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\subsubsection*{Scattering}

Consider a potential $V(r)$ such that $V(r) \to 0$
as $r \to \infty$. For a repulsive potential, the
trajectory will look something like
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/7f5533808b8f11ec.png}
\end{center}
Conservation of energy implies
\[ V_{\text{initial}} = v_{\text{final}} := v \]
In a central potential, we have conservation of
angular momentum, i.e.
\[ l = |\bf{x} \times \dot{\bf{x}}|
=\frac{|\bf{L}|}{m} \]
is conserved.

\begin{claim*}
    \[ l_{\text{initial}} =bv_{\text{initial}} \]
\end{claim*}

\begin{proof}
    Suppose $V = 0$. The particle follows a
    straight horizontal line. At the closest point
    to the origin, its angular momentum per unit
    mass is $bv_{\text{initial}}$, but by
    conservation of $l$, this must be the same as
    $l_{\text{initial}} = bv_{\text{initial}}$.
\end{proof}

\myskip
In the presence of $V$ i.e. the potential, we have
\[ l_{\text{initial}} = l_{\text{final}} =
b'v_{\text{final}} \implies b = b' \]

\subsubsection*{Rutherford Scattering}

A repulsive Coulomb potential
\[ V = \frac{qQ}{4\pi \eps_0 r} \]
(replace $k = -\frac{qQ}{4\pi \eps_0 m}$)
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/5822d1a48b9011ec.png}
\end{center}
$\phi$ is the \emph{scattering angle}. In the
short term though, it's more useful to consider
$\alpha$. ($\phi = \pi - 2\alpha$)

\myskip
\textbf{Question}: How does $\phi$ depend on $v$
and $b$? \\
Recall from last lecture for a repulsive potential
\[ r = \frac{|r_0|}{e\cos\theta - 1} \to \infty \]
when $\theta = \alpha \implies \cos\alpha =
\frac{1}{e}$. ($e > 1 \implies \alpha <
\frac{\pi}{2}$) \\
Energy of particle (last lecture)
\begin{align*}
    E
    &= \frac{mk^2}{2l^2} (e^2 - 1) \\
    &= \half mv^2 \\
    &= \frac{mk^2}{2l^2} \tan^2 \alpha
\end{align*}
sub in $l = bv$:
\[ \implies \frac{1}{\tan^2 \alpha} =
\frac{k^2}{b^2v^4} \qquad \text{or} \qquad
\frac{1}{\tan\alpha} = \frac{|k|}{bv^2} \]

\begin{note*}
    \[ \tan \frac{\phi}{2} = \tan \left(
    \frac{\pi}{2} - \alpha \right) =
    \frac{1}{\tan\alpha} \]
\end{note*}

\[ \implies \boxed{\phi = 2\tan^{-1} \left(
\frac{|k|}{bv^2} \right)} \]
This is the scattering formula.

\newpage
\section{Systems of Particles}

So far we've only covered the motion of a single
particle. We now describe $N$ interacting
particles. Put a label $i = 1, \dots, N$ on
everything. The $i$-th particle has mass $m_i$,
position $\bf{x}_i$ and momentum $\bf{p}_i$. \\
For each particle,
\begin{enumerate}[N1]
    \item[N2] is $\dot{\bf{p}}_i = \bf{F}_i$,
        force on $i$-th particle. The force can be
        split into two parts:
        \[ \bf{F}_i = \bf{F}_i^{\text{ext}} +
        \sum_{j \neq i} \bf{F}_{ij} \]
        where $\bf{F}_i^{\text{ext}}$ is an
        external force on the $i$-th particle, and
        $\bf{F}_{ij}$ is the force on the $i$-th
        particle due to the $j$-th particle.
    \item[N3] is $\bf{F}_{ij} = -\bf{F}_{ji}$.
        This holds for gravitational and Coulomb
        forces.
\end{enumerate}

\subsection{Centre of Mass Motion}

The total mass of the system $M = \sum_i m_i$. The
\emph{centre of mass} is defined as
\[ \bf{R} := \frac{1}{M} \sum_{i = 1}^N m_i
\bf{x}_i \]
The total momentum of the system is captured by
the centre of mass
\[ \ul{\bf{P}} := \sum_{i = 1}^N \bf{p}_i = M \dot{\bf{R}} \]
\begin{align*}
    \ul{\dot{\bf{P}}}
    &= \sum_i \dot{\bf{p}}_i \\
    &= \sum_i \left( \bf{F}_i^{\text{ext}} +
    \sum_{j \neq i} \bf{F}_{ij} \right) \\
    &= \sum_i \bf{F}_i^{\text{ext}} + \sum_{i < j}
    (\bf{F}_{ij} + \bf{F}_{ji}) \\
    &= \sum_i \bf{F}_i^{\text{ext}}
\end{align*}
This is important: the motion of the centre of
mass is only due to the external forces.

\subsubsection*{Conservation of Momentum}

If $\sum_i \bf{F}_i^{\text{ext}} = 0$,
$\ul{\bf{P}}$ is conserved (i.e.
$\ul{\dot{\bf{P}}} = 0$).