Waves in composites and metamaterials/Fresnel equations

The content of these notes is based on the lectures by Prof. Graeme W. Milton (University of Utah) given in a course on metamaterials in Spring 2007.

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

\nabla \times 𝐄 = i ω μ (𝐱, ω) \cdot 𝐇 (𝐱); \nabla \times 𝐇 = - i ω ϵ (𝐱, ω) \cdot 𝐄 (𝐱) .

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

ϵ \approx ϵ_{0} + \frac{i}{ω} σ

where $ϵ_{0}$ is the real part of the permittivity tensor and $σ$ 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.

File:CheckerBoardMaterial.jpg

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

ϵ_{eff} = \sqrt{ϵ_{1} ϵ_{2}} = \sqrt{1 + \frac{i σ}{ω}} \approx \frac{\sqrt{i} \sqrt{σ}}{\sqrt{ω}} (for small ω) \neq a + i b for any a, b .

Plane waves

Let us assume that the material is isotropic. Then,

𝑹^{T} \cdot ϵ \cdot 𝑹 = ϵ and 𝑹^{T} \cdot μ \cdot 𝑹 = μ \forall rotations 𝑹^{T} \cdot 𝑹 = 𝑹 \cdot 𝑹^{T} = 1 .

Therefore, we can write

ϵ = ϵ 1 and μ = μ 1 .

The Maxwell equations then take the form

(2) \nabla \times 𝐄 = i ω μ (𝐱, ω) 𝐇 (𝐱); \nabla \times 𝐇 = - i ω ϵ (𝐱, ω) 𝐄 (𝐱) .

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

\nabla \times (\nabla \times 𝐄) = i ω μ \nabla \times 𝐇 = ω^{2} ϵ μ 𝐄 (𝐱); \nabla \times (\nabla \times 𝐇) = - i ω ϵ \nabla \times 𝐄 = ω^{2} ϵ μ 𝐇 (𝐱) .

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

(3) \nabla (\nabla \cdot 𝐄) - \nabla^{2} 𝐄 = ω^{2} ϵ μ 𝐄 (𝐱); \nabla (\nabla \cdot 𝐇) - \nabla^{2} 𝐇 = ω^{2} ϵ μ 𝐇 (𝐱) .

Since

\nabla \cdot 𝐁 = 0; \nabla \cdot 𝐃 = 0; 𝐇 (𝐱) = μ^{- 1} (ω) 𝐁 (𝐱); 𝐃 (𝐱) = ϵ (ω) 𝐄 (𝐱)

we have

\nabla \cdot 𝐇 = μ^{- 1} \nabla \cdot 𝐁 = 0 and \nabla \cdot 𝐄 = ϵ^{- 1} \nabla \cdot 𝐃 = 0 .

Therefore, from equation (3), we have

\nabla^{2} 𝐄 + ω^{2} ϵ μ 𝐄 (𝐱) = 0; \nabla^{2} 𝐇 + ω^{2} ϵ μ 𝐇 (𝐱) = 0 .

We can also write the above equations in the form

(4) \nabla^{2} 𝐄 + κ^{2} 𝐄 (𝐱) = 0; \nabla^{2} 𝐇 + κ^{2} 𝐇 (𝐱) = 0 where κ^{2} = \frac{ω^{2}}{c^{2}} and c^{2} = \frac{1}{ϵ μ}

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).

File:MuEpsPlane.jpg

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

(5) 𝐄 (𝐱) = 𝐄_{0} e^{i (𝐤 \cdot 𝐱)}

where $| 𝐤 | = 1 / (2 π λ)$ and $λ$ is the wavelength. Then, using the first of equations (2) we have

(6) 𝐇 (𝐱) = - \frac{i}{ω μ} \nabla \times 𝐄 = \frac{1}{ω μ} 𝐤 \times 𝐄_{0} e^{i (𝐤 \cdot 𝐱)} = 𝐇_{0} e^{i (𝐤 \cdot 𝐱)}

where $𝐇_{0} = 1 / (ω μ) 𝐤 \times 𝐄_{0}$ .

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

\nabla \cdot 𝐄 = \frac{\partial}{\partial x_{m}} [E_{0 m} e^{i (k_{l} x_{l})}] = i k_{l} \frac{\partial x_{l}}{\partial x_{m}} E_{0 m} e^{i (k_{l} x_{l})} = i k_{m} E_{0 m} e^{i (k_{l} x_{l})} = i (𝐤 \cdot 𝐄_{0}) e^{i (𝐤 \cdot 𝐱)} = 0 .

Hence,

𝐤 \cdot 𝐄_{0} = 0 .

Similarly, since $\nabla \cdot 𝐇 = 0$ , we have

\nabla \cdot 𝐇 = \frac{1}{ω μ} [\frac{\partial}{\partial x_{m}} (ℰ_{m p q} k_{p} E_{0 q} e^{i (k_{l} x_{l})})] = \frac{i}{ω μ} [ℰ_{m p q} k_{p} E_{0 q} k_{m} e^{i (k_{l} x_{l})}] = \frac{i}{ω μ} 𝐤 \cdot (𝐤 \times 𝐄_{0}) e^{i (𝐤 \cdot 𝐱)} .

Hence,

𝐤 \cdot 𝐇_{0} = 0 .

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

\begin{matrix} {[\nabla^{2} 𝐄 + κ^{2} 𝐄 (𝐱)]}_{n} & = \frac{\partial}{\partial x_{m}} [\frac{\partial}{\partial x_{m}} (E_{0 n} e^{i k_{l} x_{l}})] + κ^{2} E_{0 n} e^{i k_{l} x_{l}} \\ = \frac{\partial}{\partial x_{m}} [E_{0 n} (i k_{l} \frac{\partial x_{l}}{\partial x_{m}}) e^{i k_{l} x_{l}}] + κ^{2} E_{0 n} e^{i k_{l} x_{l}} \\ = i E_{0 n} k_{m} \frac{\partial}{\partial x_{m}} (e^{i k_{l} x_{l}}) + κ^{2} E_{0 n} e^{i k_{l} x_{l}} \\ = i E_{0 n} k_{m} (i k_{l} \frac{\partial x_{l}}{\partial x_{m}}) e^{i k_{l} x_{l}} + κ^{2} E_{0 n} e^{i k_{l} x_{l}} \\ = - E_{0 n} k_{m} k_{m} e^{i k_{l} x_{l}} + κ^{2} E_{0 n} e^{i k_{l} x_{l}} . \end{matrix}

Reverting back to Gibbs notation, we get

- (𝐤 \cdot 𝐤) 𝐄_{0} e^{i (𝐤 \cdot 𝐱)} + κ^{2} 𝐄_{0} e^{i (𝐤 \cdot 𝐱)} = 0 .

Therefore,

𝐤 \cdot 𝐤 = κ^{2} .

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

\begin{matrix} {[\nabla^{2} 𝐇 + κ^{2} 𝐇 (𝐱)]}_{n} & = \frac{1}{ω μ} {\frac{\partial}{\partial x_{m}} [\frac{\partial}{\partial x_{m}} (ℰ_{n p q} k_{p} E_{0 q} e^{i k_{l} x_{l}})] + κ^{2} ℰ_{n p q} k_{p} E_{0 q} e^{i k_{l} x_{l}}} \\ = \frac{1}{ω μ} {\frac{\partial}{\partial x_{m}} [ℰ_{n p q} k_{p} E_{0 q} (i k_{l} \frac{\partial x_{l}}{\partial x_{m}}) e^{i k_{l} x_{l}}] + κ^{2} ℰ_{n p q} k_{p} E_{0 q} e^{i k_{l} x_{l}}} \\ = \frac{1}{ω μ} {i ℰ_{n p q} k_{p} k_{m} E_{0 q} [i k_{l} \frac{\partial x_{l}}{\partial x_{m}}] e^{i k_{l} x_{l}} + κ^{2} ℰ_{n p q} k_{p} E_{0 q} e^{i k_{l} x_{l}}} \\ = \frac{1}{ω μ} {- ℰ_{n p q} k_{p} k_{m} k_{m} E_{0 q} e^{i k_{l} x_{l}} + κ^{2} ℰ_{n p q} k_{p} E_{0 q} e^{i k_{l} x_{l}}} \end{matrix}

In Gibbs notation, we then have

\nabla^{2} 𝐇 + κ^{2} 𝐇 (𝐱) = \frac{1}{ω μ} (- 𝐤 \cdot 𝐤 + κ^{2}) 𝐤 \times 𝐄_{0} e^{i (𝐤 \cdot 𝐱)} = 0 .

Therefore, once again, we get

𝐤 \cdot 𝐤 = κ^{2} .

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.

File:ReflectInt.jpg

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{matrix} 𝐄_{i} & = 𝐄_{0 i} e^{i (𝐤_{i} \cdot 𝐱)} \\ 𝐄_{r} & = 𝐄_{0 r} e^{i (𝐤_{r} \cdot 𝐱)} \\ 𝐄_{t} & = 𝐄_{0 t} e^{i (𝐤_{t} \cdot 𝐱)} \end{matrix}

where $𝐤_{i}, 𝐤_{r}, 𝐤_{t}$ are the wave vectors.

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

𝐤_{i} \cdot 𝐱 = 𝐤_{r} \cdot 𝐱 = 𝐤_{t} \cdot 𝐱 \forall 𝐱 on the interface .

This means that the tangential components of $𝐤_{i}, 𝐤_{r}, 𝐤_{t}$ must be equal at the interface. Therefore,

| 𝐤_{i} | \sin θ_{i} = | 𝐤_{r} | \sin θ_{r} = | 𝐤_{t} | \sin θ_{t} .

Now,

| 𝐤_{i} | = κ_{i} = \frac{ω}{c_{1}}; | 𝐤_{r} | = κ_{r} = \frac{ω}{c_{1}}; | 𝐤_{t} | = κ_{t} = \frac{ω}{c_{2}}

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

\frac{ω}{c_{1}} \sin θ_{i} = \frac{ω}{c_{1}} \sin θ_{r} = \frac{ω}{c_{2}} \sin θ_{t} .

This implies that

θ_{i} = θ_{r} and \frac{\sin θ_{i}}{\sin θ_{t}} = \frac{c_{1}}{c_{2}} .

The refractive index is defined as

n : = \frac{c_{0}}{c}

where $c_{0}$ is the phase velocity is vacuum. Therefore, we get

\frac{\sin θ_{i}}{\sin θ_{t}} = \frac{n_{2}}{n_{1}} (Snells Law) .

Polarized wave with the 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 $𝐄_{i}$ vector polarized parallel to the plane of incidence. This is also called the TM (transverse magnetic) case.

File:PPolarized.jpg

Figure 4. Infinite wave polarized with the

𝐄_{i}

-vector parallel to the plane of incidence.

Let us define

κ_{1} : = | 𝐤_{i} | = | 𝐤_{r} | and κ_{2} : = | 𝐤_{t} | .

Recall that,

\begin{matrix} 𝐄_{i} (𝐱) & = 𝐄_{0 i} e^{i (𝐤_{i} \cdot 𝐱)} \\ 𝐄_{r} (𝐱) & = 𝐄_{0 r} e^{i (𝐤_{r} \cdot 𝐱)} \\ 𝐄_{t} (𝐱) & = 𝐄_{0 t} e^{i (𝐤_{t} \cdot 𝐱)} \end{matrix}

Let us choose an orthonormal basis ( $𝐞_{1}, 𝐞_{2}, 𝐞_{3}$ ) such that the $𝐞_{1}$ vector lies on the interface and is parallel the plane of incidence. The $𝐞_{2}$ vector lies on the plane of incidence and the $𝐞_{3}$ vector is normal to the interface. Then the vectors $𝐤_{i}$ , $𝐤_{r}$ , and $𝐤_{t}$ may be expressed in this basis as

\begin{matrix} 𝐤_{i} & = κ_{1} \sin θ_{i} 𝐞_{1} - κ_{1} \cos θ_{i} 𝐞_{3} \\ 𝐤_{r} & = κ_{1} \sin θ_{r} 𝐞_{1} + κ_{1} \cos θ_{r} 𝐞_{3} \\ 𝐤_{t} & = κ_{2} \sin θ_{t} 𝐞_{1} - κ_{2} \cos θ_{t} 𝐞_{3} . \end{matrix}

Similarly, defining

ℰ_{i} : = | 𝐄_{0 i} |; ℰ_{r} : = | 𝐄_{0 r} |; ℰ_{t} : = | 𝐄_{0 t} |

we get

\begin{matrix} 𝐄_{0 i} & = ℰ_{i} \cos θ_{i} 𝐞_{1} + ℰ_{i} \sin θ_{i} 𝐞_{3} \\ 𝐄_{0 r} & = - ℰ_{r} \cos θ_{r} 𝐞_{1} + ℰ_{r} \sin θ_{r} 𝐞_{3} \\ 𝐄_{0 t} & = ℰ_{t} \cos θ_{t} 𝐞_{1} + ℰ_{t} \sin θ_{t} 𝐞_{3} . \end{matrix}

Using the definition

𝐇_{0} : = \frac{1}{ω μ} 𝐤 \times 𝐄_{0}

we then get

\begin{matrix} 𝐇_{0 i} & = \frac{κ_{1}}{ω μ_{1}} ℰ_{i} 𝐞_{2} ⟹ ℋ_{i} : = | 𝐇_{0 i} | = \frac{κ_{1} ℰ_{i}}{ω μ_{1}} \\ 𝐇_{0 r} & = - \frac{κ_{1}}{ω μ_{1}} ℰ_{r} 𝐞_{2} ⟹ ℋ_{r} : = | 𝐇_{0 r} | = \frac{κ_{1} ℰ_{r}}{ω μ_{1}} \\ 𝐇_{0 t} & = \frac{κ_{2}}{ω μ_{2}} ℰ_{t} 𝐞_{2} ⟹ ℋ_{t} : = | 𝐇_{0 t} | = \frac{κ_{2} ℰ_{t}}{ω μ_{2}} . \end{matrix}

Hence, with the vector $𝐱$ expressed as $𝐱 = x_{1} 𝐞_{1} + x_{2} 𝐞_{2} + x_{3} 𝐞_{3}$ , we get

\begin{matrix} 𝐄_{i} (𝐱) & = (ℰ_{i} \cos θ_{i} 𝐞_{1} + ℰ_{i} \sin θ_{i} 𝐞_{3}) e^{i [κ_{1} (x_{1} \sin θ_{i} - x_{3} \cos θ_{i})]} \\ 𝐄_{r} (𝐱) & = (- ℰ_{r} \cos θ_{r} 𝐞_{1} + ℰ_{r} \sin θ_{r} 𝐞_{3}) e^{i [κ_{1} (x_{1} \sin θ_{r} + x_{3} \cos θ_{r})]} \\ 𝐄_{t} (𝐱) & = (ℰ_{t} \cos θ_{t} 𝐞_{1} + ℰ_{t} \sin θ_{t} 𝐞_{3}) e^{i [κ_{2} (x_{1} \sin θ_{t} - x_{3} \cos θ_{t})]} . \end{matrix}

Similarly,

\begin{matrix} 𝐇_{i} (𝐱) & = - ℋ_{i} 𝐞_{2} e^{i [κ_{1} (x_{1} \sin θ_{i} - x_{3} \cos θ_{i})]} \\ 𝐇_{r} (𝐱) & = - ℋ_{r} 𝐞_{2} e^{i [κ_{1} (x_{1} \sin θ_{r} + x_{3} \cos θ_{r})]} \\ 𝐇_{t} (𝐱) & = - ℋ_{t} 𝐞_{2} e^{i [κ_{2} (x_{1} \sin θ_{t} - x_{3} \cos θ_{t})]} . \end{matrix}

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

ℋ_{i} + ℋ_{r} = ℋ_{t} .

In terms of the electric field, we then have

\frac{κ_{1}}{ω μ_{1}} ℰ_{i} + \frac{κ_{1}}{ω μ_{1}} ℰ_{r} = \frac{κ_{2}}{ω μ_{2}} ℰ_{t} .

Recall that the refractive index is given by $n = c_{0} / c = c_{0} κ / ω$ . Therefore, we can write the above equation as

(7) \frac{n_{1}}{μ_{1}} (ℰ_{i} + ℰ_{r}) = \frac{n_{2}}{μ_{2}} ℰ_{t} .

The tangential components of the $𝐄$ vectors at the interface are given by $𝐄 \times 𝐞_{3}$ . Therefore, the tangential components of the $𝐄_{0}$ vectors at the interface are

\begin{matrix} 𝐄_{0 i} \times 𝐞_{3} & = - ℰ_{i} \cos θ_{i} 𝐞_{2} \\ 𝐄_{0 r} \times 𝐞_{3} & = ℰ_{r} \cos θ_{r} 𝐞_{2} \\ 𝐄_{0 t} \times 𝐞_{3} & = - ℰ_{t} \cos θ_{t} 𝐞_{2} . \end{matrix}

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

ℰ_{i} \cos θ_{i} - ℰ_{r} \cos θ_{r} = ℰ_{t} \cos θ_{t} .

Since $θ_{i} = θ_{r}$ , we have

(8) (ℰ_{i} - ℰ_{r}) \cos θ_{i} = ℰ_{t} \cos θ_{t} .

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

(9) \frac{ℰ_{r}}{ℰ_{i}} = \frac{\frac{n_{2}}{μ_{2}} \cos θ_{i} - \frac{n_{1}}{μ_{1}} \cos θ_{t}}{\frac{n_{2}}{μ_{2}} \cos θ_{i} + \frac{n_{1}}{μ_{1}} \cos θ_{t}}

and

(10) \frac{ℰ_{t}}{ℰ_{i}} = \frac{2 \frac{n_{1}}{μ_{1}} \cos θ_{i}}{\frac{n_{2}}{μ_{2}} \cos θ_{i} + \frac{n_{1}}{μ_{1}} \cos θ_{t}} .

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

If we define,

μ_{r 1} : = \frac{μ_{1}}{μ_{0}} and μ_{r 2} : = \frac{μ_{2}}{μ_{0}}

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

(11) \frac{ℰ_{r}}{ℰ_{i}} = \frac{\frac{n_{2}}{μ_{r 2}} \cos θ_{i} - \frac{n_{1}}{μ_{r 1}} \cos θ_{t}}{\frac{n_{2}}{μ_{r 2}} \cos θ_{i} + \frac{n_{1}}{μ_{r 1}} \cos θ_{t}}; \frac{ℰ_{t}}{ℰ_{i}} = \frac{2 \frac{n_{1}}{μ_{r 1}} \cos θ_{i}}{\frac{n_{2}}{μ_{r 2}} \cos θ_{i} + \frac{n_{1}}{μ_{r 1}} \cos θ_{t}} .

Note that

\frac{n_{2}}{μ_{r 2}} \cos θ_{i} = \frac{n_{1}}{μ_{r 1}} \cos θ_{t} ⟹ ℰ_{r} = 0 .

For non-magnetic materials we have $μ_{r 1} = μ_{r 2} = 1$ . Hence,

(12) \frac{n_{2}}{n_{1}} = \frac{\cos θ_{t}}{\cos θ_{i}} .

Also, from Snell's law

(13) \frac{n_{2}}{n_{1}} = \frac{\sin θ_{i}}{\sin θ_{t}} .

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

\cos θ_{t} \sin θ_{t} - \cos θ_{i} \sin θ_{i} = 0 ⟹ \sin (2 θ_{t}) - \sin (2 θ_{i}) = 0 ⟹ \cos (θ_{t} + θ_{i}) \sin (θ_{t} - θ_{i}) = 0 .

If $n_{1} \neq n_{2}$ we have $\sin (θ_{t} - θ_{i}) \neq 0$ . Hence,

(14) \cos (θ_{t} + θ_{i}) = 0 ⟹ θ_{t} + θ_{i} = \frac{π}{2} .

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

\frac{\sin θ_{B}}{\sin (π / 2 - θ_{B})} = \tan θ_{B} = \frac{n_{2}}{n_{1}} .

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

\frac{ℰ_{r}}{ℰ_{i}} = \frac{\frac{n_{2}}{μ_{r 2}} \cos θ_{B} - \frac{n_{1}}{μ_{r 1}} \sin θ_{B}}{\frac{n_{2}}{μ_{r 2}} \cos θ_{B} + \frac{n_{1}}{μ_{r 1}} \sin θ_{B}} = 0 .

Hence, the sign of $ℰ_{r} / ℰ_{i}$ changes at the Brewster angle.

Also, note that if $n_{2} = - n_{1}$ and $μ_{2} = - μ_{1}$ , since $n = c_{0} \sqrt{ϵ μ}$ we must have $ϵ_{2} = - ϵ_{1}$ . Then, by Snell's law

\frac{\sin θ_{t}}{\sin θ_{i}} = - 1 ⟹ θ_{t} = - θ_{i} .

Hence,

\frac{ℰ_{r}}{ℰ_{i}} = \frac{\frac{n_{1}}{μ_{r 1}} \cos θ_{i} - \frac{n_{1}}{μ_{r 1}} \cos θ_{i}}{\frac{n_{1}}{μ_{r 1}} \cos θ_{i} + \frac{n_{1}}{μ_{r 1}} \cos θ_{i}} = 0 and \frac{ℰ_{t}}{ℰ_{i}} = \frac{2 \frac{n_{1}}{μ_{r 1}} \cos θ_{i}}{\frac{n_{1}}{μ_{r 1}} \cos θ_{i} + \frac{n_{1}}{μ_{r 1}} \cos θ_{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, $𝐄$ and $𝐇$ solve Maxwell's equations

\nabla \times 𝐄 + i ω μ_{1} 𝐇 = 0; \nabla \times 𝐇 - i ω ϵ_{1} 𝐄 = 0 .

Let the solution be of the form

𝐄 (𝐱) = [E_{1} (𝐱), E_{2} (𝐱), E_{3} (𝐱)] and 𝐇 (𝐱) = [H_{1} (𝐱), H_{2} (𝐱), H_{3} (𝐱)] .

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

\begin{matrix} 𝐄 (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})] and \\ 𝐇 (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{matrix}

File:MirrorNegativeRefract.jpg

Figure 5. Reflection at an interface due to negative

ϵ

and

μ

.

Also, on the right hand side, let

\nabla \times 𝐄 = [F_{1} (𝐱), F_{2} (𝐱), F_{3} (𝐱)] .

Then, to the right of the interface, we have

\begin{matrix} \nabla \times 𝐄 = & {[\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})]], \\ [- \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})]], \\ [- \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})]]} \end{matrix}

or,

\nabla \times 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})] .

Polarized wave with the E_i perpendicular to the plane of incidence

For a plane polarized wave with the $𝐄_{i}$ vector perpendicular to the plane of incidence, we have

\begin{matrix} 𝐄_{0 i} & = ℰ_{i} 𝐞_{2} \\ 𝐄_{0 r} & = ℰ_{r} 𝐞_{2} \\ 𝐄_{0 t} & = ℰ_{t} 𝐞_{2} . \end{matrix}

Therefore,

\begin{matrix} 𝐇_{0 i} & = \frac{κ_{1}}{ω μ_{1}} (ℰ_{i} \cos θ_{i} 𝐞_{1} + ℰ_{i} \sin θ_{i} 𝐞_{3}) \\ 𝐇_{0 r} & = \frac{κ_{1}}{ω μ_{1}} (- ℰ_{r} \cos θ_{r} 𝐞_{1} + ℰ_{r} \sin θ_{r} 𝐞_{3}) \\ 𝐇_{0 t} & = \frac{κ_{2}}{ω μ_{2}} (ℰ_{t} \cos θ_{t} 𝐞_{1} + ℰ_{t} \sin θ_{t} 𝐞_{3}) . \end{matrix}

Continuity of tangential components of $𝐄$ at the interface gives

(15) ℰ_{i} + ℰ_{r} = ℰ_{t} .

The tangential components of $𝐇_{0}$ at the interface are given by

\begin{matrix} 𝐇_{0 i} \times 𝐞_{3} & = - \frac{κ_{1}}{ω μ_{1}} ℰ_{i} \cos θ_{i} 𝐞_{2} \\ 𝐇_{0 r} \times 𝐞_{3} & = \frac{κ_{1}}{ω μ_{1}} ℰ_{r} \cos θ_{r} 𝐞_{2} \\ 𝐇_{0 t} \times 𝐞_{3} & = - \frac{κ_{2}}{ω μ_{2}} ℰ_{t} \cos θ_{t} 𝐞_{2} . \end{matrix}

From continuity at the interface and using the arbitrariness of $x_{1}$ , we get (from the above equations with $θ_{i} = θ_{r}$ )

\frac{κ_{1}}{ω μ_{1}} (ℰ_{i} - ℰ_{r}) \cos θ_{i} = \frac{κ_{2}}{ω μ_{1}} ℰ_{t} \cos θ_{t} .

Using the relation $κ = n ω / c_{0}$ , we get

(16) \frac{n_{1}}{μ_{1}} (ℰ_{i} - ℰ_{r}) \cos θ_{i} = \frac{n_{2}}{μ_{2}} ℰ_{t} \cos θ_{t} .

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

(17) \frac{ℰ_{r}}{ℰ_{i}} = \frac{\frac{n_{1}}{μ_{1}} \cos θ_{i} - \frac{n_{2}}{μ_{2}} \cos θ_{t}}{\frac{n_{1}}{μ_{1}} \cos θ_{i} + \frac{n_{2}}{μ_{2}} \cos θ_{t}}

and

(18) \frac{ℰ_{t}}{ℰ_{i}} = \frac{2 \frac{n_{1}}{μ_{1}} \cos θ_{i}}{\frac{n_{1}}{μ_{1}} \cos θ_{i} + \frac{n_{2}}{μ_{2}} \cos θ_{t}} .

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

(19) \frac{ℰ_{r}}{ℰ_{i}} = \frac{\frac{n_{1}}{μ_{r 1}} \cos θ_{i} - \frac{n_{2}}{μ_{r 2}} \cos θ_{t}}{\frac{n_{1}}{μ_{r 1}} \cos θ_{i} + \frac{n_{2}}{μ_{r 2}} \cos θ_{t}}; \frac{ℰ_{t}}{ℰ_{i}} = \frac{2 \frac{n_{1}}{μ_{r 1}} \cos θ_{i}}{\frac{n_{1}}{μ_{r 1}} \cos θ_{i} + \frac{n_{2}}{μ_{r 2}} \cos θ_{t}} .

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

\tan θ_{i} = \tan θ_{t} \frac{μ_{1}}{μ_{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.

Footnotes

↑ The above relation for the permittivity tensor can be obtained as follows. Recall that
$\tilde{𝐃} (𝐱, t) = 𝐃 (𝐱, t) + \int_{- \infty}^{t} 𝐉_{f} (𝐱, τ) d τ .$
Differentiating the above relation with respect to time, we get
$\frac{\partial \tilde{𝐃}}{\partial t} = \frac{\partial 𝐃}{\partial t} + 𝐉_{f} (𝐱, t) .$
Assuming harmonic solutions of the form
$\tilde{𝐃} (𝐱, t) = \hat{\tilde{𝐃}} (𝐱) \exp (- i ω t); 𝐃 (𝐱, t) = \hat{𝐃} (𝐱) \exp (- i ω t); 𝐉_{f} (𝐱, t) = {\hat{𝐉}}_{f} (𝐱) \exp (- i ω t); 𝐄 (𝐱, t) = \hat{𝐄} (𝐱) \exp (- i ω t)$
and plugging into the differential equation above, we get
$(- i ω) \hat{\tilde{𝐃}} (𝐱) = (- i ω) \hat{𝐃} (𝐱) + {\hat{𝐉}}_{f} (𝐱) .$
Now, the free current density $𝐉_{f}$ and the electric displacement $𝐃$ are related to the electric field $𝐄$ by
$(1) 𝐉_{f} = σ \cdot 𝐄; 𝐃 = ϵ_{0} \cdot 𝐄 .$
Therefore,
$\hat{\tilde{𝐃}} (𝐱) = ϵ_{0} \cdot \hat{𝐄} (𝐱) + \frac{i}{ω} σ \cdot \hat{𝐄} (𝐱) = ϵ \cdot \hat{𝐄} (𝐱)$
where
$ϵ = ϵ_{0} + \frac{i}{ω} σ .$

References

[Lorrain88]
P. Lorrain, D. R. Corson, and F. Lorrain. Electromagnetic fields and waves: including electric circuits. Freeman, New York, 1988.

[1] The above relation for the permittivity tensor can be obtained as follows. Recall that
$\tilde{𝐃} (𝐱, t) = 𝐃 (𝐱, t) + \int_{- \infty}^{t} 𝐉_{f} (𝐱, τ) d τ .$
Differentiating the above relation with respect to time, we get
$\frac{\partial \tilde{𝐃}}{\partial t} = \frac{\partial 𝐃}{\partial t} + 𝐉_{f} (𝐱, t) .$
Assuming harmonic solutions of the form
$\tilde{𝐃} (𝐱, t) = \hat{\tilde{𝐃}} (𝐱) \exp (- i ω t); 𝐃 (𝐱, t) = \hat{𝐃} (𝐱) \exp (- i ω t); 𝐉_{f} (𝐱, t) = {\hat{𝐉}}_{f} (𝐱) \exp (- i ω t); 𝐄 (𝐱, t) = \hat{𝐄} (𝐱) \exp (- i ω t)$
and plugging into the differential equation above, we get
$(- i ω) \hat{\tilde{𝐃}} (𝐱) = (- i ω) \hat{𝐃} (𝐱) + {\hat{𝐉}}_{f} (𝐱) .$
Now, the free current density $𝐉_{f}$ and the electric displacement $𝐃$ are related to the electric field $𝐄$ by
$(1) 𝐉_{f} = σ \cdot 𝐄; 𝐃 = ϵ_{0} \cdot 𝐄 .$
Therefore,
$\hat{\tilde{𝐃}} (𝐱) = ϵ_{0} \cdot \hat{𝐄} (𝐱) + \frac{i}{ω} σ \cdot \hat{𝐄} (𝐱) = ϵ \cdot \hat{𝐄} (𝐱)$
where
$ϵ = ϵ_{0} + \frac{i}{ω} σ .$

[1]

Waves in composites and metamaterials/Fresnel equations

Contents

A brief excursion into homogenization

Plane waves

Reflection at an Interface

Polarized wave with the E_i parallel to the plane of incidence

Polarized wave with the E_i perpendicular to the plane of incidence

Footnotes

References

Navigation menu

Waves in composites and metamaterials/Fresnel equations

A brief excursion into homogenization

Plane waves

Reflection at an Interface

Polarized wave with the Ei parallel to the plane of incidence

Polarized wave with the Ei perpendicular to the plane of incidence

Footnotes

References

Navigation menu

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Polarized wave with the E_i parallel to the plane of incidence

Polarized wave with the E_i perpendicular to the plane of incidence