# Poisson brackets: prove that they are canonical invariants

EDIT: I haven't forgotten to accept answer, the question is still open..

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

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

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? Many thanks

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There are now both a 2nd and a 3rd edition of Goldstein Classical Mechanics. Could you please check if the your problem persists in the newer editions? – Qmechanic Dec 3 '12 at 23:08
@Qmechanic: 2nd and 3rd editions use symplectic approach, and so they don't answer to my question... – sunrise Dec 5 '12 at 9:28

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

Second term is zero and first term is

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

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I'm sorry, but I haven't understood. What's the meaning of this $\delta$? I know that $[P_k, F]_q, _p = \sum_j \frac {\partial F}{\partial Q_j} [P_k, Q_j]$ and I can't understand why $[F, P_k]$ is equal to $\frac {\partial F}{\partial Q_k}$. Thank you. – sunrise Nov 26 '12 at 13:53
$\delta$ is the kronecker tensor. Recall that the Poisson bracket s a summation over the generalized co-ordinates. The $\delta$ just means that only the partial derivative wrt to k will survive, all others would be zero as they are independent variables. Tell me if more clarification is needed. – user7757 Nov 26 '12 at 14:44
Thank you! So I can say that $[P_k, F]_{q,p} = \frac { \partial F}{\partial Q_k}$. But, if I have correctly understood, $[P_k, F]_{q,p}= -[F, P_k]_{q,p}$. And so I can't understand why $[F, P_k]=\frac { \partial F}{\partial Q_k}$. Where am I wronging? Thanks again – sunrise Nov 26 '12 at 14:56
Don't get confused. You are right, $[F,P_k]=-\frac{\partial F}{\partial Q_k}$. In your question, I don't see the negative of the RHS used anywhere. – user7757 Nov 26 '12 at 17:35
I had look at goldstein, and in my edition, the invariant of the poisson bracket is proved using symplectic matrices. So I couldnt see the derivation mentioned above. I guess it is a printing mistake. – user7757 Nov 27 '12 at 0:30

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 $$\left\{f,g \right\}=\frac{\partial f}{\partial y_k}I_{ki}J_{ij}I_{lj}\frac{\partial g}{\partial y_l}=\frac{\partial f}{\partial y_k}J_{kl}\frac{\partial g}{\partial y_l}$$ $\blacksquare$

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Thank you, it is very interesting! – sunrise Nov 26 '12 at 13:54
I like this answer, can you please point me to a reference which talks about the definition of a canonical transformation you mentioned (i.e I being symplectic)? I would give you 100 points of my rep if I knew how! – Antillar Maximus Dec 12 '12 at 14:36
You can find all of you need about this theme in these books: 1) David Tong: Lectures on Classical Dynamics (damtp.cam.ac.uk/user/tong/dynamics.htm) 2) V. I. Arnold, Mathematical Methods of Classical Mechanics – Oiale Dec 12 '12 at 16:03