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Network Administration

Computer networks are critical parts of almost every organization. Network and computer systems administrators are responsible for the day-to-day operation of these networks. They organize, install, and support an organization’s computer systems, including local area networks (LANs), wide area networks (WANs), network segments, intranets, and other data communication systems. Administrators manage an organization’s servers and desktop and mobile equipment. They ensure that email and data storage networks work properly. They also make sure that employees’ workstations are working efficiently and stay connected to the central computer network. In some cases, administrators help network architects design and analyze network models. They also participate in decisions about buying future hardware or software to upgrade their organization’s network. Some administrators provide technical support to computer users, and they also may supervise computer support specialists who help solve users’ problems.

As with many technical roles, network administrator positions require a breadth of technical knowledge and the ability to learn the intricacies of new networking and server software packages quickly. Within smaller organizations, the more senior role of network engineer is sometimes attached to the responsibilities of the network administrator. It is common for smaller organizations to outsource this function.

Purpose of networks
A computer network is a telecommunications network which allows computers to exchange data. In computer networks, networked computing devices exchange data with each other along network links (data connections). The connections between nodes are established using either cable media or wireless media. The best-known computer network is the Internet.

Network computer devices that originate, route and terminate the data are called network nodes. Nodes can include hosts such as personal computers, mobile phones, servers as well as networking hardware. Two such devices can be said to be networked together when one device is able to exchange information with the other device, whether or not they have a direct connection to each other.

Computer networks differ in the transmission media used to carry their signals, the communications protocols to organize network traffic, the network's size, topology and organizational intent. In most cases, communications protocols are layered on (i.e. work using) other more specific or more general communications protocols, except for the physical layer that directly deals with the transmission media.

Computer networks support an enormous number of applications such as access to the World Wide Web, video, digital audio, shared use of application and storage servers, printers, and fax machines, and use of email and instant messaging applications as well as many others.

Network Hardware
Every piece of hardware in a computer network is called a node. There are various types of nodes which may exist on a network:
 * Network Interface Controller (NIC) Card
 * Typically when you think of a NIC Card, you should think of an NIC embedded onto the motherboard of your computer or an expansion card that is installed into an ISA,PCI, or PCI-Express slot inside your computer. A NIC could be an interface for a copper (BNC, Cat5/5e, Cat6), radio or microwave, or optical/fiber connection. A NIC can have one or many ports built into the interface.

 * Repeater
 * One of the limitations of computer networking is found in the transmission medium. Certain cables are only capable of transmitting a certain distance before attenuation comes into play. If the physical distance of a cable or distance between wireless antenna exceed the physical limitations of the medium, a repeater may be placed before the limitation distance to recondition and repeat the signal.

 * Hubs
 * Hubs are very basic devices that are made up of many NIC ports. They take the electrical signals that a computer transmits into them and repeats them out every port on the device except for the one the signals arrived in. Since hubs offer no services other than repeating signals to multiple ports, they are often called multiport repeaters. A hub works at the physical layer (layer 1) of the OSI model.

 * Bridges
 * A network bridge connects multiple network segments at the data link layer (layer 2) of the OSI model, and the term layer 2 switch is often used interchangeably with bridge. Bridges are similar to repeaters or network hubs, devices that connect network segments at the physical layer, however a bridge works by connecting multiple network segments, as opposed to simply rebroadcasting to adjacent network segments. In Ethernet networks, the term "bridge" formally means a device that behaves according to the IEEE 802.1D standard—this is most often referred to as a network switch in marketing literature.

 * Switches
 * Low-end network switches appear nearly identical to network hubs, but a switch contains more "intelligence" than a network hub. Network switches are capable of inspecting data packets as they are received, determining the source and destination device of that packet, and forwarding it appropriately. By delivering each message only to the connected device it was intended for, a network switch conserves network bandwidth and offers generally better performance than a hub. It works on OSI layer 2.

 * Routers
 * A router allows connectivity to one or more computers, helping create a network. It works on OSI layer 3. For home users, these are particularly useful for taking a single broadband internet account, and spreading it to at least two or more computers. Standard routers require the internet connection from a standalone modem, but modem-routers are increasing in popularity. Modem-routers can be plugged into any broadband-enabled phone line, reducing cable clutter, and only taking up one power socket.

 * Full-featured industrial routers form the backbone of the internet. They work rather like telephone exchanges, passing data between network segments to form a connection. The rules they use for handling (routing) traffic are an important part of internet security.
 * Each router has a configuration table, or routing table, that contains information about which connections lead to certain groups of addresses, which connections have priority for usage, and rules for handling different kinds of traffic. A typical home/office router has a very small routing table, but the big routers that handle the main internet traffic can have huge complicated routing tables. Each time a router receives a packet of data it will attempt to send it along the best possible route to its destination, based on its routing table. If that connection is not currently available, it will send it along the next best route. In this way, the routers that form the internet adapt to work around any problems with the network.
 * A home/office router may have rules limiting how computers outside the network can connect to computers inside the network, as well as preventing private network traffic from spilling into the outside world. Many home routers include additional security features—they scan and filter all traffic that passes through them, usually through an integrated firewall in the hardware. Some may perform other useful roles, such as acting as a print server.
 * Wireless routers have become more common. A wireless router does exactly the same job in the home as a regular wired (Ethernet) router, except a computer can be connected to it without needing to run Ethernet cable between the computer and the router. All you need is a wireless network adapter in each PC you want to connect, usually in the form of a card in your PCI slot (or a laptop's PCMCIA card slot) or an adapter for USB. Wireless routers generally have four ports to connect Ethernet cable as well, so computers can be connected by whatever means is most convenient—you might want to use a cable for your desktop PC, which sits close to the router, but use the wireless adapter in your laptop.
 * Firewalls
 * A firewall is a dedicated appliance, or software running on another computer, which inspects network traffic passing through it, and denies or permits passage based on a set of rules. A firewall's basic task is to regulate some of the flow of traffic between computer networks of different trust levels. Typical examples are the Internet which is a zone with no trust and an internal network which is a zone of higher trust. A zone with an intermediate trust level, situated between the Internet and a trusted internal network, is often referred to as a "perimeter network" or Demilitarized zone (DMZ).

 * IP Phone
 * An IP phone uses Voice over IP technologies allowing telephone calls to be made over an IP network such as the internet instead of the ordinary PSTN system. Calls can traverse the Internet, or a private IP Network such as that of a company.

Also involved in network hardware is the medium by which the nodes are connected. This can be radio waves, infrared signals, the home power network, Ethernet cabling, USB (universal serial bus) cabling, co-ax cabling, satellite, and telephone lines. Also introduced is fiber-optic cabling, which boasts even greater increases in speed and transmission, which also requires a bit of hardware to decode the light-based signal at each node.

General

 * Open Systems Interconnection (OSI) model
 * Transmission Control Protocol (TCP) / Internet Protocol (IP) ← EXTREMELY important. Read thoroughly, including linked pages.
 * Network Address Translation (NAT)

The OSI model is the basis of networking, it attempts to describe the various networking protocols in a layered approach. This is a useful theory to read at least once in your networking life.

TCP/IP is the predominant protocol of the internet and has replaced other older protocols (see obsolete protocols).

Network Address Translation, often abbreviated NAT, is a way to get a private zone connected to a public zone by rewriting the destination or address of IP packets as they pass through a router or firewall. Its most commonly used so that multiple computers on a network can connect to the internet using one shared public IP address.

Wireless

 * Wi-Fi
 * Bluetooth

Network Services
Not only is a network administrator responsible for designing and maintaining how nodes communicate with each other on the network, there are typically several critical network services that are provided to the nodes on the network.

Domain Name Service (DNS)
DNS is a service that helps nodes resolve network names into IP addresses.

Network Attached Storage (NAS)
In a workgroup setting there may be file storage that is accessible via the network. This is important for sharing files, disaster recovery for corporate data, and providing "indefinite" storage to users.

Lightweight Directory Access Protocol
LDAP provides directory look-up in an enterprise. In a business setting, a directory may contain a record for each employee with employee id, name, organizational groups, email address, and other related information.

Network address assignment
Network addresses using the TCP/IP protocol, and additionally, IPv4, can be assigned dynamically by a Dynamic Host Configuration Protocol server or can be specified statically. DHCP addresses are best suited for networks in which clients will be connecting and disconnecting frequently, such as users on a wireless connection at an Internet cafe. Static IP addresses create a sense of liability for usage of the network, because each node is assigned a unique IP address. Static IP addresses are ideal for servers that need a dedicated route for clients to access them. Machines that do not change location or address are usually set as a static IP. Examples include: Servers, Gateways, Routers, Printers. Most networks, especially enterprise level networks, will utilize DHCP addressing for client nodes because it minimizes addressing conflicts caused by human error.

Routing table configuration
ip addressing

Maintenance work
The actual role of the Network Administrator will vary from company to company, but will commonly include activities and tasks such as network address assignment, assignment of routing protocols and routing table configuration as well as configuration of authentication and authorization – directory services. It often includes maintenance of network facilities in individual machines, such as drivers and settings of personal computers as well as printers and such. It sometimes also includes maintenance of certain network servers: file servers, VPN gateways, intrusion detection systems, etc.

The administrator is responsible for the security of the network and for assigning IP addresses to the devices connected to the networks. Assigning IP addresses gives the subnet administrator some control over the professional who connects to the subnet. It also helps to ensure that the administrator knows each system that is connected and who personally is responsible for the system.

Printers
A printer is a device in a network which is used to control a print operation or the actual physical machine used for printing. Multiple Printers can be assigned to a single print device, to allow for different privileges.

File Servers
A file server is a central storage space on a network. Advantages of using file servers include: 1) centralization of documents, 2) the ability to back up important data, and 3) the ability to control access to different resources within the company. Most commonly, a file server will simply be a Windows server with shared folders configured. From there, desktops will often be configured to connect to the share as a drive. Many different kinds of servers fall into this category: standard Windows file servers, ftp servers, or more specialized data management systems such as Open Text's Livelink (designed for the engineering industry), SANs, and NASs. These specialized file servers can often be accessed from Web browsers or other proprietary GUIs (Graphical User Interface)s.

VPN gateways
Virtual Private Networks enables a secure encrypted connection. This technology enables people to safely connect to a private network remotely. A VPN Tunnel is created using encryption algorithms, making a secure connection across the internet.

Bus
In a bus topology, computers in a data network are connected to each other in a linear fashion, or from network card to network card. This topology is the most prone to failure, as a severed link between any of the computers near the middle of the network would break the network into two segments.

Ring
In a ring topology, computers are connected in a linear fashion, but either end of the network is connected to the other. This topology provides more protection against failure than a bus topology, as a severed link would result in traffic traveling in the opposite direction around the ring.

Star
In a star topology, a computer or device with multiple network cards/ports acts as a central connection point for all other devices on the network.

Extended Star
An extended star topology functions much like a star topology, but, as the name implies, it offers a hierarchical approach to the network. The best example of an extended star topology is to visualize two or more star networks connected together.

Partial Mesh
In a partial mesh topology, almost every computer or device has at least one connection to every other device on the network. This is the next best failure resistant topology as it is not as expensive as a full mesh, but more expensive than any of the other topologies.

Full Mesh
In a full mesh topology, each computer or device has at least one connection to every other device on the network. This is the most failure resistant topology, but also the most expensive as extra network cards and cable is required as the network grows.

Security
Security in a network can be described as a two pronged approach: Hardware and Software.

Hardware security includes securing the hardware itself (Physical Security), changing vendors/ hardware between nodes (Diversity) and

The most common ways to secure a wireless network is with a WEP or WPA key. A WEP Key requires an encryption key for any network use; wires or not. A WPA (Wireless Protected Access) key only protects against Wireless Network use.

Intrusion-Detection Systems
An intrusion detection system (IDS) generally detects unwanted manipulations of computer systems, mainly through the Internet. The manipulations may take the form of attacks by crackers.

An intrusion detection system is used to detect several types of malicious behaviors that can compromise the security and trust of a computer system. This includes network attacks against vulnerable services, data driven attacks on applications, host based attacks such as privilege escalation, unauthorized logins and access to sensitive files, and malware (viruses, trojan horses, and worms).

An IDS is composed of several components: Sensors which generate security events, a Console to monitor events and alerts and control the sensors, and a central Engine that records events logged by the sensors in a database and uses a system of rules to generate alerts from security events received. There are several ways to categorize an IDS depending on the type and location of the sensors and the methodology used by the engine to generate alerts. In many simple IDS implementations all three components are combined in a single device or appliance.

Types of intrusion-detection systems
In a network-based intrusion-detection system (NIDS), the sensors are located at choke points in the network to be monitored, often in the demilitarized zone (DMZ) or at network borders. The sensor captures all network traffic and analyzes the content of individual packets for malicious traffic. In systems, PIDS and APIDS are used to monitor the transport and protocols illegal or inappropriate traffic or constructs of language (say SQL). In a host-based system, the sensor usually consists of a software agent, which monitors all activity of the host on which it is installed. Hybrids of these two systems also exist.

 * A network intrusion detection system is an independent platform which identifies intrusions by examining network traffic and monitors multiple hosts. Network Intrusion Detection Systems gain access to network traffic by connecting to a hub, network switch configured for port mirroring, or network tap. An example of a NIDS is Snort.

 * A protocol-based intrusion detection system consists of a system or agent that would typically sit at the front end of a server, monitoring and analyzing the communication protocol between a connected device (a user/PC or system). For a web server this would typically monitor the HTTPS protocol stream and understand the HTTP protocol relative to the web server/system it is trying to protect. Where HTTPS is in use then this system would need to reside in the "shim" or interface between where HTTPS is un-encrypted and immediately prior to it entering the Web presentation layer.

 * An application protocol-based intrusion detection system consists of a system or agent that would typically sit within a group of servers, monitoring and analyzing the communication on application specific protocols. For example; in a web server with database this would monitor the SQL protocol specific to the middleware/business-login as it transacts with the database.
 * A host-based intrusion detection system consists of an agent on a host which identifies intrusions by analyzing system calls, application logs, file-system modifications (binaries, password files, capability/acl databases) and other host activities and state. An example of a HIDS is OSSEC.

 * A hybrid intrusion detection system combines two or more approaches. Host agent data is combined with network information to form a comprehensive view of the network. An example of a Hybrid IDS is Prelude.

Passive systems vs. reactive systems
In a passive system, the intrusion detection system (IDS) sensor detects a potential security breach, logs the information and signals an alert on the console and or owner. In a reactive system, also known as an intrusion prevention system (IPS), the IDS responds to the suspicious activity by resetting the connection or by reprogramming the firewall to block network traffic from the suspected malicious source. This can happen automatically or at the command of an operator.

Though they both relate to network security, an intrusion detection system (IDS) differs from a firewall in that a firewall looks outwardly for intrusions in order to stop them from happening. Firewalls limit access between networks to prevent intrusion and do not signal an attack from inside the network. An IDS evaluates a suspected intrusion once it has taken place and signals an alarm. An IDS also watches for attacks that originate from within a system.

This is traditionally achieved by examining network communications, identifying heuristics and patterns (often known as signatures) of common computer attacks, and taking action to alert operators. A system which terminates connections is called an intrusion prevention system, and is another form of an application layer firewall.

IDS evasion techniques
Intrusion Detection System (IDS) evasion techniques bypass detection by creating different states on the IDS and on the targeted computer. The adversary accomplishes this by manipulating either the attack itself or the network traffic that contains the attack.

title: Network Administration


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Waves in composites and metamaterials/Elastodynamics and electrodynamics

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.

Dissipation
Recall from the previous lecture that the average rate of work done in a cycle of oscillation of material with frequency dependent mass is

\begin{align} \mathcal{P} & = \cfrac{\omega}{2~\pi} \int_{0}^{2\pi/\omega}~\mathbf{F}(t)\cdot\mathbf{V}(t)~\text{d}t \\ & = \cfrac{\text{Re}(\widehat{\mathbf{F}})\cdot\text{Re}(\widehat{\mathbf{V}}) + \text{Im}(\widehat{\mathbf{F}})\cdot\text{Im}(\widehat{\mathbf{V}})}{2} \\ & = \omega~[\text{Re}(\widehat{\mathbf{V}})\cdot\text{Im}[\boldsymbol{M}(\omega)]\cdot\text{Re}(\widehat{\mathbf{V}}) + \text{Im}(\widehat{\mathbf{V}})\cdot\text{Im}[\boldsymbol{M}(\omega)]\cdot\text{Im}(\widehat{\mathbf{V}})]. \end{align} $$ This quadratic form will be non-negative for all choices of $$\widehat{\mathbf{V}}$$ if and only if $$\text{Im}(\boldsymbol{M}(\omega))$$ is positive semidefinite for all real $$\omega>0$$. Therefore, a restriction on the behavior of such materials is that

{ \text{Im}[\boldsymbol{M}(\omega)] \ge 0 ~. } $$

Similarly, for electrodynamics, the average power dissipated into heat is given by
 * $$ \text{(1)} \qquad

\mathcal{P} = \cfrac{\omega}{2~\pi} \int_{0}^{2\pi/\omega}~\left[\mathbf{E}(t)\cdot\frac{\partial \mathbf{D}(t)}{\partial t} + \mathbf{H}(t)\cdot\frac{\partial \mathbf{B}(t)}{\partial t}\right]~dt ~. $$ In this case, the quantity $$\mathbf{E}$$ is equivalent to the voltage and the quantity rate of change of electrical displacement $$\partial\mathbf{D}/\partial t$$ is equivalent to the current (recall that in electrostatics the power is given by $$\mathcal{P} = V~I$$). In addition, we also have a contribution due to magnetic induction.

Let us assume that the fields can be expressed in harmonic form, i.e.,

\mathbf{E}(t) = \text{Re}[\widehat{\mathbf{E}}~e^{-i\omega t}] ~; \mathbf{H}(t) = \text{Re}[\widehat{\mathbf{H}}~e^{-i\omega t}] $$ or equivalently as

\mathbf{E}(t) = \text{Re}(\widehat{\mathbf{E}})~\cos(\omega t) + \text{Im}(\widehat{\mathbf{E}})~\sin(\omega t) ~; \mathbf{H}(t) = \text{Re}(\widehat{\mathbf{H}})~\cos(\omega t) + \text{Im}(\widehat{\mathbf{H}})~\sin(\omega t)~. $$ Also, recall that,

\frac{\partial \mathbf{D}}{\partial t} = \boldsymbol{\nabla} \times \mathbf{H} = -i\omega~\boldsymbol{\epsilon}\cdot\mathbf{E}(t) \qquad \text{and} \qquad \frac{\partial \mathbf{B}}{\partial t} = -\boldsymbol{\nabla} \times \mathbf{E} = -i\omega~\boldsymbol{\mu}\cdot\mathbf{H}(t) ~. $$ Therefore, for real $$\omega$$ and real $$\mathcal{P}$$, we can write equation (1) as (with the substitution $$z = \omega t$$),

\begin{align} \mathcal{P} = \cfrac{\omega}{2~\pi} \int_{0}^{2\pi}~ & \left[\text{Re}(\widehat{\mathbf{E}})~\cos z + \text{Im}(\widehat{\mathbf{E}})~\sin z\right]\cdot \left\{\text{Im}(\boldsymbol{\epsilon})\cdot \left[\text{Re}(\widehat{\mathbf{E}})~\cos z + \text{Im}(\widehat{\mathbf{E}})~\sin z\right]\right\} + \\ &      \left[\text{Re}(\widehat{\mathbf{H}})~\cos z + \text{Im}(\widehat{\mathbf{H}})~\sin z\right]\cdot \left\{\text{Im}(\boldsymbol{\mu})\cdot \left[\text{Re}(\widehat{\mathbf{H}})~\cos z + \text{Im}(\widehat{\mathbf{H}})~\sin z\right]\right\}~\text{d}z \end{align} $$ Expanding out, and using the fact that

\int_{0}^{2\pi} \cos^2 z~\text{d}z = \int_{0}^{2\pi} \sin^2 z~\text{d}z = \pi \quad \text{and} \quad \int_{0}^{2\pi} \sin(2 z)~\text{d}z = 0 $$ we have,
 * $$ \text{(2)} \qquad

\mathcal{P} = \cfrac{\omega}{2} \left[\text{Re}(\widehat{\mathbf{E}})\cdot\text{Im}(\boldsymbol{\epsilon})\cdot\text{Re}(\widehat{\mathbf{E}}) + \text{Im}(\widehat{\mathbf{E}})\cdot\text{Im}(\boldsymbol{\epsilon})\cdot\text{Im}(\widehat{\mathbf{E}}) + \text{Re}(\widehat{\mathbf{H}})\cdot\text{Im}(\boldsymbol{\mu})\cdot\text{Re}(\widehat{\mathbf{H}}) + \text{Im}(\widehat{\mathbf{H}})\cdot\text{Im}(\boldsymbol{\mu})\cdot\text{Im}(\widehat{\mathbf{H}})\right] ~. $$ Since $$\omega \ge 0$$ and the power $$\mathcal{P} \ge 0$$, the quadratic forms in equation (2) require that

\text{Im}(\boldsymbol{\epsilon}) > 0 \qquad \text{and} \qquad \text{Im}(\boldsymbol{\mu}) > 0 ~. $$ Note that if the permittivity is expressed as

\boldsymbol{\epsilon} = \boldsymbol{\epsilon}_0 + i\cfrac{\boldsymbol{\sigma}}{\omega} $$ the requirement $$\text{Im}(\boldsymbol{\epsilon}) > 0$$ implies that the conductivity $$\boldsymbol{\sigma} > 0$$. Therefore, if the conductivity is greater than zero, there will be dissipation.

Brief introduction to elastodynamics
A concise introduction to the theory of elasticity can be found in Atkin80. In this section, we consider the linear theory of elasticity for infinitesimal strains and small displacements.

Consider the body ($$\Omega$$) shown in Figure~1. Let $$\Gamma$$ be a subpart of the body (in the interior of $$\Omega$$ or sharing a part of the surface of $$\Omega$$). Postulate the existence of a force $$\mathbf{t}(\mathbf{x},\mathbf{n})$$ per unit area on the surface of $$\Gamma$$ where $$\mathbf{n}$$ is the outward unit normal to the surface of $$\Gamma$$. Then $$\mathbf{t}$$ is the force exerted on $$\Gamma$$ by the material outside $$\Gamma$$ or by surface tractions.

From the balance of forces on a small tetrahedron ($$\Gamma$$), we can show that $$\mathbf{t}(\mathbf{x},\mathbf{n})$$ is linear in $$\mathbf{n}$$. Therefore,

\mathbf{t} = \boldsymbol{\sigma}\cdot\mathbf{n} $$ where $$\boldsymbol{\sigma}$$ is a second-order tensor called the  stress tensor.

Since the tetrahedron cannot rotate at infinite velocity as its size goes to zero (conservation of angular momentum), we can show that the stress tensor is symmetric, i.e.,

\boldsymbol{\sigma} = \boldsymbol{\sigma}^T ~. $$ In particular, for a fluid,

\boldsymbol{\sigma} = - p~\boldsymbol{\mathit{1}} $$ where $$p$$ is the pressure.

Let us assume that the stress depends only on the strain (and not on strain gradients or strain rates), where the strain is defined as
 * $$ \text{(3)} \qquad

\boldsymbol{\epsilon} = \frac{1}{2}\left[\boldsymbol{\nabla}\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})^T\right] ~. $$ Here $$\mathbf{u}(\mathbf{x},t)$$ is the displacement field. Note that a gradient of the displacement field is used to define the strain because rigid body motions should not affect $$\boldsymbol{\sigma}$$ and a rigid body rotation gives zero strains (for small displacements).

Assume that $$\boldsymbol{\sigma}$$ depends linearly on $$\boldsymbol{\epsilon}$$ so that

\boldsymbol{\sigma}(\mathbf{x},t) = \int \text{d}\mathbf{x}'~ \left[\int \boldsymbol{\mathsf{K}}_{\varepsilon}(\mathbf{x},\mathbf{x}',t' - t):\boldsymbol{\epsilon}(\mathbf{x},t')~\text{d}t'\right] ~. $$ Note that this assumption ignores preexisting internal stresses such as those found in prestressed concrete. If the material can be approximated as being local, then
 * $$ \text{(4)} \qquad

\boldsymbol{\sigma}(\mathbf{x},t) = \int \boldsymbol{\mathsf{K}}_{\varepsilon}(\mathbf{x},t' - t):\boldsymbol{\epsilon}(\mathbf{x},t')~\text{d}t' ~. $$ Taking the Fourier transforms of equation (4), we get
 * $$ \text{(5)} \qquad

\widehat{\boldsymbol{\sigma}}(\mathbf{x},\omega) = \boldsymbol{\mathsf{C}}(\mathbf{x},\omega):\widehat{\boldsymbol{\epsilon}}(\mathbf{x},\omega) $$ where

\boldsymbol{\mathsf{C}}(\mathbf{x},\omega) = \int \boldsymbol{\mathsf{K}}_{\varepsilon}(\mathbf{x},\tau)~e^{-i\omega\tau}~\text{d}\tau ~; \tau = t' - t ~. $$ In index notation, equation (5) can be written as

\widehat{\sigma}_{ij} = C_{ijkl}~\widehat{\epsilon}_{kl} ~. $$ Causality implies that stresses at time $$t$$ can only depend on strains of previous times, i.e., if $$t' \le t$$ or $$\tau \le 0$$. Therefore,

\boldsymbol{\mathsf{K}}_{\varepsilon}(\mathbf{x},\tau) = \boldsymbol{\mathsf{0}} \qquad \text{if} \qquad \tau > 0 ~. $$ This in turn implies that the integral converges only if $$\text{Im}(\omega) > 0$$, i.e., $$\boldsymbol{\mathsf{C}}(\mathbf{x},\omega)$$ is analytic when $$\text{Im}(\omega) > 0$$.

In the absence of body forces, the equation of motion of the body can be written as
 * $$ \text{(6)} \qquad

\boldsymbol{\nabla} \cdot \boldsymbol{\sigma} = \rho(\mathbf{x})~\frac{\partial^2 \mathbf{u}}{\partial t^2} $$ where $$\rho$$ is the mass density, $$\boldsymbol{\nabla} \cdot \boldsymbol{\sigma}$$ is the internal force per unit volume, and $$\partial^2 \mathbf{u}/\partial t^2$$ is the acceleration. Hence, this is just the expression of Newton's second law for continuous systems.

For a material which has a frequency dependent mass, equation (6) may be written as
 * $$ \text{(7)} \qquad

\boldsymbol{\nabla} \cdot \boldsymbol{\sigma} = \int_{-\infty}^{\infty} \rho(\mathbf{x}, t - t')~\frac{\partial^2 \mathbf{u}}{\partial t^2}~\text{d}t' $$ where causality implies that if $$t' > t$$ then $$\rho = 0$$.

Taking the Fourier transform of equation (7), we get
 * $$ \text{(8)} \qquad

\boldsymbol{\nabla} \cdot \widehat{\boldsymbol{\sigma}}(\mathbf{x},\omega) = -\omega^2~\widehat{\rho}(\mathbf{x},\omega)~\widehat{\mathbf{u}}(\mathbf{x},\omega)~. $$ Substituting equation (5) into equation (8) we get
 * $$ \text{(9)} \qquad

\boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}:\widehat{\boldsymbol{\epsilon}}) + \omega^2~\widehat{\rho}~\widehat{\mathbf{u}} = 0~. $$ Also, taking the Fourier transform of equation (3), we have
 * $$ \text{(10)} \qquad

\widehat{\boldsymbol{\epsilon}} = \frac{1}{2}\left[\boldsymbol{\nabla} \widehat{\mathbf{u}} + (\boldsymbol{\nabla} \widehat{\mathbf{u}})^T\right] ~. $$ Since $$\widehat{\boldsymbol{\sigma}}$$ and $$\widehat{\boldsymbol{\epsilon}}$$ are symmetric, we must have

C_{ijkl} = C_{jikl} = C_{ijlk} ~. $$ Because of this symmetry, we can replace $$\widehat{\boldsymbol{\epsilon}}$$ by $$\boldsymbol{\nabla} \widehat{\mathbf{u}}$$ in equation (9) to get

\boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}:\boldsymbol{\nabla \widehat{\mathbf{u}})} + \omega^2~\widehat{\rho}~\widehat{\mathbf{u}} = 0~. $$ Dropping the hats, we then get the  wave equation for elastodynamics
 * $$ \text{(11)} \qquad

{ \boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}:\boldsymbol{\nabla \mathbf{u})} + \omega^2~\rho~\mathbf{u} = 0~. } $$

Antiplane shear
Let us now consider the case of antiplane shear. Assume that $$\boldsymbol{\mathsf{C}}$$ is isotropic, i.e.,

\boldsymbol{\mathsf{C}}:\boldsymbol{\nabla}\mathbf{u} = \mu~[\boldsymbol{\nabla}\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})^T] + \lambda~\text{tr}(\boldsymbol{\nabla}\mathbf{u})~\boldsymbol{\mathit{1}} $$ where $$\mu$$ is the shear modulus and $$\lambda$$ is the Lame modulus.

Let us assume that $$\mu$$ and $$\lambda$$ are independent of $$x_1$$, i.e.,

\mu \equiv \mu(x_2,x_3) \qquad \text{and} \qquad \lambda \equiv \lambda(x_2,x_3) ~. $$ Let us look for a solution with $$u_2 = u_3 = 0$$ and $$u_1$$ independent of $$x_1$$, i.e., $$u_1 \equiv u_1(x_2,x_3)$$. This is an out of plane mode of deformation.

Then, noting that $$\text{tr}(\boldsymbol{\nabla}\mathbf{u}) = \boldsymbol{\nabla} \cdot \mathbf{u}$$, we have

\text{tr}(\boldsymbol{\nabla}\mathbf{u}) = \frac{\partial u_i}{\partial x_i} = 0 ~. $$ Therefore,

[\boldsymbol{\mathsf{C}}:\boldsymbol{\nabla}\mathbf{u}]_{ij} = \mu~\left[\frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i}\right] $$ or

\boldsymbol{\mathsf{C}}:\boldsymbol{\nabla}\mathbf{u} = \mu \begin{bmatrix} 0 & \frac{\partial u_1}{\partial x_2} & \frac{\partial u_1}{\partial x_3} \\ \\      \frac{\partial u_1}{\partial x_2} & 0 & 0 \\ \\      \frac{\partial u_1}{\partial x_3} & 0 & 0 \end{bmatrix} ~. $$ Therefore

\boldsymbol{\nabla} \cdot (\boldsymbol{\mathsf{C}}:\boldsymbol{\nabla\mathbf{u})} = \begin{bmatrix} \frac{\partial }{\partial x_2}\left(\mu~ \frac{\partial u_1}{\partial x_2}\right) + \frac{\partial }{\partial x_3}\left(\mu~ \frac{\partial u_1}{\partial x_3}\right) \\ 0 \\      0    \end{bmatrix} ~. $$ Plugging into the wave equation (11) we get

\frac{\partial }{\partial x_2}\left(\mu~ \frac{\partial u_1}{\partial x_2}\right) + \frac{\partial }{\partial x_3}\left(\mu~ \frac{\partial u_1}{\partial x_3}\right) + \omega^2~\rho~u_1 = 0 $$ or (using the two-dimensional gradient operator $$\overline{\boldsymbol{\nabla}}$$)
 * $$ \text{(12)} \qquad

{  \overline{\boldsymbol{\nabla}} \cdot(\mu~\overline{\boldsymbol{\nabla} u_1)} + \omega^2~\rho~u_1 = 0 ~. } $$

TM and TE modes in electromagnetism
Let us now consider the TM (transverse magnetic field) and TE (transverse electric field) modes in electromagnetism and look for parallels with antiplane shear in elastodynamics.

Recall the Maxwell equations (with hats dropped)
 * $$ \text{(13)} \qquad

\boldsymbol{\nabla} \times \mathbf{E} = i\omega\mu\mathbf{H} ~; \boldsymbol{\nabla} \times \mathbf{H} = -i\omega\epsilon\mathbf{E} ~. $$ Assume that $$\mu$$ and $$\epsilon$$ are scalars which are independent of $$x_1$$, i.e., $$\mu \equiv \mu(x_2,x_3)$$ and $$\epsilon \equiv \epsilon(x_2,x_3)$$.

For the TE case, we look for solutions with $$E_2 = E_3 = 0$$ and $$E_1$$ independent of $$x_1$$, i.e., $$E_1 \equiv E_1(x_2, x_3)$$. Then,

\boldsymbol{\nabla} \times \mathbf{E} = \left[0, \frac{\partial E_1}{\partial x_3}, - \frac{\partial E_1}{\partial x_2}\right]~. $$ This implies that

\mathbf{H} = \left[ 0, \cfrac{1}{i\omega\mu}~\frac{\partial E_1}{\partial x_3}, - \cfrac{1}{i\omega\mu}~\frac{\partial E_1}{\partial x_2}\right]~. $$ Therefore,

\boldsymbol{\nabla} \times \mathbf{H} = \left[ -\frac{\partial }{\partial x_3}\left(\cfrac{1}{i\omega\mu}~\frac{\partial E_1}{\partial x_3}\right) -\frac{\partial }{\partial x_2}\left(\cfrac{1}{i\omega\mu}~\frac{\partial E_1}{\partial x_2}\right), 0, 0\right]~. $$ or,

\boldsymbol{\nabla} \times \mathbf{H} = \cfrac{i}{\omega}\left[ \frac{\partial }{\partial x_2}\left(\cfrac{1}{\mu}~\frac{\partial E_1}{\partial x_2}\right) + \frac{\partial }{\partial x_3}\left(\cfrac{1}{\mu}~\frac{\partial E_1}{\partial x_3}\right), 0, 0\right] = \cfrac{i}{\omega} \left[\overline{\boldsymbol{\nabla}} \cdot\left(\cfrac{1}{\mu(x_2,x_3)}       \overline{\boldsymbol{\nabla}} E_1\right), 0, 0\right] ~. $$ Plugging into equation (13) we get the TE equation

{ \overline{\boldsymbol{\nabla}} \cdot\left(\cfrac{1}{\mu(x_2,x_3)}\overline{\boldsymbol{\nabla}} E_1\right) + \omega^2~\epsilon(x_2,x_3)~E_1 = 0 ~. } $$ This equation has the same form as equation (12).

More generally, if

\boldsymbol{\mu} = \boldsymbol{\mu}(x_2,x_3) = \begin{bmatrix} \mu_{11} & 0 & 0 \\ 0 & \mu_{22} & \mu_{23} \\ 0 & \mu_{23} & \mu_{33} \end{bmatrix} = \begin{bmatrix} \mu_{11} & \left[\mathsf{0}\right] \\ \left[\mathsf{0}\right] & \left[\mathsf{M}\right] \end{bmatrix} \quad \text{where} \quad \left[\mathsf{M}\right] = \begin{bmatrix} \mu_{22} & \mu_{23} \\ \mu_{23} & \mu_{33} \end{bmatrix} \equiv \boldsymbol{M} $$ and

\boldsymbol{\epsilon} = \boldsymbol{\epsilon}(x_2,x_3) = \begin{bmatrix} \epsilon_{11} & \left[\mathsf{0}\right] \\ \left[\mathsf{0}\right] & \left[\mathsf{N}\right] \end{bmatrix} \quad \text{where} \quad \left[\mathsf{N}\right] = \begin{bmatrix} \epsilon_{22} & \epsilon_{23} \\ \epsilon_{23} & \epsilon_{33} \end{bmatrix} \equiv \boldsymbol{N} $$ we get the TE equation

{ \overline{\boldsymbol{\nabla}} \cdot\left[\left(\boldsymbol{R}_{\perp}\cdot\boldsymbol{M}^{-1}\cdot\boldsymbol{R}_{\perp}^T\right)\cdot \overline{\boldsymbol{\nabla}} E_1\right] + \omega^2~\epsilon_{11}~E_1~\mathbf{1} = \mathbf{0} } \qquad \text{where} \qquad \boldsymbol{R}_{\perp} \equiv \left[\mathsf{R}\right]_{\perp} = \begin{bmatrix} 0 & 1 \\ -1 & 0 \end{bmatrix} ~. $$

Similarly, there is a TM equation with $$H_2 = H_3 = 0$$ of the form

{ \overline{\boldsymbol{\nabla}} \cdot\left[\left(\boldsymbol{R}_{\perp}\cdot\boldsymbol{N}^{-1}\cdot\boldsymbol{R}_{\perp}^T\right) \cdot\overline{\boldsymbol{\nabla}} H_1\right] + \omega^2~\mu_{11}~H_1~\mathbf{1} = \mathbf{0} } $$ which for the isotropic case reduces to

{ \overline{\boldsymbol{\nabla}} \cdot\left(\cfrac{1}{\epsilon(x_2,x_3)}\overline{\boldsymbol{\nabla}} H_1\right) + \omega^2~\mu(x_2,x_3)~H_1 = 0 ~. } $$ The general solution independent of $$x_1$$ is a superposition of the TE and TM solutions. This can be seen by observing that the Maxwell equations decouple under these conditions and a general solution can be written as

(E_1,E_2,E_3) = (E_1, 0, 0) + (0, E_2, E_3) $$ where the first term represents the TE solution. We can show that the second term represents the TM solution by observing that

\boldsymbol{\nabla} \times (0, E_2, E_3) = \left[\frac{\partial E_3}{\partial x_2} - \frac{\partial E_2}{\partial x_3},0,0\right] $$ implying that $$H_2 = H_3 = 0$$ which is the TM solution.

A resonant structure
Consider the periodic geometry shown in Figure 2. The matrix material has a high value of shear modulus ($$\mu$$) while the split-ring shaped region has a low shear modulus or is a void. The material inside the ring has the same shear modulus as the matrix material and is connected to the matrix by a thin ligament. The system is subjected to a displacement $$u_1$$ in the $$x_1$$ direction (parallel to the axis of each cylindrical split ring).

Clearly, each periodic component of the system behaves like a mass attached to a spring. This is a resonant structure and the effective density $$\rho_{11}^{*}(\omega)$$ can be negative. A detailed treatment of the problem can be found in Movchan04. Note that the governing equation for this problem is

\overline{\boldsymbol{\nabla}} \cdot(\mu~\overline{\boldsymbol{\nabla}} u_1) + \omega^2~\rho~u_1 = 0 ~. $$

Let us compare this problem with the TM case where $$H_1$$ is the out of plane magnetic induction. The governing equation now is

\overline{\boldsymbol{\nabla}} \cdot(\cfrac{1}{\epsilon}~\overline{\boldsymbol{\nabla}} H_1) + \omega^2~\mu~H_1 = 0 ~. $$ If the value of $$1/\epsilon$$ in the region of the void (ring) is small and hence $$\epsilon$$ is large (which implies that the conductivity $$\sigma$$ is large), analogy with the equation of elastodynamics implies that the effective permeability $$\mu_{11}^{*}$$ can be negative for this material.

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


text:
Waves in composites and metamaterials/Acoustic metamaterials and negative moduli

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.

Acoustic Metamaterials
Recall that for isotropic materials, the stress is given by

\boldsymbol{\sigma} = \boldsymbol{\mathsf{C}}:\boldsymbol{\nabla}\mathbf{u} = \mu~[\boldsymbol{\nabla}\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})^T] + \lambda~(\boldsymbol{\nabla} \cdot \mathbf{u})~\boldsymbol{\mathit{1}} ~. $$

For a fluid, the shear modulus $$\mu = 0$$ and the Lame constant $$\lambda = \kappa - 2/3~\mu = \kappa$$, where $$\kappa$$ is the bulk modulus. Therefore, for fluids,
 * $$ \text{(1)} \qquad

\boldsymbol{\sigma} = \boldsymbol{\mathsf{C}}:\boldsymbol{\nabla}\mathbf{u} = \kappa~(\boldsymbol{\nabla} \cdot \mathbf{u})~\boldsymbol{\mathit{1}} = -p~\boldsymbol{\mathit{1}} ~. $$ The quantity
 * $$ \text{(2)} \qquad

p = -\kappa~(\boldsymbol{\nabla} \cdot \mathbf{u}) $$ is the pressure in the fluid.

Taking the Fourier transform of equations (1) and (2) we get
 * $$ \text{(3)} \qquad

\widehat{\boldsymbol{\sigma}} = -\widehat{p}~\boldsymbol{\mathit{1}} \qquad \text{and} \qquad \widehat{p} = -\widehat{\kappa}(\omega)~(\boldsymbol{\nabla} \cdot \widehat{\mathbf{u}}) ~. $$

Also recall that, for low frequency processes,
 * $$ \text{(4)} \qquad

\boldsymbol{\nabla} \cdot \widehat{\boldsymbol{\sigma}} + \omega^2~\widehat{\rho}(\omega)~\widehat{\mathbf{u}}(\omega) = \boldsymbol{0} ~. $$ For fluids, equations (4) becomes
 * $$ \text{(5)} \qquad

\boldsymbol{\nabla} \widehat{p} = \omega^2~\widehat{\rho}(\omega)~\widehat{\mathbf{u}}(\omega) ~. $$

From equations (3) and (5), we have

\boldsymbol{\nabla} \cdot \widehat{\mathbf{u}} = \boldsymbol{\nabla} \cdot \left(\cfrac{1}{\omega^2~\widehat{\rho}}~\boldsymbol{\nabla} \widehat{p}\right) = -\cfrac{\widehat{p}}{\widehat{\kappa}} ~. $$ Therefore, we get the  acoustic wave equation

{ \boldsymbol{\nabla} \cdot \left(\cfrac{1}{\widehat{\rho}}~\boldsymbol{\nabla} \widehat{p}\right) + \cfrac{\omega^2}{\widehat{\kappa}}~\widehat{p} = 0 ~. } $$ If $$\widehat{\rho}$$ and $$\widehat{\kappa}$$ only depend on $$x_2$$ and $$x_3$$ and $$\widehat{p}$$ is independent of $$x_1$$, then the three-dimensional gradient operator $$\boldsymbol{\nabla}$$ can be replaced with the two-dimensional gradient operator $$\overline{\boldsymbol{\nabla}}$$, and we get

{ \overline{\boldsymbol{\nabla}} \cdot\left(\cfrac{1}{\widehat{\rho}(\omega)}~\overline{\boldsymbol{\nabla}} \widehat{p}\right) + \cfrac{\omega^2}{\widehat{\kappa}(\omega)}~\widehat{p} = 0 } $$ and

\widehat{\mathbf{u}}(\omega) = \cfrac{1}{\omega^2~\widehat{\rho}(\omega)}~\overline{\boldsymbol{\nabla}} \widehat{p} ~. $$ Therfore, by analogy with the results from antiplane shear elasticity and TE electromagnetism, if $$\widehat{\rho}$$ and $$\widehat{\kappa}$$ are both negative, we get a negative refractive index material for acoustics.

The speed of sound in an acoustic medium is given by $$c = \sqrt{\kappa/\rho}$$. The sound speed is imaginary is either $$\kappa$$ or $$\rho$$ is negative and a material with these properties appears opaque to sound waves. However, if $$\kappa$$ and $$\rho$$ are both positive or both negative, the sound speed is real and the medium becomes transparent to acoustic waves, i.e., acoustic waves can propagate through the medium.

Let us now consider the propagation of acoustic waves across the interface shown in Figure 1. If the pressure is the same at points at equal distances from the interface as shown, the vectors $$\boldsymbol{\nabla} p$$ are reflections of each other. If $$\mathbf{n}$$ is the normal to the interface, the quantity $$\mathbf{n}\cdot\boldsymbol{\nabla} p$$ changes sign across the interface. Since $$\rho$$ also changes sign, $$\mathbf{n}\cdot\mathbf{u}$$ is continuous across the interface. Hence, the boundary condition across the interface is physical.

Since this situation is analgous to the one we observed earlier for electromagnetism, the effect of a slab of negative $$\kappa,\rho$$ material will be to translate a source of acoustic waves and the slab will act like a perfect lens.

Negative Elastic Moduli
Early work on negative elastic moduli can be found in Lakes01. More recent work on ultrasonic metamaterials can be found in Fang06. Consider the array of Helmholtz resonators shown in Figure 2 (an example of such a resonator in everyday life is a soda bottle which resonates when you blow over the top of the neck.)  The resonator can be thought of as a spring-mass system where the air inside the cavity acts as a spring and the water in the narrow neck acts as a mass. There is some frequency at which the spring-mass system resonates.

A model of the Helmholtz resonator
A model of the Helmholtz resonator is shown in Figure 3. For simplicity, we assume that each cavity has a square cross section as do the piston arms. The cross sectional area is assumed to be $$d^2$$. The air in each cavity is model with a spring of complex spring constant $$K$$ and the water in the neck is modeled as a rigid body of mass $$m$$. The piston is filled with a compressible fluid with complex bulk modulus $$\kappa$$.

Let the force applied on the pistons be $$F(t)$$. Then the pressure in the fluid is
 * $$ \text{(6)} \qquad

p(t) = \cfrac{F(t)}{d^2} ~. $$ Since the fluid transmits the pressure to the mass and the area of cross section of the cavity is $$d^2$$, the force applied by the fluid on the mass is also $$F(t)$$.

Let $$f(t)$$ be the force that the spring applies on the mass. By symmetry, the same forces is applied to both cavities shown in Figure 3. Let $$x(t)$$ denote the position of the piston and let $$y(t)$$ be the position of the mass.

Assume harmonic time dependence of the quantities $$F(t)$$, $$f(t)$$, $$x(t)$$, and $$y(t)$$. Then
 * $$ \text{(7)} \qquad

F(t) = \text{Re}(\widehat{F}~e^{-i\omega t}) ~; f(t) = \text{Re}(\widehat{f}~e^{-i\omega t}) ~; x(t) = \text{Re}(\widehat{x}~e^{-i\omega t}) ~; y(t) = \text{Re}(\widehat{y}~e^{-i\omega t}) ~. $$

Assume that $$p = 0$$ at $$t = 0$$. Then, we also have
 * $$ \text{(8)} \qquad

p(t) = \text{Re}(\widehat{p}~e^{-i\omega t}) ~. $$ Newton's law then implies that
 * $$ \text{(9)} \qquad

m~\cfrac{d^2 y}{d t^2} = F(t) - f(t) ~. $$ Substituting equations (7) into equation (9), we get
 * $$ \text{(10)} \qquad

-m~\omega^2~\widehat{y} = \widehat{F} - \widehat{f} ~. $$ Also, from Hooke's law
 * $$ \text{(11)} \qquad

f(y) = K~y(t) \qquad \implies \qquad \widehat{f} = K~\widehat{y} ~. $$ Combining equations (10) and (11) we get
 * $$ \text{(12)} \qquad

\widehat{F} = (K - m~\omega^2)~\widehat{y} ~. $$ Also, from equation (6) we have
 * $$ \text{(13)} \qquad

\widehat{p} = \cfrac{\widehat{F}}{d^2} ~. $$ Combining equations (13) and (12) we have
 * $$ \text{(14)} \qquad

\widehat{p} = \cfrac{(K - m~\omega^2)}{d^2}~\widehat{y} ~. $$ The change in volume of the fluid due to the motions of the pistons and the masses is given by
 * $$ \text{(15)} \qquad

\Delta V(t) = 2~d^2~[y(t) - x(t)]~. $$ Now, from the constitutive relation for the fluid,
 * $$ \text{(16)} \qquad

p(t) = -\kappa~\cfrac{\Delta V(t)}{V_0} = -\kappa~\cfrac{2~d^2~[y(t) - x(t)]}{V_0} ~. $$ where $$\kappa$$ is the complex bulk modulus of the fluid and $$V_0$$ is the initial volume.

Substituting equations (7) and (8) into equation (16) we get
 * $$ \text{(17)} \qquad

\widehat{p} = \cfrac{2~\kappa~d^2~(\widehat{x} - \widehat{y})}{V_0} ~. $$ From equations (14) and (17), eliminating $$\widehat{y}$$, we get

\widehat{p} = \cfrac{2~\kappa~d^2}{V_0} \left(\widehat{x} - \cfrac{d^2}{K - m~\omega^2}~\widehat{p}\right) ~. $$ Solving for $$\widehat{p}$$, we have

\widehat{p} = \cfrac{2~\kappa~d^2}{V_0} \left[1 + \cfrac{2~\kappa~d^4}{V_0(K - m~\omega^2)}\right]^{-1} ~\widehat{x} ~. $$ Therefore,
 * $$ \text{(18)} \qquad

\widehat{F} = d^2~\widehat{p} = \cfrac{2~\kappa~d^4}{V_0} \left[1 + \cfrac{2~\kappa~d^4}{V_0(K - m~\omega^2)}\right]^{-1} ~\widehat{x} ~. $$ By definition, the Young's modulus relates the stress to the strain. For our model this implies that

\cfrac{\widehat{F}}{d^2} = Y~\cfrac{\widehat{\Delta L}}{L_0} $$ where $$Y$$ is a complex Young's modulus, $$\Delta L(t)$$ is the change in length and $$L_0$$ is the initial length of the region between the two pistons. Therefore,
 * $$ \text{(19)} \qquad

\widehat{F} = \cfrac{d^2}{L_0}~Y~(2~\widehat{x}) ~. $$ From equations (18) and (19) we can deduce an expression for the Young's modulus of the form
 * $$ \text{(20)} \qquad

{ Y(\omega) = \cfrac{\kappa~d^2~L_0}{V_0} \left[1 + \cfrac{2~\kappa~d^4}{V_0(K - m~\omega^2)}\right]^{-1}~. } $$ If in particular $$\kappa$$ and $$K$$ are real (purely elastic springs with not damping) and positive, then a plot of $$Y$$ as a function of $$\omega$$ has the form shown in Figure 4. So it will appear that the system will have a negative Young's modulus for frequencies higher than $$\sqrt{K/m}$$ !

In fact, several other negative modulus materials can be envisaged.

Why are negative moduli a surprise?
It is common for people to use a generalized Maxwell model in linear viscoelasticity (see Figure 5). The question is: what is the relation between $$F$$ and $$u$$?

Recall that for a single Maxwell element ($$j$$), the displacements in the spring and the dashpot are given by

u_1(t) = \cfrac{f_j(t)}{K_j} \qquad \text{and} \qquad \cfrac{d^2 u_2(t)}{d t^2} = \cfrac{f_j(t)}{\eta_j} ~. $$ Fourier transforming these equations gives

\widehat{u}_1(\omega) = \cfrac{\widehat{f}_j(\omega)}{K_j} \qquad \text{and} \qquad i\omega~\widehat{u}_2(\omega) = \cfrac{\widehat{f}_j(\omega)}{\eta_j} ~. $$ The total displacement of the Maxwell element is

\widehat{u}_j(\omega) = \widehat{u}_1 + \widehat{u}_2 = \left[\cfrac{1}{K_j} -\cfrac{i}{\omega~\eta_j}\right]~\widehat{f}_j(\omega)~. $$ Dropping the hats, we have

u_j = \left(\cfrac{1}{K_j} - \cfrac{i}{\omega~\eta_j}\right)~f_j ~. $$ Therefore, from the balance of forces (for the generalized Maxwell model)

F = f_0 + \sum f_j = \left[K_0 + \sum_j \left(\cfrac{1}{K_j} - \cfrac{i}{\omega~\eta_j}\right)^{-1}\right]~u ~. $$ The quantity

K_{\text{eff}} = K_0 + \sum_j \left(\cfrac{1}{K_j} - \cfrac{i}{\omega~\eta_j}\right)^{-1} = K_0 + \sum_j \cfrac{K_j}{1 - i/\omega\tau_j} $$ is the effective spring constant for the model ($$\tau_j = \eta_j/K_j$$ is called the relaxation time).

Similarly, for the Young's modulus and converting the sum into an integral, we have

Y_{\text{eff}}(\omega) = Y_0 + \int_0^{\infty} \cfrac{H(\tau)}{1 - i/\omega\tau}~\text{d}\tau $$ where $$\tau$$ is the relaxation spectrum, $$H(\tau) \ge 0$$, and $$Y_0 \ge 0$$. Therefore,

Y_{\text{eff}}(\omega) = Y_0 + \int_0^{\infty} \cfrac{H(\tau)~(1 + i/\omega\tau)}{1 + 1/\omega^2\tau^2}~\text{d}\tau ~. $$ If we discretize the spectrum we get

Y_{\text{eff}}(\omega) = Y_0 + \sum_j \cfrac{H_j~(1 + i/\omega\tau_j)}{1 + 1/\omega^2\tau_j^2} $$ where $$H_j \ge 0$$, $$\tau_j > 0$$, and $$Y_0 \ge 0$$. This implies that $$\text{Im}(Y_{\text{eff}}(\omega)) \ge 0$$ and $$\text{Re}(Y_{\text{eff}}(\omega)) \ge 0$$ for all $$\omega$$. Clearly, viscoelastic models cannot represent negative elastic moduli and can therefore fail badly when used to model materials such as Helmholtz resonators.

In fact, for viscoelastic material models, we find that $$Y(\omega)$$ is analytic in the entire complex plane except for isolated poles (at intervals of $$\omega\tau$$) in the negative imaginary axis. This is contrary to what we observe for the frequency dependent models we have dealt with so far for which causality forces the moduli to be analytic only the upper half complex $$\omega$$ plane.


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text:
Waves in composites and metamaterials/Fading memory and waves in layered media

The content of these notes is based on the lectures by Prof. Graeme W. Milton (University of Utah) given in a course on metamaterials in Spring 2007.

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

Specifically, if

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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

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


text:
Digital libraries

Digital Libraries are collections of information that are available in digital form usually accessible through the internet.

This is part of both the Library and Information Science school and the Computer Science

Software

 * Dspace
 * Greenstone
 * Fedora
 * EPrints

Education

 * Digital Library Curriculum Project

title: Digital libraries


text:
Waves in composites and metamaterials/Airy solution and WKB solution

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
Recall from the previous lecture that we assumed that the permittivity and permeability are scalars and are locally isotropic though not globally so. Then we may write

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

\overline{\boldsymbol{\nabla}} \cdot\left(\cfrac{1}{\mu}~\overline{\boldsymbol{\nabla}} E_1\right) + \omega^2~\epsilon~E_1 = 0 $$ where $$\overline{\boldsymbol{\nabla}} $$ represents the two-dimensional gradient operator. Equation (1) can also be written as
 * $$ \text{(2)} \qquad

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

E_1(x_2,x_3) = \tilde{E}_1(x_3)~e^{\pm i~k_2~x_2} $$ and equation (2) then becomes an ODE:
 * $$ \text{(3)} \qquad

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

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

TE waves in a non-magnetic medium
For a non-magnetic medium, $$\mu$$ is constant and we can write (3) as
 * $$ \text{(4)} \qquad

\left[ \cfrac{d^2 }{d x^2} + \omega^2~\epsilon~\mu - k_2^2\right]~\tilde{E}_1 = 0 \qquad \text{where} \quad x \equiv x_3 ~. $$

Permittivity varies linearly with x
If the permittivity varies linearly with $$x$$, then we may write

\epsilon(x) = a + b~x $$ where $$a$$ and $$b$$ are constants. Plugging this into (4) we get
 * $$ \text{(5)} \qquad

\left[ \cfrac{d^2 }{d x^2} + A + B~x\right]~\tilde{E}_1 = 0 \qquad \text{where} \quad A := \omega^2~\mu~a - k_2^2~; B := \omega^2~\mu~b ~. $$ Let us assume that $$B > 0$$ (this is not strictly necessary, but simplifies things for our present analysis). Let us introduce a change of variables

\eta = B^{1/3}\left(x + \cfrac{A}{B}\right) ~. $$ Then (5) becomes
 * $$ \text{(6)} \qquad

\left[\cfrac{d^2 }{d \eta^2} + \eta\right]~\tilde{E}_1 = 0 ~. $$ Equation (6) is called the  Airy equation. The solution of this equation is

\tilde{E}_1(\eta) = C_1~\text{Ai}(-\eta) + C_2~\text{Bi}(-\eta) $$ where $$\text{Ai}$$ and $$\text{Bi}$$ are Airy functions of the first and second kind (see Abram72 for details.) A plot of the behavior of the two Airy functions as a function of real $$-\eta$$ is shown in Figure~1.

As $$x \rightarrow -\infty$$ (i.e., as $$-\eta \rightarrow \infty$$), the Airy functions asymptotically approach the values
 * $$\begin{align}

\text{Ai}(-\eta) & \sim \cfrac{1}{2}~\pi^{-1/2}~(-\eta)^{-1/4}~e^{-2/3~(-\eta)^{3/2}} \\ \text{Bi}(-\eta) & \sim \pi^{-1/2}~(-\eta)^{-1/4}~e^{2/3~(-\eta)^{3/2}} ~. \end{align}$$

So $$\text{Ai}(-\eta)$$ corresponds to an exponentially decaying wave as $$|x| \rightarrow \infty$$ and $$\text{Bi}(-\eta)$$ corresponds to an exponentially increasing waves at $$|x| \rightarrow \infty$$. A schematic of the situation is shown in Figure 2.

If there are no sources in the region $$x < 0$$ then the solution $$\text{Bi}(-\eta)$$ is unphysical which implies that $$C_2 = 0$$. Therefore,
 * $$ \text{(7)} \qquad

\tilde{E}_1(\eta) = C_1~\text{Ai}(-\eta) ~. $$

Now, as $$x \rightarrow \infty$$ (i.e., as $$-\eta \rightarrow -\infty$$), the Airy function $$\text{Ai}(-\eta)$$ takes the asymptotic form
 * $$ \text{(8)} \qquad

\text{Ai}(-\eta) \sim \pi^{-1/2}~\eta^{-1/4}~ \sin\left(\cfrac{2}{3}~\eta^{3/2}+\cfrac{\pi}{4}\right)~. $$ This is a superposition of right and left travelling waves (because the sine can be decomposed into two exponentials one of which corresponds to a wave travelling in one direction and the seconds to a wave travelling in the opposite direction.)

The Wentzel-Kramers-Brillouin (WKB) method
If we don't assume any particular linear variation of the permittivity $$\epsilon(x)$$, we can use the WKB method to arrive at a solution for high frequency waves.

The WKB method is a high frequency method for obtaining solutions to one-dimensional (time-independent) wave equations of the form
 * $$ \text{(9)} \qquad

\cfrac{d^2 \varphi}{d x^2} + k_3^2(x)~\varphi(x) = 0 ~. $$ Recall from (1) that the TE equation in a nonmagnetic medium is

\cfrac{d^2 \tilde{E}_1}{d x^2} + (\omega^2~\epsilon~\mu - k_2^2)~\tilde{E}_1 = 0 ~. $$ Clearly this equation can be written in form (9) by setting

\varphi = \tilde{E}_1 ~; k_3^2(x) = \omega^2~\epsilon~\mu - k_2^2 ~. $$ Recall also that the TM equation is
 * $$ \text{(10)} \qquad

\epsilon~\cfrac{d }{d x}\left(\epsilon^{-1}~\cfrac{d \tilde{H}_1}{d x}\right) + + k_3^2(x)~\tilde{H}_1 = 0 ~. $$ Equation (11) can also be reduced to the form (9). The procedure is as follows. Let us first set $$\psi = \tilde{H}_1$$ to get
 * $$ \text{(11)} \qquad

\epsilon~\cfrac{d }{d x}\left(\epsilon^{-1}~\cfrac{d \psi}{d x}\right) + k_3^2(x)~\psi = 0 ~. $$ After expanding (11) we get
 * $$ \text{(12)} \qquad

\cfrac{d^2 \psi}{d x^2} = \epsilon^{-1}~\cfrac{d \epsilon}{d x}~\cfrac{d \psi}{d x} - k_3^2~\psi = 0 ~. $$ Define
 * $$ \text{(13)} \qquad

\varphi(x) := \epsilon^{-1/2}(x)~\psi(x) ~. $$ Differentiating (13) twice, we get
 * $$ \text{(14)} \qquad

\cfrac{d^2 \varphi}{d x^2} = \cfrac{d^2 }{d x^2}\left(\epsilon^{-1/2}\right)~\psi + \epsilon^{-1/2}~\cfrac{d^2 \psi}{d x^2} + 2~\cfrac{d }{d x}\left(\epsilon^{-1/2}\right)~\cfrac{d \psi}{d x} ~. $$ Substituting (12), (13) into (14) we have
 * $$ \text{(15)} \qquad

\cfrac{d^2 \varphi}{d x^2} = \cfrac{d^2 }{d x^2}\left(\epsilon^{-1/2}\right)~ (\epsilon^{1/2}~\phi) + \epsilon^{-3/2}~\cfrac{d \epsilon}{d x}~\cfrac{d \psi}{d x}   - k_3^2~\epsilon^{-1/2}~\psi - \epsilon^{-3/2}~\cfrac{d \epsilon}{d x}~\cfrac{d \psi}{d x} $$ or,
 * $$ \text{(16)} \qquad

\cfrac{d^2 \varphi}{d x^2} + \left[k_3^2 - \epsilon^{1/2}~ \cfrac{d^2 }{d x^2}\left(\epsilon^{-1/2}\right)\right]~\varphi = 0 ~. $$ Equation (16) has the same form as (9).

At this stage recall that

k_3^2 = \omega^2~\epsilon~\mu - k_2^2 ~. $$ Let us assume that $$k_2$$ is proportional to $$\omega$$ which implies that $$k_3$$ is also proportional to $$\omega$$, i.e.,
 * $$ \text{(17)} \qquad

k_3^2(x) = \omega^2~s^2(x) $$ where $$s(x)$$ is independent of $$\omega$$.

In equation (16), if $$\omega$$ is large, then $$k_3$$ will dominate and we will end up with exactly the same equation as (9), provided variations in $$\epsilon$$ are smooth (and we don't get large jumps in its derivatives).

Let us now try to solve (9). When $$k_3$$ is constant, the solution of the equation is a traveling wave. If we assume that $$k_3$$ varies slowly with $$x$$, we can try to get solutions of the form
 * $$ \text{(18)} \qquad

\varphi(x) = A~e^{i\omega~\tau(x)} $$ and examine the phase $$\tau(x)$$ rather than the solution $$\varphi(x)$$. Differentiating (18), we get
 * $$ \text{(19)} \qquad

\varphi'(x) = i\omega~\tau'(x)~A~e^{i\omega~\tau(x)} ~; \varphi(x) = \left[i\omega~\tau(x) - \omega^2~(\tau'(x))^2\right]~ A~e^{i\omega~\tau(x)} ~. $$ Plugging (19) into (9), we get
 * $$ \text{(20)} \qquad

i\omega~\tau''(x) - \omega^2~(\tau'(x))^2 + k_3^2(x) = 0 ~. $$ If we assume that $$k_3^2 > 0$$ (i.e., $$k_3$$ is real) we can simplify the analysis slightly at this stage (even though this is not strictly necessary).

For large $$\omega$$, i.e., $$\omega \gg 1$$, we can seek a perturbation solution of the form
 * $$ \text{(21)} \qquad

\tau(x) = \tau_0(x) + \cfrac{1}{\omega}~\tau_1(x) + \cfrac{1}{\omega^2}~\tau_2(x) + \dots ~. $$ Plugging (21) into (20) and using (17) we get
 * $$ \text{(22)} \qquad

\left[i\omega~\tau_0(x) + i~\tau_1(x) + \cfrac{i}{\omega}~\tau_2''(x) + \dots \right] - \left[\omega~\tau_0'(x) + \tau_1'(x) + \cfrac{1}{\omega}~ \tau_2'(x) + \dots\right]^2 + \omega^2~s^2(x) = 0 $$ or,
 * $$ \text{(23)} \qquad

\left[\cfrac{i}{\omega}~\tau_0(x) + \cfrac{i}{\omega^2}~\tau_1(x) + \cfrac{i}{\omega^3}~\tau_2''(x) + \dots \right] - \left[\tau_0'(x) + \cfrac{1}{\omega}~\tau_1'(x) + \cfrac{1}{\omega^2}~\tau_2'(x) + \dots\right]^2 + s^2(x) = 0 ~. $$ For large $$\omega$$, equation (23) reduces to
 * $$ \text{(24)} \qquad

- [\tau_0'(x)]^2 + s^2(x) = 0 \qquad \text{or} \qquad [\tau_0'(x)]^2 = s^2(x) \qquad \qquad \text{(Eikonal equation)}~. $$ Therefore,
 * $$ \text{(25)} \qquad

\tau_0'(x) = \pm s(x) ~. $$ Integrating (25) from an arbitrary point $$x_0$$ to $$x$$, we get
 * $$ \text{(26)} \qquad

\tau_0(x) = \pm \int_{x_0}^x s(y)~\text{d}y + C_{0_{\pm}}. $$ where $$C_{0_{\pm}}$$ depends on the sign of the integral.

Next, collecting terms of order $$\omega$$ in equation (22), we get
 * $$ \text{(27)} \qquad

i\omega~\tau_0''(x) - 2~\omega~\tau_0'(x)~\tau_1'(x) = 0 ~. $$ Substituting (25) into (27) we get

\pm i\omega~s'(x) \mp 2~\omega~s(x)~\tau_1'(x) = 0 $$ or,
 * $$ \text{(28)} \qquad

\cfrac{i}{2}~\cfrac{s'(x)}{s(x)} = \tau_1'(x) ~. $$ Integrating (28) we get
 * $$ \text{(29)} \qquad

\tau_1(x) = \cfrac{i}{2}~\ln[s(x)] + C_1 = i\ln[\sqrt{s(x)}] + C_1 ~. $$ Plugging (26) and (29) into (21) (and ignoring terms containing powers of $$\omega^2$$ and higher) we get
 * $$ \text{(30)} \qquad

\tau(x) = \pm \int_{x_0}^x s(y)~\text{d}y + \cfrac{i}{\omega}~\ln[\sqrt{s(x)}] + C_{\pm} ~. $$ This implies that the solution (18) has the form
 * $$ \text{(31)} \qquad

{ \varphi(x) = \cfrac{A_{+}}{\sqrt{s(x)}}~ \exp\left(i\omega~\int_{x_0}^x s(y)~\text{d}y\right) + \cfrac{A_{-}}{\sqrt{s(x)}}~ \exp\left(-i\omega~\int_{x_0}^x s(y)~\text{d}y\right)~. } $$ Equation (31) is the WKB solution assuming $$k_3^2 > 0$$. Note that when $$s(x) = 0$$, a solution does not exist.

Also note that since $$k_3^2$$ is proportional to $$\omega^2$$,
 * $$ \text{(32)} \qquad

|k_3^2| \gg |i\omega~\tau_0''| \qquad \text{for large}~\omega ~. $$ Therefore,

\omega^2~s^2(x) \gg \omega~s'(x) \qquad \implies \qquad \omega~s(x) \gg \cfrac{\omega~s'(x)}{\omega~s(x)} = \cfrac{d }{d x}\left[\ln(\omega~s(x))\right] $$ or,

k_3(x) \gg \cfrac{d }{d x}\left[\ln(k_3(x))\right] ~. $$ Therefore, the restriction is that $$\omega$$ is large and that $$k_3$$ is smooth with respect to $$x$$.

Now, consider for example the profile shown in Figure 3. In region I, the WKB solution is valid since $$k_3^2 > 0$$. At the point where the profile meets the $$x$$ axis, a solution does not exist since $$s(x) = 0$$. However, if the profile is smooth enough, we can assume that $$k_3(x)$$ is linear and we can use the Airy solution for the region II around this point. When the profile goes below the $$x$$ axis, $$k_3^2 < 0$$. However, the WKB solution is valid in this region (III) too as equation 32 can still be satisfied with $$s(x) = i~\alpha(x)$$.

There is an area of overlap between the regions where the WKB solution is valid and the region where the Airy solution is valid. In fact, the unknown parameters in the two solutions can be determined by matching the solutions at points in this region of overlap.

To do this, let $$\zeta$$ be the point on the $$x$$-axis where $$s(x) = 0$$. Then, in region I, the solution is
 * $$ \text{(33)} \qquad

\varphi_I(x) = \cfrac{A_{+}}{\sqrt{s(x)}}~ \exp\left(i\omega~\int_{\zeta}^x s(y)~\text{d}y\right) + \cfrac{A_{-}}{\sqrt{s(x)}}~ \exp\left(-i\omega~\int_{\zeta}^x s(y)~\text{d}y\right)~. $$

If there are no sources in region III the solution decays exponentially in the $$-x$$ direction. Then the WKB solution with $$s(x) = i\alpha(x)$$ is
 * $$ \text{(34)} \qquad

\varphi_{III}(x) \sim \cfrac{B_{-}}{\sqrt{\alpha(x)}}~ \exp\left(\omega~\int_{\zeta}^x \alpha(y)~\text{d}y\right) $$ where the coefficient $$B_{-} = A_{-}/\sqrt{i}$$.

In region II, since $$\epsilon$$ or $$\mu$$ vary linearly with $$x$$, we may write
 * $$ \text{(35)} \qquad

k_3^2 = \omega^2~\epsilon~\mu - k_2^2 \sim \Omega(x-\zeta) $$ Then, from (7)

\varphi_{II}(x) \sim C~\text{Ai}(-\eta) \qquad \text{with} \qquad \eta = \Omega^{1/3} (x-\zeta) ~. $$ When $$\omega$$ is high, the region I, II, and III overlap. Also, from (35) we observe that $$\Omega \propto \omega^2$$. Hence, the large $$\eta$$ expansion (equation (8)) for the Airy function can be used in the overlap region, i.e.,

\varphi_{II}(x) \sim C~\pi^{-1/2}~\eta^{-1/4}~ \sin\left(\cfrac{2}{3}~\eta^{3/2}+\cfrac{\pi}{4}\right) ~. $$ Substituting for $$\eta$$ and using the identity

\sin\theta = \cfrac{1}{2i}\left(e^{i\theta} - e^{-i\theta}\right) $$ we get
 * $$ \text{(36)} \qquad

\varphi_{II}(x) \sim \cfrac{C}{2i~\pi^{1/2}~\Omega^{1/12}~(x-\zeta)^{1/4}}~ \left\{ \exp\left[\cfrac{2i}{3}~\Omega^{1/2}~(x-\zeta)^{3/2}+ \cfrac{\pi~i}{4}\right] - \exp\left[-\cfrac{2i}{3}~\Omega^{1/2}~(x-\zeta)^{3/2}- \cfrac{\pi~i}{4}\right]\right\}~. $$ Also, in the neighborhood of region II,

\omega~s(x) \sim \Omega^{1/2}~(x-\zeta)^{1/2} ~. $$ So

\omega~\int_{\zeta}^x s(y)~\text{d}y \sim \cfrac{2}{3}~\Omega^{1/2}~ (x-\zeta)^{3/2} ~, x \sim \zeta ~. $$ Therefore, $$\varphi_I$$ becomes
 * $$ \text{(37)} \qquad

\varphi_I(x) = \cfrac{A_{+}~\omega^{1/2}}{\Omega^{1/4}~(x-\zeta)^{1/4}}~ \exp\left(\cfrac{2i}{3}~\Omega^{1/2}~(x-\zeta)^{3/2}\right) + \cfrac{A_{-}~\omega^{1/2}}{\Omega^{1/4}~(x-\zeta)^{1/4}}~ \exp\left(-\cfrac{2i}{3}~\Omega^{1/2}~(x-\zeta)^{3/2}\right)~. $$ Comparing (37) with (36) we get

\cfrac{A_{+}}{A_{-}} = - i \qquad \text{and} \qquad C = 2~A_{+}~\omega^{1/2}~\pi^{1/2}~\Omega^{-1/6}~e^{i\pi/4} ~. $$ Similarly, by comparing $$\varphi_{II}$$ and $$\varphi_{III}$$ in the region of overlap, we get

C = 2~B_{-}~\omega^{1/2}~\pi^{1/2}~\Omega^{-1/6}~. $$


text:
Waves in composites and metamaterials/TE waves in multilayered media

The content of these notes is based on the lectures by Prof. Graeme W. Milton (University of Utah) given in a course on metamaterials in Spring 2007.

Review
While considering a single interface between two layers (in the previous lecture) we had used a coordinate system ($$x_1,x_2,x_3$$). In the following we switch to a system ($$x, y, z$$) to make our notation a bit less confusing for multilayered media. In the following we assume that the material properties of each layer in a multilayered material are piecewise constant.

Consider the TE wave shown in Figure 1.

Then, from the previous lecture and using the new notation ($$x, y, z$$) shown in the figure, the solution for the TE wave can be written as

E_y(x, z) = \tilde{E}_y(z)~e^{\pm i~k_x~x} $$ and the governing equation is

\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 ~. $$ Recall that plane waves propagating in the $$z$$ direction have the form

\tilde{E}_y(z) = \tilde{E}_0~\exp[i(k_z~z - \omega~t)] ~. $$ Therefore, in region 1 (see Figure~1) for fixed $$\omega$$,
 * $$ \text{(1)} \qquad

\tilde{E}_{y1}(z) = \tilde{E}_0~\left[\exp(-i~k_{z1}~z) + R^{TE}~\exp(i~k_{z1}~z) \right] ~. $$ The first term of the left hand side of (1) represents the incoming wave while the second term on represents the reflected wave (hence the difference in signs of $$z$$). The quantity $$R^{TE}$$ is a reflection coefficient.

Similarly, in region 2 (see Figure 1),
 * $$ \text{(2)} \qquad

\tilde{E}_{y2}(z) = \tilde{E}_0~T^{TE}~\exp(-i~k_{z2}~z) $$ where $$T^{TE}$$ is a transmission coefficient.

Continuity at the interface requires that the following conditions be satisfied:
 * $$ \text{(3)} \qquad

\tilde{E}_{y1} = \tilde{E}_{y2} ~; \mu_1^{-1}~\cfrac{d \tilde{E}_{y1}}{d z} = \mu_2^{-1}~\cfrac{d \tilde{E}_{y2}}{d z} ~. $$

If we choose the coordinate system such that $$z = 0$$ at the interface, substitution of (1), (2) into (3) gives
 * $$ \text{(4)} \qquad

1 + R^{TE} = T^{TE} ~; \mu_1^{-1}~k_{z1}~(1 - R^{TE}) = \mu_2^{-1}~k_{z2}~T^{TE} ~. $$ Solving for $$R^{TE}$$ and $$T^{TE}$$ from equations (4) gives
 * $$ \text{(5)} \qquad

R^{TE} = \cfrac{\mu_2~k_{z1} - \mu_1~k_{z2}}{\mu_2~k_{z1} + \mu_1~k_{z2}}~; T^{TE} = \cfrac{2~\mu_2~k_{z1}}{\mu_2~k_{z1} + \mu_1~k_{z2}}~; $$ Note that these quantities are the Fresnel coefficients of the bilayer and that the reflection and transmission coefficients may be complex.

Recall from the previous lecture that

k_z^2 = \omega^2~\mu~\epsilon - k_x^2 ~. $$ Therefore, if $$\omega^2~\mu_2~\epsilon_2 < k_{x2}^2$$, then $$k_{z2}$$ is purely imaginary. If $$k_{z1}$$ is real, then the first of equations (5) implies that the numerator and the denominator are complex conjugates. This means that

|R^{TE}| = 1 \qquad \implies \qquad \text{perfect reflection!} $$ If such a situation exists, the wave in region 2 is called  evanescent.

Multilayered Systems
Let us first examine the problem of reflection and transmission in a three layer medium (see Figure 2). Our goal is to find the effective reflection and transmission coefficients in this medium. Once we know these coefficients, we can choose the materials in the layers to achieve a desired reflectivity or transmissivity.

Let the interface between regions 1 and 2 be located at $$z = -d_1$$ and that between regions 2 and 3 be located at $$z = -d_2$$. Then, using a change of coordinates $$z \leftarrow z + d_1$$, in region 1 (from equation 1) we have

\begin{align} \tilde{E}_{y1}(z) & = \tilde{E}_0~\left[\exp[-i~k_{z1}~(z+d_1)] + \tilde{R}_{12}~\exp[i~k_{z1}~(z+d_1)] \right] \\ & =     \tilde{E}_0~\exp(-i~k_{z1}~d_1)~\left[\exp(-i~k_{z1}~z) + \tilde{R}_{12}~\exp(i~k_{z1}~z)~\exp(2~i~k_{z1}~d_1) \right] \end{align} $$ or,

\tilde{E}_{y1}(z) = A_1~\left[\exp(-i~k_{z1}~z) + \tilde{R}_{12}~\exp(i~k_{z1}~z + 2~i~k_{z1}~d_1) \right] $$ where

A_1 = \tilde{E}_0~\exp(-i~k_{z1}~d_1) $$ and $$\tilde{R}_{12}$$ is the apparent reflection coefficient at the interface between regions 1 and 2 due to the slab.

Similarly, for region 2, we have

\tilde{E}_{y2}(z) = A_2~\left[\exp(-i~k_{z2}~z) + \tilde{R}_{23}~\exp(i~k_{z2}~z + 2~i~k_{z2}~d_2) \right] $$ and $$\tilde{R}_{23}$$ is the apparent reflection coefficient for a downgoing wave at the interface between regions 2 and 3 due to the slab. However, since the wave is transmitted in region 3 and there are no further reflections, we have

\tilde{R}_{23} = R_{23} = \cfrac{\mu_3~k_{z2} - \mu_2~k_{z3}}{\mu_3~k_{z2} + \mu_2~k_{z3}}~; $$

Since the wave is only transmitted in region 3, we have

\tilde{E}_{y3}(z) = A_3~\exp(-i~k_{z3}~z) ~. $$

At this stage we don't know what $$\tilde{R}_{12}$$ is. To find this quantity, note that the downgoing wave in region 2 equals the sum of the transmission of the downgoing wave in region 1 and a reflection of the upgoing wave in region 2 (see Figure 2). Hence, at the top interface $$z = -d_1$$,
 * $$ \text{(6)} \qquad

A_2~\exp(i~k_{z2}~d_1) = T_{12}~A_1~\exp(i~k_{z1}~d_1) + R_{21}~A_2~R_{23}~\exp(-i~k_{z2}~d_1 + 2~i~k_{z2}~d_2) $$ where $$T_{12}$$ is the transmission coefficient between regions 1 and 2 and $$R_{21} = - R_{12}$$ is the reflection coefficient of waves from region 2 incident upon region 1.

Also, the upgoing wave in region 1 is the sum of the reflection of the incoming wave in region 1 and the transmission at interface 2-1 of the reflected wave at interface 2-3. Hence, at $$z = -d_1$$ we have

A_1~\tilde{R}_{12}~\exp(-i~k_{z1}~d_1 + 2~i~k_{z1}~d_1) = R_{12}~A_1~\exp(i~k_{z1}~d_1) + T_{21}~A_2~R_{23}~\exp(-i~k_{z2}~d_1 + 2~i~k_{z2}~d_2) $$ or,
 * $$ \text{(7)} \qquad

A_1~\tilde{R}_{12}~\exp(i~k_{z1}~d_1) = R_{12}~A_1~\exp(i~k_{z1}~d_1) + T_{21}~A_2~R_{23}~\exp[-i~k_{z2}~d_1 + 2~i~k_{z2}~d_2) ~. $$

Eliminating $$A_2$$ from (6) gives
 * $$ \text{(8)} \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)]} ~. $$ Plugging (8) into (7), we get
 * $$ \text{(9)} \qquad

\tilde{R}_{12} = R_{12} + \cfrac{T_{12}~T_{21}~R_{23}~\exp[2~i~k_{z2}~(d_2 - d_1)]} {1 - R_{21}~R_{23}~\exp[2~i~k_{z2}~(d_2-d_1)]} $$ which gives us an expression for the generalized reflection coefficient $$\tilde{R}_{12}$$.

We have considered only two internal reflections so far. How about further reflections? It turns out that equation (9) can be interpreted to include all possible internal reflections. To see this, let us assume that

0 \le R_{21}~R_{23}~e^{2~i~k_{z2}~(d_2-d_1)} < 1 ~. $$ Then we can expand (9) in series form to get
 * $$ \text{(10)} \qquad

\tilde{R}_{12} = R_{12} + T_{12}~T_{21}~R_{23}~e^{2ik_{z2}(d_2 - d_1)} \left[ 1 + R_{21}~R_{23}~e^{2ik_{z2}(d_2-d_1)} + R_{21}^2~R_{23}^2~e^{4ik_{z2}(d_2-d_1)} + \dots \right] ~. $$ This equation can be interpreted as shown in Figure 3. However, sometimes the series may not converge at it is preferable to use (9) for computations.

We can now generalize the above results to a medium with $$N$$ layers (see Figure 4 for a schematic of the situation). If one additional layer is added, then we only need to replace $$R_{23}$$ in equation (9) with $$\tilde{R}_{23}$$.

Therefore, in general, the wave in the $$j$$-th region takes the form

\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] ~. $$ For the last layer,

\tilde{R}_{N, N+1} = 0 ~. $$ For all other layers we get a recursion relation
 * $$ \text{(11)} \qquad

\tilde{R}_{j, j+1} = R_{j, j+1} + \cfrac{T_{j, j+1}~T_{j+1, j}~\tilde{R}_{j+1, j+2}~ \exp[2~i~k_{z, j+1}~(d_{j+1} - d_j)]} {1 - R_{j+1, j}~\tilde{R}_{j+1, j+2}~ \exp[2~i~k_{z, j+1}~(d_{j+1}-d_j)]} ~. $$ Recall from equation (4) that
 * $$ \text{(12)} \qquad

T_{ij} = 1 + R_{ij} \qquad \text{and} \qquad R_{ij} = - R_{ji} ~. $$ Using equation (12), equation (11) simplifies to
 * $$ \text{(13)} \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

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 transverse electric waves.

In the next lecture we will take the continuum limit of these equations and derive equations for the effective reflection coefficient of a smoothly graded multilayered medium with a few isolated jumps.


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Solar System, interactive

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The Solar System An interactive (audio-picture) study module for young students

When the Sun sets and people go to bed, if there are no clouds, we can see the stars in the sky. Thousands of lights hang over the Earth. Come with us on a journey through our Solar System. Let’s touch the stars. Click on the sky and enjoy this journey through space.

title: Solar System, interactive