First of all, as for etymology, estuary came from the Latin word "aestus" meaning the boiling tide.

What makes the edge of the ocean special?

1. Turbulent mixing result from the strong tidal currents

2. Form drag and hydraulic control as the rough as well as shallow bathymetry

3. Strong lateral density gradients majorly contributed by rivers

4. Coastal boundary induces Ekman transport to creates pressure gradients rapidly

5. Coastal dynamics including Kelvin waves, shelf waves, and wind-driven upwelling

6. Important but often not dominant Coriolis Force

7. Concentrated biological productivity and pollution

Let's talk about the scales!

1. Lubrication theory

Most geophysical flows have thin "aspect ratio", meaning \([\frac{\partial}{\partial x}]\sim \frac{1}{L}\) and \([\frac{\partial}{\partial z}]\sim \frac{1}{H}\), then we have \(\frac{H}{L}\ll 1\). One of the most significant consequences is \([w]\ll[u,v]\), i.e. the vertical velocity is very small compared to the horizontals. And also the pressure field is regarded as hydrostatic, except for large gradient areas such as fronts, surface, gravity waves, internal waves with \(\omega\sim N\).

2. Density variations are small

The density can be written as: \(\rho = \rho_0+\rho'(\vec{x},t)\), where \(\rho_0\) is the constant "background", \(\rho'(\vec{x},t)\) is the variation. In a typical seawater case, \([\rho_0]\sim1,000 kg m^{-3}\), \([\rho']\sim1-25 kg m^{-3}\). So it is clear that \(\frac{[\rho']}{[\rho_0]}\ll1\). This leading to an essential simplification of gthe momentum as wellas the mass conservation equations called the "Boussinesq approximation", named after mathematician Joseph Valentin Boussinesq.

3. Reynolds averaging

We average over both turbulent time and spatial scales in terms of full velocity, like: \(u_f=u+u'\). If we define \(< >\) as the average over turbulence, we will obtain \(<u_f>=u\), \(<u'>=0\). Now consider the x-momentum material derivative, the rate of change of u following a fluid parcel:

$$ \frac{Du_f}{Dt}=\frac{\partial u_f}{\partial t} + \vec{u_f}\nabla u_f. $$

Assume incompressibility works, i.e. \(\nabla \vec{u_f}=0\), then the orginal equation can be rewritten as:

$$ \frac{Du_f}{Dt}=\frac{\partial u_f}{\partial t}+\nabla(u_f\vec{u_f}), $$

and take the average:

$$ <\frac{Du_f}{Dt}>=<\frac{\partial u_f}{\partial t}>+<u_fu_f>_x+<u_fv_f>_y+<u_fw_f>_z+<uw>_z+<uw'>_z+<u'w>_z+<u'w'>_z $$

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