What does symplecticity imply?

Question

Symplectic systems are a common object of studies in classical physics and nonlinearity sciences.

At first I assumed it was just another way of saying Hamiltonian, but I also heard it in the context of dissipative systems, so I am no longer confident in my assumption.

My question now is, why do authors emphasize symplecticity and what is the property they typically imply with that? Or in other more provocative terms: Why is it worth mentioning that something is symplectic?

Arnold Neumaier · Answer 1 · 2012-07-24T11:14:17.763

Standard Hamiltonian mechanics in $N$-particle phase space $R^{6N}$ is inadequate to describe mechanical systems of interest that are not of the $N$ particle form, for example rigid bodies. However, all major techniques in classical mechanics do not depend on the specific structure of $R^{6N}$ but only on the fact that one can define on it a Poisson bracket.

Thus classical mechanics generalizes without difficulties to mechanics in Poisson manifolds. These are phase spaces on whose smooth function algebra one can define a Poisson bracket with the properties familar from $N$-particle phase space. (For example, the phase space of rigid bodies is the Lie-Poisson manifold of the Lie algebra generating the group of rigid motions.) For conservative classical mechanics in terms of Poisson manifolds see the book
J.E. Marsden and T.S. Ratiu, Introduction to mechanics and symmetry: a basic exposition of classical mechanical systems, Springer 1999.
\url{http://higherintellect.info/texts/science_and_technology/physics/Introduction to Mechanics and Symmetry.pdf}

An important class of Poisson manifolds are the symplectic manifolds, where the Poisson bracket is defined though a symplectic form. (A typical example is the cotangent space of a configuration manifold.) The importance of symplectic manifolds stems from the fact that Poisson manifolds typically foliate into symplectic leaves, and any Hamiltonian dynamics restricted to such a leaf is symplectic.

Edit: Although classical mechanics in tectbooks is usually confined to the conservative case, one can add dissipative terms to an otherwise Hamiltonian mechanics. For example, dissipative classical mechanics for realistic fluids is discussed in terms of Poisson brackets in the book
A.N. Beris and B.J. Edwards, Thermodynamics of flowing systems with internal microstructure, New York 1994

score 5 · Accepted Answer · answered Jul 24 '12 at 10:18

Classical mechanics is the study of second-order systems. The obvious geometric formulation is via semi-sprays, ie second-order vectorfields on the tangent bundle. However, that's not particularly useful as there's no natural way to derive a semi-spray from a function (ie potential).

Lagrangian and Hamiltonian mechanics are two solutions to that problem. While these formalisms are traditionally formulated on the tangent and cotangent bundles (ie velocity and momentum phase space), they were further generalized: Lagrangian mechanics led to the jet-bundle formulation of classical field theory, and Hamiltonian mechanics to the Poisson structure.

The symplectic structure is a stripped-down version of the structure of the cotangent bundle - the part that turned out to be necessary for further results, most prominently probably phase space reduction via symmetries. It doesn't feature prominently in undergraduate mechanics lecture (at least not the ones I attended) because when working in canonical coordinates, it takes a particular simple form - basically the minus in Hamilton's equations, where it's used similarly to the metric tensor in relativity, ie to make a contravariant vector field from the covariant differential of the Hamilton function.

Symplectic geometry also plays its role in thermodynamics: As I understand it, the Gibbs-Duhem relation basically tells us that we're dealing with a Lagrangian submanifold of a symplectic space, which is the reason why the thermodynamical potentials are related via Legendre transformations.

More precisely, in thermodynamics one exploits contact geometry, not symplectic geometry. — Arnold Neumaier, Jul 24 '12 at 11:16

score 3 · Answer 3 · answered Jul 24 '12 at 10:50

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Symplectic geometry is may be the cornerstone of the geometrization of physics. In addition to the very known fact that classical mechanics can be described by symplectic geometry, given some other structures, symplectic spaces can be quantized to produce quantum mechanics as well. A subclass of symplectic geometries namely Kaehler geometry is especially important to quantization problems.

Many physical theories such as Yang-Mills and gravity have descriptions in the context of symplectic geometry, please see the review: THE SYMPLECTIZATION OF SCIENCE by Gotay and Isenberg.

Also many types of dissipative systems can be treated using symplectic geometry if we allow complex Hamiltonians please see S.G. Rajeev's article.

Finally, I want to remark that in the symplectic geometry terminology there is a distinction between symplectic and Hamiltonian vector fields, while a symplectic vector field is required to leave the symplectic structure invariant, a Hamiltonian vector field is required in addition to produce an exact form upon the contraction with the symplectic form. For example the vector fields along the generators of the two-torus are symplectic but not hamiltonian. This distinction exists only if the symplectic manifold is nonsimply connected.

answered Jul 24 '12 at 10:50

David Bar Moshe

30,503

Not all of classical mechanics can be described by symplectic geometry. Rigid bodies need Poisson manifolds. – Arnold Neumaier Jul 24 '12 at 11:08
@Arnold Of course, but symplectic geometry remains important even within Poisson geometry (I think that this is the reason that they don't have an Arxiv subject classification of Poisson geometry). The rigid body example you gave can be formulated on the symplectic manifold $T^SO(3)$ (Euler angles + angular momenta), then Poisson reduced to $\mathbf{so(3)^}$ a Poisson non symplectic manifold, but the dynamics actually takes place on a single coadjoint orbit: $S^2$, agian a symplectic manifold. – David Bar Moshe Jul 24 '12 at 11:24
Then a more complex example is needed. The natural geometry of perfect fluid is also Poisson, not symplectic. Or can you describe a perfect fluid in terms of symplectic geometry? – Arnold Neumaier Jul 24 '12 at 11:32
Even on a Poisson manifold, the dynamics given any particular initial condition are always constrained to a symplectic leaf, so you will have trouble finding examples that are "genuinely Poisson". You might find more interesting examples by looking at systems with constraints, but then even Poisson geometry is not entirely adequate for this--see, e.g., Dirac geometry. – Jonathan Jul 24 '12 at 12:46
@Arnold (in addition to Jonathan's remark). The same happens for the ideal fluid example. The motion is restricted to a single coadjoint orbit of the volume preserving diffeomorphism group. Actually, the integral curves of any Hamiltonian on any Poisson manifold are restricted to a single symplectic leaf. The importance of Poisson geometry lies in that it includes the solutions of all possible initial conditions in the classical case and all inequivalent quantizations after quantization. – David Bar Moshe Jul 24 '12 at 12:56
(cont.) There are also advantages in Poisson geometry, since the symplectic leaves of a Poisson manifold may not be manifolds, and also it is more natural to perform stability analysis within Poisson geometry. – David Bar Moshe Jul 24 '12 at 12:57
David and @Jonathan: By the same reasoning, the dynamics given any particular initial condition is on a 1-dimensional curve, even for symplectic manifolds, so no symplectic geometry were needed. But we need it as we want to study a symplectic system with varying initial conditions. For the same reason, the physically and computationally relevant dynamics in my examples is Poisson and not symplectic. In the fluid case, one cannot even determine the coadjoint orbit on which the dynamics in each case would live. Would it even be a manifold? – Arnold Neumaier Jul 24 '12 at 13:35
The two non-wiki links in this answer(v1) did not work when I tried today. – Qmechanic Jan 17 '13 at 22:31

score 2 · Answer 4 · answered Jul 24 '12 at 09:45

Given a symplectic structure, some awesome results occur. This is seen most obviously in Classical Mechanics as the Wiki-site states.

For instance, in talking about particle motion, you are lead to phase-space, which is the cotangent bundle $T\approx\mathbb{R}^6$ over $\mathbb{R}^3$, and this bundle naturally carries a symplectic structure.

Once you have such a structure, then (as Wiki states verbatim):
Any real-valued differentiable function, on a symplectic manifold can serve as an energy function or Hamiltonian.
You can now discuss gradient flows (like in fluid dynamics), and some conservation statements such as Liouville's theorem.

But yes, the main good thing is that you are now able to get your hands on a differential equation which predicts the future behavior of your system.

"Given a symplectic structure, some awesome results occur.", haha. — Nikolaj-K, Jul 24 '12 at 09:55

score 2 · Answer 5 · edited May 04 '20 at 12:34

Why is it worth mentioning that something is symplectic?

This question is a little like asking why it's worth mentioning that an electric field is in the room.

As an important characteristic, I'd point out that if you have a symplectic structure, you have a Poisson algebra. That means that not only can the functions $$f:P\in \mathcal M\ \longrightarrow\ f(P)\in\mathbb{R}$$ on your manifold do things like

$$(f,g,h,P)\ \longrightarrow\ f(P)g(P)+h(P),$$

but also like

$$(f,g,P)\ \longrightarrow\ \{f,g\}(P).$$

Consequently, if you add a symplectic structure in your function algebra, some awesome results occur. Notice that the structure as well as the manifold you consider might be wild, but the Possion bracket has some qualities to it, which are true in general.

What does symplecticity imply?

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