Hi everybody! I'm currently studying Noether's theorem, but I'm a bit stuck around a stupid line of calculation for the variation of the symmetry. The script of my teacher says (roughly translated from German, equations left as he wrote them):(adsbygoogle = window.adsbygoogle || []).push({});

"V.2. Noether Theorem

How does the action change under arbitrary variations of the path and of the time?

##t' = t + \tau (t)##; ##q_a ' (t') = q_a (t) + \Delta q_a (t)##

where ##\tau## and ##\Delta q_a## are arbitrary functions, which must not be equal zero at the boundaries. For the variation of ##q_a## by a fixed time ##q_a ' (t') = q_a (t) + \delta q_a (t)##, we find a relation between the variations ##\delta q_a## and ##\Delta q_a##:

##q_a ' (t') = q_a ' (t + \tau (t) = q_a ' (t) + \tau (t) \dot{q}_a (t)##

##= q_a (t) + \delta q_a (t) + \tau (t) \dot{q}_a (t)##

##\overset{!}{=} q_a (t) + \Delta q_a (t)##

##\implies \delta q_a (t) = \Delta q_a (t) - \tau (t) \dot{q}_a (t)##"

However, here is what I got when trying to repeat his steps:

##\delta q (t) = q ' (t') - q (t)##

##= q' (t + \tau (t)) - q(t) = q'(t) + \tau (t) \dot{q} ' (t) - q(t)##

##= q(t) + \Delta q(t) + \tau (t) \dot{q} ' (t) - q(t)##

##= \Delta q(t) + \tau (t) \dot{q} ' (t)##

That's close but not quite the same. He (very!) often makes little mistakes in his script, so I wondered if that was one of them. If not, I would be very grateful if someone could explain me when I went wrong :)

Thank you very much in advance for your answers.

Julien.

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# A Noether's Theorem: mistake in my teacher's script?

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