This page was made while I was writing my M.S. thesis. It may contain typos and inaccuracies.
























Survey of Geo-Fluid Dynamics

Where to start? Physically, fluid consists of molecules. But for many geo applications, molecular interactions are approximately negligible. So in this document, molecular (discrete distributions of mass) mechanics are approximated by continuum (continuous distribution of mass) mechanics. The continuum is over the independent variables position $\vec{x}$ and time $t$. For geo applications, the dependent variables of interest are \begin{align} \vec{u}(\vec{x},t) &= \text{velocity field}\\ p(\vec{x},t) &= \text{pressure field}\\ \rho(\vec{x},t) &= \text{density field}\\ T(\vec{x},t) &= \text{temperature field}\\ \vec{S}(\vec{x},t) &= \text{chemical concentration field} \end{align} Given initial and boundary conditions, how do these dependent variables evolve in time? Equations will be derived using physical laws such as conservation of mass, momentum, and energy, and, if necessary, a closure to the system of equations. The variables are assumed to be appropriate for the context, that is, differentiable and continuous enough to allow the equations to be solved.

The survey of general equations includes: (1) Derivation of main equations, and approximations isolating (2) rotation effects, (3) stratification effects, and (4) combined rotation and stratification effects. Rotation refers to the rotation of a planet about its axis and stratification refers to a tendency toward layers due to density inhomogeneity.

General Equations and Geo Approximations

First, the general equations will be derived and discussed. Then the general equations will be restated in various reference frames. The general equations are analytically and computationally inconvenient, so approximatinos will be made to obtain more convenient equations which support geo fluid phenomena. Table 1 shows derivation of equations, solutions (if available), and discussion.
Table 1: .
Name Heuristic Derivation Equation(s) Solution(s) Properties and Notes
General Equations
in Eulerain viewpoint
Euler's viewpoint is a fixed cube with constant volume.
Physical laws are applied to conserve mass, momentum, and energy in this cube, and if necessary, close the set of equations.
  • Conservation of mass (``continuity equation'')
    Figure: flow of mass in and out of a cube
    Assuming no sources or sinks of mass, the time-change in mass equals the convergence (or negative divergence) of mass flux $\rho\vec{u}$
    In the limit as the cube shrinks to a point mass, $$ \pd{\rho}{t} = -\div\left( \rho\vec{u} \right) $$ where quantities are per volume for convenience.
    Have one equation and four unknowns: $u,v,w,\rho$. Need more equations.
  • Conservation of momentum (``Newton's second law'')
    Newton's second law states mass times acceleration equals the sum of forces acting on the mass, $$\rho \der{\vec{u}}{t}=\sum_n F_n$$ where quantities are per volume for convenience. The following is an enumeration of the forces:
    1. The pressure force $-\grad p$ is from the pressure gradient
    2. The conservative body force $\rho\grad\phi$ is represented by potential $\phi$, and is related to gravity.
    3. The nonconservative body force $\vec{F}_{\text{fric}}$ represents friction from viscosity. For Newtonian fluid, $\vec{F}_{\text{fric}}=\mu\left( \lap \vec{u} + \frac{1}{3}\grad\left( \div \vec{u} \right) \right)$
    Figure: forces on fixed cube.
    Now there are four eqns and five unknowns $u,v,w,\rho,p$. If $\rho$ is const, we are done, otherwise, need another equation.
  • Conservation of Energy (``first law of thermodynamics'')
    The first law of thermodynamics states that time-change in internal energy equals heat recieved minus mechanical work done. The following is an explanation of each term.
    1. Internal energy is the thermal agitation of the particles, and its change corresponds to the time derivative of temperature, up to scaling constant.
    2. For simplicity, consider heat gained as diffusion from neighbors, which can be generalized if desired
    3. For simplicity, consider work done only by pressure force $p\pd{}{t}\left(\frac{1}{\rho}\right)$ which equals $\frac{p}{\rho^2}\pd{\rho}{t}$ by quotient rule
    So the energy equation is $$ C_0\pd{T}{t} = \frac{k_T}{\rho}\lap T -\frac{p}{\rho^2}\pd{\rho}{t} $$ We have introduced a new variable $T$, so need another equation to close the model.
  • Closing the model:
    • Density equation (``equation of state''):
      Density $\rho$ is in general a function of time $t$, position $\vec{x}$, pressure $p$, temperature $T$, and chemical composition $S$ e.g. salinity or humidity.
      • In the ocean, density is a function of temperature, pressure, and salinity. But nearly incompressible so ignore pressure, and assume linearly dependent on temperature and salinity $ \rho = \rho_0[1-\alpha(T-T_0)+\beta(S-S_0)] $
      • In the atmosphere, dry air obeys the ideal gas law $\rho=\frac{p}{RT}$. For humid air, generalize ideal gas law to include humidity $\rho=\frac{p}{RT(1+\gamma S)}$
      Now we must resolve chemical composition $S$.
    • Equations for chemical composition:
      For simplicity, assume only diffusion $\pd{S}{t}=k_S \lap S$ of oceanic salt and atmospheric humidity
      Note $S$ is percentage by mass.
Have five eqns in five unknowns: $\vec{u},\rho, p, T,$ and $S$
$$ \left\{ \begin{array}{ll} \text{mass} & \pd{\rho}{t} = -\div\left( \rho\vec{u} \right) \\ \text{momentum} & \rho \left(\pd{\vec{u}}{t} \right) = -\grad p +\rho \vec{g} + F_{\text{shear}} \\ \text{energy} & C_0\pd{T}{t} = \frac{k_T}{\rho}\lap T -\frac{p}{\rho^2}\pd{\rho}{t} \\ \text{closure} & \text{ocean}: \rho = \rho_0[1-\alpha(T-T_0)+\beta(S-S_0)] \\ & \text{atmosphere}: \rho=\frac{p}{RT(1+\gamma S)}\\ & \pd{S}{t}=k_S S\\ \end{array} \right\} $$ See Petcua et al (2009) for analysis on a sphere.
General Equations

in Lagrangian viewpoint
Above equations derived in Eulerian viewpoint (fixed cube).
Consider moving parcel (``blob'', ``chunk'', ``element'', ``control volume'') with constant mass and trajectory $\vec{x}(t)=(x(t),y(t),z(t))$
A scalar or vector field $f(\vec{x}(t),t)$ of the parcel changes in time (using chain rule) \begin{align} \der{f}{t} &= \pd{f}{t}+\pd{x}{t}\pd{f}{x}+\pd{y}{t}\pd{f}{y}+\pd{z}{t}\pd{f}{z}\\ &= \pd{f}{t}+u\pd{f}{x}+v\pd{f}{y}+w\pd{f}{z}\\ &= \pd{f}{t}+(\vec{u}\cdot\grad) f\\ &= \mD{f} \end{align} The symbol $\mD{}$ is called the matierial (``total'', ``Lagrangian'', ``substantial'') derivative.
So replace time derivatives with $\mD{}$
  • Mass conservation:
    Expand the divergence of product and group terms to isolate material derivative of density. \begin{align} \pd{\rho}{t} + (\vec{u}\cdot \grad)\rho &= -\rho\div\vec{u}\\ \frac{D\rho}{Dt} &= -\rho\div\vec{u} \end{align}
  • Conservation of momentum:
    Simply replace time derivative with material derivative.
  • Conservation of Energy:
    Replace time derivative with material derivative
  • Model closure:
    • Equation of state:
      Keep eqns
    • Chemical equations:
      Simply replace time derivative with material derivative.
$$ \left\{ \begin{array}{ll} \text{mass} & \frac{D\rho}{Dt} = -\rho\div\vec{u} \\ \text{momentum} & \rho \left(\mD{\vec{u}} \right) = -\grad p +\rho \vec{g} + F_{\text{shear}} \\ \text{energy} & C_0\mD{T} = \frac{k_T}{\rho}\lap T -\frac{p}{\rho}\div \vec{v} \\ \text{closure} & \text{ocean}: \rho = \rho_0[1-\alpha(T-T_0)+\beta(S-S_0)] \\ & \text{atmosphere}: \rho=\frac{p}{RT(1+\gamma S)}\\ & \mD{S}=k_S S\\ \end{array} \right\} $$ Same as above This is the same equation as above, just from a moving viewpoint.
Rotating general equations
in Lagrangian viewpoint

Consider the Lagrangian equations and let the domain rotate with angular velocity $\vec{\Omega}(t)=\Omega\unitnormal{k}$. Want to adjust equations so that variables are in rotating basis.

Figure: fixed and rotating basis vectors, and vector $\vec{x}$

Want the acceleration of vector $\vec{x}(t)$ in the fixed basis in terms of its velocity in rotating basis. Derivatives will be described with subscripts $I$ and $R$ for fixed and rotating basis. A convenient operator is $\left( \der{}{t} \right)_I = \left( \der{}{t} \right)_R + \vec{\Omega}\times $. First velocity: \begin{align} \vec{u}_I = \left( \der{\vec{x}}{t} \right)_I &= \left( \der{\vec{x}}{t} \right)_R + \vec{\Omega}\times \vec{x}\\ &=\vec{u}_R + \vec{\Omega}\times \vec{x} \end{align} Next acceleration. \begin{align} \vec{a}_I = \left( \der{\vec{u}_I}{t} \right)_I &= \left( \der{\vec{u}_I}{t} \right)_R + \vec{\Omega}\times \vec{u}_I\\ &= \left( \der{( \vec{u}_R + \vec{\Omega}\times \vec{x} )}{t} \right)_R + \vec{\Omega}\times \left( \vec{u}_R + \vec{\Omega}\times \vec{x} \right)\\ &= \left( \der{ \vec{u}_R }{t} \right)_R +\left(\der{\vec{\Omega}}{t}\right)_R \times \vec{x} + \vec{\Omega}\times\left(\der{\vec{x}}{t}\right)_R + \\ &\quad+ \vec{\Omega}\times\vec{u}_R + \vec{\Omega}\times \left(\vec{\Omega}\times \vec{x} \right)\\ &= \left( \der{ \vec{u}_R }{t} \right)_R +\left(\der{\vec{\Omega}}{t}\right)_R \times \vec{x} + 2\vec{\Omega}\times\vec{u}_R + \vec{\Omega}\times \left(\vec{\Omega}\times \vec{x} \right)\\ \end{align} The first term is the acceleration in rotating basis, as desired. The second disappears for constant $\vec{\Omega}$. The third term is the ``Coriolis'' acceleration. The fourth term is the ``centripetal'' acceleration.

$$ \left\{ \begin{array}{ll} \text{mass} & \frac{D\rho}{Dt} = -\rho\div\vec{u} \\ \text{momentum} & \rho \left(\left( \der{ \vec{u}_R }{t} \right)_R +\left(\der{\vec{\Omega}}{t}\right)_R \times \vec{x} + 2\vec{\Omega}\times\vec{u}_R + \vec{\Omega}\times \left(\vec{\Omega}\times \vec{x} \right) \right)\\ &\qquad = -\grad p +\rho \vec{g} + F_{\text{shear}} \\ \text{energy} & C_0\mD{T} = \frac{k_T}{\rho}\lap T -\frac{p}{\rho}\div \vec{v} \\ \text{closure} & \text{ocean}:\\ & \rho = \rho_0[1-\alpha(T-T_0)+\beta(S-S_0)] \\ & \mD{S}=k_S\\ & \text{atmosphere}:\\ & \rho=\frac{p}{RT(1+\gamma q)}\\ & \mD{q} = k_q\lap q \end{array} \right\} $$ same as above Dynamics themselves are unchanged by rotating basis, but the equations must be adjusted.
Rotating general equations
in Lagrangian viewpoint

with geophysical simplifications
The above equations are equivalent, but in different viewpoints and bases. The following is a list of simplifications for geophysical fluids.
  • The angular velocity $\vec{\Omega}$ of Earth is assumed to be constant, so the term $\der{\vec{\Omega}}{t}$ is zero.
  • The ``centripetal'' acceleration can be rewritten as a potential, and combined with the gravity force to form a geopotential force. This force is no longer perpendicular to a spherical surface, but perpendicular to a geopotential surface (``geiod'') which resembles earth's shape. Actually, the centripetal force is small, so the geoid is nearly a sphere.
  • A basis must be chosen for the coordinates. The operators can be written in any curvilinear coordinate system. To enjoy convenience of cartesian basis, at each point $\vec{x}$ on a geopotential surface, define orthonormal basis $(\unitnormal{i}(\vec{x}),\unitnormal{j}(\vec{x}),\unitnormal{k}(\vec{x}))$ pointing eastward, northward, and outward respectively. This basis rotates with earth, so rotation terms are retained.
    Figure: basis $(\unitnormal{i}(\vec{x}),\unitnormal{j}(\vec{x}),\unitnormal{k}(\vec{x}))$ at each point of geopotential surface.
    The coriolis acceleration becomes $$ 2\vec{\Omega}\times\vec{u} = 2 \left| \matrix{ \unitnormal{i} & \unitnormal{j} & \unitnormal{k} \\ 0 & f_* & f \\ u & v & w } \right| $$ where $f=2\abs\Omega \sin\phi$, $f_*=2\abs\Omega \cos\phi$, $\phi$ is the latitude.
    The body force conveniently acts in the $\unitnormal{k}$ direction.
$$ \left\{ \begin{array}{ll} \text{mass} & \frac{D\rho}{Dt} = -\rho\div\vec{u} \\ \text{momentum} & \rho \left( \pd{u}{t} + (\vec{u}\cdot\grad)u + f_*w - fv \right) = -\frac{1}{\rho_0}\pd{p}{x}+\nu\lap u\\ & \rho \left( \pd{v}{t} + (\vec{u}\cdot\grad)v + fu \right) = -\frac{1}{\rho_0}\pd{p}{y}+\nu\lap v\\ & \rho \left( \pd{w}{t} + (\vec{u}\cdot\grad)w + f_* u \right) = -\frac{1}{\rho_0}\pd{p}{z}-\rho g+\nu\lap w\\ \text{energy} & C_0\mD{T} = \frac{k_T}{\rho}\lap T -\frac{p}{\rho}\div \vec{v} \\ \text{closure} & \text{ocean}:\\ & \rho = \rho_0[1-\alpha(T-T_0)+\beta(S-S_0)] \\ & \mD{S}=k_S\\ & \text{atmosphere}:\\ & \rho=\frac{p}{RT(1+\gamma q)}\\ & \mD{q} = k_q\lap q \end{array} \right\} $$
Boussinesq Approximation
Assume Newtonian fluid
Consider $\vec{u}=0$. Then $z$-momentum equation becomes $\pd{p_0}{z}=-\rho_0 g$ where $p_0$ and $\rho_0$ are static pressure and density.
Let density and pressure be be sum of static and perturbed. $$ \rho(t,\vec{x}) = \rho_0(z) + \rho'(t,\vec{x})\\ p(t,\vec{x})=p_0(z) + p'(t,\vec{x}) $$ Assume deviation of density is small, $\abs{\rho'} << \rho_0$ [justified since density varies 0.03% in ocean and 5% in troposphere where weather patterns occur].
The following are resulting consequences.
  • Mass conservation:
    Expand the mass conservation equation. \begin{align} &\rho_0\left( \pd{u}{x} +\pd{v}{y} +\pd{w}{z} \right) + \rho'\left( \pd{u}{x} +\pd{v}{y} +\pd{w}{z} \right) \\ &\qquad + \left( \pd{\rho'}{t} + u\pd{\rho'}{x} +v\pd{\rho'}{y} +w\pd{\rho'}{z} \right) = 0 \end{align} The third term is less than or close the the second and the second is much less than the first. So the only the first term is significant, and the incompressibility assumption is made, $$ \div \vec{u} = 0 $$ A consequence of this incompressibility is that pressure waves (e.g. sound waves) cannot exist.
  • Momentum equation
    • Incompressible Newtonian fluid so shearing is linearly proportional to velocity gradients
      So frictional force becomes laplacian of velocity
    • $\rho_0$ dominates $\rho'$, so cancel $\rho'$ except for gravity term.
    • For $z$ equation, define hydrostatic pressure linearly, $p_0(z)=P_0-\rho_0 gz$, so that $\der{p_0}{z} = -\rho_0 g$. Plug in $p$ and $\pd{p}{z}$ into $z$ equation, $p_0$ terms cancel.
      For $x,y$ equations, the $p_0(z)$ term gets killed in $x$ and $y$ derivatives.
      So everything is in terms of pressure deviation $p'$.
  • Energy equation
    incompressibility kills mechanical work term. Plug $\rho=\rho_0+\rho'$, kill $\rho'$, combine consts, end up with diffusion eqn for temperature.
  • Equation of state
    For ocean, take eqn of state, diffusion of salinity, and diffusion of temperature, assume that the diffusion coefficients are equal for large scale diffusion by eddies
    For air, more complicated, but can make it density diffusion like ocean.
From now on, replace $\rho'$ and $p'$ with $\rho$ and $p$.
$$ \left\{ \begin{array}{ll} \text{mass} & \div \vec{u} = 0\\ \text{momentum} & \pd{u}{t} + (\vec{u}\cdot\grad)u + f_*w - fv = -\frac{1}{\rho_0}\pd{p}{x}+\nu\lap u\\ & \pd{v}{t} + (\vec{u}\cdot\grad)v + fu = -\frac{1}{\rho_0}\pd{p}{y}+\nu\lap v\\ & \pd{w}{t}+(\vec{u}\cdot\grad)w -f_* u = -\frac{1}{\rho_0} \pd{p}{z} - \frac{\rho}{\rho_0}g + \nu\lap w\\ \text{energy} & \mD{T} = k\lap T \\ \text{closure} & \mD{\rho} = k\lap\rho \\ \end{array} \right\} $$ -
Geo scaled equations Consider the scale variables \begin{align} L &= \text{horizontal distance}\\ H &= \text{vertical distance}\\ T &= \text{time}\\ U &= \text{horizontal speed}\\ W &= \text{vertical speed}\\ P &= \text{pressure}\\ \Delta\rho &= \text{density change} \end{align} Below are some simplifications for geophysical fluid dynamics:
  • $T\gtrsim\frac{1}{\omega}$ since rotation is significant
  • $H\lt\lt L$ since the domain of flow is shallow
  • The continuity equation with a scale below each term. $$ \underset{\frac{U}{L}}{\pd{u}{x}} + \underset{\frac{U}{L}}{\pd{v}{y}} + \underset{\frac{W}{H}}{\pd{w}{z}} = 0 $$ \begin{align} &\pd{u}{x}&+&\pd{v}{y}&+&\pd{w}{z}& = 0 \\ &\frac{U}{L}&&\frac{U}{L}&&\frac{W}{H}& \end{align} The two complementary cases are $\frac{U}{L}\gtrsim\frac{W}{H}$ and $\frac{U}{L}\lt\lt\frac{W}{H}$. The second case must be false since the resulting equation is $\pd{w}{z}=0$, i.e. $w$ is constant, which cannot exist unless horizontal flow feeds it at the boundary. So the first case holds.
    The first case implies $W\lt\lt U$. Thus the flow is almost 2d.
  • The $x$ (and similarly $y$) momentum equation with a scale below each term. $$ \underset{\frac{U}{T}}{\pd{u}{t}} + \underset{\frac{U^2}{L}}{u\pd{u}{x}}+ \underset{\frac{U^2}{L}}{v\pd{u}{y}}+ \underset{\frac{UW}{H}}{w\pd{u}{z}} + \underset{\Omega W}{f_* w} - \underset{\Omega U}{fv} = \underset{\frac{P}{\rho_0 L}}{-\frac{1}{\rho_0}\pd{p}{x}} +\underset{\frac{\nu U}{L^2}}{\nu\pd{^2 u}{x^2}} +\underset{\frac{\nu U}{L^2}}{\nu\pd{^2 u}{y^2}} +\underset{\frac{\nu U}{H^2}}{\nu\pd{^2 u}{z^2}} $$
    • The $\Omega W$ term is killed since $W \lt\lt U$ implies $\Omega W \lt\lt \Omega U$ (except near the equator, in which case it is usually smaller than the pressure term, but further scale analysis is required)
    • The $\frac{\nu U}{L^2}$ terms are killed since $H\lt\lt L$ implies $\frac{\nu U}{L^2}\lt\lt \frac{\nu U}{H^2}$.
  • The $z$ momentum equation with a scale below each term. $$ \underset{\frac{W}{T}}{\pd{w}{t}} + \underset{\frac{UW}{L}}{u\pd{w}{x}}+ \underset{\frac{UW}{L}}{v\pd{w}{y}}+ \underset{\frac{W^2}{H}}{w\pd{w}{z}} - \underset{\Omega U}{f_* u} = \underset{\frac{P}{\rho_0}H}{-\frac{1}{\rho_0}\pd{p}{z}} -\underset{\frac{g\Delta\rho}{\rho_0}}{\frac{g\rho}{\rho_0}} +\underset{\frac{\nu W}{L^2}}{\nu\pd{^2 w}{x^2}} +\underset{\frac{\nu W}{L^2}}{\nu\pd{^2 w}{y^2}} +\underset{\frac{\nu W}{H^2}}{\nu\pd{^2 w}{z^2}} $$ The following follow from the scale requirements above:
    • The $\frac{W}{T}$ term is killed since $T\gtrsim \frac{1}{\Omega}$ implies $\frac{W}{T} \lesssim \Omega W$ and $W\lt\lt U$ implies $\Omega W \lt\lt \Omega U$
    • The $\frac{UW}{L}$ and $\frac{W^2}{H}$ are killed since $\frac{u}{L}\lesssim\Omega$, $W\lesssim\frac{H}{L}U$, and $W\lt\lt U$, imply that they are both $\lt\lt \Omega U$
    • The $\Omega U$ is itself small since for geophysical fluids, its ratio with the second term on the right, $\frac{\rho_0 \Omega U}{g\Delta \rho}$, is small for geophysical values of the various scales.
    • The $\frac{\nu U}{L^2}$ terms are killed since $H\lt\lt L$ implies $\frac{\nu W}{L^2}\lt\lt \frac{\nu W}{H^2}$.
    • The term $\frac{\nu U}{H^2}$ is killed since vertical friction cannot dominate coriolis force $\frac{\nu U}{H^2}\lesssim\Omega U$ and pluggin $W$ gives $\frac{\nu W}{H^2}\lesssim\Omega W$ which is $\lesssim \Omega U$
    Therefore, the $z$ momentum equation is hydrostatic for geostrophic flows.
  • In the density equation, $H\lt\lt L$ implies that vertical diffusion dominates horizontal diffusion.
We are left with five equations for five unknowns $u,v,w,p,\rho$
$$ \left\{ \begin{array}{ll} \text{mass} & \div \vec{u} = 0\\ \text{momentum} & \pd{u}{t} + (\vec{u}\cdot\grad)u - fv = -\frac{1}{\rho_0}\pd{p}{x}+\nu\pd{^2 u}{z^2}\\ & \pd{v}{t} + (\vec{u}\cdot\grad)v + fu = -\frac{1}{\rho_0}\pd{p}{y}+\nu\pd{^2 v}{z^2}\\ & 0 = \pd{p}{z} - \rho g \\ \text{density} & \mD{\rho} = k\pd{^2\rho}{z^2} \\ \end{array} \right\} $$ These equations will be used as a starting point for various models below. Consider the $x$ (or similarly $y$) momentum equations. Compare each term to the coriolis term by dividing. $$ \underset{\frac{1}{\Omega T}}{\pd{u}{t}} + \underset{\frac{U}{\Omega L}}{u\pd{u}{x}} + \underset{\frac{U}{\Omega L}}{v\pd{u}{y}} + \underset{\frac{WL}{UH}\frac{U}{\Omega L}}{w\pd{u}{z}} + \underset{1}{fv} = \underset{\frac{P}{\rho_0 L\Omega U}}{-\frac{1}{\rho_0}\pd{p}{x}} + \underset{\frac{\nu}{\Omega H^2}}{\nu\pd{^2 u}{z^2}} $$ \begin{align} &\pd{u}{t}& + &u\pd{u}{x}& + &v\pd{u}{y}& + &w\pd{u}{z}& - &fv& = &-\frac{1}{\rho_0}\pd{p}{x}& + &\nu\pd{^2 u}{z^2}&\\ &\frac{1}{\Omega T}& &\frac{U}{\Omega L}& &\frac{U}{\Omega L}& &\frac{W}{H\Omega}& &1& &\frac{P}{\rho_0 L\Omega U}& &\frac{\nu}{\Omega H^2}& \end{align} The following is an explanation of each term:
  • The first term is the temporal Rossby number $Ro_T$, it is usually $\lesssim 1$
  • The second term is the Rossby number $Ro$
  • The third term is the Rossby number times $\frac{WL}{UH}$, and is usually on the same scale as the Rossby number.
  • The first term on the right hand side shows the balance between pressure force and coriolis force. When it $\sim 1$, pressure counteracts some of the coriolis force, and the dynamic pressure scale is $P=\rho\Omega LU$. The dynamic pressure is usually much less than hydrostatic pressure.
  • The last term is the Eckman number $Ek$. It is usually much less than 1. But it is retained because it provides a higher order derivative, which is used to explain phenomena at the boundary like Eckman layer.
Another fluid dynamics coefficeint, the Reynolds number $Re$, compares inertial and frictional forces, and can be stated in terms of Rossby and Eckman numbers, $Re = \frac{Ro}{Ek}\left( \frac{L}{H} \right)^2$. This number is usally extremely large, so turbulence is prevelent.
Equations in density coordinates Consider
  • geophysically scaled equations
  • Stable stratification with density increasing monotincally with depth of ocean (analogous for atmosphere).
  • Density is conserved by a parcel.
  • Flow only along isopycnal (constant density) surfaces.
It is convenient to change variable $z$ to $\rho$. That is independent and dependent variables $z$ and $\rho(x,y,z,t)$ are swapped to become $z(z,y,\rho,t)$ and $\rho$.
First consider the partial derivatives of a field $f(x,y,\rho(x,y,z,t),t)$. \begin{align} \left. \pd{f}{x} \right|_{z=\text{const}} &= \left. \pd{f}{x} \right|_{\rho=\text{const}} + \pd{f}{\rho} \left. \pd{\rho}{x} \right|_{z=\text{const}}\\ &= \left. \pd{f}{x} \right|_{\rho=\text{const}} + \pd{f}{\rho} \frac{\pd{z}{x}}{\pd{z}{\rho}}\\ \end{align} since $f=z \Ra 0=\pd{z}{x}+\pd{z}{\rho}\pd{\rho}{x}$. Derivatives $\pd{}{y}$ and $\pd{}{t}$ are similar. For derivative with respect to $x$, \begin{align} \left. \pd{f}{z} \right|_{z=\text{const}} &= \pd{f}{\rho} \left. \pd{\rho}{z} \right|_{z=\text{const}} \\ &= \pd{f}{\rho} \frac{1}{\pd{z}{\rho}} \end{align} First consider the $z$ momentum equation (``hydrostatic relation'') which becomes $$ \pd{p}{\rho} = -\rho g pd{z}{\rho} $$ Want to rewrite in terms of $P(x,y,\rho,t)=p+\rho gz$, the pressure in density coordineates. \begin{align} \left. \pd{\rho}{x} \right|_{z=\text{const}} &= \left. \pd{p}{x} \right|_{\rho=\text{const}} - \frac{\pd{z}{x}}{\pd{z}{\rho}} \pd{p}{\rho}\\ &= \left. \pd{P}{x} \right|_{\rho=\text{const}} \end{align} And similarly for the $y$ derivative. So the hydrostatic relation becomes $$ \pd{P}{\rho} = gz $$ Next consider the density conservation equation in regular basis for conserved density. $$ \pd{\rho}{t} + u\pd{\rho}{x} + v\pd{\rho}{y} + w\pd{\rho}{z} = 0 $$ Rewrite in density coordinates $$ \pd{\rho}{t} + u\pd{\rho}{x} + v\pd{\rho}{y} + w\pd{\rho}{z} = 0 $$ Solve for $w$. $$ w=\pd{z}{t} + u\pd{z}{x} + v\pd{z}{y} $$ So this is the vertical velocity of the flow along isopycnal surfaces (which was assumed), and the horizontal dimensions can be treated as two dimensional.
Next consider the horizontal momentum equations. The material derivative in isobaric basis looks similar, except the vertical term is omitted since no motion across isopycnal surfaces. The equations become \begin{align} \mD{u} - fv &= -\frac{1}{\rho_0}\pd{P}{x}\\ \mD{v} + fu &= -\frac{1}{\rho_0}\pd{P}{y} \end{align} where $u,v$ are in the regular coordinate system with motion along isopynic surfaces due to corrections by $w$. This is useful since the original BC are reused.
Finally, the continuity equation. $$ \pd{h}{t} + \pd{hu}{x} + \pd{hv}{y} = 0 $$ where $h=-\Delta\rho \pd{\rho}{z}$ is the thickness of the fluid layer between densities $\rho$ and $\rho+\Delta rho$ So there are five equations in five unknowns $u,v,P,z,h$. From these, $w$ and $p$ can easily be recovered.
$$ \left\{ \begin{array}{ll} &\mD{u} - fv = -\frac{1}{\rho_0}\pd{P}{x}\\ &\mD{v} + fu = -\frac{1}{\rho_0}\pd{P}{y}\\ &\pd{P}{\rho} = gz\\ &\pd{h}{t} + \pd{hu}{x} + \pd{hv}{y} = 0\\ &h=-\Delta\rho \pd{\rho}{z} \end{array} \right\} $$ Where $u,v$ are in original basis. Transform back to original basis using $$ P=p+\rho gz\\ w=\pd{z}{t} + u\pd{z}{x} + v\pd{z}{y} $$








Rotation Effects

Table 1: .
Name Heuristic Derivation Equation(s) Solution(s) Properties and Notes
Rotation Effects
Geostrophic Flow $Ro, Ro_T \lt\lt 1$ i.e. coriolis acceleration dominates
$Ek \lt\lt 1$ i.e. negligible friction
$\rho=0$ i.e. no density variation
homogeneous fluid i.e. uniform composition
$$ \left\{ \begin{array}{ll} \text{mass} & \div \vec{u} = 0\\ \text{momentum} & -fv = -\frac{1}{\rho_0}\pd{p}{x}\\ & fu = -\frac{1}{\rho_0}\pd{p}{y}\\ & 0 = -\frac{1}{\rho_0} \pd{p}{z} \end{array} \right\} $$ Using algebra,
\begin{align} u &= \frac{1}{\rho_0 f} \pd{p}{y}\\ v &= -\frac{1}{\rho_0 f} \pd{p}{x} \end{align}
  • The word ``geostrophic'' is Greek $\gamma \eta$ meaning Earth and $\sigma \tau \rho o \phi \eta$ meaning turning
  • Horizontal velocity has no vertical shear $\pd{u}{z}=\pd{v}{z}=0$ i.e. vertical rigidity (particles on vertical line move in concert)
    This is because vertical derivative of $x$ (similarly for $y$) eqn combined with the $z$ momentum eqn
    \begin{align} -f\pd{v}{z} &=-\frac{1}{\rho_0} \pd{}{z}\pd{p}{x}\\ &=-\frac{1}{\rho_0} \pd{}{x}\pd{p}{z}\\ &= 0 \end{align} This is called Taylor Proudman theorem.
  • Balance between coriolis force and pressure forces.
    Velocity vector is perpendicular to pressure gradient. $$ \matrix{u\\v}\perp\ \matrix{\pd{p}{x}\\ \pd{p}{y} } $$ So velocity is along lines of constant pressure (isobars), not down the pressure gradient
    So isobaric flow - isobars are streamlines
    So no pressure work on or by the fluid
    So can persist without energy source
  • direction of flow:
    $f>0$ (i.e. Northern hemisphere, CCW rotation) $\Ra$ high pressure on right
    $f \lt 0$ (i.e. Southern hemisphere, CW rotation) $\Ra$ high pressure on left
  • Example: If $f$-plane approximation (i.e. $f=$const), then \begin{align} \pd{u}{x}+\pd{v}{y} &=-\pd{}{x}\left( \frac{1}{\rho_0 f} \pd{p}{y}\right) + \pd{}{y}\left( \frac{1}{\rho_0 f} \pd{p}{x}\right)\\ &= 0 \end{align} So no horizontal divergence and $\pd{w}{z}=0$
    So flow with flat boundary is 2$d$ since $w=0$ at boundary and therefore everywhere.
  • Example: irregular bottom and no surface displacements
    Figure: irregular bottom, depth $h(x,y)$ from flat surface, triangle showing u,w
    Let $h(x,y)$ be the depth of the fluid. Then when climbing the bottom slope, \begin{align} w &= -u\pd{h}{x}-v\pd{h}{y}\\ &=0 \end{align} since $w$ constant for entire depth and zero at surface.
    So flow can't climb up or down bottom slope
    So isobath flow - along contours of constant depth
    So vertical rigidity is maintained.
    e.g. fluid above bump or dip forms a permanent tube called Taylor column.
    e.g. if isobaths run from boundary to boundary, then no flow. So flow only in closed isobaths.
    e.g. If surface displacements are allowed, then still move along constant depth $h$.
Non-geostrophic Flow Non-geostrophic flow generalizes geostrophic flow to include intertial acceleration.
So consider:
$Ro, Ro_t \lesssim 1$ i.e. coriolis does NOT dominate acceleration
$Ek \lt\lt 1$ i.e. negligible friction
$\rho=0$ i.e. no density variation
homogeneous fluid
$$ \left\{ \begin{array}{ll} \text{mass} & \div \vec{u} = 0\\ \text{momentum} &\pd{u}{t}+u\pd{u}{x}+v\pd{u}{y}+w\pd{u}{z}-fv = -\frac{1}{\rho_0}\pd{p}{x}\\ &\pd{v}{t}+u\pd{v}{x}+v\pd{v}{y}+w\pd{v}{z}+fu = -\frac{1}{\rho_0}\pd{p}{y}\\ &0 = -\frac{1}{\rho_0} \pd{p}{z}\end{array} \right\} $$
Barotropic Flow Take non-geostrophic flow.
Let $u,v$ independent of depth initially. Then $u,v$ remain independent of depth for all time since $u,v$ initially depth-independent implies advection and coriolis also initially depth independent, and $p$ is also depth independent by the $z$-momentum equation. So $\pd{u}{t}$ and $\pd{v}{t}$ are initially depth-independent, so they will remain depth independent.
$$ \left\{ \begin{array}{ll} \text{mass} & \div \vec{u} = 0\\ \text{momentum} &\pd{u}{t}+u\pd{u}{x}+v\pd{u}{y}-fv = -\frac{1}{\rho_0}\pd{p}{x}\\ &\pd{v}{t}+u\pd{v}{x}+v\pd{v}{y}+fu = -\frac{1}{\rho_0}\pd{p}{y}\\ &0 = -\frac{1}{\rho_0} \pd{p}{z} \end{array} \right\} $$ The flow has no vertical structure, like geostrophic.
The flow is not required to be aligned with isobars.
The vertical velocity can vary vertically with depth, allowing 2d divergence, and flow across isobaths.
Shallow water equations Can simplify barotropic flow to three equations
Let bottom surface $b(x,y)$ and top free surface $h(x,y,t)$ above $b$
Figure: bottom surface $b(x,y)$ and top surface $h(x,y,t)$, and average height H
Find $w$ by integrating $\div \vec{u}=0$ from bottom to surface
\begin{align} \left(\pd{u}{x}+\pd{v}{y}\right)\int_b^{b+h}\upd z + w|_b^{b+h} &= 0\\ h\pd{u}{x}+h\pd{v}{y} + w(b+h)-w(b) &= 0\\ h\pd{u}{x}+h\pd{v}{y} + \mD{(b+h)}-\mD{b} &= 0\\ \end{align} Since a particle on a boundary must stay there. Continuing, $$ \begin{split} &h\pd{u}{x}+h\pd{v}{y} + \pd{(b+h)}{t}+u\pd{(b+h)}{x}\quad+\\ &\quad+v\pd{(b+h)}{y} - \left( \pd{b}{t}+u\pd{b}{x}+v\pd{b}{y} \right) = 0 \end{split}\\ \pd{h}{t}+\pd{hu}{x}+\pd{hv}{y}=0 $$ so $w$ is no longer a variable, only $u,v,h,p$
Finally, we can replace pressure with $h$. The fluid is homogeneous (and assume uniform surface pressure) so dynamic pressure $p$ is independent of depth i.e. dynamic pressure is constant for the whole depth, $$ p(x,y,t) = \rho_0 g (b+h-H) $$ Intuitively, $b+h-H$ is the surface deviation, and is greater for larger pressure.
So we have three equations in three unknowns $u,v,h$. Also note that the independent variables are $x,y,t$
$$ \left\{ \begin{array}{ll} \text{mass} & \pd{h}{t}+\pd{hu}{x}+\pd{hv}{y}=0\\ \text{momentum} & \pd{u}{t}+u\pd{u}{x}+v\pd{u}{y}-fv = -g\pd{(b+h)}{x}\\ & \pd{v}{t}+u\pd{v}{x}+v\pd{v}{y}+fu = -g\pd{(b+h)}{y} \end{array} \right\} $$ Despite its name, it applies to the atmosphere and ocean.
Vorticity dynamics Will derive three equations from barotropic and shallow water models. Will look at vorticity.
  • Take barotropic $x,y$ momentum equations. subtract the $x$-derivative of the $x$ equation from the $y$-derivative of the $y$ equation, rearrange the fourteen terms and add the term $\pd{f}{t} (=0)$. \begin{align} \pd{(y\text{ equation})}{y} - \pd{(x\text{ equation})}{x} &= 0\\ \mD{}( f+\zeta ) + \left( \pd{u}{x}+\pd{v}{y} \right) ( f+ \zeta ) &= 0 \end{align} where $\zeta = \pd{v}{x}-\pd{u}{y}$ is the realtive vorticity and is in vertical direction, $f$ is the coriolis parameter (``ambient vorticity'') and is also in the vertical direction. Note that material derivative is over time and the horizontal directions.
    Note that the pressure terms cancelled.
  • Take shallow water continuity equation, expand with product rule, combine terms to a material derivative, $$ \mD{h} + \left( \pd{u}{x}+\pd{v}{y} \right) h = 0 $$
  • Consider fluid columnar parcel with cross section $\upd s$.
    Figure: fluid column with height $h$ and cross section $\upd s$
    Incompressibility implies that fluid column volume is conserved, $$\mD{(h\upd s)}=0$$ Expand using product rule and impose above bulletpoint to get $$ \mD{\upd s} = \left( \pd{u}{x}+\pd{v}{y} \right) \upd s $$ Intuitively, horizontal divergence means $\upd s$ increases, and vice versa.
$$ \left\{ \begin{array}{ll} &\mD{}( f+\zeta ) + \left( \pd{u}{x}+\pd{v}{y} \right) ( f+ \zeta ) = 0\\ &\mD{h} + \left( \pd{u}{x}+\pd{v}{y} \right) h = 0\\ & \mD{\upd s} = \left( \pd{u}{x}+\pd{v}{y} \right) \upd s \end{array} \right\} $$
  • Expand $\mD{(f+\zeta)\upd s}$ with product rule, impose eqns \begin{align} \mD{(f+\zeta)\upd s} &= \upd s\mD{(f+\zeta)} + (f+\zeta)\mD{\upd s}\\ &=\upd s \left[ -(\pd{u}{x}+\pd{v}{y})(f+\zeta) \right] + (f+\zeta) \left[ (\pd{u}{x}+\pd{v}{y})\upd s \right]\\ &=0 \end{align} So $(f+\zeta)\upd s$ is concerved by the fluid parcel. This quantity is vorticity flux (vorticity integrated over $\upd s$) or circulation. This is the 2$d$ rotating fluid version of Kelvin's theorem that circulation is conserved by a inviscid fluid parcel.
    Figure: parcel with small $\upd s$, large $f+\zeta$, and vice versa.
    Since circulation and volume are conserved, then so is their ratio. $$ \mD{q} = 0 $$ where $q=\frac{f+\zeta}{h}$ is known as potential vorticity.
  • Example: $Ro \lt\lt 1$ i.e. geostrophic flow.
    Then $f$ dominates $\zeta$.
    Then $q=\frac{f}{h}$, which is conserved
    If $f$ constant, then $h$ constant, as desired for geostrophic flow.
  • Consider fluid with bottom $b(x,y)$ and surface $b+h(t,x,y)$.
    Consider relative height of a particle with height $z\in[b,b+h]$ $$ \sigma = \frac{z-b}{h} $$ Note $0\leq\sigma\leq 1$
    Take material derivative, use quotient rule $\mD{\sigma} = \frac{1}{h}\mD{(z-b)} - \frac{z-b}{h^2}\mD{h}$ Note that $\mD{z}=w$ which varies linearly from $\mD{b}$ at bottom $z=b$ to $\mD{b+h}$ at top $z=b+h$ $$ \mD{z} = \mD{b} + \frac{z-b}{h}\mD{h} $$ Combine above two $$ \mD{\sigma} = 0 $$ So fluid parcel retains its relative position within a fluid column.
    So layers remain stacked, and are simply squeezed and streched.
Eckman layer Vertical friction is small (e.g. ocean at midlatitudes, $Ek=10^{-4}$) and can usually be omitted. But retaining second order derivative allows us to apply boundary conditions at surface. Otherwise, accept slipping at boundary.
So consider
  • Homogeneous fluid
  • Geostrophic flow away from boundary, $z\to\infty$
  • Flat boundary at $z=0$ (generalized to non-flat in notes)
  • Boundary layer with friction bringing velocity to zero
  • $f$-plane, w.l.o.g. let $f>0$ i.e. northern hemisphere
Figure: interior flow slowed to zero in boundary layer
The equations for flow ($\vec{u}(t,\vec{x}), p(t,x,y)$) are geostrophic plus non-negligible friction due to boundary, $$ \left\{ \begin{array}{ll} \text{mass} & \div \vec{u} = 0\\ \text{momentum} & -fv = -\frac{1}{\rho_0}\pd{p}{x}+\nu\pd{^2u}{z^2}\\ & fu = -\frac{1}{\rho_0}\pd{p}{y}+\nu\pd{^2v}{z^2}\\ & 0 = -\frac{1}{\rho_0} \pd{p}{z} \end{array} \right\} $$ The equations for interior flow ($\bar{\vec{u}}(t,x,y),\bar{p}(t,x,y)$) are simply geostrophic. $$ \left\{ \begin{array}{ll} \text{mass} & \div \bar{\vec{u}} = 0\\ \text{momentum} & -f\bar{v} = -\frac{1}{\rho_0}\pd{\bar{p}}{x}\\ & f\bar{u} = -\frac{1}{\rho_0}\pd{\bar{p}}{y}\\ & 0 = -\frac{1}{\rho_0} \pd{\bar{p}}{z} \end{array} \right\} $$ Note: $p=\bar{p}$ since both $z$ momentum equations say pressure indep of depth i.e. const along each vertical line.
Subtract interior system from boundary system to isolate effects of boundary. $$ \left\{ \begin{array}{ll} & -f(v-\bar{v}) = \nu\pd{^2u}{z^2}\\ & f(u-\bar{u}) = \nu\pd{^2v}{z^2} \end{array} \right\} $$ And the boundary conditions can be for an unmoving boundary (e.g. ground) $$ \left.\matrix{u\\v}\right|_{z=0} = \matrix{0\\0}\\ \left.\matrix{u\\v}\right|_{z\to\infty} = \matrix{\bar{u}\\\bar{v}} $$ or for a sheared boundary (e.g. ocean with surface wind shear) $$ \left.\pd{}{z}\matrix{u\\v}\right|_{z=0} = \frac{1}{\rho_0 \nu}\matrix{\tau^x \\ \tau^y}\\ \left.\matrix{u\\v}\right|_{z\to\infty} = \matrix{\bar{u}\\\bar{v}} $$
For unmoving surface, $$ \left\{ \begin{array}{ll} \text{DE} & -f(v-\bar{v}) = \nu\pd{^2u}{z^2}\\ & f(u-\bar{u}) = \nu\pd{^2v}{z^2}\\ \text{BC} & \left.\matrix{u\\v}\right|_{z=0} = \matrix{0\\0}\\ & \left.\matrix{u\\v}\right|_{z\to\infty} = \matrix{\bar{u}\\\bar{v}} \end{array} \right\} $$ For sheared surface, $$ \left\{ \begin{array}{ll} \text{DE} & -f(v-\bar{v}) = \nu\pd{^2u}{z^2}\\ & f(u-\bar{u}) = \nu\pd{^2v}{z^2}\\ \text{BC} & \left.\pd{}{z}\matrix{u\\v}\right|_{z=0} = \frac{1}{\rho_0 \nu}\matrix{\tau^x \\ \tau^y}\\ & \left.\matrix{u\\v}\right|_{z\to\infty} = \matrix{\bar{u}\\\bar{v}} \end{array} \right\} $$ Try plugging solution which looks like: $$ \matrix{u-\bar{u}\\v-\bar{v}}=\matrix{a\\b}e^{\lambda z} $$ Get characteristic equation $\nu^2\lambda^4+f^2=0$
Then $\lambda = \pm\frac{1\pm i}{d}$ where $d=\sqrt{\frac{2\nu}{f}}$ is called the Eckman depth.
Boundary condition at $z\to\infty$ cancels unbounded solution. Boundary condition at $z=0$ is used to find constants $a,b$.
So solution for unmoving surface is $$ \matrix{u\\v} = \bar{u} \matrix{1-e^{-\frac{z}{d}}\cos\frac{z}{d} \\ e^{-\frac{z}{d}}\sin\frac{z}{d} } + \bar{v} \matrix{ e^{-\frac{z}{d}}\sin\frac{z}{d} \\ 1-e^{-\frac{z}{d}}\cos\frac{z}{d} } $$ And the solution for sheared surface is $$ \matrix{u\\v} = \matrix{\bar{u}\\ \bar{v}} + \frac{\sqrt{2}}{\rho_0 fd}e^{\frac{z}{d}} \matrix{ \tau^x\cos\left( \frac{z}{d}-\frac{\pi}{4} \right) - \tau^y\sin\left( \frac{z}{d}-\frac{\pi}{4} \right) \\ \tau^x\sin\left( \frac{z}{d}-\frac{\pi}{4} \right) + \tau^y\cos\left( \frac{z}{d}-\frac{\pi}{4} \right) } $$
  • The Eckman depth $d$ is the distance over which boundary layer solutions approach interior flow $(\bar{u},\bar{v})$
  • Example: unmoving flat surface
    Net transport of fluid due to boundary layer is \begin{align} \matrix{U\\V} &= \matrix{ \int_0^\infty (u-\bar{u})\upd z \\ \int_0^\infty (v-\bar{v})\upd z }\\ &= \frac{d}{2}\matrix{ \bar{u}+\bar{v} \\ \bar{u}-\bar{v} } \end{align} Divergence of transport \begin{align} \pd{U}{x}+\pd{V}{y} &= \int_0^\infty \left( \pd{u}{x} - \pd{v}{y} \right)\upd z\\ &=-\frac{d}{2} \left( \pd{\bar{v}}{x} - \pd{\bar{u}}{y} \right)\\ &= -\frac{d}{2\rho_0 f}\lap \bar{p} \end{align} So boundary flow converges or diverges if interior has relative vorticity. This fluid is coming from or going to interior via vertical velocity. Since $f$-plane, $\pd{\bar{w}}{z}=0$ so $\bar{w}$ is const along vertical column. This is called Eckman pumping.
    Compute $\bar{w}$ by integrating $\div \vec{u}=0$ \begin{align} \int_0^\infty \pd{w}{z} \upd{z} &= -\int_0^\infty \pd{u}{x} +\pd{v}{y}\upd z\\ w|_{z=\infty}-w|_{z=0} &= \frac{d}{2}\left( \pd{\bar{v}}{x} - \pd{\bar{u}}{y} \right)\\ \bar{w} &= \frac{d}{2\rho_0 f} \lap \bar{p} \end{align} So greater vorticity of mean interior flow results in greater upwelling or downwelling. Also, greater at equator since $f$ small and $d$ big.
    A cyclone (CCW in Northern hemisphere) produces upward $\bar{w}$
    An anticyclone (CW in Northern hemisphere) produces downward $\bar{w}$.
    For $f\lt 0$, adjust $d=\sqrt{\frac{2\nu}{\abs{f}}}$, but same behavior for cyclones, anticyclones.
  • Example: Simplify above to $\bar{v}=0$ i.e. uniform (indep of position) flow $\bar{u}$
    Then $v$ is nonzero, and there is sideways flow (at a 45 degree angle near the boundary).
  • Example: Generalize to non-flat bottom $b(x,y)$.
    Let bottom slope $\left( \pd{b}{x},\pd{b}{y} \right)$ be small.
    The solution is the same, with $z$ replaced with $z-b$.
    The vertical thickness of the boundary layer is still $d=\sqrt{\frac{2\nu}{f}}$ measured perpendicular to bottom.
    Get $\pd{\bar{w}}{z}$ from $\div \bar{\vec{u}}=0$ \begin{align} \pd{\bar{w}}{z} &= -\pd{\bar{u}}{x} -\pd{\bar{v}}{y}\\ &= e^{\frac{b-x}{d}}\left\{ \left( \pd{\bar{v}}{x} - \pd{\bar{u}}{y} \right) \sin\frac{z-b}{d} \right.\\ &\quad + \frac{1}{d} \pd{b}{x}\left[ (\bar{u}-\bar{v})\cos\frac{z-b}{d} + (\bar{u}+\bar{v})\sin\frac{z-b}{d} \right]\\ &\quad \left. + \frac{1}{d} \pd{b}{y}\left[ (\bar{u}+\bar{v})\cos\frac{z-b}{d} - (\bar{u}-\bar{v})\sin\frac{z-b}{d} \right] \right\} \end{align} Integrate from $b$ to $\infty$ to find $\bar{w}$ \begin{align} \bar{w} &= \left( \bar{u}\pd{b}{x} + \bar{v}\pd{b}{y} \right) + \frac{d}{2}\left( \pd{\bar{v}}{x} - \pd{\bar{u}}{y} \right) \end{align} So there is the flat-bottom Eckman pumping term plus a term accounting for $\bar{w}$ which is not perpendicular to bottom, and thus not brought to zero.
  • Example: sheared surface.
    The second term of the solution is strictly due to shear and independent of interior flow. Also, it is inversely proportional to $d$, so it can be large even for small shear stress $\tau$.
    Net transport of fluid due to boundary layer is \begin{align} \matrix{U\\V} &= \matrix{ \int_0^\infty (u-\bar{u})\upd z \\ \int_0^\infty (v-\bar{v})\upd z }\\ &= -\frac{1}{\rho_0 f} \matrix{ \tau^y \\ \tau^x } \end{align} So net velocity is perpendicular to wind stress.
    Get $\pd{\bar{w}}{z}$ from $\div \vec{u}=0$ and integration \begin{align} \pd{w}{z} &= -\pd{u}{x} -\pd{v}{y}\\ \int_0^\infty \pd{w}{z} \upd z &= -\int_0^\infty \pd{u}{x} +\pd{v}{y} \upd z\\ \bar{w} &= -\frac{1}{\rho_0}\left[ \pd{}{x}\frac{\tau^y}{f} - \pd{}{y}\frac{\tau^x}{f} \right] \end{align} So with $f$ const, horizontal divergence and vertical velocity due only to wind-stress curl. This vertical velocity is called Eckman pumping.
    For $f>0$ (northern hemisphere), a clockwise wind (negative curl) generates ocean downwelling and opposite for upwelling.
Barotropic waves Take shallow water model and suppose
  • Flat bottom $b(x,y)=0$, free surface $f(t,x,y)$
  • $f > 0$ wlog
  • $Ro\lt\lt1$, $Ro_t\sim 1$ i.e. fluid parcel speed much slower than wave traveling distance $L$ in time $T$ i.e. $U\lt\lt \frac{L}{T}$
For momentum equations, replace $b+h$ with single variable $\eta(t,x,y)=b+h-H$ representing surface displacement relative to mean fluid thickness $H$.
For continuity equations, expand using product rule $$ \pd{\eta}{t}+\left( u\pd{\eta}{x}+v\pd{\eta}{y} \right) + H\left( \pd{u}{x}+v\pd{v}{y} \right) + \eta\left( \pd{u}{x}+v\pd{v}{y} \right) = 0 $$ The scale of the terms, from left to right, are $\frac{\Delta H}{T}, U\frac{\Delta H}{L}, H\frac{U}{L}, \Delta H\frac{U}{L}$. The second and fourth term are negligible since $\frac{U}{L}\lt\lt \frac{1}{T}$ so the first term dominates them.
Independent variables are $x,y,t$ and dependent variables are $u,v,\eta$. Note: flow is independent of $z$.
$$ \left\{ \begin{array}{ll} \text{mass} & \pd{\eta}{t} + H\left( \pd{u}{x}+\pd{v}{y} \right) = 0\\ \text{momentum} & \pd{u}{t}-fv = -g\pd{\eta}{x}\\ & \pd{v}{t}+fu = -g\pd{\eta}{y} \end{array} \right\} $$ Barotropic wave equations linear so susceptible to solution methods.
Waves are small amplitude, since mass equation must balance so $\Delta H\lt\lt H$.

A brief review of 2$d$ plane waves with straight crest lines:

  • Let $a(x,y,t)=A\cos(lx+my-\omega t)$ be a wave. More complicated waves can be constructed from a superposition of waves like this.
  • Wavenumbers are $l$, $m$, and $k=\sqrt{l^2+m^2}$
    Wavelength in $x$ is $\frac{2\pi}{l}$, in $y$ is $\frac{2\pi}{m}$ and in propagation direction is $\lambda=\frac{2\pi}{k}$.
  • Crest, trough, and phase lines are wave-height contours in the $x,y$ plane. Wavenumber vector $\vec{k}=(l,m)$ is perpendicular to these lines.
  • Frequency $\omega$ is analogous to wavenumber
    Period $T=\frac{2\pi}{\omega}$ is analogous to wavelength
  • Wave (``phase'') speed is $c=\frac{\omega}{k}$
    Dispersive waves if $c$ is a function of $k$, i.e. different waves have different speeds
    Nondispersive waves if $\omega\propto k$
    Can plot $\omega$ as a function of wavenumbers. Lines through origin mean nondispersive waves.
  • Many superposed waves can each travel at different speeds.
    Group velocity $c_g = \grad_k \omega$
The following are examples of barotropic waves:
  • Geostrophic waves degenerate i.e. do not change with time.
  • Kelvin waves are non-dispersive, along a boundary or equator.
  • Poincare (Inertio-gravity) waves are dispersive, frequency $\omega\geq f$, behave as inertial oscillations and/or gravity waves.
    • Gravity waves are basic waves which are not affected by Coriolis force.
    • Inertial oscillations are waves whose frequency is a multiple of $f$ and result in a fluid parcel moving in a circle.
  • Rossby (Planetary) waves are slow
Figure: Dispersion relation $w(k)$ with geostrophic, kelvin poincare (inertial oscillations, gravity), and Rossby
Surface Kelvin wave Take Barotropic wave.
  • Consider domain of flat bottom, free surface, vertical wall (along $x=0$)
    Figure: flat bottom, free surface, vertical wall (along $x=0$)
  • $u|_{x=0}=0$ since wall.
  • velocity $v$ along wall is possible since no viscosity
Ansatz: $u=0$ everywhere.
$$ \left\{ \begin{array}{ll} \text{mass} & \pd{\eta}{t} + H\pd{v}{y} = 0\\ \text{momentum} & -fv = -g\pd{\eta}{x}\\ & \pd{v}{t} = -g\pd{\eta}{y} \end{array} \right\} $$ Cancel $\eta$ with $y$ momentum equation and mass equation to obtain $$ \pd{^2 v}{t^2} = c^2 \pd{^2 v}{y^2} $$ where $c=\sqrt{gH}$ is the wave speed.
The 1$d$ wave equation is on an infinite domain along the coast.
$$ \left\{ \begin{array}{ll} & u=0\\ & \pd{^2 v}{t^2} = c^2 \pd{^2 v}{y^2}\\ & \pd{\eta}{t} + H\pd{v}{y} = 0 \end{array} \right\} $$ Note: one can also use Fourier methods on the original system with $u=0$, but it is more economical to solve the modified system. Fourier methods will be used next for Poincare waves.
General solution to 1d wave equation: $$ v=V_1(x,y+ct)+V_2(x,y-ct) $$ where $V_1$ and $V_2$ are to be solved for. Plug into mass (or $y$-momentum) equation to find $\eta$ $$ \eta = -\sqrt{\frac{H}{g}}(V_1-V_2) $$ where the additive constant of integration can is subsumed into a properly redefined $H$.
Plug into $x$-momentum equation to find $V_1,V_2$ $$ \pd{V_1}{x} = -\frac{f}{\sqrt{gH}}V_1 ,\qquad \pd{V_2}{x} = \frac{f}{\sqrt{gH}}V_2 $$ Which superpose to make $$ v=V_1(y+ct)e^{-\frac{x}{R}}+V_2(y-ct)e^{\frac{x}{R}} $$ where $R=\frac{\sqrt{gH}}{f}=\frac{c}{f}$ is called the Rossby radius of deformation, which is the distance covered by the wave with speed $c$ in time $\frac{1}{f}$.
The second term grows exponentially away from shore, so accept the first. So the solution is \begin{align} u &= 0\\ v &= V_1(y+ct)e^{-\frac{x}{R}}\\ \eta &= -\sqrt{\frac{H}{g}}V_1(y+ct)e^{-\frac{x}{R}} \end{align} Where $V_1$ is the same shape as the initial conditions.
  • wave properties:
    speed $c=\sqrt{gH}$ which is constant
    nondispersive since wave speed $c$ independent of wave number.
  • wave exponentially decays away from boundary, so it is ``trapped'' to the boundary, and $R$ is a measure of its trapping distance.
  • Direction of wave:
    The wave travels along the boundary at speed $c$
    If $f>0$ (Northern hemisphere), the coast is on the right.
    If $f\lt0$ (Southern hemisphere), the coast is on the left.
    If $\eta>0$ (Upwelling wave), wave travels with the current along coast
    If $\eta\lt 0$ (Downwelling wave), wave travels against the current
  • Example: $f\to 0$
    Then trapping distance $R\to\infty$
    So this is gravity wave with crests and troughs perpendicular to coast.
  • Note: Internal Kelvin waves are different, require stratification
  • Surface Kelvin waves are generated by ocean tides and coastal wind.
    e.g. Storm off Northeast coast of Britain generates surface Kelvin wave of width 40m which follows North sea counterclockwise from France to Norway (2200km) in 31 hours. In English Channel, France has higher waves than England.
Poincare (inertia-gravity) wave Take barotropic wave equations
Flat bottom, free top, no side boundaries.
Assume $f$-plane.
$$ \left\{ \begin{array}{ll} \text{mass} & \pd{\eta}{t} + H\left( \pd{u}{x}+\pd{v}{y} \right) = 0\\ \text{momentum} & \pd{u}{t}-fv = -g\pd{\eta}{x}\\ & \pd{v}{t}+fu = -g\pd{\eta}{y} \end{array} \right\} $$ Take fourier transform of system i.e. integrate each term with kernel $e^{i(lx+my-\omega t)}$ where $l,m$ are wavenumbers and $\omega$ is frequency. Transformed system is: $$ -i\omega \mathscr{F}[u] -f \mathscr{F}[v] = -igl \mathscr{F}[\eta]\\ -i\omega \mathscr{F}[v] +f \mathscr{F}[u] = -igm \mathscr{F}[\eta]\\ -i\omega \mathscr{F}[\eta]+ H( il\mathscr{F}[u] + im\mathscr{F}[v] ) = 0 $$ Have a 3 by 3 algebraic system $$ \matrix{ -i\omega & -f & igl\\ f & -i\omega & igm\\ iHl & iHm & -i\omega } \matrix{\mathscr{F}[u] \\ \mathscr{F}[v] \\ \mathscr{F}[\eta]} = \matrix{0\\0\\0} $$ Need zero determinant for nontrivial (i.e. nonzero) solution. $$ \omega[\omega^2 - f^2 - gHk^2 ]=0 $$ where $k=\sqrt{l^2+m^2}$ is the wave-number magnitude.
This is called the dispersion relation. Can plot to see frequency verses wave number magnitude.
So the solution $(\mathscr{F}[u], \mathscr{F}[v], \mathscr{F}[\eta])$ is restricted to that subset of the phase space. Get back solutions $u,v,\nu$ with inverse fourier transform over that subset of the phase space. \begin{align} u(t,x,y) &= \frac{1}{2\pi} \int\int\int \mathscr{F}[u](\omega,l,m) e^{-i(lx+my-\omega t)}\upd \omega \upd l \upd m\\ v(t,x,y) &= \frac{1}{2\pi} \int\int\int \mathscr{F}[v](\omega,l,m) e^{-i(lx+my-\omega t)}\upd \omega \upd l \upd m\\ \eta(t,x,y) &= \frac{1}{2\pi} \int\int\int \mathscr{F}[\eta](\omega,l,m) e^{-i(lx+my-\omega t)}\upd \omega \upd l \upd m \end{align}
The roots to dispersion relation are $\omega=0,\pm\sqrt{f^2 + ghk^2}$
$\omega=0$ means indep of time, equations reduce to geostrophic with surface is an arrested wave
$\omega = \pm \sqrt{f^2 + gHk^2}$ results in travelling waves whose frequency $\omega \geq f$ i.e. oscillation period is at least as fast as earth rotation. These are called Poincare waves, and have many possibilities.
  • case $f=0$: this degenerate case results in frequency $\omega=k\sqrt{gH}$ and wave speed $c=\frac{\omega}{k}=\sqrt{gH}$. So waves are basic gravity waves.
  • case large wavenumbers ($k^2>>\frac{f^2}{gH}$): wavelengths much shorter than deformation radius, so waves are too short and fast to feel rotation of earth, so similar to $f=0$
  • case small wavenumbers ($k^2\lt\lt \frac{f^2}{gH}$): wavelengths much longer than deformation radius, so rotation effects dominate yielding $\omega \sim f$. Flow paterns are laterally uniform, and fluid particles move in unison, each in a circular inertial oscillation.
  • case intermediate wave numbers: waves exhibit behavior of gravity waves and inertial oscillations.
Waves are dispersive since phase speed $c=\frac{\omega}{k}$ depends on wavenumber, and different wavelengths travel at different speeds.
Rossby (planetary) wave, and a note on topographical waves Take Barotropic wave equations and let.
  • $Ro_T$ be small but not negligible. If negligible, then geostrophic flow.
  • Let beta plane i.e. coriolis parameter becomes $f+\beta y$ where $f$ and $\beta$ are constants. This accounts for phenomena which spans latitude e.g. cyclones, anticyclones, gulf stream. This is valid for only small deviations in latitude, i.e. $\frac{\beta L}{f}\lt\lt 1$ where $L$ is the distance in latitude. This ratio is called tha planetary number.
Replace coriolis parameter in the barotropic wave equations to obtain the Rossby wave equations.
The equations are nearly geostrophic, since the $Ro_T$ and $\beta$ terms are small.
$$ \left\{ \begin{array}{ll} \text{mass} & \pd{\eta}{t} + H\left( \pd{u}{x}+\pd{v}{y} \right) = 0\\ \text{momentum} & \pd{u}{t}-(f+\beta y)v = -g\pd{\eta}{x}\\ & \pd{v}{t}+(f+\beta y)u = -g\pd{\eta}{y} \end{array} \right\} $$ In scale, the large terms are $f$, $g$, and $H$, and represent geostrophic motion. The small scale terms are $\beta$ and time derivatives, and represent perturbations from geostrophic motion. The approximate geostrophic solutions are $$ u\approx -\frac{g}{f}\pd{\eta}{y}\\ v\approx \frac{g}{f}\pd{\eta}{x}\\ $$ Plug these approximations to the small terms in the $x$ and $y$ momentum equations. $$ -\frac{g}{f}\pd{^2\eta}{y \partial t}-fv-\frac{\beta g y}{f}\pd{\eta}{x} = -g\pd{\eta}{x}\\ \frac{g}{f}\pd{^2\eta}{x \partial t}+fu+\frac{\beta g y}{f}\pd{\eta}{y} = -g\pd{\eta}{y}\\ $$ Solve with simple algebra $$ u=-\frac{g}{f}\pd{\eta}{x}-\frac{g}{f^2}\pd{^2\eta}{x \partial t}+\frac{\beta g y}{f^2}\pd{\eta}{x}\\ v=\frac{g}{f}\pd{\eta}{x}-\frac{g}{f^2}\pd{^2\eta}{y \partial t}-\frac{\beta g y}{f^2}\pd{\eta}{y}\\ $$ The first terms are the geostrophic velocity, and the others are corrections in perturbation series of the velocity field. These terms are cooled ageostrophic.
Plug into the continuity equation. $$ \pd{\eta}{t}-R^2\pd{}{t}\lap\eta - \beta R^2\pd{\eta}{x}=0 $$ where $R=\frac{\sqrt{gH}}{f}$ is the Rossby radius of deformation.
Take Fourier transform i.e. integrate with kernel $e^{i(lx+my-\omega t)}$ $$ -i\omega -R^2 (-i\omega)(i^2l^2+i^2m^2)-\beta R^2 il]\mathscr{F}[\eta] = 0 $$ Solve for $\omega$ and the recurrence relation is $$ \omega = -\frac{\beta R^2 l}{1+R^2(l^2+m^2)} $$ Get back $\eta$ by taking the inverse transform over allowable values of $\omega$, l, and m (which are contours?). $$ \eta = \iiint \mathscr{F}[\eta](\omega,l,m) e^{-i(lx+my-\omega t)}\upd \omega \upd l \upd m $$
  • Rossby waves are slow compared to Kelvin and Poincare waves.
  • Rossby waves are dispersive.
  • Behavior is like goestrophic, since $\beta$ is small and $\beta=0$ corresponds to zero frequency and geostrophic flow.
  • Frequency is small.
    This is proved by scale analysis. Let $L$ be the scale of a given wavelength ($\frac{2\pi}{l}$ or $\frac{2\pi}{m}$). Then either waves are shorter $L\lesssim R$ or longer $L\gtrsim R$. In the first case, $\omega \sim \beta L$ and in the second, $\omega\sim \frac{\beta R^2}{L}\lesssim \beta L$.
    Also $\omega\lt\lt f$ since $\beta L\lt\lt f$.
  • The wave speed is frequency divided by wave number.
    Waves can't move east since $c=\frac{\omega}{l}=-\frac{\beta R^2}{1+R^2(l^2+m^2)}$ which is negative so the direction is always west.
    Waves can move north or south, depending on the sign of $l$.
    The speed is slow. The maximum speed occurs at long wavelengths ($\gt\gt R$), which results in the $R^2(l^2+m^2)$ being small, so wave speed $c$ reaches its maxium, $-\beta R^2$.
    The maximum frequency can be found by setting frequency constant and obtaining circles in the $l,m$ plane $$ \left( l+\frac{\beta}{2\omega} \right)^2 + m^2 =\frac{\beta^2}{4\omega^2} - \frac{1}{R^2} $$ The right hand side must be positive, so $\beta^2\gt \frac{4\omega^2}{R}$ so $\omega<\frac{\beta R}{2}$, the maximum frequency.
    Group velocity is $\grad_k\omega$. It can be visualized by plotting the circles of constant frequency in the $l,m$ plane, where the gradient is perpendicular to these circles.
  • Topographic waves are barotropic on $f$-plane with weak bottom slope $\alpha$. They are solved similarly and behave analogously to Planetary waves, with $\alpha$ playing a similar role to $\beta$
    There is an elegant vorticity viewpoint of planetary (Rossby) and topographic waves








Stratification Effects

In the previous section, only homogeneous fluids in a rotating basis were considered. In this section, nonhomogeneous fluids in nonrotation basis are considered. Gravitational force (per volume) $\rho g$ produces a tendency toward stratification (separation into strata (``layers'')) where denser fluid underlies lighter fluid. For example, Earth's atmosphere consists of layers troposphere, stratosphere, mesosphere, thermosphere, and exosphere.

Table * exhibits effects of stratification including (i) a tendency toward horizontal rigidity (contrasting vertical rigidity imparted by rotation), (ii) internal waves, (iii) horizontal stratification such as fronts and heat flux from lower latitudes, and (iv) discrete layers of approximately constant density.

A scale for stratification will be derived. The Froude number, $Fr=\frac{U}{NH}$, is the stratification analogue to the Rossby number $Ro=\frac{U}{\Omega L}$.

Table 1: .
Name Heuristic Derivation Equation(s) Solution(s) Properties and Notes
Stratification Effects
Hydrostatic equilibrium Consider a nonhomogeneous fluid with zero velocity, no horizontal forces.
The resulting equation is $\pd{p}{z} = -\rho g$. Define $p_0(z)$ and $\rho_0(z)$ to satisfy this situation.
$$\pd{p_0(z)}{z} = -\rho_0(z) g$$ The most basic example of stratification. The fluid has horizontal homogeneity and purely vertical density stratification.
Perturbed parcel Consider fluid in hydrostatic equilibrium
Perturb a parcel vertically from position $z$ to $z+h$ where $h$ is small.
note: if compressible fluid, use potential density instead of density
Employing Newton's second law, acceleration equals the bouyancy force \begin{align} \rho_{\text{parcel}}\pd{^2h}{t^2} &= g(\rho_{\text{ambient}}-\rho_{\text{parcel}})\\ \pd{^2h}{t^2} &= N^2 h \end{align} where $N^2 = -\frac{g}{\rho_{\text{parcel}}}\pd{\rho_{\text{parcel}}}{z}$
$$ \pd{^2h}{t^2} = N^2 h $$ where $N^2 = -\frac{g}{\rho}\pd{\rho}{z}$ case $N^2\lt 0$: Then
$$ h=a\sin(N t)+b\cos(N t) $$ where initial perterbed conditions imply $a=0$ since parcel initially at rest and $b$ is the initial $h$.
case $N^2> 0$: Then
$$ h=ae^{N t} $$ where $a$ is initial $h$.
case $N^2\lt 0$ corresponds to $\pd{\rho}{z}\lt 0$. So density decreases with height. This is considered stable. The parcel will be bouyed toward equilibrium, overshoot, and continue returning and overshooting, resulting in oscillations with frequency $N$ known as Brunt-Vaisala (``bouyancy'', ``stratification'') frequency.
case $N^2> 0$ corresponds to $\pd{\rho}{z}> 0$. So density increases with height. This is considered unstable, as higher density fluids tend to underly light density. So the parcel will exponentially run away from equilibrium.
Internal Waves Consider:
  • stratified fluid
  • infinite domain in all directions
  • no dissipation (ie no friction)
  • fluid motions and wave amplitudes are small (allowing linearization of equations)
  • reinstate vertical acceleration $\pd{w}{t}$ so no hydrostatic balance in $z$ momentum equation
Let fluid density be sum of const reference $\rho_0$, deviation of equilibrium due to stratification $\bar{\rho}(z)$, and fluctuations due to waves $\rho'(\vec{x},t)$. $$ \rho(\vec{x},t) = \rho_0+\bar{\rho}(z) + \rho'(\vec{x},t) $$ Let $\abs{\bar{\rho}}\lt\lt \rho_0$ which was assumed for Boussinesq eqn.
Let $\abs{\rho'}\lt\lt \abs{\bar{\rho}}$ so that the equations can be linearized.
Let pressure $p=p_0+\bar{p}+p'$ similarly.
The equation $$ \pd{\rho'}{t}+w\pd{\bar{\rho}}{z} = 0 $$ has term $\pd{\bar{\rho}}{z}$ which equals $-\frac{\rho_0}{g}N^2$.
Assume $N^2$ is consant. Then $\bar{\rho}$ is linear.
$$ \left\{ \begin{array}{ll} \text{momentum} &\pd{u}{t} = -\frac{1}{\rho_0}\pd{p'}{x}\\ &\pd{v}{t} = -\frac{1}{\rho_0}\pd{p'}{y}\\ &\pd{w}{t} = -\frac{1}{\rho_0}\pd{p'}{z} - \frac{1}{\rho_0}g\rho'\\ \text{density} &\pd{u}{x} + \pd{v}{y} + \pd{w}{z} = 0\\ &\pd{\rho'}{t}+w\der{\bar{\rho}}{z} = 0 \end{array} \right\} $$ Note that all coefficients are constant.
Take fourier transform with kernel $e^{i(lx+my+nz-\omega t)}$ to obtain $5\times5$ system of algebraic equations. $$ \matrix{ i\omega & & & -i\frac{l}{\rho_0} & \\ & i\omega & & -i\frac{m}{\rho_0} & \\ & & i\omega & -i\frac{n}{\rho_0} & -\frac{g}{\rho_0} \\ il & im & in & & \\ & & \der{\bar{\rho}}{z} & & i\omega } \matrix{ \mathscr{F}[u] \\ \mathscr{F}[v] \\ \mathscr{F}[w] \\ \mathscr{F}[p'] \\ \mathscr{F}[\rho'] } = \matrix{ 0\\0\\0\\0\\0 } $$ Need zero determinant for nontrivial (i.e. nonzero) solution. Then the dispersion relation is $$ \omega^2 = N^2 \frac{l^2+m^2}{l^2+m^2+n^2} $$
  • $\omega\leq N$ since $\frac{l^2+m^2}{l^2+m^2+n^2}\leq 1$. So only small wave frequencies are supported.
  • If $\pd{w}{t}$ is neglected, then the denominator becomes $n^2$. So $\pd{w}{t}$ can be neglected for phenomena where $l^2+m^2\lt\lt n^2$. Then $\omega\lt\lt N$.
  • If $l,n$ get large i.e. short waves, then $\omega$ grows until it saturates at $N$.
  • If we agitate the fluid at $\omega>N$, then there are no waves, just a local patch of turbulence.
  • Consider \begin{align} \vec{k} &= (l,m,n)^T\\ k &= \sqrt{l^2+m^2+n^2}\\ \theta &= \text{angle from }\vec{k} \text{ to the horizontal}\\ \phi &= \text{angle from }\vec{k} \text{ to the }x,z \text{ plane}\\ l &= k\cos\phi \cos\phi\\ m &= k\cos\phi\sin\phi\\ n &= k\sin\theta \end{align} Then $\omega=\pm N\cos\theta$. So $\omega$ depends on $N$ and $\theta$, not $\vec{k}$. So waves travel upward or downward along $\vec{k}$ since $\pm$?
    e.g. $\omega$ imposed (e.g. by tide), then waves of all wavelengths propogate at angle $\theta$
    e.g. $\omega\to 0$ then $\theta\to 90$ degrees i.e. phase propagation is vertical








Combined Rotation and Stratification Effects

$Fr$ vs $Ro$
Table 1: .
Name Heuristic Derivation Equation(s) Solution(s) Properties and Notes
Layered models Consider general geo fluid equation (ie stratified and rotating) in isobaric coordinates.
Assume the stratification results in layers each of constant unchanging density i.e. discretize vertical coordinate
Then each layer has its own horizontal motion which is coupled with its neighboring layers.
For each layer $k=1,...,m$ of $m$ layers, prescribe density $\rho_k$. Let $\rho_0=\rho_1$ be the reference density.
The equations of motion are modified as follows.
  • The relationship $h=-\Delta\rho \pd{\rho}{z}$ becomes $$ z_k=z_{k-1}-h_k=z_{k+1}+h_{k+1} $$ where $h_k(x,y,t)$ as the thickness of layer $k$ and $z_k(x,y,t)$ as position from mean surface $z=0$ to bottom of layer $k$.
    Note: the bottom $z_m(x,y)$ is prescribed and unchanging with time.
  • The hydrostatic relation $\pd{P}{\rho} = gz$ becomes $$ P_k=P_{k-1}+\rho_0 g_k z_{k-1} = P_{k+1}-\rho_0 g_{k+1} z_k $$ where $P_k$ is the Montgomery pressure and $g_k=\frac{\rho_{k-1}-\rho_k}{\rho_0}g$ is the reduced gravity.
    Let $P_0=p_{\text{surface}}-\rho_0 g z_0$, but the $p_{\text{surface}}$ is usually small so we will neglect.
  • The other three equations are unchanged, but use $u_k,v_k$ as the horizontal velocities.
$$ \left\{ \begin{array}{ll} &\mD{u_k} - fv_k = -\frac{1}{\rho_0}\pd{P_k}{x}\\ &\mD{v_k} + fu_k = -\frac{1}{\rho_0}\pd{P_k}{y}\\ &P_k=P_{k-1}+\rho_0 g_k z_{k-1} = P_{k+1}-\rho_0 g_{k+1} z_k\\ &\pd{h_k}{t} + \pd{h_ku_k}{x} + \pd{h_kv_k}{y} = 0\\ &h_k=z_k-1-z_k = z_k-z_{k+1} \end{array} \right\} $$ where $g_k=\frac{\rho_{k-1}-\rho_k}{\rho_0}g$
  • This discretized method allow each layer to have two dimensional flow.
  • Conservation of potential vorticity $q_k=\frac{f+\zeta_k}{h_k}$ for each level $k$.
  • The external radius of deformation $R_{\text{ext}}=\frac{NH}{\Omega}\sim\frac{\sqrt{gH}}{f}$ has an analogue for each layer.
    The internal radius of deformation is $R_{\text{int},k}=\frac{N_kH}{\Omega}\sim\frac{\sqrt{g_kH_k}}{f}$
    Note that $R_{\text{int},k}\lt\lt R_{\text{ext}}$ since $\frac{\Delta \rho}{\rho_0}$ is small
Thermal wind Consider stratification with both vertical and horizontal components e.g. cold air mass in a wedge shape between ground and warm air mass
Align with the $x,z$ axes.
Assume steady geostrophic hydrostatic flow $$ -fv = -\frac{1}{\rho_0}\pd{p}{x}\\ \pd{p}{z} = -\rho g $$ Combine equations by taking derivatives of pressure terms. $$ \pd{v}{z} = -\frac{g}{\rho_0 f}\pd{\rho}{y} $$ So vertical shear of $v$ and horizontal density gradient persist in steady state.
Similar for other horizontal direction $$ \pd{u}{z} = \frac{g}{\rho_0 f}\pd{\rho}{x} $$
$$ \pd{v}{z} = -\frac{g}{\rho_0 f}\pd{\rho}{y} \pd{u}{z} = \frac{g}{\rho_0 f}\pd{\rho}{x} $$ In vector form, $$ \pd{}{z}\matrix{u\\v} = -\frac{g}{\rho_0 f}\grad^\perp \rho $$ Coriolis force allows persistence of horizontal density gradient, no energy needed.
Quasi-Geostrophic