In a general approach derivation of Noether theorem, we have

$$ \alpha \mathcal{L} = \alpha \partial _{\mu} \left( \frac{\partial \mathcal{L}}{\partial \left( \partial _{\mu} \phi \right)} \Delta \phi \right) + \alpha \left( \frac{\partial \mathcal{L}}{\partial \phi} - \partial _{\mu} \frac{\partial \mathcal{L}}{\partial \left( \partial _{\mu} \phi \right)} \right) \Delta \phi $$

The second term vanishes because of Lagrange equation. Then we define $\partial _{\mu} \mathcal{J}^{\mu} \equiv \partial _{\mu} \left( \frac{\partial \mathcal{L}}{\partial \left( \partial _{\mu} \phi \right)} \Delta \phi \right)$. Therefore conclude $\partial _{\mu} j^{\mu} = 0$ for

$$ j^{\mu} = \frac{\partial \mathcal{L}}{\partial \left( \partial _{\mu} \phi \right)} - \mathcal{J}^{\mu} $$

It made me so confused because is $j^{\mu}$ not just identically ZERO? If it is, why do we then care about its four divergence anyway?

Any thoughts would be appreciated!


closed as off-topic by Qmechanic Oct 29 '18 at 20:18

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  • $\begingroup$ "Then we define $\partial _{\mu} \mathcal{J}^{\mu} \equiv \partial _{\mu} \left( \frac{\partial \mathcal{L}}{\partial \left( \partial _{\mu} \phi \right)} \Delta \phi \right)$." No, we don't. Where did you find this definition? $\endgroup$ – AccidentalFourierTransform Oct 29 '18 at 20:09
  • $\begingroup$ @AccidentalFourierTransform It is defined so in Peskin and Schroeder's QFT book, right above equation 2.12. $\endgroup$ – zyy Oct 29 '18 at 20:11
  • 2
    $\begingroup$ You misunderstood what P&S are saying. They do not define $\mathcal J$ as the "remaining term". Rather, they have two expressions for $\Delta\mathcal L$, one defines $\mathcal J$ and the other one is just a Taylor expansion. Maybe you should check the examples and come back to the theorem after that. $\endgroup$ – AccidentalFourierTransform Oct 29 '18 at 20:13
  • $\begingroup$ To reopen this question (v1) consider to quote the various assertions and sources. $\endgroup$ – Qmechanic Oct 29 '18 at 20:18

You're misreading the cited source. An action-preserving transformation of $\phi$ adds a total derivative to $\mathcal{L}$. In particular, the fact the $\delta S=0$ doesn't use the Euler-Lagrange equation. The trick you're missing is you should find, without using the ELE, a valid-on-shell choice of $\mathcal{J}^\mu$ for which $\delta\mathcal{L}=\alpha\mathcal{J}^\mu$. I'm sure you'll see this at work if you go through their examples carefully.


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