In Goldstein's Classical Mechanics (second edition), he derives a general expression for a Noether current in field theory. Given a general infinitesimal transformation:
$$ x^\mu \rightarrow x^\mu + \delta x^\mu$$ $$ \phi_A(x^\mu) \rightarrow \phi_A'(x^\mu+\delta x^\mu),$$
he defines the total and vertical variations of the field:
$$\tag1 \delta\phi_A \equiv \phi_A'(x^\mu+\delta x^\mu)-\phi_A(x^\mu)$$ $$\tag2\bar \delta \phi_A\equiv \phi_A'(x^\mu)-\phi_A(x^\mu).$$
His final expression for the Noether current is then, for a one-parameter transformation (equation 12-165 in his book):
$$\tag3 J^\mu=T^\mu_{\enspace\nu}X^\nu-\frac {\partial \mathcal L} {\partial(\partial_\mu\phi_A)}\Psi_A,$$ where:
$$T^\mu_{\enspace \nu}=\frac {\partial \mathcal L} {\partial (\partial_\mu \phi_A)}\partial_\nu \phi_A - \delta^\mu_\nu\mathcal L$$ $$\delta x^\mu = \epsilon X^\mu$$ $$\delta \phi_A = \epsilon\Psi_A,$$
My problem is the appearance of $\Psi_A$ in this expression. In the first example he gives (eq. 12-153), Goldstein says that in the special case of a translation of a single coordinate $x^\lambda$, we get that $$X^\nu=\delta^\nu_\lambda , \qquad\Psi_A=0,$$ so we get the expected conserved currents of the stress-energy tensor. But I don't see why $\Psi_A=0$, since $\Psi_A$ expresses the total variation of the field, and should be nonzero due to the variation in the argument of $\phi_A$. In this case we should have, according to $(1)$:
$$\delta\phi_A = \partial_\lambda \phi_A \epsilon \Rightarrow \Psi_A=\partial_\lambda \phi_A,$$
which obviously leads to the incorrect result when plugging into $(3)$. The same problem occurs with any transformation of only the coordinates. The only way I see for $\Psi_A$ to vanish in such cases is for it to be defined in terms of $\bar \delta \phi_A$ rather than $\delta \phi_A,$ which is certainly not the case in equation $(3)$.
