Integration by parts

[ditaelita]

Somebody could explain me, how of the second line arrive to the third one? in my book says that is integration by parts, please helpppp :eek:

[ditaelita]

the same is for this pleaseee :uhh:

[hunt_mat]

The first one is quite easy, you have \delta \dot{\phi}=\partial_{t} \delta \phi, now pat of the assumptions of calculus of variations is that the variation vanish at the boundaries. I think that will help you clear up your first problem.

How is quantum field theory treating you?

[ditaelita]

yes the variation vanish at the booundaries but i try and get a differente solution mmmmm

I study classical field theory from Field Quantization-Greiner and Reinhardt

[hunt_mat]

Okay you have:

\delta S = \int_{t_{1}}^{t_{2}}\frac{\delta L}{\delta \phi}\delta \phi +\frac{\delta L}{\delta \dot{\phi}}\delta \dot{\phi} d^{3} \mathbf{x}

Using the hint that I gave:

\delta S = \int_{t_{1}}^{t_{2}}\frac{\delta L}{\delta \phi}\delta \phi +\frac{\delta L}{\delta \dot{\phi}}\partial_{t}\delta \phi d^{3} \mathbf{x}

The second term can be integrated by parts to obtain:

\int_{t_{1}}^{t_{2}}\frac{\delta L}{\delta \dot{\phi}}\partial_{t}\delta \phi d^{3}\mathbf{x} =\left[ \frac{\delta L}{\delta \dot{\phi}}\delta \phi\right]_{t_{1}}^{t_{2}}-\int_{t_{1}}^{t_{2}}\frac{\partial}{\partial t}\frac{\delta L}{\delta \dot{\phi}}\delta \phi d^{3}\mathbf{x}

Putting this back in the same integral we have:

\delta S =\int_{t_{1}}^{t_{2}}\frac{\delta L}{\delta \phi}\delta \phi -\frac{\partial}{\partial t}\frac{\delta L}{\delta \dot{\phi}}\delta \phi d^{3}\mathbf{x}

So you see now?

[ditaelita]

Thank youuuuuu so much, you're the best
I'm really happy:biggrin:

[ditaelita]

Now I'll try the other one

[hunt_mat]

For that one, take a simple case of where the field has one space variable, i.e. when \phi \phi (t,x) It will be easier and give you a good feeling for the general case.

Glad to be of help...