Dynamics II

Lecture: June 12, 2023 (Monday), 14:00 Prof. Dr. Gerrit Lohmann

 
 
 

Linear stability analysis

Consider the continuous dynamical system described by \[ \dot x=f(x,\lambda)\quad \] A bifurcation occurs at \[(x_E,\lambda_0)\] if the Jacobian matrix \[ \textrm{d}f/dx (x_E,\lambda_0)\] has an Eigenvalue with zero real part.

 

Convection in the Rayleigh-Benard system

Bild

Rayleigh (1916) temperature difference between the upper- and lower-surfaces \[ T(x, y, z=H) = \, T_0 \] \[ T(x, y, z=0) \, = \, T_0 + \Delta T \]

Furthermore \[ \rho = \rho_0 = const. \] except in the buoyancy term, where:

\[ \varrho = \varrho_0 (1 - \alpha(T-T_0)) \mbox{ with } \alpha > 0 \quad . \]

common feature of geophysical flows

 
 

Vorticity in the Rayleigh-Benard system

\[ D_t u = - \frac{1}{\rho_0} \partial_x p + \nu \nabla^2 u \label{eqref:einse} \qquad \qquad \qquad \qquad \qquad \qquad (1) \]

\[ D_t w = - \frac{1}{\rho_0} \partial_z p + \nu \nabla^2 w + g (1- \alpha (T-T_0)) \qquad \qquad (2)\]

\[ \partial_x u + \partial_z w = 0 \qquad \qquad \qquad \qquad \qquad \qquad \qquad (3) \]

\[ D_t T = \kappa \nabla^2 T \qquad \qquad \qquad \qquad \qquad \qquad \qquad (4) \]

 

Introduce vorticity to have (1,2,3) in one equation:

\[ D_t \left( \nabla^2 \Psi\right) = \nu \nabla^4 \Psi - g \alpha \frac{\partial \Theta}{\partial x} \]

Temperature in the Rayleigh-Benard system

\[ T_{eq} = T_0 + \left(1 - \frac{z}{H}\right) \Delta T \]

\[ \mbox{with } \quad T = T_{eq} + \Theta \quad , \quad \mbox{where } \quad \Theta \quad \mbox{is the anomaly} \]

Non-dimensional Rayleigh-Benard system

\[ D_t \left( \nabla_d^2 \Psi_d\right) = \sigma \nabla_d^4 \Psi_d - R_a \sigma \frac{\partial \Theta_d}{\partial x_d} \]

\[ D_t \Theta_d \quad = \frac{\partial \Psi_d}{\partial x_d} + \nabla_d^2 \Theta_d \quad \]

\[ \Psi (x,z,t) \, = \, \sum_{k=1}^\infty \sum_{l=1}^\infty \Psi_{k,l} (t) \, \, \sin \left(\frac{k \pi a}{H} x \right) \, \times \, \sin \left(\frac{ l \pi}{H} z \right) \] \[ \Theta (x,z,t) \, = \, \sum_{k=1}^\infty \sum_{l=1}^\infty \Theta_{k,l} (t) \cos \left(\frac{k \pi a}{H} x \right) \, \times \, \sin \left( \frac{l \pi}{H} z \right) \]

Approximation: Just 3 Modes X(t), Y(t), Z(t)

\[ \frac{a}{1+a^2} \, \kappa \, \Psi = X \sqrt{2} \sin\left(\frac{\pi a}{H} x \right) \sin\left(\frac{\pi}{H} z \right) \]

\[ \pi \frac{R_a}{R_c} \frac{1}{\Delta T} \, \Theta = Y \sqrt{2} \cos\left(\frac{\pi a}{H} x\right) \sin\left(\frac{\pi}{H} z \right) - Z \sin\left(2 \frac{\pi}{H} z \right) \]

Rayleigh number Ra: Buoyancy & Viscosity

\[ \mbox{Motion develops if } \quad R_a = \frac{g \alpha H^3 \Delta T}{\nu \kappa} \quad \mbox{ exceeds } \quad R_c = \pi^4 \frac{(1+a^2)^3}{a^2} \mbox{ $=657.51$ occurs when $a^2 = 1/2$. }\]

\[\mbox{When } R_a < R_c,\mbox{ heat transfer is due to conduction} \]

\[\mbox{When } R_a > R_c, \mbox{ heat transfer is due to convection.} \]

Lorenz system

Bifurcation at \[ r = R_a/R_c = 1\]

Geometry constant \[b = 4(1+a^2)^{-1}\]

 
Famous low-order model:

\[ \dot X = -\sigma X + \sigma Y \]

\[ \dot Y = r X - Y - X Z \]

\[ \dot Z = -b Z + X Y \]

 

\[\mbox{dimensionless time } \quad t_d = \pi^2 H^{-2} (1+a^2) \kappa t,\]

\[ \mbox{ Prandtl number } \quad \sigma = \nu \kappa^{-1}, \]

Lorenz system r=24

left: 52%

 

r=24
s=10
b=8/3
dt=0.01
x=0.1
y=0.1
z=0.1
vx<-c(0)
vy<-c(0)
vz<-c(0)
for(i in 1:10000){
x1=x+s*(y-x)*dt
y1=y+(r*x-y-x*z)*dt
z1=z+(x*y-b*z)*dt
vx[i]=x1
vy[i]=y1
vz[i]=z1
x=x1
y=y1
z=z1}
plot(vx,vy,type="l",xlab="x",ylab="y")
plot(vy,vz,type="l",xlab="y",ylab="z")

Lorenz system r=0.9

 

r=0.9
s=10
b=8/3
dt=0.01
x=1.1
y=0.1
z=11.1
vx<-c(0)
vy<-c(0)
vz<-c(0)
for(i in 1:100){
x1=x+s*(y-x)*dt
y1=y+(r*x-y-x*z)*dt
z1=z+(x*y-b*z)*dt
vx[i]=x1
vy[i]=y1
vz[i]=z1
x=x1
y=y1
z=z1}
plot(vx,type="l",xlab="time",ylab="x")
plot(vy,type="l",xlab="time",ylab="y")

Lorenz system r=3.5

 

r=3.5
s=10
b=8/3
dt=0.01
x=1.1
y=0.1
z=11.1
vx<-c(0)
vy<-c(0)
vz<-c(0)
for(i in 1:1000){
x1=x+s*(y-x)*dt
y1=y+(r*x-y-x*z)*dt
z1=z+(x*y-b*z)*dt
vx[i]=x1
vy[i]=y1
vz[i]=z1
x=x1
y=y1
z=z1}
plot(vx,type="l",xlab="time",ylab="x")
plot(vy,type="l",xlab="time",ylab="y")

 

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

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

 

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

 

integral over the depth \[ \int_{0}^h \, dz \]

 

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

 

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

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

 

Potential vorticity is conserved (without stress)

 

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

Ocean/Atmosphere with depth h(x,y)

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

 

Couples depth, vorticity, latitude

– Changes in the depth results in change in \(\zeta\).

– Changes in latitude require a corresponding change in \(\zeta\).

Vorticity

Dietrich et al. (1980)


Vorticity

Steward, Oceanography

 

Sverdrup transport

Sverdrup

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

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

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

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

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:

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

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

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.

 

How much heat is transported northward ?

Calculation:

\[ \underbrace{ c_p}_{4.2 \cdot 10^3 Ws/(m^3 kg)} \, \cdot \, \underbrace{ \rho }_{10^3 kg/m^3 } \, \cdot \, \underbrace{\Phi}_{15 \cdot 10^6 m^3/s} \, \cdot \, \underbrace{\Delta T}_{(18-2) K } = 1 \cdot 10^{15} W \]

The flow carried by the conveyor belt loses 1 Petawatts (PW), close to estimates of Rintoul and Wunsch (1991)

The deep bottom water from the North Atlantic is mixed upward in other regions and ocean, and it makes its way back to the Gulf Stream and the North Atlantic. Thus most of the water that sinks in the North Atlantic must be replaced by water from the far South Atlantic and Pacific Ocean.

Conveyor Belt

Conveyor

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

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

=======================================================

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

  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.

Box Model

THC

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

Density gradients

THC

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

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

 

Box model

1) Boxmodel (Online Version Jupyter Notebook)

link to online tool

 

2) Boxmodel (Jupyter Notebook)

download jupyter notebook

 

3) Boxmodel (Version using R-shiny)

How to setup and run the Boxmodel R-shiny app

download

password: DynamicsII2020