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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
| + | |
| − | \( \Phi \in C^\infty(\mathbb{R}^{2n}) \)
| + | |
| − | 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) \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \)
| + | | [[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) \( \{P_i (q,p), P_j(q,p)\} = 0 \)
| + | |- |
| − | | + | | [[Aufgaben:Problem 4|Problem 4]] |
| − | | + | | 1 |
| − | | + | |- |
| − | ==Solution==
| + | | [[Aufgaben:Problem 5|Problem 5]] |
| − | | + | | 1 |
| − | ===Important equations===
| + | |- |
| − | | + | | [[Aufgaben:Problem 6|Problem 6]] |
| − | $$ \underbrace{\frac{\partial p_i}{\partial q_k}}_{0} = \frac{\partial}{\partial q_k}\left(\frac{\partial \Phi}{\partial q_i}(q, Q)\right) $$
| + | | 1 |
| − | $$ = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial q_k}}_{\delta_{sk}} + \frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) $$
| + | |- |
| − | $$ \Rightarrow 0 = \frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q) + \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) $$
| + | | [[Aufgaben:Problem 7|Problem 7]] |
| − | $$ \tag{1} \Rightarrow \boxed{ -\frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q) = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) } $$
| + | | 1 |
| − | | + | |- |
| − | | + | | [[Aufgaben:Problem 8|Problem 8]] |
| − | $$ \underbrace{\frac{\partial p_i}{\partial p_k}}_{\delta_{ik}} = \frac{\partial}{\partial p_k}\left(\frac{\partial \Phi}{\partial q_i}(q, Q)\right) $$
| + | | 2 |
| − | $$ = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial p_k}}_{0} + \frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$
| + | |- |
| − | $$ \tag{2} \boxed{ \Rightarrow \delta_{ik} = \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) } $$
| + | | [[Aufgaben:Problem 9|Problem 9]] |
| − | | + | | 1 |
| − | | + | |- |
| − | $$ \frac{\partial P_i}{\partial q_k} = \frac{\partial}{\partial q_k}\left(-\frac{\partial \Phi}{\partial Q_i}(q, Q)\right) $$
| + | | [[Aufgaben:Problem 10|Problem 10]] |
| − | $$ = \sum_{s=1}^{n}\left(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial q_k}}_{\delta_{sk}} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) $$
| + | | a by craven seems legit |
| − | $$ \tag{3} \boxed{ \Rightarrow \frac{\partial P_i}{\partial q_k} = -\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q) - \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) } $$
| + | |- |
| − | | + | | [[Aufgaben:Problem 11|Problem 11]] |
| − | | + | | 1 |
| − | $$ \frac{\partial P_i}{\partial p_k} = \frac{\partial}{\partial p_k}\left(-\frac{\partial \Phi}{\partial Q_i}(q, Q)\right) $$
| + | |- |
| − | $$ = \sum_{s=1}^{n}\left(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_s}(q, Q)\underbrace{\frac{\partial q_s}{\partial p_k}}_{0} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$
| + | | [[Aufgaben:Problem 12|Problem 12]] |
| − | $$ \tag{4} \Rightarrow \boxed{ \frac{\partial P_i}{\partial p_k} = -\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) } $$
| + | | bitch please |
| − | | + | |- |
| − | | + | | [[Aufgaben:Problem 13|Problem 13]] |
| − | ===Solution I)===
| + | | |
| − | This is the most complicated problem but we calculate it first because we use it in II) und III)
| + | |- |
| − | First we write equations (1) and (2) again an shift our eyeballs between them and look at it as matrix multiplication:
| + | | [[Aufgaben:Problem 14|Problem 14]] |
| − | $$ \underbrace{-\frac{\partial^2 \Phi}{\partial q_i \partial q_k}(q, Q)}_{D_{ik}} = \sum_{s=1}^{n}\left(\underbrace{\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)}_{A_{is}}\underbrace{\frac{\partial Q_s}{\partial q_k}}_{C_{sk}}\right) $$
| + | | |
| − | $$ \underbrace{\delta_{ik}}_{\mathbb{1}} = \sum_{s=1}^{n}\left(\underbrace{\frac{\partial^2 \Phi}{\partial q_i \partial Q_s}(q, Q)}_{A_{is}}\underbrace{\frac{\partial Q_s}{\partial p_k}}_{B_{sk}}\right) $$
| + | |- |
| − | Now let's do some linear algebra: We know that \( \mathbb{1} = AB \Rightarrow A^{-1} = B \text{ and } D = AC \)
| + | | [[Aufgaben:Problem 15|Problem 15]] |
| − | $$ \Rightarrow A^{-1}D = C $$
| + | | a): 1 |
| − | $$\Rightarrow BD = C $$
| + | |- |
| − | With that we can now calculate C which we need to calculate the Poisson bracket
| + | |} |
| − | $$ C_{sk} = - \sum_{e=1}^{n}\left(\frac{\partial Q_s}{\partial p_e}\frac{\partial^2 \Phi}{\partial q_k \partial q_e}\right) $$
| + | |
| − | Now we change the indices such that it fits into the equation
| + | |
| − | $$ \frac{\partial Q_i}{\partial q_k} = - \sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_s}\frac{\partial^2 \Phi}{\partial q_k \partial q_s}\right) $$
| + | |
| − | Finally we are prepared to calculate the Poisson bracket
| + | |
| − | $$ \{Q_i, Q_j\} = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial Q_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial q_k}\right) $$
| + | |
| − | Put in the sum from above
| + | |
| − | $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_j}{\partial p_k}\left(-\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_s}\frac{\partial^2 \Phi}{\partial q_k \partial q_s}(q, Q)\right)\right) - \frac{\partial Q_i}{\partial p_k}\left(-\sum_{s=1}^{n}\left(\frac{\partial Q_j}{\partial p_s}\frac{\partial^2 \Phi}{\partial q_k \partial q_s}(q, Q)\right)\right)\right) $$
| + | |
| − | Take the sum out
| + | |
| − | $$ \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial Q_j}{\partial p_s}\frac{\partial^2 \Phi}{\partial q_k \partial q_s}(q, Q) - \frac{\partial Q_j}{\partial p_k}\frac{\partial Q_i}{\partial p_s}\frac{\partial^2 \Phi}{\partial q_k \partial q_s}(q, Q) \right) = 0 $$
| + | |
| − | Since we sum with \(s\) and \(k\) completely from \(1\) to \(n\) the two parts are equal and therefore the difference is zero.
| + | |
| − | | + | |
| − | ===Solution II)===
| + | |
| − | | + | |
| − | $$ \{Q_i, P_j\} = \sum_{k=1}^{n}\left( \frac{\partial Q_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial Q_i}{\partial p_k}\frac{\partial P_j}{\partial q_k} \right) $$
| + | |
| − | Now we put in the results from equation (3) and (4)
| + | |
| − | $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\left(-\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right)\right) - \frac{\partial Q_i}{\partial p_k}\left(-\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q) - \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right)\right)\right) $$
| + | |
| − | Now we multiply out and separate all the sums. We can rearrange it like this because all the parts we put inside are independent on the running variable.
| + | |
| − | $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) + \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right) - \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial Q_i}{\partial q_k}\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right) $$
| + | |
| − | Now we rearrange again such that we can build a new Poisson bracket
| + | |
| − | $$ = \underbrace{\sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right)}_{\color{red}{\delta_{ij}}} + \sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q) \underbrace{\sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial Q_s}{\partial q_k} - \frac{\partial Q_i}{\partial q_k}\frac{\partial Q_s}{\partial p_k}\right)}_{=\{Q_s, Q_i\}=0}\right) $$
| + | |
| − | Now we use the definition from the problem and then the chain rule
| + | |
| − | $$ = \sum_{k=1}^{n}\left(\frac{\partial Q_i}{\partial p_k}\frac{\partial p_k}{\partial Q_j} \right) = \frac{\partial Q_i}{\partial Q_j} = \delta_{ij} $$
| + | |
| − | <span style="color:red">The chain rule is here not correct. It would imply that \( Q \) is independent of \( q \) and that is not correct. Since I got no better solution I would write \( \delta_{ij} \) directly in the sum above. </span>
| + | |
| − | | + | |
| − | ===Solution III)===
| + | |
| − | | + | |
| − | $$ \{P_i, P_j\} = \sum_{k=1}^{n}\left(\frac{\partial P_i}{\partial q_k}\frac{\partial P_j}{\partial p_k} - \frac{\partial P_i}{\partial p_k}\frac{\partial P_j}{\partial q_k}\right) $$
| + | |
| − | Now we put in the results from equation (3) and (4) in all 4 parts of the sum
| + | |
| − | $$ = \sum_{k=1}^{n}\left(\left(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q) - \underbrace{\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right)}_{(a)}\right)\left(\underbrace{-\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right)}_{(b)}\right) \\ - \left(\underbrace{-\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\right)}_{(c)}\right)\left(-\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q) - \underbrace{\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial q_k}\right)}_{(d)}\right)\right) $$
| + | |
| − | If we multiply out we see that (a) * (b) = (c) * (d) and because of the sign they cancel out.
| + | |
| − | So we multiply out the remaining parts
| + | |
| − | $$ = \sum_{k=1}^{n}\sum_{s=1}^{n}\left(\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q)\frac{\partial^2 \Phi}{\partial Q_j \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k} - \frac{\partial^2 \Phi}{\partial Q_i \partial Q_s}(q, Q)\frac{\partial Q_s}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) $$
| + | |
| − | And use the definition from the problem set
| + | |
| − | $$ = \sum_{k=1}^{n}\sum_{s=1}^{n}\left(-\frac{\partial^2 \Phi}{\partial Q_i \partial q_k}(q, Q)\underbrace{\frac{\partial P_s}{\partial Q_j}}_{=0}\frac{\partial Q_s}{\partial p_k} + \underbrace{\frac{\partial P_s}{\partial Q_i}}_{=0}\frac{\partial Q_s}{\partial p_k}\frac{\partial^2 \Phi}{\partial Q_j \partial q_k}(q, Q)\right) $$
| + | |
| − | | + | |
| − | | + | |
| − | | + | |
| − | $$ = 0 $$
| + | |
| − | | + | |
| − | | + | |
| − | ===Alternative Solution===
| + | |
| − | | + | |
| − | '''Alternative solution:'''
| + | |
| − | | + | |
| − | Here's a proof I've found on my usual source of knowledge (meaning a youtube-lecture):
| + | |
| − | | + | |
| − | '''Claim:''' For a generating function of type one (that's the smartman's name for the Phi in our problem) it holds: \( \{ Q, P \} = 1 \).
| + | |
| − | | + | |
| − | ''Proof:''
| + | |
| − | | + | |
| − | We first fix the \( q \) coordinates in the momentum equation and derive by \( p \), denoted: \( \Phi_p |_q\).
| + | |
| − | | + | |
| − | So we get:
| + | |
| − | | + | |
| − | '''Equation 1:'''
| + | |
| − | | + | |
| − | $$ 1 = \Phi_{q Q} Q_p |_q \Rightarrow Q_p |_q = \frac{1}{\Phi_{q Q}} $$
| + | |
| − | | + | |
| − | '''Equation 2:'''
| + | |
| − | | + | |
| − | Holding \( p \) fixed and derive by \( q \):
| + | |
| − | | + | |
| − | $$ 0 = \frac{\partial p}{\partial q} = \Phi_{qq} + \Phi_{qQ}Q_{q}|_p \Rightarrow Q_{q}|_p = - \frac{\Phi_{qq}}{\Phi_{qQ}} $$
| + | |
| − | | + | |
| − | '''Equation 3:'''
| + | |
| − | | + | |
| − | Holding \( q \) fixed and derive by \( p \):
| + | |
| − | | + | |
| − | $$ P_p |_q = - \Phi_{QQ}Q_q |_p \Rightarrow P_p |_q = - \frac{\Phi_{QQ}}{\Phi_{qQ}} $$
| + | |
| − | | + | |
| − | where we used equation 1 again.
| + | |
| − | | + | |
| − | '''Equation 4:'''
| + | |
| − | | + | |
| − | Holding \( p \) fixed and derive by \( q \):
| + | |
| − | | + | |
| − | $$ P_q |_p = - \Phi_{Qq} - \Phi_{QQ} Q_q |_p \Rightarrow P_q |_p = - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} $$
| + | |
| − | | + | |
| − | using equation 2.
| + | |
| − | | + | |
| − | We are now prepared to calculate:
| + | |
| − | | + | |
| − | $$ \{ Q, P \} = Q_qP_p - Q_pP_q = \frac{\Phi_{QQ}}{\Phi_{qQ}} \frac{\Phi_{qq}}{\Phi_{qQ}} - \frac{1}{\Phi_{q Q}} \left( - \Phi_{Qq} + \frac{\Phi_{QQ}\Phi_{qq}}{\Phi_{qQ}} \right) = 1 $$ by just cancelling out the terms. \( \square \)
| + | |
| − | | + | |
| − | ( ''My first idea would have been:''
| + | |
| − | | + | |
| − | By the script:
| + | |
| − | | + | |
| − | $$ \{ Q, P \} = \langle \nabla Q , J \nabla P \rangle_{\mathbb{R}^{2n}} = 1 $$
| + | |
| − | | + | |
| − | with \( \nabla = \left( \frac{\partial}{\partial q_1}, ..., \frac{\partial}{\partial p_n} \right) \) and
| + | |
| − | | + | |
| − | $$ J = \begin{bmatrix}
| + | |
| − | 0 & I_n \\
| + | |
| − | -I_n & 0 \\
| + | |
| − | \end{bmatrix} $$
| + | |
| − | | + | |
| − | and this should result in the things we want to show (by some symplectic argument or so...).)
| + | |
| − | | + | |
| − | ''But it might be better to just show this (the property is surely right, found it in a classical mechanics textbook):''
| + | |
| − | | + | |
| − | | + | |
| − | \( \vdash : \{ f, g \}_{p, q} = \{ P, Q \} \{ f, g \}_{P, Q} \)
| + | |
| − | | + | |
| − | ''Proof''
| + | |
| − | | + | |
| − | With simplified notation:
| + | |
| − | | + | |
| − | \( \{ f, g \}_{p, q} = \sum_k \left( \frac{\partial f}{\partial q_k} \frac{\partial g}{\partial p_k} - \frac{\partial f}{\partial p_k} \frac{\partial g}{\partial q_k} \right) = \sum_k \left( \frac{\partial f}{\partial Q} \frac{\partial Q}{\partial q_k} + \frac{\partial f}{\partial P} \frac{\partial P}{\partial q_k} \right) \left( \frac{\partial g}{\partial Q} \frac{\partial Q}{\partial p_k} + \frac{\partial g}{\partial P} \frac{\partial P}{\partial p_k} \right) - \sum_k \left( \frac{\partial f}{\partial Q} \frac{\partial Q}{\partial p_k} + \frac{\partial f}{\partial P} \frac{\partial P}{\partial p_k} \right) \) \(\left( \frac{\partial g}{\partial Q} \frac{\partial Q}{\partial q_k} + \frac{\partial g}{\partial P} \frac{\partial P}{\partial q_k} \right) \)
| + | |
| − | | + | |
| − | \( = \sum_k \left( \frac{\partial Q}{\partial q_k} \frac{\partial P}{\partial p_k} - \frac{\partial Q}{\partial p_k} \frac{\partial P}{\partial q_k} \right) \left( \frac{\partial f}{\partial Q} \frac{\partial g}{\partial P} - \frac{\partial f}{\partial P} \frac{\partial g}{\partial Q} \right) = \{ Q, P \}_{q, p} \cdot \{ f, g \}_{Q, P} \) \( \square \)
| + | |
| − | | + | |
| − | | + | |
| − | | + | |
| − | Now compute: \( \{ Q, P \}_{q, p} \cdot \{ Q_i, Q_j \}_{Q, P} = \{ Q, P \}_{q, p} \cdot \{ P_i, P_j \}_{Q, P} = 0 \) which follows from the properties of the Poisson-bracket.
| + | |
| − | | + | |
| − | Furthermore: \( \{ Q, P \}_{q, p} \cdot \{ Q_i, P_j \}_{Q, P} = \{ Q, P \}_{q, p} \cdot \delta_{i j} = \delta_{i j} \)
| + | |
| − | | + | |
| − | which solves the problem.
| + | |
