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When I looked at the Noether's theorem(here, we discuss the field case), there are two ways to derive it. One way is to assume that the variation of Lagrange $\delta L$ is exactly equal to derivative of some function, say $\partial_\mu F^{\mu}$, then $$j^{\mu}=\frac{\partial\mathcal L}{\partial(\partial_\mu\varphi_a)}Y_a(\varphi)-F^{\mu}(\varphi) ,$$ where $Y_a$ is a small variation in $\varphi$. On the other hand, one can start by varying in coordinates and $\varphi$. Then one can derive $$j^{\mu}=\left[\frac{\partial \mathcal L}{\partial(\partial_\mu \varphi)}\partial_\nu\varphi-\delta_\nu^\mu\mathcal L\right]\delta x^\nu-\frac{\partial \mathcal L}{\partial (\partial_\mu\varphi)}\tilde{\delta}\varphi . $$ Here $\tilde{\delta}{\varphi}:=\varphi'(x')-\varphi(x)$ the difference in both $x$ and $\varphi$.

However, I have no idea how to connect these two charges, even though they should be the same. For the first kind of derivation, see here pp.17-18. For the second derivation, see here Field-theoretical derivation.

Thanks.

flippiefanus
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  • Possible duplicates: https://physics.stackexchange.com/q/301847/2451 , https://physics.stackexchange.com/q/369729/2451 and links therein. – Qmechanic Feb 19 '18 at 06:24

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