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

Hamiltonian formalism of mechanics

The Hamiltonian formalism of mechanics is based on the definitions posed by Lagrangian mechanics and the axioms, postulates and principles posed in the Newtonian formalism.

Where the Lagrangian formalism used the 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 

$$
    \partial_{q_j} \mathcal{L} - d_t (\partial_{q_j'} \mathcal{L}) = 0.
$$