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Noether's theoem and conserved current arbitrariness

  1. Apr 20, 2015 #1
    Hello everyone,
    I have a question about conserved currents. Suppose that we have a Lagrangian which is perfectly invariant under a certain infinitesimal transformation of the field ##\phi\to\phi+\alpha\Delta\phi##, i.e. ##\mathcal{L}\to\mathcal{L}##. Now, more generally, the Lagrangian should be invariant up to a total divergence, ##\mathcal{L}\to\mathcal{L}+\alpha\partial_\mu\mathcal{J}^\mu##. This means that in our case ##\partial_\mu\mathcal{J}^\mu=0## and so ##\mathcal{J}^\mu## could be any aribtrary function with vanishing divergence.

    Now, the general form for the conserved current is:
    $$
    j_\mu(x)=\frac{\partial \mathcal{L}}{\partial(\partial^\mu\phi)}\Delta\phi-\mathcal{J}_\mu.
    $$

    Suppose now that the first term has zero divergence as well because of the equation of motion. Now, it seems to me that in this case, I could simply set the current to be zero by choosing ##\mathcal{J}^\mu=\frac{\partial \mathcal{L}}{\partial(\partial^\mu\phi)}\Delta\phi## since this indeed satisfies ##\partial^\mu\mathcal{J}_\mu=0##.

    Am I doing something wrong or is this indeed the case?

    Thanks!
     
  2. jcsd
  3. Apr 21, 2015 #2

    vanhees71

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  4. Apr 21, 2015 #3

    samalkhaiat

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    If the transformation [itex]\phi \to \phi + \delta \phi[/itex] leaves some Lagrangian invariant (off-shell): [itex]\mathcal{L} \to \mathcal{L}[/itex] (or [itex]\delta \mathcal{L} = 0[/itex]), then you have (on-shell) conserved current of the form [tex]J^{\mu} = \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \phi )} \ \delta \phi ,[/tex] and the transformation is an INTERNAL symmetry transformation. If, under some transformation [itex]\Delta \phi[/itex] the Lagrangian change (off-shell) by a total divergence of some object [itex]\Lambda^{\mu}[/itex]: [itex]\delta \mathcal{L} = \partial_{\mu} \Lambda^{\mu}[/itex], then the (on-shell) conserved current is given by [tex]J^{\mu} = \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \phi )} \ \Delta \phi - \Lambda^{\mu} ,[/tex] and the transformation [itex]\Delta \phi[/itex] is SPACE-TIME symmetry transformation. The object [itex]\Lambda^{\mu}[/itex] IS NOT arbitrary. How can it be arbitrary? It is what you get when you transform the Lagrangian. So, [itex]\Lambda^{\mu}[/itex] depends on the Lagrangian, and you can not choose it arbitrarily. For example, under the translation [itex]\Delta \phi = - a^{\mu} \partial_{\mu} \phi[/itex], the Lagrangian transforms like [itex]\delta \mathcal{L} = \partial_{\mu} ( a^{\mu} \mathcal{L} )[/itex]. So, [itex]\Lambda^{\mu} = a^{\mu} \mathcal{L}[/itex]. Is this arbitrary? You are confusing the two type of transformation with the freedom to add a divergence term into the Lagrangian. See, the discussion on pages 7-8 in the PHD below.
    Sam
     

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  5. Apr 21, 2015 #4
    Yes, I understand now. Thanks you very much for your help!
     
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