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# 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.

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# Hamiltonian formalism of mechanics
The Hamiltonian formalism of mechanics is based on the definitions posed by [Lagrangian mechanics](/en/physics/mechanics/lagrangian-mechanics/lagrangian-formalism) and the axioms, postulates and principles posed in the [Newtonian formalism](/en/physics/mechanics/newtonian-mechanics/newtonian-formalism/).
Where the Lagrangian formalism used the [principle of virtual work](/en/physics/mechanics/lagrangian-mechanics/lagrange-equations/#principle-of-virtual-work) to derive the Lagrangian equations of motion, the Hamiltonian formalism will derive the Lagrangian equations with the stationary action principle. A derivative of Fermat's principle of least time.
In Hamilton's formulation the stationary action principle is referred to as Hamilton's principle.
## Hamilton's principle
> *Principle 1*: of all the kinematically possible motions that take a mechanical system from one given configuration to another within a time interval $T \subset \mathbb{R}$, the actual motion is the stationary point of the time integral of the Lagrangian $\mathcal{L}$ of the system. Let $S$ be the functional of the trajectories of the system, then
>
> $$
> S = \int_T \mathcal{L} dt,
> $$
>
> has stationary points.
The functional $S$ is often referred to as the action of the system. With this principle the equations of Lagrange can be derived.
> *Theorem 1*: let $\mathcal{L}: (\mathbf{q}, \mathbf{q'}) \mapsto \mathcal{L}(\mathbf{q}, \mathbf{q'})$ be the Lagrangian, the equations of Lagrange are given by
>
> $$
> \partial_{q_j} \mathcal{L}(\mathbf{q}, \mathbf{q'}) - d_t \Big(\partial_{q_j'} \mathcal{L}(\mathbf{q}, \mathbf{q'}) \Big) = 0,
> $$
>
> for all $t \in \mathbb{R}$.
??? note "*Proof*:"
Let the redefined generalized coordinates $\mathbf{q}: (t,a) \mapsto \mathbf{q}(t,a)$ be given by
$$
\mathbf{q}(t,a) = \mathbf{\hat q}(t) + a \varepsilon(t),
$$
with $\mathbf{\hat q}: t \mapsto \mathbf{\hat q}(t)$ the generalized coordinates of the system and $\varepsilon: t \mapsto \varepsilon(t)$ a smooth differentiable function.
Let $S: a \mapsto S(a)$ be the action of the system and let $\mathcal{L}: (\mathbf{q}, \mathbf{q'}) \mapsto \mathcal{L}(\mathbf{q}, \mathbf{q'})$ be the Lagrangian of the system, according to Hamilton's principle
$$
S(a) = \int_T \mathcal{L}(\mathbf{q}, \mathbf{q'})dt,
$$
for all $a \in \mathbb{R}$. To determine the stationary points we must have that $S'(0) = 0$. We have that $S'$ is given by
$$
\begin{align*}
S'(a) &= \int_T \partial_a \mathcal{L}(\mathbf{q}, \mathbf{q'})dt, \\
&= \int_T \sum_{j=1}^f \bigg(\partial_{q_j} \mathcal{L} \partial_a q_j + \partial_{q_j'} \mathcal{L} \partial_a q_j'\bigg)dt, \\
&= \int_T \sum_{j=1}^f \bigg(\partial_{q_j} \mathcal{L} \varepsilon_j(t) + \partial_{q_j'} \mathcal{L} \partial_a \partial_t q_j\bigg)dt. \\
\end{align*}
$$
Partial integration may be used for the second part:
$$
\begin{align*}
\int_T \partial_{q_j'} \mathcal{L} \partial_a \partial_t q_j dt &= \Big[\partial_{q_j'} \mathcal{L} \partial_a q_j \Big]_T - \int_T \partial_a q_j d_t (\partial_{q_j'} \mathcal{L})dt, \\
&= \Big[\partial_{q_j'} \mathcal{L} \varepsilon_j(t) \Big]_T - \int_T \partial_a q_j d_t (\partial_{q_j'} \mathcal{L})dt.
\end{align*}
$$
Choose $\varepsilon_j$ such that
$$
\Big[\partial_{q_j'} \mathcal{L} \varepsilon_j(t) \Big]_T = 0.
$$
Obtains
$$
\int_T \partial_{q_j'} \mathcal{L} \partial_a \partial_t q_j dt = - \int_T \partial_a q_j d_t (\partial_{q_j'} \mathcal{L})dt.
$$
The general expression of $S'$ may now be given by
$$
\begin{align*}
S'(a) &= \int_T \sum_{j=1}^f \bigg(\partial_{q_j} \mathcal{L} \varepsilon_j(t) - \partial_a q_j d_t (\partial_{q_j'} \mathcal{L})\bigg)dt, \\
&= \int_T \sum_{j=1}^f \bigg(\partial_{q_j} \mathcal{L} \varepsilon_j(t) - \varepsilon_j(t) d_t (\partial_{q_j'} \mathcal{L})\bigg)dt, \\
&= \sum_{j=1}^f \int_T \varepsilon_j(t) \Big(\partial_{q_j} \mathcal{L} - d_t (\partial_{q_j'} \mathcal{L})\Big)dt.
\end{align*}
$$
Then
$$
S'(0) = \sum_{j=1}^f \int_T \varepsilon_j(t) \Big(\partial_{q_j} \mathcal{L} - d_t (\partial_{q_j'} \mathcal{L})\Big)dt = 0,
$$
since $\varepsilon_j$ can be chosen arbitrary this implies that
$$
\partial_{q_j} \mathcal{L} - d_t (\partial_{q_j'} \mathcal{L}) = 0.
$$