We know that the magnetic dipole moment $\mathbf{m}$ in some volume $V_0$ is: $$ \mathbf{m} := \frac{1}{2} \int_{V_0} \mathbf{r} \times \mathbf{j}\ \mathrm{d}V = \int_{V_0} \mathbf{M}\ \mathrm{d}V, \tag{1} \label{eq:1} $$ where $\mathbf{M}$ is the magnetization vector (the magnetic dipole moment per unit volume). This leads to volume and surface current densities: $$ \begin{cases} \mathbf{j}_m = \operatorname{curl} \mathbf{M} \\ \mathbf{j}_{ms} = \mathbf{M} \times \mathbf{\hat{n}} \end{cases} \tag{2} \label{eq:2} $$
Like the electric dipole, we can define a magnetization density using hypothetical magnetic charges: $$ \mathbf{p}_m := \int_{V_0} \rho_m \mathbf{r}\ \mathrm{d}V = \int_{V_0} \mathbf{P}_m\ \mathrm{d}V, \tag{3} \label{eq:3} $$ where $\rho_m$ is the magnetic charge density. This leads to volume and surface magnetic charge densities: $$ \begin{cases} \rho_m = -\operatorname{div} \mathbf{P}_m \\ \rho_{ms} = \mathbf{P}_m \cdot \mathbf{\hat{n}} \end{cases} \tag{4} \label{eq:4} $$
Question:
Does $\mathbf{m}$ and $\mathbf{p}_m$ exactly the same? (I think up to a constant: $\mathbf{p}_m = \mu_0 \mathbf{M}$, $\mathbf{p}_m$ solves the problem in this post)
What's the relation between $\mathbf{j}_m, \mathbf{j}_{ms}$ and $\rho_m, \rho_{ms}$? How to derive \eqref{eq:2} from \eqref{eq:4}?
