Past exam of the mathematics course of the University of Cambridge / 2006 / ii / Paper 2 / 35A (General Relativity)

In the equatorial plane, the Killing symmetries give conserved quantities $E=(1-2M/r)\dot{t}$ and $L=r^2\dot{\varphi }$. The null normalization givesFor $L\ne 0$, put $u=1/r$; then $du/d\varphi =-\dot{r}/L$. ConsequentlyDifferentiating gives $u^{\prime \prime }+u=3M{u}^2$, initially where $u^{\prime }\ne 0$ and then at turning points by continuity. The circular solution follows directly from the radial equation as well. The ratio $|L/E|$ is the impact parameter of the asymptotic null orbit.

The constant $u=1/(3M)$ satisfies $u=3M{u}^2$. It is the circular photon orbit at $r=3M$, the photon sphere. A perturbation $u=u_c+\eta$ satisfies ${\eta }^{\prime \prime }=\eta$ to first order, so perturbations grow exponentially with orbital angle: the circular null orbit is unstable. Substitution into the first integral gives $E^2/L^2=1/(27M^2)$.

Write $c=\cosh \varphi$. The proposed function is $u=(c-2)/(3M(c+1))$. Differentiating using $c^{\prime }=\sinh \varphi$ and $(c^{\prime }{)}^2=c^2-1$ verifies $u^{\prime \prime }+u-3M{u}^2=0$. Its physical branches have $u>0$, namely $|\varphi |>{\varphi }_0=\mathrm{arcosh}2$, andOn the positive branch it comes from infinity at ${\varphi }_0$ and spirals toward $r=3M$ from outside as $\varphi \to \infty$. The negative branch gives the reversed orbit after reflection. The interval where $u<0$ does not describe positive physical radius.

Near ${\varphi }_0$, $\cosh \varphi -2=\sqrt{3}(\varphi -{\varphi }_0)+O((\varphi -{\varphi }_0{)}^2)$. Thus $r\sim 3\sqrt{3}M/(\varphi -{\varphi }_0)$, while $\sin (\varphi -{\varphi }_0)\sim \varphi -{\varphi }_0$. HenceThis is the perpendicular offset of the incoming asymptote and agrees with $|L/E|$ from the circular orbit. The orbit is the critical separatrix between scattering and capture.