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 edit

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

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

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

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

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

  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

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