Waves in composites and metamaterials/Fresnel equations

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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[edit | edit source]

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

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

where is the real part of the permittivity tensor and 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 while the dielectric material has a permittivity of . The effective permittivity of the checkerboard is given by

Plane waves[edit | edit source]

Let us assume that the material is isotropic. Then,

Therefore, we can write

The Maxwell equations then take the form

If we assume that and do not depend upon position, i.e., and , and take the curl of equations (2), we get

Using the identity , we get


we have

Therefore, from equation (3), we have

We can also write the above equations in the form

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

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

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

where and is the wavelength. Then, using the first of equations (2) we have

where .

Since , we have (in terms of components with respect to a orthonormal Cartesian basis)


Similarly, since , we have


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

Reverting back to Gibbs notation, we get


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

In Gibbs notation, we then have

Therefore, once again, we get

Reflection at an Interface[edit | edit source]

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

where are the wave vectors.

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

This means that the tangential components of must be equal at the interface. Therefore,


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

This implies that

The refractive index is defined as

where is the phase velocity is vacuum. Therefore, we get

Polarized wave with the Ei parallel to the plane of incidence[edit | edit source]

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

Figure 4. Infinite wave polarized with the -vector parallel to the plane of incidence.

Let us define

Recall that,

Let us choose an orthonormal basis () such that the vector lies on the interface and is parallel the plane of incidence. The vector lies on the plane of incidence and the vector is normal to the interface. Then the vectors , , and may be expressed in this basis as

Similarly, defining

we get

Using the definition

we then get

Hence, with the vector expressed as , we get


At the interface, continuity requires that the tangential components of the vectors and are continuous. Clearly, from the above equations, the vectors are tangential to the interface. Also, at the interface and is arbitrary. Hence, continuity of the components of at the interface can be achieved if

In terms of the electric field, we then have

Recall that the refractive index is given by . Therefore, we can write the above equation as

The tangential components of the vectors at the interface are given by . Therefore, the tangential components of the vectors at the interface are

Using the arbitrariness of and from the continuity of the vectors at the interface, we have

Since , we have

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


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

If we define,

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

Note that

For non-magnetic materials we have . Hence,

Also, from Snell's law

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

If we have . Hence,

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

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

Hence, the sign of changes at the Brewster angle.

Also, note that if and , since we must have . Then, by Snell's law


So the radiation is transmitted at the angle and none is reflected.

More can be said about the matter. In fact, an interface separating media with and "behaves like a mirror". Consider the interface in Figure 5. Suppose that on the left side of the mirror, and solve Maxwell's equations

Let the solution be of the form

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

Figure 5. Reflection at an interface due to negative and .

Also, on the right hand side, let

Then, to the right of the interface, we have


Polarized wave with the Ei perpendicular to the plane of incidence[edit | edit source]

For a plane polarized wave with the vector perpendicular to the plane of incidence, we have


Continuity of tangential components of at the interface gives

The tangential components of at the interface are given by

From continuity at the interface and using the arbitrariness of , we get (from the above equations with )

Using the relation , we get

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


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

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

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[edit | edit source]

  1. The above relation for the permittivity tensor can be obtained as follows. Recall that
    Differentiating the above relation with respect to time, we get
    Assuming harmonic solutions of the form
    and plugging into the differential equation above, we get
    Now, the free current density and the electric displacement are related to the electric field by

References[edit | edit source]

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