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Question about noether's theorem argument

  1. Oct 16, 2008 #1
    Given a lagrangian [itex]L[\phi][/itex], where [itex]\phi[/itex] is a generic label for all the fields of the system, a transformation [itex]\phi(x) \rightarrow \phi(x) + \epsilon \delta \phi(x)[/itex] that leaves the lagrangian invariant corresponds to a conserved current by the following argument.

    If we were to send [itex]\phi(x) \rightarrow \phi(x) + \epsilon(x) \delta \phi(x) [/itex], this would not in general be a symmetry, but would be in the special case that [itex]\epsilon[/itex] is constant. Therefore*, the change in the lagrangian must be proportional to the derivative of [itex]\epsilon(x)[/itex], that is:

    [tex] \delta L = j^\mu (x) \partial_\mu \epsilon(x) [/tex]

    Now when the equations of motion are satisfied, all infinitessimal variations, symmetries or not, leave the action unchanges, so in this case we must have:

    [itex] 0 = \delta S = \int d^4x j^\mu (x) \partial_\mu \epsilon(x) [/itex]

    or, integrating by parts:

    [itex] 0 = \int d^4x \epsilon(x) \partial_\mu j^\mu (x) [/itex]

    Since [itex]\epsilon(x)[/itex] is arbitrary, this implies [itex] \partial_\mu j^\mu(x) = 0[/itex].

    My problem is with the part marked by a *. Just because something vanishes when [itex]\epsilon[/itex] is constant, why should we expect the thing to be proportional to the dderivative of [itex]\epsilon(x)[/itex]? I could imagine other dependences. For example, the following things all vanish when [itex]\epsilon(x)[/itex] is constant:

    [tex] (j^\mu \partial_\mu \epsilon(x) )^2 [/tex]

    [tex] \partial^2 \epsilon(x) [/tex]

    [tex] \epsilon(x+1) - \epsilon(x) [/tex]

    I could go on. Granted, you could eliminate these examples individually, eg, the variation should be linear and local in [itex]\epsilon(x)[/itex], and by integrating by parts we can turn the middle one into the desired form. But there are other examples, and I'm wondering how we can argue for the form [itex]\partial_\mu \epsilon[/itex] directly rather than eliminating these other possibilities one by one.
  2. jcsd
  3. Oct 16, 2008 #2

    Ben Niehoff

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    Try expanding L as a Taylor series in [itex](\epsilon \delta \phi)[/itex]. Then because [itex]\delta \phi[/itex] is by definition infinitesimal, the higher-order terms vanish.

    Now, because you are looking at the change in L under infinitesimal variations, the zeroth-order term also vanishes. What you should be left with is the first-order term.
  4. Oct 16, 2008 #3
    So the taylor expansion would have to be a sum over all terms of the form:

    [tex] \Pi_{i, m_i, n_i} ({\partial_i}^{m_i} \epsilon(x) )^{n_i}[/tex]

    where the product is over some subset of the values i, m_i, and n_i could take, and the sum would be over all such subsets. This is a pretty extreme generalization of taylors theorem. If it is true, then I guess we could argue we can ignore terms except those of the form:

    [tex] {\partial_i}^{m_i} \epsilon (x) [/tex]

    and then integrate by parts. This might be what they mean, but if so, it isn't very satisfying or intuitive, and they really skipped over a lot of details.
  5. Oct 17, 2008 #4


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    I'm actually the OP, I illegally had two user names, painfive is now gone. But I'd still like someone to tell me if this taylor's theorem argument is the way to go.
  6. Oct 17, 2008 #5


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