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

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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

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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

 

Since

 

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)

 

Hence,

 

Similarly, since  , we have

 

Hence,

 

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

 

Reverting back to Gibbs notation, we get

 

Therefore,

 

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

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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,

 

Now,

 

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

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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

 

Similarly,

 

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:

 

and

 

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

 

Hence,

 

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

 

or,

 

Polarized wave with the Ei perpendicular to the plane of incidence

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

 

Therefore,

 

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

 

and

 

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

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  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
     
    Therefore,
     
    where
     

References

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