Difference between revisions of "Talk:Aufgaben:Problem 8"
(→Suggestion for less writing work: new section) |
|||
| Line 6: | Line 6: | ||
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. | 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. | ||
| + | |||
| + | [[User:Mario|Mario]] ([[User talk:Mario|talk]]) 20:40, 29 June 2015 (CEST) | ||
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.
