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Noether current

  1. Oct 26, 2013 #1

    I read about Noether's theorem, which states that if, under a continuous transformation, the Lagrangian is changed by a total derivative
    [itex] \delta \cal L = \partial_\mu F^\mu [/itex]

    then there is a conserved current
    [tex] j^\mu = \frac{\partial \cal L}{\partial(\partial_\mu \phi)}\delta \phi - F^\mu [/tex]

    However, I have seen in a different place the formulation that if the action is invariant, then the conserved quantity is:

    [tex] \frac{\partial \cal L}{\partial(\partial_\mu \phi)}\delta \phi - T^{\mu \nu}\delta x_\nu [/tex]
    where [itex] T^{\mu \nu} [/itex] is the energy-momentum tensor.

    Is the second formulation equivalent to the first? or is it a particular case
    Last edited: Oct 26, 2013
  2. jcsd
  3. Oct 26, 2013 #2


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    Neither of both formulae is the most general case of a symmetry and Noether's theorem but special cases.

    The first case is the symmetry of the action under a variation of the field and unvaried space-time coordinates, where the Lagrangian changes by a total four-gradient, which means that the action is invariant.

    The second case is a symmetry under a more general transformation, where the space-time coordinates and fields are changed under the transformation and the Lagrangian is invariant. Of course a symmetry is still present also under such transformations, if the Lagrangian changes by a total four-gradient. Then the Noether current is
    [tex]\frac{\partial}{\partial (\partial_{\mu} \phi)} \delta \phi - T^{\mu \nu} \delta x_{\nu}-F^{\mu}.[/tex]
    You find this derived in some detail in my quantum-field theory manuscript:
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