# Noether's theoem and conserved current arbitrariness

1. Apr 20, 2015

### Einj

Hello everyone,
I have a question about conserved currents. Suppose that we have a Lagrangian which is perfectly invariant under a certain infinitesimal transformation of the field $\phi\to\phi+\alpha\Delta\phi$, i.e. $\mathcal{L}\to\mathcal{L}$. Now, more generally, the Lagrangian should be invariant up to a total divergence, $\mathcal{L}\to\mathcal{L}+\alpha\partial_\mu\mathcal{J}^\mu$. This means that in our case $\partial_\mu\mathcal{J}^\mu=0$ and so $\mathcal{J}^\mu$ could be any aribtrary function with vanishing divergence.

Now, the general form for the conserved current is:
$$j_\mu(x)=\frac{\partial \mathcal{L}}{\partial(\partial^\mu\phi)}\Delta\phi-\mathcal{J}_\mu.$$

Suppose now that the first term has zero divergence as well because of the equation of motion. Now, it seems to me that in this case, I could simply set the current to be zero by choosing $\mathcal{J}^\mu=\frac{\partial \mathcal{L}}{\partial(\partial^\mu\phi)}\Delta\phi$ since this indeed satisfies $\partial^\mu\mathcal{J}_\mu=0$.

Am I doing something wrong or is this indeed the case?

Thanks!

2. Apr 21, 2015

### vanhees71

3. Apr 21, 2015

### samalkhaiat

If the transformation $\phi \to \phi + \delta \phi$ leaves some Lagrangian invariant (off-shell): $\mathcal{L} \to \mathcal{L}$ (or $\delta \mathcal{L} = 0$), then you have (on-shell) conserved current of the form $$J^{\mu} = \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \phi )} \ \delta \phi ,$$ and the transformation is an INTERNAL symmetry transformation. If, under some transformation $\Delta \phi$ the Lagrangian change (off-shell) by a total divergence of some object $\Lambda^{\mu}$: $\delta \mathcal{L} = \partial_{\mu} \Lambda^{\mu}$, then the (on-shell) conserved current is given by $$J^{\mu} = \frac{\partial \mathcal{L}}{\partial ( \partial_{\mu} \phi )} \ \Delta \phi - \Lambda^{\mu} ,$$ and the transformation $\Delta \phi$ is SPACE-TIME symmetry transformation. The object $\Lambda^{\mu}$ IS NOT arbitrary. How can it be arbitrary? It is what you get when you transform the Lagrangian. So, $\Lambda^{\mu}$ depends on the Lagrangian, and you can not choose it arbitrarily. For example, under the translation $\Delta \phi = - a^{\mu} \partial_{\mu} \phi$, the Lagrangian transforms like $\delta \mathcal{L} = \partial_{\mu} ( a^{\mu} \mathcal{L} )$. So, $\Lambda^{\mu} = a^{\mu} \mathcal{L}$. Is this arbitrary? You are confusing the two type of transformation with the freedom to add a divergence term into the Lagrangian. See, the discussion on pages 7-8 in the PHD below.
Sam

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4. Apr 21, 2015

### Einj

Yes, I understand now. Thanks you very much for your help!