Schwarzschild null scattering and critical impact, with conditional Kruskal capture

Choose $R>r_{\mathrm{p}\mathrm{h}}$ so $b^{2}H(R)<1$. At $r=R$, set $\vartheta =\pi \mathrm{/}2$, ${\dot{x}}^0=E\mathrm{/}f(R)$, $\dot{r}=-E\sqrt{1-b^{2}H(R)}$, $\dot{\vartheta }=0$, and $\dot{\phi }=L\mathrm{/}{R}^2$, with $L=bE$. Substitution into the diagonal Schwarzschild metric gives $g(\dot{\eta },\dot{\eta })=-\frac{E^2}{f(R)}+\frac{E^2}{f(R)}(1-b^{2}H(R))+\frac{L^2}{R^2}=0.$ Also $g(\dot{\eta },{\mathrm{\partial }}_0)=-E<0$, so the initial tangent is future directed, and [F2] gives a unique local geodesic. It remains equatorial: define the sphere reflection $S(y_1,y_2,y_3)=(y_1,y_2,-y_3)$ and $\mathrm{\Psi }(x^0,r,\omega )=(x^0,r,S\omega )$. The reflection is orthogonal, so it preserves the induced round metric; [F4] therefore makes $\mathrm{\Psi }$ an isometry of the exterior metric. Pulling back the metric-selected connection by $\mathrm{\Psi }$ gives another torsion-free, metric-compatible connection, hence the same connection by uniqueness in [F4]; thus $\mathrm{\Psi }\circ \eta$ is an affinely parametrized geodesic. At the initial event $y_3=0$, and $\dot{\vartheta }=0$ makes the initial tangent lie in the fixed tangent plane of $S$, so $\mathrm{\Psi }$ fixes both initial data. Geodesic uniqueness [F2] gives $\mathrm{\Psi }\circ \eta =\eta$ on the connected maximal interval, and its image lies in the fixed set $y_3=0$, the equatorial plane. Finally [F5] preserves the initial null norm along the geodesic, so it is null throughout.

By [F1], along the equatorial null geodesic ${\dot{r}}^2=E^2(1-b^{2}H(r))$. For $r\ge R$, $H$ is decreasing and $b^{2}H(r)<1$, so the past-directed parameter branch has $\dot{r}<0$ and extends through every finite radial interval $[R,R^{\mathrm{\prime }}]$: the first integral bounds $\dot{r}$, while $f(r)\ge f(R)>0$ bounds ${\dot{x}}^0=E\mathrm{/}f(r)$ and $\dot{\phi }=L\mathrm{/}{r}^2$; compact continuation [F2] excludes a finite-parameter endpoint there. As $r\to \mathrm{\infty }$, $H(r)\to 0$ and $\dot{r}\to -E$, so the affine distance to infinity is ${\int }_R^{\mathrm{\infty }}dr\mathrm{/}[E\sqrt{1-b^{2}H(r)}]=\mathrm{\infty }$. Hence the constructed solution supplies an incoming ray from infinity for every $b\ge 0$.

By [F1], $H$ increases on $(s,r_{\mathrm{p}\mathrm{h}})$, decreases on $(r_{\mathrm{p}\mathrm{h}},\mathrm{\infty })$, and vanishes at both exterior ends. If $b>b_{\mathrm{c}}$, then $b^{2}H(r_{\mathrm{p}\mathrm{h}})>1$, so there is one root $r_{-}\in (s,r_{\mathrm{p}\mathrm{h}})$ and one simple outer root $r_{\ast }>r_{\mathrm{p}\mathrm{h}}$. The affine time from any finite large radius to $r_{\ast }$ is finite because near the simple root $[E\sqrt{1-b^{2}H(r)}{]}^{-1}=O((r-r_{\ast }{)}^{-1\mathrm{/}2})$.

If $b=b_{\mathrm{c}}$, direct factorization gives $1-b_{\mathrm{c}}^{2}H(r)=(2r-3s{)}^2(r+3s)\mathrm{/}(4r^3)$. The incoming radial speed vanishes linearly as $r\downarrow r_{\mathrm{p}\mathrm{h}}$; hence the affine integral diverges like $\int dr\mathrm{/}(r-r_{\mathrm{p}\mathrm{h}})$. The angular integrand has the same simple-pole behavior, so the ray approaches the photon sphere only at infinite affine parameter and accumulates logarithmically divergent azimuth.

If $0\le b<b_{\mathrm{c}}$, then $b^{2}H(r)<1$ for every $r>s$ by [F1]. The inward branch has no turning point and reaches the exterior boundary $r=s$ in finite affine parameter because $1-b^{2}H(r)\to 1$ there. The exterior calculation ends at that boundary. If a future continuation is placed in the Kruskal model and enters $U\ge 0$, [F3] prevents it from returning to the right exterior.

For $b>b_{\mathrm{c}}$, [F2] extends the incoming ray through the outer turn. Away from $r_{\ast }$, differentiating the radial first integral and dividing by $\dot{r}\mathrm{\ne }0$ gives $\ddot{r}=-E^2{b}^2{H}^{\mathrm{\prime }}(r)\mathrm{/}2$; smoothness of the geodesic makes $\ddot{r}$ continuous at the simple root, so this identity holds there as a limit. Since $H^{\mathrm{\prime }}(r_{\ast })<0$, the acceleration is positive and the ray leaves the turn with $\dot{r}>0$. On every finite segment with $r\ge r_{\ast }$, the conserved quantities bound $\mathrm{\mid }\dot{r}\mathrm{\mid }$, $\mathrm{\mid }{\dot{x}}^0\mathrm{\mid }=E\mathrm{/}f(r)$ and $\mathrm{\mid }\dot{\phi }\mathrm{\mid }=L\mathrm{/}{r}^2$; the state stays in a compact subset of the smooth exterior tangent domain and cannot end at a finite affine parameter by [F2]. On the outgoing branch $\dot{r}\to E>0$, while $\mathrm{\mid }\dot{r}\mathrm{\mid }\le E$, so it reaches arbitrarily large radii only at infinite affine parameter. Also $\mathrm{\mid }d\phi \mathrm{/}dr\mathrm{\mid }=b\mathrm{/}[r^2\sqrt{1-b^{2}H(r)}]$ is $O((r-r_{\ast }{)}^{-1\mathrm{/}2})$ at the turn and $O(r^{-2})$ at infinity; integrating both halves gives the convergent formula in clause 1.

The simple outer root, critical double root and no-turn regimes in 3.1–3.3 give all three clauses. Only the last clause uses the separate Kruskal continuation, and no observation or apparatus model is used.