# 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](/en/physics/mechanics/lagrangian-mechanics/lagrange-generalizations/#the-generalized-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.