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

Statement

Let μ>0, let J0⊆(0,∞) be an open interval, let U:J0→R be of class C2, let I⊆R be an interval with at least two points, and let r:I→R3 be a central-force trajectory for (μ,U) whose angular momentum j=μ r×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 (e1,e2,e3) be an oriented orthonormal triad with e3=j/J (Oriented orthonormal triads in R3: existence over any unit vector, and the orthonormal expansion with no fourth orthogonal direction).

Then there are twice differentiable ρ:I→(0,∞) and θ:I→R with

r(t)=ρ(t)(cos⁡θ(t) e1+sin⁡θ(t) e2)(t∈I),

and for every such pair (ρ,θ):

  1. ρ(t)=∣r(t)∣ for all t;
  2. ρ2θ′=J/μ on I, so θ′>0;
  3. the effective-potential equation μρ′′=−(UJeff)′(ρ) holds on I, with UJeff=U+J2/(2μρ2) (Central-force trajectories: the central-force equation, angular momentum, energy and the effective potential);
  4. E:=12μ∥r′∥22+U(∣r∣) equals 12μρ′2+UJeff(ρ) and is constant on I.

Facts & Assumptions

Given: A central-force trajectory r for (μ,U) with constant nonzero angular momentum j=Je3, J>0, an oriented orthonormal triad (e1,e2,e3), and twice differentiable ρ,θ with r=ρ(cos⁡θ e1+sin⁡θ e2).

[L1]

For the oriented orthonormal triad: every x∈R3 is x=⟨x,e1⟩e1+⟨x,e2⟩e2+⟨x,e3⟩e3 and ∣x∣2=∑i⟨x,ei⟩2 (Oriented orthonormal triads in R3: 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,e3⟩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 ρ1,ρ2:I→R with (ρ1,ρ2)≠(0,0) there is a twice differentiable angle function such that ρ1=r0cos⁡θ, ρ2=r0sin⁡θ with r0=ρ12+ρ22 (Local polar coordinates for a curve with nonvanishing position in the plane: a differentiable angle function with prescribed derivative).

[L5]

Cross product rules: e1×e2=e3, e2×e1=−e3, the product is bilinear, and ei are unit vectors (The cross product in R3, 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).

Proof

technique · direct
1.1givenL1L2L4algebra

Put ρi:=⟨r,ei⟩ for i=1,2; these are twice differentiable by [L4]. They are never both zero: by [L1] and [L2], r=ρ1e1+ρ2e2+⟨r,e3⟩e3=ρ1e1+ρ2e2, so ρ1=ρ2=0 would give r=0, contradicting ∣r∣∈J0⊆(0,∞).

1.2givenL1L5algebra

Clause 1 for an arbitrary pair: if r=ρ(cos⁡θ e1+sin⁡θ e2), then ∣r∣2=∣r∣2 is computed by [L1] and [L5] as ρ2(cos⁡2θ+sin⁡2θ)=ρ2, hence ρ=∣r∣.

1.3givenL4L5algebra

Differentiating [L5]'s expansion r=ρ n with n:=cos⁡θ e1+sin⁡θ e2 and n⊥:=−sin⁡θ e1+cos⁡θ e2 by [L4] gives r′=ρ′n+ρθ′ n⊥,r′′=(ρ′′−ρθ′2)n+(2ρ′θ′+ρθ′′)n⊥. Here n×n⊥=e3 and n⊥×n=−e3 by bilinearity of the cross product and e1×e2=e3.

2.1step 1.1L1L2L3algebra

Apply the polar lift [L3] to (ρ1,ρ2): there is a twice differentiable θ with ρ1=r0cos⁡θ, ρ2=r0sin⁡θ for r0=ρ12+ρ22; by Parseval in [L1] and ⟨r,e3⟩=0, r02=ρ12+ρ22=∣r∣2, so r0=∣r∣=:ρ is twice differentiable with values in J0⊆(0,∞). Substituting into the expansion of step 1.1 gives r=ρ(cos⁡θ e1+sin⁡θ e2), so such a pair exists.

2.2givenstep 1.3L5algebra

Clause 2: from step 1.3, r×r′=ρ n×(ρ′n+ρθ′n⊥)=ρ2θ′ e3, so j=Je3=μr×r′=μρ2θ′e3, giving J=μρ2θ′ and hence ρ2θ′=J/μ>0.

3.1givenstep 1.2step 1.3step 2.2L1L5algebra

Clause 3: the trajectory equation is μr′′=−(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 αn+βn⊥=0 forces α=β=0 by taking inner products with n and n⊥ — the radial component gives μ(ρ′′−ρθ′2)=−U′(ρ). Substituting ρ2θ′=J/μ from step 2.2 gives μρ′′=−U′(ρ)+μρθ′2=−U′(ρ)+J2μρ3=−(UJeff)′(ρ), because (UJeff)′=U′(ρ)−J2μρ3.

3.2step 1.3step 2.2L1algebra

Clause 4, the energy identity: from step 1.3 and orthonormality, ∥r′∥22=ρ′2+ρ2θ′2, so E=12μρ′2+12μρ2θ′2+U(ρ)=12μρ′2+J22μρ2+U(ρ)=12μρ′2+UJeff(ρ) by step 2.2.

4.1step 3.1step 3.2L4L6algebra∎

The right-hand side is differentiable on I and its derivative is μρ′ρ′′+(UJeff)′(ρ) ρ′=ρ′(μρ′′+(UJeff)′(ρ))=0 by step 3.1 (the equation holds at every point of I); hence E is constant on I by [L6].

Depends on

Dependency tree · 2 levels

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