# Peskin QFT - Noether's theorem

1. Jun 15, 2014

### diegzumillo

Hi all
Maybe you could help me understanding this bit from the beginning of the book (peskin - intro to QFT).

1. The problem statement, all variables and given/known data

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:
$$\phi (x) \rightarrow \phi (x) + \alpha \Delta \phi (x)$$
More specifically, he arguments that symmetries require the Lagrangian to be invariant under that transformation and writes:
$$\mathcal L (x) \rightarrow \mathcal L (x) +\alpha \partial _\mu \mathcal J ^\mu$$

Right, now let's see what this actual variation is using the field variation:

$$\alpha \Delta \mathcal L = \frac {\partial \mathcal L}{\partial \phi}(\alpha \Delta \phi) + \left( \frac{\partial \mathcal L}{\partial (\partial _\mu \phi) } \right) \partial _\mu (\alpha \Delta \phi )$$
which becomes (if anyone wants more steps, let me know)
$$\alpha \Delta \mathcal L = \alpha \partial _\mu \left( \frac{\partial \mathcal L}{\partial (\partial _\mu \phi) } \Delta \phi \right)$$

So far I follow! but then he writes "we set the remaining term equal to $\alpha \partial _\mu \mathcal J ^\mu$ and find
$$\partial _\mu j^\mu (x) = 0$$
where
$$j^\mu (x)= \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)} \Delta \phi - \mathcal J^\mu$$

I just don't get his definition of $j^\mu$.

3. The attempt at a solution

When he says "set the remaining term equal to $\alpha \partial _\mu \mathcal J ^\mu$" this is what he's saying:
$$\alpha \partial _\mu \left( \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\Delta \phi \right) = \alpha \partial _\mu \mathcal J^\mu$$
correct? so his definition of $j^\mu$ becomes
$$j^\mu = \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\Delta \phi - \frac{\partial \mathcal L}{\partial (\partial _\mu \phi)}\Delta \phi = 0$$
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 $\Delta \mathcal L = 0$ leads to $\partial _\mu \mathcal J^\mu = 0$ which is the conserved quantity. But I don't get what he is doing.

2. Jun 15, 2014

### king vitamin

I have a suspicion that this is what you're missing. We do not require $\mathcal{L}$ to be invariant, we require that the action has the same saddle point; so the action is only shifted by a constant by the transformation. Thus, the lagrangian density can only change by a total 4-divergence.

This does NOT mean $\partial_{\mu} J^{\mu} = 0$! The upper-case J parametrizes a change in the lagrangian density under the transformation. It's an input into the definition of the current lower-case j. (They really did pick a notation that could confuse, huh?)

3. Jun 15, 2014

### diegzumillo

Oh! That makes more sense.

I think I figured my other mistake too. I instinctively thought that if
$$\partial _\mu \mathcal J ^\mu = \partial _\mu \left( \frac{ \partial \mathcal L }{\partial (\partial _\mu \phi)} \Delta \phi \right)$$
then
$$\mathcal J ^\mu = \left( \frac{ \partial \mathcal L }{\partial (\partial _\mu \phi)} \Delta \phi \right)$$
which was turning that definition into nonsense, but this isn't (necessarily) true, as we could have some constant in there.

I'm starting to think I need to review my covariant notation knowledge. This stuff is really confusing me.

4. Jun 16, 2014

### CAF123

Or, for example, the divergence of an anti symmetric tensor.

5. Jun 16, 2014

### diegzumillo

Could you elaborate? I can see how a curl of something could be there, as it would vanish when applying the divergence.

6. Jun 16, 2014

### Orodruin

Staff Emeritus
If you add the divergence of an anti-symmetric tensor $\partial_\nu F^{\nu\mu}$ to $\mathcal J^\mu$, then you get
$$\partial_\mu \mathcal J^\mu \rightarrow \partial_\mu (\mathcal J^\mu + \partial_\nu F^{\nu\mu}) = \partial_\mu \mathcal J^\mu,$$
since $\partial_\mu \partial_\nu F^{\nu\mu} = 0$ - so essentially the same reason the divergence of a curl vanishes.

7. Jun 16, 2014

### diegzumillo

Thanks everyone, I think those are all my difficulties with that part :)
(the other difficulties deserve new topics)

8. Jun 16, 2014

### CAF123

I thought I had an argument, but it appears I was circling myself.
Given $$\partial_{\mu}\partial_{\nu}F^{\mu \nu} = \partial_{\nu}\partial_{\mu}F^{\mu \nu} = -\partial_{\nu}\partial_{\mu}F^{\nu \mu}$$ and then relabel ($\mu \rightarrow \nu, \nu \rightarrow \mu$) gives $$\partial_{\nu}\partial_{\mu}F^{\nu \mu} = \partial_{\mu}\partial_{\nu}F^{\nu \mu} = - \partial_{\mu}\partial_{\nu}F^{\mu \nu} = -\partial_{\nu}\partial_{\mu}F^{\mu \nu}$$

But I can't quite use this to get $\partial_{\mu}\partial_{\nu}F^{\mu \nu} = 0$. Can you see?

9. Jun 16, 2014

### WannabeNewton

$\partial_{\mu}\partial_{\nu}$ is symmetric in its indices and $F^{\mu\nu}$ is antisymmetric in its indices so the contraction vanishes. More explicitly, $\partial_{\mu}\partial_{\nu}F^{\mu\nu} = -\partial_{\mu}\partial_{\nu}F^{\nu\mu} = -\partial_{\nu}\partial_{\mu}F^{\nu\mu} = -\partial_{\mu}\partial_{\nu}F^{\mu\nu}$.

10. Jun 16, 2014

### CAF123

Hi WbN! Thanks, but what property are you using to establish the last equality?

11. Jun 16, 2014

### WannabeNewton

I relabeled the indices.

12. Jun 16, 2014

### diegzumillo

I also convinced myself with a wrong argument.

Let me see if I get this. Usually relabeling has to be done in an entire equation, but you can also relabel when the indices are being summed, which is the case there. So this equation is not valid:
$$F_a = F _y$$
but this one is
$$\partial _a F^a = \partial _b F^b$$

edit: this seems so embarassingly obvious after I wrote it down.

13. Jun 16, 2014

### CAF123

That's what I thought and indeed what you wrote was what I had in mind initially. But can you relabel like that without relabeling the 'other' side of the equation so to speak?

For example, $\partial_{\mu}\partial_{\nu}F^{\mu \nu} = \partial_{\nu}\partial_{ \mu} F^{\nu \mu} = \partial_{\mu}\partial_{\nu}F^{\nu \mu} \Rightarrow$ a solution is that $F^{\mu \nu} = F^{\nu \mu}$ which is not true. In the first equality, I relabelled and in the second used the commutivity of the partial derivatives.

14. Jun 16, 2014

### king vitamin

I'm not sure what you mean. It is true that $\partial_{\mu}\partial_{\nu}F^{\mu \nu} = \partial_{\mu}\partial_{\nu}F^{\nu \mu}$, but this doesn't mean that F is symmetric. In fact, every singe antisymmetric F satisfies both equations, since both sides equal zero by WannabeNewton's proof. EDIT: To be clear, a symmetric F would also satisfy this equation, it's an exact identity.

Last edited: Jun 16, 2014
15. Jun 16, 2014

### CAF123

Yes, it makes sense. Thanks WbN and king vitamin.

16. Jun 16, 2014

### George Jones

Staff Emeritus
Adding to what king vitamin wrote

$\partial_{\mu}\partial_{\nu}F^{\mu \nu} = \partial_{\mu}\partial_{\nu}F^{\nu \mu}$ leads to $0 = \partial_{\mu} \partial_{\nu} \left(F^{\mu \nu} - F^{\nu \mu} \right)= \partial_{\mu} \partial_{\nu} K^{\mu \nu}$, where $K^{\mu \nu} = F^{\mu \nu} - F^{\nu \mu}$ is antisymmetric for any $F^{\mu \nu}$. Consequently, any $F^{\mu \nu}$ is a solution.

CAF123 posted while I was working on my latex. Didn't see it.[/edit]