= Solution
Put $f(r)=1-2\mu/r$, where $\mu=GM/c^2$, and work in the orbital plane. A dot denotes an affine-parameter derivative; for a material particle choose that parameter to be <proper time>.
\b[First integrals and normalization.] The <geodesic Lagrangian> is
$$
L=\frac12\left(fc^2\dot t^2-f^{-1}\dot r^2-r^2\dot\phi^2\right).
$$
Time translation and axial rotation make $t$ and $\phi$ cyclic, giving
$$
\boxed{f\dot t=k,\qquad r^2\dot\phi=h.}
$$
The tangent norm is constant along an affine <geodesic> because the <Levi-Civita connection> preserves the metric. Therefore
$$
fc^2\dot t^2-f^{-1}\dot r^2-r^2\dot\phi^2=\mathcal E.
$$
For proper-time parametrization of a timelike <worldline> this norm is $c^2$. For a <null geodesic> it is zero, for every affine normalization. Thus \b[$\mathcal E=c^2$ for the chosen material-particle normalization and $\mathcal E=0$ for light]. <Proper time> cannot itself parametrize a null ray, despite the same printed symbol $\tau$ being used for its <affine parameter>.
\b[Null normalization and <impact parameter>.] Under $\tau\mapsto A\tau+B$ with $A>0$, both $k$ and $h$ scale as $1/A$, while the null path is unchanged. Their ratio is invariant. At large radius the spatial speed with respect to asymptotic time is $c$, and orbital <angular momentum> per affine tangent gives
$$
\boxed{b=\frac{|h|}{kc};\qquad k=1\ \Longrightarrow\ |h|=bc.}
$$
For an orientation chosen with $h>0$, this is $h=bc$. Setting $k=1$ is a choice of affine scale, not a claim that all <photons> have equal measured <energy>. Once a tangent is identified with a physically normalized <four-momentum>, its <energy> fixes that scale.
\b[Orbit shape.] Substituting the constants in the norm and multiplying by $f$ gives
$$
\dot r^2=k^2c^2-f\left(\mathcal E+\frac{h^2}{r^2}\right).
$$
For $u=1/r$ and $h\ne0$, $du/d\phi=-\dot r/h$. Hence
$$
\boxed{(u')^2+u^2=\frac{k^2c^2}{h^2}+2\mu u^3
-\frac{\mathcal E}{h^2}(1-2\mu u),\qquad u'=\frac{du}{d\phi}.}
$$
The $h=0$ radial orbit must be treated with the radial equation, because azimuth is then constant.
\b[Circular <photons> and stability.] For a null ray define $W(r)=f(r)/r^2$. Its radial equation becomes
$$
\dot r^2+h^2W(r)=k^2c^2,\qquad
W'(r)=-\frac2{r^3}+\frac{6\mu}{r^4}.
$$
The circular orbit requires $\dot r=0$ and $W'=0$, so
$$
\boxed{r_{\rm ph}=3\mu,\qquad W''(3\mu)=-\frac2{81\mu^4}<0.}
$$
It is at a maximum of the null <effective potential> and is unstable. Equivalently the <Binet equation> $u''+u=3\mu u^2$ linearizes about $u_0=1/(3\mu)$ to $\delta u''=\delta u$, admitting exponentially growing radial deviations in azimuth.
\b[Capture from infinity.] Divide the null radial equation by $k^2c^2$. The quantity opposing inward motion is $b^2W(r)$, whose exterior maximum is $b^2/(27\mu^2)$. There is no outer radial <turning point> precisely when
$$
\boxed{b<3\sqrt3\,\mu.}
$$
At equality an exactly critical ray from infinity asymptotes to the <photon sphere>; it does not cross the horizon at a finite <affine parameter>. Above the threshold an initially inward ray from infinity reaches an outer <turning point> and scatters. The radial case $b=0$ is also captured. These statements refer to inward rays launched from the asymptotically flat exterior, not arbitrary initial data inside the angular-momentum barrier.
\Image[/past-exam-of-the-relativity-course-of-the-university-of-cambridge/2015/paper-1-photon-capture.png]
{title=Original Schwarzschild capture calculation: the null-potential barrier and a captured equatorial ray crossing the horizon and ending at finite azimuth}
\b[Approach to the singularity.] A captured nonradial ray bends inward, crosses $r=3\mu$ and then the <event horizon> $r=2\mu$, and continues toward the <curvature singularity>. A plot in $(r\cos\phi,r\sin\phi)$ is only a coordinate projection, not a Euclidean spatial slice inside the horizon. For large $u$ the cubic term in the shape equation dominates:
$$
\frac{du}{d\phi}\simeq\sqrt{2\mu}\,u^{3/2}
$$
for the inward branch with $h>0$. Its integral is
$$
\phi_*-\phi\simeq\sqrt{\frac2\mu}\,u^{-1/2},\qquad
\boxed{r\simeq\frac\mu2(\phi_*-\phi)^2.}
$$
Thus the azimuth tends to a finite limit at the singularity, rather than winding infinitely often there. Near-critical rays can accumulate many turns near $3\mu$ before capture, which is a separate phenomenon. A strictly radial ray has constant azimuth and does not admit this azimuthal asymptotic form. The radial equation also gives $r\propto(\tau_*-\tau)^{2/5}$ for a nonradial null ray, confirming finite remaining <affine parameter>.
\b[Material capture and the condition at infinity.] To recover the quoted threshold, take the usual marginally bound orbit, initially at rest at infinity: $k=1$. With $\mathcal E=c^2$,
$$
\dot r^2=\frac{2\mu c^2}{r}-\frac{h^2}{r^2}+\frac{2\mu h^2}{r^3}
=\frac{2\mu c^2r^2-h^2r+2\mu h^2}{r^3}.
$$
The turning-point quadratic has <discriminant> $h^2(h^2-16\mu^2c^2)$. It is positive for large $r$ and has no real positive root when $0<|h|<4\mu c$, while the radial case also falls in. Therefore
$$
\boxed{|h|<4\mu c\quad\Longrightarrow\quad\text{capture from rest at infinity}.}
$$
At equality the quadratic is proportional to $(r-4\mu)^2$ and the orbit asymptotes to the unstable marginally bound circular orbit at $r=4\mu$. For larger $|h|$, an incoming marginally bound particle encounters the outer <turning point> and scatters.
The source says only “from infinity”, without explicitly fixing the asymptotic speed. For a general incoming speed $v_\infty$, $k=(1-v_\infty^2/c^2)^{-1/2}\ge1$ and
$$
\dot r^2=c^2(k^2-1)+\frac{2\mu c^2}{r}
-\frac{h^2}{r^2}+\frac{2\mu h^2}{r^3}.
$$
The same bound $|h|<4\mu c$ remains sufficient for capture, since the added constant is nonnegative. It is \b[the exact threshold only for $k=1$]. At fixed $k>1$, the critical unstable orbit satisfies
$$
k^2=\frac{(r_c-2\mu)^2}{r_c(r_c-3\mu)},\qquad
h_c^2=\frac{\mu c^2r_c^2}{r_c-3\mu},\qquad3\mu<r_c<4\mu.
$$
Then $h_c>4\mu c$. This states both the printed sufficient condition and the additional assumption needed to call it a sharp capture boundary.
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