gef 2220: dynamics · gef 2220: dynamics j. h. lacasce department for geosciences university of...
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GEF 2220: DynamicsJ. H. LaCasce
Department for Geosciences
University of Oslo
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Primitive equations
Momentum:
∂
∂tu + ~u · ∇u + fyw − fzv = −
1
ρ
∂
∂xp + ν∇2u
∂
∂tv + ~u · ∇v + fzu = −
1
ρ
∂
∂yp + ν∇2v
∂
∂tw + ~u · ∇w − fyu = −
1
ρ
∂
∂zp − g + ν∇2w
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Primitive equations
Continuity:
∂
∂tρ + ~u · ∇ρ + ρ∇ · ~u = 0
Ideal gas:
p = ρRT
Thermodynamic energy:
cvdT
dt+ p
dα
dt= cp
dT
dt− α
dp
dt=
dq
dt
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Momentum equations
Two types of forces:
1) Real 2) Apparent
Two ways to write the derivative:
1) Lagrangian 2) Eulerian
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Derivatives
Example: derive the continuity equation using the Eulerianapproach.
Then derive it using the Lagrangian approach.
Then derive it in pressure coordinates. What’s best for this,Lagrangian or Eulerian?
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Rotation
Acceleration:
(d~uR
dt)R = (
d~uF
dt)F − 2~Ω × ~uR − ~Ω × ~Ω × ~r
Two additional terms:
Coriolis acceleration → −2~Ω × ~uR
Centrifugal acceleration → −~Ω × ~Ω × ~r
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Estimating forces
1) What is the centrifugal force for a parcel at the Equator?
Show that it is small compared to g.
2) What is the Coriolis force on a parcel moving eastward at10 m/sec at 45 N?
Find the direction and magnitude. Which componentmatters?
What happens in the Southern Hemisphere?
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First law of thermodynamics
heat added = change in internal energy + work done:
dq = de + dw
At constant volume:
dq = cvdT + p dα
or, at constant pressure:
dq = cpdT − αdp
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Scaling
∂
∂tu + u
∂
∂xu + v
∂
∂yu + w
∂
∂zu + fyw − fzv = −
1
ρ
∂
∂xp
U
T
U2
L
U2
L
UW
DfW fU
HP
ρL
10−4 10−4 10−4 10−5 10−6 10−3 10−3
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Geostrophy
From scaling the horizontal momentum equations atweather scales:
fzv =1
ρ
∂
∂xp, fzu = −
1
ρ
∂
∂yp
or, in pressure coordinates:
fv =∂
∂xΦ, fu = −
∂
∂yΦ
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Geostrophy
p/ ρL
Hfu
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Geostrophy
L
fu
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Geostrophy
Given the pressure or geopotential, can derive the winds
Example: What is the pressure gradient required at theearth’s surface at 45 N to maintain a geostrophic wind of 30m/sec?
Low geopotential height to left of the wind in NorthernHemisphere
Lies to the right in Southern Hemisphere
Balance fails at equator, because fz = 2Ωsin(0) = 0
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Geostrophy
Is a diagnostic relation
• Given the pressure, can calculate the horizontal velocities
But geostrophy cannot be used for prediction
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Prediction
Must retain the 10−4 terms:
∂
∂tu + u
∂
∂xu + v
∂
∂yu − fzv = −
1
ρ
∂
∂xp
∂
∂tv + u
∂
∂xv + v
∂
∂yv + fzu = −
1
ρ
∂
∂yp
The equations are quasi-horizontal: neglect vertical motion
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Other momentum balances
Constant, circular motion:
u2
θ
r+ fuθ =
1
ρ
∂
∂rp
Scaling:
U
fR1
HP
ρfUR
First parameter is the Rossby number, ǫ
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Other momentum balances
If ǫ ≪ 1, geostrophic balance:
fuθ =1
ρ
∂
∂rp
If ǫ ≪ 1, cyclostrophic balance:
u2
θ
r=
1
ρ
∂
∂rp
or:
uθ = ±(r
ρ
∂
∂rp)1/2
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Inertial oscillations
If ∂∂rp = 0, inertial motion:
u2
θ
r+ fuθ = 0
or:
uθ = −fr
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Gradient wind
If ǫ ≈ 1, gradient wind balance:
uθ = −1
2fr ±
1
2(f2r2 + 4 r f ug)
1/2
If ug < 0 (anticyclone), require:
|ug| <fr
4
If ug > 0 (cyclone), there is no limit
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Gradient wind balance
v−r2
v−r2
L H
p
fvp
fv
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Hydrostatic balance
Scale the vertical momentum equation:
∂
∂tw + u
∂
∂xw + v
∂
∂yw + w
∂
∂zw − fyu = −
1
ρ
∂
∂zp − g
UW
L
UW
L
UW
L
W 2
DfU
V P
ρDg
10−7 10−7 10−7 10−10 10−3 10 10
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Hydrostatic balance
Remove the static contributions:
∂
∂tw + u
∂
∂xw + v
∂
∂yw + w
∂
∂zw − fyu = −
1
ρ0
∂
∂zp′ −
ρ′
ρ0
g
10−7 10−7 10−7 10−10 10−3 10−1 10−1
Vertical accelerations unimportant at synoptic scales.
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Pressure coordinates
Use the hydrostatic balance to simplify equations
∂p
∂x|z = ρg
dz
dx|p ≡ ρ
∂Φ
∂x|p
where:
Φ ≡
∫ z
0
g dz
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Vertical velocities
Different vertical velocities:
d
dt=
∂
∂t+
dx
dt
∂
∂x+
dy
dt
∂
∂y+
dp
dt
∂
∂p
=∂
∂t+ u
∂
∂x+ v
∂
∂y+ ω
∂
∂p
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Continuity
∂u
∂x+
∂v
∂y+
∂ω
∂p= 0
The flow is incompressible in pressure coordinates
ω = −
∫ p
p∗(
∂
∂xu +
∂
∂yv) dp
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Vertical motion
Showed that:
ω ≈ −ρgw
Note that upward motion corresponds to ω < 0
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Thermal wind
Geostrophy + hydrostatic balance → velocity shear
vg(p1) − vg(p0) =1
f
∂
∂x(Φ1 − Φ0) ≡
g
f
∂
∂xZ10
and:
ug(p1) − ug(p0) = −1
f
∂
∂y(Φ1 − Φ0) ≡ −
g
f
∂
∂yZ10
Shear proportional to layer thickness, Z10
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Thermal wind
Alternate expression:
vg(p1) − vg(p0) = −R
fln(
p1
p0
)∂T
∂x
ug(p1) − ug(p0) =R
fln(
p1
p0
)∂T
∂y
Shear proportional to the temperature gradient
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Thermal wind
Thus:
Z10 =R
gT ln(
p1
p0
)
Layer thickness proportional to its mean temperature
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Layer thickness
6
7
8
9
p
p2
1
v
v
1
2
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Definitions
Barotropic atmosphere: no vertical shear
Equivalent barotropic: constant direction with height
Baroclinic atmosphere: speed and direction change
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Jet stream
Example: At 30N, the zonally-averaged temperaturegradient is 0.75 Kdeg−1, and the average wind is zero at theearth’s surface. What is the mean zonal wind at the level ofthe jet stream (250 hPa)?
20
15
10
p
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Temperature advection
Z1
Z1+ Zδ
δ
ZT
ZT+ Z
v1
v2
vT
Warm
Cold
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Temperature advection
Warm advection → veering
• Anticyclonic (clockwise) rotation with height
Cold advection → backing
• Cyclonic (counter-clockwise) rotation with height
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Geostrophy vs. thermal wind
Geostrophic wind parallel to geopotential contours
• Wind with high pressure to the right (NorthHemisphere)
Thermal wind parallel to thickness (mean temperature)contours
• Wind with high thickness to the right
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Divergence and vorticity
D = ∇ · ~u
The divergence in a region is constant and positive.What happens to the density of an air parcel?
ζ = ∇× ~u
What is the vorticity of a typical tornado? Assume solidbody rotation, with a velocity of 100 m/sec, 20 m fromthe center.
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Vorticity equation
(∂
∂t+ u
∂
∂x+ v
∂
∂y+ ω
∂
∂p) ζa
= −ζa(∂u
∂x+
∂v
∂y) + (
∂u
∂p
∂ω
∂y−
∂v
∂p
∂ω
∂x) + (
∂
∂xFy −
∂
∂yFx)
where the absolute vorticity is:
ζa = ζ + f
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Scaling
∂
∂tζ + u
∂
∂xζ + v
∂
∂yζ ∝
U2
L2≈ 10−10
ω∂
∂pζ ∝
Uω
LP≈ 10−11
v∂
∂yf ∝ U
∂f
∂y≈ 10−10
(∂u
∂p
∂ω
∂y−
∂v
∂p
∂ω
∂x) ∝
Uω
LP≈ 10−11
(ζ + f) (∂u
∂x+
∂v
∂y) ≈ f (
∂u
∂x+
∂v
∂y) ∝
fU
L≈ 10−9
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Scaling
Why is the divergence term unbalanced? Argued:
u = ug + ǫua, v = vg + ǫva
with ǫ ≈ 0.1. So the divergence is:
D =∂
∂xug −
∂
∂yvg + ǫ(
∂
∂xua +
∂
∂yva)
1
f
∂
∂x(−
∂Φ
∂y) +
1
f
∂
∂y(∂Φ
∂x)
+ǫ(∂
∂xua +
∂
∂yva) = ǫ(
∂
∂xua +
∂
∂yva)
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Scaling
Thus the divergence estimate is ten times smaller
Retain only the 10−10 terms:
(∂
∂t+ u
∂
∂x+ v
∂
∂y) (ζ + f) = −f (
∂u
∂x+
∂v
∂y)
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Rossby waves
No divergence:
(∂
∂t+ u
∂
∂x+ v
∂
∂y) (ζ + f) = 0
Linearize, with a constant mean flow U :
(∂
∂t+ U
∂
∂x) ζ + v
∂
∂yf = 0
Beta-plane approximation: f = f0 + βy.
(∂
∂t+ U
∂
∂x) ζ + βv = 0
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Rossby waves
Write in terms of geopotential:
u = −1
f0
∂
∂yΦ, v =
1
f0
∂
∂xΦ
ζ =∂
∂xv −
∂
∂yu =
1
f0
∇2Φ
So:
(∂
∂t+ U
∂
∂x)∇2Φ + β
∂
∂xΦ = 0
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Rossby waves
Substitute a wave solution:
Φ = A exp(ikx + ily − iωt)
Get:
(−iω + ikU)(−k2 − l2)A + ikA = 0
or the dispersion relation:
ω = Uk −βk
k2 + l2
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Rossby waves
Has a phase speed of:
cx =ω
k= U −
β
k2 + l2
Waves move west relative to mean flow
Larger waves move faster than smaller waves (waves aredispersive)
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Westward propagation
y=0
+
−
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Divergence
Example: Consider small area of air:
δA = δx δy
Show that:
d
dtζaA = 0
This is Kelvin’s theorem
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Divergence
Important for storm development.
Example: Consider flow with constant divergence:
∂
∂xu +
∂
∂yv = D
Example: show that divgence favors anticyclonic vorticity,convergence favors cyclonic. Show why anticyclonicvorticity is bounded and cyclonic not. This is why cyclonesare more intense.
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Forecasting
Method:
Obtain Φ(x, y, t0) from measurements on p-surface
Calculate ug(t0), vg(t0), ζg(t0)
Calculate ζg(t1)
Invert ζg to get Φ(t1)
Start over
Obtain Φ(t2), Φ(t3),...
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Barotropic potential vorticity
Imagine atmosphere as a layer with constant density. Then:
∂u
∂x+
∂v
∂y= −
∂w
∂z
So:
(∂
∂t+ u
∂
∂x+ v
∂
∂y) (ζ + f) = (ζ + f)
∂w
∂z
Derive:
d
dt(ζ + f
h) = 0
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Layer potential vorticity
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Adiabat
Same idea applies to motion on an adiabatic
Flow between two isentropic surfaces trapped if zeroheating
Derive Ertel’s potential vorticity:
d
dt[(ζ + f)
∂θ
∂p] = 0
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Vorticity summary
d
dt(ζ + f) = −(ζ + f) (
∂u
∂x+
∂v
∂y)
or:
d
dt(ζ + f
h) = 0
Vorticity changes due to meridional motion
Vorticity changes due to divergence
Change in layer thickness
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Planetary boundary layer
Small scale (unresolved) eddies mix momentum vertically:
∂
∂tu + u
∂
∂xu + v
∂
∂yu + w
∂
∂zu − fv = −
1
ρ
∂
∂xp −
∂
∂zu′w′
∂
∂tv + u
∂
∂xv + v
∂
∂yv + w
∂
∂zv + fu = −
1
ρ
∂
∂yp −
∂
∂zv′w′
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PBL equations
Too many unknowns, so parameterize the eddy stresses
Two cases:
Stable boundary layer: stratified, sheared flow
Unstable boundary layer: convective mixing, no shear
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Convective boundary layer
With empirical surface stress:
fh(v − vg) = CdV u
and:
−fh(u − ug) = CdV v
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Stable boundary layer
f(v − vg) =∂
∂z(Az
∂
∂zu)
and:
−f(u − ug) =∂
∂z(Az
∂
∂zv)
Example 1: the Ekman layer (Az = const.)
Example 2: the Surface layer (Az = k2z2 ∂∂zV)
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Boundary layer effects
Most important is how PBL affects troposphere
Permits downgradient winds
Divergence/convergence causes vertical motion intotroposphere
Weakens pressure system, causing them to spin down
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Spin-down
Example: Ekman layer. Can derive:
D
Dt(ζ + f) = −
fd
2Hζ
If f = const., then:
ζ(t) = ζ(0) exp(−t/τE)
where:
τE ≡2H
fd
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General circulation
HH
H
LL
L L
HH
Ferrel
Hadley
Jet stream
Jet stream
(indirect)
(direct)
Trade winds
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