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I Derivative of a Variation vs Variation of a Derivative

  1. Jul 13, 2018 #1
    When a classical field is varied so that ##\phi ^{'}=\phi +\delta \phi## the spatial partial derivatives of the field is often written $$\partial _{\mu }\phi ^{'}=\partial _{\mu }(\phi +\delta \phi )=\partial _{\mu }\phi +\partial _{\mu }\delta \phi $$. Often times the next step is to switch the order of the variation and the partial derivative to get ##\partial _{\mu }\phi ^{'}=\partial _{\mu }\phi +\delta (\partial _{\mu }\phi )##. What justifies the replacement of ##\partial_{\mu }(\delta\phi )## by ##\delta (\partial _{\mu }\phi )##?
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  3. Jul 13, 2018 #2


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    Variations and derivatives commute if you keep your coordinates fixed during the variation. In deriving the Euler Lagrange eqns e.g. this is the case: the field variations involve functional variations.
  4. Jul 13, 2018 #3
    I guess my question is "why do variations and derivatives commute?"
  5. Jul 13, 2018 #4


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    haushofer answered your question correctly. The variation [itex]\delta[/itex] measures the change in the functional form of a field at a fixed coordinate value. So, if you define the field [itex]\psi_{\mu}(x) = \partial_{\mu}\phi (x)[/itex], then it follows from the definition of [itex]\delta[/itex] that [tex]\delta \psi_{\mu}(x) = \psi_{\mu}^{'}(x) - \psi_{\mu}(x),[/tex] or
    [tex]\delta (\partial_{\mu}\phi )(x) = \partial_{\mu}\phi^{'}(x) - \partial_{\mu}\phi(x) = \partial_{\mu}\left(\phi^{'} - \phi \right) (x) = \partial_{\mu}( \delta \phi )(x) .[/tex]
  6. Jul 13, 2018 #5
    Thinking of ##\partial _{\mu}\phi ## as an independent vector field that itself varies seemed to help! Thanks haushofer and samalkhaiat!
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