I am stuck on some details of Noether's theorem and was hoping for some further enlightenment. I want to include some details for those helping me to understand my thought process and perhaps what is tripping me up. I begin with the Lagrangian $L(\phi, \partial_{\mu}\phi)$ and vary it.
$$\delta L(\phi, \partial_{\mu}\phi) = \frac{\partial L}{\partial \phi}\delta \phi \hspace{2pt} + \hspace{2pt} \frac{\partial L}{\partial(\partial_{\mu}\phi)}\delta\partial_{\mu}\phi\tag{1}$$
Now I make the identification $\delta \partial_{\mu}\phi = \partial_{\mu}\delta \phi \hspace{1pt}$ and also use the Euler-Langrage equation $\frac{\partial L}{\partial \phi} = \partial_{\mu}\frac{\partial L}{\partial(\partial_{\mu}\phi)}$ to change first term of (1) to obtain
$\delta L(\phi, \partial_{\mu}\phi) = \partial_{\mu}\frac{\partial L}{\partial(\partial_{\mu}\phi)}\delta \phi \hspace{2pt} + \hspace{2pt} \frac{\partial L}{\partial(\partial_{\mu}\phi)}\partial_{\mu}\delta \phi$ $\hspace{20pt}$ (2)
We can then recognize this as a total derivative.
$\delta L(\phi, \partial_{\mu}\phi) = \partial_{\mu} \left(\frac{\partial L}{\partial(\partial_{\mu}\phi)}\delta \phi \hspace{2pt} \right)$ $\hspace{20pt}$ (3)
where we suppose that $\delta \phi$ is an infinitesimal field variation of a continuous symmetry. Since $\delta S = 0$, then $L$ could only change by a total derivative
$\delta L = \delta \alpha \partial_{\mu}J^{\mu}_{0}$ $\hspace{20pt}$ (4)
where $J^{\mu}_{0}$ is some function and $\alpha$ is the variation parameter. Equating (3) and (4) yields the conserved current $J^{\mu}$ where
$J^{\mu} = \frac{\partial L}{\partial(\partial_{\mu}\phi)}\frac{\delta \phi}{\delta \alpha} - J^{\mu}_{0}$ $\hspace{20pt}$ (5)
Question: This is where I have the first point of confusion. I want to derive the conserved current for a particular symmetry. Now, the way I understand this is that we:
Calculate the variance of the field under a particular symmetry
For our case, let's use $\phi \rightarrow e^{i\alpha}\phi$. The variance of this field is then $\delta \phi = \delta \alpha\hspace{1pt} i\phi$.
Given that I already derived the conserved current $J^{\mu}$ above in (5), I assume it is safe to interpret this as a sum over all types of fields present in our problem. Why can this be interpreted as a sum over all types of fields?
In the case of the symmetry that we are working with ($\phi \rightarrow e^{i\alpha}\phi$), apparently the $J^{\mu}_{0}$ term vanishes
$J^{\mu}_{0} = 0$
Why does $J^{\mu}_{0} = 0$ with a U(1) symmetry, but $J^{\mu}_{0} = \delta^{\mu}_{\nu}L$ with a translation symmetry $((\phi(x^{\mu}) \rightarrow \phi(x^{\mu}+a^{\mu}))$?
In essence, my question is asking if (5) is supposed to be interpreted as a sum over all types of fields in the given Lagrangian and why it should be interpreted that way. I am also asking how to determine the $J^{\mu}_{0}$ term in (5) will be for different symmetries. In these examples, $J^{\mu}_{0}$ is zero for the U(1) symmetry and non-zero for the translation symmetry. In the translation symmetry, it leads to the energy-momentum tensor. I want to learn how to calculate the $J^{\mu}_{0}$ term for these situations and not rely on the text just giving me the condition.
