Waves in composites and metamaterials/Effective tensors using Hilbert space formalism

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[edit] Recap

In the previous lecture we introduced the Hilbert space $\mathcal{H}$ of of square-integrable, $Q$ -periodic, complex vector fields with the inner product

$(\mathbf{a}, \mathbf{b}) = \int_Q \overline{\mathbf{a}(\mathbf{x})}\cdot\mathbf{b}(\mathbf{x})~\text{d}\mathbf{x} \equiv \int_{\hat{Q}} \widehat{\mathbf{a}}(\mathbf{k})\cdot\widehat{\mathbf{b}}(\mathbf{k})~\text{d}\mathbf{k} ~.$

We then decomposed $\mathcal{H}$ into three orthogonal subspaces.

The subspace $\mathcal{U}$ of uniform fields, i.e., $\mathbf{a}(\mathbf{x})$ is independent of $\mathbf{x}$ , or in Fourier space, $\widehat{\mathbf{a}}(\mathbf{k}) = \boldsymbol{0}$ unless $\mathbf{k} = \boldsymbol{0}$ .
The subspace $\mathcal{J}$ of zero divergence, zero average value fields, i.e., $\boldsymbol{\nabla} \cdot \mathbf{a}(\mathbf{x}) = 0$ and $\langle \mathbf{a}(\mathbf{x}) \rangle = \boldsymbol{0}$ , or in Fourier space, $\widehat{\mathbf{a}}(\boldsymbol{0}) = \boldsymbol{0}$ and $\mathbf{k}\cdot\widehat{\mathbf{a}}(\mathbf{k}) = 0$ .
The subspace $\mathcal{E}$ of zero curl, zero average value fields, i.e., $\boldsymbol{\nabla} \times \mathbf{a}(\mathbf{x}) = \boldsymbol{0}$ and $\langle \mathbf{a}(\mathbf{x}) \rangle = \boldsymbol{0}$ , or in Fourier space, $\widehat{\mathbf{a}}(\boldsymbol{0}) = \boldsymbol{0}$ and $\widehat{\mathbf{a}}(\mathbf{k}) = \alpha(\mathbf{k})~\mathbf{k}$ .

To determine the effective permittivity, we introduced the operator $\boldsymbol{\mathit{\Gamma}}$ with the properties

$\text{(1)} \qquad \mathbf{b}(\mathbf{x}) = \boldsymbol{\mathit{\Gamma}}\cdot\mathbf{a}(\mathbf{x})$

if and only if

$\mathbf{b} \in \mathcal{E} \qquad \text{and} \qquad \mathbf{a} - \boldsymbol{\epsilon}_0\cdot\mathbf{b} \in \mathcal{U} \oplus \mathcal{J} \implies \boldsymbol{\nabla} \cdot (\mathbf{a} - \boldsymbol{\epsilon}_0\cdot\mathbf{b}) = 0 ~.$

We also found that in Fourier space,

$\widehat{\mathbf{b}}(\mathbf{k}) = \boldsymbol{\mathit{\widehat{\Gamma}}}(\mathbf{k})\cdot\widehat{\mathbf{a}}(\mathbf{k})$

where

$\text{(2)} \qquad \boldsymbol{\mathit{\widehat{\Gamma}}}(\mathbf{k}) = \begin{cases} \cfrac{\mathbf{k}\otimes\mathbf{k}}{\mathbf{k}\cdot\boldsymbol{\epsilon}_0\cdot\mathbf{k}} = \boldsymbol{\mathit{\Gamma}}(\mathbf{n}) & \qquad\text{with}~~ \mathbf{n} = \cfrac{\mathbf{k}}{|\mathbf{k}|} ~~\text{if}~~ \mathbf{k} \ne 0 \\ \boldsymbol{\mathit{0}} & \qquad\text{if}~~\mathbf{k} = 0 ~. \end{cases}$

[edit] Deriving a formula for the effective permittivity

Let us now derive a formula for the effective tensor. Recall that the polarization is defined as

$\text{(3)} \qquad \mathbf{P} = (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\mathbf{E} = \mathbf{D} - \boldsymbol{\epsilon}_0\cdot\mathbf{E}$

where the permittivity tensor is being thought of as an operator that operates on $\mathbf{E}$ .

Also notice that

$\text{(4)} \qquad \langle \mathbf{E} \rangle - \mathbf{E} \in \mathcal{E} \qquad \text{(curl free)}.$

From (3) we have

$\text{(5)} \qquad \mathbf{P} - \boldsymbol{\epsilon}_0\cdot[\langle \mathbf{E} \rangle - \mathbf{E}] = \mathbf{D} - \boldsymbol{\epsilon}_0\cdot\langle \mathbf{E} \rangle \in \mathcal{U}\oplus\mathcal{J} \qquad \text{(divergence free)}.$

From the definition of $\boldsymbol{\mathit{\Gamma}}$ (equations (1) and (2)) and using (4) and (5), we can show that

$\text{(6)} \qquad \boldsymbol{\mathit{\Gamma}}\cdot\mathbf{P} = \langle \mathbf{E} \rangle - \mathbf{E} ~.$

Let $\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0$ act on both sides of (6). Then we get

$(\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\boldsymbol{\mathit{\Gamma}}\cdot\mathbf{P} = (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\langle \mathbf{E} \rangle - (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\mathbf{E} = (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\langle \mathbf{E} \rangle - \mathbf{P} ~.$

Therefore,

$\text{(7)} \qquad [\boldsymbol{\mathit{1}} + (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\boldsymbol{\mathit{\Gamma}}]\cdot\mathbf{P} = (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\langle \mathbf{E} \rangle ~.$

Inverting (7) gives

$\text{(8)} \qquad \mathbf{P} = [\boldsymbol{\mathit{1}} + (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\boldsymbol{\mathit{\Gamma}}]^{-1}\cdot (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\langle \mathbf{E} \rangle ~.$

Averaging (8) gives

$\text{(9)} \qquad \langle \mathbf{P} \rangle = \langle [\boldsymbol{\mathit{1}} + (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\boldsymbol{\mathit{\Gamma}}]^{-1} \cdot (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0) \rangle\cdot\langle \mathbf{E} \rangle ~.$

Averaging (3) leads to the relation

$\text{(10)} \qquad \langle \mathbf{P} \rangle = (\boldsymbol{\epsilon}_\text{eff} - \boldsymbol{\epsilon}_0)\cdot\langle \mathbf{E} \rangle ~.$

Comparing (9) and (10) shows us that

$\text{(11)} \qquad { \boldsymbol{\epsilon}_\text{eff} = \boldsymbol{\epsilon}_0 + \langle [\boldsymbol{\mathit{1}} + (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\boldsymbol{\mathit{\Gamma}}]^{-1} \cdot(\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\rangle~. }$

Recall from the previous lecture that, in equation (11), the operator $\boldsymbol{\epsilon}$ is local in real space while the operator $\boldsymbol{\mathit{\Gamma}}$ is local in Fourier space. It is therefore not obvious how one can invert $[\boldsymbol{\mathit{1}} + (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\boldsymbol{\mathit{\Gamma}}]$ .

Let us define

$\boldsymbol{\delta} := \boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0~.$

Then (8) can be written as

$\mathbf{P} = [\boldsymbol{\mathit{1}} + \boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}}]^{-1}\cdot \boldsymbol{\delta}\cdot\langle \mathbf{E} \rangle ~.$

Assuming that $-\boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}} < \boldsymbol{\mathit{1}}$ , we can expand the first operator in terms of an infinite series, i.e.,

$[\boldsymbol{\mathit{1}} + \boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}}]^{-1} = \sum_{n=0}^\infty (-\boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}})^n ~.$

Then we have

$\mathbf{P} = \left[\sum_{n=0}^\infty (-\boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}})^n\right]\cdot \boldsymbol{\delta}\cdot\langle \mathbf{E} \rangle ~.$

Also, from the definition of $\mathbf{P}$ , we have

$\mathbf{P} = \boldsymbol{\delta}\cdot\mathbf{E} ~.$

Hence,

$\boldsymbol{\delta}\cdot\mathbf{E} = \left[\sum_{n=0}^\infty (-\boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}})^n\right]\cdot \boldsymbol{\delta}\cdot\langle \mathbf{E} \rangle ~.$

Now,

$(-\boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}})^n\cdot\boldsymbol{\delta} = (-1)^n~ \boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}}\cdot\boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}}\dots\boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}} \cdot\boldsymbol{\delta} = \boldsymbol{\delta}\cdot(-\boldsymbol{\mathit{\Gamma}}\cdot\boldsymbol{\delta})^n ~.$

Therefore,

$\boldsymbol{\delta}\cdot\mathbf{E} = \boldsymbol{\delta}\cdot\left[\sum_{n=0}^\infty (-\boldsymbol{\mathit{\Gamma}}\cdot\boldsymbol{\delta})^n\right]\cdot \langle \mathbf{E} \rangle$

or,

$\text{(12)} \qquad \mathbf{E} = \left[\sum_{n=0}^\infty (-\boldsymbol{\mathit{\Gamma}}\cdot\boldsymbol{\delta})^n\right]\cdot \langle \mathbf{E} \rangle ~.$

Let us now define

$\text{(13)} \qquad \begin{align} \mathbf{P}^{(m)} & := \left[\sum_{n=0}^m (-\boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}})^n\right]\cdot \boldsymbol{\delta}\cdot\langle \mathbf{E} \rangle \\ \mathbf{E}^{(m)} & := \left[\sum_{n=0}^m (-\boldsymbol{\mathit{\Gamma}}\cdot\boldsymbol{\delta})^n\right]\cdot\langle \mathbf{E} \rangle ~. \end{align}$

Then we can write

$\text{(14)} \qquad \boldsymbol{\mathit{\Gamma}}\cdot\mathbf{P}^{(m)} = \langle \mathbf{E} \rangle - \mathbf{E}^{(m+1)} \qquad \text{and} \qquad \boldsymbol{\delta} \cdot \mathbf{E}^{(m)} = \mathbf{P}^{(m)} ~.$

These recurrence relations may be used to compute these fields inductively. An algorithm that may be used is outlined below:

Set $m = 0$ . Then

$\mathbf{E}^{(0)} = \langle \mathbf{E} \rangle ~.$

Compute $\mathbf{P}^{(m)}(\mathbf{x})$ in real space using the relation

$\mathbf{P}^{(m)} = \boldsymbol{\delta}\cdot\mathbf{E}^{(m)} ~.$

Take a fast Fourier transform to find $\widehat{\mathbf{P}}^{(m)}(\mathbf{k})$ .
From (14) we get

$\widehat{\mathbf{E}}^{(m+1)} = \begin{cases} - \boldsymbol{\mathit{\widehat{\Gamma}}}(\mathbf{k})\cdot\widehat{\mathbf{P}}^{(m)}(\mathbf{k}) & \text{for}~~\mathbf{k} \ne 0 \\ \langle \mathbf{E} \rangle & \text{for}~~\mathbf{k} = 0 ~. \end{cases}$

Compute $\widehat{\mathbf{E}}^{(m+1)}(\mathbf{k})$ in Fourier space.
Take an inverse fast Fourier transform to find $\mathbf{E}^{(m+1)}(\mathbf{x})$ in real space.
Increment $m$ by 1 and repeat.

This is the method of Moulinec and Suquet (Mouli94). The method also extends to nonlinear problems (Mouli98). However, there are other iterative methods that have faster convergence.

[edit] Convergence of expansions

For simplicity, let us assume that $\boldsymbol{\epsilon}_0$ is isotropic, i.e., $\boldsymbol{\epsilon}_0 = \epsilon_0~\boldsymbol{\mathit{1}}$ . Then,

$\boldsymbol{\mathit{\Gamma}} = \cfrac{\boldsymbol{\mathit{\Gamma}}_1}{\epsilon_0}$

where $\boldsymbol{\mathit{\Gamma}}_1$ is the projection from $\mathcal{H}$ onto $\mathcal{E}$ .

Define the norm of a field $\mathbf{P}$ as

$|\mathbf{P}| := (\mathbf{P}, \mathbf{P})^{1/2} = \langle \overline{\mathbf{P} \rangle\cdot\mathbf{P}}^{1/2} ~.$

Also, define the norm of a linear operator $\boldsymbol{A}$ acting on $\mathbf{P}$ as

$\lVert\boldsymbol{A}\rVert_{} := \sup_{\mathbf{P}\in\mathcal{H}~,~ |\mathbf{P}| = 1} |\boldsymbol{A}\cdot\mathbf{P}| ~.$

Therefore,

$|\boldsymbol{A}\cdot\mathbf{P}| \le \lVert\boldsymbol{A}\rVert_{}\cdot|\mathbf{P}| ~.$

Hence,

$|\boldsymbol{A}\cdot\boldsymbol{B}\cdot\mathbf{P}| \le \lVert\boldsymbol{A}\rVert_{}\cdot|\boldsymbol{B}\cdot\mathbf{P}| \le \lVert\boldsymbol{A}\rVert_{}\cdot\lVert\boldsymbol{B}\rVert\cdot|\mathbf{P}| ~.$

So

$\lVert\boldsymbol{A}\cdot\boldsymbol{B}\rVert_{} \le \lVert\boldsymbol{A}\rVert_{}\cdot\lVert\boldsymbol{B}\rVert_{} ~.$

In addition, we have the triangle inequality

$\lVert\boldsymbol{A} + \boldsymbol{B}\rVert_{} \le \lVert\boldsymbol{A}\rVert_{} + \lVert\boldsymbol{B}\rVert_{} ~.$

So from (12) and (13), we have

$\begin{align} |\mathbf{E} - \mathbf{E}^{(m)}| & = \left|\sum_{n=m+1}^\infty (-\boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}})^n\cdot\langle \mathbf{E} \rangle\right| \\ & \le \lVert\sum_{n=m+1}^\infty (-\boldsymbol{\delta}\cdot\boldsymbol{\mathit{\Gamma}})^n\rVert_{} \cdot|\langle \mathbf{E} \rangle| \\ & \le \left(\sum_{n=m+1}^\infty \lVert\cfrac{\boldsymbol{\delta}}{\epsilon_0}\rVert^n\cdot \lVert\boldsymbol{\mathit{\Gamma}}_1\rVert_{}^n\right) \cdot|\langle \mathbf{E} \rangle| ~. \end{align}$

But $\lVert\boldsymbol{\mathit{\Gamma}}_1\rVert_{} = 1$ since it is a projection onto $\mathcal{E}$ . Hence,

$|\mathbf{E} - \mathbf{E}^{(m)}| \le \left(\sum_{n=m+1}^\infty \lVert\cfrac{\boldsymbol{\delta}}{\epsilon_0}\rVert^n\right) \cdot|\langle \mathbf{E} \rangle| ~.$

Therefore the series converges provided

$\lVert\cfrac{\boldsymbol{\delta}}{\epsilon_0}\rVert < 1 ~.$

In that case

$|\mathbf{E} - \mathbf{E}^{(m)}| \le \cfrac{\gamma^{m+1}}{1-\gamma}~|\langle \mathbf{E} \rangle|$

where

$\gamma := \lVert\cfrac{\boldsymbol{\delta}}{\epsilon_0}\rVert_{}~.$

To get a better understanding of the norm $γ$ , let us consider a $n$ -phase composite with isotropic phases, i.e.,

$\boldsymbol{\epsilon} = \sum_{i=1}^n \chi_i~\epsilon_i~\boldsymbol{\mathit{1}}$

where

$\chi_i = \begin{cases} 1 & \text{in phase}~~i \\ 0 & \text{otherwise}. \end{cases}$

In this case,

$\gamma = \lVert \cfrac{\boldsymbol{\delta}}{\epsilon_0} \rVert = \sup_{\mathbf{P}\in\mathcal{H}~,~|\mathbf{P}|=1} \sum_{i=1}^n \left\langle \overline{\left(\cfrac{\epsilon_i-\epsilon_0} {\epsilon_0}~\mathbf{P}_i\right)} \cdot \left(\cfrac{\epsilon_i-\epsilon_0}{\epsilon_0}~\mathbf{P}_i\right)\right\rangle^{1/2}$

where

$\mathbf{P}_i = \chi_i~\mathbf{P} \quad \text{and} \quad |\mathbf{P}| = \left(\sum_i \overline{\mathbf{P}_i}\cdot\mathbf{P}\right)^{1/2} = 1 ~.$

Hence,

$\gamma = \sup_{\mathbf{P}\in\mathcal{H}~,~|\mathbf{P}|=1} \sum_{i=1}^n \left|\cfrac{\epsilon_i-\epsilon_0}{\epsilon_0}\right| \langle \overline{\mathbf{P}}_i \cdot\mathbf{P}_i\rangle^{1/2} ~.$

Since, $\langle \overline{\mathbf{P}}_i \cdot\mathbf{P}_i\rangle^{1/2}$ are weights, it makes sense to put the weights where $ε i$ is maximum. Hence, we can write

$\gamma = \max_i \left|\cfrac{\epsilon_i-\epsilon_0}{\epsilon_0}\right| = \max_i \cfrac{|\epsilon_i-\epsilon_0|}{|\epsilon_0|} ~.$

For $γ$ to be less than 1, we therefore require that, for all $i$ ,

$|\epsilon_i-\epsilon_0| < |\epsilon_0| ~.$

A geometrical representation of this situation is shown in Figure 1.

Figure 1. Allowable values of

ε i

for convergence of series expansion.

If the value of $ε 0$ is sufficiently large, then we get convergence if all the $ε i$ s lie in one half of the complex plane (shown by the green line in the figure).

Similarly, we can expand

$\boldsymbol{\epsilon}_\text{eff} = \boldsymbol{\epsilon}_0 + \langle [\boldsymbol{\mathit{1}} + (\boldsymbol{\epsilon} - \boldsymbol{\epsilon}_0)\cdot\boldsymbol{\mathit{\Gamma}}]^{-1} \cdot(\boldsymbol{\epsilon}-\boldsymbol{\epsilon}_0)\rangle$

in the form

$\text{(15)} \qquad \boldsymbol{\epsilon}_\text{eff} = \boldsymbol{\epsilon}_0 + \sum_{j=0}^{\infty} \boldsymbol{\mathit{\Gamma}}_0\cdot\boldsymbol{\delta} \cdot(-\boldsymbol{\mathit{\Gamma}}\cdot\boldsymbol{\delta})^j\cdot\boldsymbol{\mathit{\Gamma}}_0$

where $\boldsymbol{\mathit{\Gamma}}_0$ is a projection onto $\mathcal{U}$ , i.e.,

$\boldsymbol{\mathit{\Gamma}}_0\cdot\mathbf{P} = \langle \mathbf{P} \rangle ~.$

In this case, we find that the series converges provided

$|\epsilon_i-\epsilon_0| < |\epsilon_0| \qquad \text{for all}~~i ~.$

Note that each term in (15) is an analytic function of $ε 1$ (in fact a polynomial). So, if we truncate the series, we have an analytic function of $ε 1$ .

Since a sequence of analytic functions which is uniformly convergent in a domain converges to a function which is analytic in that domain (see, for example, Rudin76), we deduce that $\boldsymbol{\epsilon}_\text{eff}$ is an analytic function of $ε 1$ in the $ε 0$ disk (see Figure~1) with $r < | ε 0 |$ provided $| ε i - ε 0 | < ε 0$ for $i = 2, 3, \dots, n$ .

Similarly, the effective tensor is an analytic function of $ε 2$ , $ε 3$ , etc.

Since $\boldsymbol{\epsilon}_\text{eff}$ is independent of $ε 0$ , by taking the union of all such regions of analyticity, we conclude that $\boldsymbol{\epsilon}_\text{eff}$ is an analytic function of $\epsilon_1, \dots, \epsilon_n$ provided all these $ε i$ s lie inside a half-plane (see Figure~1). This means that there exists a $θ$ such that

$\text{Re}(e^{i\theta}~\epsilon_j) > 0 \quad \text{for all}~ j ~.$

[edit] Corollary:

A corollary of the above observations is the following. If each $ε i (ω)$ is an analytic function of $ω$ for $Im(ω) > 0$ (which is what one expects with $ω$ as the frequency) and $Im[ε i (ω)] > 0$ for all $ω$ with $Reω > 0$ , then $\boldsymbol{\epsilon}_\text{eff}(\omega)$ will be analytic in $ω$ .

[edit] Another interesting property:

Now, if $\mathbf{D} = \boldsymbol{\epsilon}\cdot\mathbf{E}$ , we have

$\lambda~\mathbf{D} = \lambda~\boldsymbol{\epsilon}\cdot\mathbf{E} \qquad \text{for all constants}~\lambda ~.$

Therefore,

$\langle \mathbf{D} \rangle = \boldsymbol{\epsilon}_\text{eff}\cdot\langle \mathbf{E} \rangle \qquad \implies \qquad \langle \lambda~\mathbf{D} \rangle = \lambda~\boldsymbol{\epsilon}_\text{eff}\cdot\langle \mathbf{E} \rangle ~.$

This means that

$(\lambda~\boldsymbol{\epsilon})_\text{eff} = \lambda~\boldsymbol{\epsilon}_\text{eff} ~.$

Therefore, $\boldsymbol{\epsilon}_\text{eff}(\epsilon_1,\epsilon_2, \dots, \epsilon_n)$ is homogeneous of degree one and

${ \boldsymbol{\epsilon}_\text{eff}(\lambda~\epsilon_1, \lambda~\epsilon_2, \dots, \lambda~\epsilon_n) = \lambda~\boldsymbol{\epsilon}_\text{eff}(\epsilon_1, \epsilon_2, \dots, \epsilon_n)~. }$

For a two-phase composite, if we take

$\lambda = \cfrac{1}{\epsilon_2}$

we get

$\boldsymbol{\epsilon}_\text{eff}(\epsilon_1,\epsilon_2) = \epsilon_2~\boldsymbol{\epsilon}_\text{eff}(\epsilon_1/\epsilon_2, 1) ~.$

Therefore, it suffices to study the analytic function $\boldsymbol{\epsilon}_\text{eff}(\epsilon_1, 1) = \boldsymbol{\epsilon}_\text{eff}(\epsilon_1)$ . For further details see Milton02.

[edit] References

[Milton02] G. W. Milton. Theory of Composites. Cambridge University Press, New York, 2002.
[Mouli94] H. Moulinec and P. Suquet. A fast numerical method for computing the linear and nonlinear mechanical properties of composites. Comptes rendus de l'Académie des sciences II, 318(11):1417--1423, 1994.
[Mouli98] H. Moulinec and P. Suquet. A numerical method for computing the overall response of nonlinear composites with complex microstructure. Comput. Methods Appl. Mech. Engrg., 157:69--94, 1998.
[Rudin76] W. Rudin. Principles of Mathematical Analysis. McGraw-Hill, New York, 1976.

Waves in composites and metamaterials/Effective tensors using Hilbert space formalism

From Wikiversity

Contents

[edit] Recap

[edit] Deriving a formula for the effective permittivity

[edit] Convergence of expansions

[edit] Corollary:

[edit] Another interesting property:

[edit] References

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