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\subsection{Newton's Second Law}

The \emph{equation of motion} for a particle
subjected to a force $\bf{F}$ is
\[ \dfrac{}{t} (m \dot{\bf{x}}) = \bf{F}(\bf{x},
\dot{\bf{x}}) \]
$\bf{p} = m \dot{\bf{x}}$ is \emph{momentum}, and
the force $\bf{F}$ can depend on position and
velocity. $m$ is \emph{inertial mass}. It is a
measure of the reluctance of a particle to move.
\\
When $\frac{m}{t} = 0$ (true in most situations),
we have
\[ \boxed{m \ddot{\bf{x}} = \bf{F}(\bf{x},
\dot{\bf{x}})} \]
The equation of motion is a second order
differential equation $\implies$ we need to
specify two initial conditions for each degree of
freedom. \\
For example, $\bf{x} \in \RR^3 \implies$ 3 degrees
of freedom i.e. 6 initial conditions are needed.

\bigskip

\noindent
There are two steps in any Newtonian mechanics
problem:
\begin{itemize}
    \item Write down the equation(s).
    \item Solve it.
\end{itemize}

\newpage
\section{Forces}

\subsection{Potentials in One Dimension}

Consider a particle moving in a line with position
$x(t)$. Suppose that $F = F(x)$, i.e. it depends
on position, not on velocity. We define a
\emph{potential energy} $V(x)$ by
\[ F = - \dfrac{V}{x} \qquad \text{or} \qquad V(x)
= - \int_{x_0}^x \dd x' F(x') \]
\begin{note*}
    The $x'$ does note derivative here; $x'$ is a
    dummy variable.
\end{note*}
The equation of motion is
\[ m \ddot{x} = - \dfrac{V}{x} \tag{$*$} \]

\begin{claim*}
    The energy $E = \half m\dot{x}^2 + V(x)$ is
    conserved (i.e. $\dot{E} = 0$) for any
    trajectory which obeys ($*$).
\end{claim*}

\begin{proof}
    \begin{align*}
        \dfrac{E}{t}
        &= m \dot{x} \ddot{x} + \dfrac{V}{x}
        \dot{x} \\
        &= \dot{x} \left( m \ddot{x} +
        \dfrac{V}{x} \right) \\
        &= 0
        &&\text{by ($*$)}
    \end{align*}
\end{proof}

\begin{note*}
    If $F = F(x, \dot{x})$, there is no conserved
    quantity.
\end{note*}

\begin{example*}[Harmonic Oscillator]
    \[ V = \half kx^2 \]
    Then ($*$) becomes $m\ddot{x} = -kx$. The
    general solution is
    \[ x(t) = A \cos(\omega t) + B \sin(\omega t) \]
    where $\omega := \sqrt{\frac{k}{m}}$ is the
    \emph{angular frequency}. $A$ and $B$ are
    integration constants. It's simple to show
    that
    \[ E = \half m\dot{x} + \half kx^2 \]
    is constant. The time take to complete a cycle
    is the \emph{period} $T = \frac{2
    \pi}{\omega}$.
\end{example*}

\noindent
For a general potential $V(x)$, the conserved
quantity allows use to `solve' any one-dimensional
problem.
\begin{align*}
    E
    &= \half m\dot{x}^2 + V(x) \\
    \implies \dfrac{x}{t}
    &= \pm \sqrt{\frac{2}{m}(E - V(x))} \\
    \implies t - t_0
    &= \pm \int_{x_0}^x \frac{\dd
    x'}{\sqrt{\frac{2}{m}(E - V(x'))}}
\end{align*}
which is `the solution' - we `just' need to do the
integral.

\subsubsection*{Motion in a Potential}

Sometimes even if you can't do the integral, it's
simple to get a qualitative picture of the
solution.

\begin{example*}
    \[ V(x) = m(x^3 - 3x) .\]
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/69c408047b7011ec.png}
    \end{center}
    \iffalse
    \begin{center}
    \begin{tsqx}
        (-2,0)->>(2,0)
        (0,-2)->>(0,2)
        ! real f(real x) {return 0.5*(x**3 - 3*x);}
        ! draw(graph(f, -2, 2, operator, ..));
        label $x$ @ (2.1,0)
        label $V(x)$ @ (0,2.1)
    \end{tsqx}
    \end{center}
    \fi
    Drop particle at $x = x_0 \implies E =
    V(x_0)$.
    \begin{itemize}
        \item $x_0 = \pm 1 \implies$ particle
            stays there $\forall t$ ($*$ says
            there's no force).
        \item $x_0 \in (-1, 2) \implies$ particle
            oscillates back and forth in dip.
        \item $x_0 > 2$. Particles keeps on going
            to $-\infty$.
        \item $x_0 = 2$ is a special case. It
            reaches $x = -1$ but how long does it
            take? Write $x = -1 + \eps$ and as
            $\eps \to 0 \implies V(x) \bf{u} 2m -
            3m\eps^2$
            \[ t - t_0 = \int_{x_0}^x \frac{\dd
            x'}{\sqrt{\frac{2}{m} (E - V(x'))}}
            = -\int_{\eps_0}^\eps \frac{\dd
            \eps'}{\sqrt{6} \eps'} =
            -\frac{1}{\sqrt{6}} \ln \left(
            \frac{\eps}{\eps_0} \right) \]
            so $t \to \infty$ as $\eps \to 0$,
            i.e. it takes infinite time.
    \end{itemize}
    The initial energy is all potential. It turns
    this into kinetic energy and falls down the
    dip.
\end{example*}