Current density from charge density?

1. Apr 14, 2012

jjustinn

In statistical mechanics, a probability current (http://en.wikipedia.org/wiki/Probability_current) can be derived from the probability amplitude via

ψ*ψ,i - ψψ*,i (i=x,y,z)

...and it would be nice or any number of reasons to be able to do the same thing with a non-probability "charge"/current -- to take the archetypical example, electric charge/current.

The usual electrodynamic literature defines current as the (four-) velocity of a point charge times its charge, but that obviously doesn't work if you're not dealing with localized point particles -- but it seems like given the time-dependence of the charge density across space, this could be done, just like it's done for probability current.

I thought I had a non-covariant solution -- the gradient of the time-change of the density:
Ji = (ρ,t),i

For example, take a point charge at the origin at t=0, that moves to x = 1 at t=1. So Δρ is +1 at x=1, and -1 at x=0, and the gradient points to the right, giving at least something proportional to the expected current-density.

However, this turns out to basically be a re-phrasing of the point-charge-velocity model, except even less usable--say you have a "line charge" (or a line of charges) between x=0 and x=10, moving with a constant velocity along the x-axis. In that case, the Δρ (ρ,t at the limit), and therefore the current density, will be 0 except at the ends, while this is obviously the type of current we are most familiar with.

So, is this a fool's errand, because charges are fundamentally particular? But if that's the case, what about other locally-conserved properties, like energy/momentum currents in non-particulate matter (or fields)? Perhaps the answer is that in those cases, the only measurable "current" really is at the ends (where the absolute "charge" is changing, rather than merely "moving"), because there's nothing like the magnetic field in these cases, where the "current" is the source of a field in its own right?

2. Apr 15, 2012

Staff: Mentor

What about the simple $\vec{j}=\rho \cdot \vec{v}$ with the average velocity v of your charges?
There is no classical analogon to your quantum-mechanical formula. You cannot derive the current density from the charge density alone.

3. Apr 15, 2012

jjustinn

Yeah, that works perfectly if you do have discrete / localized charges, but it just doesn't seem like it should be necessary.

Also, I think it's a mistake to relegate the probability-current to QM: if my recollection is correct, it originally came from statistical mechanics (see eg the chapters on [classical] ensembles in Dirac's _Principles of QM_).

To go back to the "line charge" -- e.g. ρ(x, t) = 1 (|x=vt| ≤ 10), 0 otherwise, it seems like one could reason out that if the ends are moving, and the "medium" is considered incompressible (and there are no sources/sinks in the region...e.g. The continuity equation holds), then the charge away from the endpoints must be moving as well (e.g. nonzero current), but the way to formalize that into a general statement about the charge density escapes me.

Conversely, if there is a similarly rigorous way to show that it's *not* possible, that would work too (though it would be infinitely more disheartening).

Last edited: Apr 15, 2012