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Question - cal of variations

  1. Jul 29, 2014 #1
    I'm reading a book - and I've been stuck for a while on the same page. This is only a calculus question. We have the action:

    [tex]S=\int d^4x \;\mathcal{L}[/tex]

    with the Lagrangian (density):

    [tex]\mathcal{L}=\mathcal{L}(\phi,\dot{\phi},\nabla\phi)[/tex]

    We then vary S:

    [tex]\delta S = \int d^4x \left[\frac{\partial \mathcal{L}}{\partial \phi}\delta\phi + \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi)} \delta(\partial_\mu \phi)\right] [/tex]

    which is all fine and dandy, but now the next line says:
    [tex] = \int d^4 x \left[\frac{\partial \mathcal{L}}{\partial \phi}-\partial_\mu \left(\frac{\partial \mathcal{L}}{\partial (\partial_\mu\phi)} \right) \right]\delta \phi + \partial_\mu \left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)} \delta \phi\right)[/tex]

    Something like integration by parts must have befallen the second term...but I don't see it. I'm very inadequate in variational calculus, the mere sight of [itex]\delta[/itex] throws me off.

    So what happened there between the two lines? Thanks
     
  2. jcsd
  3. Jul 29, 2014 #2

    Matterwave

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    Gold Member

    Indeed it is integration by parts. By the chain rule we have:

    [tex]\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial_ \mu\phi)}\delta\phi\right)=\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial( \partial_\mu\phi)}\right)\delta\phi+\frac{\partial\mathcal{L}}{ \partial(\partial _\mu\phi)}\partial_\mu(\delta\phi)[/tex]

    The only additional step you need then is to know that the partial derivative commutes with the ##\delta## so that the above equation can be rearranged to:

    [tex]\frac{\partial\mathcal{L}}{\partial(\partial_ \mu\phi)}\delta(\partial_\mu\phi)=\partial_\mu\left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi\right)-\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial _\mu\phi)}\right)\delta\phi[/tex]
     
    Last edited: Jul 29, 2014
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