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Symmetry of a lagrangian & Noether's theorem

  • Thread starter irycio
  • Start date
  • #1
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Homework Statement


Assuming that transformation q->f(q,t) is a symmetry of a lagrangian show that the quantity
[tex]f\frac{\partial L}{\partial q'} [/tex] is a constant of motion ([tex]q'=\frac{dq}{dt}[/tex]).

2. Noether's theorem
http://en.wikipedia.org/wiki/Noether's_theorem

The Attempt at a Solution



Now, what I guess is that this exercise isn't well-formulated. Noether's theorem is obviously first thing that comes to our minds. Quantity mentioned above, however, would be a constant of motion, if the transformation was q->q+f(q,t), not simply q->f(q,t), since the latter transformation leads us to q->q+(f(q,t)-q), and the conserved quantity [tex](f-q)\frac{\partial L}{\partial q'} [/tex].

Am I right and there is a tiny mistake in the exercise or there is a way to show that the given quantity is really conserved? (I tried and failed).
 

Answers and Replies

  • #2
213
8
Yeah I think they did mean q-> q+f(q,t), and it smells a lot like q->exp(iat)q
 
Last edited:
  • #3
213
8
Otherwise since q+dq=f(q,t) then dq = t*df/dt
 

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