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% 09/03/2023 10AM

\subsection{Rotating Fluid Dynamics}

Introduction: We live on a rotating Earth ($\Omega
= \frac{2\pi}{86400\mathsf{s}} \simeq 10^{-4}
\mathsf{s^{-1}}$) and the rotation has a huge
effect on the large-scale motion of the oceans and
atmosphere, hence climate. Moreover, the ocean and
atmosphere are thin ($\simeq 4\mathsf{km}$,
$\simeq 10\mathsf{km}$) compared to the Earth's
radius ($R \simeq 6000 \mathsf{km}$). Rotation of
the Earth $\implies$ speed $R\Omega$ at the
equation $\simeq 500\mathsf{ms^{-1}}$ $\gg$ winds
relative to Earth. So sensible to work in a
rotating frame.

\subsubsection{Euler Equations in a Rotating Frame}

Assume $\rho$ uniform and $\nabla \cdot \bf{u} =
0$ -- good for ocean, can make so for atmosphere
by a fix. Acceleration of a fluid particle in a
rotating frame is
\[ \Dfrac{\bf{u}}{t} + 2\bf{\Omega} \times \bf{u}
+ \bf{\Omega} \times (\bf{\Omega} \times \bf{x}) \]
($2\bf{\Omega} \times \bf{u}$ is Coriolis force
and $\bf{\Omega} \times (\bf{\Omega} \times
\bf{x})$ is centrifugal force). Now
$\rho\bf{\Omega} \times (\bf{\Omega} \times
\bf{x}) = \nabla \left( \half \rho |\bf{\Omega}
\times \bf{x}|^2 \right)$ so can combine with
$\rho\bf{g} = -\nabla \chi$ and treat them as an
``effective'' $\bf{g}$ with an effective
definition ``vertical'', which is normal to the
surface of the spheroidal Earth (Or note
$\frac{|\bf{\Omega} \times (\bf{\Omega} \times
\bf{x})|}{g} \simeq \frac{1}{300} \ll 1$). Hence
\[ \boxed{\rho \Dfrac{y}{t} + 2\rho \bf{\Omega}
\times \bf{u} = -\nabla p + \rho \bf{g}} \]
\[ \nabla \times (2\bf{\Omega} \times \bf{u}) =
2\bf{\Omega} \cancel{\nabla \cdot \bf{u}} -
2\bf{\Omega} \cdot \nabla \bf{u} \]
Hence the voriticty equation becomes
\[ \Dfrac{\bf{\omega}}{t} = (\bf{\omega} +
2\bf{\Omega}) \cdot \nabla \bf{u} \]
The vortex stretching term now involves the
\emph{total / absolute vorticity}
\[ \nabla \times (\bf{u} + \bf{\Omega} \times
\bf{x}) = \bf{w} + 2\bf{\Omega} \]
($\bf{w}$ is relative vorticity and $2\bf{\Omega}$
is planetary voriticity).

\subsubsection*{The Rossby Number}

Suppose we have a flow with a characteristic
lengthscale $L$ and velocity scale $U$ and
timescale $T = \frac{L}{U}$:
\begin{align*}
    \rho (\bf{u} \cdot \nabla) \bf{u}
    &\sim \frac{\rho U^2}{L} \\
    2\rho \bf{\Omega} \times \bf{u}
    &\sim \rho \Omega U
\end{align*}
The ratio
\[ \frac{\rho U^2 / L}{\rho U \Omega} =
\boxed{\frac{U}{\Omega L} \equiv R_0} \]
is the \emph{Rossby number} describes the relative
importance of inertia and Coriolis effects.
\begin{itemize}
    \item For example Weather system $L \sim 1000
        \mathsf{km}$, $U \sim 10\mathsf{ms^{-1}}$,
        $T\ sim 10^5 \mathsf{s} = 1 \mathsf{day}$.
        \[ \implies R_0 \sim 10^{-1} \]
        so Coriolis terms dominate.
    \item Another example would be emptying
        bathtub. Near the plughole $L \sim 30
        \mathsf{cm}$, $U \sim 3
        \mathsf{cms^{-1}}$, $T \sim 10\mathsf{s}$
        \[ \implies R_0 \sim 10^3 \]
        so Coriolis effects are a tiny
        perturbation. $R_0 \ll 1$ also implies
        $\bf{\omega} \lesssim \frac{U}{L} \ll
        \bf{\Omega}$, so relative vorticity $\ll$
        planetary.
\end{itemize}

\subsubsection*{Rotating Flow in a Shallow Layer}

\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/8e32d65ebe6611ed.png}
\end{center}
Consider flows with horizontal lengthscale $L$
such that $H \ll L \ll R$. For example
$100\mathsf{km}$ to $1000 \mathsf{km}$. Take local
cartesian coordinates $x$ eastwards, $y$
northwards and $z$ upwards, and $\bf{u} = (u, v,
w)$. If $\ol{\theta}$ is the latitude ($\subset
[-\pi/2, \pi/2]$) then $\bf{\Omega} = (0, \cos
\ol{\theta} \Omega, \sin\ol{\theta} \Omega)$ and
$2 \bf{\Omega} \times \bf{u} = 2\Omega
(\cos\ol{\theta} w - \sin\ol{\theta} v, \sin
\ol{\theta} u, -\cos\ol{\theta} u)$. We make 3
good approximations:
\begin{enumerate}[(1)]
    \item For $|\bf{u}| \sim 10\mathsf{ms^{-1}}$,
        $\frac{\Omega u}{g} \sim \frac{10^{-4}
        10}{10} = 10^{-4} \ll 1$ so we can neglect
        $-\cos\ol{\theta} u$ in the vertical
        equation.
    \item For $L \gg H$,
        \[ \pfrac{w}{z} = -\pfrac{u}{x} -
        \pfrac{v}{y} \]
        suggests $\frac{w}{H} \sim \frac{u, v}{L}
        \implies w \sim \frac{H}{L} (u, v) \ll u,
        v$ so the flow is dominantly horizontal.
        (cf velocities in shallow water waves).
        Hence:
        \begin{enumerate}[(i)]
            \item Neglect $\cos\ol{\theta} w$ in
                the $x$-direction  (except near
                the equator)
            \item Neglect the vertical
                acceleration $\rho \Dfrac{w}{t}$
                so the vertical force balance is
                hydrostatic.
        \end{enumerate}
\end{enumerate}
Hence
\begin{equation*}
    \boxed{
    \begin{aligned}
        \Dfrac{u}{t} - fv
        &= -\frac{1}{\rho} \pfrac{p}{x} &&\text{(1)} \\
        \Dfrac{v}{t} + fu
        &= -\frac{1}{\rho} \Dfrac{p}{y} &&\text{(2)} \\
        0
        &= -\frac{1}{\rho} \pfrac{p}{z} - g \\
        \pfrac{u}{x} + \pfrac{v}{y} + \pfrac{w}{z}
        &= 0
    \end{aligned}
    }
\end{equation*}
where $f = 2\Omega \sin\ol{\theta}$ is the
\emph{Coriolis parameter} (only the vertical
component of $\bf{\Omega}$ matters). Also,
$\pfrac{w}{z} \sim \frac{w}{H} \gg \pfrac{w}{x},
\pfrac{w}{x} \sim \frac{w}{L}$ so the vertical
component of the vorticity equation becomes
\begin{equation*}
    \boxed{
    \begin{aligned}
        \Dfrac{\zeta}{t}
        &= (\zeta + f) \pfrac{w}{z} \\
        &= -(\zeta + f) \left( \pfrac{u}{x} +
        \pfrac{v}{y} \right)
    \end{aligned}
    }
\end{equation*}
where $\boxed{\zeta = \omega_2}$. Consider a flow
in a shallow layer of thickness $h(x, y, t)$:
\begin{center}
    \includegraphics[width=0.6\linewidth]
    {images/1f247368be6a11ed.png}
\end{center} 
\begin{align*}
    \pfrac{p}{z} = \rho g
    &\implies p = p_0 + \rho g(h - z) \\
    &\implies \nabla_H p = \rho g \nabla_H h
\end{align*}
is independent of $z$ ($\nabla_H \equiv \left(
\pfrac{}{x}, \pfrac{}{y} \right)$)