Difference between revisions of "Talk:Aufgaben:Problem 8"

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There are two things which I'm not sure about: whether in lemma 2 we can use Lebesgue (we need a discrete sequence of dominated functions, whereas h is a continuous variable; can we change h to 1/n and let \( n \rightarrow \infty \) instead of \( h \rightarrow 0 \) without any problem?) and how lemma 2 exactly justifies the passage with \( \color{red}{*} \)
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I tried transforming to polar coordinates in an attempt to make the solution shorter but I don't think it works. I think you can abbreviate in the exam at some places and  also its all just derivatives so not very difficult, you just have to remember to invert the metric :)
  
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Some derivative signs should be partial derivatives, but I guess it's not such a problem...
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[[User:Carl|Carl]] ([[User talk:Carl|talk]]) 23:41, 11 June 2015 (CEST)
  
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Best, Nick
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== Suggestion for less writing work ==
  
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So, should be okay now... I think the induction step at \( \color{red}{*} \) is clear enough.  
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Maybe it is possible to not explicitly show that C_2 is conserved, by saying that for u->v and v->u C_1->C_2, so C_2 must be conserved if C_1 is conserved.
  
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Best, A.
 
  
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Yes, we can. If the limit of f(h) exists for h -> 0, then for every discrete sequence hn f(hn) will have the same limit for n -> infinity. So I guess you can just substitute them and use Lebesgue at will.
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== b) Poisson bracket? ==
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And \( \color{red}{*} \) isn't even an induction step... that's just applying the lemma we just proved. However, I believe in Lemma 2 the 1/sqrt(2*pi) on the right hand side of the equality is superfluous. It's not being used in the proof, as far as I can see, and it comes back to bite you in \( \color{red}{*} \).
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Instead of calculating d/dt(C) I think it would be nice to use the fact that d/dt(C) = {C,h}. Of course it's basically the same but the hamiltonian eqs wouldn't have to be substituted in but are kind of built in. However, so far I didn't manage to make this resulting in a shorter solution..
  
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Cheers, Nathalie
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Regarding your u<->v symmetry proposal, at least here you can easily argue with {,}:
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With C2 = C1(x1 <-> x2, p1 <-> p2), it obviously holds that {C1,C2}=0. Thus {C1,h}=0 ==> {C2,h}=0 by Jacobi identity.
  
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Ok, for the lemma and for the corollary it's all clear. I still can't see the induction step from the first derivative to the n-th derivative, though: the proof is for a specific function and doesn't work anymore when you have a derivative of this function inside the integral, instead of the function itself. In fact, maybe we could forget all about lemma 2 and its corollary by using something of this kind: http://en.wikipedia.org/wiki/Differentiation_under_the_integral_sign
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[[User:Mario|Mario]] ([[User talk:Mario|talk]]) 20:40, 29 June 2015 (CEST)
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But the problem is that we have \( \infty \) as integration bounds...
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I wonder whether we really have to justify all about this passage or whether we can just do it "the physician's way" (with no offense for anyone ;), because we're dealing with continuous functions with continuous derivatives and so on.
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Cheers, Nick
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Latest revision as of 18:40, 29 June 2015

I tried transforming to polar coordinates in an attempt to make the solution shorter but I don't think it works. I think you can abbreviate in the exam at some places and also its all just derivatives so not very difficult, you just have to remember to invert the metric :)

Carl (talk) 23:41, 11 June 2015 (CEST)

Suggestion for less writing work

Maybe it is possible to not explicitly show that C_2 is conserved, by saying that for u->v and v->u C_1->C_2, so C_2 must be conserved if C_1 is conserved.


b) Poisson bracket?

Instead of calculating d/dt(C) I think it would be nice to use the fact that d/dt(C) = {C,h}. Of course it's basically the same but the hamiltonian eqs wouldn't have to be substituted in but are kind of built in. However, so far I didn't manage to make this resulting in a shorter solution..

Regarding your u<->v symmetry proposal, at least here you can easily argue with {,}: With C2 = C1(x1 <-> x2, p1 <-> p2), it obviously holds that {C1,C2}=0. Thus {C1,h}=0 ==> {C2,h}=0 by Jacobi identity.

Mario (talk) 20:40, 29 June 2015 (CEST)

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