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The free particle: the complete integral S=qα−tα2/(2m), its characteristics and its canonical labels

Example

Let m>0 and let H:R×R×R→R, H(t,q,p):=p2/(2m), be the free-particle Hamiltonian (The Hamiltonian field of a C2 function on a phase domain is J∇H, and the Poisson bracket is ∑i(∂qif ∂pig−∂pif ∂qig)), and let S:R×R×R→R, S(t,q,α):=qα−tα2/(2m). Then:

  1. S solves the Hamilton-Jacobi equation St+H(t,q,Sq)=0 on R3 and is a complete integral at every point, with Sqα=1 (The Hamilton-Jacobi equation and a complete integral of it);
  2. the Jacobi constants are β:=Sα=q−tα/m, so the characteristics are q(t)=β+tα/m with momentum p=α; these satisfy q′=p/m, p′=0, and β is constant along them (A complete integral produces Hamilton characteristics, the Jacobi constants, a local converse and canonical (β,α) labels);
  3. on R3 at fixed t one has dq∧dp=dβ∧dα: the pair (β,α) is canonically conjugate to (q,p).

Facts & Assumptions

Given: The mass m>0, the free-particle Hamiltonian H=p2/(2m) and the function S=qα−tα2/(2m) on R3.

[F1]

Partial derivatives (Directional derivatives and partial derivatives of a map U⊆Rm→Rn): St=−α2/(2m), Sq=α, Sα=q−tα/m, Sqα=1.

[F2]

Complete integral and characteristics (The Hamilton-Jacobi equation and a complete integral of it, A complete integral produces Hamilton characteristics, the Jacobi constants, a local converse and canonical (β,α) labels): a C3 function with St+H(t,q,Sq)=0 and det⁡Sqα≠0 is a complete integral, and the characteristic equations q′=Hp(t,q,Sq), p=Sq then satisfy Hamilton's equations, with Sα constant.

[F3]

Two-form calculus (A phase domain, its canonical one-form and its standard symplectic two-form, The pullback of a differential form, The wedge product of differential forms): on a phase domain with coordinates (q,p) the standard form is dq∧dp, and dβ∧dα is computed from the differentials dq,dα by bilinearity and alternation.

Verification

technique · direct; compute the derivatives, check the Hamilton-Jacobi equation and the characteristics, then compare the two two-forms
1.1F1F2algebra

By [F1], St+H(t,q,Sq)=−α2/(2m)+α2/(2m)=0 on R3; since Sqα=1≠0, S is a complete integral at every point by [F2], proving clause 1.

1.2F1F2algebra

Clause 2: by [F1], β=Sα=q−tα/m, so for fixed α and β the characteristic curve is q(t)=β+tα/m with p(t)=Sq=α; then q′=α/m=Hp(t,q,p) and p′=0=−Hq, and β is constant along the curve by construction.

2.1F3step 1.2algebra∎

Clause 3: at fixed t, dβ=dq−(t/m)dα and dp=dα; hence dβ∧dα=(dq−(t/m)dα)∧dα=dq∧dα=dq∧dp, since dα∧dα=0.

Depends on

Dependency tree · 2 levels

Sources