\documentstyle[12pt,psfig]{article}
\message{
 	Copyright, 1993, all rights reserved, Charlie Ebner (Dept.of Physics, 
 The Ohio State University, Columbus OH 43210) and Mark Jarrell (Dept.of 
 Physics, The University of Cincinnati, Cincinnati, OH 45221-0011).  This 
 material may not be reproduced for profit, modified or published in any 
 form (this includes electronic redistribution) without the prior written 
 permission of the authors listed above.	
}
\title{Plane Waves and Wave Propagation}
\author{Augustin Jean Fresnel\\(1788 - 1827)}
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\tableofcontents

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In this chapter we start by considering plane waves in infinite or 
semi-infinite media. We shall look at their properties in both insulating 
and conducting materials and shall give some thought to the possible 
properties of materials of different kinds. We will also look at the 
reflection and refraction of waves at planar boundaries between different 
materials, a topic which forms the basis for much of physical optics. If 
time allows, we shall also look at some of the more abstract aspects of 
wave propagation having to do with causality and signal propagation.

\section{Plane Waves in Uniform Linear Isotropic Nonconducting Media}
\subsection{The Wave Equation}

One of the most important predictions of the Maxwell equations is the 
existence of electromagnetic waves which can transport energy. The simplest 
solutions are plane waves in infinite media, and we shall explore these 
now.

Consider a material in which
\beq
\B=\mu\H\hsph\D=\ep\E\hsph\J=\rh=0.
\eeq
Then the Maxwell equations read
\beq
\div\E=0\hsph\div\B=0\hsph\curl\E=-\frac1c\pde\B t\hsph\curl\B=\frac{\mu
\ep}c\pde\E t.
\eeq
Now we do several simple manipulations that will become second nature. 
First take the curl of one of the curl equations, e.g., Faraday's law, to 
find
\beq
\curl(\curl\E)=\grad(\div\E)-\lap\E=-\frac1c\frac{\partial}{\partial t}
(\curl\B)=-\frac{\mu\ep}{c^2}\spd\E t,
\eeq
where the generalized Amp\`ere's law was employed in the last step. Because 
the divergence of $\E$ is zero, this equation may be written as
\beq
\lep\lap-\frac{\mu\ep}{c^2}\frac{\partial^2}{\partial t^2}\rip\E=0.
\eeq
Identical manipulations starting from Amp\`ere's law rather than Faraday's 
law also lead to 
\beq
\lep\lap-\frac{\mu\ep}{c^2}\frac{\partial^2}{\partial t^2}\rip\B=0.
\eeq
Thus any Cartesian component of $\E$ or $\B$ obeys a classical wave 
equation of the form
\beq
\lep\lap-\frac1{v^2}\frac{\partial^2}{\partial t^2}\rip\psxt=0,
\eeq
where $v=c/\sqrt{\mu\ep}$.

There is a simple set of complex traveling wave solutions to this equation. 
They are of the form
\beq
\uvk(\x,t)=\ekxot
\eeq
where $\om=vk$ and $\k$ is any real vector.\footnote{This vector is real if 
$\ep$ and $\mu$ are real; they can be complex, in which case there are 
still solutions of this form with complex $\k$.} Notice that the derivatives of 
this function are 
\beqa
\grad\uvk=i\k\uvk\nonumber\\
\lap\uvk=-k^2\uvk\nonumber\\
\pde\uvk t=-i\om\uvk\nonumber\\
\spd\uvk t=-\om^2\uvk.
\eeqa
Hence
\beq
\lep\lap-\frac1{v^2}\frac{\partial^2}{\partial t^2}\rip\uvk=\lep-k^2+\frac{
\om^2}{v^2}\rip\uvk=0,
\eeq
demonstrating that we do indeed have a solution of the wave equation.

This solution is a wave ``traveling'' in the direction of $\k$ in the sense 
that a point of constant phase, meaning $\k\cdot\x-\om t=constant$, moves along 
this direction with a speed $v$ which is $\om/k$. Furthermore, 
we have a {\em plane wave}, by which we mean that a surface of constant 
phase is a plane; in particular, the surfaces of constant phase are just 
planes perpendicular to $\k$.

\centerline{\psfig{figure=fig1.ps,height=2.125in,width=4.25in}}

{\narrower{Fig.1:  A point of stationary phase moves with 
velocity $|v|=\omega/k$}}

\subsection{Conditions Imposed by Maxwell's Equations}
Next, let us see how the electromagnetic fields can be described in terms 
of these scalar plane waves. Let us look for an electric field and a 
magnetic induction with the forms
\beq
\Ext=\E_0\ekxot \hsph\Bxt=\B_0\ekxot
\eeq
with the understanding that the true fields are the real parts of these 
complex expressions. 

In addition to satisfying the wave equation, the complex fields must be 
solutions of the Maxwell equations. Let us see what additional constraints 
are thereby imposed. Consider first the divergence equations; these require that
\beq
0=\div\Bxt=\div\leb\B_0\ekxot\rib=i\k\cdot\B_0\ekxot
\eeq
and
\beq
0=\div\Ext=\div\leb\E_0\ekxot\rib=i\k\cdot\E_0\ekxot.
\eeq
Or
\beq
\k\cdot\B_0=0\hbox{\hsph and \hsph}\k\cdot\E_0=0.
\eeq
These conditions mean that $\B_0$ and $\E_0$ must be perpendicular to $\k$,
which is to say, parallel to the surfaces of constant phase and 
perpendicular to the direction in which the surface of constant phase is 
moving. Such an electromagnetic wave is called a {\em transverse wave}. 
Notice that this nomenclature is consistent with our definition in the last 
chapter of a transverse vector field as one having zero divergence.

There are further conditions on the amplitudes $\E_0$ and $\B_0$ from the 
other Maxwell equations. From the Amp\`ere law one has
\beq
\curl\Bxt=\frac{\mu\ep}c\pde\Ext t
\eeq
which leads to
\beq
i\k\times\B_0=-\frac{i\om\ep\mu}c\E_0
\eeq
or
\beq
\E_0=-\frac{\k\times\B_0}{k\sqrt{\mu\ep}}=-\frac{\nn\times\B_0}{\sqrt{\ep
\mu}}
\eeq
where $\nn=\k/k$ is a unit vector in the direction of propagation of the 
wave. From Faraday's Law and similar manipulations one finds the further, 
and final condition that
\beq
\B_0=\sqrt{\mu\ep}(\nn\times\E_0);
\eeq
however, one may also find this relation from \eq{16} and the condition
that $\nn\cdot\B_0=0$ and so it is not an additional constraint. Alternatively, 
one may derive \eq{16} from \eq{17} and the condition $\nn\cdot\E_0=0$. As a 
consequence, one may, for example, write
\beq
\Ext=\E_0\ekxot
\eeq
where the only condition on $\E_0$ is $\nn\cdot\E_0=0$. Then $\Bxt$ follows 
from \eq{17} and is
\beq
\Bxt=\sqrt{\mu\ep}(\nn\times\E_0)\ekxot.
\eeq
Alternatively, we may start by writing
\beq
\Bxt=\B_0\ekxot
\eeq
where $\B_0$ is orthogonal to $\k$, $\nn\cdot\B_0=0$. Then $\Ext$ is given 
from \eq{16} as
\beq
\Ext=-\frac{\nn\times\B_0}{\sqrt{\ep\mu}}\ekxot.
\eeq
From these conditions, and those obtained in the previous paragraph, we
may conclude that $\E$, $\B$ and $\k$ form a mutually orthogonal set.

Before leaving this section, let's look at the time-averaged energy density 
and Poynting vector in such electromagnetic waves. We shall write them in 
terms of the amplitude $\E_0$. First,
\beq
<\S>=\frac c{8\pi}\Re\leb\Ext\times\H^*(\x,t)\rib=\frac c{8\pi}\sqrt{\frac
\ep\mu}\Re\leb\E_0\times(\nn\times\E_0^*)\rib=\frac c{8\pi}\sqrt{\frac\ep
\mu}|\E_0|^2\nn.
\eeq
Similarly,
\beq
<u>=\frac1{16\pi}\Re\lep\Ext\cdot\D^*(\x,t)+\Bxt\cdot\H^*(\x,t)\rib=
\frac\ep{8\pi}|\E_0|^2.
\eeq
The time-averaged momentum density is:
\beq
<\g>=\frac1{8\pi c}\Re\leb\Ext\times\H^*(\x,t)\rib=\frac{\sqrt{\ep/\mu}}
{8\pi c}|E_0|^2\nn.
\eeq
The evaluation of the time-averaged Maxwell stress tensor is left as an 
exercise.
\section{Polarization}
In this section we address the question of the most general possible 
monochromatic plane wave, which amounts to asking what are the possible 
choices of $\E_0$. Let us specify that $\k=k\ept$ and suppose that we have 
an orthogonal right-handed set of real unit basis vectors $\epi$, $i=1,2,3$. 
Then it must be the case that $\E_0\cdot\ept=0$ which means that the most 
general amplitude $\E_0$ can be expanded as
\beq
\E_0=E_{01}\epu+E_{02}\epd.
\eeq
The scalar amplitudes in this expansion can be complex so we have in all 
four real amplitudes which we may choose with complete abandon. Let us 
write the complex scalar amplitudes in polar form,
\beq
E_{01}=E_1e^{i\ph_1}\hsp{1.0}E_{02}=E_2e^{i\ph_2} 
\eeq
where $E_i$ and $\ph_i$, $i=1,2$, are real. Further, introduce
\beq
E_0=(E_1^2+E_2^2)^{1/2}\hsph\hbox{and}\hsph\ph=\ph_2-\ph_1.
\eeq
Then the complex field becomes
\beq
\Ext=E_0\al_1\lep\epu+(\al_2/\al_1)e^{i\ph}\epd\rip e^{i(\k\ept\cdot\x-\om t)}
e^{i\ph_1}
\eeq 
where $\al_i=E_i/E_0$ and $\al_1^2+\al_2^2=1$. In this form, the wave is 
seen to have just two interesting parameters, $\al_2/\al_1$ and $\ph_2-
\ph_1$; these specify the relative phase and amplitude of the two 
components of the vector amplitude. The other two parameters simply to set 
the overall magnitude of the field and its absolute phase\footnote{These 
will, of course, be interesting if the wave meets another wave; but they 
are not interesting if there is no other wave.}. 

Look at the real part of the complex wave as a function of time at a point in 
space which is conveniently taken to be the origin. Aside from the overall 
magnitude and phase, the wave looks like
\beq
\E\sim\epu\cos(\om t)+(\al_2/\al_1)\epd\cos(\om t-\ph).
\eeq
If we map out the path traced by the tip of this vector in the space of 
$\ep_1$ and $\ep_2$, we find in general an ellipse. The ellipse is 
characterized by two parameters, equivalent to $\al_2/\al_1$ and $\ph$, these 
being its eccentricity (the ratio of the semi-minor to the semi-major axis) and
the amount by which the major axis is rotated relative to some fixed direction 
such as that of $\ep_1$. Such a wave is said to be {\em elliptically 
polarized}, the term ``polarization'' referring to the behavior of the 
electric field at a point as a function of time. There are two limiting 
special cases. One is when the eccentricity is unity in which case the 
ellipse becomes a circle and the wave is said to be {\em circularly 
polarized}; the second is when the eccentricity becomes zero so that the 
ellipse reduces to a line and the wave is {\em linearly polarized}.

\centerline{\psfig{figure=fig2.ps,height=2.0in,width=8.5in}}

{\narrower\em{Fig.2: linearly ($\al_2=0$) and circularly ($\al_2/\al_1=1\;\;\ph=\pi/2$)
polarized }}

Often one uses a set of complex basis vectors in which $\epu$ and $\epd$ 
are replaced by vectors $\eppm$ defined by 
\beq
\eppm\equiv\frac1{\sqrt2}(\epu\pm i\epd).
\eeq
These have the properties
\beq
\eppm\cdot\ept=0\hsph\eppm\cdot\epmp^*=0\hsph\eppm\cdot\eppm^*=1,
\eeq
and it is possible to write the electric field of a general plane wave as
\beq
\Ext=(E_+\epp+E_-\epm)\ekxot,
\eeq
where $E_+$ and $E_-$ are arbitrary complex constants. If just one of these 
is non-zero and is written in polar form, then, aside from phase, the 
complex electric field at a point is
\beq
\E=|E_\pm|\frac1{\sqrt2}(\epu\pm i\epd)e^{-i\om t}.
\eeq
The real part then varies as $\epu\cos(\om t)\pm\epd\sin(\om t)$ which is 
a circularly polarized wave. In the case of the upper sign, one says that 
the wave is {\em left-circularly polarized} or that it has {\em positive 
helicity}; in the case of the lower sign, it is {\em right-circularly 
polarized} or has {\em negative helicity}. In writing the general wave in 
terms of these basis vectors, we have expressed it as a superposition of 
positive and negative helicity waves with amplitudes $E_+$ and $E_-$, 
respectively.

\section{Boundary Conditions; Waves at an Interface}

In this section, we shall find out what plane waves must look like in 
semi-infinite media or when there is a planar boundary between two 
nonconducting materials such as air (or vacuum) and glass. We will need 
appropriate continuity conditions on the fields at the interface. There may 
be derived from general kinematic considerations, and from
Maxwell equations.


The basic example from which all cases may be inferred is that of a planar 
interface located at $z=0$ dividing space into two regions, $z<0$ and 
$z>0$. In the former, we assume an insulating material with dielectric 
constant $\ep$ and permeability $\mu$; in the latter there is another 
insulating material with $\ep'$ and $\mu'$. 

\centerline{\psfig{figure=fig2a.ps,height=2.0in,width=8.5in}}

\noindent Now suppose that from the 
left, or $z<0$, there is an {\em incident} wave which has 
electromagnetic fields
\beq
\Ext=\E_0\ekxot,\hsph\Bxt=\sqrt{\mu\ep}\frac{\k\times\Ext}k.
\eeq
Also, $k=\om\sqrt{\mu\ep}/c$, and $\k\cdot\zh>0$ so that the wave is 
approaching the interface. Finally, $\E_0$ is such that $\k\cdot\E_0=0$.

The incident wave is a solution of the Maxwell equations in the region $z<
0$. At the interface, however, it is not a solution; there must be other waves 
present in order to satisfy the Maxwell equations (or boundary conditions) 
here. To phrase it another way, when the incident wave hits the interface, 
additional waves, called {\em transmitted} (or {\em refracted}) and {\em 
reflected} waves must be generated. The refracted waves are the ones that 
propagate into the medium at $z>0$; the reflected waves are the ones that 
propagate back into the other medium. 

\subsection{Kinematic Conditions}
We can, from quite general considerations, learn a lot about 
the properties of the reflected and refracted waves. 

{\bf{First}}, in order that 
the continuity conditions remain satisfied at all times, given that they 
are satisfied at one instant of time, these waves must have the same time 
dependence as the incident wave. This statement follows from the linear 
nature of the field equations (each term in the equations is proportional 
to some component of one of the fields). Hence, all fields vary in time as
$\emot$. 

\centerline{\psfig{figure=fig4.ps,height=1.75in,width=4.25in}}

{\bf{Second}}, the continuity conditions must be satisfied at all points 
on the interface or $z=0$ plane. Suppose that they are satisfied at one 
particular point, such as $\x=0$. Then, in order that they remain so for 
other points on the interface, each wave must vary in the same fashion as 
each of the other waves as one moves in the plane of the interface. This 
statement follows, as does the first one, from the linearity of the field 
equations. Now, since 
the dependence of a plane wave on position is $\exp(i\k\cdot\x)$, this 
condition means that all waves (incident, reflected, and refracted) must 
have wave vectors whose components lying in the plane of the interface are 
identical. 

\centerline{\psfig{figure=fig5.ps,height=1.75in,width=4.25in}}

\noindent We can express this condition as
\beq
\nn\times\k=\nn\times\k'=\nn\times\k''
\eeq
where $\k'$ and $\k''$ are, respectively, the wave vectors of any refracted 
and reflected waves. This relation may also be written as
\beq
k\sin i=k''\sin r''=k'\sin r
\eeq
where $i$, $r''$, and $r$ are the angles between the wavevectors of the 
incident, reflected, and transmitted waves and the normal to the interface. 
They are called the {\em angle of incidence}, the {\em angle of 
reflection}, and the {\em angle of refraction}.

\centerline{\psfig{figure=fig6.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 6:  Definition of the angles $i$, $r''$, and $r$}}
\medskip


{\bf{Finally}}, any reflected wave is a solution of the same wave equation 
as the incident wave; consequently, it has a wave number $k''=k$. Any 
transmitted wave, however, has wave number $k'=\om\sqrt{\mu'\ep'}/c$, so
$k'\ne k$. If we combine these statements with \eq{36}, we can see that 
$r''=i$, {\em{the angle of incidence equals the angle of reflection}}.
For a transmitted wave, however, the wave equation is such that 
$k'=\om\sqrt{\mu'\ep'}/c$ which is not $k$; in fact, $k'/n'=k/n$ where
\beq
n\equiv\sqrt{\mu\ep} \hsph\hbox{and}\hsph n'\equiv\sqrt{\mu'\ep'}
\eeq 
are the {\em indices of refraction} in the two materials. Using these 
definitions in \eq{36} we find
\beq
n\sin i=n'\sin r
\eeq
which is known in optics as {\em Snell's Law}.

\subsection{Conditions from Maxwell's Equations}

Notice that we derived Snell's law and the statement $i=r''$ without using 
explicitly the continuity conditions; we had only to use the fact that 
there {\bf are} linear continuity conditions. Hence these properties are 
called kinematic properties (they don't depend on the particular dynamics of 
the fields which are given by the Maxwell equations) and they are 
applicable to much more than just electromagnetic phenomena.

To fully develop the rules of reflection and refraction for electromagnetic 
waves, we must use the Maxwell equations to tell us the specific relations 
among the fields and then must apply the continuity conditions at a 
specific point on the interface, such as $\x=0$, and at a specific time, 
such as $t=0$\footnote{For other points and times we know that the conditions will be satisfied by making sure the kinematic conditions derived above 
are satisfied.}. 
\beq
\div\D=4\pi\rh,\;\div\B=0,\;\curl\H=\frac{4\pi}c\J+\frac1c\pde\D t,\;
\curl\E=-\frac1c\pde\B t.
\eeq

\centerline{\psfig{figure=fig3.ps,height=1.75in,width=4.25in}}

\noindent Application of the divergence theorem to the two 
divergence equations using 
the familiar pillbox construction leads, as for the static case, to the 
continuity conditions
\beqa
(\D_1-\D_2)\cdot\nn=4\pi\si\\(\B_2-\B_1)\cdot\nn=0
\eeqa
where $\nn$ is a unit outward normal from material 1 and $\si$ is the 
macroscopic surface-charge density. Application of Stokes' theorem to the 
curl equations in the ``usual'' way leads to 
\beqa
\nn\times(\E_2-\E_1)=0\\\nn\times(\H_2-\H_1)=\frac{4\pi}c\K
\eeqa
where $\K$ is the macroscopic surface-current density lying inside of the 
loop C to which Stokes' theorem is applied. Notice that the 
time derivatives in Faraday's law and Amp\`ere's law do not contribute to 
the continuity conditions.\footnote{We have assumed that there are no 
singular parts of the time derivatives localized at the interface; were 
there any such contributions, they would show up in the continuity 
conditions.} 

For uncharged insulators, the surface sources $\si$ and $\K$ are always 
zero; then the continuity conditions are especially simple and state that 
the normal components of $\D$ and $\B$ are continuous as are the tangential 
components of $\H$ and $\E$.

At $\x=0$, $t=0$, the fields of an incident wave, a single transmitted 
wave, and a single reflected wave\footnote{ We don't know at this point 
that we need only one reflected and one transmitted wave to obtain a 
solution to the boundary value problem. By construction, we will see that 
such is the case.} may be written as follows:\newline Incident wave:
\beq
\E=\E_0\hsph\B=\frac n k(\k\times\E_0)\nonumber
\eeq
Reflected wave:
\beq
\E=\E_0''\hsph\B=\frac n k(\k''\times\E_0'')\nonumber
\eeq
Transmitted wave:
\beq
\E=\E_0'\hsph\B=\frac{n'}{k'}(\k'\times\E_0')=\frac n k(\k'\times\E_0').
\eeq
We suppose that we are given $n$, $n'$, $\k$, and $\E_0$; we need to find 
$\k'$, $\k''$, $\E_0'$, and $\E_0''$. The wave vectors follow from the 
kinematic relations; they all lie in the plane containing the normal to the 
interface and the incident wave vector, called the {\em plane of incidence}
and make angles with the normal as 
discussed above. As for the amplitudes, they are found from the continuity 
conditions: \newline
1.~$D_n$ continuous:
\beq
\ep(\E_0+\E_0'')\cdot\nn=\ep'\E_0'\cdot\nn
\eeq
2.~$B_n$ continuous:
\beq
(\k\times\E_0+\k''\times\E_0'')\cdot\nn=(\k'\times\E_0')\cdot\nn
\eeq
3.~$\E_t$ continuous:
\beq
(\E_0+\E_0'')\times\nn=\E_0'\times\nn
\eeq
4.~$\H_t$ continuous:
\beq
\frac1\mu(\k\times\E_0+\k''\times\E_0'')\times\nn=\frac1{\mu'}(\k'\times\E_0'
)\times\nn.
\eeq
It is a messy bit of algebra to solve these equations in the general case. 
The task can be made simpler by writing the incident wave's electric field 
as a linear combination of two linearly polarized waves, which is always 
possible. One solves each of these cases separately. The appropriate sum of 
the two solutions is then the solution of the original problem. Once again, 
the linearity of the field equations leads to enormous simplification of 
the algebra. The two cases that we are going to treat are
\begin{enumerate}
\item polarization of $\E_0$
perpendicular to the plane of incidence and
\item polarization parallel to 
the plane of incidence.
\end{enumerate}

\subsubsection{Polarization of $\E_0$ Perpendicular to the Plane}

\centerline{\psfig{figure=fig7.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 7: Polarization of $\E_0$ perpendicular to the plane 
of incidence }}
\medskip

The figure sets the conventions for 
the first case. They are such that $\E_0=E_0\yh$, $\E_0'=E_0'\yh$, and 
$\E_0''=E_0''\yh$. Remember also that $k''=k$, $k/n=k'/n'$, and 
$n\sin i=n'\sin r$.
Now apply the four continuity conditions. The first gives nothing because 
there is no normal component of the electric displacement or electric 
field; the second gives $E_0+E_0''=E_0'$; the third gives the same 
constraint as the second; and the fourth results in $(k/\mu)\cos i\,(E_0-
E_0'')=(k'/\mu')\cos r\,E_0'$. Since $k'=kn'/n$ and $n=\sqrt{\mu\ep}$, 
we can write the latter as $\sqrt{\ep/\mu}\cos i\,(E_0-E_0'')=\sqrt{\ep'/
\mu'}\cos r\,E_0'$. In addition, $\cos r=\sqrt{1-\sin^2r}=\sqrt{1-(n/n')^2
\sin^2i}$. Combining these relations we find the two conditions
\beq
E_0'-E_0''=E_0\hsph\sqrt{\frac{\ep'}{\mu'}}\sqrt{1-\lep\frac{n}{n'}\rip^2
\sin^2i}\,E_0'+\sqrt{\frac\ep\mu}\cos i\,E_0''=\sqrt{\frac\ep\mu}\cos i\,E_0.
\eeq
Notice that these are written entirely in terms of the angle of incidence; 
the angle of refraction does not appear. Their solution is easily shown to 
be
\beqa
E_0'=\frac{2n\cos i}{n\cos i+(\mu/\mu')\sqrt{n'^2-n^2\sin^2i}}E_0\nonumber
\eeqa
and
\beq
E_0''=\frac{n\cos i-(\mu/\mu')\sqrt{n'^2-n^2\sin^2i}}{n\cos i+(\mu/\mu')
\sqrt{n'^2-n^2\sin^2i}}E_0
\eeq

\subsubsection{Polarization of $\E_0$ Parallel to the Plane}


\centerline{\psfig{figure=fig8.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 8:  Polarization of $\E_0$ parallel to the plane 
of incidence}}
\medskip

The second case, polarization in the plane of incidence may be similarly 
analyzed.  The figure shows the conventions for this case. 
They are such that $\E_0=E_0(\sin i\,\zh-\cos i\,\xh)$, $\E_0'=E_0'(\sin r
\,\zh-\cos r\,\xh)$, and $\E_0''=E_0''(\sin i\,\zh+\cos i\,\xh)$. The first 
boundary condition implies that $\ep\sin i\,(E_0+E_0'')=\ep'\sin r\,E_0'$; the 
second gives nothing; the third gives $\cos i\,(-E_0+E_0'')=-\cos r\,E_0'$; and
the fourth gives a condition that is redundant with the first when Snell's law 
is invoked. Thus we may write the two conditions, after removing all 
occurrences of $r$ as in the first case, as
\beq
\sqrt{\frac\ep\mu}(E_0+E_0'')=\sqrt{\frac{\ep'}{\mu'}}E_0'\hsph
\cos i\,(E_0+E_0'')=\sqrt{1-(n/n')^2\sin^2 i}\,E_0'.
\eeq
Their solution is 
\beqa
E_0'=\frac{2nn'\cos i}{(\mu/\mu')n'^2\cos i+n\sqrt{n'^2-n^2\sin^2i}}E_0
\nonumber
\eeqa
and
\beq
E_0''=\frac{(\mu/\mu')n'^2\cos i-n\sqrt{n'^2-n^2\sin^2i}}{(\mu/\mu')n'^2
\cos i+n\sqrt{n'^2-n^2\sin^2i}}E_0
\eeq

Our solutions to the reflection-refraction problem have the following 
characteristics by design. First, as mentioned above, they involve only the 
angle of incidence, the angle of refraction having been removed wherever it 
appeared by using Snell's law; second, the material properties enter 
through the permeabilities and indices of refraction as opposed to the 
permeabilities and dielectric constants. The reason is that for most of the 
materials one encounters, $\mu=\mu'=1$ and so the permeabilities drop out 
of the relations. Second, one is generally more likely to be given an index 
of refraction than a dielectric constant and so expressing the amplitudes 
in terms of $n$ makes them more readily applicable.\footnote{Of course, the 
relation between $n$ and $\ep$ is sufficiently simple that there is really 
no great difference.}

Equations (52) and (54) are known as {\em Fresnel's equations}; with them 
we can calculate the reflection and transmission of a plane wave at a 
planar interface for arbitrary initial polarization. Such an incident wave 
gives rise to a single reflected plane wave and a single transmitted plane 
wave, meaning that there is just one reflected wave vector $\k''$ and one 
transmitted wave vector $\k'$. 

\subsection{Parallel Interfaces}
With a little thought we may see how to 
generalize to the case of two (or more) parallel interfaces. Consider the 
figure showing two parallel interfaces separating three 
materials. If we follow the consequences of an incident plane wave from the 
first material on one side we can see that the reflection processes within the 
middle material of the ``sandwich'' generate many plane waves in here, but that
these waves have just two distinct wave vectors. 

\centerline{\psfig{figure=fig9.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 9: Plane wave incident on a sandwich. }}
\medskip

\noindent Also, all waves
transmitted into the third material have the same wave vector, and the 
``reflected'' waves in the first medium all have a single wave vector. Hence 
one finds that in the first medium, there are just two waves with electric 
fields
\beq
\E=\E_0\ekxot\hsph\E_r=\E_{r0}e^{i(\k_r\cdot\x-\om t)};
\eeq
in the middle medium there are again just two distinct waves with fields
\beq
\E'=\E_0'e^{i(\k'\cdot\x-\om t)}\hsph\E_r'=\E_{r0}'e^{i(\k_r'\cdot\x-\om t)
}
\eeq
and in the third medium there is just one plane wave with field
\beq
\E''=\E_0''e^{i(\k''\cdot\x-\om t)}.
\eeq
To find the four amplitudes $\E_{r0}$, $\E_0'$, $\E_{r0}'$, and $\E_0''$, 
one must apply the boundary conditions at the two interfaces, leading to 
four distinct linear relations involving these amplitudes and that of the 
incident wave, $\E_0$. Solving these equations, one finds the amplitudes of 
all waves in terms of that of the incident wave.

Returning briefly to Fresnel's equations for reflection and refraction at a 
single interface, let us look at the special case of normal incidence, $i=
0$. then $r=0$ also, and the first set (polarization normal to the plane of 
incidence) of Fresnel equations tells us that \footnote{Actually, both sets 
of Fresnel equations are applicable for normal incidence. The second set, 
however, will produce a result with some signs switched as a consequence of the
different conventions used for the directions of the electric fields in the 
two cases.}
\beq
E_0'=\frac{2n}{n+(\mu/\mu')n'}E_0 \hsph E_0''=\frac{n-(\mu/\mu')n'}{n+
(\mu/\mu')n'}E_0.
\eeq
These are simple results, especially when $\mu=\mu'$. They clearly tell us 
that when the two materials have comparable indices of refraction and 
permeabilities, the wave is mostly transmitted and when they have very 
different properties (an engineer would call that impedance mismatch), 
reflection is the rule. Notice also that if $n'\mu/\mu'>n$, the reflected 
amplitude is opposite in sign to the incident one, meaning that the 
electric field of the reflected wave is phase shifted by $\pi$ radians relative
to that of the incident one under these circumstances.
\section{Reflection and Transmission Coefficients}
In this section we look at the power or energy transmitted and reflected at 
an interface between two insulators. To do so, we must evaluate the 
time-averaged power in the incident, reflected, and transmitted waves which 
is done by calculating the Poynting vector. The energy current density 
toward or away from the interface is then given by the component of the 
Poynting vector in the direction normal to the interface. In the second  
medium, where there is just a single (refracted) wave, the normal component of 
$\S$ is unambiguously the transmitted power per unit area. But in the first 
medium, the total electromagnetic field is the sum of the fields of the 
incident and reflected waves. In evaluating $\E\times\H$, one finds three 
kinds of terms. There is one which is the cross-product of the fields in 
the incident wave, and its normal component gives the incident power per 
unit area. A second is the cross-product of the fields in the reflected 
wave, giving the reflected power. But there are also two cross-terms 
involving the electric field of one of the plane waves and the magnetic 
field of the other one. It turns out that the time-average of the normal 
component of these terms is zero, so that they may be ignored in the 
present context. Bearing this in mind, we have the following quantities of 
interest: \newline The time-averaged incident power per unit area:
\beq
{\cal P}=<\S>\cdot\nn=\frac c{8\pi}\sqrt{\frac\ep\mu}|\E_0|^2\
\frac{\k\cdot\nn}{k}
\eeq
The time-averaged reflected power per unit area:
\beq
{\cal P}''=-<\S''>\cdot\nn=\frac c{8\pi}\sqrt{\frac\ep\mu}|\E_0''|^2
\frac{\k''\cdot\nn}{k}
\eeq
The time-averaged transmitted power per unit area:
\beq
{\cal P}'=<\S'>\cdot\nn=\frac c{8\pi}\sqrt{\frac{\ep'}{\mu'}}|\E_0'|^2
\frac{\k'\cdot\nn}{k'}d
\eeq
The {\em reflection coefficient} $R$ and the {\em transmission coefficient} 
$T$ are defined as the ratios of the reflected and transmitted power to the 
incident power.

We may calculate the reflection and transmission coefficients for the cases 
of polarization perpendicular and parallel to the plane of incidence by 
using the Fresnel equations. If an incident wave has general polarization so 
that its fields are linear combinations of these two special cases, then 
there is once again the possibility of cross terms in the power involving 
an electric field with one type of polarization and a magnetic field with 
the other type. Fortunately, these turn out to vanish, so that one may 
treat the two polarizations individually. 

For the case of {\bf{polarization 
perpendicular to the plane of incidence}}, we use the Fresnel equations (52) 
and (54) for the reflected and transmitted amplitudes and have
\beq
T=\frac{\sqrt{\frac{\ep'}{\mu'}}\frac{4n^2\cos^2i\cos r}{(n\cos i+(\mu/\mu'
)\sqrt{n'^2-n^2\sin^2i})^2}}{\sqrt{\frac\ep\mu}\cos i}
\eeq
Making use of the relations $n=\sqrt{\ep\mu}$, $n'=\sqrt{\ep'\mu'}$, 
$\sin r=(n/n')\sin i$, and $\cos i=\sqrt{1-\sin^2i}$, one finds that
\beq
T=\frac{4n(\mu/\mu')\cos i\sqrt{n'^2-n^2\sin^2i}}{[n\cos i+(\mu/\mu')
\sqrt{n'^2-n^2\sin^2i}\,]^2}.
\eeq
By similar means one can write the reflection coefficient as
\beq
R=\frac{[n\cos i-(\mu/\mu')\sqrt{n'^2-n^2\sin^2i}\,]^2}{[n\cos i+(\mu/\mu')
\sqrt{n'^2-n^2\sin^2i}\,]^2}
\eeq
By inspection one can see that $R+T=1$ which expresses the conservation of 
energy; what is not transmitted is reflected.

The case of {\bf{polarization in the plane of incidence}} is treated similarly.
One finds
\beq
T=\frac{4nn'^2(\mu/\mu')\cos i\sqrt{n'^2-n^2\sin^2i}}{[(\mu/\mu')n'^2\cos i
+n\sqrt{n'^2-n^2\sin^2i}\,]^2}
\eeq
and
\beq
R=\frac{[(\mu/\mu')n'^2\cos i-n\sqrt{n'^2-n^2\sin^2i}\,]^2}{[(\mu/\mu')n'^2
\cos i+n\sqrt{n'^2-n^2\sin^2i}\,]^2}.
\eeq
Once again, $R+T=1$.

\section{Examples }

\subsection{Polarization by Reflection}

From inspection of Fresnel's equations, we can see that the relative 
amounts of transmitted and reflected amplitude  depend on the state of 
polarization and are distinctly not the same for both polarizations.

\centerline{\psfig{figure=fig10.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 10: Reflection Coefficient when $n'>n$, and 
$\mu'=\mu=1$}}
\medskip

\noindent That 
means that in the general case, the polarizations of the transmitted and 
reflected waves will not be the same as that of the incident one. A very 
special case has to do with the reflected wave given incident polarization 
in the plane of incidence. We see that the reflected amplitude will vanish
if\footnote{We let $\mu=\mu'$ in this section unless explicitly stated 
otherwise; keeping the permeability around usually contributes nothing but 
extra work and obfuscation.}
\beq
n'^2\cos i=n\sqrt{n'^2-n^2\sin^2i}.
\eeq
Squaring this relation we find
\beq
n'^4\cos^2i=n^2n'^2-n^4\sin^2i=n'^4(1-\sin^2i)\hbox{ or }\sin^2i=
\frac{n'^2}{n'^2+n^2}\hbox{ or }\tan i=\frac{n'}n.
\eeq
This special angle of incidence is called the {\em Brewster angle},
\beq
i_B=\arctan(n'/n);
\eeq

\centerline{\psfig{figure=fig11.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 11: No reflected wave when $i=i_B$ and the field
is polarized in the plane.}}
\medskip

\noindent a wave polarized in the plane of incidence and incident 
on the interface at the
Brewster angel is completely transmitted with no reflected wave. If a wave 
of general polarization is incident at the Brewster angle, then the 
reflected wave is completely (linearly) polarized perpendicular to the plane
of incidence. Hence this phenomenon provides a method for obtaining a linearly 
polarized wave from an unpolarized one. More generally, if the angle of 
incidence is reasonably close to the Brewster angle, the reflected light is 
to a large degree polarized perpendicular to the plane of incidence. This 
fact is utilized by polarizing sun glasses which screen out most of the 
light polarized parallel to the surface of the earth, which is to say, most 
of the light reflected by the earth.

\centerline{\psfig{figure=fig12.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 12: Light reflected from the ocean (glare) is largely
polarized along the horizon, and may be removed with polarized sunglasses.}}
\medskip


\subsection{Total Internal Reflection}


As a second example we look at the phenomenon of total internal reflection 
which is the opposite of the one just considered in that no energy is 
transmitted across an interface under appropriate conditions. Suppose that 
$n>n'$.  As shown in the figure, this means that $r>i$.  

\centerline{\psfig{figure=fig13.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 13: A series of angles when $n>n'$.}}
\medskip

\noindent  Now consider an 
incident wave with $i$ large enough that $n\sin i>n'$. 
 How can we have a refracted wave with $r$ such that Snell's law,
$n\sin i=n'\sin r$ is satisfied?  Recall our argument for Snell's law; it  was 
based on the fact that the wave vector $\k'$ of the refracted wave had to have 
a component $k_t'$ parallel to the interface equal to the same component of the 
incident wave. Given that $n\sin i>n'$, this condition means that $k_t'$ is 
{\bf larger} than $\om n'/c$ which is supposed to be the magnitude of $\k'$,
according to the wave equation. But there is a way around this. The 
condition that comes from the wave equation is that, if $k_t'$ and $k_n'$ 
are respectively the components of $\k'$ tangential and normal to the 
interface, then $k_t'^2+k_n'^2=\om^2n'^2/c^2$. If $k_t'>\om n'/c$, we can 
satisfy this condition by having $k_n'$ be imaginary. In particular,
\beq
k_n'=\pm i\frac{n\om}c\sqrt{\sin^2i-(n'/n)^2}.
\eeq

The choice of sign has to be such as to produce a wave that damps away to 
nothing in the second medium; otherwise it becomes exceedingly large (which 
is unphysical behavior) as one moves far away from the interface. 
Now that we have figured out what is $\k'$; that is, $k_t'=(n\om/c)\sin i$ 
and $k_n'$ is given by \eq{70}, we can see the character of the transmitted 
electric field. It is
\beq
\E'\sim e^{ik_t'x}e^{-|k_n'|z}\emot
\eeq
where $\xh$ is the direction of the tangential component of $\k$.

The Poynting vector for a wave of this sort has no component directed 
normal to the interface although there is one parallel to the interface.
To see this, take $\E$ to be in the y-direction. 

\centerline{\psfig{figure=fig14.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 14: Polarization $\perp$ to the plane of incidence.}}
\medskip

\noindent Then
\[
\E'=\E_0'e^{i(k_t'x-\om t)}e^{-|k_n'| z}
\]
so that
\[
S_z'=\frac{c}{8\pi} \Re\lep \E'\times\B'^*\rip_z
=\frac{c}{8\pi} \Re\lep E'_y\times B_x'^*\rip\,.
\]
We may use Faraday's law to relate $\E$ to $\B$
\[
\curl\E=-\frac{1}{c}\pde{B}{t} \to \frac{i\om}{c} B_x' = |k_n'| E_y'
\,.\]
Thus, 
\[
S_z'=\frac{c}{8\pi} \Re\left\{-E_y'\frac{c|k_n'|}{-i\om}E_y'^*\right\}=0
\]
Thus, as shown in the figure below, when $i>i_c$, the power is totally
reflected.

\centerline{\psfig{figure=fig15.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 15: Reflection Coefficient when $n>n'$, and $\mu'=\mu=1$}}
\medskip

\noindent What we have is therefore a {\em surface wave} confined to the 
region close to the interface and transporting energy parallel to it. 
Moreover, by 
evaluating the Poynting vector of the reflected and incident waves, one 
finds that as much energy is reflected from the interface as is incident 
upon it. Hence we have the phenomenon of perfect or total reflection of 
the incident wave. This phenomenon is utilized in fiber optics; an 
electromagnetic wave is propagated inside of a thin tube of some material 
having a large index of refraction and surrounded by another material 
having a much smaller index.  Wherever the wave is incident upon the 
wall of the tube, it is completely reflected. 

\centerline{\psfig{figure=fig16.ps,height=1.75in,width=4.25in}}
{\narrower\em{Figure 16: Total internal reflection occurs within a fiber
optic tube.}}
\medskip

\noindent There is some natural 
attenuation of the wave because of imperfect dielectric properties of the 
material itself or its coating; nevertheless, a beam of light, for example, 
can be transmitted long distances and around many curves (as long as they 
aren't too sharp) in such a ``pipe.''

\section{Models of Dielectric Functions}
The dielectric ``constant'' of almost any material is in fact a function of 
frequency, meaning that it has different values for waves of different 
frequencies. 

\centerline{\psfig{figure=fig17.ps,height=1.0in,width=4.25in}}
{\narrower\em{Figure 17:  In a dispersive medium waves of different 
frequencies have different phase velocities $v=c/\sqrt{\ep(\om)\mu}$.}}
\medskip

\noindent We can make a simple model of the dielectric ``function'' of 
an insulating material as follows: Suppose that the charges which primarily 
respond to an electric field are electrons bound on atoms or molecules. Let 
one such electron be harmonically bound, meaning that the binding forces 
are treated as linear in the displacement of the charge from its 
equilibrium  position. Also, let there be a damping force proportional to 
the velocity $\v$ of the electron. Then, if the mass and charge of the 
electron are $m$ and $-e$, the equation of motion of the electron under the 
influence of an electric field $\Ext$ is
\beq
m\lep\sde\x t+\ga\der\x t+\om_0^2\x\rip=-e\Ext.
\eeq
The harmonic restoring force is expressed through a ``natural'' frequency 
of oscillation $\om_0$ of the electron. We have ignored the possible 
influence of a magnetic induction $\Bxt$ on the electron's motion. Typically 
this force is much smaller than the electric field force because the 
electron's speed is much smaller than $c$; there can be exceptions, 
however, and one of them is explored below.

Next, the typical magnitude of the electron's displacement $|\x|$ is on the 
order of an atomic size. 

\centerline{\psfig{figure=fig18.ps,height=1.5in,width=4.25in}}
{\narrower\em{Figure 18:  If the wavelength of the incident wave is
much larger than the electronic displacement, then we may neglect the
spacial dependence of $\Ext$.}}
\medskip

\noindent If the electric field $\Ext$ is that of visible or 
even ultraviolet light, then  the displacement is much smaller than 
distances over which $\Ext$ varies significantly, meaning that we can 
approximate $\Ext\approx\E(0,t)=\E_0 \exp(-i\om t)$. In this limit, the 
solution we seek is of the form $\x(t)=\x_0 \exp(-i\om t)$. Substituting into 
the equation of motion, we find that the equation for the amplitude of the 
motion is
\beq
m(-\om^2-i\om\ga+\om_0^2)\x_0=-e\E_0
\eeq
or
\beq
\x_0=\frac{-e\E_0}{m(\om_0^2-i\om\ga-\om^2)}.
\eeq

The amplitude of the dipole moment associated with the motion of this electron 
is $\edm_0=-ex_0$. To find the polarization, we need to compute the dipole 
moments of all electrons in some finite volume of material. These electrons 
will not all have the same damping or natural frequencies, so let us say that 
there are $n$ molecules per unit volume with $z$ electrons each. If $f_i$ of 
the electrons on each molecule have resonant frequency $\om_i$ and damping 
constant $\ga_i$, then we get a polarization or dipole moment per unit volume 
which varies harmonically with an amplitude
\beq
\P_0=e^2\E_0n\sum_j\lep\frac{f_j}{m(\om_j^2-i\om\ga_j-\om^2)}\rip;
\eeq
this is also the relation between $\Ext$ and $\Pxt$. If we further say that 
$\Ext$ is the macroscopic field\footnote{In this we follow Jackson, but 
remember the Clausius-Mossotti relation from last quarter; we argued that the 
electric field which produces the polarization should be the local field 
and not the macroscopic field. It is not difficult to make the necessary 
corrections to what is given here.}, then we can write $\D=\E+4\pi\P=\ep\E$
with the preceding expression for the polarization. The result is an 
expression for $\ep(\om)$:
\beq
\ep(\om)=1+\frac{4\pi ne^2}m\sum_j\lep\frac{f_j}{\om_j^2-i\om\ga_j-\om^2}\rip
\eeq 
or
\beq
\epo=1+\frac{4\pi nze^2}{m}\sum_j\lep\frac{f_j}z\rip\lep\frac{\om_j^2-\om^2
+i\om\ga_j}{(\om_j^2-\om^2)^2+\om^2\ga_j^2}\rip\equiv\ep_1+i\ep_2
\eeq
where $\ep_1$ and $\ep_2$ are real.

In a typical term of the sum, different regimes of the relative sizes of $\om$,
$\om_j$, and $\ga_j$ give rise to very different behaviors. The resonant 
frequencies are, when Planck's constant is thrown in, comparable to 
binding energies of electrons which are on the order of a few electron-volts, 
so that $\om_j$ is of order $10^{15}\,sec^{-1}$, much the same as optical 
frequencies. The damping constants tend to be somewhat smaller, perhaps of 
order $10^{12}\,sec{-1}$ (see below). Starting from low frequencies,
$\om<<\om_j^2$ and also {\bf{$\om\ga_j<<\om_j^2$}}, then we can approximate 
the dielectric function as
\beq
\ep(\om)\approx1+\frac{4\pi ne^2}m\sum\frac{f_j}{\om_j^2}
\eeq
which is a constant. 
Now, as $\om$ increases from a low 
value, the real part of $\ep$ will also increase (slowly at first); when it 
gets to within about $\ga_j$ of the smallest $\om_j$, there is a resonance 
(the electron is being ``pushed'' by the electric field at a frequency close 
to its natural frequency) which will show up in $\ep_1$ as a sudden rise, 
fall, and rise. After this, $\ep_1$ is again roughly constant. 
There are as many such resonances as there are distinct resonant 
frequencies or terms in the sum over $j$.
\begin{figure}
\vspace{3in}
\caption{Real and imaginary parts of $\ep$ near resonances}
\end{figure} 

The rapid variation of the 
dielectric function in the vicinity of a resonance also produces a rapidly 
varying index of refraction, meaning that waves with 
relatively close frequencies propagate with quite different speeds. The 
frequency regime where $\ep_1$ decreases with increasing $\om$ is known as a 
region of {\em anomalous dispersion}. 

The imaginary part of $\ep$ also behaves in an interesting 
fashion near a resonance. Because the denominator of the resonant term in 
$\epo$ gets quite small at $\om=\om_j$ while the numerator for the imaginary
part does not get small, there is a pronounced peak in $\ep_2$ here. The 
smaller the value of $\ga_j$, the bigger the peak. A large imaginary part of 
the dielectric function produces strong damping or {\em absorption} of the 
wave, so a region of anomalous dispersion is also a region of strong 
absorption, termed {\em resonant absorption}.

Finally, for $\om$ very large in comparison with any other frequency in the 
system {\bf{$\om\gg\om_j$}}, the dielectric function once again 
becomes simple and has the form
\beq
\ep(\om)=1-\frac{4\pi nze^2}{m\om^2}\equiv1-\frac{\om_p^2}{\om^2}
\eeq 
where we have introduced the {\em plasma frequency} of the electron system,
\beq
\om_p\equiv\sqrt{\frac{4\pi nze^2}m}.
\eeq
For typical values of $n$ in solids, this frequency is of order 
$10^{16}\,sec^{-1}$ which is as large as or larger than the frequency of 
visible light. Our result is interesting in that the dielectric function is 
smaller than unity in this regime of frequency, meaning that a 
point of constant phase in a harmonic wave actually travels faster than the 
speed of light $c$. Even more remarkable is the possibility that $\ep(\om)<0$
in some range of frequency. For this to occur it is necessary to have $\om<
\om_p$ but at the same time $\om$ must be considerably larger than any 
resonant frequency $\om_j$ and also larger than the damping parameters 
$\ga_j$. Such conditions can be attained in some materials; a simple example 
is a tenuous plasma, or gas of charged particles. Then the resonant frequencies
are all zero, the plasma frequency is rather low because the density of 
charges is not large, and the damping is small. See the following section.

\subsection{Dielectric Response of Free Electrons}

Some special cases are also worthy of mention. One is the case of 
free electrons. For these electrons there is no restoring force and so we may 
set the corresponding $\om_j$, called $\om_0$, to zero. This has a profound 
effect on the dielectric function at low frequencies. If we extract the 
free-electron term from the remainder of the dielectric function and regard the
latter as some constant $\ep_0$ at low frequencies (see \eq{78}), then we have
\beq
\ep=\ep_0-\frac{4\pi nf_0e^2}{m\om(\om+i\ga_0)}=\ep_0+i\frac{4\pi 
nf_0e^2}{m\om(\ga_0-i\om)}.
\eeq

This thing is singular as $\om\rightarrow0$, reflecting the fact that in 
the zero-frequency limit, the free electrons will be displaced arbitrarily 
far from their initial positions by any small electric field, producing a 
very large polarization. The singular term in $\ep$ in fact represents the 
conductivity of the free-electron material. To see how it is related to the 
conductivity, let us examine Amp\`ere's law using this dielectric function 
and {\bf no} macroscopic current $\J$, as this current will be included in the 
dielectric response (the polarization produced by the free electrons).
From $\curl\H=c^{-1}\partial\D/t$, we find
\beq
\curl\H=-i\frac\om c\lep\ep_0+i\frac{4\pi nf_0e^2}{m\om(\ga_0-i\om)}\rip\E
=\frac{4\pi}c\frac{nf_0e^2}{m(\ga_0-i\om)}\E-i\frac\om c\ep_0\E.
\eeq
By contrast, we may choose {\bf not} to include the free electrons' 
contribution to the polarization in which case $\ep=\ep_0$. Then, however, we 
have to include them as macroscopic current $\J$; assuming linear response and 
isotropy, we may write $\J=\si\E$ where $\si$ is the {\em electrical 
conductivity}. Using these relations, and Amp\`ere's law, $\curl\H=(4\pi/c)
\J+c^{-1}\partial\D/\partial t$, we find
\beq
\curl\H=\frac{4\pi}c\si\E-i\frac\om c\ep_0\E.
\eeq
Comparison of the two preceding equations shows that by including the 
contribution of the free electrons in the polarization we have actually 
derived a simple expression for the conductivity,
\beq
\si=\frac{nf_0e^2}{m(\ga_0-i\om)}\rightarrow\frac{nf_0e^2}{m\ga_0},
\eeq
the last expression holding in the zero-frequency or static limit.

Comparison of measured conductivities with this result gives one an estimate 
of the damping constant. In very good metallic conductors such as $Cu$ or 
$Ag$, $\si\sim10^{17}\,sec^{-1}$. The free-electron density is of order
$10^{22}\,cm^{-3}$ and so one is led to $\ga_0\sim10^{13}\,sec^{-1}$ which 
is considerably smaller than typical resonant frequencies (for bound 
electrons, of course).

\section{A Model for the Ionosphere}
The ionosphere is a region of the upper atmosphere which is ionized by
solar radiation (ultraviolet, x-ray, etc.).  It may be simply
described as a dilute gas of charged particles, composed of
electrons and protons or other heavy charged objects. The dielectric 
properties of this medium are mainly produced by the lighter electrons, so 
we shall include only them in our description. We then have just one kind of 
charge and it has zero resonant frequency. Because the medium is dilute, 
the damping is small; we shall ignore it.  This is the approximation of a {\em
collisionless} plasma and it leaves us with a very simple dielectric function,
\beq
\ep(\om)=1-\frac{\om_p^2}{\om^2}.
\eeq
For frequencies smaller than the plasma frequency, $\epo<0$, meaning that
the wave number is pure imaginary since $k=\om\sqrt{\ep}/c$; the 
corresponding wave will not propagate because its dependence on position is 
proportional to $\exp(i\k\cdot\x)$ or $\exp(-|\k|z)$ given that $\k\parallel
\zh$\footnote{For most laboratory plasmas, this occurs at microwave
frequencies}.

In the case of the ionosphere there is an additional complicating factor 
(which also makes the problem more interesting); the earth has a magnetic 
field which influences the motions of the electrons and hence the 
dielectric function. The equation of motion of the charges, including this 
field $\B_0$ is

\beq
m\sde\x t=-e\leb\E+\frac1c\lep\der\x t\times\B_0\rip\rib.
\eeq
We ignore the effect of the wave's magnetic induction. We shall also restrict 
(for simplicity) attention to the case of $\k\parallel\B_0$ and shall ignore 
the spatial variations of $\E$. In addition, and without loss of generality, we
can let the wave have circular polarization. Hence we write the electric field 
as $\E=E_0\eppm\emot$.

Under these conditions, $\x$ will be of the form $\x=\x_0\emot$; using this 
relation in the equation of motion, we find
\beq
-m\om^2\x_0=-e\leb E_0\eppm-\frac{i\om}cB_0(\x_0\times\zh)\rib.
\eeq
The solution of this equation is $\x_0=x_0\eppm$; one can see this easily 
if one realizes that $\eppm\times\zh=\pm i\eppm$:
\beq
\eppm\times\zh=\frac1{\sqrt2}(\xh\pm i\yh)\times\zh=\frac1{\sqrt2}(-\yh\pm i\xh
)=\pm\frac i{\sqrt2}(\xh\pm i\yh)=\pm i\eppm.
\eeq
Hence the equation of motion, using $\x_0=x_0\eppm$, is
\beq
-m\om^2x_0\eppm=-e[E_0\pm\frac{\om B_0x_0}c]\eppm.
\eeq
or
\beq
x_0=\frac{eE_0/m}{\om(\om\mp\om_B)}
\eeq
where $\om_B\equiv eB_0/mc$ is the {\em cyclotron frequency}. From this 
point we may determine the dielectric function by repeating the arguments used 
in the preceding section and find
\beq
\ep(\om)=1-\frac{\om_p^2}{\om(\om\mp\om_B)}.
\eeq

Our result tells us that waves with different polarization elicit 
different dielectric responses from the medium; such a phenomenon is known 
as {\em birefringence}. If a wave of general 
polarization is incident upon the plasma, it is in effect broken into 
its two circularly polarized components and these propagate independently. 
It is possible to have a wave with a frequency such that for one 
component $\epo<0$ and for the other, $\epo>0$. Hence, one will propagate and 
the other will not, providing a (not particularly practical) way of producing 
a circularly polarized wave.
\begin{figure}
\vspace{4.5in}
\caption{Dielectric constants vs. $\om$ for the ionosphere}
\end{figure}

In the specific case of the ionosphere, $\om_p$, $\om_B$, and $\om$ can all 
be quite comparable. The density of electrons, which varies with the time 
of day and solar activity, is typically $\sim10^5-10^6\,cm^{-3}$, leading 
to $\om_p\sim10^7\,sec^{-1}$. The earth's field $B_0\sim0.1-1.0\,gauss$, 
leading to $\om_B\sim10^7\,sec^{-1}$. A wave with $\om\sim10^7\,sec^{-1}$
is in the AM band; short-wave radio frequencies are somewhat higher, and FM 
radio or television have considerably higher frequencies. This means that 
FM and television signals are at frequencies so large that $\ep\approx1$ 
and they propagate right through the ionosphere without significant 
reflection or attenuation. For this reason, the signals can be received 
only at locations where there is a direct path through the atmosphere from 
transmitter to receiver. For the lower frequency signals (short-wave and AM), 
however, there can be strong reflection from the ionosphere, making it 
possible to receive them relatively far from the transmitter. The higher 
the point in the ionosphere where the reflection takes place, the greater 
the effective range of the signal. Because the electron density increases with 
height (and then decreases again), the higher frequencies tend to be 
reflected at greater heights (if they are reflected at all) than the lower 
ones, thereby giving greater range. That is why short-wave signals have 
longer range than AM signals, at least some of the time.
\begin{figure}
\vspace{2.0in}
\caption{Electron density vs. height in the ionosphere}
\end{figure}

What happens if $\k$ is not parallel to $\B_0$? The medium is still 
birefringent so that a wave of arbitrary polarization is broken into two 
components that propagate independently; however, the two components are 
not simple circularly polarized waves. In addition, the dielectric 
functions and hence the indices of refraction for these two waves depend on 
the angle between $\B_0$ and $\k$, so the medium is not only birefringent 
but also anisotropic.
\section{Waves in a Dissipative Medium}
We have seen in the preceding sections that the dielectric function will is 
general be complex, reflecting the fact that a wave will be dissipated or 
damped under many conditions. It therefore behooves us to learn more about 
the properties of waves when dissipation is present. As we have seen, we 
can do this by employing a complex dielectric function, and we can also 
do it, with the same basic results, by letting $\ep$ be real while introducing 
a real conductivity and thus a macroscopic current density. We shall do the 
latter, for no particular reason.

Suppose that once again we have some linear medium with $\D=\ep\E$, $\B=\mu
\H$, and $\J=\si\E$; $\ep$, $\mu$, and $\si$ are taken as real. Then the 
Maxwell equations become
\beq
\div\B=0,\hsph\div\E=0,\hsph\curl\E=-\frac1c\pde\B t,\nonumber
\eeq
and
\beq
\curl\B=\frac{4\pi\mu}c\si\E+\frac{\ep\mu}c\pde\E t.
\eeq
We have set $\rh$ equal to zero in these equations. It may be that there is 
initially some macroscopic charge density within a conductor. If this is 
the case, that density will decay to zero with a characteristic time on the 
order of $\ga^{-1}$ where $\ga$ is the damping constant introduced in the 
section on dielectric functions; see Jackson, Problem 7.7. 

Let us look for plane wave solutions to the field equations. Set $\Ext=\E_0
\ekxot$ and $\Bxt=\B_0\ekxot$. The divergence equations then tell us that 
$\B_0\cdot\k=0$ and $\E_0\cdot\k=0$ as in a nondissipative medium. From 
Faraday's law we find the familiar result 
\beq
\B_0=\frac c\om(\k\times\E_0),
\eeq
and from the Amp\`ere's law we find
\beq
i(\k\times\B_0)=\frac{4\pi\mu\si}c\E_0-i\frac{\om\mu\ep}c\E_0.
\eeq
If we take the cross product of $\k$ with \eq{94} and substitute \eq{95} 
into the result where $\k\times\B_0$ appears, we find, after using 
$\k\times(\k\times\E_0)=-k^2\E_0$, that
\beq
-i\frac{4\pi\mu\si}c\E_0-\frac{\om\mu\ep}c\E_0=-\frac{ck^2}\om\E_0
\eeq
or
\beq
k^2=\frac{\om^2\mu\ep}{c^2}+i\frac{4\pi\mu\si\om}{c^2}.
\eeq
Taking the point of view that $\om$ is some given real frequency, we can 
solve this relation for the corresponding wavenumber $k$, which is complex. 
If we write $k=k_0+i\al$, then the real and imaginary parts of \eq{97} 
give us two equations which may be solved for $k_0$ and $\al$:
\beq
k_0^2-\al^2=\frac{\om^2\ep\mu}{c^2}\hsph2k_0\al=\frac{\om^2\ep\mu}{c^2}
\lep\frac{4\pi\si}{\ep\om}\rip.
\eeq
The solution is
\beq
\lec\barr{c} k_0 \\ \al \ear\ric=\sqrt{\mu\ep}\lep\frac\om c\rip
\lec\frac{\sqrt{1+\lep\frac{4\pi\si}{\om\ep}\rip^2}\pm1}{2}\ric^{1/2}.
\eeq
where the + sign refers to $k_0$ and the - sign to $\al$.

This expression appears somewhat impenetrable although it doesn't say 
anything unexpected or remarkable. It takes on simple forms in the limits of 
high and low conductivity. The relevant dimensionless parameter is $4\pi\si
/\ep\om$. It if is much larger than unity, corresponding to a 
{\bf{good conductor}}, then 
\beq
k_0\approx\al\approx\frac{\sqrt{2\pi\om\mu\si}}c\equiv\frac1\de
\;\;\;\frac{4\pi\sigma}{\ep\om}\gg 1 \, .
\eeq
where we have introduced the {\em penetration depth} $\de$. This is the 
distance that an electromagnetic wave will penetrate into a good conductor 
before being attenuated to a fraction $1/e$ of its initial amplitude. Since 
the wavelength of the wave is $\la=2\pi/k_0$, $\de$ is also a measure of 
the wavelength in the conductor. 

For a {\bf{poor conductor}}, by which we mean $4\pi\si/\om\ep<<1$, one has
\beq
k_0+i\al\approx\sqrt{\mu\ep}\frac\om c+i\frac{2\pi}c\sqrt{\frac\mu\ep}\si.
\eeq
Notice that in the latter case, the real part of the wavenumber is the same 
as in a nonconducting medium and the imaginary part is independent of 
frequency so that waves of all frequencies are attenuated by equal amounts 
over a given distance. Also, $\al<<k_0$ which tells us that the wave 
travels many wavelengths before being attenuated significantly.

For a given $\si$, $\al$ is an increasing function of $\om$ and saturates at
high  frequencies. Therefore, if one wants a wave to travel as far as 
possible, one wants to use as low freqency a wave as possible. Then one 
should be in the good-conductor limit where the attenuation varies as 
$\sqrt{\om}$ and vanishes as $\om\rightarrow0$. A well-known application of 
this rule has to do with radio communication with submarines; sea water is a 
reasonably good conductor $\si\sim 10^{11} \,sec^{-1}$ and so to communicate 
with a submerged boat, one should send out low frequency signals which will
penetrate to greater depths in the ocean than more standard signals.

\centerline{\psfig{figure=fig22.ps,height=1.5in,width=4.25in}}
{\narrower\em{Figure 22: Low frequency waves can propagate through sea water.}}
\medskip

Given that we have found the complex wave number, and letting $\k$ point in 
the $z$-direction, we have
\beq
\Ext=\E_0e^{i(k_0z-\om t)}e^{-\al z};
\eeq
the corresponding magnetic induction is found in the usual way (take $\k
\times\E$):
\beq
\Bxt=\frac c\om(k_0+i\al)(\zh\times\E_0)e^{i(k_0z-\om t)}e^{-\al z}.
\eeq
Define the complex index of refraction 
\beq
n\equiv\frac c\om k=\frac c\om(k_0+i\al), 
\eeq
so that
\beq
\B=n(\zh\times\E).
\eeq

{\em{Notice that because $n$ is complex, $\B$ is not in phase with $\E$}}; 
to make the phase difference explicit, let us write $n$ in polar form:
\beq
n=|n|e^{i\ph}\hsph\hbox{where}\hsph\ph=\arctan\lep\frac\al{k_0}\rip.
\eeq
We can find $|n|$ and $\ph$ in terms of other parameters; let $\ga\equiv(4
\pi\si/\om\ep)^2$. Then
\beq
\ph=\arctan\leb\frac{\sqrt{1+\ga}-1}{\sqrt{1+\ga}+1}\rib^{1/2}.
\eeq
Consider $\tan(2\ph)$:
\beqa
\tan(2\ph)=\frac{2\tan\ph}{1-\tan^2\ph}
=2\frac{[(\sqrt{1+\ga}-1)/(\sqrt{1+\ga}+1)]^{1/2}}{1-\frac{\sqrt{1+
\ga}-1}{\sqrt{1+\ga}+1}} \nonumber \\
=[(\sqrt{1+\ga}-1)(\sqrt{1+\ga}+1)]^{1/2}=\sqrt{\ga}
\eeqa
Thus,
\beq
\ph=\frac12\arctan\sqrt\ga=\frac12\arctan\lep\frac{4\pi\si}{\om\ep}\rip.
\eeq
Also,
\beq
|n|=\frac c\om\sqrt{k_0^2+\al^2}=\sqrt{\mu\ep}\leb1+\lep\frac{4\pi\si}{\om
\ep}\rip^2\rib^{1/4}.
\eeq
Using these results in \eq{105}, we have
\beq
\Bxt=\sqrt{\mu\ep}\leb1+\lep\frac{4\pi\si}{\om\ep}\rip^2\rib^{1/4}
e^{\frac i2\arctan\lep\frac{4\pi\si}{\om\ep}\rip}(\zh\times\E_0).
\eeq

The amount by which $\Bxt$ is phase-shifted from $\Ext$ is easily seen from 
this expression to lie between 0 and $\pi/4$; it is zero in the small $\si
/\om$ limit and $\pi/4$ in the large $\si/\om$ limit. 
Another significant feature of the expression for $\Bxt$ is that in the 
small $\si/\om$ limit, the amplitude of $\B$ relative to that of $\E$ is 
just $\sqrt{\mu\ep}$ as for insulators. But in the opposite limit, one finds 
that the relative amplitude is $\sqrt{4\pi\si\mu/\om}$ which is much larger 
than unity. Here the wave has, relatively speaking, a much larger magnetic 
induction than electric field.

\subsection{Reflection of a Wave Normally Incident 
on a Conductor}

As an example, let us calculate the reflection of a wave normally incident 
on a conductor from vacuum. 

\centerline{\psfig{figure=fig23.ps,height=1.5in,width=4.25in}}
{\narrower\em{Figure 23: Wave normally incident on a conductor.}}
\medskip

\noindent Then
\beq
\k=\frac\om c\zh\hsph\k'=\frac\om c n\zh, n=\sqrt{\mu\ep}(1+\ga)^{1/4}e^{i
\ph}.
\eeq
The relevant boundary conditions are $\H_t$ and $\E_t$ continuous. Let
$\E_0=E_0\xh$, $\E_0''=E_0''\xh$, and $\E_0'=E_0'\xh$. The corresponding 
magnetic field amplitudes are $\H_0=E_0\yh$, $\H_0''=-\E_0''\yh$, and, for 
the transmitted wave in the conductor,
\beq
\H_0'=\sqrt{\frac\ep\mu}(1+\ga)^{1/4}e^{i\ph}E_0'\yh.
\eeq
Our boundary conditions give immediately
\beq
E_0+E_0''=E_0'\hsph E_0-E_0''=\sqrt{\frac\ep\mu}(1+\ga)^{1/4}e^{i\ph}E_0'.
\eeq
These may be combined to yield
\beq
E_0'=\frac2{1+\sqrt{\ep\mu}(1+\ga)^{1/4}e^{i\ph}}E_0
\eeq
and
\beq
E_0''=\frac{1-\sqrt{\ep/\mu}(1+\ga)^{1/4}e^{i\ph}}{1+\sqrt{\ep/\mu}(1+\ga)
^{1/4}e^{i\ph}}E_0.
\eeq

Let us calculate the Poynting vector in the conductor. Its time average is
\beq
<\S'>=\frac c{8\pi}\Re(\E'\times\H'^*)=\frac c{8\pi}\Re\lec\frac{
4|E_0|^2\sqrt{\ep/\mu}(1+\ga)^{1/4}e^{-i\ph}}{|1+\sqrt{\ep/\mu}(1+\ga)^{1/
4}e^{i\ph}|^2}\ric e^{-2\al z}\zh.
\eeq
Using the interpretation of this vector as the energy current density, we 
may find the power per unit area transmitted into the conductor by 
evaluating $<\S'>\cdot\zh$ at $z=0$,
\beq
{\cal P'}=\frac c{2\pi}|E_0|^2\sqrt{\frac\ep\mu}\,\frac{(1+\ga)^{1/4}\cos
\ph}{1+2\sqrt{\ep/\mu}\cos\ph\,(1+\ga)^{1/4}+(\ep/\mu)(1+\ga)^{1/2}}.
\eeq
The incident power per unit area is ${\cal P}=c|E_0|^2/8\pi$, so the 
fraction of the incident power which enters the conductor, where it is 
dissipated as Joule heat, is
\beq
T=\frac{{\cal P}'}{{\cal P}}=4\sqrt{\frac\ep\mu}\,\frac{(1+\ga)^{1/4}\cos\ph}
{1+2\sqrt{\ep/\mu}\cos\ph\,(1+\ga)^{1/4}+(\ep/\mu)(1+\ga)^{1/2}}.
\eeq
This expression is much simplified in the limit of a good conductor where
$\ph=\pi/4$, $\cos\ph=1/\sqrt2$, and $\ga>>1$. Then
\beq
T\approx4\sqrt{\frac\ep\mu}\,\frac{\ga^{1/4}(1/\sqrt2)}{\ep\ga^{1/2}/\mu}=2
\sqrt2\sqrt{\frac\mu\ep}\,\sqrt{\frac{\om\ep}{4\pi\si}}=\frac{2\mu\om} c
\frac c{\sqrt{2\pi\si\om\mu}}=\frac{2\mu\om}c\de.
\eeq
For a good conductor such as $Cu$, $\si\sim10^{17}\,sec^{-1}$ and so a wave 
with frequency around $10^{10}\,sec^{-1}$ will have $\de\sim10^{-4}\,cm$ or 
$1\,\mu m$. Also, the better the conductor, the smaller the fraction of the 
incident power lost in the reflection process. For the example just given, 
$T\approx10^{-4}$, meaning that the wave can be reflected some ten thousand 
times before becoming strongly attenuated.

\section{Superposition of Waves; Pulses and Packets}

No wave is truly monochromatic, although some waves, such as those produced 
by lasers, are exceedingly close to being so. Fortunately, in the case of 
linear media, the equations of motion for electromagnetic waves are completely 
linear and so any sum of harmonic solutions is also a solution. By making use 
of this superposition ``principle'' we can construct quite general solutions by 
superposing solutions of the kind we have already studied. 

\centerline{\psfig{figure=fig24.ps,height=1.5in,width=6.3875in}}
{\narrower\em{Figure 24: Any pulse in a linear media may be decomposed into 
a superposition of plane waves.}}
\medskip

This procedure amounts to making a Fourier transform of the pulse. 
For simplicity we shall 
work in one spatial dimension which simply means that we will superpose 
waves whose wave vectors are all in the same direction (the $z$-direction). 
For the same reason, we shall also employ scalar waves; these could, for 
example, be the $x$ components of the electric fields of the waves. One 
such wave has the form $\ekzokt$ where we shall not initially restrict $\om(k)$
to any particular form. Given a set of such waves, we can build a general 
solution of this kind (wave vector parallel to the $z$-axis) by integrating 
over some distribution $A(k)$ of them:
\beq
u(z,t)=\ftk\Ak\ekzokt.
\eeq
At time $t=0$, this function is simply
\beq
u(z,0)=\ftk\Ak\ekz
\eeq
and the inverse transform gives $A$ in terms of the zero-time wave:
\beq
A(k)=\ftz u(z,0)\emkz.
\eeq 
All of the standard rules of Fourier transforms are applicable to the 
functions $\Ak$ and $u(z,0)$. For example, if $\Ak$ is a sharply peaked 
function with width $\De k$, then the width of $u(z,0)$ must be of order
$1/\De k$ or larger, and conversely. One may make this statement more 
precise by defining
\beq
(\De z)^2\equiv<z^2>-<z>^2\hsph(\De k)^2\equiv<k^2>-<k>^2, 
\eeq
where
\beq
<f(k)>\equiv\frac{\int_{-\infty}^\infty dk\,f(k)|\Ak|^2}{\int_{-\infty}^
\infty dk\,|\Ak|^2}
\eeq
and
\beq
<f(z)>\equiv\frac{\int_{-\infty}^\infty dz\,f(z)|u(z,0)|^2}{\int_{-\infty}^
\infty dz\,|u(z,0)|^2}.
\eeq
The relation between these widths which must be obeyed is
\beq
\De z\De k\ge1/2.
\eeq

Now, given a ``reasonable'' initial wave form $u(z,0)$\footnote{Its time 
derivative $\partial u(z,t)/\partial t|_{t=0}$ must also be given to allow a 
unique solution of the initial value problem; our discussion is therefore 
incomplete but can be corrected easily.} with some $\De z$ 
and a Fourier transform $\Ak$ with some $\De k$, the question we seek to 
answer is what will be the nature of $u(z,t)$? The answer is simple in 
principle because all we have to do is Fourier transform to find $\Ak$ and 
then do the integral specified by \eq{121} to find $u(z,t)$. One can always 
do these integrals numerically if all else fails. Here we shall do some 
approximate calculations designed to demonstrate a few general points.

Suppose that we have found $\Ak$ and that it is some peaked function 
centered at $k_0$ with a width $\De k$. If $\om(k)$ is reasonably well 
approximated by a truncated Taylor's series expansion for $k$ within $\De 
k$ of $k_0$, then we may write
\beq
\om(k)\approx \om_0+\left.\der\om k\ril_{k_0}(k-k_0)\equiv\om_0+v_g(k-k_0)
\eeq
where
\beq
\om_0\equiv\om(k_0)\hsph\mbox{and}\hsph v_g=d\om/dk|_{k_0};
\eeq
$v_g$ is called the {\em group velocity} of the packet; notice that it can 
depend on the wave number $k_0$ which characterizes the typical wave 
numbers in the wave.
In this approximation, one finds
\beq
u(z,t)=\ftk\Ak e^{ik(z-v_gt)}e^{-i\om_0t}e^{iv_gk_0t}=e^{i(v_gk_0-\om_0)t}
u(z-v_gt,0).
\eeq
This result tells us that the wave packet retains its initial form and 
translates in space at a speed $v_g$. It does not spread (disperse) or 
distort in any way. In particular, the energy carried by the wave will move 
with a speed $v_g$. 

The group velocity is evidently an important quantity. We may write it in 
terms of the index of refraction by using the defining relation $k=\om n(
\om)/c$. Take the derivative of this with respect to $k$:
\beq
1=\lep\frac nc+\frac\om c\der n\om\rip\der\om k
\eeq
or
\beq
v_g=\frac c{n+\om\der n\om}.
\eeq
As an example consider the collisionless plasma relation $n=\sqrt{1-\om_p^2/
\om^2}$. One easily finds that
\beq
v_g=c\sqrt{1-\om_p^2/\om^2}.
\eeq
For $\om<\om_p$, the group velocity is imaginary which corresponds to a 
damped wave; for $\om>\om_p$, it is positive and increases from zero to $c$ 
as $\om$ increases.

Our calculations thus far have not resulted in any spreading or distortion 
of the wave packet because we did not include higher-order terms in the 
relation (called a {\em dispersion relation}) between $\om$ and $k$. Let's 
treat a simple example in which $\Ak$ is a gaussian function of $k-k_0$,
\beq
\Ak=\lep\frac{A_0}\de\rip e^{-(k-k_0)^2/2\de^2}.
\eeq
Further, let $\om(k)$ be approximated by 
\beq
\om(k)=\om_0+v_g(k-k_0)+\al(k-k_0)^2.
\eeq
The corresponding $u(z,t)$ is

\beqa
u(z,t)=\ftk\frac{A_0}\de e^{-(k-k_0)^2/2\de^2}e^{ikz-i[\om_0+v_g(k-k_0)+\al(k-
k_0)^2]t}\nonumber\\
=\frac{A_0}{\de\sqrt{2\pi}}e^{i(k_0z-\om_0t)}\int_{-\infty}^\infty dk\,
e^{i(k-k_0)(z-v_gt)}e^{-(1/2\de^2+i\al t)(k-k_0)^2}\nonumber\\
=\frac{A_0}{\sqrt{1+2i\al\de^2t}}e^{-(z-v_gt)^2\de^2/[2(1+2i\al\de^2t)]}
e^{i(k_0z-\om_0t)}.
\eeqa
If $\al=0$, this is a Gaussian-shaped packet which travels at speed $v_g$ 
with a constant width equal to $\de^{-1}$. If $\al\ne0$, it is still a 
Gaussian-shaped packet traveling at speed $v_g$; however, it does not have 
a constant width any longer. To make the development of the width 
completely clear, consider $|u(z,t)|^2$ which more nearly represents the 
energy density in the wave:
\beq
|u(z,t)|^2=\frac{A_0^2}{\sqrt{1+4\al^2\de^4t^2}}e^{-(z-v_gt)^2\de^2/(1+4\al^2
\de^4t^2)}.
\eeq
The width of this traveling Gaussian is easily seen to be
\beq
w(t)=\sqrt{1+4\al^2\de^4t^2}/\de.
\eeq
At short times the width increases as the square of the time, while at long 
times it becomes linear with $t$. 

When the packet spreads, or disperses, in this fashion, to what extent does 
it make sense to think about the wave as a localized object? One measure is 
the width of the packet as compared with the distance it has moved. After a 
long time the width is approximately $2\al\de t$ while the distance the 
packet has moved is $v_gt$. The ratio of these distances is $2\al\de/v_g$, 
so our condition for having a localized object is
\beq
2\al\de/v_g<<1.
\eeq

\centerline{\psfig{figure=fig25.ps,height=2.0in,width=4.25in}}
{\narrower\em{Figure 25: When $\delta$ is small, the wave is composed 
of few wavenumbers.}}
\medskip

\noindent
In addition, of course, the initial width of the packet must be small 
compared to $v_gt$ which is always possible if one waits long enough. Our 
inequality clearly puts a limit on the allowable size of $\al$, for a given 
$\de$, necessary to have a well-defined pulse. For smaller $\de$, one can 
get away with larger $\al$, a simple consequence of the fact that small 
$\de$ means the width of the packet in $k$-space is small, leading to less 
dispersion.

\subsection{A Pulse in the Ionosphere}

Let's look also at the fate of a wave packet propagating in the ionosphere; 
we found in an earlier section, treating the ionosphere as a collisionless 
plasma and with $\k$ parallel to $\B_0$, that $\ep(\om)=1+\om_p^2/\om(\om_B
-\om)$ for one particular polarization of the wave. If $\om$ is small enough 
compared to other frequencies, we may approximate in such a way that $n(\om)=
\om_p/\sqrt{\om\om_B}$, which gives rise to anomalous dispersion indeed. 
Defining $\om_0\equiv\om_p^2/\om_B$, one finds that the group velocity of a 
signal is $v_g=2c\sqrt{\om/\om_0}$.

Let us see how a pulse with the same $\Ak$ as in the previous example 
propagates. We have
\beqa
u(z,t)=\ftk\frac{A_0}\de e^{-(k-k_0)^2/2\de^2+ikz-ic^2k^2t/\om_0}
\nonumber\\=\ftk\frac{A_0}\de e^{-(k-k_0)^2/2\de^2+i(k-k_0)z+ik_0z
-ic^2t(k-k_0)^2/\om_0-i2c^2k_0t(k-k_0)/\om_0-ic^2k_0^2t/\om_0}
\nonumber\\=\frac{A_0}{1+2i\de^2c^2t/\om_0)^{1/2}}
e^{i(k_0z-c^2k_0^2t/\om_0)}e^{-\frac{(z-2c^2k_0t/\om_0)^2\de^2}{2(1+2i
\de^2c^2t/\om_0)}}
\eeqa
This is a traveling, dispersing Gaussian. Its speed is the group velocity
$v_g(k_0)$. The width of the Gaussian is
\beq
w(t)=\sqrt{1+4\de^4c^4t^2/\om_0^2}/\de\rightarrow2\de c^2t/\om_0
\eeq
at long times. The packet spreads at a rate given by $v_w=2\de c^2/\om_0$. 
The ratio of this spreading rate to the group velocity is $\de/k_0$ and so 
we retain a well-defined pulse provided the spread in wavenumber is small 
compared to the central wavenumber.

Pulses of this general type are generated in the ionosphere by 
thunderstorms. They have a very broad range of frequencies ranging from 
very low ones up into at least the AM radio range. The electromagnetic waves 
tend to be guided along lines of the earth's magnetic induction, and so, if for 
example the 
storm is in the southern hemisphere, the waves travel north in the ionosphere 
along lines of $\B$ and then come back to earth in the northern hemisphere. 

\centerline{\psfig{figure=fig26.ps,height=2.0in,width=4.25in}}
{\narrower\em{Figure 26: Lightning in the southern hemisphere yields 
wistlers in the north.}}
\medskip

\noindent By this time they are much dispersed, with the higher frequency components 
arriving well before the lower frequency ones since 
$v_g=2c\sqrt{\om/\om_0}$ for $\om<<\om_0$. Frequencies in the audible range,
$\om\sim10^2$ or $10^3\,sec^{-1}$ take one or more seconds (a long time for 
electromagnetic waves) to arrive. If one receives the signal and converts 
it directly to an audio signal at the same frequency, it sounds like a 
whistle, starting at high frequencies and continuing down to low ones over 
a time period of several seconds. This characteristic feature has caused 
such waves to be known as {\em whistlers}. 

\section{Causality and the Dielectric Function}

A linear dispersive medium is characterized by a dielectric function 
$\ep(\om)$ having physical origins that we have just finished exploring. 
One consequence of having such a relation between $\Dxo$ and $\Exo$, that 
is, 
\beq
\Dxo=\ep(\om)\Exo,
\eeq 
is that the relation between $\Dxt$ and $\Ext$ is {\bf nonlocal} in time. 
To see this we have only to look at the Fourier transforms of $\D$ and 
$\E$. One has
\beq
\Dxt=\fto\Dxo\emot
\eeq
and its inverse
\beq
\Dxo=\ftt\Dxt\emot;
\eeq
similar relations hold for $\Ext$ and $\Exo$. Using the relation $\Dxo=\epo
\Exo$, we have
\beq
\Dxt=\fto\epo\Exo\emot. 
\eeq
We can write $\Exo$ here as a Fourier integral and so have
\beqa
\Dxt=\frac1{2\pi}\int_{-\infty}^\infty d\om\,\epo\emot\int_{-\infty}^\infty
dt'\,\eotp\E(\x,t')\nonumber\\
=\frac1{2\pi}\intmp dt\,d\om\,[\epo-1+1]\E(\x,t')e^{-i\om(t-t')}=\nonumber
\\
\Ext+\frac1{2\pi}\intmp dt\,d\om\,[\epo-1]\E(\x,t')e^{-i\om(t-t')}\equiv
\Ext+4\pi\Pxt.
\eeqa
The final term, $4\pi\Pxt$, can be written in terms the Fourier 
transform\footnote{Provided the order of integration can be 
reversed and the transform exists.} of $\epo-1$; introduce the function
\beq
G(t)\equiv\frac1{2\pi}\int_{-\infty}^\infty d\om\,[\epo-1]\emot.
\eeq
Then we have
\beq
\Dxt=\Ext+\intmp dt'\,G(t-t')\E(\x,t')
\eeq
which may also be written as 
\beq
\Dxt=\Ext+\intmp d\ta\,G(\ta)\E(\x,t-\ta).
\eeq
This equation makes it clear that when the medium has a frequency-dependent
dielectric function, as all materials do, then the electric displacement 
at time $t$ depends on the electric field not only at time $t$ but also at 
times other than $t$. This is somewhat disturbing because one can see that, 
depending on the character of $G$, we could get a polarization $\Pxt$ that 
depends on values of $\E(\x,t')$ for $t'>t$, which means we get an effect 
arising from a cause that occurs at a time later than the effect. This 
behavior can be avoided if the function $G(\ta)$ vanishes when $\ta<0$, and 
that is what in fact happens.

Let's look at a simple example with 
\beq
\epo=1+\frac{\om_p^2}{\om_0^2-\om^2-i\om\ga}.
\eeq
Then
\beq
G(\tau)=\frac{\om_p^2}{2\pi}\intmp d\om\,\frac{e^{-i\om\tau}}{\om_0^2-\om^2-i\om\ga}
\eeq
This integral was made for contour integration techniques. The poles of the 
integrand are in the lower half-plane in complex frequency space at
\beq
\om_\pm=\frac12[\pm\sqrt{4\om_0^2-\ga^2}-i\ga];
\eeq
without producing a contribution to the integral, we can close the contour in 
the upper (lower) half-plane when $\ta$ is smaller (larger) than zero. Because 
there are poles only in the lower half-plane, we can see immediately that 
$G(\ta)$ will be zero for $\ta<0$. That is pleasing since we don't want 
the displacement (that is, the polarization) to respond at time $t$ to the 
electric field at times later than $t$.

\centerline{\psfig{figure=fig27.ps,height=1.5in,width=4.25in}}
{\narrower\em{Figure 27: Because 
there are poles only in the lower half-plane, we can see immediately that 
$G(\ta)$ will be zero for $\ta<0$.}}
\medskip

Applying Cauchy's theorem to the case of $\ta>0$, one finds that, for all $\ta$,
\beq
G(\ta)=\om_p^2e^{-\ga\ta/2}\frac{\sin(\nu_0\ta)}{\nu_0}\th(\ta)
\eeq
where $\th(x)$ is the step function, equal to unity for $x>0$ and to zero 
otherwise, and $\nu_0=\sqrt{\om_0^2-\ga^2/4}$. The characteristic range in 
time of this function is $\ga^{-1}$ and hence the nonlocal (in time) character 
of the response is not important for frequencies smaller than about $\ga$; it 
becomes important for larger ones. 

One may naturally wonder whether there should also be nonlocal character of 
the response in space as well as in time. In fact there should and will be 
under some conditions. If we look back at our derivation of the model 
dielectric function, we see that the equation of motion of the particle was 
solved using $\E(0,t)$ instead of $\Ext$; the latter is of course the more
correct choice. The difference is not important so long as the excursions 
of the charge from the point on which it is bound are much smaller than the 
wavelength of the radiation, which is the case for any kind of wave with 
frequencies up to those of soft X-rays. Hence the response can be expected 
to be local in space in insulating materials. However, if an electron is 
free, it can move quite far during a cycle of the field and if it does so,
the response will be nonlocal in space as well as time.
\centerline{\psfig{figure=fig28.ps,height=1.5in,width=4.25in}}
{\narrower\em{Figure 28:  $G(\tau,\x)$ will not be $\x$ dependent 
if the excursions 
of the charge from the point on which it is bound are much smaller than the 
wavelength of the radiation.}}
\medskip

Returning to the question of causality, we have seen that the simple model 
dielectric function produces a function $G(t)$ which is zero for $t<0$, as 
is necessary if ``causality'' is to be preserved, by which we mean there is 
no response in advance of the ``cause'' of that response. It is easy to see 
what are the features of the dielectric function that give rise 
to the result $G(t)=0$ for $t<0$. One is that there are no simple poles of 
the dielectric function in the upper half of the complex frequency plane. 
Another is that the dielectric function goes to zero for large $\om$ fast 
enough that we can do the contour integral as we did it. 

More generally, if 
one wants to have a function $G(t)$ which is consistent with the requirements 
of causality, this implies certain conditions on {\em{any}} $\epo$. 
Additional conditions can be extracted from such simple things as the fact 
that $G(t)$ must be real so that $\D$ is real if $\E$ is. Without going 
into the details of the matter (see Jackson) let us make some general 
statements. The reality of $G$ requires that
\beq
\ep(-\om)=\ep^*(\om^*).
\eeq
That $G$ is zero for negative times requires that $\epo$ be analytic in the 
upper half of the frequency plane. Assuming that $G(t)\rightarrow 0$ as 
$t\rightarrow\infty$, one finds that $\epo$ is analytic on the real axis. 
This last statement is actually not true for conductors which give a 
contribution to $\ep$ $\sim i\si/\om$ so that there is a pole at the 
origin. Finally, from the small-time behavior of $G(t)$, one can infer that 
at large frequencies the real part of $\epo-1$ varies as $\om^{-2}$ while the 
imaginary part varies as $\om^{-3}$.  This is accomplished by repeatedly
integrating by parts
\beq
\ep(\om)-1=\int_0^\infty d\tau G(\tau) e^{i\om\tau}
\approx \frac{iG(0^+)}{\om} - \frac{G'(0^+)}{\om^2} +\frac{iG'(0^+)}{\om^3}
+\cdots
\eeq
This series is convergent for large $\om$.  The first term vanishes
if $G(\tau)$ is continuous accross $\tau=0$.  Thus
\beq
\Re\lep \ep(\om)-1\rip \sim \frac{1}{\om^2}\;\;\;\;\;\;\;\;\;
\Im \lep \ep(\om)-1\rip \sim \frac{1}{\om^3}
\eeq
From inspection, one may see that the 
various dielectric functions we have contrived satisfy these conditions.

Given that the dielectric function has the analyticity properties described 
above, it turns out that by rather standard manipulations making use of 
Cauchy's integral theorem, one can write the imaginary part of $\epo$ in terms 
of an integral of the real part over real frequencies and conversely. That one 
can do so is important because it means, for example, that if one succeeds in 
measuring just the real (imaginary) part, the imaginary (real) part is then 
known. The downside of this apparent miracle is that one has to know the real 
or imaginary part for all real frequencies in order to obtain the other part.

To see how this works, notice that as a consequence of the analytic properties 
of the dielectric function, it obeys the relation
\beq
\ep(z)=1+\frac1{2\pi i}\inopc\frac{\epop-1}{\om'-z}
\eeq
where the contour does not enter the lower half-plane (where $\ep$ may have 
poles) anywhere and where $z$ is inside of the contour. Let C consist of 
the real axis and a large semicircle which closes the path in the upper 
half-plane. 

\centerline{\psfig{figure=fig29.ps,height=1.5in,width=4.25in}}
{\narrower\em{Figure 29:  Contour $C$: $\ep(\om)$ is analytic inside an on C..}}
\medskip

\noindent 
Then, given that $\ep$ falls off fast enough, as described 
above, at large $\om$, the semicircular part of the path does not 
contribute to the integral. Hence we find that
\beq
\ep(z)=1+\frac1{2\pi i}\intmp d\om'\,\frac{\epop-1}{\om'-z}.
\eeq
At this juncture, $z$ can be any point in the upper half-plane. Let's use
$z=\om+i\et$ and take the limit of $\et\rightarrow0$, finding
\beq
\ep(\om+i\eta)=1+\frac1{2\pi i}\intmp d\om'\frac{\epop-1}{\om'-\om-i\et}.
\eeq
The presence of the $\et$ in the denominator means that at the integration 
point $\om'=\om$, we must be careful to keep the singularity inside of, or 
above, the contour. Here we pick up $2\pi i$ times the residue, and the 
residue is just $\epo-1$. This relation shows identity but is
not useful otherwise.  However, one can also pull the following trick: If we 
integrate right across the singularity, taking the principal part (denoted $P$)
of the integral plus an infinitesmal semicircle right below the singularity
that amounts to taking $i\pi$ times the residue. Hence we can 
make the replacement
\beq
\frac1{\om'-\om-i\et}\rightarrow P\lep\frac1{\om'-\om}\rip+i\pi\de(\om'-\om)
\eeq
where $P$ stands for the principal part; this substitution leads to 
\beq
\epo=1+\frac1{\pi i}P\intmp d\om'\,\frac{\epop-1}{\om'-\om}
\eeq
Let us write separately the real and imaginary parts of this expression:
\beqa
\Re[\epo]=1+\frac1\pi P\intmp d\om'\,\frac{\Im[\epop]}{\om'-\om}\nonumber\\
\Im[\epo]=\frac1\pi P\intmp d\om'\,\frac{\Re[\epop-1]}{\om'-\om}
\eeqa
These equations are known as the {\em Kramers-Kronig relations} for the 
dielectric function. They may be written as integrals over only positive 
frequencies because of the fact that the real part of $\epo$ is an even 
function of $\om$ while the imaginary part is odd. It should also be 
pointed out that we have assumed there is no pole in $\epo$ at $\om=0$; if 
there is one (conductors have dielectric functions with this property) some
modification of these expressions will be necessary.
\section{Arrival of a Signal in a Dispersive Medium}
Most of the wave trains one receives, such as radio signals, messages from 
within or without the galaxy (sent by stars, pulsars, neutron stars, etc), and 
so on, have to traverse dispersive media to get wherever they go. 
Consequently it is of considerable importance to know how the signals are 
distorted by the intervening material. The basic idea is this: we have seen 
how a pulse centered at some particular wave number or frequency tends to 
travel with the group velocity of the central frequency and also spreads 
some as a consequence of the frequency-dependence of the index of 
refraction or dielectric function. If the dispersion is very large, as in 
regions of anomalous dispersion, the pulse will not simply spread some but 
will be distorted beyond recognition. In addition, frequency components in 
this region will be strongly attenuated and so will disappear from the wave 
train after awhile. If a signal is initially very broad in frequency, 
having components ranging from very low ones, where the group velocity is 
roughly constant and equal to $c/\sqrt{\ep(0)}$, to very high ones where 
$\epo\approx1$ and the group velocity is about $c$, then the signal that 
arrives after traveling through a significant length of medium will be very 
different indeed from the initial one. All of the frequency components 
around the regions of anomalous dispersion will be gone. There will be some 
high-frequency component which travels at a speed around $c$ and so arrives 
first; it is generally called the ``first precursor.'' Then after awhile
the remainder of the signal will arrive. The leading edge of this part is 
called the ``second precursor'' and it consists of those lower frequency 
components which have the largest group velocity and which are not appreciably 
attenuated. These are usually\footnote{But not always; the whistler 
provides a a counter example.} the very low frequency components.

It is a straightforward matter to determine what the signal will be, using 
the superposition principle. Consider a pulse in one dimension with an 
amplitude $u(z,t)$. Given that one knows the form of this pulse and its 
first space derivatives as functions of time at some initial position in 
space\footnote{Notice that instead of solving an initial value problem in time,
we here rephrase it as an initial value problem in space.}, called $z_0$, then 
one may determine by Fourier analysis the amplitude $A(\om)$ of the various 
frequency components in it. Since a frequency component $\om$ propagates 
according to $\exp[i(\om n(\om)z/c-\om t)]$, it is then easy in principle to 
find $u(z,t)$:
\beq
u(z,t)=\frac1{\sqrt{2\pi}}\intmp d\om\,A(\om)e^{i(\om n(\om)z/c-\om t)}.
\eeq
If we can do this integral for the index of refraction of our choice, we 
can find the form of the wave train at all space points at any later time.
Among other things, one can show by making use of the analyticity 
properties of the dielectric function that it is impossible for an 
electromagnetic signal to travel faster than the speed of light. See 
Jackson. 

As a very simple example, consider a single-resonance dielectric function 
with no absorption,
\beq
\epo=1+\frac{\om_p^2}{\om_0^2-\om^2}=n^2(\om)
\eeq
or
\beq
n(\om)=\lep\frac{\om_0^2-\om^2+\om_p^2}{\om_0^2-\om^2}\rip^{1/2}.
\eeq
Then 
\beq
2n\der n\om=2\om\frac{\om_p^2}{(\om_0^2-\om^2)^2}
\eeq
so
\beq
n\om\der n\om+n^2=\frac{\om^2\om_p^2}{(\om_0^2-\om^2)^2}+1+\frac{\om_p^2}{
\om_0^2-\om^2}=\frac{\om_0^4-2\om_0^2\om^2+\om^4+\om_p^2\om_0^2}{(\om_o^2
-\om^2)^2}.
\eeq
Hence
\beq
v_g=\frac c{\om\der n\om+n}=c\lep\frac{\om_0^2-\om^2+\om_p^2}{\om_0^2-
\om^2}\rip^{1/2}\frac{(\om_0^2-\om^2)^2}{(\om_0^2-2\om_0^2\om^2+\om^4+
\om_p^2\om_0^2)}
\eeq
\vspace{2.0in}

The first plot shows the character of $v_g$ and of $n(\om)$. The group 
velocity is largest for the largest 
frequencies; these will combine to provide the first precursor which may 
well be weak to the extent that the initial pulse does not contain many 
high-frequency components. The first precursor continues as lower frequency 
components (but still larger than $\sqrt{\om_0^2+
\om_p^2}$) come through. While this is going on, all of the very low 
frequency components arrive. This is the second precursor. Finally, if the 
pulse is actually a long wave train which has one predominant frequency in 
it, then after some time the received pulse settles down to something more 
or less harmonic, showing just this frequency.

%Rather than struggle with analytic approximations to the integral for $u(z,
%t)$, for which there are standard techniques (see Jackson), let us look at 
%some results obtained from numerical integration. The plots show, in suitably 
%chosen units, the amplitude as a function of time at some fixed point $z$ for 
%pulses which were located around $z=0$ at time zero. The point $z$ at which 
%the dispersed pulse is shown as a function of time is some one hundred 
%units from the initial location of the pulse. Thus there is no signal at 
%all for $t$ less than about 100 in units such that the speed of light in 
%vacuum is unity. Then the first precursor arrives. Its 
%strength depends on the amplitudes of the high frequency components of the 
%initial pulses. The cases treated and shown in the plots have, from the first 
%to the last, steadily decreasing amounts of high-frequency waves. 

%The arrival time of the second precursor is determined by the index of 
%refraction at small $\om$; the frequency $\om_p$ was chosen to make this 
%equal to two so that the second precursor should arrive around $t=200$, 
%which it clearly does in all cases. 

%Finally, if the pulse has a 
%predominant frequency in the range of real $n(\om)$, then the long-time 
%behavior of the arriving pulse will show oscillations with this frequency. 
%The fourth pulse is of this kind; the first three are not. Consequently the 
%first three simply damp out at long times while the fourth oscillates 
%harmonically with constant magnitude at long times.

\pagebreak
\appendix\section{Waves in a Conductor}

	When we discussed the propagation of waves in an ideal
dielectric, we showed that the fields were transverse to the direction
of propagation.  This corresponds to an isulating material, with a
vanishing electrical conductivity.  When we extend our discussion
to include media of finite conductivity $\sigma$, there is no
a priori reason that the fields will still be transverse to 
the direction of propagation.

Let's show that we need not worry about any longitudinal fields.
Suppose that once again we have some linear medium with $\D=\ep\E$, $\B=\mu
\H$, and $\J=\si\E$; $\ep$, $\mu$, and $\si$ are taken as real. Then the 
Maxwell equations become
\beq
\div\B=0,\hsph\div\E=0,\hsph\curl\E=-\frac1c\pde\B t,\nonumber
\eeq
and
\beq
\curl\B=\frac{4\pi\mu}c\si\E+\frac{\ep\mu}c\pde\E t.
\eeq

  Let's look for solutions to 
Maxwell's equations in the form of logitudinial waves,
\beq
\E=\zh E(z,t)\;\; ;\;\; \B=\zh B(z,t)
\eeq
Since $\div\B=\div\E=0$, $E$ and $B$ can be functions of time only.
Thus $\curl\E=\curl\B=0$, and the other two Maxwell's equations become
\beq
\pde{\B}{t}=0 \;\;\; ; \;\; 
\frac{4\pi\sigma}{c}\E + \frac{\ep}{c}\pde{\E}{t}=0
\eeq
The first says that $\B$ must be constant.  The second says that
$\E$ while uniform in space has a time dependence
\beq
\E(t)=\E(0)e^{-4\pi\sigma t/\ep}
\eeq
In a conductor, $\sigma\approx 10^{16}\;{\rm{sec}}^{-1}$.  Thus
$\E(t)$ falls off very rapidly and may be neglected.  Thus
as worst there is a constant logitudinal $\B$-field as part
of our wave in a conductor.  Since  Maxwell's equations are linear,
we may drop this trivial solution and just consider the transverse
fields.


\edo

