Weather Observation and Analysis
John Nielsen-Gammon
Course Notes
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Chapter 10. GEOSTROPHIC BALANCE
10.1 Wind and Height Gradients
You hear it frequently in weather reports, you’ve seen it on
weather maps, and it makes intuitive sense: strong pressure gradients are
associated with strong winds. That statement is a general one that applies
to all classes of fluid motion. But if you have a friend who’s an engineer
or physicist and you want to freak them out, tell them about the
atmosphere. Sure, strong gradients mean strong winds, but generally the
wind doesn’t blow from high pressure toward low pressure. Instead, it
blows sideways.
This characteristic is observed in its purest form aloft, away from
the effects of surface friction. Most of the time, meteorologists use
pressure coordinates when considering the distribution of weather
elements aloft, and on a pressure surface, gradients of geopotential height
have the same meaning and effect as gradients of pressure at a constant
altitude. So from here on, we’ll be talking about strong height gradients
and the wind not blowing from high heights toward lower heights (on a
constant pressure surface such as 500 mb).
On a map of a pressure surface, the height information is usually
depicted with contours, just like isobars depict pressure information on a
sea level map. If you examine any pressure surface, from 850 mb on up,
you will find the following general characteristics:
1. Stronger winds tend to occur where the height contours are
closer together; weaker winds where the height contours are farther apart.
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2. The winds tend to mostly blow parallel to the height contours.
3. The winds tend to be oriented so that the higher heights are 90
degrees to the right of the direction toward which the wind is blowing.
These rules aren’t followed exactly, but they are close enough that
if you just had the height contours to go by, you could make a pretty good
estimate of the wind speed and direction just about everywhere.
A hypothetical wind that follows these rules exactly is called a
geostrophic wind. Since the real wind comes close in most cases, it’s fair
to say that the actual wind is nearly geostrophic.
10.2 The Geostrophic Equation
If you were, in fact, confronted with a geopotential height analysis
and asked to estimate the wind, you could obtain an exact estimate of the
direction pretty easily, by following rules 2 and 3 above. The challenge
with estimating the speed is that you need some mathematical formula that
tells you exactly how strong a height gradient corresponds to how strong a
geostrophic wind speed.
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This formula is
r
g
vg = 0 ∇h Z
f
In words, the magnitude of the geostrophic wind is proportional to the
magnitude of the geopotential height gradient on a constant pressure
surface, with the constant of proportionality being the ratio of the
gravitational acceleration and the Coriolis parameter.
The Coriolis parameter, f, pops up frequently when dealing with
large-scale and medium-scale atmospheric motion. It is defined as
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f ≡ 2Ω sin(lat )
so it is not a constant but instead depends on latitude. The rotation rate of
the Earth, Ω, is not quite constant itself, but it’s mighty close. It has the
value of 7.292 x 10-5 s-1 (actually, radians/second, but radians are unitless).
This number must seem rather abstract to you, but if you multiply it by the
number of seconds in a day, 86,400, you get 2π, saying that the Earth
rotates 2π radians (one complete rotation) in one day. Which, after all, is
the definition of one day in the first place.
[Those science geeks among you who like to check the math found that
you get something just a little bit bigger than 2π. In fact, it’s bigger by
366.25/365.25, a correction of less than 1% that’s necessary because the
Earth is also revolving around the Sun. Once the Earth has completed one
entire rotation, it has orbited a bit around the Sun, so to get the Sun in the
same position relative to the horizon the Earth has to rotate just a little bit
more than one complete rotation.]
Defined as such, f ranges from -1.4 x 10-4 s-1 at the South Pole to
+1.4 x 10-4 s-1 at the North Pole. It is zero at the equator. At midlatitudes,
a commonly-used representative value for f is 1 x 10-4 s-1.
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Okay, so now you know f, and you already know g. Let’s plug in
some numbers to answer the question: if the height gradient is 3 decameter
(30 m) per degree of latitude, how strong would the geostrophic wind be?
r
g
9.8ms −2 1 30m
1o lat
vg = 0 ∇h Z =
×
×
= 26.4ms −1 = 51kt
f
1 × 10 −4 s −1 1o lat 1.11 × 10 5 m
The last step used the conversion factor of 1.93 from meters per
second to knots. A speed of 51 kt is a moderate wind speed, so in the
geostrophic sense, a height gradient of 3 dam per degree is a moderate
height gradient.
With the equation for the magnitude of the geostrophic wind and
our knowledge of the direction of the geostrophic wind, we can write
down equations for the individual geostrophic wind components. For
example, if we choose our x axis to be parallel to the geostrophic wind, the
height gradient must be oriented in the negative y direction, with lower
heights in the direction of positive y. So the magnitude of the height
gradient is given by the magnitude of the derivative of height with respect
to y, and to get the wind going in the positive x direction we must multiply
by -1:
ug = −
g 0 ∂Z
f ∂y
By the same reasoning, if the geostrophic wind is blowing in the
positive y direction, heights must increase in the positive x direction. So
the derivative will already be positive, and no minus sign is needed:
vg =
g 0 ∂Z
f ∂x
These are the equations defining the two horizontal components of
the geostrophic wind. There is no vertical component. They actually are
valid in all circumstances, not just when the geostrophic wind is parallel to
one of the axes. You can use them to compute the components of the
geostrophic wind separately and thereby determine the geostrophic wind
vector.
Finally, there is an equation in vector form that defines the
geostrophic wind. To determine its form, note that the equation for the x
component of the geostrophic wind depends on the y component of the
height gradient, and vice versa. Indeed, the rules describing the
geostrophic wind imply that the geostrophic wind is oriented at right
angles to the height gradient. So, our vector geostrophic wind equation
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must take care of the magnitude (by multiplying the height gradient by g/f)
and also take care of the direction (by rotating the direction of the height
gradient vector by 90o.
In our vector bag of tricks from Chapter 7, there is one vector
manipulation that can accomplish the latter task. If we take the cross
product of two vectors, the result is a vector that’s at right angles to both
the original vectors. Now then, if we take a horizontal vector and cross it
with a vertical vector, the answer will be a horizontal vector at right angles
to the original horizontal vector. And if the vertical vector is the unit
vertical vector, the magnitude of the cross product will equal the
magnitude of the original vector.
The only thing left is to get the signs right. The answer is:
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r
g
vg = o k × ∇h Z
f
Try the Gig ‘em rule: hold your arm up over your head, fist
pointing skyward, and hand in the shape of a Gig ‘em. If the palm of your
hand faces higher heights on a pressure surface, your thumb points in the
direction of the geostrophic wind. Sure enough, the higher heights would
be 90 degrees to the right of the direction the wind is blowing. So the
above equation works!
It is also a useful exercise for those who have already begun vector
calculus to verify that the two component equations can be derived from
the vector equation.
For future reference, here’s an equivalent statement of the
relationship between geostrophic wind and heights:
r
f (k × v g ) = − g o ∇ h Z
You can verify with your own hand that when you take the cross
product of the unit vertical vector and the geostrophic wind, you end up
with your thumb pointing toward lower heights 90 degrees to the left of
the geostrophic wind, as you should.
So, to summarize, we’ve got the relationship between geostrophic
wind and heights in words and in equation form both for wind components
and for the total vector geostrophic wind.
10.3 Geostrophic Divergence
One neat property of the geostrophic wind is its horizontal
divergence, or lack thereof. Recall that horizontal divergence is defined as
r ∂u ∂v
∇h • v =
+
∂x ∂y
To see what this has to do with the geostrophic wind, differentiate
the equation for the u component of the geostrophic wind with respect to
x:
∂u g
∂x
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=−
∂ ⎛ g 0 ∂Z ⎞
⎜
⎟
∂x ⎜⎝ f ∂y ⎟⎠
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and differentiate the equation for the v component of the geostrophic wind
with respect to y:
∂v g
∂y
=
∂ ⎛ g 0 ∂Z ⎞
⎜
⎟
∂y ⎜⎝ f ∂x ⎟⎠
Add the two together to get an equation for geostrophic
divergence:
∂u g
∂x
+
∂v g
∂y
=−
∂ ⎛ g 0 ∂Z ⎞ ∂ ⎛ g 0 ∂Z ⎞
⎜
⎟+ ⎜
⎟
∂x ⎜⎝ f ∂y ⎟⎠ ∂y ⎜⎝ f ∂x ⎟⎠
According to the rules of calculus, g, being a constant, can be
brought out of the differential operator. The Coriolis parameter f is only a
function of y, if we follow the convention that y is oriented toward the
north. Using the chain rule, we get
∂u g
∂x
+
∂v g
∂y
=−
g 0 ∂ ⎛ ∂Z ⎞ g 0 ∂ ⎛ ∂Z ⎞
∂Z ∂ ⎛ 1 ⎞
⎜⎜ ⎟⎟ +
⎜ ⎟
⎜ ⎟ + g0
f ∂x ⎝ ∂y ⎠ f ∂y ⎝ ∂x ⎠
∂x ∂y ⎜⎝ f ⎟⎠
Since x and y are independent variables, the order of differentiation
can be reversed. so the first two terms on the right-hand side cancel each
other, leaving
∂u g
∂x
+
∂v g
∂y
= g0
∂Z ∂ ⎛ 1 ⎞
⎜ ⎟
∂x ∂y ⎜⎝ f ⎟⎠
As noted before, f is not quite a constant, so the north-south
derivative of its reciprocal does not vanish. However, f changes so slowly
as a function of latitude that the right-hand side of the equation is typically
much smaller in magnitude than any typical values of observed divergence
of the actual wind. So even though the large-scale wind is mostly
geostrophic, only a very small part of the divergence is geostrophic. It is
so small, in fact, that the geostrophic wind can be regarded as being
nondivergent:
∂u g
∂x
+
∂v g
∂y
≅0
This statement includes the following consequences. First, the
primary wind patterns in the atmosphere are not, by themselves, associated
with upward or downward motion. It must be the departures from
geostrophic balance that make things interesting. Second, the actual wind
being mostly non-divergent, it is almost impossible to look at winds on a
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constant pressure map and eyeball convergence and divergence, unless
you have some way of isolating departures from geostrophic balance.
Now consider, once again, the relationship between the
geostrophic wind and height contours. We’ve already noted that the
geostrophic wind blows parallel to the height contours. What if you’re
following along with the height contours and the contours themselves
constrict so that they’re only half as far apart as they used to be? What
happens to all the air that used to be between the height contours?
According to the approximation that the geostrophic divergence is
zero, it must all still be there. (As in all other cases considered in this
book, the actual small divergence of the geostrophic wind can be
neglected.) So where does it go? It speeds up! After all, if the height
contours are half as far apart as before, the height gradient must be twice
as strong as before, and so the geostrophic wind must be twice as strong as
before. You’ve got air flowing twice as fast through an opening half the
size, so the same amount of air is passing between the height contours as
before.
It is a useful conceptual tool to think of the height contours as
channels for the geostrophic wind. If the wind obeys geostrophic balance,
air between two height contours at a particular level must stay between
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those two height contours and stay on that level as well. As the channel
gets narrower and the same amount of air is forced through a smaller
opening, it speeds up, and as the channel gets wider the air slows down.
It’s a lot like a river, which is broad where it is flowing slowly and is
narrow where it is flowing swiftly.
10.4 Geostrophic Advection
Advection also turns out to have some neat properties in
geostrophic-land. In Chapter 7, we expressed advection of a scalar
quantity (T there, an arbitrary scalar A here) using the standard dot product
formula for two vectors:
− v • ∇ h A = − v ∇ h A cos(θ ) This can be rewritten as
r
r
− v • ∇ h A = − ( v cos(θ ) ) ∇ h A
The interpretation is that the advection is proportional to the
projection of the wind onto the gradient, multiplied by the magnitude of
that gradient.
For geostrophic advection, it is useful to lump terms in a slightly
different way:
r
r
− v g • ∇ h A = − v g ( ∇ h A cos(θ ) )
Geostrophic advection, written this way, is described as the
projection of the gradient onto the direction of geostrophic wind,
multiplied by the magnitude of that wind. And the projection of the
gradient is just the derivative of the scalar A in the direction of the
geostrophic wind.
Now think for a moment. The magnitude of the geostrophic wind
is inversely proportional to the spacing between height contours: the
smaller the spacing, the stronger the height gradient and the stronger the
geostrophic wind. Similarly, the magnitude of the gradient of A is
inversely proportional to the distance between successive A contours in the
direction of the geostrophic wind: the closer together the A contours, the
stronger the A gradient.
Assuming we have contours close enough together, we could draw
a four-sided area (approximately a parallelogram), two of whose sides are
height contours and the other two of which are A contours. The area of
that parallelogram is inversely proportional to the spacing between the
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height contours, and is also inversely proportional to the distance between
the A contours. The area of our parallelogram has the same properties as
advection. Indeed, it turns out that (as long as the various contours are
close enough together) the magnitude of advection of any scalar quantity
by the geostrophic wind is inversely proportional to the area of the
parallelograms formed by the contours of height and of the scalar quantity.
It gets even better. Imagine a grid formed by height contours and
scalar contours such as isotherms. The closer together the contours, the
smaller the areas. Notice also that smaller areas correspond to more
intersections between height contours and isotherms. In fact, the relative
spatial density of intersection points is directly proportional to advection
too.
Advection, somewhat hard to eyeball with actual winds, is a piece
of cake with geostrophic winds. Just compare the height contours with the
contours of the field being advected. The smaller the areas bounded by
those contours (or, if you prefer, the more frequent the intersections of the
contours), the larger the advection. Identification of the area on the map
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with the strongest advection reduces to trying to find the smallest
parallelograms or the most frequent contour intersections.
10.5 Streamfunctions
It can be shown using vector calculus that any vector wind that is
nondivergent can be expressed in terms of a scalar quantity known as a
streamfunction. The definition of a streamfunction, normally written with
the Greek letter ψ, is any scalar for which
r
v = k × ∇ψ
Reexamining our vector definition of geostrophic wind, we find
that the geostrophic streamfunction, in the approximate case of constant
Coriolis parameter, is
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ψg =
g0
Z
f
All the properties of the geostrophic wind (nondivergent, channel
effect, eyeballing advection) apply to any wind field that can be defined
by a streamfunction. In turn, if the wind is nondivergent, it is not
necessary to plot the wind vectors to show the wind pattern; it suffices to
plot the contours of the streamfunction, since they carry all information
about the wind speed and direction.
Conversely, if there is any significant divergence or convergence
in the wind field, it is impossible to represent it as a streamfunction. Any
accurate attempt to draw streamlines through a divergent or convergent
wind field will inevitably lead to streamlines that appear to get too far
apart or too close together. If the wind is nondivergent, streamlines will
behave nicely, becoming closer together when the winds are strong and
farther apart when the winds are weak. Indeed, if the wind is geostrophic,
all the streamlines should be parallel to the height contours.
Let’s take that one step farther: you can actually draw streamlines
for the geostrophic wind by simply adding arrows to the height contours!
10.6 The Nature of Geostrophic Balance
Why should the wind above the ground be almost equal to the
geostrophic wind? For the wind to be blowing parallel to height or
pressure contours, there must be something holding the air parallel to
those contours. That something sure isn’t the pressure gradient force.
Pressure gradients in the atmosphere work about as you would
expect them to. The force caused by pressure gradients is a force directed
away from high pressure and toward low pressure. So the direction of the
force vector is directly opposite the pressure gradient itself, since a
gradient vector always points in the direction of higher values.
According to Newton’s second law, the acceleration of an object (or a
parcel of air, for that matter), is equal to the total force per unit mass. If
the pressure gradient force were the only horizontal force acting on the air,
we would see the air accelerate toward low pressure. Instead the air
happily blows along with low pressure always to its left. So there must be
some force opposing the pressure gradient force and preventing the air
from accelerating “down” the pressure gradient. That second force is the
Coriolis force.
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The Coriolis force has the following characteristics:
1. Its magnitude is proportional to the Coriolis parameter.
2. Its magnitude is also proportional to the horizontal wind speed.
3. In the Northern Hemisphere, its direction is 90 degrees to the
right of whichever way the wind is blowing.
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4. In the Southern Hemisphere, its direction is 90 degrees to the
left of whichever way the wind is blowing.
5. Its magnitude is so weak that it would take several hours for the
Coriolis force to cause a substantial change in the wind, if it were the only
force present.
The above statements are approximate, because they neglect the
vertical component of the Coriolis force as well as horizontal components
associated with vertical motion. But the effects of those details are so
small that we can safely ignore them here.
The Coriolis force is so weak that we can’t feel it. In fact, studying
the Coriolis force was mostly an intellectual exercise until World War I.
At that time, artillery technology had advanced to the point that shells
could be launched from several miles away with the expectation of high
accuracy. However, that accuracy wasn’t initially realized. Instead, the
shells always seemed to miss to the right by a little bit. The problem was
not in the aim but in the Coriolis force. The shells covered a long enough
distance that the Coriolis force was having a noticeable effect on their
trajectories.
As far as the shells were concerned, the Earth was moving beneath
them before they could reach their target. As far as the observers were
concerned, the shells were curving in midair as though a force was acting
on them. That force was the Coriolis force.
Now consider some air that’s moving parallel to a pressure or
height contour. The pressure gradient force is at right angles to its motion,
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toward lower pressure. The Coriolis force is at right angles to its motion,
toward higher pressure. If the speed of the air is just right, the strength of
the Coriolis force might just equal the strength of the pressure gradient
force. If air attains this happy state of Coriolis nirvana, the forces are
equal and opposite, and the acceleration is zero. The air can simply keep
moving along at the same speed with no force altering its trajectory by
causing an acceleration.
10.7 How to Attain Coriolis Nirvana
According to the above description, it sounds like the state of
geostrophic balance ought to be a pretty rare event. If air doesn’t start out
in balance, it’ll accelerate, and it doesn’t seem like it would accelerate in
the right direction. For example, if the air is going too slow for
geostrophic balance, the pressure gradient force must be stronger than the
Coriolis force, so the air would accelerate toward lower pressure. Besides,
pressure gradients are changing all the time, so even if air makes it to
geostrophic balance, what’s to keep it there when the pressure gradient
changes?
Yet, from looking at upper air maps, we know that the air actually
does stay in geostrophic balance, or pretty close to it. How is this
possible? The air can’t seek geostrophic balance on its own; all it does is
accelerate in response to the forces acting on it. So somehow, some way,
the forces must all conspire in such a way as to make geostrophic balance
stable and prevent air from getting hopelessly out of geostrophic balance.
Let’s list the four basic ways the air might not be in geostrophic
balance: (1) The component of wind across contours is zero, but the
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component along contours is too weak. (2) The component of wind across
contours is zero, but the component along contours is too strong. (3) The
component of wind along contours is just right, but there is a component
of wind across contours from high to low pressure. (4) The component of
wind along contours is just right, but there is a component of wind across
contours from low to high pressure.
We will now work out an example starting with situation (1). For
simplicity, we will imagine that the pressure gradient is uniform, that is,
all the isobars or contours are straight and evenly spaced. Okay, so our air
is moving too slowly. The Coriolis force must be too weak to balance the
pressure gradient force. The air, in response to the net force, must be
accelerating toward lower pressure. So, as time goes on, the air develops
an increasingly strong component of motion toward low pressure, in
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addition to its initial component of motion parallel to the height or
pressure contours.
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As the air drifts increasingly rapidly toward lower pressure, its
motion toward low pressure means that now there’s a component of the
Coriolis force to the right of this component of motion toward low
pressure, which, if you think about it, is the same direction as the initial
motion of the air. And this component of the Coriolis force is going to
cause its own acceleration, in this case parallel to the contours. So now
we have two things going on. In the cross-contour direction, the pressure
gradient force, not quite balanced by the Coriolis force, is accelerating the
air toward lower pressure. In the along-contour direction, the Coriolis
force, not balanced by anything, is accelerating the air in the along contour
direction.
Remember that we started out with the air going not fast enough in
the along-contour direction. If these accelerations go on long enough,
eventually the air will have sped up enough in the along-contour direction
that the cross-contour Coriolis force will be strong enough to balance the
pressure gradient force. Ta-daa, the pressure gradient and Coriolis forces
have come into balance.
But wait, there’s still the along-contour Coriolis force to worry
about. Just because the cross-contour forces balance doesn’t mean the air
is now moving parallel to the contours. And just because almost all
introductory textbooks think it does mean that, doesn’t make it so either.
When those two force components balance, that means there’s no longer
any acceleration toward lower pressure. But we’ve had an awful lot of
acceleration toward lower pressure already, so the air has built up a pretty
good velocity component in that direction. And when the forces in that
direction balance, the acceleration is zero, but there’s still nothing to
actually slow the air down in that direction. To do that, we’d need a net
force toward higher pressure, and we don’t have that.
So now we have air moving at the right speed along pressure
contours, but it’s also moving toward lower pressure. Notice that we’ve
gone from situation (1) to situation (3). What happens next? We still
have that pesky component of the Coriolis force directed along the
contours, so the air keeps speeding up in that direction. And as it
continues to speed up, the cross-contour component of the Coriolis force
must continue to strengthen too, since its magnitude is proportional to the
wind speed. Once that cross-contour balance is lost and the Coriolis force
is stronger than the pressure gradient force, the air will experience an
acceleration toward higher pressure and the cross-contour component of
wind will therefore begin to weaken.
Follow the situation a little while longer. The along-contour
component of wind is speeding up (because of the Coriolis force in that
direction), and because of what that does to the Coriolis force, the cross-
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contour component of wind is slowing down. Eventually, the crosscontour wind ends up back at zero.
So now are we in geostrophic balance? Alas, no. The wind speed
has been increasing the whole time, and it’s well beyond the geostrophic
wind speed at this point. The good news is, now that the cross-contour
component of the wind has finally decreased to zero, the along-contour
component of the Coriolis force has decreased to zero too, so the air is no
longer speeding up. Let’s see: no cross-contour flow, but wind speed too
fast ... sounds like situation (2)!
Perhaps you can imagine what happens next. With the Coriolis
force still stronger than the pressure gradient force, the air actually
develops a component of motion across contours toward higher pressure
or higher heights. As that kicks in, a component of the Coriolis force
develops that’s parallel to the height contours but pointing in the direction
opposite the geostrophic wind. Because of the acceleration caused by this
Coriolis force component, the air begins to slow down. Eventually, we
reach a point where the wind is the right speed again, but the air is moving
toward higher pressure: situation (4). And as the air continues to slow
down, eventually we end up back in situation (1) again!
What have we learned from all this? First, departures from
geostrophic balance don’t grow, they just sort of reorient themselves.
Second, look how regular the progression from (1) to (3) to (2) to (4) was.
The air spent just as much time going too fast as it did going too slow, and
just as much time going toward lower pressure as toward higher pressure.
If one figures out the average vector wind, it turns out to be exactly equal
to the geostrophic wind! So the air was almost in geostrophic balance all
along, just wobbling a bit. It’s like a marble at the bottom of a bowl. Even
if you push the marble, it will roll back. Its average position will still be at
the bottom of the bowl and its average vector motion is zero.
In the real world, our marble will eventually come to rest in the
center of the bowl again, because friction and wind resistance will steadily
act to settle it down. Similarly, although we won’t go into the details here,
energy dissipation and pressure perturbations in the atmosphere will act to
damp any departures from balance, causing air to eventually settle into
geostrophic balance as long as the pressure gradient isn’t changing too
rapidly.
10.8 The Nature of the Coriolis Force
So what is this Coriolis force that is always directed at right angles
to the wind? Saying it’s an apparent force caused by the Earth’s rotation,
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while true, doesn’t really help. To understand it, we must first understand
another apparent force we’ve discussed before: the centrifugal force.
Because of the centrifugal force, “down” is not toward the center
of the Earth. Instead, if you were digging a hole toward the center of the
Earth you’d have to tilt it a little bit toward the north. The Earth is not a
sphere, but instead it bulges at the equator. The ocean is a great “leveler”,
it settles into a “flat” shape so that neither gravity nor the centrifugal force
is causing it to move.
Okay, now back to the Coriolis force. Suppose the wind is
blowing toward the east. The Earth is also spinning toward the east, so the
air is actually rotating faster than the Earth underneath it. That means it
will experience a centrifugal force stronger than normal, and will tend to
accelerate to the right, toward the equator. Conversely, if the air is
blowing from east to west, it’s rotating slower than the Earth and will thus
experience a centrifugal force weaker than normal, accelerating it toward
the pole, which again is toward the right.
Meanwhile, if air is actually blowing directly toward the equator,
it’s moving toward a place where the Earth is rotating faster than before.
Without some force to speed it up, it will curve to the west to keep
spinning around the Earth’s axis at the same speed: thus, an apparent force
toward the right. If the air is moving toward the North Pole, it will be
moving to a place where the Earth is spinning slower than before, thus it
will curve toward the east (toward the right) unless something makes it
slow down.
On top of all this, the spinning Earth means that, relative to a fixed
frame of reference, the actual directions of north, east, south, and west are
constantly changing. Something initially heading south will start moving
west as the direction “west” replaces the direction “south” because of the
Earth’s rotation. No matter which way something is moving, the rotation
of the coordinate system will cause an apparent curve to the right in the
motion of the object.
Both of these things, the variations in the centrifugal force and the
rotation of the coordinate system, combine to produce what we call the
Coriolis force. All moving air would curve to the right unless there is
some other force pushing it to the left. In geostrophic balance, that other
force is the pressure gradient force.
10.9 Acceleration
In the previous discussions, we’ve talked about accelerations as
changes in the components of the wind. Specifically, they are changes in
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east-west and/or north-south motion of a particular air parcel. We would
write these as
Du Dv
,
Dt Dt
As noted before, Newton’s second law states that acceleration
equals the sum of the forces per unit mass. The law applies in each
direction, to each of the wind components. For horizontal motion in the
atmosphere, those forces are the Coriolis force, the pressure gradient
force, and turbulent mixing. We’ll just refer to turbulent mixing as
friction, since it’s most important in the context of turbulence generated by
friction near the Earth’s surface, and we’ll write it as Fx and Fy for the
east-west and north-south components of friction, respectively.
With that definition, the mathematical statement of Newton’s
second law for each wind component is
Du
∂Z
− fv = − g 0
+ Fx
Dt
∂x
and
Dv
∂Z
+ fu = − g 0
+ Fy
Dt
∂y
By convention, the Coriolis force is written on the left-hand side of
the equation. Notice that if the acceleration and friction are both zero, this
leaves the equations of geostrophic balance. So mathematically we see
that if there’s no acceleration and no friction, the wind has to be in
geostrophic balance.
Since the wind is a vector, we can also write Newton’s second law
in vector form:
r
r
r
Dvh
+ f (k × vh ) = − g 0 ∇ h Z + Fh
Dt
The interpretation of each of the terms is the same as in the
equations for the individual components. In particular, acceleration means
the same thing, changes in one or more of the components of velocity.
The subscript “h” means that the vectors are only horizontal (twodimensional) vectors.
It is also useful to think about acceleration in terms of the change
of a vector. Consider two vectors, sharing the same origin. Suppose one
vector represents the horizontal motion of an air parcel, and the other
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vector represents the horizontal wind one hour later. The change in the
wind vector is equal to a vector drawn from the end of the first vector to
the end of the second vector. In other words, it’s the second vector minus
the first vector. To get the acceleration, you would divide the difference
vector by the time period over which that change occurred, in this case
3600 seconds.
We can envision two prototypical situations. In the first, the two
wind speeds are the same but the directions are slightly different. In this
case, the difference vector will be roughly at right angles to the first
vector. This acceleration represents a change of direction. In the second,
the two wind directions are the same but the speeds differ. In this case,
the difference vector will be parallel to (or antiparallel to) the first vector.
This acceleration represents an increase (or decrease) of speed.
In everyday usage, for example when talking about cars, the term
“acceleration” refers to an increase of speed, and the term “deceleration”
refers to a decrease of speed. In meteorology, that’s not the case. Here,
acceleration refers to any and all changes in the wind, whether it’s an
increase or decrease of speed or just a change of direction. Any such
change in the wind vector, following an air parcel, requires that the forces
not add up to zero.
10.10 Gradient and Cyclostrophic Winds
Imagine a situation where the net force (and thus the acceleration)
is always at right angles to the wind, 90 degrees to the left of it. This
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corresponds to the acceleration that corresponds to a change of direction.
In this case, the direction of motion of the air parcel will be continuously
changing, curving to the left. With an acceleration at right angles to the
wind, there is no change in the wind speed, only in the wind direction, so
eventually, if that acceleration keeps up, the air parcel will have executed
a complete circle and ended up back where it started, ready to complete
another circle.
If we’re above the ground and can ignore friction, the only force
that can possibly do what has just been described is the horizontal pressure
gradient force. The Coriolis force is oriented in the wrong direction,
toward the right. Strictly speaking, the pressure gradient force must be
larger than the Coriolis force to produce a net acceleration to the left. And
since the acceleration is always to the left as the air parcel travels in a
circle, the isobars must also be circular. Specifically, we’re talking about
a closed low pressure center.
For atmospheric motion, we can extend the definition of balance to
include not just the situation with straight isobars, but also the situation
described above with curved isobars. In the latter case, although the air is
accelerating, it is conventional to call the winds balanced if the
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acceleration simply keeps the air moving parallel to the isobars. The
specific term for this type of balance is gradient wind balance.
Compare the wind speed in this case to the wind speed with the
same magnitude pressure gradient but straight isobars. In the latter case,
the wind speed must be just strong enough for the Coriolis force to be
equal and opposite to the horizontal pressure gradient force. In other
words, the wind is equal to the geostrophic wind. In the curved case, the
wind speed must be smaller because the Coriolis force has to be weaker
than the pressure gradient force. We say that the wind speed is
subgeostrophic. This applies to any balanced cyclonic curvature, as
around a low pressure center.
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We can also have constant acceleration to the right, as around a
high pressure center. In that case, the Coriolis force must be stronger than
the horizontal pressure gradient force, so the wind in that situation will be
stronger than geostrophic, or supergeostrophic.
Since it’s the Coriolis force that’s causing the acceleration and
rotation, the strength of an anticyclone is limited by the strength of the
Coriolis force. There’s no such limit for a cyclone, since there’s no real
limit to the strength of the pressure gradient force. Since the pressure
gradient force must be weaker than the Coriolis force, the center of a high
will almost always be broad and flat. In a low the pressure force is
dominant, so the center of a low will often be fairly tight, with strong
winds and strong pressure gradients very close to the center.
For really small vortices, such as tornadoes or severe
thunderstorms, it is also possible for the rightward acceleration to be due
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to the pressure gradient force. So even in the Northern Hemisphere, you
can have clockwise circulation around a low, but that low won’t be big
enough to show up on a standard weather map.
Once you get down to vortices that are small and intense, the
Coriolis force really starts being irrelevant because it’s so weak compared
to the pressure gradient force and to acceleration. When that happens, the
balance is called cyclostrophic.
Questions
1. Compute the height gradients corresponding to the following
geostrophic winds: (a) 25 kt from 260 degrees; (b) 45 kt from 310 degrees;
(c) 14 m/s from 190 degrees; (d) 33 m/s from 240 degrees; (e) 5 m/s from
60 degrees. For f, assume a latitude of 45 degrees.
2. Compute the geostrophic wind corresponding to the following
height gradients: (a) 10 dam/100 km toward 210 degrees; (b) 3 dam/100
km toward 180 degrees; (c) 5 dam/100 km toward 110 degrees; (d) 2
dam/100 km toward 330 degrees; (e) 14 dam/100 km toward 170 degrees.
For f, assume a latitude of 60 degrees.
3. Compute the geostrophic wind speed corresponding to the
following height gradient magnitudes: (a) 3 dam/100 km at 30N; (b) 5
dam/100 km at 70N; (c) 2 dam/100 km at 15N; (d) 1 dam/100 km at 40N.
4. Taking a map with analyzed height contours, identify the
location with apparently the largest geostrophic wind speed. Compute the
geostrophic wind speed and direction and the two horizontal geostrophic
wind components.
5. (a) Sketch a map of height contours and isotherms in which the
geostrophic temperature advection is 1 x 10-4 K/s. (b) Sketch a map of
height contours and isotherms in which the geostrophic temperature
advection is 0.5 x 10-4 K/s, but both the height gradient and temperature
gradient have the same magnitude as in 5a.
6. Specify the magnitude and direction of the Coriolis force and
pressure gradient force that would allow an air parcel to move toward the
north at 20 m/s without accelerating.
7. Specify the magnitude and direction of the Coriolis force and
pressure gradient force that would allow an air parcel moving toward the
north at 20 m/s to accelerate toward the east at 1 m/s per hour.
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