Course: Dynamics II

Planetary and relative vorticity

\[ \mbox{Absolute Vorticity }\equiv\left(\zeta+f\right) \]

\[ \frac{Du}{Dt}-f\;v = -\frac{1}{\rho}\frac{\partial p}{\partial x} \] \[ \frac{Dv}{Dt}+f\;u = -\frac{1}{\rho}\frac{\partial p}{\partial y} \]

\[ \mbox{subtract } \partial/\partial y \mbox{ of (u-equation) from } \partial /\partial x \mbox{ of (v-equation) } \]

Use \[ \frac{D}{Dt} f = v \, \partial_y f: \]

to obtain

\[ \underline{ \frac{D}{Dt}\left(\zeta+f\right) + \left(\zeta + f\right)\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=0 } \quad \]

Examples for Vorticity: Ocean/Atmosphere with depth h(x,y)

Constant density \(\rho\)

\[ \mbox{Continuity equation } \quad \partial_x \left( u h \rho \right) + \partial_y \left( v h \rho \right) \quad = \quad 0 \]

\[ \quad \quad \partial_x \left( u h \right) + \partial_y \left( v h \right) \quad = \quad 0 \]

\[ \frac{D}{Dt} h = \partial_t h + u \partial_x h + v \partial_y \quad = \quad \partial_x \left( u h \right) + \partial_y \left( v h \right) - h \left( \partial_x u + \partial_y v \right) \quad = \quad - h \left( \partial_x u + \partial_y v \right) \]

Therefore,

\[ \frac{D}{Dt}\left(\zeta +f\right)-\frac{\left(\zeta+f\right)}{h}\frac{Dh}{Dt}=0 \]

\[ \frac{1}{h} \frac{D}{Dt}\left(\zeta+f\right) - \left(\zeta + f\right) \frac{D_t h}{h^2} =0 \]

\[ \underline{ \frac{D}{Dt}\left( \frac{\zeta+f}{h}\right) = 0 } \quad \]

Potential vorticity is conserved along a fluid trajectory.

Example: Flow Tends to be Zonal

In the ocean: \[f >> \zeta \] and thus \[ \frac{D}{Dt}\left( \frac{f}{h}\right) = 0 \]

flow in an ocean of constant depth be zonal.

Contours \(f/h\): combination of latitude circles and bottom topography

Over small horizontal distances, h tends to dominate,

over longer distances, the latitude-variation of f dominates.

f/h

Potential vorticity: Example 2

Ocean/Atmosphere with depth h(x,y)

\[ \frac{D}{Dt}\left( \frac{\zeta+f}{h}\right) = 0 \quad \]

 

As the depth decreases, \(\zeta + f\) must also decrease, which requires that f decrease, and the flow is turned toward the equator.

If the change in depth is sufficiently large, no change in latitude will be sufficient to conserve potential vorticity, and the flow will be unable to cross the ridge.

Dietrich et al. (1980)

 

Taylor-Proudman Theorem

\[ \mbox{Assume constant density } \rho_0 \mbox{ on a plane with constant rotation } f=f_0 \neq 0 \]

Taylor’s lab experiments: homogeneous fluid tends to move in vertical columns

icon

Experiment on youtube

Then,

\[ \frac{\partial v}{\partial z}=\frac{\partial u}{\partial z}=\frac{\partial v}{\partial z}=0 \]

Flow is two-dimensional and does not vary in the vertical direction.

Theorem applies to slowly varying flows.

Physical origin: stiffness endowed to the fluid by rapid rotation of the Earth.

Taylor-Proudman Theorem

\[ \mbox{Assume constant density } \rho_0 \mbox{ on a plane with constant rotation } f=f_0 \] \[ -f\;v = - \frac{1}{\rho_0} \frac{\partial p}{\partial x} \] \[ f\;u= -\frac{1}{\rho_0} \frac{\partial p}{\partial y} \] \[ g= -\frac{1}{\rho_0}\frac{\partial p}{\partial z} \] \[ \mbox{and the continuity equation is:} \quad 0=\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}+\frac{\partial w}{\partial z} \]

Taking z-derivative of the first \[ -f_0\frac{\partial v}{\partial z}=-\frac{1}{\rho_0}\frac{\partial}{\partial z}\left(\frac{\partial p}{\partial x}\right)=\frac{\partial}{\partial x}\left(-\frac{1}{\rho_0}\frac{\partial p}{\partial z}\right)=\frac{\partial g}{\partial x}=0 \] \[ \mbox{ Therefore for } f_0 \neq 0 \quad \frac{\partial v}{\partial z}=0, \frac{\partial u}{\partial z}=\frac{\partial v}{\partial z}=0 \] Flow is two-dimensional and does not vary in the vertical direction.

Vertical velocity & north-south currents

Taylor-Proudman theorem: flow cannot expand or contract in the vertical

Assumption that \(f=f_0\) can not be appropiate

\[ \quad \quad \quad \quad \frac{D}{Dt}\left(\zeta+f\right) + \left(\zeta + f\right)\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)=0 \quad \]

\[ \mbox{poor man's vorticity:} \quad \boxed{\beta\;v \quad + \quad f\left(\frac{\partial u}{\partial x}+\frac{\partial v}{\partial y}\right)= 0 } \quad \quad \quad \quad \]

Using the continuity equation, we obtain \[ \beta\;v = f \frac{\partial w_g}{\partial z} \] in the ocean’s interior, geostrophic flow.

Variation of Coriolis force with latitude allows vertical velocity gradients in the interior of the ocean, and the vertical velocity leads to north-south currents.

What drives the ocean currents?

Friction: transfer of momentum from atmosphere to oceanic Ekman layer

Vorticity dynamics for the ocean and include the wind stress term

\[ D_t u - f v = - \frac{1}{\rho} \frac{\partial p}{\partial x} + \frac{1}{\rho} \partial_z \tau_{xz} \] \[ D_t v + f u = - \frac{1}{\rho} \frac{\partial p}{\partial y} + \frac{1}{\rho} \partial_z \tau_{yz} \]

\[ \frac{D}{Dt} \left( {\zeta+f}\right) - \frac{\left(\zeta+f \right)}{h} \frac{D}{Dt} h \, = \, \frac{1}{\rho} \underbrace{\left( \frac{\partial}{\partial x} \, \partial_z \tau_{yz} - \frac{\partial}{\partial y}\, \partial_z \tau_{xz} \right)}_{curl \, \partial_z \tau} \quad . \]

\[ \frac{D}{Dt} \left( \frac{\zeta+f}{h}\right) = \frac{1}{\rho \, h} \, \mbox{curl} \, \partial_z \tau \, \]

Sverdrup transport

\[ \beta v = \frac{1}{\rho } \, \mbox{curl} \, \partial_z \tau \, \]

\[ \int_{-H}^0 dz \, \beta v = \frac{1}{\rho } \, \int_{-H}^0 dz \, \mbox{curl} \, \partial_z \tau \, = \frac{1}{\rho } \, \mbox{curl} \, \tau \, \]

\[ V = \frac{1}{\rho \beta} \, \left( \frac{\partial \tau_{yz} }{\partial x} \, - \frac{\partial \tau_{xz}}{\partial y}\, \right) = \frac{1}{\rho \beta} \, \, \operatorname{curl} \, \tau \]

Sverdrup

applied globally using the wind stress from Hellerman and Rosenstein (1983). Contour interval is \(10\) Sverdrups (Tomczak and Godfrey, 1994).

\[ V = \frac{1}{\rho \beta} \, \left( \frac{\partial \tau_{yz} }{\partial x} \, - \frac{\partial \tau_{xz}}{\partial y}\, \right) = \frac{1}{\rho \beta} \, \, \operatorname{curl} \, \tau \]

Ekman Pumping in a thin Ekman layer

The vertical velocity at the surface is zero and denote \(w_E\) as the Ekman vertical velocity the bottom of the Ekman layer. \[ - \int_{-E}^0 \frac{\partial w}{\partial z} dz = w_E = \frac{\partial}{\partial x} U_E + \frac{\partial}{\partial y} V_E \]

\(\operatorname{curl} \mathbf{\tau}\) produces a divergence of the Ekman transports leading to a vertical velocity \(w_E\) at the bottom of the Ekman layer.

\[ w_E = \, \frac{\partial }{\partial x} \left( \frac{ \tau_{y}}{\rho \;f }\, \right) - \frac{\partial }{\partial y}\, \left( \frac{ \tau_{x}}{\rho \;f }\, \right) =\operatorname{curl}\left(\frac{\mathbf{\tau}}{\rho\;f}\right) \simeq \frac{1}{\rho\;f} \, \operatorname{curl} \mathbf{\tau} \]

The order of magnitude of the Ekman vertical velocity:

typical wind stress variation of \(0.2 N m^{-2}\) per 2000 km in y-direction:

\[ w_E \simeq - \frac{ \Delta \tau_{x}}{\rho \;f_0 \Delta y}\, \simeq \frac{1 }{10^3 kg m^{-3}} \frac{0.2 N m^{-2} }{10^{-4} s^{-1}\, \, 2 \cdot 10^6 m} \simeq 32 \, \, \frac{m}{yr} \]

Ekman Pumping & Sverdrup Transport

Ekman

The center of a subtropical gyre is a high pressure zone: clockwise on the Northern Hemisphere

Ekman surface currents towards the center of the gyre

The Ekman vertical velocity balanced by \[ w_E=w_g \] vertical geostrophic current in the interior

geostrophic flow towards the equator

returned flow towards the pole in western boundary currents

Ekman Pumping: vertical velocity at the bottom of the Ekman layer E

\(w_E\) as the Ekman vertical velocity the bottom of the Ekman layer \[ w_E = - \int_{-E}^0 \frac{\partial w}{\partial z} dz = \frac{\partial}{\partial x} U_E + \frac{\partial}{\partial y} V_E \]

\(\operatorname{curl} \mathbf{\tau}\) produces a divergence of the Ekman transports leading to \(w_E\) at the bottom E

\[ w_E = \, \frac{\partial }{\partial x} \left( \frac{ \tau_{y}}{\rho \;f }\, \right) - \frac{\partial }{\partial y}\, \left( \frac{ \tau_{x}}{\rho \;f }\, \right) =\operatorname{curl}\left(\frac{\mathbf{\tau}}{\rho\;f}\right) \simeq \frac{1}{\rho\;f} \, \operatorname{curl} \mathbf{\tau} \]

The order of magnitude of the Ekman vertical velocity:

typical wind stress variation of \(0.2 N m^{-2}\) per 2000 km in y-direction:

\[ w_E \simeq - \frac{ \Delta \tau_{x}}{\rho \;f_0 \Delta y}\, \simeq \frac{1 }{10^3 kg m^{-3}} \frac{0.2 N m^{-2} }{10^{-4} s^{-1}\, \, 2 \cdot 10^6 m} \simeq 32 \, \, \frac{m}{yr} \]

North Atlantic Current & Gulfstream

Gulf Stream & North Atlantic Current

Part of deep ocean

brings warm water northward where it cools.

returns southward as a cold, deep, western-boundary current.

Gulf Stream carries 40 Sv of 18°C water northward.

Of this, 15 Sv return southward in the deep western boundary current at a temperature of 2°C.

Ocean Conveyor Belt

Conveyor

Conveyor belt circulation

The the conveyor is driven by deepwater formation in the northern North Atlantic.

The conveyor belt metaphor necessarily simplifies the ocean system, it is of course not a full description of the deep ocean circulation.

Broecker’s concept provides a successful approach for global ocean circulation, although several features can be wrong like the missing Antarctic bottom water, the upwelling areas etc..

metaphor inspired new ideas of halting or reversing the ocean circulation and put it into a global climate context.

interpretation of Greenland ice core records indicating different climate states with different ocean modes of operation (like on and off states of a mechanical maschine).

Thermohaline ocean circulation

Overturning

Modelled meridional overturning streamfunction in Sv 10^6 = m^3 /s in the Atlantic Ocean. Grey areas represent zonally integrated smoothed bathymetry

Estimates of overturning ?

It is observed that water sinks in to the deep ocean in polar regions of the Atlantic basin at a rate of 15 Sv. (Atlantic basin: 80,000,000 km^2 area * 4 km depth.)

– How long would it take to ‘fill up’ the Atlantic basin?

– Supposing that the local sinking is balanced by large-scale upwelling, estimate the strength of this upwelling.

Hint: Upwelling = area * w

– Compare this number with that of the Ekman pumping!

Estimates of overturning: Solution

Timescale T to ‘fill up’ the Atlantic basin:

\[ T = \frac{ 80 \cdot 10^{12} \, m^2 \cdot 4000 \, m}{15 \cdot 10^6 \, m^3 s^{-1}} = 2.13 \cdot 10^{10} s = 676 \;years\]

Overturning is balanced by large-scale upwelling:

\[ area \cdot w = 15 \cdot 10^6 \, m^3 s^{-1}\]

\[ w = 0.1875 \cdot 10^{-6} m\;s^{-1} = 5.9 \cdot 10^{-15} m \; y^{-1}. \]

Ekman pumping \[ w_E \simeq 32 \, \, m \; y^{-1}. \]

Vorticity dynamics of meridional overturning (y,z)

\[ \frac{\partial}{\partial t} v \quad = \quad - \frac{1}{\rho_0} \frac{\partial p}{\partial y} \quad - \quad f u \quad - \quad \kappa v \]

\[ \frac{\partial}{\partial t} w \quad = \quad - \frac{1}{\rho_0} \frac{\partial p}{\partial z} \quad - \quad \frac{g}{\rho_0} (\rho -\rho_0) \quad - \quad \kappa w \] \(\kappa\) as parameter for Rayleigh friction.

Vorticity dynamics of meridional overturning (y,z)

\[ \frac{\partial}{\partial t} v \quad = \quad - \frac{1}{\rho_0} \frac{\partial p}{\partial y} \quad - \quad f u \quad - \quad \kappa v \]

\[ \frac{\partial}{\partial t} w \quad = \quad - \frac{1}{\rho_0} \frac{\partial p}{\partial z} \quad - \quad \frac{g}{\rho_0} (\rho -\rho_0) \quad - \quad \kappa w \] \(\kappa\) as parameter for Rayleigh friction.

Using the continuity equation
\[ 0 \quad = \quad \frac{\partial v}{\partial y} \quad + \quad \frac{\partial w}{\partial z} \]

\[ \mbox{ one can introduce a streamfunction } \Phi (y,z): v= \partial_z \Phi; w = - \partial_y \Phi \]

The associated vorticity equation in the (y,z)-plane is therefore

\[ \frac{\partial}{\partial t} \nabla^2 \Phi \, = \, - f \frac{\partial u}{\partial z} \quad + \quad \frac{g}{\rho_0} \frac{\partial \rho}{\partial y} \quad - \quad \kappa \nabla^2 \Phi \]

Galerkin approximation

\[ {\frac{\partial}{\partial t} \nabla^2 \Phi} \, = \, \underbrace{- f \frac{\partial u}{\partial z}}_{2.} \quad + \quad { \frac{g}{\rho_0} \frac{\partial \rho}{\partial y}} \quad - \quad \underbrace{ \kappa \nabla^2 \Phi }_{4.} \]

Term 2. is absorbed into the viscous term (4.)

\[ \Phi (y,z,t) \, = \, \sum_{k=1}^\infty \sum_{l=1}^\infty \Phi_{max}^{k,l} (t) \, \, \sin (\pi k y/L) \, \times \, \sin (\pi l z/H) \] yielding a first order differential equation in time for \(\Phi_{max}^{k,l} (t)\).

Simple ansatz satisfying that the normal velocity at the boundary vanishes \[ \Phi (y,z,t) \, = \, \Phi_{max} (t) \, \, \sin \left(\frac{\pi y}{L}\right) \, \times \, \sin \left(\frac{\pi z}{H}\right) \]

L and H dentote the meridional and depth extend: y from 0 to L, z from 0 to H

The remaining terms and integration

\[ \underbrace{\int_0^L dy \int_0^H dz \quad \frac{\partial}{\partial t} \nabla^2 \Phi}_{1.} \, = \, \underbrace{ \int_0^L dy \int_0^H dz \quad \frac{g}{\rho_0} \frac{\partial \rho}{\partial y}}_{2.} \quad - \underbrace{ \int_0^L dy \int_0^H dz \quad \kappa \nabla^2 \Phi }_{3.} \]

1. \[ \frac{d}{dt} \Phi_{max} \left(\frac{\pi^2}{L^2} + \frac{\pi^2}{H^2}\right) \int\limits_0^L dy \sin \left(\frac{\pi y}{L}\right) \int\limits_0^H dz \sin \left(\frac{\pi z}{H}\right) = 4 LH \left(\frac{1}{L^2} + \frac{1}{H^2}\right) \frac{d}{dt} \Phi_{max} \]

The remaining terms and integration

\[ \underbrace{\int_0^L dy \int_0^H dz \quad \frac{\partial}{\partial t} \nabla^2 \Phi}_{1.} \, = \, \underbrace{ \int_0^L dy \int_0^H dz \quad \frac{g}{\rho_0} \frac{\partial \rho}{\partial y}}_{2.} \quad - \underbrace{ \int_0^L dy \int_0^H dz \quad \kappa \nabla^2 \Phi }_{3.} \]

1. \[ \frac{d}{dt} \Phi_{max} \left(\frac{\pi^2}{L^2} + \frac{\pi^2}{H^2}\right) \int\limits_0^L dy \sin \left(\frac{\pi y}{L}\right) \int\limits_0^H dz \sin \left(\frac{\pi z}{H}\right) = 4 LH \left(\frac{1}{L^2} + \frac{1}{H^2}\right) \frac{d}{dt} \Phi_{max} \]

2. \[ \int\limits_0^L dy \int\limits_0^H dz \frac{g}{\rho_0} \frac{\partial \rho}{\partial y} = \frac{g}{\rho_0} \, H \, (\rho_{north} - \rho_{south}) \quad \mbox{ with } \rho_{north} = \rho (y=L) , \rho_{south} = \rho (y=0) \]

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\[ \underbrace{\int_0^L dy \int_0^H dz \quad \frac{\partial}{\partial t} \nabla^2 \Phi}_{1.} \, = \, \underbrace{ \int_0^L dy \int_0^H dz \quad \frac{g}{\rho_0} \frac{\partial \rho}{\partial y}}_{2.} \quad - \underbrace{ \int_0^L dy \int_0^H dz \quad \kappa \nabla^2 \Phi }_{3.} \]

1. \[ \frac{d}{dt} \Phi_{max} \left(\frac{\pi^2}{L^2} + \frac{\pi^2}{H^2}\right) \int\limits_0^L dy \sin \left(\frac{\pi y}{L}\right) \int\limits_0^H dz \sin \left(\frac{\pi z}{H}\right) = 4 LH \left(\frac{1}{L^2} + \frac{1}{H^2}\right) \frac{d}{dt} \Phi_{max} \]

2. \[ \int\limits_0^L dy \int\limits_0^H dz \frac{g}{\rho_0} \frac{\partial \rho}{\partial y} = \frac{g}{\rho_0} \, H \, (\rho_{north} - \rho_{south}) \quad \mbox{ with } \rho_{north} = \rho (y=L) , \rho_{south} = \rho (y=0) \]

3. \[ \kappa \Phi_{max} \left(\frac{\pi^2}{L^2} + \frac{\pi^2}{H^2}\right) \int\limits_0^L dy \sin \left(\frac{\pi y}{L}\right) \int\limits_0^H dz \sin \left(\frac{\pi z}{H}\right) = \kappa \, 4 LH \left(\frac{1}{L^2} + \frac{1}{H^2}\right) \Phi_{max} \]

Amplitude of overturning

\[ \frac{d}{dt} \Phi_{max} \, = \, \frac{a}{\rho_0} (\rho_{north} - \rho_{south}) \, \, - \, \, \kappa \Phi_{max} \]

with \[ a = g L H^2/4(L^2 + H^2) \, \]

This shows that the overturning circulation depends on the density differences on the right and left boxes.

It is simplified to a diagnostic relation

\[ \Phi_{max} = \frac{a}{\rho_0 \, \kappa} \, \, (\rho_{north} - \rho_{south}) \quad \]

because the adjustment of \(\Phi_{max}\) is quasi-instantaneous due to adjustment processes, e.g. Kelvin waves.

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THC

Schematic picture of the hemispheric two box model (a) and of the interhemispheric box model

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THC

1. The Atlantic surface density is mainly related to temperature differences.

2. But the pole-to-pole differences are caused by salinity differences. }

Meteor

Meteor Expedition, the first accurate hydrographic survey of the Atlantic from Wuest (1935).

Meteor

Temp & salinity

Meteor Expedition, the first accurate hydrographic survey of the Atlantic from Wuest (1935).

Rotation and Hadley Cell

Low Pressure: Friction on the bottom matters

left: 50%

Tea

As the ground is approached from above, and the influence of friction becomes important, this balance changes. The friction slows down the primary circular motion, and thus the Coriolis force is lower and no longer sufficient for the fluid to flow parallel to the isobars.

There is a net inward force, and the air thus moves toward the center of the low, much like the tea leaves example.

The converse case is also interesting. Around a high-pressure system, the air rotation is anticyclonic to achieve balance. In this case for the air outside the bottom boundary layer, the inward Coriolis force balances the outward radial pressure gradient.