AOSS 401, Fall 2007Lecture 2

September 7, 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)– Calendar (completed for whole semester)– Syllabus– Lectures

• Posted on day of

– Homework (and solutions)

• Homework has been posted– Under “resources” in homework folder

• Due next Wednesday (September 12, 2007)

Class news: Schedule issues

• Currently 4.5 hours are scheduled for a 4.0 hour course. (So we have some flexibility; we can “cancel” 4 classes)– There will be no class on September 14– There will be no class on October 12– There will be no class on November 21– When to schedule final exam?

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

• Pressure gradient force

• Gravitational Force

• Viscous force

• Centrifugal Force

• Coriolis ForceShould be review. So we are going fast.

You have the power to slow us down.

Some basics of the atmosphere

Earth: radius ≡ a = 6.37 x 106 m

atmosphere: depth ~ 1.0 x 105 m

Mountain: height ~ 5.0 x 103 m

Ocean Land Biosphere

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.

Newton’s Law of Motion

F = ma Force = mass x acceleration

In general we will work with force per unit mass; hence,

a = F/m

And with the definition of acceleration

Bold will represent vectors.

Newton’s Law of Motion

mdt

d/F

v

Which is the vector form of the momentum equation.(Conservation of momentum)

What are the forces?

• Pressure gradient force• Gravitational force• Viscous force• Apparent forces• Can you think of other classical forces and

would they be important in the Earth’s atmosphere?

• Total Force is the sum of all of these forces.

Newton’s Law of Motion

i

imdt

dF

v 1

Where i represents the different types of forces.

How do we express the forces?

• In general, we assume the existence of an idealized parcel or “particle” of fluid.

• We calculate the forces on this idealized parcel.• We take the limit of this parcel being

infinitesimally small.– This yields a continuous, as opposed to discrete,

expression of the force.

• Use the concept of the continuum to extend this notion to the entire fluid domain.

An intrinsic assumption

• There is an equation of state that describes the thermodynamic properties of the fluid, the air.

A particle of atmosphere

x

y

z

≡ density = mass per unit volume (V)

V = xyz

m = xyz

-------------------------------------

p ≡ pressure = force per unit area acting on the particle of atmosphere

xy

z

ij

k

Pressure gradient force (1)

x

y

z

p0 = pressure at (x0, y0, z0)

.

(x0, y0, z0)

x axis

p = p0 + (∂p/∂x)x/2 + higher order terms

Pressure gradient force (2)

x

y

z

.

x axis

p = p0 + (∂p/∂x)x/2 + higher order terms

p = p0 - (∂p/∂x)x/2 + higher order terms

Pressure gradient force (3)(ignore higher order terms)

x

y

z

.

x axis

FBx = (p0 - (∂p/∂x)x/2) (yz)

FAx = - (p0 + (∂p/∂x)x/2) (yz)

AB

Pressure gradient force (4) Total x force

= (p0 - (∂p/∂x)x/2) (yz) - (p0 + (∂p/∂x)x/2) (yz)

Fx = FBx + FAx

= - (∂p/∂x)(xyz)

We want force per unit mass

Fx/m = - 1/ (∂p/∂x)

Vector pressure gradient force

pm

z

p

y

p

x

pm

1/

)(1

/

F

kjiF

xy

z

ij

k

Viscous force (1)

• There is in a fluid friction, which resists the flow. It is dissipative, and if the fluid is not otherwise forced, It will slow the fluid and bring it to rest. Away from boundaries in the atmosphere this frictional force is often small, and it is often ignored. (We will revisit this as we learn more.)

• Close to the boundaries, however, we have to consider friction.

Derivation is at end of lecture or in the text.

Viscous force (2)

Ocean Land Biosphere

velocity ≡ u m/sec

velocity must be 0 at surface

Viscous force (3)

Ocean Land Biosphere

velocity ≡ u m/sec

Velocity is zero at the surface; hence, there is some velocity profile.

Viscous force (4)(How do we think about this?)

The drag on the moving plate is the same as the force required to keep the plate moving. It is proportional to the area (A), proportional to the velocity of the plate, and inversely proportional to the

distance between the plates; hence,

Proportional usually means we assume linear relationship. This is a model based on observation, and it is an approximation. This is often said to be “Newtonian.” The constant of proportionality assumes some physical units. What are they?

u(0) = 0

u(z)

u(h) = u0

h F = μAu0/h

Viscous force (5)

)(2 uF

m

Where Laplacian is operating on velocity vector ≡ u = (u, v, w)

≡ /kinematic viscosity coefficient

Surface forces

• Pressure gradient force and the viscous force are examples of a surface force.

• Surface forces are proportional to the area of the surface of our particle of atmosphere.

• Surface forces are independent of the mass of the particle of atmosphere.

• They depend on characteristics of the particle of atmosphere; characteristics of the flow.

Highs and Lows

Motion initiated by pressure gradient

Opposed by viscosity

Where’s the low pressure?

Geostrophic and observed wind 1000 mb (ocean)

Body forces

• Body forces act on the center of mass of the parcel of fluid.

• Magnitude of the force is proportional to the mass of the parcel.

• The body force of interest to dynamic meteorology is gravity.

Newton’s Law of Gravitation

rr

mGm rF

221

Newton’s Law of Gravitation: The force between any two particles having masses m1 and m2 separated by distance r is an attraction acting along the line joining the particles and has the magnitude proportional to G, the universal gravitation constant.

Gravitational Force

Gravitational force for dynamic meteorology

rr

GMm rF

2

Newton’s Law of Gravitation: M = mass of Earth m = mass of air parcel r = distance from center (of mass) of Earth directed down, towards Earth, hence - sign

Adaptation to dynamical meteorology

zara

GMg

20

a is radius of the Earthz is height above the Earth’s surface

Can we ignore z, the height above the surface? How would you make that argument?

Gravity for Earth

a2

=g0

a

mg0

a

Gravitational force per unit mass

rza

ag

m

rF2

2

0 )(

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

Back to Basics:Newton’s Laws of Motion

• Law 1: Bodies in motion remain in motion with the same velocity, and bodies at rest remain at rest, unless acted upon by unbalanced forces.

• Law 2: The rate of change of momentum of a body with time is equal to the vector sum of all forces acting upon the body and is the same direction.

• Law 3: For every action (force) there is and equal and opposite reaction.

Back to basics:A couple of definitions

• Newton’s laws assume we have an “inertial” coordinate system; that is, and absolute frame of reference – fixed, absolutely, in space.

• Velocity is the change in position of a particle (or parcel). It is a vector and can vary either by a change in magnitude (speed) or direction.

Apparent forces:A mathematical approach

• Non-inertial, non-absolute coordinate system

Two coordinate systems

xy

z

x’

y’

z’

Can describe the velocity and forces (acceleration) in either coordinate system.

dt

dor

dt

d 'xx

One coordinate system related to another by:

ztCytBxtAz

ztCytBxtAy

ztCytBxtAx

zzz

yyy

xxx

)()()('

)()()('

)()()('

Velocity (x’ direction)

zdt

tCdy

dt

tBdx

dt

tAd

dt

dztC

dt

dytB

dt

dxtA

dt

dx

ztCytBxtAx

xxxxxx

xxx

))(())(())(()()()(

'

)()()('

So we have the velocity 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.

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.

Acceleration (force) in rotating coordinate system

0'

))cos()sin(())sin()cos((2'

))sin()cos(())cos()sin((2'

2

2

22

2

22

2

dt

zd

tytxtdt

dyt

dt

dx

dt

yd

tytxtdt

dyt

dt

dx

dt

xd

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.

Apparent forces:A physical approach

Circle Basics

ω

θ

s = rθ

r (radius)

Arc length ≡ s = rθ

dt

dsv

dt

ddt

dr

dt

ds

... Magnitude

Centrifugal force:Treatment from Holton

rdt

d

dt

d rv

v

ω

ΔθΔv

... Magnitude

r (radius)

Centrifugal force: for our purposes

rv

v

rv

v

2

dt

d

rdt

drdt

d

dt

d

Now we are going to think about the Earth

• The preceding was a schematic to think about the centrifugal acceleration problem. Note that the r vector above and below are not the same!

What direction does the Earth’s centrifugal force point?

Ω

Ω2RR

Earth

What direction does gravity point?

Ω

R

Earth

a

rza

ag

m

rF2

2

0 )(

What direction does the Earth’s centrifugal force point?

Ω

Ω2RR

Earth

So there is a component that is in the same coordinate direction as gravity (and local vertical).

And there is a component pointing towards the equator

We are now explicitly considering a coordinate system tangent to the Earth’s surface.

What direction does the Earth’s centrifugal force point?

Ω

Ω2RR

Earth

So there is a component that is in the same coordinate direction as gravity:

~ aΩ2cos2()

And there is a component pointing towards the equator

~ - aΩ2cos()sin()

Φ = latitude

So we re-define gravity as

rg

m

ra

za

ag

m

rF

rF

))(cos)(

( 222

2

0

What direction does the Earth’s centrifugal force point?

Ω

Ω2R

R

Earth

And there is a component pointing towards the equator.

The Earth has bulged to compensate for the equatorward component.

Hence we don’t have to consider the horizontal component explicitly.

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

Next time

• Coriolis Force– Read and re-read the section in the text.

• Pressure as a vertical coordinate– Geopotential

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 viscous force

Return to lecture body.

Viscous force (1)

• There is in a fluid friction, which resists the flow. It is dissipative, and if the fluid is not otherwise forced, It will slow the fluid and bring it to rest. Away from boundaries in the atmosphere this frictional force is often small, and it is often ignored. (We will revisit this as we learn more.)

• Close to the boundaries, however, we have to consider friction.

Return to lecture body.

Viscous force (2)

Ocean Land Biosphere

velocity ≡ u m/sec

velocity must be 0 at surface

Viscous force (3)

Ocean Land Biosphere

velocity ≡ u m/sec

Velocity is zero at the surface; hence, there is some velocity profile.

Viscous force (4)(How do we think about this?)

Moving plate with velocity u0

Linear velocity profile

u(0) = 0

u(z)

u(h) = u0

h u(z) = (u0-u(0))/h × z

Viscous force (5)(How do we think about this?)

The drag on the moving plate is the same of as the force required to keep the plate moving. It is proportional to the area (A), proportional to the velocity of the plate, and inversely proportional

to the distance between the plates; hence,

Proportional usually means we assume linear relationship. This is a model based on observation, and it is an approximation. The constant of proportionality assumes some physical units. What are they?

u(0) = 0

u(z)

u(h) = u0

h F = μAu0/h

Viscous force (6)(How do we think about this?)

Recognize the u0/h can be represented by ∂u/∂z

Force per unit area is F/A is defined as shearing stress (). Like pressure the shearing stress is proportional to area.

u(0) = 0

u(z)

u(h) = u0

h F = μA(∂u/∂z)

Viscous force (7)(How do we think about this?)

zx = μ(∂u/∂z), which is the viscous force per unit area in the x direction, due to the variation of velocity in the z direction

Force per unit area; hence, like pressure.

u(0) = 0

u(z)

u(h) = u0

h Fzx/A = μ(∂u/∂z) ≡ zx

Viscous force (8)(Do the same thing we did for pressure)

x

y

z

zx0 = stress at (x0, y0, z0)

.(x0, y0, z0)

zx = zx0 + (∂zx/∂z)z/2

zx = -(zx0 - (∂zx/∂z)z/2)

Viscous force (9)

x

y

z

.

C

D

FCzx = (zx0 + (∂zx/∂z)z/2)yx

FDzx = -(zx0 - (∂zx/∂z)z/2)yx

Viscous force (10)

= (∂zx/∂z)zyx

Fzx = FCzx + FDzx

We want force per unit mass

Fzx/m = 1/ ∂zx/∂z

Viscous force (11)(using definition of )

)(1

z

u

zm

Fzx

Viscous force (12)

2

2

2

2

z

u

z

u

m

Fzx

Assume μ constant

≡ /kinematic viscosity coefficient

Viscous force (13)

)(2

2

2

2

2

2

z

u

y

u

x

u

m

Frx

Do same for other directions of shear (variation of velocity)

Viscous force (14)

)(

)vvv

(

2

2

2

2

2

2

2

2

2

2

2

2

z

w

y

w

x

w

m

F

zyxm

F

rz

ry

Do same for other directions of force

velocity vector ≡ u = (u,v,w)

Viscous force (15)

)(2 uF

m

Where Laplacian is operating on velocity vector ≡ u = (u, v, w)

Return to lecture body.

Summary

• Pressure gradient force and viscous force are examples of surface forces.

• They were proportional to the area of the surface of our particle of atmosphere.

• They are independent of the mass of the particle of atmosphere.

• They depend on characteristics of the particle of atmosphere; characteristics of the flow.

Return to lecture body.

Our surface forces

)(1 2 u

u p

dt

d

other forces

Now using the text’s convention that the velocity is u = (u, v, w).

Return to lecture body.

Highs and Lows

Motion initiated by pressure gradient

Opposed by viscosity

Return to lecture body.