mathematics-physics-wiki/docs/en/physics/classical-mechanics/hamiltonian-mechanics/equations-of-hamilton.md

Equations of Hamilton

The Hamiltonian

Definition 1: let \mathcal{L}: (\mathbf{q},\mathbf{q}',t) \mapsto \mathcal{L}(\mathbf{q},\mathbf{q}',t) be the Lagrangian of the system, suppose that the generalized momenta \mathbf{p} are defined in terms of the active variables \mathbf{q}' and the passive variables (\mathbf{q},t) such that

\mathbf{p} = \nabla_{\mathbf{q}'}\mathcal{L}(\mathbf{q},\mathbf{q}',t),

for all t \in \mathbb{R}.

We may now pose that there exists a function that meets the inverse, which can be obtained with Legendre transforms.

Theorem 1: there exists a function \mathcal{H}: (\mathbf{q},\mathbf{p},t) \mapsto \mathcal{H}(\mathbf{q},\mathbf{p},t) such that

\mathbf{q}' = \nabla_{\mathbf{p}} \mathcal{H}(\mathbf{q},\mathbf{p},t),

for all t \in \mathbb{R}. Where \mathcal{H} is the Hamiltonian of the system and is related to the Lagrangian \mathcal{L} by

\mathcal{H}(\mathbf{q},\mathbf{p},t) = \langle \mathbf{q'}, \mathbf{p} \rangle - \mathcal{L}(\mathbf{q},\mathbf{q}',t),

for all t \in \mathbb{R} with \mathcal{L} and \mathcal{H} the Legendre transforms of each other.

??? note "Proof:"

Will be added later.

The equations of Hamilton

Corollary 1: the partial derivatives of \mathcal{L} and \mathcal{H} with respect to the passive variables are related by

\begin{align*} \nabla_{\mathbf{q}} \mathcal{H}(\mathbf{q},\mathbf{p},t) &= - \nabla_{\mathbf{q}} \mathcal{L}(\mathbf{q},\mathbf{q}',t), \ \partial_t \mathcal{H}(\mathbf{q},\mathbf{p},t) &= - \partial_t \mathcal{L}(\mathbf{q},\mathbf{q}',t), \end{align*}

for all t \in \mathbb{R}.

??? note "Proof:"

Will be added later.

Obtaining the equations of Hamilton

\begin{align*} \mathbf{p}' &= -\nabla_{\mathbf{q}} \mathcal{H}(\mathbf{q},\mathbf{p},t), \ \mathbf{q}' &= \nabla_{\mathbf{p}} \mathcal{H}(\mathbf{q},\mathbf{p},t), \end{align*}

for all t \in \mathbb{R}.

Proposition 1: when the Hamiltonian \mathcal{H} has no explicit time dependence it is a constant of motion.

??? note "Proof:"

Will be added later.

To put it differently; a Hamiltonian of a conservative autonomous system is conserved.

Theorem 2: for conservative autonomous systems, the Hamiltonian \mathcal{H} may be expressed as

\mathcal{H}(\mathbf{q},\mathbf{p}) = T(\mathbf{q},\mathbf{p}) + V(\mathbf{q}),

for all t \in \mathbb{R} with T: (\mathbf{q},\mathbf{p}) \mapsto T(\mathbf{q},\mathbf{p}) and V: \mathbf{q} \mapsto V(\mathbf{q}) the kinetic and potential energy of the system.

??? note "Proof:"

Will be added later.

It may be observed that the Hamiltonian \mathcal{H} and generalised energy h are identical. Note however that \mathcal{H} must be expressed in (\mathbf{q},\mathbf{p},t) which is not the case for h.

Proposition 2: a coordinate q_j is cyclic if

\partial_{q_j} \mathcal{H}(\mathbf{q},\mathbf{p},t) = 0,

for all t \in \mathbb{R}.

??? note "Proof:"

Will be added later.

Proposition 3: the Hamiltonian is seperable if there exists two mutually independent subsystems.

??? note "Proof:"

Will be added later.

Poisson brackets

Definition 2: let G: (\mathbf{q},\mathbf{p},t) \mapsto G(\mathbf{q},\mathbf{p},t) be an arbitrary observable, its time derivative may be given by

\begin{align*} d_t G(\mathbf{q},\mathbf{p},t) &= \sum_{j=1}^f \Big(\partial_{q_j} G q_j' + \partial_{p_j} G p_j' \Big) + \partial_t G, \ &= \sum_{j=1}^f \Big(\partial_{q_j} G \partial_{p_j} \mathcal{H} - \partial_{p_j} G \partial_{q_j} \mathcal{H} \Big) + \partial_t G, \ &\overset{\mathrm{def}}= {G, \mathcal{H}} + \partial_t G. \end{align*}

for all t \in \mathbb{R} with \mathcal{H} the Hamiltonian and \{G, \mathcal{H}\} the Poisson bracket of G and \mathcal{H}.

The Poisson bracket may simplify expressions; it has distinct properties that are true for any observables. The following theorem demonstrates the usefulness even more.

Theorem 3: let f: (\mathbf{q}, \mathbf{p}, t) \mapsto f(\mathbf{q}, \mathbf{p}, t) and g: (\mathbf{q}, \mathbf{p}, t) \mapsto f(\mathbf{q}, \mathbf{p}, t) be two integrals of Hamilton's equations given by

\begin{align*} f(\mathbf{q}, \mathbf{p}, t) = c_1, \ g(\mathbf{q}, \mathbf{p}, t) = c_2, \end{align*}

for all t \in \mathbb{R} with c_{1,2} \in \mathbb{R}. Then

{f,g} = c_3

with c_3 \in \mathbb{R} for all t \in \mathbb{R}.

??? note "Proof:"

Will be added later.