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\title{Chapter Thirteen\\
Charged Particle Collisions, Energy Loss, Scattering}
\author{Niels Henrik David Bohr\\(1885 - 1962)}
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The topic of interest is a charged particle traversing a material
medium. Such a particle looses energy by scattering from the charged
particles, electrons and nuclei, in the material. 

\centerline{\psfig{figure=fig1.ps,height=2.0in,width=8.5in}}

\noindent This is an interesting
system from many points of view. Historically it was extremely important in
resolving the question of the structure of matter (the Rutherford atom), and
at present energy loss is an important phenomenon in particle physics and
is also studied in detail by nuclear engineers and by condensed matter
physicists in connection with the properties of materials and radiation damage
to materials.

The problem can be studied as a straightforward application of
electromagnetism; charged particles scatter from one another with the
result that energy and momentum are transferred. The scattering centers
in a material are of two distinct types; there are electrons of charge $-e$
and small mass $m\sim10^{-27}\,g$, and there are nuclei of charge $Ze$ with
$Z$ up to about 10$^2$ and large mass $M\sim10^{-22}\,g$.
Thus the nuclear charge is significantly larger than that of an electron,
and the nuclear mass is much larger---some $10^5$ times larger---than the
electronic mass.  It is also important to realize that there are $Z$ more
electrons than nuclei ($Z$ is the atomic number of the atoms in the
material) in a given volume of target material. Consequently the electrons
provide $Z$ times as many scattering centers as the nuclei. 

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

\noindent As we shall see,
it turns out that electrons soak up most of the energy of an incident particle
while nuclei are responsible for most of the momentum transfer in the sense
that they are more effective than electrons at deflecting the incident particle
from its initial direction of motion.

\section{Energy Transfer in Coulomb Collisions}
The general problem of energy transfer when a charged particle traverses a
material is naturally very complicated. We shall approach it a little at a
time starting with the classical impulse approximation applied to a pair of
particles. 

\subsection{Classical Impulse Approximation}
Consider a particle $(q,M)$, where $q$ is the charge and $M$,
the mass, incident with speed $v$ on a second particle $(-e,m)$ at rest in
the frame of our calculation. The incident particle has total energy
$M\ga c^2$, where $\ga=1/(1-v^2/c^2)^{1/2}$. In the impulse approximation the
incident particle is treated as undeflected by the collision. Further, the
target is approximated as stationary during the collision. Then it is easy
to calculate the momentum, or impulse, transferred from the incident particle
to the target.

Given the approximation that the incident particle's trajectory is
unaffected by the collision, it travels with constant velocity and passes the
target at some distance $b$ called the {\em impact parameter}. 

\centerline{\psfig{figure=fig3.ps,height=2.0in,width=8.5in}}

\noindent The
momentum transferred to the target can be expressed as the integral over
time of the force acting on it, and we can find that from knowledge of the
electric field produced by the incident particle at the location of the
target. From prior calculations in Chapter 11, we know that this field is
$\E=\E\per+\E\pll$ where the parallel and perpendicular components act
parallel and perpendicular to the line of motion of the incident particle.
These components are given at the target, by
\beq
E\per(b)=\frac{\ga qb}{(b^2+\ga^2v^2t^2)^{3/2}}\andh E\pll(b)=-\frac{\ga
qvt}{(b^2+\ga^2v^2t^2)^{3/2}}
\eeq
where the origin of time is chosen so that the particles are closest at
$t=0$. 

\centerline{\psfig{figure=Erel.ps,height=2.5in,width=4.5in}}

\noindent The integral over time of $E\pll$ is zero while that of $E\per$
provides the momentum transferred to the target,
\beqa
p&=&\lel\intmp dt\,(-eE\per)\ril=\lel\intmp dt\,\frac{e\ga qb}{(b^2+
\ga^2v^2t^2)^{3/2}}\ril=\lel\frac{eqb}v\ril\intmp\frac{dx}{(b^2+x^2)^{3/2}}
\nonumber\\&=&\lel\frac{qe}{bv}\ril\intmp\frac{du}{(1+u^2)^{3/2}}=
\frac{2|qe|}{bv}.
\eeqa
Next, we shall assume that $p<<mc$ so that the energy transfer to the
target may be approximated by the non-relativistic formula $p^2/2m$. This
gives us a third approximation whose validity we must scrutinize. Adopting
it, we find that the energy transfer to the target is\footnote{We shall
suppose $qe>0$ so that the notation is simplified.}
\beq
\De E=\frac{p^2}{2m}=\frac{2q^2e^2}{mb^2v^2}=\lep\frac{qe}b\rip\frac
{(qe/b)}{(mv^2/2)}\propto \frac{e^2}{m}.
\eeq

Notice that the energy transfer is proportional to the square of the charge
of the target particle and inversely proportional to its mass. Possible
targets are electrons and nuclei. A nucleus has a larger charge than an
electron by a factor of the atomic number $z$, giving the nucleus an
``advantage'' by a factor of $z^2$ when it comes to extracting energy from
the incident particle. However, nuclei are more massive than electrons by a
factor of $1836A$ where $A$ is the atomic weight which is as large as or
larger than $z$. Furthermore, there are $z$ more electrons than nuclei to
act as targets. {\em{Hence we see that the electrons are more effective than
nuclei at taking the energy of the incident particle by a factor of at
least 1836.}} For this reason, we shall henceforth suppose that the target
particle is an electron so long as we are interested in the energy transfer,
as opposed to the momentum transfer, from the incident particle to the
target.

\begin{center}
\begin{tabular}{|l|c|c|}\hline
effect	& nucleus	& electron	\\ \hline\hline
charge	& $z^2$		&   $1$		\\ \hline
mass	& $1/(1836 z)$	&    $1$	\\ \hline
number	&  $1$		&   $z$		\\ \hline \hline
total	& $z/(1836)$	&   $z$		\\ \hline
\end{tabular}
\end{center}

\subsection{Validity of Approximations}
Our simple calculation of the energy transfer contains three distinct
approximations. 
\begin{enumerate}
\item It is assumed that the incident particle is not
deflected from its straight-line path. This assumption is valid so long as
the actual angle of deflection $\th$ obeys the inequality $\th<<1$.
\item It is assumed that the target particle is at a particular point
during the entire collision. This assumption is valid provided the target
recoils a distance $d$ during the collision which is small compared to the
impact parameter $b$, $d<<b$. 
\item We assumed that the recoiling particle
is non-relativistic, $p<<mc$.
\end{enumerate}

We can determine the conditions under which the approximations are valid.
{\bf{First}}, the angle of deflection of the incident particle is of order
$p/M\ga v$,

\centerline{\psfig{figure=fig4.ps,height=1.0in,width=8.5in}}

\beq
\th\approx\frac p{M\ga v}\approx\frac{2qe}{\ga bMv^2}=\frac2\ga\lep
\frac{(qe/b)}{Mv^2}\rip.
\eeq
Thus we require that the electrostatic potential energy of interaction at a
separation of the impact parameter should be small compared to the
energy $M\ga v^2$ which is something like the energy of the incident
particle. This condition is generally met. It is also not
independent of the other conditions required for the validity of the
impulse approximation.

{\bf{Second,}} consider the requirement that the target not recoil
far in comparison with $b$ during the collision. From the form of the
electric field 
\[
E\per(b)=\frac{\ga qb}{(b^2+\ga^2v^2t^2)^{3/2}}
\]
and hence the force experienced the the particle, we can see
that the duration $\ta$ of the collision is roughly $b/\ga v$. During this time
the recoil particle moves a distance of order $(p/m)\ta$, assuming it was
initially at rest, so the requirement is
\beq
\frac b{\ga v}\lep\frac{2qe}{mbv}\rip<<b\orh\frac1\ga\frac{(qe/b)}{mv^2}<<1.
\eeq
This inequality is much like the previous one but note the replacement of
$M$ with the mass $m$. Given that the target is an electron, which has the
smallest mass of all charged particles\footnote{As far as anyone knows.},
the present condition is at least as strong as the condition $\th<<1$.
Consequently, we can forget about the latter.

Notice that the condition (5) can also be written as
\beq
\frac1\ga\frac{c^2}{v^2}\frac{r_0}b<<1
\eeq
where $r_0\equiv e^2/mc^2$ is the {\em classical radius of the electron}
which is about $2.82\times10^{-13}\,cm$. Thus, provided the factor $c^2/v^2
\ga$ is not much larger than unity, this condition is met for impact
parameters (almost) down to $r_0$ which is also about the size of a baryon
or nucleus at which point we would expect the calculation to fail for
entirely different reasons. Notice, however, that the condition becomes
much more severe if $v$ is not large, i.e., if the incident particle is not
relativistic. That is not surprising; the collision will last much longer
if the incident particle moves slowly and the target has more time to recoil
during the collision.

{\bf{Third}}, comes the condition that $p<<mc$, or $2(qe/b)/mvc<<1$. This
condition is not much different from \eq{5}.

\subsection{Energy Loss}

We have calculated, in the impulse approximation, the energy absorbed from
an incident particle by a single electron. There is never just one electron.
We have to figure out how to add up the contributions of many electrons to
determine how much energy the incident particle will lose per unit length
of its path. Given an electron density $n$, then an incident particle
having speed $v$ will pass $n(vdt)(2\pi b\,db)$ scatterers per unit time at
distances between $b$ and $b+db$. 

\centerline{\psfig{figure=diff_volume.ps,height=2.0in,width=6.0in}}

\noindent The consequent energy change of the
incident particle in time $dt$ is, from \eq{3}
\beq
d^2E=-dtdb\,v2\pi bn\lep\frac{2q^2e^2}{mb^2v^2}\rip.
\eeq
Now integrate over $b$ to get the contributions from scatterers at all
distances. This must be done with some care. Let's just integrate $b$ from
some minimum to some maximum:
\beq
\der Et=-\frac{4\pi nq^2e^2}{mv}\int_{b_{min}}^{b_{max}}\frac{db}b=-\frac{
\fpi nq^2e^2}{mv}\ln(b_{max}/b_{min}).
\eeq

The mathematical necessity of the upper and lower cutoffs on $b$ is clear;
the integral would diverge at either end without the cutoff. Physically,
what is the reason for them? We have just seen that the impulse
approximation breaks down at $b\rta0$ because the recoiling particle
recoils by a distance comparable to or larger than the impact parameter
during the collision in that limit. Referring back to the condition that
our approximation is valid
\[
\frac b{\ga v}\lep\frac{2qe}{mbv}\rip<<b\orh\frac1\ga\frac{(qe/b)}{mv^2}<<1\,,
\]
we see that
a reasonable value for the cutoff is $b_{min}=qe/m\ga v^2$. This will also
make certain that the incident particle's deflection $\th$ is small. 

What about the upper cutoff? The physical reason for the breakdown of the
impulse approximation (which then necessitates the cutoff) at large $b$ is
that when $b$ is large, the collision time $\ta=b/\ga v$ is long. When this
time is long, the natural motions of the target cannot be neglected; that
is, the electron or target is not really at rest although we treated it as
such when calculating the energy transfer. Most electrons are bound to
atoms, molecules, or ions with some binding energy $E_e$ giving them a
natural angular frequency of motion $\om_0=E_e/\hbar$. The corresponding
period is of order $1/\om_0$. The collision time must be small compared to
this time or the impulse approximation, as we have derived it, breaks down.
That suggests we choose $b_{max}$ according to $b_{max}/\ga v=1/\om_0$ or
$b_{max}=\ga v/\om_0$.

Using these  cutoffs, we find that the rate of change with time of the
incident particle's energy is
\beq
\der Et=-\frac{4\pi nq^2e^2}{mv}\ln\lep\frac{m\ga^2v^3}{qe\om_0}\rip.
\eeq
A perhaps more interesting quantity is $dE/dx=v^{-1}dE/dt$,
\beq
\der Ex=-\frac{4\pi nq^2e^2}{mv^2}\ln\lep\frac{m\ga^2v^3}{qe\om_0}\rip
\eeq

In this derivation, we have determined the lower cutoff on $b$ by looking
at the breakdown of the classical impulse approximation. There is also a
breakdown associated with quantum effects which implies a somewhat
different lower cutoff. The quantum breakdown can be understood by
appealing to the uncertainty principle. The value of $b$ is uncertain by
an amount related to the momentum of the incident particle. We claim that
it has no momentum in the direction in which the impact parameter is
measured. We can't really know this to be precisely the case and there has
to be an uncertainty in the impact parameter which is of order $\hbar/m\ga
v$. If this uncertainty is comparable to $b$ itself, then our calculation
fails. Hence the quantum mechanical cutoff is
$b_{min}^{(q)}=\hbar/m\ga v$. In any given situation, we have to use the larger
of the two lower cutoffs. The ratio of the two is
\beq
\frac{b_{min}^{(q)}}{b_{min}}=\frac\hbar{m\ga v}\frac{\ga mv^2}{qe}=\frac{\hbar
v}{qe}=\frac1{\al(q/e)}\frac vc\approx\frac{137}{(q/e)}\frac vc
\eeq
where $\al\equiv e^2/\hbar c\approx1/137$ is the {\em fine structure
constant}. If this parameter is larger than unity, the quantum cutoff should be
employed; if it is smaller than unity, the classical one is appropriate.

\section{Collisions with a Harmonically Bound Charge; Energy Loss}

One can remove the need for introducing the upper and lower cutoffs on $b$
by doing a more careful treatment of the (classical) energy transfer in the
collision. The more careful treatment needed at small $b$ is relegated to
the homework. The one needed at large $b$, which must include the natural
motion of the target particle, is given here.

\centerline{\psfig{figure=fig5.ps,height=2.0in,width=8.5in}}

Suppose that the target is bound harmonically at a point, taken as the origin
of coordinates, meaning that there is a restoring force $-m\om_0^2\x$, where
$\x$ is the particle's position and $\om_0$ is the natural frequency of the
oscillator, in the absence of damping or perturbing forces. Given that the
particle is an electron with mass $m$ and charge $-e$, its equation of
motion in the presence of an applied electric field (the one coming from
the incident particle) is
\beq
m\sde\x t=-m\om_0^2\x-m\Ga\der\x t-e\Ext
\eeq
where the term $-m\Ga d\x/dt$ is a damping force proportional to the
particle's velocity; $\Ga$ is a `damping constant'.  This term is typically
small compared to the restoring force term. 

To simplify the solution, we will make several
{\bf{Approximations}}.  
\begin{enumerate}
\item We have not included the
magnetic force which acts on the bound charge. This force is smaller than the
electric force by a factor of the recoiling particle's velocity divided by $c$
even if the incident particle is relativistic; since the recoiling particle
is not relativistic, we may ignore the magnetic force. 
\item We make one
more approximation which is to evaluate $\Ext$ at the origin or point
where the target particle is bound; this is reasonable provided $b>>|\x|$
because then the  electric field will vary but little over distances of order
$|\x|$.
\end{enumerate}

We solve \eq{12} by making a Fourier analysis of the motion. Write
\beq
\E(t)\equiv\E(0,t)=\frac1\stp\intmp d\om'\,\E(\om')e^{-i\om't}
\eeq
and
\beq
\x(t)=\frac1\stp\intmp d\om'\,\x(\om')e^{-i\om't}.
\eeq
The inverse transforms are
\beq
\E(\om)=\frac1\stp\intmp dt\,\E(t)\eot
\eeq
and
\beq
\x(\om)=\frac1\stp\intmp dt\,\x(t)\eot.
\eeq
Substitute \eqs{13}{14} directly into the equation of motion and perform
the time derivatives to find
\beq
\frac1\stp\intmp d\om'\,\leb-\om'^2-i\om'\Ga+\om_0^2\rib\x(\om')
e^{-i\om't}=-\frac{(e/m)}\stp\intmp d\om'\,\E(\om')e^{-i\om't}.
\eeq
If we multiply by $\eot$ and integrate over $t$, we obtain a delta-function,
$\de(\om-\om')$, and can then integrate trivially over $\om'$ to find a
solution for $\x(\om)$ which is
\beq
\x(\om)=-\frac{e\E(\om)}m\frac1{\om_0^2-i\om\Ga-\om^2}.
\eeq

We could now figure out what is $\E(\om)$ since we know $\E(t)$ and use it
in \eq{18} to find $\x(\om)$ and then Fourier transform the latter to
find $\x(t)$. But we aren't really interested in $\x(t)$. What we are
trying to determine is the energy transferred to the target from the
incident charge. That energy can be found as follows:
\beq
\der Et=\F\cdot\der\x t=-e\E(t)\cdot\der{\x(t)}t
\eeq
where we again approximate $\Ext$ with $\E(0,t)$. The total energy
transferred in the collision is
\beqa
\De E&=&-\intmp dt\,e\E(t)\cdot\der{\x(t)}t\nonumber\\&=&-\intmp dt\,
\frac{e}{\sqrt{2\pi}}\intmp d\om'\,\E(\om')e^{-i\om't}\cdot\der{}t
\lep\frac1{\sqrt{2\pi}}\intmp d\om\,\x(\om)\emot\rip\nonumber\\
&=&-e\intmp d\om\,(-i\om)\x(\om)\cdot\E(-\om).
\eeqa
The last step is achieved by, first, taking the time derivative; second,
integrating over $t$ to obtain a delta-function $\de(\om+\om')$; and,
finally, integrating over $\om'$.

Because the electric field is real, $\E(-\om)=\E^*(\om)$. Similarly,
$\x(-\om)=\x^*(\om)$; hence
\beq
\De E=ie\intmp d\om\,\om\,\x(\om)\cdot\E^*(\om)=\Re\,\leb2ie\int_0^\infty d\om
\,\om\,\x(\om)\cdot\E^*(\om)\rib.
\eeq
Using our solution for $\x(\om)$, we find
\beqa
\De E&=&\Re\lep-2i\frac{e^2}m\int_0^\infty d\om\,\frac{\om|\E(\om)|^2}{
\om_0^2-i\om\Ga-\om^2}\rip\nonumber\\
&=&\Re\lep-2i\frac{e^2}m\intzp d\om\,\frac{\om|\E(\om)|^2(\om_0^2-\om^2+i
\om\Ga)}{(\om_0^2-\om^2)^2+\om^2\Ga^2}\rip\nonumber\\
&=&\frac{e^2}m\intzp d\om\,\frac{2\om^2\Ga|\E(\om)|^2}{(\om_0^2-\om^2)^2+
\om^2\Ga^2}
\eeqa

Finally, consider the limit that $\Ga$ is very small (small damping). Then
the entire weight in the integrand is at $\om=\om_0$ which means that the
only part of $\E$ which contributes to the energy transfer is the part
whose frequency matches the natural frequency of the oscillator. In this
limit the integral can be done by evaluating $\E(\om)$ at $\om_0$ so that
\beqa
\De E&\approx&\frac{2e^2}m|\E(\om_0)|^2\intzp d\om\,\frac{\om^2\Ga}{(\om_0^2-
\om^2)^2+\om^2\Ga^2}\nonumber\\
&=&\frac{2e^2}m|\E(\om_0)|^2\intzp dx\,\frac{x^2}{[(\om_0/\Ga)^2-x^2]^2+x^2}
\eeqa
where $x\equiv\om/\Ga$. The remaining integral is
\beqa
I&=&\intzp\frac{dx}{[(\om_0/\Ga)^2-x^2]^2/x^2+1}\approx\int_{-\om_0/\Ga}^\infty
\frac{dy}{(\Ga/\om_0)^2[2y\om_0/\Ga+y^2]^2+1}\nonumber\\
&\approx&\intmp\frac{dy}{4y^2+1}=\frac\pi2.
\eeqa
Hence
\beq
\De E=\frac{\pi e^2}m|\E(\om_0)|^2
\eeq
in the limit of $\Ga<<\om_0$.

We still need to evaluate $\E(\om_0)$. 

\centerline{\psfig{figure=fig6.ps,height=2.0in,width=8.5in}}

\noindent If the incident particle is moving
in the $z$ direction and the target lies in the $x$ direction relative to
the track of the incident particle, then the components of the electric field
are
\beq
\E\pll(t)=-\frac{qv\ga t}{(b^2+\ga^2v^2t^2)^{3/2}}\ept\andh
\E\per(t)=\frac{\ga qb}{(b^2+\ga^2v^2t^2)^{3/2}}\epu.
\eeq
Hence
\beqa
\E\per(\om)&=&\frac{qb\ga}\stp\intmp dt\,\frac\eot{(b^2+\ga^2v^2t^2)^{3/2}}
\epu=\frac{qb\ga}\stp\frac b{\ga v}\frac1{b^3}\intmp dx\,\frac{e^{-i(\om b/\ga
v)x}}{(1+x^2)^{3/2}}\epu\nonumber\\
&=&\frac q{bv\stp}\intmp dx\,\frac{e^{-izx}}{(1+x^2)^{3/2}}\epu=\frac{2q}{
bv\stp}\intzp dx\,\frac{\cos(zx)}{(1+x^2)^{3/2}}\epu
\eeqa
where $z=\om b/\ga v$.
The integral we are contemplating is a Bessel function; that is,
\beq
K_\nu(z)=\frac{2^\nu\Ga(\nu+1/2)}{\sqrt\pi\,z^\nu}\intzp dx\,\frac{\cos(xz)
}{(1+x^2)^{\nu+1/2}};
\eeq
specifically,
\beq
K_1(z)=\frac{2\Ga(3/2)}{\sqrt\pi\,z}\intzp dx\,\frac{\cos(zx)}{(1+x^2)^{3/
2}}.
\eeq
Further, $\Ga(3/2)=\sqrt\pi/2$, so
\beq
\E\per(\om)=\frac{q}{bv}\sqrt{\frac2\pi}\,zK_1(z)\epu.
\eeq
By similar manipulations one finds that
\beq
\E\pll(\om)=-i\frac q{\ga vb}\sqrt{\frac2\pi}\,zK_0(z)\ept.
\eeq
Hence the energy transfer is, from \eq{25},
\beq
\De E=\frac{\pi e^2}m\frac{q^2}{b^2v^2}\frac2\pi\leb z^2K_1^2(z)+
\frac{z^2}{\ga^2}K_0^2(z)\rib=\frac{2q^2e^2}{mb^2v^2}\leb z^2K_1^2(z)
+\frac{z^2}{\ga^2}K_0^2(z)\rib
\eeq
where $z=\om_0b/\ga v=b/b_{max}$ using $b_{max}=\ga v/\om_0$ as per
the criterion discussed in the preceding section.

%%BoundingBox: 81 77 509 330
\centerline{\psfig{figure=fig7.ps,height=2.5in,width=4.0in}}

{\em{Cutoffs}}.  What is the qualitative behavior of this result 
as a function of $b$? For
$b<<b_{max}$, $z<<1$ and $zK_0(z)\rta0$ as $z\rta0$ while $zK_1(z)\rta1$.
Making these substitutions in \eq{32} we find that for small
$b$, the result is the same as what emerged from the
impulse approximation.\footnote{Which is almost miraculous because we
approximated $\Ext$ as $\E(0,t)$ which is not good when $b$ is small;
evidently, some cancellation of errors takes place.}  Thus we must
still insert by hand a cutoff for small $b$ ($b_{min}$).
For large $b>>b_{max}$,
however, $z>>1$ and the Bessel functions' behavior is
\beq
K_0(z)\sim K_1(z)\sim\sqrt{\frac\pi{2z}}e^{-z}
\eeq
so that in this regime of $b$,
\beq
\De E\approx\frac{2q^2e^2}{mv^2b^2}\leb\frac\pi{2z}z^2\lep1+\frac1{
\ga^2}\rip e^{-2z}\rib=\frac{q^2e^2\pi z}{mv^2b^2}\lep1+\frac1{\ga^2}
\rip e^{-2z}.
\eeq
Thus the large $b$ cutoff is automatically included in this formalism

We can find the total energy loss per unit path length by integrating over
$b$ as before.  Given an electron density $n$, then an incident particle
traversing a distance $dx$ will pass $n(dx)(2\pi b\,db)$ scatterers with
impact parameters between $b$ and $b+db$.  The integral for the energy loss 
by the incident particle can then be put in the form
\[
d^2E=-(2\pi b\, db)(v\, dt)n\,\Delta E\,,
\]
or, since $b=b_{max}z=\frac{v\gamma}{\omega_0} z$ and $dx=v\, dt$,
\[
d^2E=-2\pi\lep\frac{v\gamma}{\omega_0}\rip^2 z\, dz\, dx\, n \Delta E\,.
\]
Then, integrating on $z$, we get
\beq
\der Ex=-2\pi n\frac{2q^2e^2}{mv^2}\int_{z_{min}}^\infty\frac{dz}z\leb
z^2K_1^2(z)+\frac{z^2}{\ga^2}K_0^2(z)\rib
\eeq
where $z_{min}=b_{min}/b_{max}=qe\om_0/m\ga^2v^3$. The integral, which
is
\beq
I\equiv\int_{z_{min}}^\infty dz\,z\lep K_1^2(z)+\frac1{\ga^2}K_0^2(z)\rip=
\int_{z_{min}}^\infty dz\,z\lep K_1^2(z)+K_0^2(z)-\frac{v^2}{c^2}K_0^2(z)
\rip,
\eeq
can be done by making use of certain identities satisfied by the Bessel
functions. These identities are
\beq
K'_\nu(x)=-K_{\nu-1}(x)-\frac\nu xK_\nu(x)\andh K'_\nu(x)=-K_{\nu+1}(x)+
\frac\nu xK_\nu(x);
\eeq
from them, it follows that
\beqa
\der{}x[xK_0(x)K_1(x)]&=&-x[K_0^2(x)+K_1^2(x)]\nonumber\\
\der{}x[x^2(K_1^2(x)-K_0^2(x))]&=&-2xK_0^2(x).
\eeqa
It is now easy to do the integral; the result is
\beq
I=z_{min}K_0(z_{min})K_1(z_{min})-\frac{v^2}{2c^2}z_{min}^2[K_1^2(z_{min})
-K_0^2(z_{min})].
\eeq
Now, $z_{min}=qe\om_0/m\ga^2v^3\sim10^{-7}$ or less for a relativistic
particle, so we can expand the Bessel functions in the small argument limit:
\beq
K_1(x)\approx1/x\andh K_0(x)\approx-[\ln(x/2)+0.577]=\ln(1.123/x).
\eeq
Thus $I=\ln(1.123/z_{min})-v^2/2c^2$, and
\beq
\der Ex=-\frac{4\pi nq^2e^2}{mv^2}\leb\ln\lep\frac{1.123m\ga^2v^3}{qe
\om_0}\rip-\frac{v^2}{2c^2}\rib.
\eeq

This formula may be easily extended to a (slightly) more realistic form,
accounting for different charges with different resonant frequencies.  
Assume an elemental solid with a density of atoms $N$, each with $Z$ electrons.
The $Z$ electrons will be split into groups of $f_j$ electrons
distinguished by the resonant frequency of the group $\om_j$.
The oscillator strengths $f_j$ must satisfy the sum rule $\sum_j f_j =Z$.
The groups add linearly so that 
\beq
\der Ex=-4\pi n Z \frac{q^2e^2}{mv^2}\leb\ln{B_c}-\frac{v^2}{2c^2}\rib.
\eeq
where
\beq 
B_c=\frac{1.123m\ga^2v^3}{qe
<\om>}\;\;\;\;\; Z\ln<\om>=\sum_j f_j \ln\om_j
\eeq

This is the classical energy-loss formula derived by Bohr in 1915. It
actually works rather well despite the fact that the effects responsible
for the energy loss (scattering of small objects by other small objects)
really ought to be treated using quantum theory. The reason why the classical
theory works as well as it does is that any macroscopic energy loss is the
result of many collisions. The energy loss in each collision is not given
very accurately by the classical theory, but \eq{41} represents the energy loss
over a large number of collisions, and that is pretty close to the mark. Thus
the usefulness of the classical theory is in part a consequence of statistical
effects. Bohr's original formula was eventually superseded by a calculation
based on quantum theory and done by Bethe in 1930. Read the appropriate section
in Jackson for more details.

\section{Density Effect in Energy Loss}
A charged particle traversing a material produces a local electric polarization
of that material, as a consequence of which the electric field acting on
any given charge in the material is not the electric field that we used in
the preceding sections. 

\centerline{\psfig{figure=fig8.ps,height=3.0in,width=8.5in}}

\noindent This ``screening'' effect is especially important for collisions
of large impact parameter $b$, since then the field will be screened
by the charges closer to the path of the incident particle.

\centerline{\psfig{figure=fig10.ps,height=2.5in,width=8.5in}}

\noindent Thus the energy loss formulas we derived earlier will overestimate
the energy loss of a charged particle traversing a polarizable
medium. As we will see, this effect is most important for
fast or ultra-relativistic particles.

We can produce a calculation of the consequences of
this ``screening'' effect using the familiar formalism of macroscopic
electrodynamics. Let the material have a frequency-dependent dielectric
function $\epo$, as discussed in Chapter~7, so that the displacement and
macroscopic electric field, expressed as functions of position and
frequency, are related by
\beq
\Dxo=\epo\Exo;
\eeq
the connection between any field $F$ as a function of $\x$ and $\om$ and
the same field as a function of $\x$ and $t$ is
\beq
\Fxt=\frac1\stp\intmp d\om\,\Fxo\emot
\eeq
with the inverse transformation
\beq
\Fxo=\frac1\stp\intmp dt\,\Fxt\eot.
\eeq
Further, let us introduce Fourier transforms in space:
\beq
\Fxo=\frac1{(\stp)^3}\int d^3k\,\Fko\ekx
\eeq
with the inverse
\beq
\Fko=\frac1{(\stp)^3}\int d^3x\,\Fxo\emkx.
\eeq

We begin from the macroscopic Maxwell equations with $\B\equiv\H$, i.e.,
$\mu=1$; the inhomogeneous equations are
\beq
\curl\B=\frac{4\pi}c\J+\frac1c\pde\D t\andh\div\D=4\pi\rh,
\eeq
and the homogeneous field equations may be replace by
\beq
\B=\curl\A\andh\E=-\grad\Ph-\frac1c\pde\A t.
\eeq
which ensure that $\div\B=0$ and that 
$\curl\E=-\frac{1}{c}\partial\B/\partial t$.
Fourier transform the Maxwell equations (49) to find
\beq
i\k\times\B=\frac{4\pi}c\J-i\frac\om c\D\andh i\k\cdot\D=4\pi\rh.
\eeq
Similarly, from the Fourier transforms of Eqs.~(50) one finds
\beq
\B=i\k\times\A\andh\E=-i\k\Ph+i\frac\om c\A.
\eeq
Substitution of this pair of equations into the immediately preceding ones and
using $\D=\ep\E$, we arrive at Fourier transformed wave equations for the
potentials:
\beq
-\k(\k\cdot\A)+k^2\A=\frac{4\pi}c\J-\frac\om c\ep\leb\k\Ph-\frac\om c\A\rib
\eeq
and

\beq
\ep\leb k^2\Ph-\frac\om c\k\cdot\A\rib=4\pi\rh.
\eeq

We can make the equations for $\A$ and $\Ph$ separate by choosing an
appropriate gauge; specifically,
\beq
\k\cdot\Ako=\ep\frac\om c\Phko\,,
\eeq
which is a slightly modified form of the Lorentz gauge.
Within this gauge, the equations of motion are
\beq
(k^2-\ep\,\om^2/c^2)\A=\frac{4\pi}c\J\andh(k^2-\ep\,\om^2/c^2)\Ph=4\pi\,
\frac\rh\ep
\eeq
which are simple familiar\footnote{They may not look so familiar because
they are in wavenumber and frequency space.} wave equations.

The only macroscopic source is the incident charge\footnote{The dielectric
function accounts for any sources associated with charges in the material.}, so
\beq
\rhxt=q\de(\x-\v t)\andh\Jxt=\v\rhxt
\eeq
where we approximate $\v$ as a constant, $\v=v\ept$. The Fourier transforms
of these source densities are
\beqa
\rhko&=&\frac1{(2\pi)^2}\int d^3xdt\,\emkxot q\de(\x-\v t)\nonumber\\
&=&\frac q{(2\pi)^2}\intmp dt\,e^{-i(\k\cdot\v-\om)t}=
\frac q{2\pi}\,\de(\om-\v\cdot\k)
\eeqa
and, similarly,
\beq
\Jko=\frac{q\v}{2\pi}\,\de(\om-\v\cdot\k).
\eeq
The solutions for the Fourier-transformed potentials are trivially found:
\beqa
\Phko&=&\lep\frac{2q}\ep\rip\frac{\de(\om-\k\cdot\v)}{k^2-\ep\,\om^2/c^2}
\nonumber\\
\Ako&=&\lep\frac{2q\v}c\rip\frac{\de(\om-\k\cdot\v)}{k^2-\ep\,\om^2/c^2}
\eeqa
Now, $\Eko=-i\k\Phko+i(\om/c)\Ako$, so
\beq
\Eko=2iq\lep\frac{\om\v}{c^2}-\frac\k\ep\rip\frac{\de(\om-\k\cdot\v)}{k^2-\ep
\,\om^2/c^2}
\eeq
and
\beq
\Bko=i\k\times\Ako=i\lep\frac{2q}c\rip(\k\times\v)\frac{\de(\om-\k\cdot\v)
}{k^2-\ep\,\om^2/c^2}.
\eeq

Now let us compute the rate at which the incident particle loses energy by
finding the flow of electromagnetic energy away from the track of this
particle. Let the point at which the fields are to be evaluated be
$\x=b\epu$ and find $\Exo$ and $\Bxo$:
\beqa
\Bxo&=&\frac1{(\stp)^3}\int d^3k\,\ekx\,i\lep\frac{2q}c\rip(\k\times\v)\frac{
\de(\om-\k\cdot\v)}{k^2-\ep\,\om^2/c^2}\nonumber\\
&=&\frac{(2iq/c)}{(2\pi)^{3/2}}\int d^3k\,e^{ibk_1}(k_2\epu-k_1\epd)v
\frac{\de(\om-k_3v)}{k^2-\ep\,\om^2/c^2}\nonumber\\
&=&-\epd\frac{(2iq/c)}{(2\pi)^{3/2}}\int dk_1dk_2\,
k_1e^{ibk_1}\les\leb k_1^2+k_2^2+\frac{\om^2}{v^2}\lep1-\ep
\frac{v^2}{c^2}\rip\rib\right.
\eeqa
Set $\la^2=(\om/v)^2(1-\ep\,v^2/c^2)$ and $\be=v/c$. Then
\beqa
\Bxo&=&-i\epd\frac{2q}{c(2\pi)^{3/2}}\int dk_1dk_2\frac{k_1e^{ibk_1}}{k_1^2+
k_2^2+\la^2}=-i\epd\frac q{c\stp}\int dk_1\frac{k_1e^{ibk_1}}{\sqrt{k_1^2+
\la^2}}\nonumber\\
&=&-\epd\frac q{c\stp}\der{}b\lep\int dk_1\,\frac{e^{ibk_1}}{
\sqrt{k_1^2+\la^2}}\rip=-\epd\frac qc\sqrt{\frac2\pi}\der{}b\lep\intzp dx\,
\frac{\cos(b\la x)}{\sqrt{1+x^2}}\rip\nonumber\\
&=&-\epd\frac qc\sqrt{\frac2\pi}\,\der{}b[K_0(b\la)]=\epd\frac qc\sqrt{\frac2
\pi}\,\la K_1(b\la).
\eeqa
Similarly,
\beq
\Exo=\epu\frac qv\sqrt{\frac2\pi}\,\frac\la\ep K_1(b\la)-i\ept\frac{q\om}{v^2}
\sqrt{\frac2\pi}\lep\frac1\ep-\be^2\rip K_0(b\la).
\eeq

Next, for real $\ep$, $\la^2$ may be positive or negative depending on whether
the incident particle moves more slowly or more rapidly than the speed of light
in the medium, $c'=c/\sqrt\ep$. For $v<c'$, $\la^2>0$, $\la$ is real, and $\E$
reduces to our previous result except for the appearance of $\ep$ here and
there. It is then a straightforward matter to calculate $\De E(b)$ by the same
procedure as before, assuming\footnote{A risky assumption.} the field acting
on a target particle is the same as the macroscopic field.

Rather than reproducing the previous calculation, let's look at an alternative:
we shall calculate the radial outward part ($\rh$ component) of the Poynting
vector at $\x=\rh\rhv$.

\centerline{\psfig{figure=fig11.ps,height=2.7in,width=8.5in}}

\noindent  When this component of $\S$ is integrated over all time
and over a closed loop of radius $b$ around  the path of the particle, the
result is the total electromagnetic field energy which flows away from the
particle, per unit length of path, and at distance $b$ from the path.
Letting this energy be $E_f$, to distinguish it from 
the energy change of
anything else (such as the incident particle), we have
\beq
\lep\der {E_f}z\rip_{\rh=b}=\frac c{4\pi}2\pi b\intmp dt\,(\E\times\B)\cdot\nn
\eeq
Given the geometry introduced earlier, the quantity $(\E\times\B)\cdot\nn$ is
just $-E_3B_2$.

Let's complete the integral:
\beqa
\lep\der {E_f}x\rip_{\rh=b}&=&-\frac{cb}2\intmp dt\,B_2(t)E_3(t)=-\frac{cb}{4
\pi}\intmp dtd\om d\om'\,B_2(\om')E_3(\om)e^{-i(\om+\om')t}\nonumber\\
&=&-\frac{cb}2\intmp d\om B_2(-\om)E_3(\om)=-\frac{cb}2\intmp d\om E_3(\om)
B_2^*(\om)\nonumber\\&=&-cb\Re\leb\intzp d\om\,B_2^*(\om)E_3(\om)\rib\nonumber\\
&=&-\frac{2cbq^2}{\pi v^2}\Re\leb\intzp d\om\,(-i\om)\lep\frac1\ep-\be^2\rip K_0
(b\la)\frac1c\la^*K_1(b\la^*)\rib\nonumber\\
&=&\frac{2q^2}{\pi v^2}\Re\leb\intzp d\om\,(i\om\la^*b)\lep\frac1\ep-\be^2\rip
K_1(b\la^*)K_0(b\la)\rib,
\eeqa
an expression first derived by Enrico Fermi.

In order for the integral to have a real part, either $\la$ or $\ep$ must
be complex. If $\ep$ is real, then $\la$ can still be complex if $\ep\be^2>1$
meaning that the particle is travelling faster than the speed of light in
the material. In this case one finds the phenomenon of Cherenkov radiation
which we shall discuss presently. 

For now, let us look at the case of complex
$\ep$. Introduce the frequency-dependent polarization $\Pxo$ via the
relation
\beq
\Dxo=\Exo+4\pi\Pxo;
\eeq
In a linear medium such as we are considering, $\Pxo=\ch(\om)\Exo$ with
$\ch(\om)=(\epo-1)/4\pi$. The frequency dependent polarization is just the
Fourier transform in time of the usual polarization $\Pxt$. If we calculate
it using the damped harmonic oscillator model introduced above and in
chapter~7, we find
\beq
\Po=\frac{ne^2}{m}\frac\Eo{\om_0^2-\om^2-i\om\Ga}
\eeq
where $n$ is the electron density in the material; the corresponding
dielectric function is
\beq
\epo=1+\frac{\om_p^2}{\om_0^2-\om^2-i\om\Ga}
\eeq
where $\om_p$ is the plasma frequency, $\om_p^2=4\pi ne^2/m$.

Now we have an expression for $\epo$ based on a simple model. We need to do
the integral presented in \eq{67}. Unfortunately that cannot be done in
terms of simple functions so we shall approximate the integral in a
physically reasonable way. The important range of $\om$ should be $\om\sim
\om_0$ so that $b\la\sim b\om/v\sim b(\om_0/v)<<1$ for $b$ less than about an
atomic size and $v\sim c$; $\om_0$ is a typical atomic energy.
Thus we make the small argument approximations
\beq
b\la^*K_1(b\la^*)\approx b\la^*\frac1{b\la^*}=1
\eeq
and
\beq
K_0(b\la)\approx\ln(1.123/b\la)
\eeq
which leads to
\beq
\lep\der {E_f}x\rip_{\rh=b}=\frac{2q^2}{\pi v^2}\Re\leb\intzp d\om\,i\om\lep
\frac1\ep-\be^2\rip\ln\lep\frac{1.123}{b\la}\rip\rib.
\eeq
Because we just fouled up the integrand in the region $\om>>\om_0$, we had
best make sure that no contribution comes from this region of frequency;
physically, we believe this should be the case. Since $\ep\rta1$ sufficiently
rapidly here (something that should be checked to be sure our belief), we can
guarantee convergence of the integral by approximating $\be^2$ with 1. Then
\beq
\lep\der {E_f}x\rip_{\rh=b}=\frac{2q^2}{\pi v^2}\,\Re(I)
\eeq
where
\beq
I=\intzp d\om\,i\om\leb\ln
\lep\frac{1.123c}{\om b}\rip-\frac12\ln(1-\ep)\rib\lep\frac{1-\ep}\ep\rip.
\eeq
Using \eq{68} for $\epo$, we have
\beq
I=i\intzp d\om\,\om\lep\frac{-\om_p^2}{\om_0^2+\om_p^2-\om^2-i\om\Ga}\rip
\leb\ln\lep\frac{1.123c}{\om_pb}\rip-\ln\om+\frac12\ln(\om^2-\om_0^2+i\om\Ga)
\rib
\eeq

We can employ the Cauchy theorem to evaluate this integral by closing the
contour around the first quadrant; that is, construct a closed path by
adding a quarter-circle from a point where $\om$ is large and real to one
where it is large and imaginary and then coming down the positive
imaginary-$\om$ axis to the origin. 

\centerline{\psfig{figure=fig9.ps,height=3.0in,width=8.5in}}

\noindent The total integral around this contour is
zero because there are no poles of the integrand within it. This point is
clarified by looking for the zeroes of the integrand's denominator and by
looking for the zeroes of the logarithm's argument. They are located at points
in the lower half plane and so are well away from the interior of the contour.

The integral along the imaginary-frequency axis is, with $\om=i\Om$, $\Om$
real,
\beqa
I_3=-i\intzp id\Om\,\frac{-i\Om\om_p^2}{\Om^2+\om_0^2+\om_p^2+\Om\Ga}
\leb\ln\lep\frac{1.123c}{b\om_p}\rip-\ln(i\Om)+\frac12\ln[-
(\Om^2+\om_0^2+\Om\Ga)]\rib\nonumber\\
=i\intzp d\Om\,\frac{-\Om\om_p^2}{\Om^2+\om_0^2+\om_p^2+\Om\Ga}\leb\ln\lep
\frac{1.123c}{b\om_p}\rip-\ln\Om+\frac12\ln(\Om^2+\om_0^2+\Om\Ga)\rib
\eeqa
which is pure imaginary, meaning that $\Re(I_3)=0$.
The integral over the quarter-circle, $I_2$, is thus such that $-\Re(I_2)=
\Re(I)$, or, letting $\om=\Om\exp(i\th)$ on the quarter-circle,
\beqa
\Re(I)=-\Re\int_0^{\pi/2}i\Om e^{i\th}i\Om e^{i\th}d\th\,\lep\frac{-
\om_p^2}{\om_0^2+\om_p^2-\Om^2e^{2i\th}-i\Om e^{i\th}\Ga}\rip\nonumber\\
\times\leb\ln\lep
\frac{1.123c}{b\om_p}\rip-\ln\lep\Om e^{i\th}\rip+\frac12\ln\lep\Om^2e^{2i\th}-
\om_0^2+i\Om\Ga e^{i\th}\rip\rib\nonumber\\
=\om_p^2\Re\int_0^{\pi/2}d\th\,\leb\ln\lep\frac{1.123c}{b\om_p}\rip+{\cal
O}\lep\frac{\Ga}\Om\rip\rib=\om_p^2\frac\pi2\ln\lep\frac{1.123c}{b\om_p}
\rip.
\eeqa
Hence,
\beq
\lep\der {E_f}z\rip_{\rh=b}=\lep\frac{q^2\om_p^2}{c^2}\rip\ln\lep
\frac{1.123c}{b\om_p}\rip;
\eeq
The negative of this quantity is the energy loss of the incident particle per
unit distance traveled.

This result is to be compared with the one we found before taking screening
into account,
\beq
\lep\der Ez\rip_{\rh>b}=-\frac{q^2\om_p^2}{c^2}\leb\ln\lep\frac{1.123\ga c}{b
\om_0}\rip-\frac12\rib.
\eeq
The two differ significantly in principle, if not numerically. In
particular, the dependence of our original formula on the specific natural
frequency of the target, $\om_0$, is gone, replace by a dependence on
$\om_p$ which depends only on the density of the target electrons. Also,
a factor of $\ga$ has, in our most recent result, disappeared from the argument
of the logarithm, meaning that the energy loss by highly relativistic
charged particles is much reduced by the screening effect.

\section{Cherenkov Radiation}
We are also in a position to calculate energy loss by Cherenkov
radiation which is something that takes place when the incident particle's
speed exceeds the speed of light in the medium. We can avoid the mechanism
just discussed and so isolate the Cherenkov radiation mechanism by letting
$\ep$ be real (no damping). In this approximation we will also miss 
the attenuation of the radiation. Under these conditions, and as discussed
in the last section, the only way to get any radiation is if
\beq
\lambda=\frac{\om}{v} \sqrt{ 1-\ep(\om)\beta^2} \in {\Large{\rm C}}\,,
\eeq
or, more correctly, $\lambda$ must be imaginary.  We must 
have $v^2>c^2/\ep$ or there will be no radiation. Since $n=\sqrt{\ep}$,
then $c/\sqrt{\ep}$ is the speed of light in the medium, and thus
the condition for radiation is that the particle exceed the speed
of light in the medium.

	This will not happen for all frequencies.  By assuming a simple
model dielectric function
\beq
\epo=1+\frac{\om_p^2}{\om_0^2-\om^2}\,,
\eeq
and expressing the condition as $\ep(\om) > 1/\beta^2$ we can see
that the radiation tends to be emitted near regions of anomalous
dispersion.  

\centerline{\psfig{figure=cheren_band.ps,height=2.5in,width=6.0in}}

Under these conditions, we evaluate the fields which are
present at distance $b$ from the axis of the incident particle, using $b$ large
enough that we can make simple approximations to the Bessel functions,
$b|\la|>>1$. Then
\beq
K_0(\la b)\approx K_1(\la b)\approx\sqrt{\frac{\pi}{2\la b}}e^{-\la b}
\eeq
and so, 
\beq
\Bxo=\epd\frac{q}{c}\sqrt{\frac{\la}{b}}\,e^{-\la b}\,.
\eeq
Similarly,
\beq
\Exo=\epu\frac{q}{\ep v}\sqrt{\frac{\la}{b}}\,e^{-\la b}
-i\ept\frac{q\om}{v^2}
\frac{1}{\sqrt{\la b}}\lep\frac1\ep-\be^2\rip \,e^{-\la b}.
\eeq

\centerline{\psfig{figure=cheren2.ps,height=2.5in,width=8.5in}}

Then from \eq{67} and the equations above, the field energy passing 
through the cylinder of radius $b$ per unit length is
\beq
\lep\der Ez\rip_C=\frac{2q^2}{\pi v^2}\Re\leb\intzp d\om\,i\om\frac\pi2\sqrt{
\frac{\la^*}\la}e^{-(\la+\la^*)a}\lep\frac1\ep-\be^2\rip\rib.
\eeq
The wonderful thing that happens when  $\la$ is pure imaginary 
is that the exponential functions have imaginary arguments and will not 
become small as $b$ becomes large. Thus we find the energy given off 
as {\em Cherenkov radiation} to be
\beq
\lep\der Ez\rip_C=\frac{q^2}{v^2}\Re\leb\int d\om\,i\om\sqrt{-1}\lep\frac{1-\ep
\be^2}\ep\rip\rib=\frac{q^2}{c^2}\int d\om\,\om\lep1-\frac1{\ep\be^2}\rip
\eeq
where the integration extends over only those frequencies $\ep\be^2>1$.
One can see that this is indeed radiative energy loss because it
is independent of $b$ provided only $b$ is large enough that the Bessel
functions are well-represented by their large-argument forms. In this respect
it is quite distinct from the energy loss by transfer of energy to other
charged particles that we studied earlier (real as opposed to virtual
photons). We were able to treat that energy
loss by examining the energy carried by the electromagnetic fields because the
mechanism by which the energy is transferred from one particle to another
is by means of the fields; in effect, we did that calculation in such a way
as to ``intercept'' the energy that was on its way from one charge to another.

From the picture above it is clear that the radiation is 
completely linearly polarized
in the plane containing the observer and the path of the particle.
In addition the angle $\th_c$ of emission of Cherenkov radiation
relative to the direction $\ept$ of the particle's velocity
is given by
\beq
\cos(\th_c)=\frac{E_1}{\sqrt{E_1^2+E_3^2}}=\frac{c/n}{v}\,
\eeq
where $n=\sqrt{\ep}$.  Thus the condition that $\la$ be complex,
and thus that required for Cherenkov radiation, can be rephrased as the
requirement that $\th_c$ be a physical angle with a cosine less
than unity.

	As shown in the picture below, the emission angle
$\th_c$ can also be interpreted in terms of a shock wave angle.
\vspace{0.3in}

\centerline{\psfig{figure=cheren.ps,height=2.6in,width=8.5in}}

\section{Momentum Transfer}
The final topic we shall study in this chapter is the deflection of the
incident particle produced by scattering from the particles in the material
through which it moves. The targets mainly responsible for the deflection
turn out to be the highly charged ones---the nuclei.

We start by introducing the number of particles incident per unit time on the
target with an impact parameter between $b$ and $b+db$ and at an azimuthal
angle between $\ph$ and $\ph+d\ph$. 

\centerline{\psfig{figure=mom1.ps,height=2.0in,width=8.5in}}

\noindent If the incident beam has a particle number
density $n$ and a speed $v$, then the incident flux is $nv$ particles per unit
area per unit time, and the number incident in the area element just described
is
\beq
d^2N=nvbdbd\ph.
\eeq
Now, given a smoothly varying scattering potential, these particles will,
after scattering, show up in some element of solid angle $d\Om$. 

\centerline{\psfig{figure=mom2.ps,height=2.8in,width=8.5in}}

\noindent Hence we can write that
\beq
d^2N=N'd\Om
\eeq
where $N'$ is the number of particles scattered into unit solid angle in
unit time and $d\Om$ is the element of solid angle into which the
particular $d^2N$ particles under consideration are scattered. Using \eq{89},
we have
\beq
nvbd\ph db=N'd\Om\orh bd\ph db=\frac{N'}{nv}d\Om
\eeq
The quantity $N'$ is proportional to the incident particle flux; that is, the
number of particles per unit solid angle that come out in some given direction
is directly proportional to the incident flux. Hence a more intrinsic measure
of the scattering than $N'$ is provided by the quantity $N'/nv$, 
the {\em differential scattering cross-section} $d\si/d\Om$:
\beq
\der\si\Om\equiv\frac{N'}{nv}
\eeq
Making this substitution in \eq{91}, we get
\beq
bd\ph db=\der\si\Om d\ph \sin\th d\th
\eeq
We will also assume that the potential between the incident particle and 
the scatterer is central.   In this case we have azimuthal symmetry so
the particles incident on the target in some increment $d\ph$ of azimuthal
angle around $\ph$ are scattered into the same element of azimuthal angle, 

\centerline{\psfig{figure=mom3.ps,height=2.0in,width=8.5in}}

\noindent thus we find
\beq
b\,db=\der\si\Om\sin\th d\th\orh\der\si\Om=\frac b{\sin\th}\lel\der b\th\ril
\eeq
where $\th$ is the angle by which the particle is deflected or scattered.

The differential scattering cross-section, by its definition, has
dimensions of length squared or area. We can evaluate it if we have an
equation relating $b$ and $\th$. In the impulse approximation, the
scattering angle $\th$ is given by ratio of the momentum transfer to the
incident momentum; and that is, from \eq{2},
\beq
|\th|=\frac{p}{P}=\lel\frac{2qe}{Pvb}\ril
\eeq

\centerline{\psfig{figure=mom4.ps,height=1.5in,width=8.5in}}

\noindent where, in this equation, $\P=\gamma M\v$ is the momentum of the 
incident particle, and $p=\frac{2|qe|}{bv}$ (\eq{2}) is the momentum 
transfer from the incident particle to the target.  From this relation 
we can evaluate $|d\th/db|$ and find that the cross-section is
\beq
\der\si\Om=\frac b{\sin\th}\lel\frac{Pvb^2}{2qe}\ril=\frac{Pv}{2qe\sin\th}
\lep\frac{2qe}{Pv\th}\rip^3=\lep\frac{2qe}{Pv}\rip^2\frac1{\th^4}
\eeq
where we make the small angle approximation $\th\approx\sin\th$ which
is valid anywhere that the impulse approximation is valid. In this, the
small-angle regime, our result matches the Rutherford scattering 
cross-section.

From \eq{96} we can see that nuclei are more effective than electrons at
producing a given deflection $\th$. The charge $e$ that appears in the
cross-section is the charge of the target, a holdover from when we let the
target be an electron. More generally, replace this charge by $ze$, in case
the target is, \eg a nucleus. 
\beq
\der\si\Om=\frac b{\sin\th}\lel\frac{Pvb^2}{2qze}\ril=\frac{Pv}{2qze\sin\th}
\lep\frac{2qze}{Pv\th}\rip^3=\lep\frac{2qze}{Pv}\rip^2\frac1{\th^4}
\eeq
One can then see that the cross-section is
proportional to $z^2$, meaning that a nucleus is more effective by a
factor of $z^2$ at producing a given angle of deflection $\th$. At the same
time there are $z$ times as many electrons, leading to $z$ times as many
scattering events. This is not enough to offset the larger cross-section
produced by the nuclei, and therefore they are the dominant scatterers where
deflection of the incident particle is concerned.

\subsection{Average Angle of Deflection}
	Of course the target is rarely composed of a single atom.
Rather, we generally scatter from a molecular solid, or material.  Here,
we want to calculate a typical or average angle of deflection produced in a
scattering event. That will require integrating over $\th$ using
$d\si/d\Om$ as the distribution function. Cutoffs on the integration must
be introduced. At small $\th$, corresponding to large $b$, the cutoff is
determined by the condition $b_{max}\sim a$ where $a$ is an atomic size.

\centerline{\psfig{figure=mom5.ps,height=3.5in,width=8.5in}}

\noindent The reason is that for $b>a$, the incident particle passes completely
outside of the electronic shell surrounding the nucleus and so the
interaction between the incident particle and nucleus is almost completely
screened. Thus
\beq
\th_{min}\approx\lel\frac{qze}{Pvb_{max}}\ril\approx\lel\frac{qze}{Pva}\ril.
\eeq
This still leaves a large range of impact parameter $b$, since the nuclear
radius $\sim 10^{-13}\,{\rm cm.}$ and that of a typical atomic radius
is $\sim 10^{-8}\,{\rm cm.}$.  An alternative, quantum-based argument 
can be made for choosing $\th_{min}\sim
\hbar/pa$. There is also a maximum scattering angle which is not of much
significance in the present context; we may suppose that $\th_{max}$ is of
order one.

Given appropriate cutoffs, we can determine the mean value of $\th^2$ in
scattering events. Using the small-angle approximation for all trigonometric
functions, we have
\beqa
<\th^2>=\frac{\int d\Om\,\th^2[d\si/d\Om]}{\int d\Om\,[d\si/d\Om]}
\approx\frac{\int_{\th_{min}}^{\th_{max}}d\th/\th}{\int_{\th_{min}}^{
\th_{max}}d\th/\th^3}\nonumber\\
=\frac{2\ln(\th_{max}/\th_{min})}{1/\th_{min}^2-1/\th_{max}^2}
\approx2\th_{min}^2\ln(\th_{max}/\th_{min}).
\eeqa
This result is some not-very-large multiple\footnote{Because the cross-section
is strongly peaked at small angles.} of $\th_{min}^2$. Hence, a single
scattering event cannot be expected to deflect the incident particle very
much. 

	A sizable net deflection can be obtained in two quite different ways.
One is that a large number of small-angle scatterings can result in a large
deflection. The other is that a single large-angle scattering, though rare,
can occur. If one bombards a thin slab of target material with a beam of
particles, then what one finds is that most of the particles which come
through will have experienced a large number of small-angle scatterings and
no large-angle scatterings. These will have a distribution of net
scattering angles which reflects their experience (many small-angle
scatterings). Some particles, however, will have experienced a large-angle
scattering in addition to the many small-angle scatterings. They will have
a distribution of scattering angles which reflects their experience and
which will be quite unlike the distribution of the particles which
experience only small-angle scatterings. Let's give each of these
possibilities a little further thought.

\subsubsection{Distribution of Small Angle Scattering}
If the particle experiences only a large number of small-angle scattering
events, its deflection will resemble a random walk. \vspace{0.1in}

\centerline{\psfig{figure=mom6.ps,height=2.0in,width=8.5in}}

\noindent A collection of such
random walkers will provide a distribution of observed scattering angles
which will have approximately a Gaussian form,
\beq
P(\th)\sim e^{-(\th^2/<\Th^2>)},
\eeq
where $<\Th^2>$ is the width of the distribution. To carry the analysis
further in a quantitative manner, let's make the random walk effectively
one-dimensional by projecting it onto a plane.

\centerline{\psfig{figure=projects.ps,height=2.5in,width=8.5in}}

Consider a particle that is scattered into the direction
$(\th,\ph)$; project this direction onto the $y$-$z$~plane where it becomes
$\th'$ with $\th'=\th\sin\ph$ for $\th<<1$. Hence $\th'^2=\th^2\sin^2\ph$, and
the observed mean value of $\th'^2$ in single scattering events is
\beq
<\th'^2>=\frac{\int d\Om\,\th'^2(\th,\ph)[d\si/d\Om]}{\int d\Om\,
[d\si/d\Om]}=\frac12<\th^2>.
\eeq
Also, $<\th'>=0$. Assuming that the scattering directions produced by the
different collisions that any one particle suffers are independent, and
that there are many such collisions, then, from the theory of the
elementary one-dimensional random walk, the normalized distribution of
observed net scattering angles $\th'$ is well-approximated by a Gaussian
\beq
P_m(\th')=\frac1{\sqrt{\pi<\Th^2>}}e^{-\th'^2/<\Th^2>}
\eeq
with the random walk distribution width
\beq
<\Th^2>=N<\th'^2>,
\eeq
where $N$ is the mean number of collisions experienced by each particle in
traversing the material. If the total cross-section is $\si$, the density
of scatterers is $n$, and the thickness of the slab is $a$, then $N=n\si a$
and so
\beq
<\Th^2>=n\si a<\th'^2>.
\eeq
For our particular cross-section \eq{97}, $\si=\pi(2qze/Pv)^2/\th^2_{min}$, 
so, using also \eq{99},
\beq
<\Th^2>=2\pi n\lep\frac{2qze}{Pv}\rip^2a\ln(\th_{max}/\th_{min}).
\eeq

\subsubsection{The Distribution of Large Angle Scattering}
This distribution may be contrasted with the one that arises for particles
which undergo a single large-angle scattering and many small-angle ones. If
the net effect of the latter is less than the deflection produced by the
former, which in some sense defines what we mean by a large-angle
scattering, then we need only consider the distribution produced by a
single large-angle event. The number of such events is proportional to the
cross-section or, for\footnote{Evidently, ``large-angle'' means an angle
large compared to $\th_{min}$; it does not mean an angle so large as to be
of order one.} $\th<<1$,
\beq
d\si=\der\si\Om d\Om=\lep\frac{2qze}{pv}\rip^2\frac1{\th^4}d\ph\,\th d\th.
\eeq
We may convert $\th$ to $\th'$ using $\th=\th'/\sin\ph$,
\beq
d\si=\lep\frac{2qze}{pv}\rip^2\frac{d\th'}{\th'^3}\sin^2\ph\,d\ph.
\eeq
Now integrate $\ph$ from zero to $\pi$ to pick up all events corresponding
to $\th'>0$. The result is that\footnote{The original $d\si$ is a second-order
differential; the result of integrating over $\ph$, unfortunately still called
$d\si$, is a first-order differential.}
\beq
d\si=\frac\pi2\lep\frac{2qze}{pv}\rip^2\frac{d\th'}{\th'^3}.
\eeq
For a slab of thickness $a$ with a density $n$ of scatterers, the
probability of having a single large-angle scattering in an interval $d\th'$
around $\th'$ is
\beq
P_s(\th')d\th'=nad\si=\frac\pi2na\lep\frac{2qze}{pv}
\rip^2\frac{d\th'}{\th'^3}.
\eeq
Because this distribution falls off only as $\th'^{-3}$ while the
multiple-scattering distribution falls off exponentially as $\th'^2$, there is
some angle $\th_0$ such that the single-scattering distribution is larger than
the multiple scattering one for $\th'>\th_0$ and conversely.

Roughly speaking, the total distribution of scattered particles as a function
of $\th'$ is just $P_m$ for $\th'<\th_0$ and $P_s$ for $\th'>\th_0$. In any
given system, one can easily compute the two distributions along with $\th_0$.
It is expected that the description will work quite well for $\th'$
significantly smaller than $\th_0$ and also for $\th'$ significantly larger.
For $\th'\approx\th_0$, the actual behavior is complicated considerably by the
contribution of particles that have undergone several scatterings through
``almost-large'' angles. There are not enough such scattering events per
particle for them to be properly treated using statistical methods, and they
are not easily treated in any other way, except for numerical simulations.

\edo

