# Materials Science and Engineering/Doctoral review questions/Daily Discussion Topics/01062008
## Relationship between Bond Strength and Electronic Properties

* As the bond strength increases, the electric polarizability decreases, and the dielectric constant.
* As the bond strength increases, the Einstein temperature increases, which corresponds to a lower heat capacity.

## Types of Polarizability
### Ferroelectric

The internal electric dipoles of a ferroelectric material are physically tied to the material lattice so anything that changes the physical lattice will change the strength of the dipoles and cause a current to flow into or out of the capacitor even without the presence of an external voltage across the capacitor. Two stimuli that will change the lattice dimensions of a material are force and temperature. The generation of a current in response to the application of a force to a capacitor is called piezoelectricity. The generation of current in response to a change in temperature is called pyroelectricity. There are two main types of ferroelectrics: displacive and order-disorder. The effect in barium titanate, a typical ferroelectric of the displacive type, is due to a polarization catastrophe, in which, if an ion is displaced from equilibrium slightly, the force from the local electric fields due to the ions in the crystal increase faster than the elastic-restoring forces. This leads to an asymmetrical shift in the equilibrium ion positions and hence to a permanent dipole moment. In an order-disorder ferroelectric, there is a dipole moment in each unit cell, but at high temperatures they are pointing in random directions. Upon lowering the temperature and going through the phase transition, the dipoles order, all pointing in the same direction within a domain.
Another important ferroelectric material is lead zirconate titanate.

## Dielectric Constant

The dielectric constant is an essential piece of information when designing capacitors, and in other circumstances where a material might be expected to introduce capacitance into a circuit. If a material with a high dielectric constant is placed in an electric field, the magnitude of that field will be measurably reduced within the volume of the dielectric. This fact is commonly used to increase the capacitance of a particular capacitor design.

### Heating with Radiation

The frequency of a microwave is tuned to the peak absorption of the imaginary component
There is maximum heat generation at the peak
The static relative permittivity of a medium is related to its static electric susceptibility, $\chi _{e}$, by
    $\varepsilon _{r}=1+\chi _{e},$

## Function of a Capacitor

A capacitor is used to separate charge
$Q=CV\;$
$C={\frac {k\epsilon _{o}A}{d}}$
$E={\frac {1}{2}}CV^{2}$

### Why is a ferroelectric material not useful as a capacitor in microelectronics?

The dielectric constant is dependent on temperature

## Maxwell's Equations
### Divergence and Curl

Consider the following two operations when recalling Maxwell's equations.
* Divergence: $\nabla {\overrightarrow {\mathrm {X} }}$
** Local flow outward from a point
* Curl: $\nabla \times {\overrightarrow {\mathrm {X} }}$
** Rotational flow in a vector field

### Maxwell's Equations

| Name | Differential form | Integral form |
| Gauss's law: | $\nabla \cdot \mathbf {E} ={\frac {\rho }{\epsilon _{0}}}$ | $\oint _{S}\mathbf {E} \cdot \mathrm {d} \mathbf {A} ={\frac {Q_{S}}{\epsilon _{0}}}$ |
| Gauss' law for magnetism (absence of magnetic monopoles): | $\nabla \cdot \mathbf {B} =0$ | $\oint _{S}\mathbf {B} \cdot \mathrm {d} \mathbf {A} =0$ |
| Faraday's law of induction: | $\nabla \times \mathbf {E} =-{\frac {\partial \mathbf {B} }{\partial t}}$ | $\oint _{\partial S}\mathbf {E} \cdot \mathrm {d} \mathbf {l} =-{\frac {d\Phi _{B,S}}{dt}}$ |
| Ampère's Circuital Law (with Maxwell's correction): | $\nabla \times \mathbf {B} =\mu _{0}\mathbf {J} +\mu _{0}\epsilon _{0}{\frac {\partial \mathbf {E} }{\partial t}}$ | $\oint _{\partial S}\mathbf {B} \cdot \mathrm {d} \mathbf {l} =\mu _{0}I_{S}+\mu _{0}\epsilon _{0}{\frac {d\Phi _{E,S}}{dt}}$ |

### Derivation of Wave Equation

Electromagnetic waves as a general phenomenon were predicted by the classical laws of electricity and magnetism, known as Maxwell's equations. If you inspect Maxwell's equations without sources (charges or currents) then you will find that, along with the possibility of nothing happening, the theory will also admit nontrivial solutions of changing electric and magnetic fields. Beginning with Maxwell's equations for a vacuum:
       $\nabla \cdot \mathbf {E} =0\qquad \qquad \qquad \ \ (1)$
       $\nabla \times \mathbf {E} =-{\frac {\partial }{\partial t}}\mathbf {B} \qquad \qquad (2)$
       $\nabla \cdot \mathbf {B} =0\qquad \qquad \qquad \ \ (3)$
       $\nabla \times \mathbf {B} =\mu _{0}\epsilon _{0}{\frac {\partial }{\partial t}}\mathbf {E} \qquad \ \ \ (4)$
    where
       $\nabla$ is a vector differential operator (see Del).
One solution,
       $\mathbf {E} =\mathbf {B} =\mathbf {0}$,
is trivial.
To see the more interesting one, we utilize vector identities, which work for any vector, as follows:
       $\nabla \times \left(\nabla \times \mathbf {A} \right)=\nabla \left(\nabla \cdot \mathbf {A} \right)-\nabla ^{2}\mathbf {A}$
To see how we can use this take the curl of equation (2):
       $\nabla \times \left(\nabla \times \mathbf {E} \right)=\nabla \times \left(-{\frac {\partial \mathbf {B} }{\partial t}}\right)\qquad \qquad \qquad \quad \ \ \ (5)\,$
Evaluating the left hand side:
       $\nabla \times \left(\nabla \times \mathbf {E} \right)=\nabla \left(\nabla \cdot \mathbf {E} \right)-\nabla ^{2}\mathbf {E} =-\nabla ^{2}\mathbf {E} \qquad \quad \ (6)\,$
    where we simplified the above by using equation (1).
Evaluate the right hand side:
       $\nabla \times \left(-{\frac {\partial \mathbf {B} }{\partial t}}\right)=-{\frac {\partial }{\partial t}}\left(\nabla \times \mathbf {B} \right)=-\mu _{0}\epsilon _{0}{\frac {\partial ^{2}}{\partial ^{2}t}}\mathbf {E} \qquad (7)$
Equations (6) and (7) are equal, so this results in a vector-valued differential equation for the electric field, namely
       $\nabla ^{2}\mathbf {E} =\mu _{0}\epsilon _{0}{\frac {\partial ^{2}}{\partial t^{2}}}\mathbf {E}$
Applying a similar pattern results in similar differential equation for the magnetic field:
       $\nabla ^{2}\mathbf {B} =\mu _{0}\epsilon _{0}{\frac {\partial ^{2}}{\partial t^{2}}}\mathbf {B}$.
These differential equations are equivalent to the wave equation:
       $\nabla ^{2}f={\frac {1}{c^{2}}}{\frac {\partial ^{2}f}{\partial t^{2}}}\,$
    where
       c is the speed of the wave and
       f describes a displacement
Or more simply:
       $\Box ^{2}f=0$
    where $\Box ^{2}$ is d'Alembertian:
       $\Box ^{2}=\nabla ^{2}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}={\frac {\partial ^{2}}{\partial x^{2}}}+{\frac {\partial ^{2}}{\partial y^{2}}}+{\frac {\partial ^{2}}{\partial z^{2}}}-{\frac {1}{c^{2}}}{\frac {\partial ^{2}}{\partial t^{2}}}\$
Notice that in the case of the electric and magnetic fields, the speed is:
       $c={\frac {1}{\sqrt {\mu _{0}\epsilon _{0}}}}$
Which, as it turns out, is the speed of light. Maxwell's equations have unified the permittivity of free space $\epsilon _{0}$, the permeability of free space $\mu _{0}$, and the speed of light itself, c. Before this derivation it was not known that there was such a strong relationship between light and electricity and magnetism.

### Solutions to Wave Equation

Complex part leads to attenuation