\begin{equation}
\begin{aligned}
&W \leq \mathcal{O}(\delta U) \\
&W \leq \frac{D}{L}U\\
&WU \leq \frac{DU^2}{L}\\
&\frac{WU}{D} \leq \frac{U^2}{L}
\end{aligned}
\end{equation}
Resulting in the following:
\begin{equation}
\frac{UW}{D} = \mathcal{O}(\frac{U^2}{L})
\end{equation}
So the Pressure scale: \(P = \rho U (U,fL)_{max}\). This is so for the horizontal pressure gradient to enter as a forcing term in the horizontal momentum balance; otherwise the flow would be unaccelerated.
Rossby Number Condition
- Kelvin waves: they are affected by earth's rotation. There are equatorial kelvin waves and coastal kelvin waves.
- Rossby waves: affected by earth's rotation. There are equatorial Rossby waves and Mid-latitude Rossby waves (jet stream).
\[R_o = \frac{U}{fL}\]
The Rossby number can tell us what forces are dominant the inertial if \(U >> fL\) or Coriolis force (atmosphere forces) if \(U << fL\).
- Internal gravity waves: affected by stratification. and it happens anywhere.
Scaling p
As mentioned the Rossby number is the ratio of the relative to the coriolis accelerration. By the assumption that \(R_o << 1\). Meaning that the characteristic pressure field:\(P = \rho \frac{U^2}{R_o}\).
Proof:
\[
P = \rho U (U,fL)_{max} = \rho U f L = \rho \frac{U^2}{R_o}
\]
The Shallow-Water model condition: hydrostatic approximation
The z-component of the momentum equation:
\[\frac{\partial w}{\partial t}+u \frac{\partial w}{\partial x} + v \frac{\partial w}{\partial y} + w\frac{\partial w}{\partial z} = -\frac{1}{\rho} \frac{\partial \tilde{p}}{\partial z}\]
Now let us examine the following term (interms of the pressure gradient) from (5), (6), and (8):
\begin{equation}
\begin{aligned}
\rho \frac{\frac{\partial w}{\partial t}}{\frac{\partial \tilde{p}}{\partial z}} &= \mathcal{O}(\rho \frac{\frac{UW}{L}}{\frac{P}{D}})\\
&= \mathcal{O}(\rho \frac{\frac{UW}{L}}{\frac{\rho U^2}{R_o D}})\\
&= \mathcal{O}(\frac{R_o DW}{UL})\\
&= \mathcal{O}(\delta^2 R_o) \quad \text{$\delta = \frac{D}{L}= \frac{W}{U}$ which is the ratio of the vertical to horizontal}
\end{aligned}
\end{equation}
Hence we can conclude that that \(\frac{\partial \tilde{p}}{\partial z}\) is negligible to \(\mathcal{O}(\delta^2 R_o)\) or in terms of the total pressure:
\begin{equation}
\frac{\partial p}{\partial z} = -pg + \mathcal{O}(\delta^2 R_o)
\end{equation}
which is the hydrostatic approximation or the shallow-water model.
Horizontal Momentum Equations
Integrating (10) with respect to z:
\begin{equation}
p = \rho g(h-z) +p_o
\end{equation}
The horizontal pressure gradient is independent of z:
\begin{equation}
\begin{aligned}
&\frac{\partial p}{\partial x} = \rho g \frac{\partial h}{\partial x}\\
& \frac{\partial p }{\partial y} = \rho g \frac{\partial h}{\partial y}
\end{aligned}
\end{equation}
So the horizontal acceleration must be independent of z \(\implies\) it is consistent to consider that the horizontal velocities themselves remain z-independent if they are so initially.
Plug in the horizontal pressure gradient in (2) , to get the Horizontal momentum equations:
\begin{equation}
\begin{aligned}
&\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}- fv = -g \frac{\partial h}{\partial x} \\
& \frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+fu= -g \frac{\partial h}{\partial y}
\end{aligned}
\end{equation}
Continuity Equation
We can integrate the incompressibility equation since the horizontal velocities are independent of z:
\begin{equation}
\begin{aligned}
& \int \frac{\partial w}{\partial z} dz = \int (- \frac{\partial u}{\partial x} - \frac{\partial v}{\partial y }) dz \\
& \boxed{w(x,y,z,t) = - z (\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}) +\tilde{w}(x,y,t)}
\end{aligned}
\end{equation}
Applying boundary conditions at \(z = h_B\):
\begin{equation}
w(z=h_B) = \frac{D h_B}{Dt} = u \frac{\partial h_B}{\partial x} + v \frac{\partial h_B}{\partial y}
\label{eq: h_B}
\end{equation}
Then we can write \(\tilde{w}(x,y,t)\):
\begin{equation}
\tilde{w}(x,y,t) = u \frac{\partial h_B}{\partial x} + v \frac{\partial h_B}{\partial y} + h_B(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y})
\end{equation}
We can write \(w(x,y,z,t)\):
\begin{equation}
w(x,y,z,t) = (h_B -z) (\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}) + u \frac{\partial h_B}{\partial x} + v \frac{\partial h_B}{\partial y}
\end{equation}
Note: The corresponding kinematic condition at \(z = h\):
\begin{equation}
w|_{z=h} = \frac{Dh}{Dt} = \frac{\partial h}{\partial t}+ u \frac{\partial h}{\partial x}+v \frac{\partial h}{\partial y}
\end{equation}
Subtracting (17) from (18), knowing that \(H = h-h_B\):
\begin{equation}
\begin{aligned}
\frac{\partial h}{\partial t} +u \frac{\partial h}{\partial x} + v \frac{\partial h}{\partial y} - [(h_B -z) (\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}) + u \frac{\partial h_B}{\partial x} + v \frac{\partial h_B}{\partial y}]&= \frac{\partial h}{\partial t} + u( \frac{\partial h}{\partial x} - \frac{\partial h_B}{\partial x}) + v( \frac{\partial h}{\partial y} - \frac{\partial h_B}{\partial y})\\& - (h_B -z) (\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}) \\
&= \frac{\partial H}{\partial t} + \frac{\partial (uH)}{\partial x} + \frac{\partial(vH)}{\partial y} =0 \\
&=\frac{D H}{Dt }+H(\frac{\partial u}{\partial x}+ \frac{\partial v}{\partial y}) = 0
\end{aligned}
\end{equation}
Analysis of the continuity equation (19): Notice that if the local horizontal divergence of volume is positive, it must be balanced by decrease of the layer thickness, due to the drop in the free surface.
Hence as the cross section of A of a fluid column increases at a rate:
\begin{equation}
\frac{1}{A}\frac{DA}{Dt} = \frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}
\end{equation}
The total thickness must decrease so that the volume \(HA\) remains constant, as following:
\begin{equation}
\frac{1}{H} \frac{D H}{D t} + \frac{1}{A} \frac{DA}{Dt} =0
\end{equation}
We are done deriving the Shallow-Water equations. They are the following:
\begin{equation}
\boxed{\begin{aligned}
&\frac{\partial u}{\partial t} + u \frac{\partial u}{\partial x}+v \frac{\partial u}{\partial y}- fv = -g \frac{\partial h}{\partial x} \\
& \frac{\partial v}{\partial t} + u \frac{\partial v}{\partial x}+v \frac{\partial v}{\partial y}+fu= -g \frac{\partial h}{\partial y} \\
& \frac{\partial H}{ \partial t} + \frac{ \partial (uH)}{\partial x} + \frac{\partial (vH)}{\partial y} =0 \text{ or } \frac{D H}{Dt} + H(\frac{\partial u}{\partial x} + \frac{\partial v}{\partial y}) = 0
\end{aligned}}
\end{equation}
Reflection
- Consequence of \(\delta << 1\): have reduced the number of dynamical equations by 1, the number of dependent variables by 1 (eliminating \(w\)), and the number of independent variables by 1 (no explicit dependence on \(z\).
- The remaining variables are \(u,v,h\), meaning functions \(x,y,t\) only.
- The fact that \(w\) is linear function of \(z\) has vital implications. Using (19) and plugging in horizontal divergence in (17) we get:
\begin{equation}
\begin{aligned}
w(x,y,z,t) &= \frac{Dz}{Dt} \\
&= \frac{(h_B - z)}{H}[\frac{-DH}{Dt}] + u \frac{\partial h_B}{\partial x} +v \frac{\partial h_B}{\partial y} \\
&= \frac{(z - h_B )}{H}[\frac{DH}{Dt}] + u \frac{\partial h_B}{\partial x} +v \frac{\partial h_B}{\partial y}
\end{aligned}
\end{equation}
which can be written as:
\begin{equation}
\frac{D}{Dt} [\frac{z-h_B}{H}] =0
\end{equation}
This tells us that during stretching or contraction of each column the relative position of each column, the relative position of the fluid is unchanged.
- the independence of \(u,v\) of \(z\), means that during motion of the fluid, the fluid moves as set of columns oriented parallel to z-axis.