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Lagrangian density for fields

  1. Oct 24, 2013 #1
    This is probably a minor point, but I have seen in some QFT texts the Euler-Lagrange equation for a scalar field,

    [tex] \partial_{\mu} \left(\frac{\delta \cal{L}}{\delta (\partial_{\mu}\phi)}\right) - \frac{\delta \cal L}{\delta \phi }=0 [/tex]

    i.e. [itex] \cal L [/itex] is treated like a functional (seen from the [itex] \delta [/itex] symbol). But why would it be a functional? Functonals map functions into numbers, and in our case [itex] \cal L [/itex] is a function of the fields (and their derivatives).
     
    Last edited: Oct 24, 2013
  2. jcsd
  3. Oct 25, 2013 #2

    fzero

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    If you want to be completely rigorous, the action is the true functional. The variational derivatives of the Lagrangian (density) should be considered distributions.
     
  4. Oct 25, 2013 #3
    But why should there be a functional derivative of [itex] \cal L [/itex]? we have [itex] \cal L [/itex] which is a function of [itex] (\phi, \partial_\mu \phi) [/itex] and we differentiate (as a function) with respect to [itex] \partial_\mu \phi [/itex]
     
  5. Oct 25, 2013 #4

    Bill_K

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    If they wrote it that way it's a misprint. The derivatives should be ∂'s, not δ's.
     
  6. Oct 25, 2013 #5

    fzero

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    It's common to abuse the notation and use ##\delta## for these derivatives in order to distinguish them from the coordinate derivatives ##\partial_\mu##.
     
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