A Brief Introduction to AtmosphericDynamics

J. H. LaCasce, UiO

Contents

1 General dynamics 21.1 Derivatives . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21.2 Continuity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31.3 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41.4 Scaling the horizontal acceleration . . . . . . . . . . . . . . . .. . . . . 91.5 Geostrophic balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 121.6 Other momentum balances . . . . . . . . . . . . . . . . . . . . . . . . . 131.7 Hydrostatic balance . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 181.8 Pressure coordinates . . . . . . . . . . . . . . . . . . . . . . . . . . . . 211.9 Thermal wind . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 241.10 Thermodynamics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 291.11 Potential temperature . . . . . . . . . . . . . . . . . . . . . . . . . . .. 33

2 Waves 352.1 General waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 352.2 Sound waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 382.3 Gravity waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 402.4 The vorticity equation . . . . . . . . . . . . . . . . . . . . . . . . . . . .47

2.4.1 Example: Constant divergence . . . . . . . . . . . . . . . . . . . 482.4.2 Kelvin’s theorem . . . . . . . . . . . . . . . . . . . . . . . . . . 50

2.5 Barotropic potential vorticity . . . . . . . . . . . . . . . . . . . .. . . . 512.6 Barotropic Rossby waves . . . . . . . . . . . . . . . . . . . . . . . . . . 542.7 Quasi-geostrophic potential vorticity . . . . . . . . . . . . .. . . . . . . 572.8 Baroclinic Rossby waves . . . . . . . . . . . . . . . . . . . . . . . . . . 59

1

1 General dynamics

The motion in the atmosphere is governed by a set of equations, known as

theNavier-Stokesequations. These equations, solved numerically by com-

puters, are used to produce our weather forecasts. While there are details

about these equations which are uncertain (for example, howwe paramet-

erize processes smaller than the grid size of the models), the equations for

the most part are accepted as fact. In the following, we will consider how

these equations come about.

1.1 Derivatives

Weather forecasting specifies how fields (temperature, wind, atmospheric

moisture) change in time and space. Thus we must first specifyhow to

take derivatives.

Consider a scalar,ψ, which varies in both time and space. By the chain

rule, the total change in theψ is:

dψ =∂

∂tψ dt+

∂

∂xψ dx+

∂

∂yψ dy +

∂

∂zψ dz (1)

so

dψ

dt=

∂

∂tψ + u

∂

∂xψ + v

∂

∂yψ + w

∂

∂zψ =

∂

∂tψ + ~u · ∇ψ (2)

We refer to the left side as theLagrangianderivative. The time derivative

on the RHS is thelocal derivative, and the RHS is called theEulerianfor-

mulation. The Lagrangian formulation applies to moving measurements,

like balloons, while Eulerian applies to fixed measurements, like weather

stations.

2

1.2 Continuity equation

zδ

yδ

xδxδ

yδ δzρu ρuxδδ ρu yδ δz+ [ ]

x x + xδ

Figure 1: A infinitesimal element of air, with volumeδV .

Consider a box fixed in space, with air flowing through it. The flux of

density through the left side is:

(ρu) δy δz (3)

Through the right side the flux is:

[ρu+∂

∂x(ρu)δx] δy δz (4)

The net rate of change in mass is:

∂

∂tM =

∂

∂t(ρ δx δy δz) = (ρu) δy δz − [ρu+

∂

∂x(ρu)δx] δy δz

= −∂

∂x(ρu)δx δy δz (5)

The volume of the box is constant, so:

∂

∂tρ = −

∂

∂x(ρu) (6)

Taking into account all the other sides of the box we have:

∂ρ

∂t= −∇ · (ρ~u) (7)

3

We can rewrite the RHS and put it on the LHS:

∇ · (ρ~u) = ρ∇ · ~u+ ~u · ∇ρ (8)

Rearranging:

∂ρ

∂t+ ∇ · (ρ~u) =

∂ρ

∂t+ ~u · ∇ρ+ ρ(∇ · ~u) =

dρ

dt+ ρ(∇ · ~u) = 0 (9)

We can also obtain the same result using a Lagrangian (moving) box.

The box contains a fixed amount of air, so it conserves it mass.We can

write:

1

M

d

dtM =

1

ρδV

d

dt(ρδV ) =

1

ρ

dρ

dt+

1

δV

dδV

dt= 0 (10)

Expanding the volume term:

1

δV

dδV

dt=

1

δx

∂δx

dt+

1

δy

∂δy

dt+

1

δz

∂δz

dt=∂u

∂x+∂v

∂y+∂w

∂z(11)

So:

1

ρ

dρ

dt+ ∇ · ~u = 0 (12)

which is the same as (9). Thus the change in the relative density is propor-

tional to the velocitydivergence. If the box expands, the density decreases,

to preserve the box’s mass.

1.3 Equations of motion

The continuity equation pertains to mass. Now we consider wind velocit-

ies. We can derive expressions for these from Newton’s second law:

4

~a = ~F/m (13)

The forces acting on an air parcel (a vanishingly small box) are:

• pressure gradients

• gravity

• friction

For a parcel with densityρ, we can write:

d

dt~u = −

1

ρ∇p+ ~g + ~F (14)

This is themomentum equation. We use the Lagrangian derivative because

we are following the air parcel.

We have additional acceleration terms because the earth is rotating.

Consider first a stationary object on a rotating sphere. Eventhough the

object is not moving on the sphere, it appears to move when viewed from

space (a fixed frame) because of the rotation. Consider the vector, ~A, in

Fig. (2). In time,δt, the vector rotates through an angle:

δΘ = Ωδt (15)

We will assume the rotation rate is constant (Ω = const.), which is reas-

onable for the earth on weather time scales. The change inA is equal to

δA, the arc-length:

δ ~A = | ~A|sin(γ)δΘ = Ω| ~A|sin(γ)δt = (~Ω × ~A) δt (16)

So:

5

d ~A

dt= ~Ω × ~A (17)

δΑδΘ

γ

Ω

γ

Α

Figure 2: The effect of rotation on a vector,A, which is otherwise stationary. The vectorrotates through an angle,δΘ, in a timeδt.

If the vector is not stationary but moving in the rotating frame, one can

show that:

(d ~A

dt)F = (

d ~A

dt)R + ~Ω × ~A (18)

TheF here refers to the fixed frame andR to the rotating one. If~A = ~r,

the position vector, then:

(d~r

dt)F ≡ ~uF = ~uR + ~Ω × ~r (19)

So the acceleration is:

(d~uF

dt)F = (

d~uF

dt)R + ~Ω × ~uF = [

d

dt(uF + ~Ω × ~r)]R + ~Ω × ~uF =

(d~uR

dt)R + 2~Ω × ~uR + ~Ω × ~Ω × ~r (20)

6

We have two additional terms: theCoriolis andcentrifugalaccelerations.

Plugging these into the momentum equation, we obtain:

(d~uF

dt)F = (

d~uR

dt)R + 2~Ω × ~uR + ~Ω × ~Ω × ~r = −

1

ρ∇p+ ~g + ~F (21)

or:

(d~uR

dt)R + 2~Ω × ~uR = −

1

ρ∇p+ ~g′ + ~F (22)

where:

g′ = g − ~Ω × ~Ω × ~r (23)

We have absorbed the centrifugal force into the gravity termbecause both

are constant in time and both act in the radial direction. We can understand

this as follows. The spinning earth yields a centrifugal (negative centri-

petal) force outwards. If the spinning were too fast, we would all fly off

into space. But because we don’t fly off, gravity is strong enough to hold

us down. We define the net downward force (gravity plus centrifugal) as

g′.

There are three spatial directions and thus three momentum equations,

in the corresponding directions. In what follows, we assumethat we are in

localized region of the atmosphere, centered at a latitude,θ. Then we can

define local coordinates (x, y, z) such that:

δx = acos(θ)δφ, δy = aδθ, δz = δR

whereφ is the longitude,a is the earth’s radius andR is the radius. We

define:

7

u ≡dx

dt, v ≡

dy

dt, w ≡

dz

dt

We will write expressions fordu/dt, etc.

ΩcosθΩ sinθ

θ

Ω

Figure 3: A region of the atmosphere at latitudeθ. The earth’s rotation vector projectsonto the local latitudinal and radial coordinates.

Note first that the Coriolis term projects onto they andz directions:

2~Ω × ~u = 2Ω(w cosθ − v sinθ, u sinθ,−u cosθ) (24)

In addition, the use of a spherical coordinate system introduces some ad-

ditional “curvature terms” (see Batchelor,Fluid Dynamics). These involve

products of the velocities and can be derived using conservation of mo-

mentum arguments (see e.g. Holton,An Introduction to Dynamical Met-

eorology). Adding these, and the Coriolis terms, yields:

du

dt−uvtanθ

a+uw

a+ 2Ωw cosθ − 2Ωv sinθ = −

1

ρ

∂

∂xp+ Fx (25)

dv

dt+u2tanθ

a+vw

a+ 2Ωu sinθ = −

1

ρ

∂

∂yp+ Fy (26)

8

dw

dt−u2 + v2

a− 2Ωu cosθ = −

1

ρ

∂

∂zp− g + Fz (27)

whereFi is the frictional force acting in thei direction.

1.4 Scaling the horizontal acceleration

Not all the terms in the momentum equations are important. Take thex-

momentum equation:

∂

∂tu+u

∂

∂xu+v

∂

∂yu+w

∂

∂zu−

uvtanθ

a+uw

a+2Ωw cosθ−2Ωv sinθ = −

1

ρ

∂

∂xp

U

T

U 2

L

U 2

L

UW

D

U 2

a

UW

a2ΩW 2ΩU

p

ρL

1

2ΩT

U

2ΩL

U

2ΩL

W

2ΩD

U

2Ωa

W

2Ωa

W

U1

p

2ΩρUL

For the moment, we neglect will friction (which is generallyunimportant

outside the planetary boundary layer). In the second line wehavescaled

the equation by assuming typical values for the variables. In the third

line, we have divided through by the scaling for the scaling of the second

Coriolis acceleration,2ΩU .

Why have we done this? Note that the terms on the third line areall

dimensionlessparameters, i.e. they have no units. What we will do is

to evaluate each and see how it compares to one (the size of thesecond

Coriolis term). We have made an (educated) guess that this isone of the

largest terms. If any of the preceding terms is much less thanone, we can

neglect it. If a term is much greater than one, than our assumption that

the Coriolis term was the largest was wrong, and we have to divide again,

using another term.

9

To evaluate the terms, we use scales which are typical of weather dis-

turbances:

U ≈ 10m/sec, 2Ω =4π

86400 sec= 1.45× 10−4sec−1,

L ≈ 106m, D ≈ 104m, T = L/U ≈ 105sec a ≈ 6400km

HP/ρ ≈ 103m2/sec2, W ≈ 1cm/sec, (28)

The horizontal scale, 1000 km, is known as thesynoptic scalein the at-

mosphere. The time scale, proportional to the length scale divided by the

velocity scale, is theadvectivetime scale. This is what you’d expect, for

example, if a front were advected by the winds past an observer. With an

advective time scale, we have:

1

2ΩT=

U

2ΩL

So the first term is the same size as the second and third terms.This ratio

is theRossby number, a fundamental parameter in dynamical meteorology.

It has a value at synoptic scales of

U

2ΩL= 0.067

Thus the first three terms are smaller than the second Coriolis term.

However, the other terms are even smaller. The fourth term:

W

2ΩD= 0.0067

is about 10 times smaller than the Rossby number. The fifth andsixth

terms:

10

U

2Ωa= 0.01,

W

2Ωa= 10−5

are also smaller than the Rossby number. And the seventh term,

W

U= .001

is also small.

The pressure gradient term scales as:

p

2ΩρUL= 0.70

Thus this term is comparable in size to the second Coriolis term.

Saving only the terms which are the size of the Rossby number or larger,

thex-momentum equation is approximately:

du

dt− fv = −

1

ρ

∂

∂xp (29)

The same reasoning yields:

dv

dt+ fu = −

1

ρ

∂

∂yp (30)

where the Lagrangian derivative:

d

dt=

∂

∂t+ u

∂

∂x+ v

∂

∂y

now does not include the vertical advection term. Thus the advection in

these approximate equations isquasi-horizontal(two-dimensional). These

equations are fairly accurate at synoptic scales, and we will focus on them

from now on.

In addition, we have defined:

11

f ≡ 2Ωsinθ

This is the vertical component of the Coriolis parameter, the only one

which is important at these scales.

1.5 Geostrophic balance

The scaling suggest that the first order balance in the momentum equations

is between the Coriolis and pressure gradient forces:

−fv = −1

ρ

∂

∂xp (31)

+fu = −1

ρ

∂

∂yp (32)

This is called thegeostrophicbalance. It is one of two fundamental bal-

ances at synoptic scales. The balance implies that if we knowthe pressure,

we can deduce the velocities. So the winds can be determined from maps

of the pressure.

p/ ρL

Hfu

Figure 4: The geostrophic balance.

12

Consider the flow in Fig. (4). The pressure is high to the southand low

to the north. Left alone, this would force the air to move north. Because∂∂yp < 0, we have thatu > 0 (eastward), from (32). The Coriolis force is

acting to the right of the motion, exactly balancing the pressure gradient

force. And because the two forces balance, the motion is constant in time.

L

H

Figure 5: Geostrophic flow with non-constant pressure gradients.

If the pressure gradient changes in space, so will the geostrophic ve-

locity. In Fig. (5), the flow accelerates into a region with more closely-

packed pressure contours, then deaccelerates exiting the region.

Sincef = 2Ωsinθ, it is negativein the southern hemisphere. So the

flow in Fig. (4) would be westward, with the Coriolis force acting to the

left. In addition, the Coriolis force is identicallyzeroat the equator. So the

geostrophic balance cannot hold there.

1.6 Other momentum balances

The geostrophic balance occurs at synoptic scales, but other balances are

possible at smaller scales. To see this, consider a perfectly circular flow

(Fig. 6). The momentum equation in cylindrical coordinates(e.g. Batch-

elor,Fluid Mechanics) for the velocity in the radial direction is given by:

13

u

u

θ

r

Figure 6: Circular flow.

∂

∂tur+(ur

∂

∂r+uθ

r

∂

∂θ) ur−

u2θ

r−fuθ =

d

dtur−

u2θ

r−fuθ = −

1

ρ

∂

∂rp (33)

The termu2θ/r is called thecyclostrophicterm and is a curvature term like

those found with spherical coordinates. This is specifically related to the

centripetal acceleration. If the flow issteady(not changing in time), then

we have:

u2θ

r+ fuθ =

1

ρ

∂

∂rp (34)

U 2

R2ΩU

p

ρR

U

2ΩR1

p

2ρΩUR

We have scaled the equation as before. Note that the scale of the cyc-

lostrophic term is determined by the Rossby number. Let’s define that as

ǫ. If ǫ≪ 1, the cyclostrophic term is much smaller than the Coriolis term.

Then, we must have:

14

p

2ρΩUR≈ 1

and we have the geostrophic balance again:

fuθ =1

ρ

∂

∂rp (35)

Note that if the term on the RHS wasn’t order one, the pressuregradient

wouldn’t be large enough to balance the Coriolis force and there would be

no velocity.

Now consider ifǫ≫ 1. For example, a tornado at mid-latitudes has:

U ≈ 30m/s, f = 10−4sec−1, R ≈ 300m,

So ǫ = 1000. Then the cyclostrophic term dominates over the Coriolis

term. As we noted earlier, that means that we shouldn’t have divided the

scaling parameters by2ΩU , but ratherU 2/R. So we would have:

12ΩR

U

p

ρU 2

Now the second term, which is just one over the Rossby number,is very

small (0.001 for the tornado) and we require:

p

ρU 2≈ 1

In this case, we have thecyclostrophic wind balance:

u2θ

r=

1

ρ

∂

∂rp (36)

Notice that this is anon-rotatingbalance (becauseΩ doesn’t enter). The

pressure gradient is balanced by the centrifugal acceleration.

15

There is a third possibility, that there is no radial pressure gradient at

all. This is calledinertial flow. Then:

u2θ

r+ fuθ = 0 → uθ = −fr (37)

The velocity is negative, implying the rotation is clockwise (or “anticyc-

lonic”) in the Northern Hemisphere. The time for a parcel to complete a

full circle is:

2πr

uθ=

2π

f=

0.5 day

|sinθ|, (38)

which is known as the “inertial period”. “Inertial oscillations” are fre-

quently seen at the ocean surface, but are much rarer in the atmosphere.

The last possibility is thatǫ = 1, in which case all three terms in (34)

are important. This is thegradient wind balance. We can then solve foruθ

using the quadratic formula:

uθ = −1

2fr±

1

2(f 2r2+

4r

ρ

∂

∂rp)1/2 = −

1

2fr±

1

2(f 2r2+4 r f ug)

1/2 , (39)

after substituting in the definition of the geostrophic velocity. Note that

if the pressure gradient vanishes, we recover the inertial velocity. The

gradient wind estimate clearly differs from the geostrophic estimate.

This difference is typically about 10-20 % at mid-latitudes. To see this,

we rewrite (34) thus:

u2θ

r+ fuθ =

1

ρ

∂

∂rp = fug (40)

Then:

ug

uθ= 1 +

uθ

fr(41)

16

The last term scales as the Rossby number. So ifǫ = 0.1, the gradient

wind estimate differs from the geostrophic value by 10 %. At low latitudes,

whereǫ can be 1-10, the gradient wind estimate is more accurate.

An interesting point is that geostrophy flow issymmetric to sign changes.

If we change the sign of the pressure gradient, we change the sign, but not

the speed, of the wind. But this is not true with the gradient wind. If

ug < 0, which corresponds to anticyclonic (clockwise) rotation,we re-

quire:

|ug| <fr

4(42)

so that the root in (39) is positive. But there is no such limitwith cyclonic

(counter-clockwise) rotation. So cyclones can be much stronger under the

gradient wind approximation.

v−r2

v−r2

L H

p

fvp

fv

Figure 7: High and low pressure center flows under the gradient wind approximation.

The gradient wind balance alters the strength of the winds around high

and low pressure systems. Consider the examples in Fig. (7).For a low

pressure system (at left), the wind is somewhat weaker, because the cyc-

lostrophic term acts in the same direction as the Coriolis term, to balance

17

the pressure gradient force. With a high pressure system though, the Cori-

olis and cyclostrophic forces are opposed. So the Coriolis term must bal-

ance the other two and the wind is stronger.

If the cyclostrophic term is large enough, gradient wind vortices can ac-

tually have the pressure gradient and Coriolis forces in thesamedirection

(Fig. 8). This can only occur for cyclones. Such “anomalous lows” (low

pressure with clockwise flow) are usually found only near theequator.

v−r2

L

pfv

Figure 8: An anomalous low pressure system.

1.7 Hydrostatic balance

Now we will scale the vertical momentum equation. For this, we need an

estimate of the vertical variation in pressure:

V P/ρ ≈ 105m2/sec2

Neglecting the friction term,Fz, we have:

18

∂

∂tw+u

∂

∂xw+v

∂

∂yw+w

∂

∂zw−

u2 + v2

a−2Ωucosθ = −

1

ρ

∂

∂zp−g (43)

WU

L

UW

L

UW

L

W 2

D

U 2

a2ΩU

VP

ρDg

UW

gL

UW

gL

UW

gL

W 2

gD

U 2

ga

2ΩU

g

VP

gρD1

10−8 10−8 10−8 10−11 2 × 10−6 10−4 1 1

Notice that we used the advective time scale,L/U , for the time scaleT and

we have divided through byg, which we assume will be large. The vertical

pressure gradient and gravity terms are much larger than anyof the others.

However this is somewhat misleading because we obtain the same balance

if there isno motion at all! In particular, ifu = v = w = 0, the vertical

momentum equation is:

∂

∂zp = −ρg (44)

This is called the “hydrostatic balance” or literally the “non-moving fluid

balance”. We aren’t particularly interested in this component of the flow,

since weather comes from the dynamic (moving) component. Sowe sep-

arate the pressure and density into static and dynamic components:

p(x, y, z, t) = p0(z) + p′(x, y, z, t)

ρ(x, y, z, t) = ρ0(z) + ρ′(x, y, z, t) (45)

Generally the dynamic components are much smaller than the static com-

ponents, so that:

19

|p′| ≪ |p0| (46)

Then we can write:

−1

ρ

∂

∂zp− g = −

1

ρ0 + ρ′∂

∂z(p0 + p′)− g ≈ −

1

ρ0(1−

ρ′

ρ0)∂

∂z(p0 + p′)− g

≈ −1

ρ0

∂

∂zp′ + (

ρ′

ρ0)∂

∂zp0 = −

1

ρ0

∂

∂zp′ −

ρ′

ρ0g (47)

Note we neglect terms proportional to the product of the dynamical vari-

ables, likep′ρ′.

Now the question is: how do we scale the dynamical pressure terms?

Measurements suggest the vertical variation ofp′ is comparable to the ho-

rizontal variation, so:

1

ρ0

∂

∂zp′ ∝

HP

ρ0D≈ 10−1m/sec2 .

The perturbation density,ρ′, is roughly1/100 as large as the static density,

so:

ρ′

ρ0g ≈ 10−1m/sec2 .

To scale these, we again divide byg. So both terms are of order10−2.

So while they are smaller than the static terms, they are still two orders of

magnitude largerthan the next largest term in (43). Thus the approximate

vertical momentum equation is still the hydrostatic balance, but for the

perturbation pressure and density:

∂

∂zp′ = −ρ′g (48)

A weather model which uses this equation instead of the full vertical

momentum equation is called a "hydrostatic model"; a model which uses

20

the full vertical momentum is a "nonhydrostatic model". Notice that in

the hydrostatic model, we have no information about∂∂tw and so have lost

the ability to predict changes in the vertical velocity. Instead,w must be

estimated from the other variables. We discuss this more later on.

1.8 Pressure coordinates

p0p0+dp

x

z

dz

dx

Figure 9: Surfaces of constant pressure. Note that the pressure increases here with in-creasingx, but decreases with increasingz.

The hydrostatic balance can be exploited to simplify the momentum

and continuity equations. For this, we shift the vertical coordinate to one

defined by pressure rather than height. Consider a 2-D pressure surface.

By definition, the pressure doesn’t change along the surface, so that:

p =∂

∂xp x +

∂

∂zp z = 0 (49)

Substituting the hydrostatic relation, we get:

∂

∂xp x− ρg z = 0 (50)

so that:

21

∂p

∂x|z = −

∂p

∂z

z

x|p = ρg

z

x|p ≡ ρ

∂Φ

∂x|p (51)

where the subscripts indicatez andp coordinates, and whereΦ is thegeo-

potential:

Φ ≡∫ z

0g dz . (52)

Note that the geopotential is essentially a height, multiplied byg.

A big advantage of this coordinate change is that it removes the density

from momentum equation, because:

−1

ρ∇p = −∇Φ

So the horizontal momentum equations (29-30) can be written:

du

dt− fv = −

∂

∂xΦ (53)

and

dv

dt+ fu = −

∂

∂yΦ (54)

and the geostrophic balance is simply:

fv =∂

∂xΦ, fu = −

∂

∂yΦ (55)

So if we know the geopotential on a given pressure surface, wecan dia-

gnose the velocities—without knowing the density.

In addition, the coordinate change simplifies the continuity equation.

The easiest way to see this is to go back to our Lagrangian box.This has a

volume:

22

δV = δx δy δz = −δx δyδp

ρg(56)

after substituting from the hydrostatic balance. The mass of the box is:

ρ δV = ρ δx δy δz = −1

gδx δy δp

again using the hydrostatic balance. Note thatδp is negative here, so that

the mass is positive. Conservation of the box’s mass implies:

1

δM

d

dtδM =

g

δxδyδp

d

dt(δxδyδp

g) = 0 (57)

Rearranging:

1

δxδ(dx

dt) +

1

δyδ(dy

dt) +

1

δpδ(dp

dt) = 0 (58)

If we let δ → 0, we get:

∂u

∂x+∂v

∂y+∂ω

∂p= 0 (59)

So the flow isincompressiblein pressure coordinates. This equation is

much easier to use than the conventional continuity equation because it

doesn’t involve the density.

In pressure coordinates, the velocity normal to pressure surfaces is re-

ferred to asω. So for example, the three-dimensional Lagrangian derivat-

ive changes to:

d

dt=

∂

∂t+dx

dt

∂

∂x+dy

dt

∂

∂y+dp

dt

∂

∂p

=∂

∂t+ u

∂

∂x+ v

∂

∂y+ ω

∂

∂p(60)

23

1.9 Thermal wind

Geostrophy tells us what the velocities are if we know the geopotential on a

pressure surface. But say we want to estimate the velocitieson a different

surface. Then we need to know the velocityshear. We can obtain this

using the "thermal wind" relation.

Consider that we have two pressure surfaces, with pressuresp1 andp0.

Using the geostrophic relations (55), we can write:

vg(p1) − vg(p0) =1

f

∂

∂x(Φ1 − Φ0) ≡

g

f

∂

∂xZ10

ug(p1) − ug(p0) = −1

f

∂

∂y(Φ1 − Φ0) ≡ −

g

f

∂

∂yZ10 (61)

where:

Z10 =1

g(Φ1 − Φ0) (62)

is the thicknessof the layer betweenp0 andp1. Thus the velocity shear

is proportional to gradients in layer thickness, and the thermal wind (the

vertical shear vector) is aligned with surfaces of constantthickness.

There is another version of the thermal wind relation. From the defini-

tion of the geopotential and the hydrostatic relation, we have:

dΦ = g dz = −dp

ρ(63)

Neglecting atmospheric moisture, we can relate the densityand temperat-

ure using theideal gas law, which reads:

p = ρRT (64)

Substituting this into the expression fordΦ and rearranging, we infer:

24

∂Φ

∂p= −

1

ρ= −

RT

p(65)

We can use this, and the geostrophic relations, to determinehow the geo-

strophic velocities vary with pressure:

∂vg

∂p=

1

f

∂

∂p

∂Φ

∂x|p = −

R

pf

∂T

∂x|p (66)

or:

p∂vg

∂p=

∂vg

∂ ln(p)= −

R

f

∂T

∂x|p (67)

Similarly, we obtain:

∂ug

∂ ln(p)=R

f

∂T

∂y|p (68)

Thus the vertical shear is proportional to the gradients of temperature on

a pressure surface. Integrating thev relation between two pressure levels,

we find:

vg(p1) − vg(p0) = −R

f

∫ p1

p0

∂T

∂x|p d ln(p) (69)

and

ug(p1) − ug(p0) =R

f

∫ p1

p0

∂T

∂y|p d ln(p) (70)

So if we know the geostrophic velocities atp0, can calculate them atp1.

Think of p0 andp1 bounding a layer. We can define amean temperature

in the layer:

T ≡∫ p1

p0

T d(lnp)/∫ p1

p0

d(lnp) =∫ p1

p0

T d(lnp)/ln(p1

p0) (71)

25

6

7

8

9

p

p2

1

v

v

1

2

Figure 10:.

Then:

vg(p1) − vg(p0) =R

fln(

p0

p1) (∂T

∂x)p (72)

ug(p1) − ug(p0) = −R

fln(

p0

p1) (∂T

∂y)p (73)

If we equate this estimate with the earlier one, we get:

Z10 =R

gT ln(

p0

p1) (74)

So the layer thickness isproportional to its mean temperature.

Consider a idealized case in which the magnitude of the velocity changes

with height, but not the direction (Fig. 10). This is known asanequival-

ent barotropicflow.1 In the layer, the shear vector (which in this case is

parallel to the velocities themselves) is parallel to the temperature isolines,

which in turn are parallel to the thickness lines.1In a purely barotropic flow, the wind speed is also constant with height.

26

Using thermal wind, we can estimate the strength of the jet stream.

The zonally-averaged temperature decreases with latitudeon the earth (the

poles are colder than the equator). This means that∂T/∂y < 0, so thatu

should increase with height (Fig. 11).

20

15

10

p

Figure 11: The jet stream on a zonally-average earth.

At 30N, the zonally-averaged temperature gradient is roughly 0.75 Kdeg−1.

Assuming the average wind is zero at the earth’s surface, we can estimate

the mean zonal wind at the level of the jet stream (250 hPa), from (70):

ug(p1) − ug(p0) = ug(p1) =R

fln(

p1

p0)∂T

∂y(75)

or:

ug(250) =287

2Ωsin(30)ln(

250

1000) (−

0.75

1.11× 105m) = 36.8 m/sec (76)

This is roughly the speed of the jet at this height.

27

The equivalent barotropic assumption is often exploited insimplified

models of the atmosphere. However, the actual atmosphere isusually more

complicated, with both the wind speed and direction changing with height

(the atmosphere isbaroclinic). This implies that the temperature and geo-

potential isolines are not generally parallel, and therefore that the geo-

strophic wind can advect temperature (we speak of warm or cold winds).

δ

δ

v1

v2

vT

Warm

Cold

Φ1

ZT

Φ + Φ1

Z + ZT

Figure 12: Thermal wind between two layers (1 and 2). The geopotential height contoursfor the lower layer,Φ1, are the dashed lines and the thickness (temperature) contours arethe solid lines.

Consider the case shown in Fig. (12). The geopotential linesfor the

lower surface of a layer are indicated by dashed lines. The wind at this

level is parallel to these lines, with the larger values ofΦ1 to the right. The

thickness (temperature) contours are the solid lines, withthe temperature

increasing to the right. The thermal wind vector is parallelto these con-

28

tours, with larger temperatures on the right. We add the vectorsv1 andvT

to obtain the vectorv2, which is the wind at the upper surface. This is to

the northwest, carrying the warm air towards the colder.

Notice then that the wind vector turns clockwise with height. This is

calledveeringand is typical of warm advection. Cold advection produces

counter-clockwise turning, calledbacking.

Note too the similarity between the geostrophic and thermalwinds. The

geostrophic wind is parallel to the geopotential contours with larger values

to the right of the wind. The thermal wind is parallel to the thickness

(mean temperature) contours, with larger values to the right. But recall

that the thermal wind is not an actual wind, but adifferencein wind vectors

between the lower and upper levels.

1.10 Thermodynamics

The primary forcing agent for the atmosphere is the sun, which heats the

earth’s surface, changing the density of the overlying air.Thus thermody-

namics, which describes how gases respond to heat, plays a central role in

the dynamics. We require an additional equation, which links the temper-

ature, density and pressure and describes their dependenceon heating.

Consider a volume of air, as shown in Fig. (13). The First law of thermo-

dynamicsstates that the change in internal energy of the volume equals the

heat added minus the work done:

dq − dw = de (77)

Work is done by expanding the volume against external forces(which press

on the piston).

29

F

q

Figure 13: A volume being heated and doing work by forcing a piston.

dw = Fdx = pAdx = pdV (78)

If dV > 0, the volume is doing the work (otherwise the surroundings are

compressing the volume and doing work on it). We can assume the volume

has a unit mass, so that:

ρV = 1 (79)

so that

dV = d(1

ρ) ≡ dα (80)

whereα is thespecific volume. So:

dq = p dα + de (81)

If we add heat to the volume, its temperature rises. The extent of the

rise is determined by thespecific heat. If the volume is held constant, then:

30

cv ≡dq

dT|v (82)

Because the volume is constant, this is also the change in internal energy:

cv =de

dT|v (83)

Joule’s Lawstates thate only depends on temperature for an ideal gas,

even if the volume changes. This implies that:

cv =de

dT(84)

So we get the first law:

dq = cvdT + p dα (85)

We can also define a specific heat at constant pressure:

cp ≡dq

dT|p (86)

In this case, the volume expands in such a way to keepp constant. We

require more heat to raise the temperature, because some heat must now

be used for work. We can rewrite the first law thus:

dq = cvdT + d(pα) − αdp (87)

Recall that the ideal gas law is:

p = ρRT = α−1RT (88)

So

31

d(pα) = RdT (89)

Thus the first law can be written:

dq = (cv + R)dT − αdp (90)

At constant pressure,dp = 0, so:

dq

dT|p = cp = cv + R (91)

So the specific heat at constant pressure is greater than thatat constant

volume. For dry air:

cv = 717Jkg−1K−1, cp = 1004Jkg−1K−1 (92)

So we find that:

R = 287Jkg−1K−1 (93)

Thus an alternate version of the first law, for a dry gas is:

dq = cpdT − αdp (94)

This relation, or the alternate, constant volume version (85), can be used

to relate the temperature, density and pressure and determine how they

change with heating.

We often consideradiabaticprocesses, in which there is no heating at

all. We can derive a useful relation applicable whendq = 0, which we will

use later on. With zero heating:

cvdT + p dα = 0 (95)

32

If we eliminate thedT term using the ideal gas law, we obtain:

cvRd(pα) + p dα =

cvR

(p dα+ α dp) + p dα = 0 (96)

So:

cv α dp+ (cv +R) p dα = cvα dp+ cp p dα = 0 (97)

Thus:

dp

p+ γ

dα

α= d lnp+ γd lnα = 0 (98)

where

γ ≡cpcv

(99)

So:

pαγ = pρ−γ = const. (100)

for an adiabatic process. We will use this relation when discussing sound

waves (sec. 2.2).

1.11 Potential temperature

Heating changes both the temperature and pressure of air. But temperature

also changes with height (pressure) in the atmosphere. However, we can

redefine the temperature to remove the pressure contribution. This "poten-

tial temperature" then only changes due to heating.

Imagine a parcel with a pressurep and temperatureT . We can compress

or expand the parcel adiabatically (without heating) to a standard (surface)

33

pressure:ps = 100 kPa = 1000mb. The temperature of the parcel then

is defined to be its potential temperature. For adiabatic motion we have:

cpdT − αdp = 0 (101)

Using the ideal gas law:

α =RT

p(102)

we obtain:

cp d lnT −Rd lnp = 0 (103)

So we can define the potential temperature,θ:

cp lnT −R lnp = cp lnθ −R lnps (104)

Rearranging and taking the exponential of both sides:

θ = T (ps

p)R/cp (105)

Thusθ depends on T and p. Substituting in:

cp dlnT − Rdlnp = cp dlnθ + Rdlnp

ps− Rdlnp

= cp dlnθ + const. (106)

So:

dq = cpdlnθ + const. (107)

As promised, the heating changes only the potential temperature. Thus a

parcel on a surface with constantθ, called anisentropicsurface, can move

34

about the surface without being heated. In the language of thermodynam-

ics, such parcels conserve theirentropy.

2 Waves

Now we consider a class of dynamical phenomena in the atmosphere.

These are thewave(oscillatory) modes of motion. Waves represent solu-

tions to thelinearizedequations of motion. These waves are observable

and are fundamentally important in the changing atmosphere.

2.1 General waves

θ

θmg sin mg

l

Figure 14: The oscillating pendulum.

A simple example of a wave is the oscillation of a pendulum. Gravity

causes the pendulum mass to accelerate downward, but the rodprevents

it from falling. When the mass is not moving, the rod’s restoring force

exactly balances gravity. When disturbed from the verticalhowever, there

35

is a component of the rod’s force which acts toward the resting position,

causing the mass to accelerate that way.

We can write this mathematically as:

maθ = ml∂2

∂t2θ = −mgsinθ ≈ −mgθ (108)

if the deflection angle,θ, is small. So:

∂2

∂t2θ +

g

lθ = 0 (109)

The general solution is:

θ = ReAeiνt (110)

Whereν is thefrequency. NoteA can be a complex number. Substituting

the solution into the equation, we find:

−ν2 +g

l= 0 (111)

Thus:

ν ≡

√

g

l

Notice that the amplitude has dropped out. This is a typical occurrence in

linear wave problems. If:

A = Beiφ (112)

then:

θ = Bcos(νt+ φ) (113)

36

Thephase, φ, permits us to match initial conditions (i.e. the initial position

and velocity of the pendulum).

Wave equations generally take the form:

∂2

∂t2ψ − a2 ∂

2

∂x2ψ = 0 (114)

and have a solution:

ψ = ReAeik(x−ct) (115)

Substituting the solution into the equation yields:

k2c2 = k2a2 → c = ±a (116)

So we can write:

ψ = Bcos[k(x− at) + φ1] + Ccos[(k(x+ at) + φ2] (117)

The solution has awavelength

λ =2π

k(118)

and a frequency:

ν = −kc = ±ka (119)

The frequency is what an observer “hears”. In addition, the wave propag-

ates with aphase speed(the speed of the crests and troughs) of

c =ν

k= ±a (120)

37

Note that the first wave term propagates to the right, becausex must in-

crease ast increases. By the same reasoning, the second wave moves to

the left, becausex must decrease ast increases.

2.2 Sound waves

The first example we will consider is sound waves. For simplicity, assume

the motion is purely one-dimensional (in thex direction). It turns out that

the Coriolis term is not important for sound waves, because the spatial

scales are so small and the time scales short. We can also ignore friction.

Under these assumptions, the momentum equation is:

du

dt= −

1

ρ

∂

∂xp (121)

while the continuity equation is:

dρ

dt+ ρ

∂

∂xu = 0 (122)

or:

d ln ρ

dt+

∂

∂xu = 0 (123)

Since we are at small scales, we use thez-coordinate forms of the equa-

tions instead of those with the geopotential.

Thus we have two equations but three unknowns. For the third relation,

we use:

pρ−γ = const. (124)

which we derived in sec. (1.10). With this we can remove density from the

continuity equation:

38

1

γ

d ln p

dt+

∂

∂xu = 0 (125)

or:

dp

dt+ γp

∂

∂xu = 0 (126)

Now we have two equations with two unknowns. However, the equa-

tions arenonlinear, because they involve products of the unknowns, and

thus are difficult to solve. To proceed, welinearizeby assuming the waves

represent a weak perturbation about a background state. Specifically, we

separate the variables into constantmeanandperturbationcomponents:

u(x, y, z, t) = u+ u′(x, y, z, t), p = p+ p′, ρ = ρ+ ρ′

To linearize, we assume the perturbations are much smaller than the means:

|u′| ≪ u

We ignore products of perturbations (as we did before when discussing

the scaling of the perturbation pressure in sec. 1.7). Substituting into the

equations:

(∂

∂t+ u

∂

∂x) u′ +

1

ρ

∂

∂xρ′ = 0 (127)

(∂

∂t+ u

∂

∂x) p′ + γp

∂

∂xu′ = 0 (128)

Eliminatingu′ between the equations yields:

(∂

∂t+ u

∂

∂x)2 p′ −

γp

ρ

∂2

∂x2p′ = 0 (129)

We now substitute a wave-like solution:

39

p′ = ReAeik(x−ct) (130)

to get:

(−ikc+ iuk)2 +γp

ρk2 = 0 (131)

which yields a phase speed:

c = u±

√

√

√

√

γp

ρ(132)

What does this mean? If the background flow,u, is absent, then waves

can propagate to the left or right with a phase speed of√

γp/ρ. This speed

then is determined by the background pressure and density, as well asγ.

Increasing the pressure or decreasing the density yields faster sound waves.

The mean flow causes aDoppler shiftof this phase speed. The waves

moving in the same direction as the mean are faster while those in the

opposite direction are slower. This directly affects the wave frequency, as

ν = kc. So the sound from an approaching train’s whistle is higher when

the train is approaching us and lower when it is moving away.

If u =√

γp/ρ, the left-going wave isstationary. This occurs when a

plane reaches the so-called "sound barrier".

2.3 Gravity waves

Sound waves arelongitudinal wavesbecause the restoring force is in the

same direction as the phase velocity, as with waves moving along a coiled

spring. Gravity waves are an example of atransverse wave, where the

restoring force, gravity, is not parallel to the wave motion. The most fa-

miliar gravity waves are those at the surface of the ocean. These travel

40

perpendicular to gravity. Gravity waves also exist in the atmosphere, and

these can propagate in nearly any direction.

We begin by making a few assumptions. First, we consider two-dimensional

waves, in (x, z) plane. Second, we assume the spatial scales are small, so

that we can ignore the Coriolis force. We also use thez-coordinate ver-

sions of the equations. Third, we make theBoussinesq approximation. In

this, we replace the density by a constant value,ρ = ρ0, except in the hy-

drostatic relation. Under the Boussinesq approximation, the flow is incom-

pressible (see eqn. 9, with a constant density), and this greatly simplifies

the derivation. We also assume the motion is adiabatic, i.e.that there is no

heating.

Under these approximations, the equations are:

∂

∂tu+ u

∂

∂xu+ w

∂

∂zu = −

1

ρ

∂

∂xp (133)

∂

∂tw + u

∂

∂xw + w

∂

∂zw = −

1

ρ

∂

∂zp− g (134)

∂

∂xu+

∂

∂zw = 0 (135)

∂

∂tθ + u

∂

∂xθ + w

∂

∂zθ = 0 (136)

The last equation signifies that the potential temperature,θ, doesn’t change

following an air parcel.

As with the sound waves, we will reduce this set to one equation with

one unknown. We first linearize by assuming perturbations:

u = u+ u′, w = w′, ρ = ρ0 + ρ′, p = p(z) + p′, θ = θ(z) + θ′

41

Thus we assume a constant mean velocity which has no verticalcompon-

ent (which would otherwise require a force to counteract gravity). We also

have background temperature and pressure gradients which vary only in

the vertical direction. The background pressure must satisfy the hydro-

static relation:

∂

∂zp = −ρ0g (137)

From the definition of the potential temperature, we have:

θ = T (ps

p)R/cp =

p

ρR(ps

p)R/cp (138)

Taking the log of both sides:

ln θ = ln p1−R/cp − ln ρ+ const. = γ−1ln p− ln ρ+ const. (139)

The background state must satisfy this, so:

ln θ = γ−1ln p− ln ρ0 + const. (140)

From before we showed:

1

ρ

∂

∂zp+ g =

1

ρ0 + ρ′∂

∂z(p+ p′) + g ≈

1

ρ0

∂

∂zp′ −

ρ′g

ρ0(141)

We can eliminateρ′ by using the definition of the potential temperature:

ln(θ) = ln θ(1+θ′

θ) = γ−1ln [p(1+

p′

p)]− ln [ρ0(1+

ρ′

ρ0)]+const. (142)

Canceling the part due to the basic state, we get:

42

ln(1 +θ′

θ) = γ−1ln (1 +

p′

p) − ln (1 +

ρ′

ρ0) + const. (143)

Becauseln(1 + ǫ) ≈ ǫ, we get:

θ′

θ≈

p′

γp−ρ′

ρ0(144)

or

ρ′ =ρ0

γpp′ −

ρ0θ′

θ=p′

c2−ρ0θ

′

θ(145)

In the atmosphere, density fluctuations are dominated by temperature changes,

so we can neglect the first term on the RHS. So we have, simply:

ρ′

ρ0≈ −

θ′

θ(146)

Using these approximations, the equations become:

∂

∂tu′ + u

∂

∂xu′ = −

1

ρ0

∂

∂xp′ (147)

∂

∂tw′ + u

∂

∂xw′ = −

1

ρ0

∂

∂zp′ −

θ′g

θ(148)

∂

∂xu′ +

∂

∂zw′ = 0 (149)

∂

∂tθ′ + u

∂

∂xθ + w′ ∂

∂zθ = 0 (150)

Now we take thex-derivative of the vertical momentum equation and sub-

tract thez-derivative of the horizontal momentum equation:

43

(∂

∂t+ u

∂

∂x)(∂

∂xw′ −

∂

∂zu′) +

g

θ

∂

∂xθ′ = 0 (151)

This removesp′ from the equations. Then we use the other two equations

to eliminateu′ andθ′. This leaves an equation in terms ofw′:

(∂

∂t+ u

∂

∂x)2(

∂2

∂x2w′ +

∂2

∂z2w′) +N2 ∂

2

∂x2w′ = 0 (152)

where:

N2 = g∂

∂zln θ

is the square of theBrunt-Vaisala frequency. This is a fundamental meas-

ure of a stably-stratified environment (in which the densitydecreases going

upward).2 An air parcel displaced vertically under stable stratification will

oscillate up and down. The frequency of oscillation is the Brunt-Vaisala

frequency,N .

Now that we have a single linear equation with one unknown, wecan

substitute in a wave solution:

w′ = Rew′ei(kx+mz−νt) (153)

to get:

(ν − uk)2(k2 +m2) −N2k2 = 0 (154)

Note that the amplitude,w′, has dropped out again. So we have:

ν = uk ±Nk

(k2 +m2)1/2≡ uk ±

Nk

κ(155)

2An unstable stratification has denser fluid over lighter fluid. This tends tooverturn, resulting incon-vection.

44

This is thedispersion relation, which relates the frequency to the wavenum-

ber. The waves are calledinternal waves.

Notice that the frequency depends on the angle of propagation. If we

define:

cosχ ≡k

κ(156)

then:

ν = uk ±Ncosχ (157)

If the phase lines are parallel to the ground, thenk = 0 andχ = π/2. Then

the frequency is zero (there is no phase propagation). If on the other hand

the phase lines are perpendicular to the ground, thenm = 0 andχ = 0;

these have the maximum frequency. The more horizontal the phase lines

are, the lower the frequency. In addition, the frequency isalways less than

N without a mean flow.

The phase velocities are:

cx =ν

k= u±

N

κ, cz =

ν

m= u

k

m±Nk

mκ(158)

Notice that longer waves (waves with smallerκ) have faster phase speeds.

Thus if there is an initial disturbance which generates a range of different

size waves, the longest waves will separate out and propagate more rapidly

away. Thus gravity waves aredispersive.

We can also obtain thegroup velocities:

cgx =∂ν

∂k= u±

N(l2 − k2)

κ3/2, cgz =

∂ν

∂m= ∓

Nm

κ3/2(159)

45

One can show that the group velocity pertains to theenergy propagation.

For example, gravity waves generated by the wind blowing over mountains

carry energy upward, away from the mountains. Interestingly, the group

velocity of internal waves need not be parallel to the phase velocity. In

particular, one can show that the group velocity in the vertical direction

is in the opposite direction to that of the phase velocity. Sothe waves

carrying energy upward away from the mountains have their crests moving

downward, yielding a false impression of the movement of energy.

In addition, wind blowing over a mountain can excite stationary waves

downwind of the mountain. These are calledlee waves. They havecx = 0,

so:

u =N

κ(160)

Solving form, we get:

m2 =N2

u2− k2 (161)

If the RHS ispositive, thenm is real. This implies that the waves have a

sinusoidal structure inz. Then the waves can transport energy high into

the atmosphere. Such lee waves are important in the momentumbudget in

the upper atmosphere (above 75 km).

If the RHS of (161) isnegative, thenm is imaginary, so thatm = imi.

Then the wave’s vertical structure varies as:

exp(imz) ∝ exp(−miz) (162)

Thus the waves have an exponential structure inz rather than a sinusoidal

one. The sign of the exponential must be negative, as the waves must decay

46

going up rather than grow. In such cases, the waves, and thus the energy,

are trapped at the surface and cannot influence the upper atmosphere.

From (161), we see that weaker mean winds favor upward propagation.

So too do broader mountain ranges, which have a smaller valueof k. So

the Himalayas and Rockies have a greater effect on the atmosphere than

do smaller ranges.

Lee waves often appear as bands of clouds, because the upwardvertical

motion associated with the waves (the positivew′) cause condensation.

Upward-propagating waves can alsobreakfurther up, just as ocean gravity

waves break on a beach. This breaking generates turbulence (and bumpy

airplane rides).

2.4 The vorticity equation

Sound waves have very small spatial scales, while gravity waves have

scales of tens of meters. There are still larger waves though, which have a

scale comparable to the earth’s radius. Theseplanetaryor Rossbywaves

are important for weather and large scale adjustment. To seehow they

come about, we will need thevorticity equation.

The vorticity is the curl of the velocity. It has components in x, y andz.

But at synoptic scales, the vertical component is by far the most important.

This is:

ζ ≡∂

∂xv −

∂

∂yu (163)

To obtain an equation for the vorticity, we cross-differentiate the momentum

equations (we take∂∂x of the equation forv and subtract∂∂y of the equation

for u). Since we are at large scales, we use the pressure coordinate equa-

tions, (53) and (54) After some rearranging, we obtain:

47

(∂

∂t+ ~u · ∇) ζ + v

∂f

∂y= −(ζ + f) (

∂

∂xu+

∂

∂yv) (164)

Notice that the process of cross-differentiating has removed the geopoten-

tial. Becausef = f(y), we can rewrite the equation this way:

(∂

∂t+ u

∂

∂x+ v

∂

∂y)ζa = −ζa (

∂

∂xu+

∂

∂yv) (165)

The quantityζa = ζ+f is referred to as theabsolute vorticity. It is the sum

of therelative vorticity, ζ, and theplanetary vorticity, f . We can determine

the relative sizes of these two terms by scaling:

ζ ∝U

L, →

|ζ|

f∝

U

2ΩL= ǫ

So if the Rossby number,ǫ, is small, the relative vorticity is smaller than

the planetary vorticity. Then we can further approximate the equation as:

(∂

∂t+ u

∂

∂x+ v

∂

∂y)(ζ + f) ≈ −f (

∂

∂xu+

∂

∂yv) (166)

2.4.1 Example: Constant divergence

Equations (165) and (166) state that the absolute vorticityof an air parcel

changes in response to horizontal divergence. This processis different for

the formation of cyclones and anticyclones. Consider a region where the

divergence is constant:

∂

∂xu+

∂

∂yv = D = const. (167)

First letD > 0, which corresponds to a divergent flow (for example,

below a downdraft at the surface; Fig. 15). Then the vorticity equation

(165) is:

48

Figure 15: Divergent flow at the surface below a downdraft, producing anticyclonic cir-culation.

d

dtζa = −ζaD (168)

This implies:

ζa = ζa(0) e−Dt (169)

This implies that the absolute vorticity decays to zero, or that:

limt→∞ζ → −f (170)

This is true regardless of the initial vorticity of the air parcel. Cyclonic

and anticyclonic anomalies both become anticyclones with avorticity ap-

proaching−f .

Physically, the outward flow associated with the divergenceis diverted

to the right by the Coriolis force (Fig. 15). This produces anticyclonic

(clockwise) circulation.

49

Figure 16: Convergent flow at the surface feeding an updraft.

Now consider convergent flow, withD < 0 (Fig. 16). Now we have:

ζa = ζa(0) eDt (171)

So the vorticity increases without bound. But which sign obtains? If the

Rossby number is small, then:

ζa ≈ f > 0 (172)

So convergent flow produces intensecyclonesrather than anticyclones.

The inward flow in a convergence is steered to the right, generating cyc-

lonic flow (Fig. 16). This is a reason why intense storms are usually cyc-

lonic in the atmosphere.

2.4.2 Kelvin’s theorem

Kelvin’s theorem is a famous result due to the English scientist, Lord

Kelvin. It concerns the change in “circulation” of an area offluid. We

50

can derive it in a straightforward way from the vorticity equation.

Consider an air parcel, with sidesδx andδy. The time change in the

area is:

δA

δt= δy

δx

δt+ δx

δy

δt= δy δu+ δx δv (173)

Therelativechange in area is the divergence:

1

δA

δA

δt=δu

δx+δv

δy. (174)

So we can rewrite the divergence term (165) thus:

−(ζ + f)(∂

∂xu+

∂

∂yv) = −

ζ + f

A

dA

dt(175)

Putting this in the Lagrangian version of the vorticity equation, we get:

d

dt(ζ + f) +

ζ + f

A

dA

dt= 0 (176)

or

d

dt[(ζ + f)A] = 0 (177)

The area times the absolute vorticity is theabsolute circulation. This is

conserved for inviscid motion, meaning the quantity doesn’t change, fol-

lowing the air parcel. This is Kelvin’s theorem. If the area changes, the

vorticity must also change.

2.5 Barotropic potential vorticity

We can exploit the vorticity equation further under certainidealized condi-

tions. Consider for example that the flow is in a layer where the temperat-

51

ure,T , is constant. Because of this, there isno vertical shearin the velocity

field. To see this, take thep-derivative of the momentum equations:

d

dt

∂u

∂p+∂u

∂p· ∇u− f

∂v

∂p= −

∂

∂x

∂Φ

∂p= −

R

p

∂

∂xT (178)

d

dt

∂v

∂p+∂u

∂p· ∇v + f

∂u

∂p= −

∂

∂y

∂Φ

∂p= −

R

p

∂

∂yT (179)

after using the relation (65). WithT = const., the RHS of both equations

is zero. Thus:

d

dt

∂u

∂p= −

∂u

∂p· ∇u+ f

∂v

∂p(180)

d

dt

∂v

∂p= −

∂u

∂p· ∇v − f

∂u

∂p(181)

This implies that if:

∂u

∂p=∂v

∂p= 0 (182)

initially, there can be no shear later on. This is known as theTaylor-

Proudmanconstraint.

A simpler way to see this is from the geostrophic relations. For ex-

ample, if we take thep-derivatives of (55), we obtain:

f∂

∂pv =

∂

∂x

∂Φ

∂p= −

R

p

∂

∂xT = 0 (183)

The same result obtains for∂∂pu. Thus there is no vertical shear.

From the continuity equation (59), we have:

∂

∂xu +

∂

∂yv = −

∂

∂pω (184)

52

We can use this to replace the divergence in the vorticity equation:

(∂

∂t+ u

∂

∂x+ v

∂

∂y) (ζ + f) = (ζ + f)

∂ω

∂p(185)

Because there is no shear, the relative vorticity is also constant inz. So we

can easily integrate the equation. Say that the constant temperature layer

lies betweenp2 andp1. Then

h(∂

∂t+ u

∂

∂x+ v

∂

∂y) (ζ + f) = (ζ + f) (ω(p2) − ω(p1)) (186)

whereh = p2 − p1 is the layer thickness. Becauseω = dp/dt, we have:

hd

dt(ζ + f) = (ζ + f)

d

dt(p2 − p1) = (ζ + f)

dh

dt(187)

Then:

1

ζ + f

d

dt(ζ + f) −

1

h

dh

dt= 0 (188)

or:

d

dt[ln(ζ + f) − ln(h)] = 0 (189)

So we obtain the barotropic potential vorticity (PV) equation:

d

dt(ζ + f

h) = 0 (190)

The barotropic PV is conserved on a parcel in the layer. The PVtakes

into account changes in layerthickness. Consider the case shown in Fig.

(f:vortstretch). The vortex moving to the right is stretched as the layer

thickens, and this changes its vorticity. Ash increases, the numerator in

(190) must also increase, meaning the vorticity must becomegreater. So

an initially cyclonic vortex intensifies as it stretches.

53

δρ

Figure 17: A vortex moving between two density surfaces. Thevortex strengthens as thelayer thickness increases.

2.6 Barotropic Rossby waves

Now we can derive the relation for planetary waves. In the simplest version

of this, we assume the troposphere has constant density and aconstant

depth,h. Also, we will confine our attention to a limited range of latitudes,

centered about a latitudeθ0. We can simplify the Coriolis parameter by

making a Taylor expansion about the center latitude:

f(θ) = f(θ0) +df

dθ(θ0) (θ − θ0) +

1

2

d2f

dθ2(θ0) (θ − θ0)

2 + ... (191)

If range of latitudes is small, we can neglect the higher order terms, so that:

f ≈ f(θ0) +df

dθ(θ0) (θ − θ0) ≡ f0 + βy (192)

where

β =1

R

df

dθ(θ0) =

2Ω

Rcos(θ0)

and

y = R(θ − θ0)

Relation (192) is known as the "β-plane approximation". It simplifies life

by shifting from spherical to cartesian coordinates.

54

Under these assumptions, the barotropic PV equation is:

d

dt(ζ + f0 + βy) =

d

dtζ + βv = 0 (193)

We linearize the equation by assuming a constant mean zonal velocity:

u = u+ u′, v = v′

Then:

(∂

∂t+ u

∂

∂x)ζ ′ + βv′ = 0 (194)

With a constant density and layer depth, the continuity equation is:

∂

∂xu′ +

∂

∂yv′ = 0 (195)

This allows us to write the velocity in terms of astreamfunction:

u = −∂

∂yψ, v =

∂

∂xψ, ζ = (

∂2

∂x2+

∂2

∂y2)ψ (196)

So the vorticity equation is:

(∂

∂t+ u

∂

∂x)∇2ψ + β

∂

∂xψ = 0 (197)

Now we have a single linear equation with one unknown, as desired. We

substitute a wave-like solution:

ψ = Reψei(kx+ly−νt) (198)

and get:

(−iν + iuk)(−k2 − l2)ψ + iβkψ = 0 (199)

55

or:

ν = ku−βk

k2 + l2(200)

The corresponding zonal phase speed is:

c =ν

k= u−

β

k2 + l2(201)

This is the Rossby wave dispersion relation.

The dispersion relation has a number of interesting points.The phase

speed depends on the wavenumber, so the waves are dispersive, as with

gravity waves. The largest speeds occur with the largest waves, which

have small values ofk andl. But unlike the gravity or sound waves, which

could propagate in any direction in the absence of a mean flow,Rossby

waves propagate onlywestward. This is a signature of the waves. Satellite

observations of Rossby waves in the Pacific Ocean show that the waves,

originating off of California and Mexico, sweep westward toward Asia.

With a mean flow, the waves can be swept eastward, producing the

appearance of eastward propagation. But the waves are stillpropagating

westward relative to the mean flow. If:

u =β

k2 + l2(202)

the wave is stationary. So for a particular mean flow, we can define a

stationary wavenumber:

κ ≡ (k2 + l2)1/2 = (β/u)1/2 (203)

which determines the scale (wavelength) of the standing wave. Note that

stationary waves occur only if the mean flow is eastward.

56

Despite that the phase velocity is only westward, the group velocity can

be in any direction:

cgx = u+β(k2 − l2)

κ4, cgy =

2βkl

κ4, (204)

An interesting case is when long waves, with small values ofk, en-

counter a western boundary, like Asia in the Pacific ocean. Assumeu = 0

in the Pacific. The long waves transport energy westward, becausecgx < 0

if k < l. But the reflected waves must carry energy eastward, or else en-

ergy would accumulate at the wall. For this to happen, the reflected wave

must havek > l. So the reflected waves areshort. This effect is clearly

seen in numerical simulations.

2.7 Quasi-geostrophic potential vorticity

To consider the more realistic case with density stratification, we need to

take thermodynamics into account. With several approximations, we can

derive a single vorticity equation which does this. We return to our approx-

imate vorticity equation:

(∂

∂t+ u

∂

∂x+ v

∂

∂y) (ζ + f) = −f (

∂u

∂x+∂v

∂y) (205)

With the continuity equation, we have:

(∂

∂t+ u

∂

∂x+ v

∂

∂y) (ζ + f) − f

∂

∂pω = 0 (206)

We linearize the equation using a constant mean zonal velocity and we also

invoke theβ-plane approximation:

(∂

∂t+ u

∂

∂x) ζ ′ + βv′ − f0

∂

∂pω′ = 0 (207)

57

We neglect the termβy ∂∂pω

′, which is assumed to be small.

As we have seen, the Rossby number at synoptic scales is small, which

implies that the velocities are nearly geostrophic. We exploit this by mak-

ing thequasi-geostrophicapproximation:

u′ ≈ −1

f0

∂

∂yΦ, v′ ≈

1

f0

∂

∂xΦ, (208)

and thus that:

ζ ′ ≈1

f0∇2Φ (209)

So the vorticity equation is:

(∂

∂t+ u

∂

∂x)∇2Φ + β

∂Φ

∂x− f 2

0

∂

∂pω′ = 0 (210)

We have two unknowns,Φ andω. To close the system, we use the

thermodynamic equation. As before, we assume there is no heating, so

that the potential temperature is conserved. Thus

d

dtθ = 0 (211)

We linearize this, assuming a weak perturbation on a stable background

stratification:

θ = θ(p) + θ′(x, y, p, t) (212)

Then we obtain:

∂

∂tθ′ + u

∂

∂xθ′ + ω′ ∂

∂pθ = 0 (213)

58

Now, from (105), the potential temperature is proportionalto the tem-

perature on a surface of constant pressure. So we can write the equation in

terms of temperature:

∂

∂tT ′ + u

∂

∂xT ′ + ω′ ∂

∂pT = 0 (214)

From the hydrostatic relation (65), we have:

T = −p

R

∂Φ

∂p(215)

we can write:

(∂

∂t+ u

∂

∂x)∂Φ

∂p+ σω = 0 (216)

where:

σ ≡ −R

p

∂

∂pT (217)

is a measure of the stratification (like the Brunt-Vaisala frequency).

We can then combine the thermodynamic and vorticity equations to

eliminateω′. The result is:

(∂

∂t+ u

∂

∂x) [∇2Φ +

∂

∂z(f 2

0

σ

∂

∂zΦ)] + β

∂

∂xΦ = 0 (218)

This is thequasi-geostrophic potential vorticity equation. This takes into

account changes in vorticity due to both the Coriolis term and to changes

in temperature. This is a useful and powerful relation.

2.8 Baroclinic Rossby waves

Now we can examine baroclinic Rossby waves, or Rossby waves in the

presence of stratification. We substitute a wave solution:

59

Φ = Φei(kx+ly+mz−νt) (219)

to get:

ν = uk −βk

k2 + l2 + f 20m

2/σ(220)

with a zonal phase speed of:

cx = u−β

k2 + l2 + f 20m

2/σ(221)

We see that the relations are nearly the same as those we obtained with

barotropic waves. The difference is that there is an additional term in the

denominator of the term multiplied byβ. Like the barotropic waves, the

baroclinic waves propagate to the west relative to the mean flow and larger

waves propagate faster than smaller waves. But the additional term in

the denominator means the waves propagateslowerthan their barotropic

counterparts.

Of central interest for the large scale atmospheric circulation are the

stationary waves, generated by flow over mountains. Assuming cx = 0,

we can solve form:

m2 =σ

f0[β

u− (k2 + l2)] (222)

As with the gravity waves, we have two possibilities. Ifm2 < 0, thenm

is imaginary and the waves decay exponentially in the vertical. These are

trapped at the lower boundary. But ifm2 > 0, or:

(k2 + l2) <β

u(223)

60

the waves can propagate upward, even up into the stratosphere. The re-

lation implies that long waves (smallk, l) are more likely to propagate

upward than short waves. Also, because the LHS is positive, the relation

cannot be satisfied unless the mean flow is eastward (u > 0).

In fact, Rossby waves are found in the stratosphere, and can dramat-

ically alter the circulation (even changing the equator-to-pole temperat-

ure difference—a so-called “stratospheric warming” event). They are ob-

served there during the winter months. But they disappear during the sum-

mer, because the mean wind reverses direction (u < 0). Thus this (relat-

ively) simple theory (due to Charney and Drazin, 1961) can account for

this disappearance.

61