TheoremStatement: AI-adaptedProof: AI-generatedPipeline-generatedprecheck passjudge pass (gpt-5.6-terra)audited 2026-09-14

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Equivalence of Euler–Lagrange and Hamilton equations for hyperregular Lagrangians

Statement

Assume ${A C}_{ω}$ . Let $L : T Q \to R$ be hyperregular and $H = E_{L} \circ (F L)^{- 1}$ . The Legendre map bijects Euler–Lagrange trajectories $(q, \dot{q})$ with Hamiltonian trajectories $(q, p)$ of $H$ .

Facts & Assumptions

Given: A hyperregular $L$ and its associated $H$ .

[F1]

Euler–Lagrange equations are $\frac{d}{d t} L_{v^{i}} = L_{q^{i}}$ . Euler–Lagrange equations.

[F2]

With $p = L_{v}$ and inverse $v (q, p)$ , $H (q, p) = p_{i} v^{i} - L (q, v)$ . Energy and Hamiltonian of a hyperregular Lagrangian.

[F3]

Hamilton's equations are ${\dot{q}}^{i} = H_{p_{i}}$ and ${\dot{p}}_{i} = - H_{q^{i}}$ . Hamilton equations in canonical cotangent coordinates.

Proof

technique · direct

1.1

Differentiate the formula in [F2]. Since $p_{i} = L_{v^{i}} (q, v)$ , the $p_{i} d v^{i}$ and $L_{v^{i}} d v^{i}$ terms cancel, giving $d H = v^{i} d p_{i} - L_{q^{i}} d q^{i}$ . Hence $H_{p_{i}} = v^{i}$ and $H_{q^{i}} = - L_{q^{i}}$ .

F2algebra

2.1

If $q (t)$ satisfies [F1] and $p (t) = L_{v} (q (t), \dot{q} (t))$ , then step 1.1 gives ${\dot{q}}^{i} = v^{i} = H_{p_{i}}$ and ${\dot{p}}_{i} = \frac{d}{d t} L_{v^{i}} = L_{q^{i}} = - H_{q^{i}}$ . Thus $(q, p)$ satisfies [F3].

F1F3step 1.1

3.1

Conversely, a Hamiltonian trajectory satisfies $\dot{q} = v (q, p)$ by [F3] and step 1.1, so inverse Legendre gives $p = L_{v} (q, \dot{q})$ . Its second Hamilton equation then reads $\frac{d}{d t} L_{v^{i}} = L_{q^{i}}$ , which is [F1]. Hyperregularity makes both assignments global inverses.

F1F2F3step 1.1∎

Depends on

Used by

Geodesic flow as a Hamiltonian flow on the cotangent bundle Example

Dependency tree · two levels

14 results within two dependency steps of this one, each drawn at its shortest distance from it. An arrow runs from a result to what uses it, so the chart reads left to right and ends at this result, which carries a heavier outline. Every node is a link to that result. Click elsewhere on the chart to enlarge it.

Sources

Ana Cannas da Silva, Lectures on Symplectic Geometry (standard reference, not scraped)