 Solution

Reflection across $\theta =\pi /2$ is an isometry; a geodesic initially tangent to its fixed equatorial plane remains in that plane by uniqueness of the geodesic equation. Thus take $\theta =\pi /2$, $\dot{\theta }=0$, and $\Sigma =r^2$. Define the algebraic abbreviationsThe conserved quantities of part (b) becomeFor $\Delta \ne 0$, inversion givesThe null condition can be written without expanding a quadratic square:Substituting the preceding expressions yields the equatorial Kerr null geodesic radial equationAn equivalent useful polynomial form isThese expressions use an affine parameter $s$; a common rescaling of $E$ and $L$ accompanies its rescaling. The inversion was performed in the exterior $r>r_{+}$, where Boyer-Lindquist coordinates are regular. The polynomial radial identity has a smooth continuation to a horizon, but the formulas for $\dot{t}$ and $\dot{\varphi }$ there must be replaced by regular-coordinate expressions. In the limit $a=0$, the result reduces to the Schwarzschild spacetime equation ${\dot{r}}^2=E^2-(1-2M/r)L^2/r^2$.