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Poisson bracket of velocities

Proved
RelPartMech.poisson_velocities

by Lucas · Oct 12, 2026 · Mathlib 0df444a (Lean v4.33.1)

mathematical-physicsrelativity

With uα=1mgαβ(pβ−eAβ)u^\alpha=\frac1m g^{\alpha\beta}(p_\beta-eA_\beta)uα=m1​gαβ(pβ​−eAβ​), smooth metric ggg, smooth potential AAA and m≠0m\neq0m=0,

{uα,uβ}=1m2(Gαβγ m gγδuδ+e Fαβ),\{u^\alpha,u^\beta\} = \frac1{m^2}\Big(G^{\alpha\beta\gamma}\,m\,g_{\gamma\delta}u^\delta + e\,F^{\alpha\beta}\Big),{uα,uβ}=m21​(Gαβγmgγδ​uδ+eFαβ),

where GαβγG^{\alpha\beta\gamma}Gαβγ and FαβF^{\alpha\beta}Fαβ are the raised symbols of eq. (11). The non-commutativity of the velocities combines the force tensor with the symbol GGG.

Preamble
import Mathlib
import Definitions.Def_RelPartMech_coordinates

open scoped ContDiff
Formal statement
namespace RelPartMech

theorem poisson_velocities {n : ℕ} (g : MetricField n) (hg : IsSmoothMetric g)
    (A : CovectorField n) (hA : ∀ α, ContDiff ℝ ∞ (A α)) (e m : ℝ) (hm : m ≠ 0)
    (α β : Fin n) (x p : Coord n) :
    poisson (velocity g A e m α) (velocity g A e m β) x p =
      (1 / m ^ 2) * (∑ γ, ∑ δ, gSymbolUp g α β γ x * m * g x γ δ * velocity g A e m δ x p
        + e * forceTensorUp g A α β x) := by sorry

end RelPartMech
Source
G. Carcassi, C. A. Aidala, "Relativity and Particle Mechanics" (note, uploaded PDF), p. 6, eq. (13), third block (final line)
Human review
  • Endorsed by Shuze Chen · Oct 13, 2026

    Confirmed by the moderator at approval.

  • Endorsed by Lucas · Oct 13, 2026

    Confirmed by the mission captain (proposal self-audit).

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