|
|
| (60 intermediate revisions by 4 users not shown) |
| Line 1: |
Line 1: |
| − | == Foreword ==
| + | Niklaus Messerli |
| − | I use \(Q\:/\:P\:/\:H\) for the transformed system instead of \(\widetilde{q}\:/\:\widetilde{p}\:/\:\widetilde{h}\) 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]] |
| − | \( \{Q_i(q,p), Q_j(q,p)\} = \{P_i (q,p), P_j(q,p)\} = 0 \)
| + | | a-c):1, d):0 |
| − | \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \)
| + | |- |
| − | | + | | [[Aufgaben:Problem 3|Problem 3]] |
| − | ==Solution==
| + | | 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. |
| − | ===Assumptions===
| + | |- |
| − | From the script we use the hamilton's equations \(\dot{q_i} = \{q_i, h\} = \frac{\partial h}{\partial p_i} \text{ and } \dot{p_i} = \{p_i, h\} = -\frac{\partial h}{\partial q_i} \)
| + | | [[Aufgaben:Problem 4|Problem 4]] |
| − | | + | | 1 |
| − | | + | |- |
| − | Therefore (to show) $$ \tag{1} \dot{Q_i} = \{Q_i, h\} = \frac{\partial H}{\partial P_i} $$
| + | | [[Aufgaben:Problem 5|Problem 5]] |
| − | and $$ \tag{2} \dot{P_i} = \{P_i, h\} = -\frac{\partial H}{\partial Q_i} $$
| + | | 1 |
| − | where \( H = h\circ\Phi \)
| + | |- |
| − | | + | | [[Aufgaben:Problem 6|Problem 6]] |
| − | The question if we are allowed to use this is still not clear. Waiting for an answer of Raisa (02.01.15 - 10:00)
| + | | 1 |
| − | | + | |- |
| − | | + | | [[Aufgaben:Problem 7|Problem 7]] |
| − | | + | | 1 |
| − | We also use that $$ \tag{3} \frac{\partial h}{\partial p_l} = \sum_{k=1}^n(\frac{\partial H}{\partial Q_k}\frac{\partial Q_k}{\partial p_l} + \frac{\partial H}{\partial P_k}\frac{\partial P_k}{\partial p_l}) $$
| + | |- |
| − | and $$ \tag{4} \frac{\partial h}{\partial q_l} = \sum_{k=1}^n(\frac{\partial H}{\partial Q_k}\frac{\partial Q_k}{\partial q_l} + \frac{\partial H}{\partial P_k}\frac{\partial P_k}{\partial p_l}) $$
| + | | [[Aufgaben:Problem 8|Problem 8]] |
| − | which is just using the chain rule.
| + | | 2 |
| − | | + | |- |
| − | ===Shaking variables===
| + | | [[Aufgaben:Problem 9|Problem 9]] |
| − | With these definitions it's just about shaking and changing sums:
| + | | 1 |
| − | | + | |- |
| − | $$ \dot{Q_i} = \{Q_i, h\} = \sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\frac{\partial h}{\partial p_l} - \frac{\partial Q_i}{\partial p_l}\frac{\partial h}{\partial q_l}) $$
| + | | [[Aufgaben:Problem 10|Problem 10]] |
| − | With (3) and (4) it follows
| + | | a by craven seems legit |
| − | $$ = \sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\frac{\partial Q_j}{\partial p_l} + \frac{\partial H}{\partial P_j}\frac{\partial P_j}{\partial p_l}) - \frac{\partial Q_i}{\partial p_l}\sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\frac{\partial Q_j}{\partial q_l} + \frac{\partial H}{\partial P_j}\frac{\partial P_j}{\partial p_l})) $$
| + | |- |
| − | Now we exchange the sums
| + | | [[Aufgaben:Problem 11|Problem 11]] |
| − | $$ = \sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\frac{\partial Q_j}{\partial p_l} - \frac{\partial Q_i}{\partial p_l}\frac{\partial Q_j}{\partial q_l}) + \frac{\partial H}{\partial Q_j}\sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\frac{\partial P_j}{\partial p_l} - \frac{\partial Q_i}{\partial p_l}\frac{\partial P_j}{\partial q_l})) $$
| + | | 1 |
| − | And with the definition of the Poisson-brackets
| + | |- |
| − | $$ = \sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\{Q_i, Q_j\} + \frac{\partial H}{\partial Q_j}\{Q_i, P_j\}) $$
| + | | [[Aufgaben:Problem 12|Problem 12]] |
| − | To fulfill (1)
| + | | bitch please |
| − | $$ \implies \{Q_i, Q_j\} = 0 \text{ and } \{Q_i, P_j\} = \delta_{ij} $$
| + | |- |
| − | | + | | [[Aufgaben:Problem 13|Problem 13]] |
| − | And the same for the other equation:
| + | | |
| − | | + | |- |
| − | $$ \dot{P_i} = \{Q_i, h\} = \sum_{l=1}^n(\frac{\partial P_i}{\partial q_l}\frac{\partial h}{\partial p_l} - \frac{\partial P_i}{\partial p_l}\frac{\partial h}{\partial q_l}) $$
| + | | [[Aufgaben:Problem 14|Problem 14]] |
| − | With (3) and (4) it follows
| + | | |
| − | $$ = \sum_{l=1}^n(\frac{\partial P_i}{\partial q_l}\sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\frac{\partial Q_j}{\partial p_l} + \frac{\partial H}{\partial P_j}\frac{\partial P_j}{\partial p_l}) - \frac{\partial P_i}{\partial p_l}\sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\frac{\partial Q_j}{\partial q_l} + \frac{\partial H}{\partial P_j}\frac{\partial P_j}{\partial p_l})) $$
| + | |- |
| − | Now we exchange the sums
| + | | [[Aufgaben:Problem 15|Problem 15]] |
| − | $$ = \sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\sum_{l=1}^n(\frac{\partial P_i}{\partial q_l}\frac{\partial Q_j}{\partial p_l} - \frac{\partial P_i}{\partial p_l}\frac{\partial Q_j}{\partial q_l}) + \frac{\partial H}{\partial P_j}\sum_{l=1}^n(\frac{\partial P_i}{\partial q_l}\frac{\partial P_j}{\partial p_l} - \frac{\partial P_i}{\partial p_l}\frac{\partial P_j}{\partial q_l})) $$
| + | | a): 1 |
| − | And with the definition of the Poisson-brackets
| + | |- |
| − | $$ = \sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\{P_i, P_j\} + \frac{\partial H}{\partial P_j}\{P_i, P_j\}) $$
| + | |} |
| − | To fulfill (2)
| + | |
| − | $$ \implies \{P_i, P_j\} = 0 \text{ and } \{P_i, Q_j\} = -\delta_{ij} = -\{Q_i, P_j\} \:\:\:\blacksquare $$
| + | |
| − | | + | |
| − | ==Attempts==
| + | |
| − | - Use of hamilton's equations from script page 59. I asked Raisa if we are allowed to use things from the script without proof, no answer yet (31.12.14 - 21:00)
| + | |
| − | The problem there was, that you have to use a hamiltonian from the new coordinate system. May work, haven't checked that exactly.
| + | |
| − | | + | |
| − | - Use of simply the chain rule and insert all the things. Some solutions i saw with that said that \( \frac{\partial \Phi}{\partial p} = 0 \) but since \(\Phi \) depends on \(Q \) whicht depends on \(p \) I am not convinced with that. | + | |
| − | | + | |
| − | - Use of simply the chain rule as before. At some point of this solution you should show that \( \frac{\partial Q_a}{\partial q_j}\frac{\partial}{\partial p_j}\frac{\partial}{\partial Q_j}\Phi (q, Q) = 0 \) which nobody has until now.
| + | |
