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Confusion about Noether's theorem

  1. May 27, 2013 #1
    Hi,

    I keep running my brain in circles while trying to get a solid grip on Noether's theorem. (In Peskin and Schroeder they present this as a one-liner.) But I'm having trouble seeing the equivalence between "equations of motion are invariant" and "action is invariant (up to boundary term)". Now I know that when the equations of motion are satisfied then there is no change in the action for infinitesimal variations. More exactly for variations which are zero on the boundary. Thus, there is a solution field [itex]\phi_0[/itex], and neighboring fields [itex]\phi_0 + \delta\phi[/itex], all of which have the same boundary values. If I apply a symmetry transform [itex]U[/itex] on all these fields, then their boundary values need not all transform the same way (right?). If they don't have the same boundary values, then it doesn't feel like we should be comparing their action anymore. Or at least that we're comparing the wrong set of fields. And if that's the case, then who's to say that the transformed field [itex]U(\phi_0)[/itex] continues to be an extrema of the action (i.e. solution of equations of motion)?

    Sam
     
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  3. May 28, 2013 #2

    vanhees71

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    I'm not sure whether I understand your question correctly. Noether's theorem is the statement that if an action functional is invariant under a one-parameter Lie group then the generator of this group is conserved along the trajectory of the system which is given by a stationary point of the action functional. Here are my 2cts for a proof

    http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

    p. 53ff. I cannot provide a one-line proof as Peskin and Schroeder, but maybe the somewhat lengthier proof helps.
     
  4. May 28, 2013 #3
    Peskin is probably not the best book to learn classical field theory from. Goldstein has a good treatment of Noether's theorem.
     
  5. May 28, 2013 #4
    vanhees71 (or should I say Dr. vanhees71), That's a nice set of lecture notes. To clarify: I think I can follow along that a conserved current exists when the action is taken as invariant. In that sense I'm not confused about the formal statement of Noether's theorem.

    What I have trouble seeing is how invariance of action (up to boundary term) implies the equations of motion are also invariant (i.e. [itex]U(\phi_0)[/itex] ALSO satisfies the EOM). (I know the former can be taken as a definition of symmetry, as in your notes, but the latter viewpoint is what I'm accustomed to. Either way, the two ought be equivalent.) The reason I don't find it obvious: After symmetry transforming two fields with the same boundary value, their 'new' boundary values no longer need to agree. Thus, if some field is an extrema with respect to neighboring fields that agree on the boundary (i.e. solution), it is not clear to me that the transformed field is still an extrema with respect to a 'new' set of neighboring fields that agree on the transformed boundary. I noticed in your notes that the fields are taken to vanish at infinity of space and time. If there is no boundary, then I don't have much to quibble about. In other proofs though I typically see there is a start and end time. And then variations of the field are constrained to zero at these endpoints.

    physwizard: Actually, My hope is to learn quantum field theory well. This is one of those early waypoints.

    Thanks for responses/feedback.
     
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