In classical mechanics, we know $L(q,\dot{q},t)$ and $L(q,\dot{q},t)+\frac{d}{dt}\Lambda(q,t)$ give the same Euler-Lagrange equation $\frac{d}{dt}\frac{\partial L(q,\dot{q},t)}{\partial \dot{q}_i}=\frac{\partial L(q,\dot{q},t)}{\partial q_i}$.
What I want to know is the precise corresponding statement in (classical) field theory. Let us work in $d$ spatial dimensions and let us set $x^0=ct$. Usually one says the Lagrangian density $\mathcal{L}(\phi,\partial\phi,x)$ and $\mathcal{L}(\phi,\partial\phi,x)+\partial_\mu \Lambda^\mu(\phi,x)$ give the same Euler-Lagrange equation $ \partial_\mu\frac{\partial \mathcal{L}(\phi,\partial\phi,x)}{\partial \partial_\mu \phi_i}=\frac{\partial \mathcal{L}(\phi,\partial\phi,x)}{\partial \phi_i}$. This is fine. (Note that $\partial_\mu\Lambda^\mu$ represents $\frac{\partial\Lambda^\mu}{\partial x_\mu}+\frac{\partial\Lambda^\mu}{\partial \phi_i}\partial_\mu\phi_i+\frac{\partial\Lambda^\mu}{\partial \partial_\nu\phi_i}\partial_\mu\partial_\nu\phi_i$.)
But I also want to allow $\partial_\mu \phi_i$ dependence for $\Lambda^\mu$. For example, this situation arises when we say the term $\epsilon^{\mu\nu\rho\sigma} F_{\mu\nu} F_{\rho\sigma}=\partial_\mu \Lambda^\mu$ with $\Lambda^\mu=2\epsilon^{\mu\nu\rho\sigma}A_\nu F_{\rho\sigma}$ does not change the Maxwell equation in U(1) gauge theory.
I think the condition on $\Lambda^\mu(\phi,\partial\phi,x)$ is $$ \frac{\partial \Lambda^\mu}{\partial \partial_\nu\phi_i}+\frac{\partial \Lambda^\nu}{\partial \partial_\mu\phi_i}=0\quad...(*) $$ for all $\mu,\nu=0,1,\cdots,d$. This condition requires that the second term in $$ \delta\mathcal{L}=\partial_\mu\Lambda^\mu(\phi,\partial\phi,x)=\frac{\partial \Lambda^\mu}{\partial \phi_i}\partial_\mu\phi_i+\frac{\partial \Lambda^\mu}{\partial \partial_\nu\phi_i}\partial_\mu\partial_\nu\phi_i+\frac{\partial \Lambda^\mu}{\partial x^\mu}. $$ vanishes so that the Lagrangian does not contain derivatives higher than $\partial_\mu\phi_i$ (summation over repeated $i$ is assumed).
My question is: (1) is the above condition $(*)$ correct? I checked Weinberg and Peskin-Schroeder, but couldn't really find the same expression.
An answer in this post clearly explained that even if we do not put any condition on $\Lambda^\mu$, then the equivalent of the Euler-Lagrange equation $\frac{\delta S}{\delta\phi_i}=0$ stays unchanged at least in the "bulk". However, as far as I understand this modification produces higher derivatives in the EOM.
My question is different. I want to keep the same form of Euler-Lagrange equation. I'm checking whether my condition stated above strictly keeps the EOM invariant.
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The Euler-Lagrange equation is unchanged under the condition $(*)$ because one can show $$ \partial_\mu\frac{\partial \delta\mathcal{L}}{\partial \partial_\mu \phi_i}=\frac{\partial \delta\mathcal{L}}{\partial\phi_i}\quad...(**)$$ The calculation was lengthy but after all $(**)$ was true.
