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Momentum equation applied to
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\[ \ub{\cancel{\dfrac{}{t} \int_V \rho \bf{u} \dd
V}}_{0 \text{ because steady}} = -\int_{\partial
V} \rho \bf{u} (\bf{u} \cdot \bf{n}) + p \bf{n}
\dd A + \ub{\cancel{\int_V \bf{f} \dd V}}_{
\text{neglect}} \]
$p = p_a$ except near the impact, $\bf{u} \cdot
\bf{n} = 0$ except at the ends.

\myskip
Component parallel to the diagonal plane imply
\[ -\rho U^2 a \cos\beta + pU^a a_2 - pU^2 a_1
\tag{2} \]
(1) and (2) implies $a_2 = a \frac{1 +
\cos\beta}{2}$, $a_1 = a \frac{1 - \cos\beta}{2}$.

\myskip
Component perpendicular to plane
\[ \half \rho U^2 a \sin \beta = \int_{\partial V}
p \bf{n} \dd A = \int_{\partial V} (p - p_a)
\bf{n} \dd A = \text{Force on wall} \]
(because $p_a \int_{\partial V} \bf{n} \dd A = 0$
by divergence theorem).
Also: Aerofoil lift
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so sucked up.

\myskip
Barges in canals:
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\subsection{Hydrostatic Pressure and Archimedes Principle}

If $\bf{u} \equiv \bf{0}$ then Euler gives $0 =
-\nabla p + \rho \bf{g} = -\nabla (p + \chi)$
\[ \implies p + \chi = \text{const} \implies p =
p_0 - \rho gz \]
\emph{hydrostatic pressure}. (here gravity is in
the $z$ direction).

\myskip
Archimedes: Pressure force on a submerged body
with $\bf{u} = \bf{0}$
\begin{align*}
    \text{Force}
    &= -\int p \bf{n} \dd A
    &&(\text{fluid on body}) \\
    &= -\int (p_0 - \rho gz) \bf{n} \dd A \\
    &= -\int \nabla (p_0 - \rho gz) \dd V \\
    &= \rho_{\text{fluid}} \bf{g} \hat{\bf{z}}
    \\
    \text{upthrust / buoyancy}
    &= \text{weight of fluid displaced}
\end{align*}
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For $\bf{u} \neq \bf{0}$ and constant $\rho$
everywhere we can write $p = p_0 - \rho gz + p'$
\[ \implies \rho \Dfrac{\bf{u}}{t} = -\nabla p' \]
(no $\bf{f} = \rho \bf{g}$). $p'$ is the
\emph{dynamic} (\emph{modified}) pressure --
variation due to the motion after subtracting
hydrostatic balance. Can ignore gravity if there
are no density variations and no free surfaces.
Free surface between air and water,
$\rho_{\text{water}} \neq \rho_{\text{air}}$ hence
gravity clearly has an effect $\to$ waves (Section
5).

\subsection{Vorticity}

Vorticity defined by $\bf{\omega} = \nabla \times
\bf{u}$ (angular momentum by another name).

\begin{example*}
    $\bf{u} = \Omega \times \bf{x}$, solid body
    rotation
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    $\implies \bf{\omega} = 2\bf{\Omega}$ (check)
\end{example*}

\begin{example*}
    $\bf{u} = (0, \gamma x, 0)$, simple shear
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    $\implies \bf{\omega} = (0, 0, \gamma)$
\end{example*}

\begin{example*}
    $\bf{u} = \left( 0, \frac{R}{2\pi r}, 0
    \right)$ in cylindrical polars, ``line
    vortex''
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    \end{center}
    $\bf{\omega} = \bf{0}$ except at $r = 0$
    \[ \int_{r < a} \bf{\omega} \cdot \hat{\bf{z}}
    \dd A = \oint_{r = a} \bf{u} \cdot \dd l =
    \int_0^{2\pi} \frac{R}{2\pi r} r \dd \theta =
    R \]
    $\implies \bf{\omega} = (0, 0, R \delta(r))$.
\end{example*}

\subsubsection{Interpretation of $\omega$ as 2
$\times$ average location rate}

Consider a material line element $\delta l$ (i.e.
one moving with the fluid).
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i.e. $\bf{\delta} \bf{l} \to \bf{\delta}\bf{l} +
(\bf{\delta}\bf{l} \cdot \nabla) \bf{u} \delta t$
\[ \implies \boxed{\dfrac{}{t} \bf{\delta}\bf{l} =
(\bf{\delta}\bf{l} \cdot \nabla) \bf{u}} \]
Hence the tensor $\pfrac{u_i}{x_j}$ determines the
rate of change of $\bf{\delta}\bf{l}$. We write
\begin{align*}
    \pfrac{u_i}{x_j}
    &= \half \left( \pfrac{u_i}{x_j} +
    \pfrac{u_j}{x_i} \right) + \half \left(
    \pfrac{u_i}{x_j} - \pfrac{u_j}{x_i} \right) \\
    &= e_{ij} + \half \eps_{ijk} \omega_k
\end{align*}
The local rotation of line elements due to the
second term is $\half \eps_{jik} \omega_k \delta
k_j = \half (\bf{\omega} \times
\bf{\delta}\bf{l})_i = $ rotation at angular
velocity $\half \bf{\omega}$.

\myskip
Local rotation due to first term, called the
\emph{strain rate}, gives same angular velocity
averaged over all orientations
$\bf{\delta}\bf{l}$. $\ul{\bf{e}}$ is symmetric
(hence diagonalisable) and traceless. Therefore
$\nabla \cdot \bf{u} = 0$.
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Note $\half \bf{\omega}$ gives the rate of
rotation of blobs, not whether the blobs are going
in circles!