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Cm Math Central

1 · Prerequisites

2 · Summary

The central-force layer of the non-relativistic classical mechanics collection, all of it mathematics: the reduction of the non-radial central-force equation to polar data, the Kepler problem up to the determination of its conic, period and one-dimensional action variable, and the finite-range and Coulomb scattering layer with its cross-section densities. Every item on this page carries domain: mathematics and depends only on mathematical suppliers; nothing here asserts that the equations describe a physical interaction, that incident flux is realized by counted events, or that any empirical conclusion follows. The physical reading of masses, forces, energies and scattering measurements remains with the consuming classical-mechanics pages.

The reduction layer begins with Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential, which defines the central-force equation $μ r^{''} = - (U^{'} (∣ r ∣) / ∣ r ∣) r$ for a curve $r : I \to R^{3}$ with values in $J_{0} \subseteq (0, \infty)$ , together with the angular momentum vector $j = μ r \times r^{'}$ , the energy and the effective potential $U_{J}^{e f f}$ . Local polar coordinates for a curve with nonvanishing position in the plane: a differentiable angle function with prescribed derivative supplies local polar coordinates $ρ, σ$ of a planar curve as a twice differentiable rotation, and Conservation of the angular momentum vector along a central-force trajectory and its orthogonality to position and velocity proves that $j$ is constant and orthogonal to $r$ and $r^{'}$ . Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation then derives the polar identities $ρ^{2} θ^{'} = J / μ$ , the radial equation $μ ρ^{''} = - (U_{J}^{e f f})^{'} (ρ)$ and the conserved energy $E = \frac{1}{2} μ ρ^{' 2} + U_{J}^{e f f} (ρ)$ . Circular orbits of a non-radial central-force trajectory: the critical-point condition and the squared frequency of the linearised radial equation characterises circular orbits by $(U_{J}^{e f f})^{'} (r_{0}) = 0$ , exhibits them explicitly, and reduces the linearised radial equation to the constant-coefficient oscillator, whose solutions are classified by Solutions of $v^{''} + ω^{2} v = 0$ are exactly the functions $A \cos (ω (t - t_{0})) + B \sin (ω (t - t_{0}))$ . The Kepler specialization proceeds through Binet's equation for the attractive inverse-square field: $d^{2} u / d φ^{2} + u = μ κ / J^{2}$ for $u = 1 / ρ$ , which derives $u^{''} + u = μ κ / J^{2}$ for $u = 1 / ρ$ along the angular variable, and The Kepler orbit equation is a conic: $u = c (1 + e \cos (φ - θ_{0}))$ with $E = (μ κ^{2} / 2 J^{2}) (e^{2} - 1)$ , which proves $u = c (1 + e \cos (φ - θ_{0}))$ and the exact relation $E = \frac{μ κ^{2}}{2 J^{2}} (e^{2} - 1)$ connecting the sign of the energy with the eccentricity. The algebraic prerequisites The vector triple product identity $u \times (v \times w) = ⟨ u, w ⟩ v - ⟨ u, v ⟩ w$ and Oriented orthonormal triads in $R^{3}$ : existence over any unit vector, and the orthonormal expansion with no fourth orthogonal direction are proved locally: the first is the vector triple product identity, the second constructs an oriented orthonormal triad over any unit vector and proves that no four orthonormal vectors exist in $R^{3}$ , together with the orthonormal expansion.

The Kepler layer proper is 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 $\frac{1}{2} \int_{0}^{2 π} ϱ^{2}$ , which proves that the subcritical conic is the ellipse with centre $(- a e, 0)$ and semiaxes $a = p / (1 - e^{2})$ , $b = a \sqrt{1 - e^{2}}$ , with area $λ_{2} (E_{e l l}) = π a b = \frac{1}{2} \int_{0}^{2 π} ϱ (φ)^{2} d φ$ ; the item declares the Axiom of Countable Choice used by its affine change-of-variables and polar-decomposition arguments. Completeness and periodicity of the negative-energy Kepler orbit, with $T = 2 π μ a b / J$ and $T^{2} = 4 π^{2} μ a^{3} / κ$ proves completeness of the negative-energy orbit, the periodicity of the motion and the exact period value $T = 2 π μ a b / J = 2 π \sqrt{μ / κ} a^{3 / 2}$ , hence $T^{2} = 4 π^{2} μ a^{3} / κ$ ; it also declares the Axiom of Countable Choice. The one-dimensional action layer is The one-dimensional action variable of a family of two-turning-point energy levels, which defines $J_{a} (E) = \frac{1}{π} \int_{a (E)}^{b (E)} \sqrt{2 m (E - U (q))} d q$ for a two-simple-turning-point family, and The action variable of a two-simple-turning-point family is $C^{1}$ with derivative $T (E) / (2 π)$ , the period quadrature over $2 π$ , which proves that the action variable is $C^{1}$ with $J_{a}^{'} (E) = T (E) / (2 π)$ by the inverse function theorem at the turning points, a $C^{3}$ endpoint rescaling and differentiation under the integral sign.

The measure part of the scattering layer assumes countable choice. The scattering layer is built on Scattering data: the incident flux measure, the solid-angle element, deflection branches and the cross-section density, which defines the finite-range data, the incident flux measure $b d b d φ$ , the solid-angle measure $\sin θ d θ d φ$ , deflection branches and their cross-section density $b (θ) / ⁣ (\sin θ ∣ θ^{'} (b (θ)) ∣)$ as abstract measures and functions; no counting of events and no empirical claim belongs to the definition. Finite-range scattering: the simple outer turning point, the two global branches and the free tails assumes a simple outer turning point and proves the two unique global branches with free tails, Existence of the asymptotic velocities and the impact-parameter identity $J = μ b ∣ v_{-} ∣$ constructs the asymptotic velocities $v_{\pm}$ and the impact-parameter identity $J = μ b ∣ v_{-} ∣$ , and The scattering angle advance $Δ$ and the deflection relation: the total polar advance is $2 Δ$ and the velocity directions make the cosine $\cos (π - 2 Δ)$ proves the advance quadrature $Δ = \int_{r_{*}}^{\infty} J d r / (r^{2} \sqrt{2 μ (E - V (r))})$ and the deflection relation modulo whole revolutions. One-dimensional substitution on a deflection branch: the outgoing angular measure has density $b ∣ d b / d θ ∣$ , hence cross-section density $b ∣ d b / d θ ∣ / \sin θ$ performs the one-dimensional substitution on compact interior intervals and a Borel measure-uniqueness argument, giving the full open-branch pushforward identity for nonnegative or absolutely integrable Borel test functions and the outgoing angular measure density $b ∣ b^{'} ∣$ and the cross-section density $b ∣ b^{'} ∣ / \sin θ$ , additively over finitely many disjoint branches. Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and $Δ = π / 2 - \arctan (κ / (2 E b))$ separately treats the long-range the repulsive Coulomb potential, where the effective potential is strictly decreasing with a single simple turning point and the motion is global and non-colliding, with genuine asymptotic velocity and impact-distance limits and angle advance $Δ = π / 2 - \arctan (κ / (2 E b))$ ; finally 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 proves, for fixed energy and $J (b) = \sqrt{2 μ E} b$ , the deflection map $θ_{d e f} (b) = 2 \arctan (κ / (2 E b))$ , its inverse $b (θ) = \frac{κ}{2 E} \cot (θ / 2)$ , the density $(κ / 4 E)^{2} \csc^{4} (θ / 2)$ and the divergence of the total cross-section at forward angles together with finiteness over every compact angular interval. The companion page cm-math-central-examples records exact finite checks of the action derivative, of the Kepler ellipse data and of the Rutherford density, and a zero-angular-momentum collision counterexample that marks the excluded radial degenerate case.

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential

Definition

Let $μ > 0$ be a real number, let $J_{0} \subseteq (0, \infty)$ be an open interval, let $U : J_{0} \to R$ be of class $C^{2}$ ( $C^{k}$ Euclidean maps and diffeomorphisms), let $I \subseteq R$ be an interval with at least two points (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), and let $r : I \to R^{3}$ be twice differentiable with $∣ r (t) ∣ \in J_{0}$ for every $t \in I$ (The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral). The curve $r$ is a central-force trajectory for $(μ, U)$ when

$μ r^{''} (t) = - \frac{U^{'} (∣ r (t) ∣)}{∣ r (t) ∣} r (t) (t \in I),$

where $∣ x ∣ : = ∥ x ∥_{2}$ is the Euclidean norm (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ) and $U^{'}$ is the derivative of $U$ on the open interval $J_{0}$ .

Attached objects. The angular momentum vector of such a trajectory is the curve $j : I \to R^{3}$ ,

$j (t) : = μ r (t) \times r^{'} (t),$

with $\times$ the cross product of $R^{3}$ (The cross product in $R^{3}$ ); its energy is the function $E : I \to R$ ,

$E (t) : = \frac{1}{2} μ ∥ r^{'} (t) ∥_{2}^{2} + U (∣ r (t) ∣);$

and for every $J \geq 0$ the effective potential of momentum magnitude $J$ is the function $U_{J}^{e f f} : J_{0} \to R$ ,

$U_{J}^{e f f} (ρ) : = U (ρ) + \frac{J^{2}}{2 μ ρ^{2}} .$

Why the objects are defined. Since $ρ > 0$ throughout $J_{0}$ , the right-hand side of the central-force equation is defined, the map $ρ \mapsto J^{2} / (2 μ ρ^{2})$ is $C^{2}$ on $(0, \infty)$ (a quotient of $C^{2}$ functions with nowhere vanishing denominator), and $U_{J}^{e f f}$ is again $C^{2}$ ; in particular second derivatives of $U_{J}^{e f f}$ are available. The vector $j (t)$ is defined for every $t \in I$ because $r$ is differentiable there.

Mathematical status. The data $μ$ , $U$ and $J$ are real numbers and functions; the equation and the attached objects are mathematical statements about curves. Nothing here asserts that the equation describes a physical interaction, and the names are conventional labels for the mathematical objects only.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Local polar coordinates for a curve with nonvanishing position in the plane: a differentiable angle function with prescribed derivative

Statement

Let $I \subseteq R$ be an interval with at least two points (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), let $t_{0} \in I$ , and let $ρ, σ : I \to R$ be twice differentiable with $(ρ (t), σ (t)) \neq (0, 0)$ for every $t \in I$ . Here twice differentiable means that $f$ and its derivative $f' : I \to R$ are differentiable at every point of $I$ in the domain-relative sense of 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; $f'' = (f')'$ , including one-sided limits at included endpoints. Put

$r (t) : = \sqrt{ρ (t)^{2} + σ (t)^{2}} > 0 (t \in I) .$

Then there is a twice differentiable $θ : I \to R$ with

$ρ = r \cos θ, σ = r \sin θ on I,$

and every twice differentiable $ϑ : I \to R$ with $ρ = r \cos ϑ$ and $σ = r \sin ϑ$ on $I$ satisfies

$ϑ^{'} (t) = \frac{ρ (t) σ^{'} (t) - σ (t) ρ^{'} (t)}{r (t)^{2}} (t \in I) .$

Facts & Assumptions

Given: An interval $I \subseteq R$ with at least two points, a point $t_{0} \in I$ , and twice differentiable functions $ρ, σ : I \to R$ with $(ρ, σ) \neq (0, 0)$ on $I$ .

[L1]

The function $r = \sqrt{ρ^{2} + σ^{2}}$ is twice differentiable with $r^{'} = (ρ ρ^{'} + σ σ^{'}) / r$ : the square root is the inverse of the square on $(0, \infty)$ , whose derivative $2 y$ does not vanish there, the inverse-function rule gives $(\sqrt{})^{'} (x) = 1 / (2 \sqrt{x})$ for $x > 0$ , and the chain rule gives $r^{'} = (ρ^{2} + σ^{2})^{'} / (2 r) = (ρ ρ^{'} + σ σ^{'}) / r$ (Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ , 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 $g (c) \neq 0$ ).

[L2]

Sine and cosine are differentiable with $(\sin u)^{'} = \cos u$ and $(\cos u)^{'} = - \sin u$ , so with the chain and product rules the functions $t \mapsto ρ (t) \cos θ (t)$ , $t \mapsto σ (t) \sin θ (t)$ and their difference are differentiated by the ordinary product rule (The derivatives of sine and cosine are cosine and minus sine, 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 $g (c) \neq 0$ ).

[L3]

$t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the unit circle $S^{1} = {(x, y) : x^{2} + y^{2} = 1}$ , and $\sin^{2} + \cos^{2} = 1$ ( $t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the real unit circle, Parity and the Pythagorean identity for sine and cosine).

[L4]

A continuous function $f$ on the interval $I$ has a primitive: for fixed $c_{0} \in I$ the function $F (x) = \int_{c_{0}}^{x} f$ is differentiable on $I$ with $F^{'} = f$ (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ); sums, products and quotients with nonvanishing denominator of continuous functions are continuous (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 $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L5]

A function continuous on $I$ with vanishing derivative at every interior point of $I$ is constant on $I$ (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

Proof

technique · direct

1.1givenL1algebra

By [L1], $r$ is differentiable with $r^{'} = (ρ ρ^{'} + σ σ^{'}) / r$ . Since $ρ, σ$ are twice differentiable and $r > 0$ , this quotient is differentiable by the product and quotient rules; hence $r$ is twice differentiable.

1.2givenL3choose

The vector $(ρ (t_{0}) / r (t_{0}), σ (t_{0}) / r (t_{0}))$ satisfies $(ρ / r)^{2} + (σ / r)^{2} = 1$ by the definition of $r$ , so by [L3] there is a real $θ_{0}$ with $\cos θ_{0} = ρ (t_{0}) / r (t_{0})$ and $\sin θ_{0} = σ (t_{0}) / r (t_{0})$ .

2.1givenstep 1.1step 1.2L2L4construct

Define $g : I \to R$ by $g : = (ρ σ^{'} - σ ρ^{'}) / r^{2}$ . The numerator is continuous and $r^{2} > 0$ is continuous, so $g$ is continuous on $I$ by [L4]; hence by [L4] the function $θ (t) : = θ_{0} + \int_{t_{0}}^{t} g (s) d s$ is differentiable on $I$ with $θ^{'} = g$ . Moreover $g$ is differentiable by the product and quotient rules, since $ρ, σ$ are twice differentiable and $r^{2} > 0$ ; therefore $θ$ is twice differentiable with $θ^{'} = g$ .

2.2givenstep 1.2L3algebra

Define $A : = ρ \cos θ + σ \sin θ$ and $B : = σ \cos θ - ρ \sin θ$ on $I$ . Evaluating at $t_{0}$ with the choice of $θ_{0}$ and [L3] gives $A (t_{0}) = ρ (t_{0})^{2} / r (t_{0}) + σ (t_{0})^{2} / r (t_{0}) = r (t_{0})$ and $B (t_{0}) = σ (t_{0}) ρ (t_{0}) / r (t_{0}) - ρ (t_{0}) σ (t_{0}) / r (t_{0}) = 0$ .

3.1step 1.1step 2.1L1L2algebra

Put $c : = r^{'} / r$ on $I$ . Differentiating with [L2] and substituting $θ^{'} = g$ gives the two identities $ρ^{'} + σ θ^{'} = c ρ$ and $σ^{'} - ρ θ^{'} = c σ$ : indeed $ρ^{'} + σ θ^{'} = ρ^{'} + σ (ρ σ^{'} - σ ρ^{'}) / r^{2} = (ρ^{'} ρ^{2} + ρ σ σ^{'}) / r^{2} = ρ (ρ ρ^{'} + σ σ^{'}) / r^{2} = c ρ$ , and $σ^{'} - ρ θ^{'} = σ^{'} - ρ (ρ σ^{'} - σ ρ^{'}) / r^{2} = (σ^{'} σ^{2} + ρ σ ρ^{'}) / r^{2} = σ (ρ ρ^{'} + σ σ^{'}) / r^{2} = c σ$ , using [L1]. Hence $A^{'} = (ρ^{'} + σ θ^{'}) \cos θ + (σ^{'} - ρ θ^{'}) \sin θ = c A,$ $B^{'} = (σ^{'} - ρ θ^{'}) \cos θ - (ρ^{'} + σ θ^{'}) \sin θ = c B .$

4.1step 2.2step 3.1L1L2L5algebra

By the quotient rule of [L2] and $r^{'} = c r$ from [L1], $(A / r)^{'} = \frac{A^{'} r - A r^{'}}{r^{2}} = \frac{c A r - A c r}{r^{2}} = 0, (B / r)^{'} = \frac{c B r - B c r}{r^{2}} = 0 on I .$ The functions $A / r$ and $B / r$ are continuous on $I$ ; by [L5] and step 2.2 they are the constants $1$ and $0$ : $A = r$ and $B = 0$ on $I$ .

5.1step 4.1L3algebra

Solving the two equations $A = ρ \cos θ + σ \sin θ = r$ and $B = σ \cos θ - ρ \sin θ = 0$ for $(ρ, σ)$ gives $ρ = A \cos θ - B \sin θ = r \cos θ$ and $σ = A \sin θ + B \cos θ = r \sin θ$ on $I$ : the displayed linear system has the invertible coefficient matrix $(\begin{array}{cc} \cos θ & \sin θ \\ - \sin θ & \cos θ \end{array})$ of determinant $\cos^{2} θ + \sin^{2} θ = 1$ .

6.1givenL1L2algebra∎

Let $ϑ : I \to R$ be twice differentiable with $ρ = r \cos ϑ$ , $σ = r \sin ϑ$ . Differentiating the two identities and eliminating $r^{'}$ gives $ρ σ^{'} - σ ρ^{'} = r \cos ϑ (r^{'} \sin ϑ + r ϑ^{'} \cos ϑ) - r \sin ϑ (r^{'} \cos ϑ - r ϑ^{'} \sin ϑ) = r^{2} ϑ^{'},$ so $ϑ^{'} = (ρ σ^{'} - σ ρ^{'}) / r^{2}$ on $I$ .

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Solutions of $v^{''} + ω^{2} v = 0$ are exactly the functions $A \cos (ω (t - t_{0})) + B \sin (ω (t - t_{0}))$

Statement

Let $ω > 0$ , let $I \subseteq R$ be an interval with at least two points (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length), let $t_{0} \in I$ , and let $v : I \to R$ be twice differentiable with

$v^{''} (t) + ω^{2} v (t) = 0 (t \in I) .$

Here twice differentiable means that $v$ and $v' : I \to R$ are differentiable at every point of $I$ in the domain-relative sense of 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; $v'' = (v')'$ , with one-sided limits at included endpoints.

Then there are reals $A, B$ with

$v (t) = A \cos (ω (t - t_{0})) + B \sin (ω (t - t_{0})) (t \in I) .$

Conversely, for all reals $A, B$ the function $t \mapsto A \cos (ω (t - t_{0})) + B \sin (ω (t - t_{0}))$ is twice differentiable on $R$ and satisfies $v^{''} + ω^{2} v = 0$ there.

Facts & Assumptions

Given: A real $ω > 0$ , an interval $I \subseteq R$ with at least two points, a point $t_{0} \in I$ , and a twice differentiable $v : I \to R$ with $v^{''} + ω^{2} v = 0$ on $I$ .

[L1]

Sine and cosine are differentiable with $(\sin u)^{'} = \cos u$ and $(\cos u)^{'} = - \sin u$ ; the chain rule differentiates the compositions $t \mapsto \cos (ω (t - t_{0}))$ , $t \mapsto \sin (ω (t - t_{0}))$ with derivatives $- ω \sin (ω (t - t_{0}))$ and $ω \cos (ω (t - t_{0}))$ (The derivatives of sine and cosine are cosine and minus sine, 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L2]

Sums, products and constant multiples of differentiable functions are differentiable with the usual rules (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 $g (c) \neq 0$ ).

[L3]

A function continuous on $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant).

[L4]

$\sin^{2} + \cos^{2} = 1$ identically (Parity and the Pythagorean identity for sine and cosine).

Proof

technique · direct

1.1givenL1L2construct

Write $φ (t) : = ω (t - t_{0})$ and define $w : = v \cos φ - \frac{v^{'}}{ω} \sin φ, z : = v \sin φ + \frac{v^{'}}{ω} \cos φ$ on $I$ . Every factor is differentiable (including $v'$ , by the twice differentiable hypothesis), so $w$ and $z$ are differentiable by [L2].

1.2givenL1L2algebra

Differentiating and using $v^{''} = - ω^{2} v$ : $w^{'} = v^{'} \cos φ - ω v \sin φ - \frac{v^{''}}{ω} \sin φ - v^{'} \cos φ = - \frac{ω^{2} v + v^{''}}{ω} \sin φ = 0,$ and likewise $z^{'} = v^{'} \sin φ + ω v \cos φ + \frac{v^{''}}{ω} \cos φ - v^{'} \sin φ = \frac{ω^{2} v + v^{''}}{ω} \cos φ = 0$ on $I$ .

2.1step 1.1step 1.2L3

By [L3] and step 1.2 there are reals $A, B$ with $w = A$ and $z = B$ on $I$ .

3.1step 1.1step 2.1L4algebra

Multiplying out and using [L4], $w \cos φ + z \sin φ = v (\cos^{2} φ + \sin^{2} φ) = v$ on $I$ ; hence $v = A \cos φ + B \sin φ$ , which is the asserted representation.

4.1L1L2algebra∎

Conversely, if $u (t) : = A \cos (ω (t - t_{0})) + B \sin (ω (t - t_{0}))$ , then [L1] and [L2] give $u^{'} = - ω A \sin φ + ω B \cos φ$ and $u^{''} = - ω^{2} A \cos φ - ω^{2} B \sin φ = - ω^{2} u$ on $I$ , so $u^{''} + ω^{2} u = 0$ .

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

The vector triple product identity $u \times (v \times w) = ⟨ u, w ⟩ v - ⟨ u, v ⟩ w$

Statement

For all $u, v, w \in R^{3}$ , $u \times (v \times w) = ⟨ u, w ⟩ v - ⟨ u, v ⟩ w .$

Facts & Assumptions

Given: Vectors $u, v, w \in R^{3}$ with coordinates $u = (u_{1}, u_{2}, u_{3})$ , $v = (v_{1}, v_{2}, v_{3})$ , $w = (w_{1}, w_{2}, w_{3})$ .

[L1]

The cross product is defined coordinatewise by $u \times v = (u_{2} v_{3} - u_{3} v_{2}, u_{3} v_{1} - u_{1} v_{3}, u_{1} v_{2} - u_{2} v_{1})$ (The cross product in $R^{3}$ ).

[L2]

The Euclidean inner product is $⟨ u, w ⟩ = \sum_{k = 1}^{3} u_{k} w_{k}$ (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L3]

Finite sums may be expanded, reordered and combined (Laws of finite sums and finite products).

Proof

technique · direct

1.1L1L2L3algebra

First component. By [L1], $(u \times (v \times w))_{1} = u_{2} (v \times w)_{3} - u_{3} (v \times w)_{2} = u_{2} (v_{1} w_{2} - v_{2} w_{1}) - u_{3} (v_{3} w_{1} - v_{1} w_{3}),$ which expands to $v_{1} (u_{2} w_{2} + u_{3} w_{3}) - w_{1} (u_{2} v_{2} + u_{3} v_{3})$ ; since $u_{2} w_{2} + u_{3} w_{3} = ⟨ u, w ⟩ - u_{1} w_{1}$ and $u_{2} v_{2} + u_{3} v_{3} = ⟨ u, v ⟩ - u_{1} v_{1}$ by [L2], this equals $v_{1} (⟨ u, w ⟩ - u_{1} w_{1}) - w_{1} (⟨ u, v ⟩ - u_{1} v_{1}) = (⟨ u, w ⟩ v - ⟨ u, v ⟩ w)_{1}$ , which is the $1$ -component of the asserted identity.

2.1L1L2L3algebra

Second and third components. The same computation, with the coordinate labels cyclically relabelled and the components of the cross product read off in the corresponding cyclic order, gives $(u \times (v \times w))_{2} = (⟨ u, w ⟩ v - ⟨ u, v ⟩ w)_{2}$ and $(u \times (v \times w))_{3} = (⟨ u, w ⟩ v - ⟨ u, v ⟩ w)_{3}$ : each is the same finite expansion of [L1] as in step 1.1 with $(1, 2, 3)$ cyclically replaced, and [L2], [L3] are unaffected by the relabelling.

3.1step 1.1step 2.1L1algebra∎

Two vectors of $R^{3}$ with equal components are equal, so steps 1.1 and 2.1 prove the identity in all three coordinates.

LemmaStatement: AI-adaptedProof: AI-adaptedOpen item page →

Oriented orthonormal triads in $R^{3}$ : existence over any unit vector, and the orthonormal expansion with no fourth orthogonal direction

Statement

Call a triple $(e_{1}, e_{2}, e_{3})$ of vectors of $R^{3}$ an oriented orthonormal triad when $⟨ e_{i}, e_{j} ⟩ = δ_{i j}$ for all $i, j \in {1, 2, 3}$ and $e_{1} \times e_{2} = e_{3}$ . Then:

Existence. For every $e_{3} \in R^{3}$ with $∣ e_{3} ∣ = 1$ there are $e_{1}, e_{2} \in R^{3}$ such that $(e_{1}, e_{2}, e_{3})$ is an oriented orthonormal triad.
Dimension bound. Every orthonormal list $f_{1}, \dots, f_{m} \in R^{3}$ (pairwise orthogonal unit vectors) has $m \leq 3$ ; consequently, for every oriented orthonormal triad $(e_{1}, e_{2}, e_{3})$ and every $x \in R^{3}$ , $x = ⟨ x, e_{1} ⟩ e_{1} + ⟨ x, e_{2} ⟩ e_{2} + ⟨ x, e_{3} ⟩ e_{3}, ∣ x ∣^{2} = ⟨ x, e_{1} ⟩^{2} + ⟨ x, e_{2} ⟩^{2} + ⟨ x, e_{3} ⟩^{2} .$

Throughout, $⟨ \cdot, \cdot ⟩$ and $∣ \cdot ∣$ are the Euclidean inner product and norm (The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ), $\times$ is the cross product (The cross product in $R^{3}$ ), and $s_{1}, s_{2}, s_{3}$ is the standard basis of $R^{3}$ (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ ).

Facts & Assumptions

Given: A unit vector $e_{3} \in R^{3}$ , standard basis vectors $s_{1}, s_{2}, s_{3}$ , and arbitrary orthonormal lists as in the statement.

[L1]

The triple product identity $u \times (v \times w) = ⟨ u, w ⟩ v - ⟨ u, v ⟩ w$ holds for all $u, v, w \in R^{3}$ (The vector triple product identity $u \times (v \times w) = ⟨ u, w ⟩ v - ⟨ u, v ⟩ w$ ).

[L2]

The cross product is bilinear and alternating and is orthogonal to both factors; and $∣ u \times v ∣^{2} = ∣ u ∣^{2} ∣ v ∣^{2} - ⟨ u, v ⟩^{2}$ (The cross product is bilinear, alternating, and orthogonal to both factors, The squared cross-product norm is the Gram determinant of two vectors, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L3]

Relabel the standard basis indexed $0, 1, 2$ in the supplier as $s_{1}, s_{2}, s_{3}$ . This basis is orthonormal: $⟨ s_{i}, s_{j} ⟩ = δ_{i j}$ , and every $x \in R^{3}$ is $x = \sum_{j = 1}^{3} ⟨ x, s_{j} ⟩ s_{j}$ with $∣ x ∣^{2} = \sum_{j = 1}^{3} ⟨ x, s_{j} ⟩^{2}$ , by the coordinate definition of the Euclidean inner product and the standard basis (The standard list $e : n \to F^{n}$ with $e_{i} (i) = 1_{F}$ and $e_{i} (j) = 0_{F}$ for $j \neq i$ is an ordered basis of $F^{n}$ ; hence $\dim_{F} F^{n} = n$ , and $F^{0}$ is the zero space with basis $\emptyset$ and dimension $0$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L4]

Bessel's inequality: for an orthonormal family $(f_{1}, \dots, f_{m})$ in an inner product space (Real and complex inner product spaces, with the inner product linear in the first argument, Orthogonal vectors and subspaces, orthogonal and orthonormal sets, and orthonormal bases) and every vector $x$ , $\sum_{i = 1}^{m} ⟨ x, f_{i} ⟩^{2} \leq ∣ x ∣^{2}$ (The Bessel inequality for an arbitrary orthonormal family).

[L5]

Finite sums may be reordered and combined (Laws of finite sums and finite products); and two distinct standard basis vectors are not parallel, since they have distinct, nonvanishing coordinates.

Proof

technique · direct

1.1givenL3L5

At most one standard basis vector is parallel to $e_{3}$ : if $s_{i} = λ e_{3}$ and $s_{j} = μ e_{3}$ with $i \neq j$ and $λ, μ \neq 0$ , then $s_{i} = (λ / μ) s_{j}$ , which is false because $s_{i}$ and $s_{j}$ have $1$ in different coordinates. Hence some $k$ with $s_{k}$ not parallel to $e_{3}$ exists.

1.2givenL3L4L5

Let $f_{1}, \dots, f_{m} \in R^{3}$ be orthonormal, and suppose $m \geq 4$ . Applying Bessel's inequality [L4] to each standard basis vector $s_{j}$ gives $\sum_{i = 1}^{m} ⟨ s_{j}, f_{i} ⟩^{2} \leq ∣ s_{j} ∣^{2} = 1$ . Summing over $j = 1, 2, 3$ and exchanging the finite sums [L5] gives $\sum_{i = 1}^{m} \sum_{j = 1}^{3} ⟨ s_{j}, f_{i} ⟩^{2} \leq 3$ .

2.1step 1.1L3algebra

Put $v : = s_{k} - ⟨ s_{k}, e_{3} ⟩ e_{3}$ . Then $v \neq 0$ : otherwise $s_{k} = ⟨ s_{k}, e_{3} ⟩ e_{3}$ is parallel to $e_{3}$ , contrary to step 1.1. Set $e_{1} : = v / ∣ v ∣$ . Then $⟨ e_{1}, e_{3} ⟩ = 0$ and $∣ e_{1} ∣ = 1$ , because $⟨ v, e_{3} ⟩ = ⟨ s_{k}, e_{3} ⟩ - ⟨ s_{k}, e_{3} ⟩ ∣ e_{3} ∣^{2} = 0$ and $∣ v ∣ \neq 0$ .

2.2step 1.2L3L5

For each $i$ , the standard-basis expansion of [L3] applied to $f_{i}$ gives $\sum_{j = 1}^{3} ⟨ s_{j}, f_{i} ⟩^{2} = ∣ f_{i} ∣^{2} = 1$ . Hence the left-hand side of step 1.2 equals $\sum_{i = 1}^{m} 1 = m \geq 4$ , contradicting the bound $\leq 3$ of step 1.2. Therefore no orthonormal list in $R^{3}$ has four or more members.

3.1step 2.1L1L2algebra

Set $e_{2} : = e_{3} \times e_{1}$ . By [L2], $⟨ e_{2}, e_{3} ⟩ = 0 = ⟨ e_{2}, e_{1} ⟩$ and $∣ e_{2} ∣^{2} = ∣ e_{3} ∣^{2} ∣ e_{1} ∣^{2} - ⟨ e_{3}, e_{1} ⟩^{2} = 1$ ; and $e_{1} \times e_{2} = e_{1} \times (e_{3} \times e_{1}) = ⟨ e_{1}, e_{1} ⟩ e_{3} - ⟨ e_{1}, e_{3} ⟩ e_{1} = e_{3}$ by the triple product identity [L1]. So $(e_{1}, e_{2}, e_{3})$ is an oriented orthonormal triad, which is clause 1.

3.2givenstep 2.2L3algebra

Let $(e_{1}, e_{2}, e_{3})$ be an oriented orthonormal triad and $x \in R^{3}$ . Put $w : = x - \sum_{i = 1}^{3} ⟨ x, e_{i} ⟩ e_{i}$ . Then $⟨ w, e_{j} ⟩ = ⟨ x, e_{j} ⟩ - \sum_{i} ⟨ x, e_{i} ⟩ δ_{i j} = 0$ for each $j$ , by bilinearity of the inner product. If $w \neq 0$ , then $e_{1}, e_{2}, e_{3}, w / ∣ w ∣$ would be four pairwise orthogonal unit vectors, contradicting step 2.2; hence $w = 0$ , which is the expansion.

4.1step 3.2L3algebra∎

With the expansion in hand, $∣ x ∣^{2} = ⟨ \sum_{i} ⟨ x, e_{i} ⟩ e_{i}, \sum_{k} ⟨ x, e_{k} ⟩ e_{k} ⟩ = \sum_{i, k} ⟨ x, e_{i} ⟩ ⟨ x, e_{k} ⟩ δ_{i k} = \sum_{i} ⟨ x, e_{i} ⟩^{2},$ by bilinearity and orthonormality. This completes clause 2.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Conservation of the angular momentum vector along a central-force trajectory and its orthogonality to position and velocity

Statement

$j$ is differentiable on $I$ with $j^{'} (t) = 0$ for every $t \in I$ ; hence $j$ is constant on $I$ ;
for every $t \in I$ , $⟨ r (t), j (t) ⟩ = 0$ and $⟨ r^{'} (t), j (t) ⟩ = 0$ .

Facts & Assumptions

Given: A central-force trajectory $r$ for $(μ, U)$ on an interval $I$ with at least two points.

[L1]

For differentiable curves $u, v : I \to R^{3}$ the cross product $u \times v$ is differentiable with $(u \times v)^{'} = u^{'} \times v + u \times v^{'}$ ; moreover sums, and scalar multiples of curves, differentiate by the usual rules (Differentiation of sums, scalar multiples, inner products, cross products and squared norms of differentiable curves in $R^{3}$ ).

[L2]

The trajectory satisfies $r^{''} = - (U^{'} (∣ r ∣) / (μ ∣ r ∣)) r$ on $I$ , where $∣ x ∣ = ∥ x ∥_{2}$ (Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential, The derivative and the Riemann integral of a vector-valued function: an intrinsic derivative and a componentwise integral).

[L3]

The cross product is bilinear and alternating, so $u \times u = 0$ and $(u \times v) ⊥ u$ , $(u \times v) ⊥ v$ ; orthogonality means vanishing Euclidean inner product (The cross product is bilinear, alternating, and orthogonal to both factors, The cross product in $R^{3}$ , The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ ).

[L4]

If $f : [a, b] \to R^{3}$ is differentiable with integrable $f^{'}$ , then $\int_{a}^{b} f^{'} = f (b) - f (a)$ (If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct

1.1givenL1algebra

By [L1] the curve $j = μ (r \times r^{'})$ is differentiable with $j^{'} = μ (r^{'} \times r^{'} + r \times r^{''}) = μ r \times r^{''} .$

2.1step 1.1L2L3algebra

Substituting [L2] and using alternation [L3], $j^{'} = μ r \times (- \frac{U^{'} (∣ r ∣)}{μ ∣ r ∣} r) = - \frac{U^{'} (∣ r ∣)}{∣ r ∣} r \times r = 0 on I .$

3.1step 2.1L4

Let $a, b \in I$ with $a < b$ . Then [L4] applied to $j$ gives $\int_{a}^{b} j^{'} = j (b) - j (a)$ ; since $j^{'}$ is the zero curve by step 2.1, the integral vanishes and $j (b) = j (a)$ . As $a, b$ were arbitrary, $j$ is constant on $I$ .

4.1L3algebra∎

For every $t \in I$ , [L3] gives $⟨ r (t), j (t) ⟩ = μ ⟨ r (t), r (t) \times r^{'} (t) ⟩ = 0$ and $⟨ r^{'} (t), j (t) ⟩ = μ ⟨ r^{'} (t), r (t) \times r^{'} (t) ⟩ = 0$ : the cross product is orthogonal to each factor.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation

Statement

Let $μ > 0$ , let $J_{0} \subseteq (0, \infty)$ be an open interval, let $U : J_{0} \to R$ be of class $C^{2}$ , let $I \subseteq R$ be an interval with at least two points, and let $r : I \to R^{3}$ be a central-force trajectory for $(μ, U)$ whose angular momentum $j = μ r \times r^{'}$ is a nonzero constant vector (Conservation of the angular momentum vector along a central-force trajectory and its orthogonality to position and velocity); write $J : = ∣ j ∣ > 0$ . Let $(e_{1}, e_{2}, e_{3})$ be an oriented orthonormal triad with $e_{3} = j / J$ (Oriented orthonormal triads in $R^{3}$ : existence over any unit vector, and the orthonormal expansion with no fourth orthogonal direction).

Then there are twice differentiable $ρ : I \to (0, \infty)$ and $θ : I \to R$ with

$r (t) = ρ (t) (\cos θ (t) e_{1} + \sin θ (t) e_{2}) (t \in I),$

and for every such pair $(ρ, θ)$ :

$ρ (t) = ∣ r (t) ∣$ for all $t$ ;
$ρ^{2} θ^{'} = J / μ$ on $I$ , so $θ^{'} > 0$ ;
the effective-potential equation $μ ρ^{''} = - (U_{J}^{e f f})^{'} (ρ)$ holds on $I$ , with $U_{J}^{e f f} = U + J^{2} / (2 μ ρ^{2})$ (Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential);
$E : = \frac{1}{2} μ ∥ r^{'} ∥_{2}^{2} + U (∣ r ∣)$ equals $\frac{1}{2} μ ρ^{' 2} + U_{J}^{e f f} (ρ)$ and is constant on $I$ .

Facts & Assumptions

Given: A central-force trajectory $r$ for $(μ, U)$ with constant nonzero angular momentum $j = J e_{3}$ , $J > 0$ , an oriented orthonormal triad $(e_{1}, e_{2}, e_{3})$ , and twice differentiable $ρ, θ$ with $r = ρ (\cos θ e_{1} + \sin θ e_{2})$ .

[L1]

For the oriented orthonormal triad: every $x \in R^{3}$ is $x = ⟨ x, e_{1} ⟩ e_{1} + ⟨ x, e_{2} ⟩ e_{2} + ⟨ x, e_{3} ⟩ e_{3}$ and $∣ x ∣^{2} = \sum_{i} ⟨ x, e_{i} ⟩^{2}$ (Oriented orthonormal triads in $R^{3}$ : existence over any unit vector, and the orthonormal expansion with no fourth orthogonal direction).

[L2]

The angular momentum is orthogonal to $r$ and to $r^{'}$ : $⟨ r, e_{3} ⟩ J = ⟨ r, j ⟩ = 0$ and $⟨ r^{'}, j ⟩ = 0$ (Conservation of the angular momentum vector along a central-force trajectory and its orthogonality to position and velocity).

[L3]

Polar lift: for twice differentiable $ρ_{1}, ρ_{2} : I \to R$ with $(ρ_{1}, ρ_{2}) \neq (0, 0)$ there is a twice differentiable angle function such that $ρ_{1} = r_{0} \cos θ$ , $ρ_{2} = r_{0} \sin θ$ with $r_{0} = \sqrt{ρ_{1}^{2} + ρ_{2}^{2}}$ (Local polar coordinates for a curve with nonvanishing position in the plane: a differentiable angle function with prescribed derivative).

[L4]

Differentiability rules: inner products and cross products of differentiable curves differentiate by the product rule, $⟨ x, y ⟩^{'} = ⟨ x^{'}, y ⟩ + ⟨ x, y^{'} ⟩$ and $(x \times y)^{'} = x^{'} \times y + x \times y^{'}$ ; the chain rule differentiates $\cos θ$ and $\sin θ$ with derivatives $- θ^{'} \sin θ$ and $θ^{'} \cos θ$ ; sums, products and quotients with nonvanishing denominator differentiate by the usual rules (Differentiation of sums, scalar multiples, inner products, cross products and squared norms of differentiable curves in $R^{3}$ , The derivatives of sine and cosine are cosine and minus sine, 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 $g (c) \neq 0$ ).

[L5]

Cross product rules: $e_{1} \times e_{2} = e_{3}$ , $e_{2} \times e_{1} = - e_{3}$ , the product is bilinear, and $e_{i}$ are unit vectors (The cross product in $R^{3}$ , The cross product is bilinear, alternating, and orthogonal to both factors); $\cos^{2} + \sin^{2} = 1$ (Parity and the Pythagorean identity for sine and cosine).

[L6]

A function continuous on $I$ whose derivative vanishes at every interior point is constant on $I$ (A function continuous on an interval $I$ whose derivative vanishes at every interior point of $I$ is constant on $I$ ; consequently two such functions with the same derivative differ by a constant, $C^{k}$ Euclidean maps and diffeomorphisms, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct

1.1givenL1L2L4algebra

Put $ρ_{i} : = ⟨ r, e_{i} ⟩$ for $i = 1, 2$ ; these are twice differentiable by [L4]. They are never both zero: by [L1] and [L2], $r = ρ_{1} e_{1} + ρ_{2} e_{2} + ⟨ r, e_{3} ⟩ e_{3} = ρ_{1} e_{1} + ρ_{2} e_{2}$ , so $ρ_{1} = ρ_{2} = 0$ would give $r = 0$ , contradicting $∣ r ∣ \in J_{0} \subseteq (0, \infty)$ .

1.2givenL1L5algebra

Clause 1 for an arbitrary pair: if $r = ρ (\cos θ e_{1} + \sin θ e_{2})$ , then $∣ r ∣^{2} = ∣ r ∣^{2}$ is computed by [L1] and [L5] as $ρ^{2} (\cos^{2} θ + \sin^{2} θ) = ρ^{2}$ , hence $ρ = ∣ r ∣$ .

1.3givenL4L5algebra

Differentiating [L5]'s expansion $r = ρ n$ with $n : = \cos θ e_{1} + \sin θ e_{2}$ and $n_{⊥} : = - \sin θ e_{1} + \cos θ e_{2}$ by [L4] gives $r^{'} = ρ^{'} n + ρ θ^{'} n_{⊥}, r^{''} = (ρ^{''} - ρ θ^{' 2}) n + (2 ρ^{'} θ^{'} + ρ θ^{''}) n_{⊥} .$ Here $n \times n_{⊥} = e_{3}$ and $n_{⊥} \times n = - e_{3}$ by bilinearity of the cross product and $e_{1} \times e_{2} = e_{3}$ .

2.1step 1.1L1L2L3algebra

Apply the polar lift [L3] to $(ρ_{1}, ρ_{2})$ : there is a twice differentiable $θ$ with $ρ_{1} = r_{0} \cos θ$ , $ρ_{2} = r_{0} \sin θ$ for $r_{0} = \sqrt{ρ_{1}^{2} + ρ_{2}^{2}}$ ; by Parseval in [L1] and $⟨ r, e_{3} ⟩ = 0$ , $r_{0}^{2} = ρ_{1}^{2} + ρ_{2}^{2} = ∣ r ∣^{2}$ , so $r_{0} = ∣ r ∣ = : ρ$ is twice differentiable with values in $J_{0} \subseteq (0, \infty)$ . Substituting into the expansion of step 1.1 gives $r = ρ (\cos θ e_{1} + \sin θ e_{2})$ , so such a pair exists.

2.2givenstep 1.3L5algebra

Clause 2: from step 1.3, $r \times r^{'} = ρ n \times (ρ^{'} n + ρ θ^{'} n_{⊥}) = ρ^{2} θ^{'} e_{3},$ so $j = J e_{3} = μ r \times r^{'} = μ ρ^{2} θ^{'} e_{3}$ , giving $J = μ ρ^{2} θ^{'}$ and hence $ρ^{2} θ^{'} = J / μ > 0$ .

3.1givenstep 1.2step 1.3step 2.2L1L5algebra

Clause 3: the trajectory equation is $μ r^{''} = - (U^{'} (∣ r ∣) / ∣ r ∣) r = - U^{'} (ρ) n$ by clause 1. Comparing with the decomposition of step 1.3 and using the linear independence of $n$ and $n_{⊥}$ — their components in the orthonormal triad are distinct, so $α n + β n_{⊥} = 0$ forces $α = β = 0$ by taking inner products with $n$ and $n_{⊥}$ — the radial component gives $μ (ρ^{''} - ρ θ^{' 2}) = - U^{'} (ρ) .$ Substituting $ρ^{2} θ^{'} = J / μ$ from step 2.2 gives $μ ρ^{''} = - U^{'} (ρ) + μ ρ θ^{' 2} = - U^{'} (ρ) + \frac{J^{2}}{μ ρ^{3}} = - (U_{J}^{e f f})^{'} (ρ)$ , because $(U_{J}^{e f f})^{'} = U^{'} (ρ) - \frac{J^{2}}{μ ρ^{3}}$ .

3.2step 1.3step 2.2L1algebra

Clause 4, the energy identity: from step 1.3 and orthonormality, $∥ r^{'} ∥_{2}^{2} = ρ^{' 2} + ρ^{2} θ^{' 2}$ , so $E = \frac{1}{2} μ ρ^{' 2} + \frac{1}{2} μ ρ^{2} θ^{' 2} + U (ρ) = \frac{1}{2} μ ρ^{' 2} + \frac{J^{2}}{2 μ ρ^{2}} + U (ρ) = \frac{1}{2} μ ρ^{' 2} + U_{J}^{e f f} (ρ)$ by step 2.2.

4.1step 3.1step 3.2L4L6algebra∎

The right-hand side is differentiable on $I$ and its derivative is $μ ρ^{'} ρ^{''} + (U_{J}^{e f f})^{'} (ρ) ρ^{'} = ρ^{'} (μ ρ^{''} + (U_{J}^{e f f})^{'} (ρ)) = 0$ by step 3.1 (the equation holds at every point of $I$ ); hence $E$ is constant on $I$ by [L6].

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Circular orbits of a non-radial central-force trajectory: the critical-point condition and the squared frequency of the linearised radial equation

Statement

Let $μ > 0$ , let $J_{0} \subseteq (0, \infty)$ be an open interval, let $U : J_{0} \to R$ be of class $C^{2}$ , let $I$ be an interval with at least two points, and let $r : I \to R^{3}$ be a central-force trajectory for $(μ, U)$ with constant nonzero angular momentum $j = J e_{3}$ , $J > 0$ , and polar data $(ρ, θ)$ as in Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation, so that $ρ^{2} θ^{'} = J / μ$ and $μ ρ^{''} = - (U_{J}^{e f f})^{'} (ρ)$ on $I$ . Fix $r_{0} \in J_{0}$ . Then:

(Necessity.) If $ρ \equiv r_{0}$ is constant on $I$ , then $(U_{J}^{e f f})^{'} (r_{0}) = 0$ and $θ^{'} \equiv J / (μ r_{0}^{2})$ is constant; the trajectory is the uniform rotation $r (t) = r_{0} (\cos θ (t) e_{1} + \sin θ (t) e_{2})$ .
(Existence.) Conversely, if $(U_{J}^{e f f})^{'} (r_{0}) = 0$ , then for every $θ_{0} \in R$ the curve $r^{\circ} (t) : = r_{0} (\cos (θ_{0} + ω t) e_{1} + \sin (θ_{0} + ω t) e_{2}), ω : = \frac{J}{μ r_{0}^{2}},$ is a central-force trajectory for $(μ, U)$ defined on $R$ , with angular momentum $j = J e_{3}$ and constant energy $E = \frac{1}{2} μ r_{0}^{2} ω^{2} + U (r_{0}) = U_{J}^{e f f} (r_{0})$ .
(Linearised radial equation.) Let $ρ$ be a solution of the reduced equation $μ ρ^{''} = - (U_{J}^{e f f})^{'} (ρ)$ on $I$ with $(U_{J}^{e f f})^{'} (r_{0}) = 0$ , and put $δ : = ρ - r_{0}$ . Then $μ δ^{''} = - (U_{J}^{e f f})^{''} (r_{0}) δ + g (δ), g (x) : = (U_{J}^{e f f})^{''} (r_{0}) x - (U_{J}^{e f f})^{'} (r_{0} + x),$ where $g (x) / x \to 0$ as $x \to 0$ ; if moreover $(U_{J}^{e f f})^{''} (r_{0}) > 0$ , then the associated linearised equation $μ δ^{''} = - (U_{J}^{e f f})^{''} (r_{0}) δ$ has exactly the solutions $δ (t) = A \cos (ω_{l i n} (t - t_{0})) + B \sin (ω_{l i n} (t - t_{0}))$ with $A, B \in R$ and $ω_{l i n} : = \sqrt{(U_{J}^{e f f})^{''} (r_{0}) / μ} > 0$ , for any $t_{0} \in I$ . This is a statement about the quadratic radial approximation; no nonlinear stability of the circular orbit is asserted.

Facts & Assumptions

Given: The central-force data with constant nonzero angular momentum $j = J e_{3}$ and polar data $(ρ, θ)$ from Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation, a radius $r_{0} \in J_{0}$ , and the functions $n (θ) : = \cos θ e_{1} + \sin θ e_{2}$ , $n_{⊥} (θ) : = - \sin θ e_{1} + \cos θ e_{2}$ .

[L1]

Polar reduction: $r = ρ n (θ)$ with $ρ = ∣ r ∣$ , $ρ^{2} θ^{'} = J / μ$ , $μ ρ^{''} = - (U_{J}^{e f f})^{'} (ρ)$ , and the energy is constant (Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation).

[L2]

Derivative rules: the chain rule, the product rule and the trigonometric derivatives $(\cos)^{'} = - \sin$ , $(\sin)^{'} = \cos$ ; the cross product differentiates by the product rule; $e_{1} \times e_{2} = e_{3}$ and $e_{2} \times e_{1} = - e_{3}$ and the cross product is bilinear (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 derivatives of sine and cosine are cosine and minus sine, Differentiation of sums, scalar multiples, inner products, cross products and squared norms of differentiable curves in $R^{3}$ , Oriented orthonormal triads in $R^{3}$ : existence over any unit vector, and the orthonormal expansion with no fourth orthogonal direction).

[L3]

$\sin^{2} + \cos^{2} = 1$ ; and for the linearised equation: the solutions of $v^{''} + ω^{2} v = 0$ on an interval with $ω > 0$ are exactly $A \cos (ω (t - t_{0})) + B \sin (ω (t - t_{0}))$ (Parity and the Pythagorean identity for sine and cosine, Solutions of $v^{''} + ω^{2} v = 0$ are exactly the functions $A \cos (ω (t - t_{0})) + B \sin (ω (t - t_{0}))$ , $C^{k}$ Euclidean maps and diffeomorphisms).

[L4]

The derivative of $(U_{J}^{e f f})^{'}$ at $r_{0}$ is $(U_{J}^{e f f})^{''} (r_{0})$ : for every $ε > 0$ there is $η > 0$ with $∣ (U_{J}^{e f f})^{'} (r_{0} + x) - (U_{J}^{e f f})^{'} (r_{0}) - (U_{J}^{e f f})^{''} (r_{0}) x ∣ \leq ε ∣ x ∣$ for $∣ x ∣ < η$ (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).

Proof

technique · direct

1.1givenL1algebra

Clause 1: if $ρ \equiv r_{0}$ then $ρ^{''} = 0$ , so the reduced equation of [L1] gives $(U_{J}^{e f f})^{'} (r_{0}) = 0$ ; and $θ^{'} = J / (μ ρ^{2}) = J / (μ r_{0}^{2})$ is constant by [L1].

1.2givenL2L3algebraconstruct

Clause 2, the kinematics: for $r^{\circ} (t) = r_{0} n (θ_{0} + ω t)$ with $ω = J / (μ r_{0}^{2})$ , [L2] gives $(r^{\circ})^{'} = r_{0} ω n_{⊥} (θ_{0} + ω t)$ and $(r^{\circ})^{''} = - r_{0} ω^{2} n (θ_{0} + ω t)$ ; hence $j = μ r^{\circ} \times (r^{\circ})^{'} = μ r_{0}^{2} ω e_{3} = J e_{3}$ and $∥ (r^{\circ})^{'} ∥_{2}^{2} = r_{0}^{2} ω^{2}$ , using [L2] and [L3].

1.3givenL1L4algebra

Clause 3, the remainder: by [L4] applied to $(U_{J}^{e f f})^{'}$ at $r_{0}$ , for every $ε > 0$ there is $η > 0$ with $∣ g (x) ∣ \leq ε ∣ x ∣$ for $∣ x ∣ < η$ , where $g (x) = (U_{J}^{e f f})^{''} (r_{0}) x - (U_{J}^{e f f})^{'} (r_{0} + x)$ and $(U_{J}^{e f f})^{'} (r_{0}) = 0$ ; that is, $g (x) / x \to 0$ . Substituting $ρ = r_{0} + δ$ into the reduced equation of [L1] and expanding gives $μ δ^{''} = - (U_{J}^{e f f})^{'} (r_{0} + δ) = - (U_{J}^{e f f})^{''} (r_{0}) δ + g (δ)$ .

2.1givenstep 1.2L1algebra

Clause 2, the equation of motion: from step 1.2 and $ω^{2} = J^{2} / (μ^{2} r_{0}^{4})$ , $μ (r^{\circ})^{''} = - μ r_{0} ω^{2} n = - \frac{J^{2}}{μ r_{0}^{3}} n,$ while $- (U^{'} (r_{0}) / r_{0}) r^{\circ} = - U^{'} (r_{0}) n$ ; the two agree because $(U_{J}^{e f f})^{'} (r_{0}) = 0$ means $U^{'} (r_{0}) = J^{2} / (μ r_{0}^{3})$ . The energy is $\frac{1}{2} μ r_{0}^{2} ω^{2} + U (r_{0}) = \frac{J^{2}}{2 μ r_{0}^{2}} + U (r_{0}) = U_{J}^{e f f} (r_{0})$ , and it is constant because the equation is satisfied and the trajectory is periodic; more directly it equals the constant right-hand side of clause 4 of [L1] along this motion.

3.1step 1.3L3algebra∎

Clause 3, the linearised solutions: if $(U_{J}^{e f f})^{''} (r_{0}) > 0$ then $ω_{l i n}^{2} = (U_{J}^{e f f})^{''} (r_{0}) / μ > 0$ and the linearised equation reads $δ^{''} + ω_{l i n}^{2} δ = 0$ ; by [L3] its solutions on $I$ are exactly the displayed trigonometric functions for any $t_{0} \in I$ .

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Binet's equation for the attractive inverse-square field: $d^{2} u / d φ^{2} + u = μ κ / J^{2}$ for $u = 1 / ρ$

Statement

Let $μ > 0$ , $κ > 0$ , let $U : (0, \infty) \to R$ be $U (ρ) : = - κ / ρ$ (of class $C^{2}$ ), let $I$ be an interval with at least two points, and let $r : I \to R^{3}$ be a central-force trajectory for $(μ, U)$ with constant angular momentum $j = J e_{3}$ , $J > 0$ . Let $(ρ, θ)$ be polar data as in Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation, let $I^{\circ}$ be the interior of $I$ (a nonempty interval with at least two points, or empty if $I$ is a singleton, which is excluded), and put $W : = θ (I^{\circ})$ , an interval. Then:

$θ : I^{\circ} \to W$ is a $C^{1}$ bijection with inverse $σ : W \to I^{\circ}$ of class $C^{1}$ , and the function $u : = 1 / (ρ \circ σ) : W \to (0, \infty)$ is of class $C^{2}$ ;
on $W$ , $u^{''} + u = μ κ / J^{2}$ (derivatives with respect to the variable of $W$ );
the auxiliary identity $ρ^{'} = - \frac{J}{μ} u^{'} \circ θ$ holds on $I^{\circ}$ ;
the energy satisfies $E = \frac{J^{2}}{2 μ} ((u^{'})^{2} + u^{2}) - κ u$ on $W$ , where $E$ is the constant energy of the trajectory.

Facts & Assumptions

Given: The inverse-square central-force trajectory with $j = J e_{3}$ , $J > 0$ , and polar data $(ρ, θ)$ , the interval $W = θ (I^{\circ})$ , and the associated functions $σ$ and $u$ .

[L1]

Polar reduction: on all of $I$ , $r = ρ n (θ)$ , $ρ^{2} θ^{'} = J / μ$ (so $θ^{'} > 0$ and $θ$ is strictly increasing), $μ ρ^{''} = - (U_{J}^{e f f})^{'} (ρ)$ with $U_{J}^{e f f} (ρ) = - κ / ρ + J^{2} / (2 μ ρ^{2})$ , and the energy $E$ is constant (Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation).

[L2]

For a continuous injective strictly monotone function on an interval with nonvanishing derivative, the inverse is differentiable with $σ^{'} (φ) = 1 / θ^{'} (σ (φ))$ (Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ , Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, 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, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length); the interior $I^{\circ}$ of an interval is an interval, and $θ (I^{\circ}) = W$ is an interval with nonempty interior.

[L3]

The chain rule and the algebra of derivatives compute the derivatives of composites and quotients of $C^{k}$ functions (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 $g (c) \neq 0$ , $C^{k}$ Euclidean maps and diffeomorphisms).

Proof

technique · direct

1.1givenL1L2L3algebra

By [L1], $ρ^{''} = - (U_{J}^{e f f})^{'} (ρ) / μ$ on $I^{\circ}$ . Since $U_{J}^{e f f}$ is $C^{2}$ and $ρ$ is continuous, the right-hand side is continuous, so $ρ$ is $C^{2}$ there. Thus $θ^{'} = J / (μ ρ^{2})$ is $C^{1}$ and positive. By [L2], $θ : I^{\circ} \to W$ has a $C^{1}$ inverse $σ$ with $σ^{'} = 1 / (θ^{'} \circ σ)$ ; the right-hand side is $C^{1}$ by [L3], so $σ$ is $C^{2}$ . Therefore $u : = 1 / (ρ \circ σ)$ is a quotient of $C^{2}$ functions with nonvanishing denominator and is $C^{2}$ on $W$ .

2.1step 1.1L1L3

Chain rule along the trajectory: for $t \in I^{\circ}$ , $u (θ (t)) = 1 / ρ (t)$ , so [L3] gives $u^{'} (θ (t)) θ^{'} (t) = - ρ^{'} (t) / ρ (t)^{2}$ ; multiplying by $ρ^{2}$ and using $ρ^{2} θ^{'} = J / μ$ from [L1], $ρ^{'} (t) = - ρ (t)^{2} θ^{'} (t) u^{'} (θ (t)) = - \frac{J}{μ} u^{'} (θ (t)),$ which is clause 3.

3.1step 1.1step 2.1L1L3algebra

Differentiating clause 3 with [L3] and using $θ^{'} = J / (μ ρ^{2})$ again gives $ρ^{''} (t) = - \frac{J}{μ} u^{''} (θ (t)) θ^{'} (t) = - \frac{J^{2}}{μ^{2} ρ (t)^{2}} u^{''} (θ (t)) (t \in I^{\circ}) .$ Since $W = θ (I^{\circ})$ , every $φ \in W$ equals $θ (t)$ for some $t \in I^{\circ}$ , so this identity supplies the corresponding value of $u^{''}$ at every point of $W$ .

4.1step 1.1step 3.1L1algebra

The reduction equation of [L1] for $U (ρ) = - κ / ρ$ reads $μ ρ^{''} = - κ / ρ^{2} + J^{2} / (μ ρ^{3})$ . Substituting steps 2.1 and 3.1, with $ρ = 1 / u$ , gives $- \frac{J^{2}}{μ ρ^{2}} u^{''} = - \frac{κ}{ρ^{2}} + \frac{J^{2}}{μ ρ^{3}} ⟹ u^{''} = \frac{μ κ}{J^{2}} - u,$ that is, $u^{''} + u = μ κ / J^{2}$ on $W$ , which is clause 2.

5.1step 1.1step 2.1L1L3algebra∎

Clause 4: the energy identity of [L1] gives $E = \frac{1}{2} μ ρ^{' 2} + U_{J}^{e f f} (ρ) = \frac{1}{2} μ ρ^{' 2} - \frac{κ}{ρ} + \frac{J^{2}}{2 μ ρ^{2}}$ . Substituting $ρ = 1 / u$ and clause 3, $\frac{1}{2} μ ρ^{' 2} = \frac{J^{2}}{2 μ} (u^{'} \circ θ)^{2}$ and $\frac{J^{2}}{2 μ ρ^{2}} = \frac{J^{2}}{2 μ} u^{2}$ ; hence $E = \frac{J^{2}}{2 μ} ((u^{'} \circ θ)^{2} + u (θ)^{2}) - κ u (θ)$ along the trajectory. Since $θ$ ranges over $W$ , the displayed identity between the constant $E$ and the continuous function $\frac{J^{2}}{2 μ} ((u^{'})^{2} + u^{2}) - κ u$ on $W$ holds on $W$ .

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

The Kepler orbit equation is a conic: $u = c (1 + e \cos (φ - θ_{0}))$ with $E = (μ κ^{2} / 2 J^{2}) (e^{2} - 1)$

Statement

Let $μ > 0$ , $κ > 0$ , $U (ρ) = - κ / ρ$ on $(0, \infty)$ , let $r : I \to R^{3}$ be a central-force trajectory for $(μ, U)$ with constant angular momentum $j = J e_{3}$ , $J > 0$ and constant energy $E$ (Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential), and let $u : W \to (0, \infty)$ , $θ$ and $σ$ be as in Binet's equation for the attractive inverse-square field: $d^{2} u / d φ^{2} + u = μ κ / J^{2}$ for $u = 1 / ρ$ , with $W = θ (I^{\circ})$ . Write $c : = μ κ / J^{2} > 0$ and $p : = 1 / c = J^{2} / (μ κ)$ . Then:

there are $e \geq 0$ and $θ_{0} \in R$ with $u (φ) = c (1 + e \cos (φ - θ_{0})) (φ \in W),$ and for $t \in I^{\circ}$ accordingly $ρ (t) = p / (1 + e \cos (θ (t) - θ_{0}))$ ;
the energy and the eccentricity are related by $E = \frac{μ κ^{2}}{2 J^{2}} (e^{2} - 1);$ consequently $E < 0 ⟺ e < 1$ , $E = 0 ⟺ e = 1$ and $E > 0 ⟺ e > 1$ .

Facts & Assumptions

Given: The inverse-square central-force trajectory with constant angular momentum $j = J e_{3}$ , $J > 0$ , constant energy $E$ , the solution $u$ of Binet's equation on the interval $W$ , and the constants $c = μ κ / J^{2}$ , $p = 1 / c$ .

[L1]

Binet's equation and the auxiliary identities: $u^{''} + u = μ κ / J^{2} = c$ on $W$ , $W$ is an interval, and $E = \frac{J^{2}}{2 μ} ((u^{'})^{2} + u^{2}) - κ u$ on $W$ (Binet's equation for the attractive inverse-square field: $d^{2} u / d φ^{2} + u = μ κ / J^{2}$ for $u = 1 / ρ$ ).

[L2]

Solutions of $v^{''} + v = 0$ on an interval with at least two points are $A \cos (φ - φ_{0}) + B \sin (φ - φ_{0})$ for fixed $φ_{0}$ in that interval (Solutions of $v^{''} + ω^{2} v = 0$ are exactly the functions $A \cos (ω (t - t_{0})) + B \sin (ω (t - t_{0}))$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

Subtraction formula: $\cos (α - ψ) = \cos α \cos ψ + \sin α \sin ψ$ ; $(t \mapsto (\cos t, \sin t))$ is a bijection $[0, 2 π) \to S^{1}$ ; $\sin^{2} + \cos^{2} = 1$ (The subtraction formulas for sine and cosine, $t \mapsto (\cos t, \sin t)$ is a bijection from $[0, 2 π)$ onto the real unit circle, Parity and the Pythagorean identity for sine and cosine).

[L4]

Chain rule and algebra of derivatives for the differentiated expressions below (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 $g (c) \neq 0$ ).

Proof

technique · direct

1.1L1L2algebra

Note $u^{''} + u = c$ is equivalent to $(u - c)^{''} + (u - c) = 0$ on $W$ by linearity of the second derivative. Since $W$ is an interval and Binet's equation holds on it (it contains at least two points because $I^{\circ}$ has at least two points), [L2] gives reals $A, B$ with $u (φ) - c = A \cos (φ - φ_{0}) + B \sin (φ - φ_{0}) (φ \in W)$ for a fixed $φ_{0} \in W$ .

2.1step 1.1L3algebra

Put $R : = \sqrt{A^{2} + B^{2}} \geq 0$ . If $R > 0$ , then $(A / R, B / R)$ is a unit vector, so by [L3] there is $ψ \in [0, 2 π)$ with $\cos ψ = A / R$ , $\sin ψ = B / R$ ; if $R = 0$ take $ψ : = 0$ . Then for every $φ$ , [L3] gives $A \cos (φ - φ_{0}) + B \sin (φ - φ_{0}) = R \cos (φ - φ_{0} - ψ) .$ Setting $θ_{0} : = φ_{0} + ψ$ and $e : = R / c = R J^{2} / (μ κ) \geq 0$ , step 1.1 becomes $u (φ) = c + e c \cos (φ - θ_{0}) = c (1 + e \cos (φ - θ_{0}))$ on $W$ , which is the first assertion; the polar form $ρ = p / (1 + e \cos (θ - θ_{0}))$ follows by inverting $u = 1 / ρ$ on $I^{\circ}$ .

3.1step 2.1L1L4algebra

Energy: by [L1] and step 2.1, with $ψ (φ) : = φ - θ_{0}$ , $u = c (1 + e \cos ψ)$ and $u^{'} = - c e \sin ψ$ (differentiate with [L4]); hence $(u^{'})^{2} + u^{2} = c^{2} (e^{2} \sin^{2} ψ + 1 + 2 e \cos ψ + e^{2} \cos^{2} ψ) = c^{2} (1 + 2 e \cos ψ + e^{2}),$ and $κ u = κ c (1 + e \cos ψ)$ . Substituting into [L1]: $E = \frac{J^{2} c^{2}}{2 μ} (1 + 2 e \cos ψ + e^{2}) - κ c (1 + e \cos ψ) .$ With $c = μ κ / J^{2}$ one has $\frac{J^{2} c^{2}}{2 μ} = \frac{μ κ^{2}}{2 J^{2}}$ and $κ c = \frac{μ κ^{2}}{J^{2}}$ , so $E = \frac{μ κ^{2}}{2 J^{2}} [(1 + 2 e \cos ψ + e^{2}) - 2 (1 + e \cos ψ)] = \frac{μ κ^{2}}{2 J^{2}} (e^{2} - 1) .$

4.1step 3.1algebra∎

Classification: since $μ κ^{2} / (2 J^{2}) > 0$ , the sign of $E$ equals the sign of $e^{2} - 1$ ; with $e \geq 0$ this is negative exactly for $e < 1$ , zero exactly for $e = 1$ and positive exactly for $e > 1$ .

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

The one-dimensional action variable of a family of two-turning-point energy levels

Definition

Let $J_{0} \subseteq R$ be an open interval, let $U : J_{0} \to R$ be continuous (Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point), let $m > 0$ , let $E_{0} < E_{1}$ be reals, and let $a, b : (E_{0}, E_{1}) \to J_{0}$ be functions with

$a (E) < b (E), U (a (E)) = U (b (E)) = E, U (q) < E (a (E) < q < b (E)) (E_{0} < E < E_{1}) .$

The action variable of this datum is the function $J_{a} : (E_{0}, E_{1}) \to R$ ,

$J_{a} (E) : = \frac{1}{π} \int_{a (E)}^{b (E)} \sqrt{2 m (E - U (q))} d q,$

the integral being the Riemann integral on the compact interval $[a (E), b (E)]$ (The lower and upper Darboux integrals of a bounded $f$ on $[a, b]$ as $\sup_{P} L (f, P)$ and $\inf_{P} U (f, P)$ , Darboux integrability as their equality, and the notation $\int_{a}^{b} f$ ). It is defined because $q \mapsto \sqrt{2 m (E - U (q))}$ is continuous on $[a (E), b (E)]$ : the radicand is a continuous nonnegative function there (it vanishes exactly at the two endpoints). Existence and uniqueness of nonnegative roots follow from Square roots exist: a unique $\sqrt{a} \geq 0$ with $(\sqrt{a})^{2} = a$ ; the positives are ${x^{2} : x \neq 0}$ . For $x, y \geq 0$ , if $s = \sqrt{x} \geq t = \sqrt{y}$ , then $(s - t)^{2} \leq (s - t) (s + t) = x - y$ ; interchanging $x, y$ gives $∣ \sqrt{x} - \sqrt{y} ∣^{2} \leq ∣ x - y ∣$ . Thus $∣ x - y ∣ < ε^{2}$ implies $∣ \sqrt{x} - \sqrt{y} ∣ < ε$ , proving continuity, including at zero. The continuous integrand is Riemann integrable by A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion.

Reading of the notation. The interval $[a (E), b (E)]$ is the classically allowed segment between the two turning points of the energy $E$ , and the closed curve $p = \pm \sqrt{2 m (E - U (q))}$ encloses the area $A (E) = 2 \int_{a (E)}^{b (E)} \sqrt{2 m (E - U (q))} d q$ (the upper graph minus the lower graph), so $J_{a} (E) = A (E) / (2 π)$ ; no physical interpretation of $m$ , $U$ or $E$ is a mathematical premise here, and no period is claimed by this definition.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

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 $\frac{1}{2} \int_{0}^{2 π} ϱ^{2}$

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $p > 0$ and $0 \leq e < 1$ , and put $a : = \frac{p}{1 - e^{2}} > 0, b : = a \sqrt{1 - e^{2}}, ϱ (φ) : = \frac{p}{1 + e \cos φ} (φ \in R) .$

$ϱ$ is $C^{1}$ , $2 π$ -periodic, and $\frac{p}{1 + e} \leq ϱ \leq \frac{p}{1 - e}$ .
Let $Q (x, y) : = (1 - e^{2}) x^{2} + y^{2} + 2 p e x - p^{2}$ and $E_{e l l} : = {(x, y) \in R^{2} : Q (x, y) \leq 0}$ . Then $E_{e l l}$ is Lebesgue measurable, and for the closed unit disc $\overline{B} \subseteq R^{2}$ and $T (u, v) : = (a u - a e, b v)$ one has $E_{e l l} = T [\overline{B}]$ ; consequently $λ_{2} (E_{e l l}) = π a b = π a^{2} \sqrt{1 - e^{2}}$ .
For every $φ$ , the ray ${s (\cos φ, \sin φ) : s > 0}$ meets $\partial E_{e l l}$ in the unique point $ϱ (φ) (\cos φ, \sin φ)$ ; that point satisfies $Q = 0$ , and $λ_{2} (E_{e l l}) = \frac{1}{2} \int_{0}^{2 π} ϱ (φ)^{2} d φ$ .

Thus the conic $Q = 0$ is the ellipse with centre $(- a e, 0)$ , semiaxes $a, b$ , its right focus at the origin, and enclosed area $π a b$ .

Facts & Assumptions

Given: The constants $p > 0$ , $0 \leq e < 1$ , $a = p / (1 - e^{2})$ , $b = a \sqrt{1 - e^{2}}$ , the profile $ϱ$ , and the region $E_{e l l}$ .

[L1]

Substitution and completion of the square: $Q (T (u, v)) = Q (a u - a e, b v) = a^{2} (1 - e^{2}) (u^{2} + v^{2} - 1)$ for all $(u, v) \in R^{2}$ ; equivalently $Q (x, y) = (1 - e^{2}) (x + p e / (1 - e^{2}))^{2} + y^{2} - p^{2} / (1 - e^{2})$ . [algebra]

[L2]

Assuming countable choice, a $C^{1}$ diffeomorphism satisfies the change-of-variables formula for $L^{1}$ functions; the polar map $Φ (φ, s) = (s \cos φ, s \sin φ)$ is a $C^{1}$ diffeomorphism of $(0, 2 π) \times (0, \infty)$ onto its image (injective there, with nonsingular derivative, hence a diffeomorphism onto its open image by the injective-regular theorem); and a continuous function on a product with a nonnegative area element may be integrated iteratively by Tonelli (A C^1 diffeomorphism satisfies the change-of-variables formula for L^1 functions, An injective regular $C^{1}$ map is a diffeomorphism onto its image, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, The Axiom of Countable Choice ( ${A C}_{ω}$ )).

[L3]

A line in $R^{2}$ is Lebesgue null, and so is every subset of it (Every affine hyperplane of $R^{n}$ , and hence every proper linear subspace, is Lebesgue null); an invertible linear map scales Lebesgue measure by the absolute determinant, and the closed unit disc has measure $π$ (A linear map $T$ of $R^{n}$ sends Lebesgue measurable sets to Lebesgue measurable sets, with $λ_{n} (T [E]) = ∣ \det T ∣ λ_{n} (E)$ when $T$ is invertible and $T [E]$ Lebesgue null when it is not, The closed form for the volume of the unit $n$ -ball).

[L4]

$\cos$ is differentiable with $(\cos)^{'} = - \sin$ , is $2 π$ -periodic, and ranges over $[- 1, 1]$ ; $\sin^{2} + \cos^{2} = 1$ (The derivatives of sine and cosine are cosine and minus sine, Signs, monotonicity intervals, and ranges of sine and cosine, Parity and the Pythagorean identity for sine and cosine); sums, products and quotients of $C^{1}$ functions with nonvanishing denominator are $C^{1}$ (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 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

Proof

technique · direct

1.1givenL4algebra

Clause 1: $ϱ$ is a quotient of $C^{1}$ functions with denominator $1 + e \cos φ \geq 1 - e > 0$ , hence $C^{1}$ by [L4]; it is $2 π$ -periodic because $\cos$ is; and $p / (1 + e) \leq ϱ \leq p / (1 - e)$ because $\cos φ \in [- 1, 1]$ .

1.2givenL1algebra

Clause 2, the identity $E_{e l l} = T [\overline{B}]$ : by [L1], for $(u, v) \in R^{2}$ , $Q (T (u, v)) = a^{2} (1 - e^{2}) (u^{2} + v^{2} - 1)$ , so $Q (T (u, v)) \leq 0$ if and only if $u^{2} + v^{2} \leq 1$ , that is, $(u, v) \in \overline{B}$ ; hence $(x, y) \in E_{e l l}$ exactly when $(x, y) = T (u, v)$ for some $(u, v) \in \overline{B}$ , since $T$ is a bijection of $R^{2}$ onto itself (its linear part is invertible, $\det d i a g (a, b) = a b > 0$ ).

1.3givenL1L4algebra

Clause 3, the radial description: for $s > 0$ put $(x, y) : = s (\cos φ, \sin φ)$ . Using [L4], $Q (x, y) = s^{2} (1 - e^{2} \cos^{2} φ) + 2 p e s \cos φ - p^{2}$ , a quadratic in $s$ with positive leading coefficient (since $1 - e^{2} \cos^{2} φ \geq 1 - e^{2} > 0$ ) and negative constant term $- p^{2} < 0$ ; hence the set of $s > 0$ with $Q \leq 0$ is the interval $(0, ϱ (φ)]$ : indeed $s = ϱ (φ)$ is the unique positive root of $Q = 0$ , because substituting $ϱ = p / (1 + e \cos φ)$ and multiplying by $(1 + e \cos φ)^{2}$ gives $p^{2} (1 - e^{2} \cos^{2} φ) + 2 p^{2} e \cos φ (1 + e \cos φ) - p^{2} (1 + e \cos φ)^{2} = 0,$ a direct expansion — and $Q (0) = - p^{2} < 0$ while $Q (s) \to + \infty$ as $s \to \infty$ , so the two roots have opposite signs and the positive one is $ϱ (φ)$ . Therefore $E_{e l l} = {s (\cos φ, \sin φ) : 0 \leq φ < 2 π, 0 \leq s \leq ϱ (φ)}$ , the boundary is the displayed profile, and each ray meets it exactly once.

2.1step 1.2L2L3algebra

Clause 2, the area: $T$ is an affine map with invertible linear part, hence a $C^{1}$ diffeomorphism of $R^{2}$ with $∣ \det D T (u, v) ∣ = a b$ everywhere; $E_{e l l} = {Q \leq 0}$ is closed (so Lebesgue measurable with indicator in $L^{1}$ ), and $T [\overline{B}] = E_{e l l}$ by step 1.2. The change-of-variables formula of [L2] applied to the indicator of $E_{e l l}$ gives $λ_{2} (E_{e l l}) = \int_{\overline{B}} a b d λ_{2} = a b λ_{2} (\overline{B})$ ; the closed unit disc has $λ_{2} (\overline{B}) = π$ by [L3], so $λ_{2} (E_{e l l}) = π a b = π a^{2} \sqrt{1 - e^{2}}$ .

3.1step 1.3step 2.1L2L3algebra∎

Clause 3, the polar area formula: let $W : = {(φ, s) : 0 < φ < 2 π, 0 < s \leq ϱ (φ)} \subseteq D : = (0, 2 π) \times (0, \infty)$ and $S : = Φ (W)$ , where $Φ (φ, s) = (s \cos φ, s \sin φ)$ is the restriction of the polar map of [L2]. The set $W$ is Borel because $ϱ$ is continuous, and $S$ is Borel because $Φ$ is a homeomorphism from $D$ onto its image. Since $S \subseteq E_{e l l}$ by step 1.3 and $E_{e l l}$ has finite measure by step 2.1, $1_{S}$ is integrable. The change-of-variables formula [L2] applied on $D$ gives $λ_{2} (S) = \int_{W} s d λ_{2} (φ, s) = \int_{0}^{2 π} ⁣ ⁣ \int_{0}^{ϱ (φ)} s d s d φ = \frac{1}{2} \int_{0}^{2 π} ϱ (φ)^{2} d φ,$ the middle equality by Tonelli [L2] applied to the nonnegative function $(φ, s) \mapsto s 1_{W}$ . By step 1.3, $E_{e l l} ∖ S$ is contained in the closed positive $x$ -axis, including the origin, a subset of a line and hence null by [L3]. Therefore $λ_{2} (E_{e l l}) = λ_{2} (S) = \frac{1}{2} \int_{0}^{2 π} ϱ (φ)^{2} d φ$ , which is clause 3.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

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

Statement

Assume the Axiom of Countable Choice (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $μ > 0$ , $κ > 0$ , $U (ρ) = - κ / ρ$ on $(0, \infty)$ , and let $r : I \to R^{3}$ be a central-force trajectory for $(μ, U)$ with constant angular momentum $j = J e_{3}$ , $J > 0$ , and constant energy $E < 0$ . Let $e \in [0, 1)$ , $θ_{0} \in R$ , $p = J^{2} / (μ κ)$ , $a = p / (1 - e^{2})$ , $b = a \sqrt{1 - e^{2}}$ , and $ϱ (φ) : = p / (1 + e \cos (φ - θ_{0}))$ be 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 let $(ρ, θ)$ be polar data as in Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation. Then:

(Completeness.) $r$ extends uniquely to a central-force trajectory for $(μ, U)$ defined on all of $R$ ; along it $ρ (t) \in [p / (1 + e), p / (1 - e)]$ for every $t$ , and $ρ$ and $ρ^{'}$ are bounded. Consequently every maximal central-force trajectory with these data has domain $R$ .
(Periodicity.) $θ : R \to R$ is a strictly increasing $C^{1}$ bijection with $\frac{J}{μ} (\frac{1 - e}{p})^{2} \leq θ^{'} \leq \frac{J}{μ} (\frac{1 + e}{p})^{2}$ ; if $t_{0} \in R$ and $T > 0$ is determined by $θ (t_{0} + T) = θ (t_{0}) + 2 π$ , then $r (t + T) = r (t)$ and $r^{'} (t + T) = r^{'} (t)$ for all $t \in R$ .
(Period value.) With $r_{\max} : = p / (1 - e)$ and $r_{\min} : = p / (1 + e)$ , $T = \frac{2 π μ a b}{J} = 2 π \sqrt{\frac{μ}{κ}} a^{3 / 2}, T^{2} = \frac{4 π^{2} μ a^{3}}{κ} .$

Facts & Assumptions

Given: The inverse-square central-force trajectory with constant nonzero angular momentum and negative energy, its conic data from 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 its polar data.

[L1]

Polar reduction: $r = ρ n (θ)$ with $ρ = ∣ r ∣$ , $ρ^{2} θ^{'} = J / μ$ , $μ ρ^{''} = - (U_{J}^{e f f})^{'} (ρ)$ with $U_{J}^{e f f} (ρ) = - κ / ρ + J^{2} / (2 μ ρ^{2})$ , so $(U_{J}^{e f f})^{'} (ρ) = κ / ρ^{2} - J^{2} / (μ ρ^{3})$ is $C^{1}$ on $(0, \infty)$ ; the energy is constant (Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation).

[L2]

Conic data: for $t$ in the interior of $I$ , $ρ (t) = ϱ (θ (t))$ with $ϱ$ $C^{1}$ and $2 π$ -periodic, $ϱ \in [r_{\min}, r_{\max}]$ , and $E < 0 ⟺ e < 1$ (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 $\frac{1}{2} \int_{0}^{2 π} ϱ^{2}$ ).

[L3]

Compact continuation: for $z^{'} = F (z)$ with $F$ continuous and locally Lipschitz on an open $U \subseteq R^{n}$ , a solution whose values near a finite endpoint lie in a compact subset of $U$ extends past that endpoint; solutions agreeing at one time agree on every common interval containing it (Compact continuation for locally Lipschitz ODEs: a solution whose graph stays in a compact subset of the domain near a finite endpoint extends past it, Picard-Lindelöf local existence and uniqueness for first-order systems, First-order systems, initial value problems, and solutions on intervals, Local Lipschitz continuity in the state variable, locally uniform in time and parameters).

[L4]

Chain rule and derivative algebra: $ρ = ϱ \circ θ$ gives $ρ^{'} = ϱ^{'} (θ) θ^{'}$ on the interior of $I$ , and $ϱ^{'}$ is $2 π$ -periodic; sums, products, quotients and compositions of $C^{1}$ functions are $C^{1}$ (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 $g (c) \neq 0$ , 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, $C^{k}$ Euclidean maps and diffeomorphisms).

[L5]

The inverse of a continuous injective function with nonvanishing derivative on an interval is differentiable with $τ^{'} = 1 / (θ^{'} \circ τ)$ (Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ ); continuous functions on intervals have primitives and integrals are additive over adjacent subintervals (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ , 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 $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; with the oriented form for arbitrary $a, b, c$ ).

[L6]

$\cos$ and $\sin$ have period $2 π$ (The zero sets of sine and cosine and the least positive common period 2 pi, The derivatives of sine and cosine are cosine and minus sine, Parity and the Pythagorean identity for sine and cosine).

Proof

technique · direct

1.1givenL1L2L4algebra

By [L1] and [L2], for $t$ in the interior of $I$ the function $φ (t) : = θ (t)$ satisfies $ρ (t) = ϱ (φ (t)) \in [r_{\min}, r_{\max}]$ and, by [L4], $ρ^{'} (t) = ϱ^{'} (φ (t)) θ^{'} (t)$ with $∣ θ^{'} ∣ = J / (μ ρ^{2}) \leq J / (μ r_{\min}^{})$ ; since $ϱ^{'}$ is continuous and $2 π$ -periodic, it is bounded on $R$ by $M_{1} : = \max ∣ ϱ^{'} ∣$ , so $∣ ρ^{'} ∣ \leq M : = M_{1} J / (μ r_{\min}^{})$ on the interior of $I$ ; by continuity the bounds for $ρ$ and $ρ^{'}$ hold on all of $I$ .

1.2givenL1L3algebra

Consider the state $z : = (ρ, ρ^{'}, θ)$ in the open set $U_{0} : = (0, \infty) \times R \times R$ with right-hand side $F (x) : = (x_{2}, - μ^{- 1} (U_{J}^{e f f})^{'} (x_{1}), J / (μ x_{1}^{2}))$ ; the radial equation of [L1] and $θ^{'} = J / (μ ρ^{2})$ give $z^{'} = F (z)$ . Here $(U_{J}^{e f f})^{'} (s) = κ / s^{2} - J^{2} / (μ s^{3})$ , so $F$ is locally Lipschitz: on a state box with $x_{1} \geq δ > 0$ , each $x_{1}^{- k}$ for $k = 2, 3$ satisfies $∣ x_{1}^{- k} - y_{1}^{- k} ∣ \leq k δ^{- k - 1} ∣ x_{1} - y_{1} ∣$ by factoring the difference of reciprocal powers, while the first component is linear. Along every solution of this system, $H (ρ, ρ^{'}) : = \frac{1}{2} μ ρ^{' 2} + U_{J}^{e f f} (ρ)$ is constant, since $H^{'} = ρ^{'} (μ ρ^{''} + (U_{J}^{e f f})^{'} (ρ)) = 0$ ; its value is the given $E < 0$ . With $p = J^{2} / (μ κ)$ , direct algebra gives $E - U_{J}^{e f f} (s) = \frac{κ (1 - e^{2})}{2 p s^{2}} (s - \frac{p}{1 + e}) (\frac{p}{1 - e} - s) .$ Thus every such solution has $r_{\min} = p / (1 + e) \leq ρ \leq r_{\max} = p / (1 - e)$ ; its energy identity bounds $ρ^{'}$ , and $θ^{'} = J / (μ ρ^{2})$ is bounded as well. On a bounded time interval, $θ$ is consequently bounded, so near every finite endpoint $z$ lies in a compact subset of $U_{0}$ . Compact continuation [L3] extends it past every finite right endpoint. For a finite left endpoint apply [L3] to $\hat{z} (s) : = z (- s)$ , which solves the time-reversed locally Lipschitz system ${\hat{z}}^{'} = - F (\hat{z})$ . Hence the maximal extension of $z$ is defined on all of $R$ . For $n (θ) : = \cos θ e_{1} + \sin θ e_{2}$ and $n_{⊥} (θ) : = - \sin θ e_{1} + \cos θ e_{2}$ , differentiating $r = ρ n (θ)$ gives $r^{''} = (ρ^{''} - ρ θ^{' 2}) n (θ) + (2 ρ^{'} θ^{'} + ρ θ^{''}) n_{⊥} (θ) .$ Since $θ^{'} = J / (μ ρ^{2})$ , $θ^{''} = - 2 ρ^{'} θ^{'} / ρ$ and the second coefficient vanishes; the radial equation then gives $μ (ρ^{''} - ρ θ^{' 2}) = - U^{'} (ρ)$ . Thus the reconstructed $r$ satisfies the central-force equation. For any other continuation, the product rule and the central-force equation give $j^{'} = μ (r^{'} \times r^{'} + r \times r^{''}) = 0$ , so its angular momentum vector agrees with the one on the overlap; cross-product orthogonality places it in the same plane. By [L1] it has polar data there, and shifting its angle lift by an integer multiple of $2 π$ makes the angle agree at a common time. The resulting state solves the same locally Lipschitz system with the same initial state, so uniqueness [L3] gives the same $ρ, θ$ and hence the same trajectory. This is clause 1.

2.1step 1.1step 1.2L1L2algebra

On $R$ , $ρ \in [r_{\min}, r_{\max}]$ and $θ^{'} = J / (μ ρ^{2}) \in [J (1 - e)^{2} / (μ p^{2}), J (1 + e)^{2} / (μ p^{2})]$ , since $r_{\min} = p / (1 + e)$ and $r_{\max} = p / (1 - e)$ . Applying the conic supplier [L2] to the global extension gives the same profile $ϱ$ : its sine and cosine coefficients agree with the original ones on the nonempty open angle interval $θ (I^{\circ})$ , so their differences vanish identically (evaluate the difference and its derivative at one point). Thus $ρ = ϱ \circ θ$ on $R$ . Hence $θ$ is strictly increasing with $θ^{'} (t) \geq c_{1} > 0$ , so $θ (t) \to \pm \infty$ as $t \to \pm \infty$ and $θ (R) = R$ .

3.1step 2.1L4L6algebra

Let $t_{0} \in R$ and let $T > 0$ satisfy $φ (t_{0} + T) = φ (t_{0}) + 2 π$ , i.e. $θ (t_{0} + T) = θ (t_{0}) + 2 π$ (existence and uniqueness of $T$ come from step 2.1). Then $ρ (t_{0} + T) = ϱ (φ (t_{0}) + 2 π) = ϱ (φ (t_{0})) = ρ (t_{0})$ by the $2 π$ -periodicity of $ϱ$ , and $ρ^{'} (t_{0} + T) = ϱ^{'} (φ (t_{0}) + 2 π) θ^{'} (t_{0} + T) = ϱ^{'} (φ (t_{0})) \frac{J}{μ ρ (t_{0} + T)^{2}} = ϱ^{'} (φ (t_{0})) \frac{J}{μ ρ (t_{0})^{2}} = ρ^{'} (t_{0}),$ using [L4], the $2 π$ -periodicity of $ϱ^{'}$ , and step 2.1. Also $n (θ (t_{0} + T)) = n (θ (t_{0}))$ by the periodicity of sine and cosine [L6], so $r (t_{0} + T) = r (t_{0})$ and $r^{'} (t_{0} + T) = r^{'} (t_{0})$ (differentiate $r = ρ n (θ)$ and use the equalities just proved).

4.1step 2.1step 3.1L2L4L5algebra

Change of variables in the defining relation of $T$ : $θ : [t_{0}, t_{0} + T] \to [θ (t_{0}), θ (t_{0}) + 2 π]$ is a continuous bijection with $θ^{'} > 0$ , so its inverse $θ^{- 1}$ is differentiable with $(θ^{- 1})^{'} = 1 / θ^{'}$ composed with $θ^{- 1}$ [L5]. Hence, using $θ^{'} = J / (μ ρ^{2})$ and $ρ = ϱ (θ)$ on the interval, $T = \int_{t_{0}}^{t_{0} + T} d t = \int_{θ (t_{0})}^{θ (t_{0}) + 2 π} \frac{1}{θ^{'}} d θ = \frac{μ}{J} \int_{θ (t_{0})}^{θ (t_{0}) + 2 π} ϱ (θ)^{2} d θ = \frac{μ}{J} \int_{0}^{2 π} ϱ (ψ)^{2} d ψ,$ the last equality by the $2 π$ -periodicity of $ϱ$ .

4.2step 1.2step 3.1L3L6algebra

Define $z_{T} (t) : = (ρ (t + T), ρ^{'} (t + T), θ (t + T) - 2 π)$ . Since the state field in step 1.2 is independent of $θ$ , $z_{T}$ solves the same equation as $z (t) : = (ρ (t), ρ^{'} (t), θ (t))$ . By step 3.1 and the definition of $T$ , $z_{T} (t_{0}) = z (t_{0})$ ; uniqueness [L3] gives $z_{T} = z$ on $R$ . Periodicity of sine and cosine then yields $r (t + T) = r (t)$ , and differentiation yields $r^{'} (t + T) = r^{'} (t)$ . This is clause 2.

5.1step 4.1L2algebra

By clause 3 of 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 $\frac{1}{2} \int_{0}^{2 π} ϱ^{2}$ , applied to the unshifted profile $\tilde{ϱ} (ψ) : = p / (1 + e \cos ψ)$ , with a translation of the integration variable by $θ_{0}$ and invariance of a full-period integral under translation, $\frac{1}{2} \int_{0}^{2 π} ϱ (ψ)^{2} d ψ = λ_{2} (E_{e l l}) = π a b$ , so step 4.1 gives $T = \frac{2 π μ a b}{J}$ .

6.1step 5.1algebra∎

Algebra: $p = J^{2} / (μ κ)$ and $p = a (1 - e^{2})$ by the definitions, so $J^{2} = μ κ a (1 - e^{2})$ and $J = \sqrt{μ κ a (1 - e^{2})}$ ; with $b = a \sqrt{1 - e^{2}}$ , $\frac{2 π μ a b}{J} = \frac{2 π μ a^{2} \sqrt{1 - e^{2}}}{\sqrt{μ κ a (1 - e^{2})}} = 2 π \sqrt{\frac{μ}{κ}} a^{3 / 2},$ and squaring gives $T^{2} = 4 π^{2} μ a^{3} / κ$ . This is clause 3.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

The action variable of a two-simple-turning-point family is $C^{1}$ with derivative $T (E) / (2 π)$ , the period quadrature over $2 π$

Statement

Let $J_{0} \subseteq R$ be an open interval, let $U : J_{0} \to R$ be of class $C^{3}$ ( $C^{k}$ Euclidean maps and diffeomorphisms), let $m > 0$ , let $E_{0} < E_{1}$ , and let $a, b : (E_{0}, E_{1}) \to J_{0}$ be functions with $a (E) < b (E)$ , $U (a (E)) = U (b (E)) = E$ , $U (q) < E$ for $a (E) < q < b (E)$ , and

$U^{'} (a (E)) < 0 < U^{'} (b (E)) (E_{0} < E < E_{1})$

(the energy $E$ has exactly the two simple turning points $a (E)$ , $b (E)$ in $J_{0}$ ). Then:

$a$ and $b$ are of class $C^{1}$ on $(E_{0}, E_{1})$ ;
the action variable $J_{a}$ of The one-dimensional action variable of a family of two-turning-point energy levels is of class $C^{1}$ on $(E_{0}, E_{1})$ , and for every $E \in (E_{0}, E_{1})$ $J_{a}^{'} (E) = \frac{1}{2 π} T (E), T (E) : = \sqrt{2 m} \int_{a (E)}^{b (E)} \frac{d q}{\sqrt{E - U (q)}} \in (0, \infty),$ where the integral is the improper integral over the open interval $(a (E), b (E))$ , convergent at both endpoints.

Facts & Assumptions

Given: The data $J_{0}, U, m, E_{0}, E_{1}, a, b$ of the statement, with $U$ of class $C^{3}$ , exactly two simple turning points, and $U < E$ in between.

[L1]

The Euclidean inverse function theorem: if $F$ is $C^{1}$ near $c$ and $D F (c)$ is invertible, then $F$ restricts to a $C^{1}$ bijection from a neighbourhood of $c$ onto a neighbourhood of $F (c)$ with $C^{1}$ inverse (The Euclidean inverse function theorem). A $C^{k}$ map with invertible derivative has a $C^{k}$ local inverse by the same theorem applied to the derivative and induction on smoothness; in the one-variable case $U^{'} (ξ) \neq 0$ makes $ξ \mapsto U (ξ)$ a $C^{3}$ diffeomorphism near $ξ$ onto its image.

[L2]

Leibniz's rule on a compact rectangle: for continuous $g, h$ on $[a, b] \times [c, d]$ with $x \mapsto g (x, t)$ differentiable on $(a, b)$ with derivative $h (x, t)$ , the function $x \mapsto \int_{c}^{d} g (x, t) d t$ is differentiable with derivative $\int_{c}^{d} h (x, t) d t$ ; with the chain rule this gives the variable-limit rule $\frac{d}{d E} \int_{c}^{z (E)} G (E, z) d z = G (E, z (E)) z^{'} (E) + \int_{c}^{z (E)} \partial_{E} G (E, z) d z$ (Leibniz's rule on a compact rectangle: an interior parameter derivative with a continuous extension may be passed through a Riemann integral, 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

[L3]

Substitution: for $φ$ differentiable on $[c, d]$ with integrable $φ^{'}$ and $f$ continuous on an interval containing $φ ([c, d])$ , $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ (Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ ).

[L4]

A continuous real function on a nonempty compact metric space attains a maximum (A continuous real-valued function on a nonempty compact metric space is bounded and attains a greatest and a least value); the integral is monotone and additive, and a continuous nonnegative function with a positive value has positive integral (If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , If $m \leq f \leq M$ on $[a, b]$ then $m (b - a) \leq L (f, P) \leq \underline{\int_{a}^{b}} f \leq \overline{\int_{a}^{b}} f \leq U (f, P) \leq M (b - a)$ for every partition $P$ ; in particular every constant function is integrable, with $\int_{a}^{b} c = c (b - a)$ , A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion).

[L5]

Continuous functions on intervals have primitives, sums, products and quotients with nonvanishing denominators of $C^{k}$ functions are $C^{k}$ , and the square root is $C^{1}$ on $(0, \infty)$ with derivative $1 / (2 \sqrt{})$ (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ , 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 $g (c) \neq 0$ , 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, 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ ).

Proof

technique · direct

1.1givenL1choosealgebra

Fix $E_{*} \in (E_{0}, E_{1})$ . Since $U^{'} (a (E_{*})) \neq 0$ and $U^{'} (b (E_{*})) \neq 0$ , the inverse function theorem [L1] applied to $U$ at $a (E_{*})$ and at $b (E_{*})$ gives $δ > 0$ and $C^{3}$ (hence $C^{1}$ ) functions $A, B : (E_{*} - δ, E_{*} + δ) \to J_{0}$ with $U (A (E)) = U (B (E)) = E$ , $A (E_{*}) = a (E_{*})$ and $B (E_{*}) = b (E_{*})$ . Shrinking $δ$ , continuity gives $A < B$ on the smaller interval. For every such $E$ , the two real numbers $A (E)$ and $B (E)$ are distinct solutions of $U (q) = E$ ; by hypothesis the only two solutions in $J_{0}$ are $a (E) < b (E)$ , so ${A (E), B (E)} = {a (E), b (E)}$ ; since $A < B$ and $a < b$ , this forces $a = A$ and $b = B$ on that neighbourhood. Hence $a$ and $b$ are $C^{1}$ near the arbitrary point $E_{*}$ , which is clause 1.

2.1givenstep 1.1constructalgebra

Fix $E_{*}$ again. Since $U^{'}$ is continuous and $U^{'} (a (E_{*})) < 0 < U^{'} (b (E_{*}))$ , choose disjoint neighborhoods of $a (E_{*})$ and $b (E_{*})$ on which $U^{'} < 0$ and $U^{'} > 0$ , respectively. Choose $q_{1}, q_{2}$ in these neighborhoods with $a (E_{*}) < q_{1} < q_{2} < b (E_{*})$ . By continuity of $a$ and $b$ from step 1.1, shrink $δ > 0$ so that for $∣ E - E_{*} ∣ < δ$ , $a (E)$ and $q_{1}$ remain in the first neighborhood, $q_{2}$ and $b (E)$ remain in the second, and $a (E) < q_{1} < q_{2} < b (E)$ . Thus $U$ is strictly decreasing on $[a (E), q_{1}]$ and strictly increasing on $[q_{2}, b (E)]$ . On this energy interval write $M (E) : = \int_{q_{1}}^{q_{2}} \sqrt{2 m (E - U (q))} d q, L (E) : = \int_{a (E)}^{q_{1}} \sqrt{2 m (E - U (q))} d q, R (E) : = \int_{q_{2}}^{b (E)} \sqrt{2 m (E - U (q))} d q,$ so that $J_{a} = (L + M + R) / π$ there.

3.1step 2.1L2L4L5algebra

The function $u (E, q) : = \sqrt{2 m (E - U (q))}$ is continuous on the compact rectangle $[E_{*} - δ, E_{*} + δ] \times [q_{1}, q_{2}]$ and $\partial_{E} u (E, q) = m / \sqrt{2 m (E - U (q))}$ is continuous there, because $U (q) < E_{*}$ for every $q \in [q_{1}, q_{2}]$ and $U$ is continuous on the compact interval, so that $E - U (q) \geq ϵ > 0$ on the rectangle after shrinking $δ$ [L4, L5]. Hence by [L2] the middle term is $C^{1}$ with $M^{'} (E) = \int_{q_{1}}^{q_{2}} \frac{m d q}{\sqrt{2 m (E - U (q))}} (∣ E - E_{*} ∣ < δ) .$

3.2step 2.1L1L3algebraconstruct

For the left leg, use the open interval about $a (E_{*})$ chosen in step 2.1, containing every segment $[a (E), q_{1}]$ and having $U^{'} < 0$ . On this interval $U$ is strictly monotone; its local $C^{3}$ inverses from [L1] agree on overlaps by injectivity, giving a $C^{3}$ inverse $ψ$ on the whole image interval. For $E$ near $E_{*}$ the substitution $s = U (q)$ [L3] gives $L (E) = \int_{U (q_{1})}^{E} \sqrt{2 m (E - s)} w (s) d s, w (s) : = - \frac{1}{U^{'} (ψ (s))} > 0,$ and $w$ is $C^{2}$ on the image interval; here $U (q_{1}) < E$ for $E$ near $E_{*}$ since $U (q_{1}) < E_{*}$ .

3.3step 2.1L1L2L3L5algebra

The right leg is treated identically: near $b (E_{*})$ one has $U^{'} > 0$ , the $C^{3}$ local inverse $ψ$ of $U$ gives $R (E) = \int_{U (q_{2})}^{E} \sqrt{2 m (E - s)} w (s) d s$ with $w : = 1 / U^{'} (ψ (s)) > 0$ of class $C^{2}$ , the substitution $s = E - z^{2}$ and the variable-limit Leibniz rule give $R^{'} (E) = \int_{q_{2}}^{b (E)} \frac{m d q}{\sqrt{2 m (E - U (q))}} .$

4.1step 3.2L2L3L5algebra

Substituting $s = E - z^{2}$ in the last integral [L3] turns the left leg into $L (E) = 2 \sqrt{2 m} \int_{0}^{z (E)} z^{2} w (E - z^{2}) d z, z (E) : = \sqrt{E - U (q_{1})} > 0,$ with $z$ of class $C^{1}$ and the integrand $G (E, z) : = 2 \sqrt{2 m} z^{2} w (E - z^{2})$ of class $C^{1}$ on a compact rectangle. By the variable-limit Leibniz rule [L2], $L^{'} (E) = G (E, z (E)) z^{'} (E) + \int_{0}^{z (E)} \partial_{E} G (E, z) d z = \sqrt{2 m} z (E) w (E - z (E)^{2}) + 2 \sqrt{2 m} \int_{0}^{z (E)} z^{2} w^{'} (E - z^{2}) d z .$

5.1step 4.1L3L5algebra

Integration by parts, equivalently the product rule applied to $z \mapsto z w (E - z^{2})$ , gives $\int_{0}^{z (E)} w (E - z^{2}) d z = z (E) w (E - z (E)^{2}) + 2 \int_{0}^{z (E)} z^{2} w^{'} (E - z^{2}) d z;$ substituting this into step 4.1 yields $L^{'} (E) = \sqrt{2 m} \int_{0}^{z (E)} w (E - z^{2}) d z = \int_{a (E)}^{q_{1}} \frac{m d q}{\sqrt{2 m (E - U (q))}},$ the last equality by the same substitutions read backwards [L3]: $\sqrt{2 m} \int_{0}^{z (E)} w (E - z^{2}) d z = \int_{a (E)}^{q_{1}} u (E, q)^{- 1} m d q$ .

6.1step 3.1step 5.1step 3.3L3L4algebra

Adding the three derivatives of steps 3.1, 5.1 and 3.3 gives $J_{a}^{'} (E) = \frac{1}{π} \int_{a (E)}^{b (E)} \frac{m d q}{\sqrt{2 m (E - U (q))}} = \frac{\sqrt{2 m}}{2 π} \int_{a (E)}^{b (E)} \frac{d q}{\sqrt{E - U (q)}} = \frac{T (E)}{2 π}$ for $∣ E - E_{*} ∣ < δ$ , with the middle integral understood improperly at $a (E)$ and $b (E)$ . The endpoint pieces converge by steps 3.2–5.1 and 3.3; additivity on interior truncations and passage to the endpoint limits justify splitting this improper integral at $q_{1}, q_{2}$ . The second equality uses $m / \sqrt{2 m} = \sqrt{2 m} / 2$ , and the last is the definition of $T (E)$ .

7.1step 6.1L4algebra∎

The integrand $q \mapsto 1 / \sqrt{E - U (q)}$ is continuous and strictly positive on the nonempty open interval $(a (E), b (E))$ , so $T (E) > 0$ by [L4]; and the derivative computed in step 6.1 is independent of the auxiliary $E_{*}$ , so the formula holds on all of $(E_{0}, E_{1})$ and $J_{a}$ is $C^{1}$ there. This is clause 2.

DefinitionDefinition: AI-adaptedProof: Not applicableOpen item page →

Scattering data: the incident flux measure, the solid-angle element, deflection branches and the cross-section density

Definition

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the Borel/Lebesgue measure foundation below.

Data. Let $μ > 0$ , let $U : (0, \infty) \to R$ be of class $C^{2}$ ( $C^{k}$ Euclidean maps and diffeomorphisms), let $R > 0$ satisfy $U (r) = 0$ for every $r \geq R$ (a finite-range potential), and let $E > 0$ and $J > 0$ be reals. Write

$V (r) : = U (r) + \frac{J^{2}}{2 μ r^{2}} (r > 0) .$

Incident flux measure. Let $ν$ be the Borel measure on the open rectangle $(0, \infty) \times (0, 2 π)$ given by the density $b$ against Lebesgue measure $λ_{2}$ ,

$ν (A) : = \int_{A} b d λ_{2} (b, φ) (A \subseteq (0, \infty) \times (0, 2 π) Borel),$

the integral being that of Integral over a measurable subset; the nonnegative Borel function $(b, φ) \mapsto b$ defines a measure by The indefinite integral of a nonnegative measurable function is a measure (Borel measurable and Lebesgue measurable functions on $R^{n}$ ). In the model this measure records a uniform incident beam: an annulus of impact parameters $[b_{1}, b_{2})$ contributes $π (b_{2}^{2} - b_{1}^{2})$ over the full azimuth range, or $\frac{1}{2} (b_{2}^{2} - b_{1}^{2})$ per unit azimuth.

Solid-angle element. Let $Ω$ be the Borel measure on $(0, π) \times (0, 2 π)$ with density $\sin$ against $λ_{2}$ ,

$Ω (B) : = \int_{B} \sin θ d λ_{2} (θ, φ) (B \subseteq (0, π) \times (0, 2 π) Borel),$

where $\sin$ is the function of Sine and cosine defined by their real power series. It is continuous, being differentiable by The derivatives of sine and cosine are cosine and minus sine. Also $\sin 0 = \sin π = 0$ by The zero sets of sine and cosine and the least positive common period 2 pi; strict increase on $[0, π / 2]$ and strict decrease on $[π / 2, π]$ (Signs, monotonicity intervals, and ranges of sine and cosine) give $\sin θ > 0$ for $0 < θ < π$ . Thus $(θ, φ) \mapsto \sin θ$ is a nonnegative Borel function (continuous preimages of Borel sets are Borel, A continuous map has Borel preimages of Borel sets), and The indefinite integral of a nonnegative measurable function is a measure establishes that $Ω$ is a measure. This is the standard solid-angle element written $\sin θ d θ d φ$ (the surface measure of the unit sphere in spherical coordinates, restricted to the open chart).

Deflection branch. A deflection branch is an interval $(a_{1}, a_{2}) \subseteq (0, \infty)$ with at least two points together with a function $θ : (a_{1}, a_{2}) \to (0, π)$ that is of class $C^{1}$ and satisfies $θ^{'} (b) \neq 0$ for every $b \in (a_{1}, a_{2})$ . Such a $θ$ is strictly monotone (the continuous nonvanishing derivative has constant sign, and 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 $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ gives strict monotonicity) and its image is an interval with endpoints $θ_{1} < θ_{2}$ (Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length); its inverse $b : (θ_{1}, θ_{2}) \to (a_{1}, a_{2})$ is differentiable with $b^{'} (θ) = 1 / θ^{'} (b (θ))$ , by the derivative-of-an-inverse theorem Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ applied to the continuous injective $θ$ (every point of $(a_{1}, a_{2})$ is a limit point of it, Limit point, isolated point, adherent point, derived set, and dense subset of $R$ , 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).

Cross-section density. The cross-section density of the branch is the function

$\frac{d σ}{d Ω} (θ) : = \frac{b (θ)}{\sin θ ∣ θ^{'} (b (θ)) ∣} (θ_{1} < θ < θ_{2}) .$

Mathematical status. The measures, branches and density are abstract data independent of the finite-range trajectory data $(μ, U, R, E, J)$ . A branch is not asserted to arise from any trajectory; the scattering theorems of this page identify the deflection angle computed from the central-force equation with the values of such a function under their hypotheses, and they compute the pushforward of $ν$ under the induced map. No counting of detected particles, and no empirical claim, is part of this definition.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Finite-range scattering: the simple outer turning point, the two global branches and the free tails

Statement

Let $μ > 0$ , let $U : (0, \infty) \to R$ be of class $C^{2}$ , let $R > 0$ satisfy $U (r) = 0$ for every $r \geq R$ , let $E > 0$ and $J > 0$ , and put $V (r) : = U (r) + J^{2} / (2 μ r^{2})$ (Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential). Suppose $r_{*} > 0$ satisfies

$V (r_{*}) = E, V^{'} (r_{*}) < 0, V (r) < E (r > r_{*}) .$

Let $r : I \to R^{3}$ be a central-force trajectory for $(μ, U)$ with constant nonzero angular momentum $j = J e_{3}$ , $J > 0$ , energy $E$ , and $∣ r (t) ∣ \geq r_{*}$ for all $t \in I$ , whose radial part attains the value $r_{*}$ at some $t_{*} \in I$ (Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation). Then:

$(ρ^{'})^{2} = \frac{2}{μ} (E - V (ρ))$ on $I$ , where $ρ = ∣ r ∣$ ; $ρ^{'} (t_{*}) = 0$ and $ρ^{''} (t_{*}) = - V^{'} (r_{*}) / μ > 0$ , so $t_{*}$ is a strict minimum of $ρ$ ;
$r$ extends uniquely to a central-force trajectory on all of $R$ ; along the extension $ρ \geq r_{*}$ , $ρ (t) > ρ (t_{*}) = r_{*}$ for $t \neq t_{*}$ , and $ρ$ is strictly decreasing on $(- \infty, t_{*}]$ and strictly increasing on $[t_{*}, \infty)$ ;
the time quadrature $τ (r) : = \int_{r_{*}}^{r} \frac{d s}{\sqrt{(2 / μ) (E - V (s))}} (r \geq r_{*})$ is finite for every $r$ and satisfies $ρ (t_{*} + τ (r)) = r$ for all $r \geq r_{*}$ ;
for every time $t_{1} \geq t_{*}$ with $ρ (t_{1}) > R$ the motion is exactly free from $t_{1}$ on: $r (t) = r (t_{1}) + (t - t_{1}) r^{'} (t_{1})$ for all $t \geq t_{1}$ , and $ρ$ is strictly increasing there; symmetrically, for $t_{2} \leq t_{*}$ with $ρ (t_{2}) > R$ , $r (t) = r (t_{2}) + (t - t_{2}) r^{'} (t_{2})$ for all $t \leq t_{2}$ .

Facts & Assumptions

Given: The finite-range central-force data $(μ, U, R)$ , the energy $E > 0$ , the angular momentum $j = J e_{3}$ with $J > 0$ , the simple outer turning point $r_{*}$ of $V$ , and a trajectory with $∣ r ∣ \geq r_{*}$ attaining $r_{*}$ at $t_{*}$ .

[L1]

Polar reduction and energy: $r = ρ n (θ)$ , $ρ = ∣ r ∣$ , $ρ^{2} θ^{'} = J / μ$ , $μ ρ^{''} = - V^{'} (ρ)$ , and the constant energy is $E = \frac{1}{2} μ (ρ^{'})^{2} + V (ρ)$ (Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation; here $V = U_{J}^{e f f}$ by Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential).

[L2]

Compact continuation and uniqueness for locally Lipschitz autonomous systems: solutions trapped in a compact subset of the open state domain have no finite endpoints, and two solutions with the same state at one time agree on every common interval (Compact continuation for locally Lipschitz ODEs: a solution whose graph stays in a compact subset of the domain near a finite endpoint extends past it, Picard-Lindelöf local existence and uniqueness for first-order systems, First-order systems, initial value problems, and solutions on intervals, Local Lipschitz continuity in the state variable, locally uniform in time and parameters).

[L3]

Substitution and monotone inversion: a $C^{1}$ strictly monotone function on an interval with nonvanishing derivative has a $C^{1}$ inverse with reciprocal derivative, and substitution converts $\int d t$ into $\int d ρ / ∣ ρ^{'} ∣$ on such a branch (Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ , 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, $C^{k}$ Euclidean maps and diffeomorphisms).

[L4]

Behaviour of the quadrature: $s \mapsto E - V (s)$ is continuous and strictly positive on $(r_{*}, \infty)$ , comparable to $s - r_{*}$ near $r_{*}$ (so $1 / \sqrt{E - V (s)}$ is integrable at $r_{*}$ ), and tends to $E > 0$ as $s \to \infty$ (so $τ (r) \to \infty$ , $∣ ρ^{'} ∣ \leq \sqrt{2 E / μ}$ whenever $ρ \geq R$ , since there $V (ρ) = J^{2} / (2 μ ρ^{2}) \geq 0$ ); continuous functions on intervals have primitives and the integral is additive over adjacent intervals (Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ , Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point, 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 $g (c) \neq 0$ ).

Proof

technique · direct

1.1givenL1algebra

Square the energy identity of [L1]: $\frac{1}{2} μ (ρ^{'})^{2} = E - V (ρ)$ , which is clause 1's first display. At a time with $ρ = r_{*}$ the right side is $E - V (r_{*}) = 0$ , so $ρ^{'} = 0$ ; then $μ ρ^{''} (t_{*}) = - V^{'} (r_{*}) > 0$ , so $t_{*}$ is a strict local minimum of $ρ$ , and in particular $ρ (t) > ρ (t_{*})$ for $t$ near but different from $t_{*}$ .

2.1step 1.1L1algebra

On the open set where $ρ > r_{*}$ , the right side of the energy identity is strictly positive, so $ρ^{'} \neq 0$ ; hence the sign of $ρ^{'}$ is constant on every connected component of ${t : ρ (t) > r_{*}}$ (a union of open intervals on which the continuous nonvanishing function $ρ^{'}$ has a sign).

2.2step 1.1L1L2L4algebra

Global extension: take the state $z : = (ρ, ρ^{'}, θ)$ in $(0, \infty) \times R \times R$ with right-hand side $F (x) : = (x_{2}, - μ^{- 1} V^{'} (x_{1}), J / (μ x_{1}^{2}))$ . This field is locally Lipschitz: near any $x_{1} > 0$ , continuity of $V^{''}$ bounds it by some $M$ on a small closed interval; the fundamental theorem for continuous $V^{''}$ gives $V^{'} (a) - V^{'} (b) = \int_{b}^{a} V^{''}$ , and the Darboux definition with $∣ V^{''} ∣ \leq M$ gives $∣ V^{'} (a) - V^{'} (b) ∣ \leq M ∣ a - b ∣$ . Also $x_{1}^{- 2}$ is locally Lipschitz by factoring the difference of reciprocal powers. Along every solution, $H (ρ, ρ^{'}) : = \frac{1}{2} μ (ρ^{'})^{2} + V (ρ)$ is constant by the radial equation. Also $ρ \geq r_{*}$ on every continuation: if it entered $ρ < r_{*}$ , continuity from $t_{*}$ would give a boundary time with $ρ = r_{*}$ and values below $r_{*}$ arbitrarily nearby on one side; there $H = E$ gives $ρ^{'} = 0$ and the equation gives $ρ^{''} = - V^{'} (r_{*}) / μ > 0$ , contradicting the strict local-minimum property. Thus $V (ρ) \leq E$ and $ρ^{' 2} = (2 / μ) (E - V (ρ))$ . The function $V$ is bounded below on $[r_{*}, \infty)$ because $U$ is continuous and vanishes beyond $R$ , so $ρ^{'}$ is bounded; also $0 < θ^{'} = J / (μ ρ^{2}) \leq J / (μ r_{*}^{2})$ . On every bounded time interval, $ρ$ and $θ$ are therefore bounded, so near every finite right endpoint $z$ lies in a compact subset of the state domain and [L2] continues it. For a finite left endpoint apply [L2] to $\hat{z} (s) : = z (- s)$ , which solves ${\hat{z}}^{'} = - F (\hat{z})$ . Hence the maximal extension of $z$ is defined on $R$ . Set $n (θ) : = \cos θ e_{1} + \sin θ e_{2}$ and $n_{⊥} (θ) : = - \sin θ e_{1} + \cos θ e_{2}$ . The relation $θ^{'} = J / (μ ρ^{2})$ gives $θ^{''} = - 2 ρ^{'} θ^{'} / ρ$ ; differentiating gives $r^{''} = (ρ^{''} - ρ θ^{' 2}) n (θ) + (2 ρ^{'} θ^{'} + ρ θ^{''}) n_{⊥} (θ),$ whose second coefficient vanishes. Using $μ ρ^{''} = - V^{'} (ρ)$ then yields $μ (ρ^{''} - ρ θ^{' 2}) = - U^{'} (ρ)$ , so the reconstructed curve satisfies the central-force equation. Any other continuation has the same angular momentum on the overlap: along it $j^{'} = μ (r^{'} \times r^{'} + r \times r^{''}) = 0$ by the central-force equation, so it remains in the same plane. By [L1] it has polar data there; shift its angle lift by an integer multiple of $2 π$ to match $θ$ at a common time. The resulting state solves the same locally Lipschitz system with the same initial state, so uniqueness [L2] gives the same $ρ, θ$ and hence the same trajectory.

3.1step 1.1step 2.1algebra

Shape of the set ${ρ > r_{*}}$ . This set is open in $R$ , and on each of its connected components $ρ^{'}$ has a constant sign (step 2.1). Every bounded component $(t_{0}, t_{1})$ satisfies $ρ (t_{0}) = ρ (t_{1}) = r_{*}$ (endpoints of a component of the open set ${ρ > r_{*}}$ lie in its complement, where $ρ \leq r_{*}$ , i.e. $ρ = r_{*}$ ); at both endpoints $ρ^{''} = - V^{'} (r_{*}) / μ > 0$ by step 1.1, so $ρ^{'}$ changes sign from negative to positive at each of them, whence just inside the component the sign is positive near $t_{0}$ and negative near $t_{1}$ , contradicting constancy of the sign. Hence ${ρ > r_{*}}$ has no bounded component. Since $t_{*}$ is an isolated zero of $ρ - r_{*}$ (a strict local minimum by step 1.1) and points just left and just right of $t_{*}$ belong to ${ρ > r_{*}}$ , the two components adjacent to $t_{*}$ are $(- \infty, t_{*})$ and $(t_{*}, + \infty)$ : an open subset of $R$ has at most two unbounded components, and each of these two is unbounded on its outer side (a bounded component is impossible, and it cannot end on the inner side at a point other than $t_{*}$ , since its inner endpoint would be an endpoint of a component at which $r_{*}$ is attained, producing a second strict minimum and the same sign contradiction as above). Therefore $ρ > r_{*}$ exactly for $t \neq t_{*}$ , $ρ^{'} < 0$ on $(- \infty, t_{*})$ and $ρ^{'} > 0$ on $(t_{*}, \infty)$ .

4.1step 2.2step 3.1L3L4algebra

The quadrature: set $q (s) : = \sqrt{(2 / μ) (E - V (s))}$ for $s > r_{*}$ . The integral defining $τ$ is finite and continuous at $r_{*}$ by [L4], and $τ^{'} (s) = 1 / q (s)$ for $s > r_{*}$ . On the outgoing branch, the chain rule and $ρ^{'} = q (ρ)$ give $\frac{d}{d t} τ (ρ (t)) = 1$ for $t > t_{*}$ ; taking the limit as $t ↓ t_{*}$ yields $τ (ρ (t)) = t - t_{*}$ . On the incoming branch, $ρ^{'} = - q (ρ)$ gives $\frac{d}{d t} τ (ρ (t)) = - 1$ , hence $τ (ρ (t)) = t_{*} - t$ . The function $τ$ is strictly increasing on $[r_{*}, \infty)$ and tends to $+ \infty$ by [L4], so it maps this interval onto $[0, \infty)$ . Inverting the outgoing identity gives $ρ (t_{*} + τ (r)) = r$ for every $r \geq r_{*}$ ; the incoming identity gives the corresponding incoming branch. In particular the unique time with $ρ = r_{*}$ is $t_{*}$ , and $ρ (t) > r_{*}$ for $t \neq t_{*}$ . These are clauses 2 and 3.

5.1step 2.2step 3.1L1algebra∎

Free tails: suppose $t_{1} \geq t_{*}$ with $ρ (t_{1}) > R$ . By step 3.1, $ρ$ is strictly increasing on $[t_{*}, \infty)$ , so $ρ (t) \geq ρ (t_{1}) > R$ for all $t \geq t_{1}$ . On that regime the potential is identically zero and the Cartesian central-force equation reduces to $μ r^{''} = - (U^{'} (ρ) / ρ) r = 0$ , since $U^{'} (ρ) = 0$ for $ρ > R$ ; hence $r$ is affine on $[t_{1}, \infty)$ and $r (t) = r (t_{1}) + (t - t_{1}) r^{'} (t_{1})$ there. The incoming statement is the same argument on $(- \infty, t_{2}]$ . This is clause 4.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Existence of the asymptotic velocities and the impact-parameter identity $J = μ b ∣ v_{-} ∣$

Statement

Let the hypotheses of Finite-range scattering: the simple outer turning point, the two global branches and the free tails hold, with finite-range potential $U$ , energy $E > 0$ , angular momentum $j = J e_{3}$ , $J > 0$ , outer turning point $r_{*}$ , and let $r : R \to R^{3}$ be the extension of the trajectory. Then:

the limits $v_{\pm} : = \lim_{t \to \pm \infty} r^{'} (t)$ exist and satisfy $∣ v_{+} ∣ = ∣ v_{-} ∣ = \sqrt{2 E / μ}$ ;
for every $t_{2} \leq t_{*}$ with $ρ (t_{2}) > R$ , the incoming free tail is the half-line ${r (t_{2}) + s v_{-} : s \leq 0}$ , whose affine extension is the straight line ${r (t_{2}) + s v_{-} : s \in R}$ ; its perpendicular distance from the origin is $b : = \frac{∣ r (t_{2}) \times v_{-} ∣}{∣ v_{-} ∣} = \frac{J}{μ ∣ v_{-} ∣}, i.e. J = μ b ∣ v_{-} ∣ .$

Facts & Assumptions

Given: The finite-range scattering orbit with its simple outer turning point $r_{*}$ and its two global branches, and times $t_{1} \geq t_{*}$ , $t_{2} \leq t_{*}$ with $ρ (t_{1}), ρ (t_{2}) > R$ .

[L1]

Free tails: for $t_{1} \geq t_{*}$ with $ρ (t_{1}) > R$ , $r (t) = r (t_{1}) + (t - t_{1}) r^{'} (t_{1})$ for all $t \geq t_{1}$ ; and symmetrically for $t_{2} \leq t_{*}$ with $ρ (t_{2}) > R$ , $r (t) = r (t_{2}) + (t - t_{2}) r^{'} (t_{2})$ for all $t \leq t_{2}$ (Finite-range scattering: the simple outer turning point, the two global branches and the free tails).

[L2]

On the free regime $ρ > R$ the potential vanishes identically, so the constant energy is $\frac{1}{2} μ ∥ r^{'} ∥_{2}^{2} = E$ there (Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential, Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential).

[L3]

Angular momentum: $j = μ r \times r^{'} = J e_{3}$ is constant with $∣ j ∣ = J$ (Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential, Finite-range scattering: the simple outer turning point, the two global branches and the free tails).

[L4]

For nonzero $u, v \in R^{3}$ , the perpendicular distance from the origin to the line ${u + s v : s \in R}$ equals $∣ u \times v ∣ / ∣ v ∣$ : the closest point is $u - \frac{⟨ u, v ⟩}{∣ v ∣^{2}} v$ , whose squared norm is $∣ u ∣^{2} - \frac{⟨ u, v ⟩^{2}}{∣ v ∣^{2}} = \frac{∣ u \times v ∣^{2}}{∣ v ∣^{2}}$ by the Gram identity $∣ u \times v ∣^{2} = ∣ u ∣^{2} ∣ v ∣^{2} - ⟨ u, v ⟩^{2}$ (The squared cross-product norm is the Gram determinant of two vectors, The Euclidean inner product $⟨ x, y ⟩ = \sum_{k < n} x_{k} y_{k}$ on $R^{n}$ , The cross product in $R^{3}$ ).

Proof

technique · direct

1.1givenL1

By [L1] the velocity is constant on the outgoing tail: $r^{'} (t) = r^{'} (t_{1})$ for all $t \geq t_{1}$ , so $v_{+} : = \lim_{t \to \infty} r^{'} (t) = r^{'} (t_{1})$ exists; symmetrically $v_{-} : = r^{'} (t_{2})$ is the constant velocity on the incoming tail $t \leq t_{2}$ .

2.1step 1.1L2algebra

By [L2] applied at $t_{1}$ and $t_{2}$ , $∥ r^{'} (t_{1}) ∥_{2}^{2} = ∥ r^{'} (t_{2}) ∥_{2}^{2} = 2 E / μ$ ; hence $∣ v_{+} ∣ = ∣ v_{-} ∣ = \sqrt{2 E / μ}$ . This is clause 1.

2.2step 1.1givenL1algebra

The incoming line: by [L1] all points $r (t)$ with $t \leq t_{2}$ lie on the set ${r (t_{2}) + s v_{-} : s \in R}$ , a line through $r (t_{2})$ with direction $v_{-} \neq 0$ (it has norm $\sqrt{2 E / μ} > 0$ ).

3.1step 1.1step 2.2L3L4algebra∎

Its perpendicular distance from the origin is, by [L4] applied to $u = r (t_{2})$ , $v = v_{-}$ , $\frac{∣ r (t_{2}) \times v_{-} ∣}{∣ v_{-} ∣} = \frac{∣ r (t_{2}) \times r^{'} (t_{2}) ∣}{∣ v_{-} ∣} = \frac{∣ j ∣ / μ}{∣ v_{-} ∣} = \frac{J}{μ ∣ v_{-} ∣},$ using that $j = μ r (t_{2}) \times r^{'} (t_{2})$ and $∣ j ∣ = J$ from [L3]. Defining $b$ to be this distance gives $J = μ b ∣ v_{-} ∣$ , which is clause 2.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

The scattering angle advance $Δ$ and the deflection relation: the total polar advance is $2 Δ$ and the velocity directions make the cosine $\cos (π - 2 Δ)$

Statement

Let the hypotheses of Finite-range scattering: the simple outer turning point, the two global branches and the free tails hold, let $r : R \to R^{3}$ be the extension of the trajectory, and let $(ρ, θ)$ be polar data with $r = ρ n (θ)$ , $n (φ) = \cos φ e_{1} + \sin φ e_{2}$ (Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation). Write $t_{*}$ for the apsis time, and for $α \in R$ let $R_{α}$ be the rotation of the plane $span (e_{1}, e_{2})$ , $R_{α} (x) : = \cos α x + \sin α (e_{3} \times x)$ . Then:

the half-angle $Δ : = \int_{r_{*}}^{\infty} \frac{J d r}{r^{2} \sqrt{2 μ (E - V (r))}}$ is finite and positive, and $θ (\infty) - θ (t_{*}) = Δ = θ (t_{*}) - θ (- \infty), θ (\infty) - θ (- \infty) = 2 Δ;$
$\lim_{t \to \pm \infty} n (θ (t))$ exist and equal $v_{+} / ∣ v_{+} ∣$ (as $t \to + \infty$ ) and $- v_{-} / ∣ v_{-} ∣$ (as $t \to - \infty$ );
consequently $\frac{v_{+}}{∣ v_{+} ∣} = - R_{2 Δ} (\frac{v_{-}}{∣ v_{-} ∣}), \frac{⟨ v_{-}, v_{+} ⟩}{∣ v_{-} ∣ ∣ v_{+} ∣} = - \cos 2 Δ = \cos (π - 2 Δ);$ in particular, if $0 < Δ < π / 2$ then $π - 2 Δ \in (0, π)$ is the (undirected) angle between the asymptotic velocity directions, and the signed velocity rotation is $- (π - 2 Δ)$ in the orientation defined by $e_{3}$ .

Facts & Assumptions

Given: The finite-range scattering orbit with its apsis time $t_{*}$ , its polar data, its asymptotic velocities $v_{\pm}$ (Existence of the asymptotic velocities and the impact-parameter identity $J = μ b ∣ v_{-} ∣$ ) and the frozen angular momentum $j = J e_{3}$ , $J > 0$ .

[L1]

Branch structure and quadrature: $ρ (t_{*}) = r_{*}$ , $ρ$ is strictly decreasing on $(- \infty, t_{*}]$ and strictly increasing on $[t_{*}, \infty)$ , $ρ^{'}$ has the corresponding signs, and $(ρ^{'})^{2} = (2 / μ) (E - V (ρ))$ (Finite-range scattering: the simple outer turning point, the two global branches and the free tails).

[L2]

$ρ^{2} θ^{'} = J / μ$ and $θ$ is strictly increasing (Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation).

[L3]

Substitution on a strictly monotone branch and the conversion $d t = d ρ / ∣ ρ^{'} ∣$ (Substitution: if $φ$ is differentiable on $[c, d]$ with $φ^{'}$ integrable and $f$ is continuous on an interval containing $φ ([c, d])$ , then $\int_{φ (c)}^{φ (d)} f = \int_{c}^{d} (f \circ φ) φ^{'}$ , Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ , 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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , $C^{k}$ Euclidean maps and diffeomorphisms, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L4]

The addition formulas give $n (φ + α) = R_{α} (n (φ))$ for all $φ, α$ , and the cross product identity $⟨ u, e_{3} \times u ⟩ = 0$ for every $u$ : the mixed term of $⟨ u, R_{α} u ⟩$ vanishes because $⟨ u, e_{3} \times u ⟩ = \det [u, e_{3}, u] = 0$ under the determinant form of the scalar triple product (The addition formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, The cross product is bilinear, alternating, and orthogonal to both factors, The cross product in $R^{3}$ ).

[L5]

Free tails: for $t_{1} \geq t_{*}$ with $ρ (t_{1}) > R$ , $r (t) = r (t_{1}) + (t - t_{1}) v_{+}$ for $t \geq t_{1}$ ; symmetrically on the incoming tail with $v_{-}$ (Finite-range scattering: the simple outer turning point, the two global branches and the free tails, Existence of the asymptotic velocities and the impact-parameter identity $J = μ b ∣ v_{-} ∣$ ).

Proof

technique · direct

1.1givenL1L2L3algebra

Outgoing half-angle: on $[t_{*}, \infty)$ , $θ$ is $C^{1}$ and strictly increasing with $θ^{'} = J / (μ ρ^{2}) > 0$ , and $ρ$ is a $C^{1}$ strictly increasing bijection of $[t_{*}, \infty)$ onto $[r_{*}, \infty)$ ; substituting $ρ$ for $t$ first on compact subintervals of $(t_{*}, \infty)$ , where $ρ' > 0$ , and then taking endpoint limits [L3] gives $θ (\infty) - θ (t_{*}) = \int_{t_{*}}^{\infty} θ^{'} (t) d t = \int_{r_{*}}^{\infty} \frac{J}{μ s^{2}} \frac{d s}{ρ^{'} (ρ^{- 1} (s))} = \int_{r_{*}}^{\infty} \frac{J d s}{s^{2} \sqrt{2 μ (E - V (s))}} = Δ,$ using $ρ^{'} = \sqrt{(2 / μ) (E - V)}$ from [L1]; the integral converges at $r_{*}$ because $E - V (s)$ is comparable to $s - r_{*}$ there and diverges nowhere else, and it is positive.

2.1step 1.1L1L2L3algebra

Incoming half-angle: the same computation on the decreasing branch $(- \infty, t_{*}]$ , where $ρ^{'} = - \sqrt{(2 / μ) (E - V (ρ))}$ and the substitution reverses the sign, gives $θ (t_{*}) - θ (- \infty) = Δ$ ; adding the two halves gives $θ (\infty) - θ (- \infty) = 2 Δ$ , which is clause 1.

3.1step 1.1step 2.1L5algebra

Direction limits: on the outgoing tail [L5], $r (t) / ∥ r (t) ∥_{2} \to v_{+} / ∣ v_{+} ∣$ as $t \to \infty$ (an affine curve with nonzero velocity direction tends to its velocity direction), and $r (t) / ρ (t) = n (θ (t)) \to n (θ (\infty))$ because $θ$ has a finite limit and $n$ is continuous; hence $n (θ (\infty)) = v_{+} / ∣ v_{+} ∣$ . On the incoming tail, $t - t_{2} < 0$ tends to $- \infty$ , so $r (t) / ∥ r (t) ∥_{2} \to - v_{-} / ∣ v_{-} ∣$ , while $r (t) / ρ (t) \to n (θ (- \infty))$ ; hence $n (θ (- \infty)) = - v_{-} / ∣ v_{-} ∣$ . This is clause 2.

4.1step 2.1step 3.1L4algebra

Rotation form: by [L4], $v_{+} / ∣ v_{+} ∣ = n (θ (\infty)) = n (θ (- \infty) + 2 Δ) = R_{2 Δ} (n (θ (- \infty))) = - R_{2 Δ} (v_{-} / ∣ v_{-} ∣)$ , the first equality of clause 3.

5.1step 4.1L4algebra∎

Cosine: writing $u : = v_{-} / ∣ v_{-} ∣$ , the last identity gives $⟨ v_{-}, v_{+} ⟩ / (∣ v_{-} ∣ ∣ v_{+} ∣) = ⟨ u, - R_{2 Δ} u ⟩ = - \cos 2 Δ$ , because $⟨ u, R_{2 Δ} u ⟩ = \cos 2 Δ ∣ u ∣^{2} + \sin 2 Δ ⟨ u, e_{3} \times u ⟩ = \cos 2 Δ$ by [L4]. Since $\cos (π - 2 Δ) = - \cos 2 Δ$ , this gives the displayed cosine; if $0 < Δ < π / 2$ then $π - 2 Δ \in (0, π)$ and the identity says that the angle between the two nonzero velocity vectors (the unique angle in $[0, π]$ whose cosine is the displayed value, cosine being strictly decreasing there) is $π - 2 Δ$ .

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

One-dimensional substitution on a deflection branch: the outgoing angular measure has density $b ∣ d b / d θ ∣$ , hence cross-section density $b ∣ d b / d θ ∣ / \sin θ$

Statement

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )). Let $θ : (a_{1}, a_{2}) \to (θ_{1}, θ_{2}) \subseteq (0, π)$ be a deflection branch of Scattering data: the incident flux measure, the solid-angle element, deflection branches and the cross-section density, with inverse $b (θ)$ , and set $D = (a_{1}, a_{2}) \times (0, 2 π)$ , $H = (θ_{1}, θ_{2}) \times (0, 2 π)$ and $T (b, φ) = (θ (b), φ)$ . Restrict the incident measure $ν$ to $D$ . Its pushforward, defined by $T_{*} ν (B) = ν (T^{- 1} (B))$ for Borel $B \subseteq H$ , satisfies $T_{*} ν (B) = \int_{B} b (θ) ∣ b^{'} (θ) ∣ d λ_{2} (θ, φ) .$ For every nonnegative Borel $g : H \to [0, + \infty]$ , and for every real Borel $g$ with either side's integral of $∣ g ∣$ finite, consequently $\int_{D} g (θ (b), φ) b d λ_{2} (b, φ) = \int_{H} g (θ, φ) b (θ) ∣ b^{'} (θ) ∣ d λ_{2} (θ, φ) .$ The nonnegative equality allows $+ \infty$ ; in the real case both sides are finite and absolute integrability on one side is equivalent to that on the other. In particular it applies to continuous periodic test functions that are nonnegative or absolutely integrable, and to compactly supported continuous test functions. No equality of undefined signed improper integrals is asserted. Nonnegative double-integral notation denotes the corresponding Lebesgue integrals, with iterated integrals justified by Tonelli.

Thus the outgoing angular measure has density $b (θ) ∣ b^{'} (θ) ∣$ against $d θ d φ$ and density $\frac{d σ}{d Ω} (θ) = \frac{b (θ) ∣ b^{'} (θ) ∣}{\sin θ} = \frac{b (θ)}{\sin θ ∣ θ^{'} (b (θ)) ∣}$ against the solid-angle measure. For finitely many branches on disjoint incident intervals the measures add, including on overlapping image intervals; in particular contributions with pairwise disjoint images add.

Facts & Assumptions

Given: Countable choice, the branch, its inverse, $D, H, T, ν$ and the test function as stated.

[L1]

The inverse is $C^{1}$ , with $b^{'} = 1 / (θ^{'} \circ b)$ ; the sign of the continuous nonzero derivative is constant, so the branch is strictly monotone (Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ , 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 $c \in (a, b)$ with $f (b) - f (a) = f^{'} (c) (b - a)$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, Continuity of $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

[L2]

Compact substitution with absolute derivative holds on every closed interval lying strictly inside the branch; continuous functions there are Riemann integrable and their Riemann and Lebesgue integrals agree under countable choice (In one dimension the compact-Jordan formula is substitution over the unoriented image interval with the absolute derivative, A continuous function on $[a, b]$ is Riemann integrable, by Heine-Cantor and Riemann's criterion, A bounded Riemann integrable function on a closed bounded interval is Lebesgue measurable and has the same integral).

[L3]

On Borel subsets of the plane $λ_{2} = λ_{1} \times λ_{1}$ , and Tonelli applies to nonnegative Borel integrands. Measures agreeing on a generating pi-system with an increasing common finite-measure exhaustion agree everywhere (On Borel subsets of R^{m+n}, the product lambda_m times lambda_n agrees with lambda_{m+n}, Tonelli's theorem for nonnegative measurable functions on a sigma-finite product, Measures agreeing on a generating pi-system are equal under an increasing finite-measure exhaustion from that pi-system).

[L4]

Continuous maps have Borel preimages; a nonnegative measurable density defines a measure. Nonnegative measurable functions have increasing simple approximations and monotone convergence holds; real integrability means finite integral of the absolute value (A continuous map has Borel preimages of Borel sets, The indefinite integral of a nonnegative measurable function is a measure, Every nonnegative measurable function is the increasing limit of simple measurable functions, Monotone convergence for the integral, Integrable real and complex functions, and their integrals, Scattering data: the incident flux measure, the solid-angle element, deflection branches and the cross-section density).

[L5]

$\sin θ > 0$ on $(0, π)$ : the solid-angle definition proves this from $\sin 0 = \sin π = 0$ and strict increase/decrease on the two half intervals; its continuous density is Borel measurable (Scattering data: the incident flux measure, the solid-angle element, deflection branches and the cross-section density).

Proof

technique · direct

1.1givenL1L4

By [L1], $b$ is $C^{1}$ and $h (θ) = b (θ) ∣ b^{'} (θ) ∣$ is continuous and positive. The map $T$ is a homeomorphism $D \to H$ , so $T_{*} ν$ is a Borel measure: preimages are Borel and preserve disjoint countable unions. The nonnegative density $h$ defines another Borel measure $η (B) = \int_{B} h d λ_{2}$ by [L4].

2.1step 1.1L2L3algebra

Let $c < d$ lie in $(θ_{1}, θ_{2})$ and $0 < p < q < 2 π$ . Apply compact substitution [L2] to the inverse map $b$ on $[c, d]$ , with integrand $x \mapsto x$ : writing $u = \min (b (c), b (d))$ , $v = \max (b (c), b (d))$ , it gives $\int_{c}^{d} h (θ) d θ = \int_{u}^{v} x d x$ . Endpoints have Lebesgue measure zero, so Tonelli [L3] yields $η ((c, d) \times (p, q)) = (q - p) \int_{u}^{v} x d x = ν (T^{- 1} ((c, d) \times (p, q)))$ . These quantities are finite.

3.1step 1.1step 2.1L3

Interior open rectangles, together with the empty set, form a pi-system generating the Borel sets of $H$ : rational rectangles form a countable base, and every open set is the union of those base rectangles it contains. Choose explicit nested interior intervals exhausting each open coordinate interval; their product rectangles exhaust $H$ and have the common finite measure of step 2.1. Thus measure uniqueness [L3] gives $T_{*} ν = η$ on every Borel set of $H$ , proving the asserted pushforward identity even for unbounded open branches.

4.1step 3.1L3L4algebra

The test-function identity first holds for indicator functions by step 3.1 and the definition of pushforward, then for nonnegative simple functions by finite addition, and finally for every nonnegative Borel $g$ by increasing simple approximation and monotone convergence [L4] on both sides. Apply this to $∣ g ∣$ to obtain equivalence of absolute integrability; applying it to the positive and negative parts gives the real identity whenever that absolute integral is finite. Tonelli [L3] supplies the nonnegative iterated form.

5.1step 1.1step 3.1step 4.1L1L4L5algebra∎

Since $\sin θ > 0$ by [L5], rewriting $h d λ_{2} = (h / \sin θ) d Ω$ gives the solid-angle density; [L1] gives its reciprocal-derivative expression. Pushforward commutes with finite addition of measures by its definition, so branches on disjoint incident intervals contribute the sum of these measures, and the sum of the densities wherever their images overlap. This proves all assertions.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and $Δ = π / 2 - \arctan (κ / (2 E b))$

Statement

Let $μ > 0$ , let $κ > 0$ and $U (r) = κ / r$ on $(0, \infty)$ (the repulsive Coulomb potential, of class $C^{2}$ ), let $E > 0$ and $J > 0$ , and put $V (r) : = \frac{κ}{r} + \frac{J^{2}}{2 μ r^{2}}, b : = \frac{J}{\sqrt{2 μ E}} > 0, α : = \frac{κ}{E b} > 0.$ Then:

$V$ is strictly decreasing on $(0, \infty)$ , with $V (r) \to \infty$ as $r ↓ 0$ and $V (r) \to 0$ as $r \to \infty$ ; hence there is exactly one $r_{*} > 0$ with $V (r_{*}) = E$ , it is simple ( $V^{'} (r_{*}) < 0$ ), and $V (r) < E$ for all $r > r_{*}$ .
Let $r : I \to R^{3}$ be a central-force trajectory for $(μ, U)$ with constant angular momentum $j = J e_{3}$ , $J > 0$ , energy $E$ , with $∣ r (t) ∣ \geq r_{*}$ for all $t \in I$ and with $∣ r (t_{*}) ∣ = r_{*}$ for some $t_{*} \in I$ . Then $r$ extends uniquely to all of $R$ ; along the extension $ρ = ∣ r ∣ \geq r_{*} > 0$ (no collision), $ρ$ is strictly decreasing on $(- \infty, t_{*}]$ and strictly increasing on $[t_{*}, \infty)$ , and the asymptotic velocities $v_{\pm} = \lim_{t \to \pm \infty} r^{'} (t)$ exist with $∣ v_{+} ∣ = ∣ v_{-} ∣ = \sqrt{2 E / μ}$ . Moreover the asymptotic perpendicular distance of each branch to its limiting direction exists and equals $b = \frac{J}{\sqrt{2 μ E}} = \frac{J}{μ ∣ v_{\pm} ∣}, so that J = μ b ∣ v_{\pm} ∣ .$ The distance in clause 2 means the limit of $∣ r (t) \times v_{\pm} ∣ / ∣ v_{\pm} ∣$ as $t \to \pm \infty$ , the distance to the line through the origin in the limiting velocity direction; it does not assert a free tail or a finite longitudinal intercept.
For the polar angle, with $u_{*} : = 1 / r_{*}$ , $Δ : = \int_{r_{*}}^{\infty} \frac{J d r}{r^{2} \sqrt{2 μ (E - V (r))}} = \int_{0}^{u_{*}} \frac{b d u}{\sqrt{1 - \frac{κ}{E} u - b^{2} u^{2}}} = \frac{π}{2} - \arctan \frac{κ}{2 E b} .$
Write $n_{\pm} = \lim_{t \to \pm \infty} r (t) / ∣ r (t) ∣$ . Then $n_{+} = v_{+} / ∣ v_{+} ∣$ , $n_{-} = - v_{-} / ∣ v_{-} ∣$ , and in an oriented polar plane the outgoing velocity direction is the incoming direction rotated by $2 Δ - π \in (- π, 0)$ . In particular $\frac{⟨ v_{-}, v_{+} ⟩}{∣ v_{-} ∣ ∣ v_{+} ∣} = \cos (π - 2 Δ), θ_{s} = π - 2 Δ = 2 \arctan \frac{κ}{2 E b} .$ For fixed $μ, κ, E$ this gives an actual family for every $b > 0$ , with angular momentum $J (b) = \sqrt{2 μ E} b$ : choose its unique turning radius and initial data $r (0) = r_{*} e_{1}$ , $r^{'} (0) = J (b) e_{2} / (μ r_{*})$ . Local existence and the extension in clause 2 give a member of the family for every $b$ , unique up to apsis time and oriented plane placement.

Facts & Assumptions

Given: The repulsive Coulomb data $μ, κ, E, J$ , the effective potential $V$ , the outer turning point $r_{*}$ , and a scattering trajectory with apsis time $t_{*}$ .

[L1]

Polar reduction and energy: $r = ρ n (θ)$ , $ρ = ∣ r ∣$ , $ρ^{2} θ^{'} = J / μ$ , $μ ρ^{''} = - V^{'} (ρ)$ , $E = \frac{1}{2} μ (ρ^{'})^{2} + V (ρ)$ (Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation); here $V = U_{J}^{e f f}$ (Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential).

[L2]

A continuous real function on an interval with everywhere negative derivative is strictly decreasing; a strictly monotone continuous function assumes each intermediate value exactly once, and its inverse is differentiable at $f (c)$ with derivative $1 / f' (c)$ whenever $f$ is differentiable at $c$ with $f' (c) \neq 0$ (On an interval $I$ , for $f$ continuous on $I$ and differentiable at every interior point: $f^{'} \geq 0$ throughout gives $f$ nondecreasing, $f^{'} > 0$ gives $f$ increasing, $f^{'} \leq 0$ and $f^{'} < 0$ give the two decreasing forms; conversely a nondecreasing $f$ has $f^{'} \geq 0$ and a nonincreasing $f$ has $f^{'} \leq 0$ wherever it is differentiable, and no strict converse is claimed, Intermediate value theorem, by bisection with a canonical left-half rule: a continuous function on $[a, b]$ takes every value between $f (a)$ and $f (b)$ , Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ , Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L3]

The Cartesian field $G (x, v) : = (v, κ x / (μ ∣ x ∣^{3}))$ is defined on the open set $D : = {(x, v) : ∣ x ∣ > 0}$ and is $C^{1}$ there. Its acceleration derivative is $D_{x} (κ x / (μ ∣ x ∣^{3})) [h] = \frac{κ}{μ} (∣ x ∣^{- 3} h - 3 ∣ x ∣^{- 5} x ⟨ x, h ⟩)$ . Thus $D G$ is bounded on a sufficiently small convex ball about each point of $D$ ; along a segment in such a ball, the chain rule and the vector-valued fundamental theorem of calculus give $G (z) - G (w) = \int_{0}^{1} D G (w + s (z - w)) (z - w) d s$ , and the norm inequality for vector integrals gives a finite Lipschitz bound. Hence $G$ is locally Lipschitz, so Picard–Lindelöf uniqueness and compact continuation apply on $D$ (Compact continuation for locally Lipschitz ODEs: a solution whose graph stays in a compact subset of the domain near a finite endpoint extends past it, Picard-Lindelöf local existence and uniqueness for first-order systems, First-order systems, initial value problems, and solutions on intervals, Local Lipschitz continuity in the state variable, locally uniform in time and parameters, If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz, For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable, 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 $g (c) \neq 0$ , $C^{k}$ Euclidean maps and diffeomorphisms).

[L4]

Substitution on strictly monotone branches converts time into radial level and vice versa (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 $(f \circ g)^{'} (c) = f^{'} (g (c)) g^{'} (c)$ , Every continuous function on an interval has a primitive; two primitives differ by a constant; and $\int_{a}^{b} f = G (b) - G (a)$ for any primitive $G$ ).

[L5]

Trigonometry (The derivatives of sine and cosine are cosine and minus sine gives $\sin x / x \to 1$ from the derivative at zero): $\arctan : R \to (- π / 2, π / 2)$ is the inverse of $\tan$ ; $\arcsin$ , $\arccos$ are the principal inverses; the addition and subtraction formulas hold; $\sin^{2} + \cos^{2} = 1$ ; $\cos$ is strictly decreasing on $[0, π]$ and $\sin$ is strictly increasing on $[- π / 2, π / 2]$ (The principal inverse tangent $\arctan : R \to (- π / 2, π / 2)$ , Principal inverse sine and inverse cosine, The addition formulas for sine and cosine, The subtraction formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Signs, monotonicity intervals, and ranges of sine and cosine, Tangent, cotangent, secant, and cosecant on their exact natural domains).

[L6]

The norm bound for the vector integral and componentwise Euclidean completeness justify the velocity Cauchy criterion; the Gram identity gives $∣ x \times v ∣ / ∣ v ∣ = ∣ x ∣ \sin ψ$ when $ψ$ is the angle to the limiting direction (For $a \leq b$ and $f : [a, b] \to R^{m}$ integrable when $a < b$ , $∥ \int_{a}^{b} f ∥_{2} \leq \int_{a}^{b} ∥ f ∥_{2}$ ; for $a < b$ , $∥ f ∥_{2}$ is integrable, If $f : [a, b] \to R^{m}$ is differentiable with integrable $f^{'}$ then $\int_{a}^{b} f^{'} = f (b) - f (a)$ ; and a bounded derivative makes $f$ Lipschitz, $R$ and $R^{n}$ for $n \geq 1$ with the Euclidean metric are complete, componentwise from the Cauchy criterion in $R$ , The squared cross-product norm is the Gram determinant of two vectors).

Proof

technique · direct

1.1givenL2algebra

The derivative $V^{'} (r) = - κ / r^{2} - J^{2} / (μ r^{3})$ is negative on $(0, \infty)$ , so $V$ is strictly decreasing by [L2]; $κ / r \to \infty$ and $J^{2} / (2 μ r^{2}) \to \infty$ as $r ↓ 0$ , while both terms tend to $0$ as $r \to \infty$ . Hence $V$ is a continuous strictly decreasing bijection $(0, \infty) \to (0, \infty)$ ; since $E > 0$ lies in the range, $r_{*} = V^{- 1} (E)$ is the unique solution of $V = E$ , it is simple because $V^{'} < 0$ everywhere, and $V (r) < E$ for $r > r_{*}$ . This is clause 1.

2.1step 1.1L1L3algebra

Global continuation: energy and angular momentum are conserved by [L1], and $V (ρ) \leq E$ forces $ρ \geq r_{*}$ since $V$ is strictly decreasing. Also $∣ r^{'} ∣ \leq c : = \sqrt{2 E / μ}$ because $U > 0$ . At any finite endpoint, $∣ r ∣$ is bounded above by its value at a fixed time plus $c$ times the elapsed time, and $r$ stays at least $r_{*}$ from the origin. Thus the Cartesian state $(r, r^{'})$ stays in a compact subset of ${(x, v) : ∣ x ∣ > 0}$ near that endpoint. The Cartesian field $(v, κ x / (μ ∣ x ∣^{3}))$ is smooth there; compact continuation [L3] extends the solution past every finite right endpoint. For a finite left endpoint apply the same argument to the time-reversed state $(r (- s), - r^{'} (- s))$ , which solves the same autonomous Cartesian system; uniqueness [L3] then makes the global extension unique. The maximal extension therefore has domain $R$ .

3.1step 1.1step 2.1L1L2L4algebra

Branch shape: $ρ^{''} = - V^{'} (ρ) / μ > 0$ for all times by step 1.1. Hence $ρ^{'}$ is strictly increasing, vanishes at $t_{*}$ by the energy identity, is negative before $t_{*}$ and positive after it. Consequently the apsis is unique and $ρ > r_{*}$ elsewhere. Fixing any $t_{1} > t_{*}$ gives $ρ^{'} (t) \geq ρ^{'} (t_{1}) > 0$ for $t \geq t_{1}$ , so $ρ \to \infty$ forward; the same argument backward gives $ρ \to \infty$ there. On either open branch, compact substitution [L4] and then limits give $∣ t - t_{*} ∣ = \int_{r_{*}}^{ρ (t)} d s / \sqrt{(2 / μ) (E - V (s))}$ . Its lower endpoint is integrable since $E - V (s)$ is comparable to $s - r_{*}$ by $V^{'} (r_{*}) < 0$ ; its upper endpoint diverges since the integrand tends to $1 / c > 0$ . This proves the branch and quadrature claims without a finite-range hypothesis.

4.1step 3.1L1L4L6algebra

Asymptotic velocities and their speed: since $V (ρ (t)) \to 0$ along each branch (as $ρ (t) \to \infty$ ), $∣ ρ^{'} ∣ \to \sqrt{2 E / μ}$ . Put $c : = \sqrt{2 E / μ}$ and $c_{0} : = c / 2 > 0$ . The radial-speed limit implies $∣ ρ^{'} (t) ∣ \geq c_{0}$ for sufficiently large $∣ t ∣$ ; integrating on each monotone branch gives $ρ (t) \geq c_{0} ∣ t ∣ - C_{0}$ for a finite constant $C_{0}$ . In particular $ρ (t) \geq (c_{0} / 2) ∣ t ∣$ after increasing the tail cutoff. The acceleration of the Cartesian curve is $μ r^{''} = - (U^{'} (ρ) / ρ) r$ , so $∥ r^{''} ∥_{2} = ∣ U^{'} (ρ) ∣ / μ = κ / (μ ρ^{2}) \leq C / (1 + ∣ t ∣)^{2}$ for large $∣ t ∣$ , an integrable bound along each half-line; the Cauchy criterion for the vector integral (the integral of $r^{''}$ over $[s, t]$ is bounded by $\int_{s}^{t} ∥ r^{''} ∥_{2}$ ) therefore gives existence of $v_{\pm} = \lim_{t \to \pm \infty} r^{'} (t)$ . Passing to the limit in the energy identity $E = \frac{1}{2} μ ∥ r^{'} ∥_{2}^{2} + U (ρ (t))$ with $U (ρ (t)) \to 0$ gives $∣ v_{\pm} ∣ = \sqrt{2 E / μ}$ .

4.2step 3.1L1L4algebra

Half-angle quadrature: on compact subintervals of either open branch use $θ^{'} = J / (μ ρ^{2})$ and $∣ ρ^{'} ∣ = \sqrt{(2 / μ) (E - V)}$ in [L4]. Taking limits gives the same positive half-angle on both sides of the apsis, $Δ = \int_{r_{*}}^{\infty} \frac{J d s}{s^{2} \sqrt{2 μ (E - V (s))}} = \int_{0}^{u_{*}} \frac{b d u}{\sqrt{1 - (κ / E) u - b^{2} u^{2}}} .$ Each substitution is first on an interior compact interval. The integral converges at $r_{*}$ by the square-root bound of step 3.1 and at infinity by the bound $C / s^{2}$ , so it gives finite limits $θ_{\pm}$ with $θ_{+} - θ_{*} = θ_{*} - θ_{-} = Δ$ .

5.1step 4.2L4L5algebra

Evaluation with scaled roots: put $w = b u$ , $A = \sqrt{1 + α^{2} / 4}$ and $w_{*} = (\sqrt{α^{2} + 4} - α) / 2 = b u_{*}$ , where $α = κ / (E b)$ . Then $1 - α w - w^{2} = A^{2} - (w + α / 2)^{2}$ . Substitute $w + α / 2 = A \sin ψ$ on interior compact intervals, where $ψ \in (ψ_{0}, π / 2)$ and $ψ_{0} = \arcsin (α / (2 A)) \in (0, π / 2)$ . The derivative is $A \cos ψ > 0$ and the positive denominator is $A \cos ψ$ , so the integral equals $π / 2 - ψ_{0}$ after taking limits. Since $\sin ψ_{0} = α / \sqrt{α^{2} + 4}$ and $\cos ψ_{0} = 2 / \sqrt{α^{2} + 4}$ , $\tan ψ_{0} = α / 2$ , giving $ψ_{0} = \arctan (α / 2)$ by [L5]. Hence $Δ = π / 2 - \arctan (κ / (2 E b)) \in (0, π / 2)$ . This proves clause 3.

6.1step 4.1step 4.2step 5.1L1L5algebra

Direction limits: differentiating $r = ρ n (θ)$ gives $r^{'} = ρ^{'} n (θ) + (J / (μ ρ)) n_{⊥} (θ)$ , where $n_{⊥} (θ) = - \sin θ e_{1} + \cos θ e_{2}$ . By step 4.1, $ρ^{'} \to c$ forward and $ρ^{'} \to - c$ backward, while the tangential term tends to zero. Step 4.2 therefore gives $v_{+} = c n (θ_{+})$ and $v_{-} = - c n (θ_{-})$ . The addition formulas [L5] and $θ_{+} - θ_{-} = 2 Δ$ show the velocity rotation $2 Δ - π$ and the displayed cosine. There is no hidden extra winding: the radial coefficient $ρ^{'}$ changes sign only once, the tangential coefficient is always positive, so the velocity direction relative to $n (θ)$ varies continuously from $π$ to $0$ through the upper half-plane; its net change is $2 Δ - π$ .

7.1step 3.1step 4.1step 4.2step 6.1L1L4L5L6algebra

Impact limit: put $q (s) = \sqrt{(2 / μ) (E - V (s))}$ for $s > r_{*}$ . On the outgoing branch the remaining angle is $Ψ (t) = \int_{ρ (t)}^{\infty} J / (μ s^{2} q (s)) d s$ , so $ρ (t) Ψ (t) = \frac{J}{μ} \int_{1}^{\infty} \frac{d u}{u^{2} q (ρ (t) u)} .$ As $q (s) \to c > 0$ , for any $ε > 0$ one has $∣ 1 / q (s) - 1 / c ∣ < ε$ for every $s \geq S$ for some $S$ . Once $ρ (t) \geq S$ , the difference between the integral above and $1 / c$ is at most $ε \int_{1}^{\infty} u^{- 2} d u = ε$ . Thus $ρ Ψ \to J / (μ c)$ and $Ψ \to 0$ ; the elementary limit $\sin Ψ / Ψ \to 1$ gives $ρ \sin Ψ \to J / (μ c) = b$ . By step 6.1 this is exactly $∣ r (t) \times v_{+} ∣ / ∣ v_{+} ∣$ . The incoming computation uses $q = ∣ ρ^{'} ∣$ and the remaining angle $θ (t) - θ_{-}$ and gives $∣ r (t) \times v_{-} ∣ / ∣ v_{-} ∣ \to b$ as well.

8.1step 1.1step 2.1step 3.1step 5.1step 6.1step 7.1L1L3algebra∎

For every $b > 0$ set $J (b) = \sqrt{2 μ E} b$ and take the turning radius of clause 1. The initial data stated in clause 4 have angular momentum $J (b) e_{3}$ and energy $V (r_{*}) = E$ , radial derivative zero and positive radial second derivative. Local existence [L3] supplies a trajectory and conservation [L1] forces $ρ \geq r_{*}$ ; the preceding continuation, angle and impact-limit steps apply to it. Uniqueness [L3] determines it from its apsis state. Thus this actual family realises the deflection formula for all $b > 0$ , and all clauses are proved.

TheoremStatement: AI-adaptedProof: AI-adaptedOpen item page →

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

Statement

Assume ${A C}_{ω}$ (The Axiom of Countable Choice ( ${A C}_{ω}$ )) for the cross-section measures. Fix $μ > 0$ , $κ > 0$ and $E > 0$ . For every $b > 0$ put $J (b) = \sqrt{2 μ E} b$ and use the Coulomb trajectory family of Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and $Δ = π / 2 - \arctan (κ / (2 E b))$ clause 4. Its impact distance is $b$ and its deflection map $θ_{d e f} : (0, \infty) \to (0, π), θ_{d e f} (b) : = 2 \arctan \frac{κ}{2 E b} .$ Then:

$θ_{d e f}$ is $C^{1}$ and strictly decreasing, with $θ_{d e f} (b) \to π$ as $b ↓ 0$ and $θ_{d e f} (b) \to 0$ as $b \to \infty$ ; it is a $C^{1}$ bijection $(0, \infty) \to (0, π)$ with inverse $b (θ) = \frac{κ}{2 E} \cot \frac{θ}{2} (0 < θ < π) .$
For every interval $(θ_{1}, θ_{2}) \subseteq (0, π)$ the pair $(θ, b)$ is a deflection branch with cross-section density $\frac{d σ}{d Ω} (θ) = b (θ) ∣ \frac{d b}{d θ} (θ) ∣ \frac{1}{\sin θ} = (\frac{κ}{4 E})^{2} \csc^{4} \frac{θ}{2},$ finite and positive on $(θ_{1}, θ_{2})$ ; 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 $b ∣ d b / d θ ∣$ , hence cross-section density $b ∣ d b / d θ ∣ / \sin θ$ .
For every $θ_{0} \in (0, π)$ the truncated total cross-section $\int_{0}^{θ_{0}} ⁣ ⁣ \int_{0}^{2 π} (\frac{κ}{4 E})^{2} \csc^{4} \frac{θ}{2} \sin θ d φ d θ = + \infty,$ while for $0 < θ_{1} < θ_{2} < π$ the integral over $(θ_{1}, θ_{2}) \times (0, 2 π)$ 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 $J (b) = \sqrt{2 μ E} b$ , the deflection map $θ_{d e f}$ and its asserted inverse $b (θ)$ .

[L1]

$\arctan : R \to (- π / 2, π / 2)$ is the inverse of $\tan$ , $\arctan$ is $C^{1}$ 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 $\arctan : R \to (- π / 2, π / 2)$ , 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).

[L2]

Trigonometric calculus: derivatives of $\tan, \cot, \sec, \csc$ and their Pythagorean and parity identities, the cotangent double-angle form $\sin θ = 2 \sin (θ / 2) \cos (θ / 2)$ , and $\cos (θ / 2) > 0$ for $0 < θ < π$ (Derivatives and fundamental periods of tangent, cotangent, secant, and cosecant, Pythagorean and parity identities for all six trigonometric functions on their natural domains, The addition formulas for sine and cosine, The subtraction formulas for sine and cosine, Parity and the Pythagorean identity for sine and cosine, Signs, monotonicity intervals, and ranges of sine and cosine, Tangent, cotangent, secant, and cosecant on their exact natural domains).

[L3]

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

[L4]

Chain, quotient and inverse-function rules: $θ_{d e f}$ and $b$ are $C^{1}$ with the derivatives computed below; the inverse of a continuous injective $C^{1}$ function with nonvanishing derivative has reciprocal derivative (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 $g (c) \neq 0$ , Derivative of an inverse: if $f$ is continuous and injective on a nondegenerate interval $I$ and differentiable at $c \in I$ with $f^{'} (c) \neq 0$ , then the inverse $g$ is differentiable at $f (c)$ with $g^{'} (f (c)) = 1 / f^{'} (c)$ ; and if $f^{'} (c) = 0$ then $g$ is not differentiable at $f (c)$ , $C^{k}$ Euclidean maps and diffeomorphisms, 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, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length).

[L5]

One-dimensional substitution/change of variables with the absolute derivative and the branch density identity $d σ / d Ω = b ∣ b^{'} ∣ / \sin θ$ hold on every deflection branch (One-dimensional substitution on a deflection branch: the outgoing angular measure has density $b ∣ d b / d θ ∣$ , hence cross-section density $b ∣ d b / d θ ∣ / \sin θ$ , Scattering data: the incident flux measure, the solid-angle element, deflection branches and the cross-section density).

[L6]

A continuous nonnegative function with a positive value has positive integral, the integral is monotone, and for $0 < ε < θ_{0}$ the computed integral $\int_{ε}^{θ_{0}} θ^{- 3} d θ = \frac{1}{2} (ε^{- 2} - θ_{0}^{- 2})$ diverges as $ε ↓ 0$ ; $\sin x \leq x$ for $x \geq 0$ (mean value theorem applied to $\sin$ on $[0, x]$ , using $\cos \leq 1$ ) and $\cos (θ / 2) \geq \cos (θ_{0} / 2) > 0$ for $0 < θ \leq θ_{0} \leq π / 2$ (The second fundamental theorem: if $G$ is differentiable on $[a, b]$ with $G^{'} = f$ and $f$ is integrable, then $\int_{a}^{b} f = G (b) - G (a)$ , If $f \leq g$ on $[a, b]$ and both are integrable then $\int_{a}^{b} f \leq \int_{a}^{b} g$ ; and $m (b - a) \leq \int_{a}^{b} f \leq M (b - a)$ , 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 $\int_{a}^{b} f = G (b) - G (a)$ 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 $c \in (a, b)$ 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 $\int_{a}^{b} f = \int_{a}^{c} f + \int_{c}^{b} f$ ; 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 $f : A \to R$ at a point of $A$ and on $A$ : the $ε$ - $δ$ condition, its agreement with $\lim_{x \to c} f (x) = f (c)$ at a limit point, and continuity at an isolated point).

Proof

technique · direct

1.1givenL1L2L4algebra

Differentiation of the branch: by [L4] and [L2], $θ_{d e f}^{'} (b) = 2 \cdot \frac{1}{1 + (\frac{κ}{2 E b})^{2}} \cdot (- \frac{κ}{2 E b^{2}}) < 0 (b > 0),$ so $θ_{d e f}$ is $C^{1}$ with strictly negative derivative, hence strictly decreasing; as $b ↓ 0$ , $κ / (2 E b) \to \infty$ and $\arctan$ tends to $π / 2$ , so $θ_{d e f} \to π$ ; as $b \to \infty$ , $κ / (2 E b) \to 0$ and $θ_{d e f} \to 0$ . The image is therefore $(0, π)$ and the inverse exists, is $C^{1}$ 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 $\tan (θ / 2) = κ / (2 E b (θ))$ with $b (θ) = \frac{κ}{2 E} \cot (θ / 2)$ , because $\tan (θ / 2) \cot (θ / 2) = 1$ by [L2]; substituting shows $θ_{d e f} (b (θ)) = θ$ .

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 $π - 2 Δ = 2 \arctan (κ / (2 E b))$ ; 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], $b^{'} (θ) = - \frac{κ}{4 E} \csc^{2} (θ / 2)$ , so $∣ b^{'} (θ) ∣ = \frac{κ}{4 E} \csc^{2} (θ / 2)$ ; combining with [L5], $\frac{d σ}{d Ω} (θ) = \frac{b (θ) ∣ b^{'} (θ) ∣}{\sin θ} = \frac{κ^{2}}{8 E^{2}} \cdot \frac{\cot (θ / 2) \csc^{2} (θ / 2)}{\sin θ} .$ Using $\sin θ = 2 \sin (θ / 2) \cos (θ / 2)$ from [L2], $\cot \frac{θ}{2} \csc^{2} \frac{θ}{2} / \sin θ = \frac{\cos (θ / 2)}{\sin^{3} (θ / 2)} \cdot \frac{1}{2 \sin (θ / 2) \cos (θ / 2)} = \frac{1}{2} \csc^{4} \frac{θ}{2},$ whence $d σ / d Ω = \frac{κ^{2}}{16 E^{2}} \csc^{4} (θ / 2) = (κ / (4 E))^{2} \csc^{4} (θ / 2)$ , finite and positive on $(θ_{1}, θ_{2})$ ; this is clause 2.

3.1step 2.2L6algebra

Divergence: for $0 < θ \leq θ_{0}$ with $θ_{0} \leq π / 2$ , the integrand of clause 3 equals $(κ / (4 E))^{2} \sin θ \csc^{4} (θ / 2) = (κ / (4 E))^{2} \cdot 2 \cos (θ / 2) / \sin^{3} (θ / 2)$ by the same identities; since $\sin (x) \leq x$ for $x \geq 0$ and $\cos (θ / 2) \geq \cos (θ_{0} / 2) > 0$ by [L6], the integrand is at least $C θ^{- 3}$ with $C = 16 (κ / (4 E))^{2} \cos (θ_{0} / 2) > 0$ . Hence for $0 < ε < θ_{0}$ , by monotonicity of the integral and the $φ$ -integration of a constant, $\int_{ε}^{θ_{0}} ⁣ ⁣ \int_{0}^{2 π} (κ / (4 E))^{2} \csc^{4} (θ / 2) \sin θ d φ d θ \geq 2 π C \int_{ε}^{θ_{0}} θ^{- 3} d θ = π C (ε^{- 2} - θ_{0}^{- 2}) ⟶ + \infty$ as $ε ↓ 0$ , by the fundamental theorem and [L6]; hence the truncated integral over $(0, θ_{0})$ diverges. If instead $π / 2 < θ_{0} < π$ , 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 $(κ / (4 E))^{2} \csc^{4} (θ / 2) \sin θ$ is continuous on the compact rectangle $[θ_{1}, θ_{2}] \times [0, 2 π]$ because $\sin (θ / 2) > 0$ there, so its integral is finite by [L6]; this completes clause 3.

5 · Examples, experiments and counterexamples

None yet.

Cm Math Central

1 · Prerequisites

2 · Summary

3 · Logical flowchart

4 · Definitions, postulates, theorems and proofs

Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential

Definition

Local polar coordinates for a curve with nonvanishing position in the plane: a differentiable angle function with prescribed derivative

Statement

Facts & Assumptions

Proof

Solutions of v′′+ω2v=0v′′+ω2v=0 are exactly the functions Acos⁡(ω(t−t0))+Bsin⁡(ω(t−t0))Acos(ω(t−t0​))+Bsin(ω(t−t0​))

Statement

Facts & Assumptions

Proof

The vector triple product identity u×(v×w)=⟨u,w⟩v−⟨u,v⟩wu×(v×w)=⟨u,w⟩v−⟨u,v⟩w

Statement

Facts & Assumptions

Proof

Oriented orthonormal triads in R3R3: existence over any unit vector, and the orthonormal expansion with no fourth orthogonal direction

Statement

Facts & Assumptions

Proof

Conservation of the angular momentum vector along a central-force trajectory and its orthogonality to position and velocity

Statement

Facts & Assumptions

Proof

Polar reduction of a non-radial central-force trajectory: ρ2θ′=J/μρ2θ′=J/μ, the effective-potential equation and energy conservation

Statement

Facts & Assumptions

Proof

Circular orbits of a non-radial central-force trajectory: the critical-point condition and the squared frequency of the linearised radial equation

Statement

Facts & Assumptions

Proof

Binet's equation for the attractive inverse-square field: d2u/dφ2+u=μκ/J2d2u/dφ2+u=μκ/J2 for u=1/ρu=1/ρ

Statement

Facts & Assumptions

Proof

The Kepler orbit equation is a conic: u=c(1+ecos⁡(φ−θ0))u=c(1+ecos(φ−θ0​)) with E=(μκ2/2J2)(e2−1)E=(μκ2/2J2)(e2−1)

Statement

Facts & Assumptions

Proof

The one-dimensional action variable of a family of two-turning-point energy levels

Definition

The subcritical Kepler conic is an ellipse with semiaxes a=p/(1−e2)a=p/(1−e2), b=a1−e2b=a1−e2​, focus at the origin and area πabπab; its area is also 12∫02πϱ221​∫02π​ϱ2

Statement

Facts & Assumptions

Proof

Completeness and periodicity of the negative-energy Kepler orbit, with T=2πμab/JT=2πμab/J and T2=4π2μa3/κT2=4π2μa3/κ

Statement

Facts & Assumptions

Proof

The action variable of a two-simple-turning-point family is C1C1 with derivative T(E)/(2π)T(E)/(2π), the period quadrature over 2π2π

Statement

Facts & Assumptions

Proof

Scattering data: the incident flux measure, the solid-angle element, deflection branches and the cross-section density

Definition

Finite-range scattering: the simple outer turning point, the two global branches and the free tails

Statement

Facts & Assumptions

Proof

Existence of the asymptotic velocities and the impact-parameter identity J=μb∣v−∣J=μb∣v−​∣

Statement

Facts & Assumptions

Proof

The scattering angle advance ΔΔ and the deflection relation: the total polar advance is 2Δ2Δ and the velocity directions make the cosine cos⁡(π−2Δ)cos(π−2Δ)

Statement

Facts & Assumptions

Proof

One-dimensional substitution on a deflection branch: the outgoing angular measure has density b∣db/dθ∣b∣db/dθ∣, hence cross-section density b∣db/dθ∣/sin⁡θb∣db/dθ∣/sinθ

Statement

Facts & Assumptions

Proof

Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and Δ=π/2−arctan⁡(κ/(2Eb))Δ=π/2−arctan(κ/(2Eb))

Statement

Facts & Assumptions

Proof

The Rutherford deflection map, its inverse, the cross-section density (κ/4E)2csc⁡4(θ/2)(κ/4E)2csc4(θ/2) and the divergence of the total cross-section

Solutions of $v^{''} + ω^{2} v = 0$ are exactly the functions $A \cos (ω (t - t_{0})) + B \sin (ω (t - t_{0}))$

The vector triple product identity $u \times (v \times w) = ⟨ u, w ⟩ v - ⟨ u, v ⟩ w$

Oriented orthonormal triads in $R^{3}$ : existence over any unit vector, and the orthonormal expansion with no fourth orthogonal direction

Polar reduction of a non-radial central-force trajectory: $ρ^{2} θ^{'} = J / μ$ , the effective-potential equation and energy conservation

Binet's equation for the attractive inverse-square field: $d^{2} u / d φ^{2} + u = μ κ / J^{2}$ for $u = 1 / ρ$

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 $\frac{1}{2} \int_{0}^{2 π} ϱ^{2}$

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

The action variable of a two-simple-turning-point family is $C^{1}$ with derivative $T (E) / (2 π)$ , the period quadrature over $2 π$

Existence of the asymptotic velocities and the impact-parameter identity $J = μ b ∣ v_{-} ∣$

The scattering angle advance $Δ$ and the deflection relation: the total polar advance is $2 Δ$ and the velocity directions make the cosine $\cos (π - 2 Δ)$

One-dimensional substitution on a deflection branch: the outgoing angular measure has density $b ∣ d b / d θ ∣$ , hence cross-section density $b ∣ d b / d θ ∣ / \sin θ$

Repulsive Coulomb scattering: the unique outer turning point, global non-colliding motion, the asymptotic impact distance and $Δ = π / 2 - \arctan (κ / (2 E b))$

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