Dynamics II

Lecture: April 15, 2024 (Monday), 14:00 Prof. Dr. Gerrit Lohmann

 
 
 

Vorticity concept

 

Rayleigh-Benard convection (and Rayleigh number)

Bridge of Schneider Böck: Stability - Instability

And there he is on the bridge,
Cracks! The bridge is falling apart;

Böck
Böck

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.

 

Example: transcritical bifurcation

a fixed point interchanges its stability with another fixed point as the control parameter is varied. Bifurcation at \(r=0\).

\[ \frac{dx}{dt}=rx (1-x) \, \]

The two fixed points are 0 and 1. When r is negative, the fixed point at 0 is stable and 1 is unstable. But for \(r>0\), 0 is unstable and 1 is stable.

Saddle-node bifurcation: two fixed points collide

The normal form: \[ \frac{dx}{dt}=r+x^2 \]

\[ r<0: \mbox{ a stable equilibrium point at } -\sqrt{-r} \mbox{ and an unstable one at } \sqrt{-r} \qquad \qquad \]

\(r=0\): exactly one equilibrium point, saddle-node fixed point.

\(r>0\): no equilibrium points. Saddle-node bifurcations may be associated with hysteresis loops.

Saddle-node bifurcation: two fixed points collide

1) Calculate the equilibrium points:

\[ f(x) = r + x^2 = 0 \]

\[ x_{E1,E2} = \pm \sqrt{-r} \mbox{ for } r \le 0 \]

2) Linear stability conditions for the equilibrium points

\[ f'(x) = 2x \]

\[ f'(x_{E1}) = 2\sqrt{-r} > 0 \quad \mbox{unstable} \]

\[ f'(x_{E2}) = -2\sqrt{-r} < 0 \quad \mbox{stable} \]

For r=0: equilibrium point \(x_E=0\) which is indifferent (not stable/unstable).

3) Potential or Lyapunov Method

\[ \frac{dx}{dt}=b+x^2 = - \frac{d}{dx} \left( - b x - \frac{x^3}{3} \right) = - \frac{d}{dx} V (x) \]

Global analysis including basins of attraction for \(x_{E2}: (-\infty,3)\)

\[ x_{E1} = 2\sqrt{9} = 6 \quad \mbox{unstable} \]

\[ x_{E1} = -2\sqrt{9} = -6 \quad \mbox{stable} \]

4) Graphical method: slope at equilibrium points

\[ \frac{dx}{dt}=b+x^2 \]

Graphical Method
Graphical Method
filled points: positive slope => unstable
open points: negative slope => stable

No Convection Equilibrium: Diffusion

Bild
Bild

Diffusion: Temperature varies linearly with depth:

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

No movement of particles:

\[ u = w= 0 \]

When this solution becomes unstable, convection should develop.

No Convection Equilibrium: Diffusion

Vorticity in the Rayleigh-Benard system

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

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

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

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

 

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

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

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

In vorticity equation:

\[ \frac{\partial }{\partial x} g (1- \alpha (T_{eq} + \Theta -T_0)) = - g \alpha \frac{\partial }{\partial x} \Theta \]

Temperature equation:

\[ D_t T = D_t T_{eq} + D_t \Theta = w \cdot \frac{- \Delta T}{H} + D_t \Theta = - \frac{\Delta T}{H} \frac{\partial \Psi}{\partial x} + D_t \Theta \]

\[ D_t \Theta \quad = \frac{\Delta T}{H} \frac{\partial \Psi}{\partial x} + \kappa \nabla^2 \Theta \quad \]

Non-dimensional Rayleigh-Benard system

\[ \frac{1}{T} \frac{1}{L^2} \frac{L^2}{T} D_t \left( \nabla_d^2 \Psi_d\right) = \nu \frac{1}{L^4} \frac{L^2}{T} \nabla_d^4 \Psi_d - g \alpha \frac{ \Delta T}{L} \frac{\partial \Theta_d}{\partial x_d} \]

\[ \frac{\Delta T}{T} D_t \Theta_d \quad = \frac{\Delta T}{H} \frac{L^2}{ T L} \frac{\partial \Psi_d}{\partial x_d} + \kappa \frac{\Delta T}{L^2} \nabla_d^2 \Theta_d \quad \]

This yields (remember \(L=H\)) \[ D_t \left( \nabla_d^2 \Psi_d\right) = \nu \frac{T}{H^2} \nabla_d^4 \Psi_d - g \alpha \frac{T^2 \Delta T}{H} \frac{\partial \Theta_d}{\partial x_d} \] \[ D_t \Theta_d \quad = \frac{\partial \Psi_d}{\partial x_d} + \kappa \frac{T}{H^2} \nabla_d^2 \Theta_d \quad \]

\[ \mbox{Inserting} \quad T= H^2/\kappa \] \[ \mbox{Rayleigh number} \quad R_a = \frac{g \alpha H^3 \Delta T}{\nu \kappa}\] \[ \mbox{Prandtl number} \quad \sigma = \frac{ \nu}{ \kappa} \]

\[ D_t \left( \nabla_d^2 \Psi_d\right) = \frac{ \nu}{ \kappa} \nabla_d^4 \Psi_d - g \alpha \frac{H^3 \Delta T}{\kappa^2} \frac{\partial \Theta_d}{\partial x_d} \label{eqref:psieqnn3} \] \[ D_t \Theta_d \quad = \frac{\partial \Psi_d}{\partial x_d} + \nabla_d^2 \Theta_d \quad . \] Finally, inserting the

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

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

Galerkin approximation: Get a low-order model

Bild
Bild

\[ \mbox{ Saltzman (1962): Expand } \Psi, \Theta \mbox{ in double Fourier series in x and z: } \]

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

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

\[ \mbox{Motion develops if } \quad R_a = \frac{g \alpha H^3 \Delta T}{\nu \kappa} \quad \mbox{exceeds a critical } \quad R_c \]

As the Rayleigh number increases, the gravitational force becomes more dominant. The critical Rayleigh number can be obtained analytically for a number of different boundary conditions by doing a perturbation analysis on the linearized equations in the stable state.

\(R_c = \pi^4 \frac{\left(1+a^2\right)^3}{a^2} = \pi^4 \frac{27}{4}= 657.51\) occurs when \(a^2 = 1/2\).

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

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

Bild
Bild

Lorenz system

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

 
Famous low-order model:

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

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

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

 

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")

 

Scaling: Rotating frame of reference

The Coriolis effect is one of the dominating forces for the large-scale dynamics.

\[ \underbrace{\frac{\partial \mathbf{v}}{\partial t}}_{ U/T \sim 10^{-8} } + \underbrace{\mathbf{v} \cdot \nabla \mathbf{v}}_{ U^2/L \sim 10^{-8} }= {\underbrace{- \frac{1}{\rho} \nabla p}_{ \bf \delta P/(\rho L) \sim 10^{-5} } + \underbrace{2 \mathbf{\Omega \times v}}_{ \bf f_0 U \sim 10^{-5} } + \underbrace{fric}_{ \nu U/H^2 \sim 10^{-13}}} \quad \]

where fric is due to eddy stress divergence (\(\sim \nu \nabla^2 \mathbf{v}\)).

Values from the ocean -> exercise.

 

Because of the continuity equation \[ U/L \sim W/H \] and horizontal scales are orders of magnitude larger than the vertical ones, \[ W << U . \]

The timescales are related to \[ T \sim L/U \sim H/W \]

Exception for small scales (e.g., ocean convection or cumuls clouds): \[ H \sim L \quad \rightarrow \quad W \sim U \]

Rossby number Ro

\[ \underbrace{\frac{\partial \mathbf{v}}{\partial t}}_{ U/T \sim 10^{-8} } + \underbrace{\mathbf{v} \cdot \nabla \mathbf{v}}_{ U^2/L \sim 10^{-8} } = {\underbrace{- \frac{1}{\rho} \nabla p}_{ \bf \delta P/(\rho L) \sim 10^{-5}} + \underbrace{2 \mathbf{\Omega \times v}}_{ \bf f_0 U \sim 10^{-5} } + \underbrace{fric}_{ \nu U/H^2 \sim 10^{-13}}} \]

 

\[ Ro = \frac{ \mbox{Inertial (the left hand side)} }{ \mbox{Coriolis term } } \]

 

\[ Ro = \frac{(U^2/L)}{(f U)} = \frac{U}{f L} \quad \]

 

characterizes the importance of Coriolis acceleration

Ro is small when the flow is in a so-called geostrophic balance.

 

Vorticity is the rotation of the fluid

\[ \zeta \equiv \frac{\partial v}{\partial x}-\frac{\partial u}{\partial y} \]

or in 3D:

\[ \equiv \nabla \times \boldsymbol{ u} \]

Example: Rigid body rotating

\[ \boldsymbol{ u} = \begin{pmatrix} u \\ v \\ w \end{pmatrix}\ = \boldsymbol{ \Omega} \times \boldsymbol{ r} = \begin{pmatrix} \omega_1 \\ \omega_2 \\ \omega_3 \end{pmatrix}\ \times \ \begin{pmatrix} x \\ y \\ z \end{pmatrix}\ = \begin{pmatrix} \omega_2 z- \omega_3 y\\ \omega_3 x- \omega_1 z \\ \omega_1 y - \omega_2 x \end{pmatrix}\ \]

Rotation vector

\[ \nabla \times \boldsymbol{ u} = \begin{pmatrix} \partial_x \\ \partial_y \\ \partial_z \end{pmatrix}\ \times \ \begin{pmatrix} u \\ v \\ w \end{pmatrix}\ = \begin{pmatrix} \partial_x \\ \partial_y \\ \partial_z \end{pmatrix}\ \times \ \begin{pmatrix} \omega_2 z- \omega_3 y \\ \omega_3 x- \omega_1 z \\ \omega_1 y - \omega_2 x \end{pmatrix}\ \]

\[ = \begin{pmatrix} \partial_y (\omega_1 y - \omega_2 x) - \partial_z (\omega_3 x- \omega_1 z ) \\ \\ \\ \end{pmatrix}\ \]

\[ = \begin{pmatrix} \omega_1 + \omega_1 \\ \omega_2 + \omega_2 \\ \omega_3 + \omega_3 \end{pmatrix}\ = 2 \boldsymbol{ \Omega} \]

 

Example: Vorticity from shear

 

Cape
Cape

Tomczak & Godfrey: Regional Oceanography


 
 

\[u=0, v=v\left(x\right)\]

\[\zeta= \partial v\left(x\right)/\partial x \]

Estimate for \(\zeta\) off Cape Hatteras:

the velocity changes by \(1 \, {m}{s}^{-1}\) in 100 km

\[ \zeta= \frac{\partial v}{\partial x} = \frac{ 1 \, {m}{s}^{-1}}{100 \, {km}} = 10^{-5} \, \frac{1}{s} \]

still much smaller than

\[ f= 2 \Omega \sin \varphi = 2 \, \frac{2 \pi}{day} \sin \varphi \approx \, 10^{-4} \, \frac{1}{s} \]

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 with depth h(x,y)

\[ \mbox{Because of the continuity equation } \quad \partial_x \left( u h \right) + \partial_y \left( v h \right) \quad = \quad 0 \]

\[ \quad \frac{D}{Dt} h + h \left( \partial_x u + \partial_y v \right) = \quad 0 \]

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

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

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

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

Potential vorticity is conserved along a fluid trajectory.

Potential vorticity: Examples

 

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

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 \(\zeta\).

Cape
Cape

Dietrich et al. (1980)


Cape
Cape

Steward, Oceanography

 

Angular Momentum and Hadley Cell

 
Subtropical and polar jet

Hadely Cell
Hadely Cell

 

 

 

Tropical air rises to tropopause & moves poleward

Deflected eastward by the Coriolis force

Subtropical jet: forms at poleward limit of Hadley Cell

It tends to conserve angular momentum, friction small

equatorward moving air: westward component

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
Sverdrup

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

 

Ekman Pumping & Sverdrup Transport

 

Ekman
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
Gulf Stream & North Atlantic Current
Part of deep ocean
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.

Ocean Conveyor Belt

Conveyor
Conveyor

Conveyor Belt: Industry

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

 

End of this lecture