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Conserved charge from Lorentz symmetry

  1. Apr 11, 2014 #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. Apr 11, 2014 #2

    samalkhaiat

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    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. Apr 11, 2014 #3
  5. Apr 11, 2014 #4

    samalkhaiat

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    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]
     
  6. Apr 11, 2014 #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]?
     
  7. Apr 11, 2014 #6

    samalkhaiat

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    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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