Obtaining the conserved current of the Lagrangian making the parameter depending on $x$

Question

To calculate the conserved current due to an internal symmetry of the system (expressed by the Lagrangian density) we can proceed as follows: if it is invariant under $\delta \phi = \alpha \phi$, where $\alpha$ is constant we make $\alpha$ depending on the space-time $x$, and consequently the variation on the Lagrangian should be of the form $\partial_\mu \alpha * f^\mu(\phi)$ the argument for this is that when $\partial_\mu \alpha $ is zero ($\alpha$ is constant) we should recover that $\delta L=0$ my question is why it should have this form? I don't see why we couldn't have something like $\log (\partial_\mu \alpha f^\mu(\phi) + 1)$ or something like that.

I will finish the argument for those who are curious, The change in the action should be zero for any change of the fields, and after integrating by parts we obtain $\partial_\mu f^\mu(\phi) =0$ and therefore $f^\mu$ is the sought current.

Answer 1

When physicists write a field variation

$$ \delta\phi~=~ \alpha f(\phi) $$

it is usually implicitly implied that $\alpha$ is infinitesimally small. This in particular means that higher powers of $\alpha$ can be ignored, so that e.g. OP's suggestion

$$\ln (1+\partial_\mu \alpha f^\mu(\phi))~\approx~\partial_{\mu} \alpha f^{\mu}(\phi), $$

etc. The rule is that the variation $\delta L$ is always linear in $\alpha$.

Finally let us mention that if the Lagrangian $L$ is allowed to depend on higher derivatives, say $\partial_{\mu}\partial_{\nu}\phi$, then the variation $\delta L$ can also contain higher derivatives, say $\partial_{\mu}\partial_{\nu}\alpha$, of $\alpha$.