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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

 D.J. Griffiths section 5.2.1 paragraph above equation 5.31

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  • $\begingroup$ I'm pretty sure the charge density cannot be static in the case of nonzero currents. Where did you find this? $\endgroup$ – NDewolf Mar 4 at 21:32
  • $\begingroup$ @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). $\endgroup$ – a_point_particle Mar 5 at 4:16
  • $\begingroup$ @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) $\endgroup$ – a_point_particle Mar 5 at 4:19
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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.

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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.

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