Past exam of the relativity course of the University of Cambridge / 2021 / Section B / 6 / b / Solution

Put $r_s=2\mu$ and $f(r)=1-r_s/r$. Radial null geodesics of the Schwarzschild metric obeyNear the horizon, $\delta =r-r_s$ satisfies$C>0$ describes the exterior and $C<0$ the interior. The exponential approach in Schwarzschild coordinate time resembles part (a), but $\delta$ is not a proper radial distance.

Outside, set $\rho =2\sqrt{r_s\delta }$ and retain the leading near-horizon terms. The radial metric becomes the Rindler limit of a Schwarzschild horizon:It agrees with part (a) for $x=\rho$, $\alpha =c/(2r_s)$. Accordingly $\rho \propto e^{\pm ct/(2r_s)}$, while $\delta \propto {\rho }^2$ explains the factor two in its exponent.

Inside, write $r=r_s-{\rho }^2/(4r_s)$. The leading radial metric is insteadHere the radial coordinate is timelike and the Schwarzschild $t$ coordinate is spacelike. Thus $x<0$ in the Rindler metric does not represent the black-hole interior: $g_{tt}=(\alpha x{)}^2$ stays positive on both Rindler wedges. In the black-hole interior all future-directed radial light rays move towards decreasing $r$; the two signs of $dr/dt$ do not mean opposite future radial directions because $t$ is no longer time.

The horizon is a coordinate singularity in Schwarzschild coordinates and in the accelerated chart, not a curvature singularity. The full Schwarzschild geometry nevertheless has genuine tidal forces; the flat Rindler metric captures its leading local radial near-horizon structure.