I am currently studying the textbook Physics of Photonic Devices, second edition, by Shun Lien Chuang. Section 2.1.1 Maxwell's Equations in MKS Units says the following:
The well-known Maxwell's equations in MKS (meter, kilogram, and second) units are written as
$$\nabla \times \mathbf{E} = – \dfrac{\partial}{\partial{t}}\mathbf{B} \ \ \ \ \text{Faraday's law} \tag{2.1.1}$$
$$\nabla \times \mathbf{H} = \mathbf{J} + \dfrac{\partial{\mathbf{D}}}{\partial{t}} \ \ \ \ \text{Ampére's law} \tag{2.1.2}$$
$$\nabla \cdot \mathbf{D} = \rho \ \ \ \ \text{Gauss's law} \tag{2.1.3}$$
$$\nabla \cdot \mathbf{B} = 0 \ \ \ \ \text{Gauss's law} \tag{2.1.4}$$
where $\mathbf{E}$ is the electric field (V/m), $\mathbf{H}$ is the magnetic field (A/m), $\mathbf{D}$ is the electric displacement flux density (C/m$^2$), and $\mathbf{B}$ is the magnetic flux density (Vs/m$^2$ or Webers/m$^2$). The two source terms, the charge density $\rho$ (C/m$^3$) and the current density $\mathbf{J}$ (A/m$^2$), are related by the continuity equation
$$\nabla \cdot \mathbf{J} + \dfrac{\partial}{\partial{t}} \rho = 0 \tag{2.1.5}$$
Section 2.1.2 Boundary Conditions then says the following:
By applying the first two Maxwell's equations over a small rectangular surface with a width $\delta$ (dashed line in Fig. 2.1a) across the interface of a boundary and using Stokes' theorem between a line integral over a contour $C$ and the surface $S$ enclosed by the contour
$$\oint_C \mathbf{E} \cdot d \mathscr{l} = \int_S \nabla \times \mathbf{E} \cdot \mathbf{\hat{n}} \ dS = – \dfrac{d}{dt} \int_S \mathbf{B} \cdot \mathbf{\hat{n}} \ dS \tag{2.1.9a}$$
$$\oint_C \mathbf{H} \cdot d \mathscr{l} = \int_S \nabla \times \mathbf{H} \cdot \mathbf{\hat{n}} \ dS = \int_S \mathbf{J} \cdot \mathbf{\hat{n}} \ dS + \dfrac{d}{dt} \int_S \mathbf{D} \cdot \mathbf{\hat{n}} \ dS, \tag{2.1.9b}$$
the following boundary conditions can be derived by letting the width $\delta$ approach zero:
$$\mathbf{\hat{n}} \times (\mathbf{E}_1 – \mathbf{E}_2) = 0 \tag{2.1.10}$$
$$\mathbf{\hat{n}} \times (\mathbf{H}_1 – \mathbf{H}_2) = \mathbf{J}_s, \tag{2.1.11}$$
where $\mathbf{J}_s(= \lim\limits_{\mathbf{J} \to \infty, \ \delta \to 0} \mathbf{J} \delta)$ is the surface current density (A/m). Note that the unit normal vector $\hat{n}$ points from medium 2 to medium 1. Similarly, if we apply Gauss's laws (2.1.3) and (2.1.4) and integrate over a small volume (Fig. 2.1b) with a surface area $A$ and a thickness $\delta$ and let $\delta$ approach zero, for example,
$$\oint_S \mathbf{D} \cdot \mathbf{\hat{n}} \ dS = \int_V \nabla \cdot \mathbf{D} \ dv = \int_V \rho \ dv = \rho \delta A,$$
we obtain the following boundary conditions:
$$\mathbf{\hat{n}} \cdot (\mathbf{D}_1 – \mathbf{D}_2) = \rho_s \tag{2.1.12}$$
$$\mathbf{\hat{n}} \cdot (\mathbf{B}_1 – \mathbf{B}_2) = 0, \tag{2.1.13}$$
where $\rho_s(= \lim\limits_{\rho \to \infty, \ \delta \to 0} \rho \delta)$ is the surface charge density (C/m$^2$).
For an interface across two dielectric media, where no surface current or charge density can be supported, $\mathbf{J}_s = 0$, and $\rho_s = 0$, we have
$$\mathbf{\hat{n}} \times \mathbf{E}_1 = \mathbf{\hat{n}} \times \mathbf{E}_2 \ \ \ \ \ \ \mathbf{\hat{n}} \times \mathbf{H}_1 = \mathbf{\hat{n}} \times \mathbf{H}_2 \\ \mathbf{\hat{n}} \cdot \mathbf{D}_1 = \mathbf{\hat{n}} \cdot \mathbf{D}_2 \ \ \ \ \ \ \mathbf{\hat{n}} \cdot \mathbf{B}_1 = \mathbf{\hat{n}} \cdot \mathbf{B}_2 \tag{2.1.14}$$
For an interface between a dielectric medium and a perfect conductor,
$$\mathbf{\hat{n}} \times \mathbf{E}_1 = 0 \ \ \ \ \ \ \mathbf{\hat{n}} \times \mathbf{H}_1 = \mathbf{J}_s \\ \mathbf{\hat{n}} \cdot \mathbf{D}_1 = \rho_s \ \ \ \ \ \ \mathbf{\hat{n}} \cdot \mathbf{B}_1 = 0 \tag{2.1.15}$$
as the fields $\mathbf{E}_2$, $\mathbf{H}_2$, $\mathbf{D}_2$, and $\mathbf{B}_2$ inside the perfect conductor vanish. The surface charge density and the current density are supported by the perfect conductor surface.
I don't understand (2.1.15). How/why do the fields $\mathbf{E}_2$, $\mathbf{H}_2$, $\mathbf{D}_2$, and $\mathbf{B}_2$ inside the perfect conductor vanish, and how does this lead to (2.1.15)? My understanding is that, mathematically, "vanish" means becomes zero, but, physically, I don't understand why these fields would be zero in a "perfect conductor"; in fact, conductors, such as copper wires, are used because they do well in transmitting electromagnetic fields, right?
Best Answer
On a microscopic level, Ohm's Law is $\mathbf{J} = \sigma \mathbf{E}$. A "perfect conductor" is taken to be the $\sigma \to \infty$ limit of such a medium. In particular, this means that we must have $\mathbf{E} = 0$ inside such a medium. For most (but not all) media, the constitutive relations then tell us that $\mathbf{D} = 0$ whenever $\mathbf{E} = 0$.
I'm not so sure about the book's statement that $\mathbf{B} = 0$ in such a medium. It is true that for a superconductor we must have $\mathbf{B} \approx 0$ due to the Meissner effect; and for many (but definitely not all) media, $\mathbf{B} = 0$ implies $\mathbf{H} = 0$. Perhaps the author simply means "a superconductor" when they say "a perfect conductor". But the Meissner effect does not follow directly from taking the $\sigma \to \infty$ limit of classical electrodynamics; you have to use quantum-mechanical considerations to explain it. So (in my opinion) there's a technical distinction between a "perfect conductor" and a "superconductor" that the author is eliding here.
Finally, Eq. (2.1.15) follows from Eqs. (2.1.10–13) if you assume that all the fields in Region 2 vanish.