AOSS 401, Fall 2007 Lecture 3 September 10 , 2007
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Transcript of AOSS 401, Fall 2007 Lecture 3 September 10 , 2007
AOSS 401, Fall 2007Lecture 3
September 10, 2007
Richard B. Rood (Room 2525, SRB)[email protected]
734-647-3530Derek Posselt (Room 2517D, SRB)
Class News
• Ctools site (AOSS 401 001 F07)– A PDF of my Meeting Maker Calendar is Posted.
• It has when I am in and out of town.• It has my cell phone number.• When I am out of town, I plan to be available for my Tues-
Thursday office hours.• Write or call
• Homework has been posted– Under “resources” in homework folder
• Due Wednesday (September 12, 2007)
Weather
• National Weather Service– http://www.nws.noaa.gov/– Model forecasts:
http://www.hpc.ncep.noaa.gov/basicwx/day0-7loop.html
• Weather Underground– http://www.wunderground.com/cgi-bin/findweather/getForecast?
query=ann+arbor
– Model forecasts: http://www.wunderground.com/modelmaps/maps.asp?model=NAM&domain=US
Outline
• Review
• Coriolis Force
• Vertical structure and vertical coordinate
Should be review. So we are going fast.You have the power to slow us down.
From last time
Our momentum equation
rza
agp
dt
d ru
u2
2
02
)()(
1
+ other forces
Now using the text’s convention that the velocity is u = (u, v, w).
Apparent forces
Two coordinate systems
xy
z
x’
y’
z’
Can describe the velocity and forces (acceleration) in either coordinate system.
dt
dor
dt
d 'xx
Two coordinate systems
y
zz’ axis is the same as z, and there is rotation of the x’ and y’ axis
z’
y’
x’
x
Apparent forces
• With one coordinate system moving relative to the other, we have the velocity of a particle relative to the coordinate system and the velocity of one coordinate system relative to the other.
• This velocity of one coordinate system relative to the other leads to apparent forces. They are real, observable forces to the observer in the moving coordinate system.– The apparent forces that are proportional to rotation and the
velocities in the inertial system (x,y,z) are called the Coriolis forces.
– The apparent forces that are proportional to the square of the rotation and position are called centrifugal forces.
Centrifugal force of Earth
• Vertical component incorporated into re-definition of gravity.
• Horizontal component does not need to be considered when we consider a coordinate system tangent to the Earth’s surface, because the Earth has bulged to compensate for this force.
• Hence, centrifugal force does not appear EXPLICITLY in the equations.
Apparent forces:A physical approach
• Coriolis Force
• http://climateknowledge.org/figures/AOSS401_coriolis.mov
Two coordinate systems
y
zz’ axis is the same as z, and there is rotation of the x’ and y’ axis
z’
y’
x’
x
One coordinate system related to another by:
T
zz
tytxy
tytxx
2
'
)cos()sin('
)sin()cos('
T is time needed to complete rotation.
Circle Basics
ω
θ
s = rθ
r (radius)
Arc length ≡ s = rθ
dt
dsv
dt
ddt
dr
dt
ds
... Magnitude
Angular momentum
• Like momentum, angular momentum is conserved in the absence of torques (forces) which change the angular momentum.
• This comes from considering the conservation of momentum of a body in constant body rotation in the polar coordinate system.
• If this seems obscure or is cloudy, need to review a introductory physics text.
Angular speed
rvω
ΔθΔvr (radius)
v
v
What direction does the Earth’s centrifugal force point?
Ω
R
Earth
R Direction away from axis of rotation
Magnitude of R the axis of rotation
Ω
R
Earth
R=acos()
Φ = latitude
a
Tangential coordinate system
Ω
R
Earth
Place a coordinate system on the surface.
x = east – west (longitude)y = north – south (latitude)z = local vertical
Φ
a
Angle between R and axes
Ω
R
Earth
Φ = latitude
a
Φ
Assume magnitude of vector in direction R
Ω
R
Earth
Φ = latitude
a
Vector of magnitude B
Vertical component
Ω
R
Earth
Φ = latitude
a
z component = Bcos()
Meridional component
Ω
R
Earth
Φ = latitude
a y component = Bsin()
Earth’s angular momentum (1)
Ω
R
Earth
Φ = latitude
a
What is the speed of this point due only to the rotation of the Earth?
Rv
Earth’s angular momentum (2)
Ω
R
Earth
Φ = latitude
a
Angular momentum is
RL v
Earth’s angular momentum (3)
Ω
R
Earth
Φ = latitude
a
Angular momentum due only to rotation of Earth is
2RL
RL
v
Earth’s angular momentum (4)
Ω
R
Earth
Φ = latitude
a
Angular momentum due only to rotation of Earth is
)(cos22
2
aL
RL
Angular momentum of parcel (1)
Ω
R
Earth
Φ = latitude
a
Assume there is some x velocity, u. Angular momentum associated with this velocity is
uRLu
Total angular momentum
Ω
R
Earth
Φ = latitude
a
Angular momentum due both to rotation of Earth and relative velocity u is
)(
))cos()(cos(
)cos()(cos
2
22
2
R
uRL
uaaL
uaaL
uRRL
Displace parcel south (1)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Let’s imagine we move our parcel of air south (or north). What happens? Δy
Displace parcel south (2)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
We get some change ΔR
Displace parcel south (3)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
But if angular momentum is conserved, then u must change.
)()(
)(
2
2
RR
uuRR
R
uRL
Displace parcel south (4)(Conservation of angular momentum)
)()()( 22
RR
uuRR
R
uR
Expand right hand side, ignore squares and higher of difference terms.
RR
uRu 2
Displace parcel south (5)(Conservation of angular momentum)
yR )sin(
yR
uyu )sin()sin(2
For our southward displacement
ya
uyu )sin(
)cos()sin(2
Displace parcel south (6)(Conservation of angular momentum)
dt
dy
a
u
dt
du))sin(
)cos()sin(2(
Divide by Δt and take the limit
)tan(v
)sin(v2 a
u
dt
du
Coriolis term (check with previous mathematical derivation … what is the same? What is different?
Displace parcel south (7)(Conservation of angular momentum)
)tan(v
)sin(v2 a
u
dt
du
What’s this? “Curvature or metric term.” It takes into account that y curves, it is defined on the surface of the Earth. More later.
Remember this is ONLY FOR a NORTH-SOUTH displacement.
Coriolis Force in Three Dimensions(link to explicit derivation)
• Do a similar analysis displacing a parcel upwards and displacing a parcel east and west.
• This approach of making a small displacement of a parcel, using conversation, and exploring the behavior of the parcel is a common method of analysis.
• This usually relies on some sort of series approximation; hence, is implicitly linear. Works when we are looking at continuous limits.
Coriolis Force in 3-D
So let’s collect together today’s apparent forces.
a
uu
dt
dw
a
uu
dt
d
a
uww
a
u
dt
du
2
2
)cos(2
)tan()sin(2v
)cos(2)tan(v
)sin(v2
Definition of Coriolis parameter (f)
)tan()sin(2
)tan(v
)sin(v2
2
a
uu
dt
dv
a
u
dt
du
Consider only the horizontal equations (assume w small)
For synoptic-scale systems in middle latitudes (weather) first terms are much larger than the second terms and we have
)sin(2
)sin(2v
v)sin(v2
f
fuudt
d
fdt
du
Our momentum equation
jikuu
fufvgpdt
d )(
1 2
+ other forces that are, more often than not, ignored
Highs and Lows
Motion initiated by pressure gradient
Opposed by viscosity
In Northern Hemisphere velocity is deflected to the right by the Coriolis force
The importance of rotation
• Non-rotating fluid– http://climateknowledge.org/figures/AOSS401
_nonrot_MIT.mpg
• Rotating fluid– http://climateknowledge.org/figures/AOSS401
_rotating_MIT.mpg
Vertical StructurePressure as a vertical coordinate
Some basics of the atmosphere
Troposphere: depth ~ 1.0 x 104 m
Troposphere------------------ ~ 2Mountain
Troposphere------------------ ~ 1.6 x 10-3
Earth radius
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.
Pressure altitude
Under virtually all conditions pressure (and density) decreases with height. ∂p/∂z < 0. That’s why it is a good vertical coordinate. If ∂p/∂z = 0, then utility as a vertical coordinate falls apart.
Use pressure as a vertical coordinate?
• What do we need.– Pressure gradient force in pressure
coordinates.– Way to express derivatives in pressure
coordinates.– Way to express vertical velocity in pressure
coordinates.
Expressing pressure gradient force
Integrate in altitude
gz
p
z
z
gdzzp
gdzzpp
)(
)()(
Pressure at height z is force (weight) of air above height z.
Concept of geopotential
kF g
gdz
d
Define a variable such that the gradient of is equal to g. This is called a potential function.
We have assumed here that is a function of only z.
Integrating with height
gdz
d
z
z
gdzz
gdzz
gdzd
0
0
)(
0)0(
)0()(
What is geopotential?
• Potential energy that a parcel would have if it was lifted from surface to the height z.
• It is analogous to the height of a pressure surface.– We seek to have an analogue for pressure on
a height surface, which will be height on a pressure surface.
Linking geopotential to pressure
p
RTdpdgdz
p
RT
dpdp
gdz
dgdz
/1Definition of specific volume
Ideal gas law
Remembering some calculus
pdTRpTdRzz
pRTdd
dpp
pd
p
p
p
p
lnln)()(
ln
1ln
2
1
1
2
12
Define geopotential height(assumption of constant g = g0)
1
2
ln
)(
012
0
p
ppTd
g
RZZ
g
zZ
Z2-Z1 = ZT ≡ Thickness - is proportional to temperature is often used in weather forecasting to determine, for instance, the rain-snow transition. (We will return to this.)
Note link of thermodynamic variables, and similarity to scale heights calculated in idealized atmospheres above.
Getting pressure gradient in pressure coordinates
x
z
Constant pressure p0
Constant pressure p0+Δp
Getting pressure gradient in pressure coordinates
x
z
Constant pressure p0
Constant pressure p0+Δp
Δx
We have, for instance, ∂p/∂x on a constant z surface in our derivation of the momentum equation.((p0+Δp)-p0)/Δx
Getting pressure gradient in pressure coordinates
x
z
Constant pressure p0
Constant pressure p0+Δp
Δz
We can also calculate how pressure changes on on a z surface as we hold x constant.
(p0-(p0 +Δp))/Δz
Getting pressure gradient in pressure coordinates
x
z
Constant pressure p0
Constant pressure p0+Δp
Δz
Which we project onto the x direction by how much z changes with x on the pressure surface. Δz/Δx
(p0-(p0 + Δp))/Δz
Getting pressure gradient in pressure coordinates
x
z
Constant pressure p0
Constant pressure p0+Δp
Δz
Δx
x
zg
x
px
zg
x
px
z
z
p
x
p
1
xx
zg
x
p
x
zg
x
p
1
1
Implicit that this is on a constant z surface
Implicit that this is on a constant p surface
Horizontal pressure gradient force in pressure coordinate is the gradient of geopotential
pz
pz
yy
p
xx
p
1
1
Our momentum equation(height (z) coordinates)
jikuu
fufvgpdt
d )(
1 2
fuy
p
dt
dv
fvx
p
dt
du
zz
zz
)1()(
)1()(
Horizontal momentum equations (u, v), no viscosity
Our horizontal momentum equation(pressure coordinate)
p
pp
pp
fDt
D
fuydt
d
fxdt
du
uku
)()v
(
v)()(
Assume no viscosity
Next time
• The material derivative.
Weather
• National Weather Service– http://www.nws.noaa.gov/– Model forecasts:
http://www.hpc.ncep.noaa.gov/basicwx/day0-7loop.html
• Weather Underground– http://www.wunderground.com/cgi-bin/findweather/getForecast?
query=ann+arbor
– Model forecasts: http://www.wunderground.com/modelmaps/maps.asp?model=NAM&domain=US
Derivation of Coriolis Force (conclusion)
return to body of lecture
Displace parcel up (1)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Let’s imagine we move our parcel of air up (or down). What happens? Δz
Displace parcel up (2)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
We get some change ΔR
Displace parcel up (3)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
We get some change ΔR
zR )cos(
For our upward displacement
Displace parcel up (4)(Conservation of angular momentum)
a
uww
dt
du )cos(2
Remember this is ONLY FOR a VERTICAL displacement.
Do the same form of derivation
return to body of lecture
Displace parcel east (1)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Let’s imagine we move our parcel of air east (or west). What happens? Δx
Displace parcel east (2)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Well, there is no change of ΔR.
But remember
)(2R
uRL
Displace parcel east (3)(Conservation of angular momentum)
• So, we have changed u (=dx/dt). Hence again we have a question of conservation of angular momentum.
• We will think about this as an excess (or deficit) of centrifugal force relative to that from the Earth alone.
Displace parcel east (4)(Conservation of angular momentum)
This excess FORCE is defined as
RR
RR
2
2
22
2
)(
R
u
R
u
R
uF lcentrifugaexcess
Displace parcel east (5)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
Vector with component in north-south and vertical direction
Displace parcel east (6)(Conservation of angular momentum)
Ω
R
Earth
Φ
a
For the Coriolis component magnitude is 2Ωu. For the curvature (or metric) term the magnitude is u2/(acos())
Displace parcel east (7)(Conservation of angular momentum)
These forces in their appropriate component directions are
a
uu
dt
dw
a
uu
dt
d
2
2
)cos(2
)tan()sin(2v
return to body of lecture