text:
Waves in composites and metamaterials/Maxwell equations in media

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.

Maxwell's Equations in Media
The time-dependent Maxwell's equations in media (in the absence of any internal sources of magnetic induction) can be written as
 * $$\begin{align}

\boldsymbol{\nabla} \times \mathbf{E} & = -\frac{\partial \mathbf{B}}{\partial t} \qquad \text{(1)} \qquad \\ \boldsymbol{\nabla} \times \mathbf{H} & = \frac{\partial \mathbf{D}}{\partial t} + \mathbf{J}_f \qquad \text{(2)} \qquad \\ \boldsymbol{\nabla} \cdot \mathbf{B} & = 0 \\ \boldsymbol{\nabla} \cdot \mathbf{D} & = \rho_f \end{align}$$ where $$\mathbf{E}(\mathbf{x},t)$$ is electric field, $$\mathbf{B}(\mathbf{x},t)$$ is the magnetic induction, $$\mathbf{H}(\mathbf{x},t)$$ is the magnetic field intensity, $$\mathbf{D}(\mathbf{x},t)$$ is the electric displacement field due to the movement of bound charges, $$\mathbf{J}_f(\mathbf{x},t)$$ is the free current density, and $$\rho_f(\mathbf{x},t)$$ is the free charge density. The vector $$\mathbf{x}$$ represents the position in space and $$t$$ is the time.

We can derive the equation for the  conservation of charge by taking the divergence of equation (2) to get

\boldsymbol{\nabla} \cdot (\boldsymbol{\nabla} \times \mathbf{H}) = \frac{\partial }{\partial t}(\boldsymbol{\nabla} \cdot \mathbf{D}) + \boldsymbol{\nabla} \cdot \mathbf{J}_f \qquad \implies \qquad \frac{\partial \rho_f}{\partial t} + \boldsymbol{\nabla} \cdot \mathbf{J}_f = 0 ~. $$

The primary variables in the above equations are $$\mathbf{E}$$ and $$\mathbf{B}$$. The quantities $$\mathbf{H}$$ and $$\mathbf{D}$$ are obtained through the constitutive relations
 * $$\begin{align}

\mathbf{H} & = \boldsymbol{\mu}_0^{-1}\cdot\mathbf{B} - \mathbf{M} \\ \mathbf{D} & = \boldsymbol{\epsilon}_0\cdot\mathbf{E} + \mathbf{P} \end{align}$$ where $$\boldsymbol{\mu}_0(\mathbf{x},t)$$ is the rank 2 magnetic permeability tensor of free space, $$\boldsymbol{\epsilon}_0(\mathbf{x},t)$$ is the permittivity tensor, $$\mathbf{M}(\mathbf{x},t)$$ is the magnetization vector, and $$\mathbf{P}(\mathbf{x},t)$$ is the polarization vector. The magnetization vector $$\mathbf{M}$$ measures the net magnetic dipole moment per unit volume. This dipole is associated with electron or nuclear spins. The polarization vector $$\mathbf{P}$$ measures the net electric dipole moment per unit volume and is caused by the close proximity of two charges of opposite sign. A point electric dipole is obtained when the distance between two charges tends to zero.

Artificial Magnetic Materials (Metamaterials)
A clear definition of metamaterials does not exist yet. Some authors define metamaterials as those whose properties depend strongly on the geometry of the microstructure but appear not to depend on the properties of the constituents. This definition is not accurate because the effective properties of metamaterials do depend on the properties of the constituents as they must. Another definition is that metamaterials are those materials whose properties do not reflect everyday experience such as negative refractive indexes or negative Poisson's ratios. A more accurate definition can be based on the the fact that many of the properties of metamaterials are due to specific resonances. One such example is stained glass where the resonance of gold particles in the glass gives the glass a red tint.

The fact that artificial magnetic materials may be created from relatively non-magnetic materials was first briefly hinted by Shelkunoff and Friis ([Shelku52], pp. 584-585). The idea was developed in more detail by Pendry and coworkers [Pendry99].

In that work, split ring resonators were used to develop a magnetic material containing non-magnetic components. A schematic of the split ring resonator is shown in Figure 1.

If the magnetic field intensity $$\mathbf{H}$$ is time-dependent and the magnetization vector $$\mathbf{M}$$ is zero, then

\frac{\partial \mathbf{H}}{\partial t} \ne 0 \qquad \implies \qquad \frac{\partial \mathbf{B}}{\partial t} \ne 0 \qquad \implies \qquad \boldsymbol{\nabla} \times \mathbf{E} \ne 0 ~. $$ Therefore, there is a non-zero electric field around the loop which implies that there is a current in the split ring. Now if we place a parallel plate capacitor in the gap, charges build up in the capacitor and the current oscillates back and forth in the ring as the field $$\mathbf{H}$$ changes. The result is that the ring resonates and the net magnetic dipole moment $$\mathbf{M}$$ becomes non-zero.

It is not clear how $$\mathbf{M}$$ should be defined and whether Maxwell's equation should be modified. Avoiding these issues for the moment, we assume that

 * 1) The free current density ($$\mathbf{J}_f$$) arises only from conduction currents arising from the response of the medium and not from beams of charged particles.
 * 2) In the far distant past ($$t = -\infty$$) all field are zero.

Define
 * $$\text{(3)} \qquad

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

\frac{\partial \mathbf{\tilde{D}}}{\partial t} = \frac{\partial \mathbf{D}}{\partial t} + \mathbf{J}_f = \boldsymbol{\nabla} \times \mathbf{H} ~. $$ Taking the divergence of (3) and using the conservation of charge, we get

\boldsymbol{\nabla} \cdot \mathbf{\tilde{D}} = \boldsymbol{\nabla} \cdot \mathbf{D} + \int_{-\infty}^t \boldsymbol{\nabla} \cdot \mathbf{J}_f~\text{d}\tau = \boldsymbol{\nabla} \cdot \mathbf{D} - \int_{-\infty}^t \frac{\partial \rho_f}{\partial t}~\text{d}\tau = \boldsymbol{\nabla} \cdot \mathbf{D} - \rho_f = 0 ~. $$ Therefore, we can write Maxwell's equations in terms of $$\mathbf{\tilde{D}}$$ as
 * $$\text{(4)} \qquad

{   \boldsymbol{\nabla} \times \mathbf{E} = -\frac{\partial \mathbf{B}}{\partial t} ~; \boldsymbol{\nabla} \times \mathbf{H} = \frac{\partial \mathbf{\tilde{D}}}{\partial t} ~; \boldsymbol{\nabla} \cdot \mathbf{B} = 0 ~; \boldsymbol{\nabla} \cdot \mathbf{\tilde{D}} = 0 ~. } $$ This reduction reflects the fact that it is difficult to distinguish the free current density $$\mathbf{J}_f$$ from currents arising from the electric displacement field through $$\partial \mathbf{D}/\partial t$$.

To complete the system of equations (4), we need relations between the fields $$\mathbf{E}$$, $$\mathbf{\tilde{D}}$$, $$\mathbf{B}$$, and $$\mathbf{H}$$. Some further assumptions need to be made at this point:

 * 1) We assume that only $$\mathbf{\tilde{D}}$$ is coupled with $$\mathbf{E}$$ and that $$\mathbf{H}$$ is only coupled with $$\mathbf{B}$$. This is a good approximation for many stationary materials.  But more generally there is cross coupling between these quantities, for example, in biisotropic and bianisotropic materials.
 * 2) We assume that the relations between $$\mathbf{\tilde{D}}$$ and $$\mathbf{E}$$, and $$\mathbf{H}$$ and $$\mathbf{B}$$ are linear.
 * 3) The net magnetic dipole moment $$\mathbf{M}$$ (and hence the magnetic field $$\boldsymbol{H}$$) cannot depend on future values of $$\mathbf{B}$$. This is the principle of causality.
 * 4) The free current density $$\mathbf{J}_f$$ (and hence the electric displacement field $$\mathbf{D}$$) cannot depend on future values of $$\mathbf{E}$$.
 * 5) The materials are at rest and their properties do not depend upon time.

Therefore, using superposition, we may write
 * $$\text{(5)} \qquad

\begin{align} \mathbf{H}(\mathbf{x},t) = \int \text{d}\mathbf{x}' \int_{-\infty}^t \text{d}t'~\boldsymbol{K}_B(\mathbf{x}', \mathbf{x}, t'-t)\cdot\mathbf{B}(\mathbf{x}',t')\\ \mathbf{\tilde{D}}(\mathbf{x},t) = \int \text{d}\mathbf{x}' \int_{-\infty}^t \text{d}t'~\boldsymbol{K}_E(\mathbf{x}', \mathbf{x}, t'-t)\cdot\mathbf{E}(\mathbf{x}',t') \end{align} $$ where $$\boldsymbol{K}_B$$ and $$\boldsymbol{K}_E$$ are rank-2 tensor valued kernel functions. These kernel functions may be singular (such as delta functions) and the integrals should be interpreted in the sense of measure theory under such conditions.

We further assume that equations (5) can be approximated as being local in space (this is true for poor conductors but may fail for good conductors due to Debye screening.) This implies that the kernel functions can be chosen in such a way that the integration over space at each point evaluates to 1 and we can write
 * $$\text{(6)} \qquad

\begin{align} \mathbf{H}(\mathbf{x},t) = \int_{-\infty}^{\infty} \boldsymbol{\bar{K}}_B(\mathbf{x}, t'-t)\cdot\mathbf{B}(\mathbf{x},t')~\text{d}t'\\ \mathbf{\tilde{D}}(\mathbf{x},t) = \int_{-\infty}^{\infty} \boldsymbol{\bar{K}}_E(\mathbf{x}, t'-t)\cdot\mathbf{E}(\mathbf{x},t')~\text{d}t' ~. \end{align} $$ Note that in Fourier space the above convolutions turn into products. Also, the limits of integration have been changed to go from $$-\infty$$ to $$\infty$$ because the kernel functions have been chosen such that

\boldsymbol{\bar{K}}_B(\mathbf{x},\tau) = \boldsymbol{\bar{K}}_E(\mathbf{x},\tau) = 0 \qquad \text{if} \qquad \tau > 0 ~. $$

Next, let us assume that all the fields depend harmonically on time (we can treat more general fields by linear superposition). Then
 * $$\text{(7)} \qquad

\begin{align} \mathbf{E}(\mathbf{x},t) & = \text{Re}\{\widehat{\mathbf{E}}(\mathbf{x})~e^{-i\omega t}\} ~; \mathbf{B}(\mathbf{x},t) = \text{Re}\{\widehat{\mathbf{B}}(\mathbf{x})~e^{-i\omega t}\} \\ \mathbf{\tilde{D}}(\mathbf{x},t) & = \text{Re}\{\widehat{\mathbf{D}}(\mathbf{x})~e^{-i\omega t}\} ~; \mathbf{H}(\mathbf{x},t) = \text{Re}\{\widehat{\mathbf{H}}(\mathbf{x})~e^{-i\omega t}\} \end{align} $$ (treating $$\omega$$ as having an infinitesimally small imaginary part so that the fields are zero at $$t \rightarrow -\infty$$).

Plugging the solutions in equation (7) into equations (4), we can get new expressions for the Maxwell's equations in terms of the amplitudes of the harmonic fields. Thus, we have

\begin{align} \boldsymbol{\nabla} \times \mathbf{E} & = -\frac{\partial \mathbf{B}}{\partial t}    &\quad \implies \quad & \boldsymbol{\nabla} \times \widehat{\mathbf{E}}~e^{-i\omega t} = i\omega~\widehat{\mathbf{B}}(\mathbf{x})~e^{-i\omega t}   &\quad \implies \quad & \boldsymbol{\nabla} \times \widehat{\mathbf{E}} = i\omega~\widehat{\mathbf{B}}(\mathbf{x}) \\ \boldsymbol{\nabla} \times \mathbf{H} & = \frac{\partial \mathbf{\tilde{D}}}{\partial t}    &\quad \implies \quad & \boldsymbol{\nabla} \times \widehat{\mathbf{H}}~e^{-i\omega t} = -i\omega~\widehat{\mathbf{D}}(\mathbf{x})~e^{-i\omega t}   &\quad \implies \quad & \boldsymbol{\nabla} \times \widehat{\mathbf{H}} = -i\omega~\widehat{\mathbf{D}}(\mathbf{x}) \\ \boldsymbol{\nabla} \cdot \mathbf{B} & = 0 &\quad \implies \quad & \boldsymbol{\nabla} \cdot \widehat{\mathbf{B}}~e^{-i\omega t} = 0 &\quad \implies \quad & \boldsymbol{\nabla} \cdot \widehat{\mathbf{B}} = 0 \\ \boldsymbol{\nabla} \cdot \mathbf{\tilde{D}} & = 0 &\quad \implies \quad & \boldsymbol{\nabla} \cdot \widehat{\mathbf{D}}~e^{-i\omega t} = 0 &\quad \implies \quad & \boldsymbol{\nabla} \cdot \widehat{\mathbf{D}} = 0 \end{align} $$ or,
 * $$\text{(8)} \qquad

{ \boldsymbol{\nabla} \times \widehat{\mathbf{E}}  = i\omega~\widehat{\mathbf{B}}(\mathbf{x}) ~; \boldsymbol{\nabla} \times \widehat{\mathbf{H}} = -i\omega~\widehat{\mathbf{D}}(\mathbf{x}) ~; \boldsymbol{\nabla} \cdot \widehat{\mathbf{B}} = 0 ~; \boldsymbol{\nabla} \cdot \widehat{\mathbf{D}} = 0 ~. } $$ Similarly, plugging the equations (7) into equations (6), we get (using $$\tau = t' - t$$)

\begin{align} \widehat{\mathbf{H}}(\mathbf{x})~e^{-i\omega t} & = \int_{-\infty}^{\infty} \boldsymbol{\bar{K}}_B(\mathbf{x}, t'-t)\cdot \left[\widehat{\mathbf{B}}(\mathbf{x})~e^{-i\omega t'}\right]~\text{d}t' = \left\{ \left[\int_{-\infty}^{\infty} \boldsymbol{\bar{K}}_B(\mathbf{x}, \tau)~e^{-i\omega\tau}~\text{d}\tau\right]\cdot \widehat{\mathbf{B}}(\mathbf{x})\right\}~e^{-i\omega t} \\ \widehat{\mathbf{D}}(\mathbf{x})~e^{-i\omega t} & = \int_{-\infty}^{\infty} \boldsymbol{\bar{K}}_E(\mathbf{x}, t'-t)\cdot \left[\widehat{\mathbf{E}}(\mathbf{x})~e^{-i\omega t'}\right]~\text{d}t' = \left\{ \left[\int_{-\infty}^{\infty} \boldsymbol{\bar{K}}_E(\mathbf{x}, \tau)~e^{-i\omega\tau}~\text{d}\tau\right]\cdot \widehat{\mathbf{E}}(\mathbf{x})\right\}~e^{-i\omega t} \end{align} $$ or,
 * $$\text{(9)} \qquad

{   \widehat{\mathbf{H}}(\mathbf{x})  = [\boldsymbol{\mu}(\mathbf{x},\omega)]^{-1}\cdot\widehat{\mathbf{B}}(\mathbf{x}) ~; \widehat{\mathbf{D}}(\mathbf{x}) = \boldsymbol{\epsilon}(\mathbf{x},\omega)\cdot\widehat{\mathbf{E}}(\mathbf{x}) } $$ where
 * $$\text{(10)} \qquad

{   \left[\boldsymbol{\mu}(\mathbf{x},\omega)\right]^{-1}  = \int_{-\infty}^{\infty} \boldsymbol{\bar{K}}_B(\mathbf{x}, \tau)~e^{-i\omega\tau}~\text{d}\tau ~; \boldsymbol{\epsilon}(\mathbf{x},\omega) = \int_{-\infty}^{\infty} \boldsymbol{\bar{K}}_E(\mathbf{x}, \tau)~e^{-i\omega\tau}~\text{d}\tau ~. } $$ In general $$\boldsymbol{\mu}$$ and $$\boldsymbol{\epsilon}$$ are complex, rank-2 tensor quantities. The integrals in equations (10) converge when the imaginary part of $$\omega$$ is positive (since $$\boldsymbol{\bar{K}}_B = \boldsymbol{\bar{K}}_E = 0$$ when $$\tau > 0$$). Now, $$\exp(-i\omega\tau)$$ is an analytic function of $$\omega$$. Since a sum of analytic functions is analytic and a convergent integral of analytic functions is also analytic, the functions $$\boldsymbol{\epsilon}(\mathbf{x},\omega)$$ and $$\boldsymbol{\mu}(\mathbf{x},\omega)$$ are analytic functions of $$\omega$$ in the upper half $$\omega$$-plane, $$\text{Im}(\omega) > 0$$.

Substituting equations (9) into equations (8), and dropping the hats, we get

{ \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}) ~; \boldsymbol{\nabla} \cdot \mathbf{B} = 0 ~; \boldsymbol{\nabla} \cdot \mathbf{D} = 0 ~. } $$ These are Maxwell equations at fixed frequency.


text:
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 $$\mathbf{E}$$ 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. 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} ~. $$

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. The conducting material has a permittivity of $$\epsilon_1=1 + i\sigma/\omega$$ while the dielectric material has a permittivity of $$\epsilon_2 = 1$$. 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 ~. $$

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}) ~. $$ If we assume that $$\mu$$ and $$\epsilon$$ do not depend upon position, i.e., $$\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$$ must be real. This will be the case when $$\epsilon$$ and $$\mu$$ are both positive or both negative (see Figure 2).

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)$$ and $$\lambda$$ is the wavelength. Then, using the first of equations (2) we have
 * $$\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 ~. } $$

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. 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)} ~. $$

Polarized wave with the Ei 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.

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 ~. $$ Since $$\theta_i = \theta_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} $$ where $$\mu_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 ~. $$ For non-magnetic materials we have $$\mu_{r1} = \mu_{r2} = 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}} ~. $$ This is the condition that defines  Brewster's angle ($$\theta_i = \theta_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 $$\mu_2 = - \mu_1$$, since $$n = c_0~\sqrt{\epsilon\mu}$$ we must have $$\epsilon_2 = - \epsilon_1$$. 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~. $$ So the radiation is transmitted at the angle $$\theta_t = -\theta_i$$ and none is reflected.

More can be said about the matter. In fact, an interface separating media with $$\epsilon_2 = - \epsilon_1$$ and $$\mu_2 = -\mu_1$$ "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} $$ 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)]~. $$

Polarized wave with the Ei 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$$, we get (from the above equations with $$\theta_i = \theta_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.

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text:
Waves in composites and metamaterials/Perfect lenses and negative density materials

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.

Perfect Lenses
Recall the interface separating media with permittivities $$\epsilon_1,\epsilon_2$$ and permeabilities $$\mu_1, \mu_2$$ that "behaves like a mirror" (see Figure 1).

For the situation shown in the figure, on the left hand side (LHS) of the interface, let $$\mathbf{E}_L$$ and $$\mathbf{H}_L$$ solve Maxwell's equations

\boldsymbol{\nabla} \times \mathbf{E}_L + i\omega~\mu_1~\mathbf{H}_L = \boldsymbol{0} ~; \boldsymbol{\nabla} \times \mathbf{H}_L - i\omega~\epsilon_1~\mathbf{E}_L = \boldsymbol{0} ~. $$ Let the solutions on the LHS be of the form

\mathbf{E}_L(\mathbf{x}) = [E_1(\mathbf{x}), E_2(\mathbf{x}), E_3(\mathbf{x})] \text{and} \mathbf{H}_L(\mathbf{x}) = [H_1(\mathbf{x}), H_2(\mathbf{x}), H_3(\mathbf{x})] ~. $$ Also, on the right hand side, let

\boldsymbol{\nabla} \times \mathbf{E}_L = [F_1(\mathbf{x}), F_2(\mathbf{x}), F_3(\mathbf{x})] \text{and} \boldsymbol{\nabla} \times \mathbf{H}_L = [I_1(\mathbf{x}), I_2(\mathbf{x}), I_3(\mathbf{x})] ~. $$ On the right hand side of the interface, let $$\mathbf{E}_R$$ and $$\mathbf{H}_R$$ solve the Maxwell equations

\boldsymbol{\nabla} \times \mathbf{E}_R + i\omega~\mu_2~\mathbf{H}_R = \boldsymbol{0} ~; \boldsymbol{\nabla} \times \mathbf{H}_R - i\omega~\epsilon_2~\mathbf{E}_R = \boldsymbol{0} ~. $$ If the interface acts as a mirror such that the right hand side (RHS) of the interface has reflected fields, then

\begin{align} \mathbf{E}_R(\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}_R(\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} $$ Then, to the right of the interface, we have

\begin{align} \boldsymbol{\nabla} \times \mathbf{E}_R = & \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}_R = [F_1(-x_1,x_2,x_3), -F_2(-x_1,x_2,x_3), -F_3(x_1,x_2,x_3)]~. $$ Similarly, on the RHS, we have

\boldsymbol{\nabla} \times \mathbf{H}_R = [I_1(-x_1,x_2,x_3), -I_2(-x_1,x_2,x_3), -I_3(x_1,x_2,x_3)]~. $$ For continuity of the fields at the interface, we must have

\mu_2 = -\mu_1 \qquad \text{and} \qquad \epsilon_2 = -\epsilon_1 ~. $$ This implies that negative permeability and permittivity have the bizzare property of reflecting the fields.

A series of interfaces
Let us consider a slab of one material immersed in another medium. Let the permittivity and permeability of the surrounding medium be $$\epsilon = \mu = 1$$ (normalized with respect to the values for free space). Let the normalized permittivity and permeability of the slab be $$\epsilon = \mu = -1$$. Hence, the first interface between the medium and the slab acts as a mirror in that it reflects the electric field $$\mathbf{E}$$. The second interface also acts as a mirror and reflects the field $$\mathbf{E}$$ to the original orientation (see Figure 2).

If the source is located at a distance $$d_0$$ from the first interface, and the slab has a thickness $$d$$, we have

\mathbf{E}(x_1+2d, x_2, x_3) = \mathbf{E}(x_1, x_2, x_3) \qquad \text{for} \quad - d < x_1 < 0 ~. $$ Therefore, the effect of the slab is just a translation. The same is true for the $$\mathbf{H}$$, $$\mathbf{D}$$, and $$\mathbf{B}$$ fields.

Let $$\mathbf{E}_0$$ be the field which solves the electromagnetic problem with the slab removed for a given source. Let us now insert a negative index slab in the field. The effect of the slab is that the fields appear to move to the right of the slab, i.e., to the right of the slab it appears as if all the fields have been moved a distance $$2d$$. In other words, it appears that the source has been moved a distance $$2d$$ to the right (see Figure 3).

This implies that the slab works as a "perfect lens" in the sense that the image to the right of the slab is not diffraction limited. This observation was first made by Pendry [Pendry2000] and was a surprising result because most lenses were though to be diffraction limited.

Consider for instance the ordinary lens shown in Figure 4(a). From geometric optics we expect the rays from the source to be focussed at a point. However, if we consider the wave nature of electromagnetic radiation, several Fourier components of the wave are superimposed at the focal point and the maximum resolution of the image can never be greater than $$\lambda/2$$ where $$\lambda$$ is the wavelength.

On the other hand, the lensing effect with a slab of negative index material is expected to lead to a point source being exactly represented at the focal point. This idea dates back to Veselago [Veselago1968]. From Figure 4(b) we observe that there will appear to be sources inside the lens and at the focal point when a negative index slab is used as a lens. However, a point source leads to a singularity in the Maxwell equations and there should be no singularities where there no physical point sources. This is a paradox.

The paradox can be resolved by observing that, in fact, a solution does not exist to the time harmonic equations if $$\epsilon = \mu = -1$$ in the slab. However, if we let

\epsilon = -1 + i\delta \qquad \text{and} \qquad \mu = -1 + i\delta $$ and let $$\delta \rightarrow 0$$, then we do have a solution. This is equivalent to assuming that there some loss in the material due to the electrical conductivity of the material.

For $$d_0 > d/2$$ (see Figure 5), the fields blow up to infinity within a strip of width $$2(d-d_0)$$ starting from the focal point within the slab to the focal point outside the slab. They develop more and more oscillations (in space), i.e., at finer and finer length scales. In the remaining regions, the field converge to Pendry's solution. Therefore, the image looks like a point source only on one side of the lens if $$\delta \ne 0$$. However, in the limit that $$\delta\rightarrow 0$$, the image also looks like a point source.

If we look at the wave vectors of the electromagnetic waves, then the from the reflected direction of the wave vector inside the lens it appears that light travels backwards inside a negative refractive index lens. But one has to remember that it is the wave crests that are travelling backwards and transport of energy is in the direction propagation of the EM waves (and therefore it is not useful to think of the direction of the wave vector as the direction of the Poynting vector). Note that the phase velocity is negative if the refractive index of the material is negative. Figure 6 illustrates this point.

Negative, Complex Anisotropic Density Materials
An important question in elasticity is how can we get materials with negative moduli. The answer is "through resonance". In this section we approach the problem by exploring the difference between "dynamic" density and "static" density. The static density is defined as the mass per unit volume whereas the dynamic density is defined as the inertial density that appears in Newton's law $$F = m~a$$. Is the dynamic density the same as the static density?

This question was first addressed by Sheng and coworkers [Sheng2003]. This work was extended by Liu et al. [Liu2005] and the mathematical analysis by Avila et al. [Avila2005]. More recently, several models have been explored by Milton and Willis [Milton2007]. The following discussion is based on one of the models presented by Milton and Willis.

Consider a rigid bar with $$n$$ voids each of width $$d$$ as shown in Figure 7. Each cavity contains a spherical ball of mass $$m$$ and radius $$r$$. The ball is attached to the walls of the cavity by springs with sprint constant $$K$$ (the spring constant may be complex valued to allow for materials with viscous damping). The spring attached to the left wall of the cavity exerts a force $$f_1(t)$$ on the wall while the spring attached to the right wall exerts a force of $$f_2(t)$$ on the wall. A force $$F(t)$$ is applied on the left side of the rigid bar. Our aim is to find the response of the bar as a function of time.

Let us assume that all quantities depend harmonically on time. Let us also assume that a one-dimensional approximation of the problem is adequate. Then the forces acting on the bar are given by
 * $$ \text{(1)} \qquad

F(t) = \text{Re}(\widehat{F}~e^{-i\omega t}) ~; f_1(t) = \text{Re}(\widehat{f}_1~e^{-i\omega t}) ~; f_2(t) = \text{Re}(\widehat{f}_2~e^{-i\omega t}) $$ where the amplitudes $$\widehat{F}$$, $$\widehat{f}_1$$, and $$\widehat{f}_2$$ are generally complex.

From Newton's second law,
 * $$\text{(2)} \qquad

F(t) = \cfrac{d }{d t}[P(t)] $$ where $$P(t)$$ is the linear momentum which is of the harmonic form
 * $$\text{(3)} \qquad

P(t) = \text{Re}(\widehat{P}~e^{-i\omega t})~. $$ Therefore, plugging equations (1) and (3) into equation (2), we get

\text{Re}(\widehat{F}~e^{-i\omega t}) = \text{Re}(-i\omega~\widehat{P}~e^{-i\omega t}) $$ or

{ \widehat{F} = -i\omega~\widehat{P} ~. } $$

Let the time-dependent position of the left side of each cavity be given by

X(t) = X_0 + U(t) = X_0 + \text{Re}(\widehat{U}~e^{-i\omega t}) $$ where $$X_0$$ is the initial position and $$\widehat{U}$$ is the complex valued displacement of the bar.

Similarly, the position of each ball is assumed to be given by
 * $$ \text{(4)} \qquad

x(t) = X(t) + \cfrac{d}{2} + u(t) = X_0 + \cfrac{d}{2} + [U(t) + u(t)] = X_0 + \cfrac{d}{2} + \text{Re}(\widehat{u}~e^{-i\omega t}) $$ where $$\widehat{u}$$ is the complex valued displacement of each ball.

Then the velocity of the rigid bar is

\cfrac{d X}{d t} = \text{Re}(-i\omega~\widehat{U}~e^{-i\omega t}) = \text{Re}(\widehat{V}~e^{-i\omega t}) ~. $$ Assume that the rigid bar has mass $$M_0$$. Therefore, the linear momentum of the rigid bar is

P_{\text{bar}}(t) = M_0~\cfrac{d X}{d t} = M_0~\text{Re}(\widehat{V}~e^{-i\omega t}) ~. $$ Similarly, the velocity of each sphere is

\cfrac{d x}{d t} = \text{Re}(-i\omega~\widehat{u}~e^{-i\omega t}) = \text{Re}(\widehat{v}~e^{-i\omega t}) ~. $$ If there are $$n$$ balls, the total linear momentum of the balls is

P_{\text{ball}}(t) = n~m~\cfrac{d x}{d t} = n~m~\text{Re}(\widehat{v}~e^{-i\omega t}) ~. $$ Therefore, the total linear momentum of the system is

P(t) = M_0~\text{Re}(\widehat{V}~e^{-i\omega t}) + n~m~\text{Re}(\widehat{v}~e^{-i\omega t}) = \text{Re}(\widehat{P}~e^{-i\omega t})~. $$ From Newton's second law, the applied force equals the rate of change of linear momentum. Therefore,

F(t) = \cfrac{d }{d t}[P(t)] = M_0~\text{Re}(-i\omega~\widehat{V}~e^{-i\omega t}) + n~m~\text{Re}(-i\omega~\widehat{v}~e^{-i\omega t}) = \text{Re}(-i\omega~\widehat{P}~e^{-i\omega t})~. $$ Therefore,
 * $$ \text{(5)} \qquad

{ \widehat{P} = M_0~\widehat{V} + n~m~\widehat{v} ~. } $$ But $$\widehat{v}$$ is unobservable since it is in the hidden part of the bar and we need to relate $$\widehat{P}$$ directly to the observable velocity $$\widehat{V}$$.

Let us now consider the free-body diagram of each spring inside a cavity (see Figure 8).

Hooke's law for each spring implies that (note that $$u(t)$$ is positive in the positive $$x$$ direction)

-f_1(t) = K~u(t) = f_2(t) $$ where $$K$$ is the complex spring constant.

Recall (from equation 4) that the displacement of the spring is given by

u(t) = \text{Re}(\widehat{u}~e^{-i\omega t}) - \text{Re}(\widehat{U}~e^{-i\omega t}) ~. $$ Using the assumed harmonic forms of $$f_1(t)$$, $$f_2(t)$$ and $$u(t)$$, we then have

\text{Re}(\widehat{f}_1~e^{-i\omega t}) = - \text{Re}(\widehat{f}_2~e^{-i\omega t}) = K~\left[ \text{Re}(\widehat{U}~e^{-i\omega t}) - \text{Re}(\widehat{u}~e^{-i\omega t})\right] $$ or
 * $$ \text{(6)} \qquad

{ \widehat{f}_1 = - \widehat{f}_2 = K~(\widehat{U} - \widehat{u}) ~. } $$ Next, considering the free body diagram of the sping-mass system (see Figure 8), the balance of linear momentum for the spring-mass system implies that

f_1(t) - f_2(t) = m~\cfrac{d^2 x}{d t^2} = m~\text{Re}(-\omega^2~\widehat{u}~e^{-i\omega t})~. $$ Therefore, substituting in the harmonic forms of $$f_2(t)$$ and $$f_1(t)$$, we get
 * $$ \text{(7)} \qquad

{ \widehat{f}_1 - \widehat{f}_2 = - m~\omega^2~\widehat{u} ~. } $$ From equations (6) and (7), we have

\widehat{f}_1 - \widehat{f}_2 = - m~\omega^2~\widehat{u} = 2~K~(\widehat{U}-\widehat{u}) $$ or,

2~K~\widehat{U} = (2~K - m~\omega^2)~\widehat{u} \qquad \implies \qquad \widehat{u} = \cfrac{2~K}{2~K - m~\omega^2}~\widehat{U} ~. $$ Now $$\widehat{V} = -i\omega\widehat{U}$$ and $$\widehat{v} = -i\omega\widehat{u}$$. Hence,
 * $$ \text{(8)} \qquad

{ \widehat{v} = \cfrac{2~K}{2~K - m~\omega^2}~\widehat{V} ~. } $$ Plugging equation (8) into equation (5), we get

{ \widehat{P} = \left(M_0 + \cfrac{2~K~n~m}{2~K - m~\omega^2}\right)~\widehat{V} = M~\widehat{V} } $$ where $$M$$ is the  effective mass. Clearly, the effective mass depends on the frequency $$\omega$$ and is different from the static mass.

A normalized plot of the effective mass versus the frequency in shown in Figure 9. At $$\omega = 0$$, the effective mass is equal to the rest mass $$M_0 + n~m$$. The resonant frequency is given by

\omega_r = \sqrt{\cfrac{2~K}{m}} ~. $$ Close the resonant frequency, the effective mass can either take high positive values or  negative values.


text:
Waves in composites and metamaterials/Anisotropic mass and generalization

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.

Rigid Bar with Frequency Dependent Mass
Recall the a rigid bar with $$n$$ cavities, each containing a spring-mass system with mass $$m$$ and complex spring constant $$K$$ (see Figure 1).

The momentum of the bar ($$\widehat{P}$$) is related to its velocity ($$\widehat{V}$$) by the relation

\widehat{P} = M(\omega)~\widehat{V} $$ where $$\omega$$ is the frequency and $$M(\omega)$$ is the  effective mass which is given by

M(\omega) = M_0 + \cfrac{2~K~n~m}{2~K - m~\omega^2} $$ where $$M_0$$ is the mass of the rigid bar~.

Let us now consider a specific model for the springs. A simple model is the one-dimensional Maxwell model shown in Figure 2. In this case instead of a spring with a complex spring constant we have an elastic spring (with real spring constant $$k$$) and a dashpot (with a real viscosity $$\eta$$) in series. Let $$u_1$$ be the displacement of the elastic spring and let $$u_2$$ be that of the dashpot under the action of a force $$f$$.

For the elastic spring, we have
 * $$ \text{(1)} \qquad

u_1(t) = \cfrac{f(t)}{k} ~. $$ For the dashpot, we have
 * $$ \text{(2)} \qquad

\cfrac{d u_2}{d t} = \cfrac{f(t)}{\eta} ~. $$ Once again, assuming that the functions $$f(t)$$, $$u_1(t)$$, and $$u_2(t)$$ can be expressed as harmonic functions, we have
 * $$ \text{(3)} \qquad

u_1(t) = \widehat{u}_1~e^{-i\omega t} ~; u_2(t) = \widehat{u}_2~e^{-i\omega t} ~; f(t) = \widehat{f}~e^{-i\omega t} ~. $$ Plugging equations (3) into equations (1) and (2), we get
 * $$ \text{(4)} \qquad

\widehat{u}_1 = \cfrac{\widehat{f}}{k} \qquad \text{and} \qquad -i\omega~\widehat{u}_2 = \cfrac{\widehat{f}}{\eta} ~. $$ Recall that the displacement $$u(t)$$ of the sphere of mass $$m$$ inside the cavity is related to the applied force $$f(t)$$ by the relation

u(t) = \cfrac{f(t)}{K} \qquad \implies \qquad \widehat{u} = \cfrac{\widehat{f}}{K} ~. $$ Now, $$\widehat{u} = \widehat{u}_1 + \widehat{u}_2$$. Hence,

\cfrac{\widehat{f}}{K} = \cfrac{\widehat{f}}{k} - \cfrac{1}{i\omega}~\cfrac{\widehat{f}}{\eta} $$ or,

{   \cfrac{1}{K} = \cfrac{1}{k} + \cfrac{i}{\omega\eta} ~. } $$ So, with the Maxwell model, the effective mass is

M(\omega) = M_0 + \cfrac{2~n~m}{2 - \cfrac{1}{K}~m~\omega^2} = M_0 + \cfrac{2~n~m}{2 - \cfrac{1}{k}~m~\omega^2 - \cfrac{i}{\omega\eta}~m~\omega^2} $$ or,
 * $$ \text{(5)} \qquad

{ M(\omega) = M_0 + \cfrac{2~n~k}{\cfrac{2~k}{m} - \omega^2 - \cfrac{i\omega~k}{\eta}} ~. } $$ This model is remarkably similar to a simple model for the frequency dependent dielectric constant $$\epsilon(\omega)$$.

Comparison with a simple model for $$\epsilon(\omega)$$
Consider the following simple model of an electron bound to an atom by a harmonic force under the influence of a slowly varying electric field (a detailed description can in found in Jackson75, Sec. 7.5). A schematic of the situation is shown in Figure 3.

Let the electric field be $$E(x,t)$$ and assume that it varies slowly in space over the distance that the electron moves, i.e., $$E(x,t) \approx E(t)$$. Let $$m$$ be the mass of the electron and let $$-e$$ be its charge. Let $$\omega_0$$ be the frequency of the harmonic force binding the electron to the atom and let $$\gamma$$ be a damping coefficient due to interactions with obstacles.

Then, the equation of motion of the electron is given by
 * $$ \text{(6)} \qquad

m \left(\cfrac{d^2 x}{d t^2} + \gamma\cfrac{d x}{d t} + \omega_0^2~x\right) = -e~E(t)~. $$ Let
 * $$ \text{(7)} \qquad

x(t) = \widehat{x}~e^{-i\omega t} \text{and} E(t) = \widehat{E}~e^{-i\omega t} ~. $$ Plugging equations (7) into equation (6), we get

m (-\omega^2 - i\omega\gamma + \omega_0^2) \widehat{x} = - e~\widehat{E} $$ or

\widehat{x} = -\cfrac{e}{m} \left(\omega_0^2 - \omega^2 - i\omega\gamma\right)^{-1} ~\widehat{E} ~. $$ Let $$q(x,t)$$ be the charge density of the electron-atom system, i.e.,

q(x,t) = -\delta[x(t)-x_0]~e + \delta(x_0-x_0)~e $$ where $$\delta(x)$$ is the Dirac delta function and $$x_0$$ is the position of the atom. Then, the dipole moment (the first moment of the charge) contributed by the electron-atom system is given by

p(t) = \int q(x,t)~x(t)~\text{d}x = -e~\left[\int \delta[x(t)-x_0]~x(t)~\text{d}x + \int \delta(0)~x(t)~\text{d}x\right] ~. $$ or,

p(t) = -e~x(t) \qquad \implies \qquad \widehat{p} = -e~\widehat{x} = \cfrac{e^2}{m} \left(\omega_0^2 - \omega^2 - i\omega\gamma\right)^{-1} ~\widehat{E} ~. $$ Suppose now that instead of a binding frequency $$\omega_0$$ for all the atoms and all the electrons in a volume, there are $$N_i$$ atoms per unit volume with binding frequencies $$\omega_i$$ and damping constants $$\gamma_i$$ and. Then, we can write the polarization as
 * $$ \text{(8)} \qquad

\widehat{P} = \cfrac{e^2}{m} \left[\sum_i \cfrac{N_i}{ \omega_0^2 - \omega^2 - i\omega\gamma}\right]~\widehat{E} = \left[\sum_i \cfrac{a_i}{\omega_0^2 - \omega^2 - i\omega\gamma}\right] ~\widehat{E}~. $$ Recall that the electric displacement is related to the electric field and the polarization by
 * $$ \text{(9)} \qquad

\widehat{D} = \epsilon_0~\widehat{E} + \widehat{P} = \epsilon~\widehat{E} ~. $$ Therefore, from equations (8) and (9) we get
 * $$ \text{(10)} \qquad

{ \epsilon = \epsilon_0 + \sum_i \cfrac{a_i}{\omega_0^2 - \omega^2 - i\omega\gamma} ~. } $$ Comparing equation (5) with equation (10) we observe that they have the same form which implies that the effective permittivity is analogous to the effective mass. This also implies that the electric field is analogous to the velocity. Similarly, the polarization is analogous to the momentum.

More on frequency dependence of the mass
More generally, we get a frequency dependent density if all the constituents do not move in lock step. Lock step motion almost never occurs because there are thermal vibrations at the microscale. There are also many macroscopic situations in which lock step motion does not occur. Consider the example of a porous rock containing some water (see Figure 4). Both the rock grains and the water are connected, though this is not obvious from the figure

In this case, the water will move with a different frequency that the rock and the density of the composite will be dependent on the frequency.

At a molecular level, we can have a crystal with lead atoms attached by single bonds to the structure (see Figure 5). Presumably, the resonant frequency of such molecules is very high. so we will see the frequency dependence of the mass only at very high frequencies.

In fact, Sheng et al. Sheng03 have shown in experiments that materials indeed have frequency dependent masses. An example of such a material is shown in Figure 6.

Generalizations of the Rigid Bar Model
Consider the rigid body containing cavities shown in Figure~7. This is just a two-dimensional extension of the model shown in Figure~1. Here $$K$$ and $$L$$ are complex spring constants in the $$x$$ and $$y$$ directions.

In this case, the effective mass along the $$x$$-direction is given by
 * $$ \text{(11)} \qquad

M_x(\omega) = M_0 + \cfrac{2~K~n~m}{2~K - m~\omega^2} $$ while that along the $$y$$-direction is given by
 * $$ \text{(12)} \qquad

M_y(\omega) = M_0 + \cfrac{2~L~n~m}{2~L - m~\omega^2}~. $$ In matrix form, we then have

{ \left[\mathsf{M}\right](\omega) = M_0~\left[\mathsf{I}\right] + n~m~ \begin{bmatrix} \cfrac{2~K}{2~K - m~\omega^2} & 0 \\ 0 & \cfrac{2~L}{2~L - m~\omega^2} \end{bmatrix} ~. } $$ Hence, the effective mass is clearly  anisotropic. Note however that from a macroscopic perspective it is not the average velocity in the matrix which is important. In fact, such a quantity does not even make sense because the velocity is not defined in the void phase. Rather it is the velocity of the matrix that is the relevant quantity in this model.

One can generalize the model one step further by having the springs be oriented at different angles to each other as shown in Figure 8. Also let the springs in each cavity have different spring constants and the masses in each cavity are different.

In this case, let $$\boldsymbol{R}$$ be the rotation that is needed to orient each set of springs with the $$x$$ and $$y$$ axes. Then, if $$\left[\mathsf{R}\right]_j$$ is the rotation matrix for cavity $$j$$ containing a mass $$m_j$$, and if $$K_j$$ and $$L_j$$ are the complex spring constants for that cavity, the effective mass can be written as

{ \left[\mathsf{M}\right](\omega) = M_0~\left[\mathsf{I}\right] + \sum_{j=1}^n \left[\mathsf{R}\right]_j^T~ \begin{bmatrix} \cfrac{2~K_j~m_j}{2~K_j - m_j~\omega^2} & 0 \\ 0 & \cfrac{2~L_j~m_j}{2~L_j - m_j~\omega^2} \end{bmatrix} \left[\mathsf{R}\right]_j ~. } $$ The eigenvalues of $$\boldsymbol{M}(\omega)$$ can therefore depend on $$\omega$$.

General form of $$\boldsymbol{M}(\omega)$$
Take any body with a rigid matrix. Suppose, instead of applying harmonically varying velocities, we apply a time varying velocity $$\mathbf{V}(t)$$ and observe what the momentum $$P(t)$$ is. There will be some linear constitutive relation
 * $$ \text{(13)} \qquad

\mathbf{P}(t) = \int_{-\infty}^{\infty} \boldsymbol{H}(t - \tau)\cdot\mathbf{V}(\tau)~\text{d}\tau ~. $$ The kernel $$\boldsymbol{H}(t-\tau)$$ is second-order tensor valued and may possibly be singular. Also, since both $$\mathbf{P}$$ and $$\mathbf{V}$$ are physical and real, $$\boldsymbol{H}$$ must be real. Causality implies that $$\boldsymbol{H}(s) = \boldsymbol{\mathit{0}}$$ when $$s = t - \tau < 0$$ (or $$t < \tau$$) since the inertial force cannot depend on velocities in the future.

Taking the Fourier transform of equation (13) and using the convolution theorem, we get
 * $$ \text{(14)} \qquad

\widehat{\mathbf{P}}(\omega) = \boldsymbol{M}(\omega)\cdot\widehat{\mathbf{V}}(\omega) $$ where
 * $$ \text{(15)} \qquad

\boldsymbol{M}(\omega) := \widehat{\boldsymbol{H}}(\omega) = \int_{-\infty}^{\infty} \boldsymbol{H}(s)~e^{i\omega s}~\text{d}s ~. $$ The quantity $$\boldsymbol{M}(\omega)$$ can be shown to satisfy the Cauchy-Riemann equations only if $$\text{Im}(\omega) > 0$$. This is a consequence of causality $$s \ge 0$$ and the fact that the integral in equation (15) only converges in the upper half of the complex $$\omega$$ plane. \footnote{ Note the positive sign in the power of $$i\omega s$$ in equation (15). This occurs because we have chosen the inverse Fourier transform to be of the form

g(s) = \cfrac{1}{\sqrt{2\pi}} \int_{-\infty}^{\infty} \widehat{g}(\omega)~e^{-i\omega s}    ~\text{d}\omega ~. $$ The effect of such a choice is that the imaginary part of $$\omega$$ is positive instead of negative.} Hence, $$\boldsymbol{M}(\omega)$$ is analytic in $$\omega$$ when $$\text{Im}(\omega) > 0$$. The quantity $$\boldsymbol{H}(s)$$ is real.

Now, the complex conjugate of $$\boldsymbol{M}(\omega)$$ is given by
 * $$ \text{(16)} \qquad

\boldsymbol{M}^{*}(\omega) = \int_{-\infty}^{\infty} \boldsymbol{H}(s)~e^{-i\omega^{*} s}~\text{d}s = \boldsymbol{M}(-\omega^{*}) $$ where $$z^{*}$$ denotes the complex conjugate of $$z$$ (for any complex $$z$$).

Also assume that, for large enough frequencies, the dynamic mass tends toward the static mass, i.e.,
 * $$ \text{(17)} \qquad

\lim_{\omega\rightarrow\infty} \boldsymbol{M}(\omega) = M_0~\boldsymbol{\mathit{1}} ~. $$

Equation (15) can be used to establish the Kramers-Kronig equations for the material. To do that, recall Cauchy's formula for a function $$f$$ which is analytic on a domain that is enclosed in a piecewise smooth curve $$C$$:

f(z) = \cfrac{1}{2\pi i} \oint_C \cfrac{f(\zeta)}{\zeta - z}~\text{d}\zeta ~. $$ Since the function $$\boldsymbol{M}(\omega)$$ is analytic on the upper-half $$\omega$$ plane, for any point $$z$$ in a closed contour $$C$$ in the upper-half $$\omega$$ plane, we have
 * $$ \text{(18)} \qquad

\boldsymbol{M}(z) - M_0~\boldsymbol{\mathit{1}} = \cfrac{1}{2\pi i}       \oint_C \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' ~. $$ Let us now choose the contour $$C$$ such that it consists of the real $$\omega$$ axis and a great semicircle at infinity in the upper half plane (see Figure~9).

Also, from equation (17) we observe that

\boldsymbol{M}(\omega) - M_0~\boldsymbol{\mathit{1}} = \boldsymbol{0} \qquad \text{as} \qquad \omega \rightarrow \infty~. $$ Hence, there is no contribution to the integral in equation (18) due to the semicircular part of the contour and we just have to perform an integration only over the real line:
 * $$ \text{(19)} \qquad

\boldsymbol{M}(z) = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{2\pi i}       \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' ~. $$

Now, let us consider the integral

\boldsymbol{I} = \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' ~. $$ There is a pole at the point $$\omega' = z$$ (in the figure, the contour is shown as a semicircle of radius $$\varepsilon$$ centered at the pole). In the limit $$\varepsilon \rightarrow 0$$, this integral may be interpreted to mean the Cauchy principal value (not the same as principal values in complex analysis)

\boldsymbol{I} = P \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' = \int_{-\infty}^{z-\varepsilon} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' + \int_{z-\varepsilon}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' ~. $$ From Figure~9 we observe that this integral can be broken up into the integral over the path $$A$$ minus the sum of the integrals over the paths $$B$$ and $$D$$. From the Cauchy-Goursat theorem, the integral over the closed path $$A$$ is zero. We have also seen that since $$\boldsymbol{M}(\omega) - M_0\boldsymbol{\mathit{1}} \rightarrow 0$$ as $$\omega \rightarrow \infty$$, the integral over $$B$$ is zero. The integral over the path $$D$$ around the pole is obtained from the Residue Theorem (where the value is divided by two because the integral is over a semicircle), i.e.,

\oint_{D} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' = -\pi i~[\boldsymbol{M}(z) - M_0\boldsymbol{\mathit{1}}] ~. $$ The negative sign arises because the curve is traversed in the counter-clockwise direction.

Therefore,

\boldsymbol{I} = P \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' = \pi i~[\boldsymbol{M}(z) - M_0\boldsymbol{\mathit{1}}] ~. $$ or,

\boldsymbol{M}(z) = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi i} ~ P \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - z}~\text{d}\omega' ~. $$ Letting $$z = \omega + i\varepsilon$$ and taking the limit as $$\varepsilon \rightarrow 0$$, we get
 * $$ \text{(20)} \qquad

{ \boldsymbol{M}(\omega) = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi i}~ P \int_{-\infty}^{\infty} \cfrac{\boldsymbol{M}(\omega') - M_0~\boldsymbol{\mathit{1}}}{\omega' - \omega}~\text{d}\omega' ~. } $$ Note that, in the above equation, both $$\omega$$ and $$\omega'$$ are real. Expanding equation (20) into real and imaginary parts and collecting terms, we get the first form of the Kramers-Kronig relations
 * $$ \text{(21)} \qquad

{ \begin{align} \text{Re}[\boldsymbol{M}(\omega)] & = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi}~ P \int_{-\infty}^{\infty} \cfrac{\text{Im}[\boldsymbol{M}(\omega')]}{\omega' - \omega}~\text{d}\omega' \\ \text{Im}[\boldsymbol{M}(\omega)] & = - \cfrac{1}{\pi}~ P \int_{-\infty}^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega' - \omega}~\text{d}\omega' ~. \end{align} } $$ Therefore, the real part of the frequency dependent mass can be determined if we know the imaginary part and vice versa.

We can also eliminate the negative frequencies from equations (21). Recall from equation (16) that

\boldsymbol{M}^{*}(\omega) = \boldsymbol{M}(-\omega^{*}) ~. $$ Since, in equations (21), $$\omega$$ and $$\omega'$$ are real, we have

\boldsymbol{M}^{*}(\omega) = \boldsymbol{M}(-\omega) ~. $$ This implies that

\text{Re}[\boldsymbol{M}(\omega)] = \text{Re}[\boldsymbol{M}(-\omega)] \qquad \text{and} \qquad \text{Im}[\boldsymbol{M}(\omega)] = -\text{Im}[\boldsymbol{M}(-\omega)] ~. $$ Consider the first of equations (21). We can write this relation as

\begin{align} \text{Re}[\boldsymbol{M}(\omega)] & = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi}~\left\{ P \int_{-\infty}^{0} \cfrac{\text{Im}[\boldsymbol{M}(\omega')]}{\omega' - \omega}~\text{d}\omega' + P \int_0^{\infty} \cfrac{\text{Im}[\boldsymbol{M}(\omega')]}{\omega' - \omega}~\text{d}\omega' \right\}\\ & = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi}~\left\{ P \int_0^{\infty} \cfrac{\text{Im}[\boldsymbol{M}(-\omega')]}{-\omega' - \omega}~\text{d}\omega' + P \int_0^{\infty} \cfrac{\text{Im}[\boldsymbol{M}(\omega')]}{\omega' - \omega}~\text{d}\omega' \right\}\\ & = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi}~\left\{ -P \int_0^{\infty} \cfrac{-\text{Im}[\boldsymbol{M}(\omega')]}{\omega' + \omega}~\text{d}\omega' + P \int_0^{\infty} \cfrac{\text{Im}[\boldsymbol{M}(\omega')]}{\omega' - \omega}~\text{d}\omega' \right\}\\ & = M_0~\boldsymbol{\mathit{1}} + \cfrac{1}{\pi}~ P \int_0^{\infty} \text{Im}[\boldsymbol{M}(\omega')]\left[\cfrac{1}{\omega' + \omega}+ \cfrac{1}{\omega'-\omega}\right]~\text{d}\omega' \\ & = M_0~\boldsymbol{\mathit{1}} + \cfrac{2}{\pi}~ P \int_0^{\infty} \cfrac{\omega'~\text{Im}[\boldsymbol{M}(\omega')]}{\omega'^2 - \omega^2} ~\text{d}\omega' ~. \end{align} $$ Similarly, the second of equations (21) may be written as

\begin{align} \text{Im}[\boldsymbol{M}(\omega)] & = - \cfrac{1}{\pi}~\left\{ P \int_{-\infty}^{0} \cfrac{\text{Re}[\boldsymbol{M}(\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega' - \omega}~\text{d}\omega' + P \int_0^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega' - \omega}~\text{d}\omega' \right\} \\ & = - \cfrac{1}{\pi}~\left\{ -P \int_0^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(-\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega' + \omega}~\text{d}\omega' + P \int_0^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega' - \omega}~\text{d}\omega' \right\} \\ & = - \cfrac{1}{\pi}~\left\{ P \int_0^{\infty} (\text{Re}[\boldsymbol{M}(-\omega')] - M_0~\boldsymbol{\mathit{1}})\left[ -\cfrac{1}{\omega' + \omega} + \cfrac{1} {\omega' - \omega}\right]~\text{d}\omega' \right\} \\ & = - \cfrac{2~\omega}{\pi}~\left\{ P \int_0^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(-\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega'^2 - \omega^2}~\text{d}\omega' \right\} ~. \end{align} $$ Therefore, the alternative form of the Kramers-Kronig relations is
 * $$ \text{(22)} \qquad

{ \begin{align} \text{Re}[\boldsymbol{M}(\omega)] & = M_0~\boldsymbol{\mathit{1}} + \cfrac{2}{\pi}~ P \int_0^{\infty} \cfrac{\omega'~\text{Im}[\boldsymbol{M}(\omega')]}{\omega'^2 - \omega^2} ~\text{d}\omega' \\ \text{Im}[\boldsymbol{M}(\omega)] & = - \cfrac{2~\omega}{\pi}~\left\{ P \int_0^{\infty} \cfrac{\text{Re}[\boldsymbol{M}(-\omega')] - M_0~\boldsymbol{\mathit{1}}} {\omega'^2 - \omega^2}~\text{d}\omega' \right\} ~. \end{align} } $$

Significance of $$\text{Im}[\boldsymbol{M}(\omega)]$$
Consider harmonically varying $$\mathbf{F}(t)$$ and $$\mathbf{V}(t)$$ given by
 * $$ \text{(23)} \qquad

\mathbf{F}(t) = \text{Re}(\widehat{\mathbf{F}}~ e^{-i\omega t}) \quad \text{and} \quad \mathbf{V}(t) = \text{Re}(\widehat{\mathbf{V}}~ e^{-i\omega t})~. $$ Alternatively, we may write these as

\mathbf{F}(t) = \text{Re}(\widehat{\mathbf{F}})\cos(\omega t) + \text{Im}(\widehat{\mathbf{F}})\sin(\omega t) ~; \mathbf{V}(t) = \text{Re}(\widehat{\mathbf{V}})\cos(\omega t) + \text{Im}(\widehat{\mathbf{V}})\sin(\omega t) ~. $$ Then,

\widehat{\mathbf{F}} = -i\omega\boldsymbol{M}(\omega)\cdot\widehat{\mathbf{V}} $$ which implies that

\begin{align} \text{Re}(\widehat{\mathbf{F}}) & = \omega~[\text{Re}(\boldsymbol{M})\cdot\text{Im}(\widehat{\mathbf{V}}) + \text{Im}(\boldsymbol{M})\cdot\text{Re}(\widehat{\mathbf{V}})] \\ \text{Im}(\widehat{\mathbf{F}}) & = \omega~[\text{Im}(\boldsymbol{M})\cdot\text{Im}(\widehat{\mathbf{V}}) - \text{Re}(\boldsymbol{M})\cdot\text{Re}(\widehat{\mathbf{V}})] ~. \end{align} $$ The average rate of work done on the system in a cycle of oscillation will be

\begin{align} W & = \cfrac{\omega}{2~\pi} \int_{0}^{2\pi/\omega}~\mathbf{F}(t)\cdot\mathbf{V}(t)~\text{d}t \\ & = \cfrac{\text{Re}(\widehat{\mathbf{F}})\cdot\text{Re}(\widehat{\mathbf{V}}) + \text{Im}(\widehat{\mathbf{F}})\cdot\text{Im}(\widehat{\mathbf{V}})}{2} \\ & = \omega~[\text{Re}(\widehat{\mathbf{V}})\cdot\text{Im}[\boldsymbol{M}(\omega)]\cdot\text{Re}(\widehat{\mathbf{V}}) + \text{Im}(\widehat{\mathbf{V}})\cdot\text{Im}[\boldsymbol{M}(\omega)]\cdot\text{Im}(\widehat{\mathbf{V}})]. \end{align} $$ This quadratic form will be non-negative for all choices of $$\widehat{\mathbf{V}}$$ if and only if $$\text{Im}(\boldsymbol{M}(\omega))$$ is positive semidefinite for all real $$\omega>0$$. Note that the quadratic form does not contain $$\text{Re}{\boldsymbol{M}}$$. Since the work done in a cycle should be zero in the absence of dissipation, this implies that the imaginary part of the mass is connected to the energy dissipation (for instance, into heat).


text:
User:Wundermacht

 * LSAT Test Help
 * American J.D. Program

title: User:Wundermacht


text:
United States Law/Introduction

Lecture
This is your orientation to Wikiversity 1L Program. 1L is how lawyers and law students informally refer to the first year of the law school education. Law school education (full time) in the United States is typically 3 years. The first year courses are typically: /Contracts/, /Torts/, /Property/, /Civil Procedure/, /Criminal Law/, and /Legal Writing/.

Law school experience can be quite new to many. It has its own logic and jargons. It is your job to get used to those strange terms and rules.

Some reference wikibooks you need to get before you start school.

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 * How to read a Law School Casebook: from LexisNexis.
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Assignment
Read a list of law school terminology, e.g., http://www.law.gmu.edu/library/guides/glossary

Read the cases to be discussed and prepare to answer questions about them.

Complete the exercise. /Legal Terminology/

Pre-Law

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1L

 * /Contracts/
 * /Torts/
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2L

 * /Evidence/
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Bar Review

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Other/Miscellany
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title: United States Law/Introduction


text:
User talk:Wundermacht

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Welcome
Hello Wundermacht, and welcome to Wikiversity! If you need help, feel free to visit my talk page, or contact us and ask questions. After you leave a comment on a talk page, remember to sign and date; for it helps everyone follow the threads of the discussion. The signature icon in the edit window makes it simple. To get started, you may
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And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! --Mirwin 01:49, 4 October 2007 (UTC)

Legal Writing Course
The legal writing course you are writing looks interesting. I hope you are considering adding a section on business contracts. Mirwin 01:51, 4 October 2007 (UTC)

title: User talk:Wundermacht