Difference between revisions of "User:Nik"
From Ferienserie MMP2
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| + | === Foreword === | ||
| + | I use \(Q\:/\:P\) instead of \(\widetilde{q}\:/\:\widetilde{p}\) because it's easier to write in Latex. | ||
| + | === Problem === | ||
Let | Let | ||
\( \Phi \in C^\infty(\mathbb{R}^n) \) | \( \Phi \in C^\infty(\mathbb{R}^n) \) | ||
have the property that the system | have the property that the system | ||
| − | \( p_i = \frac{\partial}{\partial q_i} \Phi (q, | + | \( p_i = \frac{\partial}{\partial q_i} \Phi (q, Q \) |
has a unique smooth solution | has a unique smooth solution | ||
| − | \( | + | \( Q = Q(q,p) \). |
Define | Define | ||
| − | \( | + | \( P_i(q,p) = - \frac{\partial}{\partial Q_i} \Phi (q, Q) | _{Q= Q(q,p)} \) |
Let \( \{\cdot,\cdot\} \) be the Poisson bracket, such that | Let \( \{\cdot,\cdot\} \) be the Poisson bracket, such that | ||
| Line 15: | Line 18: | ||
i) | i) | ||
| − | \( \{ | + | \( \{Q_i(q,p), Q_j(q,p)\} = \{P_i (q,p), P_j(q,p)\} = 0 \) |
ii) | ii) | ||
| − | \( \{ | + | \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \) |
=== Solution for i) === | === Solution for i) === | ||
=== Solution for ii) === | === Solution for ii) === | ||
Revision as of 10:55, 30 December 2014
Foreword
I use \(Q\:/\:P\) instead of \(\widetilde{q}\:/\:\widetilde{p}\) because it's easier to write in Latex.
Problem
Let \( \Phi \in C^\infty(\mathbb{R}^n) \) have the property that the system \( p_i = \frac{\partial}{\partial q_i} \Phi (q, Q \) has a unique smooth solution \( Q = Q(q,p) \).
Define \( P_i(q,p) = - \frac{\partial}{\partial Q_i} \Phi (q, Q) | _{Q= Q(q,p)} \)
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) \( \{Q_i(q,p), Q_j(q,p)\} = \{P_i (q,p), P_j(q,p)\} = 0 \)
ii) \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \)
