# Which transformations *aren't* symmetries of a Lagrangian?

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?

• 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? – guillefix Oct 12 '14 at 12:10