Waves in composites and metamaterials/Fading memory and waves in layered 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.

Viscoelastic Materials
In the previous lecture, we discussed viscoelastic materials and wondered why the Maxwell model works even though the effective Young's modulus $$Y(\omega)$$ for such materials is analytic in the entire complex plane (except for a few isolated points).

Recall that the Maxwell model (see Figure 1) predicts that the frequency dependent Young's modulus of the system is given by

Y(\omega) = Y_e + \sum_j \cfrac{H_j}{1 + \cfrac{i}{\omega\tau_j}}~. $$

This implies that the function $$Y(\omega)$$ is analytic in the entire imaginary $$\omega$$ plane except for poles at $$\omega = -i/\tau_j$$. On the other hand, for the frequency dependent metamaterials that we have discussed earlier, the effective modulus is generally analytic only in the upper half $$\omega$$ plane (see Figure~2). Also, for such materials, $$Y^{*}(\omega) = Y(-\omega^{*})$$, where $$(.)^{*}$$ indicates the complex conjugate. Note that we do not consider the mass when we derive the modulus of the Maxwell model. The relation between viscoelastic models of the Maxwell type and general frequency dependent materials continues to be an open question.

A justification of the Maxwell model can be provided by considering the behavior of viscoelastic materials (Christ03). Consider an experiment where a bar of viscoelastic material of length $$l$$ is deformed by a fixed amount. We want to see how the stress changes with time. Recall, that if the bar is extended by an amount $$u(x)$$ where $$x = 0$$ at one end of the bar, then the one-dimensional strain is defined as

\epsilon = \cfrac{d u}{d x}~. $$ Therefore, the displacement in the bar can be expressed in terms of the strain as

u(x) = \epsilon~x \qquad \implies \qquad \epsilon = \cfrac{u(l)}{l} = \cfrac{\Delta l}{l} ~. $$ Also, if $$F$$ is the applied force on the bar and $$A$$ is its cross-sectional area, then the stress is given by

\sigma = \cfrac{F}{A} ~. $$ Let us now apply a strain to the bar at time $$t = 0$$ and hold the strain fixed. Due to the initial application of the strain, the stress reaches a value $$K_0$$ and then relaxes at time increase (due to the relaxation of polymer chains for instance). Figure 3 shows a schematic of this situation.

If the strain is applied by the superposition of a two step strains as shown in the figure, we have

\cfrac{d \epsilon}{d t} = a~\delta(t - t_1) + b~\delta(t - t_2) ~. $$ The stress is then given by

\sigma = a~K(t - t_1) + b~K(t - t_2) ~. $$ If the strain is applied by a series of infinitesimal steps, then we get a more general form for the stress:

\text{(1)} \qquad \sigma(t) = \int_{-\infty}^t K(t - t')~\cfrac{d \epsilon(t')}{d t'}~\text{d}t' $$ where the integral should be interpreted in the distributional sense. Integrating by parts (and assuming that $$\epsilon = 0$$ at $$t = -\infty$$), we get

\sigma(t) = K_0~\epsilon(t) + \int_{-\infty}^t K(t - t')~\epsilon(t')~\text{d}t' ~. $$ Now, $$\sigma(t)$$ clearly depends on past values of $$d\epsilon/dt$$. We expect $$\sigma(t)$$ should have a stronger dependence on $$d\epsilon/dt$$ in the recent past than in the distant past. More precisely, the dependence should decrease monotonically as $$\tau = t - t'$$ increases. This implies that $$K(\tau)$$ should decrease at $$\tau$$ increases, i.e.,

\cfrac{d K(\tau)}{d \tau} < 0 \quad \forall \tau > 0 \text{and} K(\tau) > 0 ~. $$ This is the assumption of  fading memmory.

From equation (1) the rate of change of $$\sigma$$ is given by

\cfrac{d \sigma(t)}{d t} = \int_{-\infty}^t \left.\cfrac{d K(\tau)}{d \tau}\right|_{\tau=t-t'}~ \cfrac{d \epsilon(t')}{d t'}~\text{d}t' ~. $$ Again, we expect $$d\sigma/dt$$ to have a stronger dependence on $$d\epsilon(t')/dt'$$ in the recent past than in the far past, i.e,.

\left|\cfrac{d K(\tau)}{d \tau}\right| \text{should decrease as}~\tau~ \text{increases.} $$ Now,

\cfrac{d K(\tau)}{d \tau} < 0 \quad \implies \quad \cfrac{d^2 K(\tau)}{d \tau^2} > 0 \quad \implies \quad \cfrac{d^3 K(\tau)}{d \tau^3} < 0 \quad \implies \quad \cfrac{d^4 K(\tau)}{d \tau^4} > 0 \quad \dots \qquad \forall \tau > 0 ~. $$ Such functions are said to be  completely monotonic. An example is

K(\tau) = e^{-\tau/\tau'} \qquad \implies \qquad \cfrac{d K(\tau)}{d \tau} < 0 ~, \cfrac{d^2 K(\tau)}{d \tau^2} > 0 ~\text{etc.} $$ More generally,

K(\tau) = K_{\infty} + \int_0^{\infty} H(\tau')~e^{-\tau/\tau'}~\text{d}\tau' $$ is completely monotonic if $$H(\tau') \ge 0$$ for all $$\tau$$ and $$K_{\infty} \ge 0$$. The function $$H(\tau')$$ is called the {\bf relaxation spectrum.}

Conversely, any completely monotonic function can be written in this form (Bernstein28).

Specifically, if

\epsilon(t) = \text{Re}(\widehat{\epsilon}~e^{-i\omega t}) $$ then

\cfrac{d \epsilon(t)}{d t} = \text{Re}(-i\omega~\widehat{\epsilon}~e^{-i\omega t})~. $$ Therefore,

\sigma(t) = \text{Re}\left(-i\omega~\widehat{\epsilon}~\int_{-\infty}^t K(t-t')~e^{-i\omega t'}~\text{d}t'\right)~. $$ Let $$\tau - t - t'$$. Then

\sigma(t) = \text{Re}\left(-i\omega~\widehat{\epsilon}~e^{-i\omega t}~      \int_0^{\infty} K(\tau)~e^{-i\omega\tau}~\text{d}\tau\right)~. $$ Define
 * $$ \text{(2)} \qquad

Y(\omega) := -i\omega~\int_0^{\infty} K(\tau)~e^{-i\omega\tau}~\text{d}\tau~. $$ Then we have

\sigma(t) = \text{Re}\left(Y(\omega)~\widehat{\epsilon}~e^{-i\omega t}\right)~. $$ If we define

\widehat{\sigma} := Y(\omega)~\widehat{\epsilon} $$ we get

\sigma(t) = \text{Re}\left(\widehat{\sigma}~e^{-i\omega t}\right)~. $$ Now, let $$K(\tau)$$ be a completely monotonic function of the form

K(\tau) = K_{\infty} + \int_0^{\infty} H(\tau')~e^{-\tau/\tau'}~\text{d}\tau'~. $$ Then from equation (2) we get

Y(\omega) = -i\omega~\int_0^{\infty} K_{\infty}~e^{-i\omega\tau}~\text{d}\tau -i\omega~\int_0^{\infty} \text{d}\tau'~H(\tau')~ \int_0^{\infty} \text{d}\tau~e^{\tau(i\omega-1/\tau')}~. $$ Assume that $$\omega$$ has a very small poistive imaginary part (which implies that $$\epsilon(t)$$ increases very slowly as $$t$$ goes to $$\infty$$). Then

Y(\omega) = -i\omega~\left(\cfrac{-K_{\infty}}{i\omega}\right) -i\omega~\int_0^{\infty} \text{d}\tau'~H(\tau')~ \left(\cfrac{-1}{i\omega-1/\tau'}\right) $$ or,

Y(\omega) = K_{\infty} + \int_0^{\infty} \cfrac{H(\tau')}{1+\cfrac{i}{\omega\tau'}}~\text{d}\tau'~. $$ This is the generalized Maxwell model.

This brings up the question: Is the assumption of fading memory always correct?

Recall the model of the Helmholtz resonator shown in Figure~4. If we apply a strain in the form of a step function to this model, the resulting stress response is not a monotonically decreasing function of time. Rather if oscillates around a certain value and may damp out over time. A similar oscillatory behavior is expected in other spring-mass systems and $$K(\tau)$$ will, in general, not be monotonic.

A short interlude: Maxwell's equations in Elasticity Form
In this section, we discuss how Maxwell's equation can be reduced to the form of the elasticity equations. Recall that, at a fixed $$\omega$$, Maxwell's equation take the form

\boldsymbol{\nabla} \times \mathbf{E} = i\omega\boldsymbol{\mu}(\mathbf{x})\cdot\mathbf{H}(\mathbf{x}) ~; \boldsymbol{\nabla} \times \mathbf{H} = -i\omega\boldsymbol{\epsilon}(\mathbf{x})\cdot\mathbf{E}(\mathbf{x}) ~. $$ Therefore,

\omega^2~\boldsymbol{\epsilon}\cdot\mathbf{E} = i\omega~\boldsymbol{\nabla} \times \mathbf{H} = \boldsymbol{\nabla} \times [\boldsymbol{\mu}^{-1}\cdot(\boldsymbol{\nabla} \times \mathbf{E})] ~. $$ Recall that, in index notation and using the summation convention, we have

[\boldsymbol{\nabla} \times \mathbf{a}]_i = \mathcal{E}_{ijk}~\frac{\partial a_k}{\partial x_j} $$ where $$\mathcal{E}_{ijk}$$ is the permutation tensor defined as

\mathcal{E}_{ijk} = \begin{cases} 1 & \text{for even permutations, i.e., 123, 231, 312 } \\ -1 & \text{for odd permutations, i.e., 132, 321, 213 } \\ 0 & \text{otherwise}. \end{cases} $$ Therefore,

\begin{align} \left[\omega^2~\boldsymbol{\epsilon}\cdot\mathbf{E}\right]_j & = \mathcal{E}_{jim}~\frac{\partial }{\partial x_i}[\boldsymbol{\mu}^{-1}\cdot(\boldsymbol{\nabla} \times \mathbf{E})]_m \\ & =    \mathcal{E}_{jim}~\frac{\partial }{\partial x_i}[(\boldsymbol{\mu}^{-1})_{mn}~(\boldsymbol{\nabla} \times \mathbf{E})_n] \\ & =    \mathcal{E}_{jim}~\frac{\partial }{\partial x_i}[(\boldsymbol{\mu}^{-1})_{mn}~\mathcal{E}_{nkl}~ \frac{\partial E_l}{\partial x_k}] \\ & =     \frac{\partial }{\partial x_i}[C_{ijkl} \frac{\partial E_l}{\partial x_k}] \quad \text{where} \quad C_{ijkl} := \mathcal{E}_{jim}~\mathcal{E}_{nkl}~[\boldsymbol{\mu}^{-1}]_{mn} \end{align} $$ or,

{   \omega^2~\boldsymbol{\epsilon}\cdot\mathbf{E} = \boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}:\boldsymbol{\nabla} \mathbf{E}) ~. } $$ This is very similar to the elasticity equation

-\omega^2~\rho~\mathbf{u} = \boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}:\boldsymbol{\nabla} \mathbf{u}) ~. $$ The permittivity is similar to a negative density and the electric field is similar to the displacement. The equations also hint at a tensorial density. However, continuity conditions are different for the two equations, i.e., at an interface, $$\mathbf{u}$$ is continuous while only the tangential component of $$\mathbf{E}$$ is continuous. Also, the tensor $$\boldsymbol{\mathsf{C}}$$ has different symmetries for the two situations. Interestingly, for Maxwell's equations

C_{ijkl} = - C_{jikl} = - C_{ijlk} = C_{klij} ~. $$

Waves in Layered Media
A detail exposition of waves in layer media can be found in Chew95. In this section we examine a few features of electromagnetic waves in layered media.

Assume that the permittivity and permeability are scalars and are locally isotropic though not globally so. Then we may write

\boldsymbol{\epsilon} = \epsilon(x_3)~\boldsymbol{\mathit{1}} \quad \text{and} \quad \boldsymbol{\mu} = \mu(x_3)~\boldsymbol{\mathit{1}} ~. $$ The TE (transverse electric field) equations are given by
 * $$ \text{(3)} \qquad

\overline{\boldsymbol{\nabla}} \cdot\left(\cfrac{1}{\mu}~\overline{\boldsymbol{\nabla}} E_1\right) + \omega^2~\epsilon~E_1 = 0 $$ where $$\overline{\boldsymbol{\nabla}} $$ represents the two-dimensional gradient operator.

Multiplying (3) by $$\mu(x_3)$$, we have
 * $$ \text{(4)} \qquad

\left[ \frac{\partial }{\partial x_2^2} + \mu(x_3)~\frac{\partial }{\partial x_3}\left(\cfrac{1}{\mu(x_3)}~\frac{\partial }{\partial x_3}\right) + \omega^2~\epsilon(x_3)~\mu(x_3)\right]~E_1 = 0 ~. $$ Equation (4) admits solutions of the form

E_1(x_2,x_3) = \tilde{E}_1(x_3)~e^{\pm i~k_2~x_2} $$ and equation (4) then becomes an ODE:
 * $$ \text{(5)} \qquad

\left[ \mu(x_3)~\cfrac{d }{d x_3}\left(\cfrac{1}{\mu(x_3)}~\cfrac{d }{d x_3}\right) + \omega^2~\epsilon(x_3)~\mu(x_3) - k_2^2\right]~\tilde{E}_1 = 0 ~. $$ The quantity

k_3^2 := \omega^2~\epsilon(x_3)~\mu(x_3) - k_2^2 $$ can be less than zero, implying that $$k_3$$ may be complex. Also, at the boundary, both $$\tilde{E}_1$$ and $$1/\mu \partial\tilde{E}_1/\partial x_3$$ must be continuous.

Similarly, for TM (transverse magnetic) waves, we have

H_1(x_2,x_3) = \tilde{H}_1(x_3)~e^{\pm i~k_2~x_2} $$ and the ODE is
 * $$ \text{(6)} \qquad

\left[ \epsilon(x_3)~\cfrac{d }{d x_3}\left(\cfrac{1}{\epsilon(x_3)}~\cfrac{d }{d x_3}  \right) + \omega^2~\epsilon(x_3)~\mu(x_3) - k_2^2\right]~\tilde{H}_1 = 0 ~. $$