Atmospheric Science 4320 / 7320

Anthony R. Lupo

Syllabus Atmospheric Dynamics ATMS 4320 / 7320 MTWR 9:00 – 9:50 / 4 credit hrs. Location: 123 Anheuser Busch Natural Resources Building Instructor: A.R. Lupo Address: 302 E ABNR Building Phone: 884-1638 Fax: 884-5070 Email: lupo@bergeron.snr.missouri.edu or LupoA@missouri.edu Homepage: www.missouri.edu/~lupoa/author.html Class Home: www.missouri.edu/~lupoa/atms4320.html Office hours: MTWR 8:00 – 8:50 or by appointment 302 E ABNR Building Grading Policy: “Straight” 97 – 100 A+ 77 – 79 C+ 92 – 97 A 72 – 77 C 89 – 92 A- 69 – 72 C- 87 – 89 B+ 67 – 69 D+ 82 – 87 B 62 – 67 D 79 – 82 B- 60 – 62 D- < 60 F Grading Distribution: Final Exam 20% 2 Tests 40% Homework/Labs 35% Class participation 5% Attendance Policy: “Shouldn’t be an issue!” (Each unexcused absence will be charged one participation point up to

5)

Syllabus

Texts: Bluestein, H.B., 1992: Synoptic-Dynamic Meteorology in the Mid-

latitudes Vol I: Priciples of Kinematics and Dynamics. Oxford University Press, 431 pp. (Required)

Holton, J.R., 2004: An Introduction to Dynamic Meteorology, 4th Inter, 535 pp.

Hess, S.L., 1959: An Introduction to Theoretical Meteorology. Robert E. Kreiger Publishing Co., Inc., 362 pp.

Zdunkowski, W., and A. Bott, 2003: Dynamics of the Atmosphere: A course in Theoretical Meteorology. Cambridge University Press, 719 pp. (a good math review)

Various relevant articles from AMS and RMS Journals. Course Prerequisites: Atmospheric Science 1050, 4320, Calculus through Calculus II, Physics

2750, or their equivalents. Senior standing or the permission of the Instructor.

Syllabus Calendar: “Wednesday is Lab exercise day” Week 1: hh 17 18L 19 January Marty King day! Intro.

to ATMS 4320. Lab 1: Real and apparent forces: Coriolis Force. Week 2: 23 24 25L 26 January Lab 2: Methodologies

for calculating derivative in the fundamental equations of hydrodynamics. Week 3: 30 31 01L 02 February Lab 3: Estimating the

geostrophic wind Week 4: 06 07 08L 09 February Lab 4: The Nocturnal

boundary layer jet and severe weather. Week 5: 13 14 15L 16 February Test 1 – 16 February

covering materials up to 15 February. Lab 5: Using the kinematic method in estimating vertical motion.

Week 6: 20 21 22L 23 February Lab 6: The Use of isentropic coordinate maps in weather forecasting I.

Week 7: 27 28 01L 02 March Lab 7: The Use of isentropic coordinate maps in weather forecasting II.

Week 8: 06 07 08L 09 March Lab 8: The Thermal wind: Forecasting problems and the analysis of fronts.

Week 9: 13 14 15L 16 March Lab 9: Computing divergence using large data sets.

Week 10: 20 21 22L 23 March Test 2 – March 31 covering material up to 29 March. Lab 10: Vorticity and Cyclone Development.

Week 11: 27 28 29L 30 March No classes! Spring Break!!

Week 12: 03 04 05L 06 April Lab 11: Sutcliffe Methodology vs. Potential Vorticity Thinking.

Week 13: 10 11 12L 13 April Lab 12: The Rossby wave equation.

Syllabus

Week 14: 17 18 19L 20 April Lab 13: The Omega Equation, a physical interpretation.

Week 15: 24 25 26L 27 April Lab 14: Convergent/Divergent patterns associated with jet maxima: Forecasting applications.

Week 16: 01 02 03L 04 May Lab 15: The practical uses of Q-G theory in daily analysis!

Finals Week: 8 - 12 May, 2006 ATMS 4320 / 7320 Final Exam The Exam will be quasi-comprehensive. Most of the material

will come from the final third of the course, however, important concepts (which I will explicitly identify) will be tested. All tests and the final exam will use materials from the Lab excercises! Thus, all material is fair game! The final date and time is:

Thursday 11 May 2006 – 8:00 am to 10:00 am in ABNR 123

Syllabus REGULAR REGISTRATION INFORMATION - WINTER SEMESTER 2006 January 9-13, 2006 WS 2006 January 9-13 Regular Registration - WS2006 January 11 Residence halls open after 1:00 p.m. January 13 Easy Access Registration - 12 noon - 6:00 p.m. January 16 Martin Luther King Holiday-University closed *January 17 WS2006 Classwork begins (8:00 a.m.) January 17-24 Late registration - Late fee assessed beginning January 17 January 24 Last day to register, add or change sections Jan. 25-Feb. 21 Drop only January 30 Last day to change grading option February 21 Last day to drop a course without a grade March 6-24 SS2006/FS2006 Early Registration Appointment Times (currently enrolled students only)* April 15 Last day to transfer divisions March 26-April 2 Spring Break (begins at 12:00 a.m. March 26 and ends 8:00 a.m. April 2) April 3 Last day to withdraw from a course - WS2006 May 5 Winter semester classwork ends May 5 Last day to withdraw from the University - WS2006 May 6 Reading Day May 8 Final Examinations begin May 10-14 Registration - Add/Drop period for SS2006 and FS2006 (currently enrolled students) May 12 Semester ends at close of day - WS2006 **May 12-13-14 Commencement Weekend

Syllabus Special Statements: ADA Statement (reference: MU sample statement) Please do not hesitate to talk to me! If you need accommodations because of a disability, if you have emergency medical information to share with me, or

if you need special arrangements in case the building must be evacuated, please inform me immediately. Please see me privately after class, or at my office.

Office location: 302 E ABNR Building Office hours : ________________ To request academic accommodations (for example, a notetaker), students must also register with Disability

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disciplinary action. Academic honesty is fundamental to the activities and principles of a university. All members of the academic

community must be confident that each person's work has been responsibly and honorably required, developed, and presented. Any effort to gain an advantage not given to all students is dishonest whether or not the effort is successful. The academic community regards academic dishonesty as an extremely serious matter, with serious consequences that range from probation to expulsion. When in doubt about plagiarism, paraphrasing, quoting, or collaboration, consult the instructor. In cases of suspected plagiarism, the instructor is required to inform the provost. The instructor does not have discretion in deciding whether to do so.

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Syllabus

Syllabus ** Introduction to the equations of motion The fundamental equations of geophysical hydrodynamics Horizontal flow and horizontal flow approximations The Isobaric coordinate system The isentropic coordinate system The vertical variation of the geostrophic wind (Thermal wind) Some fundamental kinematic concepts Vorticity and circulation the vorticity equation and vorticity theorems Inertial Instability Introduction to Quasi-geostrophic theory Fundamentals of numerical weather prediction* * These topics will be taught if there is time (also covered in

atms 4800 / 7800). All Lecture schedules are tentative! ** Students with special need are encouraged to schedule an

appointment with me as soon as possible!

Syllabus Lab Exercise Write-up Format: All lab write-ups are due at the beginning of the next ‘lab’ Wednesday.

Grading format also given. Total of 100 pts

Name

Lab #

Atms 4320 (7320) Neatness and Grammar 10 pts

Date Due Title Introduction: brief discussion of relevant background material (5 pts) Purpose: brief discussion of why performed (5 pts) Data used: brief discussion of data used if relevant (5 pts) Procedure: (15 pts) 1. 2. Results: brief discussion of results (50 pts) observations discussion (answer all relevant questions here) Summary and Conclusions (10 pts) summary conclusions Write-ups need to be the appropriate length for the exercise done. If one section does not apply, just say

so. However, one should never exceed 6 pages for a particular write – up. That’s too much! Finally, answer all questions given in the assignment.

Day One

The Equation of Motion - Newton’s Second Law

Dynamics will focus on these equations and their various incarnations!

The generalized form of Newton’s Second Law of Motion (sometimes called “Conservation of Momentum”) can be written as:

dtFVmd

)(

Day One

(Recall Velocity - V and Force - F are VECTORS)

The differential change of momentum (following a particle):

“Impulse” )( VmddtF

Day One

If there is no “force” acting, (F 0); then no change in momentum d(mV) = 0

“Conservation of Momentum”

If: 0dtVd

Day One

Then the following MUST be true

Recall the basic form of a differential equation

0F

kssourcesdtdQ

sin

Day One

The more “familiar” form of the Second Law:

dtVd

a

where

amF

,

Day One

Thus, there are several forces which can contribute to the Left-hand-Side, and when applied to geophysical fluids (or F is defined for an oceanic or atmospheric environment) these would be called the Navier - Stokes equations (published in 1901).

These (N-S) equations are partial differential equations with an infinite number of solutions. The particular solution is dependent on the specification of boundary and initial conditions.

Day One

Even though there are infinite number of solutions, we tend to see similar solutions over and over again.

Thus, while the atmosphere is a “chaotic” system, there are elements on different time and space scales which are quite predictable as solutions tend “cluster” on a time-space phase diagram.

Day One

How did we get there?

Hold mass constant (unit mass - typical atmospheric assumption)

0

0

FdtVd

m

or

Vmdtd

Day One

Q: Where is Newton’s second law valid?

A: In an Inertial frame of reference (or coordinate system) or one that is NOT being accelerated. This could be a “fixed” coordinate system, or:

This coordinate system can be moving though, it is just moving at constant V w/r/t another inertial frame.

0V

Day One

The “other” frame sometimes is called an “absolute” frame or a “fixed frame”, but may really be neither!

Thus,

describes the motion of a particle or parcel of UNIT mass in an inertial frame of reference.

amF

Day One / Two

Absolute and relative motion on a rotating system

We observe and refer motion to our position or coordinate system on the Earth’s surface which has a velocity and acceleration in “absolute” space due to the rotation of the Earth.

Thus,sysrelabs VVV

Day One / Two

(Recall V is changing directions since the coordinate system we use is attached to the rotating Earth. Thus, it is NOT an inertial coordinate system.

Now Some Additions - The coriolis acceleration: is that portion of the total absolute

acceleration which is due to the change of orientation of the relative coordinate system as it moves relative to the absolute (rotating earth).

Day Two

The coriolis acceleration exists whether you coordinate system defines the origin on the rotating earth or at the earth’s center.

The centrifugal acceleration:

the part of the total absolute acceleration which is due to a rotating earth. It is necessary that the centrifugal force keeps a surface based point of zero relative velocity (origin) in orbit about the earth’s axis in a plane normal to the axis at a constant distance normal to the axis.

Day Two

In plain English: It only exists for a coordinate system defined on the earth’s surface, and whose origin is there. It does not exist in an absolute framework, or in a coordinate system defined at the center of the earth.

Varies as R = r cos where is the latitude. R is distance from the rotation axis, r is earth radius. Thus this force is maximum at the equator R is maximum. R is 0 at the poles.

Day Two

Recall from last week we said that:

And yesterday we said that:

(1)Coriolis Centrifugal Acceleration Acceleration

sysrelabs VVV

crelrel

abs rrVdtVd

a ˆsin22

Day Two

and we could substitute (1) in the equation of motion for dV/dt:

Now let’s put in the other real forces as we did yesterday (symbolically)

mF

dtVd

abs

Day Two

Now Newton’s law with real and apparent forces:

viscgravityFricPGFm

rrVdtVd

crelrel

1

ˆsin22

Day Two

Or

This form is similar to the inertial form, but is only valid on a rotating planet (Earth)! Note there is real and the apparent force, coriolis force, which appears as a “negative” since we included it as a forcing mechanism and not an acceleration per se!

viscgravityFricPGFm

rrVdtVd

crelrel

1

ˆsin22

Day 2

Newtonian Gravitation

Newtonian gravitation (absolute gravity):

directed between the centres of mass, proportional to the product of masses and inversely proportional to the square of the distance between the centres of mass.

r

r

rGMem

gmF abs

2

Day two

Thus for a particle of unit mass

r = re or radius of earth 6371 km.

M = Me or mass of earth 6.0 x 1024 kg

G is the universal gravitation constant G: 6.6 x 10-11 Nm2kg-2

2e

abs rGMe

g

Day 2

thus: g approximately 9.8 ms-2 at the surface of the earth!!!

Apparent gravity: the centrifugal force:

Centrifugal Force The vector sum of the two is apparent gravity!

Rgg absapp

2

Day Two

Net Force = Newtonian gravity + Centrifugal force

At equator:

at 45 N

at poles.

Rgg absapp

2

Day Two

Apparent gravity only has a direction (which is always normal to or perpendicular to the surface) in k, now we can further simplify the equation of motion by stating:

gapp. = -g k = gabs. + centrifugal force

g does not appear in the horizontal equations of motion, only in the vertical!

Day Two/Three

Thus, due to centrifugal forces, and the fact that earth is an oblate spheriod, not a sphere, the surface gravity:

at the poles: g = 9.83 ms-2

at the equator: g = 9.78 ms-2

in MO: g = 9.81 ms-2

Day Two/Three

Geopotential Surfaces

geopotential surfaces are surfaces of constant gravitational geopotential energy (Potential energy).

Potential is defined as relative to some position, in this case, the earth’s surface: z = 0 (Mean Sea Level (MSL)).

z

gzgdz0

Day Three

but, g = g(,z)

Thus, surfaces of constant geometric height (z) are not also surfaces of constant geopotential (due to the variation of g with latitude)

Surfaces of constant slope very slightly towards the poles forming oblate spheriods.

Day Three

(Careful I: some atmospheric data sets give you GEOPOTENTIAL not height!)

(Careful II: MANY atmospheric scientists use the term height and geopotential interchangably! They are not!)

Day Three

at z = 0 and = 45o g = go = 9.81 ms-2

we can express:

z z

o

oo

o

z

o

dzgdzgg

ggdz0

*

Day Three

Where:

gravity is everywhere normal to surfaces of constant geopotential and is proportional in size to the vectoral gradient of geopotential!

dzgg

dzo

*

Day Three

we have many ways of expressing apparent gravity in the equation of motion:

kz

g ˆ

Rgkgkz

g abs

2ˆ

Day Three

The Pressure Gradient Force

Consider a unit volume of air with dimensions (xyz)

Consider the force on this volume due to pressure differences in the positive x direction (the x component of the total PGF) (This argument is also valid in the y and z direction!)

Day Three

Pressure: Force / Area

Force = Pressure x Area

Total force on the LHS of the box:F1 = P1 Dy Dz

Total force on the right hand side (note sign convention!)

F2 = -P2 Dy Dz

Day Three

But P2 can also be expressed as:

So F2 is:

xx

pPx

x

pP

11

zyxx

pzyP

1

Day Three

Thus, the net force in the positive x direction: F1 + F2

--F1------ -------F2-----------------

zyxx

pzyPzyP

11

Day Three

So the net force in the x direction is:

The net force per unit mass: Total Mass = Thus the x component of the Pressure

gradient force is:

dxdydzxp

zyxxp

dxdydzzyxVol

Day Three

Net Force / Unit Mass

Then it follows that the y component is:

ixp

ixp

dxdydzdxdydz

xp ˆˆ1

jyp

jyp ˆˆ1

Day Three

And finally the z component, (recall from thermo.)

Thus the three dimensional PGF is the sum of all three components:

kzp

kzp ˆˆ1

k

zp

jyp

ixp

PGF ˆˆˆ1

Day Three

or in “Vector Notation”

where is the three dimensional Pressure gradient!

pPGF 3

1

P3

Day Three/Four

We must realize that

1) the pressure gradient force is directed from high to low pressure, this means

2) air “flows” or is pushed from high to low pressure (source to sink)

We recall from thermodynamics, and our scale analyses that the vertical component dominates:

Day Four

(1) (2) (3)

Term (1) ~ 100 hPa / km

Term (2),(3) ~ 1 hPa / 100 km

but, although the horizontal components are small, they are still very important to horizontal accelerations and motions.

yp

xp

zp

Day Four

That is why it is customary to “scale” the “z” equation of motion as the hydrostatic approximation, while the horizontal equations are analyzed separately.

jyp

ixp

p

where

kzp

pp

h

h

ˆˆ

ˆ1113

Day Four

Let’s now put in the Coriolis force, the centrifugal force and the PGF into the equations of motion

Grav. Fric Visc PGF Coriolis

This is the equation of motion! We have derived CF, gapp. and PGF.

VpFgdtVd

app

2

13

Day Four

Frictional force + Viscous force (Piexoto and Oort, 1992, p. 36: The Physics of climate) in three dimensions:

divF

Day Four

X-component

/

1

where

zU

zzyxFx zzyyxx

Day Four

Kinematic Viscosity

where ( ) is a 2nd order “stress” tensor!

Recall from thermo. a tensor has magnitude, and 2 directions (Vector is 1st order tensor)!

Day Four

Stress

zzzyzx

yzyyyx

xzxyxx

Day Four

Or we can parameterize friction (e.g. Lupo et al., 1992, MWR, Aug.) (horizontal motions):

Now put in the Equations of Motion:

Fk

ˆ

Day Four

PGF CO Grav. Friction

We now have the Navier-Stoke’s (1901) equations which represents the dynamics of a geophysical fluid (Atmosphere or Ocean).

12

13 gVp

dtVd

Day Four

Again, it is common in atmospheric science to examine the horizontal and vertical components of the equation of motion separately. These equations in component form:

(Since your homework involves grinding out Coriolis force we won’t do here)

Day Four

NS equations

Fzguz

p

dt

dw

Fyuy

p

dt

dv

Fxwvx

p

dt

du

cos21

sin21

cos2sin21

Day Four

It is also VERY common to look at changes in momentum by scaling these relationships to fit the system we are trying to describe.

This was especially true in the early days of meteorology (Taylor, Richardson, Rossby, Eliassen, etc.), when there were no fancy computers to work with. These researchers had to find ways of simplifying the equations, and making approximations in order to make their life easier!

Day Four

The “Z” component (we won’t re-derive from thermo.) is the hydrostatic relation:

which can be written as (recall from thermo.):

gzp

dzdp

gzp

,0

1

Day Four

A scale analysis (using typical synoptic-scale values) of the hoizontal equations would show that:

01

gzp

oo

Day Four

10-4 ms-2 10-3 ms-2 10-3 ms-2 10-5 ms-2 10-6 ms-2

Thus, the approximate form of the horizontal, inviscid N-S equations appopriate for synoptic and plaentary-scales (tropospheric and stratospheric flow):

FrictionCoriolisPGFdtVd

Day Four

Scalar N-S equations

sin2

10

1

1

f

where

gzp

fuyp

dtdv

fvxp

dtdu

Day Four

Vector N-S equations

kgkzp

kdzdw

jfujyp

jdtdv

ifvixp

idtdu

ˆˆ1ˆ

ˆˆ1ˆ

ˆˆ1ˆ

Vkfpdt

Vdh

h

ˆ1

Day Five

The Equations of Motion (Navier Stokes Equations) in Spherical coordinates

Coordinate systems: Cartesian (x,y,z)

Polar (r,) Cylindrical (r,,z) Spherical (,,) Natural (n,s,z)

For scales of motion that are sufficently large, we must take into account changes in the orientation of earth relative to the coordinate system over the earth’s system (ie orientation of cartesian coordinates differs at poles and equator

Day Five

The acceleration vector (N-S equations), where are treated as constants

From thermo. we said that they did not have to be so:

kji ˆ,ˆ,ˆ

kdtdw

jdtdv

idtdu

dtVd ˆˆˆ

wdtkd

vdtjd

udtid

kdtdw

jdtdv

idtdu

dtVd ˆˆˆˆˆˆ

Day Five

Then the total derivative of the Unit vector is:

(An aside: can you write out the “j” and “k” equations?)

Now in spherical coords, meteorology defines:

zi

wyi

vxi

uti

dtid

ˆˆˆˆˆ

Day Five

Math meaning meteorology

radius re latitude longitude

Visualize yourself on a sphere:

Day Five

A Sphere The Earth

Day Five

Angles

Day Five

Thus, it follows that:

yk

vxk

uzk

wyk

vxk

utk

dtkd

yj

vxj

uzj

wyj

vxj

utj

dtjd

xi

uzi

wyi

vxi

uti

dtid

ˆˆˆˆˆˆˆ

ˆˆˆˆˆˆˆ

ˆˆˆˆˆˆ

Day Five

Evaluate the partial derivatives of: “i” “j” and “k” terms, all nine!?

Let’s consider (partial “i” / partial x)

as: (i/x) the variation of i with increment of longitude

Day Five

“look down” from North pole:

Day Five

Thus, we can show (with ball) i must point inward! It must have components in the y and z components:

Magnitude in “y” dir:

cos11ˆ

limˆ

erRRx

i

xi

ee rrxi

xi

tan

sincos1

sinˆˆ

Day Five

Magnitude in “z” dir:

Then:

ee rrxi

xi 1

coscos1

cosˆˆ

kru

jr

uxi

udtid

kr

jrx

i

ee

ee

ˆˆtanˆˆ

ˆ1ˆtanˆ

Day Five

Now look at: (k/x)

Buuut,

xr

rxr

rxk

xk

rr

k e

e

e

ee

e

11ˆˆ

ˆ

xRre

Day Five/Six

So,

and the direction of re is perpendicular to k in the positive x or i direction, so:

1

xRxR

xre

irx

k

e

ˆ1ˆ

Day Six

Now consider:

We can do this using geometry, or use our cross product rules:

Since:

xj

ˆ

ikj ˆˆˆ

Day Six

Then,

Then substitute our other expressions!

xi

kixk

ikxx

j

ˆˆˆ

ˆˆˆˆ

ijk

kk

ii

but

kr

jr

kiir eee

ˆˆˆ

,0ˆˆ

,0ˆˆ

ˆ1ˆtanˆˆˆ1

Шестой днљй

Так (sooo….)

Then consider….

irx

j

e

ˆtanˆ

yj

yj

ˆˆ

ery

j ,ˆ

Day Six

Then,

and the direction is in the –k

ee ryj

ryj 1ˆˆ

kry

j

e

ˆ1ˆ

Day Six

Last, consider

We can consider this via geometry again, but let’s use our cross products!

yk

yk

ˆˆ

yj

ijyi

jiyy

k

ˆ

ˆˆˆ

ˆˆˆ

Day Six

And;

substitute these in, and finally:

,0ˆ

yi

kry

j

e

ˆ1ˆ

jry

k

e

ˆ1ˆ

Day Six

Then substitute back into the derivatives of the i,j,k vectors:

jruv

iruw

yk

vxk

udtkd

w

krv

iruv

yj

vxj

uvdtjd

v

kru

jru

xi

udtid

u

ee

ee

ee

ˆˆˆˆˆ

ˆˆtanˆˆˆ

ˆˆtanˆˆ

22

222

Day Six

Now substitute back into the Navier Stoke’s equations:

FgV

pjruv

iruw

krv

iruv

kru

jru

kdtdw

jdtdv

idtdu

dtVd

eeee

ee

2

1ˆˆˆˆtan

ˆˆtanˆˆˆ

2

22

Day Six

In component form:

Component – i:

10-4 10-5 10-7 10-3 10-3 10-5 10-6

xFFwfvxp

ruw

ruv

dtdu

ee

1

tan

Day Six

In component form:

Component – j:

10-4 10-5 10-7 10-3 10-3 10-6

yFfuyp

rvw

ru

dtdv

ee

1

tan2

Day Six

In component form:

Component – k:

10-6 10-5 10-5 10 10 10-3 10-6

zFFugzp

rv

ru

dtdw

ee

122

Day Six

The Navier Stokes equations in the Natural Coordinate system

Natural Coordinates n,s,z

s is in the direction of horizontal flow (+ with flow, - against)

n is normal to the direction of flow (+ left, - right)

Day Six

z is vertical coordinate

Thus,

The acceleration term in Natural coordinates:

sVVh ˆ

dtsd

Vsdt

Vd

dtsVd

dt

Vda hh ˆ

ˆˆ

Day Six/Seven

Thus we must look at:

For , the direction is perpendicular to s and to it’s left in the n direction.

ts

dtsd

ˆˆ

xs

xs

Day Seven

So;

thus, we know x = Rc, so;

ndtd

dtsd

or

ns

ˆˆ

ˆˆ

dtd

Rdtsd

or

Rs

c

c

ˆ

ˆ

седЬмой днЬи

and in natural coordinates the definition of:

So;

dtsd

Vh

ˆ

nR

V

dtsd

R

V

dtd

c

h

c

h

ˆˆ

,

Day Seven

сейчас(Now):

(1) (2) (1) Downstream or Tangential

Acceleration (speed)

(2) Centripetal acceleration (curvature)

n

R

Vs

dt

Vd

dt

sVd

dt

Vda

c

hh ˆˆˆ 2

Day Seven

Natural Coordinates, the pressure gradient force

Natural Coordinates: the coriolis force

nnp

ssp

ph ˆ1

ˆ11

nVfVkf hh ˆˆ

Day Seven

The horizontal invicid equations of motion in Natural Coordinates (e.g. Bell and Keyser, 1993, MWR, Jan)

The s-component, in the direction of motion:

sp

dt

Vd h

1

Day Seven

The n-component

The energy equations (Bernoulli’s equations)

The 3-D Navier Stokes equation (x,y,z):

hc

Vfnp

RV f

12

Day Seven

Dot N-S equations with V:

(For horizontal equations the result is a Kinetic Energy budget equation, Smith, Kung, Orlanski - e.g. Orlanski and Sheldon, 1993, MWR, November)

rFgVpdtVd

21

Day Seven

OK, here’s the result

dt

d

dt

dzgwgkgV

VCOVV

pVpV

V

dt

dVV

dt

d

dt

VdV

ˆ

,02

11

22

2

Day Seven

Recall dot product rules?

(Note Coriolis vector disappears, dot product rule, coriolis force contributes nothing to changes in Kinetic energy)

Day Seven/Eight

so (following Orlanski and Sheldon),

(1) (2)

rFVpVgzV

dtd

rFVdtd

pVV

dtd

12

122

2

Day Seven / Eight

where,

= geopotential

V2/2 = Kinetic energy per unit mass

(1) is the work done per unit mass by PGF

(2) is the work done per unit mass by FRIC

Day Eight

Now Recall the First Law:

add KE equation and first law:

dtd

pdtdT

cQ v

rFVpVdtd

pQTcgzV

dtd

v

2

2

Day Eight

Now take definition of change of pressure with time (multiply by ):

now substitute the above into the energy equation (where class?)

pV

tp

dtdp

rFVtp

dtdp

dtd

pQTcgzV

dtd

v

2

2

Day Eight

we get (do you see the “product rule”?):

but since (using Eqn of state):

rFVtp

QpTcgzV

dtd

v

2

2

TcpTc pv

Day Eight

Generalized Bernoulli Eq.

and if we assume, adiabatic, inviscid, and steady state flow, then what?

rFVtp

QTcgzV

dtd

p

2

2

02

2

Tcgz

Vdtd

p

Day Eight

and then….

Moist Static Energy Equation

If the diabatic heating is assumed to involve condensation and evaporative processes within a parcel of air,

tConsTcgzV

p tan2

2

Day Eight

Then;

where m is mixing ratio and L is the latent heat of condensation/evaporation.

dtLmd

dtdh

Q

Day Eight

Then our generalized Bernoulli equation becomes:

Or;

rFV

tp

dtLmd

TcgzV

dtd

p

2

2

rFVtp

LmTcgzV

dtd

p

2

2

Day Eight

and, of course we can assume again, steady state and inviscid flow:

or (total energy remains constant along the trajectory)

02

2

LmTcgz

Vdtd

p

tConsLmTcgzV

p tan2

2

Day Eight

Consider these typical orders of magnitude:

V ~ 15 m/s V2/2~102 m2/s2 ~ 102 J/kg

g z ~ (3000m) g z ~3x104 m2/s2 ~ 3x104 J/kg

CpT ~ (270 K) CpT~2.7x105 m2/s2 ~ 2.7x105 J/kg

L m ~ (4 g/kg) L m~104 m2/s2 ~ 104 J/kg

Day Eight

If the total energy E is:

102 104 105 104

Moist static energy!

LmTcgz

VE pT 2

2

Day Eight

It is typical to neglect V2 /2 term, thus we could derive this quantity exclusively from the 1st law of thermodynamics for an inviscid flow.

Concept of moist static energy is important when talking about the energy balance of the general circulation, especially that of the tropics.

0

2

2

dtdE

constLmTcgzV

EE pmStatT

Day Eight

Also, it should be obvious that if we examine an adiabatic atmosphere, we’re back to dry static energy, a concept we studied in thermo. (from which we can derive dry adiabatic lapse rate).

Geopotential temperature (dynamic potential temperature):

divide all terms by Cp, and viola!:

Day Eight

This is the temperature an arbitrary sample of air would have after adiabatic decent to z=0 or sealevel!! (Similar concept to potential temperature)

Q: Is this the same as potential temperature (whose reference level is 1000 hPa)?

A: No! 1000 hPa not always at z=0 in the real atmosphere. But where 1000 hPa is at z =0, they are equal!

Day Eight

You say: Prove it! 12Z 29 Friday, 1999 data near British Columbia-Alberta border

T700 = 264K SLP = 1000 hPa Z700 = 2820 m

Now consider case where latent energy is added or lost, the non-adiabatic case is (saturated):

And,

Now we can talk about geopotential equivalent potential temperature

Day Eight

This is conserved for both saturated and unsaturated adiabatic process including air which is being evaporatively cooled by falling rain.

are similarly conservative and

porportional to their thermodynamic counter parts, but are simpler to calculate.

Day Eight

The practical Applications of Geopotential Temperature

Storm-scale --> updrafts and downdrafts

Static Energy Index ( )

Layers in which are negative, possess potential convective instability, with respect to geopotential temperature.

Day Eight

The Coupling Index (Bosart and Lackmann, 1995; Lupo et al; 2001)

Calculates a “lapse rate” using a moist low level, thus identifying regions of moist instability. Values lower than ‘10’ indicate very unstable air, and are associated with vigourous cyclogenesis. Can be identified using geopotential temperature also.

Subsynoptic and synoptic scale:

areas of maximum geopotential temperature at low levels indicate maximum updraft potential.

Day Nine

The Equation of Continuity (Hess, p212 ch 13)!!

This is an expression of the principle of the conservation of mass.

In the cartesian system:

Day Nine

A Cube:

Day Nine

Consider the mass flux though the faces of a cube.

The flux of any quantity Q:

‘flux’ = QV

Day Nine

mass flux:

the total mass flux into the left face

smkg

sm

mkg

V 23

dydzuorzyu 11

Day Nine

the total mass flux out of the right face (recall convention)

but

dydzuorzyu 22

dxxu

uxxu

uu

112

Day Nine

so the mass out of the right face:

the net inflow of mass due to the u-component of the flow equals:

mass into the left - mass out of the right

dydzdxxu

u

1

Day Nine

The result?

the net mass increment per unit volume in the x direction (u - component)

dxdydzxu

dydzudydzu 11

Day Nine

After cancellation:

where dV = dxdydz

xu

dVdxdydzxu

/)(

Day Nine

the net mass increment per unit volume in the y direction (v - component)

the net mass increment per unit volume in the z direction (w component)

yv

zw

Day Nine

the total increment in mass per unit volume per unit time:

so, t

Vzw

yv

xu

t

Day Nine

thus,

which is the equation of continuity!

This represents the local rate of change in mass inside the cube equals the net divergence or convergence of mass.

Vt

Day Nine

Mass increase mass convergence

(Hess way: more molecules stuffed in the box, which hasn’t changed size!)

Mass decrease mass divergence

Day Nine

Alternative expressions for the equation of continuity

recall

ABBABA

Day Nine

Then:

Thus,

VVVf

VVt

f

Day Nine

Or

But since:

VVt

f

dt

dV

t

Day Nine

the continuity equation may be rewritten as:

or

Vdtd

V

dtd

or

Vdtd

ln

1

Day Nine

The fractional increment in the density of air following along a parcel trajectory is due to the 3- D convergence

Specific Volume form (Oceanography)

Day Nine

Here it is!

Vdtd

V

dtd

or

Vdtd

ln

1

Day Ten

The equation of continuity from Bluestein (p 190 - 193 Bluestein) another view.

Assume constant density = o:

Vdtd

Day Ten

Then (“it’s the size of the box, stupid!”):

0,0 Vdtd

Day Ten

Convergence:

Day Ten

Divergence:

Day Ten

Water is nearly incompressible, so oceanographers can use constant density form. The assumption is also reasonable for most atmospheric applications. Thus, let’s consider the continuity equation in constant pressure coordinates.

Day Ten/11

In a hydrostatic atmosphere:

(x,y,z) (x,y,p)

0

gdV

dtd

dVdtd

Day 11

Thus, by substituting for:

it follows that;

gp

zzw

Vhh

0

pVhh

Day 11

This was derived by Sutcliffe and Godart (1942), thus making the continuity equation a Diagnostic equation.

Diagnostic No time derivatives appear in the equation! (or they approach 0 in a more formal definition)

0,0 dtdQ

dtdQ

Day 11

Prognostic Time derivatives are explicit in the equation

This reduces the problem to two dimensions, since a hydrostatically balanced atmosphere must conserve the mass between two pressure levels. So our material behaves as an incompressible fluid!

SiSodtdQ

Day 11

Continuity in isentropic coordinates:

(1) (2) (3)

(1) represents the local change in the inverse static stability. (Sz)

(2) represents the horizontal flux of inverse static stability

0

dtd

SVStS

zzz

Day 11

(3) represents the vertical flux of inverse static stability

Thus we can think of this equation

as a static stability tendency equation as well.

p

S z

Day 11

The equation of moisture Continuity

This is a statement of the conservation of water vapor mass per unit mass of air

A differential equation describing water vapor budget (coming from Geophysical Fluid dyn.):

Sdtdq

Day 11

where:

S = Sources and Sinks

S1 evaporation and sublimation,

S2 condensation and precipitation

Day 11

Where conservation of water vapor is:

0 Sdtdq

Day 11

Which is commonly examined by examining horizontal and vertical transport (advection).

pq

qVtq

hh

Day 11

Then we can rewrite in “flux” form (as is done in studies of the General Circulation budget)

pq

qVtq

hh

Day 11

So, Let’s restate the fundamental equations of geophysical fluid dynamics. In this class we’ll look at:

Equation of State (Elemental Kinetic Theory or Gasses):

RTP

Day 11

Conservation of Energy (1st Law of Thermodynamics):

dtdp

dtdT

cQ p

Day 11

Conservation of mass (Continuity):

Dry

Moist

Vdtd

pq

qVtq

hh

Day 11

Equation of Motion (also Navier Stokes, Newton’s 2nd Law, Conservation of Momentum:

Fzguz

p

dt

dw

Fyuy

p

dt

dv

Fxwvx

p

dt

du

cos21

sin21

cos2sin21

Day 11

These equations are also called the “primitive equations and represent a closed set (7 variables [u,v,w(), p, T(), , q] , 7 equations. This is a mathematicallly solvable system which, given the proper initial and boundary conditions will yield all future states of the system.

Day 11/12

Horizontal Flow

we have already derived a set of equations which represents the most rudimentary, but yet realistic, approximation of horizontal flow.

Geostrophic flow in cartesian coords (x,y,p!):

Day 12

Geostropic wind!

kf

Vgˆ1

xxfv

yyfu

g

g

1

1

Day 12

In natural coordinates:

(We need to discuss the properties of a geostrophically balanced system or geostrophic wind)

np

fV

nnp

p

VfpdtVd

g

h

geoh

1

ˆ

10

Day 12

Geostrophic Wind

horizontal and non-accelerating and can be calculated where ever pressure gradient force and Coriolis force exist.

normal to the pressure gradient force, with low (high) pressure to the left (right)

0dtVd

Day 12

magnitude is directly proportional to pressure gradient force (packing of isobars)

Virtually non-divergent (though there is a small amount of divergence and Helmholtz partitioning would demonstrate this). Simplest way to show:

Day 12

non-divergent (f = fo):

y

v

x

uV gg

hh

0

xyfxfyy

v

yxfyfxx

u

oo

g

oo

g

2

2

11

11

Day 12

And if we add this up, does it equal 0?

small divergences (f varies – div-ergence on the order of 10-4, 10-5 s-

1):

Day 12

Here:

xy

f

fxyfxfyy

v

yx

f

fyxfyfxx

u

g

g

2

2

2

2

111

111

Day 12

Adds up to be:

a term on order of 10-6 or 10-7 s-1.

f

v

xyf

fg

2

1

Day 12

Balanced system

Geostrophic Balance (Is by strictest definition, a steady state system). Thus, disturbances (energy) cannot be generated in such an atmosphere, just moved around!)

In the strictest sense, this atmosphere (if f = fo) is barotropic! (Wind speeds are the same at all heights

Day 12

Wind speed profile is constant, and there are no vertical motions!

Rossby Number (Ro)

Is a scaling parameter, or measure of geostrophic approximation validity:

Ratio of Accelerations to coriolis force.

Day 12

The N-S equations in symbolic form:

Thus, the Rossby number is a measure of the departure from geostrophic flow:

1.

.

Co

Accel

CO

PGF

AccelCOPGF

Ro = U/fL

Day 12

where U is the mean zonal wind

f is the Coriolis parameter

L is the characteristic length scale of an Atmospheric disturbance (synoptic-scale)

Day 12

If Ro is << 1 flow is nearly geostrophic

If Ro = 0.1 flow is with 10% of geostrophic balance

If Ro = 1 then geostrophic balance begins to fail

Day 12

Consequence of geostrophic balance (Pedlosky p43ff)

Taylor-Proudman theorem

If the atmosphere is in approximate geostrophic and hydrostatic balance, and flow is inviscid, and the baroclinic vector is zero!

02 p

Day 12

Then velocity is perpendicular to rotation vector () and must always be so!

Flow is non-divergent!

Vorticity lines will always be parallel to rotation vector.

V0 hh V

Day 12

Since flow is incompressible then:

Thus, the motions are COMPLETELY 2-dimensional!

0

zw

yv

xu

Day 12

And

0,,

zv

zu

zw

Day 12

Taylor’s experiment (1923):

Day 12