I'm reviewing some EM theory and I've come across something that doesn't make sense to me. My textbook states that the magnetic vector potential is defined such that \$ \textbf{B}=\nabla\times\textbf{A}\$. Then it goes on to derive \$ \textbf{A}=\int\frac{\mu\textbf{J}}{4\pi R}dV\$ using the Biot-Savart law, which (as far as I'm aware) allows one to find \$ \textbf{B}\$ given any dc current distribution (no displacement current). Then the book 'fixes' \$ \textbf{E}=-\nabla V\$ so that this equation is valid for time-varying fields: \$ \textbf{E}=-\nabla V-\frac{\partial\textbf{A}}{\partial t}\$, where \$ V=\int\frac{\rho_v}{4\pi\epsilon R}dV\$. So we've got four equations: $$ V(t)=\int\frac{\rho_v(t')}{4\pi\epsilon R}dV$$ $$ \textbf{A}(t)=\int\frac{\mu\textbf{J}(t')}{4\pi R}dV$$ $$ \textbf{E}=-\nabla V-\frac{\partial\textbf{A}}{\partial t}$$ $$ \textbf{B}=\nabla\times\textbf{A}$$
The book states that, given a (potentially time-varying) charge and current distribution, the scalar electric and vector magnetic potentials can be found and then the electric and magnetic fields can be found from these potentials (taking into account retarded time). Now \$ \textbf{E}\$ makes sense to me, since it depends on both the charge distribution (through \$ V\$) and the magnetic field (through \$ \textbf{A}\$).
But how is it that \$ \textbf{B}\$ does not depend on displacement current in these equations? Is it the case that \$ \textbf{J}\$ ought to be interpreted as the total current distribution, including the displacement current? I don't think so because later in the book it uses \$ \textbf{A}\$ to determine the fields around an antenna and only the free current density in the antenna is included in \$ \textbf{J}\$.
Best Answer
The expression for the vector potential as an integral over the current is derived for electrostatic fields, or for fields slowly varying enough to neglect the displacement current. So your observation is correct that these expressions neglect the dependence of B upon displacement current.
Edit, after dusting off my EM textbook: Although the equation for A is often derived for the quasi-static case, by using retarded time in the equation, it turns out, we end up including the dependence of A on the displacement current in the non-quasistatic case. This occurs after a lot of mathematical detail, going into the Coulomb Gauge and breaking the current into transverse (zero divergence) and and irrotational (zero curl) components. The contributions from the longitudinal current term ends up canceling with a term from the scalar potential, leaving you with the wave equation for A, with the transverse component of J as the source. The solution to this is your equation for A with retarded time. When you do this, the displacement current contribution ends up being built into the equation, for which the solution is your equation for A using the retarded time. I believe this implies that the current distribution must have zero divergence.
To me, like you, it is not at all apparent looking at the equation that the displacement current contribution to A is baked into the equation for the vector potential (with retarded time), but it ends up being true.