Current density of moving charge distribution - mobile charge density vs. "ordinary" charge density of the distribution?

Question

in Introduction to Electrodynamics by David J. Griffiths I have latched upon this definition of current density vector $\mathbf{J}$ (Chapter 5, section 5.1.3, p. 220 in 4th edition) and I would appreciate your help in getting a feel for this:

$$\mathbf{J}=\frac{d\mathbf{I}}{da_{\perp}}$$

Where $da_{\perp}$ is an infinitesimal area perpendicular to the direction of the flow of the charge carriers. Then it is introduced that $\mathbf{J}=\rho\mathbf{v}$, where $\rho$ is the mobile volume charge density. My question is what exactly is this mobile volume charge density? How is that different from the "ordinary" charge density? Can the mobile charge density vary in space and in time?

Similarly, a surface current density is defined as a current per unit width:

$$\mathbf{K}=\frac{d\mathbf{I}}{dl_{\perp}}$$

Which was again connected to the unfortunate mobile surface charge density $\sigma$ by $\mathbf{K}=\sigma\mathbf{v}$.

A problem to consider: In the $xy$ plane there is some charge distribution (an "ordinary" charge distribution) which is moving in the $x$ direction with some constant speed $v$, then the surface charge density will vary in space an in time (as an example consider $\sigma(x,y,t)=sin(x-vt)+sin(y)$). How to find the surface current density $\mathbf{K}$? How to find the mobile surface charge density $\sigma_{mobile}$? Are these concepts really different from each other?

Thanks a lot!

Answer 1

Is easier if you write the equations in integral form: $$ I = \iint_{\Omega} \mathbf J \cdot d \mathbf A $$ $$ I = \int_{C} \mathbf J_s \cdot d \mathbf \Gamma $$ As you might know, from calculus, the right hand side is a flux integral of $\mathbf J$, over an arbitrary surface $\Omega$. This Integral measures how much the field $\mathbf J$, passes through the surface $\Omega$. Same reasoning applies for the surface current density $\mathbf J_s$ but over the contour $\mathbf \Gamma$.

As you see, in the second picture the flux is less than in the first one, because it "penetrates" less onto the surface.

Why is $d\mathbf A$, a vector? Easy, because this infinitesimal area encodes, not only its size, but is normal vector. Add up all contributions and you end up with the total amount of current passing through the surface.

Your formula can interpreted from this result: $$ \frac{\partial I}{\partial \mathbf A} = \mathbf J $$ If the other was the total current density contribution along the surface, this is the other way around. The quantity of current per unit area.

Now for getting a sense of what $\mathbf J = \rho \mathbf u$ means, we have to use Maxwell's equations: $$ \nabla \cdot \mathbf E = \rho / \varepsilon_0 \\ \nabla \times \mathbf B = \mu_0 \mathbf J + \mu_0 \varepsilon_0 \frac{\partial \mathbf E}{\partial t} $$ apply the divergence onto the second equation and insert the first one onto the result. You will get: $$ \nabla \cdot \mathbf J + \frac{\partial \rho}{\partial t} = 0 $$ Plug $\mathbf J = \rho \mathbf u$: $$ \nabla \cdot (\rho \mathbf u) + \frac{\partial \rho}{\partial t} = 0 $$ So here you have an equation that describes the rate of change of your "mobile volume charge density" given a velocity $\mathbf u$. Maybe this $\mathbf u$ is the velocity of electrons inside a wire, or the speed of one particle in free space.

For example if $\nabla \cdot \mathbf u = 0$ (for simplicity): $$ \mathbf u \cdot \nabla \rho + \frac{\partial \rho}{\partial t} = 0 $$ $$ \rho(\mathbf r, t) = F(\mathbf r - \mathbf u t) $$ Which is a function which is being displaced from its original position with velocity and direction $\mathbf u$. For example:

This blob, is your charge density moving through space.

See this and this, for more information.

Answer 2

It's less complicated than you think. "Mobile charge density" refers to the portion of the charge density that is due to charge that is actually moving. If the total charge density $\rho$ at a point is made up of $\rho_s$ that is stationary and $\rho_m$ that is moving with velocity $v$, then the current density is $\rho_m v$.

This is not the first time I've seen some confusion about this. I think the issue is similar to that of Zeno's arrow paradox, in which one wonders how an arrow can be moving if, at any given instant in time, the arrow is still. The answer is that in order to believe the paradox, you have to make the faulty assumption that the system's state is fully determined by the positions of all bodies. In reality the state (at least in classical mechanics) is determined by the combination of all positions and momenta, not the positions alone. Similarly, the confusion about current density seems to arise from an assumption that the description of an electromagnetic source is made up of only the charge density distribution $\rho(x)$ and that $J(x)$ is somehow derived from that. No; you have to know both $\rho$ and $J$ in the first place.

$\rho$ and $J$ are not fundamental quantities anyhow. In classical electrodynamics, $\rho$ and $J$ are an abstraction over large numbers of electrons and protons that each individually have a position and momentum. The "mobile charge density" thus refers to the numerical density of charged particles that are literally in motion, multiplied by the average charge of such particles. In quantum electrodynamics, the quantities $\rho$ and $J$ are operators that are computed from the underlying charged fields, such as the electron field.

In the OP's example, since the surface charge density is $\sin(x - vt) + \sin(y)$ and all the charge is assumed to be moving in the x direction with speed v, the surface current density is simply $(\sin(x - vt) + \sin(y)) v \hat{x}$.

Answer 3

Perhaps considering an example will help. You apply a voltage across a metal wire. The wire itself remains neutral, so the charge density is zero inside (or very nearly so). Yet, the current density isn't zero. In writing $\mathbf J = \rho \mathbf v$, we are not including in $\rho$ the fixed positively charged nuclei or the bound electrons that are not moving, just the "mobile" electrons.

In general, when you have different "species" of charge carriers moving at different velocities, it can be more helpful to write $\mathbf J = \sum\limits_i\rho_i\mathbf v_i$, where species $i$ has density $\rho_i$ and is moving with velocity $\mathbf v_i$. Regarding the positive nuclei to be one such species, it is then clear that they don't contribute to current.

Other examples of different species of carriers include valence and conduction band electrons in semiconductors which move at different velocities, and the different types of ions in electrolytes.