As far as I understand, Noether's theorem for fields works, as explained in David Tong's QFT lecture notes (page 14) for example, by saying that a transformation $\phi(x) \mapsto \phi(x) + \delta \phi (x)$ is called a symmetry if it produces a change in the Lagrangian density which can be expressed as a four divergence, $$\delta \mathcal{L} = \partial_{\mu} F^{\mu}\tag{1.35} $$ for some 4-vector field $F^{\mu}$.

We thengo onto show that the change in this Lagrangian density may also be expressed for an arbitrary transformation as

$$\delta \mathcal{L} = \partial_{\mu}\bigg(\frac{\partial \mathcal{L}}{\partial(\partial_{\mu} \phi)}\delta \phi\bigg)\tag{1.37}.$$

Which is a 4-divergence. So how could we say any transformation is not a symmetry in the sense above?


The point is that eq. (1.35) should hold off-shell to have a symmetry, while eq. (1.37) may only hold on-shell.

[The term on-shell (in this context) means that the Euler-Lagrange equations are satisfied. See also this Phys.SE post.]

In other words: On-shell, the action will only change with at most a boundary term for any infinitesimal variation, whether or not it is a symmetry.

Phrased differently: By a symmetry is meant an off-shell symmetry. An on-shell symmetry is a vacuous notion.

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    $\begingroup$ Oh yeah, of course! And eq. (1.35) should be off-shell so that the action is still a minimum (as the nearby trajectories of the new trajectory will have their action changed by the same amount each) $\endgroup$ – guillefix Oct 11 '14 at 23:16
  • $\begingroup$ On another note, even if $\delta \mathcal{L}$ doesn't have an off-shell 4-divergence form, can it , for example, happen to still equal zero in the on-shell trajectory and thus give a conserved 4-current density, even though our transformation wasn't a symmetry? $\endgroup$ – guillefix Oct 12 '14 at 12:10

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