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mathematics-physics-wiki/docs/en/physics/classical-mechanics/hamiltonian-mechanics/hamiltonian-formalism.md

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2024-06-20 21:27:12 +02:00
# Hamiltonian formalism of mechanics
2024-04-03 19:55:17 +02:00
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.
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In Hamilton's formulation the stationary action principle is referred to as Hamilton's principle.
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## 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
$$
\partial_{q_j} \mathcal{L} - d_t (\partial_{q_j'} \mathcal{L}) = 0.
$$