Chapter 10 Geostrophic Balance. (1) Wind and Height Gradients On a contour analysis of a pressure surface: Strong winds associated with contours close.

Chapter 10

Geostrophic Balance

(1) Wind and Height GradientsOn a contour analysis of a pressure surface:Strong winds

associated with contours close together; weak winds with contours farther apart.

Winds tend to blow (nearly) parallel to height contours.

Winds tend to be oriented such that higher heights are 90o to the right (in the northern Hemisphere) of the direction toward which the winds are blowing. (left in S. H.)

Winds that follow this are called Geostrophic Winds.

(2) The Geostrophic Equation

Expresses the magnitude of the wind speed as a function of the geopotential height gradient on a constant pressure surface.

Where, go = gravitational acceleration, f = Coriolis parameter = 2Ωsin(lat).

Ω = 7.292 x 10-5 s-1

f ranges from -1.4 x 10-4 s-1 at South Pole to +1.4 x 10-4 s-1 at the North Pole.f ≈ about 1 x 10-4 s-1 for mid-latitudes.

€

rv g =

go

f∇h Z

Consider: If the height gradient is 3 decameters (30 meters) per degree of latitude, how strong would the geostrophic wind be?

Notice: 1o latitude = 1.11 x 105 mThis is an average value at mid-latitudes.

€

rv g =

go

f∇h Z =

9.8ms−2

1× 10−4s−1×

30m

1olat×

1olat

1.11 × 105 m= 26.4 m

s ×1.943kt

1ms

= 51.29kt

If the x-axis is oriented parallel to the geostrophic wind, the height gradient must be oriented in the negative y-direction with lower heights in the positive y-direction. The magnitude of the height gradient is the magnitude of: To get the wind going in the positive x-direction, it must be multiplied by -1.

€

∂Z

∂y

€

ug = −go

f

∂Z

∂y

€

vg =go

f

∂Z

∂x

These equations can be used to compute the components of the geostrophic wind in all circumstances, not just when the x component is parallel to the wind direction.

€

Magnitude = ug2 + vg

2( )

€

Direction = arctan31

36

⎛

⎝ ⎜

⎞

⎠ ⎟+ 90o + 180o = 310.7o

The u-component of the geostrophic wind is dependent on the height gradient with respect to y.The v-component is dependent on the height gradient with respect to x.

€

ug = −go

f

∂Z

∂y

€

vg =go

f

∂Z

∂x

The geostrophic wind is parallel to the height gradient; i.e., 90o to the right of the height gradient (in the northern hemisphere).

Remember cross products.

If a horizontal vector is crossed with a vertical vector, the result will be a horizontal vector at right angles with the original horizontal vector.If the vertical vector is a unit vector, the magnitude of the resulting cross product vector will equal the magnitude of the original horizontal vector.

€

a × b = a b sin angle between( )

Thus we have:

The index finger is pointing upward, parallel to the unit vector k. The palm and three fingers are pointing “curled” toward increasing height gradient (the vector pointing toward increasing height gradient).The thumb is pointing out from the page parallel to the geostrophic wind vector.

€

rv g =

go

fk × ∇h Z

Right hand Rule: All fingers point in direction of k. Curl fingers in direction of increasing height gradient vector. Thumb points in direction of resultant of cross product.

An equivalent geostrophic equation is:

Meaning, crossing the vertical unit vector with the geostrophic winds results in a vector oriented toward lower heights.

€

f k ×r v g( ) = −go∇h Z

(3) Geostrophic DivergenceRemember, divergence is defined as:

And, the two components of the Geostrophic wind are:

Then if we take the derivative of ug with respect to x and the derivative of vg with respect to y, we can get the divergence by adding them together.

€

∇h •r v =

∂u

∂x+

∂v

∂y

€

ug = −go

f

∂Z

∂y

€

vg =go

f

∂Z

∂x

€

∂ug

∂x= −

∂

∂x

go

f

∂Z

∂y

⎛

⎝ ⎜

⎞

⎠ ⎟

€

∂vg

∂y=

go

f

∂Z

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟

Adding the two together gives:

Since go is a constant, it can be brought out and since f is a function only of y (latitude), then using the chain rule gives:

Since x and y are independent variables, the order of differentiation can be reversed and the first two terms cancel each other.

€

∂ug

∂x+

∂vg

∂y= −

∂

∂x

go

f

∂Z

∂y

⎛

⎝ ⎜

⎞

⎠ ⎟+

∂

∂y

go

f

∂Z

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟

€

∂ug

∂x+

∂vg

∂y= −

go

f

∂

∂x

∂Z

∂y

⎛

⎝ ⎜

⎞

⎠ ⎟+

go

f

∂

∂y

∂Z

∂x

⎛

⎝ ⎜

⎞

⎠ ⎟+ go

∂Z

∂x

∂

∂y

1

f

⎛

⎝ ⎜

⎞

⎠ ⎟

Thus, the divergence of the geostrophic wind equation reduces to:

Since f changes so slowly with latitude, the right-hand side is typically much smaller in magnitude than any typical values of observed divergence of the actual wind. Thus, divergence of the geostrophic wind is essentially zero. It is non-divergent.

€

∂ug

∂x+

∂vg

∂y= go

∂Z

∂x

∂

∂y

1

f

⎛

⎝ ⎜

⎞

⎠ ⎟

Therefore:(1) the primary wind patterns in the atmosphere are not, by themselves, associated with upward or downward motion. (2) departures from geostrophic balance are required for strong upward and downward motion. (3) it is very difficult to see convergence and divergence on constant pressure maps. (4) Because geostrophic winds are not divergent or convergent, constrictions on contours means the air must flow faster since the air, in geostrophic balance, is essentially confined between the contours.

If the wind obeys geostrophic balance, air between two height contours at a particular level must stay between those two height contours and stay on that level as well.

(4) Geostrophic AdvectionRemember, advection can be expressed as a dot product of two vectors with one of the vectors being wind (v)and the other the “del” of a scalar quantity (A) such as temperature (the gradient of A).The result is a scalar quantity.

€

−v • ∇h A = − v ∇h A cos Θ( )

This can be written as:

Meaning, advection is proportional to the projection of the wind (⎮v⎮) onto the gradient (⎮∇hA⎮) multiplied by the magnitude of that gradient(⎮∇hA⎮).

€

− r

v • ∇h A = −r v cos Θ( )( ) ∇h A

For geostrophic wind, it is written:

Meaning, the projection of the gradient (⎮∇hA⎮)onto the direction of the geostrophic wind (⎮vg⎮) multiplied by the magnitude of that wind.

€

− r

v • ∇h A = −r v g ∇h A cos Θ( )( )

So, the advection of “A” by the geostrophic wind, vg, is directly proportional to:

(1) the magnitude of the geostrophic wind - which is inversely proportional to the spacing between height contours.(2) the magnitude of the gradient of “A” - which is inversely proportional to the spacing between successive contours (isopleths) of “A.”

€

− r

v • ∇h A = −r v g ∇h A cos Θ( )( )

Then, the smaller the areas bounded by the height contours and isopleths of “A”, the greater the advection.

Thus, we could say that the advection of “A” by the geostrophic wind is inversely proportional to the area of the parallelogram formed by the height contours and the isopleths of “A.”

We could also say that the advection is greatest when the “spatial density of intersections” of contours and isopleths of “A” is greatest.

(5) StreamfunctionsAny vector that is non-divergent can be expressed in terms of a scalar quantity known as a stream function; Ψ.

€

rv = k × ∇Ψ

€

rv g =

go

fk × ∇h Z

€

Ψg =go

fZ

The geostrophic wind vector is written as:

Therefore, the geostrophic stream function must be:

If “f” is considered constant.

Then, if the wind is non-divergent as it is for geostrophic wind:

It is not necessary to plot the wind vectors to see the wind pattern. Contours of the stream function will represent the wind.

Where the stream function contours are close together, the wind speed is strong; and farther apart where the wind is weak.All stream function contours (streamlines) should parallel the height contours.Adding arrows to the height contours produces streamlines (stream function contours).

(6) Nature of Geostrophic BalanceConsider the horizontal forces acting on a volume of air above the near surface region (where friction becomes significant).

The forces are the Pressure Gradient Force,

The Coriolis Force.

€

Fx = −m

ρ

dP

dx

€

Fy = −m

ρ

dP

dy

The Pressure Gradient Force is directed away from high pressure and toward low pressure.

In other words, it is opposite to the pressure gradient (which is directed from low pressure toward high pressure).

A gradient vector always points in the direction of thegreatest rate of increase.

The Coriolis Force - keeps the air from moving directly from high pressure to low pressure.

Its magnitude is proportional to the Coriolis parameter:Its magnitude is proportional to the horizontal wind speed.Its direction in the northern/southern hemisphere is 90o to the right/left of whatever direction the horizontal wind is blowing. Its magnitude is so weak it would take several hours for the Coriolis force to cause a substantial change in the wind direction if it were the only force present.

€

2Ω sinΦ

€

2Ωv sinΦ

Consider a parcel of air moving parallel to the contours on an upper-air chart.

The pressure gradient force is oriented roughly 90o to the direction the parcel is moving and toward lower heights.The Coriolis force is oriented 90o to the direction the parcel is moving and toward higher heights.If the speed of the wind is just right, the pressure gradient force and the Coriolis force are of the same magnitude and pointed opposite to each other.

(7) Attaining Coriolis Nirvana

If the forces are not in balance, if something changes; e.g., the pressure gradient force, the parcel will accelerate in the direction of the stronger force. How does the parcel stay in geostrophic balance?

Consider the ways the wind might be out of geostrophic balance.

(1) Component of wind across contours is zero but the component along the contours is too weak.(2) Component of wind across contours is zero but the component along the contours is too strong.

(3) Component of wind along the contours is just right but there is a component of wind across contours from high to low pressure (heights).Component of wind along contours is just right, but (4) there is a component of wind across contours from low to high pressure (heights).

Consider case (1), the wind speed is too weak.

Departures from geostrophic balance cause a reorientation of the forces which causes the air motion to speed up or slow down and move back toward geostrophic balance. The time spent going to fast is the same as going to slow, so in the average, the air is in geostrophic balance.

(8) Nature of the Coriolis ForceConsider a ball of mass, m, attached to a string and whirled through a circle of radius r at a constant angular velocity .v is the tangential velocity of the ball. ΔΘ is the angle through which the ball moves (the change in the angle measured from the zero angle position).r is the radius from the axis of rotation to the ball.

Δv is the change in velocity (acceleration - change in direction) of the ball as it moves in a circular path.v + Δv is the new velocity of the ball (I.e., because it changed direction).

From the point of view of an observer in fixed space, the speed of the ball is constant, but its direction of travel is continuously changing, so its velocity is not constant; i.e., it has an acceleration.

Δv =vΔΘ

To compute the acceleration, we consider the change in velocity, v, which occurs for a time increment, t, during which the ball rotates through an angle .

is also the angle between the vectors v and v + v.

Remember: S = rΘ

If we divide by t and note that in the limit as , v is directed toward the axis of rotation, (it is a -v), then:

Since, tangential and angular velocity are:

Then,

Δt→ 0

dvdt

=−vdΘdt

v =ωr and dΘdt

=ω

dvdt

=−ω2r

Or, since

And,

The acceleration of the ball (directed toward the axis of rotation) is equal to the square its linear velocity divided by the distance from the axis of rotation.

€

v = ωr then, ω =v

r

€

a =dv

dt= −

v 2

r 2

⎛

⎝ ⎜

⎞

⎠ ⎟r = −

v 2

r

dvdt

=−ω2r

Therefore, when viewed from fixed coordinates, the motion is one of uniform acceleration directed toward the axis of rotation and equal to the square of the angular velocity times the distance from the axis of rotation. This acceleration is the Centripetal Acceleration.The resulting force on the ball to produce this acceleration toward the axis of rotation is the Centripetal Force.

Now, suppose the motion is observed in a coordinate system rotating with the ball; e.g., the earth. Now, the ball appears stationary (if it is rotating at the same rate as the earth, but the centripetal force is still acting on the ball, namely the pull of the string.In order to apply Newton’s second law to describe the motion relative to this rotating coordinate system, we must include an additional (apparent) force, the Centrifugal Force, which just balances the force of the string on the ball. It is acting outward, away from the axis of rotation.

Thus, the Centrifugal Force is equivalent to the inertial reaction of the ball on the string and is equal but opposite to the centripetal acceleration.In a fixed system, the rotating ball undergoes constant Centripetal acceleration in response to the force exerted by the string.In a moving system, the ball is stationary and the force exerted by the string is balanced by a Centrifugal force.The Centrifugal force can be considered the force necessary to balance all the other forces acting on the ball.

Since, for air at rest on Earth’s surface, M = r, Where M represents the linear wind velocity and direction vector, (Note: here we are letting M represent the wind whereas previously, we were letting v represent the velocity of the ball. In the following, we are letting v represent a component of the wind.)Then and

Remember, for centripetal acceleration:

€

Ω = r

M

r

€

dr

M

dt=

M 2

r=

r F centrifugal

m= Ω2r

€

dv

dt= −ω 2r = −

vball2

r

€

=vball

r

To get the components in the x- and y-direction, remember that

and to get the sign right, (Centrifugal force going in the proper direction), consider the following low pressure center. €

M 2 = u2 + v 2

By writing the equations as: and using the sign below, the direction of the force is proper.

Hemisphere For flow around aLow High

1 -1-1 1

NorthernSouthern

M = +VM = -U

M = -VM = +U

F = +y centrifugalU2r

F = -y centrifugalU2r

F = +x centrifugalV2rF = -x centrifugal

V2r L

Direction and signthe Centrigugal Forcesshould have.

Fx CN

m=+s⋅

V ⋅Vr

Fy CN

m=−s⋅

U ⋅Ur

Effective GravityA particle of unit mass at rest on the surface of the earth observed in a reference frame rotating with the earth, is subject to a centrifugal force, Ω2R, where Ω is the angular speed of rotation of the earth and R is the distance of the particle from the axis of rotation.

Thus, the weight of the particle of mass, m, at rest on the Earth’s surface will generally be less than the gravitational force, mg*, because the centrifugal force partly balances the gravitational force. g=g* +Ω2RThe spheroid shape of the Earth makes g (the Effective gravity) directed normal to the level surface.

Coriolis Force: east-west motionConsider an object moving eastward in the northern hemisphere with velocity U with respect to the Earth’s surface.For an observer NOT on the Earth, its tangential (linear) velocity is:

Since it is moving in a circular path, its Centripetal acceleration is toward the axis of rotation of magnitude:

v=R(objects angular velocity) = ΩR+U( )

−ΩR+U( )

R

2

or ΩR+U( )

R

2

Expanding gives:

For an observer on the Earth, the Earth’s surface seems stationary, so there is no tangential velocity due to the Earth’s motion, only due to the object’s motion. The centripetal acceleration on the object would be just:

ΩR+U( )R

2

=Ω2R2 +2ΩRU+U2

R=Ω2R+2ΩU +

U2

R

−U2

R or magnitude

U2

R

The discrepancy between the magnitude of the two is:

The Centrifugal Force (acceleration) we know is:

The Coriolis Force (acceleration) is:

also directed outward, like the Centrifugal Force.

ΩR+U( )R

2

=Ω2R+2ΩU +U2

R

−U2

RΩ2R+2ΩU

Ω2R

2ΩU

This outward directed Coriolis Force (acceleration) can be divided into components in the vertical and parallel to the Earth’s surface along the meridional directions (North - South).

The horizontal force causes a change in the direction of movement of the object (as perceived by someone on a moving coordinate system).

That change in direction for our object moving eastward with velocity U, is toward the south, i.e., to the right of the direction of motion in the northern hemisphere. (In other words, it is causing a change in the north-south velocity of the parcel).The vertical component is much smaller than the gravitational acceleration (force) and its only effect is to cause a very minor change in the apparent weight of an object. The term is called the Coriolis parameter. It is positive in NH and negative in SH.

fc =2Ωsinφ

Objects moving north or south.Consider an object, of unit mass, initially at rest on the surface of the Earth at latitude . It starts moving southward and moves .

As the object moves equatorward, it will conserve its angular momentum in the absence of torques in the east-west direction.

Its initial angular momentum, L, is:

, where is initially equal to zero. For other than unit mass it is:

Ω +ω( )R2

L =I Ω+ω( ) =mΩ +ω( )R2

Since the distance to the axis of rotation, R, increases to R + R for the object moving equatorward, a relative westward velocity, must develop, ( becomes smaller - or negative), if the object is to conserve its angular momentum.

We can write the following showing conservation of angular momentum.

Where, is the angular momentum of the unit mass object at rest on the surface of the earth at latitude .

is the change in west-east velocity which must occur because R is increasing to . Remember that tangential velocity, U, is related to angular velocity by: , so angular velocity for the object is:

€

ΩR2 = Ω +∂U

R + ∂R

⎛

⎝ ⎜

⎞

⎠ ⎟⋅ R + ∂R( )

2

ΩR2

€

∂U

U =rω

€

=U

R + ∂R€

R + ∂R

If we expand the right side we get:

If we neglect second-order, and higher differentials, we get:

and,

€

ΩR2 = ΩR2 + 2ΩR∂R + Ω∂R2 +R2∂U

R + ∂R+

2R∂R∂U

R + ∂R+

∂R2∂U

R + ∂R

€

0 = 2ΩR2∂R + R2∂U

€

∂U = −2Ω∂R

We can see that Arc length = andSo,Dividing by a unit time increment gives the horizontal acceleration in the west-east direction.

Note: “a” is the radius of the Earth.

€

a∂Φ

€

∂R ≅ a∂Φ sinΦ

€

∂U ≅ −2Ωa∂Φ sinΦ

€

∂U

∂t≅ −2Ωa

∂Φ

∂tsinΦ

But, is just the northward velocity component, V, so:

The magnitude of the Coriolis acceleration, or Coriolis Force per unit mass, is given by which acts to the right of the direction of motion of the object. (Thus, it is causing a change in the west-east motion of the parcel.)

€

a∂Φ

∂t

€

dU

dt= −2ΩV sinΦ

2ΩVsinΦ

(9) AccelerationThe Forces acting on an air parcel are, then:

Pressure Gradient ForceCoriolis ForceFriction (turbulent mixing)

These can be expressed as accelerations (Force/unit mass).The net effect (net force/accelerations) of these expressed as components in the x- and y-direction are:

€

Du

Dt,

Dv

Dt

Since these are the result of the various forces/accelerations on the parcel, (Newton’s 2nd Law) they can be written as:

If acceleration and friction are zero, these then show geostrophic balance - Balance between the Pressure Gradient acceleration and the Coriolis acceleration.

€

Du

Dt= −go

∂Z

∂x+ 2Ωv sinΦ + Fx

Dv

Dt= −go

∂Z

∂y− 2Ωu sinΦ + Fy

Acceleration = Pressure Gradient acceleration + Coriolis acceleration + Friction

€

Du

Dt− fv = −go

∂Z

∂x+ Fx

Dv

Dt+ fu = −go

∂Z

∂y+ Fy

Since wind is a vector, the equations (Newton’s 2nd law) can be written in vector notation as one.

Where, “h” means the horizontal components of the vectors only.

€

Dr v h

Dt+ f k ×

r v h( ) = −go∇h Z +

r F h

Consider the change in the wind in time.The acceleration of the wind (change in speed or direction, or both) can be represented as the difference between two wind vectors, at different times.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 (second vector minus the first vector).

The acceleration is obtained by dividing the difference vector (blue) by the time interval.

If only the direction is changing, not the speed, the direction of the acceleration is essentially at right angles to the original direction. If only the speed is changing, the direction of the acceleration is parallel to the original direction.

(10) Gradient and Cyclostrophic Winds

If the acceleration (net force/unit mass) is always at right angles to the wind direction, (e.g., to the left), the air will constantly curve to the left.Then the Pressure Gradient force is greater than the Coriolis force (since the Coriolis force - in the northern hemisphere- tries to make the air move to the right. This is how air flows about a low pressure (low height) region in the absence of friction.

This is Gradient Wind - the air flowing parallel to curved isobars (contours).

It is called “balanced” because it flows parallel to the contours, even though the forces are not in balance.

The winds about the low are weaker than in straight isobar / contour flow.In straight flow, the wind speed must be strong enough for the Coriolis force to balance the Pressure Gradient force.

In curved flow about a low, the Coriolis Force is weaker than the Pressure Gradient force, so the winds are weaker.The winds are Subgeostrophic.The contours / isobars about a low will be close together, signifying a strong Pressure Gradient force.

About a high pressure / height region, the Coriolis force must be greater than the Pressure Gradient force.The wind speed is greater than it would be in a straight isobar / contour flow. The winds are Supergeostrophic.The center of a high will be broad and flat, signifying a weak PGF.

Homework:Do: 1, 2, 3, 6 (at latitude 37o), 7 (at latitude 37oN)Remember: Direction is from.Be careful on angles. Be careful on units.