text:
Waves in composites and metamaterials/Continuum limit and propagator matrix

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.

Previous Lecture
Recall from the previous lecture that we have been dealing with the TE equation

\left[ \mu(z)~\cfrac{d }{d z}\left(\mu(z)^{-1}~\cfrac{d }{d z}\right) + \omega^2~\epsilon(z)~\mu(z) - k_x^2\right]~\tilde{E}_y = 0 $$ where

E_y(x, z) = \tilde{E}_y(z)~e^{\pm i~k_x~x}~. $$ For a a multilayered medium with $$N$$ layers, we found that in the $$j$$-th layer
 * $$ \text{(1)} \qquad

\tilde{E}_{yj}(z) = A_j~\left[\exp(-i~k_{zj}~z) + \tilde{R}_{j, j+1}~\exp(i~k_{zj}~z + 2~i~k_{zj}~d_j) \right] $$ where $$\tilde{R}_{j, j+1}$$ is a generalized reflection coefficient. This coefficient can be obtained from a recursion relation of the form
 * $$ \text{(2)} \qquad

\tilde{R}_{j, j+1} = \cfrac{R_{j, j+1} + \tilde{R}_{j+1, j+2}~ \exp[2~i~k_{z, j+1}~(d_{j+1} - d_j)]} {1 + R_{j, j+1}~\tilde{R}_{j+1, j+2}~ \exp[2~i~k_{z, j+1}~(d_{j+1}-d_j)]} $$ where
 * $$ \text{(3)} \qquad

R_{j, j+1} = \cfrac{\mu_{j+1}~k_{z, j} - \mu_j~k_{z, j+1}} {\mu_{j+1}~k_{z, j} + \mu_j~k_{z, j+1}} $$ is the Fresnel reflection coefficient for TE waves. Equation (3) may also be written as
 * $$ \text{(4)} \qquad

R_{j, j+1} = \cfrac{\cfrac{k_{z, j}}{\mu_j} - \cfrac{k_{z, j+1}}{\mu_{j+1}}} {\cfrac{k_{z, j}}{\mu_j} + \cfrac{k_{z, j+1}}{\mu_{j+1}}} ~. $$

We will now proceed to determine the generalized reflection coefficient in the continuum limit.

Continuum Limit
Consider a medium where the layer thickness if $$\Delta$$. Denote the reflection coefficient and the generalized reflection coefficient at the interface $$z = -d_j$$ as $$R(z)$$ and $$\tilde{R}(z)$$, respectively. Also denote the phase velocity $$k_{z, j+1}$$ just below the interface as $$k_z(z - \Delta/2)$$ and the permeability $$\mu_{j+1}$$ as $$\mu(z - \Delta/2)$$. \footnote {This implies that we are measuring the phase velocity and the permeability at the center of the layer. However, this is not strictly necessary and we could alternatively measure these quantities at $$z - \Delta$$.} Then, equation (2) can be written as
 * $$ \text{(5)} \qquad

\tilde{R}(z) = \cfrac{R(z) + \tilde{R}(z-\Delta)~ \exp\left[2~i~k_z\left(z - \Delta/2\right)~\Delta\right]} {1 + R(z)~\tilde{R}(z - \Delta)~ \exp\left[2~i~k_z\left(z - \Delta/2\right)~\Delta\right]} $$ where

R(z) = \cfrac{\cfrac{k_z(z+\Delta/2)}{\mu(z+\Delta/2)} - \cfrac{k_z(z-\Delta/2)}{\mu(z-\Delta/2)}} {\cfrac{k_z(z+\Delta/2)}{\mu(z+\Delta/2)} + \cfrac{k_z(z-\Delta/2)}{\mu(z-\Delta/2)}} = \cfrac{\tilde{k}(z+\Delta/2) - \tilde{k}(z-\Delta/2)} {\tilde{k}(z+\Delta/2) + \tilde{k}(z-\Delta/2)} $$ with

\tilde{k} := \cfrac{k_z}{\mu} ~. $$ Expanding in Taylor series about $$z$$ and ignoring higher order terms, we get
 * $$ \text{(6)} \qquad

R(z) \approx \cfrac{\tilde{k}(z) + \Delta/2~\tilde{k}^{'}(z) - \tilde{k}(z) + \Delta/2~\tilde{k}^{'}(z)} {\tilde{k}(z) + \Delta/2~\tilde{k}^{'}(z) + \tilde{k}(z) - \Delta/2~\tilde{k}^{'}(z)} = \cfrac{\Delta~\tilde{k}^{'}(z)}{2~\tilde{k}(z)} ~. $$ Similarly, igoring powers $$\Delta^2$$ and higher, we get
 * $$ \text{(7)} \qquad

\exp\left[2~i~k_z\left(z - \Delta/2\right)~\Delta\right] \approx 1 + 2~\Delta~i~k_z(z) $$ and
 * $$ \text{(8)} \qquad

\tilde{R}(z - \Delta) \approx \tilde{R}(z) - \Delta~\tilde{R}^{'}(z) ~. $$ Plugging the expansions (7) and (8) into (5) gives
 * $$ \text{(9)} \qquad

\tilde{R}(z) \approx \cfrac{R(z) + \left[\tilde{R}(z) - \Delta~\tilde{R}^{'}(z)\right]~ \left[1 + 2~\Delta~i~k_z(z)\right]} {1 + R(z)~\left[\tilde{R}(z) - \Delta~\tilde{R}^{'}(z)\right]~ \left[1 + 2~\Delta~i~k_z(z)\right]} ~. $$ Substituting (6) into (9) and dropping terms containing $$\Delta^2$$ and higher gives
 * $$ \text{(10)} \qquad

\tilde{R}(z) \approx \cfrac{\cfrac{\Delta~\tilde{k}^{'}(z)}{2~\tilde{k}(z)} + \tilde{R}(z) - \Delta~\tilde{R}^{'}(z) + 2~\Delta~i~k_z(z)~\tilde{R}(z)} {1 + \cfrac{\Delta~\tilde{k}^{'}(z)}{2~\tilde{k}(z)}~\tilde{R}(z)}~. $$ If we assume that $$\Delta~\tilde{k}^{'}$$ is small such that the denominator can be expanded in a series, we get
 * $$ \text{(11)} \qquad

\tilde{R}(z) \approx \left[\cfrac{\Delta~\tilde{k}^{'}(z)}{2~\tilde{k}(z)} + \tilde{R}(z) - \Delta~\tilde{R}^{'}(z) + 2~\Delta~i~k_z(z)~\tilde{R}(z)\right] \left[1 - \cfrac{\Delta~\tilde{k}^{'}(z)}{2~\tilde{k}(z)}~\tilde{R}(z) + O(\Delta^2)\right]~. $$ After expanding and ingoring terms containing $$\Delta^2$$, we get
 * $$ \text{(12)} \qquad

\tilde{R}(z) \approx \cfrac{\Delta~\tilde{k}^{'}(z)}{2~\tilde{k}(z)} + \tilde{R}(z) - \Delta~\tilde{R}^{'}(z) + 2~\Delta~i~k_z(z)~\tilde{R}(z) - \cfrac{\Delta~\tilde{k}^{'}(z)}{2~\tilde{k}(z)}~[\tilde{R}(z)]^2 $$ or,
 * $$ \text{(13)} \qquad

{   \tilde{R}^{'}(z) = 2~i~k_z(z)~\tilde{R}(z) + \cfrac{\tilde{k}^{'}(z)}{2~\tilde{k}(z)} \left\{1-[\tilde{R}(z)]^2\right\} ~. } \qquad \text{Ricotti equation} $$ We thus get an equation that gives a continuous representation of the generalized reflection coefficient $$\tilde{R}(z)$$. Equation (13) can be solved numerically using the Runge-Kutta method.

For example, in the stuation shown in Figure 1, the generalized reflectivity coefficient at the point $$z_1$$ is $$\tilde{R}(z_1)$$ while that at point $$z_0$$ is 0. If we wish to determine the value of $$\tilde{R}(z_i)$$ at a point inside the smoothly varying layer, then one possibility is to assume that $$\mu(z)$$ and $$\epsilon(z)$$ is constant for $$z > z_i$$ and compute the value of $$\tilde{R}$$ in the usual manner.

There can also be a situation where there are a few isolated strong discontinuities inside the graded layer as shown in Figure 2. If there is a discontinuity at $$z_c$$, we can use the discrete solution with layer thickness 0 at the discontinuity.

Then, from (2), at the discontinuity
 * $$ \text{(14)} \qquad

\tilde{R}(z_c^+) = \cfrac{R(z_c) + \tilde{R}(z_c^-)} {1 + R(z_c)~\tilde{R}(z_c^-)} ~. $$ Also, from (4)
 * $$ \text{(15)} \qquad

R(z_c) = \cfrac{\cfrac{k_z^+}{\mu^+} - \cfrac{k_z^-}{\mu^-}} {\cfrac{k_z^+}{\mu^+} + \cfrac{k_z^-}{\mu^-}}~. $$ Hence we can find the generalized reflection coefficients at isolated discontinuities within the material.

Determining the coefficients Aj
Recall equation (1):

\tilde{E}_{yj}(z) = A_j~\left[\exp(-i~k_{zj}~z) + \tilde{R}_{j, j+1}~\exp(i~k_{zj}~z + 2~i~k_{zj}~d_j) \right] ~. $$ So far we have determine the value of $$\tilde{R}$$ is this equation. But how do we determine the coefficients $$A_j$$ in multilayered media?

Let us start with the coefficients for a single layer that we determined in the previous lecture. We had
 * $$ \text{(16)} \qquad

A_2 = \cfrac{T_{12}~A_1~\exp[i~(k_{z1} - k_{z2})~d_1]} {1 - R_{21}~R_{23}~\exp[2~i~k_{z2}~(d_2-d_1)]} $$ where

T_{12} = \cfrac{2~\mu_2~k_{z1}}{\mu_2~k_{z1} + \mu_1~k_{z2}} ~. $$ We can rewrite (16) as
 * $$ \text{(17)} \qquad

A_2~\exp(i~k_{z2}~d1) = \cfrac{T_{12}~A_1~\exp(i~k_{z1}~d_1)} {1 - R_{21}~R_{23}~\exp[2~i~k_{z2}~(d_2-d_1)]} $$ Using the same arguments as before, we can generalize (17) to a medium with $$N$$ layers. Thus, for the $$j$$-th layer, we have
 * $$ \text{(18)} \qquad

A_j~\exp(i~k_{z,j}~d_{j-1}) = \cfrac{T_{j-1, j}~A_{j-1}~\exp(i~k_{z, j-1}~d_{j-1})} {1 - R_{j, j-1}~\tilde{R}_{j, j+1}~ \exp[2~i~k_{z,j}~(d_j-d_{j-1})]} ~. $$ Define

S_{j-1, j} := \cfrac{T_{j-1, j}}{1 - R_{j, j-1}~\tilde{R}_{j, j+1}~ \exp[2~i~k_{z,j}~(d_j-d_{j-1})]} ~. $$ Then we can write (18) as
 * $$ \text{(19)} \qquad

\begin{align} A_j~\exp(i~k_{z,j}~d_{j-1}) & = S_{j-1, j}~A_{j-1}~\exp(i~k_{z, j-1}~d_{j-1}) \\ & =   S_{j-1, j}~A_{j-1}~\exp(i~k_{z, j-1}~d_{j-2})~ \exp[i~k_{z, j-1}~(d_{j-1} - d_{j-2})] ~. \end{align} $$ The second of equations (19) gives us a recurrence relation that can be used to compute the other $$A_j$$ s. Thus, we can write
 * $$ \text{(20)} \qquad

\begin{align} A_j~\exp(i~k_{z,j}~d_{j-1}) & =    A_{j-1}~\exp(i~k_{z, j-1}~d_{j-2})~ S_{j-1, j}~\exp[i~k_{z, j-1}~(d_{j-1} - d_{j-2})] \\ & =    A_{j-2}~\exp(i~k_{z, j-2}~d_{j-3})~ S_{j-2, j-1}~\exp[i~k_{z, j-2}~(d_{j-2} - d_{j-3})]~ S_{j-1, j}~\exp[i~k_{z, j-1}~(d_{j-1} - d_{j-2})] \\ & = \dots \\ & =    A_1~\exp(i~k_{z1}~d_{1})~ \prod_{m=1}^{j-1} S_{m, m+1}~\exp[i~k_{z, m}~(d_m - d_{m-1})]~. \end{align} $$ We can introduce a generalized transmission coefficient

\tilde{T}_{1N} := \prod_{m=1}^{N-1} S_{m, m+1}~\exp[i~k_{z, m}~(d_m - d_{m-1})]~. $$ Then,

{ A_N~\exp(i~k_{z,N}~d_{N-1})  = \tilde{T}_{1N}~A_1~\exp(i~k_{z1}~d_{1})~. } $$ So the downgoing wave amplitude in region $$N$$ at $$z = -d_{n+1}$$ is $$\tilde{T}_{1N}$$ times the downgoing amplitude in region 1 ($$z = -d_1$$). Due to the products involved in the above relation, a continuum extension of this formula is not straightforward.

State equations and Propagator matrix
The propagator matrix relates the fields at two points in a multilayered medium. This matrix is also known as the transition matrix or the transfer matrix.

Let us examine the propagation matrix for a TM wave. Recall the governing equation for a TM wave:
 * $$ \text{(21)} \qquad

\left[ \epsilon(z)~\cfrac{d }{d z}\left(\epsilon(z)^{-1}~\cfrac{d }{d z}\right) + \omega^2~\epsilon(z)~\mu(z) - k_x^2\right]~\tilde{H}_y = 0 $$ where

H_y(x, z) = \tilde{H}_y(z)~e^{\pm i~k_x~x}~. $$ Define $$\varphi := \tilde{H}_y$$. Also,

k_z^2 = \omega^2~\epsilon(z)~\mu(z) - k_x^2 ~. $$ Therefore, (21) can be written as
 * $$ \text{(22)} \qquad

\left[ \epsilon(z)~\cfrac{d }{d z}\left(\epsilon(z)^{-1}~\cfrac{d }{d z}\right) + k_z^2\right]~\varphi = 0 ~. $$ To reduce (22) to a first order differential equation, introduce the quantity
 * $$ \text{(23)} \qquad

\psi := \cfrac{1}{i\omega\epsilon}~\cfrac{d \varphi}{d z}  \qquad \implies \qquad \cfrac{d \varphi}{d z} = i\omega\epsilon~\psi ~. $$ Clearly, $$\psi$$ has to be continuous across the interface for the differential equation (22) to be satisfied.

Plugging (23) into (22) gives
 * $$ \text{(24)} \qquad

\cfrac{d \psi}{d z} = \cfrac{i~k_z^2}{\epsilon~\omega}~\varphi ~. $$ Therefore, (23)$$_2$$ and (24) for a system of differential equations which can be written as
 * $$ \text{(25)} \qquad

\cfrac{d }{d z}\begin{bmatrix} \varphi \\ \psi \end{bmatrix} = \begin{bmatrix} 0 & i\omega\epsilon \\ \cfrac{i~k_z^2}{\epsilon~\omega} & 0 \end{bmatrix}~\begin{bmatrix} \varphi \\ \psi \end{bmatrix} ~. $$ Define

\mathbf{v} := \begin{bmatrix} \varphi \\ \psi \end{bmatrix} ~; \overline{\mathbf{H}} := \begin{bmatrix} 0 & i\omega\epsilon \\ \cfrac{i~k_z^2}{\epsilon~\omega} & 0 \end{bmatrix} ~. $$ Then, equations (25) can be written as
 * $$ \text{(26)} \qquad

\cfrac{d \mathbf{v}}{d z} = \overline{\mathbf{H}}~\mathbf{v} ~. $$ If $$\overline{\mathbf{H}}$$ is constant, particular solutions to (26) can be sought of the form
 * $$ \text{(27)} \qquad

\mathbf{v} = \mathbf{v}_0~e^{\lambda~z} ~. $$ Plugging (27) into (25) leads to the eigenvalue problem

(\overline{\mathbf{H}} - \lambda~\mathbf{1})~\mathbf{v}_0 = \mathbf{0} ~. $$ Solutions exist only if

\det(\overline{\mathbf{H}} - \lambda~\mathbf{1}) = 0 \qquad \implies \qquad \lambda^2 = - k_z^2 \qquad \implies \qquad \lambda = \pm i~k_z ~. $$ Therefore, the general solution of (26) is
 * $$ \text{(28)} \qquad

\mathbf{v}(z) = A_+~e^{i~k_z~z}~\mathbf{n}_+ + A_-~e^{-i~k_z~z}~\mathbf{n}_- $$ where $$\mathbf{n}_+$$ and $$\mathbf{n}_-$$ are the eigenvectors corresponding to the eigenvalues $$i~k_z$$ and $$-i~k_z$$, respectively.

Equation (28) can be written more compactly in the form
 * $$ \text{(29)} \qquad

\mathbf{v}(z) = \begin{bmatrix} \mathbf{n}_+ & \mathbf{n}_- \end{bmatrix} \begin{bmatrix} e^{i~k_z~z} & 0 \\ 0 & e^{-i~k_z~z} \end{bmatrix} \begin{bmatrix} A_+ \\ A_- \end{bmatrix} $$ or,
 * $$ \text{(30)} \qquad

\mathbf{v}(z) = \tilde{\mathbf{n}} ~\overline{\mathbf{K}}(z)~ \tilde{\mathbf{a}} $$ where

\tilde{\mathbf{n}} := \begin{bmatrix} \mathbf{n}_+ & \mathbf{n}_- \end{bmatrix} ~; \overline{\mathbf{K}}(z) := \begin{bmatrix} e^{i~k_z~z} & 0 \\ 0 & e^{-i~k_z~z} \end{bmatrix} ~; \tilde{\mathbf{a}} := \begin{bmatrix} A_+ \\ A_- \end{bmatrix} ~. $$ Note that, for a point $$z'$$ that is different from $$z$$,

\overline{\mathbf{K}}(z - z') = \begin{bmatrix} e^{i~k_z~(z- z')} & 0 \\ 0 & e^{-i~k_z~(z-z')} \end{bmatrix} = \begin{bmatrix} e^{i~k_z~z} & 0 \\ 0 & e^{-i~k_z~z} \end{bmatrix} \begin{bmatrix} e^{-i~k_z~z'} & 0 \\ 0 & e^{i~k_z~z'} \end{bmatrix} = \overline{\mathbf{K}}(z)~\overline{\mathbf{K}}(-z') ~. $$ Also,

\overline{\mathbf{K}}(z - z')~\overline{\mathbf{K}}(z') = \begin{bmatrix} e^{i~k_z~(z- z')} & 0 \\ 0 & e^{-i~k_z~(z-z')} \end{bmatrix} \begin{bmatrix} e^{-i~k_z~z'} & 0 \\ 0 & e^{i~k_z~z'} \end{bmatrix} = \overline{\mathbf{K}}(z) ~. $$ Therefore we can write (30) in the form

\mathbf{v}(z) = \tilde{\mathbf{n}} ~\overline{\mathbf{K}}(z - z')~\overline{\mathbf{K}}(z')~ \tilde{\mathbf{a}} = \tilde{\mathbf{n}} ~\overline{\mathbf{K}}(z - z')~\tilde{\mathbf{n}}^{-1}\tilde{\mathbf{n}}~\overline{\mathbf{K}}(z')~	\tilde{\mathbf{a}} $$ or,
 * $$ \text{(31)} \qquad

{ \mathbf{v}(z) = \overline{\mathbf{P}}(z, z')~\mathbf{v}(z') } $$ where

{ \overline{\mathbf{P}}(z, z') := \tilde{\mathbf{n}} ~\overline{\mathbf{K}}(z - z')~\tilde{\mathbf{n}}^{-1} ~. } $$ The matrix $$\overline{\mathbf{P}}$$ is called the  propagator matrix or the transition matrix that related the fields at $$z$$ and $$z'$$.

In a multilayered system (see Figure~3), since the vector $$\mathbf{v}$$ is discontinuous, we can show that

\overline{\mathbf{P}}(-d_2, 0) = \overline{\mathbf{P}}(-d_2, -d_1) ~\overline{\mathbf{P}}(-d_1, 0) $$ where $$\overline{\mathbf{P}}(-d_2, -d_1)$$ depends on $$\epsilon_3, \mu_3$$ and $$\overline{\mathbf{P}}(-d_1, 0)$$ depends on $$\epsilon_2, \mu_2$$.


text:
Category:Electromagnetism

This category deals with learning materials related to electromagnetism.

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

title: Category:Electromagnetism


text:
Digital Libraries

Collaborative Research: Curriculum Development: Digital Libraries
The project was originally funded for 1/1/2006 - 12/31/2008. This was extended until 12/31/2009.

 * Project Team at VT:
 * PI: Edward A. Fox (fox@vt.edu)
 * GRA: Seungwon Yang (seungwon@vt.edu)
 * REU: John Ewers (jewers@vt.edu)

 * Project Team at UNC-CH:
 * PI: Barbara M. Wildemuth (wildem@ils.unc.edu)
 * Co-PI: Jeffrey Pomerantz (pomerantz@unc.edu)
 * GRA: Sanghee Oh (shoh@email.unc.edu)

This material is based upon work supported by the National Science Foundation under Grant Nos. IIS-0535057 (VT) and IIS-0535060 (UNC-CH).

Table 1. Curriculum Framework (Including Links to Specific Modules)
Note: Click here for a list of modules signed up for developments

Table 2. Modules for Information Retrieval Software Packages
Note: These modules can be used along with IR-related DL curriculum modules. Especially, these modules might be useful to add more technical aspect in DL classes and to provide students with realistic exercises using IR tools.

Table 3. Modules for LucidWorks Big Data Software
Note: These modules have been developed for the LucidWorks software, which is an integrated tool for processing Big Data.

Table 4. Modules for Multimedia-Related Software
Note: These modules have been developed for software tools, which are related to sound and image manipulations as well as biometric image software.

Versions of the Curriculum Framework Developed by VT/UNC Project

 * Framework for DL Curriculum (08/23/2008)
 * Framework for DL Curriculum (09/16/2007)
 * 10-Module Diagram
 * 9-Module Diagram
 * The Original 19-Module Set

Module Template

 * Template (since Aug. 27, 2007)
 * Template (before Aug. 27, 2007)

DL Syllabi Analyses

 * Computer Science DL Syllabi Reading Analysis: Resources for "Digital Library Education in Computer Science Programs, Pomerantz et al.", JCDL'07.
 * Library and Information Science DL Syllabi Reading Analysis

Others

 * DL Educational Taxonomy

Journal

 * Pomerantz, J., Abbas, J., & Mostafa, J. (2009). Teaching Digital Library Concepts Using Digital Library Applications. International Journal on Digital Libraries, 10(1), 1-13.

 * Pomerantz, J., Oh, S., Yang, S., Fox, E. A., & Wildemuth, B. (2006). The Core: Digital Library Education in Library and Information Science Programs. D-Lib Magazine, 12(11). Supplementary material: Book_list, Journal_Article_list, Journal_list.

 * Pomerantz, J., Wildemuth, B. M., Oh, S., Yang, S., & Fox, E. A. (2006). Briefing: Digital libraries curriculum development. D-Lib Magazine 12(7/8).

Conference Papers / Posters

 * Wildemuth, B.M, Pomerantz, J., Oh, S., Yang, S., Fox, E.A. (2009). I-School as a Natural Home for Digital Libraries Education. Poster presented at iConference, February 8-11, 2009.Poster

 * Wildemuth, B.M., Pomerantz, J., Oh, S., Yang, S., Fox, E.A. (2009). The Variety of Ways in Which Instructors Implement a Modular Digital Library Curriculum. Poster presented at the Joint Conference on Digital Libraries, Austin, TX, June 15-19, 2009. Poster

 * Yang, S., Levy, J., Miller, K., Pomerantz, J.P., Oh, S., Wildemuth, B.M., Fox, E.A. (2008). Two approaches to enhance the education for ETDs: Developing educational modules and migrating the ETD Guide into a community wiki. Paper presented at ETD 2008: The 11th International Symposium on Electronic Theses and Dissertations, Aberdeen Scotland, June 4-7, 2008.Paper

 * Yang, S., Wildemuth, B. Kim, S., Murthy, U., Pomerantz, J., Oh, S. & Fox, E. A. (2007). Further Development of a Digital Library Curriculum: Evaluation Approaches and New Tools. The 10th International Conference on Asian Digital Libraries, Hanoi, Vietnam, Dec. 10-13, 2007. Paper, Presentation Slides

 * Pomerantz, J., Oh, S., Wildemuth, B., Yang, S. & Fox, E. A. (2007). Digital Library Education in Computer Science Programs. The 7th ACM/IEEE-CS Joint Conference on Digital Libraries, Vancouver, British Columbia, Canada, June 18-23.Paper

 * Yang, S., Oh, S., Pomerantz, J., Wildemuth, B., & Fox, E. A. (2007). Improving Education and Understanding of NDLTD. The 10th International Symposium on Electronic Theses and Dissertations, Uppsala, Sweden, June 13-16, 2007. Paper, Presentation Slides

 * Abbas, j., Kim, K.-S., Wildemuth, B. M., Choi, Y., Mostafa, J., Brancolini, K., Pomerantz, J., Clobridge, A. (2006). Education for Digital Librarianship: Employer's Needs and How They Can Be Addressed. Panel presented at the 69th Annual Meeting of the American Society for Information Science and Technology, Austin, TX. Abstract

 * Pomerantz, J., Wildemuth, B., Fox, E. A., & Yang, S. (2006). Curriculum Development for Digital Libraries. In Proceedings of the 6th ACM/IEEE-CS Joint Conference on Digital Libraries (pp. 175-184). New York: Association for Computing Machinery. Paper in ACM DL, Powerpoint presentation, JCDL 2006 Conference site

 * Yang, S., Fox, E. A., Wildemuth, B., Pomerantz, J., & Oh, S. (2006)Interdisciplinary Curriculum Development for Digital Library Education. In Proceedings of the 9th International Conference on Asian Digital Libraries (pp. 61-70). Paper published by Springer Berlin, Slides are attached under 'Presentation slides', ICADL 2006 Conference site

Book Chapters

 * Yang, S., Fox, E., Wildemuth, B., Pomerantz, J., & Oh, S. (2009). Core Topics in Digital Library Education . In Theng, Foo, Goh, and Na (Ed.), Handbook of Research on Digital Libraries: Design, Development and Impact (pp.493-505). IGI Global.

 * Yang, S., Fox, E., Wildemuth, B., Pomerantz, J., & Oh, S. (2006). Digital library education, curricula, modules, and resources. In Encyclopedia of Digital Libraries. Proposal accepted in 2006.

Presentations

 * Pomerantz, J. (2008). Designing, Developing, and Evaluating an Interdisciplinary Digital Library Curriculum. In J. Selthofer, T. Aparac-Jelušić & M. Krtalić (Eds.), Libraries in the Digital Age (LIDA) 2008 (pp. 143-148). Dubrovnik and Mljet, Croatia: Faculty of Philosophy, University J. J. Strossmayer and Rutgers University.Presentation Slides
 * Wildemuth, B., & Fox, E. A. (2008). Preliminary Results from Field Testing of DL Curriculum Modules. JCDL Workshop: Education for Digital Stewardship: Librarians, Archivists or Curators? Pittsburge, PA, June 16-20. Presentation Slides

 * Wildemuth, B.M. (2007). Bringing your library into view. Presented at Jinonice Monday's, Institute of Information Studies and Librarianship, Charles University, October 1, 2007. Presentation Slides

Project Reports

 * NSF Reports: proposal, abstract, 1st year, 2nd year, 3rd year report, highlights (VT, UNC)

 * Project Progress Report : August 11, 2006
 * Project overview at UNC-CH : Slides for advisory board meeting at UNC-CH - June 14, 2006
 * Project introduction at ETD '06 : Presentation in Quebec, Canada - June 10, 2006
 * Project overview - VT : Slides for advisory board meeting at VT - May 1, 2006

Project Links

 * Project Main Page
 * Digital Library Curriculum WIKI homepage (password-protected)
 * Project Progress Diary
 * FAQ link
 * NSF proposal
 * NSF abstract
 * NSF 1st year report
 * NSF 2nd year report
 * NSF 3rd year report
 * NSF 4th year report

Project Evaluation Resource

 * Instructor Interview Guide

title: Digital Libraries


text:
Solar System, interactive/Earth

Earth is shaped like a ball. It is not a flat land, as we see it, when we stand on the street or in the meadow. This ball is flying through space around the Sun. You might ask, “why don’t we fall into this dark space?” The answer is because of gravity. Gravity is a power which holds us on the Earth. “And what protects us from space?” – “It is the atmosphere”. Clouds form part of this atmosphere. And now, let’s leave the Earth by clicking on the space we are flying through to have a look at Our Solar System.

Earth is a ball

title: Solar System, interactive/Earth


text:
Waves in composites and metamaterials/Waves in layered media and point sources

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.

Previous Lecture
Recall from the previous lecture the governing equation for a TM wave.
 * $$ \text{(1)} \qquad

\left[ \epsilon(z)~\cfrac{d }{d z}\left(\epsilon(z)^{-1}~\cfrac{d }{d z}\right) + k_z^2\right]~\varphi = 0 $$ where

H_y(x, z) = \varphi(z)~e^{\pm i~k_x~x} = \tilde{H}_y(z)~e^{\pm i~k_x~x} $$ and

k_z^2 = \omega^2~\epsilon(z)~\mu(z) - k_x^2 ~. $$ Also recall that we introduced a  state vector $$\mathbf{v}$$ which is continuous across boundaries. The state vector is defined as

\mathbf{v} := \begin{bmatrix} \varphi \\ \psi \end{bmatrix} $$ where

\psi := \cfrac{1}{i\omega\epsilon}~\cfrac{d \varphi}{d z} ~. $$ Then,
 * $$ \text{(2)} \qquad

\cfrac{d \mathbf{v}}{d z} = \begin{bmatrix} 0 & i\omega\epsilon \\ \cfrac{i~k_z^2}{\epsilon~\omega} & 0 \end{bmatrix}~\mathbf{v} = \overline{\mathbf{H}}~\mathbf{v} ~. $$ The general solution of (2) is
 * $$ \text{(3)} \qquad

\mathbf{v}(z) = A_+~e^{i~k_z~z}~\mathbf{n}_+ + A_-~e^{-i~k_z~z}~\mathbf{n}_- $$ where $$\mathbf{n}_+$$ and $$\mathbf{n}_-$$ are the eigenvectors corresponding to the eigenvalues $$i~k_z$$ and $$-i~k_z$$, respectively.

We also found that the solution of the differential equation (2) can be expressed in terms of a  propagator matrix $$\overline{\mathbf{P}}$$ such that
 * $$ \text{(4)} \qquad

\mathbf{v}(z) = \overline{\mathbf{P}}(z, z')~\mathbf{v}(z') $$ where

\overline{\mathbf{P}}(z, z') := \tilde{\mathbf{n}} ~\overline{\mathbf{K}}(z - z')~\tilde{\mathbf{n}}^{-1} \equiv \tilde{\mathbf{n}}~e^{i\tilde{\mathbf{K}}(z - z')}~\tilde{\mathbf{n}}^{-1} $$ where the columns of $$\tilde{\mathbf{n}}$$ are the eigenvectors of $$\overline{\mathbf{H}}$$ and

\tilde{\mathbf{K}} = \begin{bmatrix} k_z & 0 \\ 0 & - k_z \end{bmatrix} ~; e^{i\tilde{\mathbf{K}} z} = \begin{bmatrix} e^{i k_z z} & 0 \\ 0 & e^{-i k_z z} \end{bmatrix} ~. $$ The matrix $$\overline{\mathbf{P}}$$ is called the  propagator matrix or the transition matrix that related the fields at $$z$$ and $$z'$$.

Also, in a multilayered system (see Figure 1)

\overline{\mathbf{P}}(-d_2, 0) = \overline{\mathbf{P}}(-d_2, -d_1) ~\overline{\mathbf{P}}(-d_1, 0) $$ where $$\overline{\mathbf{P}}(-d_2, -d_1)$$ depends on $$\epsilon_3, \mu_3$$ and $$\overline{\mathbf{P}}(-d_1, 0)$$ depends on $$\epsilon_2, \mu_2$$.

A special case
Let us now consider the particular case shown in Figure 2. The medium consists of three layers. Regions 1 and 3 are isotropic while the sandwiched region 2 is multilayered. The interface between regions 1 and 2 is located at $$z = 0$$ while the interface between regions 2 and 3 is located at $$z = -d$$. Let the propagator matrix of region 2 be $$\overline{\mathbf{P}}(-d, 0)$$. We want to find the reflection coefficient and the transmission coefficient of the system.

In reqion 1, the state vector is given by (see equation (3))

\mathbf{v}_1(z) = A_{1_+}~e^{i~k_{z1}~z}~\mathbf{n}_{1_+} + A_{1_-}~e^{-i~k_{z1}~z}~\mathbf{n}_{1_-} ~. $$ Define

A_{1_+} := R~A_{1_-} $$ where $$R$$ is like a scalar reflection coefficient. Then
 * $$ \text{(5)} \qquad

\mathbf{v}_1(z) = R~A_{1_-}~e^{i~k_{z1}~z}~\mathbf{n}_{1_+} + A_{1_-}~e^{-i~k_{z1}~z}~\mathbf{n}_{1_-} ~. $$ Proceeding as we did in the previous lecture, let us define matrices

\tilde{\mathbf{n}}_1 := \begin{bmatrix} \mathbf{n}_{1_-} & \mathbf{n}_{1_+} \end{bmatrix} \qquad \text{and} \qquad e^{i\tilde{\mathbf{K}}_1 z} := \begin{bmatrix} e^{i k_{z1} z} & 0 \\ 0 & e^{-i k_{z1} z} \end{bmatrix} ~. $$ Then equation (5) can be written as
 * $$ \text{(6)} \qquad

{ \mathbf{v}_1(z) = \tilde{\mathbf{n}}_1~e^{i\tilde{\mathbf{K}}_1 z}~ \begin{bmatrix} R \\ 1 \end{bmatrix}~A_{1_-} ~. } $$ In region 3, there is only a transmitted wave. Therefore, the state vector is given by
 * $$ \text{(7)} \qquad

\mathbf{v}_3(z) = A_{3_-}~e^{-i~k_{z3}~(z+d)}~\mathbf{n}_{3_-} ~. $$ Define

A_{3_-} := T~A_{1_-} $$ where $$T$$ is a factor that acts like a scalar transmission coefficient. Then, equation (7) can be written in matrix form as
 * $$ \text{(8)} \qquad

{ \mathbf{v}_3(z) = \tilde{\mathbf{n}}_3~e^{i\tilde{\mathbf{K}}_3 (z+d)}~ \begin{bmatrix} 0 \\ T \end{bmatrix}~A_{1_-} } $$ where

\tilde{\mathbf{n}}_3 := \begin{bmatrix} \mathbf{n}_{3_-} & \mathbf{n}_{3_+} \end{bmatrix} \qquad \text{and} \qquad e^{i\tilde{\mathbf{K}}_3 z} := \begin{bmatrix} e^{i k_{z3} z} & 0 \\ 0 & e^{-i k_{z3} z} \end{bmatrix} ~. $$ Since we have the propagator matrix for region 2, $$\overline{\mathbf{P}}(-d,0)$$, we can use it to connect regions 1 and 3. The continuity of the state vector across the interfaces implies that
 * $$ \text{(9)} \qquad

\mathbf{v}_1(0) = \mathbf{v}_2(0) \quad \text{at}~z = 0\qquad \text{and} \qquad \mathbf{v}_2(-d) = \mathbf{v}_3(-d) \quad \text{at}~z = -d~. $$ Also, using equation (4) we have
 * $$ \text{(10)} \qquad

\mathbf{v}_2(-d) = \overline{\mathbf{P}}(-d, 0)~\mathbf{v}_2(0) ~. $$ Therefore, using equations (9), we can write (10) as
 * $$ \text{(11)} \qquad

{ \mathbf{v}_3(-d) = \overline{\mathbf{P}}(-d, 0)~\mathbf{v}_1(0) ~. } $$ From (6), at $$z = 0$$, we have
 * $$ \text{(12)} \qquad

\mathbf{v}_1(0) = \tilde{\mathbf{n}}_1~\begin{bmatrix} R \\ 1 \end{bmatrix}~A_{1_-} ~. $$ Also, from (8), at $$z = -d$$, we have
 * $$ \text{(13)} \qquad

\mathbf{v}_3(-d) = \tilde{\mathbf{n}}_3~\begin{bmatrix} 0 \\ T \end{bmatrix}~A_{1_-} ~. $$ Plugging into (11) gives

\tilde{\mathbf{n}}_3~\begin{bmatrix} 0 \\ T \end{bmatrix}~A_{1_-} = \overline{\mathbf{P}}(-d, 0)~\tilde{\mathbf{n}}_1~\begin{bmatrix} R \\ 1 \end{bmatrix}~A_{1_-} $$ or
 * $$ \text{(14)} \qquad

{ \begin{bmatrix} 0 \\ T \end{bmatrix} = \tilde{\mathbf{n}}^{-1}~\overline{\mathbf{P}}(-d, 0)~\tilde{\mathbf{n}}_1~ \begin{bmatrix} R \\ 1 \end{bmatrix} ~. } $$ Equation (14) can then be solved to find the reflection and transmission coefficients $$R$$ and $$T$$.

Extension to waves in anisotropic layered media
In a layer medium where each of the layers is isotropic, the TE and TM waves are uncoupled at the interface. However, this is not true when each of the layers is anisotropic and we have to consider the full Maxwell's equations. The state vector approach can still be used for anisotropic media by choosing the variables such that they are continuous across interfaces.

Let us start with Maxwell's equations

\boldsymbol{\nabla} \times \mathbf{E} = i \omega~\boldsymbol{\mu}\cdot\mathbf{H} ~; \boldsymbol{\nabla} \times \mathbf{H} = -i \omega~\boldsymbol{\epsilon}\cdot\mathbf{E} ~. $$ Recall that continuity of the fields requires that the tangential components of $$\mathbf{E}$$ and $$\mathbf{H}$$ be continuous across material interfaces. Therefore, an appropriate state vector for anisotropic media is

\mathbf{v} := \begin{bmatrix} \mathbf{E}_s \\ \mathbf{H}_s \end{bmatrix} $$ where $$\mathbf{E}_s$$ and $$\mathbf{H}_s$$ are the tangential components of $$\mathbf{E}$$ and $$\mathbf{H}$$ (i.e., the components on the surface normal to the $$z$$ direction).

Let us decompose the vector fields into a sum of the normal and tangential components:

\mathbf{E} = \mathbf{E}_s + \mathbf{E}_z ~; \mathbf{H} = \mathbf{H}_s + \mathbf{H}_z ~. $$ The gradient operator can also be split along the same lines, i.e.,

\boldsymbol{\nabla} = \boldsymbol{\nabla}_s + \frac{\partial }{\partial z}~\mathbf{e}_z ~; \boldsymbol{\nabla}_s := \frac{\partial }{\partial x}~\mathbf{e}_x + \frac{\partial }{\partial y}~\mathbf{e}_y $$ where $$\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$$ are the unit vectors in the $$x, y, z$$ directions, respectively. Let us express the tensors $$\boldsymbol{\mu}$$ and $$\boldsymbol{\epsilon}$$ in matrix form (with respect to the basis $$\mathbf{e}_x, \mathbf{e}_y, \mathbf{e}_z$$) as

\boldsymbol{\mu} \equiv \begin{bmatrix} \boldsymbol{\mu}_{ss} & \boldsymbol{\mu}_{sz} \\ \boldsymbol{\mu}_{zs} & \mu_{zz} \end{bmatrix} ~; \boldsymbol{\epsilon} \equiv \begin{bmatrix} \boldsymbol{\epsilon}_{ss} & \boldsymbol{\epsilon}_{sz} \\ \boldsymbol{\epsilon}_{zs} & \epsilon_{zz} \end{bmatrix} $$ where $$\boldsymbol{\mu}_{ss}, \boldsymbol{\epsilon}_{ss}$$ are $$2\times 2$$ matrices, $$\boldsymbol{\mu}_{sz}, \boldsymbol{\epsilon}_{sz}$$ are $$2\times 1$$ matrices, $$\boldsymbol{\mu}_{zs}, \boldsymbol{\epsilon}_{zs}$$ are $$1\times 2$$ matrices, and $$\mu_{zz}, \epsilon_{zz}$$ are $$1\times 1$$ matrices, i.e., scalars.

Using the splits of the various quantities and the gradient operator, we can show ({\Red work this out}) that $$\mathbf{E}_z, \mathbf{H}_z$$ can be expressed in terms of $$\mathbf{E}_s, \mathbf{H}_s$$ as

\mathbf{E}_z = -\cfrac{1}{i\omega \epsilon_{zz}}~\boldsymbol{\nabla}_s\times\mathbf{u}{\mathbf{H}_s} - \cfrac{1}{\epsilon_{zz}}~\boldsymbol{\epsilon}_{zs}~\mathbf{E}_s ~; \mathbf{H}_z = \cfrac{1}{i\omega \mu_{zz}}~\boldsymbol{\nabla}_s\times\mathbf{u}{\mathbf{E}_s} - \cfrac{1}{\mu_{zz}}~\boldsymbol{\mu}{zs}~\mathbf{H}_s ~. $$ After some further manipulations, the Maxwell equations may be expressed in matrix form as (see ~Chew95 for details)
 * $$ \text{(15)} \qquad

{ \frac{\partial }{\partial z} \begin{bmatrix} \mathbf{E}_s \\ \mathbf{H}_s \end{bmatrix} = \overline{\mathbf{H}} \begin{bmatrix} \mathbf{E}_s \\ \mathbf{H}_s \end{bmatrix} } $$ where $$\overline{\mathbf{H}}$$ is a $$4 \times 4$$ matrix. If you compare equation (15) with (2), you will see that the form of the equations is the same, except that instead of a $$2 \times 2$$ matrix $$\overline{\mathbf{H}}$$ for the isotropic case, we now have a $$4 \times 4$$ matrix.

Plane wave expansion of sources in a homogeneous medium
It is often useful to expand sources in terms of plane waves so that the results of the previous lectures may be used directly on the basis of superposition. In this section we look at the expansion of point sources in terms of plane waves (for a homogeneous medium).

Let us look at the two-dimensional scalar wave equation (which can be used for acoustics, TE and TM waves, antiplane elasticity, etc.)  In the presence of a point source, the wave equation has the form
 * $$ \text{(16)} \qquad

\left[\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + k_\rho^2 \right]~\varphi(x, y)   = -\delta(x)~\delta(y) ~. $$ Assume that $$k_\rho$$ has a small positive imaginary part (it is a slightly lossy material), i.e.,

k_\rho = k'_\rho + i k''_{\rho} ~. $$ Expressed in cylindrical coordinates, equation (16) becomes (since the equation is symmetric about the origin)
 * $$ \text{(17)} \qquad

\left[\frac{\partial^2 }{\partial \rho^2} + \cfrac{1}{\rho}~\frac{\partial^2 }{\partial \rho^2} + k_\rho^2 \right]~\varphi(\rho) = -\delta(\rho) ~; \rho = \sqrt{x^2 + y^2} ~. $$ The solution of (17) is
 * $$ \text{(18)} \qquad

{ \varphi(\rho) = \cfrac{i}{4}~H^{(1)}_0 (k_\rho~\rho) } $$ where $$H^{(1)}_0$$ is a Hankel function of the first kind. \footnote{ Recall that a Hankel function of the first kind is defined as

H^{(1)}_n(z) := J_n(z) + i~Y_n(z) $$ where $$J_n(z)$$ is a Bessel function of the first kind and $$Y_n(z)$$ is a Bessel function of the second kind. }

We can also solve (16) using Fourier transforms. Let us assume that the function $$\varphi(x, y)$$ has a Fourier transform, i.e.,
 * $$ \text{(19)} \qquad

\varphi(x,y) = \cfrac{1}{2\pi}\int_{-\infty}^{\infty} \widehat{\varphi}(k_x, y)~e^{i~k_x~x}~\text{d}k_x ~. $$ Also note that
 * $$ \text{(20)} \qquad

\delta(x) = \cfrac{1}{2\pi}\int_{-\infty}^{\infty} e^{i~k_x~x}~\text{d}k_x ~. $$ Plugging (19) and (20) into (16) gives

\cfrac{1}{2\pi}\int_{-\infty}^{\infty} \left[\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + k_\rho^2 \right]~ \left[\widehat{\varphi}(k_x, y) ~e^{i~k_x~x}\right]~\text{d}k_x = -\cfrac{1}{2\pi}\int_{-\infty}^{\infty} \delta(y)~e^{i~k_x~x}~\text{d}k_x $$ or,

\int_{-\infty}^{\infty} \left[-k_x^2 + \frac{\partial^2 }{\partial y^2} + k_\rho^2 \right]~ \widehat{\varphi}(k_x, y)~e^{i~k_x~x}~\text{d}k_x = -\int_{-\infty}^{\infty} \delta(y)~e^{i~k_x~x}~\text{d}k_x ~. $$ Since the above equation holds for all values of $$x$$, the Fourier components must agree, i.e.,

\left[\frac{\partial^2 }{\partial y^2} + k_\rho^2 - k_x^2 \right]~ \widehat{\varphi}(k_x, y) = -\delta(y) ~. $$ Defining

k_y^2 := k_\rho^2 - k_x^2 $$ we get
 * $$ \text{(21)} \qquad

{ \left[\frac{\partial^2 }{\partial y^2} + k_y^2 \right]~\widehat{\varphi}(k_x, y) = -\delta(y) ~. } $$ Note that now the equation has a source only at $$y = 0$$. Away from the source (i.e., $$y \ne 0$$), the right hand side of (21) is zero, and the solution corresponds to the homogeneous part of the equation. Therefore,
 * $$ \text{(22)} \qquad

\widehat{\varphi}(k_x, y) = A_1~e^{i~k_y~y} + A_2~e^{-i~k_y~y} \qquad y \ne 0 ~. $$ This solution must be matched with the singularity at $$y = 0$$. This can be achieved by requiring that the solution have discontinuous second derivatives at $$y = 0$$. We then have ({\Red full explanation is needed here}), considering only waves that are damping away from the source rather that those growing exponentially,
 * $$ \text{(23)} \qquad

\widehat{\varphi}(k_x, y) = \cfrac{i}{2~k_y}~e^{i~k_y~|y|} ~. $$ Plugging (23) into (19) gives
 * $$ \text{(24)} \qquad

{ \varphi(x,y) = \cfrac{i}{4\pi}\int_{-\infty}^{\infty} \cfrac{1}{k_y}~ e^{i~k_x~x + i~k_y~|y|}~\text{d}k_x ~. } $$ Equation (24) is a plane wave solution for the wave equation with a point source. So the point source has been converted into a sum of propagating plane waves and some evanescent terms.

Note that the denominator in (24) contains $$k_y = \sqrt{k_\rho^2 - k_x^2}$$. Hence, when $$\pm k_\rho = k_x$$ the solution blows up. Hence there are branch points at these locations as shown in Figure 3. In a lossless medium, $$k''_\rho \rightarrow 0$$ and these points appear as pole on the real $$k_x$$ axis. The integral in equation (24) can then be computed using the residual theorem. The region between the two poles is where waves are allowed to propagate in the $$y$$ direction while the region outside the poles is where these waves are evanescent.

If we now compare the solutions (18) and (24), we have
 * $$ \text{(25)} \qquad

H^{(1)}_0 (k_\rho~\rho) = \cfrac{1}{\pi}\int_{-\infty}^{\infty} \cfrac{1}{k_y}~ e^{i~k_x~x + i~k_y~|y|}~\text{d}k_x $$ which provides a definition for the Hankel function.

Also, differentiating (16) with respect to $$y$$ and $$x$$, we get

\left[\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + k_\rho^2 \right]~ \frac{\partial \varphi(x, y)}{\partial y}   = -\delta(x)~\frac{\partial \delta(y)}{\partial y}    \qquad \text{and} \qquad \left[\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + k_\rho^2 \right]~ \frac{\partial \varphi(x, y)}{\partial x}   = -\delta(y)~\frac{\partial \delta(x)}{\partial x}  $$ Note that the products $$\delta(x)~\frac{\partial \delta(y)}{\partial y}$$ and $$\delta(y)~\frac{\partial \delta(x)}{\partial x}$$ correspond to dipole sources in the $$y$$ and $$x$$ directions, respectively.

Define

\varphi_y(x, y) := \frac{\partial \varphi(x,y)}{\partial y}  \qquad \text{and} \qquad \varphi_x(x, y) := \frac{\partial \varphi(x,y)}{\partial x} ~. $$ Therefore, from (24), we have

{ \varphi_y(x,y) = -\cfrac{\text{sgn}(y)}{4\pi}\int_{-\infty}^{\infty} e^{i~k_x~x + i~k_y~|y|}~\text{d}k_x ~. \qquad \text{and} \qquad \varphi_x(x,y) = -\cfrac{1}{4\pi}\int_{-\infty}^{\infty} \cfrac{k_x}{k_y}~ e^{i~k_x~x + i~k_y~|y|}~\text{d}k_x ~. } $$ These are the plane wave expansions of dipoles in the $$y$$ and $$x$$ directions respectively. Taking higher derivatives gives results from quadrupoles and other multipoles.


text:
Waves in composites and metamaterials/Point sources and EM vector potentials

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.

Expanding a point source in plane waves
In the previous lecture we had determined that a two-dimensional point source could be expanded into plane waves. We may think of such a point source as a line source in three dimensions.

We can similarly try to expand true three-dimensional point sources in terms of plane waves. To do that, let us start with a three-dimensional scalar wave equation of the form
 * $$ \text{(1)} \qquad

\left[\frac{\partial^2 }{\partial x^2} + \frac{\partial^2 }{\partial y^2} + \frac{\partial^2 }{\partial z^2} + k_0^2 \right]~ \varphi(x, y, z) = -\delta(x)~\delta(y)~\delta(z) ~. $$ As before, assume that $$k_0$$ has a small positive imaginary part (it is a slightly lossy material), i.e.,

k_0 = k'_0 + i k''_0 ~. $$ If we express (1) in spherical coordinates and solve the resulting differential equation, we get
 * $$ \text{(2)} \qquad

{ \varphi(r) = \cfrac{1}{4~\pi~r}~e^{i~k_0~r} } $$ where the symmetry of the equations with respect to the $$\phi$$ and $$\theta$$ directions can be observed.

Alternatively, we can try to solve (1) using Fourier transforms. To do that, let us assume that a Fourier transform of $$\varphi(x,y,z)$$ exists and the inverse Fourier transform has the form
 * $$ \text{(3)} \qquad

\varphi(x,y,z) = \cfrac{1}{8\pi^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \widehat{\varphi}(\mathbf{k})~ e^{i~\mathbf{k}\cdot\mathbf{x}}~\text{d}\mathbf{k} $$ where $$\mathbf{k} := (k_x, k_y, k_z)$$, $$\mathbf{k}\cdot\mathbf{x} := k_x~x + k_y~y + k_z~z$$, and $$\text{d}\mathbf{k} := \text{d}k_x~\text{d}k_y~\text{d}k_z$$.

Plugging (3) into (1) and using the observation that

\delta(x)~\delta(y)~\delta(z) = \cfrac{1}{8\pi^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{i\mathbf{k}\cdot\mathbf{x}}~\text{d}\mathbf{k} $$ gives (for all $$x, y, z$$ not all zero)

\cfrac{1}{8\pi^3}~\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \left[-k_x^2 - k_y^2 - k_z^2 + k_0^2 \right]~\widehat{\varphi}(\mathbf{k})~ e^{i~\mathbf{k}\cdot\mathbf{x}} ~\text{d}\mathbf{k} = - \cfrac{1}{8\pi^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} e^{i\mathbf{k}\cdot\mathbf{x}}~\text{d}\mathbf{k} $$ Since the above equation holds for all values of $$x$$, the Fourier components must agree, i.e.,

~\left[-k_x^2 - k_y^2 - k_z^2 + k_0^2 \right]~\widehat{\varphi}(\mathbf{k}) = -1 $$ Therefore,
 * $$ \text{(4)} \qquad

\widehat{\varphi}(\mathbf{k}) = - \cfrac{1}{k_0^2 - \mathbf{k}\cdot\mathbf{k}} ~. $$ Plugging (4) into (3) gives
 * $$ \text{(5)} \qquad

{ \varphi(x,y,z) = -\cfrac{1}{8\pi^3}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \cfrac{1}{k_0^2 - \mathbf{k}\cdot\mathbf{k}}~ e^{i~\mathbf{k}\cdot\mathbf{x}}~\text{d}\mathbf{k} } $$ Let us consider the integral over $$k_z$$ first. The poles are at

k_0^2 - \mathbf{k}\cdot\mathbf{k} = 0 \qquad \implies \qquad k_z = \pm\sqrt{k_0^2 - k_x^2 - k_y^2} ~. $$ Now, for $$z > 0$$ the integral is exponentially decreasing when $$\text{Im}(k_z) \rightarrow \infty$$. Therefore, the integral over $$k_z$$ can be split into the sum of an integral along the real line + an integral over an arc of a circle of radius infinity = sum of the residues at each of the poles (see Figure 1 for a sketch of the situation).

Using the Residue theorem we can show that

\varphi(x,y,z) = \cfrac{i}{8\pi^2} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \cfrac{1}{k_{zp}}~ e^{i k_x x + i k_y y + i k_{zp} z}~\text{d}k_x~\text{d}k_y $$ where $$k_{zp}$$ is the value of $$k_z$$ at the poles, i.e.,

k_{zp} := \pm\sqrt{k_0^2 - k_x^2 - k_y^2} ~. $$ When $$z < 0$$, one takes the semicircular contour $$C$$ in the lower half plane and picks up the residue at $$-k_{zp}$$. The result for all $$z$$ can therefore be written as
 * $$ \text{(6)} \qquad

{ \varphi(x,y,z) = \cfrac{i}{8\pi^2} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \cfrac{1}{k_{zp}}~ e^{i k_x x + i k_y y + i k_{zp}|z|}~\text{d}k_x~\text{d}k_y ~. } $$ The integral is over plane waves. The waves are evanescent, i.e., $$k_{zp}$$ is imaginary when $$k_x^2 + k_y^2 > k_0^2$$.

Comparing equations (6) and (2), we get the Weyl identity Weyl19 for the solution of the wave equation in spherical coordinates
 * $$ \text{(7)} \qquad

{ \cfrac{1}{r}~e^{i~k_0~r} = \cfrac{i}{2\pi} \int_{-\infty}^{\infty}\int_{-\infty}^{\infty} \cfrac{1}{k_{z}}~ e^{i k_x x + i k_y y + i k_{z}|z|}~\text{d}k_x~\text{d}k_y ~; k_z = \sqrt{k_0^2 - k_x^2 - k_y^2} ~. } $$

Electric Dipole Fields
So far we have dealt with just planar wave equations. What about the full Maxwell's equations?

From Maxwell's equation

\boldsymbol{\nabla} \times \boldsymbol{\nabla}\times{\mathbf{E}(\mathbf{r})} - k^2~\mathbf{E}(\mathbf{r}) = i\omega\mu~\mathbf{J}(\mathbf{r}) ~. $$ Using the identity

\boldsymbol{\nabla} \times \boldsymbol{\nabla}\times{\mathbf{E}} = \boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{E}) - \nabla^2 \mathbf{E} $$ we get
 * $$ \text{(8)} \qquad

\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{E}(\mathbf{r})) - \nabla^2 \mathbf{E}(\mathbf{r}) - k^2~\mathbf{E}(\mathbf{r}) = i\omega\mu~\mathbf{J}(\mathbf{r}) ~. $$ Now, for an isotropic homogeneous medium

\boldsymbol{\nabla} \cdot \mathbf{E} = \cfrac{1}{i\omega\epsilon}~\boldsymbol{\nabla} \cdot \mathbf{J} ~. $$ Plugging this into (8) we get
 * $$ \text{(9)} \qquad

\cfrac{1}{i\omega\epsilon}~\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{J}(\mathbf{r})) - \nabla^2 \mathbf{E}(\mathbf{r}) - k^2~\mathbf{E}(\mathbf{r}) = i\omega\mu~\mathbf{J}(\mathbf{r}) ~. $$ Recall that

k^2 = \omega^2 \mu \epsilon ~. $$ Plugging this into (9) gives

-\cfrac{i\omega\mu}{k^2}~\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{J}(\mathbf{r})) - \nabla^2 \mathbf{E}(\mathbf{r}) - k^2~\mathbf{E}(\mathbf{r}) = i\omega\mu~\mathbf{J}(\mathbf{r}) $$ or,
 * $$ \text{(10)} \qquad

{ \left[\nabla^2  + k^2\right]~\mathbf{E}(\mathbf{r}) = - i\omega\mu~\left[\cfrac{1}{k^2}~\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{J}(\mathbf{r}))+\mathbf{J}(\mathbf{r})\right] ~.  } $$ This equation has the form of the scalar wave equation
 * $$ \text{(11)} \qquad

\left[\nabla^2 + k^2\right]~\varphi(\mathbf{r}) = s(\mathbf{r}) ~. $$ The only difference is that (10) consists of three scalar wave equations and the source term is given by

\mathbf{s}(\mathbf{r}) := - i\omega\mu~\left[\cfrac{1}{k^2}~\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{J}(\mathbf{r}))+\mathbf{J}(\mathbf{r})\right] ~. $$ Recall that, using the Green's function method, we can find the solution of the scalar wave equation (11) (see Chew95 p.24-28 for details) as

\varphi(\mathbf{r}) = - \cfrac{1}{4\pi} \int \cfrac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|} s(\mathbf{r}')~\text{d}r' ~. $$ In an analogous manner we can find the solution of (10), and we get
 * $$ \text{(12)} \qquad

{ \mathbf{E}(\mathbf{r}) = \cfrac{i\omega\mu}{4\pi} \int \cfrac{e^{ik|\mathbf{r}-\mathbf{r}'|}}{|\mathbf{r}-\mathbf{r}'|} ~\left[\cfrac{1}{k^2}~\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \mathbf{J}(\mathbf{r}'))+\mathbf{J}(\mathbf{r}')\right]~\text{d}\mathbf{r}' ~. } $$

For electric dipole fields, if one has a point current source directed in the $$\widehat{\boldsymbol{\alpha}}$$ direction, then the current density is given by

\mathbf{J}(\mathbf{r}, \mathbf{r}') = \widehat{\boldsymbol{\alpha}}~I~l~\delta(\mathbf{r} - \mathbf{r}') $$ where $$Il$$ is the current dipole moment, i.e., as $$l \rightarrow 0$$ and $$I \rightarrow \infty$$, $$Il$$ remains constant. If the origin is taken at the point $$\mathbf{r}'$$, we get
 * $$ \text{(13)} \qquad

\mathbf{J}(\mathbf{r}) = \widehat{\boldsymbol{\alpha}}~I~l~\delta(\mathbf{r}) ~;\qquad \delta(\mathbf{r}) := \delta(x)~\delta(y)~\delta(z) ~. $$

Plugging (13) into (12) gives

\mathbf{E}(\mathbf{r}) = \cfrac{i\omega\mu}{4\pi} \int \cfrac{e^{ikr}}{r} ~\left[{I~l}~\left\{\cfrac{1}{k^2}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \widehat{\boldsymbol{\alpha}}~\delta(\mathbf{r}'))+ \widehat{\boldsymbol{\alpha}}~\delta(\mathbf{r}')\right\} \right]~\text{d}\mathbf{r}' $$ or,
 * $$ \text{(14)} \qquad

{ \mathbf{E}(\mathbf{r}) = i\omega\mu~I~l~ \left[\cfrac{1}{k^2}\boldsymbol{\nabla} (\boldsymbol{\nabla} \cdot \widehat{\boldsymbol{\alpha}})+ \widehat{\boldsymbol{\alpha}}\right] \cfrac{e^{ikr}}{4\pi r}~. } $$ Also, from

\boldsymbol{\nabla} \times \mathbf{E} = i\omega\mu~\mathbf{H} $$ and using the identity $$\boldsymbol{\nabla} \times \boldsymbol{\nabla\mathbf{u}} = 0$$, the magnetic field is given by

{ \mathbf{H}(r) = \cfrac{I~l}{4\pi}~ \boldsymbol{\nabla} \times \left(\widehat{\boldsymbol{\alpha}}~\cfrac{e^{ikr}}{r}\right) ~. } $$ Substituting the Weyl identity (7) into these expression gives formulae for $$\mathbf{E}$$ and $$\mathbf{H}$$ in terms of plane waves.

Scattering of radiation from a sphere
Recall the Airy solution for the scattering of light by a raindrop. In the following we sketch the Mie solution which generalizes the analysis to the scattering of electromagnetic radiation by a spherical object. The problem remains similar, i.e., we wish to determine the scattering of a plane wave incident on a sphere of refractive index $$n$$. However, we now consider the case where the wavelength of the incident radiation is not necessarily much smaller than the size of the sphere.

Consider the sphere shown in Figure 2. We set up our coordinate system such that the origin is at the center of the sphere. The sphere has a magnetic permeability of $$\mu$$ and a permittivity $$\epsilon$$. The medium outside the sphere has a permittivity $$\epsilon_0$$ and a permeability $$\mu_0$$. The electric field is oriented parallel to the $$x_1$$ axis and the $$x_2$$ axis points out of the plane of the paper.

Let us now consider the situation where the material inside the sphere is non-magnetic. Then we may write

\mu = \mu_0 ~; \epsilon = \epsilon_r ~\epsilon_0 = n^2 ~\epsilon_0 $$ where $$\epsilon_r$$ is the relative permittivity of the material inside the sphere.

Also, the incident plane wave is given by

\mathbf{E} = e^{ikx_3}~\mathbf{e}_1 $$ where $$\mathbf{e}_1$$ is the unit vector in the $$x_1$$ direction.

The solution of this problem was first given by Mie Mie08. A detailed derivation is given in Kerker69. We follow the abbreviated version in Ishimaru78.

Before we can go into the details, we need to discuss vector potentials for electromagnetism.

Vector potentials for electromagnetism
Since $$\boldsymbol{\nabla} \cdot \mathbf{B} = 0$$, there exists a vector potential $$\mathbf{A}$$ such that $$\mathbf{B} = \boldsymbol{\nabla} \times \mathbf{A}$$. Hence,
 * $$ \text{(15)} \qquad

{ \mathbf{H} = \cfrac{1}{\mu}~\boldsymbol{\nabla} \times \mathbf{A} ~. } $$ Also, from Maxwell's equation

\boldsymbol{\nabla} \times \mathbf{E} + \frac{\partial \mathbf{B}}{\partial t} = 0 ~. $$ In terms of the vector potential $$\mathbf{A}$$, we then have

\boldsymbol{\nabla} \times \left(\mathbf{E} + \cfrac{\partial \mathbf{A}}{\partial t}\right) = 0 ~. $$ Therefore, there exists a scalar potential $$\phi$$ such that

\mathbf{E} + \frac{\partial \mathbf{A}}{\partial t} = -\boldsymbol{\nabla} \phi $$ i.e.,
 * $$ \text{(16)} \qquad

{ \mathbf{E} =  -\frac{\partial \mathbf{A}}{\partial t} - \boldsymbol{\nabla} \phi ~. } $$ At this stage there is some flexibility in the choice of $$\mathbf{A}$$ and $$\phi$$. A restriction that is useful is to require the potentials to satisfy the  Lorenz condition Lorenz67 (which is equivalent to requiring that the charge be conserved)

\boldsymbol{\nabla} \cdot \mathbf{A} + \epsilon~\mu~\frac{\partial \phi}{\partial t} = 0 ~. $$ Then, in the absence of free charges and currents in an isotropic homogeneous medium, both potentials satisfy the wave equation, i.e.,

\nabla^2 \phi - \epsilon~\mu~\frac{\partial^2 \phi}{\partial t^2} = 0 ~; \nabla^2 \mathbf{A} - \epsilon~\mu~\frac{\partial^2 \mathbf{A}}{\partial t^2} = \boldsymbol{0} ~. $$ Even after these restriction the potentials are not uniquely defined and one is free to make the gauge transformations

\phi' = \phi + \frac{\partial f}{\partial t} ~; \mathbf{A}' = \mathbf{A} - \boldsymbol{\nabla} f $$ to obtain new potentials $$\phi'$$, $$\mathbf{A}'$$ provide $$f$$ satisfies the wave equation

\nabla^2 f - \epsilon~\mu~\frac{\partial^2 f}{\partial t^2} = 0 ~. $$ The preceding potentials are well known. However, one can go one step further and define  superpotentials (see, for example, Bowman69).

The most widely used  superpotentials are the electric and magnetic  Hertz vector potentials $$\boldsymbol{\Pi}_e$$ and $$\boldsymbol{\Pi}_m$$ (also known as polarization potentials).

The terms of these potentials, the $$\mathbf{E}$$ and $$\mathbf{H}$$ can be expressed as
 * $$ \text{(17)} \qquad

{ \begin{align} \mathbf{E} & = \boldsymbol{\nabla} \times \boldsymbol{\nabla}\times{\boldsymbol{\Pi}_e} - \mu~\boldsymbol{\nabla} \times \cfrac{\partial \boldsymbol{\Pi}_m}{\partial t} \\ \mathbf{H} & = \boldsymbol{\nabla} \times \boldsymbol{\nabla}\times{\boldsymbol{\Pi}_m} + \epsilon~\boldsymbol{\nabla} \times \cfrac{\partial \boldsymbol{\Pi}_e}{\partial t} ~. \end{align} } $$ Comparing equations (17) with (16) and (15) one sees that the superpotentials lead to symmetric representations of $$\mathbf{E}$$ and $$\mathbf{H}$$ unlike when standard vector and scalar potentials are used.

Of course, the superpotentials $$\boldsymbol{\Pi}_e$$ and $$\boldsymbol{\Pi}_m$$ are not uniquely defined and one is free to make gauge transformations

\begin{align} \boldsymbol{\Pi}'_e & = \boldsymbol{\Pi}_e + \boldsymbol{\nabla} g_e(\mathbf{x}, t) \\ \boldsymbol{\Pi}'_m & = \boldsymbol{\Pi}_m + \boldsymbol{\nabla} g_m(\mathbf{x}, t)  \end{align} $$ where $$g_e(\mathbf{x}, t)$$ and $$g_m(\mathbf{x}, t)$$ are arbitrary scalar potential functions.

Plugging these definitions into the Maxwell's equation lead to the equations being satisfied if
 * $$ \text{(18)} \qquad

{ \begin{align} \boldsymbol{\nabla} \times \boldsymbol{\nabla}\times{\boldsymbol{\Pi}_e} + \epsilon~\mu~\frac{\partial^2 \boldsymbol{\Pi}_e}{\partial t^2} & = \boldsymbol{\nabla} f \\ \boldsymbol{\nabla} \times \boldsymbol{\nabla}\times{\boldsymbol{\Pi}_m} + \epsilon~\mu~\frac{\partial^2 \boldsymbol{\Pi}_m}{\partial t^2} & = \boldsymbol{\nabla} f \end{align} } $$ where $$f$$ is an arbitrary scalar potential which is a function of position and time.

The Lorentz condition is satisfied if

f = \boldsymbol{\nabla} \cdot \boldsymbol{\Pi}_e ~. $$

In fact, the potentials $$\mathbf{A}$$ and $$\phi$$ can be expressed in terms of $$\boldsymbol{\Pi}_e$$ and $$\boldsymbol{\Pi}_m$$ as

\begin{align} \phi & = - \boldsymbol{\nabla} \cdot \boldsymbol{\Pi}_e \\ \mathbf{A} & = \epsilon~\mu~\frac{\partial \boldsymbol{\Pi}_e}{\partial t} + \mu~\boldsymbol{\nabla} \times \boldsymbol{\Pi}_m ~. \end{align} $$

The time harmonic case
For time harmonic problems, an important class of Hertz vector potentials are those of the form (for spherical symmetry)

\boldsymbol{\Pi}_e = u~\mathbf{r} ~; \boldsymbol{\Pi}_m = v~\mathbf{r} \qquad \text{where} \quad \mathbf{r} \equiv (x_1, x_2, x_3) ~. $$ The vector $$\mathbf{r}$$ is the radial vector from the origin in a spherical coordinate system. The functions $$u$$ and $$v$$ are scalar potentials (called  Debye potentials) which satisfy the homogeneous wave equations

(\nabla^2 + k^2) u = 0 \qquad \text{and} \qquad (\nabla^2  + k^2) v = 0  ~. $$

One important result is that every electromagnetic field defined in a source-free region between two concentric spheres can be represented there by two Debye potentials Wilcox57.

In spherical coordinates, the components of the fields between two concentric spheres are given by

{ \begin{align} E_r & = \left(\frac{\partial^2 }{\partial r^2} + k^2\right)(r~u) \\ E_\theta & = \cfrac{1}{r}~\frac{\partial^2 }{\partial r \partial \theta}(r~u) + \cfrac{i k \sqrt{\mu/\epsilon}}{\sin\theta} \frac{\partial v}{\partial \phi} \\ E_\phi & = \cfrac{1}{r~\sin\theta}~\frac{\partial^2 }{\partial r \partial \phi}(r~u) - i k \sqrt{\mu/\epsilon} \frac{\partial v}{\partial \theta} \end{align} } $$ and

{ \begin{align} H_r & = \left(\frac{\partial^2 }{\partial r^2} + k^2\right)(r~v) \\ H_\theta & = \cfrac{1}{r}~\frac{\partial^2 }{\partial r \partial \theta}(r~v) - \cfrac{i k \sqrt{\mu/\epsilon}}{\sin\theta} \frac{\partial u}{\partial \phi} \\ H_\phi & = \cfrac{1}{r~\sin\theta}~\frac{\partial^2 }{\partial r \partial \phi}(r~v) + i k \sqrt{\mu/\epsilon} \frac{\partial u}{\partial \theta} ~. \end{align} } $$


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

Earth and other planets orbit around the Sun

This is the solar system. This means everything flying round the Sun. The Sun is that flaming ball in the center. Other balls are planets flying around it on their ellipses. The first four planets are near to the Sun. We call them Mercury, Venus, Earth and Mars. The small ball near to the Earth is Earth’s Moon. Further from the sun, there is a belt of asteroids. Behind it you can see other planets, such as Jupiter, Saturn, Uranus and Neptune. There are also other objects such as comets and a minor planet called Pluto. By clicking on them you can get more information.


text:
Talk:Digital Libraries

"Learning Project" is a rather general name. Any ideas for a more specific page name and maybe a category? --JWSchmidt 19:56, 4 October 2007 (UTC)
 * Where is the rest of the specific curriculum from the class documents? –SJ + > 07:17, 7 April 2010 (UTC)

title: Talk:Digital Libraries


text:
Waves in composites and metamaterials/Mie theory and Bloch theorem

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.

Scattering of radiation from a sphere
Recall the sphere shown in Figure 1. We set up our coordinate system such that the origin is at the center of the sphere. The sphere has a magnetic permeability of $$\mu$$ and a permittivity $$\epsilon$$. The medium outside the sphere has a permittivity $$\epsilon_0$$ and a permeability $$\mu_0$$. The electric field is oriented parallel to the $$x_1$$ axis and the $$x_2$$ axis points out of the plane of the paper.

Also recall that

\mu = \mu_0 ~; \epsilon = \epsilon_r ~\epsilon_0 = n^2 ~\epsilon_0 $$ where $$\epsilon_r$$ is the relative permittivity of the material inside the sphere and that the incident plane wave is given by

\mathbf{E}_i = e^{ikx_3}~\mathbf{e}_1 $$ where $$\mathbf{e}_1$$ is the unit vector in the $$x_1$$ direction.

The most widely used  superpotentials are the electric and magnetic  Hertz vector potentials $$\boldsymbol{\Pi}_e$$ and $$\boldsymbol{\Pi}_m$$ (also known as polarization potentials).

In the last lecture we discussed the Hertz vector potentials and that the $$\mathbf{E}$$ and $$\mathbf{H}$$ fields can be expressed as
 * $$ \text{(1)} \qquad

\begin{align} \mathbf{E} & = \boldsymbol{\nabla} \times \boldsymbol{\nabla}\times{\boldsymbol{\Pi}_e} - \mu~\boldsymbol{\nabla} \times \frac{\partial \boldsymbol{\Pi}_m}{\partial t} \\ \mathbf{H} & = \boldsymbol{\nabla} \times \boldsymbol{\nabla}\times{\boldsymbol{\Pi}_m} + \epsilon~\boldsymbol{\nabla} \times \frac{\partial \boldsymbol{\Pi}_e}{\partial t} ~. \end{align} $$

For spherically symmetric time harmonic problems, such as we find in the problem of scattering of EM waves by a sphere, we stated that an important class of Hertz vector potentials are the Debye potentials of the form

\boldsymbol{\Pi}_e = u~\mathbf{r} ~; \boldsymbol{\Pi}_m = v~\mathbf{r} \qquad \text{where} \quad \mathbf{r} \equiv (x_1, x_2, x_3) ~. $$

Let the time harmonic fields be given by

\mathbf{E} = \widehat{\mathbf{E}}~e^{-i\omega t} ~; \mathbf{H} = \widehat{\mathbf{H}}~e^{-i\omega t} ~; u = \hat{u}~e^{-i\omega t} ~; v = \hat{v}~e^{-i\omega t} ~. $$ Plugging these into (1) and dropping the hats gives the Maxwell equations at fixed frequency:

\begin{align} \mathbf{E} & = \boldsymbol{\nabla} \times \boldsymbol{\nabla}\times{(u~\mathbf{r})} + i\omega\mu~\boldsymbol{\nabla} \times (v~\mathbf{r}) \\ \mathbf{H} & = \boldsymbol{\nabla} \times \boldsymbol{\nabla}\times{(v~\mathbf{r})} - i\omega\epsilon~\boldsymbol{\nabla} \times (u~\mathbf{r}) ~. \end{align} $$

Recall that the Debye potentials satisfy the homogeneous wave equations
 * $$ \text{(2)} \qquad

(\nabla^2 + k^2) u = 0 \qquad \text{and} \qquad (\nabla^2  + k^2) v = 0  ~. $$

To deal with the problem of scattering by a sphere, let us split the potentials $$u$$ and $$v$$ (outside the sphere) into incident and scattered fields:

u = u_i + u_s ~; v = v_i + v_s $$ where the subscript $$i$$ indicates an incident field and the subscript $$s$$ indicates a scattered field.

Inside the sphere, the potentials are denoted by

u = u_r ~; v = v_r $$ where the subscript $$r$$ indicates a refracted + reflected field.

Let us require that these potentials satisfy wave equations of the form given in (2), i.e.,

\begin{align} (\nabla^2 + k^2) u_i &= 0 & \qquad \text{and} \qquad & (\nabla^2  + k^2) v_i = 0\\ (\nabla^2 + k^2) u_s &= 0 & \qquad \text{and} \qquad & (\nabla^2  + k^2) v_s = 0\\ (\nabla^2 + k^2 n^2) u_r &= 0 & \qquad \text{and} \qquad & (\nabla^2 + k^2 n^2) v_r = 0 ~. \end{align} $$ Since each of these satisfies a scalar wave equation, we can express each potential in terms of spherical harmonics.

In particular, the Debye potentials associated with the incident field

\mathbf{E}_i = e^{ikx_3}~\mathbf{e}_1 $$ have the expression

{ \begin{align} r u_i & = \cfrac{1}{k^2} \sum_{l=1}^{\infty} \cfrac{i^{l-1} (2l+1)}{l(l+1)}~\psi_l(k r)~P_l^1(\cos\theta)~\cos\phi \\ r v_i & = \cfrac{1}{\eta k^2} \sum_{l=1}^{\infty} \cfrac{i^{l-1} (2l+1)}{l(l+1)}~\psi_l(k r)~P_l^1(\cos\theta)~\sin\phi \end{align} } $$ where

\psi_l(\rho) = \sqrt{\cfrac{\pi~r}{2}}~J_{l+1/2}(\rho) ~; \eta = \sqrt{\cfrac{\mu_0}{\epsilon_0}} ~. $$ Here $$P_l^1(x)$$ are the Legendre polynomials which solve

\cfrac{d }{d x}\left[ (1- x^2)~\cfrac{d P^1_l}{d x}\right] + \left[l(l+1) - \cfrac{1}{1 - x^2}\right]~P^1_l = 0 $$ and $$J_\nu (\rho)$$ are the Bessel functions which solve

\cfrac{d^2 J_\nu}{d \rho^2} + \cfrac{1}{\rho}~\cfrac{d J_\nu}{d \rho} + \left[1 - \cfrac{\nu^2}{\rho^2}\right]~J_\nu = 0 ~. $$ The functions $$\psi(\rho)$$ are chosen such that

\psi_\nu(\rho) = \sqrt{\cfrac{\pi~r}{2}}~J_\nu(\rho) $$ is regular at the origin.

The scattered fields have a similar expansion

{ \begin{align} r u_s & = \cfrac{-1}{k^2} \sum_{l=1}^{\infty} \cfrac{i^{l-1} (2l+1)}{l(l+1)}~a_l~\zeta_l(k r)~P_l^1(\cos\theta)~ \cos\phi \\ r v_s & = \cfrac{-1}{\eta k^2} \sum_{l=1}^{\infty} \cfrac{i^{l-1} (2l+1)}{l(l+1)}~b_l~\zeta_l(k r)~P_l^1(\cos\theta)~ \sin\phi \end{align} } $$ where

\zeta_l(\rho) = \sqrt{\cfrac{\pi~r}{2}}~H^{(1)}_{l+1/2}(\rho) $$ and $$H^{(1)}_\nu(\rho)$$ is one of the Hankel functions solving the same equation as the Bessel function but decaying at infinity.

Inside the sphere, the expansion of the fields takes the form

{ \begin{align} r u_r & = \cfrac{1}{k^2 n^2} \sum_{l=1}^{\infty} \cfrac{i^{l-1} (2l+1)}{l(l+1)}~c_l~\psi_l(k n r)~P_l^1(\cos\theta)~ \cos\phi \\ r v_r & = \cfrac{1}{\eta k^2 n^2} \sum_{l=1}^{\infty} \cfrac{i^{l-1} (2l+1)}{l(l+1)}~b_l~\zeta_l(k n r)~P_l^1(\cos\theta)~ \sin\phi \end{align} } $$

To find the constants $$a_l, b_l, c_l, d_l$$ we need to apply continuity conditions across the boundary of the sphere.

To ensure that $$E_\theta, E_\phi, H_\theta, H_\phi$$ (tangential components of $$\mathbf{E}$$ and $$\mathbf{H}$$) are continuous across the surface of the sphere at $$r = a$$, it is sufficient that

n^2 u ~, \frac{\partial }{\partial r}(r u) ~; \frac{\partial }{\partial r}(r v) $$ are continuous.

Applying these conditions, we get

{ \begin{align} a_l & = \cfrac{\psi_l(\alpha)~\psi'_l(\beta) - n~\psi_l(\beta)~\psi'_l(\alpha)} {\zeta_l(\alpha)~\psi'_l(\beta) - n~\psi_l(\beta)~\zeta'_l(\alpha)} \\ b_l & = \cfrac{n~\psi_l(\alpha)~\psi'_l(\beta) - \psi_l(\beta)~\psi'_l(\alpha)} {n~\zeta_l(\alpha)~\psi'_l(\beta) - \psi_l(\beta)~\zeta'_l(\alpha)} \end{align} } $$ where

\alpha = k~a ~; \beta = k~n~a ~. $$ The scattered field $$E_\theta$$, $$E_\phi$$ far from the sphere are given by

\begin{align} E_\theta & = \cfrac{i~e^{i k r}}{k r}~S_1(\theta)~\cos\phi \\ E_\phi & = -\cfrac{i~e^{i k r}}{k r}~S_1(\theta)~\sin\phi \end{align} $$ where

\begin{align} S_1(\theta) & = \sum_{l=1}^\infty \cfrac{2l+1}{l(l+1)} \left[a_l~\pi_l(\cos\theta) + b_l~\tau_l(\cos\theta)\right] \\ S_2(\theta) & = \sum_{l=1}^\infty \cfrac{2l+1}{l(l+1)} \left[a_l~\tau_l(\cos\theta) + b_l~\pi_l(\cos\theta)\right] \end{align} $$ where

\pi_l(\cos\theta) = \cfrac{P^1_l(\cos\theta)}{\sin\theta} ~; \tau_l(\cos\theta) = \cfrac{d }{d \theta}~P^1_l(\cos\theta) ~. $$ Note that the tangential components of $$\mathbf{E}$$ fall off as $$1/r$$ while the radial component falls off as $$1/r^2$$.

Periodic Media and Bloch's Theorem
The following discussion is based on Ashcroft76 (p. 133-139). For a more detailed mathematical treatment see Kuchment93.

Suppose that the medium is such that the permittivity $$\epsilon(\mathbf{x})$$ and the permeability $$\mu(\mathbf{x})$$ are periodic. Recall that, at fixed frequency, the Maxwell equations are
 * $$ \text{(3)} \qquad

\boldsymbol{\nabla} \cdot (\epsilon~\mathbf{E}) = 0 ~; \boldsymbol{\nabla} \cdot (\mu~\mathbf{H}) = 0 ~; \boldsymbol{\nabla} \times \mathbf{E} + i\omega\mu~\mathbf{H} = 0 ~; \boldsymbol{\nabla} \times \mathbf{H} - i\omega\epsilon~\mathbf{E} = 0 ~. $$ Also recall the constitutive relations
 * $$ \text{(4)} \qquad

\mathbf{D} = \epsilon~\mathbf{E} ~; \mathbf{B} = \mu~\mathbf{H} ~. $$ Plugging (4) into (3), we get
 * $$ \text{(5)} \qquad

\boldsymbol{\nabla} \cdot \mathbf{D} = 0 ~; \boldsymbol{\nabla} \cdot \mathbf{B} = 0 ~; i~\boldsymbol{\nabla} \times \left(\cfrac{\mathbf{D}}{\epsilon}\right) = \omega~\mathbf{B} ~; -i~\boldsymbol{\nabla} \times \left(\cfrac{\mathbf{B}}{\mu}\right) = \omega~\mathbf{D} ~. $$ Equations (5) suggest that we should look for solutions $$\mathbf{D}$$ and $$\mathbf{B}$$ in the space of divergence-free fields such that
 * $$ \text{(6)} \qquad

\mathcal{L}\begin{bmatrix} \mathbf{D} \\ \mathbf{B} \end{bmatrix} = \omega~\begin{bmatrix} \mathbf{D} \\ \mathbf{B} \end{bmatrix} $$ where the operator $$\mathcal{L}$$ is given by
 * $$ \text{(7)} \qquad

\mathcal{L} := \begin{bmatrix} 0 & -i~\boldsymbol{\nabla} \times \mu^{-1} \\ i~\boldsymbol{\nabla} \times \epsilon^{-1} & 0 \end{bmatrix} ~. $$ Since $$\epsilon$$ and $$\mu$$ are periodic, the operator $$\mathcal{L}$$ has the same periodicity as the medium.

Clearly, equation (6) represents an eigenvalue problem where $$\omega$$ is an eigenvalue of $$\mathcal{L}$$ and $$[\mathbf{D},\mathbf{B}]$$ is the corresponding eigenvector.

Let $$\mathcal{T}_R$$ define a translation operator which, when acting upon a pair of the fields $$[\mathbf{D}, \mathbf{B}]$$ shifts the argument by a vector $$\boldsymbol{R}v$$, where $$\boldsymbol{R}v$$ is taken to be a lattice vector (see Figure~2), i.e.,

\mathcal{T}_R\begin{bmatrix} \mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x}) \end{bmatrix} = \begin{bmatrix} \mathbf{D}(\mathbf{x}+\boldsymbol{R}v) \\ \mathbf{B}(\mathbf{x}+\boldsymbol{R}v) \end{bmatrix} ~. $$

Periodicity of the medium implies that $$\mathcal{T}_R$$ commutes with $$\mathcal{L}$$, i.e.,

\mathcal{T}_R~\mathcal{L} = \mathcal{L}~\mathcal{T}_R ~. $$

Note that $$\mathcal{T}_R$$, like $$\mathcal{L}$$, maps divergence-free fields to divergence-free fields.

Now, consider the space of field pairs $$\mathbf{D}, \mathbf{B}$$ which are divergence-free and which are in the null space of $$\mathcal{L} - \omega~\boldsymbol{1}$$, i.e., they satisfy

(\mathcal{L} - \omega~\boldsymbol{1})\begin{bmatrix}\mathbf{D} \\ \mathbf{B} \end{bmatrix} = \boldsymbol{0} ~. $$               This subspace is closed under the action of $$\mathcal{T}_R$$ which is unitary, i.e.,

\mathcal{T}_R~\mathcal{T}_R^T = \mathcal{T}_R~\mathcal{T}_{-R} = \boldsymbol{1} ~. $$ Also, the translation operator commutes, i.e.,

\mathcal{T}_R~\mathcal{T}_{R'} = \mathcal{T}_{R'}~\mathcal{T}_R = \mathcal{T}_{R+R'} ~. $$ Therefore, any solution can be expressed in fields which are simultaneously eigenstates of all the $$\mathcal{T}_R$$. These eigenstates have the property

\mathcal{T}_R\begin{bmatrix} \mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x}) \end{bmatrix} = c(\boldsymbol{R}v)~\begin{bmatrix} \mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x}) \end{bmatrix} ~. $$ The Bloch condition will be discussed in the next lecture.


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

The Sun - everything orbits around the Sun in our solar system  (back to our Solar System)←The Sun →(Mercury)

The Sun is a flaming ball, a star situated in the center of Our Solar System. Planets, such as The Earth, Mars, Venus or Jupiter are flying around it on their orbits. There are also orbits of asteroids, comets and dust. The Sun consists of hydrogen, helium and other elements. Most of them are gases. In the center of the Sun, hydrogen is slowly changed into the helium. During this proces a lot of light energy is emited. This light energy makes day on the Earth and also enable the life on the Earth. Clicking on the Sun, you can get more information about this star.


text:
User talk:Jorunn

m:User talk:Jorunn

title: User talk:Jorunn


text:
User:Jorunn

m:User:Jorunn

de:Benutzer:Jorunn el:Χρήστης:Jorunn es:Usuario:Jorunn fr:Utilisateur:Jorunn it:Utente:Jorunn

title: User:Jorunn


text:
Waves in composites and metamaterials/Bloch waves and the quasistatic limit

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.

Bloch Theorem
In the previous lecture we showed that Maxwell's equations at fixed frequency can be formulated in terms of the fields $$\mathbf{D}$$ and $$\mathbf{B}$$ as
 * $$ \text{(1)} \qquad

\boldsymbol{\nabla} \cdot \mathbf{D} = 0 ~; \boldsymbol{\nabla} \cdot \mathbf{B} = 0 ~; i~\boldsymbol{\nabla} \times \left(\cfrac{\mathbf{D}}{\epsilon}\right) = \omega~\mathbf{B} ~; -i~\boldsymbol{\nabla} \times \left(\cfrac{\mathbf{B}}{\mu}\right) = \omega~\mathbf{D} ~. $$ Equations (1) suggest that we should look for solutions $$\mathbf{D}$$ and $$\mathbf{B}$$ in the space of divergence-free fields such that
 * $$ \text{(2)} \qquad

\mathcal{L}\begin{bmatrix} \mathbf{D} \\ \mathbf{B} \end{bmatrix} = \omega~\begin{bmatrix} \mathbf{D} \\ \mathbf{B} \end{bmatrix} \qquad \text{or} \qquad {  (\mathcal{L} - \omega~\boldsymbol{1})\begin{bmatrix}\mathbf{D} \\ \mathbf{B} \end{bmatrix} = \boldsymbol{0} } $$ where the operator $$\mathcal{L}$$ is given by
 * $$ \text{(3)} \qquad

{ \mathcal{L} := \begin{bmatrix} 0 & -i~\boldsymbol{\nabla} \times \mu^{-1} \\ i~\boldsymbol{\nabla} \times \epsilon^{-1} & 0 \end{bmatrix} ~. } $$

If the medium is such that the permittivity $$\epsilon(\mathbf{x})$$ and the permeability $$\mu(\mathbf{x})$$ are periodic, i.e.,

\epsilon(\mathbf{x}) = \epsilon(\mathbf{x} + \mathbf{R}) ~; \mu(\mathbf{x}) = \mu(\mathbf{x} + \mathbf{R}) $$ where $$\mathbf{R}$$ is a lattice vector (see Figure 1) then the operator $$\mathcal{L}$$ has the same periodicity as the medium.

Also recall the translation operator $$\mathcal{T}_R$$ defined as

{ \mathcal{T}_R\begin{bmatrix} \mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x}) \end{bmatrix} = \begin{bmatrix} \mathbf{D}(\mathbf{x}+\mathbf{R}) \\ \mathbf{B}(\mathbf{x}+\mathbf{R}) \end{bmatrix} ~. } $$ Periodicity of the medium implies that $$\mathcal{T}_R$$ commutes with $$\mathcal{L}$$, i.e.,

{ \mathcal{T}_R~\mathcal{L} = \mathcal{L}~\mathcal{T}_R ~. } $$ The translation operator is unitary, i.e.,

{ \mathcal{T}_R~\mathcal{T}_R^T = \mathcal{T}_R~\mathcal{T}_{-R} = \boldsymbol{1} ~. } $$ This means that the adjoint operator $$\mathcal{T}_R^T$$ is equal to the inverse operator $$\mathcal{T}_{-R}$$.

The translation operator also commutes, i.e.,

{ \mathcal{T}_R~\mathcal{T}_{R'} = \mathcal{T}_{R'}~\mathcal{T}_R = \mathcal{T}_{R+R'} ~. } $$ Also, since $$\mathcal{L}$$ and $$\mathcal{T}_R$$ commute, the operators $$\mathcal{L}-\omega\boldsymbol{1}$$ and $$\mathcal{T}_R$$ must also commute. This implies that

\mathcal{T}_R~(\mathcal{L} - \omega~\boldsymbol{1}) ~\begin{bmatrix}\mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x})\end{bmatrix} = \boldsymbol{0} = (\mathcal{L} - \omega~\boldsymbol{1})~\mathcal{T}_R ~\begin{bmatrix}\mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x})\end{bmatrix} = (\mathcal{L} - \omega~\boldsymbol{1})~ ~\begin{bmatrix}\mathbf{D}(\mathbf{x}+\mathbf{R}) \\ \mathbf{B}(\mathbf{x}+\mathbf{R})\end{bmatrix} ~. $$ Hence the eigenstates of $$\mathcal{L} - \omega~\boldsymbol{1}$$ and the eigenstates of $$\mathcal{T}_R$$ lie in the same space. Therefore, any solution can be expressed in fields which are simultaneously eigenstates of all the $$\mathcal{T}_R$$, i.e., these eigenstates have the property
 * $$ \text{(4)} \qquad

{ [\mathcal{T}_R - c(\mathbf{R})~\boldsymbol{1}]\begin{bmatrix} \mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x}) \end{bmatrix} = \boldsymbol{0} ~. } $$ Since $$\mathcal{T}_R~\mathcal{T}_{R'} = \mathcal{T}_{R+R'}$$, we have

c(\mathbf{R})~c(\mathbf{R}') = c(\mathbf{R} + \mathbf{R}') ~. $$ So it suffices to know $$c(\mathbf{R})$$ when $$\mathbf{R} = \mathbf{a}_1, \mathbf{a}_2, \mathbf{a}_3$$ where the $$\mathbf{a}_i$$'s are the primitive vectors of the lattice, i.e.,

\epsilon(\mathbf{x} + \mathbf{a}_j) = \epsilon(\mathbf{x}) ~; \mu(\mathbf{x} + \mathbf{a}_j) = \mu(\mathbf{x}) ~. $$ Let us assume that

c(\mathbf{a}_j) = e^{2\pi~i~\alpha_j} \qquad j = 1, 2, 3 $$ for a suitable choice of $$\alpha_j$$.

Then for any lattice vector

\mathbf{R} = n_1~\mathbf{a}_1 + n_2~\mathbf{a}_2 + n_3~\mathbf{a}_3 $$ we have

\begin{align} c(\mathbf{R}) & = c(n_1~\mathbf{a}_1 + n_2~\mathbf{a}_2 + n_3~\mathbf{a}_3) = c(n_1~\mathbf{a}_1)~c(n_2~\mathbf{a}_2)~c(n_3~\mathbf{a}_3) \\ & = c\left(\sum_{i=1}^{n_1}\mathbf{a}_1\right)~ c\left(\sum_{i=1}^{n_2}\mathbf{a}_2\right)~c\left(\sum_{i=1}^{n_3}\mathbf{a}_3\right) = [c(\mathbf{a}_1)]^{n_1}~[c(\mathbf{a}_2)]^{n_2}~[c(\mathbf{a}_3)]^{n_3} \\ & = e^{2\pi i \alpha_1~n_1}~e^{2\pi i \alpha_2~n_2}~e^{2\pi i \alpha_3~n_3} = e^{2\pi i (\alpha_1~n_1 + \alpha_2~n_2 + \alpha_3~n_3)} \end{align} $$ or,

c(\mathbf{R}) = e^{2\pi i (\alpha_1~n_1 + \alpha_2~n_2 + \alpha_3~n_3)} ~. $$ Define a vector

\mathbf{k} := \alpha_1~\mathbf{b}_1 + \alpha_2~\mathbf{b}_2 + \alpha_3~\mathbf{b}_3 $$ where the vectors $$\mathbf{b}_i$$ are the reciprocal lattice vectors satisfying

\mathbf{b}_i\cdot\mathbf{a}_j = 2\pi~\delta_{ij} ~. $$ Then,

\mathbf{k}\cdot\mathbf{R} = (\alpha_1~\mathbf{b}_1 + \alpha_2~\mathbf{b}_2 + \alpha_3~\mathbf{b}_3) \cdot (n_1~\mathbf{a}_1 + n_2~\mathbf{a}_2 + n_3~\mathbf{a}_3) = \alpha_1~n_1~2~\pi + \alpha_2~n_2~2~\pi + \alpha_3~n_3~2~\pi $$ or,

\mathbf{k}\cdot\mathbf{R} = 2\pi (\alpha_1~n_1 + \alpha_2~n_2 + \alpha_3~n_3) ~. $$ Therefore, we have

c(\mathbf{R}) = e^{i \mathbf{k}\cdot\mathbf{R}} ~. $$ Plugging this expression into (4), we get

[\mathcal{T}_R - e^{i \mathbf{k}\cdot\mathbf{R}}~\boldsymbol{1}] \begin{bmatrix} \mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x}) \end{bmatrix} = \boldsymbol{0} $$ or,
 * $$ \text{(5)} \qquad

{ \begin{bmatrix} \mathbf{D}(\mathbf{x}+\mathbf{R}) \\ \mathbf{B}(\mathbf{x}+\mathbf{R}) \end{bmatrix}  = e^{i \mathbf{k}\cdot\mathbf{R}}~ \begin{bmatrix} \mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x}) \end{bmatrix} ~. } $$ Equation (5) is called the  Bloch condition.

In summary, the solutions to the electromagnetic equations in a periodic medium can be expressed in  Bloch waves where each Bloch wave is a time harmonic solution to the electromagnetic equations which in addition satisfies the Bloch condition for all lattice vectors $$\mathbf{R}$$ and for some appropriate choice of $$\mathbf{k}$$.

Note that for any vector $$\mathbf{x}$$, the Bloch condition implies that

e^{-i \mathbf{k}\cdot(\mathbf{x}+\mathbf{R})} \begin{bmatrix} \mathbf{D}(\mathbf{x}+\mathbf{R}) \\ \mathbf{B}(\mathbf{x}+\mathbf{R}) \end{bmatrix} = e^{i \mathbf{k}\cdot\mathbf{R} - i \mathbf{k} \cdot\mathbf{R} - i \mathbf{k} \cdot \mathbf{x}}~ \begin{bmatrix} \mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x}) \end{bmatrix} = e^{- i \mathbf{k} \cdot \mathbf{x}}~ \begin{bmatrix} \mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x}) \end{bmatrix} ~. $$ Therefore the quantity

e^{- i \mathbf{k} \cdot \mathbf{x}}~ \begin{bmatrix} \mathbf{D}(\mathbf{x}) \\ \mathbf{B}(\mathbf{x}) \end{bmatrix} $$ is periodic.

Quasistatic Limit
Let us now consider the solution of Maxwell's equation in periodic media in the quasistatic limit. Consider the periodic medium shown in Figure 2. The lattice spacing is $$\eta$$.

Define

\epsilon_\eta(\mathbf{x}) := \epsilon\left(\cfrac{\mathbf{x}}{\eta}\right) = \epsilon(\mathbf{y})~; \mu_\eta(\mathbf{x}) := \mu\left(\cfrac{\mathbf{x}}{\eta}\right) = \mu(\mathbf{y}) ~. $$ These are periodic functions, i.e.,

\epsilon(\mathbf{y} + \mathbf{a}_i) = \epsilon(\mathbf{y}) ~; \mu(\mathbf{y} + \mathbf{a}_i) = \mu(\mathbf{y}) $$ where $$\mathbf{a}_i$$ are the primitive lattice vectors. We may also write these periodicity conditions as

\epsilon_\eta(\mathbf{x} + \eta\mathbf{a}_i) = \epsilon_\eta(\mathbf{x}) ~; \mu_\eta(\mathbf{x} + \eta\mathbf{a}_i) = \mu_\eta(\mathbf{x}) ~. $$

Similarly, define

\mathbf{D}_\eta(\mathbf{x}) := \mathbf{D}(\mathbf{y}) ~; \mathbf{B}_\eta(\mathbf{x}) := \mathbf{B}(\mathbf{y}) ~; \mathbf{E}_\eta(\mathbf{x}) := \mathbf{E}(\mathbf{y}) ~; \mathbf{H}_\eta(\mathbf{x}) := \mathbf{H}(\mathbf{y}) ~. $$ Then Maxwell's equations can be written as
 * $$ \text{(6)} \qquad

\boldsymbol{\nabla} \cdot \mathbf{D}_\eta = 0 ~; \boldsymbol{\nabla} \cdot \mathbf{B}_\eta = 0 ~; \boldsymbol{\nabla} \times \mathbf{E}_\eta - i\omega~\mathbf{B}_\eta = \boldsymbol{0} ~; \boldsymbol{\nabla} \times \mathbf{H}_\eta + i\omega~\mathbf{D}_\eta = \boldsymbol{0} ~. $$ Let us look for Bloch wave solutions of the form

\mathbf{E}_\eta(\mathbf{x}) = e^{i\mathbf{k}\cdot\mathbf{x}}~\mathbf{e}_\eta(\mathbf{x}) ~; \mathbf{D}_\eta(\mathbf{x}) = e^{i\mathbf{k}\cdot\mathbf{x}}~\mathbf{d}_\eta(\mathbf{x}) ~; \mathbf{H}_\eta(\mathbf{x}) = e^{i\mathbf{k}\cdot\mathbf{x}}~\mathbf{h}_\eta(\mathbf{x}) ~; \mathbf{B}_\eta(\mathbf{x}) = e^{i\mathbf{k}\cdot\mathbf{x}}~\mathbf{b}_\eta(\mathbf{x}) $$ where $$\mathbf{e}_\eta, \mathbf{d}_\eta, \mathbf{h}_\eta, \mathbf{b}_\eta$$ have the same periodicity as $$\epsilon$$ and $$\mu$$, i.e.,

\mathbf{e}_\eta(\mathbf{x} + \eta~\mathbf{a}_i) = \mathbf{e}(\mathbf{x}) ~; \mathbf{d}_\eta(\mathbf{x} + \eta~\mathbf{a}_i) = \mathbf{d}(\mathbf{x}) ~; \mathbf{h}_\eta(\mathbf{x} + \eta~\mathbf{a}_i) = \mathbf{h}(\mathbf{x}) ~; \mathbf{b}_\eta(\mathbf{x} + \eta~\mathbf{a}_i) = \mathbf{b}(\mathbf{x}) ~. $$ From the constitutive relations, we get

\mathbf{d}_\eta(\mathbf{x}) = \epsilon_\eta(\mathbf{x})~\mathbf{e}_\eta(\mathbf{x}) ~; \mathbf{b}_\eta(\mathbf{x}) = \mu_\eta(\mathbf{x})~\mathbf{h}_\eta(\mathbf{x}) ~. $$ Recall that, for periodic media, Maxwell's equations may be expressed as

(\mathcal{L} - \omega~\boldsymbol{1})\begin{bmatrix}\mathbf{D} \\ \mathbf{B} \end{bmatrix} = \boldsymbol{0} ~. $$ Here $$\omega$$ is an eigenvalue of $$\mathcal{L}$$. However, $$\mathcal{L}$$ depends on $$\omega$$ via $$\epsilon(\omega)$$ and $$\mu(\omega)$$. {\bf Bloch wave solutions do not exists unless $$\omega$$ takes one of a discrete set of values.}

Let these discrete values be

\omega = \omega_\eta^j(\mathbf{k}) $$ where the superscript $$j$$ labels the solution branches.

Let us see what the Bloch wave solutions reduce to as $$\eta\rightarrow 0$$. Following standard multiple scale analysis, let us assume that the periodic complex fields have the expansions
 * $$ \text{(7)} \qquad

\begin{align} \mathbf{e}_\eta(\mathbf{x}) & = \mathbf{e}_0(\mathbf{y}) + \eta~\mathbf{a}_1(\mathbf{y}) + \eta^2~\mathbf{a}_2(\mathbf{y}) + \dots \\ \mathbf{d}_\eta(\mathbf{x}) & = \mathbf{d}_0(\mathbf{y}) + \eta~\mathbf{d}_1(\mathbf{y}) + \eta^2~\mathbf{d}_2(\mathbf{y}) + \dots \\ \mathbf{h}_\eta(\mathbf{x}) & = \mathbf{h}_0(\mathbf{y}) + \eta~\mathbf{h}_1(\mathbf{y}) + \eta^2~\mathbf{h}_2(\mathbf{y}) + \dots \\ \mathbf{b}_\eta(\mathbf{x}) & = \mathbf{b}_0(\mathbf{y}) + \eta~\mathbf{b}_1(\mathbf{y}) + \eta^2~\mathbf{b}_2(\mathbf{y}) + \dots \end{align} $$ Let us also assume that the dependence of $$\omega$$ on $$\eta$$ and $$\mathbf{k}$$ has an expansion of the form
 * $$ \text{(8)} \qquad

\omega = \omega_\eta^j(\mathbf{k}) = \omega_0^j(\mathbf{y}) + \eta~\omega_1^j(\mathbf{y}) + \eta^2~\omega_2^j(\mathbf{y}) + \dots $$ Plugging (8) and (7) into (6) gives
 * $$ \text{(9)} \qquad

\begin{align} \boldsymbol{\nabla} \cdot \left(e^{i\eta~\mathbf{k}\cdot\mathbf{y}}~[\mathbf{d}_0(\mathbf{y}) + \eta~\mathbf{d}_1(\mathbf{y}) + \dots]        \right) & = 0 \\ \boldsymbol{\nabla} \cdot \left(e^{i\eta~\mathbf{k}\cdot\mathbf{y}}~[\mathbf{b}_0(\mathbf{y}) + \eta~\mathbf{b}_1(\mathbf{y}) + \dots]        \right) & = 0 \\ \boldsymbol{\nabla} \times \left(e^{i\eta~\mathbf{k}\cdot\mathbf{y}}~[\mathbf{e}_0(\mathbf{y}) + \eta~\mathbf{e}_1(\mathbf{y}) + \dots]         \right) - i [\omega_0^j + \eta~\omega_1^j + \dots] [\mathbf{b}_0(\mathbf{y}) + \eta~\mathbf{b}_1(\mathbf{y}) + \dots] & = \boldsymbol{0} \\ \boldsymbol{\nabla} \times \left(e^{i\eta~\mathbf{k}\cdot\mathbf{y}}~[\mathbf{h}_0(\mathbf{y}) + \eta~\mathbf{h}_1(\mathbf{y}) + \dots]         \right) + i [\omega_0^j + \eta~\omega_1^j + \dots] [\mathbf{d}_0(\mathbf{y}) + \eta~\mathbf{d}_1(\mathbf{y}) + \dots] & = \boldsymbol{0} ~. \end{align} $$ Define

\boldsymbol{\nabla}_y := \left(\frac{\partial }{\partial y_1}, \frac{\partial }{\partial y_2}, \frac{\partial }{\partial y_3}\right)~. $$ Then, for a vector field $$\mathbf{v}(\mathbf{y})$$, using the chain rule we get
 * $$ \text{(10)} \qquad

\boldsymbol{\nabla} \cdot \mathbf{v}(\mathbf{y}) = \cfrac{1}{\eta}~\boldsymbol{\nabla}_y \cdot\mathbf{v}(\mathbf{y}) ~; \boldsymbol{\nabla} \times \mathbf{v}(\mathbf{y}) = \cfrac{1}{\eta}~\boldsymbol{\nabla}_y\times{\mathbf{v}(\mathbf{y})} ~. $$ Using definitions (10) in (9) and collecting terms of order $$1/\eta$$ gives
 * $$ \text{(11)} \qquad

{ \begin{align} \boldsymbol{\nabla}_y \cdot\mathbf{d}_0(\mathbf{y}) = 0 \\ \boldsymbol{\nabla}_y \cdot\mathbf{b}_0(\mathbf{y}) = 0 \\ \boldsymbol{\nabla}_y\times{\mathbf{e}_0(\mathbf{y})} = \boldsymbol{0} \\ \boldsymbol{\nabla}_y\times{\mathbf{h}_0(\mathbf{y})} = \boldsymbol{0} \end{align} } $$ These are the solutions in the  quasistatic limit. Also, from the constitutive equations

\mathbf{d}_0(\mathbf{y}) = \epsilon(\mathbf{y})~\mathbf{e}_0(\mathbf{y}) ~; \mathbf{b}_0(\mathbf{y}) = \mu(\mathbf{y})~\mathbf{h}_0(\mathbf{y}) ~. $$ Similarly, collecting terms of order 1 from the expanded Maxwell's equations (9) we get
 * $$ \text{(12)} \qquad

{ \begin{align} i~\mathbf{k}\cdot\mathbf{d}_0(\mathbf{y}) + \boldsymbol{\nabla}_y \cdot\mathbf{d}_1(\mathbf{y}) = 0 \\ i~\mathbf{k}\cdot\mathbf{b}_0(\mathbf{y}) + \boldsymbol{\nabla}_y \cdot\mathbf{b}_1(\mathbf{y}) = 0 \\ i~\mathbf{k}\times\mathbf{e}_0(\mathbf{y}) + \boldsymbol{\nabla}_y\times{\mathbf{e}_1(\mathbf{y})} - i~\omega_0^j~\mathbf{b}_0(\mathbf{y}) = \boldsymbol{0} \\ i~\mathbf{k}\times\mathbf{h}_0(\mathbf{y}) + \boldsymbol{\nabla}_y\times{\mathbf{h}_1(\mathbf{y})} + i~\omega_0^j~\mathbf{d}_0(\mathbf{y}) = \boldsymbol{0} ~. \end{align} } $$ Since $$\mathbf{d}_1(\mathbf{y}), \mathbf{b}_1(\mathbf{y}), \mathbf{e}_1(\mathbf{y}), \mathbf{h}_1(\mathbf{y})$$ are periodic, this implies that
 * $$ \text{(13)} \qquad

{ \begin{align} \langle \boldsymbol{\nabla}_y \cdot\mathbf{d}_1(\mathbf{y})\rangle = 0 \\ \langle \boldsymbol{\nabla}_y \cdot\mathbf{b}_1(\mathbf{y})\rangle = 0 \\ \langle \boldsymbol{\nabla}_y \times{\mathbf{e}_1(\mathbf{y})}\rangle = \boldsymbol{0} \\ \langle \boldsymbol{\nabla}_y \times{\mathbf{h}_1(\mathbf{y})}\rangle = \boldsymbol{0} ~. \end{align} } $$ where $$\langle \bullet \rangle$$ is the volume average over the unit cell. So a necessary condition that equations (12) have a solution is that
 * $$ \text{(14)} \qquad

{ \begin{align} i~\mathbf{k}\cdot\langle \mathbf{d}_0(\mathbf{y}) \rangle = 0 \\ i~\mathbf{k}\cdot\langle \mathbf{b}_0(\mathbf{y}) \rangle = 0 \\ i~\mathbf{k}\times\langle \mathbf{e}_0(\mathbf{y}) \rangle - i~\omega_0^j~\langle \mathbf{b}_0(\mathbf{y}) \rangle = \boldsymbol{0} \\ i~\mathbf{k}\times\langle \mathbf{h}_0(\mathbf{y}) \rangle + i~\omega_0^j~\langle \mathbf{d}_0(\mathbf{y}) \rangle = \boldsymbol{0} ~. \end{align} } $$ Note that the second pair of (14) implies the first pair.


text:
User:Timwi

w:User:Timwi

title: User:Timwi


text:
User talk:Timwi

w:User talk:Timwi

title: User talk:Timwi