# Expansion of $\mathcal{L}(\phi'(x'),\partial_\mu'\phi'(x'),x')$ when looking at Noether's current

I just asked this question concerning the application of Noether's theory. Think about this got me wondering about the following. In the usual derivation of the Noether current the assumption is made that:

$$\mathcal{L}(\phi'(x'),\partial_\mu'\phi'(x'),x')=\mathcal{L}(\phi(x),\partial_\mu\phi(x),x)+\delta x^\mu\partial_\mu\mathcal{L}(\phi(x),\partial_\mu\phi(x),x).\tag{1}$$ This is usually shown by considering the Lagrangian to be a function of $x$ only then, the statement that:

$$\mathcal{L}(x')=\mathcal{L}(x)+\delta x^\mu\partial_\mu\mathcal{L}(x)\tag{2}$$

does indeed hold true by trivial Taylor expansion. But as far as I can tell this derivation is making the assumption that: $$\phi'(x')=\phi(x').\tag{3}$$ I have seen (1) used in cases where this is not the case. Thus please can someone explain why (1) holds for a general mapping $\phi(x) \mapsto \phi'(x')$

• Formula (1) does not hold for a general variation. – Qmechanic Nov 19 '17 at 16:21
• @Qmechanic, Oh ok, now I am confused. Do you know any sources that derive the Noether's current in QFT for a general case? – Quantum spaghettification Nov 19 '17 at 16:52
• @Quantumspaghettification Actually, believe it or not, the wiki page is great. Look for the fiber bundle derivation. – BB681 Nov 19 '17 at 18:31
• @Qmechanic eq (1) is actually fairly general under Dirichlet boundary conditions. Sure, it's (or looks like) a taylor expansion, but you can get to this result by formal variational techniques without expanding in small variations. – BB681 Nov 19 '17 at 18:41
• @Qmechanic If you have time I would be interested in your response to BB681 comments here. – Quantum spaghettification Nov 20 '17 at 20:13

A general variation$^1$ $$\delta {\cal L}~=~\delta_0 {\cal L} + \delta x^{\mu}~d_{\mu}{\cal L} \tag{A}$$ of the Lagrangian density ${\cal L}$ is a sum of

1. a vertical$^2$ variation $$\delta_0 {\cal L} ~=~\frac{\partial {\cal L}}{\partial \phi^{\alpha}} ~\delta_0\phi^{\alpha} +\frac{\partial {\cal L}}{\partial \phi^{\alpha}_{,\mu}} ~d_{\mu}\delta_0\phi^{\alpha}\tag{B}$$ (which OP's eqs. (1) & (2) are missing), and

2. a transport term $$\delta x^{\mu}~d_{\mu}{\cal L}\tag{C}$$ from a horizontal$^2$ variation $\delta x^{\mu}$.

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$^1$ In this answer we will for simplicity only consider infinitesimal variations/transformations.

$^2$ For terminology, see e.g. my Phys.SE answer here.

• Great thanks for your answer. Does the variation $\delta \mathcal{L}$ include the variation in the Jacobian as mentioned by Prahar in a comment to one of my previous questions – Quantum spaghettification Nov 21 '17 at 8:19
• No, the Jacobian has not yet been taken into account. – Qmechanic Nov 21 '17 at 8:33