Smooth finite-range free scattering refutes global strict decrease; as a separate contact comparison, an attractive refracting shell with index $3\mathrm{/}2$ has $\theta (b)=2(\arcsin (b\mathrm{/}a)-\arcsin (2b\mathrm{/}(3a)))$ strictly increasing, with a jump down to $0$ at grazing incidence

Motion: for $0<b<a$ the model of the Facts admits the exact broken-straight motion: an incoming straight line, a chord inside $r<a$ (the label moves freely there with speed $v_{\mathrm{i}\mathrm{n}}$), and an outgoing straight line; the entry and exit data are related by the two conservation rules as in [L1], and [L2] computes the outgoing direction. For $b>a$ the line misses the shell; at $b=a$ it has the adopted free grazing continuation. In both cases the deflection is $0$.

Genuine smooth finite-range witness and the general obstruction: fix an adopted smooth central model on time window $\mathbb{R}$ with $u(r)=0$ for $r\ge R$, and let $b>R$, $v_{\mathrm{\infty }}=\sqrt{2E\mathrm{/}\mu }$ and $r_b(t)=(v_{\mathrm{\infty }}t,b,0)$. Then $\mathrm{\mid }{r}_b(t)\mathrm{\mid }\ge b>R$, so $u=u^{\mathrm{\prime }}=0$ along it and $r_b^{\mathrm{\prime }\mathrm{\prime }}=0$ solves the reduced equation. With $M=m_1+m_2$, the curves $x_1=(m_2\mathrm{/}M)r_b$, $x_2=-(m_1\mathrm{/}M)r_b$ have separation $r_b$, zero accelerations and zero force, hence are motions of the adopted model with zero centre of mass. Their angular momentum has magnitude $J=\mu b{v}_{\mathrm{\infty }}>0$ and energy $E$; at the turning radius $b$, $V(b)=E$, $V^{\mathrm{\prime }}(b)=-J^2\mathrm{/}(\mu b^3)<0$ and $V(r)<E$ for every $r>b$, matching [L4]. Both asymptotic relative velocities equal $(v_{\mathrm{\infty }},0,0)$, so the deflection is uniquely $0$. Any incoming line with impact parameter $b>R$ stays outside the interaction region and continues freely; thus every finite-range model has this constant deflection tail. In particular $u=0$ is $C^2$ and finite-range, and the same construction works for every $b>0$, giving a unique angle $0$ for every positive impact parameter; two distinct positive parameters therefore violate the asserted strict decrease.

Deflection formula: by [L2], for $b\in (0,a)$ the angle between the incoming and outgoing directions is $\theta (b)=2(\alpha -\beta )$ with $\sin \alpha =b\mathrm{/}a$ and $\sin \beta =\sin \alpha \mathrm{/}n=b\mathrm{/}(na)$, that is $\theta (b)=2(\arcsin \frac{b}{a}-\arcsin \frac{2b}{3a})(0<b<a)\mathrm{.}$

Consistency of the contact rule: at the entry the angular momentum about the centre is $\mu b{v}_{\mathrm{\infty }}$, and inside it equals $\mu a{v}_{\mathrm{i}\mathrm{n}}\sin \beta =\mu an{v}_{\mathrm{\infty }}\sin \beta =\mu b{v}_{\mathrm{\infty }}$, so angular momentum is conserved across the interface as the radial contact force requires; the energy is conserved by construction of $v_{\mathrm{i}\mathrm{n}}$ with $U_0=\frac{5}{4}E$. This separate contact model is conservative and finite-range as an adopted radial rule; it is not a member of the smooth central-scattering class.

Inverse derivative and monotonicity: fix $x\in (-1,1)$ and put $s=\arcsin x\in (-\pi \mathrm{/}2,\pi \mathrm{/}2)$. Cosine is continuous and nonzero on that interval, since ${\cos }^{2}s=1-{\sin }^{2}s>0$ there; the intermediate value theorem and $\cos 0=1$ therefore give $\cos s=\sqrt{1-x^2}>0$. For $y\in (-1,1)$ with $y\mathrm{\ne }x$, put $q=\arcsin y$. By inverse continuity $q\to s$ as $y\to x$, and strict increase gives $q\mathrm{\ne }s$. The mean value theorem on the segment joining $q$ and $s$ gives $y-x=\cos \xi (q-s)$ for some $\xi$ between them. Consequently $\xi \to s$ and $\frac{\arcsin y-\arcsin x}{y-x}=\frac{1}{\cos \xi }⟶\frac{1}{\cos s}=\frac{1}{\sqrt{1-x^2}}\mathrm{.}$ Thus ${\arcsin }^{\mathrm{\prime }}(x)=(1-x^2{)}^{-1\mathrm{/}2}$ on $(-1,1)$. Applying the chain and algebra rules to step 2.1 gives ${\theta }^{\mathrm{\prime }}(b)=2(\frac{1}{\sqrt{a^2-b^2}}-\frac{1}{\sqrt{n^2{a}^2-b^2}})>0(0<b<a),$ because $n>1$ gives $\sqrt{n^2{a}^2-b^2}>\sqrt{a^2-b^2}>0$. The positive-derivative theorem therefore makes $\theta$ strictly increasing on $(0,a)$, the opposite of the claimed decrease.

Endpoint limits and the grazing jump: the continuous endpoint extension at $b=0$ is $\theta (0)=0$, which is not a branch value because the branch domain is $(0,a)$. As $b\uparrow a$, $\alpha \to \pi \mathrm{/}2$ and $\beta \to \arcsin (1\mathrm{/}n)=\arcsin \frac{2}{3}$, so $\theta (b)\to \pi -2\arcsin \frac{2}{3}\approx 1.682$ rad $\approx {96.4}^{\circ }$; at grazing $b=a$ and for every $b>a$ the adopted free continuation has deflection $0$. This value lies outside the branch's angular domain $(0,\pi )$. The full contact-model deflection therefore increases on $(0,a)$ and then drops to $0$, and it is not injective.

Conclusion: step 1.2 refutes the stated global strict-decrease claim with a genuine smooth central-scattering model satisfying its positive-energy, nonzero-angular-momentum and unique-angle hypotheses. More generally, any finite-range model whose angle is defined for every positive impact parameter has the constant free tail established there, so global strict decrease fails even when the interaction inside the range is nontrivial. Steps 1.1-4.1 separately retain the increasing refracting-shell branch and its adopted grazing jump; that contact comparison lies outside the smooth class and supplies no claim that $C^2$ regularity implies or is necessary for monotonicity. The central-scattering theorem itself records no global monotonicity claim and assumes a chosen nondegenerate branch for its cross-section formula.