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

$$\mathcal{L}(\phi'(x'),\partial_\mu'\phi'(x'),x')=\mathcal{L}(\phi(x),\partial_\mu\phi(x),x)+\delta x^\mu\partial_\mu\mathcal{L}(\phi(x),\partial_\mu\phi(x),x).\tag{1}$$ 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