Is magnetic moment a conserved quantity in Magnetostatics?

Question

I am interested in the equations of Magnetostatics, but without the knowledge that electric currents create magnetic fields. In other words, expressing the equations only in terms of the magnetic moment of permanent magnets, no "Steady state" currents, or any other kind of current.

These are what I found. If something is wrong, please let me know

$$\nabla \cdot \boldsymbol{B} = 0$$ $$\nabla \times\boldsymbol{B} = \nabla\times\boldsymbol{M} \mu_0$$ $$\boldsymbol{F_m} = \nabla (\boldsymbol{m}\cdot\boldsymbol{B})$$

Where $\boldsymbol{B}$ is the magnetic field, $\boldsymbol{M}$ is the magnetic moment density, $\boldsymbol{F_m}$ is the magnetostatic force on a magnetic moment $\boldsymbol{m}$, and $\mu_0$ is the vacuum permeability.

Now, since Electrostatics has analogues to those three equations, plus an equation of conservation of charge*, $\nabla \cdot \boldsymbol{J} = -\frac{d \rho}{dt}$, I'm wandering: Is there an analogous law of conservation in Magnetostatics? Is total magnetic moment conserved in Magnetostatics? How about in Electrostatics, or in general?

*From https://www.feynmanlectures.caltech.edu/II_13.html. But maybe I'm wrong in claiming it applies in electrostatics, like a commenter says. If so, I limit the question to "Is magnetic moment conserved in general?"

Answer 1

Let us write Maxwell's equations $$\nabla \cdot \mathbf E =\rho/\epsilon_0 \tag1$$ $$\nabla \cdot \mathbf B =0 \tag2$$ $$\nabla \times \mathbf E =-\frac{\partial}{\partial t} \mathbf B\tag3$$ $$\nabla \times \mathbf B =\mu_0 \mathbf J+ \epsilon_0\mu_0\frac{\partial}{\partial t} \mathbf E \tag4$$

Take the divergence of (4) and use (1) to get the continuity equation for the charge $$\nabla \cdot \mathbf J = - \frac{\partial}{\partial t}\rho \tag5$$ The continuity equation has nothing to do with statics, it holds always.

In the case of electrostatics as $\partial E/\partial t =0$, the continuity equation becomes $\partial \rho/\partial t =0$. Which in turns means $\nabla \cdot \mathbf J=0$.

Now let us replace $\mathbf B=\mu_0(\mathbf H+\mathbf M)$, we get $$\nabla \cdot \mathbf E =\rho/\epsilon_0 \tag1$$ $$\nabla \cdot \mathbf H =-\nabla \cdot \mathbf M \tag6$$ $$\nabla \times \mathbf E =-\mu_0\frac{\partial}{\partial t} \mathbf M-\mu_0\frac{\partial}{\partial t} \mathbf H \tag7$$ $$\nabla \times \mathbf H = \mathbf J-\nabla \times \mathbf M+ \epsilon_0\frac{\partial}{\partial t} \mathbf E \tag8$$ these new magnetic equations look more symmetric between $E$ and $H$, specially if we define $\rho_m=\nabla \cdot \mathbf M$ and $\mathbf J_m=\nabla \times \mathbf M$. We get $$\nabla \cdot \mathbf E =\rho/\epsilon_0 \tag1$$ $$\nabla \cdot \mathbf H =-\rho_m \tag9$$ $$\nabla \times \mathbf E =-\mu_0\frac{\partial}{\partial t} \mathbf M-\mu_0\frac{\partial}{\partial t} \mathbf H \tag7$$ $$\nabla \times \mathbf H = \mathbf J-\mathbf J_m+ \epsilon_0\frac{\partial}{\partial t} \mathbf E \tag{10}$$

Note that by construction $\nabla \cdot \mathbf J_m = \nabla \cdot (\nabla \times \mathbf M )=0$ (divergence of a curl is always 0). Which looks a lot like the electrostatic conditions, the only problem is that $\rho_m $ is not related to $\mathbf J_m$ so you cannot say anything about it. In general $\partial \rho_m / \partial t \neq 0$ even in magnetostatics.

Edit: earlier I made an argument that if you replace $\mathbf K=\mathbf \partial M/\partial t$ in equation (7), you can do a similar trick as we can do with (4), and get $\nabla \cdot \mathbf K = -\partial \rho_m /\partial t$, but if you unpack it, it is just a trivial equation.