Difference between revisions of "User:Nik"
| Line 1: | Line 1: | ||
Let | Let | ||
| − | + | \( \Phi \in C^\infty(\mathbb{R}^n) \) | |
have the property that the system | have the property that the system | ||
| − | + | \( p_i = \frac{\partial}{\partial q_i} \Phi (q, \widetilde{q}) \) | |
has a unique smooth solution | has a unique smooth solution | ||
| − | + | \( \widetilde{q} = \widetilde{q} (q,p) \). | |
| − | Define | + | Define |
| − | + | \( \widetilde{p_i}(q,p) = - \frac{\partial}{\partial \widetilde{q_i}} \Phi (q, \widetilde{q}) | _{\widetilde{q}= \widetilde{q}(q,p)} \) | |
| − | Let | + | Let \( \{\cdot,\cdot\} \) be the Poisson bracket, such that |
| − | + | \( \{f,g\} = \sum_{j=1}^n \frac{\partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j} \) | |
| + | |||
| + | Show that: | ||
| − | |||
i) | i) | ||
| − | + | \( \{\widetilde{q_i} (q,p), \widetilde{q_j} (q,p)\} = \{\widetilde{p_i} (q,p), \widetilde{p_j} (q,p)\} = 0 \) | |
ii) | ii) | ||
| − | + | \( \{\widetilde{q_i} (q,p), \widetilde{p_j} (q,p)\} = \delta_{ij} \) | |
=== Solution for i) === | === Solution for i) === | ||
=== Solution for ii) === | === Solution for ii) === | ||
Revision as of 10:50, 28 December 2014
Let \( \Phi \in C^\infty(\mathbb{R}^n) \) have the property that the system \( p_i = \frac{\partial}{\partial q_i} \Phi (q, \widetilde{q}) \) has a unique smooth solution \( \widetilde{q} = \widetilde{q} (q,p) \).
Define \( \widetilde{p_i}(q,p) = - \frac{\partial}{\partial \widetilde{q_i}} \Phi (q, \widetilde{q}) | _{\widetilde{q}= \widetilde{q}(q,p)} \)
Let \( \{\cdot,\cdot\} \) be the Poisson bracket, such that \( \{f,g\} = \sum_{j=1}^n \frac{\partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j} \)
Show that:
i) \( \{\widetilde{q_i} (q,p), \widetilde{q_j} (q,p)\} = \{\widetilde{p_i} (q,p), \widetilde{p_j} (q,p)\} = 0 \)
ii) \( \{\widetilde{q_i} (q,p), \widetilde{p_j} (q,p)\} = \delta_{ij} \)
