Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 311 / 3 / c / Solution

For the zero-angular-momentum curves, the fiber term in the induced metric tensor vanishes along the tangent, leavingfor a null geodesic. In the exterior, a future geodesic has positive conserved energy $E=(f/h)\dot{t}$. Its equations areDefine the tortoise coordinate, separately on intervals bounded by horizon radii, byFor an outgoing curve $\dot{r}=+E\sqrt{h}$,for an ingoing curve $\dot{r}=-E\sqrt{h}$, $d(t+r_{\ast })/d\lambda =0$. Thus $u=t-r_{\ast }$ and $v=t+r_{\ast }$ are the respective constant labels. The derivative $\sqrt{h}/f$, with its sign, is required for continuation; taking $\sqrt{h}/|f|$ would give the wrong labels between the horizons.

For a nonextremal horizon $r=r_H$, the tortoise coordinate has leading behaviorAt the double root of the extremal black hole it instead has a pole. These singularities are coordinate effects at a regular Killing horizon; they are not the curvature singularity at $r=0$. To cross the future horizon, the ingoing coordinatesgive the regular induced metric tensorThis also fixes the future branch used in the causal argument below.