# How/why do the fields $\mathbf{E}_2$, $\mathbf{H}_2$, $\mathbf{D}_2$, and $\mathbf{B}_2$ inside the perfect conductor vanish?

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?

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.
• Ferroelectric materials may have $D \ne 0$ for $E = 0$. Dec 16, 2021 at 21:48