What does a volume integral of $\textbf{J}$ mean?

Question

Assume stationary currents in vacuum, $\text{curl}\textbf{B} = \mu_0\textbf{J}$. With $\text{curl}\textbf{A}=\textbf{B}$ and $\text{div}\textbf{A} =0$ the vector potential $\textbf{A}$ can be written explicitly as a spatial integral over all space of the current density $\textbf{J}$, (see for example Jackson, section 5.4): $$\textbf{A(x)} = \frac{\mu_0}{4\pi}\int \textbf{J(x')}\frac{d^3x'}{\textbf{|x-x'|}} \tag{1}$$ So far so good, but what is puzzling about this integral is that the current density $\textbf{J}$ is associated with a 2-form and we are integrating it with differential 3D volume elements not 2D surface elements as I would have naively expected it.

Compare (1) with a similar formula for the scalar potential $\phi$ where $\textbf{E}=-\text{grad} \phi $, and now the charge density $\rho$ is given: $$\phi{(\textbf{x})} = \frac{1}{4\pi\epsilon_0 }\int \rho{(\textbf{x}')}\frac{d^3x'}{\textbf{|x-x'|}} \tag{2}$$ Charge density $\rho$ is a spatial density and as such is associated with a 3-form, and as I would expect it is integrated with a 3D differential volume element.

My question is what does this all mean, what happens when something that is apparently a natural 2-form is integrated if it was something else?

Answer 1

It's best to think relativisticly. We are solving the Maxwell equation $d\star (d A)=\star J$ where $$ A=-\phi dt+ A_x dx +A_ydy+A_zdz\\ J= -\rho dt + j_x dx+j_ydy+j_z dz $$ are 1-forms in four dimensions and $\star$ is the Hodge dual. So $\star J$ is a three form.

The $j_x$ appearsin $\star J$ with $dy\wedge dz \wedge dt$ so it's only a 2-form if you ignore the $dt$.

$\rho$, in the other hand, is the coefficient $dx\wedge dy\wedge dz$ and so in $\star J$ it is a three-form.

For time-dependent sources we integrate over four dimensions to get $A$ --- so the Green function must be a form which contains a delta function that only keeps contributions from the retarded time. For static current/charge distributions. The $t$ integral can be done leaving the 1-form answers you cite.

Answer 2

If one intends to calculate average value of $\mathbf j$ on a line $L$, one integrates over that line:

$$ \langle \mathbf j \rangle = \int_L \mathbf j d x' / L. $$

If one intends to calculate average value of $\mathbf j$ on a surface $S$, one integrates over that surface: $$ \langle \mathbf j \rangle = \int_S \mathbf j d^2 x' / S. $$

If one intends to calculate average value of $\mathbf j$ in a volume $V$, one integrates over that volume:

$$ \langle \mathbf j \rangle = \int_V \mathbf j d^3 x' / V. $$

The dimensionality of the integral is determined by dimensionality of the region one is considering. There is no reason one should integrate $\mathbf j$ only over two-dimensional regions. Of course, the definition of $\mathbf j$ is based on the flux integral over oriented surface $\mathbf S$:

$$ electric~current~through~\mathbf S = \int_\mathbf S \mathbf j \cdot d\mathbf S $$

and this integral is two-dimensional. But the usage of $\mathbf j$ in other integrals is not limited in any way.

Magnetic field or magnetic vector potential are given by a three-dimensional integral in general because those integrals have the right limit in case the current is limited to a narrow long tube - the limit being the integral from the Biot-Savart law for field of uni-dimensional wire:

$$ \mathbf A(\mathbf x) = \frac{\mu_0}{4\pi} \int_V \frac{\mathbf j(\mathbf x')}{|\mathbf x - \mathbf x'|}\,d^3\mathbf x' \rightarrow \frac{\mu_0}{4\pi} \int_L \frac{I(\mathbf x')}{|\mathbf x - \mathbf x'|}\,d x'~~~,x' = |\mathbf x'| $$