# Magnetic vector potential and displacement current

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}\$$.

• Biot-Savart law finds B not H. Commented Jul 23, 2020 at 15:06
• @Andyaka That's just semantics, but okay I can edit if you like. Commented Jul 23, 2020 at 15:23
• The total electric field E is composed of a conservative part (-grad V) and a solenoidal part (-dA/dt). The conservative part only depends on charge distribuition, as per your first equation, and admits a scalar potential V; the nonconservative part is related to the (time-varying) nonsolenoidal magnetic field that has a vector potential A. In the same way that V does not depend on the nonconservative part, A does not depend on the conservative part. Or, if you want to see this other way, while in time-varying conditions B is purely solenoidal, E is no longer conservative. Commented Jul 23, 2020 at 15:23
• @SredniVashtar Can you explain why B does not appear to depend on displacement current density in the above equations? But in Maxwell's equations, B clearly does depend on displacement current density. Commented Jul 23, 2020 at 15:29
• Let's get retarded. What is t' in your equations? Commented Jul 23, 2020 at 15:58

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.

• Then why is the vector potential used to perform antenna analysis where the quasi-static assumption is clearly not valid? Commented Jul 23, 2020 at 15:52
• That is a good question that I hope someone else can answer. Perhaps there is a derivation of the equation that I don't know about which does not require the assumption. Commented Jul 23, 2020 at 16:04
• Maybe the quasi-static analysis is used to find the near (non radiating) field? Could you provide us some further equations for the antenna analysis being carried out by the author? Commented Jul 23, 2020 at 17:21
• I have edited my answer now to address the question of antenna analysis. Commented Jul 23, 2020 at 17:23
• @rpm2718 I don't think it's the Coloumb gauge that makes the trick. It's the Lorenz gauge that incorporates the dE/dt term (the divergence of A depends on the derivative of V - roughly meaning that a variable E field creates a magnetic field) and allows for the potential equations to be decoupled and transformed into 'common' wave equations. The need for retarded time is a consequence of the propagation with speed c in said equations, and that's why t' appears in the determination of potentials from the sources. So to speak, it's the other way around: the formula for the potentials follow. Commented Jul 25, 2020 at 6:50

Back inn 1898 and 1900 (120 years ago), this was one of the views of E&M

https://en.wikipedia.org/wiki/Li%C3%A9nard%E2%80%93Wiechert_potential

For an more modern interpretation, consider Jefimenko's Equations

https://en.wikipedia.org/wiki/Jefimenko%27s_equations

The importance of these equations is many fold; one factor is the LACK OF CAUSALITY in Maxwell's Equations.