Integration by parts hint

  • Thread starter Ratzinger
  • Start date
  • #1
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this no homework, but nevertheless can someone hint me how this integration by parts works?
[tex]
\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)
[/tex]
where [tex]
L(\phi ,\partial _\mu \phi )
[/tex]

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
[/tex]

thanks
 
Last edited:

Answers and Replies

  • #2
George Jones
Staff Emeritus
Science Advisor
Gold Member
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1,094
This is just the product rule, i.e.,

[tex]
\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)
[/tex]

Regards,
George
 
  • #3
287
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I knew that it was going to be simple. Thanks George!
 

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