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?
#4
ditaelita
5
0
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
Okay you have:
<br />
\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} <br />
Using the hint that I gave:
<br />
\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} <br />
The second term can be integrated by parts to obtain:
<br />
\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}<br />
Putting this back in the same integral we have:
<br />
\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}<br />
So you see now?
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#6
ditaelita
5
0
Thank youuuuuu so much, you're the best
I'm really happy
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.