Hamiltonian systems as an EDSs

Coming from contact geometry#Example encompassing time-dependent Hamiltonian.
The exterior differential system generated by the 1-form $α = p_{i} d q^{i} - H d t$ contains all the necessary information to recover the full time-dependent classical Hamiltonian system.

The core idea is that the trajectories of the Hamiltonian system are precisely the integral curves of the characteristic vector field associated with the 2-form $d α$ .

Recovering Hamilton's Equations

The recovery process works by finding the vector field whose flow describes the system's evolution. This vector field is uniquely determined (up to a scalar multiple) by the condition that it lies in the kernel of the 2-form $d α$ . This is a fundamental concept in geometric mechanics.

1. Calculate the Exterior Derivative ( $d α$ )

First, we compute the exterior derivative of the contact form $α$ :

α = p_{i} d q^{i} - H (q, p, t) d t

Using the rules of exterior calculus ( $d (f g) = d f \land g + f \land d g$ and $d (d x) = 0$ ):

d α = d (p_{i} d q^{i}) - d (H d t)

d α = (d p_{i} \land d q^{i}) - (d H \land d t)

Since $H$ is a function of $q$ , $p$ , and $t$ , its differential is $d H = \frac{\partial H}{\partial q^{i}} d q^{i} + \frac{\partial H}{\partial p_{i}} d p_{i} + \frac{\partial H}{\partial t} d t$ . Substituting this in:

d α = d p_{i} \land d q^{i} - (\frac{\partial H}{\partial q^{i}} d q^{i} + \frac{\partial H}{\partial p_{i}} d p_{i} + \frac{\partial H}{\partial t} d t) \land d t

Since $d t \land d t = 0$ and using $d x \land d y = - d y \land d x$ , we get:

d α = d p_{i} \land d q^{i} + \frac{\partial H}{\partial q^{i}} d t \land d q^{i} - \frac{\partial H}{\partial p_{i}} d p_{i} \land d t

2. Find the Characteristic Vector Field ( $X_{H}$ )

The dynamics are described by a vector field $X_{H}$ whose integral curves are the trajectories of the system. We can write this vector field in the coordinates $(q^{i}, p_{i}, t)$ as:

X_{H} = {\dot{q}}^{i} \frac{\partial}{\partial q^{i}} + {\dot{p}}_{i} \frac{\partial}{\partial p_{i}} + 1 \cdot \frac{\partial}{\partial t}

Here, the coefficient of $\frac{\partial}{\partial t}$ is 1 because we parameterize the trajectories by time $t$ .

This vector field $X_{H}$ is the characteristic vector field of $d α$ , meaning it satisfies the condition $i_{X_{H}} d α = 0$ , where $i_{X}$ is the interior product. Let's compute this:

i_{X_{H}} d α = i_{X_{H}} (d p_{i} \land d q^{i}) + \frac{\partial H}{\partial q^{i}} i_{X_{H}} (d t \land d q^{i}) - \frac{\partial H}{\partial p_{i}} i_{X_{H}} (d p_{i} \land d t) = 0

Evaluating each term:

$i_{X_{H}} (d p_{i} \land d q^{i}) = {\dot{p}}_{i} d q^{i} - {\dot{q}}^{i} d p_{i}$
$i_{X_{H}} (d t \land d q^{i}) = 1 \cdot d q^{i} - {\dot{q}}^{i} d t$
$i_{X_{H}} (d p_{i} \land d t) = {\dot{p}}_{i} d t - 1 \cdot d p_{i}$

Substituting back and grouping terms by $d q^{i}$ and $d p_{i}$ :

({\dot{p}}_{i} + \frac{\partial H}{\partial q^{i}}) d q^{i} - ({\dot{q}}^{i} - \frac{\partial H}{\partial p_{i}}) d p_{i} + (- {\dot{q}}^{i} \frac{\partial H}{\partial q^{i}} - {\dot{p}}_{i} \frac{\partial H}{\partial p_{i}}) d t = 0

Wait, the calculation for the $d t$ term is not right. Let's re-calculate $i_{X_{H}} d α$ carefully.

i_{X_{H}} d α = ({\dot{p}}_{i} + \frac{\partial H}{\partial q^{i}}) d q^{i} - ({\dot{q}}^{i} - \frac{\partial H}{\partial p_{i}}) d p_{i} - ({\dot{q}}^{i} \frac{\partial H}{\partial q^{i}} + {\dot{p}}_{i} \frac{\partial H}{\partial p_{i}}) d t

Let's regroup the original expression $d α = d p_{i} \land d q^{i} + \frac{\partial H}{\partial q^{i}} d t \land d q^{i} - \frac{\partial H}{\partial p_{i}} d p_{i} \land d t$ :

i_{X_{H}} d α = ({\dot{p}}_{i} d q^{i} - {\dot{q}}^{i} d p_{i}) + \frac{\partial H}{\partial q^{i}} (1 \cdot d q^{i} - {\dot{q}}^{i} d t) - \frac{\partial H}{\partial p_{i}} ({\dot{p}}_{i} d t - 1 \cdot d p_{i})

Now, collect coefficients of the basis 1-forms $d q^{i}$ , $d p_{i}$ :

= ({\dot{p}}_{i} + \frac{\partial H}{\partial q^{i}}) d q^{i} + (- {\dot{q}}^{i} + \frac{\partial H}{\partial p_{i}}) d p_{i} = 0

For this 1-form to be zero, all of its coefficients must be zero. This gives us two sets of equations:

{\dot{p}}_{i} = - \frac{\partial H}{\partial q^{i}} and {\dot{q}}^{i} = \frac{\partial H}{\partial p_{i}}

These are precisely Hamilton's equations of motion. ✅

The Role of $α$ Itself

Evaluating the contact form $α$ along the trajectory provides insight:

α (X_{H}) = (p_{i} d q^{i} - H d t) ({\dot{q}}^{j} \frac{\partial}{\partial q^{j}} + {\dot{p}}_{j} \frac{\partial}{\partial p_{j}} + \frac{\partial}{\partial t}) = p_{i} {\dot{q}}^{i} - H

This is the definition of the Lagrangian $L$ via the Legendre transform ( $L = p_{i} {\dot{q}}^{i} - H$ ). This is a consistency condition that must hold along the physical path, connecting the Hamiltonian $H$ to the velocities ${\dot{q}}^{i}$ and momenta $p_{i}$ .

Hamiltonian systems as an EDSs

Recovering Hamilton's Equations

The Role of α Itself

The Role of $α$ Itself