Reducing relativistic effective Hamiltonian to non-relativistic limit

[Malamala]

Hello! My question is related to going from Eq. 32 to Eq. 33 in this paper (however I have seen this in other papers, too). In summary, starting with:

$$H \propto \bar{e}\gamma_\mu\gamma_5 e \bar{q}\gamma^\mu q$$
where we have the gamma matrices, e is the electron field and q is the quark/nucleon field, if we assume the nucleus is non-relativistic we end up with:

$$H \propto \gamma_5\rho(r)$$
where $\rho(r)$ is the nuclear density and this is a Hamiltonian acting in the electronic sector.

My confusions are:
1. Where did the electronic $\gamma_\mu$ go?
2. (More important) I am not sure I understand the $\rho(r)$ term. Is r referring to the nuclear or electronic position? If it is nuclear, I am confused as, given we integrated out the nuclear part, the nuclear term should be a constant, no? If it refers to electron, I am not sure how to think of the nuclear density as a function of electron coordinates and how does it even end up being a function of the electronic coordiates?

I would really appreciate some help with this. Thank you!

 

[vanhees71]

$\rho=j^{0}=\bar{q}\gamma^0 q=q^{\dagger} q$. In the non-relativistic limit, the dominating term is indeed the Coulomb-like interaction.