Past exam of the mathematics course of the University of Cambridge / 2026 / iii / Paper 311 / 4

A black-hole metric is stationary when it admits a Killing field $K$ generating time translations and timelike near spatial infinity. It is static when $K$ is also hypersurface orthogonal,In coordinates adapted to a static field, the metric coefficients are time independent and the time-space cross terms vanish.

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A stationary axisymmetric spacetime additionally admits an axial Killing field $m$ with closed orbits. The two symmetry generators commute,and preserve the black-hole exterior and horizon. Coordinates adapted to them make the metric independent of both $t$ and $\varphi$.

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The Kerr black hole coefficients are independent of $t$ and $\varphi$, so ${\partial }_t$ and ${\partial }_{\varphi }$ are commuting Killing fields; the latter has closed circular orbits. Kerr is therefore stationary and axisymmetric.

Expanding the metric gives the nonzero cross componentFor $a\ne 0$ this cannot be removed throughout the exterior by a constant redefinition of $\varphi$, and the asymptotically timelike stationary Killing field has nonzero twist. It is not hypersurface orthogonal, so Kerr is not static.

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The stationary Killing field becomes null whereSolving $r^2-2Mr+a^2{\cos }^2\theta =0$ gives the stationary-limit surfacesThe exterior Kerr ergoregion isThe event horizon is the constant-radius surface $\Delta =0$, whereas the outer stationary limit lies outside it except at the poles, where the two meet.

Inside the ergoregion ${\partial }_t$ is spacelike, so future-directed particles can have negative conserved stationary Killing energy while still following causal trajectories. Sending such a particle through the horizon permits another particle to escape with increased positive energy. This is the Penrose process, which extracts the black hole's rotational energy.

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On the north-pole surface, $\theta =0$ and $d\theta =d\varphi =0$, while $\Sigma =r^2+a^2$. Pulling back the Boyer-Lindquist coordinates metric therefore gives

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Set $F(r)=\Delta /(r^2+a^2)$. The Kerr polar tortoise coordinate obeysThen $d{s}_{(2)}^2=F(-d{t}^2+d{r}_{\ast }^2)$. Introduce retarded time $u=t-r_{\ast }$, so $dt=du+d{r}_{\ast }$ andThe determinant of this metric is $-1$, including at $\Delta =0$. These outgoing Eddington-Finkelstein-type coordinates therefore extend analytically across the past event horizon into the white-hole region of Kerr.

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At $r=0$ on the north pole,so nothing singular occurs. The Kerr curvature singularity is the ring $\Sigma =0$, which requires both $r=0$ and $\theta =\pi /2$ and is absent from this polar surface. The analytic extension passes through $r=0$ into the negative-$r$ asymptotic region, giving the maximal coordinate range

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On a constant-$t$ asymptotically flat hypersurface, transform the large-$r$ spatial metric to asymptotically Cartesian coordinates. Its leading perturbation isThe rotation parameter $a$ first affects the spatial metric at orders that vanish in the Arnowitt-Deser-Misner energy surface limit. Direct differentiation givesHenceUsing ${\int }_{S_r^2}dA=4\pi r^2$ in the ADM surface integral yieldsThus the Kerr parameter $M$ is its total mass in the rest frame.

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