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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:. Day one. We can prove that divergence is the fractional change with time of some horizontal area A. Day one/two. Then - PowerPoint PPT Presentation
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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
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