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\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.
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\title{Plane Waves and Wave Propagation}
\author{Augustin Jean Fresnel\\(1788 - 1827)}
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\maketitle
\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
__=\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<=\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-^2\hsph(\De k)^2\equiv-^2,
\eeq
where
\beq
\equiv\frac{\int_{-\infty}^\infty dk\,f(k)|\Ak|^2}{\int_{-\infty}^
\infty dk\,|\Ak|^2}
\eeq
and
\beq
\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
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