Rossby Waves
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.
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
 o   capped by a much shallower layer of density  o .
The lower layer is assumed to be motionless. This is consistent with our
observation that the deep ocean is typically relatively quiescent.
The interface between the two layers is at z=H(x,y,t) and the free surface is
at z=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 znm.
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As we have seen previously, znm must coincide with p=constant for there to
be no flow and the weight of water between z = h and z = znm 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  gH
Constant pressure at depth means this internal pressure difference due to
displacement of thermocline is compensated by the elevated sea level h.
 o g (h  H )  (  o   ) gH  const
 o gh  gH  const  0

h
H
o
Density varies by perhaps 3 kg/m3 across the thermocline, so  /  o is
around 3/1000 = 0.003.
The only way we can get this to balance is to have H and h with opposite
sign, and since the factor  /  o 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.
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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
B p1
M    v dz dx
A po
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
M /
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]
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(define local x,y coordinates on the diagram)
We can use the geostrophic relation to calculate the average velocity
between stations A and B.
u
g h
f y
where h is the steric height (of the surface relative to some depth).
The mass transport per unit depth between A and B is
B p1
M    v dz dx   o
A po
g h
g
LH   o h H
f L
f
where H is now the average thickness of the layer (it could be the depth to
the reference level znm).
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
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 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
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]
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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
h  h2  h1  

o
(H 2  H1 )


Q: The mass transport per unit depth between A and B is?
Think of the surface geostrophic current:
Slope of the sea surface is h / LAB , so the velocity is (don’t worry about
signs)
v
g
h / L AB
f
and the transport is this average velocity, times the width of the current,
times the average layer thickness, say H = ½(H1+H2)
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M AB  vLAB H

g
h H
f1
or

g 
H H
f1  o
where I’ve used
layer thickness)
h

o
H
with both h and H positive (surface height and
Units are transport in m3/s (velocity time width of current times depth)
Check signs on the basis of our rules:
h > 0, f1 < 0, get M < 0 for southward transport – GOOD.
What about at y2?
The equation for MCD at section C-D is almost the same:
 h is the same
 the distance LCD doesn’t matter
 but f2 is different
So we get
M CD 
g
h H
f2
or

g 
H H
f2 o
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The Coriolis parameter increases in magnitude with distance away from the
equator, so
| f 2 | > | f1 |
Q: What does this mean for the transport between the two isobars at CD
compared to AB?
A:
|MCD| < |MAB|
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 h 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.
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In a simple 1½ layer ocean for which geostrophic dynamics holds, the layer
interface (pycnocline) displacement is opposite in sign to the sea surface
displacement and greater in magnitude by the ratio of the relative density
difference in the layers:
h

o
H
Contours of surface dynamic height, interface displacement and geopotential
height are all parallel and represent streamlines of the geostrophic flow.
The transport between two points A and B can be calculated a number of
ways. One approach is to consider the velocity implied by the surface height
gradient:
v
g
h / L AB
f
and multiply this by the distance between A and B and the average layer
thickness to get an effective volume transport:
g
M AB  vLAB H  h H
f1
or

g 
H H
f1  o
where the ratio of sea surface and interface displacement is used to write this
as an equation in H only.
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The dependence of transport on f means that on large “planetary” scales,
variation in Coriolis causes convergence and divergence to the west and east
of eddies such that their pattern propagates westward.
This propagation leads to the accumulation of energy in the west of the
ocean gyres and produces the intensification 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 -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
 y :
i.e. f  f o  y
f  2 sin 

f

 2 cos 
y
y

1

R Earth 
where  = 7.292 x 10-5 s-1 and y
Radius of the Earth is 6371 km, so
  x 7.292 x 10-5 / 6371 x 103 cos  


x 10-11 cos 
and the units are?
… f / length = s-1 m-1
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At 20oN  x 10-11 m-1s-1
At 40oN  x 10-11 m-1s-1
What about southern hemisphere latitudes, 20oS and 40oS?
The net volume convergence between y1 and y2 is
 1 1
H H   
o
 f 2 f1 

y
 gH
H
o
f1 f 2
M CD  M AB  g

and for small y we can assume that f1 f2 = f2 where f is the Coriolis
parameter at the average latitude (where we calculate  


 1
1  f  f 2 y
    1
 2
f1 f 2
f
 f 2 f1 
This volume convergence must be balanced by a pycnocline that is being
H
driven down at vertical velocity of t in the box ABCD.
Balancing these we get:
xy
H

y
 gH
H 2
t
o
f
Check the signs here:
If the thickness of the layer H is increasing with time, the LHS is positive.
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Everything on the RHS is also positive because the way I defined H was
the difference right-left. On the east side, H changes sign and this means
H
t is negative, consistent with a thinning upper layer.
Divide through by xy 
H
 H y 1
 gH
t
 o x f 2 y

g H  H
f 2  o x

H
t
H
x is the speed c at which a line of constant H (a
Now, the ratio
wave crest for example) moves eastward.
So the planetary eddy pattern moves westward at a speed of –c, with
c
 gH
o f 2
(units are m s-1) 

[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.
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Plot this pattern offset in time, now consider what it looks like if we color in
the picture.
In an interval t , the pattern moves west a distance x , so the slope of these
H x H H
x / t 

/
 c
t H
t x
lines is
which gives the wave speed.
Let’s check how well our simple theory fits these observations in the Indian
Ocean.
c = x / t  30o 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 in
the surface is 1026.6, and at depth is 1027.8, so we get:
c
 gH
o f 2
= 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-1
(close enough for such a simple estimate)
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.
Choose some reasonable approximate values:
H = 300 m,
 /  o 3 x 10-3
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We find that
c = 1.27 m/s
at 5oS or 5oN
c = 0.08 m/s
at 20oS or 20oN
c = 0.02 m/s
at 40oS or 40oN
(=> 6 months to cross Pacific)
(=> 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 4oN 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.
http://www.po.gso.uri.edu/demos/
Suggested reading for next topic:
 Pond and Pickard section 9.5
 Tomczak and Godfrey chapter 4
 Stewart Chapter 11
[Video of Australian region]
Rossby waves are a general phenomenon of planetary scale motion of fluids
and gases, including the atmospheres of the other planets as well as the
Earth. In the Earth’s atmosphere, planetary eddies are the atmospheric highs
and lows and play a key role in determining the weather.
Atmospheric highs and lows generally move eastward because they are
carried along by the mean flow, such as the jet stream. Relative to the mean
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flow of the air however, they are going westward. So the highs and lows
move slower than the jet stream around them.
Current velocities in the ocean gyres are generally much slower than the
Rossby wave speed, except at high latitudes, so oceanic Rossby wave
movement is almost always westward and can be seen in the sea surface
height displacements observed by orbiting radar altimeters, and also in sea
surface temperature patterns.
An exception to this is the Southern Ocean, where the Antarctic Circumpolar
Current is the oceanic analogue of the atmospheric jet stream. It is able to
circumnavigate the planet without interruption, unlike the mid-latitude
oceanic gyres, and eddies and Rossby waves (which are very slow at such
high latitudes) riding on the ACC do get swept eastward.
If the ocean were purely geostrophic, then the depressions and bulges in
thermocline seen, for example, in the trans-Pacific hydrographic sections or
in horizontal maps of temperature and salinity, would all move toward the
western boundary at the Rossby wave speed. Within a few years the ocean
would come to a state of horizontally uniform stratification, and no flow.
There must be some process constantly replenishing these bulges, or eddies
Q: What is this process?
A: The winds
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