Let a Lagrangian density for a field theory of $N$ fields $\left\{\phi_i\right\}_{i=1}^N$ be given.
Assume that the Lagrangian density depends on the fields, their spacetime derivatives, and their second spacetime derivatives: $\mathcal{L}(\phi_i,\partial_\mu\phi_i,\partial_\nu\partial_\mu\phi_i)$.
Then a short derivation shows that the Euler-Lagrange equations are given by:
$$ \frac{\delta\mathcal{L}}{\delta\phi_{i}}-\partial_{\mu}\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\phi_{i}}+\partial_{\nu}\partial_{\mu}\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\partial_{\nu}\phi_{i}}=0 \,\,\, \forall i\in \{1,\dots,N\} $$
Using a similar derivation to the proof of the Noether theorem, I was able to show that the conserved Noether current is:
$$ j^\mu = \sum_{i}\left[\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\phi_{i}}\Delta\phi_{i}+\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\partial_{\nu}\phi_{i}}\partial_{\nu}\Delta\phi_{i}-\left(\partial_{\nu}\frac{\delta\mathcal{L}}{\delta\partial_{\mu}\partial_{\nu}\phi_{i}}\right)\Delta\phi_{i}\right] $$
My question is: is this correct?
It feels fishy to me because there is a term $\partial_{\nu}\Delta\phi_{i}$ and I would somehow expect all terms to be proportional to $\Delta\phi_{i}$ alone.
I'm doing this to find the conserved Noether current (see this related question and this one which unfortunately had no answers yet) of the BRST transformation.
Best Answer
Yes it is correct. I derived and used the same expression in http://vixra.org/abs/1008.0051 page 5 (with one extra term to account for space-time transformations that is not needed for internal symmetries). The dependence on the derivatives $\partial_{\nu}\Delta\phi_{i}$ is necessary and not a problem.