AOSS 401, Fall 2007Lecture 3

September 10, 2007

Richard B. Rood (Room 2525, SRB)rbrood@umich.edu

734-647-3530Derek Posselt (Room 2517D, SRB)

dposselt@umich.edu734-936-0502

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