- #1
diegzumillo
- 173
- 18
Hi all
Maybe you could help me understanding this bit from the beginning of the book (peskin - intro to QFT).
In section 2.2, subsection "Noether's theorem" he first wants to show that continuous transformations on the fields that leave the equations of motion invariant (called symmetries) will have a corresponding conserved quantity. We can write a transformation like this:
[tex]\phi (x) \rightarrow \phi (x) + \alpha \Delta \phi (x)[/tex]
More specifically, he arguments that symmetries require the Lagrangian to be invariant under that transformation and writes:
[tex] \mathcal L (x) \rightarrow \mathcal L (x) +\alpha \partial _\mu \mathcal J ^\mu[/tex]
Right, now let's see what this actual variation is using the field variation:
[tex]\alpha \Delta \mathcal L = \frac {\partial \mathcal L}{\partial \phi}(\alpha \Delta \phi) + \left( \frac{\partial \mathcal L}{\partial (\partial _\mu \phi) } \right) \partial _\mu (\alpha \Delta \phi )[/tex]
which becomes (if anyone wants more steps, let me know)
[tex]\alpha \Delta \mathcal L = \alpha \partial _\mu \left( \frac{\partial \mathcal L}{\partial (\partial _\mu \phi) } \Delta \phi \right)[/tex]
So far I follow! but then he writes "we set the remaining term equal to [itex]\alpha \partial _\mu \mathcal J ^\mu[/itex] and find
[tex] \partial _\mu j^\mu (x) = 0[/tex]
where
[tex]j^\mu (x)= \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)} \Delta \phi - \mathcal J^\mu[/tex]
I just don't get his definition of [itex]j^\mu[/itex].
When he says "set the remaining term equal to [itex]\alpha \partial _\mu \mathcal J ^\mu[/itex]" this is what he's saying:
[tex]\alpha \partial _\mu \left( \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\Delta \phi \right) = \alpha \partial _\mu \mathcal J^\mu[/tex]
correct? so his definition of [itex]j^\mu[/itex] becomes
[tex]j^\mu = \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\Delta \phi - \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\Delta \phi = 0[/tex]
wait, what? what am I missing?
I have studied Noether's theorem before and I just don't see why he introduces these two J and j. To me, simply stating that [itex]\Delta \mathcal L = 0[/itex] leads to [itex]\partial _\mu \mathcal J^\mu = 0[/itex] which is the conserved quantity. But I don't get what he is doing.
Maybe you could help me understanding this bit from the beginning of the book (peskin - intro to QFT).
Homework Statement
In section 2.2, subsection "Noether's theorem" he first wants to show that continuous transformations on the fields that leave the equations of motion invariant (called symmetries) will have a corresponding conserved quantity. We can write a transformation like this:
[tex]\phi (x) \rightarrow \phi (x) + \alpha \Delta \phi (x)[/tex]
More specifically, he arguments that symmetries require the Lagrangian to be invariant under that transformation and writes:
[tex] \mathcal L (x) \rightarrow \mathcal L (x) +\alpha \partial _\mu \mathcal J ^\mu[/tex]
Right, now let's see what this actual variation is using the field variation:
[tex]\alpha \Delta \mathcal L = \frac {\partial \mathcal L}{\partial \phi}(\alpha \Delta \phi) + \left( \frac{\partial \mathcal L}{\partial (\partial _\mu \phi) } \right) \partial _\mu (\alpha \Delta \phi )[/tex]
which becomes (if anyone wants more steps, let me know)
[tex]\alpha \Delta \mathcal L = \alpha \partial _\mu \left( \frac{\partial \mathcal L}{\partial (\partial _\mu \phi) } \Delta \phi \right)[/tex]
So far I follow! but then he writes "we set the remaining term equal to [itex]\alpha \partial _\mu \mathcal J ^\mu[/itex] and find
[tex] \partial _\mu j^\mu (x) = 0[/tex]
where
[tex]j^\mu (x)= \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)} \Delta \phi - \mathcal J^\mu[/tex]
I just don't get his definition of [itex]j^\mu[/itex].
The Attempt at a Solution
When he says "set the remaining term equal to [itex]\alpha \partial _\mu \mathcal J ^\mu[/itex]" this is what he's saying:
[tex]\alpha \partial _\mu \left( \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\Delta \phi \right) = \alpha \partial _\mu \mathcal J^\mu[/tex]
correct? so his definition of [itex]j^\mu[/itex] becomes
[tex]j^\mu = \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\Delta \phi - \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\Delta \phi = 0[/tex]
wait, what? what am I missing?
I have studied Noether's theorem before and I just don't see why he introduces these two J and j. To me, simply stating that [itex]\Delta \mathcal L = 0[/itex] leads to [itex]\partial _\mu \mathcal J^\mu = 0[/itex] which is the conserved quantity. But I don't get what he is doing.