# Why is magnetostatics defined as $\frac{\partial \rho}{\partial t} = 0$?

I don't see why the idea of steady currents (i.e. magnetostatics) implies that charge density $$\rho(\vec{r},t)$$ has no explicit time dependence.

Is it just coming from magnetostatics being defined as $$\vec{\nabla} \cdot \vec{J}:= 0$$ (I don't see why this would be true either) and because of the continuity equation $$\implies \frac{\partial\rho}{\partial t} = 0$$.

Introduction to Electrodynamics, D.J. Griffiths section 5.2.1 - Steady Currents • I'm pretty sure the charge density cannot be static in the case of nonzero currents. Where did you find this? Mar 4, 2021 at 21:32
• @NDewolf I found this is D.J. Griffiths book, Introduction to Electrodynamics section 5.2.1 in the paragraph above equation 5.31 (I will attach a picture of it in my question). Mar 5, 2021 at 4:16
• @franz I just don't see how charge density having some explicit time dependence would imply the currents would not be static. If possible, could you please elaborate? (sorry if this is a trivial question) Mar 5, 2021 at 4:19

Magnetostatics is, in some sense, a toy concept taught to students in preparation for the formal magneto-quasi-static (MQS) approximation. The purpose of the MQS approximation is to decouple the electrical field from the magnetic field. This is done by setting $$\frac{\partial}{\partial t} \vec E \approx 0$$ so that Ampere's law becomes $$\nabla \times \vec H \approx \vec J$$ (see http://web.mit.edu/6.013_book/www/chapter3/3.2.html )
Since we want $$\frac{\partial}{\partial t} \vec E \approx 0$$ and also $$\epsilon_0 \nabla \cdot \vec E = \rho$$ then that implies that $$\frac{\partial}{\partial t} \rho \approx 0$$
By decoupling the fields it becomes much easier to solve. So this approximation is very useful to make. Allowing $$\frac{\partial}{\partial t}\rho \ne 0$$ would result in $$\frac{\partial}{\partial t} \vec E \ne 0$$ and thus the fields would be coupled again. So this assumption actually turns out to be more important that the steady current assumption $$\frac{\partial}{\partial t} \vec J \approx 0$$ assumption. In fact, in the MQS the latter assumption is not made and the currents are allowed to change over time, but the equations remain decoupled and simple to solve at each time point.
I would personally define a “steady current” as one that obeys $$\frac\partial{\partial t} \mathbf J = 0$$ everywhere.
The requirement that the charge density in “magnetostatics” not change with time, $$\frac\partial{\partial t} \rho = 0$$, allows the student to use the tools developed during electrostatics to figure out what the electric fields do.
Most elementary treatments (including Griffiths) begin with a chapter or two of electrostatics with $$\frac{\partial}{\partial t}\rho =0$$ and $$\mathbf J = 0$$. Next follows a chapter or two of magnetostatics with $$\mathbf J\neq0$$ but $$\mathbf J$$ and $$\rho$$ both held constant. Then the charging-or-discharging capacitor is introduced as an example where there are regions of $$\frac\partial{\partial t}\rho \neq 0$$ and therefore $$\frac\partial{\partial t}\mathbf E \neq 0$$, motivating Maxwell’s discovery of the need for the displacement current. This pedagogical strategy mirrors the historical timeline of the development of the theory.