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Lorentz transformation of a scalar field

  1. Nov 18, 2015 #1
    Hi, the following is taken from Peskin and Schroeder page 36:

    ##\partial_{\mu}\phi(x) \rightarrow \partial_{\mu}(\phi(\Lambda^{-1}x)) = (\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x)##

    It describes the transformation law for a scalar field ##\phi(x)## for an active transformation.

    I would like to work out the intermediate steps by myself as they are missing from the textbook. Can you please correct any mistakes I make?

    ##\partial_{\mu}\phi(x) \rightarrow \partial_{\mu}(\phi(\Lambda^{-1}x)) = \frac{\partial(\phi(\Lambda^{-1}x))}{\partial x^{\mu}} = \frac{\partial (\Lambda^{-1}x)^{\nu}}{\partial x^{\mu}} \frac{\partial \phi((\Lambda^{-1}x))}{\partial (\Lambda^{-1}x)^{\nu}} = (\Lambda^{-1})^{\nu}_{\mu}(\partial_{\nu}\phi)(\Lambda^{-1}x)##.

    Am I correct?
     
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  3. Nov 18, 2015 #2

    haushofer

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    Science Advisor

    Yes. Personally I find this convention very confusing; a scalar transforms under x --> x' as

    ##
    \phi(x) \rightarrow \phi'(x')
    ##
    So
    ##
    \partial_{\mu} \phi(x) \rightarrow \partial'_{\mu} \phi'(x') = \frac{\partial x^{\rho}}{\partial x^{'\mu}} \partial_{\rho} \phi'(x')
    ##

    But whatever suits you, of course; I guess it comes down to the 'passive v.s. active'-discussion.
     
  4. Nov 18, 2015 #3
    All right, so ##\partial_{\mu}(\phi(\Lambda^{-1}x)) = \frac{\partial(\phi(\Lambda^{-1}x))}{\partial x^{\mu}}## because only ##\phi## is a function of ##\Lambda^{-1}x##.

    But then, ##(\partial_{\nu}\phi)(\Lambda^{-1}x) = \frac{\partial \phi(\Lambda^{-1}x)}{\partial (\Lambda^{-1}x)^{\nu}}##, all of ##\partial_{\nu} \phi## is a function of ##(\Lambda^{-1}x)##.

    Am I correct?
     
  5. Nov 18, 2015 #4
    Was your computation from the passive point of view?
     
  6. Nov 18, 2015 #5

    haushofer

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    Science Advisor

    I use x- primes for your Lamda-x. And yes, my approach is usually called the passive one.
     
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