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The Rutherford deflection map, its inverse, the cross-section density (κ/4E)2csc⁡4(θ/2) and the divergence of the total cross-section

Statement

Assume ACω (The Axiom of Countable Choice (ACω)) for the cross-section measures. Fix μ>0, κ>0 and E>0. For every b>0 put J(b)=2μE b and use the Coulomb trajectory family of Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and Δ=π/2−arctan⁡(κ/(2Eb)) clause 4. Its impact distance is b and its deflection map θdef:(0,∞)→(0,π),θdef(b):=2arctan⁡κ2Eb. Then:

  1. θdef is C1 and strictly decreasing, with θdef(b)→π as b↓0 and θdef(b)→0 as b→∞; it is a C1 bijection (0,∞)→(0,π) with inverse b(θ)=κ2Ecot⁡θ2(0<θ<π).
  2. For every interval (θ1,θ2)⊆(0,π) the pair (θ,b) is a deflection branch with cross-section density dσdΩ(θ)=b(θ)∣dbdθ(θ)∣1sin⁡θ=(κ4E)2csc⁡4θ2, finite and positive on (θ1,θ2); and the pushforward of the incident flux measure under the branch is described by One-dimensional substitution on a deflection branch: the outgoing angular measure has density b∣db/dθ∣, hence cross-section density b∣db/dθ∣/sin⁡θ.
  3. For every θ0∈(0,π) the truncated total cross-section ∫0θ0 ⁣ ⁣∫02π(κ4E)2csc⁡4θ2 sin⁡θ dφ dθ=+∞, while for 0<θ1<θ2<π the integral over (θ1,θ2)×(0,2π) is finite; in this sense the total cross-section of the ideal Coulomb model diverges at forward angles and every compact angular acceptance away from 0 has finite cross-section.

Facts & Assumptions

Given: Countable choice, the fixed data μ,κ,E, the family J(b)=2μE b, the deflection map θdef and its asserted inverse b(θ).

[L1]

arctan⁡:R→(−π/2,π/2) is the inverse of tan⁡, arctan⁡ is C1 and strictly increasing, and tan⁡(arctan⁡y)=y for every real y; on 0<θ/2<π/2 one has arctan⁡(tan⁡(θ/2))=θ/2 (The principal inverse tangent arctan⁡:R→(−π/2,π/2), Tangent is a continuous strictly increasing bijection from (−π/2,π/2) onto R, Principal arctangent: derivative, integral, power series, and the Gregory–Leibniz series).

[L3]

The half-angle advance for the Coulomb branch is Δ=π/2−arctan⁡(κ/(2Eb)) and the deflection magnitude is π−2Δ=2arctan⁡(κ/(2Eb)) (Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and Δ=π/2−arctan⁡(κ/(2Eb)) clauses 3-4).

[L6]

A continuous nonnegative function with a positive value has positive integral, the integral is monotone, and for 0<ε<θ0 the computed integral ∫εθ0θ−3 dθ=12(ε−2−θ0−2) diverges as ε↓0; sin⁡x≤x for x≥0 (mean value theorem applied to sin⁡ on [0,x], using cos⁡≤1) and cos⁡(θ/2)≥cos⁡(θ0/2)>0 for 0<θ≤θ0≤π/2 (The second fundamental theorem: if G is differentiable on [a,b] with G′=f and f is integrable, then ∫abf=G(b)−G(a), If f≤g on [a,b] and both are integrable then ∫abf≤∫abg; and m(b−a)≤∫abf≤M(b−a), A continuous function on [a,b] is Riemann integrable, by Heine-Cantor and Riemann's criterion, Every continuous function on an interval has a primitive; two primitives differ by a constant; and ∫abf=G(b)−G(a) for any primitive G, The mean value theorem, as the case g(x)=x of Cauchy's: for f continuous on [a,b] with a<b and differentiable on (a,b) there is c∈(a,b) with f(b)−f(a)=f′(c)(b−a), Parity and the Pythagorean identity for sine and cosine, Signs, monotonicity intervals, and ranges of sine and cosine, For a<c<b: f is integrable on [a,b] if and only if it is integrable on [a,c] and on [c,b], and then ∫abf=∫acf+∫cbf; with the oriented form for arbitrary a,b,c, Sums, scalar multiples, products, absolute values, maxima, minima and quotients with nonvanishing denominator of continuous functions are continuous, as are constants, the identity and every polynomial function, Continuity of f:A→R at a point of A and on A: the ε-δ condition, its agreement with lim⁡x→cf(x)=f(c) at a limit point, and continuity at an isolated point).

Proof

technique · direct
1.1givenL1L2L4algebra

Differentiation of the branch: by [L4] and [L2], θdef′(b)=2⋅11+(κ2Eb)2⋅(−κ2Eb2)<0(b>0), so θdef is C1 with strictly negative derivative, hence strictly decreasing; as b↓0, κ/(2Eb)→∞ and arctan⁡ tends to π/2, so θdef→π; as b→∞, κ/(2Eb)→0 and θdef→0. The image is therefore (0,π) and the inverse exists, is C1 by [L4], and satisfies the claimed formula: for 0<θ<π, arctan⁡(tan⁡(θ/2))=θ/2 by [L1] (since 0<θ/2<π/2) and tan⁡(θ/2)=κ/(2Eb(θ)) with b(θ)=κ2Ecot⁡(θ/2), because tan⁡(θ/2)cot⁡(θ/2)=1 by [L2]; substituting shows θdef(b(θ))=θ.

2.1step 1.1L3algebra

The half-angle and deflection relation of [L3] applied for each b>0 to the corresponding trajectory with J=J(b) give exactly the deflection value π−2Δ=2arctan⁡(κ/(2Eb)); thus the abstract branch formula is the computed scattering deflection of the Coulomb problem, and clause 1 is complete.

2.2step 1.1L2L5algebra

Cross-section density: by [L2], b′(θ)=−κ4Ecsc⁡2(θ/2), so ∣b′(θ)∣=κ4Ecsc⁡2(θ/2); combining with [L5], dσdΩ(θ)=b(θ)∣b′(θ)∣sin⁡θ=κ28E2⋅cot⁡(θ/2)csc⁡2(θ/2)sin⁡θ. Using sin⁡θ=2sin⁡(θ/2)cos⁡(θ/2) from [L2], cot⁡θ2csc⁡2θ2/sin⁡θ=cos⁡(θ/2)sin⁡3(θ/2)⋅12sin⁡(θ/2)cos⁡(θ/2)=12csc⁡4θ2, whence dσ/dΩ=κ216E2csc⁡4(θ/2)=(κ/(4E))2csc⁡4(θ/2), finite and positive on (θ1,θ2); this is clause 2.

3.1step 2.2L6algebra

Divergence: for 0<θ≤θ0 with θ0≤π/2, the integrand of clause 3 equals (κ/(4E))2sin⁡θ csc⁡4(θ/2)=(κ/(4E))2⋅2cos⁡(θ/2)/sin⁡3(θ/2) by the same identities; since sin⁡(x)≤x for x≥0 and cos⁡(θ/2)≥cos⁡(θ0/2)>0 by [L6], the integrand is at least Cθ−3 with C=16(κ/(4E))2cos⁡(θ0/2)>0. Hence for 0<ε<θ0, by monotonicity of the integral and the φ-integration of a constant, ∫εθ0 ⁣ ⁣∫02π(κ/(4E))2csc⁡4(θ/2)sin⁡θ dφ dθ≥2πC∫εθ0θ−3 dθ=πC(ε−2−θ0−2)⟶+∞ as ε↓0, by the fundamental theorem and [L6]; hence the truncated integral over (0,θ0) diverges. If instead π/2<θ0<π, cut the interval at π/2 and use the divergence on (0,π/2] together with additivity.

4.1step 3.1L6algebra∎

Finiteness on compact angular intervals: the integrand (κ/(4E))2csc⁡4(θ/2)sin⁡θ is continuous on the compact rectangle [θ1,θ2]×[0,2π] because sin⁡(θ/2)>0 there, so its integral is finite by [L6]; this completes clause 3.

Depends on

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