= Solution
Let $f(r)=1-2m/r$ and work in the exterior $r>2m$, with $G=c=1$. The affinely parametrized <geodesic Lagrangian> is
$$
L=\tfrac12\left[f\dot t^2-f^{-1}\dot r^2-r^2\dot\theta^2-r^2\sin^2\theta\,\dot\phi^2\right].
$$
Its polar-angle equation is
$$
\ddot\theta+2\frac{\dot r}{r}\dot\theta-\sin\theta\cos\theta\,\dot\phi^2=0.
$$
The initial data $\theta=\pi/2$, $\dot\theta=0$ therefore remain in that plane. <Rotational symmetry> allows the axes to be chosen so that any nonradial initial trajectory has those data: its conserved angular-<momentum> direction fixes an orbital plane through the centre. Radial trajectories can also be placed in an equatorial plane.
The coordinates $t,\phi$ are cyclic in $L$, so their canonical momenta are constant:
$$
p_t=f\dot t=E,\qquad p_\phi=-r^2\sin^2\theta\,\dot\phi=-J.
$$
The corresponding <Killing vector fields> are $\partial_t$ and $\partial_\phi$. With the displayed signature, $g(\partial_t,\dot x)=E$ and $g(\partial_\phi,\dot x)=-J$, explaining the sign in the second <canonical momentum>. Thus $E$ is conserved <energy> per unit <rest mass> for a massive particle parametrized by <proper time>, and $J$ is its axial <angular momentum> per unit <rest mass>. For a <photon>, affine rescaling rescales both constants; after the tangent is normalized to the physical <four-momentum> they represent its <energy> at infinity and axial <angular momentum>. Their ratio $b=J/E$ is the signed <impact parameter>.
In the equatorial plane, the <photon> satisfies the <null geodesic> constraint
$$
0=f\dot t^2-f^{-1}\dot r^2-r^2\dot\phi^2,
$$
which gives
$$
\dot r^2=E^2-\frac{J^2}{r^2}\left(1-\frac{2m}{r}\right).
$$
For $J\ne0$, put $u=m/r$ and $\alpha=Em/J$. Then
$$
\frac{du}{d\phi}=-\frac mJ\dot r,
$$
and consequently
$$
\boxed{(u')^2=\alpha^2+2u^3-u^2.}
$$
The $J=0$ radial case is not represented by $\phi$ as an orbit parameter.
There is a circular <photon> orbit at $r=3m$, but \b[it is unstable]. The radial <effective potential> is
$$
V(r)=\frac{J^2}{r^2}\left(1-\frac{2m}{r}\right),\qquad
V'(r)=\frac{2J^2(3m-r)}{r^4}.
$$
A circular orbit requires $V'(r)=0$ and $E^2=V(r)$, giving
$$
\boxed{r_c=3m,\qquad \alpha^2=\frac1{27},\qquad |b_c|=3\sqrt3\,m.}
$$
Its second derivative is
$$
V''(3m)=-\frac{2J^2}{81m^4}<0.
$$
It is a maximum, rather than a minimum, so radial perturbations depart from the orbit. Equivalently the second-order orbit equation is $u''=3u^2-u$. Linearizing about $u_c=1/3$ gives $\delta u''=\delta u$, with modes $e^{\phi}$ and $e^{-\phi}$. The existence of a growing mode directly contradicts the printed stability claim. Exact initial data can remain circular, but that does not make the orbit stable.
The supplied implicit expression describes the critical-<energy> family with $\alpha^2=1/27$, as fixed by the circular orbit. Define
$$
F(u)=\frac{1-3u}{(\sqrt3+\sqrt{1+6u})^2},\qquad s=\sqrt{1+6u}.
$$
Rationalizing once gives $F=(\sqrt3-s)/(2(\sqrt3+s))$ and
$$
F'(u)=-\frac{3\sqrt3}{s(\sqrt3+s)^2}.
$$
Differentiating $F(u)=Ae^\phi$ gives $F'u'=F$, so
$$
u'=-\frac{s(1-3u)}{3\sqrt3},\qquad
(u')^2=\frac{(1+6u)(1-3u)^2}{27}
=\frac1{27}+2u^3-u^2.
$$
This verifies the first-order equation for every allowed integration constant $A$ on its real branch. For $A=0$ the derivative argument is replaced by direct substitution: $u=1/3$ is constant and solves the critical equation. The displayed family does not solve the equation at an independently arbitrary value of $\alpha$.
For $A>0$, write $q=Ae^\phi$. On the physical exterior branch,
$$
u=\frac{1-8q+4q^2}{3(1+2q)^2},\qquad
0<q<1-\frac{\sqrt3}{2}.
$$
As $\phi\to-\infty$, $q\to0^+$ and
$$
u=\frac13-4Ae^\phi+O(e^{2\phi}),\qquad
r=3m+36mAe^\phi+O(e^{2\phi}).
$$
Thus for $A=0$ the <photon> remains exactly on the circular orbit. For $A>0$ it approaches $r=3m$ from outside through indefinitely many revolutions in the negative-$\phi$ limit. With increasing $\phi$, this same branch peels away outward and reaches infinity when $q=1-\sqrt3/2$. Reversing the direction of motion gives an incoming critical ray asymptotically approaching the <photon> circle. These are <separatrix> trajectories, not evidence of stability.
\Image[/past-exam-of-the-general-relativity-course-of-the-university-of-oxford/2011-gr-photon-orbit.png]
{title=Critical Schwarzschild photon orbit and an outward separatrix}
The potential maximum and the outward-growing radial separation display the same instability in two complementary descriptions.
Back to article page
