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In classical Mechanics, momentum and position can be paired together to form a symplectic manifold. If you have the simple harmonic oscillator with energy $H = (k/2)x^2 + (m/2)\dot{x}^2$. In this case, the orbits are ellipses. How is the vector field determined by the (symplectic) gradient, then?

Also, does anyone know an interpretation for the area inside a closed curve in phase space?

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The symplectic area contained in a closed curved, that is the boundary of map of a disc, is the "action along the curve". $$ \int_\sigma \omega = \int_\sigma d\lambda = \int_{\partial \sigma} \lambda = \int_0^{2\pi} \lambda_{\gamma(t)}(\dot \gamma(t)) dt, $$ where $\sigma$ is a smooth map from the disc to $M$, and $\gamma = \partial \sigma$. In all cases, the pullback of the 2-form $\omega$ by $\sigma$ is exact since the disc is contractible, so there exists a primitive $\lambda$, on the disc, and you apply Stokes' theorem.

[I apologize for the lengthy answer]

Let me try to elaborate a little bit on a not too complicate but not that simple example to see where the symplectic form makes sense. Let us consider a point on the sphere $S^2$, let $$ TS^2 = \{ (x,v) \in S^2 \times {\bf R}^3 \mid x \cdot v = 0 \} $$ Let $$ L : TS^2 - S^2 \to {\bf R} \quad \mbox{with} \quad L(x,v) = \Vert v \Vert $$ be the "length function" as lagrangian. And you look for the variational problem $$ \delta \int L(x(t),\dot x(t))\ dt = \delta \int \Vert \dot x(t) \Vert\ dt = 0. $$ I don't put the limits of the integral on purpose, it would lead to a too long discussion. Since the lagrangian is homogeneous of degree 1 in $v$, we have the Euler identity $$ L(x,v) = \frac{\partial L(x,v)}{\partial v}(v) $$ And the nature of the partial derivative involved above is a map from $TS^2-S^2$ to the cotangent $T^*S^2$ $$ \forall v \in T_xS^2 - \{0\}, \quad \frac{\partial L(x,v)}{\partial v} = \frac{\bar v}{\Vert v \Vert} \in T^*_xS^2 $$ where the bar denotes the transposed, that is $\bar v w = v \cdot w$. Let's call this map $P$ $$ P : TS^2 - S^2 \to T^*S^2 \quad \mbox{with} \quad P(x,v) = \left(x,\frac{\partial L(x,v)}{\partial v}\right) = \left(x, \frac{\bar v}{\Vert v \Vert}\right). $$ Now let $\lambda = pdx$ the Liouville form on $T^*S^2$, its pullback by $P$, integrated along the curve $\gamma = [t \mapsto (x(t),\dot x(t))]$ is exactly the action $$ \int \Vert \dot x(t) \Vert \ dt = \int_\gamma P^*(\lambda) = \int_{P \circ \gamma} \lambda. $$ Now, let $\tilde \gamma = P \circ \gamma$, this is a path in the image $Y$ of $P$, which is the unit-cotangent bundle $$ Y = {\rm Im}(P) = \{ (x, \bar u) \in T^*S^2 \mid \bar u u = 1 \} $$ And the variational condition becomes then $$ \delta \int_{\tilde \gamma} \lambda = \int d\lambda\left(\delta\tilde\gamma(t), \frac{d\tilde \gamma}{dt}\right)\ dt = 0. $$ But $\varpi = d\lambda$ is a 2-form on $Y \simeq US^2 \simeq SO(3)$ which is of odd dimension, actually $3= 2\times 2 -1$. Now, $\varpi$ has a kernel of dimension 1, and $\gamma$ is a solution of the variational problem if and only if $$ \frac{d\tilde \gamma}{dt} \in \ker \varpi_{\tilde \gamma(t)} $$ In this case, the kernel is given explicitly by $$ \frac{dx}{dt} = \alpha u \quad \mbox{and} \quad \frac{du}{dt}= -\alpha x. $$ The quotient space ${\cal S} = Y/\ker\varpi$, the space of solutions of the variational problem, is then equivalent to the sphere $S^2$, thanks to the (SO(3)-moment map) $$ \pi : (x,u) \mapsto x \times u. $$ By construction this space inherits a symplectic form $\omega$ such that $$ \pi^*(\omega) = \varpi. $$ And $({\cal S}, \omega)$ is the space of oriented non parametrized geodesics of the sphere $S^2$ (which by chance is also a sphere $S^2$). Finally what do we get? A space $Y \simeq US^2 \simeq SO(3)$ made of couples $(x,u)$ or matrices $y=[x\ u \ x \times u]$, a 1-form $\lambda$, the "action-form" (actually called the "Cartan 1-form"), a characteristic distribution $y \mapsto \ker(d\lambda)$ whose leaves are the pre-images of the point of the sphere $S^2$ by the moment map $\mu : (x,u) \mapsto x \times u$, and the image of $\mu$ is a symplectic manifolds for the projection $\omega$ of $d\lambda$. Note that in this case $\omega$, proportional to the standard area-form, is closed but not exact.

Now you can ask the same question as previously: "What does mean the area include in a disc $\sigma : D^2 \to {\cal S}$?"

Consider the pullback by $\sigma$ of the $S^1$-principal bundle $\pi : Y \to {\cal S}$, this is a principal bundle on $D^2$, but $D^2$ is contractible, so this fiber bundle is trivial, thus it admits a smooth section, that is a lift $\tilde \sigma : D^2 \to Y$, that is $\pi \circ \tilde \sigma = \sigma$. Now, $$ \int_\sigma \omega = \int_{\pi\circ\tilde\sigma} \omega = \int_{\tilde\sigma} \pi^*(\sigma) = \int_{\tilde\sigma} d\lambda = \int_{\tilde\gamma} \lambda \quad \mbox{with} \quad \tilde\gamma = \partial\tilde\sigma. $$ Let us write $\tilde \gamma(s) = (x_s,\bar u_s) \in Y$, and let us assume that the parameter $s$ runs over $[0,2\pi]$ to describe $\tilde \gamma = \partial \tilde \sigma$, then $$ \int_\sigma \omega = \int_{\tilde\gamma} \lambda = \int_0^{2\pi} \bar u_s \frac{dx_s}{ds} \ ds. $$ And this is the action of the unit vector $s \mapsto u_s$ distribution along the curve $s \mapsto x_s$. And let us remember that the vector $x_s \times u_s$ describes a geodesic of the sphere $S^2$ for all $s$, and $s$ is not the time parameter of this geodesic.

Note 1. that this construction can be applied to any homogeneous lagrangian, and for non-homogeneous lagrangian, first we homogenize them and after we apply this construction.

Bibliography Jean-Marie Souriau, "Structure des Systèmes Dynamiques", Dunod ed., Paris 1970

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Is that primitive known as the "Lagrangian"? (In mechanics sense.) – john mangual Jan 22 '11 at 16:36
It is related, I am afraid that giving a precise answer to your question could lead us too far :-) But in your simple case if $\omega = dp \wedge dq$ (let's say $q=x$ and $p = \dot x$), ${\bf R}^2$ being simply connected there is no major problem, you can chose once and for all $\lambda = p dq$, a standard choice. Then the action is $$\int_0^{2\pi} p(t)\frac{dq(t)}{dt} dt$$, where $\gamma : t \mapsto (p(t), q(t)$. If you choose another primitive, the difference is the differential of a real function which doesn't contribute to the integral on the loop. – Patrick I-Z Jan 22 '11 at 17:22
Wonderfully informative answer! No apologies necessary; I (and I am sure, others) appreciate the tutorial. :-) – Joseph O'Rourke Jan 24 '11 at 23:15

For one interpretation of the area inside a curve in phase space, see Arnold's Mathematical Methods of Classical Mechanics page 20. In case you do not have a copy of the book, he defines a function $S: (E_0 - \epsilon, E_0 + \epsilon) \to \mathbb{R}$ which gives the area inside the curve associated to an energy level $E$ (assuming this is well defined). The problem on page 20 asserts that $T = \frac{dS}{dE}$ where $T$ is the period of motion along the curve.

Note: The relevant page is available on google books.

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You can view $\mathbb{R}^{2n}$ as a quotient of the real Heisenberg group $\mathcal{H}^{2n+1}$ modulo its center. For a closed loop $\alpha$ in $\mathbb{R}^{2n}$ and a point in $\mathcal{H}^{2n+1}$ over $\alpha(0)$ there's unique lift $\tilde{\alpha}$ of $\alpha$ to $\mathcal{H}^{2n+1}$ going through this point. The symplectic area enclosed by $\alpha$ expresses the signed distance from $\tilde{\alpha}(0)$ to $\tilde{\alpha}(1)$ with respect to a left invariant Riemannian metric on $\mathcal{H}^{2n+1}$.

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Which is also the "action". Actually this is always true, there exists always above a symplectic manifold (or not symplectic, or not a manifold) a principal bundle with group the torus of periods of the closed 2-form, that is the quotient of the real line by the group of periods (can be not a Lie group), equipped with a connection 1-form with curvature the 2-form, and any loop can be lifted horizontally (in the kernel of the connexion), such that the area enclosed is equal to the "length" of the path along the fiber of the lift. – Patrick I-Z Jan 23 '11 at 18:47

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