1
$\begingroup$

For a steady-state current, the relation must hold $\text{div} \mathbf{J}= 0 .$

This implies $$\frac{\partial}{\partial t} \; \rho_\text{charge-density inside the volume}= 0. $$ This further implies during infinitesimal time-interval $\partial t$, there has been no net consumption or depletion of charge inside the volume.

Doesn't that mean $\bf{J}$ has the same value behind & in-front of the volume?

For if it weren't there would be a net change in charge-density inside the volume in question.

But Purcell in his book Electricity & Magnetism wrote that

A general steady current distribution is described by a volume current density $\mathbf J(\mathsf{x,y,z})$ which varies from place to place but is constant in time [...]

But how can it be so? If $\bf J$ is not constant over the volume, how could there be no net change in charge density?

Even he in a preceding chapter where he introduced $\mathbf J= \sigma \mathbf E\;,$ told that $\bf J$ must remain constant on two interfaces of the current-carrying rod.

So, I'm a bit confused.

Probably I think Purcell was talking about any arbitrarily-shaped volume so that even though $\bf J$ is not constant, there wouldn't be any net-charge increment or decrement inside the volume. Is it so?

$\endgroup$
4
$\begingroup$

No, a steady-state current is not constant throughout space, it is only constant in time. You are correct that $\vec\nabla\cdot \vec J = 0$ implies $\partial_t\rho = 0$ by the continuity equation, but that the charge inside arbitrary volumes doesn't change doesn't mean something about the current "behind & in-front" of the volume.

It means the same amount of charge flows into every volume as flows out of it, and this is precisely the natural language equivalent of $\vec\nabla\cdot\vec J = 0$, since $\vec J$ is the flow of charge, and the divergence measures the net change of charge within a volume by the continuity equation/Stokes' theorem. But that the net flow out of every volume is zero doesn't mean the vector field is constant everywhere, since every vector field that is a curl of another is divergence-less, hence an admissible steady-state current, but surely not every curl is constant throughout space.

$\endgroup$
2
  • $\begingroup$ So, my last assumption is correct, isn't it? $\endgroup$
    – user36790
    Nov 24 '15 at 14:02
  • 1
    $\begingroup$ @user36790: Yes, I think so. $\endgroup$
    – ACuriousMind
    Nov 24 '15 at 14:04

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy