate vectors alone need not be affine; the factors $f^{-1}$ are essential.

For the spherical null expansions in double-null coordinates, the cross-sectional area of a symmetry sphere is $A=4\pi r^2$. For the two affine null geodesic congruences, the definition of null expansion therefore givesThere is no extra derivative of $f$ in these formulas: the null expansion differentiates the transverse area, not the normalization of the tangent. Equivalently, the projected metric is $q_{AB}=r^2{\gamma }_{AB}$ and its optical tensor is $(k(r)/r)q_{AB}$, or $(\ell (r)/r)q_{AB}$ for the other congruence. It is purely trace, so both null shears vanish.

The null twist vanishes because the covector tangents are gradients of $-U$ and $-V$, and hence their exterior derivatives vanish. Each congruence is orthogonal to a family of null hypersurfaces. This also identifies the Frobenius theorem reason for zero null twist.

Use the Null Raychaudhuri equation for $\ell =f^{-1}{\partial }_U$. Its null shear and null twist vanish, and $R_{ab}{\ell }^a{\ell }^b=f^{-2}{R}_{UU}$. Substitution of ${\theta }_{\ell }=2r_{,U}/(fr)$ givesThe quadratic terms cancel. Thus the double-null focusing identity isHere the last equality uses the Einstein field equations in units $G=1$; terms proportional to the metric vanish in this null component. The inequality follows from the null energy condition.

For any point of the future domain of dependence $D^{+}(\Sigma )$, the past continuation of its $\ell$ generator must meet the Cauchy hypersurface $\Sigma$: every past-inextendible causal curve from that point meets $\Sigma$. Along its future continuation, $U$ increases, and $f^{-1}{r}_{,U}$ is nonincreasing by the focusing identity. Its initial value is negative, so it stays negative. Since $f>0$, $r_{,U}<0$ throughout the regular part of $D^{+}(\Sigma )$. The argument is used only where the smooth double-null chart has $r>0$; it does not extend a congruence beyond a singular endpoint.

A spherical trapped surface has both future null expansions negative. Thus the initial sphere has $r_{,U}<0$ and $r_{,V}<0$. The result of the previous part already gives $r_{,U}<0$ at every regular future sphere in $D^{+}(\Sigma )$.

Apply the same Null Raychaudhuri equation to $k=f^{-1}{\partial }_V$. The exchanged-coordinate double-null focusing identity isAlong $U=U_0$, increasing $V$ is future-directed. ThereforeBoth null expansions remain negative, proving persistence of spherical trapping along this future null direction. As usual, the conclusion concerns every $V_1>V_0$ for which the sphere exists in the regular chart in $D^{+}(\Sigma )$; the inequality does not assert existence of such spheres beyond a focusing singularity.

For the Komar integral of a five-dimensional Reissner--Nordstrom metric, use the one-form metrically dual to the stationary Killing vector field: $k^{♭}=-f(r)dt$. Its exterior derivative is $d{k}^{♭}=f^{\prime }(r)dt\wedge dr$. The stated orientation gives volume formBecause the squared norm of $dt\wedge dr$ is $g^{tt}{g}^{rr}=-1$, the defining wedge identity for the Hodge star operator givesThus the Komar integral over the positively oriented angular three-sphere issince the unit three-sphere has volume $2{\pi }^2$. Write $s=r_{+}^2+r_{-}^2$, $q=r_{+}^2{r}_{-}^2$. Then $f=1-s/r^2+q/r^4$ and $r^3{f}^{\prime }=2s-4q/r^2$. With the normalization specified for the Komar mass,The orientation makes this positive. The problem's coefficient $1/(16\pi )$ must be retained: it is not the usual dimension-adjusted five-dimensional ADM-normalized Komar coefficient. With that conventional normalization the ADM mass would instead be $3\pi s/8$.

For radial null geodesics, the metric gives $dt/dr=\pm f^{-1}$ in the exterior. Define the tortoise coordinate byAn outgoing future radial ray has $dt/dr=f^{-1}$, so $u=t-r_{\star }$ is constant; an ingoing ray has $dt/dr=-f^{-1}$, so $v=t+r_{\star }$ is constant. These describe the null trajectories independently of the choice of affine parameter.

For the tortoise coordinate of a five-dimensional charged black hole, with $a=r_{+}$, $b=r_{-}$ and $\Delta =a^2-b^2$, partial fractions giveThus a useful explicit tortoise coordinate isThe logarithms signal failure of the static coordinates at the horizons, rather than a curvature singularity there.

Since $v=t+r_{\star }$, $dt=dv-f^{-1}dr$. Substitution into the radial metric cancels the $d{r}^2/f$ terms. The ingoing Eddington-Finkelstein coordinates giveThe radial block has determinant $-1$, even when $f=0$. Its coefficients are analytic for every $r>0$, so this form extends through $r=r_{+}$ and also through $r=r_{-}$ to $0<r<r_{+}$. Ordinary angular coordinate singularities are removed by angular charts on the three-sphere. The genuine endpoint $r=0$ is not included in this extension.

The field $n=-{\partial }_r$ is an ingoing null tangent. It is future-directed in the original exterior, since its inner product with the future stationary Killing vector field is $-1$, and it remains future-directed by continuation. This fixes the time orientation used in the next part.

For an asymptotically flat spacetime with a chosen exterior component of future null infinity, the black-hole region isnamely the events from which no future-directed causal curve can reach that infinity. Here ${\mathcal{I}}^{+}$ is the infinity of the original $r>r_{+}$ exterior.

Let a future causal tangent in the ingoing Eddington-Finkelstein coordinates have components $(\dot{v},\dot{r},\dot{\Omega })$. Its inner product with the future null field $n=-{\partial }_r$ is $-\dot{v}$, so $\dot{v}\ge 0$. The causal inequality isIf $\dot{v}>0$, this impliesIn the band $r_{-}<r<r_{+}$, $f<0$, so $\dot{r}<0$. If $\dot{v}=0$, the causal inequality forces $\dot{\Omega }=0$, and future direction means the remaining radial tangent is a nonnegative multiple of $n$, again giving $\dot{r}\le 0$. This is causal trapping between two spherical horizons: $r$ is nonincreasing along every future causal curve in the between-horizon band. No event there can escape outward. An event with $0<r\le r_{-}$ would also have to cross this band outward to reach the original exterior, which is impossible. Thus all $0<r<r_{+}$ events in this extension lie in the black-hole region.

Conversely, from any $r>r_{+}$ point, the outgoing radial null ray $u=\mathrm{constant}$ satisfies $dr/dv=f/2>0$. It reaches arbitrarily large $r$ and then the original future null infinity. Therefore the exterior does not intersect the black-hole region, and $r=r_{+}$ is its event horizon. This conclusion is relative to the selected asymptotic end; the maximal charged-black-hole extension can have other asymptotic ends.

In the ingoing Eddington-Finkelstein coordinates, the stationary Killing vector field is $k={\partial }_v$ with norm $k^2=-f$. The normal vector to a level set of $r$ isAt $r=r_{+}$ it equals $k^a$, which is null there. Thus this null hypersurface is a Killing horizon of $k$. Its surface gravity is fixed by ${\nabla }_a(k^2)=-2\kappa k_a$ on the horizon. Since $k_a=(dr{)}_a$ there and ${\nabla }_a(k^2)=-f^{\prime }(r_{+})(dr{)}_a$,The positive sign follows from the future outer-horizon generator, normalized by the static time at infinity. It vanishes as the two radii approach the extremal limit.

The inner horizon in the nonextreme Reissner-Nordstrom spacetime diagram is a Cauchy horizon. Radiation from exterior perturbations can approach it at arbitrarily late advanced time and undergo unbounded blueshift. Here its positive blueshift scale is ${\kappa }_{-}=(r_{+}^2-r_{-}^2)/r_{-}^3$. The inner-horizon blueshift coordinate behaves as $V_{-}\propto -e^{-{\kappa }_{-}v}$, soEven a decaying power-law tail can therefore generate divergent local flux. With counterstreaming radiation, backreaction produces mass inflation and curvature growth, rather than the smooth inner horizon of the exact solution.

This supports the strong cosmic censorship conjecture: a generically perturbed maximal Cauchy development is expected to lose the smooth extension across its Cauchy horizon, restoring predictability in the appropriate regularity class. The exact smooth inner horizon is unstable, not a robust failure of deterministic evolution. Curvature divergence supports an obstruction to a $C^2$ metric extension; it does not by itself exclude every continuous metric extension. The regularity class is part of the conjecture, and the blueshift argument is evidence, not a general theorem proving it.

The Physical-process first law for a rotating black hole, in four-dimensional general relativity with $G=1$, isIt applies to a small neutral matter influx into an initially stationary, nonextremal black hole that settles to another stationary state. The background surface gravity and angular velocity are $\kappa$ and ${\Omega }_H$, and $\delta E,\delta J$ are the Killing energy and angular momentum delivered through the event horizon. Other conserved charges are held fixed. The perturbation must be small enough that its horizon generators do not develop caustics, and all equalities below are to first order.

Let $t^a$ be the stationary Killing vector field normalized at infinity, and ${\psi }^a$ the axial Killing vector field with $2\pi$-periodic orbits. The horizon generator is ${\xi }^a=t^a+{\Omega }_H{\psi }^a$. Choose an affine tangent $k^a$ and parameter $\lambda$ with zero at the past stationary limiting section, so affine horizon-generator scaling gives, on the background horizonThe background null expansion and null shear vanish, and null twist vanishes for the hypersurface-orthogonal horizon generators. Linearizing the Null Raychaudhuri equation and using the Einstein field equations givesThe quadratic null expansion and null shear terms are second order. With background cross-sectional measure $d{A}_0$, the first-order area change is $\delta A={\int }_{\mathcal{H}}\delta \theta d\lambda d{A}_0$. Multiplying the evolution equation by $\lambda$ and integrating by parts gives the first-order horizon-area response to matter fluxThe boundary term $[\lambda \delta \theta ]$ vanishes: $\lambda =0$ at the past limiting section, and the final stationary boundary condition gives zero expansion in the settled future. Equivalently, assume a sufficiently localized influx with the required late-time decay.

The energy and angular-momentum fluxes have signsThese correspond to the particle conventions $E=-p\cdot t$ and $J=p\cdot \psi$. Their combination isThis proves the requested Physical-process first law of black-hole mechanics. For charged infall the corresponding law contains an additional horizon-potential term ${\Phi }_H\delta Q$; the energy-minus-angular-momentum version assumes that work term is absent.

Quantize the real massless scalar field using the normalized in-modes:Use the Klein-Gordon inner product convention that is antilinear in its first argument. Mode normalization givesThe out annihilation operator is $b_i=(p_i,\hat{\varphi })$. Applying the Bogoliubov transformation and the negative norm of conjugate modes givesBoth the complex conjugation and the minus sign follow from the inner-product convention. These operators annihilate the out-vacuum, but need not annihilate the in-vacuum. The particle number operator in out-mode $i$ is $N_i^{\mathrm{out}}=b_i^{†}{b}_i$. In the in-vacuum, the only nonzero contraction is $⟨0_{\mathrm{in}}|a_j{a}_k^{†}|0_{\mathrm{in}}⟩={\delta }_{jk}$, obtained from the canonical commutation relation. ThusThis is particle number from Bogoliubov coefficients: mixing with negative-frequency in-modes produces out-particles even though no in-particles were present. For continuous mode labels the sum becomes an integral; normalized wave packets make an individual occupation number well-defined. No explicit collapse geometry or thermal spectrum is needed for this conclusion.