\chapter{Brief Review of Electromagnetics}

%\setcounter{page}{1}

\renewcommand{\thefootnote}{\fnsymbol{footnote}}

\footnotetext{Lecture notes by John Schneider. {\tt

em-review.tex}}

\section{Introduction}

The specific equations on which the finite-difference time-domain

(FDTD) method is based will be considered in some detail later. The

goal here is to remind you of the physical significance of the

equations to which you have been exposed in previous courses on

electromagnetics.

In some sense there are just a few simple premises which underlie all

electromagnetics. One could argue that electromagnetics is simply

based on the following:

\begin{enumerate}

\item Charge exerts force on other charge.

\item Charge in motion exerts a force on other charge in motion.

\item All material is made up of charged particles.

\end{enumerate}

Of course translating these premises into a corresponding mathematical

framework is not trivial. However one should not lose sight of the

fact that the math is trying to describe principles that are

conceptually rather simple.

\section{Coulomb's Law and Electric Field}

Coulomb studied the electric force on charged particles. As depicted

in Fig.\ \ref{fig:coulomb}, given two

discrete particles carrying charge $Q_1$ and $Q_2$, the force

experienced by $Q_2$ due to $Q_1$ is along the line joining $Q_1$ and

$Q_2$.

\begin{figure}

\begin{center}

\epsfig{width=4.in,file=Figures/Em-review/coulomb.eps}

\end{center}

\caption{The force experienced by charge $Q_2$ due to charge $Q_1$ is

along the line which pass through both charges. The direction of

the force is dictate by the signs of the charges. Electric field is

assumed to point radially away from positive charges as is indicated

by the lines pointing away from $Q_1$ (which is assumed here to

be positive).}

\label{fig:coulomb}

\end{figure}

The force is proportional to the charges and inversely proportional to

the square of the distance between the charges. A proportionality

constant is needed to obtain Coulomb's law which gives the equation of

the force on $Q_2$ due to $Q_1$:

\begin{equation}

\mathbf{F}_{12} = \unitvec{12}

\frac{1}{4\pi\epsilon_0}

\frac{Q_1 Q_2}{R_{12}^2}

\label{eq:coulomb}

\end{equation}

where $\unitvec{12}$ is a unit vector pointing from $Q_1$ to $Q_2$,

$R_{12}$ is the distance between the charges, and $1/4\pi\epsilon_0$

is the proportionality constant. The constant $\epsilon_0$ is known

as the permittivity of free space and equals approximately

$8.854\times 10^{-12}$ F/m. Charge is expressed in units of Coulombs

(C) and can be either negative or positive. When the two charges have

like signs, the force will be repulsive: $\mathbf{F}_{12}$ will

be parallel to $\unitvec{12}$. When the charges are of opposite sign,

the force will be attractive so that $\mathbf{F}_{12}$ will be

anti-parallel to $\unitvec{12}$.

There is a shortcoming with \refeq{eq:coulomb} in that it implies

action at a distance. It appears from this equation that the force

$\mathbf{F}_{12}$ is established instantly. From this equation one

could assume that a change in the distance $R_{12}$ results in an

instantaneous change in the force $\mathbf{F}_{12}$, but this is not

the case. A finite amount of time is required to communicate the

change in location of one charge to the other charge (similarly, it

takes a finite amount of time to communicate a change in the quantity

of one charge to the other charge). To overcome this shortcoming it

is convenient to employ the concept of fields. Instead of $Q_1$

producing a force directly on $Q_2$, $Q_1$ is said to produce a field.

This field then produces a force on $Q_2$. The field produced by

$Q_1$ is independent of $Q_2$---it exists whether or not $Q_2$ is

there to experience it.

In the static case, the field approach does not appear to have any

advantage over the direct use of Coulomb's law. This is because for

static charges Coulomb's law is correct. Fields must be time-varying

for the distinction to arise. Nevertheless, to be consistent with the

time-varying case, fields are used in the static case as well. The

electric field produced by the point charge $Q_1$ is

\begin{equation}

\Evec_1 = \unitvec{r} \frac{Q_1}{4 \pi\epsilon_0 r^2}

\label{eq:eField}

\end{equation}

where $\unitvec{r}$ is a unit vector which points radially away from

the charge and $r$ is the distance from the charge. The electric

field has units of volts per meter (V/m).

To find the force on $Q_2$, one merely takes the charge times the

electric field: $\mathbf{F}_{12}=Q_2 \Evec_1$. In general, the force

on any charge $Q$ is the product of the charge and the electric field

at which the charge is present, i.e., $\mathbf{F}=Q \Evec$.

\section{Electric Flux Density}

All material is made up of charged particles. The material may be

neutral overall because it has as many positive charges as negative

charges. Nevertheless, there are various ways in which the positive

and negative charges may shift slightly within the material, perhaps

under the influence of an electric field. The resulting

charge separation will have an effect on the overall electric field.

Because of this it is often convenient to introduce a new field known

as the electric flux density, $\Dvec$, which has units of Coulombs per

square meter (C/m$^2$).\footnote{Note that not everybody advocates

using the $\Dvec$ field. See for example Volume II of {\em The

Feynman Lectures on Physics}, R. P. Feynman, R. B. Leighton, and

M. Sands, Addison-Wesley, 1964. Feynman only uses $\Evec$ and never

resorts to $\Dvec$.} Essentially the $\Dvec$ field ignores the local

effects of charge which is bound in a material.

In free space, the electric field and the electric flux density are

related by

\begin{equation}

\Dvec = \epsilon_0\Evec.

\end{equation}

Gauss's law states that

integrating $\Dvec$ over a closed surface yields the enclosed free

charge

\begin{equation}

\oint_S \Dvec\cdot \mathbf{ds} = Q_{\mbox{\scriptsize enc}}

\label{eq:gauss}

\end{equation}

where $S$ is the closed surface, $\mathbf{ds}$ is an incremental

surface element whose normal is directed radially outward, and

$Q_{\mbox{\scriptsize enc}}$ is the enclosed charge. As an example,

consider the electric field given in \refeq{eq:eField}. Taking $S$ to

be a spherical surface with the charge at the center, it is simple to

perform the integral in \refeq{eq:gauss}:

\begin{equation}

\oint_S \Dvec\cdot \mathbf{ds} =

\int_{\theta=0}^\pi\int_{\phi=0}^{2\pi}

\epsilon_0 \frac{Q_1}{4 \pi\epsilon_0 r^2}\unitvec{r} \cdot

\unitvec{r} r^2\sin\theta \, d\phi \, d\theta = Q_1.

\label{eq:gaussExamp}

\end{equation}

The result is actually independent of the surface chosen (provided it

encloses the charge), but the integral is especially easy to perform

for a spherical surface.

We want the integral in \refeq{eq:gauss} always to equal the enclosed

charge as it does in free space. However, things are more complicated

when material is present. Consider, as shown in Fig.\

\ref{fig:platesFree}, two large parallel plates which carry uniformly

distributed charge of equal magnitude but opposite sign.

\begin{figure}

\begin{center}

\epsfig{width=4.5in,file=Figures/Em-review/parallel-plates.eps}

\end{center}

\caption{Charged parallel plates in free space. The dashed line

represents the integration surface $S$.}

\label{fig:platesFree}

\end{figure}

The dashed line represents an integration surface $S$ which is assumed

to be sufficiently far from the edges of the plate so that the field

is uniform over the top of $S$. This field is identified as

$\Evec_0$. The fields are zero outside of the plates and are

tangential to the sides of $S$ within the plates. Therefore the only

contribution to the integral would be from the top of $S$. The result

of the integral $\oint_S \epsilon_0\Evec\cdot \mathbf{ds}$

is the negative charge enclosed by the surface (i.e., the negative

charge on the bottom plate which falls within $S$).

Now consider the same plates, carrying the same charge, but with a

material present between the plates. Assume this material is

``polarizable'' such that the positive and negative charges can shift

slightly. The charges are not completely free to move---they are

bound charges. The positive charges will be repelled by the top plate

and attracted to the bottom plate. Conversely, the negative charges

will be repelled by the bottom plate and attracted to the top plate.

This scenario is depicted in Fig.\ \ref{fig:platesMat}.

\begin{figure}

\begin{center}

\epsfig{width=4.5in,file=Figures/Em-review/parallel-plates-dielectric.eps}

\end{center} \caption{Charged parallel plates with a polarizable

material present between the plates. The elongated objects

represent molecules whose charge orientation serves to produce a net

bound negative charge layer at the top plate and a bound positive

charge layer at the bottom plate. In the interior, the positive and

negative bound charges cancel each other. It is only at the surface

of the material where one must account for the bound charge. Thus,

the molecules are not drawn throughout the figure. Instead, as

shown toward the right side of the figure, merely the bound charge

layer is shown. The free charge on the plates creates the electric

field $\Evec_0$. The bound charge creates the electric field

$\Evec_m$ which opposes $\Evec_0$ and hence diminishes the total

electric field. The dashed line again represents the integration

surface $S$.} \label{fig:platesMat}

\end{figure}

With the material present the electric field due to the charge on the

plates is still $\Evec_0$, i.e., the same field as existed in Fig.\

\refeq{fig:platesFree}. However, there is another field present due

to the displacement of the bound charge in the polarizable material

between the plates. The polarized material effectively acts to

establish a layer of positive charge adjacent to the bottom plate and

a layer of negative charge adjacent to the top plate. The field due

to these layers of charge is also uniform but it is in the opposite

direction of the field caused by the ``free charge'' on the plates.

The field due to bound charge is labeled $\Evec_m$ in Fig.\

\refeq{fig:platesMat}. The total field is the sum of the fields due

to the bound and free charges, i.e., $\Evec = \Evec_0 + \Evec_m$.

Because $\Evec_0$ and $\Evec_m$ are anti-parallel, the magnitude of

the total electric field $\Evec$ will be less than $\Evec_0$.

Since the material is neutral, we would like the integral of the

electric flux over the surface $S$ to yield just the enclosed charge

on the bottom plate---not the bound charge due to the material. In

some sense this implies that the integration surface cannot separate

the positive and negative bound charge of any single molecule. Each

molecule is either entirely inside or outside the integration surface.

Since each molecule is neutral, the only contribution to the integral

will be from the free charge on the plate.

With the material present, the integral of $\oint_S

\epsilon_0\Evec\cdot \mathbf{ds}$ yields too little charge. This is

because, as stated above, the total electric field $\Evec$ is less

than it would be if only free space were present. To correct for the

reduced field and to obtain the desired result, the electric flux

density is redefined so that it accounts for the presence of the

material. The more general expression for the electric flux density

is

\begin{equation}

\Dvec = \epsilon_r\epsilon_0\Evec = \epsilon\Evec

\label{eq:eAndD}

\end{equation}

where $\epsilon_r$ is the relative permittivity and $\epsilon$ is

called simply the permittivity. By accounting for the permittivity of

a material, Gauss's law is always satisfied.

In \refeq{eq:eAndD}, $\Dvec$ and $\Evec$ are related by a scalar

constant. This implies that the $\Dvec$ and $\Evec$ fields are

related by a simple proportionality constant for all frequencies, all

orientations, and all field strengths. Unfortunately the real world

is not so simple. Clearly if the electric field is strong enough, it

would be possible to tear apart the bound positive and negative

charges. Since charges have some mass, they do not react the same way

at all frequencies. Additionally, many materials may have some

structure, such as crystals, where the response in one direction is

not the same in other directions. Nevertheless, Gauss's law is the

law and thus always holds. When things get more complicated one must

abandon a simple scalar for the permittivity and use an appropriate

form to ensure Gauss's law is satisfied. So, for example, it may be

necessary to use a tensor for permittivity that is directionally

dependent. However, with the exception of frequency-dependent

behavior (i.e., dispersive materials), we will not be pursuing those

complications. A scalar permittivity will suffice.

\section{Static Electric Fields}

Ignoring possible nonlinear behavior of material, superposition holds

for electromagnetic fields. Therefore we can think of any

distribution of charges as a collection of point charges. We can get

the total field by summing the contributions from all the charges (and

this summing will have to be in the form of an integration if the

charge is continuously distributed).

Note from \refeq{eq:eField} that the field associated with a point

charge merely points radially away from the charge. There is no

``swirling'' of the field. If we have more than a single charge, the

total field may bend, but it will not swirl. Imagine a tiny wheel

with positive charge distributed around its circumference. The wheel

hub of the wheel is held at a fixed location but the wheel is free to

spin about its hub. For static electric fields, no matter where we

put this wheel, there would be no net force on the wheel to cause it

to spin. There may be a net force pushing the entire wheel in a

particular direction (a translational force), but the forces which are

pushing the wheel to spin in the clockwise direction are balanced by

the forces pushing the wheel to spin in the counterclockwise

direction.

Another property of electrostatic fields is that the electric flux

density only begins or terminates on free charge. If there is no

charge present, the lines of flux continue.

The lack of swirl in the electric field and the source of electric

flux density are fairly simple concepts. However, to be able to

analyze the fields properly, one needs a mathematical statement of

these concepts. The appropriate statements are

\begin{equation}

\nabla\times\Evec=0

\label{eq:curlE}

\end{equation}

and

\begin{equation}

\nabla\cdot\Dvec=\rho_v

\label{eq:divD}

\end{equation}

where $\nabla$ is the del or nabla operator and $\rho_v$ is the

electric charge density (with units of C/m$^3$). Equation

\refeq{eq:curlE} is the curl of the electric field and \refeq{eq:divD}

is the divergence of the electric flux density. These two equations

are discussed further in the following section.

\section{Gradient, Divergence, and Curl}

The del operator is independent of the coordinate system

used---naturally the behavior of the fields should not depend on the

coordinate system used to describe the field. Nevertheless, the del

operator can be expressed in different coordinates systems. In

Cartesian coordinates del is

\begin{equation}

\nabla \equiv \unitvec{x} \frac{\partial}{\partial x} +

\unitvec{y} \frac{\partial}{\partial y} +

\unitvec{z} \frac{\partial}{\partial z}

\end{equation}

where the symbol $\equiv$ means ``defined as.''

Del acting on a scalar field produces the gradient of the field.

Assuming $f$ is a some scalar field, $\nabla f$ produces the vector

field given by

\begin{equation}

\nabla f = \unitvec{x} \frac{\partial f}{\partial x} +

\unitvec{y} \frac{\partial f}{\partial y} +

\unitvec{z} \frac{\partial f}{\partial z}.

\end{equation}

The gradient of $f$ points in the direction of greatest change and is

proportional to the rate of change. Assume we wish to find the amount

of change in $f$ for a small movement $dx$ in the $x$ direction. This

can be obtained via $\nabla f \cdot \unitvec{x} dx$, to wit

\begin{equation}

\nabla f \cdot \unitvec{x} dx = \frac{\partial f}{\partial x} dx

= \mbox{(rate of change in $x$ direction)}\times

\mbox{(movement in $x$ direction)}.

\end{equation}

This can be generalized for movement in an arbitrary direction.

Letting an incremental small length be given by

\begin{equation}

\mathbf{d\ell} = \unitvec{x} dx + \unitvec{y} dy + \unitvec{z} dz,

\end{equation}

the change in the field realized by moving an amount $\mathbf{d\ell}$

is

\begin{equation}

\nabla f\cdot\mathbf{d\ell} =

\frac{\partial f}{\partial x} dx +

\frac{\partial f}{\partial y} dy +

\frac{\partial f}{\partial z} dz.

\end{equation}

Returning to \refeq{eq:divD}, when the del operator is dotted with a

vector field, one obtains the divergence of that field. Divergence

can be thought of as a measure of ``source'' or ``sink'' strength of the

field at a given point. The divergence of a vector field is a scalar

field given by

\begin{equation}

\nabla \cdot \Dvec = \frac{\partial D_x}{\partial x} +

\frac{\partial D_y}{\partial y} +

\frac{\partial D_z}{\partial z}.

\end{equation}

Let us consider a finite-difference approximation of this divergence

in the $xy$-plane as shown in Fig.\ \ref{fig:div}.

\begin{figure}

\begin{center}

\epsfig{width=3.5in,file=Figures/Em-review/discrete-divergence.eps}

\end{center}

\caption{Discrete approximation to the divergence taken in the

$xy$-plane. \label{fig:div}}

\end{figure}

Here the divergence is measured over a small box where the field is

assumed to be constant over each edge of the box. The derivatives can

be approximated by central differences:

\begin{equation}

\frac{\partial D_x}{\partial x} +

\frac{\partial D_y}{\partial y} \approx

\frac{D_x\left(x+\frac{\Delx}{2},y\right)

-D_x\left(x-\frac{\Delx}{2},y\right)}{\Delx} +

\frac{D_y\left(x,y+\frac{\Dely}{2}\right)

-D_y\left(x,y-\frac{\Dely}{2}\right)}{\Dely}

\label{eq:divII}

\end{equation}

where this is exact as $\Delx$ and $\Dely$ go to zero.

Letting $\Delx=\Dely=\delta$, \refeq{eq:divII} can be written

\begin{equation}

\frac{\partial D_x}{\partial x} +

\frac{\partial D_y}{\partial y} \approx

\frac{1}{\delta}

\left(D_x\left(x+\frac{\delta}{2},y\right) -

D_x\left(x-\frac{\delta}{2},y\right) +

D_y\left(x,y+\frac{\delta}{2}\right) -

D_y\left(x,y-\frac{\delta}{2}\right)\right).

\label{eq:divIII}

\end{equation}

Inspection of \refeq{eq:divIII} reveals that the divergence is

essentially a sum of the field over the faces with the appropriate

sign changes. Positive signs are used if the field is assumed to

point out of the box and negative signs are used when the field is

assumed to point into the box. If the sum of these values is

positive, that implies there is more flux out of the box than into it.

Conversely, if the sum is negative, that means more flux is flowing

into the box than out. If the sum is zero, there must be as much flux

flowing into the box as out of it (that does not imply necessarily that,

for instance, $D_x\left(x+\delta/2,y\right)$ is equal to

$D_x\left(x-\delta/2,y\right)$, but rather that the sum of all four

fluxes must be zero).

Equation \refeq{eq:divD} tells us that the electric flux density has

zero divergence except where there is charge present (as specified by

the charge-density term $\rho_v$). If the charge density is zero, the

total flux entering some small enclosure must also leave it. If the

charge density is positive at some point, more flux will leave a small

enclosure surrounding that point than will enter it. On the other

hand, if the charge density is negative, more flux will enter the

enclosure surrounding that point than will leave it.

Finally, let us consider \refeq{eq:curlE} which is the curl of the

electric field. In Cartesian coordinates it is possible to treat this

operation as simply the cross product between the vector operator

$\nabla$ and the vector field $\Evec$:

\begin{equation}

\nabla\times\Evec =

\left|

\begin{array}{ccc}

\unitvec{x} & \unitvec{y} & \unitvec{z} \\

\frac{\partial}{\partial x} &

\frac{\partial}{\partial y} &

\frac{\partial}{\partial z} \\

E_x & E_y & E_z

\end{array}

\right|

=

\unitvec{x}\left(\frac{\partial E_z}{\partial y} -

\frac{\partial E_y}{\partial z}\right) +

\unitvec{y}\left(\frac{\partial E_x}{\partial z} -

\frac{\partial E_z}{\partial x}\right) +

\unitvec{z}\left(\frac{\partial E_y}{\partial x} -

\frac{\partial E_x}{\partial y}\right).

\label{eq:curl}

\end{equation}

Let us consider the behavior of only the $z$ component of this

operator which is dictated by the field in the $xy$-plane as shown in

Fig.\ \ref{fig:curl}.

\begin{figure}

\begin{center}

\epsfig{width=3.5in,file=Figures/Em-review/discrete-curl.eps}

\end{center}

\caption{Discrete approximation to the curl taken in the

$xy$-plane. \label{fig:curl}}

\end{figure}

The $z$-component of $\nabla\times\Evec$ can be written as

\begin{equation}

\frac{\partial E_y}{\partial x} -

\frac{\partial E_x}{\partial y} \approx

\frac{E_y\left(x+\frac{\Delx}{2},y\right)

-E_y\left(x-\frac{\Delx}{2},y\right)}{\Delx} -

\frac{E_x\left(x,y+\frac{\Dely}{2}\right)

-E_x\left(x,y-\frac{\Dely}{2}\right)}{\Dely}.

\label{eq:curlII}

\end{equation}

The finite-difference approximations of the derivatives are again

based on the fields on the edges of a box surrounding the point of

interest. However, in this case the relevant fields are tangential to

the edges rather than normal to them. Again letting

$\Delx=\Dely=\delta$, \refeq{eq:curlII} can be written

\begin{equation}

\frac{\partial E_y}{\partial x} -

\frac{\partial E_x}{\partial y} \approx

\frac{1}{\delta}

\left(E_y\left(x+\frac{\delta}{2},y\right) -

E_y\left(x-\frac{\delta}{2},y\right) -

E_x\left(x,y+\frac{\delta}{2}\right) +

E_x\left(x,y-\frac{\delta}{2}\right)\right).

\label{eq:curlIII}

\end{equation}

In the sum on the right side the sign is positive if the vector

component points in the counterclockwise direction (relative to

rotations about the center of the box) and is negative if the vector

points in the clockwise direction. Thus, if the sum of these vector

components is positive, that implies that the net effect of these

electric field vectors is to tend to push a positive charge in the

counterclockwise direction. If the sum were negative, the vectors

would tend to push a positive charge in the clockwise direction. If

the sum is zero, there is no tendency to push a positive charge around

the center of the square (which is not to say there would not be a

translation force on the charge---indeed, if the electric field is

non-zero, there has to be some force on the charge).

\section{Laplacian}

In addition to the gradient, divergence, and curl, there is one more

vector operator to consider. There is a vector identity that the curl

of the gradient of any function is identically zero

\begin{equation}

\nabla\times\nabla f = 0.

\label{eq:vecIdent}

\end{equation}

This is simple to prove by merely performing the operations in

Cartesian coordinates. One obtains several second-order partial

derivatives which cancel if the order of differentiation is switched.

Recall that for a static distribution of charges, $\nabla\times\Evec=0$.

Since the curl of the electric field is zero, it should be possible to

represent the electric field as the gradient of some scalar function

\begin{equation}

\Evec = -\nabla V.

\label{eq:potential}

\end{equation}

The scalar function $V$ is the electric potential and the negative

sign is used to make the electric field point from higher potential to

lower potential (by historic convention the electric field points away

from positive charge and toward negative charge). By expressing the

electric field this way, the curl of the electric field is guaranteed

to be zero.

Another way to express the relationship between the electric field and

the potential is via integration. Consider movement from an arbitrary

point $a$ to an arbitrary point $b$. The change in potential between

these two points can be expressed as

\begin{equation}

V_b-V_a = \int_a^b \nabla V \cdot \mathbf{d\ell}.

\label{eq:potentialInt}

\end{equation}

The integrand represent the change in the potential for a movement

$\mathbf{d\ell}$ and the integral merely sums the changes over the

path from $a$ to $b$. However, the change in potential must also be

commensurate with the movement in the direction of, or against, the

electric field. If we move against the electric field, potential

should go up. If we move along the electric field, the potential

should go down. In other words, the incremental change in potential

for a movement $\mathbf{d\ell}$ should be

$dV=-\Evec\cdot\mathbf{d\ell}$ (if the movement $\mathbf{d\ell}$ is

orthogonal to the electric field, there should be no change in the

potential). Summing change in potential over the entire path yields

\begin{equation}

V_b-V_a = - \int_a^b \Evec \cdot \mathbf{d\ell}.

\label{eq:potentialIntII}

\end{equation}

The integrals in \refeq{eq:potentialInt} and \refeq{eq:potentialIntII}

can be equated. Since the equality holds for any two arbitrary points,

the integrands must be equal and we are again left with $\Evec =

-\nabla V$.

The electric flux density can be related to the electric field via

$\Dvec=\epsilon\Evec$ and the behavior of the flux density $\Dvec$ is

dictated by $\nabla\cdot\Dvec=\rho_v$. Combining these with

\refeq{eq:potential} yields

\begin{equation}

\Evec = \frac{1}{\epsilon}\Dvec = -\nabla V.

\end{equation}

Taking the divergence of both sides yields

\begin{equation}

\frac{1}{\epsilon}\nabla\cdot\Dvec =

\frac{1}{\epsilon} \rho_v =

-\nabla\cdot\nabla V.

\end{equation}

Rearranging this yields Poisson's equation given by

\begin{equation}

\nabla^2 V = -\frac{\rho_v}{\epsilon}

\label{eq:poisson}

\end{equation}

where $\nabla^2$ is the Laplacian operator

\begin{equation}

\nabla^2 \equiv \nabla\cdot\nabla =

\frac{\partial^2}{\partial x^2} +

\frac{\partial^2}{\partial y^2} +

\frac{\partial^2}{\partial z^2}.

\end{equation}

Note that the Laplacian is a scalar operator. It can act on a scalar

field (such as the potential $V$ as shown above) or it can act on a

vector field as we will see later. When it acts on a vector field,

the Laplacian acts on each component of the field.

In the case of zero charge density, \refeq{eq:poisson} reduces to

Laplace's equation

\begin{equation}

\nabla^2 V = 0.

\end{equation}

We have a physical intuition about what gradient, divergence, and curl

are telling us, but what about the Laplacian? To answer this,

consider a function of a single variable.

Given the function $V(x)$, we can ask if the function at some point is

greater than, equal to, or less than the average of its neighboring

values. The answer can be expressed in terms of the value of the

function at the point of interest and the average of samples to either

side of that central point:

\begin{equation}

\frac{V(x+\delta) + V(x-\delta)}{2} - V(x) =

\left\{

\begin{array}{ll}

\mbox{positive}&\mbox{if center point less than average of neighbors} \\

\mbox{zero} &\mbox{if center point equals average of neighbors} \\

\mbox{negative}&\mbox{if center point greater than average of neighbors}

\end{array}

\right.

\label{eq:avg}

\end{equation}

Here the left-most term represents the average of the neighboring

values and $\delta$ is some displacement from the central point.

Equation \refeq{eq:avg} can be normalized by $\delta^2/2$ without changing

the basic tenants of this equation. Performing that normalization and

rearranging yields

\begin{eqnarray}

\frac{1}{\delta^2/2}

\left\{

\frac{V(x+\delta) + V(x-\delta)}{2} - V(x)

\right\}

&=&

\frac{1}{\delta^2}

\left\{

(V(x+\delta) - V(x)) - (V(x)-V(x-\delta))

\right\} \nonumber

\\

&=&

\frac{

\frac{V(x+\delta) - V(x)}{\delta} -

\frac{V(x)-V(x-\delta)}{\delta}}{\delta} \nonumber

\\

&\approx&

\frac{

\frac{\partial V(x+\delta/2)}{\partial x} -

\frac{\partial V(x-\delta/2)}{\partial x}}{\delta} \nonumber

\\

&\approx&

\frac{\partial^2 V(x)}{\partial x^2}.

\end{eqnarray}

Thus the second partial derivative can be thought of as a way of

measuring the field at a point relative to its neighboring points.

You should already have in mind that if the second derivative is

negative, a function is tending to curve downward. Second derivatives

are usually discussed in the context of curvature. However, you

should also think in terms of the field at a point and its neighbors.

At points where the second derivative is negative those points are

higher than the average of their neighboring points (at least if the

neighbors are taken to be an infinitesimally small distance away).

In lieu of these arguments, Poisson's equation \refeq{eq:poisson}

should have physical significance. Where the charge density is zero,

the potential cannot have a local minima or maxima. The potential is

always equal to the average of the neighboring points. If one

neighbor is higher, the other must be lower (and this concept easily

generalizes to higher dimensions). Conversely, if the charge density

is positive over some region, the potential should increase as one

moves deeper into that region but the rate of increase must be such

that at any point the average of the neighbors is less than the center

point. This behavior is illustrated in Fig.\ \ref{fig:chargeSphere}

which depicts the potential along a path through the center of a

uniform sphere of charge.

\begin{figure}

\begin{center}

\epsfig{width=4in,file=Figures/Em-review/sphere-potential.eps}

\end{center}

\caption{Potential along a path which passes through a uniform

sphere of positive charge (arbitrary units). \label{fig:chargeSphere}}

\end{figure}

\section{Gauss's and Stokes' Theorems}

Equation \refeq{eq:gauss} presented Gauss's law which stated the flux

of $\Dvec$ through a closed surface $S$ is equal to the enclosed

charge. There is an identity in vector calculus, known as Gauss's

theorem, which states that the integral of the flux of any vector

field through a closed surface equals the integral of the divergence

of the field over the volume enclosed by the surface. This holds for

any vector field, but using the $\Dvec$ field, Gauss's theorem states

\begin{equation}

\oint_S \Dvec\cdot\mathbf{ds} =

\int_V \nabla\cdot \Dvec dv

\label{eq:gaussTheorem}

\end{equation}

where $V$ is the enclosed volume and $dv$ is a differential volume

element. Note that the left-hand side of \refeq{eq:gaussTheorem} is

the left-hand side of \refeq{eq:gauss}.

The right-hand side of \refeq{eq:gauss} is the enclosed charge

$Q_{\mbox{\scriptsize enc}}$ which could be determined either by

evaluating the left-hand side of \refeq{eq:gauss} or by integration of

the charge density $\rho_v$ over the volume enclosed by $S$. (This is

similar to determining the mass of an object by integrating its mass

density over its volume.) Thus,

\begin{equation}

Q_{\mbox{\scriptsize enc}} =

\int_V \rho_v dv.

\label{eq:chargeIntegral}

\end{equation}

Equating the right-hand sides of \refeq{eq:gaussTheorem} and

\refeq{eq:chargeIntegral} yields

\begin{equation}

\int_V \nabla\cdot \Dvec dv =

\int_V \rho_v dv.

\end{equation}

Since this must hold over an arbitrary volume, the integrands must be

equal which yields \refeq{eq:divD}.

Another useful identity from vector calculus is Stokes' theorem which

states that the integral of a vector field over any closed path is

equal to the integral of the curl of that field over a surface which

has that path as its border. Again, this holds for any vector field,

but using the electric field as an example one can write

\begin{equation}

\oint_L \Evec\cdot\mathbf{d\ell} =

\int_S \nabla\times\Evec \cdot \mathbf{ds}.

\label{eq:stokes}

\end{equation}

The surface normal is assumed to follow the right-hand convention so

that when the fingers of the right hand are oriented along the path of

the loop, the thumb points in the positive direction of the surface

normal.

Static electric fields are conservative which means that the net work

required to move a charge in a closed path is always zero. Along some

portion of the path positive work would have to be done to push the

charge against the field, but this amount of work would be given back

by the field as the charge travels along the remaining portions of the

path. The integrand on the left-hand side in \refeq{eq:stokes} is the

field dotted with an incremental length. If the integrand were

multiplied by a unit positive charge, the integrand would represent

work, since charge times field is force and force times distance is

work. Because the electric field is conservative, the integral on the

left-hand side of \refeq{eq:stokes} must be zero. Naturally this

implies that the integral on the right-hand side must also be zero.

Since this holds for any loop $L$ (or, similarly, any surface $S$), the

integrand itself must be zero. Equating the integrand to zero yields

\refeq{eq:curlE}.

\section{Electric Field Boundary Conditions}

Consider an interface between two homogeneous regions. Because

electric flux density only begins or ends on charge, the normal

component of $\Dvec$ can only change at the interface if there is

charge on the interface, i.e., surface charge is present. This can be

stated mathematically as

\begin{equation}

\hat{\mathbf{n}} \cdot(\Dvec_1 - \Dvec_2) = \rho_s

\end{equation}

where $\rho_s$ is a surface charge density (C/m$^2$),

$\hat{\mathbf{n}}$ is a unit vector normal to the surface, and

$\Dvec_1$ and $\Dvec_2$ are the field to either side of the interface.

One should properly argue this boundary condition by an application of

Gauss's law for a small volume surrounding the surface, but such

details are left to other classes (this is just a review!). If no

charge is present, the normal components must be equal

\begin{equation}

\hat{\mathbf{n}} \cdot \Dvec_1 = \hat{\mathbf{n}} \cdot \Dvec_2.

\end{equation}

The boundary conditions on the tangential component of the electric

field can be determined by integrating the electric field over a

closed loop which is essentially a rectangle which encloses a portion

of the interface. By letting the sides shrink to zero and keeping the

``top'' and ``bottom'' of the rectangle small but finite (so that they

are tangential to the surface), one essentially has that the field

over the top must be the same as the field over the bottom (owing to

the fact that total integral must be zero since the field is

conservative). Stated mathematical, the boundary condition is

\begin{equation}

\hat{\mathbf{n}} \times (\Evec_1 - \Evec_2) = 0.

\end{equation}

\section{Conductivity and Perfect Electric Conductors}

It is possible for the charge in materials to move under the influence

of an electric field such that currents flow. If the material has a

non-zero conductivity $\sigma$, the current density is given by

\begin{equation}

\Jvec = \sigma \Evec.

\end{equation}

The current density has units of A/m$^2$ and the conductivity has

units of S/m.

If charge is building up or decaying in a particular region, the

divergence of the current density must be non-zero. If the divergence

is zero, that implies as much current leaves a point as enters it and

there is no build-up or decay of charge. This can be stated as

\begin{equation}

\nabla \cdot \Jvec = -\frac{\partial \rho_v}{\partial t}.

\label{eq:chargeConservation}

\end{equation}

If the divergence is positive, the charge density must be decreasing

with time (so the negative sign will bring the two into agreement).

This equation is a statement of charge conservation.

Perfect electric conductors (PECs) are materials where it is assumed

that the conductivity approaches infinity. If the fields were

non-zero in a PEC, that would imply the current was infinite. Since

infinite currents are not allowed, the fields inside a PEC are

required to be zero. This subsequently requires that the tangential

electric field at the surface of a PEC is zero (since tangential

fields are continuous across an interface and the fields inside the

PEC are zero). Correspondingly, the normal component of the electric

flux density $\Dvec$ at the surface of a PEC must equal the charge

density at the surface of the PEC. Since the fields inside a PEC are

zero, all points of the PEC must be at the same potential.

\section{Magnetic Fields}

Magnetic fields circulate around, but they do not terminate on

anything---there is no (known) magnetic charge. Nevertheless, it is

often convenient to define magnetic charge and magnetic current.

These fictions allow one to simplify various problems such as integral

formulations of scattering problems. However for now we will stick to

reality and say they do not exist.

The magnetic flux density $\Bvec$ is somewhat akin to the electric

field in that the force on a charge in motion is related to $\Bvec$.

If a charge $Q$ is moving with velocity $\mathbf{u}$ in a field

$\Bvec$, it experiences a force

\begin{equation}

\mathbf{F} = Q \mathbf{u}\times\Bvec.

\end{equation}

Because $\Bvec$ determines the force on a charge, it must account for

all sources of magnetic field. When material is present, the charge

in the material can have motion (or rotation) which influences the

magnetic flux density.

Alternatively, similar to the electric flux density, we define the

magnetic field $\Hvec$ which ignores the local effects of material.

These fields are related by

\begin{equation}

\Bvec = \mu_r\mu_0 \Hvec = \mu\Hvec

\end{equation}

where $\mu_r$ is the relatively permeability, $\mu_0$ is the

permeability of free space equal to $4\pi\times 10^{-7}$ H/m, and

$\mu$ is simply the permeability. Typically the relative permeability

is greater than unity (although usually only by a small amount) which

implies that when a material is present the magnetic flux density is

larger than when there is only free space.

Charge in motion is the source of magnetic fields. If a current $I$

flows over an incremental distance $\mathbf{d\ell}$, it will produce a

magnetic field given by:

\begin{equation}

\Hvec = \frac{I\mathbf{d\ell}\times \mathbf{a}_r}

{4 \pi r^2}

\label{eq:biot}

\end{equation}

where $\mathbf{a}_r$ points from the location of the filament of

current to the observation point and $r$ is the distance between the

filament and the observation point. Equation \refeq{eq:biot} is known

as the Biot-Savart equation. Of course, because of the conservation

of charge, a current cannot flow over just a filament and then

disappear. It must flow along some path. Thus, the magnetic field

due to a loop of current would be given by

\begin{equation}

\Hvec = \oint_L\frac{I\mathbf{d\ell}\times \mathbf{a}_r}

{4 \pi r^2}.

\label{eq:biotIntegral}

\end{equation}

If the current was flowing throughout a volume or over a surface, the

integral would be correspondingly changed to account for the

current wherever it flowed.

From \refeq{eq:biotIntegral} one sees that currents (which are just

another way of saying charge in motion) are the source of magnetic

fields. Because of the cross-product in \refeq{eq:biot} and

\refeq{eq:biotIntegral}, the magnetic field essentially swirls around

the current. If one integrates the magnetic field over a closed path,

the result is the current enclosed by that path

\begin{equation}

\oint_L \Hvec \cdot \mathbf{d\ell} = I_{\mbox{\scriptsize enc}}.

\label{eq:hIntegral}

\end{equation}

The enclosed current $I_{\mbox{\scriptsize enc}}$ is the current that

passes through the surface $S$ which is bound by the loop $L$.

The left-hand side of \refeq{eq:hIntegral} can be converted to a

surface integral by employing Stokes' theorem while the right-hand

side can be related to the current density by integrating over the

surface of the loop. Thus,

\begin{equation}

\oint_L \Hvec\cdot \mathbf{d\ell} =

\int_S \nabla\times\Hvec \cdot \mathbf{ds} =

I_{\mbox{\scriptsize enc}} =

\int_S \Jvec \cdot \mathbf{ds}.

\label{eq:hIntegralII}

\end{equation}

Since this must be true for every loop (and surface), the integrands

of the second and fourth terms can be equated. This yields

\begin{equation}

\nabla\times\Hvec = \Jvec.

\end{equation}

The last equation needed to characterize static fields is

\begin{equation}

\nabla\cdot\Bvec=0.

\end{equation}

This is the mathematical equivalent of saying there is no magnetic

charge.

\section{Magnetic Field Boundary Conditions}

Note that the equation governing $\Bvec$ is similar to the equation

which governed $\Dvec$. In fact, since the right-hand side is always

zero, the equation for $\Bvec$ is simpler. The arguments used to

obtain the boundary condition for the normal component of the $\Dvec$

field can be applied directly to the $\Bvec$ field. Thus,

\begin{equation}

\hat{\mathbf{n}} \cdot(\Bvec_1 - \Bvec_2) = 0.

\end{equation}

For the magnetic field, an integration path is constructed along

the same lines as the one used to determine the boundary condition on

the electric field. Note that the equations governing $\Evec$ and

$\Hvec$ are similar except that the one for $\Hvec$ has a non-zero

right-hand side. If the current density is zero over the region of

interest, then there is really no distinction between the two and one

can say that the tangential magnetic fields must be equal across a

boundary. However, if a surface current exists on the interface,

there may be a discontinuity in the tangential fields. The boundary

condition is given by

\begin{equation}

\hat{\mathbf{n}} \times (\Hvec_1 - \Hvec_2)

= \mathbf{K}

\end{equation}

where $\mathbf{K}$ is the surface current density (with units of A/m).

\section{Summary of Static Fields}

When a system is not changing with respect to time, the governing

equations are

\begin{eqnarray}

\nabla \cdot \Dvec &=& \rho_v, \\

\nabla \cdot \Bvec &=& 0, \\

\nabla \times \Evec &=& 0, \\

\nabla \times \Hvec &=& \Jvec. \label{eq:curlHdc}

\end{eqnarray}

If a loop carries a current but is otherwise neutral, it will produce

a magnetic field and only a magnetic field. If a charge is

stationary, it will produce an electric field and only an electric

field. The charge will not ``feel'' the loop current and the current

loop will not feel the stationary charge (at least approximately).

The magnetic field and electric field are decoupled. If a charge $Q$

moves with velocity $\mathbf{u}$ in the presence of both an electric

field and a magnetic field, the force on the charge is the sum of the

forces due to the electric and magnetic fields

\begin{equation}

\mathbf{F} = Q(\Evec + \mathbf{u}\times\Bvec).

\end{equation}

\section{Time Varying Fields}

What happens when a point charge moves? We know that charge in motion

gives rise to a magnetic field, but if the charge is moving, its

associated electric field must also be changing. Thus, when a system

is time-varying the electric and magnetic fields must be coupled.

There is a vector identity that the divergence of the curl of any

vector field is identically zero. Taking the divergence of both sides

of \refeq{eq:curlHdc} yields

\begin{equation}

\nabla\cdot \nabla \times \Hvec =

\nabla\cdot \Jvec =

-\frac{\partial\rho_v}{\partial t}

\end{equation}

where the conservation of charge equation was used to write the last

equality. Since the first term must be zero, this implies that

$\partial\rho_v/\partial t$ must also be zero. However, that

is overly restrictive. In general, for a time-varying system, the

charge density will change with respect to time. Therefore something

must be wrong with \refeq{eq:curlHdc} as it pertains to time-varying

fields. It was Maxwell who recognized that by adding the temporal

derivative of the electric flux density to the right-hand side of

\refeq{eq:curlHdc} the equation would still be valid for the

time-varying case. The correct equation is given by

\begin{equation}

\nabla \times \Hvec = \Jvec + \frac{\partial \Dvec}{\partial t}

\label{eq:curlHac}

\end{equation}

The term $\partial \Dvec/\partial t$ is known as the displacement

current while $\Jvec$ is typically called the conduction current.

Equation \refeq{eq:curlHac} is known as Ampere's law.

Taking the divergence of the right-hand side of \refeq{eq:curlHac}

yields

\begin{equation}

\nabla\cdot\Jvec + \frac{\partial \nabla\cdot\Dvec}{\partial t} =

-\frac{\partial \rho_v}{\partial t} +

\frac{\partial \nabla\cdot\Dvec}{\partial t} =

-\frac{\partial \rho_v}{\partial t} +

\frac{\partial \rho_v}{\partial t}

= 0

\end{equation}

where use was made of \refeq{eq:divD} and the conservation of charge

equation \refeq{eq:chargeConservation}.

The electromotive force (EMF) is the change in potential over some

path. It has been observed experimentally that when a magnetic field

is time-varying there is a non-zero EMF over a closed path which

encloses the varying field (i.e., the electric field is no longer

conservative).

The symbol $\lambda$ is often use to represent total magnetic flux

through a given surface, i.e.,

\begin{equation}

\lambda = \int_S \Bvec\cdot\mathbf{ds}.

\end{equation}

For time-varying fields, the EMF over a closed path $L$ can be written

\begin{eqnarray}

V_{\mbox{\scriptsize emf}} &=& \frac{d\lambda}{dt}, \\

-\oint_L \Evec\cdot\mathbf{d\ell} &=&

\frac{d}{dt}\int_S \Bvec\cdot\mathbf{ds}, \\

-\int_S \nabla\times\Evec\cdot\mathbf{ds} &=&