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Why does field Lagrangian depend on four-derivative?

  1. Jan 27, 2015 #1
    Hi guys, so this is a pretty generic question.

    Starting off with the classical Lagrangian in a case where there is no interaction or explicit time dependence, the functional form is

    [itex]L=L(x,\dot{x})=L(x,\partial_{t}x)[/itex].

    Now when we look at the Lagrangian density in field theory, the functional dependence is

    [itex]\mathcal{L}=\mathcal{L}(\varphi,\partial_{\mu}\varphi)[/itex].

    And there's my question. Why does the Lagrangian density depend on [itex]\partial_{\mu}\varphi[/itex] and not only [itex]\partial_{t}\varphi[/itex]? i mean why the four-derivative?

    I have a few ideas but I'm not sure if they are correct. I'm thinking along the lines that since [itex]\partial_{\mu} = \nabla + \partial_{t}[/itex], maybe the [itex]\partial_{\mu}[/itex] just appears to merge the two together and make it more compact as this one term includes possible interactions and also the field velocity?

    Please help! thank you :)
     
  2. jcsd
  3. Jan 27, 2015 #2
    That's not true.
    The field is assumed to be lorentz invariant, hence derivative terms must be built out of scalars involving the whole four-gradient. (e.g. box operator).
     
  4. Jan 27, 2015 #3
    Okay thank you :) I do have a question though - why is it wrong? is the sign of the nabla meant to be negative or is it just completely wrong?
     
  5. Jan 31, 2015 #4
    Didn't see this sorry.
    The partial derivative with the greek subscript just refers a derivative with respect to x,y,z or t.
     
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