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Explanation for term in formula, QFT

  1. Jul 19, 2006 #1
    Hi

    I've got a question about a term in a formula I've found in Mandl&Shaw's QFT book

    It's about equation 2.18 on page 31

    [tex] L(t) = {\sum_i \delta \bf{x}_i {\cal L}_i \ ... [/tex]

    Why is there a delta x_i when summing over all lagrangians for getting the lagrange-function for the whole system?
    And what operator is that delta in this particular equation?
    The difference between two different points? (wouldn't that rather be [tex] \Delta \bf{x} [/tex] ?)
    [tex] \delta [/tex] serves only as functional derivative AFAIK
     
  2. jcsd
  3. Jul 19, 2006 #2
    I don't have Mandl & Shaw, this looks like a derivation of a field Lagrangian from a system of finite oscillators? The continuum limit [tex]\delta x_i \rightarrow 0[/tex] means you're spacing the components closer and closer together, which means their masses (and thus Lagrangians) must be scaled down as [tex]\delta x_i[/tex] if the mass density is to remain unchanged. There's a nice discussion in the final chapter of Goldstein, I assume a QFT book would not go into much detail about this.

    As for the notation, it looks like he's summing over infinitesimals [tex]\delta x_i[/tex], which is a physicist's way of doing calculus. :rolleyes: It essentially means [tex]\int dx \, {\cal L} (x)[/tex].
     
    Last edited: Jul 19, 2006
  4. Jul 19, 2006 #3
    yep, exactly, that's what is described in the book


    oh, IC now the reason of my wrong understanding. He uses [tex] \delta x_i [/tex] because he wants to show explicitly that he's talking about seperate cells (with the index i as one of the arguments of the field operator). On the page before that he declares explicitly his -in my opinion, awkward - notation.

    Thanks ;)
     
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