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\begin{center}
\LARGE\bf Physics \course \\ Lecture \lect \\
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\section{Introduction}
In the last few lectures a method for calculating differential
and total cross sections has been developed. The method was
developed for the case of the electromagnetic interaction, but in fact
is more general, and can be applied to almost any interaction as long
as the coupling constant is small. The method is based several approximation:
\begin{enumerate}
  \item That the beam is prepared a large distance from the scattering
        center, so that a plane wave solution can be used;
  \item That the outgoing particles are measured a large distance from the
        scattering center, so that a plane wave solution can be used;
  \item The time variations of the potential are large compared to the
        interaction time;
  \item That only terms to first order in the charge are kept in the
        transition amplitude.
\end{enumerate}
Based on these approximations, the general form of the
differential cross section is given by:
\begin{equation}
  d\sigma = \frac{|\mathcal{M}|^2}{F}{dQ}
\end{equation}
where the flux factor $F$ is given by:
\begin{equation}
  F=|\vv_A-\vv_B|2E_A2E_B
\end{equation}
and the phase space factor $dQ$ is given by:
\begin{equation}
  dQ = (2\pi)^4 \delta^4(P_B+P_D-P_A-P_C)
       \frac{d^3p_C}{(2\pi)^3 2E_C}\frac{d^3p_D}{(2\pi)^3 2E_D}
\end{equation}
where this describes two outgoing particles (for the general, there is
one density of states factor per outgoing particle, and the delta function
describes energy momentum conservation).

In this lecture, the Rutherford cross section will be derived. This cross 
section describes the scattering of a very light particle (small compared 
to the energy) off a very heavy particle that is initially at rest. 
The derivation will not assume
this but will put it in as a last step, so that the general form of the
cross section can be examined.
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\section{The Rutherford Cross Section}
The calculation starts with writing down the Feynman diagram that describes 
the process (see fig. \ref{fig:fyn}). First write down the invariant
amplitude (matrix element):
\begin{equation}
  -i\mathcal{M}=[ie(P_A + P_B)^\mu] \left(-i \frac{g_{\mu\nu}}{q^2}\right)
             [ie(P_C+P_D)^\nu] \label{eq:M}
\end{equation}
where:
\begin{equation}
  q^2 = (P_B - P_A)^\mu(P_B - P_A)_\mu=(P_C - P_D)^\mu(P_C - P_D)_\mu
\end{equation}
Next comes the flux factor, it is given by:
\begin{equation}
  F=|v_A|2E_A2E_C=2 |\pv_A|2m_C \qquad\text{where}\qquad 
    |\vv|=\frac{|\pv_A|}{E_A}
\end{equation}
Finally the phase space factor is given by:
\begin{equation}
  dQ = (2\pi)^4\delta^4(P_B+P_D-P_A-P_C) \frac{d^3p_B}{(2\pi)^3E_B}
        \frac{d^3p_D}{(2\pi)^3E_D}
\end{equation}
\begin{figure}
  \begin{center}
    \epsfig{file=em_vert_2.eps}
    \caption{This is the Feynman diagram that is being calculated in this
             lecture note.\label{fig:fyn}}
  \end{center}
\end{figure}

With all the factors written out explicitly, the differential cross section
can be calculated. But before proceeding, it must be decided what 
quantity is to be measured. In this problem, particle $A$ is incident on
particle $C$, which is at rest. After the scatter particle $A$ becomes
particle $B$, while particle $C$ becomes particle $D$; note that $A$ and $B$
are the same particle, the label is used to distinguish before and
after, with the same holding for particles $C$ and $D$. Since particle
$D$ is assumed to be very heavy, it will not travel far so will most
likely be unmeasurable, therefore the only quantity that will be measurable
will be particle $B$. The cross section will then be given in terms of
the angular distribution of particle $B$ relative to the incident particle
$A$.

Given the quantity that is to be measured, the first step is to do the
integral over the 3-momenta of particle $D$. This integral impose momentum
conservation, through the delta-function; the condition imposed is
$\pv_D=\pv_A+\pv_B-\pv_C$. At this point in the calculation, the 
differential cross section is in the form:
\begin{equation}
  d\sigma = \frac{\delta(E_B+E_D-E_A-E_C) |\mathcal{M}_{if}|^2}
                 {64\pi^2 m_C^2 |\pv_A|E_B}|\pv_B|^2dp_Bd\Omega
\end{equation}
where the relation $d^3p_B=|\pv_B|^2dp_Bd\Omega$ has been used. Since the
delta-function involves the energy, and there is an energy ($E_B$) 
in the equation (note that the energy depends on the momentum), 
the integral should be converted from a momentum integral
to one over the energy. The change of variables can be achieved through the
relation:
\begin{equation}
  |\pv_B|^2 = E_B^2 - m_B^2 \quad\Rightarrow\quad
  2 |\pv_B| dp_B = 2E_B dE_B
\end{equation}
With this change of variables, the delta function can be integrated out.
This yields:
\begin{equation}
  d\sigma = \frac{|\mathcal{M}_{fi}|^2}
                 {64\pi^2m_C^2}\frac{|\pv_B|}{|\pv_A|}d\Omega
\end{equation}
notice that the invariant amplitude is just carried along, since the
it contains only 4-momenta, and the
integrals over the delta function impose energy-momentum conservation.

The final step is to simplify the matrix element. First of all simplify
$q^2$. Since the final result is to be written in terms of the kinematic
variables of particle $A$-$B$, $q^2$ is given by:
\begin{align}
  q^2 = &(P_B - P_A)^\mu(P_B - P_A)_\mu = 2m^2-2E_iE_f+2|\pv_i||\pv_f|
        \cos \theta \\\quad\Rightarrow\quad &4E_iE_f\sin^2\theta/2
        \qquad \text{for } E\gg m \notag
\end{align}
note at this point the subscripts are changed to indicate that 
particles $A$ and $B$ are the same, and define the initial and final
kinematic quantities. The numerator, which is the sum of 4-momenta,
as given in equation \ref{eq:M}, is simplified by using energy-momentum
conservation to remove particle $D$ the final state of particle $C$:
\begin{align}
  e^2(P_A + P_B)^\mu
             (P_C+P_D)_\mu &= e^2(P_A + P_B)^\mu 
             (2P_C+P_A-P_B)_\mu\\ &= e^2 (2P_A\cdot P_C+m^2 - P_A\cdot P_B+
             2P_B\cdot P_C+P_B\cdot P_A -m^2)\notag \\
             &= 2e^2 (E_i+E_f) m_C \notag
\end{align}
Finally, from these expressions, the
differential cross section is given by:
\begin{equation}
  \frac{d\sigma}{d\Omega}=\frac{\alpha^2}{4}\frac{|\pv_f|}{|\pv_i|}
       \frac{(E_i+E_f)^2}{(m^2-E_iE_f+|\pv_i||\pv_f|\cos\theta)^2}
\end{equation}
where the relation $\alpha = e^2/4\pi$ was used. In the limit where the
energy of particle $A$ is much larger than its mass, and the mass
of the target particle is extremely large, the differential
cross section is given by:
\begin{equation}
  \frac{d\sigma}{d\Omega}=\frac{\alpha^2}{4}\frac{1}{E^2\sin^4\theta/2}
\end{equation}
where the kinematics yields the relations $E_i=E_f$, and
$|\pv_i|=|\pv_f|$.
This equation is referred to as the Rutherford cross section.
\end{document}