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$^2$). enter image description here 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?


1 Answer 1


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.

  • $\begingroup$ Ferroelectric materials may have $D \ne 0$ for $E = 0$. $\endgroup$ Dec 16, 2021 at 21:48
  • 1
    $\begingroup$ @SebastianRiese: Thanks for the link; I've added it to the answer. $\endgroup$ Dec 16, 2021 at 21:49
  • $\begingroup$ For the meister effect. As I understand it, it works due to any changing magnetic field inside it will cause an induced electric field ,such that because there are is no resistance, this magnetic field perfectly counteracts the external magnetic field. But what if I cool the conductor down to super conducting temperatures WHILST EXPOSED to a magnetic field? Wouldnt the magnetic field be frozen inside the superconductor and any change to that initial magnetic field would result in an emf that wants to keep it frozen? Is this a classic analogue of "flux pinning"? $\endgroup$ Dec 17, 2021 at 0:38
  • $\begingroup$ And if this last idea is true . I have no idea why in flux pinning the magnetic field squeezes into bunches. And not just resembles the external magnetic field $\endgroup$ Dec 17, 2021 at 0:40
  • $\begingroup$ @jensenpaull: Those sound like good questions to ask in a new post. $\endgroup$ Dec 17, 2021 at 3:55

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service and acknowledge that you have read and understand our privacy policy and code of conduct.

Not the answer you're looking for? Browse other questions tagged or ask your own question.