Physical interpretation of Poisson bracket properties

Question

In classical Hamiltonian mechanics evolution of any observable (scalar function on a manifold in hand) is given as $$\frac{dA}{dt} = \{A,H\}+\frac{\partial A}{\partial t}$$

So Poisson bracket is a binary, skew-symmetric operation $$\{f,g\} = - \{f,g\}$$ which is bilinear $$\{\alpha f+ \beta g,h\} = \alpha \{f, g\}+ \beta \{g,h\}$$ satisfies Leibniz rule: $$\{fg,h\} = f\{g,h\} + g\{f,h\}$$ and Jacobi identity: $$\{f,\{g,h\}\} + \{g,\{h,f\}\} + \{h,\{f,g\}\} = 0$$

How to physically interpret these properties in classical mechanics? What physical characteristic each of them is connected to? For example I suppose the anticommutativity has to do something with energy conservation since because of it $\{H,H\} = 0$.

Answer 1

Let's assume no explicit time dependence and that our Poisson bracket $\{,\}$ - I prefer curly brackets so square ones $[,]$ can be used to denote the commutator of vector fields - is non-singular, ie there's a corresponding symplectic product $\omega$.

The time derivative $$ \frac{\mathrm d}{\mathrm dt}=\{\,\cdot\,,H\} $$ is actually the Lie derivative with respect to the Hamiltonian vector field $X_H$ given by $$ X_H\rfloor\omega \equiv \mathrm dH $$ in disguise as can be seen by $$ \{f,H\} \equiv \omega(X_f,X_H)=(X_f\rfloor\omega)(X_H)=\mathrm df(X_H)=\mathcal{L}_{X_H}f $$ As $\mathcal{L}_{X_H}$ is a linear differential operator respecting the Leibniz rule, so is $\{\,\cdot\,,H\}$.

Antisymmetry translates to $$ \mathcal{L}_{X_f}g = -\mathcal{L}_{X_g}f $$ ie the change of $g$ with respect to the Hamiltonian flow induced by $f$ is the negative of the change in $f$ with respect to the Hamiltonian flow induced by $g$.

Rewriting the Jacobi identity as $$ \{f ,\{g,h\}\} = \{\{f,g\},h\} - \{\{f,h\},g\} $$ we see that $$ \mathcal{L}_{X_{\{g,h\}}}f=\left(\mathcal{L}_{X_h}\mathcal{L}_{X_g} - \mathcal{L}_{X_g}\mathcal{L}_{X_h}\right)f = \mathcal{L}_{[X_h,X_g]}f $$ ie $f\mapsto X_f$ is a Lie-algebra homomorphism.

Answer 2

The physical interpretation is integrability conditions being satisfied on the manifold. From the first equation, if you would take A not depending on 't' explicitly then dA/dt = [A,H]. The Poisson bracket contains in it the dynamics involved in canonically conjugate variables and in classical mechanics, we can measure them simultaneously. Apart from this, laws of conservation can be explicitly seen in this representation.

One important factor to note is that, Poisson brackets are valid only for exact differentials and they follow the canonical transformations. In fact, canonical transformations are nothing but invariance of Poisson brackets.

Answer 3

If we consider for simplicity a 2d phase space (q,p), then we can interpretate the poisson bracket between two functions f(q,p) and g(q,p) as the vector product of their gradients, which are vector fields in this plane:

$[f,g]=(\nabla f\times \nabla g)\cdot \mathbf{e}_z$

where $e_z$ is a unit vector perpendicular to the plane.

From that definition all the properties are obvious.

We can imagine the following physical analogy for the equation of motion, the gradient of the hamiltonian act like magnetic field $B$ and the gradient of the function is the velocity $v$, in formulas:

$\partial_tf\, \mathbf{e}_z= \nabla f\times \nabla H = \mathbf{v} \times \mathbf{B}$

which is the expression of the Lorentz force.