Past exam of the mathematics course of the University of Cambridge / 2009 / iii / Paper 56 / 2

An asymptotically flat spacetime is a stationary spacetime if it admits a complete stationary Killing vector field $K$, with ${\mathcal{L}}_{K}g=0$, timelike near infinity and normalized there to a unit time translation. It need not remain timelike in an ergoregion of a stationary spacetime. It is axisymmetric if it admits an axial Killing vector field $\Phi$, whose flow is a rotational $U(1)$ action: its spacelike orbits away from the axis are closed with period $2\pi$, and it vanishes on the regular rotation axis. For a stationary axisymmetric spacetime, the generators can be chosen to commute, $[K,\Phi ]=0$. These are geometric symmetries, not merely time-independent or angle-independent coordinate labels.

During gravitational collapse, nonspherical time-dependent multipoles can be lost in gravitational radiation or absorbed by the forming black hole. Matter remaining outside must ultimately escape or fall in if the final object is to be an isolated vacuum black hole. Assume that the classical exterior settles to a regular stationary asymptotically flat vacuum geometry, with a connected horizon and the regularity conditions of the black-hole uniqueness theorem. This settling assumption is physical input, rather than a consequence of charge conservation alone.

For a rotating remnant, the black-hole rigidity theorem supplies axial symmetry; the uniqueness theorem then identifies the exterior as Kerr spacetime. The nonrotating equilibrium case is Schwarzschild spacetime. Gravitational radiation carries energy and angular momentum but no net electric charge. In the intended neutral collapse, with no net charge carried away by expelled matter, charge conservation therefore leaves zero black-hole charge. Initial total neutrality alone would not prove this if oppositely charged matter were expelled; the two-parameter conclusion assumes the uncharged vacuum remnant just described. This leavesas its two intrinsic continuous parameters. The final $M,J$ are what remain after escaping matter and radiation have carried off their shares; they need not equal the star's initial values. A spatial rotation may align the spin axis, and a translation or boost chooses the position and rest frame, so these do not add intrinsic parameters. All higher equilibrium multipoles are fixed by $M,J$; for example Kerr's mass quadrupole is $-J^2/M$. A regular Kerr black hole obeys $|J|\le M^2$ in geometrized units. This is the content of the black-hole no-hair theorem in vacuum general relativity, with its stated hypotheses, rather than a statement about arbitrary matter theories or an unproved guarantee of dynamical settling. Hawking evaporation is ignored in this classical late-time equilibrium description.

The conserved Killing energy of a geodesic tangent $p^a$ is $E=-K_a{p}^a$, where $K={\partial }_t$. Conservation follows from $p^a{\nabla }_a{p}^b=0$ and the Killing equation. For the Kerr metric, $K^2=g_{tt}=-(1-2Mr/\Sigma )$. Outside the outer Kerr ergoregion, $K$ is timelike. Its orthogonal complement is positive definite, so a nonzero causal vector cannot have $K\cdot p=0$. Equivalently, if both are future-directed, $-K\cdot p>0$. HenceThis excludes the open exterior of the stationary-limit surface; on the surface itself a zero-energy null tangent can be proportional to the null stationary vector.

A Killing horizon must be a null hypersurface whose normal is a Killing vector field. The outer stationary-limit surface instead has $s=r-r_e(\theta )=0$ and, in Boyer-Lindquist coordinates,On it, $r_e^2-2M{r}_e+a^2{\cos }^2\theta =0$, so $\Delta (r_e)=a^2{\sin }^2\theta >0$ away from the rotation axis. Its normal is spacelike and the surface timelike. Thus the Kerr stationary-limit surface is generically timelike and is not a Killing horizon, despite $K^2=0$ there. For example its stationary null vector is not normal, since $g(K,{\partial }_{\varphi })=-2Mar{\sin }^2\theta /\Sigma \ne 0$ away from the axis. The actual outer Killing horizon is $r=r_{+}=M+\sqrt{M^2-a^2}$, generated by $K+{\Omega }_H\Phi$, with Kerr horizon angular velocity ${\Omega }_H=a/(r_{+}^2+a^2)$. The two surfaces meet on the axis, which does not make the entire stationary-limit surface null.

For the superextreme Kerr metric,for all real $r$, so there are no coordinate singularities at horizon roots. The only stipulated curvature singularity is $\Sigma =0$, which forces $r=0$, $\theta =\pi /2$. In the supplied Cartesian coordinates this is the Kerr ring singularity $x^2+y^2=a^2$, $z=0$.

Write ${\rho }^2=x^2+y^2$. The coordinate transformation impliesand henceThe positive-radius Cartesian sheet usesIts zero set is the disk $z=0$, $\rho \le a$. At an interior disk point, $\rho <a$, one has $\cos \theta \ne 0$, so $\Sigma =a^2{\cos }^2\theta >0$. The disk interior is therefore not a singularity: in signed $(r,\theta ,\varphi )$ coordinates, the metric is analytic through $r=0$. In particular, $g_{rr}=\Sigma /\Delta$ is finite and the metric determinant $-{\Sigma }^2{\sin }^2\theta$ is nonzero away from the ordinary polar coordinate axis. The polar axes are regular using Cartesian angular charts and $2\pi$ periodicity.

The required two-sheeted extension of superextremal Kerr spacetime is constructed by taking a second Cartesian sheet with the negative root $r\le 0$, cutting both sheets along the disk, and gluing the upper bank of one to the lower bank of the other, and conversely. Signed $r$ then passes continuously through zero as the disk is crossed. The ring is omitted. A useful check near an interior disk point isso signed $r$ is a smooth transverse coordinate there. The apparent factors $z/r$ in the Cartesian metric represent $\cos \theta$ and have regular signed-coordinate limits; they do not justify deleting the whole disk. Identifying the two signs of $r$ on a single Cartesian copy would also identify two physically different metric values, since the Kerr-Schild coefficient changes sign.

Continue the two sheets out to $r\to \pm \infty$. Both ends are asymptotically flat; at the negative-radius end the mass term has the opposite sign, as $g_{tt}=-1+2M/r+O(r^{-2})$. There are no further finite-radius horizon boundaries requiring additional blocks. Removing only the curvature-singular ring, and removing coordinate artifacts with the charts described above, gives the standard maximal analytic extension. The ring cannot be filled by an analytic nondegenerate metric because its curvature diverges. This extension is not globally hyperbolic; analyticity of an extension should not be confused with unique evolution from Cauchy data.

On the equatorial plane, the Kerr metric coefficient simplifies toPut $r=-\epsilon$ on the negative-radius sheet. Then $g_{\varphi \varphi }={\epsilon }^2+a^2-2M{a}^2/\epsilon <0$ for all sufficiently small $\epsilon >0$. The curveis a closed timelike curve, since its tangent is ${\partial }_{\varphi }$ and the angular coordinate is periodic. It never meets the ring at $r=0$. These are the closed timelike circles in negative-radius Kerr spacetime; they need not be geodesics. Along them $\tilde{t}$ is also constant, because its difference from $t$ depends only on $r$.

For $0<M<a$, the ring is a naked singularity and there is no event horizon shielding the extension's closed timelike curves. The weak cosmic censorship conjecture predicts that such superextreme Kerr geometries are not formed by generic collapse of regular isolated initial data satisfying the appropriate energy conditions. This is a conjectural physical restriction, not a proof from the exact metric alone.

For $M>a>0$, the closed timelike curves lie deep inside the ideal analytic extension, beyond the inner Cauchy horizon, so weak cosmic censorship does not itself exclude them. However, that inner horizon is unstable to perturbations, with mass inflation expected in realistic collapse. The strong cosmic censorship conjecture is the relevant obstruction to a regular continuation into the ideal chronology-violating region. Thus the exact Kerr extension exhibits such curves mathematically, but neither regular collapse nor the uniqueness theorem establishes that they occur in Nature.