lagrangian formalism - Noether's Theorem in Classical Field theory Confusion

Consider $N$ independent scalar fields $φ_i (x)$ in 4D space. Also consider a lagrangian density $$\mathcal{L} = \mathcal{L}(φ_i, \partial_μφ_i).$$

Suppose we perform the following infinitesimal transformations: $$x'^μ = x^μ + ε^α Χ^μ_α \tag{1} $$ $$φ_i '(x')=φ_i(x)+ ε^α Ψ_{iα} .\tag{2} $$

Let us denote $δφ = φ'(x')-φ(x)$ and $\barδφ = φ'(x)-φ(x)$. It's easy to see that $$ \barδφ_i= δφ_i - δx^ν \partial_ν φ_i. \tag{3}$$ Now doing some calculations which can be found in many books (I'm following D. Gross' lecture notes, also Peskin's and weinberg introductory QFT books), we see that under the above transformations the action changes as: $$ δS=\int d^4x \frac{δS}{δφ_i} \barδφ_i + \partial_μ[\mathcal{L}δx^μ +\frac{\partial \mathcal{L}}{\partial(\partial_μ φ)} \barδφ_i] \tag{4}$$ Where $\frac{δS}{δφ_i} =0 $ are the equations of motion (EOM).

1. My first point of confusion is the following: What do we want from this transformation so as to get Noether's Theorem: do we want the Lagrangian to change up to a 4-divergence; the Lagrangian to be invariant and with the equations of motions satisfied or not; or the action to be invariant i.e. to remain unchanged no matter if the equations of motions are satisfied or not?

Supposing the EOM satisfied then from $(4)$ we get only a 4-divergence in the integral of the action.

2. Second point:If we allow the Lagrangian to change by a 4-divergence, as Peskin says in chapter 2.2, then every infinitesimal transformation would be a symmetry and give a conserved quantity, right? (Peskin only deals with field transformations I think, but still what he says should be a special case and thus generalize to what we're doing here) On the other hand, upon getting to eq $(4)$ Gross says that if S is to be invariant under those transformations (for arbitrary volumes, he notes- is this important?) then that 4-divergence should be 0, thus we have our conserved current. Indeed I know that the general Noether current is exactly what we see from (4): $$J^μ_α = \mathcal{L}X^μ_α +\frac{\partial \mathcal{L}}{\partial(\partial_μ φ)}Ψ_{ia}-\frac{\partial \mathcal{L}}{\partial(\partial_μ φ)} \partial_ν φ_i X^ν_α. $$ So when I have ANY transformation I can just close my eyes, get this "formula" and find a conserved current?

Moreover, doesn't "invariant action" mean that it doesn't matter if equations of motions are satisfied or not, action still doesn't change? So Gross means something else?

3. In this question Energy momentum tensor from Noether's theorem the accepted answer says "For actions that only depend on first derivatives of the fields, the variation of the action will inevitably have the form $$ S = \int (\partial_\mu a) j^\mu d^d x $$ where $j^\mu$ is some particular function of the fields or other degrees of freedom (and their derivatives)". Does $(4)$ assume this form? Should it? Note that I haven't specified if $ε^α$ are constant or not, in getting $(4)$ we've been as general as possible on this matter (right?).

Answer

1. 
   

   The assumption in Noether's (first) theorem is that the action $S$ should be invariant off-shell$^1$ up to a boundary term under the transformations (1) & (2). It is not enough to know this only on-shell.

   
   
   


2. 
   

   No. Knowing only that there is an on-shell quasi-symmetry (which is a tautology), there is not enough information to conclude an on-shell continuum equation/conservation law. There is no free lunch.

   
   

   (Concerning a subquestion in a parenthesis: Knowing an off-shell quasi-symmetry for arbitrary spacetime volumes means that one can use localization, cf. e.g. my Phys.SE answer here.)

   
   


3. 
   

   Yes, eq. (4) also holds for non-constant (=$x$-dependent) $\epsilon\equiv a$.

   
   

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$^1$The words on-shell and off-shell refer to whether the Euler-Lagrange (EL) equations (=EOM) are satisfied or not.