Noether's theoem and conserved current arbitrariness

[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!

 

[vanhees71]

Have look at

http://fias.uni-frankfurt.de/~hees/publ/lect.pdf

Sect. 3.3. I hope this answers your question.

 

[samalkhaiat]

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!

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

 

[Einj]

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