# How can we obtain equations of motion from $\iota_{X_{H}}\omega =dH$?

I don't know if this is an obvious result and I am just missing a trick, so please forgive me, but how do we obtain equations of motion from the following equation. \begin{equation} \iota _{X_{H}}\omega =dH \end{equation} Where $\omega$ is a non-degenerate, closed symplectic 2-form, $X_H$ is a Hamiltonian vector field and $H$ is a Hamiltonian function. I can see how Hamilton's ODEs or the Hamilton-Jacobi equation can be solved to give the equations of motion, but not in this case. I am assuming that this equation can actually be used to find equations of motion?

A great example would be a simple system like the harmonic oscillator or even a free particle! :)

• I don't think you can obtain the equations of motion from that. The symplectic form is a nondegenerate bilinear function, and thus it creates a canonical isomorphism between the tangent and cotangent spaces, similar to a symmetric metric. The equation you stated is, in index notation $\omega_{\mu\nu}(X_H)^\mu=(dH)_\nu$, this is basically an equivalent of "lowering" and "raising" in GR. Sep 2 '15 at 20:11
• Echoing @Uldreth's comment: $\iota _{X_{H}}\omega =dH$ is an identity, or alternatively, a definition of the Hamiltonian vector field $X_H$. It contains no further information. Sep 2 '15 at 20:28

This equation reduces to the Hamiltonian-Jacobi equations in certain specific examples.

The Hamiltonian formalism in symplectic geometry, which you've written above, is dependent on the fact that for a symplectic manifold $(M, \omega)$ and a smooth function $H : M \rightarrow \mathbb{R}$, there exists a unique vector field $X_H$ satisfying

$$\iota_{X_H} \omega = d H$$,

which is the equation you've written. The Hamiltonian dynamics on $(M, \omega)$ associated with the Hamiltonian function $H : M \rightarrow \mathbb{R}$ is the dynamical system where all points follow integral curves of the vector field $X_H.$ What makes this interesting is that the value of the function $H$ is conserved along the integral curves of $X_H$—corresponding to the idea that $H$ is conserved along the dynamical paths. This corresponds to the way that conservation of energy gives rise to your standard Hamilton-Jacobi equations.

This is all a bit abstract, but if we reduce it to the example of phase space, the Hamiltonian-Jacobi equations arise naturally.

To describe a physical system with $n$ degrees of freedom, we let our manifold be $M = \mathbb{R}^{2n}.$ We label $M$ with the coordinates $(q^1, \dots, q^n, p^1, \dots, p^n)$—the standard position-momentum phase space coordinates of the Hamiltonian formalism with which you're already familiar.

We define the symplectic form on $M$ to be $\omega = \sum\limits_{i=1}^n d q^{i} \wedge d p^{i}.$ You can verify for yourself that this form is symplectic. The idea is that the symplectic form tells us how the position and momentum coordinates should be paired with each other.

Now suppose we have a smooth function $H : M \rightarrow \mathbb{R}$ (this is the standard Hamiltonian with which you are familiar—it takes in a configuration in phase space and spits out the associated energy).

Now, our claim is that the dynamics should follow the integral curves of the vector field $X_H.$ So we write the vector field $X_H$ as $\left( \frac{d q^1}{dt}, \dots, \frac{d q^n}{dt}, \frac{d p^1}{dt}, \dots, \frac{d p^n}{dt}\right)$, since it should be the "time derivative" of our desired curve. We also know that $X_H$ is the unique curve satisfying $\iota_{X_H} \omega = d H.$ Well, let's evaluate each side of this separately.

$$\iota_{X_H} \omega = \iota_{X_H} \sum\limits_{i=1}^n d q^{i} \wedge d p^{i} = \sum\limits_{i=1}^n \left[ (d q^{i} \cdot X_H) d p^{i} - (d p^{i} \cdot X_H) d q^{i} \right] = \sum\limits_{i=1}^n \left[ \frac{d q^i}{dt} d p^i - \frac{d p^i}{dt} d q^{i} \right].$$

$$d H = \sum\limits_{i=1}^n \left[ \frac{\partial H}{\partial q^i} d q^i + \frac{\partial H}{\partial p^i} d p^i \right]$$

Equating the two sides (that is, demanding that $X_H$ satisfies our original equation), we find

$$\sum\limits_{i=1}^n \left[ \frac{\partial H}{\partial q^i} d q^i + \frac{\partial H}{\partial p^i} d p^i \right] = \sum\limits_{i=1}^n \left[ \frac{d q^i}{dt} d p^i - \frac{d p^i}{dt} d q^{i} \right],$$

or, equivalently,

$$\frac{\partial H}{\partial q^i} = - \frac{d p^i}{d t}$$

and

$$\frac{\partial H}{\partial p^i} = \frac{d q^i}{d t}.$$

So we've recovered the Hamilton-Jacobi equations using the formula you originally gave.

So the formula $\iota_{X_H} \omega = dH$ is really just a more general form of the Hamilton-Jacobi equation. It is powerful in that it can be used to define dynamics that are similar to our standard physical Hamiltonian dynamics on more complicated, non-trivially curved manifolds. It also gives us room to use some really neat geometric tools (for example, Lie derivatives) in discussing Hamiltonian systems.

EDIT:

How do we guarantee the uniqueness of the vector field for a given Hamiltonian?

This follows almost immediately from the nondegeneracy of the symplectic form $\omega$, along with some basic linear algebra. The fact that $\omega$ is nondegenerate means that there is no nonzero vector field $Y$ such that $\iota_Y \omega = 0.$

We may treat $\omega$ as a map between vector fields and dual vector fields. For any vector field $Y$, $\iota_Y \omega$ is a dual vector field. For example, we have $\iota_{x_H} \omega = d H$ (and $dH$ is a dual vector field).

So given any vector field, $\omega$ gives us a corresponding dual vector field. The nondegeneracy condition on $\omega$ amounts to telling us that this map has trivial kernel and is this injective. That is, $\iota_Y \omega = 0$ if and only if $Y = 0.$

A basic theorem in linear algebra states that if a map between vector spaces of the same dimension is injective, then it is surjective. Since a vector space and its dual vector space have the same dimension, the map given by $Y \mapsto \iota_Y \omega$ is bijective. It therefore has a well-defined inverse. This amounts to saying that for any dual vector field $\xi$, there exists a unique vector field $Y$ satisfying $\iota_Y = \xi.$

The uniqueness of the vector field $X_H$ for a given Hamiltonian is a special case of this result.

• This is a really good answer, and you cleared up a lot for me! I do have one follow up question, how do we guarantee the uniqueness of the vector field for a given Hamiltonian? Many thanks once again :) Sep 2 '15 at 21:08
• Actually giving this some more thought, it be from the Cauchy-Lipschitz theorem for ODEs? Sep 2 '15 at 21:19
• Way easier than that---it comes from the fact that the symplectic form is nondegenerate. I'll modify my answer to answer this question, since I need some TeX to answer it fully. Sep 3 '15 at 1:18