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\subsubsection*{Conservation of Angular Momentum}

Total angular momentum
\[ \bf{L} = \sum_i \bf{x}_i \times \bf{p}_i \]
\begin{align*}
    \implies \dfrac{\bf{L}}{t}
    &= \sum_i \bf{x}_i \times \dot{\bf{p}}_i \\
    &= \sum_i \bf{x}_i \times \left(
    \bf{F}_i^{\text{ext}} + \sum_{j \neq i}
    \bf{F}_{ij} \right) \\
    &= \bf{\tau} + \sum_i \sum_{j \neq i} \bf{x}_i
    \times \bf{F}_{ij}
\end{align*}
where
\[ \bf{\tau} := \sum_i \bf{x}_i \times
\bf{F}_i^{\text{ext}} \]
is the \emph{total external torque}. \\
We write the second term as
\begin{align*}
    \sum_i \sum_{j \neq i} \bf{x}_i \times
    \bf{F}_{ij}
    &= \sum_{i < j} \bf{x}_i \times \bf{F}_{ij} +
    \sum_{j < i} \bf{x}_i \times \bf{F}_{ij} \\
    &= \sum_{i < j} \bf{x}_i \times \bf{F}_{ij} +
    \sum_{j < i} \bf{x}_i \times (-\bf{F}_{ji}) \\
    &= \sum_{i < j} \bf{x}_i \times \bf{F}_{ij} +
    \sum_{i < j} \bf{x}_j \times (-\bf{F}_{ij}) \\
    &= \sum_{i < j} (\bf{x}_i - \bf{x}_j) \times
    \bf{F}_{ij}
\end{align*}
which is 0 if $\bf{F}_{ij}$ is parallel to
$(\bf{x} - \bf{x}_j)$, for example in gravity and
electromagnetism. \myskip
This is a stronger version of Newton's third law:
\begin{enumerate}[N1]
    \item[N3'] $\bf{F}_{ij} = -\bf{F}_{ji}$ and is
        parallel to $(\bf{x}_i - \bf{x}_j)$.
\end{enumerate}
If N3' holds, then $\dfrac{\bf{L}}{t} =
\bf{\tau}$.

\subsubsection*{Conservation of Energy}

The kinetic energy
\[ T = \half \sum_i m_i \dot{\bf{x}}_i \cdot
\dot{\bf{x}}_i \]
We write
\[ \bf{x}_i = \bf{R} + \bf{y}_i \]
where $\bf{R}$ is the position of the centre of
mass. Since
\[ \sum_i m_i \bf{x}_i = M \bf{R} \]
\[ \implies \sum_i m_i \bf{y}_i = 0 \]
Then
\begin{align*}
    T
    &= \half \sum_i m_i (\dot{\bf{R}} +
    \dot{\bf{y}}_i)^2 \\
    &= \half \sum_i m_i \dot{\bf{R}}^2 +
    \dot{\bf{R}} \cdot \cancel{\sum_i m_i
    \bf{y}_i} + \half \sum_i m_i \dot{\bf{y}}_i
    \cdot \dot{\bf{y}}_i \\
    &= \half \sum_i m_i \dot{\bf{R}}^2 + \half
    \sum_i m_i \dot{\bf{y}}_i \cdot \dot{\bf{y}}
    \\
    &= \half M \dot{\bf{R}}^2 + \half \sum_i m_i
    \dot{\bf{y}}^2
\end{align*}
where $\half M \dot{\bf{R}}^2$ is the energy of
the centre of mass, and $\half \sum_i m_i
\dot{\bf{y}}_i^2$ is the energy due to motion
around the centre of mass. \\
We repeat the analysis of work done in 3D (section
2.2). The $i$-th particle moves on a trajectory
$C_i$. Then
\[ T(t_2) - T(t_1) = \sum_i \int_{C_i}
\bf{F}_i^{\text{ext}} \cdot \dd \bf{x}_i + \sum_i
\sum_{i \neq j} \int_{C_i} \bf{F}_{ij} \cdot \dd
\bf{x}_i \]
We can define a potential energy if
\begin{itemize}
    \item $\bf{F}_i^{\text{ext}} = -\nabla V_i
        (\bf{x}_i)$ (summation convention is
        \emph{not} used here)
    \item $\bf{F}_{ij} = -\nabla V_{ij} (|\bf{x}_i
        - \bf{x}_j|)$ (summation convention is
        \emph{not} used here)
\end{itemize}
Remember $\nabla = \left( \pfrac{}{x},
\pfrac{}{y}, \pfrac{}{z} \right)$. The energy
\[ E = T + \sum_i V_i(\bf{x}_i) + \sum_i \sum_{j
\neq i} V_{ij}(|x_i - x_j|) \]
is conserved.

\subsubsection*{The 2 Body Problem as a 1 Body Problem}

Consider 2 particles with $\bf{F}_i^{\text{ext}} =
0$. The centre of mass $M \bf{R} = m_1 \bf{x}_1 +
m_2 \bf{x}_2$ and the relative separation is
\[ \bf{r} = \bf{x}_1 - \bf{x}_2 \]
\[ \bf{x}_1 = \bf{R} + \frac{m_2}{M} \bf{r} \]
\[ \bf{x}_2 = \bf{R} - \frac{m_1}{M} \bf{r} .\]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/2ecb94b08f3e11ec.png}
\end{center}
Since $\bf{F}_i^{\text{ext}} = 0$, $\ddot{\bf{R}}
= 0$ (see last lecture). Meanwhile
\begin{align*}
    \ddot{\bf{r}}
    &= \ddot{\bf{x}_1} - \ddot{\bf{x}_2} \\
    &= \frac{1}{m_1} \bf{F}_{12} - \frac{1}{m_2}
    \bf{F}_{21} \\
    &= \frac{m_1 + m_2}{m_1m_2} \bf{F}_{12} \\
    \implies \mu \ddot{\bf{r}}
    &= \bf{F}_{12}(r)
\end{align*}
where $\mu$ is the \emph{reduced mass}
\[ \mu := \frac{m_1m_2}{m_1 + m_2} \]
This is the same as the equation for one particle
of mass $\mu$ and position $\bf{r}$.

\subsubsection*{Collisions}

\emph{elastic collisions} are ones in which both
kinetic energy and momentum are conserved - these
come from conservative inter-particle forces
between the two colliding particles. Take
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/4ac7e4b08f3f11ec.png}
\end{center}
conservation of energy
\[ \half m \bf{v}^2 = \half m \bf{v}_1^2 + \half m
\bf{v}_2^2 \]
conservation of momentum
\[ m \bf{v} = m \bf{v}_1 + m \bf{v}_2 \]
hence
\[ \bf{v}^2 = \bf{v}_1^2 + \bf{v}_2^2 + 2\bf{v}_1
\cdot \bf{v}_2 \]
Then combining with conservation of energy gives
\[ \bf{v}_1 \cdot \bf{v}_2 = 0 .\]
i.e. one of the final state particles is
stationary or they travel at right angles.

\begin{note*}
    We have some information about the final state
    from conservation laws but not a unique
    outcome. \\
    Not surprising: $\bf{v}_1, \bf{v}_2$ have 6
    parameters, but 4 constraint equations: 1
    energy, 3 momentum.
\end{note*}

\subsubsection*{Impulse}

When particles are subject to short, sharp shocks
(for example in collisions) one talks of impulse
rather than force. If a force acts for a short
time $\Delta t$, the \emph{impulse} experienced by
the particle is
\[ \bf{I} = \int_t^{t + \Delta t} \bf{F} \dd t
\stackrel{(N2)}{=} \Delta \bf{p} \]

\subsubsection*{Bouncing Balls}

2 equations, 2 unknowns: after collision, small
ball has speed $v$ and big ball has speed $V$.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/60693a088f3f11ec.png}
\end{center}
Best to think of balls as slightly separated.
Conservation of $E$:
\[ mu^2 + Mu^2 = mv^2 + MV^2 \]
Conservation of $p$:
\[ Mu - mu = mv + MV \]
One can check David Tong's note for the full
solution, but one can solve to get
\[ \boxed{v = \frac{3M - m}{M + m} u} \]