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.
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.
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