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Trouble with noether's theorem

  1. Jun 17, 2012 #1
    If you can think of an infinitismal transformation of fields that vanishes at the endpoints, then doesn't the action automatically vanish by the Euler-Lagrange equations?

    For example take the Lagrangian:

    L=.5 m v2

    and the transformation:


    At t±∞, x'(±∞)=x(±∞), so the field x(t) doesn't change at the endpoints t=±∞ under this transformation. Since the transformation is infinitismal and the endpoints don't change, then the change in the action should be zero: hence there should be a conserved quantity, namely:


    But this quantity Q is not conserved for a free particle.
  2. jcsd
  3. Jun 17, 2012 #2
    Well, this is really a classical mechanics question, but I'll answer it and leave it to mods to move the thread if they desire.

    You are confusing a couple things. Noether's Theorem deals with the consequences of transformations that leave the action invariant along ALL paths, and in particular what this symmetry causes along a stationary path. So the transformation you described does not result in a conserved quantity because it is not a symmetry of the Lagrangian (you wind up with an extra term -2mεv/t^3 ).

    If we were only concerned about transformations that leave stationary paths invariant, then there would be an infinite number of (independent) conserved quantities since any transformation at all would do!

    Hope that helps.
  4. Jun 17, 2012 #3
    Right, but mathematically I don't see why it's untrue.

    The variation in the action would be:

    [tex]\delta S=\int dt \frac{\delta L}{\delta x} \delta x + \frac{\delta L}{\delta v} \delta v [/tex]

    Along the classical trajectory, the RHS transforms:

    [tex]\delta S=\int dt \frac{d}{dt}\left( \frac{\delta L}{\delta v} \delta x\right) [/tex]

    Also on a classical trajectory, the LHS is zero (since the transformation left the endpoints intact):

    [tex]0=\int dt \frac{d}{dt}\left( \frac{\delta L}{\delta v} \delta x\right) [/tex]


    [tex]\left( \frac{\delta L}{\delta v} \delta x\right)=const. [/tex]

    I must have made a mistake logically somewhere, but I can't figure out where.
  5. Jun 17, 2012 #4
    Never mind. I'm an idiot. The argumentation is sound, but the final result is of course trivial:

    [tex]0=\int dt \frac{d}{dt}\left( \frac{\delta L}{\delta v} \delta x\right) [/tex]

    is true, and this implies:

    [tex]\left( \frac{\delta L}{\delta v} \delta x\right)=const. *\mbox{at endpoints}*

    The endpoints are at t= ±∞, but at this point δx was *constructed* to be zero, so you basically have 0=0. Basically this is empty of content:

    [tex]\left( \frac{\delta L}{\delta v} \delta x\right)=\left( \frac{\delta L}{\delta v} 0 \right)=0 *\mbox{at endpoints}*

    So basically if you specify that your arbitrary transformation vanishes at the endpoints, then you'll get a conserved quantity, but it's trivial because it'll involve δx which was zero by fiat.
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