An Introduction to Dynamic Meteorology PDF

January 24, 2004 12:0 Elsevier/AID aid An Introduction to Dynamic Meteorology FOURTH EDITION January 24, 2004 12:0 Elsevier/AID aid This is Volume 88 in the...

January 24, 2004 12:0 Elsevier/AID aid An Introduction to Dynamic Meteorology FOURTH EDITION January 24, 2004 12:0 Elsevier/AID aid This is Volume 88 in the INTERNATIONAL GEOPHYSICS SERIES A series of monographs and textbooks Edited by RENATA DMOWSKA, JAMES R. HOLTON and H. THOMAS ROSSBY A complete list of books in this series appears at the end of this volume. January 24, 2004 12:0 Elsevier/AID aid AN INTRODUCTION TO DYNAMIC METEOROLOGY Fourth Edition JAMES R. HOLTON Department of Atmospheric Sciences University of Washington Seattle,Washington Amsterdam Boston Heidelberg London New York Oxford Paris San Diego San Francisco Singapore Sydney Tokyo January 24, 2004 12:0 Elsevier/AID aid Senior Editor, Earth Sciences Frank Cynar Editorial Coordinator Jennifer Helé Senior Marketing Manager Linda Beattie Cover Design Eric DeCicco Composition Integra Software Services Printer/Binder The Maple-Vail Manufacturing Group Cover printer Phoenix Color Corp Elsevier Academic Press 200 Wheeler Road, Burlington, MA 01803, USA 525 B Street, Suite 1900, San Diego, California 92101-4495, USA 84 Theobald’s Road, London WC1X 8RR, UK This book is printed on acid-free paper. Copyright
c 2004, Elsevier Inc. All rights reserved. No part of this publication may be reproduced or transmitted in any form or by any means, electronic or mechanical, including photocopy, recording, or any information storage and retrieval system, without permission in writing from the publisher. Permissions may be sought directly from Elsevier’s Science & Technology Rights Department in Oxford, UK: phone: (+44) 1865 843830, fax: (+44) 1865 853333, e-mail: [email protected]. You may also complete your request on-line via the Elsevier homepage (http://elsevier.com), by selecting “Customer Support” and then “Obtaining Permissions.” Library of Congress Cataloging-in-Publication Data Application submitted British Library Cataloguing in Publication Data A catalogue record for this book is available from the British Library ISBN: 0-12-354015-1 CD-Rom ISBN: 0-12-354016-X For all information on all Academic Press publications visit our Web site at www.academicpressbooks.com PRINTED IN THE UNITED STATES OF AMERICA 04 05 06 07 08 9 8 7 6 5 4 3 2 1 January 24, 2004 12:0 Elsevier/AID aid C O N T E N T S Preface xi Chapter 1 Introduction 1.1 The Atmospheric Continuum 1 1.2 Physical Dimensions and Units 2 1.3 Scale Analysis 4 1.4 Fundamental Forces 4 1.5 Noninertial Reference Frames and “Apparent” Forces 10 1.6 Structure of the Static Atmosphere 19 Problems 24 MATLAB Exercises 26 Suggested References 27 Chapter 2 Basic Conservation Laws 2.1 Total Differentiation 29 2.2 The Vectorial Form of the Momentum Equation in Rotating Coordinates 33 2.3 Component Equations in Spherical Coordinates 34 2.4 Scale Analysis of the Equations of Motion 38 2.5 The Continuity Equation 42 2.6 The Thermodynamic Energy Equation 46 2.7 Thermodynamics of the Dry Atmosphere 49 v January 24, 2004 12:0 Elsevier/AID aid vi contents Problems 54 MATLAB Exercises 55 Suggested References 56 Chapter 3 Elementary Applications of the Basic Equations 3.1 Basic Equations in Isobaric Coordinates 57 3.2 Balanced Flow 60 3.3 Trajectories and Streamlines 68 3.4 The Thermal Wind 70 3.5 Vertical Motion 75 3.6 Surface Pressure Tendency 77 Problems 79 MATLAB Exercises 83 Chapter 4 Circulation and Vorticity 4.1 The Circulation Theorem 86 4.2 Vorticity 91 4.3 Potential Vorticity 95 4.4 The Vorticity Equation 100 4.5 Vorticity in Barotropic Fluids 106 4.6 The Baroclinic (Ertel) Potential Vorticity Equation 108 Problems 111 MATLAB Exercises 113 Suggested References 114 Chapter 5 The Planetary Boundary Layer 5.1 Atmospheric Turbulence 116 5.2 Turbulent Kinetic Energy 120 5.3 Planetary Boundary Layer Momentum Equations 122 5.4 Secondary Circulations and Spin Down 131 Problems 136 MATLAB Exercises 137 Suggested References 138 Chapter 6 Synoptic-Scale Motions I: Quasi-geostrophic Analysis 6.1 The Observed Structure of Extratropical Circulations 140 6.2 The Quasi-Geostrophic Approximation 146 6.3 Quasi-geostrophic Prediction 155 6.4 Diagnosis of the Vertical Motion 164 6.5 Idealized Model of a Baroclinic Disturbance 174 Problems 176 MATLAB Exercises 178 Suggested References 180 January 24, 2004 12:0 Elsevier/AID aid contents vii Chapter 7 Atmospheric Oscillations: Linear Perturbation Theory 7.1 The Perturbation Method 183 7.2 Properties of Waves 183 7.3 Simple Wave Types 188 7.4 Internal Gravity (Buoyancy) Waves 196 7.5 Gravity Waves Modified by Rotation 204 7.6 Adjustment to Geostrophic Balance 208 7.7 Rossby Waves 213 Problems 220 MATLAB Exercises 224 Suggested References 226 Chapter 8 Synoptic-Scale Motions II: Baroclinic Instability 8.1 Hydrodynamic Instability 229 8.2 Normal Mode Baroclinic Instability: A Two-Layer Model 230 8.3 The Energetics of Baroclinic Waves 242 8.4 Baroclinic Instability of a Continuously Stratified Atmosphere 250 8.5 Growth and Propagation of Neutral Modes 260 Problems 264 MATLAB Exercises 266 Suggested References 267 Chapter 9 Mesoscale Circulations 9.1 Energy Sources for Mesoscale Circulations 269 9.2 Fronts and Frontogenesis 269 9.3 Symmetric Baroclinic Instability 279 9.4 Mountain Waves 284 9.5 Cumulus Convection 289 9.6 Convective Storms 298 9.7 Hurricanes 304 Problems 309 MATLAB Exercises 310 Suggested References 311 Chapter 10 The General Circulation 10.1 The Nature of the Problem 314 10.2 The Zonally Averaged Circulation 316 10.3 The Angular Momentum Budget 329 10.4 The Lorenz Energy Cycle 337 10.5 Longitudinally Dependent Time-Averaged Flow 343 10.6 Low-Frequency Variability 349 10.7 Laboratory Simulation of the General Circulation 354 10.8 Numerical Simulation of the General Circulation 360 Problems 366 January 24, 2004 12:0 Elsevier/AID aid viii contents MATLAB Exercises 368 Suggested References 369 Chapter 11 Tropical Dynamics 11.1 The Observed Structure of Large-Scale Tropical Circulations 371 11.2 Scale Analysis of Large-Scale Tropical Motions 387 11.3 Condensation Heating 391 11.4 Equatorial Wave Theory 394 11.5 Steady Forced Equatorial Motions 400 Problems 403 MATLAB Exercises 404 Suggested References 406 Chapter 12 Middle Atmosphere Dynamics 12.1 Structure and Circulation of the Middle Atmosphere 408 12.2 The Zonal-Mean Circulation of the Middle Atmosphere 411 12.3 Vertically Propagating Planetary Waves 421 12.4 Sudden Stratospheric Warmings 424 12.5 Waves in the Equatorial Stratosphere 429 12.6 The Quasi-biennial Oscillation 435 12.7 Trace Constituent Transport 440 Problems 445 MATLAB Exercises 446 Suggested References 447 Chapter 13 Numerical Modeling and Prediction 13.1 Historical Background 449 13.2 Filtering Meteorological Noise 450 13.3 Numerical Approximation of the Equations of Motion 452 13.4 The Barotropic Vorticity Equation in Finite Differences 462 13.5 The Spectral Method 464 13.6 Primitive Equation Models 470 13.7 Data Assimilation 475 13.8 Predictability and Ensemble Prediction Systems 481 Problems 485 MATLAB Exercises 487 Suggested References 490 January 24, 2004 12:0 Elsevier/AID aid contents ix Appendix A Useful Constants and Parameters 491 Appendix B List of Symbols 493 Appendix C Vector Analysis 498 Appendix D Moisture Variables 501 Appendix E Standard Atmosphere Data 504 Appendix F Symmetric Baroclinic Oscillations 506 Answers to Selected Problems 509 Bibliography 513 Index 519 International Geophysics Series 531 January 24, 2004 12:0 Elsevier/AID aid January 24, 2004 12:0 Elsevier/AID aid P R E F A C E In this fourth edition of An Introduction to Dynamic Meteorology I have retained the basic structure of all chapters of the previous edition. A number of minor correc- tions, pedagogical improvements, and updates of material are included throughout. The major departure from previous editions, however, is inclusion of a variety of computer-based exercises and demonstrations utilizing the MATLAB
r program-
r ming language at the end of each chapter (MATLAB is a registered trademark of The MathWorks, Inc.). These will, I hope, provide students with an opportunity to visualize and experiment with various aspects of dynamics not readily accessible through analytic problem solving. I have chosen MATLAB because it is a high-level language with excellent graphing capabilities and is readily available to most university students. The ability within MATLAB to animate wave fields is particularly valuable as a learning aid in dynamic meteorology. It is not necessary to have much experience with MATLAB to solve most of the problems provided. In most cases MATLAB scripts (M-files) are provided on the accompanying CD, and the student need only run the scripts for various parameter choices or make minor revisions. Through studying the various examples, students should gradually be able to gain the confidence to program their own MATLAB scripts. xi January 24, 2004 12:0 Elsevier/AID aid xii preface Much of the material included in this text is based on a two-term course sequence for seniors majoring in atmospheric sciences at the University of Washington. It would also be suitable for first-year graduate students with little previous back- ground in meteorology. As in the previous editions the emphasis in the text is on physical principles rather than mathematical elegance. It is assumed that the reader has mastered the fundamentals of classical physics and has a thorough knowledge of elementary calculus. Some use is made of vector calculus. In most cases, how- ever, the vector operations are elementary in nature so that the reader with little background in vector operations should not experience undue difficulties. The fundamentals of fluid dynamics necessary for understanding large-scale atmospheric motions are presented in Chapters 1–5. These have undergone only minor revisions from the previous edition. The development of the Coriolis force in Section 1.5 has been substantially improved from previous editions. The dis- cussion of the barotropic vorticity equation in Section 4.5 now introduces the streamfunction. As in previous editions, Chapter 6 is devoted to quasi-geostrophic theory, which is still fundamental to the understanding of large-scale extratropical motions. This chapter has been revised to provide increased emphasis on the role of potential vorticity and potential vorticity inversion. The presentation of the omega equation and the Q vector has been revised and improved. In Chapter 9, the discussions of fronts, symmetric instability, and hurricanes have all been expanded and improved. Chapter 10 now includes a discussion of annular modes of variability, and the discussion of general circulation models has been rewritten. Chapter 11 has an improved discussion of El Niño and of steady equatorial circulations. Chapter 12 presents a revised discussion of the general circulation of the stratosphere, including discussions of the residual circulation and trace constituent transport. Finally, Chapter 13 has been updated to briefly summarize modern data assimilation techniques and ensemble forecasting. Acknowledgments: I am indebted to a large number of colleagues and students for their continuing interest, suggestions, and help with various figures. I am partic- ularly grateful to Drs. Dale Durran, Greg Hakim, Todd Mitchell, Adrian Simmons, David Thompson, and John Wallace for various suggestions and figures. January 27, 2004 13:54 Elsevier/AID aid C H A P T E R 1 Introduction 1.1 THE ATMOSPHERIC CONTINUUM Dynamic meteorology is the study of those motions of the atmosphere that are associated with weather and climate. For all such motions the discrete molecular nature of the atmosphere can be ignored, and the atmosphere can be regarded as a continuous fluid medium, or continuum. A “point” in the continuum is regarded as a volume element that is very small compared with the volume of atmosphere under consideration, but still contains a large number of molecules. The expres- sions air parcel and air particle are both commonly used to refer to such a point. The various physical quantities that characterize the state of the atmosphere (e.g., pressure, density, temperature) are assumed to have unique values at each point in the atmospheric continuum. Moreover, these field variables and their derivatives are assumed to be continuous functions of space and time. The fundamental laws of fluid mechanics and thermodynamics, which govern the motions of the atmo- sphere, may then be expressed in terms of partial differential equations involv- ing the field variables as dependent variables and space and time as independent variables. 1 January 27, 2004 13:54 Elsevier/AID aid 2 1 introduction The general set of partial differential equations governing the motions of the atmosphere is extremely complex; no general solutions are known to exist. To acquire an understanding of the physical role of atmospheric motions in determin- ing the observed weather and climate, it is necessary to develop models based on systematic simplification of the fundamental governing equations. As shown in later chapters, the development of models appropriate to particular atmospheric motion systems requires careful consideration of the scales of motion involved. 1.2 PHYSICAL DIMENSIONS AND UNITS The fundamental laws that govern the motions of the atmosphere satisfy the princi- ple of dimensional homogeneity. That is, all terms in the equations expressing these laws must have the same physical dimensions. These dimensions can be expressed in terms of multiples and ratios of four dimensionally independent properties: length, time, mass, and thermodynamic temperature. To measure and compare the scales of terms in the laws of motion, a set of units of measure must be defined for these four fundamental properties. In this text the international system of units (SI) will be used almost exclusively. The four fundamental properties are measured in terms of the SI base units shown in Table 1.1. All other properties are measured in terms of SI derived units, which are units formed from products or ratios of the base units. For example, velocity has the derived units of meter per second (m s−1 ). A number of important derived units have special names and symbols. Those that are commonly used in dynamic meteorology are indicated in Table 1.2. In addition, the supplementary unit desig- nating a plane angle, the radian (rad), is required for expressing angular velocity (rad s−1 ) in the SI system.1 In order to keep numerical values within convenient limits, it is conventional to use decimal multiples and submultiples of SI units. Prefixes used to indicate such multiples and submultiples are given in Table 1.3. The prefixes of Table 1.3 may be affixed to any of the basic or derived SI units except the kilogram. Because the Table 1.1 SI Base Units Property Name Symbol Length Meter (meter) m Mass Kilogram kg Time Second s Temperature Kelvin K 1 Note that Hertz measures frequency in cycles per second, not in radians per second. January 27, 2004 13:54 Elsevier/AID aid 1.2 physical dimensions and units 3 Table 1.2 SI Derived Units with Special Names Property Name Symbol Frequency Hertz Hz (s−1 ) Force Newton N (kg m s−2 ) Pressure Pascal Pa (N m−2 ) Energy Joule J (N m) Power Watt W (J s−1 ) kilogram already is a prefixed unit, decimal multiples and submultiples of mass are formed by prefixing the gram (g), not the kilogram (kg). Although the use of non-SI units will generally be avoided in this text, there are a few exceptions worth mentioning: 1. In some contexts, the time units minute (min), hour (h), and day (d) may be used in preference to the second in order to express quantities in convenient numerical values. 2. The hectopascal (hPa) is the preferred SI unit for pressure. Many meteo- rologists, however, are still accustomed to using the millibar (mb), which is numerically equivalent to 1 hPa. For conformity with current best prac- tice, pressures in this text will generally be expressed in hectopascals (e.g., standard surface pressure is 1013.25 hPa). 3. Observed temperatures will generally be expressed using the Celsius tem- perature scale, which is related to the thermodynamic temperature scale as follows: T C = T − T0 where TC is expressed in degrees Celsius (◦ C), T is the thermodynamic temperature in Kelvins (K), and T0 = 273.15 K is the freezing point of water on the Kelvin scale. From this relationship it is clear that one Kelvin unit equals one degree Celsius. Table 1.3 Prefixes for Decimal Multiples and Submulti- ples of SI Units Multiple Prefix Symbol 106 Mega M 103 Kilo k 102 Hecto h 101 Deka da 10−1 Deci d 10−2 Centi c 10−3 Milli m 10−6 Micro µ January 27, 2004 13:54 Elsevier/AID aid 4 1 introduction 1.3 SCALE ANALYSIS Scale analysis, or scaling, is a convenient technique for estimating the magnitudes of various terms in the governing equations for a particular type of motion. In scal- ing, typical expected values of the following quantities are specified: (1) magnitudes of the field variables; (2) amplitudes of fluctuations in the field variables; and (3) the characteristic length, depth, and time scales on which these fluctuations occur. These typical values are then used to compare the magnitudes of various terms in the governing equations. For example, in a typical midlatitude synoptic2 cyclone the surface pressure might fluctuate by 10 hPa over a horizontal distance of 1000 km. Designating the amplitude of the horizontal pressure fluctu- ation by δp, the horizontal coordinates by x and y, and the horizontal scale by L, the magnitude of the horizontal pressure gradient may be estimated by dividing δp by the length L to get
   ∂p ∂p δp , ∼ = 10 hpa/103 km 10−3 Pa m−1 ∂x ∂y L Pressure fluctuations of similar magnitudes occur in other motion systems of vastly different scale such as tornadoes, squall lines, and hurricanes. Thus, the horizon- tal pressure gradient can range over several orders of magnitude for systems of meteorological interest. Similar considerations are also valid for derivative terms involving other field variables. Therefore, the nature of the dominant terms in the governing equations is crucially dependent on the horizontal scale of the motions. In particular, motions with horizontal scales of a few kilometers or less tend to have short time scales so that terms involving the rotation of the earth are negli- gible, while for larger scales they become very important. Because the character of atmospheric motions depends so strongly on the horizontal scale, this scale provides a convenient method for the classification of motion systems. Table 1.4 classifies examples of various types of motions by horizontal scale for the spectral region from 10−7 to 107 m. In the following chapters, scaling arguments are used extensively in developing simplifications of the governing equations for use in modeling various types of motion systems. 1.4 FUNDAMENTAL FORCES The motions of the atmosphere are governed by the fundamental physical laws of conservation of mass, momentum, and energy. In Chapter 2, these principles are applied to a small volume element of the atmosphere in order to obtain the 2 The term synoptic designates the branch of meteorology that deals with the analysis of observations taken over a wide area at or near the same time. This term is commonly used (as here) to designate the characteristic scale of the disturbances that are depicted on weather maps. January 27, 2004 13:54 Elsevier/AID aid 1.4 fundamental forces 5 Table 1.4 Scales of Atmospheric Motions Type of motion Horizontal scale (m) Molecular mean free path 10−7 Minute turbulent eddies 10−2 – 10−1 Small eddies 10−1 – 1 Dust devils 1 – 10 Gusts 10 – 102 Tornadoes 102 Cumulonimbus clouds 103 Fronts, squall lines 10 – 105 4 Hurricanes 105 Synoptic cyclones 106 Planetary waves 107 governing equations. However, before deriving the complete momentum equation it is useful to discuss the nature of the forces that influence atmospheric motions. These forces can be classified as either body forces or surface forces. Body forces act on the center of mass of a fluid parcel; they have magnitudes proportional to the mass of the parcel. Gravity is an example of a body force. Surface forces act across the boundary surface separating a fluid parcel from its surroundings; their magni- tudes are independent of the mass of the parcel. The pressure force is an example. Newton’s second law of motion states that the rate of change of momentum (i.e., the acceleration) of an object, as measured relative to coordinates fixed in space, equals the sum of all the forces acting. For atmospheric motions of meteorological interest, the forces that are of primary concern are the pressure gradient force, the gravitational force, and friction. These fundamental forces are the subject of the present section. If, as is the usual case, the motion is referred to a coordinate system rotating with the earth, Newton’s second law may still be applied provided that certain apparent forces, the centrifugal force and the Coriolis force, are included among the forces acting. The nature of these apparent forces is discussed in Section 1.5. 1.4.1 Pressure Gradient Force We consider an infinitesimal volume element of air, δV = δxδyδz, centered at the point x0 , y0 , z0 as illustrated in Fig. 1.1. Due to random molecular motions, momentum is continually imparted to the walls of the volume element by the surrounding air. This momentum transfer per unit time per unit area is just the pressure exerted on the walls of the volume element by the surrounding air. If the pressure at the center of the volume element is designated by p0 , then the pressure on the wall labeled A in Fig. 1.1 can be expressed in a Taylor series expansion as ∂p δx p0 + + higher order terms ∂x 2 January 27, 2004 13:54 Elsevier/AID aid 6 1 introduction Fig. 1.1 The x component of the pressure gradient force acting on a fluid element. Neglecting the higher order terms in this expansion, the pressure force acting on the volume element at wall A is
 ∂p δx FAx = − p0 + δy δz ∂x 2 where δyδz is the area of wall A. Similarly, the pressure force acting on the volume element at wall B is just
 ∂p δx FBx = + p0 − δy δz ∂x 2 Therefore, the net x component of this force acting on the volume is ∂p Fx = FAx + FBx = − δx δy δz ∂x Because the net force is proportional to the derivative of pressure in the direction of the force, it is referred to as the pressure gradient force.The mass m of the dif- ferential volume element is simply the density ρ times the volume: m = ρδxδyδz. Thus, the x component of the pressure gradient force per unit mass is Fx 1 ∂p =− m ρ ∂x Similarly, it can easily be shown that the y and z components of the pressure gradient force per unit mass are Fy 1 ∂p Fz 1 ∂p =− and =− m ρ ∂y m ρ ∂z January 27, 2004 13:54 Elsevier/AID aid 1.4 fundamental forces 7 so that the total pressure gradient force per unit mass is F 1 = − ∇p (1.1) m ρ It is important to note that this force is proportional to the gradient of the pressure field, not to the pressure itself. 1.4.2 Gravitational Force Newton’s law of universal gravitation states that any two elements of mass in the universe attract each other with a force proportional to their masses and inversely proportional to the square of the distance separating them. Thus, if two mass elements M and m are separated by a distance r ≡ |r| (with the vector r directed toward m as shown in Fig. 1.2), then the force exerted by mass M on mass m due to gravitation is GMm  r  Fg = − 2 (1.2) r r where G is a universal constant called the gravitational constant. The law of grav- itation as expressed in (1.2) actually applies only to hypothetical “point” masses since for objects of finite extent r will vary from one part of the object to another. However, for finite bodies, (1.2) may still be applied if |r| is interpreted as the distance between the centers of mass of the bodies. Thus, if the earth is designated as mass M and m is a mass element of the atmosphere, then the force per unit mass exerted on the atmosphere by the gravitational attraction of the earth is Fg GM  r  ≡ g∗ = − 2 (1.3) m r r Fig. 1.2 Two spherical masses whose centers are separated by a distance r. January 27, 2004 13:54 Elsevier/AID aid 8 1 introduction In dynamic meteorology it is customary to use the height above mean sea level as a vertical coordinate. If the mean radius of the earth is designated by a and the distance above mean sea level is designated by z, then neglecting the small departure of the shape of the earth from sphericity, r = a + z. Therefore, (1.3) can be rewritten as g0∗ g∗ = (1.4) (1 + z/a)2 where g0∗ = −(GM/a 2 )(r/r) is the gravitational force at mean sea level. For meteorological applications, z  a so that with negligible error we can let g∗ = g0∗ and simply treat the gravitational force as a constant. 1.4.3 Viscous Force Any real fluid is subject to internal friction (viscosity), which causes it to resist the tendency to flow. Although a complete discussion of the resulting viscous force would be rather complicated, the basic physical concept can be illustrated by a simple experiment. A layer of incompressible fluid is confined between two horizontal plates separated by a distance l as shown in Fig. 1.3. The lower plate is fixed and the upper plate is placed into motion in the x direction at a speed u0. Viscosity forces the fluid particles in the layer in contact with the plate to move at the velocity of the plate. Thus, at z = l the fluid moves at speed u(l) = u0 , and at z = 0 the fluid is motionless. The force tangential to the upper plate required to keep it in uniform motion turns out to be proportional to the area of the plate, the velocity, and the inverse of the distance separating the plates. Thus, we may write F = µAu0 / l where µ is a constant of proportionality, the dynamic viscosity coefficient. This force must just equal the force exerted by the upper plate on the fluid immediately below it. For a state of uniform motion, every horizontal layer of fluid of depth δz must exert the same force F on the fluid below. This may be Fig. 1.3 One-dimensional steady-state viscous shear flow. January 27, 2004 13:54 Elsevier/AID aid 1.4 fundamental forces 9 expressed in the form F = µAδu/δz where δu = u0 δz/ l is the velocity shear across the layer δz. The viscous force per unit area, or shearing stress, can then be defined as δu ∂u τzx = lim µ =µ δz→0 δz ∂z where subscripts indicate that τzx is the component of the shearing stress in the x direction due to vertical shear of the x velocity component. From the molecular viewpoint, this shearing stress results from a net downward transport of momentum by the random motion of the molecules. Because the mean x momentum increases with height, molecules passing downward through a horizontal plane at any instant carry more momentum than those passing upward through the plane. Thus, there is a net downward transport of x momentum. This downward momentum transport per unit time per unit area is simply the shearing stress. In a similar fashion, random molecular motions will transport heat down a mean temperature gradient and trace constituents down mean mixing ratio gradients. In these cases the transport is referred to as molecular diffusion. Molecular diffusion always acts to reduce irregularities in the field being diffused. In the simple two-dimensional steady-state motion example given above there is no net viscous force acting on the elements of fluid, as the shearing stress acting across the top boundary of each fluid element is just equal and opposite to that acting across the lower boundary. For the more general case of nonsteady two-dimensional shear flow in an incompressible fluid, we may calculate the viscous force by again considering a differential volume element of fluid centered at (x, y, z) with sides δxδyδz as shown in Fig. 1.4. If the shearing stress in the x direction acting through the center of the element is designated τzx , then the stress acting across the upper boundary on the fluid below may be written approximately as ∂τzx δz τzx + ∂z 2 while the stress acting across the lower boundary on the fluid above is   ∂τzx δz − τzx − ∂z 2 (This is just equal and opposite to the stress acting across the lower boundary on the fluid below.) The net viscous force on the volume element acting in the x direction is then given by the sum of the stresses acting across the upper boundary on the fluid below and across the lower boundary on the fluid above:

 ∂τzx δz ∂τzx δz τzx + δx δy − τzx − δx δy ∂z 2 ∂z 2 January 27, 2004 13:54 Elsevier/AID aid 10 1 introduction Fig. 1.4 The x component of the vertical shearing stress on a fluid element. Dividing this expression by the mass ρδxδyδz, we find that the viscous force per unit mass due to vertical shear of the component of motion in the x direction is
 1 ∂τzx 1 ∂ ∂u = µ ρ ∂z ρ ∂z ∂z For constant µ, the right-hand side just given above may be simplified to ν∂ 2 u/∂z2 , where ν = µ/ρ is the kinematic viscosity coefficient. For standard atmosphere conditions at sea level, ν = 1.46 × 10−5 m2 s −1. Derivations anal- ogous to that shown in Fig. 1.4 can be carried out for viscous stresses acting in other directions. The resulting frictional force components per unit mass in the three Cartesian coordinate directions are  2  ∂ u ∂ 2u ∂ 2u Frx = ν + + ∂x 2 ∂y 2 ∂z2  2  ∂ v ∂ 2v ∂ 2v Fry = ν + + (1.5) ∂x 2 ∂y 2 ∂z2  2  ∂ w ∂ 2w ∂ 2w Frz = ν + + 2 ∂x 2 ∂y 2 ∂z For the atmosphere below 100 km, ν is so small that molecular viscosity is negligible except in a thin layer within a few centimeters of the earth’s surface where the vertical shear is very large. Away from this surface molecular boundary layer, momentum is transferred primarily by turbulent eddy motions. These are discussed in Chapter 5. 1.5 NONINERTIAL REFERENCE FRAMESAND “APPARENT” FORCES In formulating the laws of atmospheric dynamics it is natural to use a geocentric reference frame, that is, a frame of reference at rest with respect to the rotating earth. Newton’s first law of motion states that a mass in uniform motion relative to January 27, 2004 13:54 Elsevier/AID aid 1.5 noninertial reference frames and “apparent” forces 11 a coordinate system fixed in space will remain in uniform motion in the absence of any forces. Such motion is referred to as inertial motion; and the fixed reference frame is an inertial, or absolute, frame of reference. It is clear, however, that an object at rest or in uniform motion with respect to the rotating earth is not at rest or in uniform motion relative to a coordinate system fixed in space. Therefore, motion that appears to be inertial motion to an observer in a geocentric reference frame is really accelerated motion. Hence, a geocentric reference frame is a noninertial reference frame. Newton’s laws of motion can only be applied in such a frame if the acceleration of the coordinates is taken into account. The most satisfactory way of including the effects of coordinate acceleration is to introduce “apparent” forces in the statement of Newton’s second law. These apparent forces are the inertial reaction terms that arise because of the coordinate acceleration. For a coordinate system in uniform rotation, two such apparent forces are required: the centrifugal force and the Coriolis force. 1.5.1 Centripetal Acceleration and Centrifugal Force A ball of mass m is attached to a string and whirled through a circle of radius r at a constant angular velocity ω. From the point of view of an observer in inertial space the speed of the ball is constant, but its direction of travel is continuously changing so that its velocity is not constant. To compute the acceleration we consider the change in velocity δV that occurs for a time increment δt during which the ball rotates through an angle δθ as shown in Fig. 1.5. Because δθ is also the angle between the vectors V and V + δV, the magnitude of δV is just |δV| = |V| δθ. If we divide by δt and note that in the limit δt → 0, δV is directed toward the axis of rotation, we obtain DV Dθ  r  = |V| − Dt Dt r Fig. 1.5 Centripetal acceleration is given by the rate of change of the direction of the velocity vector, which is directed toward the axis of rotation, as illustrated here by δV. January 27, 2004 13:54 Elsevier/AID aid 12 1 introduction However, |V| = ωr and Dθ/Dt = ω, so that DV = −ω2 r (1.6) Dt Therefore, viewed from fixed coordinates the motion is one of uniform accel- eration directed toward the axis of rotation and equal to the square of the angular velocity times the distance from the axis of rotation. This acceleration is called centripetal acceleration. It is caused by the force of the string pulling the ball. Now suppose that we observe the motion in a coordinate system rotating with the ball. In this rotating system the ball is stationary, but there is still a force acting on the ball, namely the pull of the string. Therefore, in order to apply Newton’s second law to describe the motion relative to this rotating coordinate system, we must include an additional apparent force, the centrifugal force, which just balances the force of the string on the ball. Thus, the centrifugal force is equivalent to the inertial reaction of the ball on the string and just equal and opposite to the centripetal acceleration. To summarize, observed from a fixed system the rotating ball undergoes a uniform centripetal acceleration in response to the force exerted by the string. Observed from a system rotating along with it, the ball is stationary and the force exerted by the string is balanced by a centrifugal force. 1.5.2 Gravity Force An object at rest on the surface of the earth is not at rest or in uniform motion relative to an inertial reference frame except at the poles. Rather, an object of unit mass at rest on the surface of the earth is subject to a centripetal acceleration directed toward the axis of rotation of the earth given by −2 R, where R is the position vector from the axis of rotation to the object and  = 7.292 × 10−5 rad s−1 is the angular speed of rotation of the earth.3 Since except at the equator and poles the centripetal acceleration has a component directed poleward along the horizontal surface of the earth (i.e., along a surface of constant geopotential), there must be a net horizontal force directed poleward along the horizontal to sustain the horizontal component of the centripetal acceleration. This force arises because the rotating earth is not a sphere, but has assumed the shape of an oblate spheroid in which there is a poleward component of gravitation along a constant geopotential surface just sufficient to account for the poleward component of the centripetal acceleration at each latitude for an object at rest on the surface of the earth. In other words, from the point of view of an observer in an inertial reference frame, geopotential 3 The earth revolves around its axis once every sidereal day, which is equal to 23 h 56 min 4 s (86,164 s). Thus,  = 2π/(86, 164 s) = 7.292 × 10−5 rad s−1. January 27, 2004 13:54 Elsevier/AID aid 1.5 noninertial reference frames and “apparent” forces 13 Fig. 1.6 Relationship between the true gravitation vector g* and gravity g. For an idealized homogeneous spherical earth, g* would be directed toward the center of the earth. In reality, g* does not point exactly to the center except at the equator and the poles. Gravity, g, is the vector sum of g* and the centrifugal force and is perpendicular to the level surface of the earth, which approximates an oblate spheroid. surfaces slope upward toward the equator (see Fig. 1.6). As a consequence, the equatorial radius of the earth is about 21 km larger than the polar radius. Viewed from a frame of reference rotating with the earth, however, a geopo- tential surface is everywhere normal to the sum of the true force of gravity, g∗ , and the centrifugal force 2 R (which is just the reaction force of the centripetal acceleration). A geopotential surface is thus experienced as a level surface by an object at rest on the rotating earth. Except at the poles, the weight of an object of mass m at rest on such a surface, which is just the reaction force of the earth on the object, will be slightly less than the gravitational force mg∗ because, as illustrated in Fig. 1.6, the centrifugal force partly balances the gravitational force. It is, therefore, convenient to combine the effects of the gravitational force and centrifugal force by defining gravity g such that g ≡ −gk ≡ g∗ + 2 R (1.7) where k designates a unit vector parallel to the local vertical. Gravity, g, sometimes referred to as “apparent gravity,” will here be taken as a constant (g = 9.81 m s −2 ). Except at the poles and the equator, g is not directed toward the center of the earth, but is perpendicular to a geopotential surface as indicated by Fig. 1.6. True gravity g∗ , however, is not perpendicular to a geopotential surface, but has a horizontal component just large enough to balance the horizontal component of 2 R. Gravity can be represented in terms of the gradient of a potential function , which is just the geopotential referred to above: ∇ = −g However, because g = −gk where g ≡ |g|, it is clear that = (z) and d /dz = g. Thus horizontal surfaces on the earth are surfaces of constant geopo- tential. If the value of geopotential is set to zero at mean sea level, the geopotential January 27, 2004 13:54 Elsevier/AID aid 14 1 introduction (z) at height z is just the work required to raise a unit mass to height z from mean sea level:  z = gdz (1.8) 0 Despite the fact that the surface of the earth bulges toward the equator, an object at rest on the surface of the rotating earth does not slide “downhill” toward the pole because, as indicated above, the poleward component of gravitation is balanced by the equatorward component of the centrifugal force. However, if the object is put into motion relative to the earth, this balance will be disrupted. Consider a frictionless object located initially at the North pole. Such an object has zero angular momentum about the axis of the earth. If it is displaced away from the pole in the absence of a zonal torque, it will not acquire rotation and hence will feel a restoring force due to the horizontal component of true gravity, which, as indicated above is equal and opposite to the horizontal component of the centrifugal force for an object at rest on the surface of the earth. Letting R be the distance from the pole, the horizontal restoring force for a small displacement is thus −2 R, and the object’s acceleration viewed in the inertial coordinate system satisfies the equation for a simple harmonic oscillator: d 2R + 2 R = 0 (1.9) dt 2 The object will undergo an oscillation of period 2π/  along a path that will appear as a straight line passing through the pole to an observer in a fixed coordinate system, but will appear as a closed circle traversed in 1/2 day to an observer rotating with the earth (Fig. 1.7). From the point of view of an earthbound observer, there is an apparent deflection force that causes the object to deviate to the right of its direction of motion at a fixed rate. 1.5.3 The Coriolis Force and the Curvature Effect Newton’s second law of motion expressed in coordinates rotating with the earth can be used to describe the force balance for an object at rest on the surface of the earth, provided that an apparent force, the centrifugal force, is included among the forces acting on the object. If, however, the object is in motion along the surface of the earth, additional apparent forces are required in the statement of Newton’s second law. Suppose that an object of unit mass, initially at latitude φ moving zonally at speed u, relative to the surface of the earth, is displaced in latitude or in altitude by an impulsive force.As the object is displaced it will conserve its angular momentum in the absence of a torque in the east–west direction. Because the distance R to the axis of rotation changes for a displacement in latitude or altitude, the absolute angular velocity,  + u/R, must change if the object is to conserve its absolute January 27, 2004 13:54 Elsevier/AID aid 1.5 noninertial reference frames and “apparent” forces 15 0° 0° 6 hr 3 hr A A B 0° 9 hr B 12 hr A B A C C 0° Fig. 1.7 Motion of a frictionless object launched from the north pole along the 0˚ longitude meridian at t = 0, as viewed in fixed and rotating reference frames at 3, 6, 9, and 12 h after launch. The horizontal dashed line marks the position that the 0˚ longitude meridian had at t = 0, and short dashed lines show its position in the fixed reference frame at subsequent 3 h intervals. Horizontal arrows show 3 h displacement vectors as seen by an observer in the fixed reference frame. Heavy curved arrows show the trajectory of the object as viewed by an observer in the rotating system. Labels A, B and C show the position of the object relative to the rotating coordinates at 3 h intervals. In the fixed coordinate frame the object oscillates back and forth along a straight line under the influence of the restoring force provided by the horizontal component of gravitation. The period for a complete oscillation is 24 h (only 1/2 period is shown). To an observer in rotating coordinates, however, the motion appears to be at constant speed and describes a complete circle in a clockwise direction in 12 h. angular momentum. Because  is constant, the relative zonal velocity must change. Thus, the object behaves as though a zonally directed deflection force were acting on it. The form of the zonal deflection force can be obtained by equating the total angular momentum at the initial distance R to the total angular momentum at the displaced distance R + δR: 
 u 2 u + δu + R = + (R + δR)2 R R + δR where δu is the change in eastward relative velocity after displacement. Expand- ing the right-hand side, neglecting second-order differentials, and solving for δu gives u δu = −2δR − δR R January 27, 2004 13:54 Elsevier/AID aid 16 1 introduction Noting that R = a cos φ, where a is the radius of the earth and φ is latitude, dividing through by the time increment δt and taking the limit as δt → 0, gives in the case of a meridional displacement in which δR = − sin φδy (see Fig. 1.8):
   Dy Du u uv = 2 sin φ + tan φ = 2v sin φ + tan φ (1.10a) Dt a Dt a and for a vertical displacement in which δR = + cos φδz:
  Du u  Dz uw = − 2 cos φ + = −2w cos φ − (1.10b) Dt a Dt a where v = Dy/Ddt and w = Dz/Dt are the northward and upward velocity components, respectively. The first terms on the right in (1.10a) and (1.10b) are the zonal components of the Coriolis force for meridional and vertical motions, respectively. The second terms on the right are referred to as metric terms or curvature effects. These arise from the curvature of the earth’s surface. A similar argument can be used to obtain the meridional component of the Coriolis force. Suppose now that the object is set in motion in the eastward direction by an impulsive force. Because the object is now rotating faster than the earth, the centrifugal force on the object will be increased. Letting R be the position vector Ω R0 φ0 δR φ0 a ius ad δy r rth R0+δR Ea us a h radi Eart φ0 φ0+δφ Fig. 1.8 Relationship of δR and δy = aδφ for an equatorward displacement. January 27, 2004 13:54 Elsevier/AID aid 1.5 noninertial reference frames and “apparent” forces 17 from the axis of rotation to the object, the excess of the centrifugal force over that for an object at rest is  u 2 2uR u2 R + R − 2 R = + 2 R R R The terms on the right represent deflecting forces, which act outward along the vector R (i.e., perpendicular to the axis of rotation). The meridional and vertical components of these forces are obtained by taking meridional and ver- tical components of R as shown in Fig. 1.9 to yield
 Dv u2 = −2u sin φ − tan φ (1.11a) Dt a
 Dw u2 = 2u cos φ + (1.11b) Dt a The first terms on the right are the meridional and vertical components, respec- tively, of the Coriolis forces for zonal motion; the second terms on the right are again the curvature effects. For synoptic scale motions |u|  R, the last terms in (1.10a) and (1.11a) can be neglected in a first approximation. Therefore, relative horizontal motion produces a horizontal acceleration perpendicular to the direction of motion given by
 Du = 2v sin φ = f v (1.12a) Dt Co
 Dv = −2u sin φ = −f u (1.12b) Dt Co where f ≡ 2 sin φ is the Coriolis parameter. Fig. 1.9 Components of the Coriolis force due to relative motion along a latitude circle. January 27, 2004 13:54 Elsevier/AID aid 18 1 introduction The subscript Co indicates that the acceleration is the part of the total acceleration due only to the Coriolis force. Thus, for example, an object moving eastward in the horizontal is deflected equatorward by the Coriolis force, whereas a westward moving object is deflected poleward. In either case the deflection is to the right of the direction of motion in the Northern Hemisphere and to the left in the Southern Hemisphere. The vertical component of the Coriolis force in (1.11b) is ordinarily much smaller than the gravitational force so that its only effect is to cause a very minor change in the apparent weight of an object depending on whether the object is moving eastward or westward. The Coriolis force is negligible for motions with time scales that are very short compared to the period of the earth’s rotation (a point that is illustrated by several problems at the end of the chapter). Thus, the Coriolis force is not important for the dynamics of individual cumulus clouds, but is essential to the understanding of longer time scale phenomena such as synoptic scale systems. The Coriolis force must also be taken into account when computing long-range missile or artillery trajectories. As an example, suppose that a ballistic missile is fired due eastward at 43◦ N latitude (f = 10−4 s −1 at 43◦ N). If the missile travels 1000 km at a horizontal speed u0 = 1000 m s −1 , by how much is the missile deflected from its eastward path by the Coriolis force? Integrating (1.12b) with respect to time we find that v = −f u0 t (1.13) where it is assumed that the deflection is sufficiently small so that we may let f and u0 be constants. To find the total displacement we must integrate (1.13) with respect to time:  t  y0 +δy  t vdt = dy = −f u0 tdt 0 y0 0 Thus, the total displacement is δy = −f u0 t 2 /2 = −50 km Therefore, the missile is deflected southward by 50 km due to the Coriolis effect. Further examples of the deflection of objects by the Coriolis force are given in some of the problems at the end of the chapter. The x and y components given in (1.12a) and (1.12b) can be combined in vector form as
 DV = −f k × V (1.14) Dt Co where V ≡ (u, v) is the horizontal velocity, k is a vertical unit vector, and the subscript Co indicates that the acceleration is due solely to the Coriolis force. Since −k × V is a vector rotated 90˚ to the right of V, (1.14) clearly shows the January 27, 2004 13:54 Elsevier/AID aid 1.6 structure of the static atmosphere 19 deflection character of the Coriolis force. The Coriolis force can only change the direction of motion, not the speed of motion. 1.5.4 Constant Angular Momentum Oscillations Suppose an object initially at rest on the earth at the point (x0 , y0 ) is impulsively propelled along the x axis with a speed V at time t = 0. Then from (1.12a) and (1.12b), the time evolution of the velocity is given by u = V cos f t and v = −V sin f t. However, because u = Dx/Dt and v = Dy/Dt, integration with respect to time gives the position of the object at time t as V V x − x0 = sin f t and y − y0 = (cos f t − 1) (1.15a,b) f f where the variation of f with latitude is here neglected. Equations (1.15a) and (1.15b) show that in the Northern Hemisphere, where f is positive, the object orbits clockwise (anticyclonically) in a circle of radius R = V /f about the point (x0 , y0 − V /f ) with a period given by τ = 2π R/V = 2π/f = π/( sin φ) (1.16) Thus, an object displaced horizontally from its equilibrium position on the sur- face of the earth under the influence of the force of gravity will oscillate about its equilibrium position with a period that depends on latitude and is equal to one sidereal day at 30˚ latitude and 1/2 sidereal day at the pole. Constant angular momentum oscillations (often referred to misleadingly as “inertial oscillations”) are commonly observed in the oceans, but are apparently not of importance in the atmosphere. 1.6 STRUCTURE OF THE STATIC ATMOSPHERE The thermodynamic state of the atmosphere at any point is determined by the values of pressure, temperature, and density (or specific volume) at that point. These field variables are related to each other by the equation of state for an ideal gas. Letting p, T ρ, and α(≡ ρ −1 ) denote pressure, temperature, density, and specific volume, respectively, we can express the equation of state for dry air as pα = RT or p = ρRT (1.17) where R is the gas constant for dry air (R = 287 J kg−1 K −1 ). January 27, 2004 13:54 Elsevier/AID aid 20 1 introduction Fig. 1.10 Balance of forces for hydrostatic equi- librium. Small arrows show the upward and downward forces exerted by air pressure on the air mass in the shaded block. The downward force exerted by gravity on the air in the block is given by ρgdz, whereas the net pressure force given by the difference between the upward force across the lower surface and the downward force across the upper surface is −dp. Note that dp is negative, as pressure decreases with height. (After Wallace and Hobbs, 1977.) 1.6.1 The Hydrostatic Equation In the absence of atmospheric motions the gravity force must be exactly balanced by the vertical component of the pressure gradient force. Thus, as illustrated in Fig. 1.10, dp/dz = −ρg (1.18) This condition of hydrostatic balance provides an excellent approximation for the vertical dependence of the pressure field in the real atmosphere. Only for intense small-scale systems such as squall lines and tornadoes is it necessary to consider departures from hydrostatic balance. Integrating (1.18) from a height z to the top of the atmosphere we find that  ∞ p(z) = ρgdz (1.19) z so that the pressure at any point is simply equal to the weight of the unit cross section column of air overlying the point. Thus, mean sea level pressure p(0) = 1013.25 hPa is simply the average weight per square meter of the total atmospheric column.4 It is often useful to express the hydrostatic equation in terms of the geopotential rather than the geometric height. Noting from (1.8) that d = g dz and from, (1.17) that α = RT /p, we can express the hydrostatic equation in the form gdz = d = −(RT /p)dp = −RT d ln p (1.20) Thus, the variation of geopotential with respect to pressure depends only on tem- perature. Integration of (1.20) in the vertical yields a form of the hypsometric equation:  p1 (z2 ) − (z1 ) = g0 (Z2 − Z1 ) = R T d ln p (1.21) p2 4 For computational convenience, the mean surface pressure is often assumed to equal 1000 hPa. January 27, 2004 13:54 Elsevier/AID aid 1.6 structure of the static atmosphere 21 Here Z ≡ (z)/g0 , is the geopotential height, where g0 ≡ 9.80665 m s−2 is the global average of gravity at mean sea level. Thus in the troposphere and lower stratosphere, Z is numerically almost identical to the geometric height z. In terms of Z the hypsometric equation becomes  p1 R ZT ≡ Z2 − Z1 = T d ln p (1.22) g0 p2 where ZT is the thickness of the atmospheric layer between the pressure surfaces p2 and p1. Defining a layer mean temperature  p1  p1 −1 T = T d ln p d ln p p2 p2 and a layer mean scale height H ≡ R T /g0 we have from (1.22) ZT = H ln(p1 /p2 ) (1.23) Thus the thickness of a layer bounded by isobaric surfaces is proportional to the mean temperature of the layer. Pressure decreases more rapidly with height in a cold layer than in a warm layer. It also follows immediately from (1.23) that in an isothermal atmosphere of temperature T, the geopotential height is proportional to the natural logarithm of pressure normalized by the surface pressure, Z = −H ln(p/p0 ) (1.24) where p0 is the pressure at Z = 0. Thus, in an isothermal atmosphere the pressure decreases exponentially with geopotential height by a factor of e−1 per scale height, p(Z) = p(0)e−Z/H 1.6.2 Pressure as a Vertical Coordinate From the hydrostatic equation (1.18), it is clear that a single valued monotonic relationship exists between pressure and height in each vertical column of the atmosphere. Thus we may use pressure as the independent vertical coordinate and height (or geopotential) as a dependent variable. The thermodynamic state of the atmosphere is then specified by the fields of (x, y, p, t) and T (x, y, p, t). Now the horizontal components of the pressure gradient force given by (1.1) are evaluated by partial differentiation holding z constant. However, when pres- sure is used as the vertical coordinate, horizontal partial derivatives must be January 27, 2004 13:54 Elsevier/AID aid 22 1 introduction Fig. 1.11 Slope of pressure surfaces in the x, z plane. evaluated holding p constant. Transformation of the horizontal pressure gradi- ent force from height to pressure coordinates may be carried out with the aid of Fig. 1.11. Considering only the x, z plane, we see from Fig. 1.11 that    
 (p0 + δp) − p0 (p0 + δp) − p0 δz = δx z δz x δx p where subscripts indicate variables that remain constant in evaluating the differ- entials. Thus, for example, in the limit δz → 0  
 (p0 + δp) − p0 ∂p → − δz x ∂z x where the minus sign is included because δz < 0 for δp > 0. Taking the limits δx, δz → 0 we obtain5


 ∂p ∂p ∂z = − ∂x z ∂z x ∂x p which after substitution from the hydrostatic equation (1.18) yields


 1 ∂p ∂z ∂ − = −g =− (1.25) ρ ∂x z ∂x p ∂x p Similarly, it is easy to show that

 1 ∂p ∂ − =− (1.26) ρ ∂y z ∂y p 5 It is important to note the minus sign on the right in this expression! January 27, 2004 13:54 Elsevier/AID aid 1.6 structure of the static atmosphere 23 Thus in the isobaric coordinate system the horizontal pressure gradient force is measured by the gradient of geopotential at constant pressure. Density no longer appears explicitly in the pressure gradient force; this is a distinct advantage of the isobaric system. 1.6.3 A Generalized Vertical Coordinate Any single-valued monotonic function of pressure or height may be used as the independent vertical coordinate. For example, in many numerical weather predic- tion models, pressure normalized by the pressure at the ground [σ ≡ p(x, y, z, t)/ ps (x, y, t)] is used as a vertical coordinate. This choice guarantees that the ground is a coordinate surface (σ ≡ 1) even in the presence of spatial and temporal surface pressure variations. Thus, this so-called σ coordinate system is particularly useful in regions of strong topographic variations. We now obtain a general expression for the horizontal pressure gradient, which is applicable to any vertical coordinate s = s(x, y, z, t) that is a single-valued monotonic function of height. Referring to Fig. 1.12 we see that for a horizontal distance δx the pressure difference evaluated along a surface of constant s is related to that evaluated at constant z by the relationship pC − pA pC − pB δz p B − pA = + δx δz δx δx Taking the limits as δx, δz → 0 we obtain


 ∂p ∂p ∂z ∂p = + (1.27) ∂x s ∂z ∂x s ∂x z Fig. 1.12 Transformation of the pressure gradient force to s coordinates. January 27, 2004 13:54 Elsevier/AID aid 24 1 introduction Using the identity ∂p/∂z = (∂s/∂z)(∂p/∂s), we can express (1.27) in the alternate form



 ∂p ∂p ∂s ∂z ∂p = + (1.28) ∂x s ∂x z ∂z ∂x s ∂s In later chapters we will apply (1.27) or (1.28) and similar expressions for other fields to transform the dynamical equations to several different vertical coordinate systems. PROBLEMS 1.1. Neglecting the latitudinal variation in the radius of the earth, calculate the angle between the gravitational force and gravity vectors at the surface of the earth as a function of latitude. What is the maximum value of this angle? 1.2. Calculate the altitude at which an artificial satellite orbiting in the equatorial plane can be a synchronous satellite (i.e., can remain above the same spot on the surface of the earth). 1.3. An artificial satellite is placed into a natural synchronous orbit above the equator and is attached to the earth below by a wire. A second satellite is attached to the first by a wire of the same length and is placed in orbit directly above the first at the same angular velocity. Assuming that the wires have zero mass, calculate the tension in the wires per unit mass of satellite. Could this tension be used to lift objects into orbit with no additional expenditure of energy? 1.4. A train is running smoothly along a curved track at the rate of 50 m s −1. A passenger standing on a set of scales observes that his weight is 10% greater than when the train is at rest. The track is banked so that the force acting on the passenger is normal to the floor of the train. What is the radius of curvature of the track? 1.5. If a baseball player throws a ball a horizontal distance of 100 m at 30◦ latitude in 4 s, by how much is it deflected laterally as a result of the rotation of the earth? 1.6. Two balls 4 cm in diameter are placed 100 m apart on a frictionless horizontal plane at 43◦ N. If the balls are impulsively propelled directly at each other with equal speeds, at what speed must they travel so that they just miss each other? 1.7. A locomotive of 2 × 105 kg mass travels 50 m s −1 along a straight horiz