# Lernkarten

Karten 35 Karten 1 Lernende English Universität 21.07.2018 / 14.08.2018 Keine Angabe
0 Exakte Antworten 35 Text Antworten 0 Multiple Choice Antworten

Give the equation for the quasi-geostrophic potential Vorticity!

$$\psi$$ is the quasi-geostrophic stream function.

$$\frac{D_g}{Dt}(\nabla^2\psi +\beta y-\frac{f_0^2}{c^2}\psi )=\frac{1}{\rho_0H_1}(\frac{\partial }{\partial x}\tau_s^y-\frac{\partial}{\partial y} \tau_s^x)$$

$$\nabla ^2 \psi$$... is the relative vorticity

$$f_0+\beta y$$... is the planetary vorticity

$$\frac{f_0^2}{c^2 }\psi$$... is the stretching term

$$\frac{\partial }{\partial x}\tau_s^y-\frac{\partial}{\partial y}\tau_s^x$$... is the wind stress curl

If the wind stress curl is 0 then:

$$\frac{D_q}{Dt}=0\;\;where\;\;q=\nabla^2\psi+\beta y-\frac{f_0^2}{c^2}\psi$$

q is the quasi-geostrophic potential vorticity;

$$\frac{Dq}{Dt}=0$$ states the Conservation of Quasi-Geostrophic Potential Vorticity

If q is conserved, then changes of the planetary vorticity are balanced bx changes of the relative vorticity and vortex stretching.

What important assumption is made in the quasi-geostrophic theory?

It is assumed, that vertical displacements are not so large as to disturb the background stratification. In the $$1\frac{1}{2}$$ layer model, this corresponds to the assumption that the interface displacements are small compared to the undisturbed depth $$H_1$$.

Give the dispersion relation of linear Rossby Waves and explain the difference between long and short waves!

Lizenzierung: Keine Angabe

Under the assumption of conservation of the quasi-geostrophic potential vorticity $$Dq/Dt=0$$ and a solution of the form:

$$\psi=\psi_0e^{i(kx+ly-\omega t)}$$

We find the dispersion relation:

$$\omega=-\frac{\beta k }{(k^2+l^2+\frac{f_0^2}{c^2})}$$

For fixed l the solution has the following form as shown in the graphic below.

long waves are: $$\frac{\partial \omega}{\partial k}<0$$ -> westward group velocity

short waves are: $$\frac{\partial \omega}{\partial k }>0$$ -> eastward group velocity

Western intensification: short waves trapped near western boundary; long waves escape westwards.

Long Waves

The limit for the long waves is:

$$k^2+l^2<<\frac{f_0^2}{c^2}$$

... horizontal length scales are much larger than the radius of deformation

• typical for baroclinic modes in the ocean. For example, for the first baroclinic mode ($$c=1-3ms^{-1}$$)

for long waves the dispersion relation reduces to:

$$\omega =-\beta\frac{c^2}{f_0^2}k$$

-> These waves are non-dispersive with westward phase and group velocity given by $$\beta\frac{c^2}{f_0^2}$$.

thinking of the shallow water equations:

$$u_t-fv=-g'h_x\\ v_t+fu=-g'h_y\\ h_t+H_1(u_x+v_y)=0$$

The long wave limit is equivalent to neglecting $$u_t,v_t$$ and so:

$$h_t-\frac{\beta g'H_1}{f^2}h_x=0\;\;with\;\;\beta=\frac{df}{dy}$$

assuming wave solution $$h=h_0e^{i(\omega t-kx-ly)}$$ we get to the dispersion relation (see above).

Divergence of the ageostrophic flow due to $$\beta$$ is balanced by "vortex stretching", i.e. pumps the thermocline up and down via the $$h_t$$ term.

To cross the North Atlantic Ocean, these waves need roughly 25 years! Although the propagation speed increases toward equator, where it takes only 1 year.

Short Waves

Lizenzierung: Keine Angabe

The limit of short waves is:

$$k^2+l^2>>\frac{f_0^2}{c^2}$$

-> length scales are much smaller than the Radius of Deformation!

Usually appropriate for the barotropic mode in the ocean: $$c=200m/s$$

The dispersion relation reduces to:

$$\omega=-\frac{\beta k}{(k^2+l^2)}$$

These waves are highly dispersive (the phase and group velocity are far from being equal).

The time to cross the North Atlantic for a barotropic wave is typically about one week. So, much fast than a baroclinic wave.

The shallow water equations are:

$$u_t-fv=-g\eta_x\\ v_t+fu=-g\eta_y\\ \eta_t+H(u_x+v_y)=0$$

The short wave limit is equivalent to negelcting the $$\eta_t$$ term in the continuity equation.

This enables us to work with a streamfunction for which: $$u=-\psi_y,\;v=\psi_x$$:

$$\frac{\partial }{\partial t}(\nabla ^2\psi )+\beta\psi_x=0$$

The horizontal divergence is zero. This means that the dvergence flow caused by $$\beta$$ must be absorbed by the convergence of the isallobaric flow arising from the $$u_t,\;v_t$$terms.

Note that the isallobaric part of the ageostrophic flow flows towards the isallobaric low, where pressure drops most rapidly. The only way the resulting convergence can be absorbed by the $$\beta$$ part is when the waves propagate westward! (is unclear for me)

Explain the concept of normal modes!

Lizenzierung: Keine Angabe

The gorverning equations and boundary conditions from a Sturm-Liouville problem so that there is a complete set of (orthogonal) vertical structure functions $$\widehat{p}_n(z),\;\widehat{w}_n(z)$$ called vertical normal modes, enabling us to write:

$$u(x,y,z,t)=\sum^\infty_{n=0}\tilde{u_n}(x,y,t)\widehat{p}_n(z)$$

the horizontal structure associated with each mode satisfies the shallow water equations!

$$c_n$$ist the wave speed associated with the n'th mode and $$H_n$$ is the equivalent depth.

$$c_0>c_1>c_2>...$$

n=0 is called the barotropic mode. For this mode, $$c=\sqrt{gH}$$ (H is the total depth), $$\widehat{p}_0(z)=1$$ and $$\widehat{w}_0(z)=\frac{z+H}{H}$$ are a good approximation. for the barotropic mode, the vertical velocity varies essentially linearly with depth from zero at the bottom to a maximum at the top and, to good appr., the barotropic velocities $$\tilde{u_0}$$ and $$\tilde{v_0}$$ are the vertically averaged horizontal velocities.

For $$n\geq1$$we have the baroclinic modes; typically $$c_1=1-3ms^{-1}$$(the first baroclinic mode). For these modes the rigid-lid approximation can be applied, i.e. we can put $$\widehat{w}_n(0)=0$$ (i.e. w=0 at Z=0) to a good approximation and all displacements are internal. For the baroclinic modes $$\int^0_{-H}udz=0$$ (to a good approximation) and all displacements are internal. For the first mode, $$\widehat{p}_1(z)$$ has one zero crossing, for the n'th mode $$\widehat{p}_n(z)$$ has n zero crossings.

Note that all modes are "orthogonal", with the implication that each mode varies independently of the others.

the shallow water equations can be obtained from the 3-D governing equations by projecting the 3-D governing equations on to each vertical mode.

Spin-up of an ocean basin by an applied wind stress without boundaries.

Lizenzierung: Keine Angabe
• Equastion linearised about a state of rest for a stratified ocean.
• Seperation into vertical modes, since the bottom is flat.

$$u_t-fv=-g\eta_x+\frac{\tau_s^x}{\rho_0H}\\ v_t+fu=-g\eta_y+\frac{\tau_s^y}{\rho_0H}\\ \eta_t+H(u_x+v_y)=0$$

For a quasi-geostrophic flow these equations reduce to:

$$\frac{\partial }{\partial t}(\nabla^2\psi-\frac{f_0^2}{c^2}\psi)+\beta\psi_x=\frac{1}{\rho_0H}(\frac{\partial }{\partial x}\tau^y_s-\frac{\partial }{\partial y}\tau_s^x)\;where\;\psi=\frac{g\eta}{f_0 }$$

take $$\tau_s^y=0\;and\;\frac{\tau_s^x}{\rho_0H}=X_0sin(ly)$$

look for solutions of the form: $$\psi=\psi'cos(ly)$$

we become:

$$\frac{\partial }{\partial t}(\frac{\partial ^2\psi'}{\partial x^2}-(l^2+\frac{f_0^2}{c^2})\psi')+\beta\psi'_x=-lX_0$$

first we discuss the situation for no boundaries. Since the wind stress has no dependence on the zonal coordinate, x, neither will the solution.

$$\frac{\partial }{\partial t}(-(l^2+\frac{f_0^2}{c^2})\psi')=-lX_0$$

for baroclinic modes: $$l^2<< f_0^2/c^2$$ it follows:

$$\frac{\partial }{\partial t}(-\frac{f_0^2}{c^2}\psi')=-lX_0$$ or

$$\frac{\partial \eta}{\partial t}=\frac{1}{f_0\rho_0}(-\frac{\partial \tau_s^x}{\partial y})$$

-> Vertical displacement of the thermocline by Ekman pumping.

Ekman transport is part of the ageostrophic flow. The dominant part is the geostrophic flow. When there are no boundaries there is no x-dependency and $$v_g=0$$. The geostrophic flow is zonal and in the same direction as the wind. Because the thermocline is being continually pumped, the zonal geostrophic flow increases with time.

Spin-up of an ocean basin by an applied wind stress. Add the eastern boundary!

Lizenzierung: Keine Angabe

Rossby waves:

1. long waves -> westward group velocity
2. short waves -> eastward group velocity

-> need only to consider long Rossby Waves - the waves with westward group velocity.

-> Work with the baroclinic mode for which: $$k^2+l^2 << \frac{f_0^2}{c^2}$$ (horizontal length scales are long compared to the radius of deformation.

Returning to the quasi-geostrophic equation, using the long wave limit and putting $$\psi=\psi'cos(ly)$$ we come to the same solution as for no boundaries:

$$-\frac{f_0^2}{c^2}\frac{\partial \psi'}{\partial t}=-lX_0$$

This means that the solution is as for the case with no boundaries, until the (long Rossby) wave arrives from the eastern boundary and introduces x-dependences to the solution.

1. In the interior of the ocean, first there is the Ekman pumping of the thermocline by the wind stress curl. This is the same as the solution with no boundaries. -> the thermocline moves uniformly downward through Ekman pumping.
1. the balance is: $$-\frac{f_0^2}{c^2}\frac{\partial \psi'}{\partial t}=-lX_0$$
2. When the waves from the eastern boundary arrive, the solution switches to the Sverdrup balance:
1. $$\beta\psi'_x=-lX_0$$