lagrangian formalism - What is the actual form of Noether current in field theory?

Let us consider $N$ independent scalar fields which satisfy the Euler-Lagrange equations of motion and are denoted by $\phi^{(i)}(x) \ ( i = 1,...,N)$, and are extended in a region $\Omega$ in a $D$-dimensional model spacetime $\mathcal{M}_D$. Now consider the classical Lagrangian density, $\mathcal{L}(\phi^{(i)}, \partial_\mu \phi^{(i)}, x^\mu)$. We apply the following infintesimal fixed-boundary transformation to $\mathcal{M}_D$. \begin{align*} x \to \widetilde{x}^\mu &\equiv x^\mu + \delta x^\mu (x), \tag{1} \\ \text{such that, }\ \delta x^\mu\Big{|}_{\partial\Omega}&=0, \tag{2} \\ \text{and the fields transform as: }\ \phi^{(i)}(x) &\to \widetilde{\phi}^{(i)}(\widetilde{x}) \equiv \phi^{(i)} (x) + \delta\phi^{(i)} (x). \tag{3} \\ \end{align*}

According to my calculations, up to first order in the variation, the Lagrangian density is given by: $$ \boxed{ \delta \mathcal{L} = \partial_\mu \Big( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\delta\phi^{(i)} - \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\partial_\nu \phi^{(i)} \delta x^\nu + \mathcal{L} \delta x^\mu \Big) - \mathcal{L} \partial_\mu (\delta x^\mu) }\tag{4} $$

Therefore, the conserved current is $$ \boxed{ J^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\delta\phi^{(i)} - \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\partial_\nu \phi^{(i)} \delta x^\nu + \mathcal{L} \delta x^\mu - F^\mu } \tag{5}$$ where $F^\mu$ is some arbitrary field that vanishes on $ \partial \Omega$.

However, most textbooks ignore the second and the third terms in the above expression. Compare, for example, with Peskin and Schroeder (p.18) which sets:

$$ J^{\mu} = \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\delta\phi^{(i)} - F^\mu. \tag{6} $$

For another example, Schweber (p. 208) ignores all terms but the first in the variation of the Lagrangian density, and writes:

$$ \delta \mathcal{L} = \partial_\mu \Big( \frac{\partial \mathcal{L}}{\partial (\partial_\mu \phi^{(i)} )}\delta\phi^{(i)} \Big).\tag{7} $$

So what is going on here? Am I missing something? We seem to have set the same assumptions, but get different results. Am I wrong, or are they?

EDIT: Condition (2) is unnecessary, as it was never used in the derivation of the current. Please ignore its presence in the above text.

Answer

1. 
   

   Eq. (5) is (up to factors of the infinitesimal parameter $\varepsilon$) the standard expression for the full Noether current. Here:

   
   
   
   
   • $\delta x^{\mu}$ is the so-called horizontal component of the infinitesimal variation;
   
   
   • $\delta \phi -\frac{\partial \phi}{\partial x^{\mu}} \delta x^{\mu} $ is the so-called vertical component of the infinitesimal variation;
   
   
   • $F^{\mu}$ is an improvement term in case of quasisymmetry.
   
   
   
   


2. 
   

   The main point is that Schweber (7), Peskin & Schroeder (6) are only considering situations with purely vertical transformations, i.e. situations where $\delta x^{\mu}=0$.

   
   
   


3. 
   

   Let us mention that the last term in eq. (4) gets cancelled by the Jacobian contributions from the integration measure. Hence it is not present in eq. (5).

   
   


4. 
   

   Finally, it seems relevant to mention that OP's boundary condition (2) is often not fulfilled in important applications, such as the canonical stress-energy-momentum (SEM) tensor, which is the Noether current for spacetime translations. See e.g. this Phys.SE post. Therefore the boundary condition (2) should be relaxed appropriately. Similarly, the improvement term $F^{\mu}$ is not some arbitrary field that vanishes on the boundary, as OP claims (v3) under eq. (5). Instead the improvement term $F^{\mu}$ is dictated by the quasisymmetry, which fixes $F^{\mu}$ up to a divergence-free term.