# Question - cal of variations

1. Jul 29, 2014

### Travis091

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

2. Jul 29, 2014

### Matterwave

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:

$$\frac{\partial\mathcal{L}}{\partial(\partial_ \mu\phi)}\delta(\partial_\mu\phi)=\partial_\mu\left(\frac{\partial \mathcal{L}}{\partial(\partial_\mu\phi)}\delta\phi\right)-\partial_\mu\left(\frac{\partial\mathcal{L}}{\partial(\partial _\mu\phi)}\right)\delta\phi$$

Last edited: Jul 29, 2014