[proofplan] We write the scattering orbit in polar coordinates on its plane of motion and use conservation of angular momentum and energy to express $\dot\theta$ and $|\dot r|$ in terms of $r$. Dividing the angular equation by the radial equation gives the differential relation between angle and radius on each monotone half-orbit. Integrating this relation from the turning point $r_{\min}$ to infinity gives the half-orbit angle $\Phi(E,\ell)$. Finally, the incoming and outgoing asymptotic velocity directions are compared to the radial directions at infinity, producing the signed deflection formula $\Theta(E,\ell)=\pi-2\Phi(E,\ell)$. [/proofplan]

[step:Write the orbit in polar coordinates and identify the conserved quantities] Let $I\subset\mathbb{R}$ be the time interval of the scattering orbit. Since the orbit is non-colliding, define \begin{align*} r:I\to(0,\infty), \qquad r(t)=|q(t)|. \end{align*} Let $I_0\subset I$ be any open subinterval on which a continuous angular coordinate can be chosen. Define the angular coordinate map \begin{align*} \theta:I_0\to\mathbb{R} \end{align*} by \begin{align*} q(t)=r(t)(\cos\theta(t),\sin\theta(t)). \end{align*} The velocity is \begin{align*} \dot q(t)=\dot r(t)e_r(t)+r(t)\dot\theta(t)e_\theta(t), \end{align*} where $e_r(t)=(\cos\theta(t),\sin\theta(t))$ and $e_\theta(t)=(-\sin\theta(t),\cos\theta(t))$. The scalar angular momentum is \begin{align*} L=m r(t)^2\dot\theta(t). \end{align*} After reversing the orientation of $\theta$ if necessary, we may assume $L=\ell>0$. Hence \begin{align*} \dot\theta(t)=\frac{\ell}{m r(t)^2}. \end{align*} The conserved energy identity is \begin{align*} E=\frac{m}{2}|\dot q(t)|^2+V(r(t)). \end{align*} Using the polar velocity decomposition, this becomes \begin{align*} E=\frac{m}{2}\dot r(t)^2+\frac{m}{2}r(t)^2\dot\theta(t)^2+V(r(t)). \end{align*} Substituting $\dot\theta(t)=\ell/(m r(t)^2)$ gives \begin{align*} E=\frac{m}{2}\dot r(t)^2+\frac{\ell^2}{2m r(t)^2}+V(r(t)). \end{align*} Therefore \begin{align*} \dot r(t)^2=\frac{1}{m^2}\left(2m(E-V(r(t)))-\frac{\ell^2}{r(t)^2}\right). \end{align*} [/step]

[step:Convert angular change into an integral over the radius]Let \begin{align*} A:[r_{\min},\infty)\to[0,\infty) \end{align*} be the positive function \begin{align*} A(r)=2m(E-V(r))-\frac{\ell^2}{r^2}. \end{align*} By hypothesis, $A(r)>0$ for every $r>r_{\min}$. Since $r_{\min}$ is a simple turning point of $E=V_{\mathrm{eff}}(r)$, $A(r_{\min})=0$ and $A'(r_{\min})\neq 0$. The radial energy equation is \begin{align*} \dot r(t)^2=\frac{A(r(t))}{m^2}. \end{align*} The theorem statement assumes that the particular scattering orbit reaches this turning point at $t_0$, has $r(t_0)=r_{\min}$, has an incoming branch on which $r$ decreases from infinity to $r_{\min}$, and has an outgoing branch on which $r$ increases from $r_{\min}$ to infinity. On either open half-orbit away from $t_0$, $A(r(t))>0$, so $\dot r(t)\neq0$ and $r$ is monotone. Moreover, \begin{align*} |\dot r(t)|=\frac{1}{m}\sqrt{A(r(t))}. \end{align*} Combining this with $\dot\theta(t)=\ell/(m r(t)^2)$ gives \begin{align*} \left|\frac{d\theta}{dr}\right|=\frac{|\dot\theta(t)|}{|\dot r(t)|}=\frac{\ell}{r(t)^2\sqrt{A(r(t))}}. \end{align*} Thus the angular change accumulated by either half-orbit between radius $R>r_{\min}$ and the turning point $r_{\min}$ is \begin{align*} \int_{r_{\min}}^R \frac{\ell}{r^2\sqrt{2m(E-V(r))-\ell^2/r^2}}\,d\mathcal L^1(r). \end{align*} Letting $R\to\infty$ and using the assumed convergence of the improper integral yields \begin{align*} \Phi(E,\ell)=\int_{r_{\min}}^\infty \frac{\ell}{r^2\sqrt{2m(E-V(r))-\ell^2/r^2}}\,d\mathcal L^1(r). \end{align*}[/step]

[guided]The goal is to replace time by radius as the independent variable. This is valid only on a half-orbit where the radius is monotone. We recall the radial energy identity and angular momentum identity from the preceding step: for every time at which the polar coordinate is defined, \begin{align*} \dot r(t)^2=\frac{A(r(t))}{m^2} \end{align*} and \begin{align*} \dot\theta(t)=\frac{\ell}{m r(t)^2}. \end{align*} Here the radial kinetic factor is the map $A:[r_{\min},\infty)\to[0,\infty)$ defined by \begin{align*} A(r)=2m(E-V(r))-\frac{\ell^2}{r^2}. \end{align*} The hypotheses give $A(r)>0$ for $r>r_{\min}$. The additional assumption that the turning point is simple means $A(r_{\min})=0$ and $A'(r_{\min})\neq0$. The statement also specifies that this orbit actually has $r(t_0)=r_{\min}$, with an incoming branch decreasing from infinity to $r_{\min}$ and an outgoing branch increasing from $r_{\min}$ to infinity. Away from the turning point, $A(r)>0$ implies that the radial velocity cannot vanish. Hence each incoming or outgoing branch can be parametrized by $r$. From the radial energy identity, we have \begin{align*} \dot r(t)^2=\frac{A(r(t))}{m^2}. \end{align*} Taking the positive square root for speed, rather than signed radial velocity, gives \begin{align*} |\dot r(t)|=\frac{1}{m}\sqrt{A(r(t))}. \end{align*} The angular momentum identity gives \begin{align*} \dot\theta(t)=\frac{\ell}{m r(t)^2}. \end{align*} Dividing angular speed by radial speed therefore gives the absolute angular change per unit radial change: \begin{align*} \left|\frac{d\theta}{dr}\right|=\frac{|\dot\theta(t)|}{|\dot r(t)|}=\frac{\ell}{r(t)^2\sqrt{A(r(t))}}. \end{align*} Now integrate this relation over the radius. For a finite cutoff $R>r_{\min}$, the angular change from radius $R$ down to $r_{\min}$ on the incoming branch, or from $r_{\min}$ up to $R$ on the outgoing branch, is \begin{align*} \int_{r_{\min}}^R \frac{\ell}{r^2\sqrt{2m(E-V(r))-\ell^2/r^2}}\,d\mathcal L^1(r). \end{align*} The measure is explicitly $d\mathcal L^1(r)$ because this is a one-dimensional [Lebesgue integral](/page/Lebesgue%20Integral) in the radial variable. The theorem assumes that the corresponding improper integral converges as $R\to\infty$, so the limiting half-orbit angular change is exactly \begin{align*} \Phi(E,\ell)=\int_{r_{\min}}^\infty \frac{\ell}{r^2\sqrt{2m(E-V(r))-\ell^2/r^2}}\,d\mathcal L^1(r). \end{align*}[/guided]

[step:Relate the half-orbit angle to the asymptotic velocity directions] Let $\theta_0$ denote the polar angle at the turning point. By the explicit scattering-branch hypothesis, the incoming branch has $r(t)\to\infty$ at one end of $I$, the outgoing branch has $r(t)\to\infty$ at the other end of $I$, and both branches have radial range $(r_{\min},\infty)$. The convergence of the improper integral defining $\Phi(E,\ell)$ implies that the total angular variation on each half-orbit is finite; hence the polar angle has a finite limit at each end. Let $\theta_-$ denote the limiting polar angle of the incoming end of the orbit as $r\to\infty$, and let $\theta_+$ denote the limiting polar angle of the outgoing end as $r\to\infty$. The preceding step gives \begin{align*} \theta_0-\theta_-=\Phi(E,\ell) \end{align*} and \begin{align*} \theta_+-\theta_0=\Phi(E,\ell). \end{align*} Hence \begin{align*} \theta_+-\theta_-=2\Phi(E,\ell). \end{align*} At the incoming end, the velocity points opposite to the outward radial direction. Indeed, $r(t)\to\infty$ while $\dot r(t)<0$ on the incoming branch, and the tangential velocity component satisfies \begin{align*} r(t)\dot\theta(t)=\frac{\ell}{m r(t)}\to 0 \end{align*} as $r(t)\to\infty$. Also, since $V(r)\to0$ and $\ell^2/r^2\to0$ as $r\to\infty$, the radial energy identity gives \begin{align*} |\dot r(t)|=\frac{1}{m}\sqrt{2m(E-V(r(t)))-\frac{\ell^2}{r(t)^2}}\to\sqrt{\frac{2E}{m}}. \end{align*} Thus the radial component has a nonzero limiting magnitude while the tangential component vanishes. The incoming velocity direction therefore has angle $\theta_-+\pi$. At the outgoing end, $\dot r(t)>0$, the same estimate $r(t)\dot\theta(t)=\ell/(m r(t))\to0$ holds, and the same radial-speed limit gives $|\dot r(t)|\to\sqrt{2E/m}$. Hence the velocity points in the outward radial direction, so the outgoing velocity direction has angle $\theta_+$. Therefore the signed change from incoming velocity direction to outgoing velocity direction is \begin{align*} \theta_+-(\theta_-+\pi)=2\Phi(E,\ell)-\pi. \end{align*} With the convention that the deflection is the negative of this signed rotation from the incoming direction to the outgoing direction, equivalently that free straight-line motion has deflection $0$, we obtain \begin{align*} \Theta(E,\ell)=\pi-2\Phi(E,\ell). \end{align*} This is the claimed scattering angle formula. [/step]