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h1. Optical Properties
h2. Maxwell's Equations

In electromagnetics, Maxwell's equations are a set of four equations, compiled by James Clerk Maxwell, that describe the behavior of both the electric and magnetic fields, as well as their interactions with matter.

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Maxwell's four equations express, respectively, how electric charges produce electric fields (Gauss' law), the experimental absence of magnetic monopoles, how currents and changing electric fields produce magnetic fields (the Ampere-Maxwell law), and how changing magnetic fields produce electric fields (Faraday's law of induction).  Refer to [article in wikipedia|http://en.wikipedia.org/wiki/Maxwell_equations] to see in depth discussion.


||! Name! Differential form! Integral form
| Gauss's law: How electric charges produce electric fields |
| <math>\vec{\nabla} \cdot \vec{D} = \rho</math>  |
| <math>\oint_S  \vec{D} \cdot \mathrm{d}\vec{A} = \int_V \rho\, \mathrm{d}V</math> |
| Gauss' law for magnetism <br /> (absence of magnetic monopoles): |
| <math>\vec{\nabla} \cdot \vec{B} = 0</math> |
| <math>\oint_S \vec{B} \cdot \mathrm{d}\vec{A} = 0</math> |
| Faraday's law of induction: How currents and changing electric fields produce magnetic fields |
| <math>\vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}} {\partial t}</math>  |
| <math>\oint_C \vec{E} \cdot \mathrm{d}\vec{l} - \oint_C \vec{B} \times \vec{v} \cdot \mathrm{d}\vec{l} = - \ { \mathrm{d} \over \mathrm{d}t }   \int_S   \vec{B} \cdot \mathrm{d}\vec{A}</math> |
| Amp��re's law<br /> (with Maxwell's extension): How changing magnetic fields produce electric fields |
| <math>\vec{\nabla} \times \vec{H} = \vec{J} + \frac{\partial \vec{D}} {\partial t}</math> |
| <math>\oint_C \vec{H} \cdot \mathrm{d}\vec{l} = \int_S \vec{J} \cdot \mathrm{d} \vec{A} + \int_S \frac{\partial\vec{D}}{\partial t} \cdot \mathrm{d} \vec{A}</math> |

h2. Index of Refraction

The refractive index (or index of refraction) of a material is the factor by which the phase velocity of electromagnetic radiation is slowed in that material, relative to its velocity in a vacuum. It is usually given the symbol n, and defined for a material by:

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<math> n=\sqrt{\epsilon_r\mu_r}</math>

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where ��r is the material's relative permittivity, and ��r is its relative permeability. For a non-magnetic material, <math>\mu_r</math> is very close to 1, therefore n is approximately <math>\sqrt{\epsilon_r}</math>.

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The phase velocity is defined as the rate at which the crests of the waveform propagate; that is, the rate at which the phase of the waveform is moving. The group velocity is the rate that the envelope of the waveform is propagating; that is, the rate of variation of the amplitude of the waveform. It is the group velocity that (almost always) represents the rate that information (and energy) may be transmitted by the wave, for example the velocity at which a pulse of light travels down an optical fiber.

h2. Stoke's Theorem

h2. Divergence Theorem

h2. Wave Equation

The electromagnetic wave equation is a second-order partial differential equation that describes the propagation of electromagnetic waves through a medium or in a vacuum. The homogeneous form of the equation, written in terms of either the electric field <math>E</math> or the magnetic field <math>H</math>, takes the form:

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<math>(\nabla^2-{1\over c^2}{\partial^2 \over \partial t^2})\mathbf{E} = 0</math>

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<math>(\nabla^2-{1 \over c^2}{\partial^2 \over \partial t^2} ) \mathbf{H} = 0</math>

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</center>

where c is the speed of light in the medium. In a vacuum, <math>c = 2.998 x 10^8 \frac{m}{s}</math>, which is the speed of light in free space.  The electromagnetic wave equation is derived from Maxwell's equations.

h3. Derivation

To obtain electromagnetic waves in a vacuum, note that Maxwell's equations (SI units) in a vacuum are

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<math> \nabla \cdot \mathbf{E} = 0</math>

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<math> \nabla \times \mathbf{E} = -\mu_o \frac{\partial \mathbf{H}} {\partial t}</math>

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<math> \nabla \cdot \mathbf{H} = 0</math>

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<math> \nabla \times \mathbf{H} =\varepsilon_o \frac{ \partial \mathbf{E}} {\partial t}</math>

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Take the curl of the curl equations we obtain

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<math> \nabla \times \nabla \times \mathbf{E} = -\mu_o \frac{\partial } {\partial t} \nabla \times \mathbf{H} = -\mu_o \varepsilon_o \frac{\partial^2 \mathbf{E} }  {\partial t^2} </math>

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<math> \nabla \times \nabla \times \mathbf{H} = \varepsilon_o \frac{\partial } {\partial t} \nabla \times \mathbf{E} = -\mu_o \varepsilon_o \frac{\partial^2 \mathbf{H} }  {\partial t^2} </math>

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Note the vector identity

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<math>\nabla \times \left( \nabla \times \mathbf{V} \right) = \nabla \left( \nabla \cdot \mathbf{V} \right) - \nabla^2 \mathbf{V}</math>

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where <math> \mathbf{V} </math> is any vector function of space.  Recover the wave equations

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<math> {\partial^2 \mathbf{E} \over \partial t^2} \ - \  c^2 \cdot \nabla^2 \mathbf{E}  \ \ = \ \ 0</math>

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<math> {\partial^2 \mathbf{H} \over \partial t^2} \ - \  c^2 \cdot \nabla^2 \mathbf{H}  \ \ = \ \ 0</math>

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</center>

where <math>c</math> is the speed of light in free space.

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<math>c = { 1 \over \sqrt{ \mu_o \varepsilon_o } } = 2.998 \times 10^8 \frac{m}{s} </math> 

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</center>

h3. Solutions

See [article in wikipedia|http://en.wikipedia.org/wiki/Electromagnetic_wave_equation].
h2. EM Fields at Interfaces
h3. Boundary Conditions

The tangential components of <math>\vec B / \mu</math> can be discontinuous at an interface if there is a layer of surface current confined there.  In the equation below, <math>\mu_1</math> and <math>\mu_2</math> are the magnetic permeabilities of the two media and <math>\vec j_s</math> is the surface current density.  If both materials behave as perfect insulators, no surface currents can exist, and the tangential components of \vec B / \mu are continuous across the interface.

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<math>\frac{1}{\mu_1} \vec B_1 \times \vec n - \frac{1}{\mu_2} \vec B_2 \times \vec n = \vec j_s</math>

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</center>

The tangential components of the electric field are continuous at an interface.

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<math>\vec E_1 \times \vec n = \vec E_2 \times \vec n</math>

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</center>


h2. Transfer Matrix

If the electric and magnetic fields in the <math>x-y</math> plane at <math>z=0</math> are known, use Maxwell's equations at a fixed frequency to integrate the wavefield and find electric and magnetic fields in the <math>x-y</math> plane at <math>z=c</math>.  Assume that there is zero divergence of <math>\vec B</math> and <math>D</math>.  It is only necessary to know two components of each field.

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<math>\vec F(z = 0) = \left [E_x (z=0), E_y (z=0), H_x (z=0), E_y (z=0) \right]</math>

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An equation below defines the transfer matrix, <math>\vec T</math>.

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<math>\vec F (z = c) = \vec T (c, 0) \vec F(z=0)</math>

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The transfer matrix defines how waves cross a slab of material.  It is closely related to the transmission and reflection coefficients of the slab.  If the slab is infinitely repeated thourout the material in a periodic array, it is possible to impose the Bloch condition and calculate the band structure, <math>K(\omega)</math>, from the eigenvalues of <math>T</math>.

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<math>e^{iKc} \vec{F}(z=0) = \vec F (z=c)</math>

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<math>e^{iKc} \vec{F}(z=0) = \vec T (c,0) \vec F (z=0)</math>

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</center>

** Reflection
** Multiplicity of interfaces
** Periodic Medium
* Photonic Band Diagrams
* Damped Harmonic Oscillator
h2. References
* [Transfer Matrix Techniques for Electromagnetic Waves|http://www.sst.ph.ic.ac.uk/photonics/pdf/crete95.pdf]

h2. Index of Refraction, Band Gap, and Wavelength

Index of refraction measure of how much light interacts with material.  Closer to one, more transparent.

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Check graph in notes.  <math>Ge</math> above <math>Si</math> above <math>SiO</math>.  Index of refraction of oxide less than index of non-oxide.  The insulator lies below semiconductor materials.  <math>Ge</math> smaller band gap.  Band gap goes down with increasing index of refraction.

h2. Definition and Examples of Isotropic, Uniaxial and Biaxial Optical Materials

Define coordinate axes (principal axes) such that the dielectric tensor is diagonal.

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<math>\epsilon = \begin{bmatrix} \epsilon_1& 0& 0\\ 0& \epsilon_2& 0\\ 0& 0& \epsilon_3 \end{bmatrix}</math>

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</center>

Isotropic
* Cubic or amorphous materials
* <math>\epsilon_1 = \epsilon_2 = \epsilon_3</math>
* <math>Y_3Al_5O_12</math> (Yttrium Aluminum Garnet)

Uniaxial
* Trigonal, tetragonal, hexagonal
* <math>\epsilon_1 \ne \epsilon_2 = \epsilon_3</math>
* <math>LiNbO_3</math>

Biaxial
* Triclinic, monoclinic, orthorhombic
* <math>\epsilon_1 \ne \epsilon_2 \ne \epsilon_3</math>

h2. Maxwell's Equations and the Constitutive Equation



||! Name! Differential form! Integral form
| Gauss's law: How electric charges produce electric fields |
| <math>\vec{\nabla} \cdot \vec{D} = \rho</math>  |
| <math>\oint_S  \vec{D} \cdot \mathrm{d}\vec{A} = \int_V \rho\, \mathrm{d}V</math> |
| Gauss' law for magnetism <br /> (absence of magnetic monopoles): |
| <math>\vec{\nabla} \cdot \vec{B} = 0</math> |
| <math>\oint_S \vec{B} \cdot \mathrm{d}\vec{A} = 0</math> |
| Faraday's law of induction: How changing magnetic fields produce electric fields |
| <math>\vec{\nabla} \times \vec{E} = -\frac{\partial \vec{B}} {\partial t}</math>  |
| <math>\oint_C \vec{E} \cdot \mathrm{d}\vec{l} - \oint_C \vec{B} \times \vec{v} \cdot \mathrm{d}\vec{l} = - \ { \mathrm{d} \over \mathrm{d}t }   \int_S   \vec{B} \cdot \mathrm{d}\vec{A}</math> |
| Amp��re's law<br /> (with Maxwell's extension): How currents and changing electric fields produce magnetic fields |
| <math>\vec{\nabla} \times \vec{H} = \vec{J} + \frac{\partial \vec{D}} {\partial t}</math> |
| <math>\oint_C \vec{H} \cdot \mathrm{d}\vec{l} = \int_S \vec{J} \cdot \mathrm{d} \vec{A} + \int_S \frac{\partial\vec{D}}{\partial t} \cdot \mathrm{d} \vec{A}</math> |


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<math>\vec D = \epsilon \vec E = \vec E + 4 \pi \vec P</math>

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<math>\vec D = \mu \vec H = \vec H + 4 \pi \vec M</math>

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<math>\vec J = \sigma \vec E</math>

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</center>

h2. The Wave Equation and the Conditions for its Applicability

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<math> {\partial^2 \mathbf{E} \over \partial t^2} \ - \  c^2 \cdot \nabla^2 \mathbf{E}=0</math>

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<math> {\partial^2 \mathbf{H} \over \partial t^2} \ - \  c^2 \cdot \nabla^2 \mathbf{H}=0</math>

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</center>

Solutions to the wave equations can take on different forms.  Examples include plane waves, cylindrical waves, or spherical waves.  These vector equations are satisfied by the following solutions.

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<math>\vec E(x,y,z) = \vec E_o e^{i \omega t - \vec k r}</math>

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<math>\frac{c}{\sqrt{\mu \epsilon}} |\vec k| = \omega</math>

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</center>

h2. The Properties of Plane Wave Solutions, Wave Vector, Transversality

h3. Phase velocity

The solution is a complex valued function.  An expression of the phase is below.  Consider the fronts where the phase is constant.

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<math>\vec E (x,y,z) = \vec E_o e^{i \omega t - \vec k r}</math>

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<math>\phi = \omega t - \vec k \cdot \vec r</math>

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<math>\omega t - \vec k \cdot \vec r = const</math>

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</center>

Below is an expression of velocity of the constant fronts.

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<math>\vec v_{phase} = \frac{\omega}{|k|} \hat k</math>

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<math>\vec v_{phase} = \frac{c}{\sqrt{\epsilon \mu}}</math>

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The direction is defined by the wavevector's direction.  The phase velocity is changed in a material compared with its value in vacuum.

h3. Transversality of E-M Fields

There are three Cartesian terms of the solutions of the wave equations.

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<math>\vec E(x,y,z) = E_x \hat x + E_y \hat y + E_z \hat z</math>

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Consider a medium that is homogeneous and charge free.  Two equations below imply the transversality of E-M fields.

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<math>\vec \nabla \cdot \vec E = 0</math>

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<math>\vec \nabla \cdot \vec H = 0</math>

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<math>\vec E \cdot \vec k = 0</math>

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<math>\vec H \cdot \vec k = 0</math>

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</center>

h2. Dispersion Relations

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<math>\epsilon ( \omega ) \omega^2 = c^2 k^2</math>

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The term <math>k</math> is the wavevector or propagation vector and it can be related to the wavelength, as shown below.

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<math>|k| = \frac{2 \pi}{\lambda}</math>

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<math>|k| = \frac{n}{c} \lambda</math>

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</center>

h2. How are Fields on one Side of an Interface Related to the Fields on the Other Side

To calculate the behavior of the normal components of the field with respect to the interface unit vector Gauss' theorem

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<math>\int_{volume} \vec \nabla \cdot \vec B dv = \int_{surface} \vec B \cdot \hat n dS = 0</math>

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<math>\int_{volume} \vec \nabla \cdot \vec D dv = \int_{surface} \vec D \cdot \hat n dS = 4 \pi \int_{volume} \rho dv</math>

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</center>

* <math>\hat n \cdot ( \vec B_2 - \vec B_1 ) = 0</math>
** The normal component of the magnetic induction is continuous across the surface of discontinuity
* <math>\hat n \cdot ( \vec D_2 - \vec D_1 ) = 4 \pi \rho</math>
** In the presence of a layer of surface charge, the normal component of the electric displacement changes abruptly by an amount equal to <math>4 \pi \sigma</math>.  The tangential components are found through the application of Stokes Theorem.

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<math>\int_{surface} \vec \nabla \times \vec E \cdot \hat n dS = \int_{line} \vec E \cdot d \vec r</math>

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</center>

* <math>\hat n \times (\vec E_2 - \vec E_1) = 0</math>
** The tangential component of the electric field is continuous across the interface.
* <math>\vec n \times (\vec H_2 - \vec H_1 = \frac{4 \pi}{c} \vec K</math>
** In the presence of a surface current the tangential component of the magnetic induction changes abruptly.

h2. What is the Energy Law of EM Fields, Significance of Poynting Vector

h3. Energy Law of Electromagnetic Fields

Electromagnetic theory interprets the light intensity as the energy flux of the field.  The presence of a field implies the existence of energy associated with that field.

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<math>\vec E \cdot \vec \nabla \times \vec H - \vec H \cdot \vec \nabla \times \vec E = \frac{4 \pi}{c} \vec J \cdot \vec E + \frac{1}{c} \vec E \cdot \frac{\partial \vec D}{\partial t} + \frac{1}{c} \vec H \cdot \frac{\partial \vec B}{\partial t}</math>

<br>

<math>\int \vec J \cdot \vec E dv + \frac{\partial}{\partial t} \int ( \vec E \cdot \vec D + \vec H \cdot \vec B ) dv + \int ( \vec E \times \vec H ) \cdot \hat n dS = 0</math>

<br>

<math>\int \vec J \cdot \vec E dv + \frac{\partial}{\partial t} \int ( \vec E \cdot \vec D + \vec H \cdot \vec B ) dv + \int ( \vec E \times \vec H ) \cdot \hat n dS = - \vec \nabla \cdot ( \vec E \times \vec H)</math>

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<math>\frac{4 \pi}{c} \vec J \cdot \vec E + \frac{1}{c} \vec E \cdot \frac{\partial \vec D}{\partial t} + \frac{1}{c} \vec H \cdot \frac{\partial \vec B}{\partial t} + \vec \nabla \cdot ( \vec E \times \vec H) = 0</math>

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Apply Gauss's theorem.

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<math>\int \frac{4 \pi}{c} \vec J \cdot \vec E dv + \int \left ( \frac{1}{c} \vec E \cdot \frac{\partial \vec D}{\partial t} \right ) dv + \frac{1}{c} \vec H \cdot \frac{\partial \vec B}{\partial t} + \int \left ( \vec E \times \vec H \right ) \cdot \hat n dS = 0</math>

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</center>

Use materials equations

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<math>\frac{1}{4 \pi} \vec E \cdot \frac{\partial \vec D}{\partial t} = \frac{1}{4 \pi} \vec E \cdot \frac{\partial \epsilon \vec E}{\partial t}</math>

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<math>\frac{1}{4 \pi} \vec E \cdot \frac{\partial \vec D}{\partial t} = \frac{1}{8 \pi} \frac{\partial \epsilon \vec E^2}{\partial t}</math>

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<math>\frac{1}{4 \pi} \vec E \cdot \frac{\partial \vec D}{\partial t} = \frac{1}{8 \pi} \frac{\partial (\vec E \cdot \vec D)}{\partial t}</math>

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<math>\frac{1}{4 \pi} \cdot \frac{\partial \vec B}{\partial t} = \frac{1}{8 \pi} \frac{\partial (\vec H \cdot \vec B)}{\partial t}</math>

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h3. Poynting Vector

The Poynting vector describes the energy flux <math>\left ( \frac{J}{m^2s} \right )</math> of an electromagnetic field. It is named after its inventor John Henry Poynting. Oliver Heaviside independently co-discovered the Poynting vector.

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It points in the direction of energy flow and its magnitude is the power per unit area crossing a surface which is normal to it. (The fact that it points perhaps contributes to the frequency with which its name is misspelled.) It is derived by considering the conservation of energy and taking into account that the magnetic field can do no work. It is given the symbol <math>S</math> (in bold because it is a vector) and, in SI units, is given by:

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<math>\mathbf{S} = \vec{E} \times \vec{H} = \frac{1}{\mu} \vec{E} \times \vec{B}</math>

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where <math>E</math> is the electric field, <math>H</math> and <math>B</math> are the magnetic field and magnetic flux density respectively, and <math>\mu</math> is the permeability of the surrounding medium.

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For example, the Poynting vector near an ideally conducting wire is parallel to the wire axis - so electric energy is flowing in space outside of the wire. The Poynting vector becomes tilted toward wire for a resistive wire, indicating that energy flows from the e/m field into the wire, producing resistive Joule heating in the wire.

h3. Snell's Law and its Derivation

See below

h3. Conserved E-M Quantities at Interfaces (Parallel Component of Wave Vector, Frequency) and the Consequence of their Conservation

Consider a wave incident from the left onto an interface.

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<math>\mbox{Incident Wave}</math>

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<math>\vec E_i e^{\omega t - i \vec k_i \cdot \vec r</math>

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<math>\mbox{Reflected Wave}</math>

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<math>\vec E_r e^{\omega t - i \vec k_r \cdot \vec r</math>

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<math>\mbox{Transmitted Wave}</math>

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<math>\vec E_t e^{\omega t - i \vec k_t \cdot \vec r</math>

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Consider a homogeneous medium and dispersion relations.

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<math>|\vec k_i| = |\vec k_r| = \frac{\omega n_1}{c}</math>

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<math>\vec k_t = \frac{\omega n_2}{c}</math>

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</center>

Focus on the interface plane, <math>x=0</math>.  Boundary conditions dictate relations between the field components on both sides of the interface.  The phase of the fields must be equal.

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<math>( \vec k_1 \cdot \vec r )_{x=0} = ( \vec k'_1 \cdot \vec r )_{x=0} = ( \vec k_2 \cdot \vec r )_{x=0}</math>

<br>

<math>( \vec k_{1y} y + \vec k_{1z} z) = ( \vec k'_{1y} y + \vec k'_{1z} z) = ( \vec k_{2y} y + \vec k_{2z} z)</math>

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<math>k_{1y} = k'_{1y} = k_{2y}</math>

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<math>k_{1z} = k'_{1z} = k_{2z}</math>

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</center>

Consider an arbitrary vector that lies in the <math>z-y</math> plane.  in this case, the following is true.

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<math>\vec r = (x=0,y,z) = \vec r_{z-y}</math>

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<math>(\vec k_{1t} \cdot \vec r_t ) = (\vec k'_{1t} \cdot \vec r_t ) = (\vec k_{2t} \cdot \vec r_t )</math>

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</center>

The vectors <math>\vec k_1</math>, <math>\vec k'_1</math>, <math>\vec k_2</math> all lie in a place called the plane of incidence.  Orient the coordinate system so that the plane of incidence coincides with the <math>x-z</math> plane.

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<math>\vec E = \vec E e^{i(\omega t - k_x x - k_z z)}</math>

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</center>

The tangential components of the wavevector are all identical regardless of the medium that they are in and regardless of whether a medium is lossless or absorbing.

<center>

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<math>k_{1z} = k_{1'z} = k_{2z} = \beta</math>

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<math>|\vec k_1| = |\vec k'_1| = n_1 \vec {\omega}{c}</math>

<br>

<math>|\vec k_2| = n_2 \vec {\omega}{c}</math>

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<math>|\vec k_i| \sin \theta_1 = |\vec k_t| \sin \theta_2</math>

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<math>\frac{\omega n_1}{c} \sin \theta_1 = \frac{\omega n_2}{c} \sin \theta_2</math>

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</center>

This leads to two important consequences.
* Angle of reflection is equal to the angle of incidence

<center>

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<math>\theta_i = \theta_r</math>

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</center>

* Snell's law:

<center>

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<math>k_{1z} = |\vec k_1| \sin \theta_1 = n_1 \frac{\omega}{c} \sin \theta_1</math>

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<math>k_{2z} = |\vec k_2| \sin \theta_2 = n_2 \frac{\omega}{c} \sin \theta_2</math>

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<math>n_1 \sin \theta_1 = n_2 \sin \theta_2</math>

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</center>

h3. Analysis of the Reflection from a Dielectric Slab (Reflectivity for s and p Polarized Waves, Field Directions, Poynting Vector, Magnitude of Wavevectors)

Find out the direction of fields from the relation below.

<center>

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<math>\vec S = \frac{c}{4 \pi} \vec E \times \vec H</math>

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</center>

Express the fields on both sides of the interface in the following way.

<center>

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<math>\mbox{x<0}</math>

<br>

<math>\vec E = \vec E_1 e^{-i \vec k_1 \vec r} + \vec E'_1 e^{-i \vec k'_1 \vec r}</math>

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<math>\mbox{x>0}</math>

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<math>\vec E = \vec E_2 e^{-i \vec k_2 \vec r} + \vec E'_2 e^{-i \vec k'_2 \vec r}</math>

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</center>

Seprate the electric field into components that are either in the plane of incidence (p) or perpendicular to it (s), in the <math>y</math> direction.  Begin with the s-polarized electric field.

<center>

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<math>\vec E_s (x, y, z) = (0, E_s (x, z), 0 )</math>

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</center>

Find the electric field in the <math>i</math>-th medium.

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<math>\vec E_{1s} = \hat y \left ( E_{1s} e^{- i k_{ix} x} + E'_{1s} e^{i k_{ix} x} \right ) e^{i \omega t - i \beta z}</math>

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</center>

The dependence of the field on the variables <math>x</math> and <math>z</math> is a result of the selection of the <math>x-z</math> plane as the plane of incidence.  The components of <math>k</math> are in the <math>x</math> and <math>z</math> directions.  Therefore, there is no dependence on <math>y</math>.

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</p>

The component is tangential to the interface and is continuous across the interface.

<center>

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<math>E_{1y} + E'_{1y} = E_{2y} + E'_{2y}</math>

<br>

</center>

Recall the boundary condition on the components of <math>H</math> that are tangential to the interface.  Consider an insulating material in whch surface currents are not supported.  The tangential component is continuous.

<center>

<br>

<math>\hat n \times ( \vec H_2 - \vec H_1 ) = \frac{4 \pi}{c} \vec K</math>

<br>

<math>H_{1z} + H'_{1z} = H_{2z} + H'_{2z}</math>

<br>

<math>\vec \nabla \times \vec E + \frac{1}{c} \frac{\partial \vec B}{\partial t} = 0</math>

<br>

<math>\vec B = \mu \vec H</math>

<br>

</center>

Assume a harmonic time variation of <math>H</math> of the form <math>e^{i \omega t}</math>.  Assume that the material is non-magnetic, <math>\mu=1</math>, and use the angle convention in the notes.

<center>

<br>

<math>\vec \nabla \times \vec E + \frac{i \omega \mu}{c} \vec H = 0</math>

<br>

<math>H_z = \frac{c}{i \omega \mu} \frac{\partial}{\partial x} E_y</math>

<br>

<math>H_z = \frac{\partial}{\partial x} E_1 e^{-i (k_x x + k_z z)}</math>

<br>

<math>H_z = -k_x \frac{c}{i \omega \mu} E_1 e^{-i (k_x x + k_z z)}</math>

<br>

<math>k_{1x} = |k| \cos \theta_1 = \frac{n_1 \omega}{c} \cos \theta_1</math>

<br>

<math>H_{1z} = -n_1 \cos \theta_1 E_1 e^{-i(k_x x + k_z z)}</math>

<br>

<math>H'_{1z} = n_1 \cos \theta_1 E'_1 e^{-i(k'_x x + k_z z)}</math>

<br>

</center>

This leads to two equations for the unknown field amplitudes.

<center>

<br>

<math>n_1 (E_{1y} - E'_{1y} \cos \theta_1 )= n_2 (E_{2y} - E'_{2y} \cos \theta_2 )</math>

<br>

</center>

Write in matrix form.  These matrices are dynamic matrices of s polarized wave.

<center>

<br>

<math>\begin{bmatrix} 1 & 1\\ n_1 \cos \theta_1 & -n_1 \cos \theta_1\end{bmatrix} \begin{bmatrix} E_{1y}\\ E'_{1y}\end{bmatrix} = \begin{bmatrix} 1 & 1\\ n_2 \cos \theta_2 & -n_2 \cos \theta_2\end{bmatrix} \begin{bmatrix} E_{2y}\\ E'_{2y}\end{bmatrix}</math>

<br>

<math>D_2^{-1} D_1 = D_{12}</math>

<br>

<math>D_{12} \begin{bmatrix} E_{1y}\\ E'_{1y}\end{bmatrix} =  \begin{bmatrix} E_{2y}\\ E'_{2y}\end{bmatrix}</math>

<br>

</center>

Consider where the wave is incident from medium 1 moving from left to right and set <math>E'_2 = 0</math>.  The reflection and transmission coefficients.

<center>

<br>

<math>r_s \left ( \frac{E'_1}{E_1} \right ) = \frac{n_1 \cos \theta_1 - n_2 \cos \theta_2}{n_1 \cos \theta_1 + n_2 \cos \theta_2</math>

<br>

<math>r_s \left ( \frac{E_2}{E_1} \right ) = \frac{2n_1 \cos \theta_1}{n_1 \cos \theta_1 + n_2 \cos \theta_2</math>

<br>

<math>R = |r_s|^2</math>

<br>

</center>

The last term above is the reflectivity.

h3. How to Derive the Transfer Matrix that Describes the Field Transformation at Interfaces

Consider medium with the following index of refraction.  An objective is to relate any two pairs of amplitudes to any other two pairs.

<center>

<br>

<table cellpadding=10>
<tr>

<td>
<math>x<0</math>
</td>

<td>
<math>0<x<d</math>
</td>

<td>
<math>d<x</math>
</td>

</tr>

<tr>

<td>
<math>n_1</math>
</td>

<td>
<math>n_2</math>
</td>

<td>
<math>n_3</math>
</td>

</tr>

</table>

</center>

The <math>z</math> component of the wavevector does not change throughout the problem.  It leads to a simple dependence on the <math>z</math> direction evolution.

<center>

<br>

<math>\vec E = \vec E(x) e^{i(\omega t - \beta z)}</math>

<br>

</center>

Define amplitudes on both sides of each interface for an s polarized field.

<center>

<br>

<math>x<0</math>

<br>

<math>E_1 e^{-ik_{1x} x + E'_1 e^{-ik_{1x} x</math>

<br>

<math>0<x<d</math>

<br>

<math>E_2 e^{-ik_{2x} x + E'_2 e^{-ik_{2x} x</math>

<br>

<math>d<x</math>

<br>

<math>E_3 e^{-ik_{3x} x + E'_3 e^{-ik_{3x} x</math>

<br>

</center>

The propagation matrix captures the effect of propagation through a medium of index <math>n_2</math> and thickness <math>d</math>.


<center>

<br>

<math>P_2 = \begin{bmatrix} e^{ik_{2x}d}& 0\\ 0& e^{ik_{2x}d}\end{bmatrix}</math>

<br>

<math>D_{23} P_2 D_{12} \begin{bmatrix} E_1\\ E'_1\end{bmatrix} = \begin{bmatrix} E_3\\ E'_3\end{bmatrix}</math>

<br>

</center>

h3. Transfer Matrix Approach to Solving Periodic Systems of Dielectric Materials

<center>

<br>

<table cellpadding=10>
<tr>

<td>
<math>na<x<na+d_2</math>
</td>

<td>
<math>na-d_1<x<na</math>
</td>

</tr>

<tr>

<td>
<center>
<math>n_1</math>
</center>
</td>

<td>
<center>
<math>n_2</math>
</center>
</td>

</tr>

</table>

<br>

<math>n(x+na) = n(x)</math>

<br>

</center>

The solution in each medium is of the form below.

<center>

<br>

<math>E = E(x) e^{i(\omega t - \beta z)}</math>

<br>

</center>

Focus on the <math>x</math> dependence of the solutions.

<center>

<br>

<math>na -d_1 <x <na</math>

<br>

<math>E(x) = A_n e^{-ik_{1x} (x-na)} + B_n e^{ik_{1x} (x-na)}</math>

<br>

<math>(na-1) -d_1 <x <na - d_1</math>

<br>

<math>E(x) = A_n e^{-ik_{2x} (x-na+d_1)} + B_n e^{ik_{2x} (x-na+d_1)}</math>

<br>

</center>

The field notation has been changed in order to simplify the labeling of the fields in these problems which involve a multiplicity of interfaces.

<center>

<br>

<math>k_{1x} = \sqrt{ \left ( \frac{ \omega n_1}{c} \right )^2 - k_{1x}^2} = \sqrt{ \left ( \frac{ \omega n_1}{c} \right )^2 - \beta^2} = \frac{\omega n_1}{c} \cos \theta_1</math>

<br>

<math>k_{2x} = \sqrt{ \left ( \frac{ \omega n_2}{c} \right )^2 - \beta^2} = \frac{\omega n_2}{c} \cos \theta_2</math>

<br>

</center>

Dynamic matrices are below.

<center>

<br>

<math>D_l^s = \begin{bmatrix} 1& 1\\ n_1 \cos \theta_1& -n_1 \cos \theta_1\end{bmatrix}</math>

<br>

<math>D_{l-1,l} = D_l^{-1}D_{l-1}</math>

<br>

</center>

Consider the solution form.  Relate the fields across the interface.

<center>

<br>

<math>E(x)|_{x=na^{(1)}} = A_n + B_n</math>

<br>

<math>E(x)|_{x=na-d_1^{(1)}} = A_ne^{-ik_{1x}(na-d_1-na)} + B_ne^{-ik_{1x}(na-d_1-na)}</math>

<br>

<math>E(x)|_{x=na-d_1^{(1)}} = A_ne^{ik_{1x}d_1} + B_ne^{-ik_{1x}d_1}</math>

<br>

<math>E(x)|_{x=na-d_1^{(1)}} = C_n + D_n</math>

<br>

<math>P_1 = \begin{bmatrix} e^{ik_{1x}d_1} & 0\\ 0 & e^{-ik_{1x}d_1}\end{bmatrix}</math>

<br>

<math>D_2 \begin{bmatrix} C_n\\ D_n\end{bmatrix} = D_1 P_1 \begin{bmatrix} A_n\\ B_n\end{bmatrix}</math>

<br>

<math>\begin{bmatrix} C_n\\ D_n\end{bmatrix} = D_2^{-1} D_1 P_1 \begin{bmatrix} A_n\\ B_n\end{bmatrix}</math>

<br>

</center>

The solution brings one to the LHS of the <math>x=na-d_1</math> interface (i.e. in medium 2).  Propagate the wave in medium 2 back to the RHS of interface <math>x = (n-1)a</math>.

<center>

<br>

<math>E(x)|_{x=na-d_1^{(1)} = C_n +D_n</math>

<br>

<math>E(x)|_{x=(n-1)d^{(2)} = C_ne^{ik_{2x}d_2 +D_ne^{-ik_{2x}d_2</math>

<br>

<math>P_2 = \begin{bmatrix} e^{ik_{2x}d_2} & 0\\ 0 & e^{-ik_{2x}d_2}\end{bmatrix}</math>

<br>

</center>

Match fields across the interface to extract <math>A_{n-1}</math> and <math>B_{n-1}</math>.


<center>

<br>

<math>E(x)|_{x=na-d^{(2)} = C_ne^{ik_{2x}d_2 +D_ne^{-ik_{2x}d_2</math>

<br>

<math>E(x)|_{x=na-d^{(1)} = A_{n-1} + B_{n-1}</math>

<br>

<math>D_1 \begin{bmatrix} A_{n-1}\\ B_{n-1}\end{bmatrix} = D_2 P_2 \begin{bmatrix} C_{n}\\ D_{n}\end{bmatrix}</math>

<br>

<math>\begin{bmatrix} A_{n-1}\\ B_{n-1}\end{bmatrix} = D_1^{-1} D_2 P_2 \begin{bmatrix} C_{n}\\ D_{n}\end{bmatrix}</math>

<br>

<math>\begin{bmatrix} A_{n-1}\\ B_{n-1}\end{bmatrix} = D_1^{-1} D_2 P_2 D_2^{-1} D_1 P_1 \begin{bmatrix} A_{n}\\ B_{n}\end{bmatrix}</math>

<br>

</center>

h3. Eigenvalue Problem and its Solutions

The following is true from Bloch's Theorem.

<center>

<br>

<math>\begin{bmatrix} A_{n-1}\\ B_{n-1}\end{bmatrix} = e^{iKa} \begin{bmatrix} A_{n}\\ B_{n}\end{bmatrix}</math>

<br>

<math>D_1^{-1} D_2 P_2 D_2^{-1} D_1 P_1 \begin{bmatrix} A_{n}\\ B_{n}\end{bmatrix} = e^{iKa} \begin{bmatrix} A_{n}\\ B_{n}\end{bmatrix}</math>

<br>

<math>D_1^{-1} D_2 P_2 D_2^{-1} D_1 P_1 = M</math>

<br>

<math>M = \begin{bmatrix} M_{11} & M_{12}\\ M_{21} & M_{22}\end{bmatrix}</math>

<br>

</center>

Each one of the matrix elements depends on <math>\omega</math> and <math>\beta</math>.  The eigenvalues of the 2x2 matrix are below.

<center>

<br>

<math>e^{iKa} = \frac{1}{2} (M_{11} + M_{22} ) \pm \left ( \frac{1}{4} (M_{11} + M_{22})^2 - 1 \right )^{1/2}</math>

<br>

</center>

The equation defines the dispersion relations for the Bloch wavenumber <math>K</math> and <math>\omega</math> and <math>\beta</math>.  The determinant of the <math>M</math> matrix is unity, and e^{iKa} and e^{-iKa} are two eigenvalues.

<center>

<br>

<math>\frac{e^{iKa} + e^{-iKa}}{2} = \cos Ka</math>

<br>

<math>K(\beta, \omega) = \frac{1}{a} \cos^{-1} \left ( \frac{M_{11} + M_{22}}{2} \right )</math>

<br>

</center>

The eigenvectors are below

<center>

<br>

<math>\begin{bmatrix} A_{0}\\ B_{0}\end{bmatrix} = e^{iKa} \begin{bmatrix} M_{12}\\ e^{iKa} - M_{11}\end{bmatrix}</math>

<br>

</center>

h3. Photonic Band Diagrams and Bandgaps

Consider the case where <math>\beta=0</math> which corresponds to normal incidence.  The dispersion relation <math>\omega</math> vs. <math>K</math> can be expressed as below.

<center>

<br>

<math>\cos Ka = \frac{M_{11} + M_{22}}{2} = \cos k_1 d_1 \cos k_2 d_2 - \frac{1}{2} \left  ( \frac{n_2}{n_1} + \frac{n_1}{n_2} \right ) \sin k_1 d_1 \sin k_2 d_2</math>

<br>

<math>k_1 = \frac{\omega n_1}{c}</math>

<br>

<math>k_2 = \frac{\omega n_2}{c}</math>

<br>

</center>

The width of the normal incidence bandgap is given approximately by the expression below.

<center>

<br>

<math>\frac{\Delta \omega}{\omega_o} \approx \frac{2}{\pi} \frac{\Delta n}{n}</math>

<br>

</center>