1
$\begingroup$

One of the textbooks I’m reading states that an electric current is said to be steady if $\nabla \cdot \mathbf J = 0$, which is equivalent to $\frac{\partial \rho}{\partial t} = 0$ by the continuity equation, and later claims that a steady current need not be time-independent. This makes sense to me, since $\mathbf J = \rho \mathbf v$ and so $$ \frac{\partial \mathbf J}{\partial t} = \frac{\partial \rho}{\partial t} \mathbf v + \rho \frac{\partial \mathbf v}{\partial t}$$ and I don’t see any reason why $\frac{\partial \mathbf v}{\partial t}$ should equal $\mathbf 0$ in general. For this same reason, I wouldn’t say that a time-independent current is necessarily steady. However, in Purcell’s Electricity and Magnetism, steady current is defined by the condition $\frac{\partial \mathbf J}{\partial t} = \mathbf 0$, and the following explanation for why this implies that $\nabla \cdot \mathbf J = 0$ is given:

We speak of a steady or stationary current system when the current density vector J remains constant in time everywhere. Steady currents have to obey the law of charge conservation. Consider some region of space completely enclosed by the balloonlike surface S. The surface integral of J over all of S gives the rate at which charge is leaving the volume enclosed. Now if charge forever pours out of, or into, a fixed volume, the charge density inside must grow infinite, unless some compensating charge is continually being created there. But charge creation is just what never happens. Therefore, for a truly time-independent current distribution, the surface integral of J over any closed surface must be zero. This is completely equivalent to the statement that, at every point in space, div J = 0.

It seems to me that this argument not only assumes that $\frac{\partial \mathbf J}{\partial t} = \mathbf 0$, but most importantly that $\frac{\partial \rho}{\partial t} = 0$. Am I missing something? In other words, is the implication $$\frac{\partial \mathbf J}{\partial t} = \mathbf 0 \implies \nabla \cdot \mathbf J = 0$$ actually correct?

$\endgroup$
0
$\begingroup$

Purcell's argument assumes that the charge density must remain bounded. This is reasonable for physical systems (or mathematically when you take into account the full Maxwell equations and the equation of motion of matter).

Without this, the implication $\frac{\partial \mathbf{J}}{\partial t}=\mathbf{0} \Longrightarrow \nabla \cdot \mathbf{J}=0$ is false.

Indeed, take $\mathbf X$ any (time-independent) vector field, and set $\mathbf v = \frac{1}{\rho} \mathbf X$. (Again, this makes no sense physically, but is mathematically consistent.) Then $\mathbf J = \mathbf X$ so $\partial_t \mathbf J = 0$. However $\partial_t \rho = - \mathbf{\nabla \cdot X}$ is not necessarily $0$

$\endgroup$

Your Answer

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

Not the answer you're looking for? Browse other questions tagged or ask your own question.