Atmospheric Science 4320 /

7320

Anthony R. Lupo

Day one Let’s talk about fundamental

Kinematic Concepts

In lab, we talked about divergence, which is a scalar quantity:

hh V

Day one We can prove that divergence is the

fractional change with time of some horizontal area A.

dtdA

AVhh

1

dtyd

xdt

xdy

yxdtAd

A

and

dtyxd

yxdtAd

A

11

11

Day one/two Then

finally:

yv

xu

dtAd

A1

0lim

hh VdtdA

Ayv

xu

1

Day two We could extend the concept to 3-D and

get:

33

1V

dtdV

Vzw

yv

xu

Day two Horizontal divergence in terms of a

line integral, invoking Green’s theorem (2-D) and Gauss’ (3-D) theorem

Define area A (with tangential wind vectors)

Day two We must also assume Green’s theorem holds

defining a line integral:

A x y

FdxdyFdAdsnF ˆ

Day two Green’s theorem (where P = u and Q = v)

S is the oriented surface, or the position vector on the curve is R

thus ds = dr

F is a vector field in the normal direction on S, in our case V (which is tangent to the curve) where we consider:

Day two So F is:

The normal component is:

nV

hn VkV

ˆ

Day two And we invoke Green’s theorem;

Now recall vector identity: Ax(BxC) = (AdotC)B - C(AdotB)

A

hn

A

dAVkdrnV

FdAdsnF

ˆˆ

ˆ

Day two And see that:

And what is the second term on the RHS equal to????

dAkVdAVkdAVkA

hh

A

hh

A

hh ˆˆˆ

hh

A

hh VdAVk

ˆ

Day two In 3-D we invoke Gauss’s theorem:

Stoke’s Theorem

s V

VdVFFdsdsF

33

Day two Recall Green’s theorem:

And,

kjNiMyxF ˆ0ˆˆ),(

R

dAy

M

x

NNdyMdx

Day two Then

R

R

dAx

NNdy

dAy

MMdx

Day two And;

dyy

MMdx

dxxgxMxgxMMdx

dxxgxMdxxgxMMdx

dxxgxMdxxgxMMdx

dxyxMdxyxMMdx

b

a

xg

xg

b

a

b

a

b

a

C C

C C

2

1

1 3

1 3

2,1,

2,1,

2,1,

,,

Day two And if you don’t understand this:

You might have a ….little …trouble…

Day two Horizontal divergence in Natural

Coordinates:

(s,n,z,t)

The Velocity in Natural Components:

and,

sVV hh ˆ

nn

ssh ˆˆ

Day two so, the horizontal divergence is:

aha, product rule! (which terms will drop

out as 0?

sVV hhhh ˆ

nns

VsnnV

sss

VsssV

V hhhh ˆˆ

ˆˆˆˆ

ˆˆ

Day two Recall:

And;

s

nnnn

Rnsn

Rc

c

ˆˆˆ1

,

Day two Sooo,

A B

nV

sV

V hhh

Day two In the Above relationship

Term A refers to the speed divergence:

Term B is the directional divergence (Diffluence)

Day two Speed Div

Day two Diffluence (directional div)

Day two In a typical synoptic situation, these terms

tend to act opposite each other:

Confluence and speed increasing:

Diffluence and speed decreasing:

00

nsV

00

nsV

Day two Alternative derivation of horizontal

divergence in natural coordinates:

Then take derivative (product rule again:

sin

cos

h

h

Vv

Vu

yV

y

V

yv

xV

x

V

xu

hh

hh

cossin

sincos

Day two/three If we “rotate” i and j (x and y) to coincide

with s and n (s and n) then:

Thus,

1cos,0sin,0 o

s

V

x

V

xu hh

Day two/three and,

Then, the celebrated result!

nV

yV

yv

hh

nV

s

V

yv

xu

V hh

hh

Day three We can also perform our simple-minded

area analysis along the same lines:

dtdA

AVhh

1

dtsd

sdt

sdn

nsdtAd

A

anddt

nsdnsdt

AdA

11

11

Day three Then

finally:

ts

ns

V

dtAd

Ah 1

0lim

hhhh V

dtdA

AnV

s

V

1

Day three Divergence in the Large-Scale

Meteorological Coordinate System:

The divergence refers to the cartesian coordinate which is an invariant coordinate.

On large-scales we need to take into account the curvature of the earth’s surface.

Day three However, the earth is curved, thus all else

being equal, if we move a large airmass (approximated as 2 - D)northward (southward), the area gets smaller (larger) implying convergence (divergence).

We can consider the parcel moving

upward or downward also (the area or volume):

Day three These discrepencies arise from the fact

that the Earth is a sphere, and thus we cannot hold i, j, and k constant.

Recall we re-worked the Navier - Stokes equations to be valid on a spherical surface:

Day three The equation

FgV

pjruv

iruw

krv

iruv

kru

jru

kdtdw

jdtdv

idtdu

dtVd

eeee

ee

2

1ˆˆˆˆtan

ˆˆtanˆˆˆ

2

22

Day three We can also define divergence:

Thus, for example (can you do the rest?)

kwjviukz

jy

ix

V ˆˆˆˆˆˆ33

i

xk

wixj

vxu

ix

kwjviu ˆˆ

ˆˆ

ˆˆˆˆ

Day three then;

Recall, we defined expressions for:

.....ˆˆ

ˆˆ

33 ixk

wixj

vzw

yv

xu

V

jruv

iruw

yk

vxk

udtkd

w

krv

iruv

yj

vxj

uvdtjd

v

kru

jru

xi

udtid

u

ee

ee

ee

ˆˆˆˆˆ

ˆˆtanˆˆˆ

ˆˆtanˆˆ

22

222

Day three thus, we get, for the divergence:

ehh

ee

rv

yv

xu

V

rw

rv

zw

yv

xu

V

tan

2tan33

Day three How important are these “correction terms” for

each scale?

Phenomena Scale Observed

Planetary () 6,000 – 30,000 km

10-6 s-1 - 10-8 s-1

Synoptic (L) 1,000 – 6,000 km

10-5 s-1

Mesoscale 10 – 1,000 km 10-4 s-1 - 10-3 s-1

Day three And;

Then (this approximation is fine too!),

86 102

,10tan

ee rw

rv

ehh

ee

rv

yv

xu

V

rw

rv

zw

yv

xu

V

tan

2tan33

Day three Orders of magnitude of Horizontal divergence and

vertical motions:

Scale Divergence (s-1)

w (cm s-1) b s-1

Planetary 10-6 0.5 0.5

Synoptic 10-5 5 5

Mesoscale 10-4 50 50

Microscale 10-3 500 500

Day three/four Vorticity and Circulation/ unit area

Vorticity is a vector whose magnitude is directly proportional to the circulation/unit area of a plane normal to the vorticity vector.

Vorticity = Curl(V), ROT(V), or

33 V

Day four We are primarily interested in the vertical

component of vorticity due to circulations in the horizontal plane:

(xi) (eta) (zeta)

kji

kyu

xv

jzu

xw

izv

yw

V

ˆˆˆ

ˆˆˆ33

Day four The vertical component:

This is called relative vorticity:

kyu

xv

k ˆˆ

yu

xv

Day four Circulation (Kelvin’s Theorem):

consider a closed curve S, and by definition:

or

R = the position vector dr = change in the position vector

isi sVdsVC

drVC h

Day four from Green’s Theorem:

then:

A x y

FdxdyFdAdsnF ˆ

A x y

hhh dxdyVdAVdrVC

Day four Thus, circulation per unit area:

We need to show that:

Recall, line integral definition:

yxA

A

Cyu

xv

A 0lim

CrVdrVC h

Day four Then,

yvxyyu

uyxxv

vxuC

pathpathpathpathC

4321

Day four And theeen,

xyyu

yxxv

C

reducing

yvxyyu

xuyxxv

yvxuC

Day four So,

Areancirculatio

yu

xv

dadc

and

yu

xv

yxyx

yu

xv

AC

0lim

Day four Thus, we get the vertical component of

Vorticity:

Let’s Examine the vorticity on a sphere:

Vorticity:

kyu

xv ˆ

kwjviukz

jy

ix

V ˆˆˆˆˆˆ33

Day four Look at using ‘foil’, u-comp only:

zi

kuikzu

yi

juijyu

xi

iuiixu

iux

i

ˆˆˆˆ

ˆˆˆˆ

ˆˆˆˆˆˆ

Day four recall cross product rules:

and:

..,ˆˆˆ,ˆˆˆ,ˆˆˆ,0ˆˆ etcjikjkikjiii

jruv

iruw

yk

vxk

udtkd

w

krv

iruv

yj

vxj

uvdtjd

v

kru

jru

xi

udtid

u

ee

ee

ee

ˆˆˆˆˆ

ˆˆtanˆˆˆ

ˆˆtanˆˆ

22

222

Day four Then Vorticity reduces to:

jz

ui

z

v

y

kjwi

y

w

y

jjvk

y

u

x

kiw

jx

w

x

jivk

x

v

x

iiuV

ˆˆ

ˆˆˆ

ˆˆˆ

ˆ

ˆˆ

ˆˆˆˆ

33

Day four but:

thus,

0ˆ

,ˆ1ˆ

0ˆ

ˆ,0ˆ

ˆ,ˆ1ˆtanˆ

ˆ

yk

jiry

jj

xk

ixj

ijr

krx

ii

e

ee

iz

vj

z

ui

y

wi

r

u

ky

uj

x

wk

x

vj

r

uk

r

uV

e

ee

ˆˆˆˆ

ˆˆˆˆˆtan33

Day four/five Now each component:

e

e

e

ru

xw

zu

j

rv

zv

yw

i

ru

yu

xv

Vk

ˆ

ˆ

tanˆ33

Day five Compare orders of magnitude of the

curvature term:

eru

yu

xv tan

55 10,10

yu

xv

610tan

eru

Day five Now look at orders of magnitude of the

other components:

ee ru

xw

zu

rv

zv

yw

,

13

17

102

10,

sxzv

zu

and

syw

xw

1610 sr

u

r

v

ee

Day five And

Relative vorticity in the Natural coordinate system:

Recall:

1310,,

s

zu

zv

dAdC

yu

xv

A 0lim

Day five And recall

sRS

A

FdAdsF

0

0

sss

s

hhsh RRR

RV

VRVC

Day five Then,

2s

s

h

ss

hsshshsh

RR

V

RR

VRRVRVRVC

Day five Since:

sss RRRsA

ss

ss

hs

s

hssh

RR

RR

VR

R

VRRV

AC

2

s

ss

h

s

hsh

R

RR

V

R

VRV

AC

Day five Then apply:

and get :

(A) (B)

s

h

s

h

s

hA R

VRV

RV

dAdC

0lim

nRs

nV

RV h

s

h

Day five Term B, Shear Vorticity (Thanks to C.

Doswell)

Day five Term A; Curvature Vorticity (Thanks C.

Doswell)

Day five Thus, vorticity has two terms (B) shear

term, and (A) curvature term. (e.g. Bell and Keyser, 1993 (Jan MWR).

Thus each can exist independent of the other or together.

Shearing and curvature Vorticity can cancel out even within a rotating fluid!!

Day five In a vortex:

If the air speed increases as we go toward the center such that shear vorticity cancels curvature.

curvshs

h

curvshs

h

curvshs

h

nV

R

V

nV

R

V

nV

R

V

0

Day five Circulation can be irrotational!! This is the

Rankine Vortex model.

An Alternative Vorticity Derviation

Recall:

sVVVVk hhhh ˆ,ˆ33

Day five Then:

And

sVnn

ss

V hhh ˆˆˆ

ns

nVsnn

V

ss

sVsss

VV

hh

hh

hh

ˆˆˆˆ

ˆˆˆˆ

Day five Then,

And

and

ssRs s ˆˆ,

nRRs

sss

ss

ˆ1ˆˆ

nnnn

sns

ˆˆˆ

Day five So then,

and

n

nnVk

nV

nR

sVV hh

shhh ˆˆˆˆ

1ˆ

nV

RV

Vkkn

Vk

R

V

h

s

h

hhh

s

h

ˆˆˆ

Day five Absolute Vorticity:

Is the sum of relative vorticity + Planetary vorticity where,

Planetary vorticity:

fa

sin2f

dtd

dtd

Rdtds

,

Day five Continued

Angular acceleration

Rdtd

Rdtds

V

RV

RV

,

Day five And

Thus, the vorticity associated with solid rotation is a vector in the direction of the axis of rotation with a magnitude of twice the angular velocity.

2p

p

andRV

RV

Day five The earth’s vorticity vector:

thus,

Absolute vorticity:

kfkkpˆˆsin22ˆ

frpra

frpra

Day five Geostrophic relative vorticity:

y

u

x

vVV gg

ggg

,

xv

yu

g

g

Day five If f = fo:

If f varies, then:

zf

g

y

z

x

z

f

g

oog

22

2

2

2

f

uz

f

g

y

z

fy

f

y

z

x

z

f

g g

og

222

2

2

2 1

Day five/Six The Vorticity Equation and vorticity

Theorems (Bluestein, p 242 – 271)

We can calculate from three approaches:

del cross 3-d equation of motion (vector math)

obtain from circulation theorems Brute force

Day six Derive vorticity equation:

Vorticity:

Equations of motion, how do we get?

fyu

xv

rar

,

21

11

y

x

Ffuyp

zv

wyv

vxv

utv

dtdv

Ffvxp

zu

wyu

vxu

utu

dtdu

Day six After differentiation (How do we do this)? This will give us equation 3) and 4):

411

,311

2

2

22

2

22

2

2

2

2

222

x

F

xu

fyp

xxyp

xzv

wzv

xw

xyv

vyv

xv

xv

uxv

xu

xtv

y

F

yv

fvyf

xp

yyxp

yzu

wzu

yw

yu

vyu

yv

yxu

uxu

yu

ytu

y

x

Day sixs

Day six Then subtract equation 3) from 4) to get:

yv

xu

fyx

pxy

pzu

yw

zv

xw

yu

xv

yv

yu

xv

xu

yu

xv

zw

yu

xv

yv

yu

xv

xu

yu

xv

t

2

1

y

F

x

F

yf

v xy

Day six Use some old friends:

Why??

fyu

xv

rar

,

yv

xu

Vhh

dtdf

yf

v

Day six We get:

y

F

x

F

dt

df

Vfyx

p

xy

p

z

u

y

w

z

v

x

w

y

v

x

u

zw

yv

xu

t

xy

hh

rrrrrr

2

1

Day six Combine:

y

F

x

FVf

yxp

xyp

zu

yw

zv

xw

Vdtdf

dt

d

xyhh

hhrr

2

1

Day six We get (using vector and Jacobean

notation) (x,y,z,t) coords!:

This is the traditional Vorticity equation.

Fk

pJwzV

kVdt

dhha

a

ˆ

,1ˆ

2

Day six But we have the vorticity equation, w/ five

terms! (A) (B) (C)

(D) (E)

FkpJ

wzV

kVVt hhaaa

ˆ,1

ˆ

2

Day six Q: What does each term represent?

Which are sources and sinks? Which strictly move Vorticity around?

A1: Vorticity Advection, Divergence, Tilting, Solenoidal (baroclinic), and Friction.

A2: D,E A3 The rest of ‘em!

Day six/seven The vorticity advection term (A) :

CVA (PVA) AVA (NVA)

aV

0

0

V 0

0

V

Day six/seven A map (term A) (Thanks Dr. Martin!):

Day seven The divergence term (B)

Div.

Conv.

hha V

.,,, dect

BVhha

.,,, incrt

BVhha

Day seven Obviously, the first two cases are more

prevalent!

.,,, incrt

BVhha

.,,, dect

BVhha

Day seven The tilting term (C) (From C. Doswell)

wzV

Day seven The solenoidal term (D):

Generation:

Then the left hand side gets a positive

contribution.

22ˆ,

1

p

kpJ

02 p

Day seven Destruction:

Then the left hand side gets a negative contribution!

02 p

Day seven Frictional term (E) :

Increase friction, increase convergence, and the LHS decreases!

Decrease friction, decrease convergence, and the LHS increases!

Day seven The typical orders of magnitude of each term:

Term Synoptic Scale Planetary Scale

1 / s2 1 / s2

Vorticity Advection 1 x 10-9 1 x 10-10

Divergence 1 x 10-10 1 x 10-11

Solenoidal 1 x 10-10 1 x 10-11

Tilting 1 x 10-11 1 x 10-12

Frictional 1 x 10-11 1 x 10-12

Day seven In the free atmosphere, we can

approximate as:

wzV

kVVt hhaaa

ˆ

Day seven The vorticity equation in isobaric

coords

Take

2,

1,

y

x

Ffuydt

dv

Ffvxdt

du

!,2,1 subtractthenx

andy

Day seven We get:

(A) (B) (C) (D)

Note! there is no Solenoidal term here!

FkpV

kp

Vt aaa

ˆˆ

Day seven Vorticity equation in (x,y,,t)

coordinates

Then the (Frictionless) Equation of motion is;

fV a ,

hh VkfM

dtVd

ˆ

Day seven Where

get a vorticity equation that includes a tilting term,

but if motion is adiabatic, the no vertical motions

gzTcM p

VkVV

t haaa ˆ

haaa VV

t

Day seven Vorticity equation in - coordinates;

Thus, there is a solenoidal term!

FkpJ

VkVV

t haaa

ˆ,1

ˆ

2

Day eight Vorticity Theorems: Manipulation and

approximations of the basic vorticity equations.

Constant Absolute Vorticity Trajectories (CAVT).

Consider the isobaric vorticity equation:

Fkp

Vk

pdt

da

a

ˆˆ

Day eight Assume invicid, and no gradient in the

vertical motion:

Assume we are at the non-divergence level:

Thus, we reduce to the “barotropic” vorticity equation!

pdt

da

a

0dt

d a

Day eight (Obviously, vorticity is constant following

lines of absolute vorticity)

Q: Which term survives? What can we do with Vorticity?

A: Term A (advection) and we can move it around!

tconsfr tan

Day eight Assume relative vorticity is all in the curvature

term:

Thus, along the trajectory, with a southerly comp, f is increasing, thus vorticity is decreasing (becoming more anticyclonic),

Then, after some time, we get a northerly component, f is decreasing thus vorticity must increase (becoming more cyclonic).

constfR

V

R

V ha

hr

Day eight Conservation of Potential Vorticity in

a horizontal flow (quasi- geostrophic PV).

Recall earlier, we derived from the 3-D vorticity equation in x,y,z,t:

Fp

uudtd

aa

2

Day eight The vorticity equation for horizontal flow

(x,y,,t):

and rewrite:

haa V

dt

d

ha

ha

a

Vdt

d

then

Vdt

d

ln

1

Day eight We can substitute for the divergence term

by substituting 2-D continuity equation for divergence term:

Couple them:

hVdt

pd

ln

dt

ddt

pd

a lnln

Day eight Thus we can solve to show that:

0

0ln

0ln

ln

pdtd

then

pdtd

anddt

pd

dt

d

a

a

a

Day eight So…..

Thus, we get a 2-D PV (Quasi-geostrophic PV) that is conserved following a constant absolute vorticity trajectory (CAVT).

This relationship can show important concepts w/r/t PV, even though this relationship is antiquated and is not used (in favor of IPV or EPV).

constp

f

Day eight This shows that if:

This relationship can teach us about lee troughing:

gincreagdecreap

gdecreagincreap

elveryadiabaticp

sin,sin0

sin,sin

arg,0

Day eight Consider a North-South mountain chain

and assume adiabatic:

Assume that d is constant. Then as air climbs mountain, if dp decreases, so must the absolute vorticity (become anticyclonic).

constdp

f

Day eight Now, f is decreasing, and as we cross the

top of the barrier, dp increases, thus to retain constant values of vorticity, we must get a sharp increase in relative vorticity.

set up a wavelike disturbance that evanesces along the CAVT.

Day eight Flow through an absolute vorticity

maximum, or advection of cyclonic vorticity

We can use the vorticity equation to deduce these patterns.

We know that upper air flow moves much faster than (or flows through) upper air vorticity maxima and upper air waves.

Day six/seven A map (Thanks Dr. Martin!):

Day eight Let’s consider the isobaric form of the

vorticity equation (we’ll only retain the divergence term).

Assume is small or non-existent (steady state vortex, or coherent structure)

hhaa V

dtd

ta

hhaah VV

Day eight Thus Divergence will be associated with

Cyclonic Vorticity Advection (where higher values of vorticity are carried toward initially lower values of vorticity)

Convergence will be associated with Anticyclonic Vorticity advection (where lower values of vorticity are carried toward initially higher values of vorticity).

Day eight Again divergence aloft will be maximized

and provide maximum upper level support when it is superimposed over low-level convergence, inducing strong upward motions.

Day nine Divergence / Convergence and

secondary circulations associated with jet streaks: (Uccellini et al, 1984, 1985, 1987, and Uccellini and Kocin, 1987, all MWR)

Let’s first consider the straight line jet:

Divergence/convergence patterns, why there?

Day nine Thus, quadrant by quadrant, div CVA / Conv --

AVA.

Entrance Exit speed divergence speed

convergence directional conv. directional div. Geostrophic adj. Geostrophic adj. PGF > CO PGF < CO ageo. Comp. plwrd ageo. Comp. eqtrwd.

Day nine Secondary circulations induced by

jet/streaks:

Day nine Q-G perspective

Day nine Consider cyclonically and anticyclonically

curved jets: Keyser and Bell, 1993, MWR.

Day nine Superimpose high and low level jets.

Day nine A bit of lightness…

Day nine Inertial Instability (Hydrodynamic

Instability)

The inertial Stability Criteria of Zonal Geostrophic Flow:

Consider a steady zonal geostrophic current, which varies only in the y - direction. ug = ug(y).

Day nine The horizontal frictionless eq. of motion:

fuyp

dtdv

fvxp

dtdu

1

1

Day nine thus, there is no variation of pressure in x:

and the equations become:

yp

fuyp

fu

xp

gg

11

,0

uufdtdv

fvdtdu

g

Day nine We want to determine if a parcel displaced

northward or southward will be forced back to or accelerate away from the original position. (will north-south KE change w/time)?

Day nine Thus, we want KE, or more appropriately

the sign of KE:

!,02

!,02

2

2

stableV

dtd

unstableV

dtd

Day nine Displaced parcel (will experience some

change in speed):

thus substituting (from eqns of motion):

(1)

dtdtdu

uduuu ggo

fvdtuu go

Day nine at this same point ug is (recall is a function

of y):

(2)

vdtdydtdy

v ,

vdty

uuu

and

dyy

uuu

ggog

ggog

Day nine but, the second EOM, the v equation, let’s

subtract,

y

uffvdt

dtdv

fvdtuvdty

uuf

dtdv

g

gog

go

Day nine multiply by v:

y

ufdtfv

vdtd

and

y

ufdtfv

dtdv

v

g

g

22

2

2

Day nine In the NH f > 0, v2 > 0, dt > 0

Q: What’s different about the SH?

A: f < 0

Day nine Thus, the stability criteria depends on:

y

uf

y

uf

vdtd

Unstable

y

uf

y

uf

vdtd

Neutral

y

uf

y

uf

vdtd

Stable

gg

gg

gg

,0,02

:

,0,02

:

,0,02

:

2

2

2

Day nine Interpretation:

If the increase in ug to the north (anticyclonic shear), is greater than f (instability - disturbances amplify and reduce the shear to the value of f)

There are no limits in terms of cyclonic shear (stable).

Day nine is rarely but sometimes

observed, usually equatorward of the jet axis (the anticyclonic side). (Why?)

Another way to interpret (vorticity):

0

y

uf g

fyu

xv

a

Day nine Here:

stable

neutral

unstable

a

a

a

,0

,0

,0

Day nine Interpret in terms of absolute

vorticity (Barotropic instability)

Barotropic instability is important in the terrestrial tropical atmosphere, but is also important in other atmospheres (e.g., Jupiter, Saturn, etc.).

Day nine The absolute vorticity gradient must be

negative for barotropic instability! (e.g., Rayliegh, 1880; Kuo, 1949, JAM, Mudrick, 1974, JAS)

0

0

y

uf

y

or

g

a

Day nine Then:

Then Barotropic instability depends on the curvature of the geostrophic wind!

Barotropic instability should be favored in “narrow, pointy” jets.

0,0 2

2

2

2

y

u

y

u

yf gg

Day ten Introduction to Q-G Theory:

Recall what we mean by a geostrophic system:

2-D system, no divergence or vertical motion no variation in f incompressible flow steady state barotropic (constant wind profile)

Day ten We once again start with our fundamental

equations of geophysical hydrodynamics:

(4 ind. variables, seven dependent variables, 7 equations)

x,y,z,t u,v,w or ,q,p,T or ,

kssourcesdtdm

Vdtd

dtdp

dtdT

cQRTP p

sin,

,

Day ten More…….

z

y

x

Fgzp

zw

wyw

vxw

utw

dtdw

Ffuxp

zv

wyv

vxv

utv

dtdv

Ffvxp

zu

wyu

vxu

utu

dtdu

1

1

1

Day ten Our observation network is in (x,y,p,t).

We’ll ignore curvature of earth:

Our first basic assumption: We are working in a dry adiabatic atmopshere, thus no Eq. of water mass cont. Also, we assume that g, Rd, Cp are constants. We assume Po = a refernce level (1000 hPa), and atmosphere is hydrostatically balanced.

Day ten Eqns become:

p

TR

p

Ffuyp

v

y

vv

x

vu

t

v

dt

dv

Ffvxp

u

y

uv

x

uu

t

u

dt

du

d

y

x

p

c

R

o

c

Q

pV

t

p

pT

V

p

d

lnlnln

,033

Day ten Now to solve these equations, we need to

specify the initial state and boundary conditions to solve. This represents a closed set of equations, ie the set of equations is solvable, and given the above we can solve for all future states of the system.

Thus, as V. Bjerknes (1903) realizes, weather forecasting becomes an initial value problem.

Day ten These (non-linear partial differential equations)

equations should yield all future states of the system provided the proper initial and boundary conditions.

However, as we know, the solutions of these equations are sensitive to the initial cond. (solutions are chaotic).

Thus, there are no obvious analytical solutions, unless we make some gross simplifications.

Day ten So we solve these using numerical techniques.

One of the largest problems: inherent uncertainty in specifying (measuring) the true state of the atmosphere, given the observation network. This is especially true of the wind data.

So our goal is to come up with a system that is somewhere between the full equations and pure geostrophic flow.

Day ten We can start by scaling the terms:

1) f = fo = 10-4 s-1 (except where it appears in a differential)

2) We will allow for small divergences, and small vertical, and ageostrophic motions. Roughly 1 b/s

3) We will assume that are small in the du/dt and dv/dt terms of the equations of motion.

pv

pu

,

Day ten 4) Thus, assume the flow is still 2 - D.

5) We assume synoptic motions are fairly weak (u = v = 10 m/s).

Also, flow heavily influenced by CO thus ( <<< f).

Day ten 7) Replace winds (u,v,) by their

geostrophic values

8) Assume a Frictionless AND adiabatic atmosphere.

Day ten/11 The Equations of motion and

continutity

So, we could show that the RHS, will nearly cancel each other in these three equations.

hh

ohh

ohh

Vp

ufy

vVt

v

dt

dv

vfx

uVt

u

dt

du

Day 11 TIME OUT!

Still have the problem that we need to use height data (measured to 2% uncertainty), and wind data (5-10% uncertainty). Thus we still have a problem!

Much of the development of modern meteorology was built on Q-G theory. (In some places it’s still used heavily). Q-G theory was developed to simplify and get around the problems of the Equations of motion.

Day 11 Why is QG theory important?

1) It’s a practical approach we eliminate the use of wind data, and use more “accurate’ height data. Thus we need to calculate geopotential for ug and vg. Use these simpler equations in place of Primitive equations.

Day 11 2) Use QG theory to balance and replace

initial wind data (PGF = CO) using geostrophic values. Thus, understanding and using QG theory (a simpler problem) will lead to an understanding of fundamental physical process, and in the case of forecasts identifying mechanisms that aren’t well understood.

Day 11 3) QG theory provides us with a

reasonable conceptual framework for understanding the behavior of synoptic scale, mid-latitude features. PE equations may me too complex, and pure geostrophy too simple. QG dynamics retains the presence of convergence divergence patterns and vertical motions (secondary circulations), which are all important for the understanding of mid-latitude dynamics.

Day 11 So Remember……

“P-S-R”

Day 11 Informal Scale analysis derivation of

the Quasi - Geostrophic Equations (QG’s)

We’ll work with geopotential (gz):

Rewrite (back to) equations of motion:(We’ll reduce these for now!)

Day 11 Here they are;

Then, let’s reformulate the thermodynamic equation:

ufy

vVtv

dtdv

vfx

uVtu

dtdu

ohh

ohh

TcQ

Vtdt

d

p

lnlnln

Day 11 Then we have:

on p-sfc, of course:

which we can re-write as:

TRpp

pT d

c

R

o p

d

,

TccTc 231 ,

TcQ

pV

t p

ln

lnln

Day 11 Then multiply by Specifc Volume and carry

out derivative in two LHS terms:

Then from hydrostatic balance

TcQ

pV

t ph

ln

p

TcQ

ppV

pt ph

ln

Day 11 Define static stability (in Q.G. Theory, = (p)):

Thus, we have reworked the first law of thermodynamics. Then applying Q-G theory:

ppT

pR

ln

TcQ

pV

pt ph

0

ogh p

Vpt

Day 11 Next, let’s rework the vorticity

equation:

In isobaric coordinates:

Fk

pV

kVp

Vt

hhha

aah

a

ˆ

ˆ

Day 11 Let’s start applying some of the approximations:

1) Vh = Vg

2) Vorticity is it’s geostrophic value 3) assume zeta is much smaller than f = fo

except where differentiable. 4) Neglect vertical advection 5) neglect tilting term 6) Invicid flow

Day 11 Introduce each assumption:

And we get the vorticity equation. We can substitute for r:

fVV

zfg

ff

rgh

ogro

,

, 2

hhoaga

hhaaga

VfVt

VVt

Day 11 to get (recall on LHS ):

and then multiply by fo:

Thus, we are left with two equations in an adiabatic, invicid, Q-G framework.

0

tf

pff

fV

ft oo

go

22 11

pff

fVf

t oo

go

222 1

Day 11 Now let’s derive the height tendency

equation from this set

We will get another “Sutcliffe-type” equation, like the Z-O equation, the omega equation, the vorticity equation.

Like the others before them, they seek to describe height tendency, as a function of dynamic and thermodynamic forcing!

Day 11 Take the thermodynamic equation and:

1) Introduce:

2) switch: 3) apply:

t

pt

,

p

f

o

o

2

Day 11 And get:

0

0

22

2

22

pf

pV

pf

pf

and

pV

p

o

ogh

o

o

o

o

ogh

Day 11 Now add the Q-G vorticity and

thermodynamic equation (where ) and we don’t have to manipulate it:

t

pff

fVf o

ogo

222 1

Day 11/12 The result

becomes after addition:

(Dynamics – Vorticity eqn, vort adv)

(Thermodynamics – 1st Law, temp adv)

pV

p

f

ff

Vfp

f

gho

o

ogo

o

o

2

22

222 1

Day 11/12 This is the original height tendency

equation!

pV

pf

ff

Vfp

f

gho

o

ogo

o

o

2

22

222 1

Day 12 The Omega Equation (Q-G Form)

We could derive this equation by taking of the thermodynamic equation, and of the vorticity equation (similar to the original derivation). However, let’s just apply our assumptions to the full omega equation, which we have derived already. (I’ll expect on test)

Day 12 The full omega equation (The Beast!):

Fkp

Vk

tpV

pf

ppf

c

QTV

p

R

pf

h

agaah

a

pha

ˆˆ

22

22

Day 12 Apply our Q-G assumptions (round 1):

Assume:

Vh = Vgeo, = g, and r <<< fo f= fo, except where differentiable frictionless, adiabatic = (p) = const.

Day 12 Here we go:

p

Vk

pV

pf

ppfTV

p

R

pf

ghaagho

aoghoo

ˆ

22

222

Day 12 Then let’s assume:

1) vertical derivatives times omega are small, or vertical derivatives of omega, or horizontal gradients of omega are small.

2) substitute: 21

or f

Day 12 3) Use hydrostatic balance in temp

advection term.

4) divide through by sigma (oops equation too big, next page)

pRT

p

Day 12 Here we go;

pV

ff

Vp

f

p

f

and

pV

ff

Vp

fp

f

gho

ogh

o

o

o

o

gh

oghooo

2

22

222

2

22

222

1

1

1

Day 12 Of course there are dynamics and

thermodynamics there, can you pick them out?

Q-G form of the Z-O equation (Zwack and Okossi, 1986, Vasilj and Smith, 1997, Lupo and Bosart, 1999)

We will not derive this, we’ll just start with full version and give final version. Good test question on you getting there!

Day 12 Full version:

dpp

dpcQ

STVfR

Pd

dp

Fkp

Vk

t

ppV

Pdt

po

pt

p

po ph

po

pt hag

aa

ah

p

go

2

ˆˆ

tppPd

1

Day 12 Q-G version #1 (From Lupo and Bosart,

1999):

dpp

dpSTV

fR

Pd

dpff

VPdft

po

pt

p

po

og

po

pt og

po

2

22 11

Day 12 Q-G Form #2 (Zwack and Okossi, 1986;

and others)

dpp

dpTV

fR

Pd

dpff

VPdft

po

pt

p

po

g

po

pt og

po

2

22 11

Day 12 Q-G Form #3!

dpdpp

Vf

Pd

dpff

VPdft

po

pt

p

po

g

po

pt og

po

2

22

1

11

Day 12 Quasi - Geostropic potential Vorticity

We can start with the Q-G height tendency, with no assumption that static stability is not constant.

pV

pf

ff

Vftpp

f

gho

ogoo

1

11

2

222

Day 12 Multiply by 1/fo:

pV

pf

ff

Vtpp

ff

gho

ogo

o

1

111 22

Day 12 consider the temperature (thickness)

advection term:

pp

Vf

ppfV

pV

pf

go

oggho

1

11

Day 12 we get two terms (from the product rule!).

But, the second term goes to zero! Why?

pf

kp

fpp

Vf

oo

go

1ˆ1

fk

Vg

ˆ

Day 12 And:

(A) (B)

Since:

01ˆ

ppk

TermBTermA

Day 12 So, the equation becomes:

(1) (2)

(3)

01

111 22

ppfV

ff

Vpp

ffft

og

ogo

o

Day 12 or terms 2 and 3 can mesh: (1)

(2)

or (term 1 – eulerian, term 2 advective!) (hey, I slipped an “f” in there! Why OK?):

011

11

2

2

ppff

fV

ppff

ft

oo

g

oo

Day 12

Vorticity Stability

This is quasi-geostropic potential vorticity! (See Hakim, 1995, 1996, MWR; Henderson, 1999, MWR, March)

Note that we combined dynamic and thermodynamic forcing!!

011 2

ppff

fV

t oo

g

Day 13 So,

Also, you could start from our EPV expression from earlier this year:

ppff

fQGPV o

o

11 2

aEPV

Day 13 Or in (x,y,p,t) coordinates:

In two dimensions:

agEPV

constp

gPV a

Day 13 We assume that:

Thus (recall, this was an “ln” form, so we need to multiply by 1/PV):

pPV

wheredtPVd

a

,0

0

1 dtPVd

PV

Day 13 so,

and

0lnln

pdt

ddtd

a

01

pdtdp

dtd

aa

Day 13 And “QG”

Then

01 2

pdtdp

ffdt

da

o

pp

ppRT

foa

1

,

011 2

pdt

d

pff

fdt

do

o

Day 13 we get QGPV!

Again, we have both thermodynamic and dynamic forcing tied up in one variable QGPV (as was the case for EPV)!

011 2

ppff

fdtd

oo

Day 13 Thus, QGPV can also be tied to one

variable, the height field, thus we can invert QGPV field and recover the height field.

We can also “linearize” this equation, dividing the height field into a mean and perturbation height fields, then:

Day 13 Then….

then

ppf

ffQGPVq

where

ppff

fQGPV

and

QGPVdtd

oo

oo

11

11

,0

2*

2

Day 13 Thus, when we invert the PV fields; we get

the perturbation potential vorticity fields. Ostensibly, we can recover all fields (Temperature, heights, winds, etc. from one variable, Potential Vorticity, subject to the prescribed balance condition (QG)).

pp

ff

q oo

11 2*

Day 13 We’ve boiled down all the physics into one

equation! Impressive development! Thus, we don’t have to worry about non-linear interactions between forcing mechanisms, it’s all there, simple and elegant!

Disadvantage: we cannot isolate individual forcing mechanisms. We must also calculate PV to begin with! Also, does this really give us anything new?

Day 13 Forecasting using QGPV or EPV

Local tendency just equal to the advection (see Lupo and Bosart, 1999; Atallah and Bosart, 2003).

EPVVEPVt

QGPVVQGPVt

EPVdtd

QGPVdtd

0,0

Day 13 EPV and QGPV NOT conserved in a

diabatically driven event. Diabatic heating is a source or sink of vorticity or Potential Vorticity.

Potential Vorticity Generation:

.sin FricDiabaticskssourcesEPVdtd

Day 13 Generation:

Day 13/14 The Q - Vector approach (Hoskins et

al., 1978, QJRMS) Bluestein, pp. 350 - 370.

Start w/ “Q-G” Equations of motion:

gygog

gyagog

uufdt

dv

vvfdt

du

Day 14 Differentiate both w/r/t p:

20

10

p

u

p

uf

p

v

yv

y

v

p

v

p

v

xu

x

v

p

u

p

v

t

p

v

p

vf

p

u

yv

y

u

p

v

p

u

xu

x

u

p

u

p

u

t

gy

ao

gg

gggg

ggg

gy

ago

gg

gggg

ggg

Day 14 Substitute the thermal wind relationship:

,, xT

fR

p

v

yT

fR

p

u

o

g

o

g

Day 14 And substitute

0

0

2

2

yT

p

u

R

pf

xT

yv

y

v

xT

xT

xu

x

v

yT

xT

t

fR

xT

p

v

R

pf

yT

yv

y

u

xT

yT

xu

x

u

yT

yT

t

fR

yao

gg

gg

o

yago

gg

gg

o

Day 14 Here is the adiabatic form of the Q-G

thermodynamic equation:

0

pgg c

Q

R

P

y

Tv

x

Tu

t

T

Day 14 Differentiate the 1st Law first w/r/t x then

w/r/t y:

40

30

p

gg

gg

p

gg

gg

c

Q

yR

P

y

y

T

yv

y

T

y

v

x

T

yu

x

T

y

u

y

T

t

c

Q

xR

P

x

y

T

xv

y

T

x

v

x

T

xu

x

T

x

u

x

T

t

Day 14 Then, multiply Q-G equation of motion

(Eqs. 1 and 2) by:

and add to thermodynamic equations (Eqs. 3 and 4)

R

pfo

Day 14 This gives us Q1 and Q2:

jQiQQ

where

c

Q

yp

R

x

T

p

RT

y

V

p

R

p

vf

yQ

c

Q

xp

R

y

T

p

RT

x

V

p

R

p

uf

xQ

py

gao

py

gao

ˆ2ˆ1

22

21

2

2

Day 14 Then differentiate Q1 and Q2, w/r/t x and

y, respectively (in other words, take divergence).

Q1 Q2

p

gg

aao

c

Q

p

R

x

T

p

RT

y

V

yT

x

V

xp

R

y

v

x

u

p

f

2

22

2

Day 14 Then use continuity:

This give us the omega equation in Q-vector format!

py

v

x

uV aa

Ty

V

p

RQ

Tx

V

p

RQ

where

c

Q

p

R

x

T

p

RQ

p

f

g

g

p

o

2

1

2 22

222

Day 14 Then since :

This give us a Q-vector with diabatics!

2

p

g

p

g

p

o

cQ

yT

y

V

pR

Q

cQ

xT

x

V

pR

Q

where

cQ

pR

xT

pR

Qp

f

2

1

2 22

222

Day 14 Note that on the RHS, we have the

dynamic and thermodynamic forcing combined into one term.

Also, note that we can calculate these on p -surfaces (no vertical derivatives). The forcing function is exact differential (ie, not path dependent), and dynamics or thermodynamics not neglected.

Day 14 This form also gives a clear picture of

omega on a 2-D plot:

Div. Q is sinking motion:

Day 14 Conv. Q is rising motion:

Day 14 Forcing function is “Galilean Invariant”

which simply means that the forcing function is the same in a fixed coordinate system as it is in a moving one (i.e., no explicit advection terms!)

And this is the end of Dynamics! See you next year in 9350?

Day 15 Symmetric Instability

Convective instability:

Measure based on the slope of Γe to Γd or Γe to Γd. Static stability.

It’s an instability in the vertical.

Day 15 Then the slope of the observed potential

temperatures (θ) are important.

If the slope of the dry adiabats is 0, then a “negative” represents stability or “flatter” insentropes. This is a smaller dT/dz.

The opposite is true of dry adiabats which are “rotated” 90 degrees!

Day 15 With Inertial instability, recall:

Stable when absolute vorticity is less than zero: ζa < 0 which was

Could also use:

0

y

uf g

0

x

vf g

Day 15 Thus geostrophic momentum is:

(Draw typical N-S):

gyg Mvfx

gxg Mufy

Day 15 Horizonal instability! Then if the lines of

Mg are “flatter”, then neutral. When slanted, parcels can be displace “slantwise”.

So, for “symmetric” instability, we’re going to combine the two situations.

(Draw)

Day 15 Thus, when the Mg lines are sloped more

steeply than θe, then we have symmetric stability. (stable to both vertical and horizontal conditions)

When the θe lines are sloped more than the Mg lines, then we have “symmetric instability”. (unstable for both conditions).

Day 15 Symmetric instability, but the nature of

the presence of temperature gradients, is baroclinic in nature.

General Instability Problem.

We want a statement that might include both dynamic (barotropic) and thermodynamic (convective / baroclinic)instability.

Day15 Start with QGPV:

Then (in QG form):

Day 15 If we only look in the y-z plane

Then take partial / partial-y:

Day 15 And Finally

Barotropic BaroclinicInstability Instability

Day 15

Day 14 Turn out the lights!

Day 14