Conflicting definitions of Current Density J? I am reading up on dipole/multipole radiation in Zangwill's Modern Electrodynamics (2013) and I am getting super confused about the units and convention of current density. On Page 735, equation 20.111 reads:
$$ \vec{j}(\vec{r}, t) = q\vec{v}(t)\delta(\vec{r}-\vec{r_0}(t)) $$
BUT if we perform dimensional analysis, for the left hand side we have:
$$ \vec{j} = \frac{dI}{dA} = \frac{dq}{dtdA} = \frac{dq}{dtdx^2}$$
which makes sense because the integration over a surface $dA$ should give us the current. And, for the right hand side we have:
$$ \vec{j} = q\vec{v} = dq\frac{dx}{dt} = dIdx$$
Can anyone help clear my confusion? Thanks in advance!
 A: A Dirac delta-function effectively has dimensions that are the inverse of its argument.  For example, if we integrate
$$
\int \delta(x) dx,
$$
we get 1, which is dimensionless.  But if $x$ has dimensions of length, then the integral of any function $f(x)$ with respect to $x$ should have units of $[f] \cdot [\text{Length}]$, where $[f]$ are the dimensions of $f(x)$.  In the case of $f(x) = \delta(x)$, this implies that
$$
[\delta(x)] = [\text{Length}]^{-1}.
$$
A three-dimensional delta function $\delta^{(3)}(\vec{r}) = \delta(x) \delta(y) \delta(z)$ therefore has dimensions of $\text{[Length]}^{-3}$, which resolves the inconsistency in the dimensions.
A: The actual definition of $\mathbf{J}$ is that it's the charge desnity-weighted real space velocity. The usual way it's defined is by
$$\mathbf{J}(\mathbf{x})=\rho(\mathbf{x})\,\mathbf{v}(\mathbf{x}),$$
where $\rho(\mathbf{x})$ is the usual charge density and $\mathbf{v}(\mathbf{x})$ is the velocity field that describes how the charges are moving at that point. 
The definition given in the question is just a special case of this where $\rho(\mathbf{x}) = q\delta(\mathbf{x}-\mathbf{x}_q(t))$ with $\mathbf{v}=\dot{\mathbf{x}}_q$; the commentary on the unit analysis of a delta function in @MichaelSeifert's answer settles those issues. In principle, when you're dealing with a system where no two charged particles can occupy the same point, this is enough. When the particles can overlap, or when your definition of a "point" is incapable of resolving the individual point charges, you need some slightly more sophisticated machinery that is an elaboration of the same concept.
The full definition of $\mathbf{J}$ that is unambiguous requires turning to the phase space. Suppose $\mathcal{N}(\mathbf{x},\mathbf{p},t)$ is the density of particles per unit real space volume per unit momentum space volume (i.e. $\frac{\mathrm{d}N}{\mathrm{d}x^3\,\mathrm{d}p^3}$). Because the movement of a particle through phase space can be determined entirely from its position in phase space. That is, every particle at a given $(\mathbf{x},\,\mathbf{p})$ at a time $t$ has the same phase space velocity $(\dot{\mathbf{x}},\,\dot{\mathbf{p}})$, so we can describe both of them as functions of position and time (e.g. $\dot{\mathbf{x}}(\mathbf{x},\mathbf{p},t)$). Thus we can define the phase space current density
$$\vec{\mathcal{J}}(\mathbf{x},\,\mathbf{p})=q\mathcal{N}(\mathbf{x},\mathbf{p},t)\,\left(\dot{\mathbf{x}},\,\dot{\mathbf{p}}\right),$$
if all of the particles carry identical charge $q$. 
The rate at which charge is flowing through any phase space surface is then given by the integral of the phase space current density dotted with the surface normal, giving a current through that surface.
The ordinary current density then takes the form
$$\mathbf{J}(\mathbf{x}) = \int_{\text{all }\mathbf{p}} q\,\dot{\mathbf{x}}(\mathbf{x},\mathbf{p},t)\,\mathcal{N}(\mathbf{x},\mathbf{p},t)\,\mathrm{d}^3p.$$
