1

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 $\phi(x) \mapsto \phi'(x')$

Qmechanic
  • 201,751
Quantum spaghettification
  • 7,585

1 Answers1

1

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 $\delta x^{\mu}$.

--

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

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

Qmechanic
  • 201,751