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An exact Rutherford check: $κ = E = 1$ , $b = 1$ give $θ = 2 \arctan \frac{1}{2}$ and $d σ / d Ω = 25 / 16$

Example

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the cited scattering measures and vocabulary.

Take the repulsive Coulomb data $κ = 1$ and $E = 1$ (with any $μ > 0$ and $J = \sqrt{2 μ}$ , so that the asymptotic impact distance of Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and $Δ = π / 2 - \arctan (κ / (2 E b))$ is $b = 1$ ). Then the deflection map of The Rutherford deflection map, its inverse, the cross-section density $(κ / 4 E)^{2} \csc^{4} (θ / 2)$ and the divergence of the total cross-section has the exact values $θ_{d e f} (1) = 2 \arctan \frac{1}{2} \in (0, π), \frac{π}{2} - Δ = \arctan \frac{1}{2}, Δ = \frac{π}{2} - \arctan \frac{1}{2},$ the inverse branch satisfies $b (θ_{d e f} (1)) = 1$ , and $\frac{d σ}{d Ω} (θ_{d e f} (1)) = (\frac{κ}{4 E})^{2} \csc^{4} (\frac{θ_{d e f} (1)}{2}) = \frac{25}{16},$ because $θ_{d e f} (1) / 2 = \arctan \frac{1}{2}$ and $\csc^{2} (\arctan \frac{1}{2}) = 1 + \cot^{2} (\arctan \frac{1}{2}) = 1 + 4 = 5$ .

Facts & Assumptions

Given: Countable choice for the scattering measures. $κ = 1$ , $E = 1$ , $b = 1$ , the deflection map $θ_{d e f} (b) = 2 \arctan (κ / (2 E b))$ , the inverse branch $b (θ) = \frac{κ}{2 E} \cot (θ / 2)$ , and the half-angle $Δ = π / 2 - \arctan (κ / (2 E b))$ .

[L1]

Deflection and half-angle: $θ_{d e f} (b) = 2 \arctan (κ / (2 E b))$ and $Δ = π / 2 - \arctan (κ / (2 E b))$ , with $θ_{d e f} = π - 2 Δ$ (The Rutherford deflection map, its inverse, the cross-section density $(κ / 4 E)^{2} \csc^{4} (θ / 2)$ and the divergence of the total cross-section, Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and $Δ = π / 2 - \arctan (κ / (2 E b))$ , Scattering data: the incident flux measure, the solid-angle element, deflection branches and the cross-section density).

[L2]

$\tan (\arctan y) = y$ and $\cot (\arctan y) = 1 / y$ for $y > 0$ ; $\csc^{2} u = 1 + \cot^{2} u$ (The principal inverse tangent $\arctan : R \to (- π / 2, π / 2)$ , Tangent, cotangent, secant, and cosecant on their exact natural domains, Pythagorean and parity identities for all six trigonometric functions on their natural domains).

[L3]

The cross-section density is $(κ / (4 E))^{2} \csc^{4} (θ / 2)$ on the branch (The Rutherford deflection map, its inverse, the cross-section density $(κ / 4 E)^{2} \csc^{4} (θ / 2)$ and the divergence of the total cross-section).

[L4]

$π - 2 (π / 2 - \arctan \frac{1}{2}) = 2 \arctan \frac{1}{2}$ by the subtraction formula and elementary algebra (The subtraction formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Verification

technique · direct

1.1givenL1L4algebra

Substituting $κ = E = b = 1$ into [L1]: $θ_{d e f} (1) = 2 \arctan \frac{1}{2}$ and $Δ = \frac{π}{2} - \arctan \frac{1}{2}$ ; and $π - 2 Δ = 2 \arctan \frac{1}{2} = θ_{d e f} (1)$ by [L4].

2.1step 1.1L2algebra

Inverse check: $b (θ_{d e f} (1)) = \frac{1}{2} \cot (\arctan \frac{1}{2}) = \frac{1}{2} \cdot 2 = 1$ by [L2].

3.1step 1.1L2L3algebra∎

Density: $\frac{θ_{d e f} (1)}{2} = \arctan \frac{1}{2}$ , so by [L2] $\csc^{2} (\arctan \frac{1}{2}) = 1 + \cot^{2} (\arctan \frac{1}{2}) = 1 + 4 = 5$ ; hence $\csc^{4} = 25$ and, by [L3] with $κ = E = 1$ , $d σ / d Ω = (\frac{1}{4})^{2} \cdot 25 = \frac{25}{16}$ .

Depends on

Dependency tree · 2 levels

Sources

Richard Fitzpatrick, Newtonian Dynamics (University of Texas at Austin), §6.4, Eqs. (358)-(365): the Rutherford angle and density, checked here at exact rational data

An exact Rutherford check: κ=E=1κ=E=1, b=1b=1 give θ=2arctan⁡12θ=2arctan21​ and dσ/dΩ=25/16dσ/dΩ=25/16