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I am reading about retarded potentials in Griffiths' Introduction to Electrodynamics, and came across the particular case of wires. Griffiths takes wires to be electrically neutral, therefore the retarded scalar potential is zero, which has an integrand proportional to charge density, which he has taken to be zero.

The retarded vector potential depends on the current density in the wire, which is non-zero. However, I recall that the current density $\textbf{J} = \rho \textbf{v}$.

Hence, my question is, how can we have a non-zero current density, but a zero charge density? Are the $\rho$'s different? The current density being constant in say the z direction, and zero charge density does not violate conservation of charge, but how do we reconcile this fact with the relationship between $\rho$ and $\textbf{J}$?

Is the charge density, found for instance in the Maxwell equation $\nabla \cdot \textbf{E} = \frac{\rho}{\epsilon_0}$ an 'overall' charge density which takes every charge into account, while the $\rho$ found in the expression for $\textbf{J}$ a 'mobile' charge density?

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    $\begingroup$ If you have a neutral wire and get a current to start so that the negative charges are moving, the wire is still neutral even though the negative charges are moving. The motion of the negative charges doesn't suddenly add or remove charges from the wire. $\endgroup$ – Aaron Stevens Sep 16 '18 at 13:23
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If the positive and negative charges are moving at different velocities, then their contributions to the current density will be different, even if their charge densities are equal and opposite.

For example, in a normal wire, we can usually model the electrons as moving and the positive charges as stationary (though this is material-dependent and in particular doesn't hold for semiconductors). The charge density of positive charges is $\rho$ and the charge density of electrons is $-\rho$, so the total charge density is $\rho-\rho=0$. But the electrons are moving at velocity $-\vec{v}$ (where $\vec{v}$ points in the direction of the electric field) and the protons are effectively stationary, so the total current density is $\rho\vec{0}-\rho(-\vec{v})=\rho\vec{v}$, which is nonzero. You can add the current densities like this because Maxwell's equations are linear, and therefore obey the principle of superposition.

For another example, suppose we somehow had a wire made out of electrons and positrons of equal charge density (neglecting annihilation or any of the more interesting physics that this would imply). When an electric field is applied to this wire, the positrons will move in the direction of the electric field and the electrons will move in the opposite direction. Since the electron and positron charge have the same absolute value, the magnitude of the force on each species will be the same, and they will move at the same speed. Once again, the charge density is zero, since the positrons have charge density $\rho$ and the electrons have charge density $-\rho$. For the current density, we define $\vec{v}$ to be pointing in the same direction as the electric field; then the positrons have charge density $\rho$ and are moving with velocity $\vec{v}$, and the electrons have charge density $-\rho$ and are moving with velocity $-\vec{v}$, so the total current density is $\rho\vec{v}-\rho(-\vec{v})=2\rho\vec{v}$. So even when the charge carriers move at the same speed, the current density can still be nonzero if their velocity vectors are pointing in different directions.

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From a microscopic perspective $J=\rho v$ holds, but from a macroscopic perspective it can fail.

Consider two microscopic charge distributions, one with charge density and no velocity and the other with the opposite charge density and some velocity. The current of the first is 0, but the current of the second is non zero.

Maxwell’s equations are linear, so to get the overall picture we can simply add the contributions from both charge distributions to get a macroscopic 0 charge density and nonzero current density.

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Is the charge density, found for instance in the Maxwell >equation ∇⋅E=ρ/ϵ0 an 'overall' charge density which takes >every charge into account, while the ρ found in the >expression for J a 'mobile' charge density?

In short, yes.

In the Gauss's law formula, we count all charges. The electric field divergence will be zero only if the net change is zero.

In the current density formula, we count all charges that will move with a non-zero average velocity, when an electric field is applied, and thus will contribute to the electric current.

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