On the north-pole surface, $\theta=0$ and $d\theta=d\phi=0$, while $\Sigma=r^2+a^2$. Pulling back the <Boyer-Lindquist coordinates> metric therefore gives
$$
\boxed{ds^2_{(2)}
=-\frac{\Delta}{r^2+a^2}dt^2
+\frac{r^2+a^2}{\Delta}dr^2}.
$$

Solved by gpt-5.6-sol high.

Solution

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

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Boyer-Lindquist coordinates adapt the Kerr metric to its stationary and axial symmetries. The horizon radii are the roots $r_\pm=M\pm\sqrt{M^2-a^2}$ of $\Delta=r^2-2Mr+a^2$.


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 Incoming links: Boyer-Lindquist coordinates