I've seen 2 (or 3) definitions for stationary current.

Definition 1:$\quad\frac{\partial}{\partial t}\rho = 0 $ or $\nabla\cdot\mathbf{J} = 0 $

This means, as expected, that the current throught an arbitrary closed surface is $I = \oint \mathbf{J}\,d\mathbf{S} = 0$. If we consider an infinity cylinder (a wire) we conclude that the current throught any perpendicular cross-section is the same. If $\hat{x}$ is the direction along the cylinder, $\frac{\partial}{\partial x} I = 0$. This seems, at least, strange to me. Since usually "stationary" means constant wrt time. Wikipedia (in Spanish) and Mildford-Reitz's Fundamentals of electromagnetic theory support this definition.

Definition 2: $\quad\frac{\partial}{\partial t}\mathbf{J} = \vec{0}$

which would imply $\frac{\partial}{\partial t} I = 0$ for any surface. However, if we consider a closed surface and $I\neq 0$, there will be an acumulation ($I<0$) or a ¿dissipation? ($I>0$) of charges. This is a "problem" unavoidable from this definition, isn't it?

There is a third definition. In paragraph 5.2.1 of Introduction to Electrodynamics by David Griffiths, the author states

Stationary charges ⇒ constant electric fields: electrostatics.
Steady currents ⇒ constant magnetic fields: magnetostatics.
By steady current I mean a continuous flow that has been going on forever, without change and without charge piling up anywhere. (Some people call them “stationary currents”; to my ear, that’s a contradiction in terms.) Formally, electro/magnetostatics is the régime
∂ρ/∂t = 0, ∂J/∂t = 0, (5.32)
at all places and all times.

From there it seems that the autor defines steady current as both definitions together. "without charge piling up anywhere" (definition 1) and "∂J/∂t = 0" (definition 2).

I know that definition 2 is a sufficient condition for magnetostatics, definition 1 for electrostatics and both are needed for electromagnetostatics. These seems to me to be completely different conditions, not related in any ways, with very different implications. Taking both definitions together could be a solution, but I haven't seen it anywhere. I'm sure I'm missing something here.


1 Answer 1


I would normally assume steady currents mean both $\frac{\partial \vec J}{\partial t} = 0$ and $\frac{\partial \rho}{\partial t}=0$, unless it is clear from the context that one of these (probably the second) isn't valid.

However I don't know of a common situation where a current that is static ($\frac{\partial \vec J}{\partial t} = 0$) for an extended period of time is not also divergence-free: as Griffiths also indicates this would cause charge to pile up, which would cause a time-varying electric field. This would act to prevent charge from piling up.

  • $\begingroup$ Thanks for answering! What does "This" on "This would act to prevent charge from piling up." refers to? $\frac{\partial}{\partial t}\rho = 0$? $\endgroup$
    – Gilgamesh
    Commented Jun 23, 2020 at 20:49
  • $\begingroup$ It means "the time-varying electric field". If positive charge is accumulating at a certain point, the increasing E-field pointing outward around this region would act to slow down this accumulation. $\endgroup$
    – Puk
    Commented Jun 23, 2020 at 20:51
  • $\begingroup$ Oh, your're saying that, informally, $\frac{\partial}{\partial t}\mathbf{J} = 0$ and $t_0\rightarrow -\infty$ implies $\frac{\partial}{\partial t}\rho=0$. Thanks. $\endgroup$
    – Gilgamesh
    Commented Jun 23, 2020 at 21:02

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