# What are Hamilton's equations with respect to a nonstandard symplectic form?

Hamilton's equations for a Hamiltonian $H(q,p)$ w.r.t. to a standard symplectic from $\omega = dq \wedge dp$ are $$\dot{q} = \partial H_{p}, \quad \dot{p} = - \partial H_{q}$$

How do Hamilton's equations write w.r.t. a nonstandard symplectic form $F(q,p) dq \wedge dp$, where $F(q,p)$ is some smooth function?

## 2 Answers

1. More generally, let there be given a Poisson manifold $$(M,\pi)$$, where $$\pi ~=~ \frac{1}{2} \pi^{IJ} \frac{\partial}{\partial z^I} \wedge \frac{\partial}{\partial z^J} \tag{1}$$ is a Poisson bi-vector, and $$\{ f, g\}_{PB}~=~\frac{\partial f}{\partial z^I}\pi^{IJ}\frac{\partial g}{\partial z^J} \tag{2}$$ is the corresponding Poisson bracket. Let the Hamiltonian $$H$$ be a globally defined function on $$M$$. Then Hamilton's equations read $$\dot{z}^{I}~=~\{ z^I, H\}_{PB},\tag{3}$$ i.e. time-evolution is given by (minus) the Hamiltonian vector field $$X_H~=~\{H,\cdot\}_{PB}.\tag{4}$$

2. If the Poisson structure is invertible, then $$M$$ is a symplectic manifold with symplectic 2-form $$\omega ~=~\frac{1}{2} \omega_{IJ}~ \mathrm{d}z^I \wedge \mathrm{d}z^J,\tag{5}$$ where $$\omega_{IJ}$$ is the inverse matrix: $$\pi^{IJ}\omega_{JK}~=~\delta^I_K. \tag{6}$$

3. In canonical/Darboux coordinates $$(z^1, \ldots, z^{2n})~=~(q^1, \ldots, q^n,p_1,\ldots, p_n) ,\tag{7}$$ the above construction reduces to the standard Poisson bi-vector $$\pi~=~\frac{\partial}{\partial q^i} \wedge \frac{\partial}{\partial p_i},\tag{8}$$ and the standard symplectic 2-form $$\omega ~=~ \mathrm{d}p_i \wedge \mathrm{d}q^i.\tag{9}$$

• Commented Jan 2, 2017 at 20:03

A Hamiltonian $H:M\rightarrow \mathbb{R}$ defines a vector field $X_H$ through the equation $$\omega(X_H,\cdot)=dH.$$ For $\omega=F(q,p)dq\wedge dp$ and substituting the components $X_H=X_{Hq}\partial_q+X_{Hp}\partial_p$ we get $$F(q,p)(X_{Hq}dp-X_{Hp}dq)=(\partial_qH)dq+(\partial_pH)dp.$$ The integral curves $t\mapsto(q(t),p(t))$ of the vector field $X_H$ represent the Hamiltonian flow of the system. Therefore, we have \begin{align} \dot{q}=\frac{\partial_qH}{F(q,p)};\;\;\; \dot{p}=-\frac{\partial_pH}{F(q,p)}; \end{align}