What is the relationship between variational calculus and integration by parts?

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SUMMARY

The relationship between variational calculus and integration by parts is illustrated through the action integral S=\int d^4x \;\mathcal{L}, where the Lagrangian density is defined as \mathcal{L}=\mathcal{L}(\phi,\dot{\phi},\nabla\phi). The variation of S leads to the expression \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]. The integration by parts is applied to the second term, resulting in the rearrangement that allows for the application of the chain rule, confirming the commutation of the partial derivative with the variation operator δ.

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Travis091
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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:

S=\int d^4x \;\mathcal{L}

with the Lagrangian (density):

\mathcal{L}=\mathcal{L}(\phi,\dot{\phi},\nabla\phi)

We then vary S:

\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]

which is all fine and dandy, but now the next line says:
= \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)

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 \delta throws me off.

So what happened there between the two lines? Thanks
 
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Indeed it is integration by parts. By the chain rule we have:

\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)

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:

\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
 
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