classical mechanics - Meaning of the Poisson bracket as a coordinate transformation

Well, the Possion bracket:

$\{ A(q,p),B(q,p) \} \equiv \sum_{s} \left( \dfrac{\partial A}{\partial q_{s}} \dfrac{\partial B}{\partial p_{s}} - \dfrac{\partial A}{\partial p_{s}} \dfrac{\partial B}{\partial q_{s}} \right)$

is a coordinate transformation according to the words of this Wikipedia page: (link: http://en.wikipedia.org/wiki/Poisson_bracket)

...".It places mechanics and dynamics in the context of coordinate-transformations: specifically in coordinate planes such as canonical position/momentum, or canonical-position/canonical transformation."...

The Jacobian of, for example, $A(q,p)$ and $B(q,p)$ has the same form as the terms in the parenthesis. From what to what coordinates does the bracket transform? Does it make any sense to make comparison (Poisson bracket and the Jacobian)?

Answer

Poisson brackets are closely related also to transformations of a system.

Consider a "generator" $\delta G$ and some quantity $A$. (All the quantities mentioned are functions of $p_i$, $q_i$, and do not depend on time explicitly. So I'll not write the arguments unless it's necessary). The small transformation, generated by $\delta G$ is:

$A \to A+\delta A,\quad\quad \delta A = -\{\delta G, A\}$

One have to be careful with the signs in the definition. I use the one from Wikipedia.

Example 1 -- momentum
$\delta G = \epsilon_i p_i \quad \delta A = -\epsilon_i\{p_i, A\} = \epsilon_i\frac{\partial A}{\partial q_i},$
$\quad \Rightarrow \quad A(p_i,q_i)\to A(p_i,q_i)+\epsilon_i\frac{\partial A}{\partial q_i} = A(p_i,q_i+\epsilon_i)$
The momentum is the generator of translations.

Example 2 -- angular momentum
$\delta G = \epsilon_i e_{ijk} p_jq_k \quad \delta A = -\epsilon_i e_{ijk} \{p_jq_k, A\}$

Here $e_{ijk}$ -- is a Levi-Civita symbol. Expanding:
$\delta A = -\epsilon_i e_{ijk} \sum_\alpha \left(\frac{\partial p_jq_k}{\partial q_\alpha}\frac{\partial A}{\partial p_\alpha}-\frac{\partial p_jq_k}{\partial p_\alpha}\frac{\partial A}{\partial q_\alpha}\right) = -\epsilon_i e_{ijk}\left(p_j\frac{\partial A}{\partial p_k}+q_j\frac{\partial A}{\partial q_k}\right)$

$\quad \Rightarrow \quad A(p_i,q_i)\to A(p_i,q_i)-\epsilon_i e_{ijk}\left(p_j\frac{\partial A}{\partial p_k} + q_j\frac{\partial A}{\partial q_k}\right) = A(R_{ij}p_j,R_{ij}q_j)$
For infinitesimal rotations $R_{ij}$

So the angular momentum is the generator of rotations.

Example 3 -- energy
By using Hamilton equation one can calculate to which transformations the energy corresponds to (it is essentially a "reverse" of what is written here):

$\delta G = \epsilon H \quad \delta A = -\epsilon\{H, A\} = \epsilon\frac{d A}{d t},$

$\quad \Rightarrow \quad A(p_i(t),q_i(t))\to A(p_i(t),q_i(t))+\epsilon\frac{d A}{d t} = A(p_i(t+\epsilon),q_i(t+\epsilon))$

The energy is the generator of time evolution.

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From here one can directly see the relation between symmetries and conservation laws.

If a Hamiltonian is symmetric under certain transformation (and, again, doesn't depend on time explicitly), then the bracket with the corresponding "generator" must vanish $\{\delta G, H\} = 0$. But this also means that the value of the "generator" doesn't change with time.