Difference between revisions of "User:Nik"

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Let
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Niklaus Messerli
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$$ \Phi \in C^\infty(\mathbb{R}^n) $$
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nik@student.ethz.ch
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have the property that the system
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$$p_i = \frac{\partial}{\partial q_i} \Phi (q, \widetilde{q}) $$
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has a unique smooth solution
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$$ \widetilde{q} = \widetilde{q} (q,p) $$
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Define
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{| class="wikitable"
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$$ \widetilde{p_i}(q,p) = - \frac{\partial}{\partial \widetilde{q_i}} \Phi (q, \widetilde{q}) | _{\widetilde{q}= \widetilde{q}(q,p)} $$
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!Article
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!Check
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Let $$ \{\cdot,\cdot\} $$ be the Poisson bracket, such that
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|-
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$$ \{f,g\} = \sum_{j=1}^n \frac{\partial f}{\partial q_j} \frac{\partial g}{\partial p_j} - \frac{\partial f}{\partial p_j} \frac{\partial g}{\partial q_j} $$
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| [[Aufgaben:Problem 1|Problem 1]]
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| 1
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Show that: $$\\$$
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|-
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i)
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| [[Aufgaben:Problem 2|Problem 2]]
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$$ \{\widetilde{q_i} (q,p), \widetilde{q_j} (q,p)\} = \{\widetilde{p_i} (q,p), \widetilde{p_j} (q,p)\} = 0 $$
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| a-c):1, d):0
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|-
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ii)
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| [[Aufgaben:Problem 3|Problem 3]]
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$$ \{\widetilde{q_i} (q,p), \widetilde{p_j} (q,p)\} = \delta_{ij} $$
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| 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.
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|-
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| [[Aufgaben:Problem 4|Problem 4]]
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=== Solution for i) ===
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| 1
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=== Solution for ii) ===
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|-
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| [[Aufgaben:Problem 5|Problem 5]]
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| 1
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|-
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| [[Aufgaben:Problem 6|Problem 6]]
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| 1
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|-
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| [[Aufgaben:Problem 7|Problem 7]]
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| 1
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|-
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| [[Aufgaben:Problem 8|Problem 8]]
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| 2
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|-
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| [[Aufgaben:Problem 9|Problem 9]]
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| 1
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|-
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| [[Aufgaben:Problem 10|Problem 10]]
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| a by craven seems legit
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|-
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| [[Aufgaben:Problem 11|Problem 11]]
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| 1
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|-
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| [[Aufgaben:Problem 12|Problem 12]]
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| bitch please
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|-
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| [[Aufgaben:Problem 13|Problem 13]]
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|
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|-
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| [[Aufgaben:Problem 14|Problem 14]]
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|
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|-
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| [[Aufgaben:Problem 15|Problem 15]]
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| a): 1
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|-
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|}

Latest revision as of 14:13, 3 August 2015

Niklaus Messerli nik@student.ethz.ch

Article Check
Problem 1 1
Problem 2 a-c):1, d):0
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.
Problem 4 1
Problem 5 1
Problem 6 1
Problem 7 1
Problem 8 2
Problem 9 1
Problem 10 a by craven seems legit
Problem 11 1
Problem 12 bitch please
Problem 13
Problem 14
Problem 15 a): 1
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