Past exam of the mathematics course of the University of Cambridge / 2001 / iii / Paper 69 / 3

Take $M>0$, set $b=Q^2/M$, and denote the squared areal radius by $C(r)=r(r-b)$. The positive-area exterior has $r>b$. The stationary norm is $-f(r)$, where $f(r)=1-2M/r$, so its candidate Killing horizon is at $r=2M$. There it has areaA regular horizon must have $b<2M$. The curvature singularity then lies strictly inside the horizon, and ingoing coordinates extend smoothly across it because $C(2M)>0$. In the future interior $f<0$, the gradient of $r$ is timelike and future causal curves move to decreasing $r$, rather than back to infinity. Thus the regular black hole condition isThe equality case has zero horizon area and a curvature-singular null limiting surface. It is often included in the family as a singular extremal solution, but it is not a regular black hole. If that limiting nomenclature is used, the family condition is $Q^2\le 2M^2$ with this essential qualification. This geometry has the same spherical form as the magnetically charged dilaton black hole; no unprinted field equation is needed for the horizon comparison.

For $M>0$ and $Q^2>2M^2$, the physical exterior starts at $r=b>2M$. Throughout it, $f>0$, so the putative zero of $f$ is not in the exterior and no regular horizon shields the curvature singularity. The normal to $r=b$ has positive limiting squared norm $g^{rr}=f(b)>0$, identifying a timelike singular boundary. Outgoing radial null geodesics obey $dr/dt=f(r)>0$ and escape from arbitrarily close to that boundary. HenceAt $Q^2=2M^2$, the singular boundary coincides with $f=0$ and is null. There is no regular shielding horizon of positive area; this marginal singular extremal case must be separated from both the regular black hole and the supercritical timelike naked singularity. If “naked” is used simply to mean unshielded by a regular horizon, equality belongs on that side of the distinction, but its causal boundary is different.

The usual positive-mass restriction matters. For $M<0$, $f(r)>0$ at every positive radius and there is no positive-radius horizon; the central zero-area boundary at $r=0$ is singular and naked. For $Q=0$ its angular sectional curvature is $2M/r^3$, which diverges there. For $Q\ne 0$, $C=r(r-b)$ with $b<0$, and the angular sectional curvature $[1-f(C^{\prime }{)}^2/(4C)]/C$ also diverges as $r\downarrow 0$. The displayed family is undefined at $M=0$ with nonzero $Q$; its uncharged zero-mass limit is flat.

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 $E=f\dot{t}$, $L=C\dot{\varphi }$, and null normalization givesHere $B$ 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 $Q=0$, $H=(r-2M)/r^3$ andThe maximum is at the unstable photon sphere $r=3M$, with $H=1/(27M^2)$. Therefore

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 isThe 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.