In the paper by Littlejohn, 1983, the canonical Hamiltonian $h_c$ of a charged particle in electromagnetic field is given by,
$$ h_c (\vec{q}, \vec{p}, t) = \frac{1}{2m} \left[ \vec{p} - \frac{e}{c} \vec{A}(\vec{q},t) \right]^2 + e\phi(\vec{q},t) \tag{1} $$
Here, $(\vec{q},\vec{p})$ represents the canonical position and momentum of the charged particle.
Let $z = z(\vec{q},\vec{p},t)$ represent an arbitrary coordinate system on phase space. The Poincare-Cartan differential one-form is given by,
$$ \gamma = \sum_{i=1}^{6} \gamma_i \mathrm{d}z_i - h\mathrm{d}t \tag{2} $$
where
$$ \gamma_i(z,t) = \vec{p} \cdot \frac{\partial \vec{q}}{\partial z_i} \tag{3} $$
$$ h(z,t) = h_c - \vec{p} \cdot \frac{\partial \vec{q}}{\partial t} \tag{4} $$
Let us now write $z = z(\vec{x},\vec{v})$, where the $\vec{x}$ is the particle position and $\vec{v}$ is its velocity. They are related to canonical coordinates as, $\vec{x} = \vec{q}$ and $\vec{v} = \frac{1}{m} \left( \vec{p} - \frac{e}{c} \vec{A}(\vec{q},t) \right)$.
Using (2), (3) and (4), then Poincare-Cartan form for charged particle is (given in reference Eq(18)),
$$ \gamma = \left[m\vec{v} + \frac{e}{c} \vec{A}(\vec{x},t) \right] \cdot \mathrm{d} \vec{x} - \left( \frac{1}{2}m \vec{v}^2 + e\phi(\vec{x},t) \right) \mathrm{d}t \tag{5} $$
Question?
What happened to the second term in (4), i.e. $\vec{p} \cdot \frac{\partial \vec{q}}{\partial t}$ Am I missing anything here?
