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Introduction to Ocean Dynamics Lie-Yauw Oey

COVER ILLUSTRATION

A three-dimensional surface of near-inertial energy = 0.03 m2s-2 on Sep/03/12:00, approximately one week after the passage of the disastrous hurricane Katrina, Aug/25-30/2005, in the Gulf of Mexico, USA, simulated by the Princeton Ocean Model. This shows penetration of intense energies to deep layers due to the presence of the warm-core ring and the Loop Current, both represented by the dark contours in the three cut-away xy-planes. The location of an observational mooring where extensive model-observational analyses have been conducted is shown as vertical dashed line [from: Oey et al. 2008: “Stalled inertial currents in a cyclone,” Geophysical Res. Lett., 35, L12604, doi:10.1029/2008GL034273; with permission to reproduce].

“.. I have no special talents. I am only passionately curious..” (Albert Einstein)

CHAPTER 1: Wind-Driven Ocean CurrentsWe begin with the near-surface ocean response to wind. We will

for the moment assume that the ocean density ( 1035 kg m-3) is constant. Throughout this book, unless explicitly stated, we will use the SI units. Most of us experience the ocean perhaps through a summer swim in a coastal sea or in a lake. Wind-driven surface motions in the form of waves and surfs are familiar. However, these motions can involve large vertical accelerations and are of very small scales (order of centimeters to meters, O(cm-m)) which we will not model for now, at least not directly. Instead, we focus on motions which are of sufficiently

large scales in the horizontal (scale L ~ O(1-1000 km)) compared to

vertical (scale D ~ O(10-1000 m)), D/L << 1, that the pressure field

satisfies the hydrostatic equation:∂ p∂ z

=−ρg, (1-1)

where a Cartesian coordinate system xyz is used such that, conventionally, x and y are the west-east and south-north axes respectively, and z is the vertical axis with z = 0 placed at the mean sea level (MSL; Figure 1-1); x, y and z are measured in m. The pressure p is in N m-2 (also called Pascal, Pa), and g ( 9.806 m s-2) is the acceleration due to gravity. In other words, the motions of interest occur in such thin

fluid layers (with D/L << 1) that as far as the vertical force balance is

concerned the fluid may be treated as being static.1 For such an atmosphere-ocean system, the pressure at some depth z = z’ in the water is just the weight per unit area of fluid (water and air) above, or

p = [mass per area].g = [ (-z’) + <eazea> ].g (1-2)

1 The thinness of the oceanic (or atmospheric) layer relative to its horizontal extent is comparable to that of the thickness of paper of this book relative to its width.

Figure 1-1. A sketch of water of mean depth H(x,y) and free surface (x,y,t).

Here, z = (x,y,t) is the free surface; and/or z’ can be positive or negative, and z’ ( 0) represents the distance from the sea-surface to the

point z = z’ in the water. The term <eazea> represents∇ the mass per area

of the atmosphere, it is given by

<eazea> = ∫❑

∞

❑adz ∫0

∞

❑adz (1-3)

where a is the air density. Thus ea and zea may be thought of as the effective density and height of the “air-containing” atmosphere. The main bulk (about 80%) of the atmosphere’s mass is within the lowest layer (the troposphere) of about 10 km thick such that the atmospheric pressure at the sea surface, pa = <eazea>.g 105 N m-2 = 1 bar.2 Equation (1-2) can be written as

p = g(-z’) + pa(x,y,t) (1-4)which is also obtained by integrating (1-1) from z = z’ to z = ; in general, the pa is a function of the horizontal position and time.

If both a and pa do not vary with (x, y) and either are steady or at most only vary so slowly with time that the fluid remains in hydrostatic equilibrium, i.e. (1-1) is satisfied, then the fluid remains motionless provided that initially it is also at rest. A simple system to imagine is water in an annulus channel (Figure 1-2). If the axis of the annulus is far from the center, so that r/Ranu << 1, where r is the width of the channel,

2 An easy way to remember this is to use ea 1 kg m-3, zea 10 km and g 10 m s-2. Note also the familiar weather reporting of “1000 millibar” etc, which is 1 bar, the approximate sea-level pressure.

and Ranu is the radius of curvature of the annulus, then one can approximate the (motionless) water to be in a straight channel with x directed along the channel axis, y across the channel (Figure 1-2); a vertical slice along the axis then gives Figure 1-1. Modelers refer to this kind of channel a “periodic channel” since, starting from a yz-section at any x-location and going around along the axis, the field variables return to the same values. When conducting geophysical fluid experiments, a periodic channel is a convenient configuration to use because it often allows the extraction of essential flow physics while at the same time alleviates the modeler from having to formulate more complicated boundary conditions.

Figure 1-2. A circular annulus channel of radius Ranu much larger than the width of the annulus r. This is used to illustrate along-channel periodic fluid motion within the annulus.

1-1: A Simple Shearing Flow by the WindConsider therefore the straight channel (Figure 1-1) in which we

will further assume that all variables are independent of the cross-channel axis “y,” and that the earth’s rotation is nil. The flow is described by the three components of velocity (u, v, w) and the pressure p. Since there can be no flow across the channel wall, i.e.

v = 0, at y = 0 and y = r, t. (1-5)the cross-channel velocity v must be zero everywhere; the symbol t means “at all time.” The channel is further idealized by letting its depth H = constant. Initially, at t = 0, the water is at rest. A wind stress (N m-

2) is then applied uniformly at the surface, so that is a function of time only:

= (t) (1-6)Similarly, we could stipulate that pa is also a function of time only; but for simplicity we will set

pa = 0. (1-7)

It follows that, the along-channel velocity u cannot vary with x. Therefore, at any point, there can be no accumulation (convergence or divergence) of mass, and since w = 0 at z = H, the vertical velocity is nil everywhere, and the sea-surface remains flat. Thus,

u = u(z, t), w = 0 = /x = p/x. (1-8)

Under these very specialized conditions, a parcel of fluid experiences acceleration only due to the vertical shear stress. The momentum balance is then:

DuDt

=∂❑zx

∂ z (1-9)

where zx denotes shear stress (force per unit area) in the x-direction acting on the fluid elemental face that is perpendicular to the z-axis, and D(.)/Dt is the material derivative which for a fluid property S is given by

DSDt

=∂ S∂ t

+u ∂ S∂ x

+v ∂ S∂ y

+w ∂ S∂ z . (1-10)

For the present specialized case, the last three terms are nil, and with S =

u, we have DuDt = ∂u

∂ t . A loose analogy (of equation 1-9 is a stack of poker

cards which are glued with a special adhesive that is never dry and thus remains sticky. The top card is then “pulled” parallel to the card’s surface a small distance and it drags upon the card below it. In our fluid system, the wind is doing the “pulling” by transferring air momentum onto the water’s surface (and we assume that no ripples or wind waves are produced!). Momentum is vertically transferred from the surface to fluid layers below – upper fluid drags upon the lower fluid which in turn drags upon the layer further below, and so on (Figure 1-3).

Figure 1-3. The ocean is treated as a (vertical) stack of thin layers each of thickness dz : shown are layers “1” and “2;” layer “0” is air. The Fmn is (shearing) force per unit (horizontal) area due to layer “m” on layer “n.” The formulae show that the net force per unit area in layer “1” is F1net = F01 + F21 =dz.¶tzx/¶z. Therefore, the net shearing force per unit mass is F1net dx dy/(rdxdydz) = (¶tzx/¶z)/r. This, in the idealized case described in the text, is = ¶u/¶t, the acceleration in the x-direction.

For the so called Newtonian fluid such as water, the shear stress zx is proportional to the vertical rate of change of fluid velocity (i.e. the vertical “rate of strain”):

zx = ∂u∂ z (1-11)

where is the viscosity which to a good approximation may be taken as a constant and moreover is rather small; it is 10-3 kg m-1 for water and 1.7×10-5 kg m-1 for air. Equation (1-11) is valid for laminar or slowly-moving viscous flows void of turbulent movements we usually see in say, a mountain stream or in swirling air vortices around a house. However, as will be seen below, one often model oceanic and atmospheric flows using a similar formula, except that an “eddy viscosity” e is used to represent the aggregated effects of small eddies on the large-scale flows. The eddy viscosity is much larger than the molecular viscosity and is moreover not a constant – it depends on the flow. The “e” will be ascribed below but for now we will proceed with our model formulation using equation (1-11). Equation (1-9) then becomes:

∂u∂ t= ∂

∂ z(❑e

∂ u∂ z

) (1-12)

where e = e/, the kinematic eddy viscosity. This is a simple “heat-diffusion” equation that can be easily solved subject to some initial conditions and boundary conditions at z = 0 and z =H; for example:

u = 0, t = 0 (1-13)u = 0, z = H (1-14a)

❑e∂u∂ z

=❑o

❑ , z = 0, (1-14b)

where to is a constant wind stress.Exercise 1-1A : Use POM to solve (1-12) subject to (1-13) and (1-14) in a periodic straight channel. Compare your solution with exact (analytical) solution. Experiment with different grid sizes as well as with more complicated initial and/or boundary conditions. Analyze and discuss your results (e.g. use Matlab or IDL etc).Exercise 1-1B : Repeat Exercise 1-1A using a e-profile that decays with depth, e.g. e = eo exp(z), where eo and are constants.Exercise 1-1C : Repeat Exercise 1-1A using an annulus channel (Figure 1-2) instead of the straight (periodic) channel. Use a curvilinear grid (in POM) for this exercise. Experiment with various values of Ranu and r. Compare and discuss your results. Are the results substantially different from the straight-channel case when Ranu is “not too large” (define what this means), why? Is equation (1-12) still valid then? Why (or why not)?

Chapter 1.A: Equations of Motion

The material derivative (1-10), DS/Dt = ∂S/∂t + u.∇S, has two parts:

∂S/∂t = differentiation of S with time keeping space fixed;

But since S(x,t), x = (x,y,z) = (x1,x2,x3), we have:

δS = ∂S/∂t.δt + ∂S/∂x.δx + ∂S/∂y.δy + ∂S/∂z.δz

so that following a fixed particle (the "material"):

DS/Dt = ∂S/∂t + ∂S/∂x.u + ∂S/∂y.v + ∂S/∂z.w

= ∂S/∂t + u.∇S (1.A-1)

Conservation of Mass

Consider a fixed elemental volume (fig.1.A-1), δV=δxδyδz, its mass is ρδV, then

Fig. 1.A-1

∂(ρδV)/∂t = (∂ρ/∂t)δV = rate of increase of mass in δV

= [mass flux in] - [mass flux out]

= [ρu(x-δx/2,y,z).δyδz + ρv(x,y-δy/2,z).δxδz + ρw(x,y,z-δz/2).δxδy]

δx

δy

δz(x,y,z) ρu(x+δx/2,y,z).δyδz ρu(x-δx/2,y,z).δyδz

- [ρu(x+δx/2,y,z).δyδz + ρv(x,y+δy/2,z).δxδz + ρw(x,y,z+δz/2).δxδy]

so that ∂ρ/∂t = -[∂(ρu)/∂x + ∂(ρv)/∂y + ∂(ρw)/∂z] = -∇.(ρu), i.e.

∂ρ/∂t + ∇.(ρu) = 0

(1.A-2a)

or Dρ/Dt + ρ∇.u = 0 (1.A-2b)

Incompressibility

∇.u = 0 (1.A-3)

This should not be misunderstood to mean that Dρ/Dt = 0! The ρ can in fact change

because of e.g. heating. These density changes can affect buoyancy - hence forces

on fluid parcels, but the changes are too small to affect mass (continuity) balance.

For example, the thermal expansion coefficient = -(∂ρ/∂T)/ρ ≈ 2×10-4 K-1 for water,

so that a 1K change in T only results in δρ/ρ ≈ 0.02%.

Forces on a Fluid Parcel

Mass × Acceleration = Summed Forces

ρ.δV × Du/Dt = Fgravity + Fpressure + Ffriction

= -gρδVk - [i∂p/∂x + j∂p/∂y + k∂p/∂z]δV + ρFfric.δV

so that Du/Dt = -∇p/ρ - gk + Ffric (1.A-4)

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

Problem Set#2: Derive the momentum-balance equation (1.A-4) following the fixed-control volume method used for deriving the mass-balance equation (1.A-2).

We consider each direction, x, y and z separately.

The rate of change of x-momentum within the elemental volume δV is ∂(ρu)/∂t.δV. By Newton's Law, this must equal to the summed forces that act in the same x-direction: gravity, pressure and frictional. Since we have taken 'z' to be oppositely-directed to gravity, the x-gravity force = 0. Let the net frictional force per unit mass in x-direction be Ffricx, so that the net x-directed frictional force is simply ρδVFfricx. The net x-directed pressure force acting on δV is the difference of pressure forces acting on the faces δyδz to the left x-δx/2 and right x+δx/2 of the center of δV (see fig. 1.A-2). By Taylor's expansion:

p(x-δx/2,y,z) = p(x) - (δx/2)(∂p/∂x) + O(δx)2, and p(x+δx/2,y,z) = p(x) + (δx/2)(∂p/∂x) + O(δx)2.

Therefore, net x-directed pressure force acting on δV = p(x-δx/2,y,z).δyδz - p(x+δx/2,y,z).δyδz = -∂p/∂x.δxδyδz = -∂p/∂x.δ V

Fig. 1.A-2

Therefore, ∂(ρu)/∂t.δV = 0 -∂p/∂x.δV + ρδVFfricx, i.e. ∂u/∂t = -(∂p/∂x)/ρ + Ffricx. A similar formula is valid for the y-momentum with "v" and "y" replacing "u" and "x" respectively. The z-momentum is also similar (with "w" and "z") but additionally the z-directed gravity force is non-zero, and is equal to -gρδVk (acting in the opposite direction of the upward positive "z"). The 3 (x, y & z) equations in vector form become:Du/Dt = -∇p/ρ - gk + Ffricwhere u = (u,v,w), ∇p = (∂p/∂x,∂p/∂y,∂p/∂z), -gk = (0,0,-g) and Ffric = (Ffricx, Ffricy, Ffricz).

δxδy

δz(x,y,z) p(x+δx/2,y,z).δyδz p(x-δx/2,y,z).δyδz

Chapter 1.B: Wind Waves Over a Flat-Bottom Ocean

For wind waves - waves at the ocean's surface that we see when visiting the beach or

looking out from a cruise boat in the open seas, the horizontal scales L are generally

of the order of or smaller than the water depth-scale D , or L /D ~ O(1) or << 1. In

most cases, the phase speeds of wind waves are much larger than the speeds of the water parcels: |u|/|c| << 1, where c = phase velocity of the wave and u = (u, v, w) is the fluid parcel velocity, which means that the waves may be approximately described by the linearized equations of motion, and equation (1.A-4) becomes:

∂u/∂t = -∇(p/ρ) - gk (1.B-1abc)

Note that equation (1.B-1abc) has 3 equations and, since |u| is small, terms O(|u|2) on

the LHS are dropped because of linearization.3 We again assume that the density ρ is

constant. Take the curl, ∇×(1.B-1abc):

∂(∇×u)/∂t = 0 (1.B-2abc)

which states that if the vorticity (∇×u) is zero initially or at any instant of time,4 then

it remains zero at all time. Then, there exists a scalar function called the "velocity

potential" ϕ, so that:

u = ∇ϕ (1.B-3abc)

The fluid is incompressible, ∇.u = 0 (see 1.A-3), so that:

∇2ϕ = 0 (1.B-4)

This is a Laplace equation which is hardly an equation that describes "waves"! The wave part as we will now show comes from the free-surface boundary condition. The surface-wave problem serves as an excellent illustration of the importance of the boundary condition in determining the physics of the problem. The Laplace equation merely describes the distribution of the currents and mass field within the water.

The boundary condition at the ocean's bottom is:3 At this point we do not quite know that |u| << |c|, so we will check later, a priori.4 For periodic wave motion, ∇×u (indeed every periodic field) must go through zero periodically.

w = ∂ϕ/∂z = 0, z = -H (1.B-5)The boundary condition at the ocean's surface is:

w = ∂ϕ/∂z = ∂η/∂t, z = η (1.B-6)From (1.B-1) and (1.B-3), since gk = ∇(gz),

∇(∂ϕ/∂t + p/ρ + gz) = 0, i.e. ∂ϕ/∂t + p/ρ + gz = F'(t)(1.B-7)

where F'(t) = dF/dt is an arbitrary function of time. However we can always define a new ϕnew = ϕ + F(t) without changing the velocity field defined by (1.B-3), i.e. u = ∇ϕnew = ∇ϕ. Therefore, the RHS of the second equation in (1.B-7) can be set to zero:

∂ϕ/∂t + p/ρ + gz = 0(1.B-8)

where we have used the same symbol ϕ (instead of calling it ϕnew). Applying this equation at z = η, then ∂/∂t the resulting equation and using (1.B-6) to replace ∂η/∂t by ∂ϕ/∂z:

∂2ϕ/∂t2 + g∂ϕ/∂z = -ρ-1.(∂pa/∂t), z = η(1.B-9)

where pa is the atmospheric pressure, which is assumed known. This is an immensely complicated equation because it is applied at the z-position z = η which is one of the unknowns (the others are u)

we are trying to solve! To simplify, since we expect that |η| ~ |u|, we may apply (1.B-9) at z = 0. To check that this is consistent with our previously assumed linearization, i.e. that terms O(|u|2) and higher

may be neglected, we can expand the LHS of (1.B-9), which we shall call G(x,η),

about z = 0:

G(x,η) = G(x,0) + η.∂G/∂z|z=0 + ... = G(x,0) + O(η.ϕ) + ...Therefore, by replacing G(x,η) by G(x,0), we are commiting error no larger

than O(η.ϕ) ≈ O(|u|2) that we already committed in the governing equation (1.B-1).

To be consistent, we must also neglect O(η.ϕ). Therefore, (1.B-9) can indeed be

applied at z = 0:∂2ϕ/∂t2 + g∂ϕ/∂z = -ρ-1.(∂pa/∂t), z = 0 (1.B-10)To summarize, the equations that govern the small-amplitude (i.e. linear) surface gravity waves in an inviscid, incompressible and irrotational fluid are (1.B-4), (1.B-6) and (1.B-10):

∇2ϕ = 0

∂ϕ/∂z = 0, z = -H∂2ϕ/∂t2 + g∂ϕ/∂z = -ρ-1.(∂pa/∂t), z = 0 (1.B-

11abc)

The last equation containing ∂2ϕ/∂t2 is the mathematical source of the wave. Physically, the equation describes the interplay between the restoring gravitational force at the upper boundary and the

relation between the free surface elevation and the vertical velocity at the upper boundary.We seek a free (i.e. pa = 0) plane-wave solution for (1.B-11) of the form:

ϕ(x,t) = Re{A.ei(K.x-ωt)} (1.B-12)

where A = wave amplitudeK = wave vector = (k1, k2, k3) or (k, l, m)ω = frequencyRe = Real part of

Then the Laplace equation (1.B-11a) gives:

k12 + k2

2 + k32 = 0 (1.B-13)

which is not possible if all k's are real. For large x and y, the solution needs to remain finite (i.e. does not "blow up"). Therefore, k1 and k2 must be real, which means that k3 must be purely imaginary. These reasonings suggest that we look for solutions that are periodic in x, y, and t, of the following form:

ϕ(x,t) = A(z).ei(kx+ly-ωt) (1.B-14)

where again the Real part is what we want. Substitute into the Laplace equation:

d2A/dz2 - Kh2A = 0, (1.B-15a)

where Kh2 = k2 + l2 (1.B-15b)

is the squared amplitude of the horizontal wave vector. We see that the Laplace equation simply gives us the vertical structure of the plane-wave solution. The solution to (1.B-15a) is:

A(z) = a.exp(Khz) + b.exp(-Khz)

The boundary condition (1.B-11b) gives dA/dz|z=-H = 0, i.e.

a.exp(-KhH) - b.exp(KhH) = 0; so that b = a.exp(-2KhH)

Then, A(z) = a.{exp(Khz) + exp[-Kh(z+H)].exp[-KhH]}= 2a.exp[-KhH].{exp[Kh(z+H)] + exp[-Kh(z+H)]}/2

Therefore, A(z) = .cosh[Kh(z+H)] (1.B-16)

where "" may be taken as a constant. Substitute this "A(z)" and (1.B-14), i.e. ϕ(x,t)

= A(z).ei(kx+ly-ωt) into the boundary condition (1.B-11c) at z=0 now gives:

-ω2.cosh(KhH) + gKh.sinh(KhH) = 0,

i.e. ω2 = gKh.tanh(KhH),

i.e. ω = {gKh.tanh(KhH)}1/2 (1.B-

17a)

and phase speed c = ω/Kh = (gH)1/2.{tanh(KhH)/(KhH)}1/2 (1.B-17b)

Small & Large KhH limits:We will now examine (1.B-17) for cases when wavelengths 2/Kh are long compared to depth "H", i.e. when KhH << 1, as well as when KhH >> 1 (i.e. short wavelengths). Since

tanh(q) = (eq - e-q)/ (eq + e-q),

we have for the small and large argument "q", the following expansions:

tanh(q) ~ q, |q| << 1and tanh(q) ~ 1 |q| >> 1

Equation (1.B-17) then becomes, for "long" waves:ω ~ Kh(gH)1/2, for KhH << 1 (1.B-18a)

c ~ (gH)1/2 (1.B-18b)and, for "short" waves:ω ~ (gKh)1/2, for KhH >> 1 (1.B-19a)

and c ~ (g/Kh)1/2 (1.B-19b)

See Fig.1.B-1.

Fig.1.B-1. Long or shallow-water waves are non-dispersive and have wavelengths (h) larger than the water depth (H). Short or deep-water waves are dispersive and have wavelengths smaller than the water depth.

Note that:

1. the sign in (1.B-17 to 1.B-19) means that for each Kh (horizontal wave-vector amplitude or wave number), there exist 2 waves propagating in opposite direction, one parallel to Kh = (k,l) and the other antiparallel;

2. the frequency ω and phase speed c depend only on the wave-vector amplitude Kh =

|Kh|, and not on the direction of Kh;3. the "c" is different for different wavelengths (=2/Kh). In other words, the waves

disperse. The relation that connects "ω" to "Kh", eqation (1.B-17a), is therefore

called the dispersion relation.

That phase speed (and later we learn wave energy propagation velocity) depends on wave number is very interesting. From (1.B-17b), one can see that "c" is maximum when KhH is << 1 (i.e. for long waves). This can be shown by differentiating "c" with respect to "KhH" etc to find the KhH that makes "c" a maximum. However, a graphical construction also shows that KhH (= q, say) ~ 0 gives maximum "c", since

Wavelength = h = 2/Kh >> H

H

Shallow-water wave

h << H

H

Deep-water wave

d(tanh(q))/dq ~ 1 - q2 for |q| << 1, so that the graph (plot) of F = tanh(q) always falls below the graph of the line function "F = q", and {tanh(q)/q} is always < 1, and is a maximum at q = 0 (see fig.1.B-2). This maximum phase speed is given by (1.B-18b), i.e. by c ~ (gH)1/2. In this long-wave limit, the waves become non-dispersive. On the other hand, the phase speed for short waves (KhH >> 1) is small and becomes smaller with increasing Kh, given by equation (1.B-19b).

Fig.1.B-2. A sketch showing that {tanh(q)/q} is always < 1.

The solution: surface elevation, velocity & pressure fields

From (1.B-14) & (1.B-16), the solution for the velocity potential ϕ is:

ϕ(x,t) = .cosh[Kh(z+H)].ei(kx+ly-ωt) (1.B-20)

From the surface condition (1.B-8), the surface elevation η (for pa=0) is:

η = [-∂ϕ/∂t|z=0]/g = i(ω/g).cosh(KhH).ei(kx+ly-ωt) = ηo.ei(kx+ly-ωt) (1.B-21a)

where ηo = i(ω/g).cosh(KhH) (1.B-21b)

which is taken as a real-number amplitude (i.e. by appropriately defining the spatial

point where η is a maximum). Then, taking the real part of (1.B-21a):

η = ηo.cos(kx+ly-ωt) (1.B-22)

q

F

Slope = dF/dq = 1-q2

F = tanh(q)

F = qSlope = 1

+1

-1

Using ηo in place of (from 1.B-21b), i.e. = ηog/[iω.cosh(KhD)], the ϕ (eqn.(1.B-

20)) then becomes:

ϕ(x,t) = ηog/[ω.cosh(KhH)].cosh[Kh(z+H)].sin(kx+ly-ωt)

= ηoω/[Kh.sinh(KhH)]. cosh[Kh(z+H)].sin(kx+ly-ωt) (1.B-23)

The dispersion relation ω2 = gKh.tanh(KhH) has been used on the second line above.

The velocity (u,v,w) = ∇ϕ can now be obtained:

uh = ηoω(Kh/Kh).cos(Kh.xh-ωt).cosh[Kh(z+H)]/ sinh(KhH) (1.B-24a,b)

w = ηoω.sin(Kh.xh-ωt). sinh[Kh(z+H)]/ sinh(KhH) (1.B-24c)

= ∂η/∂t. sinh[Kh(z+H)]/ sinh(KhH)

the last line is obtained after using (1.B-22). In above, uh = (u,v) is the horizontal velocity vector and xh = (x,y) is the horizontal position vector.

The pressure field "p" may be obtained from the Bernoulli equation (1.B-8):

p = -ρgz - ρ∂ϕ/∂t (1.B-25)

The first part, -ρgz, is just the hydraustatic part which has nothing to do with the

wave. Substituting ∂ϕ/∂t from (1.B-23):

p = -ρgz + ρgηo.cos(Kh.xh-ωt). cosh[Kh(z+H)]/cosh[KhH]

= ρg{η. cosh[Kh(z+H)]/cosh[KhH] - z} (1.B-26)

Note:

(1). The horizontal velocity uh is in the direction of the wave vector Kh;(2). Each perturbation variable in u and p (equations 1.B-24 & 26) is proportional to

the amplitude ηo of the free surface elevation;

(3). If the water is deep, i.e. KhH >> 1, then since both

cosh[Kh(z+H)]/sinh(KhH) and sinh[Kh(z+H)]/ sinh(KhH) ~ exp(Khz),

all perturbations (u and p) decay exponentially from the free surface, with a depth scale ~ 1/Kh ~ h (the wavelength):

u and p ~ exp(Khz), KhH >> 1 (1.B-27)

Physically, the waves do not "feel" the bottom when the latter is deep. That is why the wave frequency/wave numer relation (1.B-19) for deep-water waves (hence the phace speed) do not depend on water depth H in this limit;

(4). On the other hand, for very long waves, i.e. KhH << 1, cosh[Kh(z+H)]/sinh(KhH) ~ (HKh)-1 and sinh[Kh(z+H)]/ sinh(KhH) ~ 1+z/H, so that (using also 1.B-18):

uh = (η/H)c(Kh/Kh) (1.B-

28a)

w = (∂η/∂t)(1+z/H) (1.B-28b)

p = ρg(η - z) (1.B-28c)

The horizontal velocity uh has a magnitude (speed) that is less (much less) than the

wave phase speed c, by a factor of | η/H| << 1. The vertical velocity w is linear with

depth. The pressure is simply hydrostatic - see equation (1-4).Trajectories of Fluid Parcels in the Plane Wave

Let x be the wave direction, so that i = Kh/Kh, and let us assume that the subsequent evolution remains in the xz-plane. Then let (X,Z) be the x and z displacements of the fluid parcels about some original position which for simplicity is taken to be at the origin (0,0). Then for small displacements, from (1.B-24):

dX/dt = u = ηoω.cos(kx-ωt).cosh[k(z+H)]/sinh(kH) (1.B-29a)

dZ/dt = w = ηoω.sin(kx-ωt).sinh[k(z+H)]/sinh(kH) (1.B-

29b)

Integrating ∫dt gives:

X = - ηo.sin(kx-ωt).cosh[k(z+H)]/sinh(kH) (1.B-30a)

Z = + ηo.cos(kx-ωt).sinh[k(z+H)]/sinh(kH) (1.B-30b)

Therefore (since sin2+cos2 = 1),

X2/Lx2 + Z2/Lz

2 = 1 (1.B-31a)

where Lx = ηo.cosh[k(z+H)]/sinh(kH), Lz = ηo.sinh[k(z+H)]/sinh(kH)

(1.B-31b)The trajectories are ellipses and Lx and Ly are their semi-major and semi-minor axes respectively. Note that Z = 0 at z = -H. For deep-water waves (i.e. when

wavelengths << H, or kH >> 1), Lx ~ Lz ~ ηo.ekz so that the trajectories are circles

which shrink with depth as the radii decrease exponentially with z (<0). For shallow-

water waves (i.e. when wavelengths >> H, or kH << 1), Lx ~ ηo/(Hk) and Lz ~

ηo(1+z/H); therefore the ratio of z-length to x-length of the ellipse is Lz/Lx ~ (kH)

(1+z/H) << 1, and the trajectories are essentially horizontal lines parallel to the bottom.

Visualizing the wave

Since η = ηo.cos(kx-ωt) (see 1.B-22), the u-velocity is in phase with the wave's free

surface elevation, i.e. at any z-level, the u is positive and has the maximum speed

under a wave-crest where the phase = kx-ωt = 2n (n=0, 1, 2...), and it is negative

and the maximum speed under a trough where kx-ωt = (2n+1). (n=0, 1, 2...) - see

fig.1.B-3.

Fig.1.B-3. A schematic showing deep-water wave with a given wave number "k"

propagating from left to right in an xz-plane: η = ηo.cos(kx+ly-ωt). In the linear

theory assumed in the text, the free-surface elevation is of small amplitude; in the sketch here, the amplitude has been exagerated. Circles (ellipses) are trajectories traced by fluid parcels as the wave passes. As the wave passes, the u-velocity at a crest (trough) is positive (negative) and the parcel experiences (i) a forward (backward) movement, then (ii) downward (upward), then (iii) backward (forward) - i.e. at the trough (crest), then (iv) upward (downward), and finally (v) (=(i)) forward

(backward) again at the next crest (trough). At the surface, the orbital diameter = 2ηo

(see eqn.1.B-31b), i.e. twice the wave amplitude, but it decreases exponentially with depth. The sketch also shows that the forward velocity at the crest is a little stronger than the backward velocity at the trough (why?), and this small difference actually shows up in the linear theory even though the free-surface condition is applied at z=0!

Stokes Drift:Imagine yourself shrinked to the size of a small fluid parcel, bobbing (floating or 漂浮) on the surface of the water as a wave with fixed "k" passes (see fig.1.B-3). You are carried forward at the wave crest, then downward and then backward when the wave trough arrives. Assuming deep-water wave in an xz-plane, we have,

ucrest ≈ ηoω.exp(kηo) ≈ ηoω.(1+ kηo+...) at the wave crest, and

kx-ωt = 0 η = ηo.cos(kx-ωt) /2 3/2 2

u < 0

u > 0Wave

propagating

utrough ≈ ηoω.exp(-kηo) ≈ ηoω.(1- kηo+...) at the wave trough,

so that the net forward movement, or "drift" is, very roughly, ucrest - utrough ~ 2ωkηo2.

This is actually close to the more exact value that we now derive. As you follow the fluid parcel being carried forward and backward by the passing wave, your position vector = (1, 3) keeps changing, and at any time your x-velocity u(,t) deviates from the velocity u(x,t) at the initial fixed point x = (x, z), where | - x| is small compared to the wavelength. In fact, we see from (1.B-29 & 30) that:

1 = x + ∫udt = x - ηo.sin(kx-ωt).ekz (1.B-32a)

3 = z + ∫wdt = x + ηo.cos(kx-ωt).ekz (1.B-32b)

Then u(,t) can be expanded in Taylor series about x:u(,t) = u(x,t) + (∂u/∂x).(1-x) + (∂u/∂z).(3-z) + O(|-x|2) (1.B-33)

Since u(,t) = x-velocity of the parcel riding on the wave, while u(x,t) = x-velocity at a fixed point, their difference u(,t) - u(x,t), when averaged over the wave cycle (i.e.

when averaged over the phase of the wave "x-ωt") denoted by <.>, is then the drift

mentioned above. An estimate of this Stokes drift (after Stokes (1847) who first discovered and described it) can be obtained from (1.B-33).

<(∂u/∂x).(1-x)> = ∫(ηo2ω.k.sin2(kx-ωt).e2kz).d(kx-ωt)/(2) (1.B-34a)

<(∂u/∂z).(3-z)> = ∫(ηo2ω.k.cos2(kx-ωt).e2kz).d(kx-ωt)/(2) (1.B-34b)

where the integral is from 0 to 2. Therefore,

uStokes = < u(,t) - u(x,t)> ≈ ηo2ω.k.e2kz (1.B-35)

Fig.1.B-4. Surface Stokes drift and e-folding depth (m) in the North Pacific Ocean computed from a wave-hindcast model. Courtesey of Dr. H. Tamura (JAMSTEC).

Langmuir Cells

Notes on Langmuir circulation by Leibovich (Ann. Rev. Fluid Mech, 1983, vol15, 391-427)http://princetonjournalclub.pbworks.com/w/file/25027348/Leibovich_ARFM_1983_Langmuir_circulation.pdf

The Craik-Leibovich equation can be (heuristically) derived by using the following identity for the material derivative:

Du/Dt = u/t + x u +(u2/2). (1.B-36)

Then writing u = v + us, where v = current, us = Stokes vel, and = xu xv = (exact if us(z)). Then substituting Du/Dt into the full (i.e. Navier-Stokes) equation:

Dv/Dt + x us = -(us2/2 + p) + … (1.B-37)

The author discussed 2 mechanisms for generating Langmuir circulation: CL1 (Fig.1.B-5) is direct-drive mechanism and CL2 (Fig.1.B-6) is instability mechanism.

The author explains the mechanism (below) by assuming wind blowing in x-direction which then generates waves and Stokes velocity Us.

Fig.1.B-5: CL1-mechanism (direct drive):

http://princetonjournalclub.pbworks.com/w/file/25027348/Leibovich_ARFM_1983_Langmuir_circulation.pdf

http://princetonjournalclub.pbworks.com/w/file/25027348/Leibovich_ARFM_1983_Langmuir_circulation.pdf

In this case, Us is a function of both z and y (cross-wind direction) – it is weaker at some y and stronger at other y as shown in Fig.1.B-5. (If Us is a function of z only, i.e. the Stokes velocity is uniform horizontally (i.e. the waves are uniform also), then the vortex force is balanced by wave-radiation-stress term in the vertical (see ftp://aden.princeton.edu/pub/lyo/lyo/research_group/journals/longuet-higgins-

stewart-dsr64_lyonotes.pdf)). Because of U(z) shear of the current (i.e. U/z > 0),

we produce vorticity in y-direction y = U/z directed in positive y-direction. The

vortex force fvortex = fv = us x is still upward everywhere, but now it is stronger where Us is stronger and weaker where Us is weaker, as shown by the vertical arrows in Fig.1.B-5. The differential strengths then produce the roll as shown with downwelling where Us is weakest (Fig.1.B-5). Therefore the “rolls” are directly driven. The CL1-mechanism is discussed further on pages 403-405.

Fig.1.B-6:CL2-mechanism (instability):

ftp://aden.princeton.edu/pub/lyo/lyo/research_group/journals/longuet-higgins-stewart-dsr64_lyonotes.pdf

ftp://aden.princeton.edu/pub/lyo/lyo/research_group/journals/longuet-higgins-stewart-dsr64_lyonotes.pdf

The CL1 depends on special wave structure in which the Us(y,z) can vary with “y.” A more general case is when Us(z) – i.e. function of “z” only(!) and the basic flow is still U(z), but we wish to find out how a small current irregularity (i.e. small perturbation) u(y,z,t) can grow and what its structure is. Suppose that the u varies with “y” and has a small maximum at “y=0” (Fig.1.B-6). Then = kz (where z = -u/y) points upward for y>0 and it points downward for y<0. Therefore, the vortex force fv points to negative-y direction in y>0 region, and it points to positive-y direction in y<0 region. Fluid is pushed to the center (y=0) where u is a maximum, and where there is then downwelling. This then further increases “u” and it therefore amplifies (i.e. instability: (u2)/t ~ -<uw>U/z etc; also -(uw)/z & -(uv)/y>0 which lead to u/t>0 – i.e. convergence of x-momentum flux at y=0 increases u, which increases z, hence fv, hence more momentum flux convergence etc).

Kinematically, one sees from Fig.1.B-6 that the kz is tilted by Us/z, and again produces roll with downwelling at y=0. CL2-mechanism is more general and the author discussed further on pages 405-411 which are very interesting.

The Group Velocity:

Let us write the dispersion relation in the following general form:ω = Ω(k; x,t) (1.B-38)

In our case of surface gravity waves, Ω = {gKh.tanh(KhH)}1/2 (eqn.1.B-17a). The

group velocity is then defined as the gradient in the wavenumber-space of the dispersion relation:

cg = ∇kω; or cgi = ∂ω/∂ki (1.B-39)

For wave propagating in x-direction, the group velocity cgx for deep-water waves is derived as follows.The dispersion relation is2 = gKh.tanh(KhH). For deep-water waves KhH >> 1, this is:

2 = gKh = g(k2 + l2)1/2

So, 2.¶/¶k = gk/Kh ¶/¶k = (g/Kh)1/2(k/Kh)/2 = (c/2).(k/Kh); where c = /Kh (phase speed).Similarly, ¶/¶l = = (c/2).(l/Kh)So that, cg = (cgx, cgy) = (¶/¶k, ¶/¶l) = (c/2).(k, l)/Kh (1.B-40)

Since (k, l)/Kh is just the unit-wave vector (i.e. a unit vector in the direction of the wave), (1.B-40) says that the group velocity is in the wave direction with a magnitude that is half of the wave phase speed “c”.

The corresponding group velocity cgx for shallow-water waves is derived as follows.For shallow-water waves, the dispersion relation is = (gH)1/2.Kh; so that c = /Kh = (gH)1/2, and

cg = (cgx, cgy) = (¶/¶k, ¶/¶l) = c.(k, l)/Kh (1.B-41)In this case, the group velocity is also in the wave direction but with a magnitude that is the same as the phase speed “c”.

Chapter 1.C: Equation of motion, inertial oscillation & Ekman layers

The equation of motion (1.A-4) derived in Chapter 1.A, copied here:

Du/Dt = -∇p/ρ - gk + Ffric (1.C-1)

is an accurate application of Newton’s law in a fixed (inertial) frame of reference. However, we are earth-bound, and therefore observe motion in a rotating frame of refrence. In this frame, Coriolis acceleration and other features unique to rotating fluid motions can arise, and they are most easily described in the rotating frame. Since Newton’s law applies only to the fixed frame, our goal is then to transform (1.C-1) to one that is valid for a rotating frame. Before doing that, let us study a simple case to see how Coriolis acceleration can arise,

The Coriolis Acceleration & Geostrophic Balance

To see how Coriolis acceleration arises, consider a circular dishpan half-filled with water of constant density ρ placed on a turntable rotating at a constant rate = Ω (radians per second) counter-clockwise (Fig.A-1).

Fig.1.C-1. Illustrating the effects of rotation.After many, many rotations (which may take hours) and for a sufficiently fast Ω, the water level is then raised around the rim wall of the dishpan, and is symmetrical

around the center of the circular pan. The centrifugal force per unit volume directed outward on a fluid particle is then balanced by an inward pressure gradient:

Vθ2/r = (∂p/∂r)/ρ = g∂h/∂r, (1.C-2)

where Vθ = azimuthal velocity at radius r, positive counter-clockwise (i.e. same sense as Ω), h = depth of water, and we have used the hydrostatic relation that p = ρg(h-z) with z=0 at bottom of dishpan. If we define vθ to be the azimuthal velocity relative to the rotating dishpan, then:

Vθ = vθ + Ωr (1.C-3)

Then, (vθ + Ωr)2/r = g∂h/∂r. Simplifying:

vθ2/r + 2Ωvθ = g∂ηp/∂r, (1.C-4a)

where ηp = h - Ω2r2/(2g) (1.C-4b)

is the water surface referenced to the parabolic surface Ω2r2/(2g). The term "2Ωvθ" is the Coriolis acceleration term. The "2Ωvθ = g∂ηp/∂r" in fact is the geostrophic relation valid for large-scale flows. The centrifugal acceleration term " vθ

2/r" becomes important, e.g., in a tornado which is "small" and "fast". One may think of the rotating turntable as the "earth," vθ = zonal velocity measured on earth, and Vθ = zonal velocity measured in an absolute fixed frame.

Suppose the dishpan has a tiny hole in the center so that water is drained through the hole. Then a particle placed at the rim initially has zero velocity but begins to move inward. As it does so it conserves angular momentum:

Vθ.r = Ω.r12 (the initial angular momentum at the rim; where r1 = dishpan's

radius).

Then (vθ + Ωr).r = Ω.r12; i.e. vθ = Ω(r1

2 - r2)/r (1.C-5)

Thus the particle spirals inward in an anticlockwise sense (same sense as Ω) into the center of the dishpan, and its azimuthal speed increases as the particle nears the center (Fig.1.C-2).

Fig.1.C-2. Laboratory experiments: effects of rotation on the trajectory of a particle placed initially near the rim of the rotating circular dishpan for slow (left; GFD3_slow_rotation.mpeg) and fast (right; GFD3_fast_rotation.mpeg) rotations.

Fig.1.C-3. Particle trajectories in the ridial inflow experiment viewed in the rotating frame – i.e. the camera is co-rotating also (so that the parcel at the outer edge is at rest: v = 0, see (1.C-5)). Left: Ω = 5 rotations per minute (rpm); right: Ω = 10 rpm. In the limit Ω 0, the particle goes straight from edge to center hole, and when Ω >> 1, then the particle circles around the center without “falling” into the hole.Transforming the Equation of Motion to the Rotating Frame of Reference

From the rotating dishpan idea, eqn(1.C-3), we write more generally:

ufix = urot + Ω × r (1.C-6)

In this equation, ufix is the velocity as viewed from the fixed (or so called “inertial”) frame – by someone in the outer space looking down onto the earth. The observer

Ω Ω

sees ufix as the sum of 2 vectors. The first is urot which is the velocity in the torating

frame, and the second one is Ω × r which is the velocity of the fixed vector r as it is

being “carried” by the rotating frame. Consider any vector A = (Ax, Ay, Az):

A = iAx + jAy + kAz (1.C-7)

defined in the rotating frame. Here, (i, j, k) are the 3 unit vectors fixed in the rotating frame – i.e. they are just unit vectors defined on our usual (x, y, z) coordinate axes here on earth. The material derivative of A viewed from the rotating frame is then (since (i, j, k) are fixed):

(DA/Dt)rot = i DAx/Dt + j DAy/Dt + k DAz/Dt (1.C-8)

But, to the fixed-frame, outer-space observer, (i, j, k) are rotating and their velocities are:

(Di/Dt)fix = Ω × i, (Dj/Dt)fix = Ω × j, (Dk/Dt)fix = Ω × k (1.C-9)

Therefore,

(DA/Dt)fix = (DA/Dt)rot + Ω × A (1.C-10)

Fig.1.C-3. The left panel shows the velocity uin = ufix (e.g. of a fluid parcel on earth) as seen from the inertial-frame or fixed-frame observer. The right panel shows the same velocity uin which now comprises of 2 parts: the urot as seen (and measured) by us on earth, plus the velocity Ω × r caused by the fact that the parcel (which is located at r) is also being carried by the solid rotation of the earth. (Note that i=x, j= y, and k=z).Putting A = r in (1.C-10) gives (1.C-6), since

ufix = (Dr/Dt)fix and urot = (Dr/Dt)rot (1.C-11)

Putting A = ufix, we get:

(Duin/Dt)fix = [(D/Dt)rot + Ω ×] ufix = [(D/Dt)rot + Ω ×] (urot + Ω × r)

= (Durot/Dt)rot + 2Ω × urot + Ω × Ω × r (1.C-12)

Finally, then, the equation of motion in inertial frame, eqn(1.C-1) becomes:

(Durot/Dt)rot = -2Ω × urot - Ω × Ω × r -p/ - kg + Ffric (1.C-13)

Homework:1.C-1: (a) By considering first the direction of the vector (Ω × r), show (explain) that the vector Ω×Ω×r is directed in the opposite direction to the direction of r – hence - Ω×Ω×r is in the same direction of r (see sketch below).

Fig.1.C-4. Sketch to illustrate the centrifugal acceleration -Ω×Ω×r due to Ω.

(b) Then show also that the magnitude of -Ω×Ω×r is |Ω|2r.(c) Finally, show that -Ω×Ω×r = (|Ω|2r2/2).

1.C-2: What is the value of Ω on earth? What is the radius of the earth?

1.C-3: The vector - Ω×Ω×r of problem 1.C-1 is in fact the centrifugal acceleration due to the rotating frame Ω. Calculate the centrifugal acceleration of a particle fixed to the earth at the equator, and compare it with the acceleration due to gravity g, is it much larger or smaller than g?

Geopotential Surface:

In (1.C-13), the term -Ω×Ω×r is a vector that is directed outward from the center of

rotation – i.e. it is a centrifugal acceleration term. It can be written as (HW1.C-1):

- Ω × Ω × r = (Ω2.r2/2) (1.C-14)

Then (1.C-13) becomes (dropping subscript “rot”):

Du/Dt = -2Ω×u - p/ - (gz - Ω2.r2/2) + Ffric (1.C-15)

We see that, if Ω = 0, then the geopotential surfaces are just z = constant as gravity is

directed radially inward towards the center of the earth. But because of rotation, the centrifugal acceleration modified this direction of gravity to give g* which is “deflected” from the radial direction, as shown in Fig.1.C-5. The “g*” is what we measure.

Fig.1.C-5. The gravity vector g = -gk = -(gz) is directed towards the earth’s center. Note that k is directed radially outward. Centrifugal acceleration due to the earth’s rotation Ω is directed outwards as the vector (Ω2r2/2). The vector sum g* = -gk + (Ω2r2/2) is then as shown. Over millions of years, the earth’s (solid!) surface has adjusted to become an ellipsoid.The surface

= gz - Ω2.r2/2 (1.C-16)

is the modified gravitational potential, and geopotential surfaces are:

z* = z + Ω2.r2/(2g) = z + Ω2a2cos2/(2g) (1.C-17)

where a = radius of earth (see fig.1.C-4).

Modified Hydrostatic Balance:

If there is no motion, u = 0, then from (1.C-15) and using (1.C-16), we obtain,p/ = -; p/ + = constant (assume = constant) (1.C-18)

So that instead of p-surfaces coinciding with z-surfaces when fluid is in hydrostatic balance, they now coincide with -surfaces.Coriolis Acceleration:

Also from (1.C-15), in the absence of forces acting on it, a fluid parcel experiences an acceleration to the right of its velocity:

Du/Dt = -2Ω×u (1.C-19)

Inertial Oscillations:

Imagine a fluid parcel that is given an initial “push” with a velocity

uo = u(x,t=0) = (uo, vo) (1.C-20)

We assume that there are no other forces acting on the parcel, so that (1.C-19) is satisfied (note that we are following the parcel, so write D/Dt d/dt):

du/dt – fv = 0, where f = 2Ω; (1.C-21a)

dv/dt + fu = 0 (1.C-21b)Write = u + iv, where here i = (-1)1/2. Then [(1.C-21a)+i*(1.C-21b)] gives:

d/dt + if = 0, (1.C-22) = oe-ift = (uo+ivo).[cos(ft)-i.sin(ft)]; where o = (t=0) = (uo+ivo).

Therefore,u + iv = [uo.cos(ft) + vo.sin(ft)] + i.[vo.cos(ft) – uo.sin(ft)]

Hence, u = uo.cos(ft) + vo.sin(ft) & v = vo.cos(ft) – uo.sin(ft) (1.C-23)

The trajectory of the parcel is:

dx/dt = u, and dy/dt = v (1.C-24)

so that after integration, from t’ = 0 to t’ = t, (1.C-24) with (1.C-23) give:

x = xo + (uo/f).sin(ft) – (vo/f).cos(ft) + (vo/f) (1.C-25a)y = yo + (vo/f).sin(ft) + (uo/f).cos(ft) – (uo/f) (1.C-25b)

Suppose uo = 0, and also (xo,yo) = (0,0), then (1.C-23) and (1.C-25) give:

u = vo.sin(ft) and v = vo.cos(ft) (1.C-26)

x = (vo/f).(1 – cos(ft)) and y = (vo/f).sin(ft)

i.e. (x – vo/f)2 + y2 = (vo/f)2 (1.C-27)

Eqn.(1.C-26) shows that the velocity vector moves in a clockwise sense with time “t”, repeating every 2/f – see figure below.

Fig.1.C-6. Sketch showing clockwise rotating velocity vectors with time for f > 0: eqn.(1.C-26).Eqn.(1.C-27) says that the trajectory of the parcel is a circle centered at (vo/f,0) and with radius vo/f, as shown in the figure below.

Fig.1.C-7. The trajectory of a parcel executing inertial oscillation described by eqn.(1.C-27). Initially, the parcel is at (0,0) and has a northward velocity = vo.A Laboratory Experiment to Understand Inertial Oscillation:

Imagine a rotating solid parabolic surface with Ω = rotation rate (see Fig.1.C-1), and

the parabolic surface is described by (see eqn.1.C-4b):

Zparabolic = h = Ω2r2/(2g) (1.C-28)

measured from some reference level. A ball carefully placed on this surface will remain stationary since the surface is a “geopotential surface” and v = 0 by eqn.(1.C-4a). The force balance is given below:

Fig.1.C-8. The force balance for a ball placed on a rotating (Ω) parabolic surface Ω2r2/(2g).

The parabolic surface may be imagined to be made of smooth ice. Then a puck that is initially at the center and given a velocity vo approximately executes inertial oscillation described by fig.1.C-7 (approximate because there is actually a small friction acting on the puck) – see fig.1.C-9 below.

Fig.1.C-9. Trajectories of a puck that initially is at (0,0) (at t=0) and given an initial velocity (0,vo), as seen in (a) an inertial frame and (b) a rotating frame.Effects of Friction on Inertial Oscillation:

We can modify (1.C-21) to:

du/dt – fv = -ru, (1.C-29a)

dv/dt + fu = -rv (1.C-29b)

where r = friction coefficient = positive constant.In terms of = (u+iv), we then have,

d/dt + (if+r). = 0, (1.C-30)so that

= oe-(if+r)t = (uo+ivo).e-rt.[cos(ft)-i.sin(ft)]. (1.C-31)Therefore (see 1.C-23, replaing uo by uoe-rt and vo by voe-rt),

u = [uo.cos(ft) + vo.sin(ft)].e-rt, v = [vo.cos(ft) – uo.sin(ft)].e-rt (1.C-32)

The special case of uo = 0 then gives (c.f. 1.C-26):

u = vo.sin(ft) .e-rt and v = vo.cos(ft) .e-rt (1.C-33)

The trajectory equation is:

dz/dt = ivoe-(if+r)t where z = x+iy (1.C-34)

Integrating:x-xo = [(vo/f)/(1+2)].[1 – e-ftcos(ft) - . e-ftsin(ft)] (1.C-35a)y-yo = [(vo/f)/(1+2)].[e-ftsin(ft) + .(1 - e-ftcos(ft)] (1.C-35b)

where = r/f.

Fig.1.C-10. Inertial trajectories for particles that initially are at (xo,yo) = (0,0) for various friction parameter = r/f. Left is for various rather large after 5 nertial cycles (5×2/f). Right is for a small =0.02 after 8 cycles. The x and y are multiples of vo/f (= inertial radius when =0).

Homework:1.C-2A: Examine (1.C-35) for large time; also (a) small and (b) large . (You may assume xo = yo = 0).For finite , but large time , ft >> 1, x ~ (vo/f)/(1+2) and y ~ (vo/f)/(1+2). So friction eventually produces a steady state with the final resting particle shifted to the “east” and “north” of (0,0).For small , x (vo/f).[1-cos(ft)-.sin(ft)]/(1+2) & y (vo/f).[sin(ft)+.(1-cos(ft))]/(1+2). Friction makes the inertial radius smaller by 1/(1+2).For large , x ~ 1/2 and y ~ 1/. The particle comes to rest near the origin before it has a chance to execute any inertial oscillation.

Homework 1.C-2B: Forced Inertial Oscillations - Resonance:Suppose the RHS of (1.C-22) has a forcing term with amplitude “o” oscillating at the inertial frequency:

d/dt + if = o.exp(-ift), (1.C-2B-1)Physically, the inertial-oscillation system (1.C-21) is constantly being “pushed” at a frequency that matches the system’s natural oscillation which is the inertial oscillation at frequency ‘f.’ We will shortly give an example of this for the case of ocean’s response to a propagating typhoon.

The solution is: = o t.exp(-ift); (1.C-2B-2)

We can check that this is the solution by direct calculation using the solution (1.C-2B-2):

d/dt = o.exp(-ift) – if.o.t.exp(-ift) = o.exp(-ift) - if;where we have used (1.C-2B-2) in the second term.Therefore, d/dt + if = o.exp(-ift), and equation (1.C-2B-1) is satisfied.Therefore, (1.C-2B-2) is the solution and it shows an inertial oscillation that grows linearly with time. Compare it to the “free” solution without forcing just before equation (1.C-23), we see that the difference is only in the linear “t” term.

It is often observed that when a tropical cyclone (typhoon or hurricane) propagates over the ocean, SST on the right side of the storm is cooler than on the left. The reason is because as the storm is passing, the surface current on the right side is being blown by a wind (vector) that is turning in the same clockwise sense as its own tendency for natural inertial oscillation. In fig.1.C-2B, the storm with “eye” diameter ~ L is moving westward. The blue point on right side then first feels a southward wind (at “0”, i.e. when the storm is still far but approaching), then southwestward (at “/4” as the storm approaches), then westward (at “/2”, i.e. when the storm is directly south), then northwestward (at “3/4” as the storm is leaving), and finally northward (“” when the storm is again far, and leaving). So the wind turns

clockwise. How fast is this turning (i.e. frequency S)? That depends on “L” and also the propagation speed U of the storm. Clearly, the turning is fast (i.e. S is high) if U is large and/or L is small. In fact, S = U/L, and therefore the period TS = 2L/U. A typical U 5 m/s and L 200km. Then TS 1.45 days. The inertial period T = 0.5day/sin(), where = latitude. Therefore, at a latitude of 20o where typically typhoon passes, we have TS T and the inertial oscillation amplitude increases linearly with time. The strong currents then tend to produce more mixing which brings cooler deeper waters to the surface.

By contrast, wind on the left side turns anti-clockwise, which therefore tends to damp the clockwise-turning tendency of inertial oscillation, and inertial currents there become weaker.

Fig.1.C-2B.

FIG. 1.C-2C Surface inertial amplitudes (m s-1) from 27 to 30 Aug during passage of Katrina (2005) in the Gulf of Mexico. Only the amplitudes >0.5 m s-1 are shown. Note larger amplitudes right of the storm path (solid line). From Wang and Oey [2008; BAMS, 487-495]. For an animation of hurricane’s path, wind and ocean responses, see: (http://www.aos.princeton.edu/WWWPUBLIC/PROFS/ANIMATION/katrina/anim_foa19b7DayConf20050825_0

831_ssh.gif).

Ekman Transport:

We add the shear stress term to (1.C-21):¶u/¶t – fv = ¶X/¶z (1.C-36a)¶v/¶t + fu = ¶Y/¶z (1.C-36b)

where (X,Y) = (tx,ty). In vector form, the above is:

¶u/¶t + fk×u = ¶X/¶z (1.C-37)

Integrating ∫dz from z=-∞ to 0 (ocean) or z=0 to ∞ (for atmosphere) and noting that |X| 0 for large |z| [we use ∞ anticipating that the effects of X will be confined near z = 0]:¶M/¶t + fk×M = Xo, where + is for ocean and – for atmosphere (1.C-

38a,b)

and where M = ∫u.dz, z=-∞ to 0 (ocean) or z=0 to ∞ (for atmosphere); also Xo = surface stress. In steady state, ¶/¶t = 0, (1.C-38) becomes:

http://www.aos.princeton.edu/WWWPUBLIC/PROFS/ANIMATION/katrina/anim_foa19b7DayConf20050825_0831_ssh.gif

http://www.aos.princeton.edu/WWWPUBLIC/PROFS/ANIMATION/katrina/anim_foa19b7DayConf20050825_0831_ssh.gif

-fMy = Xo and fMx = Yo (1.C-39)

so that the ocean’s surface Ekman (mass) transport is to the right of the wind stress and the atmospheric planetary Ekman (mass) transport is also to the right of the frictional stress that acts on the wind.

Summing (1.C-38a) for ocean and (1.C-38b) for atmosphere, we see that the total mass transport (from deep ocean to top of atmosphere) satisfies:¶Mtot/¶t + fk×Mtot = 0 (1.C-40)

Physically, this means that for a given wind, the Ekman mass transports in both the oceanic and atmospheric laters are opposite, and cancel out: if Mtot = 0 initially, then it remains zero at all time.

The Ekman Spiral

The above involves vertical integrals of u the detailed structure (z-profile) of which is therefore not required. We now examine these details, first for the wind-driven surface layer in the ocean, then for the bottom boundary layer in ocean or atmosphere.In terms of the kinematic stress, t, (1.C-36) becomes:

¶u/¶t – fv = ¶tx/¶z = ¶2u/¶z2 (1.C-41a)¶v/¶t + fu = ¶ty/¶z = ¶2v/¶z2 (1.C-41b)

The second form uses the fact that for a Newtonian fluid (water and air),

t = ¶u/¶z (1.C-42)

where = kinematic (molecular) viscosity is assumed to be constant. For turbulent flows (ocean and atmosphere), a gross assumption is made that (1.C-42) is also true with = eddy viscosity. This assumption is highly inaccurate but is made in order to derive analytical solution which we can examine to gain insight into the structure of the surface layer. The boundary conditions are:

¶u/¶z = to at z = 0for ocean (1.C-43a)or u = 0 at z = 0for atmosphere or ocean’s bottom (1.C-43b)

Note that for ocean bottom boundary layer, we have used z = 0 instead of z = -H.

It is convenient to rewrite (1.C-41) and (1.C-43) in non-dimensionalized forms. Define,

tn = ft, un = u/U, and = z/dE, dE = (2/f)1/2 (1.C-44)

where 1/f is the time scale, U is the velocity scale, and dE is the z-scale (the reason for the factor 21/2 will become clear below); the tn, un and are non-dimensionalized time, velocity and z. Then, [1/(Uf)]×(1.C-41) gives:

¶un/¶tn – vn = [¶2un/¶2]/2 (1.C-45a)¶vn/¶tn + un = [¶2vn/¶2]/2 (1.C-45b)

Similarly, 2/(Uf) × boundary conditions (1.C-43) gives:

¶un/¶ = 21/2to/[U(f)1/2] = tno at = 0, for ocean (1.C-46a)

or un = 0 at = 0, for atmosphere or ocean’s bottom (1.C-46b)

Steady Solution to (1.C-45 & 46):

We drop the subscript “n” so that variables will now be non-dimensionalized. As in previous section, let = u + iv, so that (1.C-45a) + i×(1.C-45b) gives:

¶/¶t + i = [¶2/¶2]/2 (1.C-47)

When the RHS = 0, this yileds the equation for inertial oscillation (see 1.C-22). Here we will find the steady-state solution in which ¶/¶t = 0. Assuming a solution of the form = em gives m2 = 2ei/2, i.e. m = 21/2.ei/4 = (1+i). Therefore,

= A.e(1+i) + B.e-(1+i) (1.C-48)

where A and B are constants to be determined.

Oceanic Surface Ekman Layer:

For eddy viscosity 2.5×10-3 m2 s-1, and f 5×10-5 s-1, typical dE 10 m. We assume that the ocean’s depth is much deeper than this. Therefore, since < 0, we set B = 0. Then apply the surface condition (1.C-46a):

d/d|=0 = A.(1+i) = (tox + i.toy);so, A = [(tox + toy) + i(-tox + toy)]/2 (1.C-49)

Substitute this into (1.C-48):

2e-. = [cos + i.sin].[(tox + toy) + i(-tox + toy)]= [(tox + toy).cos - (-tox + toy).sin] + i[(tox + toy).sin + (-tox + toy).cos]

= [tox.(cos+sin) + toy.(cos-sin)] + i[tox.(sin-cos) + toy.(sin+cos)]so, 21/2e-. = [tox.sin(+/4) + toy.cos(+/4)] + i[-tox.cos(+/4) + toy.sin(+/4)];

Finally,

= 2-1/2e.{[tox.sin(+/4)+toy.cos(+/4)] + i[-tox.cos(+/4)+toy.sin(+/4)]} (1.C-50)

At the surface, = 0,

uo = 0.5×[(tox+toy).i + (-tox+toy).j] (1.C-51)

From the definition of vector product (a × b = |a|.|b|.sin().k):

uo × to = |uo|.|to|.sin(ut).k = |to|2. sin(ut)/21/2.k (1.C-52a)

where ut is the angle between the surface current and wind stress vectors. On the other hand, direct calculation using determinant gives

uo × to = (|to|2/2).k (1.C-52b)

So, sin(ut) = 1/21/2, and ut = /4. (1.C-53)

Homework:1.C-3a: Let toy = 0 in eqn.(1.C-50), then sketch the current profile from surface to deep.Eqn.(1.C-50) gives for toy = 0:

U’ = 21/2(u/tox) = e.sin(+/4) and V’ = 21/2(v/tox) = -e.cos(+/4) U’ V’0 sin(/4) -cos(/4)-/4 0 -e-/4

-/2 -e-/2/21/2 - e-/2/21/2

.1.C-3b: Then, show by integration by parts that ∫U’.d=0 and ∫V’.d=-1/21/2, i.e. ∫v.d=-tox/2, which in dimensional form (using 1.C-44 & 1.C-46a) is: ∫v*dz*=-t*ox/f. Here ∫d means integration from =-∞ to zero (i.e. surface) and the asterisk * denote dimensional quantities.

Fig.1.C-11. Ekman transport due to (kinematic) wind stress in x-direction, t*ox.

Atmospheric ( or Oceanic Bottom ) Boundary Layer :

The oceanic surface Ekman layer is driven by wind stress. The atmospheric boundary layer (oceanic bottom boundary layer) depends on the wind (ocean’s current) high above the ground (ocean’s bottom). If there is no wind (current), then there is no boundary layer. This is clear by applying the no-slip condition (1.C-46b) to the solution (1.C-48) which since 0 ∞ becomes:

= B.e-(1+i) (1.C-54)

Then applying u = 0 at = 0 gives B = 0, the trivial no-wind solution.The wind high above the ground is driven by pressure gradients, and we therefore include these into our momentum equations (1.C-41) which in vector form are:

¶u*/¶t* + fk×u* = -p*/* + ¶2u*/¶z*2 (1.C-55)

This is written with the * to denote dimensional form. Away from the boundary, we expect frictional shear stress term to be small. The time scale (tS) it takes for a parcel with velocity ~ U to move through a horizontal length scale L is:

tS ~ L/U (1.C-56)

If this time scale is much longer than the inertial time scale, i.e. if L/U >> 1/f,

= U/(fL) << 1 (1.C-57)

then the ¶/¶t* term is small compared to the “f” term. This can be seen by non-dimensionalizing (1.C-55), using the scale tS (instead of 1/f) and L:

.¶u/¶t + k×u = -p/ + (Ev/2)¶2u/¶z2 (1.C-58)

where the non-dimensional pressure ‘p’, density ‘’ and depth ‘z’ are

p = p*/(oUfL), = */o and z = z*/D (1.C-59)and Ev = 2/(fD2) = (dE/D)2 (1.C-60)

is the (vertical) Ekman number. Note that

z2/Ev = D2z*2/(D2.dE2) = (z*/dE)2 = 2 (1.C-61)

see (1.C-44). In equation (1.C-59), the vertical scale D ~ H is used anticipating that the flow field being considred is high above the ground away from the boundary layer.

From (1.C-58), it is clear that the time-rate of change term is much smaller than the Coriolis term. Also, the Ekman number Ev << 1, which means that away from boundary layers, the frictional shear stress term is also small compared to the Coriolis term. Therefore, the main balance in (1.C-58) is:

k×ug = -p/ (1.C-62)

which is called the geostrophic balance, and we put u = ug; see Fig.1.C-12.

Fig.1.C-12. Geostrophic flows u =ug around high and low pressures in N.Hemisphere; here z=k.

Some of us who watch a lot of TV should be familiar with weather charts (Fig.1.C-13) which actually provide wonderful examples of geostrophic flows.

Fig.1.C-13. Surface pressure (mbar; top) and winds (bottom) on Nov/21 2011 in NW Pacific.

The geostrophic velocity vector, ug, always points in the direction parallel to the p-contours (for constant ) and, in the northern hemisphere, also points to the right as one looks ‘down’ from high to low. This is seen by k×(1.C-62) which gives:

ug = k×p/ (1.C-63)Now that the geostrophic wind (ug) above the ground is known, we can now proceed to examine the atmospheric boundary layer. We will again set the ¶/¶t term = 0. Substitute the pressure term in (1.C-58) by ug from (1.C-62), and using (1.C-61), we obtain:

k×(u-ug) = (1/2)¶2u/¶2 (1/2)¶2(u-ug)/¶2 (1.C-64)the second form uses the fact that the pressure gradients (hence the ug) vary very little across the thin boundary layer, i.e. ¶2ug/¶2 0. Let

uE = u - ug (1.C-65)then k×uE = (1/2)¶2uE/¶2 (1.C-66)

which is exactly the same as the steady-state form of (1.C-45), but now written in terms of uE instead of the total velocity u. In terms of this new uE, the boundary condition at = 0, (see 1.C-46b) is:

uE = -ug (1.C-67)

Therefore, the solution from (1.C-48) is (with E = uE + ivE etc):

E = -g.e-(1+i) (1.C-68)i.e. uE = -e-.[ug.cos + vg.sin], vE = e-.[ug.sin - vg.cos] (1.C-69a,b)

Homework:1.C-4a: Let vg = 0 in eqn.(1.C-69), then sketch the current profile from ground to atmosphere above.

Eqn.(1.C-69) gives for vg = 0:uE = -e-.ug.cos and vE = e-.ug.sin uE vE

0 -ug 0/4 -e-/4ug.2-1/2 e-/4ug.2-1/2

/2 0 e-/2ug 3/4 +e-3/4ug.2-1/2 e-3/4ug.2-1/2

.1.C-4b: Then, calculate the corresponding bottom Ekman transports.Let TrE = (I,J) = ∫(uE,vE).d be the transport due to uE, then:

-I/ug = ∫e-.cos.d = ∫e-.dsin = 0 + ∫e-.sin.d = -∫e-.dcos = +1 - ∫e-.cos.d;i.e. ∫e-.cos.d = ½, therefore I = ∫uE.d = -ug/2

J/ug = ∫ e-.sin.d = -∫e-.dcos = +1 - ∫e-.cos.d = 1 - ∫e-.sin.d; (see 2 lines above);i.e. ∫e-.sin.d = ½, therefore J = ∫vE.d = ug/2The Ekman transport is: TrE = (-ug, ug)/2. Thus for an eastward geostrophic current ug, the Ekman transport is directed towards the northwest.

Structure of Atmospheric Boundary Layer (ABL) over an SST Front:

What is the structure of the ABL when (geostrophic) wind ug blows over an oceanic SST (sea-surface temperature; T*(x,y)) front, e.g. the Kuroshio or the Gulf Stream? The SST gradient is (as before, asterisk * denotes dimensional, unless otherwise obvious):

T* = (¶T*/¶x*, ¶T*/¶y*) ( & ¶ are understood to be dimensional) (1.C-70)

Assume that the ABL is well-mixed, so that its temperature * follows that of the underlying SST, but the ocean’s influence decays with height above the sea-surface:

* = T*.e-k, where = z*/dE, and k > 0. (1.C-71)

For k 1 or larger, the influence of the ocean is confined near the sea-surface. For smaller k, the ocean’s influence extends to greater heights. By hydrostatic equation, the pressure in the ABL is:

(¶p*/¶z*)/o = -g*/o = ’g* = ’gT*.e-k (1.C-72)

To get the 2nd form, we have used the relation

*/o = -’.*, where ’ = -(¶*/¶*)/o; (1.C-73)

note that > 0 since density decreases (d*<0) as air warms up (d*>0). The 3rd form in (1.C-72) then uses (1.C-71). Integrating (1.C-72) from z’=z* to z’=z*n ndE, where n >> 1 (say ~10):

[p*(zn) – p*]/o = -dE[’gT*/k].[e-kn – e-k]

or, p*/o p*(z*n)/o – [dE’g/k].T*.e-k (1.C-74)

We now non-dimensionalize p* by dividing by (UfL) as in (1.C-59):

p p(=n) – c.T.e-k (T = T*/To) (1.C-75a)

where c = [’TodE.g/(k.UfL)] = [’To./Fr].k-1 (1.C-75b)

is the coupling coefficient; here, Fr = U2/(g.dE) is the Froude number and = U/(fL) is the Rossby number (see equation 1.C-57). Typical values are: Fr ~ 100/(10.102) 0.1, ~ 10/(10-4.106) 0.1, ’To 1, so that c ~ 1/k. Smaller “k” means greater coupling. The horizontal pressure gradient:

p = p(=n) – c.T.e-k ( is now understood to be dimensionless) (1.C-76)

consists of 2 parts: a part that involves the pressure high above the ocean’s surface (p(=n)), and another part that involves the SST-gradient (T) near the ocean’s surface. The former part simply drives the geostrophic wind (see 1.C-62; note that the “” there is really “1” in dimensionaless form):

k × ug = -p(=n) (1.C-77)

This together with (1.C-76) can be substituted into the governing equation (1.C-58) in steady state:

k×(u – ug) = c.T.e-k + (1/2)¶2u/¶2; or since ¶ug/¶ 0, can be rewritten as:

(1/2)¶2uE/¶2 - k×uE = -c.T.e-k; uE = u – ug; (1.C-78a)

with uE = –ug at = 0. (1.C-78b)

In component form, (1.C-78a) becomes:

(1/2)¶2uE/¶2 + vE = -c.Tx.e-k (1.C-79a)(1/2)¶2vE/¶2 - uE = -c.Ty.e-k (1.C-79b)

Thus, [¶2E/¶2]/2 - iE = -c.e-k.G ; G = Tx + iTy (1.C-80)

where E = uE + ivE, i = (-1)1/2. This is the same as (1.C-47) except now there is the forcing term on the RHS due to the SST front. The solution satisfying also E = -g at =0 (i.e. 1.C-78b) is:

E = B.e-(1+i) + C.e-k, (1.C-81a)with C = -cG(k2/2+i)/(k4/4+1) = -[c/(k4/4+1)].[(Txk2/2-Ty)+i(Tyk2/2+Tx)] (1.C-81b)and B = -g – C (1.C-81c)

After some algebra, we obtain the solution as:

uE = uE0 + {c.e-/(1+k4/4)}{Tx.1() + Ty.2()} (1.C-82a)vE = vE0 + {c.e-/(1+k4/4)}{Ty.1() – Tx.2()} (1.C-82b)

where 1() = [sin+(k2/2).(cos-e(1-k))] and 2() = [(k2/2)sin-(cos-e(1-k))] (1.C-82c)

and (uE0,vE0) is the Ekman solution in the absence of SST-gradient (c.f. 1.C-69a,b):

uE0 = -e-.[ug.cos + vg.sin], vE0 = e-.[ug.sin - vg.cos] (1.C-83a,b)

Interior Geostrophic Flow:

The ABL Ekman solution found above produces vertical velocity (Ekman pumping) at the top of the ABL, to which then the interior flow responds. To obtain the vertical velocity, we use the continuity equation (1.A-3):

¶w*/¶z* = -(¶uE*/¶x*+¶vE*/¶y*) (since ¶ug*/¶x*+¶vg*/¶y* = 0 why?).

Non-dimensionalize this: u = u*/U, x = x*/L, z = z*/D, w = w*L/(DU), etc., we obtain:

¶w/¶z = -(¶uE/¶x+¶vE/¶y) ¶w/¶ = -(dE/D).(¶uE/¶x+¶vE/¶y) (1.C-84)

From (1.C-82 & 83) (learn Greek: = Zeta; = Xi):

¶uE/¶x + ¶vE/¶y = -e-.sin.g + [c.e-/(1+k4/4)].[sin+(k2/2).(cos-e(1-k))].2Twhere g = (¶vg/¶x - ¶ug/¶y) is the relative vorticity of the geostrophic flow. (1.C-85a,b)

Using this in (1.C-84), integrating from = 0 to = ∞, and noting that w(=0) 0:

(D/dE).w(=∞) = g/2 – 2T.c’/2 (1.C-86a)where c’ = [c/(1+k4/4)][1+k+k2/2] (1.C-86b)

The first term on RHS is Ekman pumping induced by friction caused by the geostrophic wind rubbing against the ocean surface, while the second term is that produced by the SST gradient. Note that the latter term is = 0 for T = constant, but it is also zero when T is linear with x and y, so that 2T = 0. Thus the existence of an SST-gradient alone is insufficient to produce a response in the atmosphere above – the SST must have some curvature. The physics will be explained below.

Imagine a case when the atmosphere is initially quiet (no wind). We then let the SST-gradient “suddenly appears” – i.e. over a short time period of say a few days. What is the resulting wind? Since at t = 0, the w = ¶w/¶z = 0, we expect that approximately, the vertical velocity remains zero, so that from (1.C-86):

g c’.2T (1.C-87)

which gives information on the geostrophic wind ug produced by 2T.

Consider the case when ¶/¶x = 0, so the SST front is from west to east aligned with the x-axis:

T = (0, Ty) (1.C-88)

Then (1.C-87) gives -¶ug/¶y = c’.¶2T/¶y2; integrating from y’=-∞ to y’=y, and noting that ug(-∞) = 0 (wind remains quiet far from the front):

ug(y) = -c’.Ty(y) (1.C-88)

which like (1.C-87) is an elegant and exceedingly simple result. It says that westerly geostrophic wind (ug > 0; hence also the wind high above the ABL) is produced by a front for which the SST decreases northward, as for example northward across the Gulf Stream east of US, or the Kuroshio east of Japan. Also, vg is trivially zero (i.e. x-parallel geostrophic flow at x = ∞). Equations (1.C-82) and (1.C-83) then give

u(y,) = ug(y).{1 – e-.[cos + 2(;k)/(1+k+k2/2)]} (1.C-89a)v(y,) = ug(y). e-.[sin - 1(;k)/(1+k+k2/2)] (1.C-89b)

Equation (1.C-89a) shows that, since the quantity in the curly parentheses “{..}” is always positive (this can be proven analytically; one can also plot the function as in Fig.1.C-14), the zonal velocity u < 0 for Ty > 0.

As an example, let:

T(y) = -T.tanh(y/d), (1.C-90a)then Ty = -(T/d).cosh-2(y/d) = -(T/d).sech2(y/d) (1.C-90b)

where T = T*/To (>0; T* = temperature change across the front) and d = d*/L (d* = cross-front scale). Therefore, (1.C-88) gives:

ug(y) = (c’T/d).sech2(y/d) = (c’T/d)/[(ey/d + e-y/d)/2] (1.C-91)

Fig.1.C-14. Contours showing {1 – e-.[cos + 2(;k)/(1+k+k2/2)]} in (1.C-89a) is always positive.

Fig.1.C-15. A satellite-derived Sea Surface Temperature map, constructed from space-time interpolation of three weeks

of images and in situ data, for July 14, 1995 showing the Gulf Stream ring and meander region. (From

http://oceancurrents.rsmas.miami.edu/atlantic/gulf-stream_2.html).

Choose L = 6×105 m (broad scale across the Gulf Stream or Kuroshio), f = 8.5×10-5 s-1 (at 37oN), dE 500 m, and V = 1 m s-1, then 0.02 and Fr 2×10-4. With k =2 (so that c’ = c), d = 1/3, and T* 4 oC (Fig.1.C-15), we obtain the maximum ug(0) as:

c’T/d 100.(T*/o)/(k.d) (4/3)(3/2), i.e. Max|u*| 2 m s-1 (1.C-92)

Physical Interpretations:

http://oceancurrents.rsmas.miami.edu/atlantic/gulf-stream_2.html

The vertical motion (1.C-86) has 2 parts: the 1st part, WM, is the familiar Ekman pumping or suction due to friction, while the 2nd part, WT, is the SST-gradient which we will now examine more closely. We again use (1.C-90) as the frontal temperature profile, for which

Tyy = (2T/d2).tanh(y/d)/cosh2(y/d) (1.C-93)

Then, (1.C-86) gives:

(D/dE).WT = –2T.c’/2 = –(c’/2).(2T/d2).tanh(y/d)/cosh2(y/d) (1.C-94)

Fig.1.C-16. Sketches of the “tanh” and “cosh-2” functions (blue and red in (a)) and their product (green in (b)) which has

maximum (minimum) at y/d = +ln2/2 (-ln2/2). The SST-gradient induced WT is then seen to be downward (upward) in

cold (warm) side of the front, and the cross-frontal velocity VT is from cold to warm, maximum at y = 0.

The function “tanh(y/d)/cosh2(y/d)” is positive (negative) for y > 0 (< 0) (see Fig.1.C-16b – green line), so that air moves downward (upward) on the cold (warm) side of the front. From (1.C-90b), we see that |Ty| is maximum at y = 0, so that ug (hence “u”) is strongest (and positive as mentioned above) at y=0 also (see 1.C-88 and 1.C-89a). The cross-frontal motion induced by the SST-gradient is from (1.C-89b) (again take k =2):

VT(y,) = -ug(y). e-.1()/5; (k=2) (1.C-95)

The 1() (see 1.C-82c) may be seen to be positive throughout the ABL (for small , it is ~ 3 + 2/2), so that VT < 0, and air in ABL is moving from cold to warm side of the front, strongest at y = 0 as sketched in Fig.1.C-16b. There is divergence on the cold side of the front where the air moves downward, and convergence on the warm side where the air moves upward. Minobe et al. [2008] have confirmed these types of motion over the Gulf Stream – see Figs.1.C-17, 18 & 19 below.

Fig.1.C-17. Annual-mean climatology of various surface parameters over the Gulf Stream illustrating the dominance of

the SST-gradient induced winds. Contours are SST (2oC interval; dashed are 10oC & 20oC). (a) & (b) are 10-m wind

convergence (-.u). (c) and (d) are 2SLP and -2T (see 1.C-76).

Homework: 1.C-5: Are these plots consistent w/Fig.1.C-16 & discussions? Describe why they are or are not

consistent?

Fig.1.C-18. Annual-mean climatology of rain rate. (AGCM = Atmospheric General Circulation Model). Contours are

SST as in Fig.1.C-17. Observation (a) shows a narrow rain band confined to the Gulf Stream’s warmer flank with SSTs greater than 16oC. AGCM specified with the observed SST reproduces quite well the observed rain rate

(b). But when the SST is smoothed, the result is poor (c).

Homework 1.C-6: discuss why these rain band structures exist (c.f. Fig.1.C-17)?

Fig.1.C-19. Annual-mean climatology from ECMWF of (a) vertical wind velocity (color) and dE (black) and wind

convergence (-.u; color contours) and (b) upper-tropospheric wind divergence (.u; color) showing divergence and

the tropopause acting virtually as a lid for the mean circulation. (c) Satellite’s outgoing long-wave radiation (OLR) as a measure of occurrence of high clouds. Lower OLR indicates lower temperatures and higher altitudes of cloud tops. Values < 160 W m-2 roughly corresponds to 300 hPa (z 7km). Contours in b and c are SST’s as before.Homework 1.C-7: A rough way to represent ABL bottom friction is to use r*:

-fv*=-¶p*/¶x*/o-r*u* & +fu*=-¶p*/¶y*/o-r*v*.Nondimensionalize r* by f r=r*/f, show that:

(1) ru-v=-px & rv+u=-py;then (2) u=(-rpx-py)/(r2+1) & v=(+px-rpy)/(r2+1);and (3) (ux+vy)=-r2p/(r2+1).Finally, use (1.C-76): p = p(=n) – c.T.e-k to show that

(4) (ux+vy)=[r/(r2+1)].[-g+c. e-k 2T]; which should be compared to (1.C-85) etc., and discuss.

The above gives a simple solution for ug high above the surface. A more complete solution must however consider the beta-effect, which will be done after we have learned about it in more details.Chapter 1.D: Large-Scale Homogeneous Circulation

The ocean or atmosphere is confined in the vertical by z = ztop (x,y,t) and a bottom z = zbot(x,y,t). The governing equations in dimensional form are:

∂ u∂ x

+ ∂ v∂ y⏟

(1)U /L

+ ∂ w∂ z⏟

(2 )W /D

=0(1.D-1a)

∂ u∂ t+u ∂u

∂ x+v ∂ u

∂ y⏟(1)U 2 /L

+ w ∂u∂ z⏟

(2)WU /D

=+ fv⏟(3 )f o U

− 1❑o

∂ p∂ x⏟

(4 )P /❑o L

+ ∂∂ z

(K M∂u∂ z

)⏟

(5)KU /D 2

(1.D-1b)

∂ v∂t+u ∂ v

∂ x+v ∂ v

∂ y⏟(1)U 2 /L

+ w ∂ v∂ z⏟

(2)WU /D

=−fu⏟(3)f o U

− 1❑o

∂ p∂ y⏟

(4) P/❑o L

+ ∂∂ z

(K M∂ v∂ z

)⏟

(5)KU /D2

(1.D-1c)

∂ p∂ z

=−' g (1.D-1d)

where all variables are dimensional (for the time being; they should really have “*” but for simplicity the “*’s” are omitted. Under each term, we indicate its order of magnitude by using the following scale for each variable:

(x,y) L; z D; (u,v) U; KM K; (1.D-2a)t L/U; w W ~ UD/L; (1.D-2b)p P ~ oLfoU; f f*o (1.D-2c)

where “” means “have/has the scale of.” The Coroilis parameter “f” and eddy viscosity e = KM are now more general, and may depend of (x,y,z,t). For homogeneous flow, ’ = 0, and after simplifying etc as before, we obtain the following non-dimensionalized set of equations:

∂ u∂ x

+ ∂ v∂ y

+ ∂ w∂ z

=0 (1.D-3a)

( ∂ u∂ t+u ∂ u

∂ x+v ∂ u

∂ y+w ∂ u

∂ z )=fv−∂ p∂ x

+E v

2∂

∂ z(KM

∂u∂ z

) (1.D-3b)

( ∂ v∂ t

+u ∂ v∂ x

+v ∂ v∂ y

+w ∂ v∂ z )=−fu−∂ p

∂ y+

Ev

2∂

∂ z(K M

∂ v∂ z

) (1.D-3c)

∂ p∂ z

=0 (1.D-3d)

where now all variables are dimensionless, f = ±1 for constant Coriolis and

¿ Uf o¿ L=Rossb y number , Ev=

2 Kf o¿ D2=Ekman number (1.D-4)

For small , and away from surface or bottom boundaries:

fk×ug = -p (1.D-5)

where as before the subscript “g” denotes geostrophy, and = (¶/¶x, ¶/¶y). Therefore,∂ug

∂ x+

∂ vg

∂ y=0, (1.D-6)

For homogeneous flow, p is independent of “z”, so that ug is also independent of “z”. Let u and v be expanded as:

u = ug + u1 + O(2); v = vg + v1 + O(2); w = wg + w1 + O(2); etc. (1.D-7)Then, substituting into the continuity equation (1.D-3a), we obtain:

∂ wg

∂ z=0 (1.D-8a)

Figure 1.D-1. Nearly-steady homogeneous (constant-density) ocean flow in an x-periodic channel, 600km by

300km, of depth 200m except at the channel’s center where a cylinder rises 50m above the bottom. Color is

sea-surface height in meters and vectors are velocity at (A) z=0m (i.e. surface), (B) z=90m and (C) z=180m.

This model calculation was carried out for 100 days when the flow has reached a nearly steady state. The

experiment is meant to illustrate Taylor-Proudman theorem: flow below the cylinder’s height goes around the

cylinder while above it the flow also tends to go around as if the cylinder extends to the surface. The velocity

does not vary with “z” and the vertical velocity (which is not shown) is nearly zero.

Equations (1.D-7 & 8a) show that all three components of the geostrophic velocity (ug, vg, wg) are independent of z, the rotation axis. This is the Taylor-Proudman theorem which is valid for inviscid homogeneous slow (small ) flow. If the upper or lower boundary is flat, so that wg = 0 there, then wg = 0 through the water column. The motion is then two dimensional, (ug, vg) 0, but wg = 0. Figure 1.D-1 shows an example of this type of flow. Actually, the requirement that upper or lower boundary being flat is not necessary, provided that << 1. That is because the vertical velocity is very small. For example, near z* 0, the vertical velocity due to the bottom Ekman layer is, from (1.C-86a):

w* ~ (UD/L).[g*/(U/L)].Ev1/2 = Dg*.Ev

1/2 ~ O(W). Ev1/2

where W ~ UD/L is the scale for the interior vertical velocity (see 1.D-2b). Therefore, since Ev

1/2 is very small ~ O() (at most), the leading w* term “wg*” in (1.D-7) is zero:

wg* = 0 = wg (1.D-8b)

The -plane & the QG-Potential Vorticity Equation

We now consider the case when the Coriolis parameter is a function of latitude :

f* = 2.sin(), /2 +/2 (1.D-9)

Suppose we focus on a certain latitude, = o where f* = fo*, and let “y*” be the northward (distance) coordinate measured from this latitude (so that y* = yo* at = o), then for “small” distances away from this latitude, we have:

f* fo* + [df*/dy*][y*yo*] = fo* + [(df*/d)/ro][y*yo*] = fo* + [2.cos(o)/ro][y*yo*] (1.D-10)

where ro = earth’s radius. Therefore,

f* fo* + o*(y*yo*) (1.D-11a)

where

o* = 2.cos(o)/ro (1.D-11b)

Equation (1.D-11) constitutes the so called “-plane,” and is valid for |yyo| not too large (i.e. a few thousands of kilometers, for o 2×10-11 m-1s-1 near mid-latitude.) We now derive a vorticity equation taking into account that f is a function of y, i.e. is given by (1.D-11a). In terms of dimensionless variables, (1.D-11) is:

f fo + (yyo), = (o*L/|fo*|) << 1. (1.D-12)

where fo = fo*/|fo*| = ±1 denotes northern or southern hemisphere. Note that is small; we will in fact let , and let:

= / = o*L2/U O(1). (1.D-13)

The is also:

= [o*]/[(U/L)/L] = [Planetary vorticity gradient]/[Flow vorticity gradient]so that ~ O(1) means that we are considering slowly-varying large-scale flows when the flow vorticity gradients are comparable to the planetary vorticity gradient.

With (1.D-12), the Coriolis terms in equation (1.D-3b,c) become:

fv = [fo + (yyo)][vg + v1 + O(2)]= fovg + fov1 + (yyo).vg + O(, 2) (1.D-14a)

fu = foug fou1 (yyo) ug + O(, 2)] (1.D-14b)

With these expressions, substituting (1.D-7) into (1.D-3b,c), and collecting and equating terms that are of O(1) and O(, ) separately we obtain for the O(1) terms the geostrophic relation as before, while the O(, ) terms give:

∂u1

∂ x+

∂ v1

∂ y+

∂ w1

∂ z=0 (1.D-15a)

∂ug

∂ t+ug

∂ug

∂ x+v g

∂ ug

∂ y=f o v1+❑❑( y− yo)vg−

∂ p1❑

∂ x(1.D-15b)

∂ vg

∂ t+ug

∂ v g

∂ x+vg

∂ vg

∂ y=− f o u1−❑❑( y− yo)ug−

∂ p1❑

∂ y(1.D-15c)

∂ p1❑

∂ z=0 (1.D-15d)

where fo = ±1, = /. Taking the curl, i.e. [(1.D-15c)/x(1.D-15b)/y], and using (1.D-15a) give:d ξg

dt+❑❑v g= f o

∂ w1

∂ z(1.D-16a)

whereddt= ∂

∂ t+ug

∂∂ x

+vg∂

∂ y (1.D-16b)

To understand the physical meaning contained in (1.D-16a), we write it in dimensional form (using 1.D-2, 1.D-13 and fo defined after 1.D-12):

(L/U)2.d(*g)/dt* + (o*L2/U2).v*g = fo*/|fo*|.(¶w*1/¶z*).[DL/(UD)]

i.e. d(*g)/dt* + o*.v*g = .fo*(¶w*1/¶z*)

where = U/(|fo*|L). Now, w* w*1 since (from 1.D-7),

w* = w*g + w*1 + O(2) = w*1 + O(2) (since w*g = 0), (1.D-17a)and df*/dt* = v*¶f*/¶y* = o*.v* = o*.v*g.[1+O()] o*.v*g (1.D-17b)

so that the previous equation becomes:

d(f*+*g)/dt* = fo*.(¶w*/¶z*) (1.D-18a)

where d/dt* = ¶/¶t* + u*g.¶/¶x* + v*g.¶/¶y* (1.D-18b)

This is the dimensional form of equation (1.D-16). It says that in an inviscid fluid, for a fluid parcel following the geostrophic velocity, stretching of the parcel ¶w*/¶z* in the same direction as the background spin (z-directed vector) of the earth’s (planetary) vorticity field fo* will produce a change in the total vorticity (f*+*g) that is equal to the product of the background spin and stretching: i.e. = fo*.(¶w*/¶z*).

Since (ug, vg) hence g are independent of z, equation (1.D-16) can be integrated ∫zbot

ztop

dz to yield, with

ztop – zbot = (z*top – z*bot)/D = 1 for constant-depth ocean or atmospheric layer,

d ξg

dt+❑❑v g= f o[w1Top−w1Bot ] (1.D-19)

SST-gradient-induced motion (northern hemisphere f o = +1):

For the case of the SST-gradient-induced motion, the w1Bot is given by (1.C-86) which was derived for northern hemisphere, and which together with (1.D-7) (i.e. w w1 ) gives:

w1Bot = w(z/dE=∞)/ = (g – 2T.c’).Ev1/2/(2) = r.(g – 2T.c’)/2 (1.D-20a)

where r = Ev1/2/ (1.D-20b)

is a non-dimensionalized parameter measuring the effect of friction (which will be further discussed below) and Ev

1/2 = dE/D (from 1.C-60). Finally, (1.D-19) becomes, for w1Top = 0:

dg/dt + vg = -r.(g – 2T.c’)/2 (1.D-21)

To understand this equation, consider the case = 0 = c’, then dg/dt = -r.g and the solution is:

g = g0.exp(-r.t) (1.D-22)

expressing, because of friction, a decaying with time of the interior geostrophic vorticity field from its initial state g0. The effects of friction on ug is described in Fig.1.D-2. In (a) the flow is geostrphic u* = ug* so that –f*k×u* - p*/* = 0, i.e. the dashed (southward) vector balances solid (northward) vector, and u is westward parallel to the p-contours ( = constant). In (b), the solid vector “-p*/*” is still southward directed from high to low pressure. It must still be balanced by a northward dashed vector which is now the sum of “–f*k×u*” acting perpendicular and to the right of u and fritciotnal vector Fric acting opposite to u. In order to accomplish this, i.e. in order that:

Northward Dashed Vector = –f*k×u* + Fric

the u must be directed slightly inward into the low pressure as shown in the figure.

Fig.1.D-2. Balance of forces in flow (a) without and (b) with friction; shown for f > 0.

A real-case example of this phenomenon is given in Fig.1.D-3. The lower panel is same as Fig.1.C-13 showing pressure contours high above which therefore indicate the geostrophic flow. Over Taiwan, this geostrophic flow is generally westward with high pressure to the north (c.f. Fig.1.D-2a). The lower panel shows flows near the surface. The near-surface wind over Taiwan may be seen to be directed southwestward, i.e. leaning towards the low pressure region (c.f. Fig.1.D-2b).

Fig.1.D-3. Surface pressure (mbar; top) and near-surface winds (bottom) on Nov/21 2011 in NW Pacific. East of Taiwan, the p-contours are nearly zonal, so that geostrophic winds are easterly. But the (near-surface) wind vectors are actually from the northeast.

Homework 1.D-1:

Explain using simple drawings and in terms of Ekman and geostrophic dynamics why air diverges (converges) under a high- (low-) pressure center; and why therefore center of low (high) is generally associated with cloudy (sunny) days.

The meaning of r:

Let us analyze more exactly the above “competition” between pressure and frictional forces that causes the flow to ‘lean’ towards the low pressure region as the flow approaches the surface. At the same time, we then can interpret r = Ev

1/2/ in (1.D-20b). Suppose the geostrophic flow is purely zonal: ug = (ug, 0); this is Fig.1.D-2a with ug < 0. The solution is then ug + uE which from (1.C-69) (see homework 1.C-4a) is

u = ug.(1 - e-.cos) & v = ug.e-.sin, where = z/dE (1.D-23)

We will assume ug < 0 (Fig.1.D-2a). High above, ~ ∞, u ~ ug and v ~ 0, i.e. the flow is purely westward and the resulting Coriolis force balances pressure gradient force. As the surface is approached from above, the westward wind weakens, the corresponding Coriolis force also weakens, so that since the pressure gradient force is independent of z (i.e. it penetrates into the Ekman layer), it must force a flow from high to low pressure; in other words, southward wind is produced and the total wind spirals towards the low pressure region. The southward flow may be seen from expanding (1.D-23) for small : u ~ ug. and v ~ ug., so that the flow points towards the southwest, at exactly 45o anticlockwise from due west, i.e. the v-component of the near-surface wind points towards the low pressure region.

The existence of vg crossing isobars (pressure contours) from high to low means that work is being done on the fluid in the Ekman layer by the pressure-gradient force. In general, this work per unit horizontal area of the Ekman layer is (from u dot the momentum equation):

dW*/dt* = -∫u*.p*.dz*, where ∫ is from 0 to D (depth of atmospheric layer).But, u*.p* = v*.¶p*/¶y*, since ¶p*/¶x* = 0 in our case of a purely zonal flow.So, dW*/dt* = -∫v*.¶p*/¶y*.dz* = (ug*)2.f**dE. ∫e-.sin.d, where ∫ is from 0 to ∞, and the geostrophic balance f**ug* = -¶p*/¶y* has been used. Thus,

dW*/dt* = (ug*)2.f**dE/2 (1.D-24)

The geostrophic wind u* = (ug*, 0) which has a kinetic energy per unit area of:KEg* = *(ug*)2D/2 (1.D-25)

is therefore supplying (and looses its) energy to the Ekman layer at the rate of dW*/dt*, and therefore decays with a time scale of:

tE* = KEg*/(dW*/dt*) = D/(f*.dE) = .(L/U)/Ev1/2 = (L/U)/r (1.D-26a)

where the last step uses (1.D-20b): r = Ev1/2/. Therefore,

r = (L/U)/tE* = (Flow Advection Time Scale)/(Frictional Decay Time Scale) (1.D-26b)

When studying the atmospheric response to SST-gradient in Chapter 1.C, we made the simplifying assumption (without justification) and set w(=∞) = 0 in equation (1.C-86a), thereby arriving at the simple solutions (1.C-89). From the potential-vorticity equation (1.D-21), we now see that the assumption is equivalent to setting r >> 1 in that equation, or, from (1.D-26b), that friction (or mixing) is so strong that the decay (i.e. response) time scale in ABL is much shorter than the time scale for flow (vorticity) advection.

Wind Stress Curl Induced Vertical Velocity in the Ocean:

In the potential-vorticity equation (1.D-19), in addition to the w1Bot due to friction or SST-gradients in the bottom Ekman layer, stretching can also be caused by vertical velocity w1Top produced at the top of the ocean. The w1Top is produced if there are convergences and/or divergences in the surface Ekman layer, and one way these may be produced are by spatially different strengths and/or directions of the wind stress. From (1.C-39) and assume that * = constant (which is a good approximation for the ocean):

M/ = (UE, VE) = (toy/f, -tox/f) = TrE (1.D-27a)*

where the ‘*’ on the equation# indicates that all variables in this equation are dimensional, and

(tox, toy) = (Xo, Yo)/, and TrE = ∫uE.dz, integration from z=-∞ to 0(1.D-27b)*Thus UE* is the wind-stress induced Ekman volume transport in x*-direction per unit y* and it has a unit of m2.s-1; similarly for VE*.

Fig.1.D-4. (a) An xz-plane view of x-directed Ekman transport and vertical transport due to toy; (b) the corresponding yz-plane view due to tox. All variables are dimensional.

Consider Fig.1.D-4a, wind stress toy > 0 is northward, in general varying with x, and the Ekman transport is eastward (assume f > 0). Mass balance in the dashed control box gives:w.x + (toy/f)- = (toy/f)+; (i.e. inflow = outflow)

i.e. w = [(toy/f)+ - (toy/f)-]/x = ¶(toy/f)/¶x (1.D-28a)*Similarly, for Fig.1.D-4b with a wind stress tox > 0, Ekman transport is southward:

w.y + [-(tox/f)-] = -(tox/f)+;

i.e. w = -[(tox/f)+ - (tox/f)-]/y = -¶(tox/f)/¶y (1.D-28b)*So in general: w = wTop = ¶(toy/f)/¶x - ¶(tox/f)/¶y = ×(to/f) (1.D-28c)*Bottom Stress Curl Induced Vertical Velocity in the Atmosphere or Ocean:

A similar calculation can be done for the vertical motion that is produced by spatially varying bottom stress. In this case, from (1.C-39), the Ekman transport due to bottom stress is:M/ = (UE, VE) = (-tby/f, tbx/f) = TrE (1.D-29)*

Note that the signs are reversed from those in (1.D-27a). Consider Fig.1.D-5a (all variables are again dimensional), wind velocity is northward (va > 0) and in general varying with x. The resulting bottom stress is -tby (< 0) which also can vary with x. Mass balance in the dashed control box gives:-(tby/f)- = -(tby/f)+ + w.x; (i.e. inflow = outflow)i.e. w = [(tby/f)+ - (tby/f)-]/x = ¶(tby/f)/¶x .¶(va/f)/¶x (1.D-30a)*where we have approximated the bottom stress as:

tb .ua i.e. (tbx,tby) = .(ua,va), = dEf/(2) (1.D-31)*Similarly, for eastward wind (Fig.1.D-5b):

(tbx/f)- = (tbx/f)+ + w.y;i.e. w = -[(tbx/f)+ - (tbx/f)-]/y = -¶(tbx/f)/¶y -.¶(ua/f)/¶y (1.D-30b)*Combining (1.D-30a & b), and for simplicity consider f = constant, we obtain:

w = wBot = (dE/2).×ua = (dE/2).g (1.D-30c)*which is the same as (1.D-20a) (with 2T=0 and w = w1) if written in non-dimensional form. For example, for a cyclonic geostrophic wind, ×ua > 0, and there is upward movement of air inside the cyclone, and convergence of air from surrounding high to low pressure in the center of the cyclone. This high-to-low movement of air eventually “kills” (i.e. dissipates) the cyclone.

Fig.1.D-5. (a) An xz-plane view of x-directed Ekman transport and vertical transport due to the geostrophic wind va > 0; (b) the corresponding yz-plane view due to ua > 0. All variables are dimensional.

Homework 1.D-2:Rederive (1.D-28a) and (1.D-30a) by applying the continuity in xz-plane: ¶u/¶x+¶w/¶z = 0, so that w.x = -u.z = -U, where U = Ekman transport.

Then also rederive (1.D-28b) and (1.D-30b) using ¶v/¶y+¶w/¶z = 0.

The Potential Vorticity Equation with Wind Stress Curl, Bottom Friction & SST-Gradient:

We are now ready to complete equation (1.D-21) by adding at the top the wind stress curl contribution which produces upwelling and downwelling as in (1.D-28c)* which gives (in dimensional form with *):w*Top = *×(t*o/f*), which becomes in non-dimensionalized form:(UD/L).wTop = ×(to/fo).ts

o/(|fo*|L) where tso is the scale of the wind stress, so:

wTop =×(to/fo).tso/(|fo*|UD)=w1Top where w = w1 is used from (1.D-17a). Therefore,

w1Top = wTop/ = ×(to/fo).tso/(|fo*|UD) (1.D-31a)

or, w1Top = (Ev1/2/).×(to/fo)/2 = r ×(to/fo)/2, where r = Ev

1/2/ (1.D-31b)

and in which for (1.D-31b) the scale for wind stress tso from (1.C-46a) is used:

tso = U(|fo*| /2)1/2 = UdE|fo*|/2 (1.D-32)

Restricting to the northern hemisphere, fo = +1, we substitute (1.D-31a) into (1.D-19):

dg/dt + vg = [tso/(|fo*|UD)].×to - r.(g – 2T.c’)/2 (1.D-33a)

or if instead (1.D-31b) is substituted into (1.D-19), we get a more compact form:

dg/dt + vg = r.{×to - (g – 2T.c’)}/2, (r = Ev1/2/ & = /) (1.D-33b)

Sverdrup Balance in the Atmosphere:

Before we study Sverdrup flows and ocean gyres based on (1.D-33), we digress first to atmosphere. In atmosphere, we use the pressure coordinate in the vertical instead of ‘z*’. The reason is because in pressure coordinate, the “incompressibility” condition similar to (1.D-3a) is also satisfied. To show, consider an elemental mass in pressure coordinate:

dM* = * dx* dy* dz* = * dx* dy* (-dp*/*).g-1 = g-1. dx* dy* (-dp*) (1.D-34) I II

Comparing “I” and “II”, we see that in the pressure coordinate (x*, y*, p*), the effective density is actually “1/g” = constant. So that (see derivation in Chapter 1.A):

¶u*/¶x* + ¶v*/¶y* + ¶*/¶p* = 0 (1.D-35)

where

* = Dp*/Dt* = ¶p*/¶t* + uH*.p* + w*¶p*/¶z* -g*w* (1.D-36)

is the effective vertical velocity in the p*-coordinate (c.f. w* = Dz*/Dt*); here, uH* = (u*, v*) is the horizontal velocity, and in the last step, the hydrostatic equation ¶p*/¶z* = -g* has been used Then, by applying (1.D-35) to the derivation of the potential-vorticity (PV) equation (starting from the continuity and momentum equations (1.D-1)), we see that the same PV equation (1.D-18a) is obtained which in dimensional form is:

d(*g)/dt* + o*.v*g = fo*(¶*/¶p*) (1.D-37)

For steady state or slow motion, the d/dt* can be dropped, so that:

o*.v*g fo*(¶*/¶p*) (1.D-38)

which is the atmospheric version of the Sverdrup balance. Excellent examples are in Rodwell and Hoskins [2001] where the authors describe subtropical anticyclones and summer monsoons. Some of their figures are illustrated below.

Fig.1.D-6. Idealized simulation for Jun–Aug in which monsoon heating is applied at 25N, 90E (red dashed vertical line)

and 400 hPa: (a) Vertical velocity with contour interval 0.25 hPa h-1 and (b) meridional wind v with contour interval 1

m s-1. Positive contours are solid, the zero contour is dotted, and negative contours are dashed.

Figure 1.D-6 shows a vertical zonal (xz) slice of the atmosphere from 1000 hPa at the gound to the top of the atmosphere 0 hPa. Note that 1 hPa (read as hectopascal; hecto comes from the Greek hekaton which means hundred) is equivalent to a column of approximately 10 m of air, since 1 hPa = 100 N/m2 gHe, so that He 10 m; therefore 1000 hPa corresponds to approximately 10 km column of air or roughly the height of the troposphere. Rising air is seen in (a) where heating is, where * < 0 (i.e. w* = -*/(*g) > 0). The strongest rising velocity is at the core of the heating at p 400 hPa where is negative and has the largest magnitude. Below this core, ¶*/¶p* is therefore positive (as shown) and according to the Sverdrup balance (1.D-38) the meridional

velocity v*g is also positive or poleward, as seen in Fig.1.D-6b. West of the heating, the air is descending or sinking, ¶*/¶p* < 0 and the flow is therefore equatorward.

Fig.1.D-7. (a) Jun–Aug mean vertical velocity at 674 hPa, with contour interval 0.5 hPa h-1. The following panels

show terms in the thermodynamic energy equation calculated: (b) diabatic heating, Q/Cp at 674 hPa with contour

interval 0.5 K day-1. The data are taken from analyses for 1990–94 from ECMWF. Positive contours are solid, the zero

contour is dotted, and negative contours are dashed.

In the atmosphere, regions of monsoon heating occur e.g. over southwest Mexico in the eastern N-Pacific warm pool north of the equator and over south Asian continent (India) (Fig.1.D-7); here air rises and produces the Great-Plain low-level jet over the central and southeastern USA and the southwesterly summer monsoon in South China Sea, respectively (Fig.1.D-8). The polward flow in the latter is modified and steered by high mountains, but the concept of Sverdrup flow discussed above still applies. In Fig.1.D-8, the polward Sverdrup flow to the east of the heating forms the western portion of the great anticyclonic wind circulation over the North Atlantic and North Pacific Oceans. Similarly, the equatorward Sverdrup flow to the west of the heating forms the eastern portion of the anticyclonic wind circulation. The poleward flows form the Low-Level Jets (LLJ’s) which are very important to transporting heat and moistures to the northern regions. The equatorward flows are broader and brings dry cool air from north to the equator over the eastern side of the N-Atlantic or N-Pacific Oceans.

Fig.1.D-8. Mean 850-hPa horizontal winds and streamfunction for 1990–94 Jun–Aug. Data are taken from analyses

from the European Centre for Medium-Range Weather Forecasts (ECMWF). The contour interval is 2 × 106 m2 s-1.

Positive contours are solid, the zero contour is dotted, and negative contours are dashed.

Steady Oceanic Cyclonic Flow on an f-Plane Spun up by Wind Stress Curl:

On an f-plane, = 0, and also ¶/¶t = 0 for steady flow. Consider a constant depth (= H*) ocean extended to x = ∞, and no variation in y, so that ¶/¶y = 0. The wind stress acts in the y-direction only to = (0, toy). Therefore, as there is no prossure gradient acting in the y-direction, we expect that ug = 0 also. Then dg/dt = 0, and equation (1.D-33b) becomes:

g = ×to, or vg = toy (1.D-39a)

In dimensional form using the wind stress scale tso = UdE|fo*|/2 (see 1.D-32), we obtain:

vg* = [2/(|fo*|)]1/2.toy* (1.D-39b)

Typical values: 2×10-3 m2 s-1, |fo*| 4×10-5 s-1 (lat 18o), |toy*| = 10-4 m2 s-2, then |vg| =0.5 m s-1; or 5×10-3 m2 s-1, |fo*| 6×10-5 s-1 (lat 27o), |toy*| =10-4 m2 s-2, then |vg| =0.25 m s-1.

Note that vg*(x*) takes on whatever function of x* the wind stress toy* takes. For example, if toy* varies like “tanh”, i.e. toy* ~ tanh(x*/L), then v* ~ tanh(x*/L). In Fig.1.D-9, the POM is used with toy* varying linearly with x* as described in the Figure caption, then v* also varies like x*.

Fig.1.D-9. Nearly-steady homogeneous (constant-density) wind-driven rotational flow in a channel of depth 200m and

theoretically unbounded in x*, but in practice an x-length = 1000 km is used in this numerical solution using POM. The

flow is assumed independent of y*. The wind stress is specified in y-direction only: (0, 10-4).(x*/500km - 1) m2 s-2.

Panel (A) is for the y-directed velocity v*which varies nearly linearly with x* just like the wind stress, maximum |v*|

0.25 m s-1, and panel (B) the x-directed velocity u* which is seen to be nearly zero in the interior but non-zero in the

surface and bottom boundary layers. In (B), profiles of u* are also plotted at the four indicated x*-locations with scales

0.1 m/s given along the bottom for the first and fourth locations. Detailed plots comparing the profile at x*=205 km

(i.e. the first x*-location) with analytical solution are given in Fig. 1.D-10. The Coriolis parameter is constant, fo = 6

´10-5 s-1, and the vertical eddy viscosity is also constant, = 5´10-3 m2 s-1. This numerical solution is at time = 100

days, and differs slightly from the analytical one discussed in the text; instead of the no-slip condition at the bottom, a

matching of the velocity to the law-of-the-wall log layer is used [see the POM manual by Mellor, 2004].

Fig.1.D-10. Profiles of (A) u*, (B) v* and (C) w* at the x* = 205 km location described in Figure 1.D-9b. Solid line

indicates analytical solution assuming no-slip condition at the bottom, while dashed line the corresponding numerical

(i.e. Figure 1.D-9) solution assuming log-layer at the bottom. Black line indicates the Ekman velocity while red (in

panels (A) and (B)) indicates the total (Ekman + geostrophic) velocity. Note that in panel (B), total v*/4 is plotted (so

the profile fits in the same panel.

Chapter 1.E: A Model of Ocean Gyre & Western Boundary Current

In this chapter, we formulate a simple model of the ocean gyre and western boundary current (e.g. the Kuroshio). The fluid will be assumed homogeneous of constant density . Unless otherwise stated, all variables in this Chapter 1.E will be dimensional. At the same time, we will rederive the PV-equation (1.D-33) but in a different way. Consider a “box” ocean with sides Lx and Ly on a -plane (Fig. 1.E-1). A zonal wind stress is applied:

τ w=(−τwo cos ( y /Ly ) ,0) (1.E-1)so that curl z(τ ¿¿w)=−(τ¿¿wo / Ly)sin( y /Ly )¿¿ (1.E-2)

where wo (which will be set = 10-4 m2s-2 below) is the magnitude of the (kinematic) wind stress, and curlz = ×. Initially, the fluid is at rest:

(u, v, w) = (0, 0, 0), and also, free surface = 0 (1.E-3)

At the four side walls, both the normal and tangential components of the horizontal velocities are zero:

(u, v) = (0, 0) at x = 0, x = Lx and also at y = 0, y = Ly. (1.E-4)At the ocean bottom and at the free surface, the following equations are satisfied:

w= ∂∂ t+u ∂

∂ x+v ∂

∂ y,at z=( x , y , t ) , (1.E-5a)

w=−u ∂ H∂ x

−v ∂ H∂ y

,at z=−H ( x , y ) . (1.E-5b)

The continuity equation and the linearized x and y momentum equations are then:∂∂ t=−∂ U

∂ x−∂V

∂ y (1.E-6a)

( ∂ u∂ t )=+ fv−g

∂(−❑a)∂ x

+ ∂∂ z

(K M∂u∂ z) (1.E-6b)

( ∂ v∂ t )=−fu−g

∂(−❑a)∂ y

+ ∂∂ z

(K M∂ v∂ z

). (1.E-6c)

where

(U ,V )=∫−H

❑

(u ,v ) dz, depth-integrated transport (per unit length) (1.E-7a)

and a = -pa/(g) (1.E-7b)

is the surface elevation due to the atmospheric pressure pa.

Homework 1.E-1. Derive (1.E-6a) by integrating the continuity equation ¶u/¶x + ¶v/¶y + ¶w/¶z = 0 with respect to z from z=-H to z=. (Hint: use (1.E-5) and the formula for integration by parts).

Homework 1.E-2. Integrate the hydrostatic equation ¶p/¶z = -g with respect to z’ from any point z’=z to z’=, and show that: -p = -pa -g(-z) where pa = atmospheric pressure. Then write a = -pa/(g), and show that –p = -g(-a-z).

Homework 1.E-3. Using pa = 1 pascal = 1 N m-2 = 10-5 bar = 0.01 mbar, show that a decrease of 1 mbar (= 100 N m-2) of atmospheric pressure will result in approximately an increase in a of about 0.01 m. (You may assume that 103 kg/m3 and g 10 m/s2).

Homework 1.E-4. Derive (1.E-6a,b) by dropping the nonlinear advective terms in the momentum equations (1.D-1b,c), and then by using the “p” obtained from Homework 1.E-2.

Fig. 1.E-1. A “box” ocean circulation driven by eastward (northern half of basin) and westward (southern half) wind

stress distribution.

Depth-integrate (1.E-6b,c), and again drop nonlinear terms such as u etc:

( ∂ U∂ t )=+ fV−gH ∂

∂ x+τw

x−τbx (1.E-8a)

( ∂ V∂ t )=− fU−gH ∂

∂ y+τw

y−τby (1.E-8b)

where tw and tb are wind and bottom stress vectors respectively. These equations together with equation (1.E-6a) constitute there equation for three unknowns (, U, V). For our purpose, it is more convenient to write them in terms of the depth-averaged velocity:

(u , v )=∫−H

❑

(u , v )dz /D∫−H

0

(u , v ) dz /H=(U , V )/H , (1.E-9)

so that (1.E-6a) and (1.E-8) become:

∂∂ t+ ∂u H

∂ x+ ∂ v H

∂ y=Q (1.E-10a)

( ∂ u∂ t )=+ f v−g ∂

∂ x+(τ ¿¿w x−r u) /H ¿ (1.E-10b)

( ∂ v∂ t )=−f u−g ∂

∂ y+(τ ¿¿w y−r v )/H ¿ (1.E-10c)

where r is the bottom friction coefficient with dimension m/s, we have modeled the bottom friction as:

(τ bx , τb

y¿=r (u , v ), (1.E-11)

and we have also added a source term Q to the right hand side of (1.E-10a). This we will use later to model the net effect of precipitation (P) minus evaporation (E), i.e.

Q = (P – E)/o (unit is m/s). (1.E-1)

Let H = constant, a vorticity equation similar to (1.E-10) may now be derived from (1.E-10). Taking the curl of (1.E-10b,c), we obtain:

∂❑∂t

+❑o v−f ∂∂ t/H=[curl¿¿ z (τ ¿¿w)−Qf−r❑]/H ¿¿ (1.E-2a)

Apart from the terms involving Q and ¶/¶t, this equation is equivalent to (1.D-33). The term ¶/¶t does not appear in (1.D-33) because it is neglected when we derived the vertical velocity at the surface. We will explain in the next section when this term may be deleted; assuming that it can be, then (1.E-2a) becomes:

∂❑∂t

+❑o v=[curl¿¿ z (τ ¿¿w)−Qf−r❑]/H ¿¿ (1.E-2b)

The appearance of -Qf together with curlz(tw) means that as far as the large-scale ocean circulation is concerned, effects of wind stress curl and net precipitation minus evaporation are equivalent. The reason is that in both cases, the ocean layer beneath the surface boundary layer (where wind and precipitation etc act) ‘sees’ only vertical mass flux coming in (or out) of the surface layer. In the case of wind stress curl, this is “Ekman pumping” (Chapter 1.D) caused by convergence or divergence in the surface layer, while in the case of precipitation etc, it is more direct – due to the flux of water from the atmosphere. We will model these effects and give a physical interpretation below. Also, while H is the ocean depth, it may be interpreted as the depth of the main thermocline in which the predominant large-scale currents reside. In that case, the friction r u is then considered as the friction at the base of the main thermocline.

Sverdrup Relation:

In the open ocean away from side boundaries, the relative vorticity ❑ is assumed weak, in fact ❑0, and the main vorticity balance in steady state is, from (1.E-2b) (omitting Q for simplicity):

❑o v H=❑o V=curl z(τ ¿¿w)¿ (1.E-3)

Figure 1.E-2. A schematic to explain the equatorward Sverdrup transport V caused by a negative wind stress curl,

curlz(w) < 0.

Equation (1.E-3) is the Sverdrup relation; it shows that in steady motion, a negative curlz(w) produces southward transport. Figure 1.E-2 explains the physics involved. The negative curlz(w) produces convergent flow in the thin surface (Ekman) layer (Chapter 1.D) and hence a downward

pumping that squashes the vortex tube in the deeper interior; this is equivalent to a divergent subsurface flow. Squashing increases the cross-sectional area of the tube, and if it stays at the same latitude a negative relative vorticity (i.e. a reduced spin) would be produced, ❑<0. To keep the same ❑ 0, and assuming that the tube remains approximately vertical (and it does), the tube must move south along the curved earth surface where it can stretch and retain its initial zero spin. The curved surface in this case is the physical equivalence of the -effect, i.e. the Coriolis parameter f varies with latitude. If the earth’s surface is not curved, then = 0. Another way to understand the Sverdrup relation (1.E-3) is that, since the motion is slow, steady and inviscid, Taylor-Proudman theorem (Fig.1.D-1) says that the fluid column can neither change its length nor be tilted. Therefore, the column must move equatorward following the curved surface of the earth. Gill [1982; p.465] gives yet another, equivalent explanation. The divergent subsurface flow beneath the negative curlz(w) increases the cross-sectional area of the vortex tube, so the absolute vorticity (i.e. f + ) decreases (because the circulation around the tube must remain the same assuming an inviscid fluid). But since f >> , the only way that this decrease can be accomplished is that the tube moves southward where f is smaller.

Figure 1.E-2 illustrates two other features. One is that the effect of (positive) Q [= (P–E)/o] is similar to (negative) curlz(w): this is because Q ‘pumps’ water into the subsurface layer and again squashes vortex tube. The other one is that while the surface is slightly elevated ( > 0) it is not important in the argument given above; in steady state, it is in fact = 0. If as shown in Figure 1.E -1 the Sverdrup transport occurs over most portion of the ocean basin, then the total southward transport is:

SV ∫0

Lx

curlz (τ ¿¿w) /❑odx ¿ (m3 s-1). (1.E-4)

Figure 1.E-3. Annual mean distribution of Ekman pumping, positive = upwelling, in unit of 10-6 m s-1 (or 0.1 m/day).

From Tomczak and Godfrey.

Figure 1.E-3 shows a global map of Ekman pumping – note large areas of downwelling in the subtropical regions, suggesting equatorward Sverdrup transports there. From (1.E-4), we have

roughly SV curlz(w).Lx/o 2¿ τ wx∨¿Lx/(Lyo) (2×10-4/10-11)(Lx/Ly) m3 s-1 20 Sv (1 Sv = 106 m3 s-

1) for Lx Ly, ¿ τ wx∨¿ 10-4 m2 s-2 and o 10-11 m-1s-1. The corresponding Ekman pumping

(downwelling) is from (1.D-31) (in dimensional form) = curlz(w)/fo 2¿ τ wx∨¿/(Lyfo) 1~2×10-6 m

s-1 for Ly 2000 km and fo 6×10-5 s-1, roughly equal to the values shown in Figure 1.E-3. This Ekman velocity (or Figure 1.E-3) may be compared with contours of Q [= (P–E)/o] shown in Figure 1.E-4, which gives local maxima of roughly 10-7 m s-1, about 10 times weaker than the Ekman pumping.

Figure 1.E-4. Annual mean distribution of precipitation minus evaporation, Q = (P–E)/o, in m yr-1 ( 3×10-8 m s-1);

shaded regions are positive.

Stommel’s Model of the Western Boundary Current (WBC):

In our idealized closed basin (Figure 1.E-1), the southward Sverdrup transport must be returned northward. Since this return cannot be in the mid-ocean (i.e. open ocean) because of the Sverdrup constraint (as seen above), it must occur near the western and/or eastern boundaries where the Sverdrup balance is broken. In other words, other terms in the potential vorticity equation (1.E-2b), such as friction and/or nonlinearity (which is neglected in the Sverdrup balance 1.E-3), become important near the boundary. Stommel [1948] assumes that (bottom) friction is important so that (1.E-2b) becomes, in steady-state:

∂ v∂ x

+( H❑o

r )v=0. (1.E-5)

Including curlz(w) (or Q) adds unnecessary complexity without adding physical insight so it is omitted. A physical reason is that we expect the western boundary layer to be thin so that the effects of these terms in a small area are small. Also, near the western or eastern boundary, ❑∂ v /∂ x . The solution is:

v=C ( y ) exp (−H❑o x /r ) (1.E-6)

where C(y) is an arbitrary function of y to be determined. Equation (1.E-6) shows a northward jet which is maximum at the boundary (for an eastern boundary, the “x” is replaced by “xLx”) and either decays (if western boundary) or grow (if eastern boundary) exponentially away from the boundary. Therefore, the northward return flow must be along the western boundary and crucially depends on a non-zero o; the width of this return jet may be estimated as the e-1 x-scale:

xwbc = r/(Ho) (1.E-7)

This is 50 km for r 10-4 m s-1, H 200 m and o 10-11 m-1 s-1. The C(y) is determined by requiring that the total northward transport carried by the WBC is equal to the southward Sverdrup transport

at the same latitude (i.e. same “y”). Thus, ∫0

∞

v H .dx=C ( y )r /❑o=−¿SV ¿, i.e.

C ( y )=−¿o SV /r=∫0

Lx

curl z(τ ¿¿w)dx /r ¿¿ (1.E-8)

With the aforementioned values of SV 20 Sv, o 10-11 m-1 s-1 and r 10-4 m s-1, C 2 m s-1.

The Stommel WBC solution (1.E-6) therefore consists of a jet concentrated against the western boundary. Since this jet has to return northward all the Sverdrup transport across a narrow width xwbc, the jet’s velocity is large. The Sverdrup velocity scale is, from (1.E-3), curlz(w)/(Ho)

2¿ τ wx∨¿/(HLyo) (2×10-4/10-11)/(HLy) m s-1 0.05 m s-1 for H = 200 m and Ly = 2000 km; while the

Stommel velocity scale is much larger, C 2 m s-1. Stommel’s jet is maximum at the coast x = 0 and decays exponentially offshore to become very weak in the ocean’s interior where the southward Sverdrup transport prevails. Note that the Stommel’s jet, being wholly positive, cannot join smoothly to the Sverdrup flow, nor is it expected to since it is based on a frictional equation that

ignores the Ekman pumping by the wind; but this is not important. Nor is it important that the Stommel’s jet cannot satisfy the no-slip condition at the coast. The important thing is that: (1) beta (o) is crucial for the existence of the western-intensified boundary current, and (2) the WBC jet provides a mechanism that closes the ocean-basin mass balance.

Munk ’s Model of the Western Boundary Current (WBC) :

A slightly more complicated model can be constructed to satisfy the no-lip condition at x = 0. Instead of the bottom friction we include horizontal viscosity terms on the RHS of the momentum equations:

∂∂ t+ ∂u H

∂ x+ ∂ v H

∂ y=Q (1.E-9a)

( ∂ u∂ t )=+ f v−g ∂

∂ x+ τw

x /H+A H2 u (1.E-9b)

( ∂ v∂ t )=− f u−g ∂

∂ y+τw

y /H+AH2 v (1.E-9c)

where AH is the horizontal (eddy) viscosity coefficient (unit is m2 s-1), assumed a constant. If we now take the curl etc as in the derivation of (1.E-2b) we obtain:

∂❑∂t

+❑o v=[curl¿¿ z (τ ¿¿w)−Qf ] /H+AH2❑¿¿ (1.E-10)

Define a stream function (unit m2 s-1) such that:

v=❑x∧u=−¿y ¿ (1.E-11)

and therefore

¿∂ v /∂ x−∂u /∂ y=¿2 ¿ (1.E-12)

In the WBC, we now let the beta term balances the horizontal viscosity term, equation (1.E-10) in steady state is then (c.f. 2-27):

AH ∂4/∂ x4−❑o ∂/∂ x=∂4/∂❑4−∂ /∂=0 (1.E-13a)

where (let = ax, then substitute in the 1st form of above equation, etc):

= (o/AH)1/3x. (1.E-13b)

Equation (1.E-13a) is a fourth-order ordinary differential equation (with “y” as the ‘floating’ parameter not directly involved), and a solution of the form exp(m) maybe assumed. Four roots are (m4-m = 0; so that m = 0 and m = exp(i2n/3), n = 0, 1, 2; also use cos(/3)=1/2 & sin(/3)=31/2/2):

m1 = 0, m2 = 1, m3 = exp(i/3)=(1+i3)/2, m4 = exp(i2/3) =(1i3)/2 (1.E-14)

The solution then is of the form

= C1 + C2exp(m2) + C3exp(m3) + C4exp(m4) (1.E-15)

where the C’s are arbitrary function of “y” yet to be determined. The m2 root must be deleted - by choosing C2 = 0, because it leads to a solution which exponentially grows with (or x). The root m1

leads to C1 – which is chosen to be C1 = I(y), where the subscript “I” denotes “interior” (of the ocean away from the boundary) and from the interior Sverdrup solution (2-25) with v I=∂I /∂ x,

❑I ( x , y )=∫0

x

curl z(¿ τw )/(H ¿¿ o)dx '+❑o( y)¿¿ (1.E-16)

is the stream function of the interior’s Sverdrup flow; o(y) is an arbitrary function. This choice is so that ~ I for ~ , since m3 and m4 give exponentially decaying solutions for large .

Substituting m3 and m4 from (1.E-14) to (1.E-15), we obtain then:

= I + e−¿2[C3cos (√32 )+ C4sin(√3

2 )] (1.E-17)

the C3 and C4 here differ from those in (1.E-15) but the detail is not important; they are still arbitrary functions of “y.” The corresponding velocity components from (1.E-11) are:

u=−∂I

∂ y−e−¿ 2[

∂ C3

∂ ycos(√3

2 )+ ∂ C4

∂ ysin(√3

2 )] (1.E-18a)

(AH/o)1/3.v=e−¿2[(−C3

2+ √3 C4

2 )cos( √32 )−(√3 C3

2+

C4

2 )sin(√32 )] (1.E-18b)

The conditions that there can be no normal flow (u=0) and no slip (v=0) at = 0 give:

C3 = -I(0,y) + K, where K = constant, and (1.E-19a)

C4=C3/√3. (1.E-19b)

Therefore, (1.E-17) gives:

= I(x, y) - I(0,y)e−¿2[cos (√32 )+ 1

√3sin(√3

2 )]+ Ke−¿2[cos (√3

2 )+ 1√3

sin(√32 )] (1.E-20)

From equation (1.E-13b), we note that (AH/o)1/3 has the unit of length (m). Its value is 100 km (for AH = 103 m2 s-1 and o 10 -11 m-1 s-1) which is thin compared to the width of the basin 2000 km (say); it is the e-folding decay scale of the western boundary current near x 0. We are therefore making only a small error if I(0,y) in (1.E-20) is replaced by I(x,y), or vice versa. Another way of saying the same thing is that, since I varies over a large scale >> (AH/o)1/3, I(x, y) I(0,y) near the western boundary. To determine K, we note that, since our basin is closed, the stream function around the boundary is zero. Therefore, for to be also = 0 at x = 0, (1.E-20) shows that K must also = 0. Therefore, (1.E-20) becomes:

= I(0, y) {1 - e−¿2[cos (√32 )+ 1

√3sin(√3

2 )]} (1.E-21)

The northward velocity is ¶/¶x:

v=(¿o/ AH)13

I (0 , y ) 2√3

sin(√32 )e

−¿2

¿, = (o/AH)1/3x, (1.E-22)

in which we have neglected the interior velocity contribution ¶I/¶x, which is much smaller. Note that the jet form is again apparent by the exponential decay term. But in contrast to the Stommel’s solution equation (2-28), because of the sine function, the northward velocity is zero at x = 0. Also because of the sine function the velocity becomes slightly negative (i.e. reversed flow) at the offshore edge of the jet.

Class: estimate the magnitude of v from (1.E-22).

Two-Layer Ocean and the Reduced-Gravity Formulation

We temporarily denote non-dimensional variables with subscripts “n.” The above model has one layer and involves the free surface . We now show that with slight reinterpretations of the variables, the equations of that model are exactly the same as those governing the motion of a two-layer ocean in which one of the layers is infinitely thick and is quiescent. Before we do this, we find out first the condition when the term involving /t in equation (1.E-2a) may be neglected. This is necessary in order to understand the restrictions on the spatial and temporal scales under which the resulting equation(s) apply. Instead of (1.E-2a), we will non-dimensionalize the vorticity equation that uses the horizontal viscosity as in the Munk’s model instead of the bottom friction, i.e. we use (1.E-10) with f(/t)/H retained:

∂❑∂t

+❑o v−f ∂∂ t/H=[curl¿¿ z (τ ¿¿w)−Qf ]/H+A H

2❑¿¿ (1.E-23)

Define the following non-dimensional variables:

❑=❑n(UL ), o=❑on(U

L2 ) , v=vn U ,=❑n( fULg ) , t= tn( L

U ) ,

w = wno, Q = Qn Qo, AH = AHn AHo. (1.E-24)

Note that the scale for , i.e. fUL/g, is determined from balancing the pressure gradient and Coriolis terms in equation (1.E-9b,c). Substitute (1.E-24) into (1.E-23):

∂❑n

∂t n+❑onvn−( L2

R2 ) ∂❑n

∂ t n=[( Lo

HU 2 )curl ¿¿ zn(τ ¿¿wn)−( L2 fQo

HU 2 )Qn]+( AHo

UL )AHn❑n

2❑n ¿¿(1.E-25)

where R = (gH)1/2/f is the Rossby radius of deformation. We see that dropping the term involving n/tn amounts to saying that R >> L, i.e. the spatial scale of motion is much smaller than the Rossby radius. For H 1000 m and f 6×10-5 s-1, R 1500 km, so that motions with scales smaller than about 1000 km will not be much affected by the production of vorticity due to stretching associated with the motion of the free-surface. This kind of motion is called a barotropic motion. On the other hand, as we will now show, R can be quite small for the reduced-gravity model, in which case the stretching term becomes important.

The Reduced-Gravity Model from the Two-Layer Model:

We now derive the reduced-gravity (or 1.5-layer) equations from a two-layer model as sketched schematically in Figure 1.E-5. The caption of Figure 1.E-5 explains all the variables to be used. Since in each layer the density is constant, the continuity (conservation of volume) equation takes exactly the same form as that for a homogeneous fluid (see equation 1.E-9a):

∂ hi

∂t+

∂ ui hi

∂ x+

∂ v i hi

∂ y=0, i = 1, 2. (1.E-26)

Figure 1.E-5. An illustration in the xz-plane of an ocean model with two (inmiscible fluid) layers with subscripts “1” and “2” denoting the upper and lower layers respectively. Symbols are: r1 and r2 = constant densities of the upper and lower layers respectively, r2 > r1; H1 = constant undisturbed upper-layer depth; H2(x,y) = undisturbed lower-layer depth which may be a function of (x,y) for variable bottom topography; h(x,y,t) = layer depth; u(x,y,t) = the x-component velocity; v(x,y,t) = the y-component velocity; p(x,y,z,t) = pressure; h1(x,y,t) = disturbed free-surface height measured from z = 0; h2(x,y,t) = disturbed inter-layer height measured from z = -H1. Note that h1 and h2 can be positive (upward) or negative (downward), and that h1 = H1 + h1 - h2.

where i = 1 is for layer-1 and i = 2 is for layer-2, and for simplicity the source/sink term Q is omitted. (The subscripts ‘i’ in equation 1.E-26 are not to be confused with x and y components, and repeated subscripts do not mean summation!). Except for the pressure gradient terms, the momentum equations (1.E-9b,c) will also remain essentially the same. To evaluate the pressure gradients, we will assume as before that the horizontal scales are much larger than the vertical scales, so that the pressure is hydrostatic, ¶p/¶z = -g. Thus the pressure at some depth z = z’ in the water is just the weight per unit area of fluid above. For a point in layer 1, the pressure p1(x,y,z,t) is:

p1(x,y,z,t) = pa(x,y,t) + g1(1 z), for 1 z H1+2, (1.E-27a)

where pa = atmospheric pressure (i.e. = weight per unit area of air above z = ). Similarly, for a point in layer 2, the pressure p2(x,y,z,t) is:

p2(x,y,z,t) = p1(at z =H1+2) + g2(H1 + 2 z), for H1+2 z (H1+H2), (1.E-27b)

Then the horizontal pressure gradient in layer 1 is (for example /x):

p1/x = pa/x + g11/x, for 1 z H1+2, (1.E-28a)

From (1.E-27a), pressure at the interface z = -H1 + 2 is:

p1 = pa(x,y,t) + g1h1, (use 2 = H1 + h1 - h1 from Fig.2-5).

Then the horizontal pressure gradient in layer 2 is:

p2/x = pa/x + g1h1/x + g2 (1h1)/x,

so that, p2/x = p1/x + g(1h1)/x, for H1+2 z (H1+H2), (1.E-28b)

where = (2 1).

The momentum equations (2-31b,c) applied to layers 1 and 2 separately lead to (1.E-29b,c) and (1.E-30b,c) below:

∂ h1

∂t+

∂u1 h1

∂ x+

∂v1 h1

∂ y=0 (1.E-29a)

( Du1

Dt )=+ f v1−1ρo

∂ pa

∂ x−g

∂❑1

∂ x+ τw

x /H+AH ∂2u1/∂ x2 (1.E-29b)

(D v1

Dt )=−f u1−1ρo

∂ pa

∂ y−g

∂❑1

∂ y+τw

y /H+AH ∂2 v1/∂ y2 (1.E-29c)

∂ h2

∂t+

∂ u2 h2

∂ x+

∂ v2 h2

∂ y=0 (1.E-30a)

( Du2

Dt )=+ f v2−1ρo

∂ pa

∂ x−g

∂❑1

∂ x−g ρ

ρo∂(¿1−h1)

∂ x+A H ∂2 u2/∂ x2 ¿ (1.E-30b)

( D v2

Dt )=−f u2−1ρo

∂ pa

∂ y−g

∂❑1

∂ y− g ρ

ρo∂(¿1−h1)

∂ y+A H ∂2 v2/∂ y2¿ (1.E-30c)

Note that we have generalized /t to D/Dt in the momentum equations. These are the equations for our two-layer ocean. More complicated forms with additional terms exist in the literature; for our purpose, the above suffices. Note that the lower layer experiences both the layer-1 pressure gradient (p1/x)/o as well as the additional force (per unit mass) (g/o)(1h1)/x; and similarly for the y-component.

In the reduced-gravity model, we let one of the layers to be infinitely thick. Supposing we let layer-2 to be infinitely thick, then from the continuity equation (1.E-30a) the deep-layer currents (u2, v2) must approach zero, and h2/t ~ 0 also. Subtracting (1.E-30) from (1.E-29) then gives:

∂ h1

∂t+

∂ u1 h1

∂ x+

∂ v1 h1

∂ y=Q (1.E-31a)

( Du1

Dt )=+ f v1−g ρρo

∂ h1

∂ x+τ w

x /H+AH ∂2u1/∂ x2 (1.E-31b)

( D v1

Dt )=−f u1−g ρρo

∂ h1

∂ y+ τw

y /H+AH ∂2 v1/∂ y2 (1.E-31c)

in which we have put back the “Q” on the RHS of the continuity equation, and have made a further approximation that 1x << h1/x and similarly 1y << h1/y, which means that the variations of the free surface is negligible in comparison to the variations of the upper-layer depth. Equations (1.E-31) are identical to the equations that govern the motion of a single-layer (homogeneous) fluid of constant depth H1 (i.e. the topography) with a free surface except that the acceleration due to gravity g is reduced to:

g’ = g/o (1.E-32)

which appropriately is called the “reduced gravity.” To see the equivalence, set

h1 = H1 + , (1.E-33)

where now the “” is interpreted as the departure of the upper-layer depth from its ‘mean’ of H1. For typical /o 10-4~10-2, the corresponding shallow-water baroclinic phase speed Ci = (g’H1)1/2, hence also the baroclinic Rossby radius of deformation Ri = (g’H)1/2/f, is some 10 to 100 times smaller. Therefore, for reduced-gravity (baroclinic) motions, the time-dependent stretching term

(L/Ri)2 h1n/tn that produces vorticity in equation (1.E-25) can be significant. In fact, this term overwhelms the other time-dependent term 1n/t in the limit of

(L/Ri)2 >> 1. (1.E-34)

The reduced-gravity equivalence of (1.E-23) is:

∂❑1

∂ t+u1 .∇❑1+❑o v1−f

∂h1

∂ t/H 1=[curl¿¿ z (τ ¿¿w)−Qf ] /H 1+AH

2❑1¿¿(1.E-35)

in which the nonlinear advection term u1.1 (from the D/Dt terms in 1.E-31b,c) is included. But, consistent with the quesi-geostrophic approximations of Chapter 1, we may replace the u1 in this nonlinear term by its geostrophic values:

H1v1 = +(Ci2/f)h1/x, H1u1 = (Ci

2/f)h1/y (1.E-36)

Equation (1.E-36) suggests a stream function 1 such that:

1/x = H1v1 and 1/y = H1u1 (1.E-37)

By comparing (1.E-36) and (1.E-37), we then have:

1 = (Ci2/f)h1 = fRi

2 h1 (1.E-38)

Equation (1.E-35) then becomes:

−∂❑1

∂ t/Ri

2+H1

∂❑1

∂t+J (❑1 ,❑1 )+❑o

∂❑1

∂ x=[curl¿¿ z (τ ¿¿w)−Qf ]+H 1 AH

2❑1¿¿

(1.E-39)

where J(A,B) = A/x.B/yB/x.A/y is the Jacobian (of the functions A and B).

The Island Rule

Consider a single layer of constant depth (=H) of homogeneous fluid:

¶u/¶t + (f+)k×u = -(p/ + |u|2/2) + (tw - tb)/H (1.E-40a)

where the u is non-divergent, i.e. it satisfies .u = 0, so that a stream function can be defined:

Hu = k × (i.e. Hu = -¶/¶y, Hv = ¶/¶x) (1.E-40b)

Integrate “C∫” (1.E-40a) cyclonically around the circuit “C” (see Fig.1.E-6):

¶/¶t[C∫ Hu.dl] - ∫IES fHv.dl + ∫EIN fHv.dl = C∫(tw - tb). dl - C∫u.n dl (1.E-41)

where ∫IES = integration from xI(yS) to xE(yS), eastward along the southern latitude line, and∫EIN = integration from xE(yN) to xI(yN), westward along the northern latitude line.

Fig.1.E-6.

But, ∫IES fHv.dl = ∫IES fHv.dx = fS.(E - I), since dl = dx on “ys”, and∫EIN fHv.dl = -∫EIN fHv.dx = ∫IEN fHv.dx = fN.(E - I), since dl = -dx on “yN”.

Equation (1.E-41) then becomes:

¶/¶t[C∫ Hu.dl] - fS.(E - I) + fN.(E - I) = C∫(tw - tb). dl - C∫u.n dl

i.e. I.(fN-fS) = -C∫(tw - tb). dl + C∫u.n dl + ¶/¶t[C∫ Hu.dl] (1.E-42)where E has been arbitrarily set = 0. Physically, (E - I) = transport between the Island and the eastern boundary. If ¶/¶t = 0 and low Rossby number so that = |/f| << 1, then the last 2 terms can be neglected:

I = -C∫(tw - tb). dl /(fN-fS), which is the Island rule [Godfrey, 1989]. (1.E-43)

Final Exam Review

1. For each of the terms in the table below, write down where appropriate the formula, example(s) and/or brief explanation(s):

Term Formulae, Examples &/or Brief Explanations

Rossby number

Ekman number

Ekman depth

Coriolis (i.e. inertial) frequency at

lat=

Coriolis period at =90oN

Geostrophic equation in vector form

Geostrophic equations in x & y

component form

Steady Ekman-layer equation in

vector form

Steady Ekman-layer equations in x &

y component form

Ocean’s surface Ekman transport by

wind stress to

References

Batchelor, G.K., 1967: Fluid Dynamics.Baumert et al., 2005:Feliks, Y., H. Ghil and E. Simonnet, 2004: Low-frequency variability in the midlatitude atmosphere induced by an oceanic thermal front. J. Atmos. Sci., 61, 961-981.Gill, A., 1982: Atmospheric & Oceanic DynamicsGodfrey, J.S. 1989: A Sverdrup model of the depth-integrated flow from the world ocean allowing for island circulations. Geophys. Astrophys. Fluid Dyn., 45, 89-112.Mellor, G.L., 2004: POM User Guide.Mellor G.L. and Yamada, T., 1982: Review of Turbulence Models.Minobe et al. 2008: Influence of the Gulf Stream on the troposphere. Nature, 452, 206-210.Pedlosky, J., 1979: Geophysical Fluid Dynamics.Roach, P., 1972: Computational Fluid Dynamics.G.G. Stokes (1847). "On the theory of oscillatory waves". Transactions of the Cambridge Philosophical Society 8: 441–455. Reprinted in: G.G. Stokes (1880). Mathematical and Physical Papers, Volume I. Cambridge University Press. pp. 197–229.Van Dyke, M., 1964: Perturbation Methods in Fluid Mechanics.Stewart, Robert H., 2000: Introduction to Physical oceanography. Columbia University Press, http://www.earthscape.org/r3/stewart/index.html.

http://www.earthscape.org/r3/stewart/index.html

http://www.archive.org/details/mathphyspapers01stokrich

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