Past exam of the general relativity II course of the University of Oxford / 2017 / 3 / d / Solution

The assertion about every exterior radius is false for geodesics. A circular orbit here means an equatorial null geodesic at constant $r$, rather than a genuinely radial trajectory. A zero of the radial effective potential alone gives a turning point, not necessarily a circular orbit. The radial geodesic equation must hold as well.

Write $q=L-aE$ andFor an exterior circular null geodesic, the two conditions are $f(r)=f^{\prime }(r)=0$, equivalently $\mathcal{R}(r)={\mathcal{R}}^{\prime }(r)=0$ with $E,L$ fixed. One way to justify the derivative condition without dividing by $\dot{r}=0$ is to use the radial Euler-Lagrange equation. At constant $r$ it givesTogether with $g_{AB}{\dot{x}}^A{\dot{x}}^B=0$, and the fixed conserved $p_A$, this is precisely the double-root condition for $\mathcal{R}$.

NowSetting this to zero and using $f=0$ givesA nontrivial exterior null geodesic cannot have $E=0$ here: the second equation would force $q=L=0$, and then every tangent component would vanish. ThereforeBoth equations, with the same constants $E,L$, are required. They give the requested radius relations in terms of the conserved quantities, while showing why an arbitrary exterior radius is not permitted.

For a more explicit classification let $b=L/E$ be the impact parameter. The second relation gives $b-a=s{r}^{3/2}/\sqrt{M}$ with $s=\pm 1$. Substituting into the first relation yieldsFor $0<a<M$, $s=+1$ gives the prograde equatorial circular Kerr photon orbit and $s=-1$ the retrograde one. The two allowed exterior radii areBoth are strictly greater than $r_{+}$ in the subextremal range. Squaring the signed radius equation gives $r(r-3M{)}^2=4M{a}^2$, but introduces an additional inner root and loses the sign distinguishing the two senses of rotation. At $a=0$ the two exterior radii coincide at the photon sphere $r=3M$, with $b=\pm 3\sqrt{3}M$.

For a direct counterexample to the all-radii claim take $a=0$ and $r_0=4M>r_{+}=2M$. Choose $E=1$ and $L=4\sqrt{2}M$ so that $f(r_0)=0$. The radial geodesic equation nevertheless givesIt is a turning point, not a circular orbit. Alternatively, constant-$r$ null directions exist throughout the exterior because the two angular-velocity roots of $g_{tt}+2\Omega g_{t\varphi }+{\Omega }^2{g}_{\varphi \varphi }=0$ are real; almost all of their integral curves require acceleration and fail the radial geodesic equation. This distinction explains how an all-radii assertion can arise.

Finally, $r=r_{+}$ itself has the null geodesic generators of its Killing horizon, proportional to ${\chi }_{+}={\partial }_v+{\Omega }_{+}{\partial }_{\psi }$. They are a separate horizon case. On the horizon ${\chi }_{+}$ is normal to every tangent, so its contractions with ${\partial }_v$ and ${\partial }_{\psi }$ both vanish: $E=L=0$ for a generator tangent. Its affine parameter can be chosen by rescaling that tangent. The exterior calculation dividing by $E$ does not include it. The presence of these generators does not supply circular null geodesics at the intervening arbitrary exterior radii.