Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 58 / 2

Axisymmetry preserves the horizon, so the axial Killing vector field $m$ is tangent to it, so its null normal $\xi =k+\Omega m$ obeys $\xi \cdot m=0$ and ${\xi }^2=0$. Therefore $k\cdot m=-\Omega m^2$ andThus $\boxed{k^2={\Omega }^2{m}^2}$ there. A generator has $dt/d\alpha =1$, $d\phi /d\alpha =\Omega$, making $\Omega =d\phi /dt$ its angular velocity relative to the nonrotating stationary frame at infinity. For Kerr it is the Kerr horizon angular velocity $a/(r_{+}^2+a^2)$; the argument itself is not Kerr-specific.

For nonzero $a$, $\Sigma =0$ requires $r=0$, $\theta =\pi /2$. Oblate Cartesian coordinates satisfy $x^2+y^2=(r^2+a^2){\sin }^2\theta$, $z=r\cos \theta$, so this is the Kerr ring singularity of radius $|a|$. Curvature components and generic invariants diverge there. The whole $r=0$ surface is not singular: away from the ring it is a disk through which the extension continues to a sheet with $r<0$. This oblate radial parameter is not a nonnegative Euclidean distance.

In $G=c=1$ units, $\boxed{J=Ma}$. Horizon candidates solve $\Delta =0$, giving $r_{\pm }=M\pm \sqrt{M^2-a^2}$. For $J^2>M^4$, equivalently $|a|>M$, there are no real roots. The over-rotating Kerr solution has no event horizon hiding its naked ring singularity. The term black hole in this regime names the Kerr family, not a hidden-singularity black hole.

The periodic axial orbits have norm $m^2=g_{\phi \phi }$. On the equator,Thus a neighborhood on the negative sheet has $m^2<0$. A fixed-$(t,r,\theta )$ curve running once around periodic $\phi$ is a closed timelike curve: its timelike tangent returns to the same event.

No horizon in the over-rotating solution prevents passage from positive-$r$ infinity through the nonsingular disk to this region. Timelike periodic axial orbits provide the time-machine mechanism. This property of the analytic solution does not establish that such a naked geometry can be manufactured stably from generic regular collapse, as envisaged by the weak cosmic censorship conjecture.

An ergoregion of a stationary spacetime is where the stationary Killing vector field, normalized as time translation at infinity, is spacelike; a fixed-coordinate stationary observer cannot remain timelike there. For the Kerr metric, $k^2=-(1-2Mr/\Sigma )$, so the outer stationary-limit surface isFor a subextremal rotating hole $r_{\mathrm{E}}\ge r_{+}=M+\sqrt{M^2-a^2}$, with equality only at the poles. At the equator $r_{\mathrm{E}}=2M>r_{+}$. The ergosphere touches the horizon on the rotation axis and is outside it elsewhere. Its boundary is not the horizon, whose generator is $k+{\Omega }_{H}m$.

The drawing uses oblate Cartesian coordinates to display the intersection and is not an isometric embedding of the spatial geometry.

Choose spin orientation so that $0<a<M$; the printed $M/a$ threshold assumes nonzero positive $a$. Put $H=2Mr/(r^2+a^2)$ and $F=1-H$. The axial metric has components $(-F,H,1+H)$ and determinant $-1$. Conserved energy and timelike normalization giveEliminating $\dot{\tilde{t}}$ gives the Kerr axial timelike effective potentialIt extends through the zeros of $F$ in the regular original coordinates. Since $V_{\mathrm{eff}}(0)=1$, reaching zero requires ${\epsilon }^2\ge 1$. At equality the infalling orbit reaches zero with zero radial velocity and turns back, because the nearby negative sheet is forbidden. Along the axis zero is regular, unlike the off-axis ring.

For $r<0$, $V_{\mathrm{eff}}^{\prime }=2M(r^2-a^2)/(r^2+a^2{)}^2$ and the global maximum is $V_{\mathrm{eff}}(-a)=1+M/a$. If ${\epsilon }^2>1+M/a$, an inward particle has no negative-sheet turning point. It passes through zero and continues toward $\boxed{r\to -\infty }$, another asymptotically flat end, with $\dot{r}\to -\sqrt{{\epsilon }^2-1}$ and infinite proper time to infinity. It first crosses the outer and inner horizons. Perturbative Cauchy-horizon instability can invalidate this ideal Kerr axial analytic extension physically.

This is the two-dimensional axis restriction. The dashed $r=0$ curve is regular; the four-dimensional ring lies off-axis and is not an axial singular boundary.