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Existence of the asymptotic velocities and the impact-parameter identity J=μb∣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=Je3, J>0, outer turning point r∗, and let r:R→R3 be the extension of the trajectory. Then:

  1. the limits v±:=lim⁡t→±∞r′(t) exist and satisfy ∣v+∣=∣v−∣=2E/μ;
  2. for every t2≤t∗ with ρ(t2)>R, the incoming free tail is the half-line {r(t2)+s v−:s≤0}, whose affine extension is the straight line {r(t2)+s v−:s∈R}; its perpendicular distance from the origin is b:=∣r(t2)×v−∣∣v−∣=Jμ∣v−∣,i.e.J=μb∣v−∣.

Facts & Assumptions

Given: The finite-range scattering orbit with its simple outer turning point r∗ and its two global branches, and times t1≥t∗, t2≤t∗ with ρ(t1),ρ(t2)>R.

[L1]

Free tails: for t1≥t∗ with ρ(t1)>R, r(t)=r(t1)+(t−t1)r′(t1) for all t≥t1; and symmetrically for t2≤t∗ with ρ(t2)>R, r(t)=r(t2)+(t−t2)r′(t2) for all t≤t2 (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 12μ∥r′∥22=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).

[L4]

For nonzero u,v∈R3, the perpendicular distance from the origin to the line {u+sv:s∈R} equals ∣u×v∣/∣v∣: the closest point is u−⟨u,v⟩∣v∣2v, whose squared norm is ∣u∣2−⟨u,v⟩2∣v∣2=∣u×v∣2∣v∣2 by the Gram identity ∣u×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⟩=∑k<nxkyk on Rn, The cross product in R3).

Proof

technique · direct
1.1givenL1

By [L1] the velocity is constant on the outgoing tail: r′(t)=r′(t1) for all t≥t1, so v+:=lim⁡t→∞r′(t)=r′(t1) exists; symmetrically v−:=r′(t2) is the constant velocity on the incoming tail t≤t2.

2.1step 1.1L2algebra

By [L2] applied at t1 and t2, ∥r′(t1)∥22=∥r′(t2)∥22=2E/μ; hence ∣v+∣=∣v−∣=2E/μ. This is clause 1.

2.2step 1.1givenL1algebra

The incoming line: by [L1] all points r(t) with t≤t2 lie on the set {r(t2)+s v−:s∈R}, a line through r(t2) with direction v−≠0 (it has norm 2E/μ>0).

3.1step 1.1step 2.2L3L4algebra∎

Its perpendicular distance from the origin is, by [L4] applied to u=r(t2), v=v−, ∣r(t2)×v−∣∣v−∣=∣r(t2)×r′(t2)∣∣v−∣=∣j∣/μ∣v−∣=Jμ∣v−∣, using that j=μ r(t2)×r′(t2) and ∣j∣=J from [L3]. Defining b to be this distance gives J=μb∣v−∣, which is clause 2.

Depends on

Dependency tree · 2 levels

Sources