Exact repulsive relativistic Coulomb scattering branch

Along any solution of [F1], $\dot{H}={\mathrm{\nabla }}_{x}H\cdot \dot{x}+{\mathrm{\nabla }}_{p}H\cdot \dot{p}={\mathrm{\nabla }}_{x}H\cdot {\mathrm{\nabla }}_{p}H-{\mathrm{\nabla }}_{p}H\cdot {\mathrm{\nabla }}_{x}H=0,$ and $\dot{J}=\dot{x}\times p+x\times \dot{p}=0$ because $\dot{x}$ is parallel to $p$ and $\dot{p}$ is parallel to $x$. Thus $W$ and $J$ are constant. With $u=1\mathrm{/}r$, the energy relation gives $\mathcal{E}=W-\kappa u$, so $c^2\mathrm{\mid }p{\mathrm{\mid }}^2=(W-\kappa u{)}^2-m^2{c}^4,$ and $\mathrm{\mid }p{\mathrm{\mid }}^2=p_r^2+J^2\mathrm{/}{r}^2$. Rearranging yields $p_r^2=A-2du-C{u}^2=Q(u)$.

Since $Q(0)=A>0$ and $Q((W-m{c}^2)\mathrm{/}\kappa )=-J^2((W-m{c}^2)\mathrm{/}\kappa {)}^2<0$, continuity gives a positive root below $(W-m{c}^2)\mathrm{/}\kappa$. The other root is negative because the product of the roots is $-A\mathrm{/}C<0$, so the positive root is unique. The quadratic formula gives the displayed $u_{\ast }$; $Q^{\mathrm{\prime }}(u_{\ast })=-2(d+C{u}_{\ast })<0$, hence the root is simple. At this root $\mathcal{E}=W-\kappa u_{\ast }>m{c}^2$, so the momentum is real and the point lies in $D$.

At any turning state choose orthonormal axes $e_1=x\mathrm{/}{r}_{\ast }$ and $e_2=p\mathrm{/}\mathrm{\mid }p\mathrm{\mid }$, and translate its time to $0$; then $x=r_{\ast }{e}_1$ and $p=(J\mathrm{/}{r}_{\ast })e_2$. The root equation in 2.1 gives $H(x,p)=W$, and the smooth field in [F1] satisfies [F2], giving a local solution. By [F2], all solutions through this full initial state agree on their overlapping intervals; their union is a solution on the union interval, is locally a solution near every interior point, and is the unique maximal extension. Its interval is open, since local existence at an included finite endpoint would extend it. Every trajectory through the state is a restriction of this extension by uniqueness. Along it, direct differentiation gives ${\dot{p}}_r=c^2{J}^2\mathrm{/}(\mathcal{E}{r}^3)+\kappa \mathrm{/}{r}^2>0$ everywhere, so [F5] makes $p_r$ strictly increasing with its unique zero at $0$; consequently $r$ decreases before the turn and increases after it. On every allowed state $0<u\le u_{\ast }<(W-m{c}^2)\mathrm{/}\kappa$, hence $r\ge r_{\ast }$ and $m{c}^2<\mathcal{E}=W-\kappa u<W$, which gives $\mathrm{\mid }p\mathrm{\mid }<\sqrt{A}$. Also $\mathrm{\mid }\dot{x}\mathrm{\mid }<c$, so near any finite interval endpoint $x$ remains bounded and $\mathrm{\mid }x\mathrm{\mid }\ge r_{\ast }$; together with the momentum bound this gives compact containment in $D$. Applying [F2] at a finite right endpoint, or to the reversed-time field at a finite left endpoint, would properly extend the maximal solution, a contradiction. Its interval is therefore $\mathbb{R}$. Each monotone radial tail must be unbounded: if it had a finite limit greater than $r_{\ast }$, then $Q(1\mathrm{/}r)$ and $\mathcal{E}$ would give $\mathrm{\mid }\dot{r}\mathrm{\mid }$ a positive lower bound, contradicting that finite limit. Thus $r\to \mathrm{\infty }$ on both tails and $\dot{r}=c^2{p}_r\mathrm{/}\mathcal{E}\to \pm c^2\sqrt{A}\mathrm{/}W$, so $r$ grows linearly there. Then $\mathrm{\mid }\dot{p}\mathrm{\mid }=\kappa \mathrm{/}{r}^2$ is integrable, and $p$ converges to incoming and outgoing limits of magnitude $\sqrt{A}$. Since the tangential momentum has magnitude $J\mathrm{/}r\to 0$, the limiting momenta point along the limiting radial directions, giving the two asymptotic velocity directions.

Orient the orbital plane so its polar angle increases with $J=\mathrm{\mid }x\times p\mathrm{\mid }$. Then $\dot{\phi }=c^{2}J\mathrm{/}(\mathcal{E}{r}^2)$ and $\dot{r}=c^2{p}_r\mathrm{/}\mathcal{E}$, so $\mathrm{\mid }d\phi \mathrm{/}du\mathrm{\mid }=J\mathrm{/}\sqrt{Q(u)}$ and the half-angle is $\mathrm{\Delta }={\int }_0^{u_{\ast }}Jdu\mathrm{/}\sqrt{A-2du-C{u}^2}$. Completing the square gives $Q=(d^2+CA)\mathrm{/}C-C(u+d\mathrm{/}C{)}^2$. With $y=(Cu+d)\mathrm{/}\sqrt{d^2+CA}$, the root relation in 2.1 makes the endpoints $d\mathrm{/}\sqrt{d^2+CA}$ and $1$, and the integral becomes $(J\mathrm{/}\sqrt{C})\int dy\mathrm{/}\sqrt{1-y^2}=(J\mathrm{/}\sqrt{C})[\arcsin y{]}_{d\mathrm{/}\sqrt{d^2+CA}}^1$. Put $\alpha =\arctan (\sqrt{CA}\mathrm{/}d)\in (0,\pi \mathrm{/}2)$. Since $\tan \alpha =\sqrt{CA}\mathrm{/}d$, the Pythagorean identity and positivity of sine and cosine in [F3] give $\cos \alpha =d\mathrm{/}\sqrt{d^2+CA}$; the quarter-turn identity then gives $\arcsin (d\mathrm{/}\sqrt{d^2+CA})=\pi \mathrm{/}2-\alpha$ because both angles lie in $[0,\pi \mathrm{/}2]$. The evaluated integral is therefore $\mathrm{\Delta }=(J\mathrm{/}\sqrt{C})\arctan (\sqrt{CA}\mathrm{/}d)$.

Put $\alpha =\kappa \mathrm{/}(cJ)\in (0,1)$, $q=\sqrt{1-{\alpha }^2}\in (0,1)$, and $\beta =\sqrt{1-m^2{c}^4\mathrm{/}{W}^2}\in (0,1)$. If $a=\arcsin q$, then [F3] gives $a\in (0,\pi \mathrm{/}2)$, $\sin a=q$, and $\cos a=\sqrt{1-q^2}>0$; hence $\tan a=q\mathrm{/}\sqrt{1-q^2}$ and the principal inverse identity gives $a=\arctan (q\mathrm{/}\sqrt{1-q^2})$. The formula from 3.2 becomes $\mathrm{\Delta }=q^{-1}\arctan (\beta q\mathrm{/}\sqrt{1-q^2})$; since $0<\beta <1$ and $\arctan$ is strictly increasing, $0<\mathrm{\Delta }<\arcsin (q)\mathrm{/}q$.

Define $g(x)=\sin x-2x\mathrm{/}\pi$ on $[0,\pi \mathrm{/}2]$. By [F3], $g(0)=g(\pi \mathrm{/}2)=0$ and $g^{\mathrm{\prime }\mathrm{\prime }}(x)=-\sin x\le 0$. Since $g^{\mathrm{\prime }}$ is differentiable and $g^{\mathrm{\prime }\mathrm{\prime }}\le 0$, [F5] makes $g^{\mathrm{\prime }}$ nonincreasing. For any $0<x<\pi \mathrm{/}2$, [F5] applied to $g$ on $[0,x]$ and $[x,\pi \mathrm{/}2]$ gives $g(x)\mathrm{/}x=g^{\mathrm{\prime }}({\xi }_1)$ and $-g(x)\mathrm{/}(\pi \mathrm{/}2-x)=g^{\mathrm{\prime }}({\xi }_2)$ for some $0<{\xi }_1<x<{\xi }_2<\pi \mathrm{/}2$. Thus $g(x)\mathrm{/}x\ge -g(x)\mathrm{/}(\pi \mathrm{/}2-x)$; multiplying by the positive $x(\pi \mathrm{/}2-x)$ yields $g(x)\ge 0$. Taking $x=\arcsin q$ gives $\arcsin q\le \pi q\mathrm{/}2$, hence step 4.1 gives $\mathrm{\Delta }<\pi \mathrm{/}2$.

On either asymptotic tail, $\mathrm{\mid }d\phi \mathrm{/}dr\mathrm{\mid }=J\mathrm{/}(r^2\sqrt{Q(1\mathrm{/}r)})\sim J\mathrm{/}(\sqrt{A}{r}^2)$. For every $0<ϵ<1$, the definition of this limit bounds the integrand between $(1-ϵ)J\mathrm{/}(\sqrt{A}{r}^2)$ and $(1+ϵ)J\mathrm{/}(\sqrt{A}{r}^2)$ for all sufficiently large radii; integration from $r$ to infinity gives the angular gap $\delta (r)\sim J\mathrm{/}(\sqrt{A}r)$. Step 3.1's asymptotic momentum and [F6] now give $r\sin \delta (r)\to J\mathrm{/}\sqrt{A}$, the limiting perpendicular distance of each asymptotic line from the force center; this is $b$. The incoming momentum points opposite its limiting radial direction and the outgoing momentum along its own, so their angular separation is $\theta =\pi -2\mathrm{\Delta }\in (0,\pi )$ by step 5.1.

The unique noncollision turning branch, impact parameter and exact deflection angle follow from steps 1.1–6.1 under the stated hypotheses $W>m{c}^2$ and $J>\kappa \mathrm{/}c$. This proves the selected $C>0$ branch, without asserting that $J>\kappa \mathrm{/}c$ is necessary for a positive turning root.