1A caused by half the charge density at twice the velocity

[particlezoo]

In the situation consisting of a steady current of 1A in an arbitrary closed path, what would the consequences be for the electric field if the drift velocity was non-uniform along the path due to non-uniform carrier density?

This would be a case of a "uniform" 1 amp, but where the charge carrier density varies along the conductor which could be due to wire of varying thickness or use of different conductor materials along the path.

Now I understand that, ignoring resistive heating, all the electric fields due to these "idealized" steady currents would be radial to their sources, and so we would not expect the emission of electromagnetic waves.

Now, just because the electric fields are radial to their respective sources does not mean that the curl of those fields are all zero, for after all an isolated charge moving at constant velocity will cause changes in magnetic fields that we know generate the curl of electric fields per the Maxwell-Faraday equation.

If I understand correctly, if I compare two current elements with the only difference being that the second current element has half the charge carrier density at twice the drift velocity, then the difference made on the electric field due to length contraction is not the same and is approximately double compared to that of the first current element. If that is true, then shouldn't we expect that the curl of the electric field would not vanish in the region surrounding a point where such two current elements meet each other?
Kevin M.

 

[Dale]

In such cases there will be a surface charge density at the interface. If it is a steady state field then, by definition all of the time varying terms will be zero and therefore the curl of the E field will also be 0.

 

[Delta2]

If the current density J(r,t)=J(r) does not depend on time then the vector potential A(r,t) will be A(r) that is it will not depend on time because it is

$\vec A(r,t)=\int \frac{\vec J(r',t-\frac{|r-r'|}{c})}{|r-r'|}d^3r'=\int \frac{\vec J(r')}{|r-r'|}d^3r'=\vec A(r)$

And since A(r) doesn't depend on time then
$B=\nabla \times A$ doesn't depend on time either, hence

$Curl(E)=\nabla \times E=-\frac{\partial B}{\partial t}=0$.

 

[particlezoo]

After some thought, it now appears to me that in the case of static charge/current densities that the curl due to the length contraction due to non-uniform drift velocity of the charge carriers is canceled exactly by the curl of electric field due to the convective acceleration of said charge.