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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

How does letting $\delta$ approach zero get us 2.1.12 and 2.1.13? And why do we have $\mathbf{D}_1 - \mathbf{D}_2$ and $\mathbf{B}_1 - \mathbf{B}_2$ instead of $\mathbf{D}_2 - \mathbf{D}_1$ and $\mathbf{B}_2 - \mathbf{B}_1$?

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Consider the Gaussian 'pillbox' in figure (b). Assume the curved surface is $S_3$ while the cross-sectional areas of area $A$ are $S_1$ and $S_2$. Also, we assume that the cross-section chosen is arbitrarily small, such that $\mathbf{D}_2,\mathbf{D}_1$ are consant across them respectively. $$ \oint _{S} \textbf{D} \cdot \mathbf{\hat n} \,dS = \iint _{S_1} \textbf{D} \cdot \mathbf{\hat n_1} \,dS_1 + \iint _{S_2} \textbf{D} \cdot \mathbf{\hat n_2}\,dS_2 + \iint _{S_3} \textbf{D} \cdot \mathbf{\hat n_3}\,dS_3 $$

As $\delta$ goes to zero, the surface integral through $S_3$ becomes zero. We note that $\mathbf{\hat n_1}=\mathbf{\hat n},\mathbf{\hat n_2}=-\mathbf{\hat n}$ $$ \lim \limits_{\delta \to 0} \oint _{S} \textbf{D} \cdot \mathbf{\hat n}\,dS = \iint_{S_1} \mathbf{D_1} \cdot \mathbf{\hat n}\,dS_1 + \iint_{S_2} \mathbf{D_2} \cdot (-\mathbf{\hat n})\,dS_2 $$ $$ = (\mathbf{D_1} \cdot \mathbf{\hat n})A - (\mathbf{D_2} \cdot \mathbf{\hat n})A =A\mathbf{\hat n} \cdot (\mathbf{D_1} - \mathbf{D_2})$$

Applying the divergence theorem on the LHS, we get $$\lim \limits_{\delta \to 0} \oint _{S} \textbf{D} \cdot \mathbf{\hat n} \,dS = \lim \limits_{\delta \to 0} \iiint_V (\nabla\cdot \textbf{D}) \,dV $$ $$ = \lim \limits_{\delta \to 0} \iiint_V \rho \,dV \text{(applying Gauss Law)}$$ $$ =\lim \limits_{\delta \to 0} \rho \delta A = \rho_sA $$

The final step is justified by the following argument. $$ \rho_s = \frac{\,dQ}{\,dS} = \frac{\,dQ}{{}^{\,dV}{\mskip -5mu/\mskip -3mu}_{\delta}} = \rho \delta $$ Equating them, we get the desired boundary condition: $ \mathbf{\hat n} \cdot (\mathbf{D_1} - \mathbf{D_2}) = \rho_s $. A similar argument works for the second boundary condition too.

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