Chapter 8: Problem 3
Show that the map \(A: \mathbb{R}^{2 n} \rightarrow \mathbb{R}^{2 n}\) sending \((\mathbf{p}, \mathbf{q}) \rightarrow(\mathbf{P}(\mathbf{p}, \mathbf{q}), \mathbf{Q}(\mathbf{p}, \mathbf{q}))\) is canonical if and only if the Poisson brackets of any two functions in the variables \((\mathbf{p}, \mathbf{q})\) and \((\mathbf{P}, \mathbf{Q})\) coincide: $$ (F, H)_{\mathbf{p}, \mathbf{q}}=\frac{\partial H}{\partial \mathbf{p}} \frac{\partial F}{\partial \mathbf{q}}-\frac{\partial H}{\partial \mathbf{q}} \frac{\partial F}{\partial \mathbf{p}}=\frac{\partial H}{\partial \mathbf{P}} \frac{\partial F}{\partial \mathbf{Q}}-\frac{\partial H}{\partial \mathbf{Q}} \frac{\partial F}{\partial \mathbf{P}}=(F, H)_{\mathbf{p}, \mathbf{q}} $$ Solution. Let \(A\) be canonical. Then the symplectic structures \(d \mathbf{p} \wedge d \mathbf{q}\) and \(d \mathbf{P} \wedge d \mathbf{Q}\) coincide. But the definition of the Poisson bracket \((F, H)\) was given invariantly in terms of the symplectic structure; it did not involve the coordinates. Therefore, $$ (F, H)_{\mathbf{p}, \mathbf{q}}=(F, H)=(F, H)_{\mathbf{P}, \mathbf{Q}} . $$ Conversely, suppose that the Poisson brackets \(\left(P_{i}, Q_{j}\right)_{\mathbf{p}, \mathbf{q}}\) have the standard form of Problem \(2 .\) Then, clearly, \(d \mathbf{P} \wedge d \mathbf{Q}=d \mathbf{p} \wedge d \mathbf{q}\), i.e., the map \(A\) is canonical.
Short Answer
Step by step solution
Key Concepts
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