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?"

  • 1
    $\begingroup$ Does electrostatics have $\nabla\cdot J = -\partial_t\rho$? Surely in electrostatics both sides of this are zero by definition. $\endgroup$
    – jacob1729
    Commented Nov 24, 2021 at 11:53
  • $\begingroup$ @jacob1729 Yes, that makes sense... I read it from feynmanlectures.caltech.edu/II_13.html, which uses it when explaining magnetostatics with electric currents. But maybe I'm wrong in thinking it how I do in my question... I edited it $\endgroup$
    – Juan Perez
    Commented Nov 24, 2021 at 12:07
  • $\begingroup$ Your second equation cannot be right, the curl of a vector gives you a vector but the divergence of a vector is a scalar. $\endgroup$
    – Mauricio
    Commented Nov 24, 2021 at 12:26
  • $\begingroup$ Also the equation cannot be as simple as the continuity equation for the electric charge as the magnetic moment density is a vector not a scalar. $\endgroup$
    – Mauricio
    Commented Nov 24, 2021 at 12:57
  • $\begingroup$ @Mauricio Yes, I typed dot instead of cross, edited now. I took it from Wikipedia en.wikipedia.org/wiki/Magnetization#Magnetostatics, which states Curl(H) = 0, and since the definition of H is = 1/mu.B - M (en.wikipedia.org/wiki/…), then that equation follows, I think... $\endgroup$
    – Juan Perez
    Commented Nov 24, 2021 at 13:01

1 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.


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