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

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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