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"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?
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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 $$ \{q,p\}=\{Q,P\}=0\, , $$ 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.

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  • $\begingroup$ Great answer! so saying that $f$ is "Poisson-orthogonal" to $H$ is equivalent to saying that $f$ and $H$ are in mutual involution w (the PB is null). It makes sense to me. $\endgroup$ – Samà Jul 23 '17 at 20:07
  • $\begingroup$ Since you talked about canonical transformations in your answer, is there any relation between the Poisson bracket and the Jacobian of the transformation? $\endgroup$ – Samà Jul 23 '17 at 20:13
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    $\begingroup$ @Samà Yes. In 2d systems, with $(p,q)$ as coordinates, a necessary and sufficient condition for a transformation to be canonical is that its Jacobian be 1. Unfortunately, this condition does not hold for higher dimensional system: one must verify that full set of canonical bracket relations is preserved. $\endgroup$ – ZeroTheHero Jul 23 '17 at 20:15
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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.

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    $\begingroup$ This is in the direction of what I was thinking about writing here. The one big thing that is still missing is an explanation of why there is a symplectic structure (i.e. Why are we on the cotangent bundle and why is that symplectic?). $\endgroup$ – Danu Jul 24 '17 at 6:56
  • $\begingroup$ @Danu the answers to all such questions are in Arnold's book. Heuristically, we are on a cotangent bundle because $p$, the canonical momentum, is a one-form in a natural way. John Baez has some notes about thats here: math.ucr.edu/home/baez/classical/cm05week06.pdf The cotangent bundle is symplectic because, as one can convince oneself, $p,q$ are Darboux coordinates for any coordinates $q$ on the base manifold. $\endgroup$ – Robin Ekman Jul 24 '17 at 9:48
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    $\begingroup$ Sure, I was just suggesting that you add something to this effect to the answer to complete it ;) $\endgroup$ – Danu Jul 24 '17 at 10:24
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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.

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    $\begingroup$ I'm not quite familiar with differential geometry but I think I got your point. A follow up question: can I say that the PB is, in a sense, the metric of the symplectic manifold?! $\endgroup$ – Samà Jul 23 '17 at 19:51
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    $\begingroup$ @Samà the uptick is from me, I want to know the answer to that good question in your comment as well. +1 earlier for the question $\endgroup$ – user163104 Jul 23 '17 at 19:54
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    $\begingroup$ in a sense, yes. With the symplectic two-form you can raise and lower indices just like you can with the metric. However the two-form is skew-symmetric where the metric is symmetric and it measures areas as opposed to angles. $\endgroup$ – maze-cooperation Jul 23 '17 at 20:02
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    $\begingroup$ @Samà for a question related to your comment, see here. $\endgroup$ – Danu Jul 23 '17 at 21:47
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    $\begingroup$ @Samà Due to skew-symmetry it has more in common with $3$D cross products, but an even better way to think of it is a commutator. $\endgroup$ – J.G. Jul 23 '17 at 22:25

protected by Qmechanic Jul 24 '17 at 3:24

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