Poisson brackets: prove that they are canonical invariants

Question

I need a clarification about Poisson brackets.

I'm studying on Goldstein's Classical Mechanics (1 ed.).

Goldstein proves that Poisson brackets are canonical invariants for any functions F and G.

But there is a step that I can't understand.

After some steps, he says that:

$$ \tag{1} [F, G]_q,_p = \sum_k ( \frac { \partial G}{\partial Q_k} [F,Q_k]_q, _p +\frac {\partial G}{\partial P_k}[F, P_k]_q, _p)$$

After other steps, he writes:

$$ \tag{2}[F,Q_k]= - \frac {\partial F}{\partial P_k}$$

and $$ \tag{3}[P_k, F]_q, _p = \sum_j \frac {\partial F}{\partial Q_j} [P_k, Q_j] + \sum_j \frac {\partial F}{\partial P_j}[P_k, P_j]$$

$$\Rightarrow \tag {4} [F,P_k]=\frac {\partial F}{\partial Q_k}$$

and now he replaces these relations in the first expression I have written, obtaining:

$$\tag {5}[F, G]_q, _p=[F, G]_Q, _P$$

Why does he obtain in the second last step $\frac {\partial F}{\partial Q_k}$ and not $-\frac {\partial F}{\partial Q_k}$? $[P_k, F]=-[F, P_k]$ isn't it?

EDIT: Golstein starts from (1) and substituites $Q_k$ to $F$ and $F$ to $G$ and so he obtains (2).

Then he substitutes $P_k$ to $F$ and $F$ to $G$ and obtains (3).

Immediately after he writes (4), that according to me is opposite to (3).

And so I have thought to a printing error. I have tried to substitute $-P_k$ to $F$ and I have obtained

$$[-P_k, F]=\frac {\partial F}{\partial Q_k}$$

Then, as $[-P_k, F]=[F,P_k]$, I can say that $$ [F,P_k]=\frac {\partial F}{\partial Q_k}$$

And so I can obtain (5).

Could you confirm that my argumentation is correct?

Answer 1

Maybe it will be interesting to you to see another proof of this fact.

Firstly, define $2n$ vector $\textbf{x}=(q_1,...,q_n,p_1,...,p_n)^T$ and $2n\times2n$ matrix $J$

$$J=\begin{pmatrix} 0 & 1\\ -1 & 0 \end{pmatrix}$$

With this notation we can write Hamiltonian equation in this way: $$\dot{\textbf{x}}=J\frac{\partial H}{\partial \textbf{x}}$$.

Write transformation in the form $x_i \rightarrow y_i(x)$ where $i=1,...,2n$.

Then we have $$\dot{y_i}=\frac{\partial y_i}{\partial x_j}\dot{x_j}=\frac{\partial y_i}{\partial x_j}J_{jk}\frac{\partial H}{\partial x_k}=\frac{\partial y_i}{\partial x_j}J_{jk}\frac{\partial H}{\partial y_l}\frac{\partial y_l}{\partial x_k}$$ If we define $I_{ij}=\frac{\partial y_i}{\partial x_j}$ (Jacobian) then

$$\textbf{y}=IJI^T\frac{\partial H}{\partial \textbf{y}}$$

So Hamilton's equation are left invariant if

$$IJI^T=J$$

If this holds, the Jacobian $I$ is said to be symplectic. And a change variables with a

symplectic Jacobian is said to be canonical transformation.

In this notation we can define Poisson bracket as $$\left\{f,g \right\}=\frac{\partial f}{\partial x_i}J_{ij}\frac{\partial g}{\partial x_j}$$

Now theorem.

Theorem: The poisson bracket is invariant under canonical transformations.

Proof: if $x_i \rightarrow y_i(x)$- is canonical transformation, we have $\frac{\partial f}{\partial x_i}=\frac{\partial f}{\partial y_k}I_{ki}$, so $$\left\{f,g \right\}=\frac{\partial f}{\partial y_k}I_{ki}J_{ij}I_{lj}\frac{\partial g}{\partial y_l}=\frac{\partial f}{\partial y_k}J_{kl}\frac{\partial g}{\partial y_l}$$ $\blacksquare$

Answer 2

$[F,P_k]=\frac{\partial F}{\partial Q_i}\frac{\partial P_k}{\partial P_i}-\frac{\partial F}{\partial P_i}\frac{\partial P_k}{\partial Q_i}$.

Second term is zero and first term is

$\frac{\partial F}{\partial Q_i }\delta_{ik}=\frac{\partial F}{\partial Q_k}$