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A curve solves Hamilton's equations if and only if its velocity annihilates $d (p \cdot d q - H d t)$ , with the three coordinate components computed

Statement

Let $J \subseteq R$ be a nonempty open interval, let $W \subseteq R^{2 n}$ be open and nonempty with coordinates $z = (q, p) = (q^{1}, \dots, q^{n}, p_{1}, \dots, p_{n})$ , and let $H : J \times W \to R$ be of class $C^{1}$ ( $C^{k}$ Euclidean maps and diffeomorphisms, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ ). On $J \times W$ let $α : = \sum_{i = 1}^{n} p_{i} d q^{i} - H d t$ , a coordinate one-form with $C^{1}$ coefficients. Define its coordinate exterior derivative by $d α : = \sum_{a < b} (\partial_{a} α_{b} - \partial_{b} α_{a}) d y^{a} \land d y^{b}$ in the ordered coordinates $y = (t, q^{1}, \dots, q^{n}, p_{1}, \dots, p_{n})$ ; this is a continuous coordinate two-form, using the same alternating multilinear convention as A smooth differential $k$ -form but making no smooth-form claim. In expanded form, set

$σ : = d α = \sum_{i = 1}^{n} d p_{i} \land d q^{i} + \sum_{i = 1}^{n} H_{q^{i}} d t \land d q^{i} + \sum_{i = 1}^{n} H_{p_{i}} d t \land d p_{i} .$

Every coefficient of $σ$ is continuous because $H$ is $C^{1}$ . Let $J^{'} \subseteq J$ be a nonempty open interval and let $z : J^{'} \to W$ be a $C^{1}$ curve. Then the following are equivalent:

$z$ satisfies Hamilton's equations on $J^{'}$ : ${\dot{q}}^{i} (t) = H_{p_{i}} (t, z (t))$ and ${\dot{p}}_{i} (t) = - H_{q^{i}} (t, z (t))$ for $i = 1, \dots, n$ and every $t \in J^{'}$ ;
for every $t \in J^{'}$ the covector $ι_{(1, \dot{z} (t))} σ$ vanishes, that is $σ_{(t, z (t))} ((1, \dot{z} (t)), w) = 0$ for every $w \in R \times R^{2 n}$ .

Facts & Assumptions

Given: The open sets $J \subseteq R$ and $W \subseteq R^{2 n}$ , the $C^{1}$ function $H$ , the two-form $σ$ , and a $C^{1}$ curve $z : J^{'} \to W$ .

[F1]

Coordinates and forms (A smooth differential $k$ -form, Directional derivatives and partial derivatives of a map $U \subseteq R^{m} \to R^{n}$ , $C^{k}$ Euclidean maps and diffeomorphisms): the coordinate exterior derivative of a $C^{1}$ one-form is defined by antisymmetrizing its first coefficient derivatives; applied to $α = \sum_{i} p_{i} d q^{i} - H d t$ , it gives the continuous two-form $σ = \sum_{i} d p_{i} \land d q^{i} + \sum_{i} H_{q^{i}} d t \land d q^{i} + \sum_{i} H_{p_{i}} d t \land d p_{i}$ , and $d H = H_{t} d t + \sum_{i} (H_{q^{i}} d q^{i} + H_{p_{i}} d p_{i})$ . The standard basis at a point is $\partial_{t}, \partial_{q^{1}}, \dots, \partial_{q^{n}}, \partial_{p_{1}}, \dots, \partial_{p_{n}}$ .

[F2]

Curves (First-order systems, initial value problems, and solutions on intervals, Intervals of $R$ : the nine order-convex forms, nondegeneracy, and length, The Jacobian matrix of partial derivatives and the gradient in the scalar-valued case): $z$ is differentiable on the interval $J^{'}$ with derivative $\dot{z}$ ; the tangent vector $(1, \dot{z} (t))$ of the graph curve $t \mapsto (t, z (t))$ has components $(1, {\dot{q}}^{1}, \dots, {\dot{q}}^{n}, {\dot{p}}_{1}, \dots, {\dot{p}}_{n})$ .

Proof

technique · direct. Compute the two-form in coordinates, evaluate its contraction with the graph velocity on the standard basis vectors, and read off Hamilton's equations

1.1F1algebra

By [F1] and the defining coordinate formula, $d α = \sum_{i} d p_{i} \land d q^{i} - d H \land d t = \sum_{i} d p_{i} \land d q^{i} - H_{t} d t \land d t - \sum_{i} H_{q^{i}} d q^{i} \land d t - \sum_{i} H_{p_{i}} d p_{i} \land d t$ ; since $d t \land d t = 0$ , $d q^{i} \land d t = - d t \land d q^{i}$ and $d p_{i} \land d t = - d t \land d p_{i}$ , this is the continuous two-form $σ$ displayed in the Statement.

2.1F1F2algebra

Write $v : = (1, \dot{z} (t))$ . Evaluating the three types of basis vectors with the coordinate expression of step 1.1: $σ (v, \partial_{q^{i}}) = d p_{i} (v) d q^{i} (\partial_{q^{i}}) - d H \land d t (v, \partial_{q^{i}}) = {\dot{p}}_{i} + H_{q^{i}}$ , because $d p_{i} (v) = {\dot{p}}_{i}$ , $d q^{i} (\partial_{q^{i}}) = 1$ and $(d H \land d t) (v, \partial_{q^{i}}) = d H (v) \cdot 0 - 1 \cdot H_{q^{i}}$ ; $σ (v, \partial_{p_{i}}) = d p_{i} (v) d q^{i} (\partial_{p_{i}}) - (d H \land d t) (v, \partial_{p_{i}}) = 0 - {\dot{q}}^{i} \cdot 1 + H_{p_{i}} = H_{p_{i}} - {\dot{q}}^{i}$ ; and $σ (v, \partial_{t}) = \sum_{j} d p_{j} (v) d q^{j} (\partial_{t}) - \sum_{j} d q^{j} (v) d p_{j} (\partial_{t}) - (d H \land d t) (v, \partial_{t}) = - (H_{q^{i}} {\dot{q}}^{i} + H_{p_{i}} {\dot{p}}_{i})$ summed over $i$ , the $d q, d p$ terms vanishing because $\partial_{t}$ is annihilated by $d q^{j}, d p_{j}$ .

3.1step 2.1algebra

Equivalence: if $z$ satisfies Hamilton's equations, then step 2.1 gives $σ (v, \partial_{q^{i}}) = 0$ , $σ (v, \partial_{p_{i}}) = 0$ and $σ (v, \partial_{t}) = 0$ , so the covector vanishes on the whole basis and (2) holds; conversely if (2) holds, then $σ (v, \partial_{q^{i}}) = 0$ and $σ (v, \partial_{p_{i}}) = 0$ give ${\dot{p}}_{i} = - H_{q^{i}}$ and ${\dot{q}}^{i} = H_{p_{i}}$ for every $i$ and $t$ , which is (1).

4.1step 1.1step 2.1step 3.1∎

Collecting step 1.1, step 2.1 and step 3.1 proves the two-form identity, the component computations and the equivalence.

Remarks

Sign conventions. The two-form is $σ = \sum_{i} d p_{i} \land d q_{i} - d H \land d t$ , the exterior derivative of the Poincare-Cartan one-form $\sum_{i} p_{i} d q^{i} - H d t$ ; the component signs above match the convention in which the one-form is the phase-space action integrand, as in the cited treatments. No symplectic form $ω$ is used in the statement or the proof, so this lemma is independent of the sign convention $ι_{X} ω = d H$ .

A curve solves Hamilton's equations if and only if its velocity annihilates $d (p \cdot d q - H d t)$ , with the three coordinate components computed

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A curve solves Hamilton's equations if and only if its velocity annihilates d(p⋅dq−H dt)d(p⋅dq−Hdt), with the three coordinate components computed

Statement

Facts & Assumptions

Proof

Remarks

Depends on

Dependency tree · 2 levels

Sources

A curve solves Hamilton's equations if and only if its velocity annihilates $d (p \cdot d q - H d t)$ , with the three coordinate components computed