Finite-range central scattering in the model class: the scattering motion on all of time, the asymptotic velocities and impact parameter $b=J\mathrm{/}\sqrt{2\mu E}$, the advance $\mathrm{\Delta }$ with principal deflection $\arccos (\cos (\pi -2\mathrm{\Delta }))$, and the cross-section of a deflection branch

By [L1], the relative coordinate is a $C^2$ central-force trajectory for $(\mu ,u)$ with attached angular momentum $j=J{e}_3$ and energy $E$, and $\mathrm{\mid }r(t)\mathrm{\mid }\in (0,\mathrm{\infty })$. The potential and turning-point conditions are exactly (H1)-(H2), and the trajectory attains $r_{\ast }$ at $t_{\ast }$ while satisfying $\mathrm{\mid }r\mathrm{\mid }\ge r_{\ast }$.

Global branches and extension: the first-order Cartesian central-force field of [L6] has a unique maximal solution through $(r(t_{\ast }),r^{\mathrm{\prime }}(t_{\ast }))$ by [L7], and the given trajectory on $I$ is its restriction; write the maximal curve as $\bar{r}$ and $\rho =\mathrm{\mid }\bar{r}\mathrm{\mid }$. By [L2], ${\rho }^{\mathrm{\prime }}(t_{\ast })=0$ and ${\rho }^{\mathrm{\prime }\mathrm{\prime }}(t_{\ast })=-V^{\mathrm{\prime }}(r_{\ast })\mathrm{/}\mu >0$, so $\rho >r_{\ast }$ immediately on both sides of $t_{\ast }$. Whenever $\rho >r_{\ast }$, the energy identity and (H2) give ${\rho }^{\mathrm{\prime }}\mathrm{\ne }0$, hence ${\rho }^{\mathrm{\prime }}$ has constant sign on each connected interval where $\rho >r_{\ast }$. On the outgoing interval adjacent to $t_{\ast }$ that sign is positive. If the interval had a finite right endpoint $T$ inside the maximal domain, continuity would give $\rho (T)=r_{\ast }$; then the radial equation in [L2] gives ${\rho }^{\mathrm{\prime }\mathrm{\prime }}(T)>0$, so ${\rho }^{\mathrm{\prime }}<0$ immediately before $T$, contradicting the positive sign on that interval. Thus the outgoing interval reaches the maximal right endpoint, with $\rho >r_{\ast }$ and ${\rho }^{\mathrm{\prime }}>0$ throughout. The symmetric argument gives $\rho >r_{\ast }$ and ${\rho }^{\mathrm{\prime }}<0$ on the incoming interval. In particular $\rho \ge r_{\ast }$ throughout the maximal interval. Since $u$ is continuous on $[r_{\ast },R]$ when this interval is nonempty and vanishes for $r\ge R$, it is bounded below on $[r_{\ast },\mathrm{\infty })$; the energy identity therefore bounds $\mathrm{\mid }{\bar{r}}^{\mathrm{\prime }}\mathrm{\mid }$. On any finite endpoint, $\bar{r}$ is bounded by this speed bound and remains at distance at least $r_{\ast }>0$ from the origin. Its state consequently lies in a compact subset of $\{(r,v):r\mathrm{\ne }0\}$ near that endpoint, so [L7] continues it past every finite endpoint. The maximal interval is therefore all of $\mathbb{R}$, and uniqueness is from [L6]. The global branches and radial energy identity established here supply the hypotheses for the direct quadrature and angle calculation below. For the model statement, let $X$ be the centre of mass of the original motion, affine on $I$ with $X^{\mathrm{\prime }\mathrm{\prime }}=0$ by [L1] clause 2, and let $\bar{X}$ be its affine extension to $\mathbb{R}$; then by [L1] clause 4 the pair $(\bar{X},\bar{r})$ reconstructs a motion of the model with its autonomous force law extended to time window $\mathbb{R}$, because $\mathrm{\mid }\bar{r}(t)\mathrm{\mid }\in (0,\mathrm{\infty })$ keeps the configuration admissible. This is clause 1.

Time quadrature in clause 1: for $s>r_{\ast }$ the integrand in the displayed $\tau (s)$ is positive and continuous. The derivative definition gives $(E-V(s))\mathrm{/}(s-r_{\ast })\to -V^{\mathrm{\prime }}(r_{\ast })>0$ as $s\downarrow r_{\ast }$, so near $r_{\ast }$ the integrand is bounded by a constant times $(s-r_{\ast }{)}^{-1\mathrm{/}2}$, which has a finite integral on every bounded interval starting at $r_{\ast }$. Thus $\tau (s)<\mathrm{\infty }$ for every finite $s\ge r_{\ast }$. Since $V(s)=J^2\mathrm{/}(2\mu s^2)\to 0$ for $s\ge R$, the integrand tends to $1\mathrm{/}\sqrt{2E\mathrm{/}\mu }>0$ and $\tau (s)\to \mathrm{\infty }$; it is strictly increasing from $\tau (r_{\ast })=0$. On either monotone branch, the fundamental theorem gives ${\tau }^{\mathrm{\prime }}(s)=1\mathrm{/}\sqrt{(2\mathrm{/}\mu )(E-V(s))}$ for $s>r_{\ast }$. By the chain rule and the radial energy identity, $(\tau \circ \rho {)}^{\mathrm{\prime }}=1$ on the outgoing branch and $(\tau \circ \rho {)}^{\mathrm{\prime }}=-1$ on the incoming branch. Taking the limits at $t_{\ast }$ gives $\tau (\rho (t))=\mathrm{\mid }t-t_{\ast }\mathrm{\mid }$, and strict increase and unboundedness of $\tau$ yield $\rho (t_{\ast }\pm \tau (s))=s$ for every $s\ge r_{\ast }$. This proves the quadrature assertion and also $\rho (t)\to \mathrm{\infty }$ on both tails.

Free tails and asymptotic velocities: on the outgoing branch $\rho$ is increasing. It is unbounded: if it had a finite limit $L>r_{\ast }$, the energy identity and (H2) would imply ${\rho }^{\mathrm{\prime }}(t)\to \sqrt{(2\mathrm{/}\mu )(E-V(L))}>0$, contradicting convergence to $L$. Thus $\rho (t)\to \mathrm{\infty }$, and after it passes $R$ it remains there; since $u^{\mathrm{\prime }}(r)=0$ for $r\ge R$, the Cartesian equation gives ${\bar{r}}^{\mathrm{\prime }\mathrm{\prime }}=0$ on that tail. The incoming branch is symmetric. Now let $t_1\ge t_{\ast }$ with $\rho (t_1)>R$; the motion is free for all later times and its constant velocity is $v_{+}:={\lim }_{t\to +\mathrm{\infty }}{\bar{r}}^{\mathrm{\prime }}(t)$. At any such free point $u=0$, so energy gives $\mathrm{\mid }{v}_{+}\mathrm{\mid }=\sqrt{2E\mathrm{/}\mu }=:v_{\mathrm{\infty }}>0$. If instead $\rho (t_1)=R$, define $q_{+}(t):=\bar{r}(t_1)+(t-t_1){\bar{r}}^{\mathrm{\prime }}(t_1)$ for $t\ge t_1$. Differentiating ${\rho }^2=⟨\bar{r},\bar{r}⟩$ gives $⟨\bar{r},{\bar{r}}^{\mathrm{\prime }}⟩=\rho {\rho }^{\mathrm{\prime }}$; since $\rho$ is nondecreasing on this branch, $⟨\bar{r}(t_1),{\bar{r}}^{\mathrm{\prime }}(t_1)⟩=R{\rho }^{\mathrm{\prime }}(t_1)\ge 0$. Also $\mathrm{\mid }{\bar{r}}^{\mathrm{\prime }}(t_1){\mathrm{\mid }}^2=2E\mathrm{/}\mu >0$ because $u(R)=0$, so $\mathrm{\mid }{q}_{+}(t){\mathrm{\mid }}^2=R^2+2(t-t_1)⟨\bar{r}(t_1),{\bar{r}}^{\mathrm{\prime }}(t_1)⟩+(t-t_1{)}^2\mathrm{\mid }{\bar{r}}^{\mathrm{\prime }}(t_1){\mathrm{\mid }}^2\ge R^2$. Since $u^{\mathrm{\prime }}(r)=0$ for all $r\ge R$, including $R$, $q_{+}$ and $\bar{r}$ solve the same Cartesian equation with the same state at $t_1$; [L6] gives $\bar{r}=q_{+}$ for all $t\ge t_1$. Thus its constant velocity is again $v_{+}$ and has norm $v_{\mathrm{\infty }}$. The incoming equality case is identical with $q_{-}(t):=\bar{r}(t_2)+(t-t_2){\bar{r}}^{\mathrm{\prime }}(t_2)$ for $t\le t_2$: ${\rho }^{\mathrm{\prime }}(t_2)\le 0$ and $t-t_2\le 0$ give $\mathrm{\mid }{q}_{-}(t)\mathrm{\mid }\ge R$, so [L6] gives the incoming free formula and $v_{-}:={\lim }_{t\to -\mathrm{\infty }}{\bar{r}}^{\mathrm{\prime }}(t)$ with $\mathrm{\mid }{v}_{-}\mathrm{\mid }=v_{\mathrm{\infty }}$. This proves clause 2's free-motion and velocity claims.

Clause 3, polar advance: apply Polar reduction of a non-radial central-force trajectory: ${\rho }^2{\theta }^{\mathrm{\prime }}=J\mathrm{/}\mu$, the effective-potential equation and energy conservation to the global trajectory $\bar{r}$ of step 2.1, giving polar data $(\rho ,\theta )$ on all of $\mathbb{R}$. Let $n(\phi )=\cos \phi e_1+\sin \phi e_2$ be its unit radial direction. On the outgoing branch, ${\theta }^{\mathrm{\prime }}=J\mathrm{/}(\mu {\rho }^2)>0$ and ${\rho }^{\mathrm{\prime }}=\sqrt{(2\mathrm{/}\mu )(E-V(\rho ))}$; substitution on compact subintervals away from $t_{\ast }$ gives $\theta (+\mathrm{\infty })-\theta (t_{\ast })={\int }_{r_{\ast }}^{\mathrm{\infty }}\frac{J\mathrm{d}r}{r^2\sqrt{2\mu (E-V(r))}}=:\mathrm{\Delta }\mathrm{.}$ The integral converges at $r_{\ast }$ because $E-V(r)\sim -V^{\mathrm{\prime }}(r_{\ast })(r-r_{\ast })$ as $r\downarrow r_{\ast }$, and at infinity because $V(r)=J^2\mathrm{/}(2\mu r^2)$ for $r\ge R$; it is positive. The same substitution on the incoming branch, where ${\rho }^{\mathrm{\prime }}<0$ and ${\theta }^{\mathrm{\prime }}>0$, gives $\theta (t_{\ast })-\theta (-\mathrm{\infty })=\mathrm{\Delta }$. Thus ${\theta }_{\pm }:={\lim }_{t\to \pm \mathrm{\infty }}\theta (t)$ exist and ${\theta }_{+}-{\theta }_{-}=2\mathrm{\Delta }$. By the free tails of step 3.2, $\bar{r}(t)\mathrm{/}\rho (t)\to v_{+}\mathrm{/}\mathrm{\mid }{v}_{+}\mathrm{\mid }$ as $t\to +\mathrm{\infty }$ and $\bar{r}(t)\mathrm{/}\rho (t)\to -v_{-}\mathrm{/}\mathrm{\mid }{v}_{-}\mathrm{\mid }$ as $t\to -\mathrm{\infty }$. Since $\bar{r}\mathrm{/}\rho =n(\theta )$, continuity gives $n({\theta }_{+})=v_{+}\mathrm{/}\mathrm{\mid }{v}_{+}\mathrm{\mid }$ and $n({\theta }_{-})=-v_{-}\mathrm{/}\mathrm{\mid }{v}_{-}\mathrm{\mid }$. The addition formulas imply $n({\theta }_{+})=R_{2\mathrm{\Delta }}n({\theta }_{-})$, hence $\frac{v_{+}}{\mathrm{\mid }{v}_{+}\mathrm{\mid }}=-R_{2\mathrm{\Delta }}\left(\frac{v_{-}}{\mathrm{\mid }{v}_{-}\mathrm{\mid }}\right)\mathrm{.}$ For any unit $q$ in the plane, $⟨q,e_3\times q⟩=0$, so this identity gives $⟨v_{-},v_{+}⟩\mathrm{/}(\mathrm{\mid }{v}_{-}\mathrm{\mid }\mathrm{\mid }{v}_{+}\mathrm{\mid })=-\cos (2\mathrm{\Delta })=\cos (\pi -2\mathrm{\Delta })$. Therefore ${\theta }_{\mathrm{s}}=\arccos (\cos (\pi -2\mathrm{\Delta }))$ by clause (iii) of Scattering data of the central model class: the scattering motion, the impact parameter, the asymptotic directions, the deflection conventions and the uniform-incident preparation; when $0<\mathrm{\Delta }<\pi \mathrm{/}2$, $\pi -2\mathrm{\Delta }\in (0,\pi )$ is that angle. To identify the continuous signed rotation, differentiate $\bar{r}=\rho n(\theta )$: ${\bar{r}}^{\mathrm{\prime }}={\rho }^{\mathrm{\prime }}n(\theta )+J\mathrm{/}(\mu \rho )(e_3\times n(\theta ))$, whose positive tangential component ensures $\mathrm{\mid }{\bar{r}}^{\mathrm{\prime }}\mathrm{\mid }>0$. Put $\psi (t)=\arccos ({\rho }^{\mathrm{\prime }}(t)\mathrm{/}\mathrm{\mid }{\bar{r}}^{\mathrm{\prime }}(t)\mathrm{\mid })\in (0,\pi )$ using [L9]. The Pythagorean identity and positivity give ${\bar{r}}^{\mathrm{\prime }}\mathrm{/}\mathrm{\mid }{\bar{r}}^{\mathrm{\prime }}\mathrm{\mid }=n(\theta +\psi )$, so $\theta +\psi$ is a continuous real angle for the velocity direction. On the free tails, $\mathrm{\mid }{\bar{r}}^{\mathrm{\prime }}\mathrm{\mid }=v_{\mathrm{\infty }}$ and the radial energy equation gives ${\rho }^{\mathrm{\prime }}\to -v_{\mathrm{\infty }}$ incoming and ${\rho }^{\mathrm{\prime }}\to v_{\mathrm{\infty }}$ outgoing. Thus $\psi \to \pi$ incoming and $\psi \to 0$ outgoing; the net continuous rotation is ${\theta }_{+}-({\theta }_{-}+\pi )=2\mathrm{\Delta }-\pi$. This is clause 3.

Clause 2, impact parameter: on the incoming branch choose $t_2\le t_{\ast }$ with $\rho (t_2)\ge R$; the affine extension of the incoming free tail is $\{\bar{r}(t_2)+s{v}_{-}:s\in \mathbb{R}\}$ and its perpendicular distance from the origin is $\mathrm{\mid }\bar{r}(t_2)\times v_{-}\mathrm{\mid }\mathrm{/}\mathrm{\mid }{v}_{-}\mathrm{\mid }=\mathrm{\mid }j\mathrm{\mid }\mathrm{/}(\mu \mathrm{\mid }{v}_{-}\mathrm{\mid })=J\mathrm{/}(\mu v_{\mathrm{\infty }})$ by step 3.2 and [L4]. For the outgoing line $\{\bar{r}(t_1)+s{v}_{+}:s\in \mathbb{R}\}$, the same computation with [L4] gives distance $\mathrm{\mid }\bar{r}(t_1)\times v_{+}\mathrm{\mid }\mathrm{/}\mathrm{\mid }{v}_{+}\mathrm{\mid }=\mathrm{\mid }j\mathrm{\mid }\mathrm{/}(\mu \mathrm{\mid }{v}_{+}\mathrm{\mid })=J\mathrm{/}(\mu v_{\mathrm{\infty }})$; since $J\mathrm{/}(\mu \sqrt{2E\mathrm{/}\mu })=J\mathrm{/}\sqrt{2\mu E}$, both distances equal $b=J\mathrm{/}\sqrt{2\mu E}$ as displayed, and multiplying by $v_{\mathrm{\infty }}$ gives $J=\mu b{v}_{\mathrm{\infty }}$.

Clause 4: the assumed family uses the same $\mu ,u,E$ and varying $J(b)=\sqrt{2\mu E}b$, with (H2) and the trajectory conditions for each member. By the branch hypothesis its scattering angle is $\theta (b)$; [L3] gives the measure and test-function identity on the entire open branch with the stated nonnegative or absolute-integrability condition. Under the preparation the probability law is $P=\nu {\mathrm{\mid }}_{D_{\max }}\mathrm{/}(\pi b_{\max }^{})$ and the total arrival rate is $\mathrm{\Phi }\pi b_{\max }^{}$. Hence the branch portion $(a_1,c)\times (0,2\pi )$ has expected rate $\mathrm{\Phi }\nu ((a_1,c)\times (0,2\pi ))$. Its pushforward is supported in $Y=\theta ((a_1,c))$, so [L3] identifies that area with this branch's contribution ${\sigma }_{\mathrm{b}\mathrm{r}}(Y)=2\pi {\int }_{a_1}^{c}bdb=\pi (c^2-a_1^2)$. Other branches with overlapping images contribute separately by finite addition as defined in The differential and truncated total cross-sections of a deflection branch: densities, units, branch sums and the empirical qualifications of the uniform-incident reading clause (ii). This proves the rate regardless of whether the branch increases or decreases.

Clause 5 is the statement of the scope: the branch hypothesis of clause 4 is an explicit assumption, not a consequence of (H1)-(H2); no monotonicity or nonvanishing of ${\theta }^{\mathrm{\prime }}$ off the branch is used or claimed; and the exclusions are those of Adopted central-force model class: static central pair potentials, the reduced one-label problem, the fixed-centre idealization, reading conventions and exclusions clause (v) together with (H1). This completes the proof.