Set $F(r)=\Delta/(r^2+a^2)$. The <Kerr polar tortoise coordinate> obeys
$$
\boxed{\frac{dr_*}{dr}=\frac1F
=\frac{r^2+a^2}{\Delta}}.
$$
Then $ds^2_{(2)}=F(-dt^2+dr_*^2)$. Introduce retarded time $u=t-r_*$, so $dt=du+dr_*$ and
$$
\boxed{ds^2_{(2)}=-F(r)du^2-2\,du\,dr}.
$$
The 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.

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 / ii / Solution

{c}

On the Kerr symmetry axis, the radial metric has $F=\Delta/(r^2+a^2)$ and the tortoise coordinate satisfies $dr_*/dr=F^{-1}=(r^2+a^2)/\Delta$.


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