4 Instability of the zonal �ow; genesis of mid-latitude storms

4.1 Wave conservation properties and the stability ofzonal �ows (QG)

4.1.1 The QG potential enstrophy budget

The conservation law for QGPV is

Dgq = Q

whereDg is the time derivative following the geostrophic �ow (a nonconserva-tive termQ has been added to allow for sources and sinks of PV). We assumea purely zonal basic state, with mean �ow �u(y; z), �v = 01, and mean QGPV�q(y; z); and look at small amplitude perturbations to this state. Followingour usual procedures, we linearize the QGPV equation to get

@q0

@t+ �u

@q0

@x+ v0

@�q

@y= Q0 :

Multiplying by q0 and averaging, we get

@

@t

�1

2q02�+ v0q0

@�q

@y= q0Q0 : (1)

The quantity q02 is known as the eddy potential enstrophy2, and (1) describesthe budget of this quantity: eddy potential enstrophy changes because ofinteraction with the mean state, and nonconservative sources and/or sinks..So for conservative (Q0 = 0) eddies of steady (@=@t = 0) amplitude, theQGPV �ux v0q0 must vanish.

4.1.2 PV �uxes and the Eliassen-Palm theorem

Now consider the eddy �ux itself, �v0q0. The the QGPV perturbation is

q0 =@2 0

@x2+@2 0

@y2+1

�

@

@z

��f 20N2

@ 0

@z

�: (2)

Now,v0 = @ 0=@x and:

1The geostrophic mean northward �ow must be zero, since v = @ =@x.2Integrated vorticity-squared is known as enstrophy.

1

1.@ 0

@x

@2 0

@x2=1

2

@

@x

"�@ 0

@x

�2#= 0 ;

2.

@ 0

@x

@2 0

@y2=

@

@y

�@ 0

@x

@ 0

@y

�� @ 0

@y

@2 0

@x@y

=@

@y

�@ 0

@x

@ 0

@y

�� 12

@

@x

"�@ 0

@y

�2#

=@

@y

@ 0

@x

@ 0

@y

!;

and

1

�

@ 0

@x

@

@z

��f 20N2

@ 0

@z

�=1

�

@

@z

��f 20N2

@ 0

@x

@ 0

@z

�� f 20N2

@ 0

@z

@2 0

@x@z

=1

�

@

@z

"�f 20N2

@ 0

@x

@ 0

@z

#� f 202N2

@

@x

"�@ 0

@z

�2#

=1

�

@

@z

"�f 20N2

@ 0

@x

@ 0

@z

#:

Therefore, from (2),�v0q0 = r � F . (3)

where

F =

�FyFz

�=

�@

0

@x@ 0

@y

�f20N2

@ 0

@x@ 0

@z

!=

��u0v0�f0

v0�0

d�=dz

!: (4)

F is known as the ELIASSEN-PALM �ux. Note that the northward compo-nent of F is (minus) the northward �ux of zonal momentum by the waves,while the vertical component is proportional to the northward �ux of poten-tial temperature, v0�0:

2

4.1.3 Wave activity conservation

Now, (1) gave us@

@t

�1

2q02�+ v0q0

@�q

@y= v0Q0 :

from which, if we de�ne

A =1

2�q02=

�@�q

@y

�and D = �v0Q0=

�@�q

@y

�;

then, provided @�@�q@y

�=@t = 0,

@A

@t+r � F = D . (5)

Eq. (5) is the Eliassen-Palm relation. It is a conservation law for zonally-averaged wave activity whose density is A. Note that D ! 0 for conservative�ow.The signi�cance of this relation is that it gives us a measure of the �ux

of wave activity through wave propagation. For example, if the waves areconservative (D = 0) then A must increase with time wherever F is conver-gent and decrease wherever it is divergent. Thus F is a meaningful measureof the propagation of wave activity from one place to another. This becomesmost obvious for plane, vertically propagating, Rossby waves on a uniform�ow U and @�q=@y = �, for which the geostrophic streamfunction is

0 = Re0ez=2Hei(kx+ly+mz�!t) :

when it can be shown that the �uxF =(Fy; Fz) and densityA of wave activityare related by

F = cgA :

In such plane or almost-plane wave situtations where we can de�ne groupvelocity, then, F is parallel to group velocity. However, note that F remainsvalid as a measure of the �ux of wave activity even in general situations inwhich we cannot even de�ne group velocity.So the EP�ux gives us a meaningful and straightforward way of displaying

and quantifying wave activity propagation. E. g., from a localized source,the EP �uxes will necessarily look as if the wave is indeed emanating fromthe source, as illustrated in Fig. 1.

3

y

z

Figure 1: Schematic of F for waves propagating away from a localized source(shaded)

4.1.4 The Eliassen-Palm theorem

For waves which are steady (@A=@t = 0), of small amplitude and conserv-ative (D = 0) , the �ux F is nondivergent. This is, through (3), the samething as saying that the northward �ux of quasigeostrophic potential vortic-ity vanishes under these conditions [a result we could of course have obtainedfrom (1) without involving F]:

4.1.5 Observed EP �uxes in the troposphere

(Edmon et al., J. Atmos. Sci., 37, 2600-2616, 1980.)Examples of F and r � F for stationary eddies from two (old) winter cli-

matologies are shown in Fig. 4.1.5. As we might expect, we see F indicatingupward propagation away from near-surface sources (such as topography)up to the upper troposphere, and then spreading out laterally, mostly equa-torward, in the upper troposphere. The transient component, comprising�uctuations in the quasi-stationary eddy �eld, but predominantly the mo-bile, synoptic scale eddies of middle latitudes, shows quite similar behavior,as shown in Fig. 4.1.5.Note from (4) that upward wave activity propagation implies a poleward

eddy heat �ux (f v0�0 > 0), so that upward propagating QG waves mustbe associated with a poleward heat �ux irrespective of the mean latitudi-nal temperature gradient. Similarly, equatorward wave activity propagationcorresponds to poleward eddy momentum �ux (�u0v0 > 0).

4

5

4.2 Stability of zonal �ows to QG perturbations

4.2.1 The Charney-Stern theorem

This result is due to Charney & Stern, J. Atmos. Sci., 19, 159-172, (1962):they did not frame their discussion in terms of EP wave activity (developmentof which came later) but the mathematics of the derivation is essentially thesame.Consider now the globally-integrated budget of wave activity. From (5),

@

@t

ZZRA dy dz +

ICF � n dl =

ZZRD dy dz :

where the domain R is the whole region of interest (which may be the wholeglobe), and C the boundary of the domain. For conservative �ow (D = 0)integrated wave activity can change only if there is a �ux of wave activitythrough the boundaries. For a closed system, one might imagine that thewave activity �ux through the boundaries is necessarily zero. This may notbe true: side boundaries are usually straightforward, but bottom (and top)boundaries can be surprising. At a rigid, �at, lateral boundary, where v = 0,it follows directly that Fy = ��u0v0 = 0, and hence that F � n = 0 there. Attop and bottom boundaries, however, there is in general no reason to assumethat v0�0 = 0, and so there may be a vertical �ux of wave activity (apparently)through the boundaries. However, if these boundaries are isentropic, �0 = 0and then Fz = �f0v0�

0= (d�=dz) = 0, whence F � n = 0 there also. Underthese circumstances, then, for conservative (inviscid, adiabatic) �ow

@

@t

ZZRA dy dz = 0 :

So net (globally integrated) wave activity cannot then grow or decay� it

can merely be redistributed. Now, since A = 12q02=

�@�q@y

�, if the PV gradi-

ent is single-signed we obviously have a meaningful constraint on how adisturbance can grow. In fact, if we look for normal mode growth such thatq02 = B(t)C(y; z) (where we may de�ne both B and C to be positive de�nite),then

dB

dt

ZZR

1

2

C(y; z)

@�q=@ydy dz = 0 :

Then, if the PV gradient is single-signed, the integral cannot vanish, sodB=dt = 0: the mean state is stable to normal mode disturbances. Thefull statement is then that:

6

A zonal �ow is stable to inviscid, adiabatic, quasigeostrophic normalmode perturbations if there is no change of sign of PV gradient withinthe �uid and if the system is bounded above and below by isentropicboundaries.

This statement is known as the Charney-Stern theorem. It follows from waveactivity conservation simply because a normal mode must simultaneouslyincrease in magnitude everywhere. But, if there is no wave activity �uxthrough the boundaries, wave activity cannot increase everywhere (since it isglobally conserved): the only way to get such growth is for the wave activitydensity to be positive in some places and negative in others, so both positiveand negative regions can increase in magnitude while conserving the globalintegral. Thus, instability can occur only if at least one of the assumptionswe have made here is violated (i.e., change of sign in PV gradient, or non-isentropic upper or lower boundaries)Note that this constraint does not apply to non-normal-mode growth,

wherein, for example wave amplitudes may increase (and measures such aseddy energy may increase everywhere) in some places but decrease in others.

4.2.2 Violating the stability constraint: Barotropic and baroclinicinstability

(Holton, Ch. 8; a detailed exposition of the theory is given in Vallis, Ch. 6.)As we have seen zonal �ows are stable to, inviscid, adiabatic, normal

mode, QG disturbances if the PV gradient is single signed and the upperand lower boundaries are isentropic. Since the QGPV is

q = f +@2

@y2+1

�

@

@z

��f 20N2

@

@z

�and for a zonal �ow U = �@ =@y, the PV gradient is

@q

@y= � � @2U

@y2� 1�

@

@z

��f 20N2

@U

@z

�: (6)

For a barotropic �ow (no T gradients, whence @U=@z = 0), non-isentropicboundaries are not an issue, and the PV gradient is�

@q

@y

�barotropic

= � � @2U

@y2(7)

7

In the case � = 0, such a zonal �ow is necessarily stable unless the curvatureterm has both signs within the �uid. If the curvature is everywhere �nite,this yields the in�ection point theorem: the �ow is stable unless U(y) hasan in�ection point where @2U=@y2 changes sign. Introduction of � > 0is a stabilizing in�uence: barotropic instability is then possible only if thecurvature term is of the correct sign and of su¢ cient magnitude to overcome� in (7), somewhere in the �ow.In the absence of barotropic curvature,

@q

@y= � � 1

�

@

@z

��f 20N2

@U

@z

�(8)

and baroclinic instability is possible if the upper and lower boundaries arenot isentropic, or if the vertical �curvature�of the �ow (modi�ed by �f 20 =N

2)is su¢ cient to overcome � in (8). In the extratropical troposphere, thecurvature is not usually su¢ cient (nor usually the correct sign) to do so inthe free atmosphere, but the stability condition is violated by the presenceof temperature gradients on the lower boundary.In the presence of both barotropic and baroclinic curvature, both curva-

ture terms in (6) may contribute to changing the sign of @q=@y, resulting inmixed barotropic-baroclinic instability.

4.3 Baroclinic instability: The Eady problem

The simplest example of baroclinic instability (and one that is actually morerelevant to the real atmosphere than it might appear) in a continuously strati-�ed �uid is the Eady problem. The problem has the following characteristics:

1. Rigid horizontal upper and lower boundaries at z = �12D, on which

w = 0:

2. The �uid is Boussinesq (� = constant in inertial terms)

3. Inviscid, adiabatic �ow on an f� plane (f = f0 is constant: � = 0)

4. Uniform buoyancy frequency: N2 = �g��100 (@�0=@z) is constant

5. Basic state comprises a zonal �ow that increases linearly with height:u0 = �z

8

6. Basic state density in thermal wind balance with the wind:

�0 = �00

�1 +

1

g

�f�y �N2z

��where �00 is constant, so @�0=@y = f��00=g: �0 increases uniformlywith latitude everywhere.

The basic state geostrophic streamfunction is

= �ZU(z) dy = ��zy + constant

and the basic state QGPV gradient is

@Q

@y= �@

2U

@y2� f 20N2

@2U

@z2= 0

so the basic state has no PV gradient� this is the de�ning characteristic ofthe Eady problem. It then follows from the stability criterion that this �owmust be stable unless there are density gradients on the boundaries, whichof course there are.The perturbation QGPV equation is�

@

@t+ U

@

@x

�q0 + v0

@Q

@y= 0 :

But since @Q=@y = 0, if q0 = 0 everywhere at some initial time, then

q0 =@2 0

@x2+@2 0

@y2+f 20N2

@2 0

@z2= 0 :

If we look for separable modal solutions, wave-like in the horizontal, of theform

0 = Re��(z)ei(kx+ly�kct)

�(9)

thend2�

dz2� N2

f 20�2� = 0

where � =pk2 + l2. Then � � exp (�N�z=f0), or

�(z) = A cosh

�N�

f0z

�+B sinh

�N�

f0z

�: (10)

9

Thus, we can regard this as the sum of two exponentials, decaying away fromthe lower and upper boundaries.To close the problem, we need to invoke the upper and lower boundary

conditions w0 = 0. To do so, consider the thermodynamic equation, whichfor this Boussinesq system is just d�=dt = 0, which yields the linearizedperturbation equation�

@

@t+ U

@

@x

��0 + v0

@�0@y

+ w0@�0@z

= 0 :

Since �0 = � (f0�00=g) @ 0=@z, and @�0=@y = f��00=g, and w0 = 0 on the

boundaries, we have

(U � c)d�

dz� �� = 0 (11)

on each boundary.It is convenient here to de�ne the length scale, L = ND=f0, the internal

radius of deformation. This compares with the external radius of deforma-tion, Le =

pgD=f0:

L =ND

f0=D

f0

sg

�00

����d�0dz���� = D

f0

sg

D

��0�00

= Le

s��0�00

:

Then N�z=f0 = �Lz=D.Now, applying (10) to each boundary condition (11) in turn, noting that

U = �D=2, ��D=2, on the upper and lower boundaries respectively,

�L

D

��D

2� c

��A sinh

�1

2�L

�+B cosh

�1

2�L

����

�A cosh

�1

2�L

�+B sinh

�1

2�L

��= 0

(12)

�� LD

��D

2+ c

���A sinh

�1

2�L

�+B cosh

�1

2�L

����

�A cosh

�1

2�L

��B sinh

�1

2�L

��= 0

Eq. (12) represents an eigenvalue problem for c. After a good deal ofmanipulation (see the Appendix for details) we �nd

c = ��D�L

s��L

2� tanh

�1

2�L

����L

2� coth

�1

2�L

��: (13)

10

The function (x� tanh x) (x� cothx) is plotted in Fig. 2. When x <1:1997, the function is negative3, and c is then purely imaginary; when x >

2.51.250-1.25-2.5

2.5

2

1.5

1

0.5

0

-0.5

-1

x

y

x

y

Figure 2: The function y = (x� tanh x) (x� cothx).

1:1997, c is purely real. Note that since our solution (9) depends on timeas exp (�ikct), we have propagating waves, without growth or decay, forIm (c) = 0, and growing4 waves for Im (c) > 0. Since, from (13), we have

c

�D= � 1

�L

s��L

2� tanh

�1

2�L

����L

2� coth

�1

2�L

��(14)

we can plot c=�D vs. �L; this is shown in Fig. 3.The gross characteristics of the solutions, therefore, depend solely on

whether or not �L = N�D=f0 exceeds the value 0 = 2:3994. For given Nand D, the long waves grow:

� < 0L�1 ; Im (c) 6= 0

� > 0L�1 ; Im (c) = 0

For very short waves, �L� 1 and (13) gives us

c! �12�D :

3In fact, the zero of the function occurs where x = cothx.4Since (13) has two solutions of opposite signs, whenever there is a growing solution,

there is also a corresponding decaying solution. But the growing solution with positive Impart is the interesting one.

11

0 1 2 3 4NDκ/f0

-0.2

-0.1

0.0

0.1

0.2

0.3

Im(c)

Re(c)Re(c)c/(ΛD)

Figure 3: Real and imaginary parts of c=�D, plotted against �L, for theEady problem.

In this limit, the sinh and cosh functions just decay exponentially awayfrom the boundaries (and have little amplitude at the opposite boundary).These �Eady edge waves� (which are formally equivalent to Rossby waves,owing their existence to the temperature gradients there) are trapped ateach boundary, and each is simply advected by the local �ow. For smaller �,the two boundary waves decay less rapidly with height and interact, slowingeach other�s propagation. Eventually, when � exceeds the critical value, thisinteraction stalls the waves (Re (c)! 0, so the waves propagate at the speedof the mid-level �ow, which happens to be zero in this case) and they beginto reinforce each other, causing growth of the coupled boundary waves.Even though c depends on wavenumber only through its magnitude �,

the growth rate of growing waves, � = k Im (c), is a function of k and l.Writing dimensionless wavenumbers k0 = kL, l0 = lL; �0 =

pk02 + l02 = �L;

we have�L

�D= k0 Im

� c

�D

�

12

and so from (14) we have

�N

f0�= �

�k0

�0

�s��0

2� tanh

�1

2�0���

�0

2� coth

�1

2�0��

:

The dependence of �N=f0� on k0 and l0 for the growing wave is plotted in Fig.4. The maximum growth rate, �N=f0� = 0:31, is found at k = 1:61L�1, l =

Figure 4: [James]

0. Note that the growth rate � depends on the ratio �=f0N , and therefore,

13

as one might expect, increases with increasing baroclinic shear �, but thatthe wavelength of the fastest growing wave is independent of �.Is this instability relevant to the real world? In the midlatitude tro-

posphere, D ' 10km, N ' 1:2 � 10�2s�1, f0 ' 1:0 � 10�4s�1, and �is typically 25ms�1=10km ' 2:5 � 10�3s�1. So the fastest growth rate is0:31 � 2:5 � 10�7=1:2 � 10�2 ' 6:5 � 10�6s�1, or an e-folding time scale of1:5�105s' 1:8 days. This is comparable with what is seen in a strongly devel-oping storm. The wavenumber of the fastest growing wave is 1:61f0=ND =1:61�10�4= (120)m�1 ' 1:34�10�6m�1, giving a wavelength of 2�=k ' 4700km. (At 450, where a latitude circle measures 28000 km, this corresponds tozonal wavenumber 6.)The longitude-height structure of the most rapidly growing mode is shown

in Fig. 5. Note:

1. The geopotential perturbation maximizes at the upper and lower bound-aries

2. The geopotential perturbation tilts westward with height

3. w and T are positively correlated

4. Poleward �ow (@�=@x > 0) is positively correlated with T

4.4 Energetics of the Eady problem

[Holton, Ch. 8 discusses energetics in a di¤erent situation. Lorenz (TheNature and Theory of the General Circulation of the Atmosphere, WMO,Geneva, 1967) discusses atmospheric energetics in some detail. When appliedlocally, details can be over-interpreted (Plumb, J. Atmos. Sci., 40, 1670-1688,1983) but we avoid these pitfalls here by focusing exclusively on integratedbudgets.]

4.4.1 Available potential energy

For a Boussinseq �uid (as in the problem at hand) the potential energy (PE)of an elementary �uid volume dV = dx dy dz is �gz dV , so the total amountin the �uid is

P = g

ZZZ�z dx dy dz : (15)

14

Figure 5: [Holton Fig 8.10]

15

Note that the origin of z is arbitrary: one is only ever interested in changesof PE, However, this is not a very useful expression for a �uid, since thereare other constraints, such as conservation of total mass. Moreover, if thedynamics are adiabatic, density can be changed only by redistribution, andif the �ow satisi�es QG assumptions then, to leading order in Rossby num-ber, that redistribution takes place (approximately) horizontally. Given ourgoverning QG equations

Dgu� fv = �@~�

@x

Dgv + fu = �@~�

@y(16)

d�

dt= 0

Here we have subtracted the horizontally averaged geopotential ��100 p0(z) from�, so that

@~�

@z= � g

�00(�� �0(z)) = �

g

�00~� ;

where ~� is the departure of density from its horizontal average. Taking u�the �rst of (16) plus v� the second gives

Dg

�1

2

�u2 + v2

��= �u@

~�

@x� v

@~�

@y

= �r ��u~��+ w

@~�

@z

= �r ��u~��� g

�00w~� : (17)

But the QG version of the density equation can be written as

Dg~�+ wd�0dz

= 0

so that, multiplying by ~�,

Dg

�1

2~�2�+ w~�

d�0dz

= 0 : (18)

16

So (17) and (18) together give

Dg

"1

2

�u2 + v2

�+1

2

�g

�00

�2~�2

N2

#= �r �

�u~��: (19)

where we have used N2 = � (g=�00) d�0=dz:Eq. (19) expresses conservation of energy in this system. The term

on the RHS, the divergence of what is often (and sometimes misleadingly)referred to as the energy �ux, obviously vanishes on integration if there is no�ow through the boundaries. The �rst term inside the bracket on the leftis clearly identi�able as the kinetic energy per unit mass; the second term,rather less obviously, is a measure of the potential energy per unit mass, suchthat

A =1

2

g2

�00

ZZZ~�2

N2dx dy dz (20)

is referred to as the available potential energy (APE). It is clearly not thesame thing as (15), but the di¤erence,

P � A =

ZZZ ��z � 1

2

g2

�00

~�2

N2

�dx dy dz

is evidently something that is unavailable for conversion to kinetic energy byadiabatic QG motions.Note from the form of (20) that APE depends only on departures of

density from its horizontal average, i.e., on horizontal gradients of density.A system with no horizontal gradients of density has no APE: the PE of sucha state cannot be changed by adiabatic QG motions5. The minimum APE(and, by (19), the maximum KE) is achieved by reducing ~�2 to zero, i.e., by�attening the density surfaces.

4.4.2 Eddy energies and growth

Now let�s consider the eddies separately. As before, we de�ne eddy �uctua-tions to be the departures from the zonally averaged state:

�a =1

L

Z L

0

a dx ; a0 = a� �a .

5It can, of course, be changed by non-QG motions: convective available potential energy(CAPE) is a measure of how much PE can be released by vertical motions.

17

Then the total KE is the sum of mean KE (KM) and eddy KE (KE) where6

KM =1

2L

ZZ�00�u

2dy dz ; KE =1

2L

ZZ�00

�u02 + v02

�dy dz : (21)

Similarly, the mean and eddy APEs are

AM =1

2

g2L

�00

ZZ~�2

N2dy dz ; AE =

1

2

g2L

�00

ZZ�02

N2dy dz : (22)

Now, going back to our normal mode stability problem, an initially in�n-itessimal disturbance, which by (21) and (22) has in�nitessimally small KE

and AE, grows exponentially with time. As it does so, both KE and AE mustincrease with time. Under what conditions can this happen?The perturbation equations, linearized about the zonal mean, are

@u0

@t+ �u

@u0

@x� fv0 = �@�

0

@x@v0

@t+ �u

@v0

@x+ fu0 = �@�

0

@y(23)

@�0

@t+ �u

@�0

@x+ v0

@��

@y+ w0

@��

@z= 0

Take u0� the �rst + v0� the second, average, and manipulate as before:

@

@t

�1

2

�u02 + v02

��= �r �

�u0�0

�� g

�00w0�0 ;

which integrates (assuming no �ow through the boundaries) to give simply

dKE

dt= �gL

ZZw0�0 dy dz : (24)

Thus, eddy KE grows if, on average, dense �uid is sinking and buoyant �uidrising within the eddy �eld.

6From the de�nition of mean and eddy, for any variable a,

a2 = (�a+ a0)2= �a2 + a02 :

Note also that the mean geostrophic v vanishes by de�nition: �v = f�1@�=@x = 0.

18

To get the eddy APE equation, take �0� the third of (23) and average togive

@

@t

�1

2�02�+ v0�0

@��

@y+ w0�0

@��

@z= 0:

Then from (22) we have

dAedt

= �g2L

�00

ZZ1

N2u0�0 � r�� dy dz ; (25)

where u0�0 =�u0�0; v0�0; w0�0

�is a measure of the eddy mass �ux. (Note that

the zonal component is irrelevant as �� has no zonal gradient.) What (25)implies is just a little more subtle than (24): in order to eddy APE to grow,projection of the eddy mass �ux onto the mean density gradient must bedowngradient.How the two conditions for growth can be satis�ed simultaneously can

be seen from Fig. 6. In the mean state, density increases downward and

z

y -->EQ POLE

ρ

ρu'ρ'

Figure 6: Wedge of instability in a Boussinesq �uid.

poleward. Growth of eddy KE requires, from (24), that w0�0 be downward,which is actually against the mean vertical gradient (toward higher density)in order that, simultaneously, the projection of the �ux onto r�� be neg-ative, the �ux must lie within the �wedge of instability� bounded by the

19

density surface and the horizontal, as illustrated by the shading. (Note thatthe wedge of instability is often described in terms of parcel displacements,which can be problematic: the pitfalls are avoided by phrasing it in termsof buoyancy �uxes, as done here.) So, although neither (24) nor (25) alonestates this directly, together they prescribe that the eddy mass �ux musthave a latitudinal component directed toward the lower mean density. Thus,in the usual midlatitude situation of density increasing poleward, the eddymass �ux must be downward and equatorward.In an atmospheric context, the same principle holds, though we have to

talk about the structure of � in p coordinates. In this case, eddy growthrequires that the eddy �ux of potential temperature, u0�0, must lie within thewedge de�ned by the contours of �� and pressure surfaces, as shown in Fig.7. Thus, the eddy heat �ux must be upward and poleward in a developing

Figure 7: Wedge of instability in the atmospheric context.

baroclinic wave.

4.4.3 Implications for baroclinic wave structure

We saw that w0 and T 0 are positively correlated in a growing Eady wave,as (24) implies they must, in order for eddy KE to grow. Something elseevident from the structure is that the wave tilts westward with height. Thisin fact follows directly from the result that, on average, v0�0 be directed

20

equatorward, i.e.,fv0�0 < 0 :

Since geostrophic and hydrostatic balance give us

fv0 =@�0

@x

and

�0 = ��00g

@�0

@z;

we have

fv0�0 = ��00g

@�0

@x

@�0

@z:

Now, if (Fig. 8) lines of constant �0 slope at an angle � to the horizontal,

α

φ' =

cons

tant

y

z

Figure 8: Slope (in y � z space) of a phase line.

then

tan� = �@�0=@x

@�0=@z;

whence

fv0�0 =�00g

�@�0

@x

�2tan�

and so the requirement that, on average, fv0�0 < 0 implies that tan� < 0 onaverage: the phase lines (of constant �0) must slope westward with height ina developing disturbance.An example (taken from Wallace and Hobbs) of such tilt leading to rapid

development is shown in Fig. 9. On Nov 19, 12Z, a cold front lies across

21

Figure 9: Rapid development, 1964 Nov 19-20. Note that the upper leveltrough is westward of the surface low.

22

the eastern half of the continent, with a weak surface low (central pressure alittle less than 1004 hPa) lying on the front, a little south of the Ohio valley.At 500hPa, a major trough, propagating eastward, is situated some way tothe west of the surface low. These two couple together and grow such that,within only 12 hrs, the surface low has deepened to 994 hPa, with the troughstill to its west.Note that the initial disturbance aloft was hardly in�nitessimal (it rarely

is): it was a large amplitude, pre-existing disturbance. So the initial growthwas hardly normal-mode-like. Once the upper trough and developing surfacelow begin to interact, however, the disturbance gains more similarity to anEady mode. When the disturbance matures, the upper and lower lowsbecome aligned, and growth stops.

23

Appendix to Section 4: Solution of (12).

To clean up the eqs. a bit, temporarily use the shorthand S = sinh�12�L�,

C = cosh�12�L�. Then rewrite (12) as

A

��L

D

��D

2� c

�S � �C

�+B

��L

D

��D

2� c

�C � �S

�= 0

(26)

A

��L

D

��D

2+ c

�S � �C

�+B

��� L

D

��D

2+ c

�C + �S

�= 0

Setting the determinant of coe¢ cients to zero gives us��L

D

��D

2� c

�S � �C

� ��� L

D

��D

2+ c

�C + �S

����L

D

��D

2� c

�C � �S

� ��L

D

��D

2+ c

�S � �C

�= 0

Reorganizing,

��2 L2

D2

"��D

2

�2� c2

#SC + �� L

D

��D

2+ c

�C2 + �� L

D

��D

2� c

�S2 � �2SC

��2 L2

D2

"��D

2

�2� c2

#SC + �� L

D

��D

2+ c

�S2 + �� L

D

��D

2� c

�C2 � �2SC = 0 :

! 2�2L2

D2

"c2 �

��D

2

�2#SC � 2�2SC+�2�L

�C2 + S2

�= 0 :

Now, using the identity

cosh2 x+ sinh2 x

coshx sinh x= 2 coth 2x = tanhx+ coth x

we arrive at

c2 =

��D

2

�2+D2�2

L2�2� �

2D2

2�L

�tanh

�1

2�L

�+ coth

�1

2�L

��=

��D

�L

�2 ��2L2

4� �L

2

�tanh

�1

2�L

�+ coth

�1

2�L

��+ 1

�=

��D

�L

�2 ��L

2� tanh

�1

2�L

����L

2� coth

�1

2�L

��: (27)

This takes us directly to (13).

24