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A Generalising the Euler-Lagrange equation for scalar fields

  1. Sep 18, 2016 #1
    The Euler-Lagrange equation obtained from the action ##S=\int\ d^{4}x\ \mathcal{L}(\phi,\partial_{\mu}\phi)## is ##\frac{\partial\mathcal{L}}{\partial\phi}-\partial_{\mu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\phi)}\big)=0##.

    My goal is to generalise the Euler-Lagrange equation for the action ##S=\int\ d^{4}x\ \mathcal{L}(\phi,\partial_{\mu}\phi, \partial_{\mu}\partial_{\nu}\phi)##.

    The variation of the Lagrangian density this time contains the extra term ##\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta(\partial_{\mu}\partial_{\nu}\phi)##. Now,

    ##\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta(\partial_{\mu}\partial_{\nu}\phi)##

    ##=\partial_{\mu}\Big[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\partial_{\nu}(\delta\phi)\Big]-\partial_{\mu}\Big[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\Big]\partial_{\nu}(\delta\phi)##

    ##=\partial_{\mu}\Big[\partial_{\nu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta\phi\big)-\partial_{\nu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\big)\delta\phi\Big]-\Big[\partial_{\nu}\Big(\partial_{\mu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\big)\delta\phi\Big)-\partial_{\nu}\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\Big)\delta\phi\Big]##

    Now, in the fourth term, ##\partial_{\nu}\partial_{\mu}\Big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\Big)=0## for arbitrary ##\delta\phi## - this is the new term in the Euler-Lagrange equation.

    The second and third terms must give zero, by Gauss' theorem and under the assumption that the field vanishes at spatial or temporal infinity.

    What do I do about the first term? Should I have to introduce a constraint on the derivative ##\partial_{\nu}\big(\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\delta\phi\big)## - I don't want to, because it's not physical.
     
  2. jcsd
  3. Sep 18, 2016 #2
    You don't do anything about the first term. It becomes part of the new Euler-Lagrange equations. Every time you make the Lagrangian depend on a higher order derivative, the Euler-Lagrange equations increase in order.
     
  4. Sep 18, 2016 #3
    Well, at least I need to factor out ##\delta\phi## from the first term ##\partial_{\mu}\Big[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\partial_{\nu}(\delta\phi)\Big]##, don't I?

    Could you suggest a way in which this could be done?
     
  5. Sep 18, 2016 #4
    Sorry I got it wrong, the first term doesn't remain. The fourth one does, since it is the only term that does not go away with integration by parts. The first three terms are boundary terms, since they are integrals of a divergence.
     
  6. Sep 18, 2016 #5
    Hmm...

    Well, if I integrate the term ##\partial_{\mu}\Big[\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\partial_{\nu}(\delta\phi)\Big]## by parts, then the boundary term is ##\frac{\partial\mathcal{L}}{\partial(\partial_{\mu}\partial_{\nu}\phi)}\partial_{\nu}(\delta\phi)##, but to set this boundary term to zero, I need to assume that ##\delta(\partial_{\nu}\phi)## is zero at temporal and spatial infinity. Isn't this a little unphysical?
     
  7. Sep 18, 2016 #6
    Let us make the usual assumption that ##\delta \phi## vanishes at infinity. That means ##\delta \phi## gets closer and closer to ##0## as the point in question gets closer and closer to infinity (this can be easily formulated more rigorously). Now, I do believe there is an extra assumption that ##\partial_\nu(\delta\phi)## also converges to ##0## as you go to infinity. This assumption actually seems very reasonable to me, because otherwise you would have a field that is rapidly oscillating as you go to infinity.
     
  8. Sep 18, 2016 #7
    Hmm...

    Would you also expect all the higher-order derivatives of the field to vanish at the boundary?
     
    Last edited: Sep 18, 2016
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