Difference between revisions of "User:Nik"
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| − | + | == Foreword == | |
| − | I use \(Q\:/\:P\) instead of \(\widetilde{q}\:/\:\widetilde{p}\) because it's easier to write in Latex. | + | 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. |
| − | + | == Problem == | |
Let | Let | ||
\( \Phi \in C^\infty(\mathbb{R}^n) \) | \( \Phi \in C^\infty(\mathbb{R}^n) \) | ||
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Show that: | Show that: | ||
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\( \{Q_i(q,p), Q_j(q,p)\} = \{P_i (q,p), P_j(q,p)\} = 0 \) | \( \{Q_i(q,p), Q_j(q,p)\} = \{P_i (q,p), P_j(q,p)\} = 0 \) | ||
| − | |||
| − | |||
\( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \) | \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \) | ||
| − | === | + | |
| + | ==Solution== | ||
| + | ===Assumptions=== | ||
| + | From the script we use the hamilton's equations \(\dot{q_i} = \{q_i, h\} = \frac{\partial h}{\partial p_i} \text{ and } \dot{p_i} = \{p_i, h\} = -\frac{\partial h}{\partial q_i} \) | ||
| + | |||
| + | |||
| + | Therefore $$ \tag{1} \dot{Q_i} = \{Q_i, h\} = \frac{\partial H}{\partial P_i} $$ | ||
| + | and $$ \tag{2} \dot{P_i} = \{P_i, h\} = -\frac{\partial H}{\partial Q_i} $$ | ||
| + | where \( H = h\circ\Phi \) | ||
| + | |||
| + | The question if we are allowed to use this is still not clear. Waiting for an answer of Raisa (02.01.15 - 10:00) | ||
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| + | 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 $$ \tag{4} \frac{\partial h}{\partial q_l} = \sum_{k=1}^n(\frac{\partial H}{\partial Q_k}\frac{\partial Q_k}{\partial q_l} + \frac{\partial H}{\partial P_k}\frac{\partial P_k}{\partial p_l}) $$ | ||
| + | which is just using the chain rule. | ||
| + | |||
| + | ===Shaking variables=== | ||
| + | With these definitions it's just about shaking and changing sums: | ||
| + | |||
| + | $$ \dot{Q_i} = \{Q_i, h\} = \sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\frac{\partial h}{\partial p_l} - \frac{\partial Q_i}{\partial p_l}\frac{\partial h}{\partial q_l}) $$ | ||
| + | With (3) and (4) it follows | ||
| + | $$ = \sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\frac{\partial Q_j}{\partial p_l} + \frac{\partial H}{\partial P_j}\frac{\partial P_j}{\partial p_l}) - \frac{\partial Q_i}{\partial p_l}\sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\frac{\partial Q_j}{\partial q_l} + \frac{\partial H}{\partial P_j}\frac{\partial P_j}{\partial p_l})) $$ | ||
| + | Now we exchange the sums | ||
| + | $$ = \sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\frac{\partial Q_j}{\partial p_l} - \frac{\partial Q_i}{\partial p_l}\frac{\partial Q_j}{\partial q_l}) + \frac{\partial H}{\partial Q_j}\sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\frac{\partial P_j}{\partial p_l} - \frac{\partial Q_i}{\partial p_l}\frac{\partial P_j}{\partial q_l})) $$ | ||
| + | And with the definition of the Poisson-brackets | ||
| + | $$ = \sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\{Q_i, Q_j\} + \frac{\partial H}{\partial Q_j}\{Q_i, P_j\}) $$ | ||
| + | To fulfill (1) | ||
| + | $$ \implies \{Q_i, Q_j\} = 0 \text{ and } \{Q_i, P_j\} = \delta_{ij} $$ | ||
| + | |||
| + | And the same for the other equation: | ||
| + | |||
| + | $$ \dot{P_i} = \{Q_i, h\} = \sum_{l=1}^n(\frac{\partial P_i}{\partial q_l}\frac{\partial h}{\partial p_l} - \frac{\partial P_i}{\partial p_l}\frac{\partial h}{\partial q_l}) $$ | ||
| + | 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 | ||
| + | $$ = \sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\sum_{l=1}^n(\frac{\partial P_i}{\partial q_l}\frac{\partial Q_j}{\partial p_l} - \frac{\partial P_i}{\partial p_l}\frac{\partial Q_j}{\partial q_l}) + \frac{\partial H}{\partial P_j}\sum_{l=1}^n(\frac{\partial P_i}{\partial q_l}\frac{\partial P_j}{\partial p_l} - \frac{\partial P_i}{\partial p_l}\frac{\partial P_j}{\partial q_l})) $$ | ||
| + | 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) | - 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. | The problem there was, that you have to use a hamiltonian from the new coordinate system. May work, haven't checked that exactly. | ||
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- 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 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. | + | - 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. |
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Revision as of 09:40, 2 January 2015
Foreword
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.
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:
\( \{Q_i(q,p), Q_j(q,p)\} = \{P_i (q,p), P_j(q,p)\} = 0 \) \( \{Q_i(q,p), P_j(q,p)\} = \delta_{ij} \)
Solution
Assumptions
From the script we use the hamilton's equations \(\dot{q_i} = \{q_i, h\} = \frac{\partial h}{\partial p_i} \text{ and } \dot{p_i} = \{p_i, h\} = -\frac{\partial h}{\partial q_i} \)
Therefore $$ \tag{1} \dot{Q_i} = \{Q_i, h\} = \frac{\partial H}{\partial P_i} $$
and $$ \tag{2} \dot{P_i} = \{P_i, h\} = -\frac{\partial H}{\partial Q_i} $$
where \( H = h\circ\Phi \)
The question if we are allowed to use this is still not clear. Waiting for an answer of Raisa (02.01.15 - 10:00)
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 $$ \tag{4} \frac{\partial h}{\partial q_l} = \sum_{k=1}^n(\frac{\partial H}{\partial Q_k}\frac{\partial Q_k}{\partial q_l} + \frac{\partial H}{\partial P_k}\frac{\partial P_k}{\partial p_l}) $$ which is just using the chain rule.
Shaking variables
With these definitions it's just about shaking and changing sums:
$$ \dot{Q_i} = \{Q_i, h\} = \sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\frac{\partial h}{\partial p_l} - \frac{\partial Q_i}{\partial p_l}\frac{\partial h}{\partial q_l}) $$ With (3) and (4) it follows $$ = \sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\frac{\partial Q_j}{\partial p_l} + \frac{\partial H}{\partial P_j}\frac{\partial P_j}{\partial p_l}) - \frac{\partial Q_i}{\partial p_l}\sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\frac{\partial Q_j}{\partial q_l} + \frac{\partial H}{\partial P_j}\frac{\partial P_j}{\partial p_l})) $$ Now we exchange the sums $$ = \sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\frac{\partial Q_j}{\partial p_l} - \frac{\partial Q_i}{\partial p_l}\frac{\partial Q_j}{\partial q_l}) + \frac{\partial H}{\partial Q_j}\sum_{l=1}^n(\frac{\partial Q_i}{\partial q_l}\frac{\partial P_j}{\partial p_l} - \frac{\partial Q_i}{\partial p_l}\frac{\partial P_j}{\partial q_l})) $$ And with the definition of the Poisson-brackets $$ = \sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\{Q_i, Q_j\} + \frac{\partial H}{\partial Q_j}\{Q_i, P_j\}) $$ To fulfill (1) $$ \implies \{Q_i, Q_j\} = 0 \text{ and } \{Q_i, P_j\} = \delta_{ij} $$
And the same for the other equation:
$$ \dot{P_i} = \{Q_i, h\} = \sum_{l=1}^n(\frac{\partial P_i}{\partial q_l}\frac{\partial h}{\partial p_l} - \frac{\partial P_i}{\partial p_l}\frac{\partial h}{\partial q_l}) $$ 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 $$ = \sum_{j=1}^n(\frac{\partial H}{\partial Q_j}\sum_{l=1}^n(\frac{\partial P_i}{\partial q_l}\frac{\partial Q_j}{\partial p_l} - \frac{\partial P_i}{\partial p_l}\frac{\partial Q_j}{\partial q_l}) + \frac{\partial H}{\partial P_j}\sum_{l=1}^n(\frac{\partial P_i}{\partial q_l}\frac{\partial P_j}{\partial p_l} - \frac{\partial P_i}{\partial p_l}\frac{\partial P_j}{\partial q_l})) $$ 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.
