Classical Mechanics – Is Understanding of Canonical Transformations Flawed?

classical-mechanicscoordinate systemshamiltonian-formalismphase-spacepoisson-brackets

Consider a system described by Hamilton's equations
$$\dot{q}_i=\frac{\partial H}{\partial p_i}=\{q_i,H\}, \quad \dot{p}_i=-\frac{\partial H}{\partial q_i}=\{p_i,H\}.\tag{1}$$
I want to prove that a time-independent transformation of the form $$q_i\to Q_i(q,p), \quad p_i\to P_i(q,p)\tag{2}$$ preserves Hamilton's equations provided
$$\{Q_i,Q_j\}=0,\qquad \{Q_i,P_j\}=\delta_{ij},\qquad\{P_i,P_j\}=0, \tag{3}$$
i.e. if the transformation is canonical.

However, I find that
$$\dot{Q}_i=\sum_k\left[\frac{\partial Q_i}{\partial q_k}\dot{q}_k+\frac{\partial Q_i}{\partial p_k}\dot{p}_k\right]=\{Q_i,H\},\tag{4}$$ and similarly,
$$\dot{P}_i=\sum_k\left[\frac{\partial P_i}{\partial q_k}\dot{q}_k+\frac{\partial P_i}{\partial p_k}\dot{p}_k\right]=\{P_i,H\}\tag{5}$$
without using (3)!

Does it mean that any transformation of the form (2) is canonical?

But as far as I knew, only a subset of transformations of type (1) which satisfies (3) preserve the form of Hamilton's equations i.e. are canonical. But it turns out that I am able to reproduce Hamilton's equations in the transformed variables without using the restriction that (3), rather trivially. Did I make any mistakes?

Best Answer

Hint: In symplectic notation $$z^I~=~(q^i,p_i)\qquad\text{and}\qquad Z^I~=~(Q^i,P_i),$$ and assuming no explicit time dependence in the transformation $Z^I=f^I(z)$, OP has shown that $$\begin{align}\dot{Z}^I~\stackrel{\text{chain rule}}{=}~&~ \frac{\partial Z^I}{\partial z^J}\dot{z}^J\cr ~\stackrel{\text{old Ham. eqs.}}{=}&~ \frac{\partial Z^I}{\partial z^J}J^{JK}\frac{\partial H}{\partial z^K}\cr ~=~~~~~~&\{Z^I,H\}_z,\end{align}\tag{4/5}$$ where $J^{JK}$ is the symplectic unit. However the task is instead to show the new Hamilton's equations $$\dot{Z}^I~=~ J^{IJ}\frac{\partial H}{\partial Z^J}~=~\{Z^I,H\}_{\color{red}{Z}}\tag{new Ham. eqs}$$ with the help of the symplectic condition $$ J^{IL}~=~\{Z^I,Z^L\}_z~=~\frac{\partial Z^I}{\partial z^J}J^{JK}\frac{\partial Z^L}{\partial z^K}.\tag{3}$$

References:

  1. H. Goldstein, Classical Mechanics, 2nd eds.; Section 9.3.

  2. H. Goldstein, Classical Mechanics, 3rd eds.; Section 9.4.

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