Dismiss Notice
Join Physics Forums Today!
The friendliest, high quality science and math community on the planet! Everyone who loves science is here!

Explanation for term in formula, QFT

  1. Jul 19, 2006 #1

    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 by a moderator: 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 ;)
Share this great discussion with others via Reddit, Google+, Twitter, or Facebook