The geometrical interpretation of the Poisson bracket

Question

"Hamiltonian mechanics is geometry in phase spase."

The Poisson bracket arises naturally in Hamiltonian mechanics, and since this theory has an elegant geometric interpretation, I'm interested in knowing the geometrical interpretation of the Poisson bracket.

I've read somewhere that the Poisson bracket of two functions $f$ and $g$ of the dynamical variables $(q_i,p_i,t)$ given by:
$$\{f,g\}=\sum_i \left(\frac{\partial f}{\partial q_i}\frac{\partial g}{\partial p_i}-\frac{\partial f}{\partial p_i}\frac{\partial g}{\partial q_i}\right)$$

is the dot product in phase space of the 'ordinary' gradient of $f$ and the symplectic gradient of $g$. Let me illustrate this interpretation..

As I know, the gradient in phase space is $(\partial_q,\partial_p)$, and the symplectic gradient is $(\partial_p,-\partial_q)$ (i.e is just the gradient rotated by $90^\circ$ clockwise). I think the dot product is apparent now.

Questions.

• What is the meaning/significance of the dot product interpretation?
• Is there any other geometrical interpretation of the Poisson bracket?

Answer 1

The importance comes from the equations of motions: $$ \dot{q}=\frac{\partial H}{\partial p}\, ,\qquad \dot{p}=-\frac{\partial H}{\partial q} $$ which can be rewritten as $$ \dot{q}=\{q,H\}\, ,\qquad \dot{p}=\{p,H\} $$ In particular, for an arbitrary function of $f(p,q)$, we have $$ \frac{d}{dt}f(p,q,t)=\{f,H\}+\frac{\partial f}{\partial t}\, . $$ Geometrically, changes of coordinates in phase space (i.e. canonical transformation) preserve the Poisson bracket, i.e. the transformation $(q,p)\to (Q(q,p),P(q,p)$ is such that $$ \{q,p\}=\{Q,P\}=1\, , $$ and so in this sense, and thinking of $q$ and $p$ as "basis vectors", canonical transformation preserve the Poisson bracket much like rotation preserve the dot product between two vectors. In this interpretation a quantity is conserved (i.e. $\dot f(q,p,t)=0$) when it is Poisson-orthogonal to the Hamiltonian, for instance.

Answer 2

The Poisson bracket is the bracket of a Lie algebra defined by the symplectic 2-form.

That's a lot to unpack so let's go through it slowly. A 2-form $\omega$ is an anti-symmetric two-tensor $\omega_{\mu\nu}$. If $\omega_{\mu\nu}x^\nu \neq 0$ at points where $x^\nu \neq 0$, then $\omega$ is said to be non-degenerate. So $\omega$ is like the metric tensor, except it's anti-symmetric instead of symmetric. To be symplectic, the "curl" ("exterior derivative") of $\omega$ should also be zero, $(d\omega)_{\mu\nu\rho} = \partial_{[\mu} \omega_{\nu\rho]} = 0$ where the brackets indicate complete anti-symmetrization.

Since $\omega_{\mu\nu}$ is non-degenerate, like the usual metric tensor, it defines an isomorphism between vectors (index up) and one-forms (index down). If $f$ is a scalar function, then $\partial_\mu f$ is naturally a one-form. With this isomorphism, we can define an associated vector field $X_f^\mu$. Note that in terms of concrete components, what this is does is quite different from the usual operation of raising and lowering indices in relativity. Viz., in 2 dimensions, any symplectic 2-form can be represented by the matrix $\begin{pmatrix}0 & 1 \\ -1 & 0 \end{pmatrix}$ so that if $\partial_\mu f$ has components $(a, b)$, $X_f^\mu$ has components $(b,-a)$.

Because we can get one-forms from scalars by taking the gradient, we can define an operation on scalars $f,g$ as $(f,g) \mapsto \omega(X_f, X_g)$. One can verify that this operation is linear in both arguments, anti-symmetric, and satisfies the Jacobi idenitity (because of the requirement that the exterior derivative $d\omega$ vanishes), so it defines a Lie algebra.

If you use the matrix representation of $\omega$ above and that in coordinates $p,q$, the components of $\partial_\mu f$ are $(\partial f/\partial p, \partial f/\partial q)$, then you can work out that this coincides with the usual definition of the Poisson bracket in coordinates. The extension to $2n$ dimensions with coordinates $p_i, q_i,\, i = 1,\ldots,n$ is found by replacing $1$ by the $n\times n$ idenity matrix in the matrix above. A theorem by Darboux says that this can always be done locally.

The canonical reference for this is V. I. Arnold, Mathematical Methods of Classical Mechanics.

Answer 3

Combine $q,\,p$ into a single vector $y$, said to live in phase space (which of course is even-dimensional). The PB is $\partial_i f\omega^{ij}\partial_j g$ where $y_i$ is either a component of $q$ or $p$ and $\omega^{ij}\partial_j g$ is the symplectic gradient, and the matrix/$2$-form $\omega$ is left as an index to the reader. The geometric interpretation is that phase space is a symplectic manifold, a manifold equipped with a certain kind of $2$-form.