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

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.

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.

 * 1) 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 $$\mathbf{k} = \boldsymbol{0}$$.
 * 2) 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 $$\mathbf{k}\cdot\widehat{\mathbf{a}}(\mathbf{k}) = 0$$.
 * 3) 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} $$

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.

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_{}~. $$ To get a better understanding of the norm $$\gamma$$, let us consider a $$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}$$ are weights, it makes sense to put the weights where $$\epsilon_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|} ~. $$ For $$\gamma$$ to be less than 1, we therefore require that, for all $$i$$,

|\epsilon_i-\epsilon_0| < |\epsilon_0| ~. $$ A geometrical representation of this situation is shown in Figure 1. If the value of $$\epsilon_0$$ is sufficiently large, then we get convergence if all the $$\epsilon_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 ~. $$ Note that each term in (15) is an analytic function of $$\epsilon_1$$ (in fact a polynomial). So, if we truncate the series, we have an analytic function of $$\epsilon_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 $$\epsilon_1$$ in the $$\epsilon_0$$ disk (see Figure~1) with $$r < |\epsilon_0|$$ provided $$|\epsilon_i - \epsilon_0| < \epsilon_0$$ for $$i = 2, 3, \dots, n$$.

Similarly, the effective tensor is an analytic function of $$\epsilon_2$$, $$\epsilon_3$$, etc.

Since $$\boldsymbol{\epsilon}_\text{eff}$$ is independent of $$\epsilon_0$$, 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 $$\epsilon_i$$ s lie inside a half-plane (see Figure~1). This means that there exists a $$\theta$$ such that

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

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

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.


text:
Waves in composites and metamaterials/Transformation-based cloaking in electromagnetism

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.

Introduction
In this lecture we will give a brief description of cloaking in the context of conductivity. It is useful to start off with a desciption of some variational principles for electrical conductivity at this stage.

Variational principle
Suppose that the electrical conductivity $$\boldsymbol{\sigma}(\mathbf{x})$$ is real and symmetric. Also assume that

\alpha~\boldsymbol{\mathit{1}} \ge \boldsymbol{\sigma}(\mathbf{x}) \ge \beta~\boldsymbol{\mathit{1}} \qquad\text{for all}\alpha,\beta > 0 ~. $$ Consider the body ($$\Omega$$) with boundary ($$\partial\Omega$$) shown in Figure 1. We would like to minimize the power dissipation into heat inside the body. This statement can be expressed as

\min_{u ~,~ u=u_0~\rm{on}~\partial\Omega} W(u) $$ where

W(u) = \int_{\Omega} \boldsymbol{\nabla} u\cdot\boldsymbol{\sigma}\cdot\boldsymbol{\nabla} u~\text{d}\Omega ~. $$ Now consider a variation $$v$$ where $$v = 0$$ on $$\partial\Omega$$ and let $$\delta$$ be a small parameter. Then

\begin{align} W(u + \delta~v) & = \int_{\Omega} \boldsymbol{\nabla} (u+\delta~v)\cdot\boldsymbol{\sigma}\cdot\boldsymbol{\nabla} (u+\delta~v)~\text{d}\Omega \\ & =   \int_{\Omega} \boldsymbol{\nabla} u\cdot\boldsymbol{\sigma}\cdot\boldsymbol{\nabla} u~\text{d}\Omega + 2~\delta~\int_{\Omega} \boldsymbol{\nabla} v\cdot\boldsymbol{\sigma}\cdot\boldsymbol{\nabla} u~\text{d}\Omega + \delta^2~\int_{\Omega} \boldsymbol{\nabla} v\cdot\boldsymbol{\sigma}\cdot\boldsymbol{\nabla} v~\text{d}\Omega > 0 ~. \end{align} $$ Using the identity

\mathbf{a}\cdot\boldsymbol{\nabla} b = \boldsymbol{\nabla} \cdot (b~\mathbf{a}) - b~\boldsymbol{\nabla} \cdot \mathbf{a} $$ in the middle term on the right hand side leads to

W(u + \delta~v) = \int_{\Omega} \boldsymbol{\nabla} u\cdot\boldsymbol{\sigma}\cdot\boldsymbol{\nabla} u~\text{d}\Omega + 2~\delta~\int_{\Omega} \boldsymbol{\nabla} \cdot (v~\boldsymbol{\sigma}\cdot\boldsymbol{\nabla u)}~\text{d}\Omega - 2~\delta~\int_{\Omega} v~\boldsymbol{\nabla} \cdot (\boldsymbol{\sigma}\cdot\boldsymbol{\nabla u)}~\text{d}\Omega + \delta^2~\int_{\Omega} \boldsymbol{\nabla} v\cdot\boldsymbol{\sigma}\cdot\boldsymbol{\nabla} v~\text{d}\Omega ~. $$ From the divergence theorem, we have

\int_{\Omega} \boldsymbol{\nabla} \cdot (v~\boldsymbol{\sigma}\cdot\boldsymbol{\nabla u)}~\text{d}\Omega = \int_{\partial\Omega} \mathbf{n}\cdot(v~\boldsymbol{\sigma}\cdot\boldsymbol{\nabla} u)~\text{d}\Gamma $$ where $$\mathbf{n}$$ is the outward unit normal to the surface $$\partial\Omega$$ and $$\Gamma \equiv \partial\Omega$$. Since $$v = 0$$ on $$\partial\Omega$$, we have

\int_{\Omega} \boldsymbol{\nabla} \cdot (v~\boldsymbol{\sigma}\cdot\boldsymbol{\nabla u)}~\text{d}\Omega = 0 ~. $$ Therefore,

W(u + \delta~v) = \int_{\Omega} \boldsymbol{\nabla} u\cdot\boldsymbol{\sigma}\cdot\boldsymbol{\nabla} u~\text{d}\Omega - 2~\delta~\int_{\Omega} v~\boldsymbol{\nabla} \cdot (\boldsymbol{\sigma}\cdot\boldsymbol{\nabla u)}~\text{d}\Omega + \delta^2~\int_{\Omega} \boldsymbol{\nabla} v\cdot\boldsymbol{\sigma}\cdot\boldsymbol{\nabla} v~\text{d}\Omega ~. $$ For $$W(u+\delta v)$$ to be positive for all $$v$$, it is sufficient to have

\int_{\Omega} v~\boldsymbol{\nabla} \cdot (\boldsymbol{\sigma}\cdot\boldsymbol{\nabla u)}~\text{d}\Omega = 0 ~. $$ If this is to be true for all $$v$$, then

{  \boldsymbol{\nabla} \cdot (\boldsymbol{\sigma}\cdot\boldsymbol{\nabla u)} = 0 ~. } $$ If we define the flux as

\mathbf{J}(\mathbf{x}) := \boldsymbol{\sigma}(\mathbf{x})\cdot\boldsymbol{\nabla} u $$ then we have

{ \boldsymbol{\nabla} \cdot \mathbf{J}(\mathbf{x}) = 0 ~. } $$

Coordinate transformation equations for currents
Let us take new curvilinear coordinates $$\mathbf{x}'(\mathbf{x})$$ as shown in Figure 2. The new coordinates are material coordinates.

The Jacobian of the transformation $$\mathbf{x}' \rightarrow \mathbf{x}$$ is given by

J = \det(\boldsymbol{A}) ~; A_{ij} := \frac{\partial x'_i}{\partial x_j} ~. $$ Then an infinitesimal volume $$\text{d}\Omega$$ of the body transforms as

\text{d}\Omega' \rightarrow J~\text{d}\Omega ~. $$ Recall that

W(u) = \int_{\Omega} \frac{\partial u}{\partial x_i}~\sigma_{ij}~\frac{\partial u}{\partial x_j}~\text{d}\Omega ~. $$ Then, using the chain rule, we get

W(u) = \int_{\Omega'} \left(\frac{\partial u}{\partial x'_m}~\frac{\partial x'_m}{\partial x_i}\right) ~\sigma_{ij}~\left(\frac{\partial x'_l}{\partial x_j}~\frac{\partial u}{\partial x'_l}\right) ~\cfrac{1}{J}~\text{d}\Omega' $$ or,

W(u) = \int_{\Omega'} \frac{\partial u}{\partial x'_m}~\sigma'_{ml}~\frac{\partial u}{\partial x'_l} ~\text{d}\Omega' $$ where

\sigma'_{ml} = \cfrac{1}{J}~\frac{\partial x'_m}{\partial x_i}~\sigma_{ij}~ \frac{\partial x'_l}{\partial x_j} = \cfrac{1}{J}~A_{mi}~\sigma_{ij}~A_{lj} ~. $$ Hence, in the transformed coordinates, the functional $$W(u)$$ takes the form

W(u) = \int_{\Omega'} \boldsymbol{\nabla}' u\cdot\boldsymbol{\sigma}'\cdot\boldsymbol{\nabla}' u~\text{d}\Omega' $$ where $$\boldsymbol{\nabla}' (\bullet)$$ denotes a gradient with respect to the $$\mathbf{x}'$$ coordinates and the conductivity transforms as

\boldsymbol{\sigma}'(\mathbf{x}') = \cfrac{1}{J}~\boldsymbol{A}(\mathbf{x})\cdot\boldsymbol{\sigma}(\mathbf{x})\cdot\boldsymbol{A}^T(\mathbf{x}) ~. $$

Interpretation
We can now interpret the minimization problem in the transformed coordinates as follows:
 * The function $$u'(x') = u(x')$$ minimizes $$W$$ in a body $$\Omega'$$     filled with material with conductivity $$\boldsymbol{\sigma}'(\mathbf{x}')$$ with $$x'_1,x'_2,x'_3$$      as Cartesian coordinates in $$x'$$ space.

Therefore, for $$W$$ to remain positive, we must have

\mathbf{J}'(\mathbf{x}') = \boldsymbol{\sigma}'(\mathbf{x}') \cdot \boldsymbol{\nabla}' u(\mathbf{x}') = \cfrac{1}{J}~[\boldsymbol{A}(\mathbf{x})\cdot\boldsymbol{\sigma}(\mathbf{x})\cdot\boldsymbol{A}^T(\mathbf{x})] \cdot \boldsymbol{\nabla}' u(\mathbf{x}')~. $$ Now,

[\boldsymbol{\nabla} u]_i = \frac{\partial u}{\partial x_i} = \frac{\partial x'_m}{\partial x_i}~\frac{\partial u}{\partial x'_m} = A_{mi}~\frac{\partial u}{\partial x'_m} = [\boldsymbol{A}^T\cdot\boldsymbol{\nabla}' u]_i ~. $$ Hence,

\mathbf{J}'(\mathbf{x}') = \cfrac{1}{J}~[\boldsymbol{A}(\mathbf{x})\cdot\boldsymbol{\sigma}(\mathbf{x})\cdot\boldsymbol{A}^T(\mathbf{x})] \cdot [\boldsymbol{A}^T(\mathbf{x})]^{-1}\cdot\boldsymbol{\nabla} u(\mathbf{x}) $$ or,

{ \mathbf{J}'(\mathbf{x}') = \cfrac{1}{J}~\boldsymbol{A}(\mathbf{x})\cdot\boldsymbol{\sigma}(\mathbf{x})\cdot\boldsymbol{\nabla} u(\mathbf{x}) = \cfrac{\boldsymbol{A}\cdot\mathbf{J}(\mathbf{x})}{\det(\boldsymbol{A})} ~. } $$ This is the transformation law for currents. Using the same arguments as before, we can show that

{ \boldsymbol{\nabla}'\cdot\mathbf{J}'(\mathbf{x}') = 0 ~. } $$ Let the electric field $$\mathbf{E}$$ be derived from the potential $$u$$. Then the fields

\mathbf{E}(\mathbf{x}) = \boldsymbol{\nabla} u(\mathbf{x}) \qquad \text{and} \qquad \mathbf{E}'(\mathbf{x}') = \boldsymbol{\nabla}' u(\mathbf{x}') $$ are related via

{  \mathbf{E}'(\mathbf{x}') = (\boldsymbol{A}^T)^{-1}\cdot\mathbf{E}(\mathbf{x}) ~. } $$ Therefore, there are two transformations which are equivalent. However, an isotropic material transforms to an anisotropic material via the transformation equation for conductivity.

Electrical tomography
Consider the situation shown in Figure 1. Let the conductivity of the body be $$\boldsymbol{\sigma}(\mathbf{x})$$ and let us require that $$\boldsymbol{\nabla} \cdot (\boldsymbol{\sigma}\cdot\boldsymbol{\nabla u)} = 0$$ inside the body. In electrical tomography one measures the current flux $$\mathbf{n}\cdot\mathbf{J}(\mathbf{x})$$ at the surface for all choices of the potential $$u_0$$.

Suppose one knows the Dirchlet to Neumann map ($$\varphi$$)

\varphi : u_0 \rightarrow \mathbf{n}\cdot\mathbf{J}(\mathbf{x}) = g(\mathbf{x}) ~. $$ Can one find $$\boldsymbol{\sigma}(\mathbf{x})$$? No, not generally. Figure 3 illustrates why that is the case. For the body in the figure, the transformation is $$\mathbf{x}' = \mathbf{x}$$ outside the blue region while inside the blue region $$\mathbf{x}' \ne \mathbf{x}$$. Also, outside the blue region, $$\Omega' = \Omega$$, $$\mathbf{J}' = \mathbf{J}$$, and $$u' = u$$. Inside the blue region $$\Omega' \ne \Omega$$ and $$\mathbf{J}'$$ is obtained via the transformation rule.

From the figure we can see that the Dirichlet-Neumman map will remain unchanged on $$\partial\Omega$$. Hence, the body appears to be exactly the same in $$\mathbf{x}'$$-space but has a different conductivity.

Even though this fact has been known for a while, there was still hope that you could determine $$\boldsymbol{\sigma}(\mathbf{x})$$ uniquely, modulo a coordinate transformation. However, such hopes were dashed when Greenleaf, Lassas, and Uhlmann provided a counterexample in 2003 (Greenleaf03).

First transformation based example of cloaking
Greenleaf et al. (Greenleaf03) provided the first example of transformation based cloaking. They considered a singular transformation

\mathbf{x}'(\mathbf{x}) = \begin{cases} \left(\cfrac{|\mathbf{x}|}{2} + 1\right)~\cfrac{\mathbf{x}}{|\mathbf{x}|} & \text{if}~|\mathbf{x}| < 2 \\ \mathbf{x} & \text{if}~|\mathbf{x}| > 2 ~. \end{cases} $$ The effect of this mapping is shown in the schematic in Figure 4. An epsilon ball at the center of $$\Omega$$ is mapped into a sphere of radius 1 in $$\Omega'$$. The value of $$\boldsymbol{\sigma}'(\mathbf{x}')$$ is singular at the boundary of this sphere. Inside the sphere of radius 1, the transformed conductivity has the form $$\boldsymbol{\sigma}'(\mathbf{x}') = h(\mathbf{x}')$$.

Therefore we can put a small body inside and the potential outside will be undisturbed by the presence of the body in the cloaking region.

Cloaking for Electromagnetism
Pendry, Schurig, and Smith (Pendry06) showed in 2006 that cloaking could be achieved for electromagnetic waves. The concept of cloaking follows from the observation that Maxwell's equations keep their form under coordinate transformations. The Maxwell's equations at fixed frequency $$\omega$$ are

\begin{align} \boldsymbol{\nabla} \times \mathbf{E} + i\omega\boldsymbol{\mu}\cdot\mathbf{H} &= \boldsymbol{0} \\ \boldsymbol{\nabla} \times \mathbf{H} - i\omega\boldsymbol{\epsilon}\cdot\mathbf{E} &= \boldsymbol{0} ~. \end{align} $$ A coordinate transformation ($$\mathbf{x} \rightarrow \mathbf{x}'$$) gives us the equivalent relations

\begin{align} \boldsymbol{\nabla}'\times\mathbf{E}' + i\omega\boldsymbol{\mu}'\cdot\mathbf{H}' &= \boldsymbol{0} \\ \boldsymbol{\nabla}'\times\mathbf{H}' - i\omega\boldsymbol{\epsilon}'\cdot\mathbf{E}' &= \boldsymbol{0} \end{align} $$ with

\mathbf{x}' = \mathbf{x}'(\mathbf{x}) ~; \mathbf{E}'(\mathbf{x}') = (\boldsymbol{A}^T)^{-1}~\mathbf{E}(\mathbf{x}) ~; \mathbf{H}'(\mathbf{x}') = (\boldsymbol{A}^T)^{-1}~\mathbf{H}(\mathbf{x}) $$ where

A_{ki} = \frac{\partial x'_k}{\partial x_i} ~; [\boldsymbol{A}^{-1}]_{ij} = \frac{\partial x_i}{\partial x'_j} $$ and

\boldsymbol{\mu}'(\mathbf{x}') = \cfrac{\boldsymbol{A}\cdot\boldsymbol{\mu}(\mathbf{x})\cdot\boldsymbol{A}^T}{\det(\boldsymbol{A})} ~; \boldsymbol{\epsilon}'(\mathbf{x}') = \cfrac{\boldsymbol{A}\cdot\boldsymbol{\epsilon}(\mathbf{x})\cdot\boldsymbol{A}^T}{\det(\boldsymbol{A})} ~. $$ To see that this invariance of form under coordinate transformations does indeed hold, observe that
 * $$ \text{(1)} \qquad

-i\omega\boldsymbol{\mu}'\cdot\mathbf{H}' = \cfrac{-i\omega\boldsymbol{A}\cdot\boldsymbol{\mu}\cdot\mathbf{H}}{\det(\boldsymbol{A})} = \cfrac{\boldsymbol{A}\cdot(\boldsymbol{\nabla} \times \mathbf{E})}{\det(\boldsymbol{A})} ~. $$ We want to show that this equals $$\boldsymbol{\nabla}'\times\mathbf{E}'$$.

In index notation, (1) can be written as

\begin{align} \left[-i\omega\boldsymbol{\mu}^{'}\cdot\mathbf{H}^{'}\right]_h & = \cfrac{\left[\boldsymbol{A}\cdot\left(\boldsymbol{\nabla} \times \mathbf{E}\right)\right]_h}{\det(\boldsymbol{A})} \\ & = \cfrac{1}{\det(\boldsymbol{A})}~\frac{\partial x'_h}{\partial x_j}~\mathcal{E}_{jmk}~\frac{\partial E_k}{\partial x_m} \\ & = \cfrac{1}{\det(\boldsymbol{A})}~\frac{\partial x'_h}{\partial x_j}~\mathcal{E}_{jmk}~\frac{\partial x'_l}{\partial x_m} ~\frac{\partial E_k}{\partial x'_l} ~. \end{align} $$ On the other hand,

\begin{align} \left[\boldsymbol{\nabla}'\times\mathbf{E}'\right]_h & = \left[\boldsymbol{\nabla}'\times(\boldsymbol{A}^{-T}\cdot\mathbf{E})\right]_h \\ & = \mathcal{E}_{hlm}~\frac{\partial }{\partial x_i}\left(\frac{\partial x_k}{\partial x'_m}~E_k\right) \\ & = \mathcal{E}_{hlm}~\frac{\partial^2 x_k}{\partial x'_l \partial x'_m}~E_k + \mathcal{E}_{hlm}~\frac{\partial x_k}{\partial x'_m}~\frac{\partial E_k}{\partial x'_l} \\ & = \mathcal{E}_{hlm}~\frac{\partial x_k}{\partial x'_m}~\frac{\partial E_k}{\partial x'_l} ~. \end{align} $$ The first term above evaluates to zero because of $$\text{tr}(\boldsymbol{A}\cdot\boldsymbol{B}) = 0$$ if $$\boldsymbol{A}$$ is skew and $$\boldsymbol{B}$$ is symmetric.

So we now need to show that

\mathcal{E}_{hlm}~\frac{\partial x_k}{\partial x'_m}~\frac{\partial E_k}{\partial x'_l} = \cfrac{1}{\det(\boldsymbol{A})}~\frac{\partial x'_h}{\partial x_j}~\mathcal{E}_{jmk}~\frac{\partial x'_l}{\partial x_m} ~\frac{\partial E_k}{\partial x'_l} $$ or that,
 * $$ \text{(2)} \qquad

\det(\boldsymbol{A})~\mathcal{E}_{hlm}~\frac{\partial x_k}{\partial x'_m} = \frac{\partial x'_h}{\partial x_j}~\mathcal{E}_{jmk}~\frac{\partial x'_l}{\partial x_m} ~. $$ Multiply both sides of (2) by $$A_{pk}$$ and sum over $$k$$, (i.e., multiply by $$\boldsymbol{A}$$ which is non-singular). Then we get

\det(\boldsymbol{A})~\mathcal{E}_{hlm}~\frac{\partial x_k}{\partial x'_m}~\frac{\partial x'_p}{\partial x_k} = \mathcal{E}_{jmk}~\frac{\partial x'_h}{\partial x_j}~\frac{\partial x'_l}{\partial x_m}~\frac{\partial x'_p}{\partial x_k} $$ or,

\det(\boldsymbol{A})~\mathcal{E}_{hlm}~\frac{\partial x'_p}{\partial x'_m} = \det(\boldsymbol{A})~\mathcal{E}_{hlm}~\delta_{pm} = \det(\boldsymbol{A})~\mathcal{E}_{hlp} = \mathcal{E}_{jmk}~\frac{\partial x'_h}{\partial x_j}~\frac{\partial x'_l}{\partial x_m}~\frac{\partial x'_p}{\partial x_k} ~. $$ Both sides are completely antisymmetric with respect o $$h,l,p$$. So it suffices to take $$h = 1$$, $$l = 2$$, $$p = 3$$ and we can write

\det(\boldsymbol{A})~\mathcal{E}_{123} = \det(\boldsymbol{A}) = \mathcal{E}_{jmk}~\frac{\partial x'_1}{\partial x_j}~\frac{\partial x'_2}{\partial x_m}~\frac{\partial x'_3}{\partial x_k} ~. $$ The right hand side above is the well known formula for the determinant of the Jacobian. Hence the first of the transformed Maxwell equations holds. We can follow the same procedure to show that the second Maxwell's equation also maintains its form under coordinate transformations. Hence Maxwell's equations are invariant with respect to coordinate transformations.

license: Creative Commons - Attribution Share-Alike - https://creativecommons.org/licenses/by-sa/4.0/


text:
Waves in composites and metamaterials/Transformation-based cloaking in mechanics

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.

Recap
In the previous lecture we showed that for a material with a real, symmetric, and positive definite conductivity tensor ($$\boldsymbol{\sigma}$$), we can set up a variational principle for minimal power dissipation. We also showed that this variational principle is satisfied if

\boldsymbol{\nabla} \cdot (\boldsymbol{\sigma}\cdot\boldsymbol{\nabla u)} = \boldsymbol{\nabla} \cdot \mathbf{J}(\mathbf{x}) = 0 ~. $$ Next we showed that under a coordinate transformation from $$\mathbf{x} \rightarrow \mathbf{x}'$$, the conductivity tensor transforms as

\boldsymbol{\sigma}'(\mathbf{x}') = \cfrac{1}{J}~\boldsymbol{A}(\mathbf{x})\cdot\boldsymbol{\sigma}(\mathbf{x})\cdot\boldsymbol{A}^T(\mathbf{x}) $$ where

J = \det(\boldsymbol{A}) ~; A_{ij} := \frac{\partial x'_i}{\partial x_j} ~. $$ We also found that the variational principle in transformed coordinates has the alternative interpretation that { the function $$u'(x') = u(x')$$ minimizes $$W$$ in a body $$\Omega'$$ filled with material with conductivity $$\boldsymbol{\sigma}'(\mathbf{x}')$$ with $$x'_1,x'_2,x'_3$$ as Cartesian coordinates in $$x'$$ space.}

Next we derived the transformation rule for currents:

\mathbf{J}'(\mathbf{x}') = \cfrac{1}{J}~\boldsymbol{A}(\mathbf{x})\cdot\boldsymbol{\sigma}(\mathbf{x})\cdot\boldsymbol{\nabla} u(\mathbf{x}) = \cfrac{\boldsymbol{A}\cdot\mathbf{J}(\mathbf{x})}{\det(\boldsymbol{A})} ~; \boldsymbol{\nabla}'\cdot\mathbf{J}'(\mathbf{x}') = 0 ~. $$ We also saw that the transformation rule for the electric field could be written as

\mathbf{E}'(\mathbf{x}') = (\boldsymbol{A}^T)^{-1}\cdot\mathbf{E}(\mathbf{x}) ~. $$ Next, based on insights obtained from electric tomography, we found that a transformation-based cloaking effect could be obtained using the Greenleaf-Lassas-Uhlmann mapping (Greenleaf03). This mapping is singular and of the form

\mathbf{x}'(\mathbf{x}) = \begin{cases} \left(\cfrac{|\mathbf{x}|}{2} + 1\right)~\cfrac{\mathbf{x}}{|\mathbf{x}|} & \text{if}~|\mathbf{x}| < 2 \\ \mathbf{x} & \text{if}~|\mathbf{x}| > 2 ~. \end{cases} $$ The effect of this mapping is shown in Figure 1.

Some other unusual transformations
One unusual mapping that can be used to achieve cloaking is to fold back space upon itself (Leonhardt06,Pendry06). An example of such a mapping is

\mathbf{x}' = \begin{cases} \mathbf{x} & \text{if}x_1 < 0 \\ (-x_1, x_2, x_3) & \text{if} d > x_1 > 0 \\ \mathbf{x} - (2d, 0, 0) & \text{if} x_1 > d ~. \end{cases} $$ The effect of this transformation is shown in Figure~2. Note that there is a sharp (discontinuous) fold and the separation shown in the thickness direction is simply for the purpose of illustration. In reality, space is folded upon itself and the determination of the Jacobian inside the fold is -1.

Transformations of Maxwell's equations
We also showed in the previous lecture that Maxwell's equations at fixed frequency are invariant with respect to coordinate transformations. Thus the equations in $$\mathbf{x}$$-space

\begin{align} \boldsymbol{\nabla} \times \mathbf{E} + i\omega\boldsymbol{\mu}\cdot\mathbf{H} &= \boldsymbol{0} \\ \boldsymbol{\nabla} \times \mathbf{H} - i\omega\boldsymbol{\epsilon}\cdot\mathbf{E} &= \boldsymbol{0} ~. \end{align} $$ transform, in $$\mathbf{x}'$$-space, to

\begin{align} \boldsymbol{\nabla}'\times\mathbf{E}' + i\omega\boldsymbol{\mu}'\cdot\mathbf{H}' &= \boldsymbol{0} \\ \boldsymbol{\nabla}'\times\mathbf{H}' - i\omega\boldsymbol{\epsilon}'\cdot\mathbf{E}' &= \boldsymbol{0} ~. \end{align} $$ In the transformed equations,

\mathbf{E}'(\mathbf{x}') = (\boldsymbol{A}^T)^{-1}~\mathbf{E}(\mathbf{x}) ~; \mathbf{H}'(\mathbf{x}') = (\boldsymbol{A}^T)^{-1}~\mathbf{H}(\mathbf{x}) $$ and
 * $$ \text{(1)} \qquad

\boldsymbol{\mu}'(\mathbf{x}') = \cfrac{\boldsymbol{A}\cdot\boldsymbol{\mu}(\mathbf{x})\cdot\boldsymbol{A}^T}{\det(\boldsymbol{A})} ~; \boldsymbol{\epsilon}'(\mathbf{x}') = \cfrac{\boldsymbol{A}\cdot\boldsymbol{\epsilon}(\mathbf{x})\cdot\boldsymbol{A}^T}{\det(\boldsymbol{A})} ~. $$ Let us consider the effect of the fold-back transformation shown in Figure~2 on Maxwell's equations.

Recall that this transformation has the form

\mathbf{x}' = \begin{cases} \mathbf{x} & \text{if}x_1 < 0 \\ (-x_1, x_2, x_3) & \text{if} d > x_1 > 0 \\ \mathbf{x} - (2d, 0, 0) & \text{if} x_1 > d ~. \end{cases} $$

The perfect lens
Now, let us suppose that $$\boldsymbol{\epsilon}' = \boldsymbol{\mu}' = \boldsymbol{\mathit{1}}$$ everywhere in the region. Since the Jacobian of the transformation is

A_{ki} = \frac{\partial x'_k}{\partial x_i} $$ we have

\boldsymbol{A} = \begin{cases} \boldsymbol{\mathit{1}} & \text{if} x_1 < 0 \text{or} x_1 > d \\ \begin{bmatrix} -1 & 0 & 0 \\ 0 & 1 & 0 \\ 0 & 0 & 1 \end{bmatrix} & \text{if} d > x_1 > 0 ~. \end{cases} $$ This implies that in the region $$d > x_1 > 0$$ we have

\boldsymbol{A}\cdot\boldsymbol{A}^T = \boldsymbol{\mathit{1}} ~; \det(\boldsymbol{A}) = -1 ~. $$ Since the materials in the region are isotropic, i.e., $$\boldsymbol{\mu} = \boldsymbol{\mathit{1}}$$ and $$\boldsymbol{\epsilon} = \boldsymbol{\mathit{1}}$$, then from equation (1) we see that in the region $$d > x_1 > 0$$,

\mu'(\mathbf{x}') = -1 ~; \epsilon'(\mathbf{x}') = -1 ~. $$ Therefore the fold back transformation is realized in a geometry that is equivalent to the perfect lens that we discussed earlier. Figure 3 shows the geometry involved. For a source that is less than a distance $$d$$ from the first interface, the fields blow up to infinity and there is no solution. For a solution to exist, we need to regularize the problem and add a small loss $$\delta$$, i.e., $$\epsilon=\mu = -1 + i\delta$$ in the lens. In that case, the fields blow up to infinity in two strips of length $$d - d_0$$ where $$d_0$$ is the distance of the source from the first interface. Outside this region, the fields converge to that expected by the Pendry solution (see Figure 4 for a schematic.)

If $$d - d_0 > d_0$$, i.e., $$d_0 < d/2$$, then the sources will be in a region of enormous fields. In fact, the source produces infinite energy per unit time in such regions as the loss $$\delta \rightarrow 0$$. This is clearly unphysical. So any realistic point or line source with finite energy such as a polarizable particle must have an amplitude which goes to zero as $$\delta \rightarrow 0$$. This means that the particle will have become cloaked!

Figures 5(a) and (b) show the cloaking caused by a cylindrical perfect lens with a small loss (Milton06). When a polarizable diople is located close to the lens, the field is barely perturbed. However, when the dipole is at a distance from the lens, the field shows significant perturbations.

Magnification
So far we have not dealt with the issue of magnification. Is there a coordinate transformation that leads to magnification? One such possible transformation is illustrated in Figure 6.

In this case, in the region of dilation, the transformation is

\mathbf{x}' = \beta~\mathbf{x} ~; \beta > 1 ~. $$ Therefore, the Jacobian of the transformation is

\boldsymbol{A} = \beta~\boldsymbol{\mathit{1}} ~; \det(\boldsymbol{A}) = \beta^3 ~. $$ Hence the material tensors in the region of dilation transform as

\boldsymbol{\epsilon}' = \cfrac{\boldsymbol{\epsilon}}{\beta} ~; \boldsymbol{\mu}' = \cfrac{\boldsymbol{\mu}}{\beta} ~. $$ However, in the folded region, the $$\boldsymbol{\epsilon}$$ and $$\boldsymbol{\mu}$$ tensors are anisotropic and negative. Such a transformation therefore acts like a magnifying lens.

Transformation-based cloaking in elasticity
It turns out the Willis equations in elastodynamics also transform in a manner that is very similar to the Maxwell equation in electromagnetism. Before we describe the Willis equations, let us get into a brief description of ensemble averaging (a opposed to volume averaging). The hope is that the ensemble average is a good descriptor of behavior in individual realizations.

Examples of ensembles
Some examples of ensembles are:
 * Periodic media with a period $$\delta$$ where the fields are not necessarily periodic (see Figure 7(a)). The ensemble is the material and all translations of it.  Of course,        a translation that is equal to the period gives back the same        material.
 * Media generated by some translation invariant statistical process.       This means that a particular realization and its translations are        equally likely to occur (roughly speaking).  An example is a medium        generated by a Poisson process.  We can represent the ensemble by         constructing a Voronoi tessellation and assigning constants to each        cell at random (see Figure 7(b)).
 * Media generated by some statistical process where the statistics       depend slowly on position.

Willis' equations
Recall that the equations governing the motion of a linear elastic body are
 * $$ \text{(2)} \qquad

\begin{align} & \boldsymbol{\nabla} \cdot \boldsymbol{\sigma} + \mathbf{f} = \dot{\mathbf{p}} \\ & \boldsymbol{\varepsilon} = \frac{1}{2}~[\boldsymbol{\nabla}\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})^T] \\ & \dot{\mathbf{u}} = \frac{\partial \mathbf{u}}{\partial t} ~; \dot{\mathbf{p}} = \frac{\partial \mathbf{p}}{\partial t}  \end{align} $$ where $$\mathbf{p}$$ is the momentum, $$\boldsymbol{\sigma}$$ is the stress, and $$f$$ is the body force. We assume that the body force is independent of the realization. The microscopic constitutive relations are assumed to be
 * $$ \text{(3)} \qquad

\boldsymbol{\sigma} = \boldsymbol{\mathsf{C}}\star\boldsymbol{\varepsilon} ~; \mathbf{p} = \rho~\dot{\mathbf{u}} ~. $$ Here,

\boldsymbol{\mathsf{C}}\star\boldsymbol{\varepsilon} \equiv \int_{-\infty}^t \boldsymbol{\mathsf{C}}(t-\tau) : \boldsymbol{\varepsilon}(\tau)~\text{d}\tau ~. $$ By ensemble averaging (2) we get

\begin{align} & \boldsymbol{\nabla} \cdot \left\langle \boldsymbol{\sigma} \right\rangle + \mathbf{f} = \left\langle \dot{\mathbf{p}} \right\rangle \\ & \left\langle \boldsymbol{\varepsilon} \right\rangle = \frac{1}{2}~[\boldsymbol{\nabla} \left\langle \mathbf{u} \right\rangle + (\boldsymbol{\nabla} \left\langle \mathbf{u} \right\rangle)^T] \end{align} $$ where $$\left\langle (\bullet) \right\rangle$$ is the ensemble average over realizations and not a volume average.

However, we cannot ensemble average (3) since

\left\langle \boldsymbol{\sigma} \right\rangle \ne \left\langle \boldsymbol{\mathsf{C}} \right\rangle\star\left\langle \boldsymbol{\varepsilon} \right\rangle \quad \text{and} \quad \left\langle \mathbf{p} \right\rangle \ne \left\langle \rho \right\rangle~\left\langle \dot{\mathbf{u}} \right\rangle ~. $$ We therefore need some effective constitutive relation. Willis (Willis81,Willis81a,Willis83,Willis97,Milton07) found that

\begin{align} \left\langle \boldsymbol{\sigma} \right\rangle & = \boldsymbol{\mathsf{C}}_\text{eff}\star\left\langle \boldsymbol{\varepsilon} \right\rangle + \boldsymbol{\mathsf{S}}_\text{eff}\star\left\langle \dot{\mathbf{u}} \right\rangle\\ \left\langle \mathbf{p} \right\rangle & = \boldsymbol{\mathsf{S}}_\text{eff}^\dagger \star\left\langle \boldsymbol{\varepsilon} \right\rangle + \rho_\text{eff}\star\left\langle \dot{\mathbf{u}} \right\rangle \end{align} $$ where all the operators are nonlocal in time (and in general also nonlocal in space).

By the adjoint operator (represented by the superscript $$\dagger$$), we mean

\int \boldsymbol{\pi} \star (\boldsymbol{\mathsf{S}}^\dagger_\text{eff} \star \boldsymbol{\tau})~\text{d}\mathbf{x} = \int \boldsymbol{\tau} \star (\boldsymbol{\mathsf{S}}_\text{eff} \star \boldsymbol{\pi})~\text{d}\mathbf{x} $$ for all fields $$\boldsymbol{\pi}$$ and $$\boldsymbol{\tau}$$ and at time $$t$$.

In the next lecture we will show how the Willis' equations are derived.


text:
User:Eiskalteschatten

English
Hello!

I am a graduate student and am studying the German language. I am very interested in languages and have studied German, Japanese, Spanish, Russian and Latin.

Deutsch
Hallo!

Ich bin Student im Aufbaustudium und ich studiere die deutsche Sprache. Ich interessiere mich für Sprachen und ich habe Deutsch, Japanisch, Spanisch, Russisch und Lateinisch studiert.

日本語
こんにちは！

私はドイツ語を勉強している大学院生です. 言語が大好きですので、ドイツ語や日本語やスペイン語やロシア語やラチン語を勉強しました.

Русский
Здравствуйте!

Я аспирант и изучаю немецкий язык. Я интересоваюсь языками и изучал немецкий, японский, испанский, русский и латинский.

title: User:Eiskalteschatten


text:
Waves in composites and metamaterials/Willis equations for elastodynamics

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.

Recap
In the previous lecture introduced the Willis equations (Willis81,Willis81a,Willis83,Willis97,Milton07). In this lecture we will discuss how those equations are derived.

Recall that by ensemble averaging the governing equations of elastodynamics we get

\begin{align} & \boldsymbol{\nabla} \cdot \left\langle \boldsymbol{\sigma} \right\rangle + \mathbf{f} = \left\langle \dot{\mathbf{p}} \right\rangle \\ & \left\langle \boldsymbol{\varepsilon} \right\rangle = \frac{1}{2}~[\boldsymbol{\nabla} \left\langle \mathbf{u} \right\rangle + (\boldsymbol{\nabla} \left\langle \mathbf{u} \right\rangle)^T] \end{align} $$ where $$\left\langle (\bullet) \right\rangle$$ is the ensemble average over realizations and not a volume average.

We need to derive the effective constitutive relations

\begin{align} \left\langle \boldsymbol{\sigma} \right\rangle & = \boldsymbol{\mathsf{C}}_\text{eff}\star\left\langle \boldsymbol{\varepsilon} \right\rangle +\boldsymbol{\mathcal{S}}_\text{eff}\star\left\langle \dot{\mathbf{u}} \right\rangle\\ \left\langle \mathbf{p} \right\rangle & = \boldsymbol{\mathcal{S}}_\text{eff}^\dagger \star\left\langle \boldsymbol{\varepsilon} \right\rangle + \boldsymbol{\rho}_\text{eff}\star\left\langle \dot{\mathbf{u}} \right\rangle \end{align} $$ where the operator $$\star$$ represents a convolution over time, i.e.,

\begin{align} &\boldsymbol{\mathsf{C}}\star\boldsymbol{\varepsilon} \equiv \int_{-\infty}^t \boldsymbol{\mathsf{C}}(t-\tau) : \boldsymbol{\varepsilon}(\tau)~\text{d}\tau ~; \boldsymbol{\mathcal{S}}\star\dot{\mathbf{u}} \equiv \int_{-\infty}^t \boldsymbol{\mathcal{S}}(t-\tau) \cdot \dot{\mathbf{u}}(\tau)~\text{d}\tau \\ &\boldsymbol{\mathcal{S}}^\dagger\star\boldsymbol{\varepsilon} \equiv \int_{-\infty}^t \boldsymbol{\mathcal{S}}^\dagger(t-\tau) \cdot \boldsymbol{\varepsilon}(\tau)~\text{d}\tau ~; \boldsymbol{\rho}\star\dot{\mathbf{u}} \equiv \int_{-\infty}^t \boldsymbol{\rho}(t-\tau)\cdot\dot{\mathbf{u}}(\tau)~\text{d}\tau ~. \end{align} $$ and the adjoint operator (represented by the superscript $$\dagger$$) is defined via

\int \mathbf{a} \star (\boldsymbol{\mathcal{S}}^\dagger_\text{eff} \star \boldsymbol{A})~\text{d}\mathbf{x} = \int \boldsymbol{A} \star (\boldsymbol{\mathcal{S}}_\text{eff} \star \mathbf{a})~\text{d}\mathbf{x} $$ for all vector fields $$\mathbf{a}$$ and second order tensor fields $$\boldsymbol{A}$$ and at time $$t$$. Note that the quantities $$\boldsymbol{\mathcal{S}}_\text{eff}$$ and $$\boldsymbol{\mathcal{S}}^\dagger_\text{eff}$$ are third-order tensors. In the above definition the convolutions are defined as

\mathbf{a} \star \mathbf{b} = \int_{-\infty}^t \mathbf{a}(t-\tau) \cdot \mathbf{b}(\tau)~\text{d}\tau ~; \boldsymbol{A} \star \boldsymbol{B} = \int_{-\infty}^t \boldsymbol{A}(t-\tau) : \boldsymbol{B}(\tau)~\text{d}\tau $$ where $$\mathbf{a}, \mathbf{b}$$ are vectors and $$\boldsymbol{A}, \boldsymbol{B}$$ are second-order tensors.

Derivation of Willis' equations
Let us introduce a homogeneous reference medium with properties $$\boldsymbol{\mathsf{C}}_0$$ and $$\rho_0$$ (constant). The polarization fields are defined as
 * $$ \text{(1)} \qquad

\begin{align} \boldsymbol{\tau} & := (\boldsymbol{\mathsf{C}} - \boldsymbol{\mathsf{C}}_0) \star \boldsymbol{\varepsilon} & = \boldsymbol{\sigma} - \boldsymbol{\mathsf{C}}_0\star\boldsymbol{\varepsilon} \\ \boldsymbol{\pi} & := (\rho - \rho_0)~\dot{\mathbf{u}} & = \mathbf{p} - \rho_0~\dot{\mathbf{u}} ~. \end{align} $$ Then,
 * $$ \text{(2)} \qquad

\begin{align} \boldsymbol{\sigma} & = \boldsymbol{\tau} + \boldsymbol{\mathsf{C}}_0\star\boldsymbol{\varepsilon} \\ \mathbf{p} & = \boldsymbol{\pi} + \rho_0~\dot{\mathbf{u}} ~. \end{align} $$ Taking the divergence of the equation (2)$$_1$$, we get
 * $$ \text{(3)} \qquad

\boldsymbol{\nabla} \cdot \boldsymbol{\sigma} = \boldsymbol{\nabla} \cdot \boldsymbol{\tau} + \boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}_0\star\boldsymbol{\varepsilon}) ~. $$ Also, taking the time derivative of equation (2)$$_2$$, we have
 * $$ \text{(4)} \qquad

\dot{\mathbf{p}} = \dot{\boldsymbol{\pi}} + \rho_0~\ddot{\mathbf{u}} ~. $$ Recall that the equation of motion is
 * $$ \text{(5)} \qquad

\boldsymbol{\nabla} \cdot \boldsymbol{\sigma} + \mathbf{f} = \dot{\mathbf{p}} ~. $$ Plugging (3) and (4) into (5) gives

\boldsymbol{\nabla} \cdot \boldsymbol{\tau} + \boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}_0\star\boldsymbol{\varepsilon}) + \mathbf{f} = \dot{\boldsymbol{\pi}} + \rho_0~\ddot{\mathbf{u}} $$ or,
 * $$ \text{(6)} \qquad

\boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}_0\star\boldsymbol{\varepsilon}) + \mathbf{f} + \boldsymbol{\nabla} \cdot \boldsymbol{\tau} - \dot{\boldsymbol{\pi}} = \rho_0~\ddot{\mathbf{u}} ~. $$ In the reference medium, $$\boldsymbol{\tau} = \boldsymbol{0}$$ and $$\boldsymbol{\pi} = \boldsymbol{0}$$. Let $$\mathbf{u}_0$$ be the solution in the reference medium in the presence of the body force $$\mathbf{f}$$ and with the same boundary conditions and initial conditions. For example, if the actual body has $$\mathbf{u} \rightarrow \boldsymbol{0}$$ as $$t \rightarrow -\infty$$, then $$\mathbf{u}_0 \rightarrow \boldsymbol{0}$$ as $$t \rightarrow -\infty$$. Then, in the reference medium, we have
 * $$ \text{(7)} \qquad

\begin{align} & \boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}_0\star\boldsymbol{\varepsilon}_0) + \mathbf{f} = \rho_0~\ddot{\mathbf{u}}_0\\ & \boldsymbol{\varepsilon}_0 = \frac{1}{2} [\boldsymbol{\nabla} \mathbf{u}_0 + (\boldsymbol{\nabla} \mathbf{u}_0)^T] ~. \end{align} $$ Remember that we want our effective stress-strain relations to be independent of the body force $$\mathbf{f}$$. So all we have to do is subtract (7)$$_1$$ from (6). Then we get

\boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}_0\star\boldsymbol{\varepsilon}) - \boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}_0\star\boldsymbol{\varepsilon}_0) + \boldsymbol{\nabla} \cdot \boldsymbol{\tau} - \dot{\boldsymbol{\pi}} = \rho_0~\left[\ddot{\mathbf{u}} - \ddot{\mathbf{u}}_0\right] $$ or,

\boldsymbol{\nabla} \cdot [\boldsymbol{\mathsf{C}}_0\star(\boldsymbol{\varepsilon}-\boldsymbol{\varepsilon}_0)] + \boldsymbol{\nabla} \cdot \boldsymbol{\tau} - \dot{\boldsymbol{\pi}} = \rho_0~\left[\ddot{\mathbf{u}} - \ddot{\mathbf{u}}_0\right] ~. $$ Define

\mathbf{u}' := \mathbf{u} - \mathbf{u}_0 ~; \boldsymbol{\varepsilon}' := \frac{1}{2}[\boldsymbol{\nabla} \mathbf{u}' + (\boldsymbol{\nabla} \mathbf{u}')^T] = \boldsymbol{\varepsilon} - \boldsymbol{\varepsilon}_0 ~; \mathbf{h} := \boldsymbol{\nabla} \cdot \boldsymbol{\tau} - \dot{\boldsymbol{\pi}} ~. $$ Then,
 * $$ \text{(8)} \qquad

{ -\boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}_0\star\boldsymbol{\varepsilon}') + \rho_0~\ddot{\mathbf{u}}'  = \mathbf{h} ~. } $$ If we assume that $$\mathbf{h}$$ is fixed, then (8) can be written as

\mathcal{L}~\mathbf{u}' = \mathbf{h} $$ where $$\mathcal{L}$$ is a linear operator. The solution of this equation is

\mathbf{u}' = \boldsymbol{G} \star \mathbf{h} $$ where $$\boldsymbol{G}$$ is the Green's function associated with the operator $$\mathcal{L}$$. Plugging back our definitions of $$\mathbf{u}'$$ and $$\mathbf{h}$$, we get
 * $$ \text{(9)} \qquad

{ \mathbf{u} = \mathbf{u}_0 + \boldsymbol{G} \star (\boldsymbol{\nabla} \cdot \boldsymbol{\tau} - \dot{\boldsymbol{\pi}}) = \mathbf{u}_0 + \boldsymbol{G} \star (\boldsymbol{\nabla} \cdot \boldsymbol{\tau}) - \boldsymbol{G}\star\dot{\boldsymbol{\pi}} ~. } $$ The strain-displacement relation is

\boldsymbol{\varepsilon} = \frac{1}{2}[\boldsymbol{\nabla}\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})^T] ~. $$ Plugging the solution (9) into the strain-displacement relation gives
 * $$ \text{(10)} \qquad

\boldsymbol{\varepsilon} = \boldsymbol{\varepsilon}_0 + \frac{1}{2}~\boldsymbol{\nabla} [\boldsymbol{G} \star (\boldsymbol{\nabla} \cdot \boldsymbol{\tau})] + \frac{1}{2}~[\boldsymbol{\nabla} [\boldsymbol{G}\star(\boldsymbol{\nabla} \cdot \boldsymbol{\tau})]]^T - \frac{1}{2}~\boldsymbol{\nabla} (\boldsymbol{G}\star\dot{\boldsymbol{\pi})} - \frac{1}{2}~[\boldsymbol{\nabla} (\boldsymbol{G}\star\dot{\boldsymbol{\pi})}]^T ~. $$ Define $$\boldsymbol{\mathsf{S}}_x$$ and $$\boldsymbol{\mathcal{M}}_x$$ via

\begin{align} \boldsymbol{\mathsf{S}}_x\star\boldsymbol{\tau} & = -\frac{1}{2}~\left\{\boldsymbol{\nabla} [\boldsymbol{G} \star (\boldsymbol{\nabla} \cdot \boldsymbol{\tau})] + [\boldsymbol{\nabla} [\boldsymbol{G}\star(\boldsymbol{\nabla} \cdot \boldsymbol{\tau})]]^T\right\} \\ \boldsymbol{\mathcal{M}}_x\star\boldsymbol{\pi} & = \frac{1}{2}~\left\{\boldsymbol{\nabla} (\boldsymbol{G}\star\dot{\boldsymbol{\pi})} + \frac{1}{2}~[\boldsymbol{\nabla} (\boldsymbol{G}\star\dot{\boldsymbol{\pi})}]^T\right\} ~. \end{align} $$ Then we can write (10) as
 * $$ \text{(11)} \qquad

{ \boldsymbol{\varepsilon} = \boldsymbol{\varepsilon}_0 - \boldsymbol{\mathsf{S}}_x\star\boldsymbol{\tau} - \boldsymbol{\mathcal{M}}_x\star\boldsymbol{\pi} ~. } $$ Also, taking the time derivative of (9), we get
 * $$ \text{(12)} \qquad

\dot{\mathbf{u}} = \dot{\mathbf{u}}_0 + \cfrac{d }{d t}\left[\boldsymbol{G} \star (\boldsymbol{\nabla} \cdot \boldsymbol{\tau})\right] - \cfrac{d }{d t}\left[\boldsymbol{G}\star\dot{\boldsymbol{\pi}}\right] ~. $$ Define $$\boldsymbol{\mathcal{S}}_t$$ and $$\boldsymbol{M}_t$$ via

\begin{align} \boldsymbol{\mathcal{S}}_t\star\boldsymbol{\tau} & = -\cfrac{d }{d t}\left[\boldsymbol{G} \star (\boldsymbol{\nabla} \cdot \boldsymbol{\tau})\right] \\ \boldsymbol{M}_t\star\boldsymbol{\pi} & = \cfrac{d }{d t}\left[\boldsymbol{G}\star\dot{\boldsymbol{\pi}}\right] ~. \end{align} $$ Then we can write (12) as
 * $$ \text{(13)} \qquad

{ \dot{\mathbf{u}} = \dot{\mathbf{u}}_0 - \boldsymbol{\mathcal{S}}_t\star\boldsymbol{\tau} - \boldsymbol{M}_t\star\boldsymbol{\pi} ~. } $$ Willis (Willis81a) has shown that $$\boldsymbol{\mathcal{S}}_t$$ and $$\boldsymbol{\mathcal{M}}_x$$ are formal adjoints, i.e., $$\boldsymbol{\mathcal{S}}_t = \boldsymbol{\mathcal{M}}_x^\dagger$$, in the sense that

\int \boldsymbol{\pi} \star (\boldsymbol{\mathcal{S}}_t \star \boldsymbol{\tau})~\text{d}\mathbf{x} = \int \boldsymbol{\tau} \star (\boldsymbol{\mathcal{M}}_x \star \boldsymbol{\pi})~\text{d}\mathbf{x} ~\qquad \forall~\boldsymbol{\pi},\boldsymbol{\tau},t ~. $$ From (11) and (13), eliminating $$\boldsymbol{\epsilon}$$ and $$\dot{\mathbf{u}}$$ via equations (1), we have
 * $$ \text{(14)} \qquad

\begin{align} (\boldsymbol{\mathsf{C}} - \boldsymbol{\mathsf{C}}_0)^{-1}\star\boldsymbol{\tau} + \boldsymbol{\mathsf{S}}_x\star\boldsymbol{\tau} + \boldsymbol{\mathcal{M}}_x\star\boldsymbol{\pi} & = \boldsymbol{\varepsilon}_0 \\ (\rho - \rho_0)^{-1}~\boldsymbol{\pi} + \boldsymbol{\mathcal{S}}_t\star\boldsymbol{\tau} + \boldsymbol{M}_t\star\boldsymbol{\pi} & = \dot{\mathbf{u}}_0 ~. \end{align} $$ Also, ensemble averaging equations (11) and (13), we have
 * $$ \text{(15)} \qquad

\begin{align} \left\langle \boldsymbol{\varepsilon} \right\rangle &= \boldsymbol{\varepsilon}_0 - \boldsymbol{\mathsf{S}}_x\star\left\langle \boldsymbol{\tau} \right\rangle - \boldsymbol{\mathcal{M}}_x\star\left\langle \boldsymbol{\pi} \right\rangle\\ \left\langle \dot{\mathbf{u}} \right\rangle &= \dot{\mathbf{u}}_0 - \boldsymbol{\mathcal{S}}_t\star\left\langle \boldsymbol{\tau} \right\rangle - \boldsymbol{M}_t\star\left\langle \boldsymbol{\pi} \right\rangle~. \end{align} $$ From (14) and (15), eliminating $$\boldsymbol{\varepsilon}_0$$ and $$\dot{\mathbf{u}}_0$$, we get

\begin{align} (\boldsymbol{\mathsf{C}} - \boldsymbol{\mathsf{C}}_0)^{-1}\star\boldsymbol{\tau} + \boldsymbol{\mathsf{S}}_x\star\boldsymbol{\tau} + \boldsymbol{\mathcal{M}}_x\star\boldsymbol{\pi} & = \left\langle \boldsymbol{\varepsilon} \right\rangle + \boldsymbol{\mathsf{S}}_x\star\left\langle \boldsymbol{\tau} \right\rangle + \boldsymbol{\mathcal{M}}_x\star\left\langle \boldsymbol{\pi} \right\rangle\\ (\rho - \rho_0)^{-1}~\boldsymbol{\pi} + \boldsymbol{\mathcal{S}}_t\star\boldsymbol{\tau} + \boldsymbol{M}_t\star\boldsymbol{\pi} & = \left\langle \dot{\mathbf{u}} \right\rangle + \boldsymbol{\mathcal{S}}_t\star\left\langle \boldsymbol{\tau} \right\rangle + \boldsymbol{M}_t\star\left\langle \boldsymbol{\pi} \right\rangle \end{align} $$ or,
 * $$ \text{(16)} \qquad

\begin{align} (\boldsymbol{\mathsf{C}} - \boldsymbol{\mathsf{C}}_0)^{-1}\star\boldsymbol{\tau} + \boldsymbol{\mathsf{S}}_x\star(\boldsymbol{\tau}-\left\langle \boldsymbol{\tau} \right\rangle) + \boldsymbol{\mathcal{M}}_x\star(\boldsymbol{\pi} -\left\langle \boldsymbol{\pi} \right\rangle) & = \left\langle \boldsymbol{\varepsilon} \right\rangle\\ (\rho - \rho_0)^{-1}~\boldsymbol{\pi} + \boldsymbol{\mathcal{S}}_t\star(\boldsymbol{\tau}-\left\langle \boldsymbol{\tau} \right\rangle) + \boldsymbol{M}_t\star(\boldsymbol{\pi} -\left\langle \boldsymbol{\pi} \right\rangle) & = \left\langle \dot{\mathbf{u}} \right\rangle ~. \end{align} $$ Equations (16) are linear in $$\boldsymbol{\tau}$$ and $$\boldsymbol{\pi}$$. Therefore, formally these equations have the form
 * $$ \text{(17)} \qquad

\begin{bmatrix} \boldsymbol{\tau} \\ \boldsymbol{\pi} \end{bmatrix} = \mathcal{T} \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon} \right\rangle \\ \left\langle \dot{\mathbf{u}} \right\rangle \end{bmatrix} ~. $$ That such an argument can be made has been rigorously shown for low contrast media but not for high contrast media. Hence, these ideas work for composites that are close to homogeneous.

From the definition of $$\boldsymbol{\tau}$$ and $$\boldsymbol{\pi}$$, taking the ensemble average gives us
 * $$ \text{(18)} \qquad

\left\langle \boldsymbol{\tau} \right\rangle = \left\langle \boldsymbol{\sigma} \right\rangle - \boldsymbol{\mathsf{C}}_0\star\left\langle \boldsymbol{\varepsilon} \right\rangle ~; \left\langle \boldsymbol{\pi} \right\rangle = \left\langle \mathbf{p} \right\rangle - \rho_0~\left\langle \dot{\mathbf{u}} \right\rangle ~. $$ Also, from (17), taking the ensemble average leads to
 * $$ \text{(19)} \qquad

\begin{bmatrix} \left\langle \boldsymbol{\tau} \right\rangle \\ \left\langle \boldsymbol{\pi} \right\rangle \end{bmatrix} = \left\langle \mathcal{T} \right\rangle \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon} \right\rangle \\ \left\langle \dot{\mathbf{u}} \right\rangle \end{bmatrix} ~. $$ Plugging in the relations (18) in these equations gives us

\begin{bmatrix} \left\langle \boldsymbol{\sigma} \right\rangle - \boldsymbol{\mathsf{C}}_0\star\left\langle \boldsymbol{\varepsilon} \right\rangle\\ \left\langle \mathbf{p} \right\rangle - \rho_0~\left\langle \dot{\mathbf{u}} \right\rangle \end{bmatrix} = \left\langle \mathcal{T} \right\rangle \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon} \right\rangle \\ \left\langle \dot{\mathbf{u}} \right\rangle \end{bmatrix} $$ or,

\begin{bmatrix} \left\langle \boldsymbol{\sigma} \right\rangle \\ \left\langle \mathbf{p} \right\rangle \end{bmatrix} = \left\langle \mathcal{T} \right\rangle \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon} \right\rangle\\\left\langle \dot{\mathbf{u}} \right\rangle \end{bmatrix} + \begin{bmatrix} \boldsymbol{\mathsf{C}}_0 & 0 \\ 0 & \rho_0~\boldsymbol{\mathit{1}}\end{bmatrix} \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon} \right\rangle\\\left\langle \dot{\mathbf{u}} \right\rangle \end{bmatrix} $$ or,
 * $$ \text{(20)} \qquad

{ \begin{bmatrix} \left\langle \boldsymbol{\sigma} \right\rangle \\ \left\langle \mathbf{p} \right\rangle \end{bmatrix} = \begin{bmatrix} \boldsymbol{\mathsf{C}}_\text{eff} & \boldsymbol{\mathcal{S}}_\text{eff} \\ \boldsymbol{\mathcal{S}}^\dagger_\text{eff} & \boldsymbol{\rho}_\text{eff} \end{bmatrix} \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon} \right\rangle\\\left\langle \dot{\mathbf{u}} \right\rangle \end{bmatrix} ~. } $$ These are the Willis equations.

Willis equations for electromagnetism
For electromagnetism, we can use similar arguments to obtain

\begin{align} \left\langle \mathbf{D} \right\rangle & = \boldsymbol{\epsilon}_\text{eff} \star \left\langle \mathbf{E} \right\rangle + \boldsymbol{\alpha}_\text{eff} \star \left\langle \mathbf{B} \right\rangle\\ \left\langle \mathbf{H} \right\rangle & = \boldsymbol{\alpha}_\text{eff} \star \left\langle \mathbf{E} \right\rangle + (\boldsymbol{\mu}_\text{eff})^{-1} \star \left\langle \mathbf{B} \right\rangle \end{align} $$ where $$\boldsymbol{\alpha}_\text{eff}$$ is a coupling term.

In particular, if the fields are time harmonic with non-local operators being approximated by local ones, then

\begin{align} \left\langle \widehat{\mathbf{D}} \right\rangle & = \boldsymbol{\epsilon}_\text{eff} \cdot \left\langle \widehat{\mathbf{E}} \right\rangle + \boldsymbol{\lambda}_\text{eff} \cdot \left\langle \widehat{\mathbf{H}} \right\rangle\\ \left\langle \widehat{\mathbf{B}} \right\rangle & = \overline{\boldsymbol{\lambda}_\text{eff}} \cdot \left\langle \widehat{\mathbf{E}} \right\rangle + \boldsymbol{\mu}_\text{eff} \cdot \left\langle \widehat{\mathbf{H}} \right\rangle ~. \end{align} $$ If the operators are local, then $$\boldsymbol{\epsilon}_\text{eff}, \boldsymbol{\lambda}_\text{eff}, \boldsymbol{\mu}_\text{eff}$$ will just be matrices that depend on the frequency $$\omega$$.

If the composite material is isotropic, then

\boldsymbol{\epsilon}_\text{eff} = \epsilon_\text{eff}~\boldsymbol{\mathit{1}} ~; \boldsymbol{\lambda}_\text{eff} = \lambda_\text{eff}~\boldsymbol{\mathit{1}} ~; \boldsymbol{\mu}_\text{eff} = \mu_\text{eff}~\boldsymbol{\mathit{1}} ~. $$ Under reflection, $$\left\langle \widehat{\mathbf{E}} \right\rangle$$ reflects like a normal vector. However, $$\left\langle \widehat{\mathbf{H}} \right\rangle$$ reflects like an axial vector (i.e., it changes direction). Hence $$\lambda_\text{eff}$$ would have to change sign under a reflection. Therefore, with $$\lambda_\text{eff}$$ fixed, the constitutive relations are not invariant with respect to reflections! This means that if $$\lambda_\text{eff} \ne 0$$ the medium has a certain handedness and is called a  chiral medium.

Extension of the Willis approach to composites with voids
Sometimes the quantity $$\left\langle \mathbf{u} \right\rangle$$ is not an appropriate macroscopic variable. For example, in materials with voids $$\mathbf{u}$$ is undefined inside the voids. Even if the voids are filled with an elastic material with modulus tending to zero, the value of $$\left\langle \mathbf{u} \right\rangle$$ will depend on the way this limit is taken. Also, for materials such as the rigid matrix filled with rubber and lead (see Figure 1), it makes senses to average $$\mathbf{u}$$ only over the deformable material phase.

Therefore it makes sense to look for equations for $$\left\langle \mathbf{u}_w \right\rangle$$ where
 * $$ \text{(21)} \qquad

\mathbf{u}_w(\mathbf{x}, t) = w(\mathbf{x})~\mathbf{u}(\mathbf{x}, t) $$ where $$w(\mathbf{x})$$ is a weight which could be zero in the region where there are voids. Also, the weights could vary from realization to realization.

Also, if we have $$\dot{\mathbf{u}}$$ we can recover $$\mathbf{u}$$ by integrating over time, i.e.,

\mathbf{u}(t) = \int_{-\infty}^t \dot{\mathbf{u}}(\tau)~\text{d}\tau = \int_{-\infty}^{\infty} H(t-\tau)~\dot{\mathbf{u}}(\tau)~\text{d}\tau $$ where

H(v) = \begin{cases} 1 & \text{for}v > 0 \\ 0 & \text{for}v < 0 \end{cases} $$ Hence we can write
 * $$ \text{(22)} \qquad

\mathbf{u} = H \star \dot{\mathbf{u}} ~. $$ So, from the definitions of $$\boldsymbol{\tau}$$ and $$\boldsymbol{\pi}$$ and using the relation (22), we have

\begin{bmatrix} \boldsymbol{\varepsilon} \\ \mathbf{u} \end{bmatrix} = \begin{bmatrix} (\boldsymbol{\mathsf{C}} - \boldsymbol{\mathsf{C}}_0)^{-1} & 0 \\ 0 & H\star(\rho-\rho_0)^{-1} \end{bmatrix} \star \begin{bmatrix} \boldsymbol{\tau} \\ \boldsymbol{\pi} \end{bmatrix} ~. $$ Form the Willis equations (17) we have

\begin{bmatrix} \boldsymbol{\tau} \\ \boldsymbol{\pi} \end{bmatrix} = \mathcal{T} \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon} \right\rangle \\ \left\langle \dot{\mathbf{u}} \right\rangle \end{bmatrix} ~. $$ Therefore,
 * $$ \text{(23)} \qquad

\begin{bmatrix} \boldsymbol{\varepsilon} \\ \mathbf{u} \end{bmatrix} = \begin{bmatrix} (\boldsymbol{\mathsf{C}} - \boldsymbol{\mathsf{C}}_0)^{-1} & 0 \\ 0 & H\star(\rho-\rho_0)^{-1} \end{bmatrix} \star \mathcal{T} \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon} \right\rangle \\ \left\langle \dot{\mathbf{u}} \right\rangle \end{bmatrix} ~. $$ Now, if the weighted strain is defined as

\boldsymbol{\varepsilon}_w = \frac{1}{2}~[\boldsymbol{\nabla}\mathbf{u}_w + (\boldsymbol{\nabla}\mathbf{u}_w)^T] $$ then, taking the ensemble average, we have

\left\langle \boldsymbol{\varepsilon}_w \right\rangle = \frac{1}{2}~\left\langle [\boldsymbol{\nabla}\mathbf{u}_w + (\boldsymbol{\nabla}\mathbf{u}_w)^T] \right\rangle ~. $$ Using equation (21) we can show that
 * $$ \text{(24)} \qquad

\left\langle \boldsymbol{\varepsilon}_w \right\rangle = \left\langle w~\boldsymbol{\varepsilon} \right\rangle + \frac{1}{2}~\left\langle \boldsymbol{\nabla} w\otimes\mathbf{u}+ \mathbf{u}\otimes\boldsymbol{\nabla} w\right\rangle ~. $$ Using (23) we can express (24) in terms of $$\left\langle \boldsymbol{\varepsilon} \right\rangle$$ and $$\dot{\mathbf{u}}$$, and hence also in terms of $$\dot{\mathbf{u}}_w$$. After some algebra (see Milton07 for details), we can show that

\begin{bmatrix} \left\langle \boldsymbol{\varepsilon}_w \right\rangle \\ \left\langle \dot{\mathbf{u}} \right\rangle_w \end{bmatrix} = \boldsymbol{R}_w \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon} \right\rangle \\ \left\langle \dot{\mathbf{u}} \right\rangle \end{bmatrix} $$ where $$\boldsymbol{R}_w = \boldsymbol{\mathit{1}}$$ when $$w(\mathbf{x}) = 1$$.

Taking the inverse, we can express the Willis equations (20) in terms of $$\left\langle \boldsymbol{\varepsilon}_w \right\rangle$$ and $$\left\langle \dot{\mathbf{u}} \right\rangle_w $$ as

\begin{bmatrix} \left\langle \boldsymbol{\sigma} \right\rangle \\ \left\langle \mathbf{p} \right\rangle \end{bmatrix} = \begin{bmatrix} \boldsymbol{\mathsf{C}}_\text{eff} & \boldsymbol{\mathcal{S}}_\text{eff} \\ \boldsymbol{\mathcal{S}}^\dagger_\text{eff} & \boldsymbol{\rho}_\text{eff} \end{bmatrix} \star \boldsymbol{R}_w^{-1} \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon}_w \right\rangle \\ \left\langle \dot{\mathbf{u}} \right\rangle_w \end{bmatrix} $$ or,

{ \begin{bmatrix} \left\langle \boldsymbol{\sigma} \right\rangle \\ \left\langle \mathbf{p} \right\rangle \end{bmatrix} = \begin{bmatrix} \boldsymbol{\mathsf{C}}^w_\text{eff} & \boldsymbol{\mathcal{S}}^w_\text{eff} \\ \boldsymbol{\mathcal{D}}^w_\text{eff} & \boldsymbol{\rho}^w_\text{eff} \end{bmatrix} \star \begin{bmatrix} \left\langle \boldsymbol{\varepsilon}_w \right\rangle \\ \left\langle \dot{\mathbf{u}} \right\rangle_w \end{bmatrix}~. } $$ These equations have the same form as the Willis equations. However, $$\boldsymbol{\mathcal{D}}^w_\text{eff} \ne (\boldsymbol{\mathcal{S}}^w_\text{eff})^\dagger$$. We now have a means of using the Willis equations even in the case where there are voids.


text:
Offshore engineering

Part of the Wikiversity Department of Marine Engineering, School of Engineering and the Engineering and Technology Portal

Content summary
Offshore Engineering applies ocean engineering studies to the engineering of devices within the benthic environment, such as Marine Foundations, Mooring Systems, Ocean Mining, Offshore Disposal, Offshore Pipelines and Cables, Offshore Structures, Submersible Vehicles, Submarine Habitats and Salvage Operations.

Goals
To further knowledge of the deep ocean as an engineering environment and to enable the practitioner to effectively design within the parameters of such.

Texts

 * Introduction to Oceanography; Wikibooks, the open-content textbooks collection
 * Marine Hydrodynamics; Wikibooks, the open-content textbooks collection

Offshore Engineering Lessons

 * The Benthic Environment
 * 1) Temperature
 * 2) Density
 * 3) Salinity
 * 4) Chemistry
 * Deep Water Engineering
 * 1) Anchors
 * 2) Mining & Oil Exploration
 * 3) Biological Investigation
 * Ocean engineering design
 * 1) Models & Scaling
 * 2) Resistance
 * 3) Propulsion
 * 4) Corrosion
 * 5) Galvanic Protection
 * 6) Arctic Engineering

Activities:

 * Physical Properties of the Oceans

Readings:

 * Physical Properties of the Oceans
 * 1) Peruse the appropriate sections of Introduction to Oceanography

Study guide:

 * Physical Properties of the Oceans
 * 1) Wikipedia article:Temperature
 * 2) Wikipedia article:Density
 * 3) Wikipedia article:Salinity
 * 4) Wikipedia article:Chemistry
 * 5) Wikipedia article:Marine geology
 * 6) Wikipedia article:Oceanography
 * 7) Wikipedia article:Chemical oceanography
 * 8) Wikipedia article:Physical oceanography

Active participants
Please sign up!!

Active participants in this Learning Group

title: Offshore engineering


text:
Portal talk:Learning German

I have listed your resources in the School of Language and Literature resource catalog--please list any new resources there in the future! Thanks! --Trinity507 21:52, 22 February 2009 (UTC)

Site Structure

 * German redirects to Topic:German
 * Topic:German is the top level for German
 * Topic:Learning German is the top level for the German Language course.
 * Lesson:German Introduction is the only content page at the moment.

Ideas and Contributions
Let's use the talk page instead of the German I course page for planning. That way we can actually add plans to the German I page as needed. --Alex 01:00, 5 October 2007 (UTC)

I just got rid of the redirect from the German I page. Hope you don't mind it. My ideas of this course and wikiversity in general would be a very strong interactivity (as a contrast to wikibooks). Some audio-files and tests for each lesson would be useful (as an example). If I can help you anyhow I'll do my best. --Edmond Dantes AUT 19:42, 6 October 2007 (UTC)

http://en.wikipedia.org/wiki/German_grammar is a good link to post for beginners to the german language, and it stays within the wiki sites as well -Tereasa

I suggest Lesson 2 to be "Was ist das?". It is a standard first lesson in many languages books. A dialog asking about things and animals. Was ist das? Das ist ein Hund, etc., to introduce ein, eine and new vocabulary. --Davichito 19:05, 6 October 2008 (UTC)

If anyone is still working on this, I'm fairly likely to just start an entirely different branch based on a project page. The goal of my page is going to be to help learners develop skills they need to teach themselves, so it will come with a lot more questions, guidelines and exercises and less curriculum. I might nest more curriculum within the current structure, but it depends how compatible I find the Lehrplan I develop. I'd love to chat with some other teachers about this before I do anything too crazy so HMU Jon michael swift (discuss • contribs) 17:29, 18 May 2016 (UTC)

Changes
I fixed up the introduction page which I don't think should have any actual lessons on it. I'm thinking we should use it more as a syllabus. --Alex 01:01, 5 October 2007 (UTC)

I went directly and edited Lesson 1. I thought the lesson was only a dialog; it should have an explanation of it. I added without proposing here, but I think it is how wikis are.

Davichito 18:31, 6 October 2008 (UTC)

Participants
If you're interested in helping instruct or just want to learn, add your name in the appropriate list below! There are lots of materials we can use to help each other learn German. Feel free to share your ideas here by editing this page! A more updated list can be found here: German stream

Complete (or near complete) Beginners
Hi, it is me. --Davichito 19:04, 6 October 2008 (UTC) Zim 16:09, 27 May 2007 (UTC) hey im mark from san francisco and id like to learn german Hello, I'm Rashmi from India. I respect the German Technology and the R&D carried out in the county. I'm considering Germany for my higher studies and also my fiance lives there. I love German. I know a little bit of the language and would love to speak fluently in it.
 * Hi, My name is Ramya. I live in India. I have visited Germany found it to be a great place with very friendly people...inspiring me to learn the language.--203.99.195.235 (discuss) 10:06, 9 July 2014 (UTC)
 * Hi, My name is Fernanda. My husband lived 13 years in Germany and I'd love to be able to speak this language with him. He's trying really hard to learn my language too.
 * Hello, My Name is Pat and I would like to learn German because I just moved to Wolfsburg.
 * hallo! i am tenzin, and i started learing germen this month. For me, the most easiest way to learn a language is to read easy to understand comics. If anyone has online resource like that it would be helpful to link it here.193.206.186.101 08:26, 22 August 2008 (UTC)tenzin
 * Hallo!Ich bin George und ich komme aus Rumanian und ich bin sechzhen!Hett dich zu treffen!!Well i realy like do study german becousei think that is a cool language and a very nice one even that it's very very hard!i do some self-studies and i want to do more becouse i realy wanty to know this languages!i'm also fascinating by the northen languages and i know also a little bit of finnish(finland languages).for me it's hard do learn this languagesbecouse my language is romanian-=a latin lkanguage like french, italiano, spanish and portuguese-but i try to do hte best and to learn it!i know already english, french, moldavian. a little of spanish and portuguese and i start the study for german and finnis!in my free time i study also japanese and turkish-this is only a hobby-so...if u can help me with the german and even finnish please contact me on e-mail..thanksGeorge-Cristian Potrivitu -13.06.2007
 * Hi! My name is Aena and I would really love to learn German. I'm just doing some self-study on the language so I really need help from lots of people. Thank you! ^^
 * Hi, I'm Christine. I'd like to learn German as well. I am not trying to send people to other sites (I love the Wiki!) but I found a BBC site to help people learn German for free. Maybe someone from there can help with this project. I did see a "Contact Us" page. Just a thought... They offer several languages but I can't say if the info is good or not; I don't know. I am looking for a more complete course.
 * Edvin
 * hallo! my name is Louisa, i love German and love to learn more about it... it's a cool language.. anyway, i got so many source about German.. if u wanna just send me an email..
 * Hi. I'm Nour Eddine Yaghmour from Palestine and i'd like to learn too
 * Hi, I'm Michael and I'd like to learn the German language.
 * I'd like to learn too! --HappyCamper 01:36, 12 April 2007 (UTC)
 * Hi I'm Bojana from Macedonia and I like to learn German. I started on my own but it's not very motivating.
 * I'm James and I also want to learn German
 * My name is David and I lived in Germany for three years and would like to relearn the language.
 * Hi, I´m Eli and a German wikiclass would help when I am self-teaching German this summer.
 * Hi, I'm Adam and i would also like to learn German as a second language.
 * Hi my name's Tracey and I'd love to learn German too, but it's hard to stay motivated learning on your own
 * Hi, my name is Syd and I learned German in high school but would like to refresh myself and become fluent.--SydHeresy 21:57, 27 April 2007 (UTC)
 * How ya doin,Im Jim, i would like to learn german to help me with my studies.
 * Hello, my name is Lissa and I would love to learn German.
 * craasnibooi-i'd like to learn to speak German.
 * Hi, I'm Marcus and I'm also interested.
 * Hey, I'm José, and I'd also like to learn German. 0
 * I'm Ben and I want to learn German
 * I'm Jenn from Las Vegas, USA and would love to learn the basics of German. Thanks to any possible teachers!
 * I have very little experience with German, but I'd love to learn. Blast 01.07.07 0719 (UTC)
 * I'm Cory, from Melbourne, Australia. I'd love to learn German as a fourth language, preferably before I go over to Europe =p Anyone who can help?
 * G'day, I'm Leigh. I wouldn't mind having a crack at learning German!
 * Hallo! I'm Rubèn Roca and I would like to learn German. Thanks!
 * hi,there.I'm Fatih from Turkey and would like to learn German.Thanks in advance!!
 * Ich bin's Desi. Ja, Ich will auch Deutsch lernen.
 * My name is Bryan and I live in USA, I had a German Exchange Student and I would like to learn German so that I can speak to his family
 * my name is juan, im from colombia and id like to learn some german
 * Hi, I'm Alexander. I'm lithuanian. Although I have had some experience with German in high school, I'd like to continue learning.
 * Hi, I'm Cabo from Guyana and would like to learn German.
 * Hi, I'm Alyssa and I would love to learn German.
 * Hi, I'm Denes from Hungary and I'd love to learn German.
 * Hi, I'm James from Melbourne, FL USA and I'm interested in learning German. 01:47, 18 April 2008 (UTC)
 * Hallo, Ich heiBe Abhishek. I want to learn german because i've started liking the langauge after some self learning, apart from this i've great admiration for Germany as a country. I Hope i've a good experience learning Deutche. Tschuss!! 24 Dec 08 Abhi.danej 12:21, 24 December 2008 (UTC)

Advanced Beginners
Jjkr 02:17, 7 January 2008 (UTC)
 * Hi, i am not very "advanced" but I have started already German "alleine" (alone, at home) a month already. I am interested into this project:)--Annalemesos 02:21, 12 April 2007 (UTC)
 * -Ich bin in meine dritte Jahr von Deutsch und will mehr lernen. Natürlich will ich mit die Klasse hilfen wenn eure Lehrer(in) möchtet. Kyran_Tyr 21:03 EST, 02:03 GMT, 30 April 2007
 * The Jade Knight I've studied Old English (very similar to German), and a little German on my own.  The Jade Knight 08:13, 8 June 2007 (UTC)
 * Hello! My name is Przemek. I've got a degree in linguistics applied and learned German ages ago as a kid. I have got some idea on German phonetics, basic syntax and so on. Hope to learn, help and have fun! I have got some experience as a teacher of English and Russian so I guess that could a kind of help. As it is my first edit at Wikiversity I hope I won't spoil something. :-) Stiopa 19:24, 17 June 2007 (UTC)
 * Hiya! I'm Will and I'm from Baltimore.  I've taken German 101 and am looking to get a German minor (at least), but the community college that I'm at right now keeps canceling any German classes beyond just the basics.  I'm mostly interested in keeping what little German I know fresh until I transfer to a university that offers a decent German program.  I'm also not too clear on how this whole Wikiversity thing works, though it looks promising. Willickers 05:06, 29 June 2007 (UTC)
 * Hey! My name's JJ and I took a year of German studies in High School (I remember very little, unfortunately). I'd really like to learn german.
 * Hallo. My name is Tereasa and I have self-taught myself german for years, though i am now taking a course in college. I think this project could help a lot of people. ^_^
 * Hi, I am Matt and I really want to learn Deutsch (German). I already know some, und ich bin gut (and I am good).
 * Hallo, ich heiße Adam. Ich bin amerikanisch und ich möchte lernen Deutsch. Agomulka 01:30, 16 February 2009 (UTC)

Intermediate Students
-I am also interested. My family comes from Germany and I have take some courses in Deutch. Jawohl!

-I am very interested in the German language. I had a bit of practice...years ago.

-Jetzt ich studiere Deutsch in der Universität von den Philippinen aber ich möchte mein Deutsch verbessern. Stickbreaker 20:24, 17 April 2007 (UTC)

- I can also help out. I took a total of 6 years of German throughout High School and College. Schulb72 12:39, 20 June 2007 (UTC)

- I would be pleased to assist in any way that I am able. I have been studying German for several years now at the University level. I have assisted with and have taught several subjects previously and have always loved helping others learn.

- Hallo! Ich heiße Guerric, und ich fliege in Oktober zu Dresde. Ich werde dort Deutsch lernen; aber, ich mag gern Deutsch, und ich will jetzt lernen beginnen! Ich lerne während mein Arbeit ein wenig, weil ich bin Lebensretter und es gibt an mein Schwimmbecken niemand. I'm good at discovering new vocabulary (which I've never got enough of), but word order gives me a great deal of trouble, as do prepositions. German is, chronologically, my fifth language - though better then my neglected Mandarin, and soon to beat my somewhat-neglected Spanish. --Lytyr 21:18, 5 July 2007 (UTC)

- Ich habe deutsch seit 4 Jahren studiert. Ich hoffe, dass ich helfen kann =] 70.111.230.78 21:51, 5 August 2007 (UTC)

- I used to study German a few years ago, at a German school in my native country, but haven't practiced it at all in about 2 years. I'd like to start from where i left off if possible, because although it's a rather difficult language, I think it's cool. And I'd like to travel to Germany some time. EmiLiuS 01:53, 10 August 2007 (UTC)

Proficient Students

 * Dallin.m 17:50, 2 May 2007 (UTC) Hello, people: I'm a not-quite-fluent German student who has just discovered Wikiversity.  I've studied the language for a number of years in high school, university, and I spent a brief time living in Germany.  I was hoping to use Wikiversity to keep my German sharp over the summer, but seeing as there isn't much (anything?) here, I wouldn't mind being an instructor for beginning students.

 * My name is Jeffrey Mason and I have studied German for 9 years (8th-12th grade, four years in college). My German is fairly good, I've visited Germany three times, one of those times was a job which involved translating, and I would love to help as much as I can in teaching to keep up with and improve my skills.

Native Speakers

 * Sundance Raphael can a bit help, but my English is not very well. Please be not upset if i make mistakes in grammar and spelling.
 * Michael Reschke, please contact me by Mail if necessary. I will support your project.
 * I'm willing to help where I can. Get in contact with me if necessary. --Edmond Dantes AUT 10:04, 28 July 2007 (UTC)
 * I'm no linguist but I'm a native speaker and I kind of know my language. Therefore, I suppose I would be able to help (and I would like too) Saschm 13:13, 29 March 2008 (UTC)

Interested Teachers

 * 1) --AaronRichard 11:35, 10 April 2007 (UTC) Hi, I'm a German teacher at a high school in Australia. I studied German in Australia and Germany, and have lived in Germany.

 * 1) --Alex 23:52, 4 October 2007 (UTC) Hello. I am a graduate student working on my PhD in German. I am currently student teaching with the ultimate goal of becoming a university professor in German.

 * 1) --Guten Tag! I'm starting up a project I call a German Home Immersion School. Any interested students or teachers feel free to ping me! Jon michael swift (discuss • contribs) 17:24, 18 May 2016 (UTC)

title: Portal talk:Learning German


text:
Talk:The Question

I'm curious why the original article was deleted. It references a copyright violation. I wrote that page originally and am pretty sure it was all original at that time...except perhaps for the phrase "the ultimate question of life, the universe and everything" the article currently completely misses the point of the original article. Originally it was supposed to be a collaborative PROJECT exploring an unanswerable question using a recursive method of question asking. As it is, it is now a rather pointless article talking about the ultimate question and should be removed from the projects section of philosophy. --71.222.47.189 01:59, 5 October 2007 (UTC)

 * Hello, I looked and saw that the page was deleted 12:32, 2 January 2007 with the note: copyvio from the Hitchhiker's guide to the galaxy book 2 page 37 Does that ring a bell perhaps ? Best is to ask the persons deleting that page.

 * "the article currently completely misses the point of the original article": well different people have different views - but you can still either include your ideas into the existing page now or create a new one stating at the begin of the article the goal. There are also other ways to help in the Wikiversity project. If you want to continue your engagement here I am sure we can help you while doing this. What do you say ? Erkan Yilmaz ( my talk page, wiki blog ) 06:38, 5 October 2007 (UTC)

Yeah what happened? This isn't a project...

A qusetionless world
when you have no question to ask in your life it means you know everything and in reality you know nothing.

==

title: Talk:The Question


text: User:Mike's bot account/Header

title: User:Mike's bot account/Header


text: Template:Tlxs

title: Template:Tlxs


text:
Template talk:Tlxs

en.wb edit history
* (cur) (last) 21:40, September 20, 2007 Darklama (Talk | contribs | block) m (707 bytes) (+up to 10 optional arguments) (rollback | undo) * (cur) (last) 20:37, September 20, 2007 Mike.lifeguard (Talk | contribs | block) m (127 bytes) (tweak) (undo) * (cur) (last) 20:35, September 20, 2007 Mike.lifeguard (Talk | contribs | block) m (119 bytes) (tweak) (undo) * (cur) (last) 20:34, September 20, 2007 Mike.lifeguard (Talk | contribs | block) m (102 bytes) (New page:  )

title: Template talk:Tlxs


text:
User talk:Eiskalteschatten

Welcome
Hello Eiskalteschatten, 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
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other wikiversiters on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! Erkan Yilmaz ( my talk page, wiki blog ) 05:54, 5 October 2007 (UTC)

title: User talk:Eiskalteschatten


text:
Template:EE DMM

X 100K X 100 X 100m X 100μ
 * }}

A   DC     AC     C L
 * }}

title: Template:EE DMM


text: Template:EE 7seg

title: Template:EE 7seg


text:
Template talk:EE DMM

HOW TO USE
MAXIMUM DISPLAY POSABLE LETTERS ARE 1 2 3 4 5 6 7 8 9 0 . + - first 2 letters are the switches "M" "K " " "m" "u" "r" "a "dc" "ac" "cl"

Eadthem 09:44, 5 October 2007 (UTC)

title: Template talk:EE DMM


text: Category:Solar System

title: Category:Solar System


text:
Template:Bcp/sepa/ip

(invasive or naturalized)

This template adds plant pages to Category:BCP/Flora of Pennsylvania and Category:BCP/SEPA/IP (it automatically adds ), for plants that are invasive in Southeast Pennsylvania. Later it can also be used to create a flora category for Pennsylvania, the Mid-Atlantic region, etc.

title: Template:Bcp/sepa/ip


text:
Template:Bcp/sepa/np

(native)

This template adds plant pages to Category:BCP/Flora of Pennsylvania and Category:BCP/SEPA/NP (it automatically adds ), for plants that are native in Southeast Pennsylvania. Later it can also be used to create a flora category for Pennsylvania, the Mid-Atlantic region, etc.

title: Template:Bcp/sepa/np


text:
Template:Bcp/sepa/gp

(cultivated)

This template adds plant pages to Category:BCP/Flora of Pennsylvania and Category:BCP/SEPA/GP (it automatically adds ), for plants that are cultivated in Southeast Pennsylvania. Later it can also be used to create a flora category for Pennsylvania, the Mid-Atlantic region, etc.

title: Template:Bcp/sepa/gp


text:
User:Mickyfitz13

Heres my sandbox for the time being, very busy at moment.

title: User:Mickyfitz13


text:
User:Mickyfitz13/sandbox

Introduction
The centre of mass is simply where the mass is pin-pointed on a single location on an object

title: User:Mickyfitz13/sandbox


text: User talk:85.31.137.11

title: User talk:85.31.137.11


text:
User talk:Wal S. Trebmal

Hello also from me. Have fun here. Below you find some info. Erkan Yilmaz ( my talk page, wiki blog ) 15:18, 5 October 2007 (UTC)

Welcome
Hello Wal S. Trebmal, 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
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other wikiversiters on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! Erkan Yilmaz ( my talk page, wiki blog ) 15:18, 5 October 2007 (UTC)

title: User talk:Wal S. Trebmal


text:
User talk:Benjaminmyklebust

Welcome
Hello Benjaminmyklebust, 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
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other wikiversiters on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! Erkan Yilmaz ( my talk page, wiki blog ) 15:19, 5 October 2007 (UTC)

French Mentoring
Salut, Benjaminmyklebust. You are listed at the French stream as being a beginning student of French with prior French-language experience, who is interested in learning French at Wikiversity. The French Mentoring project has recently been launched, and if you add your name to that list, I will do my best to put you in contact with an appropriate Mentor who would be glad to help you. The Jade Knight (d'viser) 08:37, 22 November 2008 (UTC)

title: User talk:Benjaminmyklebust


text:
Solar System, interactive/Earth/Solar System overview/Mercury

Mercury  (Mercury orbits close to the Sun)←Mercury→(Venus)

Mercury is the smallest planet in our solar system and it also is the closest planet to the sun.


text:
User talk:Onecardaway

Welcome
Hello Onecardaway, 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
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other wikiversiters on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! Erkan Yilmaz ( my talk page, wiki blog ) 16:02, 5 October 2007 (UTC)

title: User talk:Onecardaway


text:
User:JarrahTree

User JarrahTree - mainly at english wikipedia

title: User:JarrahTree


text:
Economic Classroom Experiments/Login Instructions

Login Instructions
Handing out printed Login Instructions is a good way of speeding up the lengthy process of getting everyone logged in at the start of a session. This is particularly so at the start of a new academic year, when the students are not yet familiar with the software.

It is possible to run the Exeter Games and Veconlab sites in kiosk mode provided you have Internet Explorer installed. The kiosk prevents the students from switching away from the experiment to check email or browse the web. Ctrl+Alt+Windows+F4 closes Internet Explorer and exits the kiosk.

Example: Exeter Games via Internet Explorer Kiosk
Start your computer and log on.

Start Internet Explorer.

Do a Google search for the words "feele kiosk".

Click on the first link: "Feele Kiosk: Internet Explorer Kiosk Mode".

Click on the upper of the two bullet-pointed links: "Log in to Feele (Exeter)". A "File Download - Security Warning" may appear, asking you, "Do you want to run or save this file?" If so, click Run (not Save). An "Internet Explorer - Security Warning" may then appear, asking you, "The publisher could not be verified. Are you sure you want to run this software?" If so, click Run (not Don't Run).

You should now see the "Feele Lab Participant Access" screen. Click on the top button, "Login" (Log in here to participate in an experiment) to bring up the "Participant Login" screen. Enter your first name, your last name and the Access Code for this experiment, which is [to be announced], then click on "Login".

Note that you do not need to write down your username and password.

You may start to read the instructions as soon as they appear; you might first have to wait for other participants to log in. When you reach the end of the instructions, you may start round 1 of the experiment.

Example: Veconlab via Internet Explorer Kiosk
Start your computer and log on.

Start Internet Explorer.

Do a Google search for the words "feele kiosk".

Click on the first link: "Feele Kiosk: Internet Explorer Kiosk Mode".

Click on the lower of the two bullet-pointed links: "Log in to Veconlab (Virginia – Charlie Holt)". A "File Download - Security Warning" may appear, asking you, "Do you want to run or save this file?" If so, click Run (not Save). An "Internet Explorer - Security Warning" may then appear, asking you, "The publisher could not be verified. Are you sure you want to run this software?" If so, click Run (not Don't Run).

You should now see the "Veconlab Participant Login Screen". Click on the top button, "Login" (Initial Login for all Programs) to bring up the "Veconlab: Enter Session Name" screen. Enter the session name for this experiment, which is [to be announced] and click on "Submit".

Enter your First Name and your Last Name; you do not need to enter an Optional Password.

Note that you do not need to write down your ID number or password.

You may start to read the instructions as soon as they appear; you might first have to wait for other participants to log in. When you reach the end of the instructions, you may start round 1 of the experiment.

Example: Exeter Games (non-kiosk)
Start your computer and log on.

Do a Google search for the word "feele".

Click on the first link: "FEELE Laboratory".

Click on the large link "Students: Log In here" in the section entitled "Students: Log In to a Teaching Experiment".

Click on the Login button, at the top of the "Feele Lab Participant Access" screen.

Enter your first name, your last name and the Access Code for this experiment, which is to be announced, then click on the Login button.

You do not need to write down your username and password, so you may now start reading the instructions.

When you reach the end of the instructions, you may begin. You will have to wait until the person with whom you are paired has also read the instructions and made their first decision.

Example: Veconlab (non-kiosk)
Start your computer and log on.

Do a Google search for "veconlab login".

Click on the first link: "Veconlab Login for all Programs".

Click on the Login button (Initial Login for All Programs).

Enter the session name for this experiment, which is to be announced, then click on the Submit button.

Enter just your first name and your last name (you do not need to enter a password), then click on the Continue button.

You do not need to write down your ID number, so you may now start reading the instructions.

When you reach the end of the instructions, you may begin. You will have to wait until the person with whom you are paired has also read the instructions and made their first decision.


text:
User talk:JarrahTree

JarrahTree talk page

title: User talk:JarrahTree


text: User:Whiteknight/Count Ticker/2007/October


text:
User:Whiteknight/Count Ticker

= About This Page =

This page is my counter ticker. Every day I query the Wikicharts page counter, analyze the data per-base page (not subpages) and post the total results here. For more information about this effort, check out my page on Wikibooks: b:User:Whiteknight/Count Ticker.

The counter is scheduled to be updated every night at 9PM EST. It may get updated more frequently if I do it manually (such as for testing).

Archives

 * 2007/October

 * 2007/December

 * 2008/January

 * 2008/February

= Reports =

title: User:Whiteknight/Count Ticker


text:
User:Whiteknight/HitsTotal

Month:   — Total Hits:   — Threshold:   — Sections Listed:   — Posted:   (About)

title: User:Whiteknight/HitsTotal


text: User:Whiteknight/Hits

title: User:Whiteknight/Hits


text: Category:User:Whiteknight/Count Ticker

title: Category:User:Whiteknight/Count Ticker


text:
Category:User:Whiteknight

These pages belong to User:Whiteknight

title: Category:User:Whiteknight


text: Template:Wikicite

title: Template:Wikicite


title: File:Lec24Fig1.jpg


text:
User:Anthrcer

Hello! Wikipedia users may recognise me as w:User:Anthrcer. Just look at that page!

title: User:Anthrcer


title: File:Lec23Fig1.jpg


title: File:Lec23Fig2.jpg


title: File:Lec23Fig3.jpg


title: File:Lec23Fig4a.jpg


title: File:Lec23Fig5.jpg


text:
User talk:Meldonium

Welcome
Hello Meldonium, 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
 * Take a guided tour and learn to edit;
 * Explore our learning projects;
 * Browse our portals, schools, and research activities;
 * Read and help develop our community policies；or
 * Chat with other wikiversiters on #wikiversity-en.

And don't forget to explore Wikiversity with the links to your left. Be bold, and see you around Wikiversity! Erkan Yilmaz ( my talk page, wiki blog ) 20:42, 5 October 2007 (UTC)

title: User talk:Meldonium


title: File:Lec23Fig6.jpg