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

  1. Sep 8, 2016 #1
    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):

    "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.
     
  2. jcsd
  3. Sep 8, 2016 #2

    Orodruin

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    These two definitions are incompatible (or equivalent - meaning ##\Delta q_a = \delta q_a## by definition). I assume that you intend for one to have ##q_a'(t)## on the LHS (or ##t'## instead of ##t## on the RHS). Without knowing which is which, it will be difficult to help you.
     
  4. Sep 8, 2016 #3
    Hi @Orodruin and thank you for your answer. Those definitions unfortunately come straight out of my teacher's script... I would assume that my teacher made a mistake and -without much certainty- that ##q_a ' (t) = q_a (t) + \Delta q_a (t)## (then it would be the definition of a translation) and that ##\delta q_a (t) = q_a ' (t') - q_a(t)## (the symmetry variation). Does that seem correct to you? Unfortunately, I already used the known definitions of translation and time translation in my calculations, and I got a ##+## instead of a ##-##.

    Thanks a lot in advance.


    Julien.
     
  5. Sep 8, 2016 #4

    Orodruin

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    Well, then the result depends on which of the ##\Delta q## and ##\delta q## you change the definition for.
     
  6. Sep 8, 2016 #5

    Orodruin

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    I mean, judging from your teacher's script, he has ##q'(t) = q(t) + \delta q## and ##q'(t') = q(t) + \Delta q##. This also rhymes with the statement that ##\delta q## is the fixed time variation. You are using the opposite definition.
     
  7. Sep 8, 2016 #6
    @Orodruin In my teacher's script, both definitions are written as ##q' (t')##... Okay, I rewrite my whole development following your advice:

    We consider the following transformation:
    ##t' = t + \tau (t)##, ##q' (t) = q(t) + \delta q(t)##

    We want to know what is the symmetry variation, defined as:
    ##\Delta q = q' (t') - q(t)##
    (transformed coordinate - original coordinate)

    So:

    ##\Delta q = q' (t + \tau) - q(t)##
    ##= q'(t) + \tau \cdot \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)##

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

    Indeed, that works now.. Raah sorry for that, I get it now.


    Thanks for your help!


    Julien.
     
  8. Sep 8, 2016 #7

    Orodruin

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    Yes, but the text itself "For the variation of qa by a fixed time" and the way he is using it in the equations suggests the definition in my previous post is what is intended.
     
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