Conserved charge from Lorentz symmetry

  1. I am trying to derive the conserved charge from the symmetry of the action under Lorentz transformations, but I am doing something wrong.
    Noether's theorem states that the current is
    [tex]
    J^\mu = \frac{\partial \cal L}{\partial(\partial_\mu \phi)} \delta\phi - T^{\mu \nu}\delta x_\nu
    [/tex]
    For an infinitesimal Lorentz transformations
    [tex]
    \Lambda^\mu _\nu = \delta^\mu_\nu +\omega ^\mu_\nu
    [/tex]
    I get
    [tex]\delta x_\nu = \omega_{\nu \sigma} x^\sigma, \quad \delta \phi = -\omega^\mu _\nu x^\nu \partial_\mu \phi [/tex]
    This gives
    [tex]
    J^\mu = \frac{\partial \cal L}{\partial(\partial_\mu \phi)}(-\omega^\nu _\sigma x^\sigma \partial_\nu \phi) -T^{\mu \nu}\omega_{\nu \sigma} x^\sigma
    [/tex]
    where
    [tex]
    T^{\mu \nu} = \frac{\partial \cal L}{\partial(\partial_\mu \phi)} \partial^\nu \phi -\cal L \eta^{\mu \nu}
    [/tex]
    so
    [tex]
    J^\mu = -\omega_{\nu \sigma} x^\sigma \left(2\frac{\partial \cal L}{\partial(\partial_\mu \phi)}\partial^\nu \phi - \cal L \eta^{\mu \nu} \right)
    [/tex]
    this is "almost" the correct form but I have this factor of 2 which is wrong. I will be happy to know where is my mistake
     
  2. jcsd
  3. samalkhaiat

    samalkhaiat 1,101
    Science Advisor

    Exactly what is it that you are doing here? You want to substitute stuff in a formula to get the same formula back?
    The general form of Noether current is
    [tex]J^{ \mu } = \frac{ \partial \mathcal{L} }{ \partial ( \partial_{ \mu } \phi ) } \delta \phi + \delta x^{ \mu } \mathcal{L} ,[/tex]
    where [itex]\delta \phi ( x ) = \bar{ \phi } ( x ) - \phi ( x )[/itex]. When you rewrite the current in terms of the energy momentum tensor, you get
    [tex]J^{ \mu } = \frac{ \partial \mathcal{L} }{ \partial ( \partial_{ \mu } \phi ) } \delta^{ * } \phi ( x ) - T^{ \mu }{}_{ \nu } \delta x^{ \nu } ,[/tex]
    where now the variation in the field is given by
    [tex]\delta^{ * } \phi ( x ) = \bar{ \phi } ( \bar{ x } ) - \phi ( x ) = \bar{ \phi } ( x + \delta x ) - \phi ( x ) .[/tex]
    When you expand to 1st order in [itex]\delta x[/itex] you find that the two variations are related by
    [tex]\delta^{ * } \phi ( x ) = \delta \phi ( x ) + \delta x^{ \mu } \partial_{ \mu } \phi ( x ) .[/tex]
    For scalar field [itex]\delta^{ * } \phi ( x ) = 0[/itex], this implies [itex]J^{ \mu } = -\delta x^{ \nu } T^{ \mu }{}_{ \nu }[/itex].
     
    Last edited: Apr 11, 2014
  4. samalkhaiat

    samalkhaiat 1,101
    Science Advisor

    Very consistent. The field in (1.51) is a scalar field.
    [tex]\bar{ \phi } ( \bar{ x } ) = \phi ( x ) , \ \ \Rightarrow \ \ \delta^{*} \phi = 0 .[/tex]
    Use [itex]x = \Lambda^{ - 1 } \bar{ x }[/itex] and rename the coordinates [itex]x[/itex], (i.e. drop the bar from [itex]\bar{ x }[/itex]), you get [itex]\bar{ \phi } ( x ) = \phi ( \Lambda^{ - 1 } x )[/itex]. Now, if you expand this, as they did, you find that [itex]\delta \phi = - \delta x^{ \mu } \partial_{ \mu } \phi[/itex]
     
  5. but why in my formula you use [itex] \delta^* \phi \equiv \phi'(x')-\phi(x) [/itex] and in the article it is [itex] \delta \phi = \phi'(x) -\phi(x) [/itex]?
     
  6. samalkhaiat

    samalkhaiat 1,101
    Science Advisor

    Look at it again. In the article, he uses the current which does not involve T, this is why he used [itex]\delta \phi[/itex]. If you want to start with the form of [itex]J^{ \mu }[/itex] which involves the tensor [itex]T^{ \mu \nu }[/itex], then you must use [itex]\delta^{ * }\phi[/itex], as I explained in my first post.
     
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