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Noether's currents under dilatations (scaling transformations)

  1. Nov 27, 2015 #1
    Hello,
    1. The problem statement, all variables and given/known data

    Suppose we have the following Lagrangian density, in ## 3 + 1## dimensions:
    $$L = \frac{1}{2}\partial_{\mu}\phi \partial^{\mu}\phi - g \phi^4$$
    Under the dilatation (scaling transformation): ##x \rightarrow \lambda x^{\mu}, \phi (x) \rightarrow \lambda^{-1} \phi(\lambda^{-1} x)##, we should calculate the resulting conserved currents.

    2. Relevant equations
    Noether's currents' definition formula:
    $$D^{\mu} = \frac{\partial L}{\partial (\partial_{\mu} \phi)}\delta\phi - F^{\mu}$$
    $$\partial_{\mu} F^{\mu} = \delta L$$

    3. The attempt at a solution
    The answer would be a simple application of the general Noether's currents' definition formula, if I knew how to calculate: ##\delta\phi## and ##F^{\mu} ##.

    Thank you in advance,
    askalot.
     
  2. jcsd
  3. Nov 30, 2015 #2

    DrDu

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    Science Advisor

    I guess ##\delta \phi= (\partial \phi/\partial \lambda) d\lambda##.
     
  4. Nov 30, 2015 #3

    samalkhaiat

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    If the coordinates transform like
    [tex]\bar{x}^{\mu} = e^{-\epsilon} x^{\mu} \approx (1-\epsilon) x^{\mu} ,[/tex]
    i.e.,
    [tex]\delta x^{\mu} = -\epsilon \ x^{\mu} ,[/tex] the fields transform according to
    [tex]\bar{\varphi}(\bar{x}) = e^{\epsilon \Delta} \ \varphi(x) ,[/tex] where [itex]\Delta[/itex] is the scaling dimension of the field ([itex]\Delta = 1[/itex] for scalar field). If you expand both sides to first order in [itex]\epsilon[/itex], you find
    [tex]\delta \varphi (x) \equiv \bar{\varphi}(x) - \varphi (x) = \epsilon \left( \Delta + x^{\mu} \partial_{\mu} \right) \varphi (x) .[/tex]
    Okay, now you do the substitution into the Noether current of dilatation [tex]D^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_{\mu} \varphi)} \delta \varphi + \delta x^{\mu} \mathcal{L} .[/tex]
     
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