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Integration by parts hint

  1. Dec 7, 2005 #1
    this no homework, but nevertheless can someone hint me how this integration by parts works?
    \int {d^4 } x\frac{{\partial L}}{{\partial \left( {\partial _\mu \phi } \right)}}\partial _\mu (\delta \phi ) = {\rm{ }} - \int {d^4 } x\partial _\mu \left( {\frac{{\partial L}}{{\partial (\partial _\mu \phi )}}} \right)\delta \phi {\rm{ }} + {\rm{ }}\int {d^4 } x{\rm{ }}\partial _\mu \left( {\frac{{\partial L}}{{\partial (\partial _\mu \phi )}}\delta \phi } \right)
    where [tex]
    L(\phi ,\partial _\mu \phi )

    I don't understand where the second term on the RHS comes from. I thought the second term should be [tex]
    \frac{{\partial L}}{{\partial (\partial _\mu \phi )}}\delta \phi \left| {^b _a } \right. = 0

    Last edited: Dec 7, 2005
  2. jcsd
  3. Dec 7, 2005 #2

    George Jones

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    This is just the product rule, i.e.,

    \frac{{\partial L}}{{\partial \left( {\partial _\mu \phi } \right)}}\partial _\mu (\delta \phi ) + \partial _\mu \left( {\frac{{\partial L}}{{\partial (\partial _\mu \phi )}}} \right)\delta \phi {\rm{ }} = \partial _\mu \left( {\frac{{\partial L}}{{\partial (\partial _\mu \phi )}}\delta \phi } \right)

  4. Dec 7, 2005 #3
    I knew that it was going to be simple. Thanks George!
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