Waves in composites and metamaterials/Fresnel equations

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[edit] A brief excursion into homogenization

One of the first questions that arise in the homogenization of composites is whether determining the effective behavior of the composite by some averaging process is the right thing to do. At this stage we ignore such questions and assume that there is a representative volume element (RVE) over which such an average can be obtained.

Let $\mathbf{E}$ be an average over some RVE of the $\mathbf{E}$ -field at an atomic scale. Similarly, let $\mathbf{B}$ be the average of the $\mathbf{B}$ -field. Recall, from the previous lecture, that the Maxwell equations have the form (we have dropped the hats over the field quantities)

$\boldsymbol{\nabla} \times \mathbf{E} = i\omega~\boldsymbol{\mu}(\mathbf{x},\omega)\cdot\mathbf{H}(\mathbf{x}) ~;~~ \boldsymbol{\nabla} \times \mathbf{H} = -i\omega~\boldsymbol{\epsilon}(\mathbf{x},\omega)\cdot\mathbf{E}(\mathbf{x}) ~.$

For some conductors, at low frequencies, the permittivity tensor is given by

$\boldsymbol{\epsilon} \approx \boldsymbol{\epsilon}_0 + \cfrac{i}{\omega}~\boldsymbol{\sigma}$

where $\boldsymbol{\epsilon}_0$ is the real part of the permittivity tensor and $\boldsymbol{\sigma}$ is the electrical conductivity tensor. ^[1]

A mixture of conductors and dielectric materials may have properties which are quite different from those of the constituents. For example, consider the checkerboard material shown in Figure 1 containing an isotropic conducting material and an isotropic dielectric material.

Figure 1. Checkerboard material containing conducting and dielectric phases.

The conducting material has a permittivity of $ε 1 = 1 + i σ / ω$ while the dielectric material has a permittivity of $ε 2 = 1$ . The effective permittivity of the checkerboard is given by

$\epsilon_{\text{eff}} = \sqrt{\epsilon_1~\epsilon_2} = \sqrt{1 + \cfrac{i~\sigma}{\omega}} \approx \cfrac{\sqrt{i}\sqrt{\sigma}}{\sqrt{\omega}} ~~~(\text{for small}~\omega) \ne a + i~b ~~\text{for any}~~ a, b ~.$

[edit] Plane waves

Let us assume that the material is isotropic. Then,

$\boldsymbol{R}^T\cdot\boldsymbol{\epsilon}\cdot\boldsymbol{R} = \boldsymbol{\epsilon} \qquad \text{and} \qquad \boldsymbol{R}^T\cdot\boldsymbol{\mu}\cdot\boldsymbol{R} = \boldsymbol{\mu} \qquad \forall ~~\text{rotations}~~\boldsymbol{R}^T\cdot\boldsymbol{R} = \boldsymbol{R}\cdot\boldsymbol{R}^T = \boldsymbol{\mathit{1}} ~.$

Therefore, we can write

$\boldsymbol{\epsilon} = \epsilon~\boldsymbol{\mathit{1}} \qquad \text{and} \qquad \boldsymbol{\mu} = \mu~\boldsymbol{\mathit{1}} ~.$

The Maxwell equations then take the form

$\text{(2)} \qquad \boldsymbol{\nabla} \times \mathbf{E} = i\omega~\mu(\mathbf{x},\omega)~\mathbf{H}(\mathbf{x}) ~;~~ \boldsymbol{\nabla} \times \mathbf{H} = -i\omega~\epsilon(\mathbf{x},\omega)~\mathbf{E}(\mathbf{x}) ~.$

If we assume that $μ$ and $ε$ do not depend upon position, i.e., $\mu \equiv \mu(\omega)$ and $\epsilon \equiv \epsilon(\omega)$ , and take the curl of equations (2), we get

$\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\times\mathbf{E}) = i\omega~\mu~\boldsymbol{\nabla} \times \mathbf{H} = \omega^2~\epsilon~\mu~\mathbf{E}(\mathbf{x})~;~~ \boldsymbol{\nabla} \times (\boldsymbol{\nabla}\times\mathbf{H}) = -i\omega~\epsilon~\boldsymbol{\nabla} \times \mathbf{E} = \omega^2~\epsilon~\mu~\mathbf{H}(\mathbf{x}) ~.$

Using the identity $\boldsymbol{\nabla} \times (\boldsymbol{\nabla}\times\mathbf{A}) = \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{A}) - \nabla^2 \mathbf{A}$ , we get

$\text{(3)} \qquad \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{E}) - \nabla^2 \mathbf{E} = \omega^2~\epsilon~\mu~\mathbf{E}(\mathbf{x})~;~~ \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{H}) - \nabla^2 \mathbf{H} = \omega^2~\epsilon~\mu~\mathbf{H}(\mathbf{x})~.$

Since

$\boldsymbol{\nabla} \cdot \mathbf{B} = 0 ~; \boldsymbol{\nabla} \cdot \mathbf{D} = 0 ~;~~ \mathbf{H}(\mathbf{x}) = \mu^{-1}(\omega)~\mathbf{B}(\mathbf{x}) ~;~ \mathbf{D}(\mathbf{x}) = \epsilon(\omega)~\mathbf{E}(\mathbf{x})$

we have

$\boldsymbol{\nabla} \cdot \mathbf{H} = \mu^{-1}~\boldsymbol{\nabla} \cdot \mathbf{B} = 0 \qquad \text{and} \qquad \boldsymbol{\nabla} \cdot \mathbf{E} = \epsilon^{-1}~\boldsymbol{\nabla} \cdot \mathbf{D} = 0 ~.$

Therefore, from equation (3), we have

$\nabla^2 \mathbf{E} + \omega^2~\epsilon~\mu~\mathbf{E}(\mathbf{x}) = \boldsymbol{0} ~;~~ \nabla^2 \mathbf{H} + \omega^2~\epsilon~\mu~\mathbf{H}(\mathbf{x}) = \boldsymbol{0} ~.$

We can also write the above equations in the form

$\text{(4)} \qquad { \nabla^2 \mathbf{E} + \kappa^2~\mathbf{E}(\mathbf{x}) = \boldsymbol{0} ~;~~ \nabla^2 \mathbf{H} + \kappa^2~\mathbf{H}(\mathbf{x}) = \boldsymbol{0} \qquad \text{where} \quad \kappa^2 = \cfrac{\omega^2}{c^2} ~~\text{and}~~ c^2 = \cfrac{1}{\epsilon~\mu} }$

where $c$ is the phase velocity (the velocity at which the wave crests travel). To have a propagating wave, $c$ must be real. This will be the case when $ε$ and $μ$ are both positive or both negative (see Figure 2).

Figure 2. Transparency and opacity of a material as a function of

μ

and

ε

.

Let us look for plane wave solutions to the equations (4) of the form

$\text{(5)} \qquad \mathbf{E}(\mathbf{x}) = \mathbf{E}_0~e^{i~(\mathbf{k}\cdot\mathbf{x})}$

where $|\mathbf{k}| = 1/(2~\pi~\lambda)$ and $λ$ is the wavelength. Then, using the first of equations (2) we have

$\text{(6)} \qquad \mathbf{H}(\mathbf{x}) = -\cfrac{i}{\omega~\mu}~\boldsymbol{\nabla} \times \mathbf{E} = \cfrac{1}{\omega\mu}~ \mathbf{k}\times\mathbf{E}_0~e^{i~(\mathbf{k}\cdot\mathbf{x})} = \mathbf{H}_0~e^{i~(\mathbf{k}\cdot\mathbf{x})}$

where $\mathbf{H}_0 = 1/(\omega\mu)~\mathbf{k}\times\mathbf{E}_0$ .

Since $\boldsymbol{\nabla} \cdot \mathbf{E} = 0$ , we have (in terms of components with respect to a orthonormal Cartesian basis)

$\boldsymbol{\nabla} \cdot \mathbf{E} = \frac{\partial }{\partial x_m}\left[ E_{0m}~e^{i(k_l~x_l)}\right] = i~k_l~\frac{\partial x_l}{\partial x_m}~E_{0m}~e^{i(k_l~x_l)} = i~k_m~E_{0m}~e^{i(k_l~x_l)} = i~(\mathbf{k}\cdot\mathbf{E}_0)~e^{i(\mathbf{k}\cdot\mathbf{x})} = 0 ~.$

Hence,

${ \mathbf{k}\cdot\mathbf{E}_0 = 0 ~. }$

Similarly, since $\boldsymbol{\nabla} \cdot \mathbf{H} = 0$ , we have

$\boldsymbol{\nabla} \cdot \mathbf{H} = \cfrac{1}{\omega\mu}\left[ \frac{\partial }{\partial x_m}\left(\mathcal{E}_{mpq}~k_p~E_{0q}~e^{i(k_l~x_l)}\right) \right] = \cfrac{i}{\omega\mu}\left[\mathcal{E}_{mpq}~k_p~E_{0q}~k_m~e^{i(k_l~x_l)} \right] = \cfrac{i}{\omega\mu}~\mathbf{k}\cdot(\mathbf{k}\times\mathbf{E}_0)~e^{i(\mathbf{k}\cdot\mathbf{x})}~.$

Hence,

${ \mathbf{k}\cdot\mathbf{H}_0 = 0 ~. }$

Plugging equation (5) into the first of equations (4) we get

$\begin{align} \left[\nabla^2 \mathbf{E} + \kappa^2~\mathbf{E}(\mathbf{x})\right]_n & = \frac{\partial }{\partial x_m}\left[\frac{\partial }{\partial x_m}\left( E_{0n}~e^{i k_l x_l}\right)\right] + \kappa^2~E_{0n}~ e^{i k_l x_l} \\ & = \frac{\partial }{\partial x_m}\left[E_{0n}~\left(i~k_l~\frac{\partial x_l}{\partial x_m}\right)~ e^{i k_l x_l}\right] + \kappa^2~E_{0n}~ e^{i k_l x_l} \\ & = i~E_{0n}~k_m~\frac{\partial }{\partial x_m}\left(e^{i k_l x_l}\right) + \kappa^2~E_{0n}~ e^{i k_l x_l} \\ & = i~E_{0n}~k_m~\left(i~k_l~\frac{\partial x_l}{\partial x_m}\right)~e^{i k_l x_l} + \kappa^2~E_{0n}~ e^{i k_l x_l} \\ & = - E_{0n}~k_m~k_m~e^{i k_l x_l} + \kappa^2~E_{0n}~e^{i k_l x_l} ~. \end{align}$

Reverting back to Gibbs notation, we get

$-(\mathbf{k}\cdot\mathbf{k})~\mathbf{E}_0~e^{i~(\mathbf{k}\cdot\mathbf{x})} + \kappa^2~\mathbf{E}_0~e^{i~(\mathbf{k}\cdot\mathbf{x})} = \boldsymbol{0} ~.$

Therefore,

${ \mathbf{k}\cdot\mathbf{k} = \kappa^2~. }$

Plugging the solution (6) into the second of equations (4) (and using index notation as before) we get

$\begin{align} \left[\nabla^2 \mathbf{H} + \kappa^2~\mathbf{H}(\mathbf{x})\right]_n & = \cfrac{1}{\omega\mu}\left\{ \frac{\partial }{\partial x_m}\left[\frac{\partial }{\partial x_m}\left( \mathcal{E}_{npq}~k_p~E_{0q}~e^{i k_l x_l}\right)\right] + \kappa^2~ \mathcal{E}_{npq}~k_p~E_{0q}~e^{i k_l x_l}\right\} \\ & = \cfrac{1}{\omega\mu}\left\{ \frac{\partial }{\partial x_m} \left[\mathcal{E}_{npq}~k_p~E_{0q}~\left(i~k_l~\frac{\partial x_l}{\partial x_m}\right)~ e^{i k_l x_l}\right] + \kappa^2~ \mathcal{E}_{npq}~k_p~E_{0q}~e^{i k_l x_l}\right\} \\ & = \cfrac{1}{\omega\mu}\left\{ i~\mathcal{E}_{npq}~k_p~k_m~E_{0q}~\left[i~k_l~\frac{\partial x_l}{\partial x_m}\right]~ e^{i k_l x_l} + \kappa^2~ \mathcal{E}_{npq}~k_p~E_{0q}~e^{i k_l x_l}\right\} \\ & = \cfrac{1}{\omega\mu}\left\{ - \mathcal{E}_{npq}~k_p~k_m~k_m~E_{0q}~e^{i k_l x_l} + \kappa^2~ \mathcal{E}_{npq}~k_p~E_{0q}~e^{i k_l x_l}\right\} \end{align}$

In Gibbs notation, we then have

$\nabla^2 \mathbf{H} + \kappa^2~\mathbf{H}(\mathbf{x}) = \cfrac{1}{\omega\mu}~(- \mathbf{k}\cdot\mathbf{k} + \kappa^2)~ \mathbf{k}\times\mathbf{E}_0~e^{i(\mathbf{k}\cdot\mathbf{x})} = \boldsymbol{0} ~.$

Therefore, once again, we get

${ \mathbf{k}\cdot\mathbf{k} = \kappa^2 ~. }$

[edit] Reflection at an Interface

The following is based on the description given in [Lorrain88]. Figure 3 shows an electromagnetic wave that is incident upon an interface separating two mediums.

Figure 3. Reflection of an electromagnetic wave at an interface.

We ignore the time-dependent component of the electric fields and assume that we can express the waves shown in Figure 3 in the form

$\begin{align} \mathbf{E}_i & = \mathbf{E}_{0i}~e^{i(\mathbf{k}_i\cdot\mathbf{x})} \\ \mathbf{E}_r & = \mathbf{E}_{0r}~e^{i(\mathbf{k}_r\cdot\mathbf{x})} \\ \mathbf{E}_t & = \mathbf{E}_{0t}~e^{i(\mathbf{k}_t\cdot\mathbf{x})} \end{align}$

where $\mathbf{k}_i, \mathbf{k}_r, \mathbf{k}_t$ are the wave vectors.

Since the oscillations at the interface must have the same period (the requirement of continuity), we must have

$\mathbf{k}_i\cdot\mathbf{x} = \mathbf{k}_r\cdot\mathbf{x} = \mathbf{k}_t\cdot\mathbf{x} \qquad \forall \mathbf{x} ~~ \text{on the interface} ~.$

This means that the tangential components of $\mathbf{k}_i, \mathbf{k}_r, \mathbf{k}_t$ must be equal at the interface. Therefore,

$|\mathbf{k}_i|~\sin\theta_i = |\mathbf{k}_r|~\sin\theta_r = |\mathbf{k}_t|~\sin\theta_t~.$

Now,

$|\mathbf{k}_i| = \kappa_i = \cfrac{\omega}{c_1} ~;~~ |\mathbf{k}_r| = \kappa_r = \cfrac{\omega}{c_1} ~;~~ |\mathbf{k}_t| = \kappa_t = \cfrac{\omega}{c_2}$

where $c 1$ and $c 2$ are the phase velocities in medium 1 and medium 2, respectively. Hence we have,

$\cfrac{\omega}{c_1}~\sin\theta_i = \cfrac{\omega}{c_1}~\sin\theta_r = \cfrac{\omega}{c_2}~\sin\theta_t~.$

This implies that

$\theta_i = \theta_r \qquad \text{and} \qquad \cfrac{\sin\theta_i}{\sin\theta_t} = \cfrac{c_1}{c_2} ~.$

The refractive index is defined as

$n := \cfrac{c_0}{c}$

where $c 0$ is the phase velocity is vacuum. Therefore, we get

$\cfrac{\sin\theta_i}{\sin\theta_t} = \cfrac{n_2}{n_1} \qquad\mbox{(Snells Law)} ~.$

[edit] Polarized wave with the $\mathbf{E}_i$ parallel to the plane of incidence

Consider the $p$ -polarized wave shown in Figure 4. The figure represents an infinite wave polarized with the $\mathbf{E}_i$ vector polarized parallel to the plane of incidence. This is also called the TM (transverse magnetic) case.

Figure 4. Infinite wave polarized with the $\mathbf{E}_i$ -vector parallel to the plane of incidence.

Let us define

$\kappa_1 := |\mathbf{k}_i| = |\mathbf{k}_r| \qquad\text{and}\qquad \kappa_2 := |\mathbf{k}_t| ~.$

Recall that,

$\begin{align} \mathbf{E}_i(\mathbf{x}) & = \mathbf{E}_{0i}~e^{i(\mathbf{k}_i\cdot\mathbf{x})} \\ \mathbf{E}_r(\mathbf{x}) & = \mathbf{E}_{0r}~e^{i(\mathbf{k}_r\cdot\mathbf{x})} \\ \mathbf{E}_t(\mathbf{x}) & = \mathbf{E}_{0t}~e^{i(\mathbf{k}_t\cdot\mathbf{x})} \end{align}$

Let us choose an orthonormal basis ( $\mathbf{e}_1,\mathbf{e}_2,\mathbf{e}_3$ ) such that the $\mathbf{e}_1$ vector lies on the interface and is parallel the plane of incidence. The $\mathbf{e}_2$ vector lies on the plane of incidence and the $\mathbf{e}_3$ vector is normal to the interface. Then the vectors $\mathbf{k}_i$ , $\mathbf{k}_r$ , and $\mathbf{k}_t$ may be expressed in this basis as

$\begin{align} \mathbf{k}_i & = \kappa_1~\sin\theta_i~\mathbf{e}_1 - \kappa_1~\cos\theta_i~\mathbf{e}_3 \\ \mathbf{k}_r & = \kappa_1~\sin\theta_r~\mathbf{e}_1 + \kappa_1~\cos\theta_r~\mathbf{e}_3 \\ \mathbf{k}_t & = \kappa_2~\sin\theta_t~\mathbf{e}_1 - \kappa_2~\cos\theta_t~\mathbf{e}_3 ~. \end{align}$

Similarly, defining

$\mathcal{E}_i := |\mathbf{E}_{0i}| ~;~~ \mathcal{E}_r := |\mathbf{E}_{0r}| ~;~~ \mathcal{E}_t := |\mathbf{E}_{0t}|$

we get

$\begin{align} \mathbf{E}_{0i} & = \mathcal{E}_i~\cos\theta_i~\mathbf{e}_1 + \mathcal{E}_i~\sin\theta_i~\mathbf{e}_3 \\ \mathbf{E}_{0r} & = -\mathcal{E}_r~\cos\theta_r~\mathbf{e}_1 + \mathcal{E}_r~\sin\theta_r~\mathbf{e}_3 \\ \mathbf{E}_{0t} & = \mathcal{E}_t~\cos\theta_t~\mathbf{e}_1 + \mathcal{E}_t~\sin\theta_t~\mathbf{e}_3 ~. \end{align}$

Using the definition

$\mathbf{H}_0 := \cfrac{1}{\omega\mu}~\mathbf{k}\times\mathbf{E}_0$

we then get

$\begin{align} \mathbf{H}_{0i} & = \cfrac{\kappa_1}{\omega\mu_1}~\mathcal{E}_i~\mathbf{e}_2 \qquad \implies \qquad \mathcal{H}_i := |\mathbf{H}_{0i}| = \cfrac{\kappa_1~\mathcal{E}_i}{\omega\mu_1}\\ \mathbf{H}_{0r} & = - \cfrac{\kappa_1}{\omega\mu_1}~\mathcal{E}_r~\mathbf{e}_2 \qquad \implies \qquad \mathcal{H}_r := |\mathbf{H}_{0r}| = \cfrac{\kappa_1~\mathcal{E}_r}{\omega\mu_1}\\ \mathbf{H}_{0t} & = \cfrac{\kappa_2}{\omega\mu_2}~\mathcal{E}_t~\mathbf{e}_2 \qquad \implies \qquad \mathcal{H}_t := |\mathbf{H}_{0t}| = \cfrac{\kappa_2~\mathcal{E}_t}{\omega\mu_2}~. \end{align}$

Hence, with the vector $\mathbf{x}$ expressed as $\mathbf{x} = x_1~\mathbf{e}_1 + x_2~\mathbf{e}_2 + x_3~\mathbf{e}_3$ , we get

$\begin{align} \mathbf{E}_i(\mathbf{x}) & = (\mathcal{E}_i~\cos\theta_i~\mathbf{e}_1 + \mathcal{E}_i~\sin\theta_i~\mathbf{e}_3) ~e^{i[\kappa_1(x_1\sin\theta_i - x_3\cos\theta_i)]} \\ \mathbf{E}_r(\mathbf{x}) & = (-\mathcal{E}_r~\cos\theta_r~\mathbf{e}_1 + \mathcal{E}_r~\sin\theta_r~\mathbf{e}_3) ~e^{i[\kappa_1(x_1\sin\theta_r + x_3\cos\theta_r)]} \\ \mathbf{E}_t(\mathbf{x}) & = (\mathcal{E}_t~\cos\theta_t~\mathbf{e}_1 + \mathcal{E}_t~\sin\theta_t~\mathbf{e}_3) ~e^{i[\kappa_2(x_1\sin\theta_t - x_3\cos\theta_t)]}~. \end{align}$

Similarly,

$\begin{align} \mathbf{H}_i(\mathbf{x}) & = -\mathcal{H}_i~\mathbf{e}_2~ ~e^{i[\kappa_1(x_1\sin\theta_i - x_3\cos\theta_i)]} \\ \mathbf{H}_r(\mathbf{x}) & = -\mathcal{H}_r~\mathbf{e}_2~ ~e^{i[\kappa_1(x_1\sin\theta_r + x_3\cos\theta_r)]} \\ \mathbf{H}_t(\mathbf{x}) & = -\mathcal{H}_t~\mathbf{e}_2~ ~e^{i[\kappa_2(x_1\sin\theta_t - x_3\cos\theta_t)]}~. \end{align}$

At the interface, continuity requires that the tangential components of the vectors $\mathbf{E}$ and $\mathbf{H}$ are continuous. Clearly, from the above equations, the $\mathbf{H}_0$ vectors are tangential to the interface. Also, at the interface $x 3 = 0$ and $x 1$ is arbitrary. Hence, continuity of the components of $\mathbf{H}$ at the interface can be achieved if

$\mathcal{H}_i + \mathcal{H}_r = \mathcal{H}_t ~.$

In terms of the electric field, we then have

$\cfrac{\kappa_1}{\omega\mu_1}~\mathcal{E}_i + \cfrac{\kappa_1}{\omega\mu_1}~\mathcal{E}_r = \cfrac{\kappa_2}{\omega\mu_2}~\mathcal{E}_t ~.$

Recall that the refractive index is given by $n = c_0/c = c_0~\kappa/\omega$ . Therefore, we can write the above equation as

$\text{(7)} \qquad { \cfrac{n_1}{\mu_1}~(\mathcal{E}_i + \mathcal{E}_r) = \cfrac{n_2}{\mu_2}~\mathcal{E}_t ~. }$

The tangential components of the $\mathbf{E}$ vectors at the interface are given by $\mathbf{E}\times\mathbf{e}_3$ . Therefore, the tangential components of the $\mathbf{E}_0$ vectors at the interface are

$\begin{align} \mathbf{E}_{0i}\times\mathbf{e}_3 & = -\mathcal{E}_i~\cos\theta_i~\mathbf{e}_2 \\ \mathbf{E}_{0r}\times\mathbf{e}_3 & = \mathcal{E}_r~\cos\theta_r~\mathbf{e}_2 \\ \mathbf{E}_{0t}\times\mathbf{e}_3 & = -\mathcal{E}_t~\cos\theta_t~\mathbf{e}_2 ~. \end{align}$

Using the arbitrariness of $x 1$ and from the continuity of the $\mathbf{E}$ vectors at the interface, we have

$\mathcal{E}_i~\cos\theta_i - \mathcal{E}_r~\cos\theta_r = \mathcal{E}_t~\cos\theta_t ~.$

Since $θ i = θ r$ , we have

$\text{(8)} \qquad { (\mathcal{E}_i - \mathcal{E}_r)~\cos\theta_i = \mathcal{E}_t~\cos\theta_t ~. }$

From equations (7) and (8), we get two more relations:

$\text{(9)} \qquad { \cfrac{\mathcal{E}_r}{\mathcal{E}_i} = \cfrac {\cfrac{n_2}{\mu_2}~\cos\theta_i - \cfrac{n_1}{\mu_1}~\cos\theta_t} {\cfrac{n_2}{\mu_2}~\cos\theta_i + \cfrac{n_1}{\mu_1}~\cos\theta_t} }$

and

$\text{(10)} \qquad { \cfrac{\mathcal{E}_t}{\mathcal{E}_i} = \cfrac {2~\cfrac{n_1}{\mu_1}~\cos\theta_i} {\cfrac{n_2}{\mu_2}~\cos\theta_i + \cfrac{n_1}{\mu_1}~\cos\theta_t}~. }$

Equations (7), (8), (9), and (10) are the Fresnel equations for $p$ -polarized electromagnetic waves.

If we define,

$\mu_{r1} := \cfrac{\mu_1}{\mu_0} \qquad \text{and} \qquad \mu_{r2} := \cfrac{\mu_2}{\mu_0}$

where $μ 0$ is the permeability of vacuum, then we can write equations (9) and (10) as

$\text{(11)} \qquad \cfrac{\mathcal{E}_r}{\mathcal{E}_i} = \cfrac {\cfrac{n_2}{\mu_{r2}}~\cos\theta_i - \cfrac{n_1}{\mu_{r1}}~\cos\theta_t} {\cfrac{n_2}{\mu_{r2}}~\cos\theta_i + \cfrac{n_1}{\mu_{r1}}~\cos\theta_t} ~;~~ \cfrac{\mathcal{E}_t}{\mathcal{E}_i} = \cfrac {2~\cfrac{n_1}{\mu_{r1}}~\cos\theta_i} {\cfrac{n_2}{\mu_{r2}}~\cos\theta_i + \cfrac{n_1}{\mu_{r1}}~\cos\theta_t}~.$

Note that

$\cfrac{n_2}{\mu_{r2}}~\cos\theta_i = \cfrac{n_1}{\mu_{r1}}~\cos\theta_t \qquad \implies \qquad \mathcal{E}_r = 0 ~.$

For non-magnetic materials we have $μ r 1 = μ r 2 = 1$ . Hence,

$\text{(12)} \qquad \cfrac{n_2}{n_1} = \cfrac{\cos\theta_t}{\cos\theta_i} ~.$

Also, from Snell's law

$\text{(13)} \qquad \cfrac{n_2}{n_1} = \cfrac{\sin\theta_i}{\sin\theta_t} ~.$

Combining equations (12) and (13), we get

$\cos\theta_t~\sin\theta_t - \cos\theta_i~\sin\theta_i = 0 \quad \implies \quad \sin(2~\theta_t) - \sin(2~\theta_i) = 0 \quad \implies \quad \cos(\theta_t+\theta_i)~\sin(\theta_t-\theta_i) = 0 ~.$

If $n_1 \ne n_2$ we have $\sin(\theta_t - \theta_i) \ne 0$ . Hence,

$\text{(14)} \qquad \cos(\theta_t+\theta_i) = 0 \qquad \implies \qquad { \theta_t + \theta_i = \cfrac{\pi}{2}} ~.$

This is the condition that defines Brewster's angle ( $θ i = θ B$ ). Plugging equation (14) into equation (13), we get

$\cfrac{\sin\theta_B}{\sin(\pi/2-\theta_B)} = \tan\theta_B = \cfrac{n_2}{n_1} ~.$

This relation can be used to solve for Brewster's angle for various media. At Brewster's angle, we have

$\cfrac{\mathcal{E}_r}{\mathcal{E}_i} = \cfrac {\cfrac{n_2}{\mu_{r2}}~\cos\theta_B - \cfrac{n_1}{\mu_{r1}}~\sin\theta_B} {\cfrac{n_2}{\mu_{r2}}~\cos\theta_B + \cfrac{n_1}{\mu_{r1}}~\sin\theta_B} = 0~.$

Hence, the sign of $\mathcal{E}_r/\mathcal{E}_i$ changes at the Brewster angle.

Also, note that if $n 2 = - n 1$ and $μ 2 = - μ 1$ , since $n = c_0~\sqrt{\epsilon\mu}$ we must have $ε 2 = - ε 1$ . Then, by Snell's law

$\cfrac{\sin\theta_t}{\sin\theta_i} = -1 \qquad \implies \qquad \theta_t = - \theta_i ~.$

Hence,

$\cfrac{\mathcal{E}_r}{\mathcal{E}_i} = \cfrac {\cfrac{n_1}{\mu_{r1}}~\cos\theta_i - \cfrac{n_1}{\mu_{r1}}~\cos\theta_i} {\cfrac{n_1}{\mu_{r1}}~\cos\theta_i + \cfrac{n_1}{\mu_{r1}}~\cos\theta_i} = 0 \qquad\text{and}\qquad \cfrac{\mathcal{E}_t}{\mathcal{E}_i} = \cfrac {2~\cfrac{n_1}{\mu_{r1}}~\cos\theta_i} {\cfrac{n_1}{\mu_{r1}}~\cos\theta_i + \cfrac{n_1}{\mu_{r1}}~\cos\theta_i} = 1~.$

So the radiation is transmitted at the angle $θ t = - θ i$ and none is reflected.

More can be said about the matter. In fact, an interface separating media with $ε 2 = - ε 1$ and $μ 2 = - μ 1$ "behaves like a mirror". Consider the interface in Figure 5. Suppose that on the left side of the mirror, $\mathbf{E}$ and $\mathbf{H}$ solve Maxwell's equations

$\boldsymbol{\nabla} \times \mathbf{E} + i\omega~\mu_1~\mathbf{H} = \boldsymbol{0} ~;~~ \boldsymbol{\nabla} \times \mathbf{H} - i\omega~\epsilon_1~\mathbf{E} = \boldsymbol{0} ~.$

Let the solution be of the form

$\mathbf{E}(\mathbf{x}) = [E_1(\mathbf{x}), E_2(\mathbf{x}), E_3(\mathbf{x})] ~~\text{and}~~ \mathbf{H}(\mathbf{x}) = [H_1(\mathbf{x}), H_2(\mathbf{x}), H_3(\mathbf{x})] ~.$

Suppose that the right hand side of the interface has reflected fields, i.e.,

$\begin{align} \mathbf{E}(\mathbf{x}) & = [-E_1(-x_1,x_2,x_3), E_2(-x_1,x_2,x_3), E_3(-x_1,x_2,x_3)] ~~\text{and}~~ \\ \mathbf{H}(\mathbf{x}) & = [-H_1(-x_1,x_2,x_3), H_2(-x_1,x_2,x_3), H_3(-x_1,x_2,x_3)] ~. \end{align}$

Figure 5. Reflection at an interface due to negative

ε

and

μ

.

Also, on the right hand side, let

$\boldsymbol{\nabla} \times \mathbf{E} = [F_1(\mathbf{x}), F_2(\mathbf{x}), F_3(\mathbf{x})] ~.$

Then, to the right of the interface, we have

$\begin{align} \boldsymbol{\nabla} \times \mathbf{E} = & \left\{ \left[\frac{\partial }{\partial x_2}[E_3(-x_1,x_2,x_3)] - \frac{\partial }{\partial x_3}[E_2(-x_1,x_2,x_3)]\right],\right. \\ & \left[-\frac{\partial }{\partial x_3}[E_1(-x_1,x_2,x_3)] + \frac{\partial }{\partial x_1}[E_3(-x_1,x_2,x_3)]\right], \\ & \left. \left[-\frac{\partial }{\partial x_1}[E_2(-x_1,x_2,x_3)] + \frac{\partial }{\partial x_2}[E_1(-x_1,x_2,x_3)]\right]\right\} \end{align}$

or,

$\boldsymbol{\nabla} \times \mathbf{E} = [F_1(-x_1,x_2,x_3), -F_2(-x_1,x_2,x_3), -F_3(x_1,x_2,x_3)]~.$

[edit] Polarized wave with the $\mathbf{E}_i$ perpendicular to the plane of incidence

For a plane polarized wave with the $\mathbf{E}_i$ vector perpendicular to the plane of incidence, we have

$\begin{align} \mathbf{E}_{0i} & = \mathcal{E}_i~\mathbf{e}_2 \\ \mathbf{E}_{0r} & = \mathcal{E}_r~\mathbf{e}_2 \\ \mathbf{E}_{0t} & = \mathcal{E}_t~\mathbf{e}_2 ~. \end{align}$

Therefore,

$\begin{align} \mathbf{H}_{0i} & = \cfrac{\kappa_1}{\omega\mu_1}~(\mathcal{E}_i~\cos\theta_i~\mathbf{e}_1 + \mathcal{E}_i~\sin\theta_i~\mathbf{e}_3) \\ \mathbf{H}_{0r} & = \cfrac{\kappa_1}{\omega\mu_1}~(-\mathcal{E}_r~\cos\theta_r~\mathbf{e}_1 + \mathcal{E}_r~\sin\theta_r~\mathbf{e}_3) \\ \mathbf{H}_{0t} & = \cfrac{\kappa_2}{\omega\mu_2}~(\mathcal{E}_t~\cos\theta_t~\mathbf{e}_1 + \mathcal{E}_t~\sin\theta_t~\mathbf{e}_3) ~. \end{align}$

Continuity of tangential components of $\mathbf{E}$ at the interface gives

$\text{(15)} \qquad { \mathcal{E}_i + \mathcal{E}_r = \mathcal{E}_t ~. }$

The tangential components of $\mathbf{H}_0$ at the interface are given by

$\begin{align} \mathbf{H}_{0i}\times\mathbf{e}_3 & = -\cfrac{\kappa_1}{\omega\mu_1}~ \mathcal{E}_i~\cos\theta_i~\mathbf{e}_2 \\ \mathbf{H}_{0r}\times\mathbf{e}_3 & = \cfrac{\kappa_1}{\omega\mu_1}~ \mathcal{E}_r~\cos\theta_r~\mathbf{e}_2 \\ \mathbf{H}_{0t}\times\mathbf{e}_3 & = -\cfrac{\kappa_2}{\omega\mu_2}~ \mathcal{E}_t~\cos\theta_t~\mathbf{e}_2 ~. \end{align}$

From continuity at the interface and using the arbitrariness of $x 1$ , we get (from the above equations with $θ i = θ r$ )

$\cfrac{\kappa_1}{\omega\mu_1}~(\mathcal{E}_i - \mathcal{E}_r)~\cos\theta_i = \cfrac{\kappa_2}{\omega\mu_1}~\mathcal{E}_t~\cos\theta_t ~.$

Using the relation $\kappa = n~\omega/c_0$ , we get

$\text{(16)} \qquad { \cfrac{n_1}{\mu_1}~(\mathcal{E}_i - \mathcal{E}_r)~\cos\theta_i = \cfrac{n_2}{\mu_2}~\mathcal{E}_t~\cos\theta_t ~. }$

From equations (15) and (16), we get

$\text{(17)} \qquad { \cfrac{\mathcal{E}_r}{\mathcal{E}_i} = \cfrac {\cfrac{n_1}{\mu_1}~\cos\theta_i - \cfrac{n_2}{\mu_2}~\cos\theta_t} {\cfrac{n_1}{\mu_1}~\cos\theta_i + \cfrac{n_2}{\mu_2}~\cos\theta_t} }$

and

$\text{(18)} \qquad { \cfrac{\mathcal{E}_t}{\mathcal{E}_i} = \cfrac {2~\cfrac{n_1}{\mu_1}~\cos\theta_i} {\cfrac{n_1}{\mu_1}~\cos\theta_i + \cfrac{n_2}{\mu_2}~\cos\theta_t} ~. }$

Equations (15), (16), (17), and (18) are the Fresnel equations a wave polarized with the $\mathbf{E}_i$ vector perpendicular to the plane of incidence. We may also write the last two equations as

$\text{(19)} \qquad \cfrac{\mathcal{E}_r}{\mathcal{E}_i} = \cfrac {\cfrac{n_1}{\mu_{r1}}~\cos\theta_i - \cfrac{n_2}{\mu_{r2}}~\cos\theta_t} {\cfrac{n_1}{\mu_{r1}}~\cos\theta_i + \cfrac{n_2}{\mu_{r2}}~\cos\theta_t} ~;~~ \cfrac{\mathcal{E}_t}{\mathcal{E}_i} = \cfrac {2~\cfrac{n_1}{\mu_{r1}}~\cos\theta_i} {\cfrac{n_1}{\mu_{r1}}~\cos\theta_i + \cfrac{n_2}{\mu_{r2}}~\cos\theta_t} ~.$

From the above equations, there is no reflected wave only if

$\tan\theta_i = \tan\theta_t~\cfrac{\mu_1}{\mu_2} ~.$

This is only possible if there is no interface. Therefore, in the presence of a interface, there is always a reflected wave for this situation.

[edit] Footnotes

↑ The above relation for the permittivity tensor can be obtained as follows. Recall that

$\mathbf{\tilde{D}}(\mathbf{x},t) = \mathbf{D}(\mathbf{x},t) + \int_{-\infty}^t \mathbf{J}_f(\mathbf{x},\tau)~\text{d}\tau ~.$

Differentiating the above relation with respect to time, we get

$\frac{\partial \mathbf{\tilde{D}}}{\partial t} = \frac{\partial \mathbf{D}}{\partial t} + \mathbf{J}_f(\mathbf{x},t) ~.$

Assuming harmonic solutions of the form

$\mathbf{\tilde{D}}(\mathbf{x},t) = \widehat{\mathbf{\tilde{D}}}(\mathbf{x})~\exp(-i\omega t) ~;~~ \mathbf{D}(\mathbf{x},t) = \widehat{\mathbf{D}}(\mathbf{x})~\exp(-i\omega t) ~;~~ \mathbf{J}_f(\mathbf{x},t) = \widehat{\mathbf{J}}_f(\mathbf{x})~\exp(-i\omega t) ~;~~ \mathbf{E}(\mathbf{x},t) = \widehat{\mathbf{E}}(\mathbf{x})~\exp(-i\omega t)$

and plugging into the differential equation above, we get

$(-i\omega)\widehat{\mathbf{\tilde{D}}}(\mathbf{x}) = (-i\omega)\widehat{\mathbf{D}}(\mathbf{x}) + \widehat{\mathbf{J}}_f(\mathbf{x}) ~.$

Now, the free current density $\mathbf{J}_f$ and the electric displacement $\mathbf{D}$ are related to the electric field $\mathbf{E}$ by

$\text{(1)} \qquad \mathbf{J}_f = \boldsymbol{\sigma}\cdot\mathbf{E} ~;~~ \mathbf{D} = \boldsymbol{\epsilon}_0\cdot\mathbf{E} ~.$

Therefore,

$\widehat{\mathbf{\tilde{D}}}(\mathbf{x}) = \boldsymbol{\epsilon}_0\cdot\widehat{\mathbf{E}}(\mathbf{x}) + \cfrac{i}{\omega}~\boldsymbol{\sigma}\cdot\widehat{\mathbf{E}}(\mathbf{x}) = \boldsymbol{\epsilon}\cdot\widehat{\mathbf{E}}(\mathbf{x})$

where

$\boldsymbol{\epsilon} = \boldsymbol{\epsilon}_0 + \cfrac{i}{\omega}~\boldsymbol{\sigma} ~.$