AOSS 321, Winter 2009
Earth System Dynamics
Lecture 9
2/5/2009
Christiane Jablonowski
cjablono@umich.edu
734-763-6238
Eric Hetland
ehetland@umich.edu
734-615-3177
Today’s class
•
•
•
•
•
Discussion of the three momentum equations
Scale analysis for the midlatitudes
Geostrophic balance / geostrophic wind
Ageostrophic wind
Hydrostatic balance
Full momentum equation:
in component form
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
Let’s think about this equation (1)
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
The equations are explicitly non-linear.
Let’s think about this equation (2)
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
These are the curvature terms that arise because the
tangential coordinate system curves with the surface
of the Earth.
Let’s think about this equation (3)
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
These are the Coriolis terms that arise because the
tangential coordinate system rotates with the Earth.
Let’s think about this equation (4)
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
These are the friction terms, the viscous forces, that
arise because the atmosphere is in motion. They
oppose the motion.
Let’s think about this equation (5)
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
These are the pressure gradient terms. They initiate
the motion.
Let’s think about this equation (6)
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
This is gravity (combined with the apparent centrifugal force).
Gravity stratifies the atmosphere, with pressure decreasing
with height. The motion of the atmosphere is largely
horizontal in the x-y plane.
Let’s think about this equation (7)
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
We are mixing (x, y, z) with (λ, Φ, z).
Let’s think about this equation (8)
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
The equations are coupled.
Let’s think about this equation (9)
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
We have u, v, w, ρ, p which depend on (x, y, z, t).
Can we solve this system of equations?
Consider x and y components
of the momentum equation
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
What are the units? Do they check out?
Time scales
D ( )/Dt looks like 1/T (with T: time scale)
We recognize that time can be characterized
by
a distance (length scale L) divided by a velocity
(with velocity scale U): T = L/U
This gives us the unit ‘s’ for the time scale.
D
Dt
 1 /( L / U )  U / L
Scale Analysis:
Let us define
U  horizontal velocity scale
W  vertical velocity scale
L  horizontal length scale
H  vertical length scale
P /   horizontal pressure fluctation scale
L / U  time scale for accelerati on
What are the scales of the terms?
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
U*U/L
U*W/a
U*U/a
P
L
Uf
Wf
U
2
H
Scales of atmospheric phenomena:
Tornadoes
• Funnel cloud
that emerges
from a
thunderstorm
• Scale of the
motion?
Scales of atmospheric phenomena:
Hurricanes
• Satellite image
• Tropical storm
that originates
over warm
ocean water
• Scale of the
motion?
Scales of atmospheric phenomena:
Extratropical storm systems
• Satellite image
• Storm system in
the Gulf of Alaska
• Scale of the
motion ?
Scales for large mid-latitude systems
U
W 
L
length
H
height
a
radius
g
gravity
pressure fluctuations P /  
T  L /U 
time
viscosity

horizontal wind
vertical wind
Coriolis term
Typical scales ?
at  0  45 , f  2sin( 0 ) 
 2cos( 0 ) 
Scales for “large-scale” mid-latitude
weather systems
U  10 m s-1
W  1 cm s-1 (units!)  102 m s-1
L  10 6 m
H  10 4 m
a  10 7 m
g  10 m s2
P /   10 3 m 2 s-2
T  L /U  10 5 s
  105 m 2 s1
at  0  45 , f  2sin( 0 )  2cos( 0 )  10 -4 s1
What are the scales of the terms?
Class exercise
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
What are the scales of the terms?
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
U*U/L
10-4
U*U/a U*W/a
10-5
10-8
P
L
Uf
Wf
10-3
10-3
10-6
U
H2
10-12
What are the scales of the terms?
Du uv tan(  ) uw
1 p



 2v sin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan(  ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
U*U/L
10-4
U*U/a U*W/a
10-5
10-8
P
L
Uf
Wf
10-3
10-3
10-6
Largest Terms
U
H2
10-12
Consider only the largest terms
1 p

 2v sin(  )  0
 x
1 p

 2usin(  )  0
 y
This is a dominant balance between the
• pressure gradient force
• Coriolis force (the dominant sin() components of
the Coriolis force)
This is the geostrophic balance.
Consider only the largest terms
1 p
 2v sin(  )
 x
1 p
 2usin(  )
 y
Note: There is no D( )/Dt term. Hence, no
acceleration, no change with time. This is a balance.

This is the geostrophic balance.
Geostrophic balance
Check out:
http://itg1.meteor.wisc.edu/wxwise/AckermanKnox/chap6/balanced_flow.html
Pressure gradient
force (PGF)
Low Pressure
High Pressure
Coriolis force
(Northern Hemisphere)
In Northern Hemisphere: Coriolis force points to the right
(perdendicular) relative to the path of the air parcel
Geostrophic Wind
Component form:
1 p
ug  
f y
1 p
vg 
f x

Vector form:
1
vg 
k  p
f
with the Coriolis parameter f = 2 sin
 D( )/Dt term.
Note: There is no
Hence, no acceleration, no change with time.
The geostrophic wind describes the dominant balance
between the pressure gradient force and the Coriois force.
Geostrophic wind 300 mb
Geostrophic & observed wind 300 mb
Geostrophic and observed wind 1000 mb (land)
Geostrophic and observed wind 1000 mb (ocean)
Geostrophic wind
• The direction of the geostrophic wind is parallel to
the isobars.
• The geostrophic wind vg is a good approximation
of the real horizontal wind vector, especially over
oceans and at upper levels. Why?
• The closer the isobars are together, the stronger
the magnitude of the geostrophic wind | vg |
(isotachs increase).
Geostrophic Wind
Component form:
1 p
ug  
f y
1 p
vg 
f x

Vector form:
1
vg 
k  p
f
with the Coriolis parameter f = 2 sin
Note: There is 
no D( )/Dt term.
Hence, no acceleration, no change with time.
This is a DIAGNOSTIC equation
Geostrophic Wind
1 p
1 p
ug  
, vg 
f y
f x
f = 2  sin(),   7.292  10-5 s-1
MSLP (hPa) and 10 m wind (knots)
45º N

30º N
1026 hPa
15º N
1013 hPa
135º W
105º W
75º W
Geostrophic
Wind
MSLP (hPa) and 10 m wind (knots)
  7°
58º N
1 p
ug  
f y
1 p
vg 
f x
990 hPa
980 hPa
f = 2  sin()
  7.292  10-5 s-1
14º W
0º
Rossby number
• Rossby number: A dimensionless number that
indicates the importance of the Coriolis force
• Defined as the ratio of the characteristic scales for the
acceleration and the Coriolis force
U2
U
L
Ro 

f 0U
f 0L
• Small Rossby numbers (≈ 0.1 and smaller): Coriolis
force is important, the geostrophic relationship (in
midlatitudes)
is valid.

• What are typical Rossby numbers of midlatitudinal
cyclones and tornadoes?
Impact of friction
• http://ww2010.atmos.uiuc.edu/(Gh)/guides/mtr/fw/fric.rxml
• Friction (especially near the Earth’s surface)
changes the wind direction and slows the wind
down, thereby reducing the Coriolis force.
• The pressure gradient force
becomes more dominant.
As a result, the total wind
deflects slightly towards
lower pressure.
• The (ageostrophic) wind
reduces the differences
between high and low pressure!
Diagnostic and Prognostic
• In order to predict what is going to happen next, we need
to know the rate of change with time. We need D ( )/Dt.
This is known as a prognostic equation.
• Scale analysis shows that for middle latitude large scale
motion (1000 km spatial), 1 day (temporal)) the
acceleration term is an order of magnitude smaller than
the geostrophic terms.
• Hence, for much of the atmosphere we can think of the
geostrophic balance as some notion of an equilibrium
state, and we are interested in determining the difference
from this equilibrium state.
• Ageostrophic is this difference from geostrophic.
What are the scales of the terms?
Du uvtan( ) uw
1 p



 2vsin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan( ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
U*U/L U*U/a U*W/a
P
L
10-5
10-3
10-4
Acceleration
10-8
Uf
Wf
10-3
10-6
Geostrophic terms
U
H2
10-12
In the free
atmosphere

A simple prognostic equation
Du
1 p
 fv 
 f (v  v g )  fva
Dt
 x
Dv
1 p
  fu 
  f (u  ug )   fua
Dt
 y
vg: geostrophic wind vector, not the real wind.
Within 10-15% of real wind in middle latitudes,
large-scale.
va: ageostrophic wind vector, difference between
actual (real) wind and geostrophic wind.
A simple prognostic equation
Du
 f (v  v g )
Dt
Dv
  f (u  ug )
Dt
This shows, explicitly, that the acceleration, the
prognostic attribute of the atmosphere, is related to
the difference from the geostrophic balance.

A simple prognostic equation
Du
 f (v  v g )
Dt
Dv
  f (u  u g )
Dt
Looks like we are getting towards two variables u
and v, but we have buried the pressure and density
in geostrophic balance. This links, directly, the
mass field and velocity field. And what about w?
Geostrophic and observed wind 1000 mb (land)
Scale analysis of the vertical
momentum equation
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
Psfc
W*U/L
U*U/a
H
g
Uf
W
2
H
Scale analysis of the vertical
momentum equation
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
W*U/L
U*U/a
Psfc
H
10-7
10-5
10

g
Uf
10
10-3
Largest Terms
W
2
H
10-15
How did we get that vertical scale for
pressure?
Troposphere: depth
~ H = 1.0 x 104 m
ΔP ~ 900 mb
This scale analysis tells us that the troposphere is thin relative to the size of
the Earth and that mountains extend half way through the troposphere.
Hydrostatic relation
1 p
 g
 z
And the vertical acceleration Dw/Dt is 8 orders of
magnitude smaller than this balance.
So the ability to use the vertical momentum
equation to estimate w is essentially nonexistent.
Hydrostatic relation
1 p
 g
 z
Hence:
w must be “diagnosed” from some balance.
exception: small scales: thunderstorms, tornadoes
Non-hydrostatic scale of ~ 10000 m, 10 km.
Momentum equation
Du uvtan( ) uw
1 p



 2vsin(  )  2w cos( )   2 (u)
Dt
a
a
 x
Dv u 2 tan( ) vw
1 p



 2usin(  )   2 (v)
Dt
a
a
 y
Dw u 2  v 2
1 p


 g  2ucos( )   2 (w)
Dt
a
 z
D() ()
()
()
()
()

 v  () 
u
v
w
Dt
t
t
x
y
z
Currently, we have 5 independent variables
u, v, w, ρ, p and 3 momentum equations.
Can we solve this system of equations? Are we done?
Are we done?
• Currently, 5 unknowns and 3 equations
• But wait, we also discussed the ideal gas law (also
called Equation of State)
1
p  Rd T  Rd T

1
with   (specific volume)

• This adds another unknown ‘T’ to the system
• Now we have 6 unknowns (u, v, w, , p, T) and
4 equations. Are we done?
• Recall what we said about conservation laws
– Momentum
– Energy
– Mass