Alphabeta Math

1 · Prerequisites

2 · Summary

The ellipse/period and cross-section checks inherit countable choice from their mathematical suppliers. The companion page to cm-math-central records explicit worked checks of its reduction, Kepler and scattering results. Everything here is mathematics: the examples and the counterexample carry domain: mathematics, depend only on the mathematical items of the parent page and on mathematical suppliers, and make no physical or empirical claim.

The harmonic oscillator action variable $J_{a} (E) = E / ω$ and $J_{a}^{'} (E) = T / (2 π)$ instantiates The action variable of a two-simple-turning-point family is $C^{1}$ with derivative $T (E) / (2 π)$ , the period quadrature over $2 π$ at $U (q) = \frac{1}{2} k q^{2}$ : the substitution $q = \sqrt{2 E / k} \sin ϑ$ gives $J_{a} (E) = \frac{\sqrt{2 m E}}{π} \sqrt{2 E / k} \int_{- π / 2}^{π / 2} \cos^{2} ϑ d ϑ = E / ω$ with $ω = \sqrt{k / m}$ , the time quadrature gives $T (E) = 2 π / ω$ , and differentiation confirms $J_{a}^{'} (E) = 1 / ω = T (E) / (2 π)$ exactly. An exact Kepler ellipse check: $e = 1 / 2$ , $J = 1$ , $μ = κ = 1$ give $a = 4 / 3$ , $b = 2 \sqrt{3} / 3$ , $E = - 3 / 8$ and $T^{2} = 256 π^{2} / 27$ evaluates The Kepler orbit equation is a conic: $u = c (1 + e \cos (φ - θ_{0}))$ with $E = (μ κ^{2} / 2 J^{2}) (e^{2} - 1)$ and Completeness and periodicity of the negative-energy Kepler orbit, with $T = 2 π μ a b / J$ and $T^{2} = 4 π^{2} μ a^{3} / κ$ at $μ = κ = J = 1$ , $e = \frac{1}{2}$ , obtaining $p = 1$ , $1 - e^{2} = \frac{3}{4}$ , $a = \frac{4}{3}$ , $b = \frac{2 \sqrt{3}}{3}$ , $E = - \frac{3}{8}$ , and the same value $T^{2} = \frac{256 π^{2}}{27}$ from both $T = 2 π μ a b / J$ and $T^{2} = 4 π^{2} μ a^{3} / κ$ . An exact Rutherford check: $κ = E = 1$ , $b = 1$ give $θ = 2 \arctan \frac{1}{2}$ and $d σ / d Ω = 25 / 16$ evaluates 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 at $κ = E = b = 1$ : $θ_{d e f} (1) = 2 \arctan \frac{1}{2}$ , $b (θ_{d e f} (1)) = \frac{1}{2} \cot (\arctan \frac{1}{2}) = 1$ , and $d σ / d Ω = (1 / 4)^{2} \csc^{4} (\arctan (1 / 2)) = \frac{25}{16}$ , using $\csc^{2} (\arctan \frac{1}{2}) = 1 + \cot^{2} (\arctan \frac{1}{2}) = 5$ .

A zero-angular-momentum Kepler trajectory reaches the force centre in finite time, refuting unconditional global existence is the counterexample that marks the excluded radial degenerate case of the Kepler layer: the curve $r (t) = A (1 - t)^{2 / 3} e_{1}$ on $(- \infty, 1)$ solves the attractive inverse-square equation $μ r^{''} = - κ r / ∣ r ∣^{3}$ with angular momentum $j \equiv 0$ , reaches the force centre at $t = 1$ in finite time and has no extension to an interval containing $1$ ; it refutes the claim that every attractive inverse-square trajectory is global, collision-free and a conic, and it shows why the non-radial hypothesis $J > 0$ of the positive results is not removable.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

None yet.

5 · Examples, experiments and counterexamples

ExampleConstruction: AI-generatedVerification: AI-generatedOpen item page →

The harmonic oscillator action variable $J_{a} (E) = E / ω$ and $J_{a}^{'} (E) = T / (2 π)$

Example

Let $m > 0$ , $k > 0$ , $U (q) = \frac{1}{2} k q^{2}$ on $J_{0} = R$ , and for $E > 0$ set $a (E) : = - \sqrt{\frac{2 E}{k}}, b (E) : = \sqrt{\frac{2 E}{k}}, ω : = \sqrt{\frac{k}{m}}, T : = \frac{2 π}{ω} .$ Then the datum $(J_{0}, U, m, a, b)$ satisfies the hypotheses of The action variable of a two-simple-turning-point family is $C^{1}$ with derivative $T (E) / (2 π)$ , the period quadrature over $2 π$ on every bounded open interval $(E_{0}, E_{1})$ with $0 < E_{0} < E_{1} < \infty$ . These restrictions define the same action on overlaps, hence together give a $C^{1}$ function on $(0, \infty)$ ; the action variable of The one-dimensional action variable of a family of two-turning-point energy levels is $J_{a} (E) = \frac{E}{ω} (E > 0),$ and $J_{a}^{'} (E) = \frac{1}{ω} = \frac{T}{2 π} (E > 0),$ in agreement with the general theorem, whose period quadrature is $T (E) = \sqrt{2 m} \int_{a (E)}^{b (E)} d q / \sqrt{E - U (q)} = 2 π / ω$ .

Facts & Assumptions

Given: Positive $m, k$ , the quadratic potential $U (q) = \frac{1}{2} k q^{2}$ , the turning-point functions $\pm \sqrt{2 E / k}$ , and the frequency $ω = \sqrt{k / m}$ .

[L1]

Action variable and theorem: $J_{a} (E) = \frac{1}{π} \int_{a (E)}^{b (E)} \sqrt{2 m (E - U (q))} d q$ , and under the two simple turning points of this datum, $J_{a}^{'} (E) = T (E) / (2 π)$ with $T (E) = \sqrt{2 m} \int_{a (E)}^{b (E)} d q / \sqrt{E - U (q)}$ (The one-dimensional action variable of a family of two-turning-point energy levels, The action variable of a two-simple-turning-point family is $C^{1}$ with derivative $T (E) / (2 π)$ , the period quadrature over $2 π$ ).

[L2]

Substitution with $q = \sqrt{2 E / k} \sin ϑ$ on $[- π / 2, π / 2]$ , where $\cos ϑ \geq 0$ , is a differentiable monotone change of variable, and $\cos^{2} ϑ = \frac{1}{2} (1 + \cos 2 ϑ)$ from the addition formula; the Taylor-free identities $\sin^{2} + \cos^{2} = 1$ and the endpoint values $\sin (\pm π / 2) = \pm 1$ , $\cos (\pm π / 2) = 0$ hold (Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ , The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Signs, monotonicity intervals, and ranges of sine and cosine).

[L3]

$\sqrt{2 E / k} \cdot \sqrt{k / 2} = \sqrt{E}$ and $\sqrt{2 m E} \cdot \sqrt{E} = \sqrt{2 m} E$ ; for $E > 0$ the square roots are the positive ones (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ).

Verification

technique · direct

1.1givenL1algebra

Fix any bounded energy interval $(E_{0}, E_{1})$ with $0 < E_{0} < E_{1} < \infty$ . Restrict $a, b$ to that interval. The datum satisfies the hypotheses of the theorem: $U$ is $C^{3}$ with $U^{'} (q) = k q$ , so $U^{'} (a (E)) = - k \sqrt{2 E / k} < 0 < k \sqrt{2 E / k} = U^{'} (b (E))$ , and $a (E) < b (E)$ ; moreover $\frac{1}{2} k q^{2} < E$ exactly for $a (E) < q < b (E)$ , and $U (a (E)) = U (b (E)) = E$ . Every $E > 0$ lies in such an interval, for example $(E / 2, 2 E)$ , and the defining integrals agree on overlaps; thus the theorem gives a $C^{1}$ action on all of $(0, \infty)$ .

2.1step 1.1L2L3algebra

Computation of the action variable: substituting $q = \sqrt{2 E / k} \sin ϑ$ by [L2] and using $\frac{1}{2} k q^{2} = E \sin^{2} ϑ$ , $J_{a} (E) = \frac{1}{π} \int_{- π / 2}^{π / 2} \sqrt{2 m E \cos^{2} ϑ} \sqrt{\frac{2 E}{k}} \cos ϑ d ϑ = \frac{\sqrt{2 m E}}{π} \sqrt{\frac{2 E}{k}} \int_{- π / 2}^{π / 2} \cos^{2} ϑ d ϑ,$ because $\sqrt{\cos^{2} ϑ} = \cos ϑ \geq 0$ on the interval. With $\cos^{2} ϑ = \frac{1}{2} (1 + \cos 2 ϑ)$ the integral is $π / 2$ , so by [L3] $J_{a} (E) = \frac{\sqrt{2 m E} \sqrt{2 E / k}}{2} = \frac{2 E \sqrt{m / k}}{2} = \frac{E}{ω} .$

2.2step 1.1L2L3algebra

Period quadrature: put $A = \sqrt{2 E / k}$ . For $- π / 2 < c < d < π / 2$ the substitution $q = A \sin ϑ$ lies strictly between the turning points, where the integrand is continuous and $\cos ϑ > 0$ . Hence [L2] gives $\int_{A \sin c}^{A \sin d} d q / \sqrt{E - k q^{2} / 2} = \sqrt{2 / k} (d - c)$ , since $\sqrt{E \cos^{2} ϑ} = \sqrt{E} \cos ϑ$ there. Taking the independent limits $c ↓ - π / 2$ , $d ↑ π / 2$ proves convergence at both endpoints and gives $\int_{a (E)}^{b (E)} d q / \sqrt{E - k q^{2} / 2} = π \sqrt{2 / k}$ . Thus $T (E) = \sqrt{2 m} π \sqrt{2 / k} = 2 π / ω$ .

3.1step 1.1step 2.1step 2.2L1algebra∎

Differentiating the closed form of step 2.1, $J_{a}^{'} (E) = 1 / ω$ for every $E > 0$ ; by step 2.2 this equals $T (E) / (2 π)$ , consistent with the general theorem [L1], which the datum satisfies by step 1.1.

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An exact Kepler ellipse check: $e = 1 / 2$ , $J = 1$ , $μ = κ = 1$ give $a = 4 / 3$ , $b = 2 \sqrt{3} / 3$ , $E = - 3 / 8$ and $T^{2} = 256 π^{2} / 27$

Example

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )), as required by the ellipse-area and period suppliers; the conic-and-eccentricity theorem does not impose this hypothesis.

Take the attractive Kepler data $μ = κ = 1$ and angular momentum magnitude $J = 1$ , and consider the negative-energy orbit with eccentricity $e = \frac{1}{2}$ . Then the conic data of The Kepler orbit equation is a conic: $u = c (1 + e \cos (φ - θ_{0}))$ with $E = (μ κ^{2} / 2 J^{2}) (e^{2} - 1)$ and The subcritical Kepler conic is an ellipse with semiaxes $a = p / (1 - e^{2})$ , $b = a \sqrt{1 - e^{2}}$ , focus at the origin and area $π a b$ ; its area is also $\frac{1}{2} \int_{0}^{2 π} ϱ^{2}$ are $p = \frac{J^{2}}{μ κ} = 1, a = \frac{p}{1 - e^{2}} = \frac{4}{3}, b = a \sqrt{1 - e^{2}} = \frac{2 \sqrt{3}}{3},$ the energy is $E = \frac{μ κ^{2}}{2 J^{2}} (e^{2} - 1) = - \frac{3}{8}$ , and the period of Completeness and periodicity of the negative-energy Kepler orbit, with $T = 2 π μ a b / J$ and $T^{2} = 4 π^{2} μ a^{3} / κ$ satisfies $T = \frac{2 π μ a b}{J} = \frac{16 π \sqrt{3}}{9}, T^{2} = \frac{256 π^{2}}{27} = \frac{4 π^{2} μ a^{3}}{κ} .$ All numbers are exact; the example checks algebraic consistency of the three theorems and the sign criterion $E < 0 ⟺ e < 1$ , and makes no empirical claim.

Facts & Assumptions

Given: Assume ${A C}_{ω}$ . $μ = κ = 1$ , $J = 1$ , $e = \frac{1}{2}$ and the derived constants $p = 1$ , $a = \frac{4}{3}$ , $b = \frac{2 \sqrt{3}}{3}$ .

[L1]

Conic and eccentricity: for a non-radial attractive Kepler trajectory with $E < 0$ , $p = J^{2} / (μ κ)$ , $a = p / (1 - e^{2})$ , $b = a \sqrt{1 - e^{2}}$ and $E = (μ κ^{2} / 2 J^{2}) (e^{2} - 1)$ (The Kepler orbit equation is a conic: $u = c (1 + e \cos (φ - θ_{0}))$ with $E = (μ κ^{2} / 2 J^{2}) (e^{2} - 1)$ , The subcritical Kepler conic is an ellipse with semiaxes $a = p / (1 - e^{2})$ , $b = a \sqrt{1 - e^{2}}$ , focus at the origin and area $π a b$ ; its area is also $\frac{1}{2} \int_{0}^{2 π} ϱ^{2}$ , Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential).

[L2]

Period: $T = 2 π μ a b / J$ and $T^{2} = 4 π^{2} μ a^{3} / κ$ (Completeness and periodicity of the negative-energy Kepler orbit, with $T = 2 π μ a b / J$ and $T^{2} = 4 π^{2} μ a^{3} / κ$ ).

[L3]

Square roots: $\sqrt{1 - \frac{1}{4}} = \frac{\sqrt{3}}{2}$ and $(\sqrt{3})^{2} = 3$ (Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ ); $\cos^{2} + \sin^{2} = 1$ provides the value $\sqrt{1 - e^{2}}$ used below via $1 - e^{2} > 0$ (Parity and the Pythagorean identity for sine and cosine).

Verification

technique · direct

1.1givenL1L3algebra

With $μ = κ = J = 1$ : $p = 1$ , $1 - e^{2} = \frac{3}{4}$ , $a = 1 / (\frac{3}{4}) = \frac{4}{3}$ , $\sqrt{1 - e^{2}} = \frac{\sqrt{3}}{2}$ and $b = \frac{4}{3} \cdot \frac{\sqrt{3}}{2} = \frac{2 \sqrt{3}}{3}$ .

1.2givenL1algebra

Energy: $e^{2} - 1 = \frac{1}{4} - 1 = - \frac{3}{4}$ , so $E = \frac{1 \cdot 1}{2 \cdot 1} \cdot (- \frac{3}{4}) = - \frac{3}{8} < 0$ , consistent with $e = \frac{1}{2} < 1$ by the equivalence of [L1].

2.1step 1.1L2algebra

Period: $μ a b = \frac{4}{3} \cdot \frac{2 \sqrt{3}}{3} = \frac{8 \sqrt{3}}{9}$ , so $T = 2 π \cdot \frac{8 \sqrt{3}}{9} = \frac{16 π \sqrt{3}}{9}$ ; squaring, $T^{2} = \frac{256 π^{2} \cdot 3}{81} = \frac{256 π^{2}}{27}$ .

3.1step 1.1step 2.1L2algebra∎

Independent check of the second period formula: $a^{3} = \frac{64}{27}$ , so $4 π^{2} μ a^{3} / κ = 4 π^{2} \cdot \frac{64}{27} = \frac{256 π^{2}}{27} = T^{2}$ , in agreement with step 2.1 and with [L2].

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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 $κ = E = 1$ , $d σ / d Ω = (\frac{1}{4})^{2} \cdot 25 = \frac{25}{16}$ .

CounterexampleConstruction: AI-generatedVerification: AI-adaptedOpen item page →

A zero-angular-momentum Kepler trajectory reaches the force centre in finite time, refuting unconditional global existence

Statement refuted

False statement. Every solution of the attractive inverse-square equation $μ r^{''} = - κ r / ∣ r ∣^{3}$ ( $μ, κ > 0$ ) on its maximal interval of definition is defined for all real times and satisfies $∣ r (t) ∣ \geq c$ for some $c > 0$ .

The witness has zero angular momentum and reaches the excluded force centre, so the non-radial positive results do not apply. Its inverse-square potential is singular and is not finite-range; this example does not establish the necessity of the simple-turning-point hypothesis in Finite-range scattering: the simple outer turning point, the two global branches and the free tails.

Facts & Assumptions

Given: Parameters $μ > 0$ , $κ > 0$ , the constant $A : = (9 κ / (2 μ))^{1 / 3} > 0$ , and the curve $r : (- \infty, 1) \to R^{3}$ , $r (t) : = A (1 - t)^{2 / 3} e_{1}$ , where $e_{1}$ is a unit vector.

[L1]

The central-force equation in the radial case: for a curve $r = ρ e_{1}$ with $ρ > 0$ of class $C^{2}$ , the equation $μ r^{''} = - κ r / ∣ r ∣^{3}$ is equivalent to $μ ρ^{''} = - κ / ρ^{2}$ , and the angular momentum $μ r \times r^{'}$ vanishes identically (Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential, $C^{k}$ Euclidean maps and diffeomorphisms).

[L2]

Power derivatives: for $t < 1$ , $\frac{d}{d t} (1 - t)^{2 / 3} = - \frac{2}{3} (1 - t)^{- 1 / 3}$ and $\frac{d}{d t} (- \frac{2}{3} (1 - t)^{- 1 / 3}) = - \frac{2}{9} (1 - t)^{- 4 / 3}$ , by the real-power derivative and the chain rule (Continuity and derivatives of positive-base real powers, The chain rule, in one line from Carathéodory: if $g$ is differentiable at $c$ and $f$ is differentiable at $g (c)$ , then $f \circ g$ is differentiable at $c$ with $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Sums, scalar multiples, products and quotients: $(f + g)^{'} (c) = f^{'} (c) + g^{'} (c)$ , $(α f)^{'} (c) = α f^{'} (c)$ , $(f g)^{'} (c) = f^{'} (c) g (c) + f (c) g^{'} (c)$ , and $(f / g)^{'} (c) = (f^{'} (c) g (c) - f (c) g^{'} (c)) / g (c)^{2}$ when $g (c) \neq 0$ , The derivative $f^{'} (c) = \lim_{x \to c} \frac{f (x) - f (c)}{x - c}$ of $f : A \to R$ at a point $c \in A$ that is a limit point of $A$ , and differentiability on a set, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

For $t < 1$ , $A^{3} (1 - t)^{2} = \frac{9 κ}{2 μ} (1 - t)^{2}$ and, equivalently, $\frac{2 μ A}{9} = \frac{κ}{A^{2}}$ : this is the defining relation of $A$ , and it is an equality of positive real numbers (multiply $A^{3} = 9 κ / (2 μ)$ by $2 μ / (9 A^{2})$ ).

Counterexample

technique · direct

1.1givenL1L2L3algebra

The curve solves the equation: by [L2], $r^{''} (t) = - \frac{2 A}{9} (1 - t)^{- 4 / 3} e_{1}$ , so $μ r^{''} = - \frac{2 μ A}{9} (1 - t)^{- 4 / 3} e_{1}$ , while $- κ r / ∣ r ∣^{3} = - κ ρ^{- 2} e_{1} = - \frac{κ}{A^{2}} (1 - t)^{- 4 / 3} e_{1}$ because $ρ (t) = A (1 - t)^{2 / 3}$ . By [L3] the two coefficients agree, so $μ r^{''} = - κ r / ∣ r ∣^{3}$ for every $t < 1$ ; and the angular momentum vanishes identically.

2.1givenstep 1.1L1algebra

Failure of global existence: $ρ (t) = A (1 - t)^{2 / 3} \to 0$ as $t ↑ 1$ , and $ρ$ is strictly positive and $C^{2}$ on $(- \infty, 1)$ . There is no solution of $μ r^{''} = - κ r / ∣ r ∣^{3}$ on an interval containing $1$ whose restriction to $(- \infty, 1)$ equals $r$ : such a solution would be continuous at $1$ with $r (1) = \lim_{t ↑ 1} r (t) = 0$ , and the equation is undefined at the origin (the right-hand side contains $r / ∣ r ∣^{3}$ ). Hence the maximal interval of $r$ is $(- \infty, 1)$ , not $R$ .

3.1step 1.1step 2.1algebra∎

Failure of the lower bound: $\inf_{t < 1} ρ (t) = 0$ , so there is no $c > 0$ with $∣ r (t) ∣ \geq c$ on the domain. Together with step 2.1, this refutes the stated universal claim.

Cm Math Central — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

The harmonic oscillator action variable $J_{a} (E) = E / ω$ and $J_{a}^{'} (E) = T / (2 π)$

Example

Facts & Assumptions

Verification

An exact Kepler ellipse check: $e = 1 / 2$ , $J = 1$ , $μ = κ = 1$ give $a = 4 / 3$ , $b = 2 \sqrt{3} / 3$ , $E = - 3 / 8$ and $T^{2} = 256 π^{2} / 27$

Example

Facts & Assumptions

Verification

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

Example

Facts & Assumptions

Verification

A zero-angular-momentum Kepler trajectory reaches the force centre in finite time, refuting unconditional global existence

Statement refuted

Facts & Assumptions

Counterexample

Sources

Cm Math Central — Examples

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

5 · Examples, experiments and counterexamples

The harmonic oscillator action variable Ja(E)=E/ωJa​(E)=E/ω and Ja′(E)=T/(2π)Ja′​(E)=T/(2π)

Example

Facts & Assumptions

Verification

An exact Kepler ellipse check: e=1/2e=1/2, J=1J=1, μ=κ=1μ=κ=1 give a=4/3a=4/3, b=23/3b=23​/3, E=−3/8E=−3/8 and T2=256π2/27T2=256π2/27

Example

Facts & Assumptions

Verification

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

Example

Facts & Assumptions

Verification

A zero-angular-momentum Kepler trajectory reaches the force centre in finite time, refuting unconditional global existence

Statement refuted

Facts & Assumptions

Counterexample

Sources

The harmonic oscillator action variable $J_{a} (E) = E / ω$ and $J_{a}^{'} (E) = T / (2 π)$

An exact Kepler ellipse check: $e = 1 / 2$ , $J = 1$ , $μ = κ = 1$ give $a = 4 / 3$ , $b = 2 \sqrt{3} / 3$ , $E = - 3 / 8$ and $T^{2} = 256 π^{2} / 27$

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