Nov 14: Rossby waves and westward intensification

Recommended reading:

Tomczak and Godfrey “Introduction to regional Oceanography” Chapter 3

East-west hydrographic sections in the sub-tropical Pacific showed a thermocline that deepens toward the west. The dynamic height of the surface therefore increases toward the west.

This implies there is geostrophic flow out of the page all across that section (except for the western boundary current).

Deepening of the thermocline moving westward is a feature of all the subtropical gyres. The dynamical reason for this is related to the large scale planetary waves, called Rossby waves, which determine how the subtropical ocean adjusts to variations in wind and buoyancy forcing.

The essence of the dynamics of Rossby wave propagation (Rossby waves always travel westward) can be captured in a very simple dynamic model, one consisting of a simple 1½ layer ocean not unlike the tank experiment Bob Chant conducted, only much much bigger!

1½ layer approximation

To understand the mechanism that gives rise to Rossby waves it is useful to consider an idealized approximation to the oceanic density structure known as the 1½ layer ocean.

[Tomczak and Godfrey, fig. 3.3 - 1½ layer ocean]

In such a model, the ocean is represented as a very deep layer of density capped by a much shallower layer of density.

The lower layer is assumed to be motionless. This is consistent with our observation that the deep ocean is typically relatively quiescent.

The displacement of the interface between the two layers is H(x,y,t) and the displacement of the free surface is h(x,y,t).

The shape of the free surface corresponds to the steric height of the p=0 surface relative to a level of no motion in the lower layer at z_nm.

As we have seen previously, z_nm must coincide with p=constant for there to be no flow and the weight of water between z = h and z = z_nm must be constant. This gives us information on the relationship of the thermocline and sea surface displacements.

The pressure difference over the depth range associated with the displacement of the layer interface (the “thermocline”) is less in the eddy center because of the less dense water. The difference in the weight of water inside compared to outside the eddy is

Constant pressure at depth means this internal pressure difference is compensated by the elevated sea level h.

Density varies by perhaps 3 kg/m³ across the thermocline, so is around 3/1000 = 0.003.

The only way we can get this to balance is to have H >> h as we have seen before.

Since the factor is so small, we see that displacement of the thermocline H(x,y) must be much larger than displacement of the ocean surface h(x,y), and in the opposite direction.

Sometimes oceanographers will be a bit cavalier about signs here, changing the sign of H for convenience. Then h and H always have the same sign. In effect this is re-interpreting H as the thickness of the upper layer. I actually prefer to think of H and h this way. My advice is don’t be dependent on getting the math right – develop intuition for what ought to happen and work from that – this will get you out of sign troubles.

Where h slopes upward, H slopes steeply downward, and vice versa.

Rule: In most ocean regions (where the 1½ layer model is a good approximation) the thermocline slopes opposite to the sea surface, and at an angle usually 100-300 times larger than the sea surface.

This simple rule allows a very easy check on the current direction from hydrographic measurements. You may want to verify this by looking at hydrographic sections across currents for which we know the direction of flow by independent measure, such as current meters or drifters.

Return to the Pacific dynamic topography map…

Notice that the dynamic topography of the ocean sea surface has a maximum in the west of every oceanic basin.

To understand how this arises we need to examine how the mass transport of geostrophic flow is modified by the variation of the Coriolis parameter with latitude.

So first we look at geostrophic currents and mass transport that goes along with the steric height map we have computed from observations of temperature and salinity.

The mass transport between two stations A and B on a steric height map is

in kg s^-1

This is mass x velocity x area kg/m³.m/s.m² = kg s^-1

The volume transport is simply mass transport divided by density

in m³/s

Typical values for a current such as the Gulf Stream would be of the order of millions of m³ s^-1.

Recall that oceanographers use a unit called the Sverdrup; 1 Sv = 10⁶ m³/s

Consider a map of steric height of the ocean surface relative to some depth of no motion:

[Tomczak and Godfrey, fig. 3.2 - transport and dynamic topography]

(define local x,y coordinates on the diagram)

We can use the geostrophic relation to calculate the average velocity between stations A and B.

where h is the steric height (of the surface relative to some depth).

The mass transport per unit depth between A and B is

where H is now the average thickness of the layer (it could be the depth to the reference level z_nm).

Note that the total transport does not depend on how far apart the points A and B are.

If the contours of steric height converge:

· the velocity between them increases

· the separation decreases

· the net transport remains fixed

The geostrophic current remains confined between the contours of steric height, so the height contours are effectively streamlines of the geostrophic flow. This is why mapping steric height is such an effective way of visualizing the flow field.

Variation in f

An important thing to notice is that the transport M depends on f, and f varies with latitude. This gives rise to waves of very long wavelength (1000s of km) that we call Rossby waves.

Rossby waves and westward intensification

Equipped with this knowledge about how the pressure field, thermocline displacement, and mass transport interact through geostrophic dynamics, we are set to consider the properties of Rossby waves.

Consider an eddy in a 1½ layer ocean.

To keep you on your toes, it’s a Southern Hemisphere anticyclonic eddy.

Is the sea level high or low?

Which way do the currents go?

What is the displacement of the thermocline?

[Tomczak and Godfrey, fig. 3.4 – Ekman convergence/divergence for eddy]

Sketch the layer depth (or thickness, since h<<H)

Isobars are parallel to the interface displacement, so we consider two isobars, H₁ and H₂, which will be streamlines of the geostrophic current.

I want to look at the mass transport between these two isobars at different latitudes y₁ and y₂.

The difference in steric height between the two contours is

Q: The mass transport per unit depth between A and B is?

Think of the surface geostrophic current:

Slope of the sea surface is , so the velocity is (don’t worry about signs)

and the transport is this average velocity, times the width of the current, times the average layer thickness, say = ½(H₁+H₂)

where I’ve used with both h and H positive (surface height and layer thickness)

Units are transport in m³/s (velocity time width of current times depth)

Check signs on the basis of our rules:

> 0, f₁ < 0, get M < 0 for southward transport – GOOD.

What about at y₂?

The equation for M_CD at section C-D is almost the same:

· is the same

· the distance L_CD doesn’t matter

· but f₂ is different

So we get

The Coriolis parameter increases in magnitude with distance away from the equator, so

| f₂ | > | f₁ |

Q: What does this mean for the transport between the two isobars at CD compared to AB?

A: |M_CD| < |M_AB|

because of the change in f.

Q: Where does the water go?

A: It pushes the thermocline down, so H increases.

On the western side of the eddy, the variation of f causes convergence of flow and pushes the thermocline down.

Q: What happens on the eastern side of the anticyclone?

A: Transport magnitudes are the same because Dh is the same

and f₁ and f₂ are the same … only the direction is reversed.

This causes divergence.

On the eastern side of the eddy, the variation of f causes divergence of flow and pushes the thermocline? … UP.

Q: What happens to the eddy because of this convergence and divergence?

A: It goes west

The westward movement of such planetary eddies is known as Rossby wave propagation.

Rossby waves move thermocline displacements westward, though they don’t actually move the water.

But they are moving energy westward, and accumulation of energy in the west leads to the intensification of currents on the western side of the ocean basins.

We can estimate the speed that the Rossby wave moves by considering these convergence and divergence processes and the rate at which they displace the thermocline.

First we introduce the b-plane approximation which is just a convenient way of representing the variation of f with latitude.

Between two latitudes y₁ and y₂, f changes by an amount :

i.e.

where = 7.292 x 10^-5 s^-1 and

Radius of the Earth is 6371 km, so

= 2 x 7.292 x 10^-5 / 6371 x 10³

= 2.28 x 10^-11

and the units are?

… f/length = s^-1 m^-1

At 20^oN = 2.15 x 10^-11

At 40^oN = 1.75 x 10^-11

What about southern hemisphere latitudes, 20^oS and 40^oS?

The net volume convergence between y₁ and y₂ is

and for small we can assume that f₁ f₂ = f² where f is the Coriolis parameter at the average latitude (where we calculate ).

1/f₂ - 1/f₁ = ( f₁ – f₂ )/(f₁ f₂) = /(f₁ f₂)

This volume convergence must be balanced by a deepening thermocline that is being driven down at vertical velocity of in the box ABCD.

Balancing these we get:

Check the signs here. If the thickness of the layer H is increasing with time, the LHS is positive. Everything on the RHS is also positive because the way I defined was the difference right-left. On the east side, changes sign and this means is negative, consistent with a thinning upper layer.

Divide through by :

Now, the ratio is the speed c at which a line of constant H (a wave crest for example) moves eastward.

So the planetary eddy pattern moves westward at a speed of –c, with

(units are m/s)

[Indian Ocean Hovmueller diagram of Rossby wave crests]

http://www.soc.soton.ac.uk/JRD/SAT/Rossby/ltplotprod_largerfont.gif

Thermocline displacements have small sea surface displacements associated with them and we can observe these from space with a satellite altimeter.

Imagine a snapshot in time of a series of wave crests and troughs across the ocean at some latitude.

A short while later, the pattern has moved westward by a fixed amount that is roughly the same at every longitude.

Plot this pattern offset in time, now consider what it looks like if we color in the picture.

In an interval , the pattern moves west a distance , so the slope of these lines is which gives the wave speed.

Let’s check how well our simple theory fits these observations in the Indian Ocean.

c = / 30^o lon * 111 km * cos(25)

60 cycles * 10 days * 86400 sec

= 5.8 x 10^-2 m s^-1

Indian Ocean thermocline is at around 1000m depth, and the average r in the surface is 1026.6, and at depth is 1027.8, so we get:

= 2.1 x 10^-11 * (27.8-26.6)/1027.8 * 9.81 * 1000 m / (6.16 x 10^-5)²

= 6.3 x 10^-2 m/s (close enough).

Things to notice about the Rossby wave speed:

· gets larger approaching the equator

· always positive (i.e. westward propagation in our sign convention)

This is an approximate equation for very long wavelength, long period (many months) Rossby waves, for the idealized 1½ layer ocean.

Take typical values:

H = 300 m, = 3 x 10^-3

we find that

c = 1.27 m/s at 5^oS or 5^oN (=> 6 months to cross Pacific)

c = 0.08 m/s at 20^oS or 20^oN

c = 0.02 m/s at 40^oS or 40^oN (=> 20 years to cross Pacific)

[Chelton and Schlax - TOPEX Rossby wave propagation across Pacific]

Hovmueller diagrams at different latitudes show different speeds.

Pacific transit times at 4^oN are only a year, compared to many years at higher latitudes.

At the equator, there is no obvious westward propagation. As we will see when we consider ENSO, there is another class of planetary waves (Kelvin waves) with quite different features that propagate eastward along the equator.