= Solution
The absorption question concerns the geometric-optics capture cross-section, assuming rays reaching the central singular boundary are absorbed. <Null geodesic> motion alone does not specify a wave boundary condition at a singularity.
By spherical symmetry choose an equatorial ray. Conservation of energy and <angular momentum> gives $E=f\dot t$, $L=C\dot\phi$, and null normalization gives
$$
\dot r^2=E^2-\frac{fL^2}{C}
=E^2\{1-B^2H(r)\},\qquad
B=\frac LE,\qquad
H(r)=\frac{f}{C}=\frac{r-2M}{r^2(r-b)}.
$$
Here $B$ is the <impact parameter> measured at infinity. A returning ray has an exterior turning point; capture occurs when none exists. The critical impact parameter obeys $B_c^2=1/\sup H$, and the cross-section is the area $\pi B_c^2$ in the incident impact-parameter plane.
For $Q=0$, $H=(r-2M)/r^3$ and
$$
H'(r)=-\frac{2(r-3M)}{r^4}.
$$
The maximum is at the unstable <photon sphere> $r=3M$, with $H=1/(27M^2)$. Therefore
$$
\boxed{B_c=3\sqrt3\,M,\qquad \sigma_{\rm abs}=27\pi M^2\quad(Q=0).}
$$
For $Q^2=2M^2$, cancellation for $r>2M$ gives $H=1/r^2$. There is no regular exterior circular photon orbit: the supremum occurs at the singular limiting surface, with $\sup H=1/(4M^2)$. Rays with $B>2M$ turn at $r=B$; those with $B<2M$ reach the central boundary. Thus the <extremal dilaton metric photon capture> result is
$$
\boxed{B_c=2M,\qquad \sigma_{\rm abs}=4\pi M^2\quad(Q^2=2M^2).}
$$
The critical ray is a measure-zero boundary of the capture set. The finite capture cross-section is not the horizon area, which vanishes in this limit. Nor is it a claim about a low-frequency quantum absorption cross-section or an unspecified reflection law at the singularity.
Back to article page
