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| − | == Foreword ==
| + | Niklaus Messerli |
| − | I use \(Q\:/\:P\) for the transformed system instead of \(\widetilde{q}\:/\:\widetilde{p}\) because it's easier to write in Latex.
| + | nik@student.ethz.ch |
| − | == 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
| + | {| class="wikitable" |
| − | \( P_i(q,p) = - \frac{\partial}{\partial Q_i} \Phi (q, Q) | _{Q= Q(q,p)} \)
| + | !Article |
| − | | + | !Check |
| − | Let \( \{\cdot,\cdot\} \) be the Poisson bracket, such that
| + | |- |
| − | \( \{f,g\} = \sum_{j=1}^n \frac{\partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j} \)
| + | | [[Aufgaben:Problem 1|Problem 1]] |
| − | | + | | 1 |
| − | Show that:
| + | |- |
| − | | + | | [[Aufgaben:Problem 2|Problem 2]] |
| − | I) \( \{Q_i(q,p), Q_j(q,p)\} = 0 \)
| + | | a-c):1, d):0 |
| − | | + | |- |
| − | II) \( \{P_i (q,p), P_j(q,p)\} = 0 \)
| + | | [[Aufgaben:Problem 3|Problem 3]] |
| − | | + | | d) Wieso müssen wir zeigen dass "\(\{L_g: g \in G\}\) is a group under composition"? Wir wissen doch schon, dass \(L: G \rightarrow \mathrm{Sym}G\) und dass \(G\) und \(\mathrm{Sym}G\) Gruppen sind. |
| − | III) \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \)
| + | |- |
| − | | + | | [[Aufgaben:Problem 4|Problem 4]] |
| − | ==Solution==
| + | | 1 |
| − | | + | |- |
| − | ===Important equations===
| + | | [[Aufgaben:Problem 5|Problem 5]] |
| − | | + | | 1 |
| − | $$ \underbrace{\frac{\partial p_i}{\partial q_k}}_{0} = \frac{\partial}{\partial q_k}(\frac{\partial \Phi}{\partial q_i}(q, Q)) $$
| + | |- |
| − | $$ = \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial q_k}}_{\delta_{sk}} + \frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}) $$
| + | | [[Aufgaben:Problem 6|Problem 6]] |
| − | $$ \Rightarrow 0 = \frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q) + \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}) $$
| + | | 1 |
| − | $$ \tag{1} \Rightarrow \boxed{ -\frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q) = \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}) } $$
| + | |- |
| − | | + | | [[Aufgaben:Problem 7|Problem 7]] |
| − | | + | | 1 |
| − | $$ \underbrace{\frac{\partial p_i}{\partial p_k}}_{\delta_{ik}} = \frac{\partial}{\partial p_k}(\frac{\partial \Phi}{\partial q_i}(q, Q)) $$
| + | |- |
| − | $$ = \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial p_k}}_{0} + \frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}) $$
| + | | [[Aufgaben:Problem 8|Problem 8]] |
| − | $$ \tag{2} \boxed{ \Rightarrow \delta_{ik} = \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}) } $$
| + | | 2 |
| − | | + | |- |
| − | | + | | [[Aufgaben:Problem 9|Problem 9]] |
| − | $$ \frac{\partial P_i}{\partial q_k} = \frac{\partial}{\partial q_k}(-\frac{\partial \Phi}{\partial Q_i}(q, Q)) $$
| + | | 1 |
| − | $$ = \sum_{s=1}^{n}(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial q_k}}_{\delta_{sk}} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}) $$
| + | |- |
| − | $$ \tag{3} \boxed{ \Rightarrow \frac{\partial P_i}{\partial q_k} = -\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q) - \sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}) } $$
| + | | [[Aufgaben:Problem 10|Problem 10]] |
| − | | + | | a by craven seems legit |
| − | | + | |- |
| − | $$ \frac{\partial P_i}{\partial p_k} = \frac{\partial}{\partial p_k}(-\frac{\partial \Phi}{\partial Q_i}(q, Q)) $$
| + | | [[Aufgaben:Problem 11|Problem 11]] |
| − | $$ = \sum_{s=1}^{n}(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial p_k}}_{0} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}) $$
| + | | 1 |
| − | $$ \tag{4} \Rightarrow \boxed{ \frac{\partial P_i}{\partial p_k} = -\sum_{s=1}^{n}(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}) } $$
| + | |- |
| − | | + | | [[Aufgaben:Problem 12|Problem 12]] |
| − | | + | | bitch please |
| − | ===Solution I)===
| + | |- |
| − | | + | | [[Aufgaben:Problem 13|Problem 13]] |
| − | $$ \{Q_i, Q_j\} = \sum_{k=1}^{n}(\frac{\partial Q_i}{\partial q_k}\frac{\partial Q_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial q_k}) $$
| + | | |
| − | | + | |- |
| − | | + | | [[Aufgaben:Problem 14|Problem 14]] |
| − | ===Solution II)===
| + | | |
| − | | + | |- |
| − | $$ \{P_i, P_j\} = \sum_{k=1}^{n}(\frac{\partial P_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial P_i}{\partial p_k}\frac{\partial P_j}{\partial q_k}) $$
| + | | [[Aufgaben:Problem 15|Problem 15]] |
| − | | + | | a): 1 |
| − | | + | |- |
| − | ===Solution III)===
| + | |} |
| − | | + | |
| − | $$ \{Q_i, P_j\} = \sum_{k=1}^{n}(\frac{\partial Q_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial P_j}{\partial q_k}) $$
| + | |
| − | $$ = \sum_{k=1}^{n}(\frac{\partial Q_i}{\partial q_k}\frac{\partial}{\partial p_k}(-\frac{\partial \Phi}{\partial Q_j}(q, Q)) - \frac{\partial Q_i}{\partial p_k}\frac{\partial}{\partial q_k}(-\frac{\partial \Phi}{\partial Q_j}(q, Q))) $$
| + | |
| − | $$ = \sum_{k=1}^{n}(\frac{\partial Q_i}{\partial q_k}\frac{\partial}{\partial p_k}(-\frac{\partial \Phi}{\partial Q_j}(q, Q)) - \frac{\partial Q_i}{\partial p_k}\frac{\partial}{\partial q_k}(-\frac{\partial \Phi}{\partial Q_j}(q, Q))) $$
| + | |
