Consider an inhomogeneous dielectric medium, i.e., one for which the dielectric constant is a function of position, \(\kappa_{e}=\kappa_{e}(x, y, z)\). Show that the fields obey the wave equations $$ \begin{aligned} &\nabla^{2} \mathbf{E}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}}=-\nabla\left(\frac{\nabla \kappa_{e}}{\kappa_{\theta}} \cdot \mathbf{E}\right) \\ &\nabla^{2} \mathbf{B}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{B}}{\partial t^{2}}=-\frac{\nabla \kappa_{e}}{\kappa_{e}} \times(\nabla \times \mathbf{B}) \end{aligned} $$ where, in general, the terms on the right-hand sides couple the cartesian components of the fields. Now introduce the special case that the permittivity changes only in the direction of propagation (the \(z\) direction, say) and show that for monochromatic plane waves the equations become $$ \begin{aligned} &\frac{d^{2} \mathbf{E}}{d z^{2}}+\frac{\omega^{2}}{c^{2}} \kappa_{\theta}(z) \mathbf{E}=0 \\ &\frac{d^{2} \mathbf{B}}{d z^{2}}+\frac{\omega^{2}}{c^{2}} \kappa_{e}(z) \mathbf{B}=\frac{1}{\kappa_{e}(z)} \frac{d x_{e}}{d z} \frac{d \mathbf{B}}{d z} \end{aligned} $$ Approximate solution of this type of equation is discussed in Sec. \(9.1 .\)

We will start by writing down the basic Maxwell's equations in the inhomogeneous dielectric medium. $$ \begin{aligned} \nabla \cdot \mathbf{D} &= \rho_{free} \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t} \\ \nabla \times \mathbf{H} &= \mathbf{J}_{free} + \frac{\partial \mathbf{D}}{\partial t} \end{aligned} $$ where \(\mathbf{D}\) is the displacement field and \(\mathbf{H}\) is the magnetic field.

In this step, we will establish the relations between \(\mathbf{D}\) and \(\mathbf{E}\), \(\mathbf{H}\) and \(\mathbf{B}\), as well as the free charge and current densities. $$ \begin{aligned} \mathbf{D} &= \kappa_e \mathbf{E} \\ \mathbf{H} &= \frac{1}{\mu}\mathbf{B} \\ \rho_{free} &= 0 \\ \mathbf{J}_{free} &= 0 \end{aligned} $$ We will use these relations to rewrite Maxwell's equations.

Now, we substitute the relations we have established in Step 2 into Maxwell's equations: $$ \begin{aligned} \nabla \cdot (\kappa_e \mathbf{E}) &= 0 \\ \nabla \cdot \mathbf{B} &= 0 \\ \nabla \times \mathbf{E} &= -\frac{\partial \mathbf{B}}{\partial t} \\ \nabla \times \frac{1}{\mu}\mathbf{B} &= \frac{\partial (\kappa_e \mathbf{E})}{\partial t} \end{aligned} $$

In order to derive the wave equations, we take the curl of both sides of the third and fourth equations: $$ \begin{aligned} \nabla \times (\nabla \times \mathbf{E}) &= -\frac{\partial}{\partial t}(\nabla \times \mathbf{B}) \\ \nabla \times (\nabla \times \frac{1}{\mu}\mathbf{B}) &= \frac{\partial}{\partial t}\left(\nabla(\kappa_e) \times \mathbf{E} + \kappa_e\nabla \times \mathbf{E}\right) \end{aligned} $$

Now, we use the vector identity \(\nabla \times (\nabla \times \mathbf{A}) = \nabla(\nabla \cdot \mathbf{A}) - \nabla^2\mathbf{A}\) and simplify the equations from Step 4: $$ \begin{aligned} \nabla(\nabla \cdot \mathbf{E}) - \nabla^2\mathbf{E} &= -\frac{\partial}{\partial t}\left(-\frac{\partial \mathbf{E}}{\partial t}\right) \\ \nabla(\nabla \cdot \frac{1}{\mu}\mathbf{B}) - \nabla^2\frac{1}{\mu}\mathbf{B} &= \frac{\partial}{\partial t}\left(\frac{\nabla \kappa_e}{\kappa_e} \cdot \mathbf{E}\right) \end{aligned} $$ Considering that \(\nabla \cdot \mathbf{E} = 0\) and \(\nabla \cdot \frac{1}{\mu}\mathbf{B} = 0\), the equations become: $$ \begin{aligned} \nabla^2\mathbf{E} &= \frac{\partial^2\mathbf{E}}{\partial t^2} \\ \nabla^2\frac{1}{\mu}\mathbf{B} &= \frac{\nabla \kappa_e}{\kappa_e} \cdot \mathbf{E} \end{aligned} $$ Finally, we multiply the first equation by \(\kappa_e\) and the second equation by \(\mu\), and then we get the desired wave equations: $$ \begin{aligned} \nabla^2 \mathbf{E}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{E}}{\partial t^{2}} &= -\nabla\left(\frac{\nabla \kappa_{e}}{\kappa_{\theta}} \cdot \mathbf{E}\right) \\ \nabla^{2} \mathbf{B}-\frac{\kappa_{e}}{c^{2}} \frac{\partial^{2} \mathbf{B}}{\partial t^{2}} &= -\frac{\nabla \kappa_{e}}{\kappa_{e}} \times(\nabla \times \mathbf{B}) \end{aligned} $$

Now, let's consider the special case where the permittivity only changes in the \(z\) direction, and we have monochromatic plane waves: First, rewrite both sides of the equations in terms of \(z\) only: $$ \begin{aligned} \frac{d^2 \mathbf{E}}{dz^2} + \frac{\omega^2}{c^2}\kappa_e(z)\mathbf{E} &= 0 \\ \frac{d^2 \mathbf{B}}{dz^2} + \frac{\omega^2}{c^2}\kappa_e(z)\mathbf{B} &= \frac{1}{\kappa_e(z)}\frac{d\kappa_e}{dz}\frac{d\mathbf{B}}{dz} \end{aligned} $$ These equations represent the wave equations for the given conditions.