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\subsection{Coriolis Force}

The Coriolis force is
\[ \bf{F}_{\text{cor}} = -2m \bf{\omega} \times
\bf{v} \]
where
\[ \bf{v} = \left( \dfrac{\bf{r}}{t} \right)_{S'} \]
($\bf{v} = 0$ for the water in the bucket). Let's
ignore the centrifugal force for now.
\begin{itemize}
    \item It's velocity dependent and independent
        of position.
    \item It's mathematically related to the
        Lorentz force for a constant magnetic
        field $\implies$ particles move in
        circles.
\end{itemize}
$\bf{\omega}$ is out of the page.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/709008b49a2811ec.png}
\end{center}
$\implies$ a free particle moves in a clockwise
direction.
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/9f212f789a2811ec.png}
\end{center}
This is a similar effect that leads to hurricanes.
Suppose there's a low pressure region. Particles
in the fluid move radially towards the low
pressure and are deflected by the Coriolis force:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/056755289a2911ec.png}
\end{center}
end result is that swirls in an anti-clockwise
direction (opposite to a free particle).
Hurricanes in the Northern hemisphere rotate in an
anticlockwise direction. (Southern hemisphere -
clockwise).

\begin{note*}
    Our discussion assumed that particles move in
    a plane perpendicular to $\bf{\omega}$. This
    isn't true on the surface of the earth.
\end{note*}

For a particle travelling due North, the Coriolis
force points East with size
\[ |\bf{F}_{\text{cor}}| = 2mv \omega \sin\theta \]
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/6ad17efc9a2911ec.png}
\end{center}
For a particle travelling East, it's a little
trickier: $\bf{F}_{\text{cir}}$ has a component
pointing upward. Looking at the component tangent
to Earth,
\[ |\bf{F}_{\text{cor}} \cdot \hat{\bf{\theta}}| =
2mv \omega \sin\theta \]
again.

\begin{note*}
    $\theta = 0 \implies \bf{F}_{\text{cor}}$ and
    indeed, there are no hurricanes within about
    500 miles of the equator.
\end{note*}

\begin{example*}[Balls and Tower]
    Climb up a tower of height $h$ and drop a
    ball. \\
    \ul{Question}: Does it land in front or behind
    the tower? \\
    \ul{Answer}: In front - a bit counter
    intuitive.
    \[ \ddot{\bf{r}} = \bf{g} - \bf{\omega} \times
    (\bf{\omega} \times \bf{r}) - 2\bf{\omega}
    \times \dot{\bf{r}} \approx \bf{g} - 2
    \bf{\omega} \times \dot{\bf{r}} \tag{$*$} \]
    (we neglect $\bf{\omega} \times (\bf{\omega}
    \times \bf{r})$ because it is $O(\omega^2) +
    \text{very small}$, so it's fine after
    $\dot{\bf{r}}$ is above a certain value).
    Integrating once:
    \[ \implies \dot{\bf{r}} = \bf{g}t -
    2\bf{\omega} \times (\bf{r} - \bf{r}_0) .\]
    ($\bf{r}_0$ is an integration constant - the
    initial position). Substitute into ($*$) to
    get:
    \[ \ddot{\bf{r}} = \bf{g} - 2\bf{\omega}
    \times \bf{g} t \]
    Now integrate twice:
    \[ \implies \bf{r} = \bf{r}_0 + \half gt^2 -
    \frac{1}{3} \bf{\omega} \times \bf{g} t^3 \]
    We'll look at this effect on the equator
    \begin{center}
        \includegraphics[width=0.6\linewidth]
        {images/69dafedc9a2a11ec.png}
    \end{center}
    \begin{align*}
        \bf{g}
        &= -g \bf{e}_3 \\
        \bf{\omega}
        &= \omega \bf{e}_1 \\
        \bf{r}_0
        &= (R + h) \bf{e}_3 \\
        \implies \bf{r}
        &= \left( R + h - \half gt^2 \right)
        \bf{e}_3 - \frac{1}{3} \omega gt^3
        \bf{e}_2
    \end{align*}
    $\bf{e}_2$ is Westerley but displacement is in
    the Easterley direction. Particle hits ground
    when $t^2 = \frac{2h}{g}$, as usual.
    \[ \implies \text{displacement } = \frac{1}{3}
    \omega g \left( \frac{2h}{g} \right)^{3/2}
    \text{ East} \]
    (ball drops in \emph{front} of tower).

    \begin{note*}
        This is really conservation of angular
        momentum. When dropped, $l = (R + h)^2
        \omega$ so the speed close to the ground
        is $v$:
        \[ Rv = (R + h)^2 \omega \implies v =
        \frac{(R + h)^2}{R} \omega > R \omega =
        v_{\text{earth}} \]
    \end{note*}
\end{example*}

\subsubsection*{Brief Discussion of Foucault's Pendulum}

A pendulum at the North pole stays aligned with
its inertial force while the Earth rotates
underneath. An observer on the earth sees the
swing process throughout the day (see David Tong's
notes for algebra)
\[ \zeta := x + iy \qquad \zeta = e^{-i \omega t
\sin\theta} \left( a \cos \sqrt{\frac{g}{l}} t + B
\sin \sqrt{\frac{g}{l}} t \right) \]
Pendulum rotates with period $\frac{1 \text{
day}}{\sin\theta}$.

\subsubsection*{Larmor Precession}

Consider a charged particle orbiting around a
second, fixed particle, under the influence of the
Coulomb force. Add this to a constant magnetic
field $\bf{B}$. Equation of motion:
\[ m\ddot{\bf{r}} = -\frac{k}{r^2} + q
\dot{\bf{r}} \times \bf{B}; \qquad k =
\frac{qQ}{4\pi \eps_0} \]
In a rotating frame ($\dot{\bf{\omega}} = 0$):
\begin{align*}
    m \left( \dfrac[2]{\bf{r}}{t} \right)_{S'}
    &= -\frac{k}{r^2} \hat{\bf{r}} + q
    \dot{\bf{r}} \times \bf{B} - 2m \bf{\omega}
    \left( \dfrac{\bf{r}}{t} \right)_{S'} -
    m(\bf{\omega} \times (\bf{\omega} \times
    \bf{r})) \\
    &= -2m \bf{\omega} \left( \dfrac{\bf{r}}{t}
    \right)_{S'} - m (\bf{\omega} \times
    (\bf{\omega} \times \bf{r})) - \frac{k}{r^2}
    \hat{\bf{r}} + q \left[ \left(
    \dfrac{\bf{r}}{t} \right)_{S'} + \bf{\omega}
    \times \bf{r} \right] \times \bf{B}
\end{align*}
Pick $\bf{\omega}$ to cancel $\left(
\dfrac{\bf{r}}{t} \right)_{S'}$ terms, i.e.
$\bf{\omega} = -\frac{q \bf{B}}{2m}$
\[ \implies m \ddot{\bf{r}} = -\frac{k}{r^2}
\hat{\bf{r}} + \frac{q^2}{4m} \bf{B} \times
(\bf{B} \times \bf{r}) \]
(when $B^2 \ll \frac{4mk}{q^2r^2}$ we can neglect
the last term). We go back to solutions with
elliptic motion. Transform back out of rotating
frame to show that the ellipses precess with
angular speed $\omega = \frac{qB}{2m}$, the Larmor
frequency ($\half$ the cyclotron frequency).