Atmospheric Science 4310 /
7310
Atmospheric Thermodynamics
By
Anthony R. Lupo
Syllabus
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Atmospheric Thermodynamics
ATMS 4310
MTWR 9:00 – 9:50 / 4 credit hrs.
Location: 1-120 Agruculture Building
Class Ref#: 15505
Instructor:
A.R. Lupo
Address:
302 E ABNR Building
Phone:
88-41638
Fax:
88-45070
Email:
lupo@bergeron.snr.missouri.edu or LupoA@missouri.edu
Homepage:
www.missouri.edu/~lupoa/author.html
Class Homepage:
www.missouri.edu/~lupoa/atms4310.html
Office hours: MTWR 10:00 – 10:50
302 E ABNR Building
Syllabus
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Grading Policy: “Straight”
97 – 100
A+
92 – 97
A
89 – 92
A87 – 89
B+
82 – 87
B
79 – 82
B< 60
F
77 – 79
72 – 77
69 – 72
67 – 69
62 – 67
60 – 62
C+
C
CD+
D
D-
Grading Distribution:
Final Exam
20%
2 Tests
40%
Homework/Labs
35%
Class participation
5% (Note, you WILL lose 1 point for each unexcused
absence, up to 5 points. This IS a half-letter grade, keep that in mind!)
Attendance Policy: “Shouldn’t be an issue!”
Syllabus
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Texts:
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Holton, J.R., 2004: An Introduction to Dynamic Meteorology, 4th Inter, 535 pp.
(Required)
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Bluestein, H.B., 1992: Synoptic-Dynamic Meteorology in the Mid-latitudes Vol I:
Priciples of Kinematics and Dynamics. Oxford University Press, 431 pp.
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Hess, S.L., 1959: An Introduction to Theoretical Meteorology. Robert E. Kreiger
Publishing Co., Inc., 362 pp.
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Zdunkowski, W., and A. Bott, 2003: Dynamics of the Atmosphere: A course in
Theoretical
Meteorology. Cambridge University Press, 719 pp. (a good math review)
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Zdunkowski, W., and A. Bott, 2004: Thermodynamics of the Atmosphere: A course in
Theoretical Meteorology. Cambridge University Press, 251 pp.
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Various relevant articles from AMS and RMS Journals.
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Course Prerequisites:
Atmospheric Science 1050, Calculus through Math 1700, Physics 2750, or their
equivalents. Senior standing or the permission of the Instructor.
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Syllabus
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Calendar: “Wednesday is Lab exercise day”
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Week 1:
21
22
23
24
August  Introduction and
Friday makeup arrangements. Intro. To Atms. 4310. Lab 1: The Thermodynamic diagram and
upper air information.
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Week 2:
28
29
30
31
August / September  Fri.
makeup 1, 1 September. Lab 2: Adiabatic Motions in the Atmosphere.
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Week 3:
hh
5
6
7
September  Mon., Labour Day
Holiday / Fri makeup 2, 8 September. Lab 3: The Thermodynamic Diagram: Examining Moist
Processes.
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Week 4:
11
12
13
14
September  Fri. makeup 3, 15
September, Lab 5 Lab 4: The Thickness Equation and it’s Uses in Operational Meteorology. (move
up other labs)
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Week 5:
18
19
20
21
September  Friday make up 4,
22 September, Lab 5 the Lapse Rates of Special Atmospheres. Test 1 22 Sept., covering material
to 19 Sept.?
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Week 6:
25
26
27
28
September  Friday 29
September makeup 5.
Lab 6: Using Thermodynamic diagrams to Determine Water Vapor Variables.
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Week 7:
2
3
4
5
October  Friday makeup 6, 6
October. Lab 7: Estimating Vertical Motions Using the First Law of Thermodynamics.
Syllabus
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Week 8: nn
nn
nn
nn
October  No Class, UCAR-NCAR member rep meetings
and Heads and Chairs. Lab 8: Atmospheric Stability I: Special Forecasting Problems: Fog Formation.
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Week 9: nn
nn
nn
nn
October  Gone to Cleveland, OH – NWA meet.
Lab 9: Atmospheric Stability II: Special Forecasting Problems: Air Pollution.
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Week 10: 23
24
25
26
Lab 10: Severe Weather: The Synoptic-Scale sets the table.
October  Makeup 7, 27 October
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Week 11: 30
31
1
2
October / November  Makeup 8, 3 November Test covering
material to 1 November. Lab 11: Using Thermodynamic diagrams in forecasting Convective Outbreaks.
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Week 12: nn
nn
nn
nn
Louis, MO. Lab 12: Estimating Various Stability Indicies in real-time.
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Week 13: 13
14
15
16
November  Makeup number 9, 17 November Lab 13:
Severe Weather I: Using thermodynamic diagrams: Super Cell Formation and Wind Gust Estimation.
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Week 14: hh
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Week 15: 27
28
Lab 14: Severe Weather II:
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Week 16: 4
5
6
7
December  Makeup 11, 8 Dec., Final 8 Dec.?
Lab 15: Severe Weather III: Using thermodynamic diagrams: Maximum Windgust and Microburst.
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Finals Week:
hh
hh
hh
November  Severe and Local Storms Conference in Saint,
November  No classes Turkey day week!
29
30
November / December  Make up number 10, 1 December.
Using thermodynamic diagrams: Hail Formation
11 – 15 December
Syllabus
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ATMS 4310 Final Exam
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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:
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Friday, 15 December 2006 – 10:30 am to 12:30 pm in ABNR 1-120
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University Important Dates Calendar
August 14-18 FS2006 Regular Registration
August 16 Residence Halls open 9:00 a.m.
August 18 Easy Access registration - noon - 6:00 p.m.
August 21 Classwork begins 8:00 a.m.
August 21 Late Registration and Add/Drop - Late fee assessed
beginning August 21
August 28 Last day to register, add, or change sections
August 29-Sept. 25 Drop Only
September 4 Labor day Holiday
September 5 Last day to change grading option
September 18 (Census Day) - Last day to register for CDIS courses
for Fall
September 25 Last day to drop course without grade
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Syllabus
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TBA WS2007 Early Registration Appointments
October 30 Last day to withdraw from a course - FS2006
November 15 Last day to change divisions
November 18 Thanksgiving recess begins, close of day
November 27 Classwork resumes, 8:00 a.m.
December 8 Fall semester classwork ends
December 8 Last day to withdraw from University
December 9 Reading Day
December 11 Final examinations begin
December 15 Fall semester ends at close of day
December 15-16 Commencement Weekend
*Please note: This calendar is subject to change
Syllabus
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Syllabus **
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Introductory and Background Material, including a math review (Calculus III)
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The Thermodynamics of Dry Air
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Hydrostatics
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The Thermodynamics of Moist Air
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Static Stability and Convection
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Vertical Stability, Instability, and Convection*
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Cloud Microphysics *
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The Thunderstorm and Non-hydrostatic Pressure *
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*
These topics will be taught if there is time. All Lecture schedules are tentative!
**
Students with special need are encouraged to schedule an appointment with
me as soon as possible!
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Syllabus
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Special Statements:
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ADA Statement (reference: MU sample statement)
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Please do not hesitate to talk to me!
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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 Services, AO38 Brady Commons, 882-4696. It is the campus office responsible for reviewing
documentation provided by students requesting academic accommodations, and for accommodations
planning in cooperation with students and instructors, as needed and consistent with course
requirements. Another resource, MU's Adaptive Computing Technology Center, 884-2828, is available to
provide computing assistance to students with disabilities.
Academic Dishonesty (Reference: MU sample statement and policy guidelines)
Any student who commits an act of academic dishonesty is subject to disciplinary action.
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Syllabus
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The procedures for disciplinary action will be in accordance with the rules and regulations
of the University governing disciplinary action.
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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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It is the duty of any instructor who is aware of an incident of academic dishonesty in his/her
course to report the incident to the provost and to inform his/her own department
chairperson of the incident. Such report should be made as soon as possible and should
contain a detailed account of the incident (with supporting evidence if appropriate) and
indicate any action taken by the instructor with regard to the student's grade. The instructor
may include an opinion of the seriousness of the incident and whether or not he/she
considers disciplinary action to be appropriate. The decision as to whether disciplinary
proceedings are instituted is made by the provost. It is the duty of the provost to report the
disposition of such cases to the instructor concerned.
Syllabus
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Lab Exercise Write-up Format: All lab write-ups are due at the beginning of the next ‘lab’
Wednesday. Grading format also given.
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Total of 100 pts
Name
Lab #
Atms 4310
Date Due
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Neatness and Grammar 10 pts
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Title
Introduction: brief discussion of relevant background material (5 pts)
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Purpose: brief discussion of why performed (5 pts)
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Data used: brief discussion of data used if relevant (5 pts)
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Procedure: (15 pts)
1.
2.
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Syllabus
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Results: brief discussion of results (50 pts)
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observations
discussion (answer all relevant questions here)
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Summary and Conclusions (10 pts)
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summary
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conclusions
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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 1
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Thermodynamics  the study of initial and final
equilibrium states of a "system" which has been
subjected to a specified energy process or
transformation.
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“System”  a specific sample of matter (air parcels)
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We will concentrate on the “macroscale” or parcel
properties only! We will not look at the microscale
(molecular level) that’s atmospheric physics (Dr.
George, Dr. Fox).
Day 1
Day 1
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Variables of state: (thermodynamic variables)
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pressure (hPa, mb) (Force Area-1)
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Temperature (oC, K, oF)
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Volume (typically m3 [kg-1]), but typically
assume “unit” mass)
Day 1
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Laws of thermodynamics:
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Equation of State
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1st law of thermodynamics (conservation of energy)
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2nd law of thermodynamics (entropy) (direction of
heat flow) (warm to cold)
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We will review Dimensions and Units, and
conventions.
Day 1
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Atmospheric science derives a set of standard
measurements or unit system, such that everyone
everywhere will be on the same page.
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AMS endorsed the SI (Systeme International) or
International system of Units (BAMS, 1974, Aug.)
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The basic units are: Length, Mass, Time (meter, m;
kilo, kg; second, s)
Day 1
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A derived unit combines basic units: example,
pressure:
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Force /Area = kg m s-2 / m2 = kg m-1 s-2 = Pascals
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1000 Pa = 1 kPa = 10 hPa = 10 mb
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Temperature (Kelvin, or absolute scale; Celsius
(1742) Farenheit (1714)).
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Coordinate System: (Cartesian)
Day 1
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Coordinate system: tangent to Earth’s surface which is really a
sphere (curvature for most applications and approximations can
be neglected).
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Cartesian coordinates:
x,y,z,t => x,y,p,t, or x,y,q,t
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Could also use natural coordinates:
s(treamline),n(normal),z,t
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Spherical coordinates
r(adius),q(longitude),f(latitude)
Day 2
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Wind
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Wind direction: direction from which the wind blows,
and compass direction, not Cartesian!
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West wind: is blowing from 270o.
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Wind direction: increasing with time or height:
veering: decreasing with time or height: backing
Day 2
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Remember:
Direction of math = 270 – compass
(meteorology) direction.
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Vector representation in Geophysical Fluid
Dynamics
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Remember the atmosphere is a fluid, and a fluid is
liquid or gas. Thus, the primitive equations will be
valid in any atmosphere, terrestrial (extraterrestrial).
Day 2
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Scalar quantity  A quantity with magnitude
only (e.g., wind
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speed has units m s-1) (zero order tensor)
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Vector  (first order Tensor) A quantity with
magnitude and direction (e.g., wind velocity)
Day 2
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Wind…
V  uiˆ  vˆj  wkˆ ( x, y, z, t )
V  uiˆ  vˆj  kˆ ( x, y, p, t )
u 0

V 0 v
0
0
0 0 w
Day 2
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Dyadic (2nd order Tensor) has a
magnitude and two directions!
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Example: stress (Force per unit area), where
A is the vector of some magnitude equal to
the area and in the direction of the normal.
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In English: Magnitude, direction (1) of the
force, and (2) on which surface applied
Day 2
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An example…
Aii
Aij
Aik
S  A ji
A jj
A jk
Aki
Akj
Akk
Example Stress: Force = Area times Stress
(has same units as pressure)
Day 2
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Vector Analysis:
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Vector Notations:
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A, aˆ
A A
|A| = magnitude of A
Day 2
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Vectors are equal if they have equal magnitude and
directions!!
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The unit vector: any vector of unit length!
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A  Aaˆ
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where a is a vector of unit length
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A  Axiˆ  Ay ˆj
Cartesian Unit vectors: iˆ, ˆj , kˆ vectors of unit length
in the positive x,y,z direction, respectively.
Day 2
sˆ, nˆ, kˆ
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Natural coordinates are:
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Vector components (2 dimensions), but we
can extend to infinite number of directions:
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A  Axiˆ  Ay ˆj  ....  AN nˆ
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A  As sˆ  An nˆ  ....  AZ zˆ
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Magnitude of A  |A| =  (Ax2 + Ay2)
Day 2
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Vector addition and subtraction:
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1) A + B = C
2) A+B = B + A = C
3) A - B = C
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Day 2
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Addition
Subtraction
Day 2
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Here’s how:
 
A  B  ( Ax  Bx)iˆ  ( Ay  By) ˆj
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Associative rule:
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(A+B) + C = A + (B+C)
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Negative Vector: Is a vector of the same
magnitude, but opposite direction.
Day 2
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Vector multiplication:
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Scalar x Vector:
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A(V )  Auiˆ  Avˆj
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In the atmospheric sciences: The wind vector
(2-D 3 – D):
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V  uiˆ  vˆj  ( wkˆ, or kˆ)
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V  Vs sˆ  Vn nˆ  ( wkˆ, or kˆ)
Day 2 /3
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Vector products
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The dot product (also the “scalar” or “inner”) product:
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A dot B = |A||B|cos(q)
Physically: The dot product is the PROJECTION of
vector B onto Vector A in direction of A!
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Day 2/3
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Projection (mathworld.wolfram.com)
(excellent math site):
Day 3
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Properties of the Dot Product:
commutative
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A dot B = B dot A
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associative
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A dot (B dot C) = (A dot B) dot C
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distributive
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A dot (B + C) = A dot B + A dot C
Day 3
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Dot product of perpendicular vector = 0
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In order for the dot product to have a value, the B
vector must have a component parallel to vector A!
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Recall:
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Thus, i dot j, and j dot k, etc… = 0, and i dot i = 1,
etc…
cos(0o) = 1
and
cos(90o) = 0
Day 3
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Orthogonal Vectors (Orthogonality property):
When the angle between two vectors is 90o,
or the dot product is zero, two vectors are
said to be “orthogonal”.
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Other Dot Product Rules:
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1) A dot A = |A||A| cos(0) = A2
2) A dot mA = mA2
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Day 3
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Dot product of two vectors (here’s how):
 
A  B  ( Axiˆ  Ay ˆj )  ( Bxiˆ  By ˆj )  ( Ax Bx  Ay By )
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Remember “Foil”?
“AxBx (i dot i) + Ax By ( i dot j) + Ay Bx (j dot i)
+ Ay By (j dot j)”
Day 3
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Ok, now you try iˆ  Axiˆ = ?
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Answer?????
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Cross Product (or vector product) =
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A x B = |A||B|sin(q)
Ax
Day 3
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What is it (again courtesy of Mathworld site)?
Day 3
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Einstien notation – permutation:
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Q: "What do you get when you cross a mountain-climber with a
mosquito?"
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A: "Nothing: you can't cross a scaler with a vector,"
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Q: "What do you get when you cross an elephant and a grape?"
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A: "Elephant grape sine-of-theta."
Day 3
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The cross product of two vectors is a third vector
that is mutually perpendicular to the two vectors and
the plane containing these vectors.
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Q: Remember your Physics?
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The positive direction of A x B may be determined
by the “right hand (or corkscrew)” rule. Just curl your
fingers from A to B, and your right thumb (vector C)
is the result! (“ayyyy”)
Day 3
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Evaluating the cross product of A and B in the
cartesian coordinate system.
iˆ
ˆj
kˆ
 
A  B  Ax
Ay
Az 
Bx
By
Bz
( AyBz  AzBy )iˆ  ( AxBz  AzBx ) ˆj  ( AxBy  AyBx)kˆ
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The vector or cross product is NOT
  
commutative!
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A B  B  A
Day 3
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Try the right hand rule to show that we
cannot switch order
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Also, remember sin(-90o) = -1.0
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The cross product is distributive:
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A X (B +C) = (A x B) + (A x C)
Day 3/4
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The cross product of a vector with itself equals 0!!
q = 0 so, sin(0) = 0, or A x A = |A||A|sin(0) = 0
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Cross products of unit vectors and the “cyclical”
property of the cross product:
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1) i x i = 0, i x j = k, i x k = -j
2) j x i = -k, j x j = 0, j x k = i
3) k x i = j, k x j = -i, k x k = 0
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Day 4
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A x B = C , B x C = A, C x A = B

Another “trick” or property:

Unit vector k x Ah = Horizontal vector of length |A|
turned 90o to the left of A (try it with right hand rule horizontal vector!!!)

-k x Ah = A vector of length A turned 90o to the right
of A
Day 4
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Multiple Vector Products
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Scalar Triple product:

A dot (B x C) = a scalar value

This is also “cyclical”
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A dot (Bx C) = B dot (C x A) = C dot (A x B)
Day 4
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Or dot and cross product may be interchanged:
A dot ( B x C) = (A x B) dot C
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Triple vector product:

(A x (B xC)) = Vector quantity

A x (B xC) = (A dot C ) B – (A dot B) C

The result is a third vector in the plane of B and C!!!
Day 4
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But,

(A x B) x C not equal to A x (B x C)

since for former result is in the plane of A and
B!!
Day 4/5
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The mathematical description of the Atmosphere:

We must eventually develop from fundamental physical laws and
concepts (first principles), the 5 vector (7 scalar) equations of
geophysical fluid dynamics, and describe and understand the
behavior of the atmosphere through the manipulation of these
equations.

Describing the atmosphere in terms of the distribution in space
and time of certain properties of the atmosphere.
Day 5
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Independent variables  these are the basis
of (or describe) our coordinate system.
We’ll use Cartesian system (x,y,z,t)
and a right handed coordinate system
Day 5
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Dependent variables  depend on your
position in space and time, and can be
described as a function of the independent
variables.
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In atmospheric science: u,v,T,r,q,P,w,or 

Example: u(x,y,z,t)
Day 5
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Invariance  A quantity that does not change if measured in a
different coordinate system

e.g., “rotationally” invariant  quantities that do not change
even if coordinate system rotates

Q: Which variable might be rotationally invariant?
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A:
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Galilean invariant  quantities that do not change even if
coordinate system is moving horizontally
T(x,y,z,t) = T(x’,y’z’,t)
Day 5
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Important Definintion!

Conserved  a quantity that does not
change with time. (e.g., Potential
Temperature and adiabatic motions)

Conservation  the change in some quantity
with time equals 0!
Day 5

Conservation = steady state = balance
(between sources and sinks!)
dQ
 0   source  sin ks
dt

where Q = any quantity
Day 5

The derivative (A review)

Let take a quantity: Q(x,y,z,t)

Now one needs to take the Total Derivative.
In order to do this, we must use the “Chain
Rule!” Remember this?
Day 5

Total derivative is (in x,y,z,t,):
dQ Q dx Q dy Q dz Q dt




dt
x dt y dt z dt t dt

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|advective derivative | + |Eulerian|
(in x,y,p,t)
dQ Q dx Q dy Q dp Q dt




dt
x dt y dt p dt t dt
Day 5

in (x,y,q,t)
dQ Q dx Q dy Q dq Q dt




dt x dt y dt q dt t dt

in natural coordinates
dQ Q ds Q dn Q dz Q dt




dt
s dt n dt z dt t dt
Day 5




now let’s give our pens a break:
1)
2)
3)
u = dx /dt,
v = dy / dt,
w= dz / dt

In Mathematics:

The total derivative (“heavy” D) (substantial,
individual, material) is exact, thus the derivative not
path dependent!
Day 5

Exactness!
Day 5

But, if path dependent, total derivative has no
meaning, and we write with a small “d”.

If path dependent, then the process is
sensitive to the initial starting place and it
corresponds to (generally) one outcome.

But I, like most atmospheric scientists use the
notation “d” for a total derivative regardless.
Day 5

The partial derivative is:

- the change in one variable or coordinate w/out
regard to the other components.


looks like…..

Q  x 2 y 2 z 2t 2
x




Q  Cx 2  2Cx
x

in the eyes of the partial derivative here, where C =
z2y2t2.
Day 5/6

Differentiation of Vectors

The normal rules of differentiation apply, but you
must preserve the order when cross product is
applied.

Let vector A be time dependent, i.e., vector A is
changing size and/or direction with time and space:

A  Axiˆ  Ay ˆj  Az kˆ
Day 6



If our coordinate system is not changing (i.e.,
i,j,k = constant)….
then Ax,Ay,Az change (of course!)

DA DAx ˆ

i  ... (oh, you fill in the rest!)
Dt
Dt
Day 6

Some more “fun” rules:



DmA
DA  Dm
m
A
Dt
Dt
Dt
1)

2)

3)
 


D( A  B )  DA  DB
 B
 A
Dt
Dt
Dt
 


D( A  B)  DB  DA
 A
 B
Dt
Dt
Dt
Day 6

but…. if i,j,k are changing….
ˆ
D
DAx
D
i
Axiˆ 
iˆ  Ax
Dt
Dt
Dt

Position, Velocity, Acceleration

The position vector R, the velocity vector V
and the Acceleration vector A
Day 6


Position vector:

r  xiˆ  yˆj  zkˆ
The velocity vector:
 dr dx
dy ˆ dz ˆ
ˆ
V
 i
j  k  uiˆ  vˆj  wkˆ
dt dt
dt
dt

The acceleration vector:

 dV d 2 r du
dv ˆ dw ˆ
A
 2  iˆ 
j
k  axiˆ  ayˆj  azkˆ
dt
dt
dt
dt
dt
Day 6

The “del” operator

Also known as the; “Hamiltonian”, “gradient”,
or “nabla” operator.

Let us define a differential operator with
vector properties
 ˆ  ˆ  ˆ
 i 
j k
x
y
z
Day 6

The del operator has no physical meaning
until it operates on another quantity such as a
scalar or another vector! (A ghost vector)

Operating on a scalar:
Q ˆ Q ˆ Q ˆ
Q 
i
j
k
x
y
z
Day 6

Del Q: is now a 3-D vector whose direction is
in the direction of the maximum increase of Q
and whose magnitude is equal to the rate of
change of Q per unit distance in that
direction.
Day 6

Del Q in normal (perpendicular to lines of Q).
On a 2 – D surface is perpendicular to Q
isolines.

Then in “plane” English: delQ is simply the
slope of Q on some planar surface. The first
derivative in space (slope) is analogous to
the first derivative in time (velocity).
Day 6/7

Now a proof! (show velocity vector is perpendicular
to gradient vector)
Q
Q
dQ 
dx 
dy
x
y

Step 1:

But, dQ = 0 on a line of Q (correct?)
Day 7

Now we know that;

a) each point of a surface of constant Q can
be defined by the position vector.

b) then on a Q surface dr (or V), must be on
the surface of Q, so dr must lie on the Q
surface.
Day 7

c) then dr dot del Q on Q surface;

 Q ˆ Q ˆ
dr  Q  dr 
i
j  scalar
x
y

So dr dot delQ = dQ, this is the definition of the
total derivative.

but dQ = 0 on Q surface as discussed above.
Day 7

Therefore since dr and delQ separately are
not 0, but their dot product IS 0, they must be
perpendicular (or orthogonal)!!!

Furthermore, we know that delQ was
perpendicular to lines of Q, thus dr or
Velocity, must be parallel to lines of Q!

Point proved!
Day 7

Advection of a scalar quantity

3-D transport of some quantity:

In Atms Sci, in (x,y,p,t) coordinates, advection
and flux are equivalent.
Day 7

A 2-D example of advection:

 V  Q
 or 

 V  T


Some other names for advective quantity:
Convective derivative or Lagrangian!
Day 7

Two definitions:

Lagrangian measurement  measurement that
moves with the flow (“goes with the flow”)

Eulerian  measurement at a stationary point

Important Concept!!

 If Q is conserved or a conserved property
(invariant with time) time derivative = 0. We also
call this “steady state”.
Day 7

Thus, if this is true either the source-sinks are
zero, or Equal and opposite. But it also
implies that advection equals the time rate of
change.
dQ
dt
0
Re call
dQ
  sources  sin ks
dt
Day 7


- or –

Q
 V  Q
t
where Q = Any variable, vector or scalar!
Day 7

The Laplacian operator

 Del dot Del: It’s a scalar operator! It typically
changes the sign of a function. A measure of the
curvature in a function.



    2  2  2
x y z
2
2
2
2

 as acceleration is to velocity, so is the
gradient operator (slope) to the Laplacian
(curvature)!
Day 7

The divergence of a vector (del dot V) (a
scalar)!
 u v w
 V   
x y z

Velocity divergence:
 
Седьмой Дни

The curl of the vector V:
 
 3  V3    iˆ  ˆj  kˆ

The curl of the velocity or vector vorticity:
Day 7

Vertical component of Vorticity (zeta)

v u  v u  ˆ
    k
x y  x y 

Magnitude in vertical is entirely dependent on
horizontal spatial variations or shears

Consider a case where all the terms
contribute positively:
v u  v u 

x

   kˆ  
y  x y 
Day 7/8

Then;

 Then, the vertical component is POSITIVE for
cyclonic circulations (shear) and negative for
anticylonic circulations. It is the opposite in the SH.
Day 8



Rotational Vectors and vectors in Rotation
Definition of the rotational vector (omega - ) 
rotational vector has a direction along the axis,
positive in the sense of the Right hand rule, and it’s
magnitude, omega ||, is porportional to the angular
velocity of the rotating system. (ang. Vel.=
radians/sec)
The rate of change of a vector A of constant
magnitude due to it’s changing direction produced
by rotation (omega - W)
Day 8

Now look down from above at the plane in
which the vector A is rotating.
Day 8

The magn. Or length of DA = DA and for small
angle Dq

Recall from Geometery:

DA = A sin(f) Dq

 since for small angles tan q = q
adj. Or opp. = adj. tan q
opp. /
Day 8

So, (now include delta t)

DA/ Dt = A sin(f) (Dq / D t)

And as Dt goes to 0 

dA/dt = A sin(f)(dq /dt)
Day 8

But dq /dt = omega  (the angular velocity)  So;

dA/dt =  A sin (f)

Then we need to;

 redefine  as W since we’re talking about earth!

 recall our definition of the cross product!!! (W x A)
Day 7


Thus the magnitude of DA/dt = DA/dt equals
the mangitudes of W x A!!
dA/dt = WxA!
Day 8



Direction of dA/dt and W x A
 Since (W x A is mutually perpendicular to W
and A as is DA/dt, and is positive in the same
direction the two vectors DA/dt and W x A are
equal vectors!
Thus for A of constant magnitude dA/dt = W x A
Day 8



 The rate of change of vector A in a
fixed (absolute) coordinate system vs. a
rotating coordinate system. (“Fixed” and
“absolute” are not good news – we can’t
do this in practice (in real atms.)).
Atms example: coriolis force!!
dV/dt = W x V
Day 8

OK, there’s more than one way to skin a cat ……..

Consider (X,Y,Z) w/ unit vectors (I,J,K)

Consider another (x,y,z) w/unit vectors (i,j,k). Allow
this one to rotate w/angluar velocity (omega).

A = AXI + AYJ + AZK = Axi + Ayj +Azk
Day 8

Now differentiate A w/r/t time (sytem 1 is
const.! System 2 is rotating)

DA/dt = AX/dt I + dAY/dt J + dAZ/dt K =
dAx/dt i + Ax di/dt + etc…..

i,j,k are unit vectors.

Di/dt = W x i and dj/dt = W x j and….
Day 8

(DA/dt)abs = dAx/dt i + dAy/dt j + dAz/dt k +
(Ax (W x i) + Ay (W x j) + Az (W x k) )

DA/dt abs = (dA/dt) (relative to rotating coord.
System) + (W x (Axi + Ayj + Azk)

DA/dt = dA/dt (relative to rot.) + W x A (rot of
coord system w/r/t vertical)
Day 8/9

The rate of change of a vector in an absolute
(inertial) frame of ref. Is equal to rate of
chage observed in the rotating system + a
term = to the cross product of the rotational
vector and the arbitrary vector A!

 Thus, you have just derived the
expression for the coriolos force! 
Day 9

Dimensional Analysis (The rules!)

1. All terms of an equation must have the same
dimensions! e.g. potential temp relationship;
Po 
q  T( )
P

2) All exponents are non – dimensional

3) All log and trig functions are also non-dimentional.
Day 9

4. The dimensions of differentials are the same
as the dimensions of a differentiated quantity.

e.g., we can say d (ln (T))

 Note: “specific” as a prefix implies the quantity
is per unit mass and has the dimensions of (Q x M1) e.g., specific volume = vol / unit mass.
Day 9

 Valid physical relationships must be
dimensionally consistent with eachother, in
other words X = Y must have the same units.
Otherwise, we say X proportional to Y. Or X =
AY where the units of AY are the same as X.

Caveat: Some proportionalities have
consistent units.
Day 9

Non-dimensional analysis (Scale analysis)

 (Mathematical formalization) Theoreticians like to look
at equations in a non-dimensional sense, that is we
choose characteristic time and space scales for some
phenomena to be studied.

We essentially change coordinate systems:

(x,y) = L(x,y) where L is 2000 km
t = Tt’
where T = 100000 sec
(u,v) = U(u’,v’) U = 10 m/s


Day 9

 In performing this type of analysis we can determine
what processes are important for du/dt, for example, in
the horizontal equation of motion on the desired scale
(cyclone), we can perform this type of analysis for
particular phenomena.

“Informal” Scale analysis

 Similar to that of non-dimensionalization. Scale (size)
analysis is also a powerful tool often used in
meteorological derivations & the analysis of physical
processes. This is a more “informal” method than nondimensional analysis.
Day 9

 Suppose we have an equation based on a
physical law or principle (e.g. Newton’s 2nd
law, 3rd eqn. of motion). This equation is a
generalized equation, valid for many
atmospheric phenomena.
dw
1 p

 g  2W cos(f )  Visc  Fric  other
dt
r z
10 6
10
10
10 3
10 7
10 7
10 7
Day 9

We choose the scale we are interested in, the
consider the order of magnitude, (e.g. the size and
space scale) each term in the equation would have,
for that particular scale of motion (again typical
values, or estimates).

Then, we can neglect the smaller terms and simplify
the equation. We’ll also know the error introduced in
doing this (ratio of neglected to retained terms).

The equation is simplified, but now less general, it’s
the trade-off for a simpler relationship.
Day 9

Hydrostatic balance:
p
  rg
z

This equation governs the movement of
synoptic-scale systems (highs/lows (waves in
the westerlies)) and fronts.
Day 9



Hydrostatic balance: the error 
Error = Neglected Terms / Retained terms
Error = 10-3 / 10 = 10-4 = 0.0001 = 0.01%

 Thus this is a darned good estimate for synoptic
and meso alpha scales.

 Important Point! The equation is much simpler,
but it’s only valid for these scales and has now lost
its generality!
Day 9/10

Scales of Atmospheric motions

Scale
Horiz Dimension

Planetary
Synoptic
Meso
Micro
10,000 km
weeks - 1 month
2,000 – 6000 km
1 to 7 days
10 km – 2,000 km
1h – 1 day
< 10 km
< 1h




Time
 Can you think of examples of real phenomena
that fit each category?
Day 10

Fundamental equations of geophysical hydrodynamics

 Seven dependent variables, four independent variables, seven
independent equations:

p(x,y,z,t)
r(x,y,z,t)
T(x,y,z,t) or q(x,y,z,t)
M(x,y,z,t)
U(x,y,z,t)
V(x,y,z,t)
W(x,y,z,t)






Pressure (mass)
density (mass, thermal)
(potential) Temperature (thermal)
Mixing Ratio (mass, thermal)
zonal wind (mass)
meridional wind (mass)
vertical wind (mass)
Day 10

Concept
Name

Elemental Kinetic TheoryEqn. Of state
P  rRT
 or 
P  RT
Day 10

Cons. of Energy
1st Law of Thermo.
dh
dT
d

dT
dp
Q 
 Cv
p
 Cp

dt
dt
dt
dt
dt

Cons. of Mass
Eq. of continuity

1 dr
  V
r dt
Day 10

Cons. of mass
Eq. of water mass
dms
dm
w
 Pr
dt
dz
 or 
dm
0
dt
Day 10

Cons. of momentum
Eq. of motion

du u
u
u
u
1 p
 u v w  
 fv  Fu  Fx
dt t
x
y
z
r x

dv v
v
v
v
1 p
 u v  w  
 fu  Fy
dt t
x
y
z
r y

dw w
w
w
w
1 p

u
v
w

 g  Fu  Fx
dt t
x
y
z
r z
 or 


dV
1
   3 p  2W  V  g  Fric
dt
r

A.K.A Navier – Stokes Equation, Newton’s 2nd
Law, etc.
Day 10

Von Helmholtz 1858: In principle this is a mathematically solvable
system (closed) given observed initial state and proper BC’s, the
solution should yield all future states of the system. (The “rub”
IC’s and BC’s).

 Thus, forecasting is an initial value problem (Bjerknes, 1903)

 These eqns. Are what will be studied in Atms 4310, 4320.
These describe behavior of the atmosphere.

 Solving these equations (Numerical methods and modeling
classes – Atms 4800)
Day 10

The Thermodynamics of Dry Air (Holton Ch 24)

Reminder:
moisture)!

Moist air: means water vapor present.

Dry air: (a homogeneous mixture of gasses from 0 –
80 km up – Homosphere)
Dry air means DRY air (no
Day 10/11

The Heterosphere is above that, gasses separate by
weight (mass)

The Atmosphere: its makeup:

Gas (atomic weight)
% by Vol
Nitrogen (N2) 28.02
78.1
Oxygen (O2) 32.00
20.9
Argon (Ar) 39.94
0.93
Carbon Dioxide (CO2) 44.01 0.036




% by mass
75.5
23.1
1.3
0.05
Day 10/11

 Many other gases present in very small quantities
(Ne, He, H, O3) they are called: Trace Gases

 Thus if we calculate the atomic weight of air:

28.97 kg mol-1

 Three of these are very important because
despite the small quantities, they help determine the
temperature structure of the troposphere and
stratosphere: H2O, CO2, and O3
Day 11

H2O – is important within the hydrologic cycle,
clouds, rain etc.. Water is the only substance in
earth atmosphere that exists in all three phase at
terrestrial pressures and temperatures.

 Water Vapor and clouds are important in
determining atmospheric structure due to their
radiative properties (albedo, infrared).

 Residence time 1 – 10 days.
Day 11

 It is the most important and potent greenhouse gas,
but its not homogeneously distributed!

CO2 – has homogeneous concentration. It is important
because of it’s radiative properties in infrared. Its
residence time near 100 years, thus important in longer
term climate change. But, the CO2 cycle is not well
known yet.

O3 – concentrated at 32km up (in stratosphere) due to
solar and chemical reactions. It absorbs UV and emits
infrared and responsible for the Stratosphere’s inversion.
Day 11

 Near surface it is present in small amounts due to
pollution, but it is highly poisonous.

Moist air

 Water vapor extremely variable near 0 – near 4%

That’s 0 - 40 g/kg!

More typical:
1% 10 g/kg (Td = 57F)
Day 11

The variables of state

Mass (M)
 Density (mass/unit vol) (r)
 Specific volume (vol / unit mass) 



 Pressure: (Force/Area) is due to molecular
collisions and the associated momentum changes
independent of direction (a scalar). The force is
normal to gas container walls. N m-2 = 1 Pa and
1mb = 102 Pa - or - kg m-1s-1
Day 11

 Temperature: (T) a measure of the average internal energy of
the molecules obtained during a state of equilibrium.

 To read temperature there needs to be equilibrium
established between the system and a temp sensor. Temp.
determines direction of heat flow. (Kelvin, Celsius) 0o C =
273.15 K.

Ideal Gas: (Kinetic theory of gasses) is a collection of molecules
that are
completely elastic spheres
with no attractive or repulsive forces, and
occupying no volume.
1)
2)
3)
Day 11

Heat: is a form of energy, which can be transferred
from a warmer to colder aubstance. Heat transfer by
(really kinetic energy transfer):

Radiation  Transfer of Electro- and Magnetic
Conduction  Transfer by molecular motions
(Contact)
Convection  Transfer by turbulent mixing (parcels,
bulk transport, advection)
Latent (phase changes)  "hidden heat"



Day 11

Relationships between our “state”
variables, r, , T, and P.

Boyle’s law:

Robert (Bob) Boyle (1600) said

if T is constant, then;
Day 11
- or –

P1 x V1 = P2 x V2

P x Volume = Constant

Recall, this is possible since Torricelli (1543)
invented the barometer!

So as a result of Boyle’s Law:
Day 11


As P increases, Volume decreases:

Day 11


Or as Volume increase, then P decreases
 
Day 11

Charles’s Law:

Jaques Charles (1787) but stated formally J. GayLussac (1802)

(1787 Bonus question: what else important
happened in Sept. 1787?)

“Jack” said if pressure is kept constant then;

Volume1/Volume2 = Temperature1/Temperature2
Day 11

Or:

Volume/Temperature = Constant

Thus, when gas is heated, volume goes up,
when gas is cooled, volume decreases!
Day 11/12

Combined or Ideal gas law

Constant = Pressure x Volume / Temperature

Derivation of Ideal Gas law for Atmosphere

 Nee Avagadro’s hypothesis (derived
experimentally, and derivable from kinetic theory of
gasses)
Day 12

His hypothesis: Different gasses, each containing
the same number of molecules, occupy the same
volume at the same temperature and pressure.

Kilogram molecular weight: a kmol of material is it’s
molecular weight expressed in kg. Thus, one kmol of
water is 18.016 kg!

Number of molecules is Avagadro’s number (Av):
6.022x 1026

Day 12

Universal gas constant:

 if we have a mixture of gases, each with it’s own
ideal gas law (which is what Dalton’s Law implies!)

Po Volume(gas) = m(gas)R(gas)To where gas = 1,2,3,
etc.

where m is the number of moles of each gas!
Day 12

Divide each equation by ng where ng is weight of a
kmol of gas.

Po Vg/ng = (mg/ng) RgTo

If each sample consists of 1 mol they have same
number of molecules. Avagadro’s hypotheses all
have same Volume. For each gas we can write;

Po (V1/ng) / To = Po (Vo/ng) / To
Day 12

or
[Po(Vg/ng) / To ] ng = mgRg

Since mgRg will be the same by Avagadro’s
hypothesis:

Then: mgRg = R* (Universal Gas constant)

R* = 8314.3 J K-1 kmol-1
Day 12

Thus, we must find the apparent weight of air, or take a weighted
average of the gasses per mol.

Molecular weight = 28.97 kg / kmol = n

So then: P = R*/n T

P = RT


Or
P = rRT (R is a constant depending on individual Gas)
Day 12





 Or to include the effects of moisture:
P = r Rd Tv (Rd = 287.04 J K-1 kg-1)
 So let’s go back to:
Po (Vg/ng) / To = mg/ng R*
Can alternatively express RHS as R*/ng = R
Day 12

Thus in ideal gas law we must express as:

PV = R*/n (air) T where air = 28.97

V is volume, where Volume could be anything. Well we’ll specify
some “specific volume”. ( = vol/unit mass) which equals 1 kg.
This is still volume, we are not changing variables here or
“playing fast and loose” with the math.

Thus:

P = R*/ (n air) T = RdT
(- or P = r R*/n T = rRdT)

Day 12

That’s for dry air. In accounting for moisture: ng =
18.016. The amount of water is very variable, and
we could be specifying a different R for air every
time amount of moisture changes.

That’s where concept of “virtual temperature”
comes in. Thus,

P  = Rd Tv
-or-
P = r Rd Tv
Day 12

Q: Which is more dense, dry air or moist air?

Ever heard a baseball announcer talk about
the ball carrying on a warm humid night?

Dry air weights 28.97, throw in 18.016 for
moist air.
Day 12

In dry air, 1.00 * 28.97 for dry air = 28.97

Now, say the air was 4% vapor:




%-age x
mol. wt.
Contribution
0.96
28.97 = 27.80
+ 0.04
18.016 = 0.72
______________________________
28.53 (molecular weightof air + vapor)
Day 12/13

Application: Let’s derive expression for Virtual
temperature (Wallace and Hobbs p51 – 52)
Md  Mv
r
 rd  rv
Vol

Ideal Gas Laws, for water vapor and dry air:
e  Rv  rv  T
Pd  rd  Rd  T
Day 13

And use Dalton’s Law

P = Pd + e

r = rd + rv

so, substitute ideal gas laws into (1) to get:

Pd / (Rd T) + (e / (Rv T))
(1)
Day 13

And then:

(P – e) / (Rd T) + (e / (Rv T))

use “strategic” multiplications of each term by
1:

1st term multiply by: P / P
2nd term multiply by: P Rd / P Rd

Day 13



then we’ll have a common factor to pull out:
(P / (Rd T))
and we get:
Then rearrange:
P  P  e eRd 



RdT  P
PRv 
P  e e Rd 
1  

RdT  P P Rv 
 or 
r
P  e

1


1

0
.
622



RdT  P

Day 13

And finally
P
rRdT
 e

1  0.378
 p

Day 13

Celsius (1742) – Temperature scale (Centigrade)

Define: at P = Po = 1000 mb at a state of thermal
equilibruim

Pure Ice and water mixture  temp = 0o C

Pure water and steam mixture at equilibrium  100
o C
Day 13

If P = Po alpha varies linearly with temp. Thus:

Y = mx + b

T in oC is the slope:
T (Celsius) = {(t – o) / (100 – o)} 100

This is the defining expression for Centigrade scale!
Day 13



The Absolute or Kelvin Temp Scale:
T oC = [100 t / (100 –o)] – [100o / (100 – o)]
VRBL
–
CONST

 Or we can use this relationship to re-define the
temperature scale by extrapolating to the point
where all molecular motion stops and Specific
Volume goes to 0!

T (Absolute) = VRBL = T oC + CONST
Day 13

CONST = 273.16

So,

K = C + 273.16

Show Absolute zero = -273.16
Day 13




Solve for () at t:
t = o + T oC / [(100 – o) 100]
t = o ([1 + 1 / 273.16]ToC)
so if t = 0 (or all molecular motion ceases),
Day 13
[1+1/273.16] ToC = 0

then

 Solve T oC = -273.16

and then –273.16 oC = 0 Absolute

or 0 K (Volume of Ideal gas goes to 0)!!
Day 14

The work done by an expanding gas

Let’s draw a piston:
Day 13

 Consider a mass of gas at Pressure P in a
cylinder of Cross section A

Now, Recall from Calc III or Physics:

Work = force x distance or Work = Force dot
distance
W  F  ds

So only forces parallel to the distance travelled do
work!
Day 13

Then,
dW  F  ds
 or 
dW
 F  ds
dt
Day 13


But, we know that:
Pressure = Force / Unit Area

So then,

Force = P x Area
Day 13

Total work increment now:
dW   P  Area   ds



Well,
Area x length = Volume
soo………………… A * ds = dVol
Day 13

Then we get the result:

 Work :

Let’s “work” with Work per unit mass:

dW  PdV
dW
dV
P
m
m
 Thus, we can start out with volume of only
one 1 kg of gas!!!
Day 13
Day 13
Day 13