Fluid Dynamics - Fundamentals and Applications [1 ed.] 9783030495619, 9783030495626

Table of contents :
Preface
Contents
Nomenclature
1 Fundamentals 1
1.1 Hydrostatics
1.2 Some Applications
1.2.1 Atmospheric Pressure
1.2.2 A First Glimpse at Stability
1.2.3 Archimedes' Law
1.2.4 Models of Earth's Interior
1.2.5 Shape of the Earth
1.3 Pressure Distribution in Rigid Body Motion
References
2 Fundamentals 2
2.1 The Continuity Equation
2.2 The Equations of Motion
2.3 The Bernoulli Theorem
2.4 Simple Applications
2.4.1 Flow from a Tank
2.4.2 The Pitot Tube
2.5 Thrust for a Propeller
2.6 Circulation and Vorticity
2.6.1 Kelvin Circulation Theorem
References
3 Fundamentals 3
3.1 Viscosity
3.2 The Equation of Navier Stokes
3.2.1 The Diffusion Equation
3.3 The Reynolds Number
3.3.1 The Blasius Boundary Layer
3.4 Plane Poiseuille Flow and Pipe Flow: A Plumber Application …
3.5 Flow Past a Sphere at Low Reynolds Number
3.6 Life at Low Reynolds Number
References
4 Aerodynamics and All That
4.1 Velocity Potential
4.2 Elementary Potential Flows
4.2.1 Source and Sink of a Fluid
4.2.2 Line Vortex
4.2.3 Plane Doublet
4.2.4 Non-lifting Flow over a Circular Cylinder
4.2.5 Lifting Flow over a Rotating Circular Cylinder
4.2.6 The Kutta-Zhukhovsky Lift Theorem
4.2.7 Lift from the Deflection of Air Stream
References
5 Waves
5.1 Sound Waves
5.2 Supersonic Flow Past Thin Plates
5.3 Wave Velocities and Energy
5.3.1 Energy and Intensity of an Acoustic Wave
5.4 Gravity Waves
5.4.1 Deep Water Waves
5.4.2 Shallow Water Waves
5.4.3 Capillary Waves
5.5 Energy Flux in Water Gravity Waves
5.6 Ship Waves
5.7 Wave Drag on Ships
5.8 Non-linear Shallow Water Waves: Solitons
References
6 Instabilities
6.1 Stability, Instability and Bifurcations
6.2 Gravitational (Jeans) Instability
6.3 Instability of Shear Flows
6.3.1 Rayleigh–Taylor Instability
6.3.2 Kelvin–Helmholtz Instability
6.4 Stratified Shear Flows
6.4.1 The Richardson Number
6.4.2 A Useful Example
References
7 Non-linearities, Randomness and Chaos
7.1 Non-linear Behaviour and Navier–Stokes Equation
7.2 Stochastic Behaviour
7.2.1 White or Gaussian Noise
7.3 A Couple of Examples of Randomness
7.3.1 The Brownian Motion
7.3.2 Climate Variability
7.4 Chaotic Behaviour
7.4.1 Generating Chaos
7.4.2 Lyapunov Exponents
7.4.3 Visualization of Chaos: Attractors, Poincarè Sections and Fractals
7.5 Chaos in a Fluid: The Loop Oscillator
7.6 The Lorenz System
7.7 ENSO: A Very Complex System
References
8 Turbulence
8.1 Some General Matter
8.2 Statistical Description of Turbulence
8.2.1 Examples of Probability Distribution
8.3 Smoke Plumes as an Application of Pdf
8.4 The Atmospheric Boundary Layer: The Mixing Length
8.5 Two-Dimensional Turbulence
8.5.1 Energy and Enstrophy Transfer
References
9 Magnetohydrodynamics
9.1 Some Preliminaries
9.1.1 Particles Motion
9.1.2 Plasma Oscillations
9.1.3 Debye Length
9.2 The MHD Equations
9.3 Vorticity and Magnetic Field
9.4 The Frozen Field
9.5 The Planetary Dynamos
9.5.1 The Faraday Self-Excited Dynamo
9.6 The Turbulent Dynamo
9.7 Magnetohydrodynamic Waves
9.7.1 Dispersion Relations for MHD Waves
References
Index

Citation preview

Guido Visconti Paolo Ruggieri

Fluid Dynamics Fundamentals and Applications

Fluid Dynamics

Guido Visconti Paolo Ruggieri •

Fluid Dynamics Fundamentals and Applications

123

Guido Visconti Scienze Fisiche e Chimiche Università dell'Aquila Coppito, AQ, Italy

Paolo Ruggieri Department of Physics and Astronomy University of Bologna Bologna, Italy Centro Euro-Mediterraneo sui Cambiamenti Climatici Bologna, Italy

ISBN 978-3-030-49561-9 ISBN 978-3-030-49562-6 https://doi.org/10.1007/978-3-030-49562-6

(eBook)

© Springer Nature Switzerland AG 2020 This work is subject to copyright. All rights are reserved by the Publisher, whether the whole or part of the material is concerned, specifically the rights of translation, reprinting, reuse of illustrations, recitation, broadcasting, reproduction on microfilms or in any other physical way, and transmission or information storage and retrieval, electronic adaptation, computer software, or by similar or dissimilar methodology now known or hereafter developed. The use of general descriptive names, registered names, trademarks, service marks, etc. in this publication does not imply, even in the absence of a specific statement, that such names are exempt from the relevant protective laws and regulations and therefore free for general use. The publisher, the authors and the editors are safe to assume that the advice and information in this book are believed to be true and accurate at the date of publication. Neither the publisher nor the authors or the editors give a warranty, expressed or implied, with respect to the material contained herein or for any errors or omissions that may have been made. The publisher remains neutral with regard to jurisdictional claims in published maps and institutional affiliations. This Springer imprint is published by the registered company Springer Nature Switzerland AG The registered company address is: Gewerbestrasse 11, 6330 Cham, Switzerland

This book is dedicated to our past, present and future students, hoping it will help them to live on the street.

Preface

Overview As happens in almost all the physics sectors today, there are plenty of books and online material and whatever on fluid dynamics. As hinted in the subtitle of the Eckert’s book The dawn of fluid dynamics, this discipline is between science and technology and in this textbook, we decided to combine science with a proper emphasis on the applications. As a matter of fact, the book draws heavily on a course that one of us taught for a few years at an undergraduate level. The preparation of the lectures required a quite important commitment, in particular, to find insightful examples that could raise the students’ interest. Hence, we provide all along examples and applications from various fields such as aerodynamics, geophysics, life sciences, sports and so on, but often related also to everyday life. Some of us learned physics on the Feynman Lectures books and we were particularly impressed by the chapters dealing with the flow of dry water and the flow of wet water. The present book will start out from those chapters to provide the basic knowledge and we will show that even with such elements, it is possible to find some interesting applications. We very much endorse T. E. Faber’s statement in the preface of his book on Fluid Dynamics: The notion that the only way to arouse the enthusiasm of physicists is to teach them about quarks and black holes is in my view a myth.

Organization The book is organized in nine chapters and each of them has an appendix where more detailed calculations are reported (if needed) or specific examples are illustrated. The Chaps. 1–3 deal with introductory materials and, therefore, they start from hydrostatics, then continue with the most fundamental equations in fluid dynamics and end up with viscous flow. These first rudiments are enough to

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illustrate theories on the shape of the Earth and its internal structure or some most elementary model of the stellar structure. In the second chapter, we introduce some of the most important equations of fluid dynamics. These are enough not only to discuss some elementary concepts of atmospheric dynamics but also to mention the hidden relations between fluid dynamics and quantum mechanics. Viscosity requires the introduction of the Navier–Stokes equation and this is a good chance to illustrate and explain the Benard convection or the intensification of the westward current in the ocean. Starting from Chap. 4, we introduce aerodynamics and all its connections. So we deal with elementary potential flow and introduce the interactions of solid bodies with a fluid and the different theories about lift. The appendix not only reports the interesting applications to sports (golf, sailing and frisbee) but we also found some insights of the potential flows for the Great Red Spot on Jupiter. Waves (Chap. 5) have a very important role in fluid dynamics, thus we illustrate gravity waves, capillary waves up to ship waves with the associated drag. Solitons have also been mentioned. The appendix has to deal with waves in the atmosphere and the oceans with a quick look at the physics of Tsunamis. Instabilities are another fundamental topic and they are treated in Chap. 6. After discussing the most important instabilities (gravitational, Rayleigh–Taylor, Kelvin– Helmholtz), we add details in the appendix with the long list of atmospheric instabilities. We also report some applications of the Rayleigh Taylor instability to the problem of supernova remnants and to some features of the solid Earth. Chapter 7 studies non-linearities, randomness and chaos, and presents a short introduction to the stochastic physics giving a few examples. The same thing is done to introduce chaos and related concepts. Then we give some applications for the loop oscillator, the Malkus waterwheel and the ENSO phenomenon. Somewhat related to this chapter is the next one on turbulence where again there is an introduction to the statistical description before discussing Kolmogorov classical approach. Here, the appendix gives mostly details on what is done in the chapter and application to the atmospheric boundary layer. We close with Chap. 9 on magnetohydrodynamics where the main interest is the generation of planetary magnetic fields with didactic tools like the Faraday self-excited dynamos and their variation (Rikitake). We also illustrate some elementary concepts on magnetic confinement of hot plasma. Our main target is to tickle the interest of the student, giving him the awareness that complex problems can be at hand even in an introductory course.

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Acknowledgements The text is based on a course taught to third-year undergraduate physics students (one of the authors was among them) so we acknowledge their interests and stimulating interactions. A very helpful and illuminating contribution to Chap. 7 came from Prof. Raffaele D’Ambrosio of the Department of Information Engineering and Computer Science and Mathematics at the University of L’Aquila. L’Aquila, Italy Bologna, Italy

Guido Visconti Paolo Ruggieri

Contents

1 Fundamentals 1 . . . . . . . . . . . . . . . . . . . . . . . . 1.1 Hydrostatics . . . . . . . . . . . . . . . . . . . . . . . . 1.2 Some Applications . . . . . . . . . . . . . . . . . . . 1.2.1 Atmospheric Pressure . . . . . . . . . . . . 1.2.2 A First Glimpse at Stability . . . . . . . 1.2.3 Archimedes’ Law . . . . . . . . . . . . . . 1.2.4 Models of Earth’s Interior . . . . . . . . 1.2.5 Shape of the Earth . . . . . . . . . . . . . . 1.3 Pressure Distribution in Rigid Body Motion . Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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2 Fundamentals 2 . . . . . . . . . . 2.1 The Continuity Equation . 2.2 The Equations of Motion 2.3 The Bernoulli Theorem . . 2.4 Simple Applications . . . . 2.4.1 Flow from a Tank 2.4.2 The Pitot Tube . . 2.5 Thrust for a Propeller . . . 2.6 Circulation and Vorticity . 2.6.1 Kelvin Circulation Appendix . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . .

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3 Fundamentals 3 . . . . . . . . . . . . . . 3.1 Viscosity . . . . . . . . . . . . . . . . 3.2 The Equation of Navier Stokes 3.2.1 The Diffusion Equation

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3.3 The Reynolds Number . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3.3.1 The Blasius Boundary Layer . . . . . . . . . . . . . . . . . . . 3.4 Plane Poiseuille Flow and Pipe Flow: A Plumber Application of Navier Stokes Equation . . . . . . . . . . . . . . . . . . . . . . . . . . 3.5 Flow Past a Sphere at Low Reynolds Number . . . . . . . . . . . . 3.6 Life at Low Reynolds Number . . . . . . . . . . . . . . . . . . . . . . . Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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4 Aerodynamics and All That . . . . . . . . . . . . . . . . . . . . . . 4.1 Velocity Potential . . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2 Elementary Potential Flows . . . . . . . . . . . . . . . . . . . . 4.2.1 Source and Sink of a Fluid . . . . . . . . . . . . . . . 4.2.2 Line Vortex . . . . . . . . . . . . . . . . . . . . . . . . . . 4.2.3 Plane Doublet . . . . . . . . . . . . . . . . . . . . . . . . 4.2.4 Non-lifting Flow over a Circular Cylinder . . . . 4.2.5 Lifting Flow over a Rotating Circular Cylinder 4.2.6 The Kutta-Zhukhovsky Lift Theorem . . . . . . . 4.2.7 Lift from the Deflection of Air Stream . . . . . . Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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5 Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5.1 Sound Waves . . . . . . . . . . . . . . . . . . . . . . 5.2 Supersonic Flow Past Thin Plates . . . . . . . 5.3 Wave Velocities and Energy . . . . . . . . . . . 5.3.1 Energy and Intensity of an Acoustic 5.4 Gravity Waves . . . . . . . . . . . . . . . . . . . . . 5.4.1 Deep Water Waves . . . . . . . . . . . . 5.4.2 Shallow Water Waves . . . . . . . . . . 5.4.3 Capillary Waves . . . . . . . . . . . . . . 5.5 Energy Flux in Water Gravity Waves . . . . 5.6 Ship Waves . . . . . . . . . . . . . . . . . . . . . . . 5.7 Wave Drag on Ships . . . . . . . . . . . . . . . . . 5.8 Non-linear Shallow Water Waves: Solitons Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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6 Instabilities . . . . . . . . . . . . . . . . . . . . . . 6.1 Stability, Instability and Bifurcations 6.2 Gravitational (Jeans) Instability . . . . 6.3 Instability of Shear Flows . . . . . . . . 6.3.1 Rayleigh–Taylor Instability . 6.3.2 Kelvin–Helmholtz Instability

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6.4 Stratified Shear Flows . . . . . . . . 6.4.1 The Richardson Number 6.4.2 A Useful Example . . . . . Appendix . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . .

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174 175 177 179 203

7 Non-linearities, Randomness and Chaos . . . . . . . . . . . . . . . . . . 7.1 Non-linear Behaviour and Navier–Stokes Equation . . . . . . . . 7.2 Stochastic Behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.2.1 White or Gaussian Noise . . . . . . . . . . . . . . . . . . . . . 7.3 A Couple of Examples of Randomness . . . . . . . . . . . . . . . . 7.3.1 The Brownian Motion . . . . . . . . . . . . . . . . . . . . . . . 7.3.2 Climate Variability . . . . . . . . . . . . . . . . . . . . . . . . . 7.4 Chaotic Behaviour . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.1 Generating Chaos . . . . . . . . . . . . . . . . . . . . . . . . . . 7.4.2 Lyapunov Exponents . . . . . . . . . . . . . . . . . . . . . . . . 7.4.3 Visualization of Chaos: Attractors, Poincarè Sections and Fractals . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.5 Chaos in a Fluid: The Loop Oscillator . . . . . . . . . . . . . . . . . 7.6 The Lorenz System . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 7.7 ENSO: A Very Complex System . . . . . . . . . . . . . . . . . . . . . Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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8 Turbulence . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8.1 Some General Matter . . . . . . . . . . . . . . . . . . . . . . . . . 8.2 Statistical Description of Turbulence . . . . . . . . . . . . . . 8.2.1 Examples of Probability Distribution . . . . . . . . . 8.3 Smoke Plumes as an Application of Pdf . . . . . . . . . . . 8.4 The Atmospheric Boundary Layer: The Mixing Length 8.5 Two-Dimensional Turbulence . . . . . . . . . . . . . . . . . . . 8.5.1 Energy and Enstrophy Transfer . . . . . . . . . . . . Appendix . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . .

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9 Magnetohydrodynamics . . . . . . . . . . . . . . . . 9.1 Some Preliminaries . . . . . . . . . . . . . . . . . 9.1.1 Particles Motion . . . . . . . . . . . . . 9.1.2 Plasma Oscillations . . . . . . . . . . . 9.1.3 Debye Length . . . . . . . . . . . . . . . 9.2 The MHD Equations . . . . . . . . . . . . . . . . 9.3 Vorticity and Magnetic Field . . . . . . . . . . 9.4 The Frozen Field . . . . . . . . . . . . . . . . . . 9.5 The Planetary Dynamos . . . . . . . . . . . . . 9.5.1 The Faraday Self-Excited Dynamo

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9.6 The Turbulent Dynamo . . . . . . . . . . . 9.7 Magnetohydrodynamic Waves . . . . . . 9.7.1 Dispersion Relations for MHD Appendix . . . . . . . . . . . . . . . . . . . . . . . . . References . . . . . . . . . . . . . . . . . . . . . . . .

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Index . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 323

Nomenclature

a b C, Cd k l m x,X / w q r s h n, f cp cv lm Ri Si;j VA vg ,vp A a B Bf C CD CL D E

Specific volume, seismic velocity (P waves) Seismic velocity (S waves), Coriolis parameter gradient Lapse rate Wavelength Coefficient of viscosity Kinematic viscosity Angular velocity, vorticity, wave frequency Potential, velocity potential Streamfunction, wave function Density Surface tension, growth rate, standard deviation Stress Potential temperature Relative vorticity Specific heat at constant pressure Specific heat at constant volume Mixing length Richardson number Stress tensor Alfven wave velocity Group and phase velocities Moment of inertia Radius of the Earth Magnetic field, probability density Bernoulli function Moment of inertia, circulation Drag coefficient Lift coefficient Diffusion coefficient Energy, electric field

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f Fr G g j K k k L M N P p Pr Q R R Ra Re S s T U W Z z

Nomenclature

Oblateness, Coriolis parameter Froude number Gravitational constant Acceleration of gravity Current density Compressibility modulus, kinetic energy Kolmogorov wave number Wave vector Monin–Obukhov length Mach number Buoyancy frequency Palinstrophy Pressure Prandtl number Source strength Autocorrelation function Gas constant Rayleigh number Reynolds number Spectral density, salinity Entropy Temperature Potential energy Wave energy Enstrophy Altitude

Chapter 1

Fundamentals 1

1.1 Hydrostatics The main property of a fluid with respect to a solid is that the former cannot maintain a shear stress. As a result, the fluid will move under the stress with thicker fluid (like honey) moving more slowly due to the higher viscosity. For the moment, we will neglect the viscosity to respect the dry water approximation but we will talk about it extensively later. The simplest approach to study fluids is to consider the properties of a fluid at rest which correspond to the hydrostatics. In the hydrostatic approximation, the stress is always perpendicular to any surface inside the fluid and the normal force per unit area is called pressure. It is easy to show (see Appendix) that if there are no shears in the fluid, then pressure should be the same in any direction. This could be another way to state what is known as Pascal’s principle. The pressure within a fluid may change from place to place and we may consider a small portion of the fluid (an elementary parallelepiped) as in Fig. 1.1. Then we can evaluate the net force in the y direction as difference between forces on opposite faces  ∂p y xz pxz − p + ∂y 

and the net force in the y direction is then −(∂ p/∂ y)yxz. If we work out in the same way the net force on the remaining faces of the cube, we get that the net force per unit volume simply must equal ∇ p. Consider now the fluid to be in a gravitational field with potential φ, then the force per unit volume is just −ρ∇φ. At equilibrium, this force added to the pressure force must give zero − ∇ p − ρ∇φ = 0 (1.1) Equation 1.1 is the hydrostatic equation that can be solved only if the density is constant. If ρ is not constant the equation cannot be solved because while the pressure © Springer Nature Switzerland AG 2020 G. Visconti and P. Ruggieri, Fluid Dynamics, https://doi.org/10.1007/978-3-030-49562-6_1

1

2

1 Fundamentals 1

term is a pure gradient, the other gradient (∇φ) is a function of ρ and p. In this case, a convection current will arise. For constant density, the solution is simply p + ρφ = const

(1.2)

In the specific case of the gravitational potential for a stratified fluid, we have p = p0 − ρgz where g is the gravity acceleration and p0 is the pressure at z = 0. This relation signifies the fact that pressure at a reference height in a hydrostatic fluid is determined by the weight per unit surface of the fluid above it. Taking the curl of Eq. 1.1, using the rule ∇ × ( f A) = f ∇ × A + ∇f × A and remembering that the curl of the gradient is zero, we have ∇ρ × ∇φ = 0

(1.3)

This means that the gradients of ρ and φ are parallel and their level surfaces coincide at the equilibrium. In particular if ρ is only a function of pressure, the pressure surfaces (isobars) are parallel to the density surfaces and in this case the fluid is called barotropic. In meteorology, this is a quite simplifying assumption. The most general case is when the two surfaces are not parallel, and the fluid in this case is called baroclinic.

1.2 Some Applications 1.2.1 Atmospheric Pressure We can formulate an insightful example using the differential form of the hydrostatic balance (Eq. 1.1) ∂p = −ρg (1.4) ∂z if we consider a planetary atmosphere where the density is determined by the local temperature, and we assume that temperature changes linearly with height according to a law T = T0 − z where  = −∂ T /∂z is the temperature gradient. For the case of an ideal gas, the density can be expressed as a function of temperature and pressure ρ = pM/RT

1.2 Some Applications

3

where M is the molecular mass and R, the gas constant. Substituting in Eq. 1.4 and integrating, we get    β T p (1.5) = p0 T0 where β = Mg/R. It can be easily shown that for the case of constant temperature, the pressure will change according to the relation p = p0 e−z/H = RT /Mg where H is the so-called scale height and gives an indication on how rapidly the pressure changes with height. The above example gives a clear idea of the importance of thermodynamics in determining the properties of the dynamics of a fluid. We will see more examples of this aspect in the chapters ahead.

1.2.2 A First Glimpse at Stability We have mentioned pressure and density and noticed that temperature can be obtained from them. A case of special relevance is the so-called stratified fluid in which density is independent of the horizontal coordinates if the fluid is stable. Temperature differences at the same height would produce a pressure difference and consequently horizontal motions. We will have many occasions to talk about that, and limiting ourselves to the vertical coordinate we could consider convective motions. These originate when there is a negative temperature gradient. Suppose we have a temperature profile T (z) and consider a parcel of air which is moved adiabatically from z to z + dz. By definition of adiabatic motion, the parcel will conserve its entropy s(z) but it will adjust to the new pressure p  = p(z + dz) with a new density ρ(s, p  ). In order for the parcel to be stable, this density must be higher than that of the surrounding air with entropy s  = s(z + dz). The condition for stability is then ρ( p  , s) > ρ( p  , s  ) ⇒



∂ρ ∂s

 p

ds 0 that is     ∂s ∂s ds dT dp = + (1.8) dz ∂ T p dz ∂ p T dz This equation using (1.7) reduces to ds dp dT = cp −α dz dz dz Now for the hydrostatic balance αdp/dz = −g, and consequently a criterion for stability is   g dT (1.9) < − dz cp For the Earth’s atmosphere c p ≈= 103 J Kg−1 K−1 so that the convection starts at a gradient larger that 10 K/km. We will see later how to treat, in a more general fashion, convective instability in a fluid but this gives just an idea how thermodynamics is important in fluid dynamics.

1.2.3 Archimedes’ Law Another application of the hydrostatic equilibrium has to do with what we know as Archimedes’ law. Consider a floating body completely submerged as in Fig. 1.1 where Fv (1) and Fv (2) are the net forces on the upper surface (1) and lower surface (2), respectively. The net force on the body (buoyancy) will be Fb = Fv (2) − Fv (1)

(1.10)

where Fv (1) is determined by the weight of the fluid above (1) and Fv (2) is the weight of the fluid above (2). The two forces can be expressed as an integral

Fig. 1.1 Pressure forces on a fluid element Δz

Δ

Δx Δy

1.2 Some Applications

5

 Fb =

 ( p2 − p1 )d A = −ρg

(z 2 − z 1 )d A

(1.11)

The integral represents the volume of the body so that the right-hand side is just the weight of the body if it would be filled with the fluid in which it is submerged. Equation 1.6 is one of the Archimedes’ law: A body immersed in fluid experiences a vertical buoyant force equal to the weight of the fluid it displaces. The buoyant force is applied to the centre of volume of the displaced body and is called centre of buoyancy. If our body is a ship, we know that only part of it is submerged such that the buoyant force must balance the entire weight of the ship (Fig. 1.2). An interesting application of the Archimedes’ law is found in geophysics. Think about a mountain chain as a block of continental crust immersed in the mantle. The crust has a density around 2800 kg m−3 while the mantle has a density around 3300 kg m−3 . Hydrostatic equilibrium requires that (see Fig. 1.3 left). ρc h = ρm b

(1.12)

so that the mountain block will emerge by a height   ρc h−b = 1− ρm

(1.13)

Using a crust thickness of 35 km, the mountain will emerge by 5.3 km. The application of the hydrostatic equilibrium to the continental crust is known as isostasy and is treated in some more detail in the examples at the end of the chapter. We can introduce what is the effect of the ocean on the continental crust following Fig. 1.3. In this case, the continental crust has thickness h cc with density ρcc . Its upper surface is at sea level and it is covered with water of depth h w and density ρw . The oceanic crust has thickness h oc and density ρoc and the mantle density as ρm . Application of hydrostatic equilibrium at the base of the continental crust gives ρcc h cc = ρw h w + ρoc h oc + ρm (h cc − h w − h oc )

(1.14)

The depth of the ocean basin relative to the continent is given by hw =

(ρw − ρcc ) (ρm − ρoc ) h cc − h oc (ρm − ρw ) (ρm − ρw )

(1.15)

Using h cc = 35 km, h oc = 6 km, ρm = 3300 Kg m−3 , ρw = 1000 Kg m−3 , ρoc = 2900 kg m−3 and ρcc = 2800 Kg m−3 , we find h w = 6.6 km. If the continental crust thins with respect to its previous value, the hydrostatic equilibrium will require the crust to sink below the sea level creating a sedimentary basin.

6

1 Fundamentals 1 p1

dA

CG

z1-z2 B displaced volume

p2

Fig. 1.2 The pressure forces on a submerged body (left), the centre of gravity (CG) and the buoyancy centre (right) for a ship. This situation corresponds in principle to an unstable case Continental crust ρc

h

b

Ocean Oceanic crust

hw, ρw

Continental crust ρcc

hoc, ρoc hcc

Mantle ρm Mantle ρm

Fig. 1.3 Archimedes’ law applied to the floating of Earth’s crust in the mantle (left) and the mantle under the oceanic crust (right)

1.2.4 Models of Earth’s Interior The hydrostatic equilibrium can be applied to find the density change at the Earth’ interior. We will anticipate now some information on the elastic waves that we will treat later. In particular, we will concentrate on the propagation velocities of seismic waves: P waves (longitudinal) with a velocity of α and S waves (transverse) with a velocity of β. These velocities are  α=

K + 23 μ ρ 

β=

μ ρ

(1.16)

(1.17)

where μ is a Lamé constant and K is the compressibility modulus. Then is easy to show that K = α 2 − 43 β 2 (1.18) ρ

1.2 Some Applications

7

Now we consider a spherical shell of thickness dr with density ρ(r ) at the radius r. For this layer, the hydrostatic equilibrium requires dP = −g(r )ρ(r ) dr

(1.19)

where g(r ) is the local acceleration of gravity given by g(r ) =

G M(r ) r2

(1.20)

where M(r ) is the mass within the radius r . The density of the Earth will change with depth using (1.20) dp dρ G M(r )ρ(r ) dρ dρ = =− dr dr dp r2 dp

(1.21)

On the other hand, the bulk modulus can be expressed as K =ρ

dp dρ

(1.22)

Equation (1.21) can then be written as G M(r )ρ(r )2 dρ =− dr r2K

(1.23)

And substituting (1.18), we have dρ G M(r )ρ(r ) =− 2 2 4 2 dr r (α − 3 β )

(1.24)

This is known as Adams Williamson equation and, in principle, it could be used to determine the density of the Earth if the profile of seismic velocity is known. The method is to start from the surface and to work inwards, applying the equation iteratively to shells of uniform composition. If M E is the mass of the Earth, the mass M(r ) within radius r is given by  Mr = M E −

R

ρ(r  )r  dr  2

(1.25)

r

The profile of seismic waves can be obtained with rather ingenious inversion methods based on the analysis of signals generated during earthquakes (see Chapter on waves).

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1 Fundamentals 1

Ω

Fig. 1.4 Forces acting on a point P on a rotating planet with co-latitude θ

P acf θ

r

1.2.5 Shape of the Earth Planets with an appreciable rotation rate (like the Earth) experience a centrifugal acceleration so that the total force acting on a body results from the sum of the gravitational force and the centrifugal force. If the θ is the co-latitude of the point P (Fig. 1.4) in terms of centrifugal potential we have 1 Vc f = − 2 r 2 sin2 θ 2

(1.26)

where is the angular velocity. The total potential is then Vtotal (r, θ ) = −

GM + Vc f (r, θ ) r

(1.27)

At equilibrium, the surface must be perpendicular to the sum of the gravitational and centrifugal acceleration vectors. If the deviation of the ocean surface δr (θ ) is small, we can write (1.28) rocean = a + δr (θ ) where a is the equatorial radius. The potential describing the surface is then Vsur f ace (r, θ ) = −

GM GM 1 + 2 δr − 2 a 2 sin2 θ − 2 a sin2 θ δr a a 2

(1.29)

In the Earth’s case, 2 a  G M/a 2 so that we can neglect the last term. Setting the equipotential surface to zero, we have δr ≈ a +

2 a 4 sin2 θ 2G M

this would give putting θ = π/2 and θ = 0

(1.30)

1.2 Some Applications

9

requ = a +

2 a 4 2G M

r pol = a

(1.31)

with the flattening or oblateness defined as f =

requ − r pol

2 a 3 = requ 2G M

(1.32)

Now the potential theory shows that the external gravitational potential of any body with an axis of symmetry can be written as Vgrav (r, θ ) = −

  2  4  a a GM 1 − J2 P2 (cosθ ) − J4 P4 (cosθ ).... r r r

(1.33)

where Jn are dimensionless constants related to the inertia moments and Pn is the Legendre polynomial of degree n. In our case, it can be shown that J2 ≈

C−A Ma 2

(1.34)

where C is the moment of inertia around the rotation axis and A = B, the moments of inertia around the remaining orthogonal axis. If we now express Eq. (1.26) in terms of Legendre polynomials, we have as total potential   G Ma 2 GM 1 2 2 + J2 + r P2 (cosθ ) Vtot (r, θ ) = r r3 3

(1.35)

We have neglected higher order terms like J4 , J6 and so on while for a symmetric body, the odd terms are zero. The reason is that it can be shown that if the flattening is small, the higher order terms go like Jn ∝ f n/2 . Again if we consider small deviations from a, r = a + δr and assume that the corresponding potential to be zero, use of (1.35) gives   1 2 a 3 a P2 (θ ) (1.36) δr = constant − J2 + 3 GM Again we can use the definition for the flattening to obtain f =

1 2 r 3 3 J2 + 2 2 GM

(1.37)

This means that if the planet is deformed due to the hydrostatic adjustment, the flattening is increased by 3J2 /2. The developed theory is accurate enough when we compare the observed value of the flattening (0.003353) with the calculated one (0.003349). If we consider the flattening without deformation Eq. (1.32), we have f = 0.0017306, that is, a much smaller value. This corresponds to a planet, not in

10

1 Fundamentals 1

hydrostatic equilibrium. In terms of distance, a flattening of this kind corresponds to a difference between the equatorial and polar radius of roughly 21 Km. These numbers could be guessed with a simpler argument. The difference in centrifugal 2 /2, and the difference potential between the equator and the poles is simply 2 requ in gravitational potential is g(requ − r pol ) so these differences must compensate and we have

2 R 2 requ − r pol = (1.38) 2g This difference should be of the same order of magnitude of the tallest mountain ≈10 km because higher deformation could not be supported by the strain of rocks. If we substitute for 2 ≈ 5.3 × 10−9 s−2 and R 2 ≈ 4.2 × 1013 m2 , we obtain a difference of 11 Km roughly half the observed value. In part, this difference is due to the fact we have assumed implicitly a spherical Earth while the centrifugal force produces an additional flattening.

1.3 Pressure Distribution in Rigid Body Motion For a fluid, we intend rigid body motion when there is no relative motion between the particles that constitute the fluid. We will see later that this implies absence of viscosity effect so that Eq. (1.1) can be generalized to include particle acceleration ∇ p = ρ(g − a)

(1.39)

The pressure gradient has the same direction as the vector g − a so that the free surfaces are perpendicular to the same vector. It is quite obvious that we need to talk about a fluid that is confined, for example, in a tank and also, in this case, this discussion, it is rather academic because you need to consider that the tank moves with constant acceleration so that its velocity could become quite high. As the first example consider the case of a fluid that moves with constant acceleration a in the positive direction x as in Fig. 1.5a. The difference g − a results in a vector that makes an angle θ with the horizontal such that tan θ =

ax g

(1.40)

A little bit more complex is the case of a rotating fluid shown in Fig. 1.5b. In this case, we need to write the radial (r ) and vertical component (z) of the equation such that ∂p ∂p (1.41) = ρr 2 = −ρg ∂r ∂z where the first equation equates the horizontal gradient to the centrifugal acceleration and the second just fixes the hydrostatic equilibrium. The pressure p will be a function

1.3 Pressure Distribution in Rigid Body Motion

z

11

(a)

(b) h

Fluid at rest -ax

Ω

θ g R

R

x

Fig. 1.5 In a, we show a fluid subject to a horizontal acceleration while in (b), a rotating fluid is shown

of both z and r but the first equation can be integrated with respect to r and gives p=

1 2 2 ρr + f (z) 2

(1.42)

where f (z) has to be determined. Substituting this solution in the second (1.41), we have ∂f ∂p = = −ρg ∂z ∂z so that Eq. (1.42) becomes 1 p = p0 − ρgz + ρr 2 2 2

(1.43)

with the requirement that pressure being p = p0 at (z, r ) = (0, 0). To find the shape of an isobaric surface at pressure p1 , we obtain from (1.43) z=

p0 − p1 r 2 2 + ρg 2g

(1.44)

The surfaces are paraboloids of revolution concave upward with the minimum on the axis of rotation. If h is the depth of the fluid, we have from (1.44) h=

ω2 R 2 2g

where R is the radius of the containing vessel. The volume of the concavity is just π R 2 h/2 so that the still water level is halfway between the high and low points of the free surface, that is h/2.

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1 Fundamentals 1

Fig. 1.6 The pressure forces on an element of submerged fluid

z p1 D

E ps

p2

p3 O

x

A

C

y

B p4

Appendix Why Pressure Does Not Depend on Direction Consider an elementary volume as depicted in Fig. 1.6. The element must be at rest in the hydrostatic approximation such that the sum of all forces on it must be zero. In the y direction 1 xz( p2 − p3 ) = 0 (1.45) 2 which implies p2 = p3 . In the z direction, we have ps yl cos α = p4 yx = 0

(1.46)

because l cos α = x. The same happens for the x direction ps yl sin α = p4 yz = 0

(1.47)

where, in this case l sin α = z. We conclude then ps = p1 = p2 = p3 = p4

(1.48)

Appendix

13

More on Isostasy We have considered isostasy before and made the case of a single block of crust floating on the molten mantle. However, the real situation is quite different because a mountain chain can be composed of several blocks with different densities. In that case, the problems can be reduced to a series of blocks with the same density (Fig. 1.6) and this corresponds to the so-called Airy model. On the other hand, the blocks can be of different densities and that corresponds to Pratt’s model (Fig. 1.7b). In the first case, the calculation, if we define the topography with respect to the sea level as b, we have ρcr ust h (1.49) b= ρm − ρcr ust While in case of negative topography (below the sea level), we have  b=

ρcr ust − ρw ρm − ρcr ust

 h

(1.50)

In the case of the Pratt model, the blocks have different densities and the hypothesis is that they will float in such a way that they have a common base level. Referring to Fig. 1.7b, the variable density ρ p is related to the elevation above the sea level by  ρ p = ρ0

W W +h

 (1.51)

where ρ0 is the reference density corresponding to zero elevation while W is called the depth of compensation. For topography below the sea level (h negative), the variable density is given by ρ0 W + ρw h (1.52) ρp = W +h These anomalies in density can be used to interpret the gravitational anomalies (Fig. 1.7).

Stellar Structure and Lane–Emden Equation The same equations we have used to determine the structure of a planet could be used to find how pressure changes within a star. We consider the equation of continuity dM = 4πr 2 ρ(r ) dr and the hydrostatic equation

(1.53)

14

1 Fundamentals 1

(a)

h

(b) sea level ρw

h

ρcrust

sea level ρw

ρp

ρ0 W

b

ρm

ρm

Fig. 1.7 The pressure forces on an element of submerged fluid

G M(r )ρ(r ) dp =− dr r2

(1.54)

Using these two equations, we can eliminate d M/dr to get 1 d r 2 dr



r 2 dp ρ dr

 = −4π Gρ

(1.55)

We assume the pressure to change according to a polytrope p = Kρ γ . Notice that the general expression for a polytrope is pV n = C that coincides with an isothermal for n = 1 and an adiabatic for n = c p /cv . To be more general, we assume that γ = (n + 1)/n so that (1.55) becomes 1 d r 2 dr



r 2 K γ −1 dρ γρ ρ dr

 = −4π Gρ

(1.56)

We now introduce a new variable α such that r = αξ where  α=

n+1 Kρc(1−n)/n 4π G

1/2 (1.57)

we then assume ρ(r ) = ρc θ n (ξ ). This substitution assures that at the centre of the star where ξ = 0, θ (0) = 1. Also considering that dp/dr goes to zero at the centre, we need dθ/dr = 0 for ξ = 0. These are the boundary conditions for the solution. With these substitutions, we have what is called the Lane–Emden equation 1 d ξ 2 dξ

  dθ ξ2 = −θ n dξ

(1.58)

Appendix

15

Analytic Solution for n = 1 Substituting in Eq. (1.58) n = 1, we have d dξ

  2 dθ ξ = −ξ 2 θ dξ

(1.59)

We now introduce another variable χ such that χ = ξ θ so that d dθ = dχ dξ Substitution in Eq. (1.59) gives

  χ ξχ − χ = ξ ξ2

(1.60)

χ  + χ = 0

(1.61)

The general solution reads θ (ξ ) = A sin ξ + Bcosξ ⇒ θ (ξ ) = A

sinξ cos ξ +B ξ ξ

(1.62)

Using the boundary conditions cos ξ =∞ ξ →0 ξ

θ (0) = 1 ⇒ B = 0, because lim

(1.63)

The same condition gives θ (0) = 1 ⇒ A = 1, because lim (sin ξ/ξ ) = 1 ξ →0

The solution of the equation is then θ (ξ ) =

sinξ ξ

(1.64)

At this point we can make some practical case. We can evaluate the mean density ρ¯ inside a certain radius r M(ξ ) ρ¯ = 4 3 3 (1.65) πα ξ 3 where the mass is obtained from   αξ 4πρr 2 dr = 4π α 3 ρc M(ξ ) = 0

That is

0

ξ

ξ 2 θ dξ

(1.66)

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1 Fundamentals 1

M(ξ ) = −4π α 3 ρc ξ 2

dθ dξ

(1.67)

The mean density can be expressed by 3 dθ ρ¯ = −ρc ξ dξ ξ =ξ1

(1.68)

using the solution (1.68) and considering that ξ1 = π , we get ρ¯ = 0.29499 ρc

(1.69)

The radius of the star according to (1.65) is then 

K R= 2π G

1/2 π

(1.70)

The central pressure according to the initial polytrope pc = Kρc2

(1.71)

We can easily find the pressure at the centre pc =

G M2 8R 4

(1.72)

It is interesting to use these relations for a gaseous giant like Jupiter or Saturn that are mostly made up of a mixture of hydrogen and helium in solar proportion. Jupiter has a mass of 2 × 1027 kg and a radius of 7 × 104 m and its average density can be easily found to be 1.39 × 103 kg/m3 . The average density is related to the central density by Eq. (1.69). For a polytrope with n = 1, the central density will be ρc = 4.57 × 103 Kg/m3 . We can find a relationship between M and R using (1.67) and (1.71) to get  3 M R 1 (1.73) = 2 π 4π ρc ξ dθ/dξ This relation can be used to find the radius of another gaseous planet like Saturn once the mass is known. For Saturn the mass is 6 × 1026 kg and using the same value for ρc found for Jupiter (it only depends according to (1.73) on the average density), we obtain R S = 4.7 × 104 km. This is quite different from the observed, 5.2 × 104 km. The difference could be attributed to the assumption of the same average density.

References

17

References Textbooks Chandrasekhar S (1967) An introduction to the study of stellar structure. Dover Publications, New York Eckert M (2003) The dawn of fluid dynamics. Wiley-VCH, New York Feynman RP, Leighton R, Sands M (2011) Lectures on physics, vols 1, 2. Basic Books, New York Fowler CRM (2004) The solid earth: an introduction to global geophysics. Cambridge University Press, Cambridge Murray CD, Dermott SF (2008) Solar system dynamics. Cambridge University Press, Cambridge Thorne KS, Blandford RD (2017) Modern classical physics: optics, fluids, plasmas, elasticity, relativity, and statistical physics. Princeton University Press, Princeton Turcotte DL, Schubert G (2002) Geodynamics. Cambridge University Press, Cambridge White FM (2015) Fluid mechanics. Mc Graw Hill, New York

Chapter 2

Fundamentals 2

In the previous chapter, we considered the fluid at rest, in what we termed hydrostatics, and the fluid in motion as a whole with the example of a rotating planet. In this chapter, we treat what happens when the particles within the fluid move, and for the time being we will consider the absence of viscous forces. This corresponds to what John von Neumann called the dry water (as reported by Richard Feynman) to stress the fact that this is a rather impossible idealization. We will see later that we pretend to apply the Bernoulli equation (valid in the above approximation) also to a very complicated problem like the calculation of the lift for an airfoil. The same ideal fluid approximation predicts that a symmetric body placed in fluid flowing past it does not experience any drag (D’Alembert paradox) contrary to the experiment. We know that this terminology may not be appropriate because it gives the impression that viscosity is related to some kind of wetting process which is not (consider that viscosity exists also for a gas). On the other hand, we would like to maintain a light approach.

2.1 The Continuity Equation Consider an elementary fluid volume as depicted in Fig. 2.1 submerged in a fluid in motions with an assigned vector velocity V with components u, v, w along the three directions x, y, z. The net flux entering the volume element must be equal to the accumulation of mass within the volume. The inflow rate per unit area through the left-hand side face is given by  ρv −

∂ δy (ρv) ∂y 2



while the rate of outflow per unit area through the right-hand side face is © Springer Nature Switzerland AG 2020 G. Visconti and P. Ruggieri, Fluid Dynamics, https://doi.org/10.1007/978-3-030-49562-6_2

19

20

2 Fundamentals 2

Fig. 2.1 The calculation of the mass flux for an elementary volume δz 6 (ρv) δy 2 6y

ρ

ρ

δy + 6 (ρv) 2 6y

δx δy

 ρv +

∂ δy (ρv) ∂y 2



The net inflow in the y direction is then the difference −

∂ (ρv)δxδyδz ∂y

Similar expressions are found for the x and z directions so that the net rate of mass inflow   ∂ ∂ ∂ − (ρu) + (ρv) + (ρw) δxδyδz ∂x ∂y ∂z The net mass inflow per unit volume is just −∇ · (ρV) and this must be equal to the net change of mass per unit volume that corresponds to the density change ∂ρ/∂t. We get ∂ρ + ∇ · (ρV) = 0 (2.1) ∂t This is the mass divergence form of the continuity equation. An alternative form of this equation could be found using the vector identity ∇ · (ρV) = ρ∇ · V + V · ∇ρ if we substitute in (2.1), we have ∂ρ + ρ∇ · V + V · ∇ρ = 0 ∂t We define substantial derivative or Lagrangian derivative as the operator d ∂ = +V·∇ dt ∂t

(2.2)

2.1 The Continuity Equation COLD

21 COLD

T

T

T+ΔT

T+ΔT

......

......

T+3ΔT

T+3ΔT

WARM

WARM

Fig. 2.2 The advection of warm air (left figure) and cold air (right figure)

There are several ways to derive (2.2) but it is rather interesting to give it a physical interpretation. A possible way is to think of a balloon that is transported by the wind in the atmosphere. A thermometer inside the balloon will register the local change of the temperature as contributed by the local change ∂ T /∂t and the convective change V · ∇T . The convective term has a simple interpretation if we refer to Fig. 2.2 where we have drawn the isotherms (lines connecting points at the same temperature) and the generic wind direction. The local change of temperature based on (2.2) is given by dT ∂T = − V · ∇T ∂t dt If the wind is parallel to the temperature gradient, cold air will be advected towards warm air that will tend to be cool. Vice versa, if the wind is opposite to the gradient, warm air will be advected towards cold air that will tend to be warm. In the first case, we will talk about cold advection while in the second case we will talk about warm advection. For this reason advection is defined with the minus sign −V · ∇T . The mass divergent form (2.1) is another way to interpret the divergence if we remember the divergence theorem that relates the flux of a vector A to the volume integral of the divergence   A · nd S = S

∇ · Ad V V

The integral of the divergence of the flux is just the rate of change of the mass inside the volume. Again Eq. (2.1) is simply another way to write the conservation law (in this case for the mass) as it is illustrated at length in the appendix. Notice also that if the fluid is incompressible, then dρ/dt = 0 and the divergence of the velocity is zero, ∇ · v = 0.

22

2 Fundamentals 2

2.2 The Equations of Motion The Lagrangian derivative can be also applied to calculate the acceleration of a fluid particle dv ∂v = + (v · ∇)v (2.3) dt ∂t The acceleration is determined by the forces per unit mass so we get ∇p ∂v + (v · ∇)v = − − ∇φ ∂t ρ

(2.4)

As you may notice, we have not considered the viscous forces because we are still in the dry water approximation. Equation (2.4) can be rearranged using the vector identity (v · ∇)v = (∇ × v) × v + 21 ∇(v · v) We now define a new vector =∇ ×v

(2.5)

that we will treat later as vorticity, and the equation of motion becomes ∇p ∂v +  × v + 21 ∇(v · v) = − − ∇φ ∂t ρ

(2.6)

The vorticity can be easily related to the circulation defined as  circulation =

v · dl

And using the Stokes’ theorem, we have 





(∇ × v) · nd S = S

 · nd S =

v · dl

S

so that the vorticity is the circulation per unit area. This means that if we put a little piece of dirt in the fluid with vorticity , it will have a circulation C = 2π vr and a vorticity  = C/πr 2 = 2v/r and so will rotate with an angular velocity /2. Equation (2.6) can be further simplified if we are only interested in the velocity field. We take the curl of both sides of (2.6) and remember that the curl of the gradient is zero. We have ∂ + ∇ × ( × v) = 0 (2.7) ∂t This equation together with the equations

2.2 The Equations of Motion

23

=∇ ×v

(2.8)

∇ ·v =0

(2.9)

and

describes completely the velocity field v. The situation is somewhat similar to the magnetism when we have the equation ∇ · B = 0 and the magnetic field is determined by the current density j, ∇ × B = μ0 j. In this case if we know the current, we can determine the magnetic field. In the same fashion if we know , we can determine v. Then when v is known using (2.7), we can determine how  changes in time and so we may know the field at any other time.

2.3 The Bernoulli Theorem It is time to define what intends for streamlines. These are lines drawn always tangent to the velocity of the fluid as shown in Fig. 2.3a. These lines again are analogous to the “field lines” of an electric or magnetic field that we have known also as “lines of force”. If the streamlines are fixed in space, we call the flow steady. This does not mean that nothing is happening in the fluid but rather that the fluid particles move always on the “trajectories” so that ∂v/∂t = 0. Then if we multiply Eq. (2.6) by v considering that v · (∇ × ) = 0, we obtain   1 2 p (2.10) +φ+ v =0 v·∇ ρ 2 This equation tells that for small displacements in the direction of the fluid velocity, the quantity inside the bracket does not change. Because in steady flow all the movements are along the streamlines, for all the points along the same streamline we have p 1 + φ + v2 = cost (streamline) (2.11) ρ 2 v2Δt

V2

ΔM v1

A2

v1Δt ΔM

A1 (a)

(b)

Fig. 2.3 The streamlines (a), a flow tube (b) and the conservation of mass (c)

(c)

24

2 Fundamentals 2

The constant is different for different streamlines while if the motion is irrotational, that is  = 0, then Eq. (2.6) gives for steady flow  1 2 p +φ+ v =0 ∇ ρ 2 

so that

p 1 + φ + v2 = cost (everywhere) ρ 2

(2.12)

This looks similar to (2.11) but now the constant is the same everywhere. The Bernoulli theorem has been derived formally using the equation of motion but there is a more simple and intuitive way to arrive at the same conclusions. Consider now what we call a stream tube that is a tube formed by adjacent streamlines as shown in Fig. 2.3b. Consider now two sections at some distance where the velocities are v1 and v2 while the cross-section areas are A1 and A2 . At the same points, we assign the densities ρ1 , ρ2 and the potential φ1 , φ2 . After a short interval t, the fluid at the two extremities has moved the distances v1 t and v2 t, respectively (Fig. 2.3c) and the continuity requires that the mass entering A1 is the same as the mass leaving A2 M = ρ1 A1 v1 t = ρ2 A2 v2 t so that ρ1 A1 v1 = ρ2 A2 v2 As expected, if the density stays the same, the velocity will change inversely with the area of the stream tube. Now we calculate the work done by the pressure forces in the fluid. This can be easily calculated to be p1 A1 v1 t − p2 A2 v2 t This difference must be equal to the gain of energy going from A1 to A2 p1 A1 v1 t − p2 A2 v2 t = M(E 2 − E 1 )

(2.13)

The energy per unit mass of the fluid can be written as E=

1 2 v +φ +U 2

where U is the internal energy of the fluid like the thermal energy if the fluid is compressible, or the chemical energy. From (2.13), we have p2 A2 v2 t 1 p1 A1 v1 t 1 − = v2 2 + φ2 + U2 − v1 2 + φ1 + U1 M M 2 2

2.3 The Bernoulli Theorem

25

And we get 1 p2 1 p1 + v1 2 + φ1 + U1 = + v2 2 + φ2 + U2 ρ1 2 ρ2 2

(2.14)

which is the Bernoulli theorem with the additional term of the internal energy that is zero in the case of incompressible fluid.

2.4 Simple Applications In this paragraph, we will report not only some of the most simple applications of the Bernoulli theorem but also some less-known and more interesting applications like those dealing with a simple helicopter or the propeller thrust.

2.4.1 Flow from a Tank Suppose we have a tank where the water level is maintained at a height H above some orifice as shown in Fig. 2.4a. The atmospheric pressure is pa and is the same at the top of the tank and near the hole. If we apply the Bernoulli theorem at the streamline shown in the figure and assume that the velocity of the fluid is negligible at the top, we have pa = pa + 21 ρvout 2 − ρg H vout =

 2g H

(2.15)

If we look carefully at the shape of the jet from the hole, we noticed that it contracts until all the streamlines are parallel. The area of the jet at this point is less than the

pa

H

ps = pa

Stream Line

Vs=0

H

pa

pa

vout Fig. 2.4 Flow from a tank. The figure on the right shows the Borda mouthpiece and the velocity

26

2 Fundamentals 2

area of the hole (the so-called vena contracta), and the ratio of these two areas is what is called efflux coefficient. It is very instructive to proceed to calculate this coefficient for the special case of the Borda mouthpiece. This is a cylinder that actually goes inside the tank as shown in Fig. 2.4b. We have used just the conservation of energy to calculate the efflux velocity but the jet leaving the tank has some kind of momentum and we need to consider also the momentum conservation. We call αc the area of the vena contracta, α the area of the mouthpiece and V the flow velocity. To evaluate the rate of changing for the momentum (i.e. the force), we simply observe that the force is the mass flowing per unit time times the change in velocity, that is rate of change of momentum = αc ρV × V This force must be equal to the pressure and we have ρα H = αc ρV 2 We now assume that the efflux velocity is proportional to the velocity (2.15), V = Cv vout so we get 1 αc = (2.16) α 2Cv 2 In the case Cv = 1, the efflux coefficient is just 0.5. Now the scope of the Borda mouthpiece is to increase the efflux and this happens because as shown in Fig. 2.4b, the jet first contracts and then expands before exiting from the tube. We indicate Vs and ps as velocity and pressure at point S and p0 the pressure at point 0. We have Vs 2 p0 V2 ps +H+ = + + losses ρ 2g ρ 2g where the losses are due to the decrease in the exiting velocity losses = (Vc − V )2 /2g where Vc is the velocity at the contraction. Considering that ps = p0 and Vs = 0, we have  2 V 2 Vc V2 + −1 H= 2g 2g V The ratio Vc /V can be easily obtained from continuity Vc α = V α so we have V /Vc = 0.5 and consequently H = V 2 /g, and from the definition of coefficient of velocity H=

√ Cv2 2g H → Cv = 1/ 2 = 0.707 g

(2.17)

2.4 Simple Applications

27

Pressure gage

C

static pressure

A

B

Fig. 2.5 The Pitot tube. The inset shows the detail of the static pressure hole

2.4.2 The Pitot Tube Most of the examples you see on the Pitot tube are very general and do not refer to the real measurement on an aircraft. Figure 2.5 gives a more precise idea of how a Pitot tube works. The tube immersed in the flow has two openings: one in front (point A) and the other just on the surface of the tube (point B). The tube is closed on one end (point C) where a gauge measures the pressure. While point A registers just the static pressure p0 , the gauge in C registers the dynamic pressure because the tube is rapidly filled with air. At point B, the velocity of the fluid is zero and this is called stagnation point. The Bernoulli equation applied to points A and B reads 1 1 p A + ρV A 2 = p B + ρVB 2 2 2 Considering that we have p A = p0 , V A = V , VB = 0 and p B = p, we get for the velocity of the fluid V 2( p − p0 ) (2.18) V = ρ The Pitot tube depicted in Fig. 2.5 is called Pitot static probe and reads both the total pressure p and the static pressure p0 .

2.5 Thrust for a Propeller We associate the thrust that equips an aircraft to a turbojet engine but actually, the more classical propeller can also be seen in terms of thrust. The propeller blades

28

2 Fundamentals 2

can be regarded as wings (we will talk extensively about them in the book) and each blade as it rotates will produce a decrease in pressure ahead of the propeller and an overpressure behind. The velocity of the air on the other hand will increase continuously going from well ahead of the plane to the back. This is shown in Fig. 2.6 where according to the Bernoulli Equation, the pressure decreases from the value p0 well ahead of the plane to some value p1 at the propeller. Far away and behind the plane, the pressure will be again p0 while at the propeller, there will be a pressure increase of p. The simple picture shown in Fig. 2.6 corresponds to what is called the actuator disc model and assumes that the propeller role is just to create a pressure increase. We also call V0 the velocity far away ahead of the plane and u the velocity at the propeller site. In the same fashion, Ve is the velocity far away behind the plane. We apply the Bernoulli equation to the points p0 and p1 and to p1 and p0 behind the propeller. We have 1 1 p0 + ρV0 2 = p1 + ρu 2 2 2 1 1 ( p1 + p) + ρu 2 = p0 + ρVe 2 2 2 We can eliminate u between these two equations and obtain p =

 1 2 ρ Ve − V0 2 2

At this point, we argue that the thrust T must not only be the product of the mass flux m˙ times the change in velocity Ve − V0 but also the product of the pressure difference times the area. So we have with m˙ = ρu A T = p A = ρu A(Ve − V0 ) → u =

1 (Ve + V0 ) 2

(2.19)

We can also evaluate the power as P=

1 m(V ˙ e − V0 )2 2

(2.20)

As possible numbers, we could mention that one of the last propeller-driven fighter (the Grumman F8F Bearcat) with an engine of 2000 CV produced a thrust of roughly 70 kN. We can reverse the propeller problem and treat the works of a windmill (or as it is called today wind generator) but in a somewhat simpler way. The windmill will slow down the entrant air that initially has velocity U and invests an area C A smaller than the area A of the blade so that C < 1. The current behind the blades will have velocity f U smaller than the entrant velocity so that the area involved will be larger, C A/ f . The thrust of the windmill can be evaluated from the pressure drop across the blade

2.5 Thrust for a Propeller

29

(a) V0 p1 u

Ve

p1+Δp

p0 p0

pressure

velocity

actuator disk

Ve

V0

(b)

p1+Δp p0

(c) p0

p1

A B (d) R

Fig. 2.6 The propeller problem. The top drawing shows the flux tube with the actuator disc. In the middle, qualitative graphs are shown for the velocity and pressure inside the tube. At the bottom, the helicopter situation is shown

30

2 Fundamentals 2

1 1 1 ρU 2 + p0 = f U 2 + p0 + p → p = ρU 2 (1 − f 2 ) 2 2 2 The thrust on the windmill will be Ap, that is T =

1 ρ AU 2 (1 − f 2 ) 2

The constant C can now be evaluated from the thrust expressed as the change of momentum T = mU ˙ (1 − f ) = ρU 2 C A(1 − f ) So we obtain for C C=

1 (1 + f ) 2

The kinetic energy removed from the air per unit time can be obtained by multiplying the thrust by the velocity U 1 1 ρU 3 C A(1 − f 2 ) = ρU 3 A(1 − f 2 )(1 + f ) 2 4 The optimal value for f can be obtained by maximizing the quantity (1 − f 2 )(1 + f ) and that turns out to be 1/3. The maximum power generated must be 8ρ AU 3 /27 Actually, there is a very simple argument to show that the power extracted by a windmill is proportional to the cubic power of wind. The kinetic energy density is simply ρu 2 /2 and the volume spanned by the wind around the blades is Aut with A, the area of the blades and t the time interval. Then the power is 1 2 1 ρu Au = ρu 3 A 2 2 T. E. Faber in his textbook gives an example for the rotor of a helicopter as shown in Fig. 2.6d. The volume is divided above the rotor (point A) and below the rotor (point B). The action of the rotor is such that it decreases the pressure at point A with respect to the pressure at point B while the pressure at the rotor level is the atmospheric pressure pa . Neglecting the change in pa due to the altitude changes we have applying Bernoulli 1 p A + ρu A 2 = pa 2 The same equation applied between B and a point R in the vena contracta gives

2.5 Thrust for a Propeller

31

1 1 1 p B + ρu B 2 = p B + ρu A 2 = pa + ρu R 2 2 2 2 where it is assumed that velocity does not change between A and B. The pressure difference is 1 p B − p A = ρu R 2 2 The upward thrust is then ( p B − p A )A =

1 ρ Au R 2 = ρu R 2 C A 2

where the last term is the downward momentum per unit time.

2.6 Circulation and Vorticity We have already defined both circulation and vorticity. However, there is a quite instructive way to define vorticity that emphasizes how it is related to the fluid behaviour. Consider a portion of fluid as in Fig. 2.7 where the initial perpendicular lines AB and BC after some time dt deform into A B  and B  C  . These lines have apparently rotated to an angle dα with respect to the x-axis and of dβ with respect to the y-axis. We define the angular velocity component ωz along the z-axis perpendicular to x − y as the average rate of counter clockwise turning of the two lines   dβ 1 dα − (2.21) ωz = 2 dt dt We can approximate in the limit of small dt dα ≈

∂v dt ∂x

dβ ≈

∂u dt ∂y

substituted in the (2.21) give the z component of the vorticity 1 ωz = 2



∂v ∂u − ∂x ∂y

 (2.22)

In the same way, we can determine the other two components ωx =

1 2



∂w ∂v − ∂y ∂z

 ωy =

1 2



∂u ∂w − ∂z ∂x

 (2.23)

32

2 Fundamentals 2

du dy dt dy A’

A

Time t

dβ

Time t + Δt C’

dα B

C

dv dx dt dx

B’

Fig. 2.7 The deformation of two linear elements of fluid to calculate the rotation and the z component of the vorticity

The vector is just the curl of the velocity ω = 21 ∇ × v. To make this consistent with our previous definition of vorticity, we just note that  = 2ω. We now summarizes the three Eqs. (2.7–2.8) written in the inverse order ∇·v =0 =∇×v ∂ω + ∇ × (ω × v) = 0 ∂t The first two of these equations simply say that the fluid is incompressible (velocity divergence is zero) and that the so-called vortex lines are always closed lines like the magnetic induction B. This statement is more evident if we notice that a vortex line has the same direction of ω and has a density proportional to the magnitude of ω. At this point, we need to show that the vortex lines move with the fluid that is as if they were frozen in the fluid. To this end, consider an infinitesimal line element δl as shown in Fig. 2.8 that at time t + t is at a different position as δl + δl. The rate of change following the motion 1 dδl = [δl(t + δt) − δl(t)] dt δt where we can write δl(t + δt) = δl + δvδt Substituting this into (2.24) and noting that δv = (δl · ∇)v, we have

(2.24)

2.6 Circulation and Vorticity

33

Fig. 2.8 The calculation of the incremental change to the element l moving with velocity v

dδl = δv = (δl · ∇)v dt

(2.25)

We can now identify δl as a vortex line element δl = Aω with A as constant. The total derivative d dω dδl (ω × δl) = × δl − ×ω (2.26) dt dt dt Substituting for dω = (ω · ∇)v dt

dδl = (δl · ∇)v dt

we have that the left-hand side of (2.26) is zero so that the vortex line continues to behave as a material line. We can go back to the interpretation of the second and third equations summarized at the beginning of the paragraph. Once v is assigned, the vorticity ω can be calculated while the third equation can be used to evaluate a new value of the same ω at some later time and so on. The classical example is to consider a small cylinder whose axis is parallel to the vortex lines as in Fig. 2.9. At some time later, the same fluid element will occupy a different position in space while its diameter and length may have changed or, for example, if the cylinder has decreased its diameter, its length must have increased because we have assumed an incompressible fluid such that its volume must remain constant. The third equation however tells us that the vortex lines stick with the fluid so in this shrinking process, their density must have increased and so its vorticity. In practice, the product of the vorticity ω and the area A of the cylinder must be constant. That means ω2 A2 = ω1 A1

(2.27)

This equation is just another way to express the conservation of angular momentum. As a matter of fact, if the fluid is with no viscosity, the pressure forces are always perpendicular to the surfaces. In this case, there are no tangential forces that are needed to change the angular momentum that in turn must be conserved. The angular momentum for a cylinder is just the product of the moment of inertia (I = M R 2 ) times the angular velocity of the fluid that we may assume proportional to the vorticity.

34

2 Fundamentals 2

Area A2 Area A2 time t ω1

ω2

time t’ > t

Fig. 2.9 The elementary vortex tube with vorticity ω1 evolving in a vortex tube with vorticity ω2

Then we have (M1 R1 2 )1 = (M2 R2 2 )2 where M1 = M2 and the area are proportional to R 2 and so we get essentially the previous equality. The vortex ring has been used in the Feynman book to show that to generate vorticity from an initial state where the same vorticity is zero, you “need” a viscous fluid. We will work out this problem later on.

2.6.1 Kelvin Circulation Theorem Now we can show more formally that in certain circumstances also, the circulation is conserved “moving with the flow”. The conditions are that the forces on the fluid are conservative (derived from a potential) and the fluid must be barotropic, that is ρ = ρ( p). We start by calculating the material derivative to the circulation defined in paragraph (2.2) for a curve that follows the flow (often called a material curve) d dC = dt dt

 

 v · dr =

dv · dr + v · dv dt

 (2.28)

We substitute Eq. (2.4) to obtain dC = dt

    1 − ∇ p − ∇φ · dr + v · dv ρ

Now the circulation of the ∇φ is zero while the term v · dv can be written as the differential 21 d(v · v) and again the circulation is zero. We are left with dC = dt



1 − ∇ p · dr ρ

(2.29)

2.6 Circulation and Vorticity

35

This integral is zero if the density is constant or if the density is just a function of p. To see this we just apply the Stokes’ theorem 

1 ∇ p · dr = ρ



 ∇× S

∇p ρ



 · dS =

− S

∇ρ × ∇p · dS ρ2

Now if p is a function of ρ alone, the gradients in the last integral are parallel and the integral is zero, and the final result reads for a barotropic fluid d dt

 v · dr = 0

(2.30)

If we again apply the Stokes’ theorem, we obtain (see paragraph 2) d dt

 v · dr =

d dt

  · dS = 0

(2.31)

S

This last property could be obtained by considering an elementary volume δV = δlδ A that is of area δ A and length δl. The vorticity crossing the area is proportional to the length of the segment (if the element shrinks, the length increases as the vorticity) so we have δV ∝ ωδ A The volume of the element remains constant so that the product on the right-hand side remains constant. When the element is stretched, the vortex lines get closer together so that the product ωδ A remains constant conserving the circulation.

Appendix Quantum Mechanics and Hydrodynamics The behaviour of atomic particles is described by the Schrödinger equation i

2 2 ∂ψ(r, t) =− ∇ ψ(r, t) + V (r)ψ(r, t) ∂t 2m

(2.32)

We can show that this equation is equivalent to the conservation of mass and momentum for a fluid. We write the wave function ψ as ψ=

√

A exp iφ/

(2.33)

So we simply have decomposed a complex function in its absolute value A and its phase ϕ. We insert Eq. (2.33) in (2.32) and separate the result into real and imaginary

36

2 Fundamentals 2

parts after dividing by

√

A exp iφ/. We obtain ∂A +∇ · ∂t

∂ϕ 1 2 + | ∇ϕ |2 + ∂t 2m 8m





A ∇ϕ m

 =0

 1 2 2 2 | ∇ A | − ∇ A = −V A2 A

(2.34)

(2.35)

We now define as velocity the quantity v=

∇ϕ m

(2.36)

that substituted in (2.34) gives the continuity equation for the amplitude A ∂A + ∇ · (Av) = 0 ∂t

(2.37)

Now to see that Eq. (2.35) is equivalent to the momentum equation (i.e. Newton’s second law), we remember that for a fluid we have ρ

dv ∂v =ρ + ρv · ∇v = F dt ∂t

And using the continuity equation, we arrive at what is called the Euler equation ∂(ρv) + ∇ · (ρvv) = F ∂t

(2.38)

which is the equivalent of the continuity equation for the momentum (ρv). To get something equivalent to this from (2.35), we derive it with respect to xi , then remembering (2.36) and that for a conservative force F = −∇V we obtain 1 ∂(v j v j ) 2 ∂vi + + ∂t 2 j ∂ xi 8m



∂ ∂ xi



   1 ∂ 1 2 1 2 − 2 ∇ | ∇ A | A = Fi A2 ∂ xi A m

(2.39) Now it is evident that if the velocity can be obtained as a gradient of a scalar field (as in (2.36)), we have ∂ 1 ∂ (v j v j ) = (v j vi ) 2 ∂x j ∂x j j j At this point, we introduce the quantity

Appendix

37

Qi = −

     1 ∂ 1 2 1 ∂ 2 − 2 ∇ | ∇ A | A 8 ∂ xi A2 ∂ xi A

(2.40)

Equation (2.39) could be written as  ∂(Av) A

F + 2 Q + ∇ · (Avv) = ∂t m

(2.41)

It is evident that such equation is equivalent to (2.38) and in the limit of  → 0 coincides with the Euler equation for a fluid. We conclude then that the Schrödinger equation is equivalent to the continuity equation (2.37) and Eq. (2.41), and this led in the 30s to the hydrodynamic formulation of quantum mechanics which is illustrated in a very clear paper on Physical Review by David Bohm in 1952.

Some Applications to the Atmospheric Dynamics Baroclinic Flow We have shown how the circulation vanishes if the fluid is barotropic but then what happens when the pressure is also a function of temperature? In this case, the pressure gradient and the density gradient are no longer parallel and Eq. (2.29) must be reconsidered. We just express ρ as the inverse of the specific volume α to get dC =− dt



dp =− ρ



∇p · dl ρ

(2.42)

Using the Stokes’ theorem the integral becomes, remembering the rule ∇× we obtain

∇p 1 = (∇ × ∇ p) + ∇ p × ∇ ρ ρ dC =− dt

 S

∇ p × ∇ρ · dS ρ2

  1 ρ

(2.43)

(2.44)

This equation tells us that circulation can be generated in a baroclinic fluid and we can consider a very illuminating example in the sea breeze problem. In Fig. 2.10, we show a simplified vertical section above the boundary sea land. During the day, the land is warmer than the sea and we assume that the column of air above the land is isothermal at temperature T2 while above the sea the temperature is T1 with T2 > T1 . The gradients for pressure and density are shown in the figure and according to (2.33) the circulation is anticlockwise: We can estimate the wind intensity if we rewrite (2.32) as

38

2 Fundamentals 2

p-2Δp

Δ

p-Δp

p

Δ

ρ

p

Fig. 2.10 The breeze circulation produced by the temperature difference above land and water. During the day the land heats up more rapidly than the seaside

dC =− dt

 RT d ln p

where R is the gas constant divided by the molecular weight of air (≈29). Solving the integral along the path shown in the figure, we have dC = R ln dt



p p0

 (T2 − T1 )

Assuming an average velocity v, we have dv R ln( p/ p0 ) = (T2 − T1 ) dt 2(h + L) Holton derives this formula and uses values like p0 = 1000 hPa, p1 = 900 hPa, T2 − T1 = 10 C, L = 20 km and h = 1 km to find an acceleration dv/dt ≈ 0.685 cm s−2 that is quite unrealistic. What is missing among other things is the frictional drag.

Geostrophic Motion If we want to apply some of the concepts we have illustrated so far to the Earth’ atmosphere, we need to talk a little about the reference system. The Earth rotates with an angular velocity of  E so that the absolute velocity of any object on the surface of the planet is given by Va = V +  E × r

(2.45)

where r is the vectorial distance. The absolute acceleration is then dVa dr Va = +  E × Va dt dt

(2.46)

Appendix

39

where the subscript r indicating the derivative is performed in the relative system. Substituting the value for Va from (2.34), we have dr dVa = (V +  E × r ) +  E × (V +  E × r) dt dt dr V + 2 E × V − 2E R = dt

(2.47)

where R is the vector perpendicular to the rotation axis. The second term on the extreme right represents the Coriolis acceleration while the last term is the centrifugal acceleration. The absolute acceleration must be zero so we have 1 dV = −2 E × V − − ∇ p + g + F dt ρ

(2.48)

where the centrifugal term has been included in the generalization of the acceleration of gravity and F includes all the other forces. At this point, we make a scale analysis of the above equation where we compare the order of magnitude of the acceleration with the Coriolis term. We assume a characteristic length L and a characteristic velocity U so that we have a reference time T = L/U . The ratio between the acceleration dV/dt and the Coriolis acceleration is then dV ≈ O(U 2 /L) dt

2 E × V ≈ O(Ω E U )   U R0 ≈ O E L

(2.49)

The ratio Ro is called Rossby number and is a very important parameter in the atmospheric circulation theory. At this point if R O  1, we can neglect the acceleration term and establish the balance between the pressure and Coriolis term in the horizontal plane 1 ∂p (2.50) − (2 E sin φ)v = − f v = − ρ ∂x + (2 E sin φ)u = + f u = −

1 ∂p ρ ∂y

(2.51)

The quantity f = 2 E sin φ is the component of the vector product on the horizontal plane and it is called Coriolis parameter while φ is the latitude. Notice that  E sin φ is the local vertical component of the angular velocity. u and v are the horizontal velocity components along x and y, respectively. Equations (2.39–2.40) represent the so-called geostrophic approximation which can be used when the Rossby number is much less than 1. For example, in the case of the Earth, we have f ≈ 10−4 s−1 , L ≈ 106 m and U ≈ 10 ms−1 so that R O = U/ f L ≈ 0.1. The Rossby number can be small either if the rotation is rapid (like

40

2 Fundamentals 2

Fp Fco

v

L

Fp

Fco

H

v

Fig. 2.11 Geostrophic circulation around a high-pressure centre (H) and low-pressure centre (L). Also indicated are the pressure force F p , Coriolis, FC O and the geostrophic velocity, v

Earth, Mars, Jupiter) or if the velocities are small. For example, for the Earth’s ocean, U ≈ 0.1 ms−1 and Ro is even smaller than the value of the atmosphere. Equations (2.39–2.40) have a very simple interpretation as it is illustrated in Fig. 2.11. The Coriolis acceleration according to the vector product rule is always to the right of the velocity. The general rule is then that the wind is counterclockwise around a high-pressure centre and clockwise around low pressure.

Streamlines and Stream Function This is the right place where we introduce the concept of streamlines and stream function. Streamline is an imaginary line within the fluid that connects all the tangents to the local velocity vector. This requirement implies in the most simple twodimensional case that v dy = (2.52) dx u where again u and v are the velocity components along x and y, respectively. In general, the streamline can be found by solving the differential equation for each time t0 dy dz dx = = (2.53) u (r, t0 ) v (r, t0 ) w (r, t0 ) Streamlines cannot cross because if they did, the velocity at the intersection point would have multiple values. Streamlines extend to infinity or are closed loops. We will see, talking extensively on kinematics, that there are exceptions to this rule. An immediate example of a streamline can be obtained from the geostrophic motion. As a matter of fact, Eqs. (2.39–2.40) can be written synthetically

Appendix

41

Vg = k ×

1 ∇p ρf

(2.54)

so that the geostrophic vector is always perpendicular to the pressure gradient as already shown in Fig. 2.11. Another important concept to be introduced is the so-called stream function. If we consider a non-divergent fluid, that is ∂u ∂v + =0 ∂x ∂y then we can argue that the velocity components u and v could be obtained from the stream function ψ ∂ψ ∂ψ u=− v= (2.55) ∂y ∂x so that the divergence is zero. It is obvious that the geostrophic motion given by Eqs. (2.39)–(2.40) has zero divergence and in this case, the stream function can be easily found to be, with the assumption that ρ = cost and f do not change with latitude, 1 ∂p ∂ψ p u=− = ⇒ψ = (2.56) ρ f ∂y ∂y ρf The dimensions of the stream function are m2 s−1 and this suggests another meaning to it. Consider as in Fig. 2.12 a two-dimensional fluid, and we want to calculate the flow between point A and B. We can use two virtual “doors”, the first along x which goes from A to C and the other along y going from C to B. The net flow is then v AC − uC B = vx − uy Substituting for the velocities ∂ψ ∂ψ x + y ∂x ∂y and the elementary flux is then dψ =

∂ψ ∂ψ dx + dy ⇒ ∂x ∂y



B

dψ = ψ(B) − ψ(A)

(2.57)

A

The stream function gives directly the flux between two points in a two-dimensional fluid. If ψ is multiplied by ρ, it gives the mass flux directly in kg m−1 s−1 .

42

2 Fundamentals 2

Fig. 2.12 Calculation of the flow between point A and B. Two “bulkheads” are chosen oriented along AC and C A that intercept fluid moving upward and along the x direction

B u

Δy

v C

A Δx

The Taylor Proudman Theorem Again consider the geostrophic motion described by Eqs. (2.39–2.40) and derive them with respect to the altitude z. We have 1 ∂ ∂p g ∂ρ ∂v = = =0 ∂z fρ ∂ x ∂z fρ ∂ x

(2.58)

where we have substituted the hydrostatic equilibrium and assumed that the density does not change with x. The result is that the geostrophic wind is constant with altitude. This result is valid for any rapidly rotating fluid. In this case “rapidly” means that the Rossby number is small. For such a fluid the equation to be used is (2.46). The equation can be simplified if we assume (a) there are not viscous forces; (b) the density is constant (incompressible). When the Rossby number is small the acceleration can be neglected and we have 1 0 = −2 × V − − ∇ p + ∇ ρ If we take the curl of this equation and consider that the curl of the gradients are zero, we have ∇ × ( × V) = 0 (2.59) We now apply the rule ∇ × ( × V) = (V · ∇) − V(∇ · ) − ( · ∇)V + (∇ · V )

(2.60)

Because of the incompressibility (∇ · V) = 0) and the constancy of , we have ( · ∇)V

(2.61)

This the Taylor–Proudman theorem that states that the velocity does not change in the direction parallel to the rotation axis. Equation (2.50) can be split into the vertical

Appendix

43 Ω

Ω

Fig. 2.13 The Taylor–Proudman theorem states that the circulation stays constant in the direction parallel to the angular velocity (left). An obstacle at the bottom (the dashed cylinder) creates a vertical column (right)

and horizontal components to read ∂w =0 ∂z

(2.62)

∂Vh =0 ∂z

(2.63)

 

which implies that if w is zero at the bottom, it must be zero everywhere. As for the horizontal components, their value does not change with altitude. For example, if there is an obstacle at the bottom of the fluid which deforms the flow, this deformation will be conserved at all altitudes. It is as if the rapid rotation imparts some degree of rigidity of the flow (see Fig. 2.13). The result is that the fluid moves in coherent vertical columns. An example of such columns may be found in the ocean while the most spectacular example could be the “Red Spot” on Jupiter. This may originate deeply in the atmosphere of the planet.

Vorticity Equation We have defined earlier vorticity and now we could find a connection with the equation of motion (2.17). Again we write the horizontal component not neglecting the acceleration so we have 1 ∂p du − fv = − (2.64) dt ρ ∂x dv 1 ∂p + fu = − dt ρ ∂y

(2.65)

44

2 Fundamentals 2

Then we derive the first with respect to y and the second with respect to x and subtract the first from the second: we get d dt



∂v ∂u − ∂x ∂y







+ f +

∂v ∂u − ∂x ∂y

 

∂v ∂u + ∂x ∂y

 +v

∂f =0 ∂y

The first parenthesis contains the vertical (∇ × v)z component of the relative vorticity ζ and the last term represents the derivative of f with respect to time d f /dt = ∂ f /∂t + v∂ f /∂ y. Notice that f is a function of latitude. Then our equation becomes   ∂u ∂v d (ζ + f ) = −( f + ζ) + dt ∂x ∂y

(2.66)

In writing this equation, we have neglected higher order terms involving pressure derivative and also the vertical velocity w. The quantity f is known as planetary vorticity and as we have seen earlier, it is just twice the local angular velocity f = 2 sin φ. The quantity ζ is known as relative vorticity and together with f , it makes up the absolute vorticity. Equation (2.66) is the simplest form of the vorticity equation and is the equivalent of the conservation of angular momentum for a fluid. Imagine a very thin disc rotating with angular velocity 2 sin φ and having a positive divergence: this means that the disc is expanding in its radial dimension. In this case, the absolute vorticity tends to decrease (d(ζ + f )/dt < 0). Vice versa, if the disc contracts with negative divergence, the absolute vorticity will increase. In average atmospheric conditions ζ  f so that it can be neglected in (2.66) and when the divergence is zero (as in the geostrophic approximation), the equation reduces to  d

ζg + f = 0 dt

(2.67)

That represents the conservation of vorticity. In Eq. (2.67), ζg indicates the geostrophic relative vorticity defined as ∂u g ∂vg − (2.68) ζg = ∂x ∂y where u g and vg are given by (2.54). When we introduce the stream function ψg , we have ∂ψg ∂ψg vg = ug = − ∂y ∂x And the geostrophic vorticity is just ζg = ∇ 2 ψg

(2.69)

This is a very interesting conclusion because now we can relate the stream function to a quantity very much used in meteorology and that is the geopotential  defined as  = gz where z is the altitude of a pressure surface and g the acceleration of

Appendix

45

z1

z2 > z1

ζ>0

vg

ζ< 0

vg

vg

ζ< 0

Fig. 2.14 The geostrophic vorticity ζg around low- pressure and high-pressure centre. The geostrophic wind (vg ) follows the geopotential contours z 1 and z 2

gravity. It is very easy to show that d = −dp/ρ so that according to (2.64) and (2.65) and with f constant ψ=

 f

⇒ ζg = ∇ 2 ψg =

1 2 ∇  f

(2.70)

Now we know from analysis that a maximum value of the Laplacian ∇ 2 ψg corresponds to a minimum in the function ψg and then . A positive value for the vorticity corresponds to a counterclockwise movement and that happens around low-pressure centre and vice versa. This is schematically illustrated in Fig. 2.14. The consequences of this will be seen later when we explore how the vorticity conservation gives rise to the Rossby atmospheric waves.

References Textbooks Anderson J (2012) Fundamentals of aerodynamics. Mc Graw Hill, New York Faber T (2010) Fluid dynamics for physicist. Cambridge University Press, Cambridge Feynman RP, Leighton R, Sands M (2011) Lectures on physics, vols 1, 2. Basic Books, New York Holton JR (2012) An introduction to dynamic meteorology. Academic Press, New York Hoskins BJ, James IN (2014) Fluid dynamics of midlatitude atmosphere. Wiley Blackwell, New York Snieder R (2001) Mathematical methods for the physical sciences. Cambridge University Press, Cambridge

Chapter 3

Fundamentals 3

We have stressed several times that all the previous matters are about a somewhat ideal fluid that does not know viscosity. But we all know from everyday life that viscosity is alive and, well, everywhere from social life (bureaucracy) or when we fill out the oil tank of our car. In fluid mechanics, viscosity complicates, a little, all the equations and requires the introduction of a new interesting subject. In most cases, things get really interesting when we consider heat transport within the fluid. Again, for our car, the oil that looks so “viscous” at the gas station looks much more “watery” with the rising temperature inside the engine. Besides heat conduction and transport, viscosity is not a familiar subject among students. In this chapter, we will illustrate most of these matters.

3.1 Viscosity We have not very much treated the interaction of an ideal fluid with the wall of a pipe or a tank. When we have done that (efflux from a tank), it was assumed that the fluid in contact with the wall has a velocity determined, for example, by the height of the fluid. Actually, experimentally it can be seen that the velocity of the fluid near the wall is zero. The assumption we made that a shearing stress in a point of the fluid would disappear must be revised if we take into account viscosity. We can devise a simple experiment as shown in Fig. 3.1; a viscous fluid is comprised between two flat plates at some distance d. The upper plate is put in motion with a force F until the plate moves with a constant velocity v0 . Then the fluid in contact with such a plate will move with the same velocity v0 and we can determine experimentally the force needed to maintain the uniform motion. It results to be v0 F =μ A d © Springer Nature Switzerland AG 2020 G. Visconti and P. Ruggieri, Fluid Dynamics, https://doi.org/10.1007/978-3-030-49562-6_3

(3.1) 47

48

3 Fundamentals 3

v0

F

vx+Δvx Δy

d

vx

(a) (b)

Fig. 3.1 In (a, left), the viscous drag between two parallel plates. In (b, right), the viscous stress within the fluid

where A is the area of the plate and μ is what is called coefficient of viscosity. Notice that the ratio F/A has the dimension of a pressure so that we call it shear stress because it is the pressure we have to exert to move one layer of the fluid with respect to the other. As a matter of fact we can go inside the fluid (Fig. 3.1b) and a portion of thickness y with velocity that changes from vx + v to vx so that we have vx ∂vx F =μ =μ A y ∂y

(3.2)

We could name this shear stress with S yx with the index indicating that is the pressure we have to exert in the direction x to produce the gradient of the velocity in the y direction. However, in order to avoid rotations, the stress is most properly defined as  Sx y = μ

∂v y ∂vx + ∂x ∂y

 (3.3)

As a matter of fact for a rotation, we should have ∂vx /∂ y = −∂v y /∂ x so that the stress is zero. In the same way, we can define 

Sx z

∂vx ∂vz + =μ ∂x ∂z



In general, we have

S yz 

Si j = μ



∂v y ∂vz + =μ ∂y ∂z

∂vi ∂v J + ∂x j ∂ xi

 (3.4)

 (3.5)

with the consequence that Sx x = 2μ

∂vx , ∂x

S yy = 2μ

∂v y , ∂y

Szz = 2μ

∂vz , ∂z

Si j = S ji

(3.6)

3.1 Viscosity

49

( Szy +

z

6Szx 6z

) ΔxΔy

Δz

( Syy +

Syy ΔxΔz

Sxy

6Syy 6y

) ΔxΔz

y Δx x

( Sxy +

6Sxy 6x

) ΔxΔy

Δy

Szy ΔxΔy

Fig. 3.2 The viscous stresses in a fluid element

In the most general case of a compressible fluid, we have to add another term to Eq. (3.5) so that the complete expression reads 

∂vi ∂v J + Si j = μ ∂x j ∂ xi



+ μ δi j (∇ · v)

(3.7)

In this equation, δi j is the Kronecker symbol that is δi j = 0 for i = j so that this term contributes only to the diagonal terms. Also, the additional term goes to zero for incompressible fluid (∇ · v = 0). We see that we need two constants to describe the viscous fluid, μ, that is, the “ordinary” coefficient of viscosity and μ that we will call second coefficient of viscosity. Now we are ready to calculate the viscous force per unit volume Fvisc by referring to the elementary volume shown in Fig. 3.2 where we have also represented the stresses on the different faces. Consider now the opposite faces normal to the y direction. The elements that contribute to the net force in the y direction are the stresses tangent to the faces d x − dy, dy − dz and normal to the face d x − dz. For the faces d x − dy, we have as the net force  Szy +

 ∂ Szy ∂ Szy dz d xd y − Szy d xd y = d xd ydz ∂z ∂z

and a similar expression for the faces dy − dz ∂ Sx y d xd ydz ∂x

50

3 Fundamentals 3

while for the faces x, z   ∂ S yy ∂ S yy dy d xdz − S yy d xdz = d xd ydz S yy + ∂y ∂y The net viscous force per unit volume in the y direction is then (Fvisc ) y =

∂ Sx y ∂ S yy ∂ Szy + + ∂x ∂y ∂z

(3.8)

Similar expressions exist for the other components and, in general, we can write for the ith component (Fvisc )i =

3  ∂ Si j j=1

∂ xi

=

   3  ∂v j ∂vi ∂ ∂ μ + + (μ ∇ · v) ∂ x ∂ x ∂ x ∂ x i j i i j=1

(3.9)

In vectorial form this is equivalent to Fvisc = μ∇ 2 + (μ + μ )∇(∇ · v)

(3.10)

In the incompressible case ∇ · v = 0 and the viscous force is just μ∇ 2 v, but we will mostly use the simplest case.

3.2 The Equation of Navier Stokes We now have an expression for the viscous force that we can add to the equation of motion (2.4) to read   ∂v − (v · ∇)v = −∇ p − ρ∇φ + μ∇ 2 v (3.11) ρ ∂t This is known as equation of Navier–Stokes and it is one of the most popular equations in physics that we will see has many different applications. Of particular interest for what we will study in a while is the introduction of the vorticity vector ω = ∇ × v so that we get (compare with Eq. (2.6))  ρ

 1 ∂v + ω × v + ∇v2 = −∇ p − ρ∇φ + μ∇ 2 v ∂t 2

(3.12)

Now apply the curl to this equation to have μ ∂ω + ∇ × (ω × v) = ∇ 2 ω ∂t ρ

(3.13)

3.2 The Equation of Navier Stokes

51

This equation is particularly interesting because without the term proportional to ∇ 2 ω, we are back to the “frozen” vorticity while taking the curl to be zero, we are left with what is known as the diffusion equation for the vorticity μ ∂ω = ∇2ω ∂t ρ

(3.14)

And this means that the vorticity “diffuses” into the fluid or it could dissipate it in case there is a source for the vorticity. We will make some examples on that but now we can justify in general, the diffusion equation.

3.2.1 The Diffusion Equation Consider an element of fluid as depicted in Fig. 3.3 and imagine that the element is invested by some heat flux measured in watt m−2 . The heat entering face 1 at the coordinate x is q(x, y, z)xz (in watts) while the flux leaving the face parallel but at distance [x is [q y (x, y, z) + (∂q/∂ y)y]xz so that the power remaining (or leaving) the elementary cube is given by  −

∂q y ∂qx ∂qz + + ∂x ∂y ∂z

 xyz = −∇ · qV

(3.15)

where V is the elementary volume. If the thermal capacity of the fluid is C in joule kg−1 K−1 , the rate of temperature change will be

z

qy ΔxΔz

( qy +

Δz

y Δx x

Δy

Fig. 3.3 The calculation of the heat fluxes through an elementary volume

6qy 6y

) ΔxΔz

52

3 Fundamentals 3

C

V 1 ∂T = ∇·q=− ∇·q ∂t m ρ

(3.16)

where m is the mass of the elementary volume while ρ is the density. We only need at this point, to express the heat flux in terms of the temperature. For a diffusive flux, such relation is of the type q = −κ∇T where k is the thermal conductivity in w K−1 m−1 . Substituting in (3.15), we have κ 2 ∂T = ∇ T = D∇ 2 T ∂t ρC

(3.17)

The quantity D = κ/ρC is the diffusion coefficient and it is measured in m2 s−1 . Equation (3.17) is known as diffusion equation and is very popular not only in physics. This equation is a good excuse to make some dimensional analysis. If T is a typical temperature difference and L is a typical length, we can evaluate a diffusion time as TD ≈ L 2 /D. Now we all know molecular diffusion from earlier physics and, for example, for air the corresponding molecular diffusion coefficient is of the order of 1 cm2 s−1 that is 10−4 m2 s−1 so if molecular diffusion should be responsible for the heat transport in a room of let’s say 5 m, it will take 25/10−4 ≈ 2.5 days to heat up the room. The reason why the heating is much faster is that the convection mechanism sets in when the air above the radiator warms up and triggers a circulation within the room that is much more efficient in transporting heat. In this case we need to generalize (3.17) by writing ∂T + v · ∇T = D∇ 2 T ∂t

(3.18)

where we have added the advection term v · ∇T . However, the situation is more complicated than that because the advection implies a change in the diffusion mechanism and we will talk about that later. The diffusion is not limited to scalar quantities like temperature. As a matter of fact, we can propose two interesting examples that will be discussed at length later. One has to do with the magnetic diffusion and the other we have already seen with the diffusion of vorticity. The first example is obtained from the Maxwell equations that relate the magnetic induction B, the electric field E and the current density j. We have the equations ∇ × B = μ0 j ∇×E=−

∂B ∂t

∇·B=0

(3.19)

j = σ (E + u × B)

(3.20)

where μ0 is the magnetic permeability and σ is the conductivity.We now obtain E from the second of (3.20) and then take the curl, we have

3.2 The Equation of Navier Stokes

∇×E=

53

∂B ∇×j − ∇ × (u × B) = − σ ∂t

Then substituting for j from the first of (3.19) and remembering the rule ∇ × (∇ × B) = ∇(∇ · B) − ∇ 2 B, we have ∂B 1 = ∇ × (u × B) + ∇2 B ∂t μ0 σ

(3.21)

The generation of the magnetic field is due to a quite complicated term (the first on the right-hand side) while the same field is dissipated by a diffusion term (the second on the right) from which we can estimate the lifetime of the magnetic field. It is possible to think that the Earth’s magnetic field originates in the nucleus of the planet where the conductivity is estimated to be 3 × 105 ohm−1 m−1 while the permeability can be assumed to be 4π 10−7 N A−2 . The quantity 1/μ0 σ is the equivalent of a diffusion coefficient so that the diffusion time is of the order Td ≈ μσ Rn2 ≈ 1.5 × 104 year s

(3.22)

where with Rn , we have indicated the radius of the nucleus. This simple approach means that the planetary magnetic field could not be the remnant of a “permanent” magnetic field but instead it must be generated by the same mechanism (the term ∇ × (u × B) that generically is indicated as dynamo mechanism). We will talk about this later in the book in the magnetohydrodynamic section. As for the diffusion of vorticity as described by Eq. (3.24), we have an equivalent diffusion coefficient of the or μ/ρ. This could have an interesting application for the mantle of the Earth.

3.3 The Reynolds Number We start by rewriting the equation of Navier Stokes in a slightly different form 

 ∂v 1 μ + (v · ∇)v = − ∇ p − g + ∇ 2 v ∂t ρ ρ

(3.23)

The quantity ν = μ/ρ is called specific viscosity or kinematic viscosity and has the dimension of a diffusion coefficient. Now it is very common in fluid dynamics (for reasons that will be more evident later) to normalize or make dimensionless the equations. In the case of Navier–Stokes, we just decide to measure the length with L ∗ , the velocity with U ∗ and the pressure with some P ∗ . The unit for the time is simply T ∗ = L ∗ /U ∗ . Based on these units the new variables will be x  = x/L ∗

y  = y/L ∗

z  = z/L ∗

v = v/U ∗

t  = t/T ∗

p  = p/P ∗

54

3 Fundamentals 3

The derivative and the operators will be 1 ∂ ∂ = ∗  ∂t T ∂t

1 ∂ ∂ = ∗  ∂x L ∂x

∇=

1  ∇ L∗

∇2 =

1 2 ∇ L ∗2

Also we make the position P ∗ = ρU ∗ 2 . Once we make the substitution and simplify, we obtain    ∂v gL ∗ ν      = −∇ + (v · ∇ )v p − + ∗ ∗ ∇ 2 v (3.24) 2 ∗ ∂t  U L U This equation is now dimensionless and that means that its solution does not depend on the dimension of the object we are studying, for example, the flux around a boat or an airplane. This is true as long as two numbers appearing in the same equation are the same. These numbers are the following: Re = Reynolds number =

U ∗ L∗ ν

U∗ Fr = Froude number = √ ∗ gL

(3.25)

(3.26)

We will talk later, in detail, of the Froude number but for the moment, it is sufficient to say that is the ratio between a typical velocity of the fluid U ∗ and something related to a “gravitational speed” that a fluid would acquire after falling from a height L ∗ . The Reynolds number can be obtained as the ratio between the inertial term (v · ∇v ) and the viscous term (ν∇ 2 v ) with the results Re =

U ∗2 L∗ νU ∗ L∗2

=

U ∗ L∗ ν

(3.27)

Considering these simplifications and dropping the primes, now Eq. (3.24) can be written as   ∂v 1 1 2 + (v · ∇)v = −∇ p − + ∇ v (3.28) ∂t Fr Re Objects with the same Reynolds number will produce a similar flow. This result is used to obtain from models of the characteristics of large objects like boats or aircraft. The simplest case is referred to a sphere of diameter d, and for it we can introduce a drag coefficient, C D , defined as the ratio of the viscous (drag) force experienced in the fluid and the quantity 21 ρu 2 A. This is called dynamic pressure. Remember that pressure has the dimension of energy density. CD =

f or ce 1 ρu 2 A 2

(3.29)

3.3 The Reynolds Number

55

where u is the velocity of the fluid and A is the section of the sphere π d 2 /4. The drag coefficient is a complicated function of the Reynolds number (as we will see in a while) and helps to determine different fluid regimes. At this point, we could only say that small Reynolds number (≈10−2 ) corresponds to a regular flow around the sphere while as the Reynolds number increases (above 20), the flow gets very complicated. We need to learn some more to discuss the case but we can say that also the drag force depends on the Reynolds number. For viscous flow (low Reynolds number), the drag turns out to be according to the definition of stress (remember A goes as the square of d μ=

1 f or ce ⇒ D R AG = μdu = ρu 2 A Lu Re

(3.30)

while at high Reynolds number, the drag does not depend any longer on Re and can be taken as ≈ ρu 2 A. We can now make a couple of examples to explain a little better this point. Suppose, we have two circular cylinders, one having four times the diameter of the other. The flow over the smaller cylinder has a free stream density, velocity and temperature given by ρ1 , v1 and T1 . If we assume that the viscosity μ is proportional to the square root of the temperature, we can calculate the density and velocity around the larger cylinder in order to have the same Reynolds number. We have T2 μ2 = μ1 T1 and the two Reynolds numbers

Re1

ρ1 v1 d1 = μ1

Re2

ρ2 v2 d2 ρ2 v2 d2 = = μ2 μ1

T1 T2

Then a possible choice to make the numbers equal is T2 = 4T1 , V2 = 2V1 , ρ2 = ρ1 /4. Another example (as the previous one taken from the Andersen’s book) bears more closely the similarity used to evaluate the characteristic of the flow. The example takes into consideration a Boeing 747 flying at a pressure of 200 hPa (about 12.000 m) at a temperature of 215 K and a speed of 880 km/h. We would like to test a model plane (in scale 1/50) on a wind tunnel at the temperature of 240 K in such a way that the drag and lift coefficient are the same for the full plane and its model. Again, we assume that both μ and the sound speed are proportional to the square root of temperature. The lift coefficient C L is a proportionality factor between lift L and the kinetic energy (3.31) L = 21 C L ρV 2 A where A is the wing area and V the velocity. The index 1 and 2 refer to the free-flight and the model, respectively. We must be assured that the Mach number is the same.

56

3 Fundamentals 3

The Mach number M is the ratio between the velocity of the plane and the local sound velocity. We must have M1 =

V1 V1 ∝√ s1 T1

V2 V1 V2 M2 ∝ √ ⇒ √ = √ T2 T1 T2

From this we can get V2 = 924 Km/h. From the equality of the two Reynolds number, we have ρ2 V2 l2 ρ1 V1 l1 = μ1 μ2 Because of the equality of the Mach numbers, we have V1 = V2

T1 ρ2 l2 ⇒ = = 50 T2 ρ1 l1

And from the gas equation p = ρ RT , we have   p2 ρ2 T2 240 = 55.1 = = 50 p1 ρ1 T1 215 So the wind tunnel should be pressurized at about 11 atm. It is worth mentioning at this time that the Rayleigh number is related to the occurrence of turbulence (something that we will explore later). The number should exceed 106 and we can see that this is quite common. For example, if we consider a sailboat, we find easily that both the boat and the sail are in this regime. The viscosity of water is about 10−3 N sm−2 so that considering the density of water 103 kg m−3 we have a kinematic viscosity of 10−6 m2 s−1 , so even a boat moving at 5 knots (2.6 ms−1 ) will have a Reynolds number greater than 106 even for a dimension of the order of a meter. The sail will be in a turbulent regime as well, even if the dynamic viscosity of air is much smaller than water (1.8 × 10−5 N sm−2 ) which corresponds to a kinematic value of 1.5 × 10−5 ms−2 .

3.3.1 The Blasius Boundary Layer There is another interpretation of the Rayleigh number that gives some further insight into its significance. Consider a uniform steady flow past a thin flat plate as shown in Fig. 3.4. In the upstream of the plate, the flow is uniform with a velocity U parallel to the x-axis. At the surface of the plate, the molecules of the fluid “stick” to the surface so that their velocity is zero (no-slip condition). If we consider a streamline at distance X from the leading edge and at distance δ above the surface, such that just above it the velocity is U . The average acceleration term will be

3.3 The Reynolds Number

57 X

δ

L

Fig. 3.4 The Blasius boundary layer produced on a plate in a moving viscous fluid

 u · ∇u O



U2 X

At the equilibrium, this acceleration is balanced by the viscous stress assuming that the flux changes from U to 0 over the distance δ 

νU ν∇ u O δ2



2

Balancing these two terms, we have for the depth of the boundary layer  δ≈

νX U

1/2 (3.32)

So the boundary thickness increases with the distance and reaches the maximum after a length L at the trailing edge of the plate  δ≈

νL U

1/2 =

L 1/2 Re

(3.33)

The velocity changes from 0 to U and this implies the generation of vorticity of magnitude    1/2 U L U ξ ≈ ∓ = Re1/2 (3.34) δ L X where the minus sign refers to the top of the plate. Notice that the vorticity is infinite at the leading edge and becomes Re1/2 (U/L) at the end of the plate. The vorticity created in this way is shed into the wake behind the plate. The most interesting conclusion of this situation (also called Blasius boundary layer) is that there are two scales of motions, L and δ, and their ratio is just the Reynolds number.

58

3 Fundamentals 3

 2 L Re = δ

(3.35)

Notice, however, that viscous stresses do not generate vorticity but rather their effect is to diffuse vorticity as can be clearly seen in the diffusion term of Eq. (3.23). Equation (3.24) tells us that vorticity increases with increasing Reynolds number. In the boundary layer of the Earth’s atmosphere Re ≈ 109 and this guarantees a fully developed turbulence that we need to examine later. The boundary layer model gives another possible interpretation of the Reynolds number. In the boundary, the change in velocity produces vortices that are either advected by the flow with characteristic time ta = L/U (with L typical length and U typical velocity) or are diffused in the flow with characteristic time td = L 2 /ν. The flow will be regular (we will call it differently later) when ta  td because in this case the vortices may diffuse in the surrounding medium. This will happen when Re 1 and so in case of viscous flow.

3.4 Plane Poiseuille Flow and Pipe Flow: A Plumber Application of Navier Stokes Equation The simplest case is to study the flow of a fluid between two parallel plates at distance h with the flow parallel to the plates in the x direction. We assume that there are no components of the velocity in the y direction and at the same time, the existence of a pressure gradient p along the x direction across some distance L (see Fig. 3.5). We also require steady state conditions so that Eq. (3.21) reduces only to the x component 1 ∂p ∂u =− +ν u ∂x ρ ∂x



∂ 2u ∂ 2u + ∂x2 ∂ y2

 (3.36)

z p2

p1 x z

L Fig. 3.5 The Poiseuille flow between plates

h

3.4 Plane Poiseuille Flow and Pipe Flow: A Plumber Application …

59

The velocity u does not depend on x and so the above equation simplifies to 1 ∂p ∂ 2u = 2 ∂y μ ∂x Substitute now the pressure gradient with p/L and the equation can be easily integrated with the boundary conditions that u(0) = u(h) = 0 u(y) =

1 p y(y − h) 2μ L

(3.37)

The velocity assumes a parabolic behaviour as depicted in Fig. 3.5. A more interesting case is when the flow happens in a pipe. In this case, it is convenient to introduce a cylindrical coordinate system with the axis oriented in the x direction. In this system, we can write the continuity and the momentum equation along x and in the radial direction. We have 1 ∂ ∂u + (r v) = 0 ∂x r ∂r   2    ∂u ∂p ∂ u ∂u ∂u 1 ∂ +v =− +μ r ρ u + ∂x ∂r ∂x ∂x2 r ∂r ∂r    2   ∂v ∂v ∂p ∂ v ∂v 1 ∂ ρ u +v =− +μ r + ∂x ∂r ∂r ∂x2 r ∂r ∂r

(3.38)

(3.39)

(3.40)

where u is the component along the cylinder axis and v is along the radius. The assumption of fully developed flow implies the constancy of the axial velocity (∂u/∂ x = 0) while v = 0. Equations (3.38–3.40) simplify as ∂u =0 ∂x   ∂u ∂p μ ∂ r = r ∂r ∂r ∂x ∂p =0 ∂r

(3.41)

(3.42)

(3.43)

To solve these equations, we assume that pressure varies linearly along x so that ∂ p/∂ x = p/L. Also, we require the no-slip condition u(R) = 0 where R is the radius of the pipe. Knowing what we expect for the solution (the velocity should have a maximum at the axis of the pipe), we also require ∂u/∂r = 0 at r = 0. We can then integrate (3.42) and obtain

60

3 Fundamentals 3

u(r ) =

p 2 r +C 4μL

(3.44)

where the constant is determined by the no-slip condition. 

r 2  p R 2 p 2 2 1− u(r ) = (r − R ) = 4μL 4μL R

(3.45)

Once the velocity is determined, the total flow Q can be evaluated using the integral Q = 2π

R

r u(r )dr =

0

π R 4 p S 2 p = 8μL 8π μL

(3.46)

This is known as Poiseuille formula and at this point, we may introduce a hydraulic resistance as the ratio between p and Q so that R=

8π μL p = Q S2

(3.47)

This equation is useful for plumbers to evaluate the loss along hydraulic pipes.

3.5 Flow Past a Sphere at Low Reynolds Number At very low Reynolds number in Eq. (3.28), the viscous term will be the most important and the inertial term (i.e. the acceleration) will be negligible so that the relevant equations for the fluid will be ∇·v =0 (3.48) −

∇p + ν∇ 2 v = 0 ρ

(3.49)

Now we assume to have a little sphere of radius a in the flow directed upward along the z-axis (see Fig. 3.6). The flux velocity field will have only the upward component W while the flux perturbed by the presence of the sphere will have components v = [vr (r, θ ), vθ (r.θ ), 0]

(3.50)

Axial symmetry requires that ∂/∂φ = 0. In polar coordinates, the equation of continuity becomes 1 ∂ 1 ∂ 2 (r vr ) + (vθ sin θ ) = 0 (3.51) r 2 ∂r r ∂θ As for the case of rectangular coordinates, we can define a stream function ψ satisfying identically (3.51)

3.5 Flow Past a Sphere at Low Reynolds Number

61

w

z vθ

θ

vr

r y

φ

x Fig. 3.6 The flow past a sphere (left). On the right is the spherical coordinate system used

vr =

r2

1 ∂ψ sin θ ∂θ

vθ = −

1 ∂ψ r sin θ ∂r

(3.52)

We can now write the components for the curl of the velocity as show in the appendix. We have   1 Qψ (3.53) ∇ × v = 0, 0, − r sin θ where Q is the differential operator ∂2 sin θ ∂ Q= 2+ 2 ∂r r ∂θ



1 ∂ sin θ ∂θ



At this point, we remember that ∇ 2 v = −∇ × (∇ × v) so that Eq. (3.37) becomes ∇ p = −μ∇ × (∇ × v) we get the components for the pressure gradient μ ∂(Qψ) ∂p = 2 ∂r r sin θ ∂θ

(3.54)

1 ∂p μ ∂(Qψ) =− r ∂θ r sin θ ∂r

(3.55)

By cross differentiation, we can eliminate the pressure and obtain 

∂2 sin θ ∂ + 2 2 ∂r r ∂θ



1 ∂ sinθ ∂θ

2 ψ =0

(3.56)

62

3 Fundamentals 3

On the surface of the sphere (r = a) the velocity is zero so we have as boundary condition 1 ∂ψ ∂ψ = =0 (3.57) ∂r r ∂θ The condition as r → ∞ is that the velocity must coincide with W so we have vr ≈ W cos θ

vθ ≈ −W sin θ

and this corresponds to a stream function ψ ≈ 21 W r 2 sin2 θ which suggests, in general, a stream function of the form ψ = f (r ) sin2 θ Substituted in Eq. (3.56), we have a 4th order ODE of the standard Euler form r 4 f  − 4r 2 f  + 8r f − 8 f = 0 This equation has solutions of the form f (r ) ∝ r n . It is easy to verify that n = −1, 1, 2, 4 so that the solution looks like   A A 2 4 2 2 4 + Br + Cr + Dr f (r ) = + Br + Cr + Dr ⇒ ψ = sin θ r r To satisfy the conditions (3.57) and the stream function at infinity, we must have D=0 vr =

A W B + 3 + = 0, 2 a a

so that A=

1 W a3, 4

C = W/2 vθ = W −

A B + =0 2 a a

3 B = − Wa 4

and the stream function ψ=

W 2

  3ar a3 r3 + − sin2 θ 2r 2

(3.58)

To obtain the drag D on the sphere, we need to find the stresses normal and tangent to the surface of the sphere. The normal stress is given by the pressure that could be found using (3.54) and then integrating from a to infinity. We get p = p∞ −

3 μW a cos θ 2 r2

(3.59)

3.5 Flow Past a Sphere at Low Reynolds Number

63

This is the normal stress while the shear stress has the form    ∂ vθ  1 ∂vr + tr θ = −μ r ∂r r r ∂θ

(3.60)

We have then the force in the z direction due to the normal stress with θ angle between the z-axis and r π D pr ess = 2πa 2 p sin θ cos θ dθ = 2πaμW 0

while the drag due to the shear stress Dshear = 2πa

π

2

tr θ sin2 θ dθ = 4πaμW

0

And the total is just the Stokes famous result Dstokes = 6πaμW

(3.61)

One of the first encounters of the student with the Stokes law is when discussing the Millikan experiment to measure the charge of the electron. The falling of the oil drops in the electric field is determined by the gravity and the Stokes drag. So we have 4 πρ p a 3 g = 6πaμW 3 where ρ p is the density of the oil. The radius of the drops can be determined as a=3

μW 2gρ p

Another relation is obtained for the charge when the falling of the drops is balanced with an electric field V /d determined by the voltage V across the distance d of the condenser plates. 4 V q = πa 3 gρ d 3 It is easy to obtain the charge q as a function of known quantities 18π d q= V

μ3 W 3 2ρg

Today, there are other ways to measure the charge of the electrons, not to mention the fact that Millikan did cut some corners in reporting his experiment.

64

3 Fundamentals 3

Before leaving this topic, we would like to calculate the so-called drag coefficient, that is the ratio between the drag force D and the force due to the dynamic pressure 1 ρv2 2 24 2D 12π μav 12μ = CD = = = (3.62) Aρv2 πa 2 ρv2 aρv Re where A is the cross section of the sphere and Reynolds number is defined as Re = 2aρv/μ. We can see that the drag coefficient decrease as Re increases. Actually, this happens only up to Re ≈ 20 but then the behaviour is quite different. We will find some initial remedy in a while because the real solution must wait for a theory based on turbulence.

3.6 Life at Low Reynolds Number In 1977, the Nobel Prize winner Edward Purcell published a paper on the American Journal of Physics with the same title of this paragraph. Actually, the paper came out of a Symposium dedicated to Victor Weisskopf who, among other things, wrote another paper on Science explaining why mountains on Earth cannot be higher than 10 km. Purcell noted that when the viscosity increases and the Reynolds number gets low, the motion of the familiar objects of everyday life is in trouble. Think about a swimmer in a pool of water that has a kinematic viscosity of roughly ν ≈ 1 × 10−6 m2 s−1 , even the faster one (100 m in 1 min) will have a Reynolds number of 3 × 106 so he can swim with no problems. Reynolds number can get low not only by increasing viscosity but also by decreasing dimension. For example, if you consider a goldfish, its Reynolds number will get down to about 102 , and if you consider a bacterium with dimensions of 1µm, its Reynolds number can be as low as 10−4 . Purcell introduced a very useful concept that can be obtained from the Stokes drag law D = 6π μva. If we consider the sphere moving in such a way that the Reynolds number is 1/6π , we have va = ν/6π so that D = μν = μ2 /ρ where ρ is the density. Purcell concluded that for a Reynolds number of the order of 1, the force μν will tow anything, even a submarine, because for water this force is roughly 10−9 N. If one is interested in low Reynolds number, he/she is also interested in small forces. The other way around works for the Earth’s mantle with a viscosity of 1022 that gives you a force of 1040 N so that it is clear why the movements of solid Earth are so slow. Now suppose you have a swimmer in a pool filled with corn syrup that has a density of 103 kg m−3 and a dynamic viscosity of 5 Pa s. The requirement for low Reynolds number would be not to move faster than 1 cm/min and he will not be able to move at all. Another problem at low Reynolds number is the negligible role of inertia. If you consider a “bug” in water with a dimension of 1µm moving at 30µs−1 , it is possible to evaluate the time it takes to stop it by considering the acceleration given by the Navier Stokes equation

Appendix

65

dV V ≈ ν∇ 2 v ≈ ν 2 dt L So it will take about L 2 /ν to stop which corresponds roughly to 1μs and a deceleration ◦ of 30 ms−2 that will bring it to a complete stop in about 0.15 A. The reason why inertia plays no role at low Reynolds number is that the dominant term in Eq. (3.28) is just the right-hand term, so we have what it is called Stokes or creeping flow ∇ p = μ∇ 2 v

(3.63)

This equation together with the continuity ∇ · v = 0 has some interesting consequences. Take the curl of (3.63) and obtain (with ω the vorticity) ∇2 ω = 0

(3.64)

Then again take to divergence of the same equation ∇2 p = 0

(3.65)

Now we know that if ψ is the stream function ω = −∇ 2 ψ, that substituted in (3.64) gives (3.66) ∇4 ψ = 0 This is exactly the same as Eq. (3.56) that was solved in that case in spherical coordinates. The absence of time in the creeping flow means, among other things, that such a motion is reversible in terms of t (change t in −t). Now, this has some very important implications for the movements of microorganisms in a fluid. As we have seen, the dimensions and velocity of small “bugs” is compatible with very low Reynolds number. These microorganisms use a complicated way of moving using flagella or cilia. Purcell invented for such movements the so-called scallop theorem that sounds like this. Suppose that a small swimming body in an infinite fluid is observed to execute a periodic cycle of configurations, relative to a coordinate system moving with constant velocity U relative to the fluid. Suppose that the fluid dynamics is that of the creeping flow. If the sequence of configurations is indistinguishable from the time of reversed sequence, then U = 0 and the body does not locomote. As a matter of fact, consider a scallop as shown in Fig. 3.7. When the scallop opens its valves, (a) it pushes the fluid to the right and so moves to the left. When completing the cycle by closing the valves (b) it moves the fluid to the left and goes back to the initial position with the results of not moving at all. This logically derives from the fact the time reversal would lead to a motion with velocity −U but since the motion is indistinguishable, then U = −U = 0. Such problems are discussed at length in the very nice book by Childress, which suggests the forms shown in Fig. 3.7. Which ones will move?

66

3 Fundamentals 3

Fluid Motion

Fluid Motion

Fig. 3.7 The scallop theorem. On the left is shown the complete cycle of a scallop while on the right are two possible propulsion techniques

Appendix Curl in Spherical Coordinate Write the elementary distance ds in Cartesian coordinates and spherical coordinates (Fig. 3.8) (3.67) ds 2 = d x 2 + dy 2 + dz 2 = dρ 2 + ρ 2 sin2 θ dϕ 2 + ρ 2 dθ 2 So we can establish that d x = dρ

dy = ρ sin θ dϕ

dz = ρdθ

(3.68)

The components of the gradient will be ∇ψ(ρ, ϕ, z) = ρ 0

ρdθ ds

θ ρ

(3.69)

3

ρ sinθ dφ dρ

∂ψ 1 ∂ψ 1 ∂ψ + θ0 + ϕ0 ∂ρ ρ ∂θ ρ sin θ ∂ϕ

4 q2 (q2,q3 ) ds2=h2dq2

ds3=h3dq3 2

1 q3

φ

Fig. 3.8 The calculation of the curl in spherical coordinates. On the left is the elementary volume in spherical coordinates while on the right a surface element with q1 = cost is shown

Appendix

67

where ρ 0 , ϕ 0 and θ 0 are the unit vectors in the respective directions. It is useful to define the so-called scale factors between the coordinate system x, y, z and, in general, q1 , q2 , q3 (in our case ρ, ϕ, θ ) is such a way that ds1 = h 1 dq1 (d x = dρ)

ds2 = h 2 dq2 (dy = ρdθ )

ds3 = h 3 dq3 (dz = ρ sin θdϕ)

(3.70) In the present case then h 1 = h ρ = 1, h 2 = h θ = ρ, h 3 = h ϕ = ρ sin θ . To evaluate the curl, we just invoke the Stokes theorem applied to a generic vector A

∇ × A · dσ =

A · dλ

(3.71)

S

Working on one component at a time, we consider an elementary surface (Fig. 3.8) on the surface q1 = cost to get

∇ × A|1 h2 h3 dq2 dq3 =

A · dλ

(3.72)

The integral on the right-hand side reads (considering the direction of integration)   ∂ A · dλ = A2 h2 dq2 − A2 h2 + (A2 h2 )dq3 dq2 ∂q3     ∂ ∂ ∂ + A3 h 3 + (A3 h 3 )dq2 dq3 − A3 h 3 dq3 = (h 3 A3 ) − (h 2 A2 ) ∂q2 ∂q2 ∂q3

Comparing with (3.72), one gets 1 ∇ × A|1 = h2h3



∂ ∂ (h 3 A3 ) − (h 2 A2 ) ∂q2 ∂q3

 (3.73)

The remaining components are obtained by rotating the index and the general expression will be ⎡ ⎤ a1 h 1 a2 h 2 a3 h 3 ⎢ ⎥ ⎥ 1 ⎢ ∂ ∂ ∂ ⎢ ⎥ ∇×A= ∂q1 ∂q2 ∂q3 ⎥ h1h2h3 ⎢ ⎣ ⎦ h 1 A1 h 2 A2 h 3 A3 And we get the curl in spherical coordinate ∇×A=

ρ0 ρ sin θ



∂ ∂ Aθ (Aϕ sin θ ) − ∂θ ∂ϕ

 +

θ ρ sin θ



 ∂ Aρ ∂ − sin θ (ρ Aρ ) + ∂ϕ ∂ρ   ϕ ∂ ∂ Aρ (ρ Aθ ) − ρ ∂ρ ∂θ (3.74)

68

3 Fundamentals 3

The Oseen Correction to the Stokes Drag and the Stokes Paradox Carl Oseen observed that far away from the sphere, the advection term in the Navier Stokes equation cannot be neglected. As noted by Kundo and Cohen, the viscous force per unit volume is given by μU a r3 To evaluate the advection terms, we need the velocity components along r, u r and along θ, u θ that can be calculated from the stream function ur =

  3a a3 1 ∂ψ = U cos θ 1 − + r 2 sin θ ∂θ 2r 2r 3

(3.75)

uθ =

  1 ∂ψ 3a a3 = −U sin θ 1 − − 3 r sin θ ∂r 4r 4r

(3.76)

where a is the radius of the sphere. Then the largest inertial term is given by ρu r

ρU 2 a ∂u θ ≈ ∂r r2

Then the ratio of the inertial term to the viscous term is r ρU a r = Re μ a a This means that when r/a is of the order of 1/Re , the viscous term is no longer negligible. As a matter of fact, if we consider x˜ = x/Re then ∂ ∂ = Re ∂ x˜ ∂x And Re (u · ∇u) ≈ Re ∂u/∂ x so we have Re ∂/∂ x ≈ Re 2 ∂/∂ x. ˜ And the viscous term 

∂2 ∂2 ∂2 + + ∂x2 ∂ y2 ∂z 2



 u≈

Re2

∂2 ∂2 ∂2 + + ∂ x˜ 2 ∂ y˜ 2 ∂ z˜ 2

 u˜

At this point ,the advection term can no longer be neglected and the solution is found as a perturbation to the basic velocity U . Our velocity fields becomes u = U + u,

v = v ,

w = w

and the advective term becomes for the x component

(3.77)

Appendix

69

u

  ∂u ∂u ∂u  ∂u  ∂u  ∂u  ∂u +v +w =U + u + u + w ∂x ∂y ∂z ∂x ∂x ∂y ∂z

(3.78)

Neglecting the quadratic terms, the equation of motion becomes ρU

∂u i ∂p =− + μ∇ 2 u i ∂x ∂ xi

(3.79)

In this equation u i represents U  , v or w . The boundary conditions for a moving sphere are u  = v = w = 0, at infinity u  = −U, v = w = 0, at the surface of the sphere The solution found for the stream function is   2    ψ r a 3 Re r 2 sin = + θ − (1 + cos θ ) − exp − (1 − cos θ ) U a2 2a 2 4r Re 4 a (3.80) where in this case Re = 2aU/ν is the Reynolds number based on dynamics. The drag coefficient calculated in this case 24 CD = Re

  3 1+ Re 16

(3.81)

Actually, this formula could include other terms, for example, the next would be 19Re2 (ln Re )/160 and so on. Even this term, however, would diverge for Re > 3 so that the idea of expanding the Stokes drag in terms of power of Re does not seem successful. Following White Fluid Viscous Flow, in Fig. 3.9 we compare some solutions to the Stokes drag with the empirical formula reported in that book in Fig. 3.B.1 that reads CD =

24 6 + √ + 0.4 Re 1 + Re

0 ≤ Re ≤ 2 × 105

(3.82)

Above the upper limit, the boundary layer on the sphere becomes turbulent. Now Eq. (3.82) is somewhat related to the Stokes paradox. We have obtained the drag formula in the hypothesis that inertia effects can be neglected. If this is true, then the force must depend only on the velocity U , the fluid viscosity μ and the dimension L so that the force per unit length would be F = force per unit length = f (U, μ, L) ⇒

F = const μU

The paradox is that the force is independent on the size of the object and this suggests the idea that there must be a dependence on density, that is

70

3 Fundamentals 3

Fig. 3.9 The drag coefficient as a function of the Reynolds number for three different approximations. Equation (3.87) reproduces well the experimental data up to Re ≈ 105

Equation (3.62)

100

Equation (3.87) Equation (3.81) 10 CD

1

10-1

101

1

10 2

10 3

104

105

Numero di Reynolds

F = f (ρ, U, μ, L) ⇒

F = f μU



ρU L μ



That means that drag depends on the Reynolds number.

The Valley Wind In a valley due to the solar heating during the day, a mountain slope may be warmer than the surrounding atmosphere (Fig. 3.10). The air so heated may have some positive buoyancy that generates wind with a diurnal behaviour. There are very complex models (numerical) but a quite old theory due to the pioneeristic work of Ludwig Prandtl is still valid and can be constructed from the few things we have learned so far.

n

z

n T1,ρ1

T2,ρ2

gΔρ/ρ

DAY

z T1,ρ1

s

T2,ρ2

NIGHT −

α

−

1 6p ρ 6x

1 6p ρ 6x α

x

s

x

gΔρ/ρ

Fig. 3.10 The forces acting on an air parcel on a mountain slope. On the left, the day situation is shown with the buoyancy force much larger than the pressure force. On the right, the night situation is shown. The grey arrow indicates the direction of motion

Appendix

71

We assume a coordinate system with the axis s along the slope direction and the other axis n perpendicular to the slope. The usual coordinate system x, z (horizontal and vertical) is such that the relation between the two is the following: z = s sin α + n cos α

(3.83)

We also assume that temperature changes with altitude with gradient  = −dT /dz. Then if the deviation from the basic temperature is T  along the slope, we have the equilibrium between the buoyant acceleration and the viscous term. Notice that during the day, the temperature down the slope is higher than the temperature up the slope (T1 > T2 ) while at night the opposite happens. At the equilibrium, we have g

∂ 2u T sin α + ν 2 = 0 T ∂n

(3.84)

Notice that we have neglected the acceleration due to the pressure gradient that as shown in Fig. 3.10 is negligible. A similar thermal equilibrium can be established between the advection term u(∂ T  /∂s) and the conduction (diffusion) of heat u sin α − D

∂2T  =0 ∂n 2

(3.85)

where D is the coefficient of diffusion of air. Deriving two terms of Eq. (3.84) and substituting (3.85), we obtain ∂ 4u + ∂n 4



g sin2 α T Dν

 u=0

(3.86)

T = 0

(3.87)

4T Dν and n = lη g sin2 α

(3.88)

And a similar equation for T  ∂4T  + ∂n 4 We now let l4 = so that

∂ 4u + 4u = 0 ∂η4



g sin2 α T Dν

and



∂4T  + 4T  = 0 ∂η4

(3.89)

The solution of these equations should depend on four constants. However, they must be finite for n → ∞ while the no-slip condition requires u = 0 for n = 0. The solution for u is then (3.90) u = U e−η sin η

72

3 Fundamentals 3

while for the temperature, it is T  = T e−η cos η

(3.91)

In these equations, U gives the amplitude of the velocity perturbation while T is the difference between the air temperature and the slope temperature. You notice that l has dimension of length and so may be the order of the thickness of the boundary layer. The constant could be obtained by rewriting Eq. (3.85) U e−η sin α sin η = 2D So we get

 U = T

T −η e sin η l2

g D νT

 21 (3.92)

It is quite simple to evaluate the total mass flux rate M= 0

∞

D ρudn =  sin α



∞ 0

 ρTnn dn

 ∂ T   D ρ0 ≈  sin α ∂n n=0

(3.93)

Now T is the temperature difference between the slope and the atmosphere and the first derivative of T  calculated at n = 0 is just T /l. During the night, the temperature of the slope will be higher than the atmosphere (T > 0) and the air will move down the slope. The opposite will happen during the day. In Fig. 3.11, the qualitative profiles of temperature and velocity are shown. A last point of the Prandtl theory concerns the amplitude of the temperature difference. This can be found if we know the sensible heat H leaving (or entering) the surface. Assuming the usual Fourier law, we have

200

Altitude (m)

160

120

80

40

0 -0.5

0

1.0

2.0

Wind velocity (m/s)

3.0

-1

0

1

2

3

4

5

Temperature (C)

Fig. 3.11 On the left, the qualitative profiles for the wind and temperature as a function of height are shown. On the right, the motions along the valley are illustrated during the day (top) and the night (bottom)

Appendix

73

H = −k T

 ∂ T   ∂n n=0

(3.94)

where k T is the thermal conductivity W m−1 K−1 . Again substituting for the derivative, we have (3.95) T = l H/k T It is worth mentioning that with T = 288 K, D = ν = 1 m2 s−1 ,  = 6.5 × 10−3 Km−1 , we get l ≈ 30 m. The calculation for (3.95) would need some more complex explanation.

The Bénard Convection Consider a fluid between two infinite flat plates kept at fixed temperatures T1 and T2 (see Fig. 3.12) with the lower plate at higher temperature. The heated fluid will rise towards the upper plate where it will cool and descend towards the lower plate. The resulting motion will be organized in cells (known as convection or Bénard cells). The problem in two-dimensional convection can be reduced to the transport of two quantities, heat and vorticity. The generation of vorticity can be traced to the positive buoyancy acquired by the fluid particles that produce an acceleration given by  a=

δρ ρ

 g = αgδT

(3.96)

where ρ is the density, g is the acceleration of gravity, α is the coefficient of thermal expansion and T the temperature. The vorticity δζ produced across a distance D will be given by

T2

D

Critical Rayleigh number

10

5

10 4

10

3

T1 10

2

0

1

2

3

4

5

6

7

8

q

Fig. 3.12 The Bénard convection. Periodic convection cells are produced between two flat plates kept at fixed temperatures (T1 > T2 ) (left). On the right, the critical Rayleigh number as a function of q is shown

74

3 Fundamentals 3

1 δv gαδT δζ ≈ ≈ δt D δt D

(3.97)

where we have assumed that the acceleration will produce in time δt, a change in velocity aδt. We assume a coordinate system with the horizontal axis x so that the temperature gradient δT /D ≈ ∂ T /∂ x. This term will be the source of vorticity that will be dissipated by the diffusion. Our equation will be ∂ζ ∂T = ν∇ 2 ζ + gα ∂t ∂x

(3.98)

A similar equation could be found for the temperature where the advection is balanced by the diffusion ∂T T (3.99) +v = k∇ 2 T ∂t D where v is the horizontal velocity and k the diffusion coefficient. Notice that we assumed a constant velocity v along the cell. In terms of the stream function ∂T ∂ 2 ∇ ψ = ν∇ 4 ψ + gα ∂t ∂x

(3.100)

∂ψ T ∂T + = k∇ 2 T ∂t ∂x D

(3.101)

As we did before, we normalize the variables in Eqs. (3.98–3.99) according to t ⇒ t

D2 , k

x ⇒ x  D, ψ ⇒ ψ  k,

θ⇒T

kν αg D 3

where the “prime” are non dimensional. The two equations now become 1 Pr



∂ 2 ∇ ψ ∂t

 = ∇ 4ψ +

∂θ ∂x

∂θ ∂ψ + Ra = ∇2θ ∂t ∂x

(3.102)

(3.103)

In these equations, we have dropped the prime and have introduced two more fluid dynamics numbers. The Prandtl number Pr = ν/k as the ratio between kinematic viscosity and diffusion coefficient, and the Rayleigh number Ra = (ρgα D 3 /μk)T . We will explore at length the significance of these two numbers and for the time being we just assume they are useful to simplify the equations. Solutions to (3.102–3.103) are sought in the form ψ(x, z, t) = ψ1 (t)cos(π z) sin(q x)

(3.104)

Appendix

75

θ (x, z, t) = θ1 (t)cos(π z) cos(q x)

(3.105)

These equations satisfy the conditions that vertical velocity is zero at the two plates  w=

∂ψ ∂x

 = qψ1 (t)cos(π z) cos(q x) = 0

(3.106)

z±1/2

while the temperature must be zero for z = ±1/2. (Notice that we are talking about perturbations of temperature). Substituting (3.104) and (3.105) in Eqs. (3.102–3.103) gives qθ1 − (π 2 + q 2 )ψ1 (3.107) Pr−1 ψ˙1 = 2 (π + q 2 θ˙1 = q Ra ψ1 − (π 2 + q 2 )θ1

(3.108)

To simplify further the equation, we put X=

q ψ1 , π 2 + q2

Y =

so that the equation reduce to

q2 θ1 , (π 2 + q 2 )3

t  = (π 2 + q 2 )t

(3.109)

X˙ = Pr (Y − X )

(3.110)

Y˙ = r X − Y

(3.111)

with r=

q2 Ra (q 2 + p 2 )3

(3.112)

An obvious equilibrium point of the system is X = Y = 0 but the interesting question is to know if this is a stable point. To ascertain that we assume small perturbation around this point of the form δ X = δ X 0 exp(σ t),

δY = δY0 exp(σ t)

when substituted in (3.110), (3.111), we obtain (σ + Pr )δ X 0 + Pr δY0 = 0 r δ X 0 − (σ + 1)δY0 = 0 This system will admit a solution if (σ + Pr )(σ + 1) − r Pr = 0 ⇒ σ 2 + (Pr + 1)σ + Pr (1 − r ) = 0

(3.113)

76

3 Fundamentals 3

One of the real roots of this equation becomes positive when r ≥ 1 so that from the definition of r , we get π 2 + q2 (3.114) Racrit ≥ q2 When this condition is satisfied, an initial perturbation at the equilibrium will grow with time establishing a convective regime. This is also known as Rayleigh–Bénard instability and will be discussed at length in the chapter on instabilities. By the way, studying this problem Edward Lorenz discovered deterministic chaos. It is interesting to find the minimum Rayleigh number that gives convection that is found minimizing (3.114) and we find that q 2 = π 2 /2, so that the minimum Rayleigh number that gives convection is (3.115) Racrit = 27π 2 /4 = 657.5 Figure 3.12 shows the critical Rayleigh number as a function of the parameter q. Let us see if this conclusion helps for the Earth’s case. First of all, we need to define a Rayleigh number for the mantle situation that can be considered a layer heated from within. Starting from the Fourier conduction law, if we have a source of heat H in w kg−1 , the equilibrium between production and conduction requires KT

∂2T = ρH ∂z 2

(3.116)

where K T is the thermal conductivity in Wm−1 K−1 . Integrating this simple equation twice over the thickness D, we have for the temperature difference T = ρ H/K T D 2 that substituted in the expression for the Rayleigh number Ra = (ρgα D 3 /μk)T gives ρ 2 αg H D 5 ραg D 3 Ra = T = (3.117) μk K T μk For the Earth’s mantle, we have μ = 1021 Pa s, K T = 4 w m−1 K−1 , k = 1 mm s−2 , α = 3 × 10−5 K−1 , ρ = 4000 kg m−3 and H = 9 × 10−12 w kg−1 . For the upper mantle D = 700 km and the Rayleigh number would be 2 × 105 well above the critical number.

Vorticity and Friction in the Boundary Layer In the atmospheric boundary layer, the turbulence is such that the friction between the atmosphere and the surface cannot be neglected. This friction will have an important influence over the geostrophic equilibrium we have seen before. Figure 3.13 exemplifies such difference. On the left (a), we see the geostrophic equilibrium with the wind velocity along the isobars. In the same figure, the situation within the boundary layer is quite different. In this case, the geostrophic wind is decreased by the friction and

Appendix

77 (c)

(a)

H

Fp

V

L

L

p

Fco

p+

Δp

(d) Vg

H

Fp Fco (b)

Fig. 3.13 The effect of friction on the geostrophic equilibrium (a) determines a secondary circulation between high pressure and low pressure (b) and (c). In d, the secondary circulation in a cup of tea is shown

the diminished Coriolis acceleration could balance only part of the pressure gradient. As a result, the wind will be oriented towards the low-pressure border. Figure 3.13b represents the possible situation of a high-pressure centre near a low-pressure centre. The deviation will generate a mass flux in the boundary layer between the high pressure and low pressure and this mass exchange will be responsible for a secondary circulation that will result in the rapid destruction of the pressure differences. This secondary circulation is similar to what happens (or happened) in a cup of tea. Many years ago, you did not have tea bags and people did use directly the tea leaves that were filtered out of when the tea was poured into the cup. Then after stirring, it was noticed that leaves would gather at the centre of the cup. Something similar happens in a cup of coffee with cream or foam. A possible explanation was given nothing less than Albert Einstein in a 1926 paper published on Die Naturwissenschaften. The paper had to explain how the erosion in the river bends changes the profile of the river bed. He noted that when you stir tea in a cup at the bottom, the friction slows down the fluid rotation in such a way that the centrifugal force no longer equilibrates the pressure of the fluid so that a secondary circulation is produced (see Fig. 3.13d) so that fluid with high angular momentum is transported to a region with low angular momentum (at the centre), and the rotation is rapidly destroyed. Noticing that friction does not slow down directly the rotation, but it is rather does through this secondary circulation. A very nice treatment of this problem is given in Baker (1968). To apply the same argument to the atmospheric boundary layer, we need to find how the wind changes with altitude. To simplify things, we assume a basic flow in the x direction and include it in the momentum equation. We have ∂ 2u + f v = 0, ∂z 2 1 ∂p ∂ 2v ν 2 + fu = − ∂z ρ ∂y ν

(3.118)

There is a simple justification for the first left-hand terms. We imagine that the geostrophic wind at the top of the boundary layer forces the lower layers so that there

78

3 Fundamentals 3

is transport of momentum from the top layer to the surface. The momentum flux being diffusive can be assumed to be ν∂u/∂z and the acceleration is just proportional to the derivative of that. Then we assume geostrophic equilibrium −∂ p/∂ y = ρ f u g , so we get for the second ∂ 2v ν 2 + f (u − u g ) = 0 (3.119) ∂z We then define a complex variable U = u + iv and the system (3.118) reduces to ν

∂ 2U − i f U = −i f Ug ∂z 2

(3.120)

The boundary conditions are U = 0 for z = 0 and U → Ug = u g + i0 for z → ∞. The solution is then      πz πz cos , u = u g 1 − exp He He     (3.121) πz πz v = u g exp sin He He √ where He = 2ν/ f is the thickness of the boundary layer. As a matter of fact, the v component become zero when z/He = 1. At this point, the basic flow only in the x direction generates a component of the flow in the y direction (v = 0), and we can evaluate the flux in the y direction from M=

He



He

ρvdz =

0

 ρu g exp

0

πz He



 sin

πz He

 dz

(3.122)

Then the vertical velocity could be obtained from the continuity equation integrating ∂w/∂z from the ground to the top of the boundary layer to get

He

ρwe = − 0



 ∂ ∂ (ρu) + (ρv) dz ∂x ∂y

(3.123)

where it is assumed that w = 0 for z = 0. Substituting (3.121), we notice that the derivative with respect to x is zero and we remain with ∂ ρwe = − ∂y

0

He



πz ρu g exp He





πz sin He

 dz

(3.124)

Comparing this equation with (3.122), we notice that vertical flux at the top of the layer is equal to the horizontal convergence of mass in the boundary layer −∂ M/∂ y. At this point, we notice that −∂u g /∂ y is just the geostrophic vorticity ζg so we have

Appendix

79

 we = ζg

ν 2f

(3.125)

Now it is possible to relate the vertical velocity to the rate of change of vorticity. From the vorticity equation d (ζ + f ) = − f dt



∂v ∂u + ∂x ∂y

 = f

∂w ∂z

Integrate this equation from the boundary layer to the top of the tropopause to obtain

H He

dζ dz = f dt



wH

dw

(3.126)

we

Assuming w = 0 for z = H and remembering that the geostrophic vorticity does not depend on z, we get  dζg we fν =−f ≈ −ζg (3.127) dt H − De 2H 2 where (3.125) has been used for we and H  He . The time it takes for the vortex to spin down is simply   2 1 dζg −1 (3.128) =H τe = ζg dt νf Using H = 10 km, f = 10−4 s−1 , ν = 10 m2 s−1 , we have a spin down time of roughly 4 days. This time can be compared with time it takes for viscous diffusion of the order of H 2 /ν which is about 100 days. So the secondary circulation is a very efficient way to destroy pressure dipoles.

Intensification of the Western Boundary Current We all know about the Gulf Current that is supposed to make the climate in the North Atlantic more bearable. However, a few may know the origin of such a current (or similar phenomena) and even less know that mild winter in England may be due to other causes. From the point of view of fluid dynamics, the Gulf Current is the manifestation of what wind stresses can produce on the ocean and, in turn, how stresses on the western boundaries may change the character of currents. For simplicity, we will reduce the Atlantic Ocean to a rectangular basin as in Fig. 3.14 of dimension L in the x direction and M in the y direction. Over the southern part of this basin, the trade winds will blow from east to the west (the so-called easterlies) while westerlies (from west to the east) will blow in the mid-latitude sector. These winds will produce stresses on the surface of the

80

3 Fundamentals 3

wind stress = -1.0 frictional = +0.1 planetary = -1.0

wind stress = -1.0 frictional = +0.1 planetary = +1.0

Total

Total

= -1.9

= +0.1

wind stress

Ekman drift

M

western side

L

eastern side

Ekman suction

Ekman pumping

Fig. 3.14 On the left side, the circulation induced in a rectangular basin by the winds’ stress is shown. On each side of the basin, a rough budget for the vorticity is reported. On the right side, it is shown how the torque associated with the stress can produce a vertical motion. The water below the thermocline is dark grey

ocean that will drive a clockwise circulation as shown in the figure. The generation of negative vorticity at the basin level must be balanced by an appropriate sink. Besides this requirement, we must invoke the conservation of absolute vorticity ζ + f . On the western side of the basin (i.e. the North American coast), the current has the south to north direction so that the Coriolis parameter f will increase and consequently ζ must decrease while the wind stress provides negative vorticity. On the eastern side of the basin (the African coast), the current will move from north to south so that f decreases, and ζ must increase while the input from the wind stress provides negative vorticity. Using the data in the original paper by Henry Stommel, we assume the contribution from the wind stress to be −1 so that the other components (friction and planetary) are relative. The frictional stress assumes the current to be zero near the coast and to increase to the “open sea” value so its contribution will be positive on either side. The net balance of vorticity over the basin must be zero and in these conditions, the western side will have an excess of negative vorticity while the eastern side will be almost neutral. To have a better idea of what is going on, we need some more information and may start from the momentum equations 1 ∂p 1 ∂τx ∂u − fv = − + , ∂t ρ ∂x ρ ∂z 1 ∂p 1 ∂τ y ∂v + fu = − + ∂t ρ ∂y ρ ∂z

(3.129)

where τx and τ y are the wind stresses in the x and y directions. We now assume that the steady state is reached in (3.129) when the wind is made up of the geostrophic terms that equilibrate the pressure gradients and some residual values that equilibrate the stress term. Then (3.129) reduces to

Appendix

81

1 ∂τx , ρ ∂z 1 ∂τ y f uE = − ρ ∂z

− f vE = −

(3.130)

Integrating these equations over z from the bottom of the basin z = −h to the surface z = 0, we have −f

0

−h 0

f −h

ρv E dz = τx (0) − τx (−h), (3.131) ρu E dz = τ y (0) − τ y (−h)

If we define the total mass flux vector  M = (Mx , My , 0) =

0

−h

ρudz,

0 −h

 ρvdz, 0

(3.132)

that could also be written as − f M = k × τ . Then taking the divergence of this vector, we obtain  0 ∂ρu ∂ρv − f + dz + ρw(0) − ρw(−h) = ∇ · M + ρw(0) − ρw(−h) = 0 ∂x ∂y −h (3.133) And we can get the vertical velocity at the bottom assuming that w(0) = 0 and   1 τ w(−h) = k · ∇ × ρ f

(3.134)

This formula has a very simple interpretation and corresponds to the Ekman pumping or Ekman suction as shown in Fig. 3.14. On the right of the figure, the fluid column rotates clockwise so the drift due to the Coriolis terms drives the fluid at the centre generating a downward motion (Ekman pumping). On the left, the fluid column rotates in anticlockwise fashion creating an upward motion Ekman suction. The result is a deformation not only of the surface of the ocean but also of the first layer down to the so-called thermocline (in dark grey in the figure). From these considerations, we can obtain another important diagnostic tool for the oceans. Again at the equilibrium, if we integrate Eq. (3.129) we obtain

∂p dz + τx (0), ∂ −h x 0 ∂p dz + τ y (0) f Mx = − −h ∂ y

− f My = −

0

(3.135)

82

3 Fundamentals 3

We can eliminate the pressure from these by deriving the first with respect to y and the second with x to obtain (3.136) β M y = k · (∇ × τ ) where β = ∂ f /∂ y = 2 cos φ/a. In this case,  is the angular velocity of the Earth, a its radius and φ the latitude. Equation (3.136) is known as Sverdrup balance. It is to notice that the meridional flux M y is proportional to the torque on the ocean water produced by the wind stress. This can be understood if we refer to Fig. 3.14 where we have a wind field that exemplifies the transition between the westerlies and the trade winds. In the northern part, the westerlies produce an almost zonal current that is deviated south by the Coriolis acceleration. At some point, this deviation is such that it produces a meridional flux that may be further deviated until the motion in inverted with respect to the initial motion. In the basin, we then have essentially a meridional motion but continuity requires that somewhere water must return north and this could happen only at the two boundaries at x = 0 or x = L. However, a return on the eastern boundary would produce negative vorticity and could not dissipate the vorticity created by the wind stress while a return to the western boundary could produce positive vorticity with an appropriate shear. As mentioned before in 1948 Henry Stommel formulated a quite simple theory to explain why currents strengthen on the western boundary. The theory starts from Eq. (3.129) where we add a simple term due to what is called Rayleigh friction, where the acceleration is proportional to the velocity of the type Ru for the x direction and Rv for the y direction. Then we apply cross differentiation to obtain   ∂ My ∂ Mx ∂τx (0) ∂τ y (0) − +R − (3.137) − β My = ∂y ∂x ∂x ∂y Let us define a mass stream function  such that Mx = −

∂ , ∂y

My =

∂ ∂x

(3.138)

And (3.137) becomes −β

∂τ y ∂ ∂τx = − +R ∂x ∂y ∂x



∂ 2 ∂ 2 + 2 ∂x ∂ y2

 (3.139)

Provided the boundary condition of no mass flux at the boundary  = 0 at x = 0 and x = L. Following Stommel, we use a wind stress purely zonal given by τ y = 0,

τx = −T cos

πy  M

(3.140)

Appendix

83

We solve the equation by separating the variables (x, y) = ψ(x) sin

πy 

(3.141)

M

and get a differential equation in ψ R

π 2 d 2ψ dψ Tπ − R + β ψ =− 2 dx dx M M

(3.142)

In the particular case of no friction (R = 0), we have the solution ψ(x) =

Tπ (L − x) βM

(3.143)

that cannot be accepted as a solution because for x = 0, ψ(0) = 0. Then we consider that the Coriolis parameter does not change with latitude β = 0. A solution in this case is given by TM ψ(x) = Rπ

  sinh(π(L − x)/M) + sinh(π x/M) 1− sinh(π L/M)

(3.144)

This expression is symmetrical with respect to x = L/2 so that cannot provide the western intensification. The complete solution could be found as (x, y) =

 

πy  TM (1 − e D2 L )e D1 x − (1 − e D1 L )e D2 x sin 1− D L D L Rπ e 1 −e 2 M

where β D1 , D2 = − ± 2R



β 2R

2 +

π 2 M

(3.145)

(3.146)

This solution is shown in Fig. 3.15 and shows a typical length scale on the western boundary given by δ≈

10−6 s−1 R ≈ ≈ 50 km β 2 × 10−11 m−1 s−1

(3.147)

The solutions (3.144) and (3.145) are shown in Fig. 3.15. The Rayleigh friction coefficient has been increased to R = 5 × 10−5 in order to make the streamlines more distinct. Following the previous consideration on the Ekman pumping, we may notice that the anticyclonic motion will produce a positive bulge at the surface of the ocean and a depression of the thermocline.

84

3 Fundamentals 3

f = cost

β=cost

Fig. 3.15 The intensification of the westward current. In the figure on the left, the solution (3.144) is shown with the Coriolis parameter constant. On the right f , it changes linearly according to f = f 0 + βy. The Rayleigh friction assumes R = 5 × 10−5 five times the values used in the text to make more clear the figure

References Textbooks Childress S (1981) Mechanics of swimming and flying. Cambridge University Press, Cambridge Feynman RP, Leighton R, Sands M (2011) Lectures on physics, vol 2. Basic Books, New York Kundu PK, Cohen IM (2015) Fluid mechanics. Academic Press, New York White F (2011) Viscous fluid flow. Tata Mc Graw Hill, New York

Articles Baker DJ (1968) Demonstrations of fluid flows in a rotating system II. The spin up problem. Am J Phys 36:980 Baker DJ (1966) Demonstrations of fluid flows in a rotating system. Am J Phys 34:647 Lauga E, Powers TR (2009) The hydrodynamics of swimming microorganisms. Rep Prog Phys 72:1 Purcell EM (1977) Life at low Reynolds number. Am J Phys 45:3 Stocker T (1996) The ocean in the climate system. In: Boutron CF (ed) Topics in atmospheric and interstellar physics and chemistry, European research course on atmospheres, vol 2. France, Les Editions de Physique, Les Ulis Zardi D, Serafin S (2015) An analytic solution for time-periodic thermally driven slope flows. Q J R Meteorol Soc 141:1968

Chapter 4

Aerodynamics and All That

What we have learned so far is enough to start dealing with practical matters and nothing in fluid dynamics is more suitable than aerodynamics. Here we do not refer only to the interaction of air with bodies but to more general phenomena including the motion of boats or fishes. A major limitation of our approach is that viscosity is neglected but we will see this simplification does not prevent a good qualitative description of the phenomena. Plenty of books about flight and aerodynamics are published and many of them written by engineers that will explain the concepts like lift or drag that have numerous applications. In the title of the chapter, we have added an “all that” which refers to a class of examples that refer to less practical issues that are treated in this chapter. We will talk about the motion of a golf ball and we will see how complicated the flight of a frisbee can be. We hope you will enjoy all that.

4.1 Velocity Potential We will follow mostly the Kundu Cohen approach so that in two dimensional frame we know that the continuity equation for an incompressible flow ∂v ∂u + =0 ∂x ∂y

(4.1)

implies the existence of a stream function ψ such that the velocity component can be obtained ∂ψ ∂ψ u= v=− (4.2) ∂y ∂x We now refer to an irrotational flow when the z component of the vorticity is zero

© Springer Nature Switzerland AG 2020 G. Visconti and P. Ruggieri, Fluid Dynamics, https://doi.org/10.1007/978-3-030-49562-6_4

85

86

4 Aerodynamics and All that

∂u ∂v − =0 ∂x ∂y

(4.3)

In this case we can also introduce a scalar function φ called velocity potential such that ∂φ ∂φ u= v= (4.4) ∂x ∂y It is easy to show that equipotential lines (where φ is constant) and streamlines are orthogonal because     ∂φ ∂φ ∂ψ ∂ψ ∂φ ∂ψ ∂φ ∂ψ ∇φ · ∇ψ = i +j · i +j = + =0 ∂x ∂y ∂x ∂y ∂x ∂x ∂y ∂y

(4.5)

Based on the equalities ∂φ ∂ψ =− ∂y ∂x

∂φ ∂ψ = ∂x ∂y

(4.6)

These are called the Cauchy-Riemann equations in complex variable theory and provided that the partial derivatives of (4.6) are continuous it follows that w = φ + iψ

(4.7)

is an analytic function of the complex variable z = x + i y and w(z) is called the complex potential. Beside, the functions ψ and φ satisfy the Laplace equations ∇2φ = 0

∇ 2 ψ = 0.

(4.8)

4.2 Elementary Potential Flows 4.2.1 Source and Sink of a Fluid We start by considering an axisymmetric potential φ ≡ φ(r ). From the Laplace equation in polar coordinate we have 1 d ∇ φ= r dr 2

  dφ r =0 dr

(4.9)

And a simple integration gives m dφ = ⇔ φ(r ) = m ln r + C dr r

(4.10)

4.2 Elementary Potential Flows

87

φ = const

t

st co n ψ=

ψ

co

φ=

con st

ns

=

Plane Source

Line Vortex Plane Doublet

Fig. 4.1 Plane source (left), line vortex (center) and plane doublet (right). The velocity potential φ and the stream function ψ are indicated

With m and C constants to be determined. The velocity component could be found using the (4.4) and the result is only the radial component u = ∇φ =

m r0 r

(4.11)

where r0 is the unit vector along r . m > 0 represents a source while m < 0 is a sink of fluid of strength m. If Q is the flow rate it is found that in two dimensions m = Q/2π . Notice that the two dimensional case is quite different from the spherical case when the velocity is u = (m/r 2 )r0 . The different result is due to the fact that in three dimensions ∇φ = m/r 2 . It is interesting to find in this case the potential ψ. We use the relations (4.6) in plane polar coordinates 1 ∂ψ ∂φ = ∂r r ∂θ

1 ∂φ ∂ψ =− r ∂θ ∂r

(4.12)

From the first of these relation we get ψ=

m θ 2π

(4.13)

An example of a plane source is shown in Fig. 4.1.

4.2.2 Line Vortex With a potential like φ(θ ) = kθ the solution of Laplace equation in plane polar coordinate is k ∂φ 1 ∂φ =0 uθ = = (4.14) ur = ∂r r ∂θ r

88

4 Aerodynamics and All that

This represents a rotating fluid around a line vortex with k = /2π if  is the circulation of the flow. It is interesting to verify that the vorticity is zero in this case and this can be done by evaluating the curl in cylindrical plane coordinates 





∂u r ∂u z − ∂z ∂r





 1 ∂(r u φ ) 1 ∂u r + φ0 + z0 − ∇ × v = r0 r ∂r r ∂φ (4.15) where the unit vector in the three directions are r0 , φ 0 and z0 . It is clear that ∇ × v = 0. Again it is easy to show, using (4.12) and (4.14), that the stream function for the vortex is given by ∂ψ  1 ∂φ =− ⇔ψ =− log r (4.16) r ∂θ ∂r 2π 1 ∂u z ∂u φ − r ∂φ ∂z

The vortex it is shown in Fig. 4.1b.

4.2.3 Plane Doublet We can use the potential for a source or sink to find the potential for a doublet considering that the Laplace equation is linear so that different solution can be superimposed. We then consider a source at point y = δ/2 and a sink at point y = −δ/2. We have for the fields φ = m log r = m log



x 2 + (y − δ/2

2 1/2

=

 2 m log x 2 + (y − δ/2 2

(4.17)

An the total field would be φ=



2   m x + (y − δ/2)2 m  2 log x + (y − δ/2)2 − log x 2 + (y + δ/2)2 = log 2 2 2 x + (y + δ/2)2

(4.18) The equipotential lines are similar to the field lines of an electric dipole. It can be shown that in the limit δ → 0 with μ = mδ fixed we have φ=−

μ·r = μ · ∇ ln r r2

(4.19)

In our case μ is oriented along y so we have φ = −μy/r 2 and using the first of (4.12) we have μx (4.20) ψ =− 2 r There is a more intuitive way to present the doublet. As a matter of fact we can obtain a doublet by placing two point sources at distance d as shown in Fig. 4.2a. In this case our stream function is

4.2 Elementary Potential Flows

89

Δθ

dθ

m 2πc

b

source +m

θ2

θ1

θ

a

sink -m d

Fig. 4.2 The doublet constructed from a source and sink

ψ=

m m (θ1 − θ2 ) = − θ 2π 2π

(4.21)

Now we want d to go to zero while keeping the strength of the doublet μ = md fixed. As shown in Fig. 4.2b, when d decreases we have  m ψ = lim − dθ d→0 2π

(4.22)

dθ = d sin θ/(r − d cos θ ) And as consequence ψ =−

μ sin θ 2π r

(4.23)

The streamlines are obtained putting this expression to some constant value c so we have μ r =− sin θ (4.24) 2π c In plane polar coordinate this represents a circle with radius μ/2π c with the center above the origin. The streamlines for a doublet are then circles tangent to x axis for different values of c as shown in Fig. 4.2c.

4.2.4 Non-lifting Flow over a Circular Cylinder We now demonstrate that a combination of a uniform flow and a doublet produces a flow over a circular cylinder. The uniform flow with velocity V∞ along the x direction is simply (4.25) ψ = V∞ y = V∞r sin θ

90

4 Aerodynamics and All that Lifting flow over a cilinder

A,B stagnation points

A

B

PLUS

Non lifting flow over a cilinder

EQUAL

Vortex of strength Γ

Fig. 4.3 The lifting flow over a rotating cylinder as a sum of the flow over a cylinder plus a vortex of strength 

So that the combination with (4.23) gives     μ sin θ μ R2 = V∞r sin θ 1 − 2 ψ = V∞r sin θ − = V∞r sin θ 1 − 2π r 2π V∞r 2 r (4.26) where we have put R 2 = μ/2π V∞ that actually represents the radius of the cylinder. The velocity components are   1 ∂ψ R2 Vr = = 1 − 2 V∞ cos θ r ∂θ r Vθ = −

  R2 ∂ψ = − 1 + 2 V∞ sin θ ∂r r

(4.27)

(4.28)

We now introduce the concept of stagnation point where the velocity is zero. Putting to zero both Eqs. (4.27–4.28) we find that such point is located in (r, θ ) = (R, 0) and (R, π ). These two points are indicated by A and B in Fig. 4.3. Inserting these values in the equation for the stream function (4.26) we obtain ψ = 0 that implies   R2 (4.29) V∞r sin θ 1 − 2 = 0 r Which is satisfied for r = R that is the circle with center in the origin shown in Fig. 4.3. The radius of the cylinder is related to the velocity of the flow and the doublet strength μ. μ R= (4.30) 2π V∞ Notice that the entire flow field is symmetrical about the vertical and horizontal axis, so that the pressure fields are symmetrical and there is no net force on the cylinder. Now it is easy to accept that there is no lift on the cylinder while it is harder to recognize there is no drag. We all know that a body in moving fluid experiences a

4.2 Elementary Potential Flows

91

drag. This is also known as D’Alembert paradox and we know that at that time they simply neglected the viscous effect as we have already studied for the Stokes drag. On the other hand. there is a way to obtain the lift from a cylinder if you just make it spinning around its axis: it is what is known as Magnus effect.

4.2.5 Lifting Flow over a Rotating Circular Cylinder Consider now, as shown in Fig. 4.3, to add a non-lifting flow over a cylinder to a vortex of strength . Just sum up the stream function (4.26) with the (4.16). We have 

R2 ψ = V∞r sin θ 1 − 2 r

 +

 ln r + const 2π

(4.31)

The constant derives from the integration of the vortex stream function and we may choose it in such a way that ψ = 0 for r = R, that is const = −(/2π ) ln R    R2 r ψ = V∞r sin θ 1 − 2 + ln r 2π R

(4.32)

From this we can obtain the velocity components   R2 Vr = V∞ 1 − 2 cos θ r

  R2  Vθ = −V∞ 1 + 2 sin θ − r 2πr

(4.33)

We can find the stagnation point by imposing these to be zero at r = R. From the second equation we have 

 θ = arcsin − 4π V∞ R

 = arcsin(B)

(4.34)

where B = − (/2π V∞ R). We see that if B < 2 there are two solutions θ = − arcsin(B/2) and as B increases the solution move around and coalesce for B = 2 and θ = 3π/2: When B > 2 there is only one stagnation point that lies outside the surface of the cylinder at B r = + R 2



B2 −1 4

θ=

3π 2

(4.35)

These different cases are illustrated in Fig. 4.4 where we see that streamlines are no longer symmetrical with respect the horizontal axis. The net force in the y direction results from the pressure on the surface of the cylinder that coincides with a streamline. So we have from the Bernoulli theorem

92

4 Aerodynamics and All that

Γ 2 2 π VR

Fig. 4.4 The stream function over a rotating cylinder as a function of the parameter B = /2π V R. Notice the position of the stagnation points

p + 21 ρu 2 = const,

on r = R

(4.36)

therefore   V sin θ + const p = ρ −2V 2 sin2 θ + πR

on r = R

(4.37)

where V∞ is now simply V . As we have observed the stream function is symmetrical to the y axis (changes θ in π − θ so that the net force must be perpendicular to the horizontal axis. The elementary force on the cylinder is simply p Rdθ and the vertical components is − p R sin θ dθ . The net force on the cylinder is then

ρ 0

2π

  V 2 2 −2V sin θ + sin θ R sin θ dθ = −ρV  πR

(4.38)

This result shows that there is a positive lift if  < 0 (that is the cylinder rotates in clockwise fashion), and there are obvious reasons for that. On top of the cylinder the rotation velocity sums up with the incoming stream producing high speeds and low pressures. Beneath the cylinder, the circulatory flow opposes the incoming stream leading to lower velocity and higher pressure. This result is also called Magnus effect and preludes to a what is called Kutta-Zhukovsky theorem that gives the formula for the lift of a wing. To arrive to that we need to make some more progress.

4.2.6 The Kutta-Zhukhovsky Lift Theorem In the previous paragraph we have shown that the drag on a circular cylinder is zero, while the lift is −ρV . These results are valid for a cylinder of any section and this is known as Blasius theorem that is the basis to obtain also the lift for a “cylindrical” wing.

4.2 Elementary Potential Flows

93

Consider a cylinder with arbitrary section surrounded by a fluid. Let us call L the lift and D the drag, as for an inviscid flow only the normal pressures are exerted, we could write these two elementary forces as d D = − pdy

d L = pd x

(4.39)

We can form the complex quantity d D − id L = − pdy − i pd x = −i p(d x − idy) = i pdz ∗

(4.40)

with complex variable z ∗ = x − i x. The total force on the body is then 

pdz ∗

D − i L = −i

(4.41)

C

In this case C represents the contour that coincides with the section of the cylinder. The pressure can be found from the Bernoulli equation 1 p∞ + ρV 2 = p + 21 ρ(u 2 + v2 ) = p + 21 ρ(u + iv)(u − iv) 2

(4.42)

where now u and v are the velocity components along x and y. Substituting for the pressure in (4.41), we have  D − i L = −i C



 p∞ + 21 ρV 2 − 21 ρ(u + iv)(u − iv) dz ∗

(4.43)

Now the integral of constant terms is zero while the complex number u + iv can be written as  u + iv = u 2 + v2 eiθ dz =| dz | eiθ Notice that this element dz is parallel to the contour and the quantity (u + iv)dz ∗ is real, so we can equate it to its complex conjugate (u + iv)dz ∗ = (u − iv)dz. Equation (4.43), becomes D − iL =

i ρ 2

  C

dw dz

2 dz

(4.44)

where we have introduced the complex velocity dw/dz = u − iv. Equation (4.44), constitutes the Blasius theorem and applies to any plane irrotational flow. We now apply this theorem to a steady flow around the airfoil assimilated to a cylindrical body with a clockwise circulation  with the velocity at infinity with magnitude V and the flow parallel to the x direction. Then the flow can be considered a superposition of a uniform stream and a set of singularities like vortex, doublet, source and sink. Taking the contour C at large distance from the body all the singularities seem to be

94

4 Aerodynamics and All that

C

dθ

V dz

θ

θ

(b)

p dy v

p dx

(a) Fig. 4.5 The Blasius theorem for a generic cylinder (a) and the calculation of the lift produced on an airfoil (b)

located in the origin z = 0. We can now switch to the complex representation and write the potential in the form w = Vz +

i μ m ln z + ln z + + · · · 2π 2π z

(4.45)

We recognize the uniform flow, the source, a clockwise vortex and a doublet. Because the contour is closed the net mass efflux must be zero and so m = 0. Then applying the Blasius theorem, we have iρ D − iL = 2



C

i μ V+ ln z − 2 + · · · 2π z

2 dz

(4.46)

From the theory of complex variables the integral can be found to be (see Appendix)

  iV iρ 2π D − iL = 2 pi

(4.47)

which implies D=0

L = ρV 

(4.48)

The same results could be obtained without recurring to the complex variables. Consider then an airfoil as depicted in Fig. 4.5b, which generates a circulation  when immersed in a fluid with density ρ and uniform flow V . Consider a circuit of radius r large with respect to the dimension of the airfoil. The airfoil lift upward must be equal to the sum of the pressure force on the whole periphery of the circuit. For the moment we will neglect the reaction to the change of downward momentum of the air that we will see in a while. Actually, given the dimension of r , large with respect to the airfoil, this effect can be neglected. On the circular boundary, the local pressure is p and the velocity is the sum of the free stream velocity (V ) and the

4.2 Elementary Potential Flows

95

one induced by the circulation v. So, if we apply the Bernoulli equation between the location of the airfoil and the circular path, we have  1 1  p0 + ρV 2 = ρ V 2 + v2 + 2V v sin θ 2 2 So that (neglecting v2 with respect to V 2 ) p = p0 − ρV v sin θ On the element subtended by the angle δθ the pressure force is − pr sin θ δθ so that substituting for p the total contribution to lift is given by

2π

L1 = −

( p0 − ρV v sin θ )r sin θ dθ = ρπ V vr

0

Now, we must take into account the contribution to the lift due to the increase of the mass flow through the boundary. The mass flow is ρV r cos θ δθ and due to the velocity increase v cos θ the rate of change of downward momentum is −ρV r cos θ δθ (v cos θ ) = −ρV vr cos2 θ δθ and the inertial contribution to the lift will be

2π ρV vr cos2 θ dθ = ρπ V vr L2 = 0

so the total lift L = L 1 + L 2 = 2ρπ V vr. Substituting to the change of velocity expressed in terms of the circulation v=

 2πr

we get the result of (4.48).

4.2.7 Lift from the Deflection of Air Stream There is a very simple argument to show that the deflection of the air stream produced by the airfoil can produce lift. Consider the wing constituted by a simple thin plate inclined of some angle α (as in Fig. 4.6), with respect to a uniform air stream. When leaving the wing the flux is deviated by the same angle α and the lift equals the change

96

4 Aerodynamics and All that 2

D

L

C

D

V

a

1

α

3 A

α

4 B

Fig. 4.6 The lifting from the deviation of the flow. On the left, the airfoil is reduced to a thin plate inclined to an angle α. 

of momentum in the vertical direction L = dp/dt where p is the momentum. Now consider a wing of span l in a stream of thickness a (Fig. 4.6b). The mass flow is simply dm = alρV dt The vertical component of the velocity after leaving the wing is Vv = V sin α so that the upward force is dm F= V sin α = alρV 2 sin α dt Referring now to Fig. 4.5 we can evaluate the circulation on the path shown, where the only contribution is from path 3 that is equal to V sin α

a ≈ V a sin α cos α

and the circulation  ≈ V a sin α Substituting the circulation in the expression for the force we have for the lift (remember is force per unit length) L=

F = ρV  l

(4.49)

that is the Kutta-Zhukhovsky result. We now derive the same result in a more detailed way. First of all, we consider the equation of motion in a steady condition and in absence of body force, that is ρ(v · ∇)v = ∇ p Following Acheson, we write this equation in the form   ∂u ∂p ∂u +v =− ρ u ∂x ∂y ∂x where u and v are the x and y components of the velocity. We now neglect ρ and take the volume integral of this term

4.2 Elementary Potential Flows

97

 



 ∂u ∂(uu) ∂(uv) ∂u u +v dV = + dV ∂x ∂y ∂x ∂y V V This equality can be easily verified considering that the ∇ · v = 0. The integrand on the left is the divergence of the vector uv, and using the divergence theorem we have

 V



∂(uu) ∂(uv) + d V = (v · n)ud S ∂x ∂y S

where n is the normal to the surface S enclosing the volume V . Therefore, we have

ρ(v · n)ud S = − S

V

( pn x )d S S



In vectorial form

∂p dV = − ∂x

( pn)d S = S

ρv(v · n)d S.

(4.50)

S

In the relation above ρu is the momentum per unit volume of the fluid element and (u · n)d S is the volume rate at which the fluid is leaving a small portion d S of the surface S. The equation states that the total force (integral on the left) on S is equal to the rate at which the momentum is leaving S.

Appendix Complex Variables Integration A complex number z is defined as z = x + i y or in polar coordinates

z = r eiθ

(4.51)

where r = (x 2 + y 2 )1/2

tan θ = y/x

(4.52)

A function of complex variable f (z) can be written in terms of real and imaginary parts f (z) = g(x, y) + i h(x, y) (4.53) And the derivative can be made in the x or y direction

98

4 Aerodynamics and All that y

y C2

C C1

C1

Z0

C2 C0

Z0

Cn

x

x

(a)

(b)

(c)

Fig. 4.7 The integration in a multiply connected region, (a). The region of Laurent series expansion, (b) and (c) the integration of a function with singularities

∂g ∂h df = +i dz ∂x ∂x

df ∂g ∂h = +i dz ∂y ∂y

(4.54)

Because the derivative must be independent of the direction of differentiation we must have ∂g ∂h ∂g ∂h = =− (4.55) ∂x ∂y ∂y ∂x These equalities are called Cauchy-Riemann conditions, and guaranty the differentiability of the complex function. The function is analytic if the real and imaginary parts satisfy the Laplace’s equations. The points of a region where the f (z) is analytic are called regular or singular where the function is not. We now consider the integral of complex function

B f (z)dz A

If the function f (z) is analytic the integral is the same independent of the path connecting A to B. This implies that the integral on a closed path is zero. This is known as Cauchy integral theorem. Consider now a curve C that surrounds n curves C1 , C2 , . . . Cn ad in Fig. 4.7. The application of the Cauchy theorem to this region for a function which is analytic inside C and outside the n curves gives the result 

 f (z) dz = C

 f (z) dz +

C1

 f (z) dz + · · · +

C2

f (z) dz

(4.56)

Cn

Consider now different cases. In the first one, the function is analytic at all points within a circle C0 with center in z 0 , then the function can be represented by the Taylor series f n (z 0 ) (z − z 0 )n (4.57) f (z) = f (z 0 ) + f (z 0 )(z − z 0 ) + · · · + n!

Appendix

99

Consider now as in Fig. 4.7b the circle C1 and C2 with center in z 0 and a function f (z) analytic between these two circles. Then the function can be represented by the Laurent series ∞  f (z) = An (z − z 0 )n (4.58) −∞

Consider now the integration of a function f (z) with singularities. As shown in Fig. 4.7c, the function is analytic inside C except for z 0 . Then surround the singularity with a circle C0 and write the function as a Laurent power series. The integral around C is then   ∞  f (z) dz = An (z − z 0 )n = 2πi A−1 (4.59) C

−∞

C0

where A−1 is the coefficient of the term (z − z 0 )−1 and is called the residue of the function f (z) at the point z 0 . As generalization if f (z) is analytic inside C except for a number of singularities N then (4.59) becomes  f (z) dz = 2πi C

N 

A−1 z( j)

(4.60)

j=1

Conformal Mapping Consider the function f (z) = x + ı y that generates a pair of values (u, v) in the w plane. Then each value of f represents a point in the z plane and a corresponding value in the w plane. The function f (z) then transforms points or curves from the z plane to the w plane and when the function is analytic the mapping is called conformal, and has interesting properties that can be illustrated with examples. Consider for example the simplest case of the complex function φ + ıψ = x + ı y The function may represent a fluid flow ad shown in Fig. 4.8, with horizontal streamlines ψ = ı y and equipotential lines φ = x. We now apply a transformation that maps √ the z plane in the w plane with the function w = u + ıv = z. The transformation reads 2uv = y = ψ (u + ıv)2 = x + ı y ⇒ u 2 − v2 = x = φ So at each value of ψ corresponds the hyperbola uv = ψ/2 while for each value of φ we have an hyperbola rotated of an angle π/4. Notice that the lines that were perpendicular in the z plane remain as such in the w plane. Another example refer to the properties of the Cauchy-Riemann relations (4.6). We consider the complex potential

100

4 Aerodynamics and All that 1 1 0.8

0.8 0.6

ψ

0.6

v 0.4

mapping

ψ=0.2 0.4

0.2

Φ=0 0.2

0 -1

-0.8

-0.6

-0.4

-0.2

0.0

0.2

0.4

0.6

0.8

1.0

Φ

Z=x+iy PLANE

0

0.2

0.4

Fig. 4.8 The mapping of a uniform flow (left) to a right angle corner flow, w =

0.6

u

√

0.8

1.0

z

2 1.8

w=z π/3

1.6 1.4 1.2 1 0.8 0.6 0.4 0.2 0 0

0.2

0.4

0.6

0.8

1.0

Fig. 4.9 Same as Fig. 4.8 for a different angle. The arrows indicate the direction of flow

φ + ıψ = Az π/α where

φ = Ar π/α cos(π θ/α)

ψ = Ar π/α sin(π θ/α)

(4.61)

We can obtain the velocity components (real and imaginary) as Vr =

∂ψ ∂φ = ∂x ∂y

Vi =

∂ψ ∂φ =− ∂y ∂x

(4.62)

As consequence the complex velocity W = Vr + i Vi =

π (π−α)/α Az α

(4.63)

Appendix

101

From (4.61), we see that ψ = 0 for θ = 0 and θ = α and the stream function represents the flow in a corner of angle α as in Fig. 4.8b, drawn for α = π/3. Notice that for y ≈ 0 the velocity is parallel to the horizontal axis and for θ = α the velocity is again constant (Fig. 4.9).

Application to the Kutta-Zhukhovsky Lift First of all we have to reformulate our potential flow theory in terms of the complex variables and we have used this concept already when we have written the generalized complex potential as F = φ + ıψ. So, we will start to generalize the potential of a rotating cylinder in uniform flow with the center of the cylinder located at the point z 0 . The potential will be Fz = V e−iα (z − z 0 ) +

me−iα  z − z0 +i z − z0 2π a

(4.64)

where the symbols are the same we have used for the paragraph (4.2.4), except for R substituted by a. Now we have to require that the flow around the circle z − z 0 = aeıθ to be zero, This implies that for z on the circle the imaginary part of F be zero and from (4.64), we must have m = V a 2 . Then the velocity can be calculated 1 d Fz V a 2 e−iα  = V e−iα − +i dz (z − z 0 )2 2π z − z 0

(4.65)

This is our complex velocity in the z plane, and our plan is to map the cylinder in this plane to an airfoil in the w plane (Fig. 4.10). The transformation is the following known as Zhukhovsky transformation

zt wt

Fig. 4.10 The mapping of a rotating cylinder into an airfoil. Notice the stagnation points on the cylinder correspond to the leading and trailing edges of the airfoil

102

4 Aerodynamics and All that

w=z+

b2 z

(4.66)

Before discussing the implications of such transformation we need to define the socalled Kutta condition that states that portions of fluid flowing above and below the surface of an airfoil must meet at the trailing edge. This requirement has originated a debate on the real cause of the lifting. If the Kutta condition is met then the fluid flowing above must have a greater velocity than the fluid flowing below, so that for the Bernoulli equation the pressure above the airfoil is less than the pressure below. Then people ask how the first planes could fly having as wing just curved canvas or how a plane can fly upside down. Refer now to Fig. 4.10, and we see that the trailing edge wt on the airfoil will map to the point z t on the cylinder and we have z t = z 0 + aeıβ with 0 < β < π/2. The Kutta condition means that (4.65) be put to zero so we have  d Fz   iβ e =0 = V e−iα − V e−i(α+2β) + i  dz zt 2πa so that =

 2πaV  i(α+β) e − e−i(α+β) = 4πaV sin(α + β) i

(4.67)

(4.68)

And following the Kutta-Zhukhovsky theorem the lift is given by L = ρV = 4πρaV 2 sin(α + β)

(4.69)

We now can derive the velocity in the mapped space. In the original z space we have for the velocity ∂φ ∂ψ dF = +i = Vr − i Vi (4.70) dz ∂x ∂x In the same way we can obtain by chain rule d Fw dz d Fw = dw dz dw And in the case of the Kutta-Zhukhovsky transform becomes d Fw Vz = dw (1 − b2 /z 2 Now go back to the transform (4.66) to make an example. We use the simplest form in which b = 1, then we have   1 x y (4.71) u + ıv = x + ı y + =x+ 2 +ı y− 2 (x + ı y) (x + y 2 ) (x + y 2 )

Appendix

103

Separating real and imaginary parts we get u=x+

(x 2

x + y2)

v=y−

(x 2

y + y2)

(4.72)

Now the module | z |= 1 that implies x 2 + y 2 = 1 and the final results is u = 2x

v=0

(4.73)

So the circle is mapped into a flat plate on the real number line from −2 to 2.

The Physics of Sailing Sailing is perhaps the oldest way of moving across long distances. Today, it is mainly practiced for fun although in many places around the world is still used as transportation or fishing. We have now the basics to study sailing and we can start from the examination of the forces acting on a boat. In Fig. 4.11, we see that in the atmosphere the force developed from the sail Fs is summed to Fk that is the force developed underwater by the keel. The resultant is the net force (Fn ) that propels the boat. When the velocity is constant Fn is balanced by the drag, Fd . Both the sail and the keel work on the same principle as the wing. The curved surface of the sail presents some kind of attack angle with respect to wind while the keel is inclined with respect to the water current. We can see that actually the boat moves “against” the wind and this is quite different from what happened for the old sailing ships that were “pushed” by the wind. Read the interesting paper by V. Radhakrishnan (1997) which discusses all the details. About the drag, we will neglect for the moment all the problems created by the waves generated by the motion of the boat. Following what we have learned so far, we can write the expression for the lift (L) and the drag (D) as L = 21 Cl ρVa2 A

D = 21 Cd ρVa2 A

(4.74)

where Va is the velocity of the boat and A is the projected area of the sail (or keel). Cl or Cd are the lift and drag coefficients that are the proportionality constants with respect to the dynamic pressure (1/2)ρV 2 . To obtain the equation of motion, we have to define the apparent fluid velocity Va , which is the velocity of the fluid in the boat rest frame. The other important velocity components are the velocity of the boat (V) and the fluid velocity Vt which is measured in the frame of reference attached to Earth. It is quite evident that Vt = V + Va Then we have to define a few more angles in the atmosphere and in the water. In the atmosphere the angle between the apparent velocity Vaa is called θa while the angle

104

4 Aerodynamics and All that direction of motion

wind

Fk

water

Fs

Vt

V

Va,a

Fs

Fk

γ

y

Fn

Fd

β−γ

π-(α−γ) V a,a β

π/2-α

Vt,a

V γ

x

α

Ls

β

y y

y

Va,a

Vt,a

Va,a

β

θa

β Lh

θcl

α

θsb

x V

x

γ

Ls

Dh

θsb

γ

x

V

Da

Fig. 4.11 The top row: the composition of the forces acting on the sailboat, the angles and velocity discussed in the text. The composition of the apparent velocity with the boat velocity. Bottom row: the first two figures angles and forces discussed in the text, and finally the case worked out analytically

of the sail boom θsb determines the angle θcl of the axis of the boat with respect our x axis. Finally, β is the angle with respect to the same x axis with of the apparent velocity. Referring to Fig. 4.11 we have θa = β − θcl − θsb The attack angle for the keel θh will be simply the difference between the angle of the centerline θcl and the angle of the vector V with the x axis that we will call γ . so that we have θh = θcl − γ To write down the equation of motion for the boat refer to Fig. 4.10a d Vx = L s sin β − Ds cos β − L h sin γ − Dh cos γ dt d Vy = L s cos β − Ds sin β + L h cos γ − Dh sin γ m dt m

(4.75) (4.76)

Some simplification is possible, if we assume there are no currents in the water, so that the fluid velocity that determines L h and Dh is just the negative of the boat velocity that can be related to the “true” air speed Vta so that referring to Fig. 4.11,

Appendix

105

and using the law of sines we have V = Vta sin(α − β)/sin(β − γ )

(4.77)

The forces L s and Ds depend on the apparent air velocity Vaa that can be expressed as a function of Vta as Vaa = Vta sin(α − γ )/sin(β − γ )

(4.78)

The equations of motion (4.75) and (4.76), can be solved analytically only in a very simple case when the sail is normal to the wind. In this case, the problem becomes one-dimensional and the equations reduce to just one in the direction of the wind. We have 1 1 dV = Cds ρ As (V − Vta )2 − Cdh ρw Ah V 2 (4.79) m dt 2 2 where ρ and ρw are the densities of air and water, respectively, As and Ah are the surface of the sail and of the immersed body, and Cds and Cdh the coefficient for the sail and the boat. We put K a = 21 Cds ρ As /m and K h = 21 Cdh ρ Ah /m and solve the equation. dV = K a (V − Vta )2 − K h V 2 (4.80) dt Which has the solution  −1 V = Vta 1 + (K h /K a )1/2 coth(t/τ )

(4.81)

where τ −1 = (K h K a )1/2 Vta . As t → ∞ the velocity approaches the equilibrium speed for the downwind situation  −1 V (∞) = Vta 1 + (K h /K a )1/2 The conclusion is that if the hull drag is negligible (K h ≈ 0) the limit velocity coincides with the wind velocity. The range of values for the ratio K h /K a can be found from the definition ρw Cdh Ah Kh = Ka ρa Cds Aa Considering that ρw /ρa ≈ 1000, Cdh ≈ 0.01 − 0.1, Cds ≈ 1 we obtain 10

Ah Kh Ah < < 100 Aa Ka Aa

The real value K h /K a ≈ 6 so the ratio V (∞)/Vta ≈ 0.3

106

4 Aerodynamics and All that

The Flight of Golf Balls The physics of golf ball has been discussed more than a century ago (1893) on paper by P.G. Tait on Nature. We present here a more simple treatment as an application of the Magnus effect. We then consider that the golf ball is subjected to the force of gravity, the air resistance (drag) and the Magnus force. The drag resistance follows the usual expression D = 21 C D ρ AV 2 where C D is the drag coefficient, ρ is the density of air, A is the surface area and V the velocity. The real news, in this case, is how to express the Magnus force that we could write as FM = S(ω × V) where ω is the angular velocity and S is a proportionality constant. As a consequence, the components of the last force can be written as FM x = S(ω y w − ωz v) FM y = S(ωz u − ωx w) FM z = S(ωx v − ω y u)

(4.82)

where u, v, w are the components of the velocity in the three directions and the same for the angular velocity. We assume z to be the vertical axis and x and y the horizontal axis. The equations of motion in the tree directions are then d2x S = − (ω y w − ωz v) dt 2 m d2 y S = − (ωz u − ωx w) dt 2 m d2z S = −g − (ωx v − ω y u) 2 dt m

(4.83)

These equations are easily integrated, if the orientation of ω is specified beside the characteristic of the ball (radius and mass). Also, the angle of attack must be specified, that is the angles the initial direction makes with the axes. There is an additional variable specific for golf balls that influence the drag coefficient. Golf ball has a dimpled surface, that is, the surface is not smooth but presents some rounded small hollows. This feature makes the air surrounding the ball more turbulent decreasing the low pressure that is generated behind a smooth ball. This makes a stronger Magnus effect and decreases the drag coefficient. We have integrated equations (4.83), in a number of circumstances, and the result is shown in Fig. 4.12. All the cases must be compared with the ball flying in a medium where the drag is negligible.

Appendix

107

80 no drag

50

60

Cd=0.15

40

“dimple” ball z(m)

40

z (m)

40

30 20

30 20

Cd=0.25

10

2

v drag

20 0 0

10

150

0

0

0

100

20

40

60

200

80

-10

100

300 X (m)

400

100

Y (m)

-20

50 -30

X (m)

0

Fig. 4.12 On the left, the results of the integration are shown. The plot shows the no drag cases, the cases with the dimple ball and the cases with the drag proportional to v 2 . Notice the different scales for z and x. On the right, the case for the dimple ball is shown with a three-dimensional plot. The initial angle was 60◦ and the ω components were different from zero only for ωx and ωz . The trajectories for Cd = 0.15 and 0.25 are reported

Then we move to the case where a drag is introduced proportional to the square of the velocity. Then we make the case when the Magnus effects is considered both on a smooth and a dimple ball. We have considered a drag coefficient of 0.25 for a smooth ball and 0.15 for a dimpled ball, while for S we have used a value of 5 × 10−5 kg. The ball is a standard ball with m = 41 g and radius 21 mm.

The Flight of a Frisbee The name “frisbee” originates from the pies shape of the Frisbee Pie Company of Bransford, Connecticut in the late 1800. Possibly, the Cookie tin lids or Pie-tins were the first prototypes for the frisbee. The invention of plastics in the 1940, was crucial for the marketing of the frisbee whose detail was important for a successful flight of the object. An essential feature for the flight of the frisbee is its rotation and its shape. As shown in Fig. 4.12a, above the leading edge the trajectories of the air molecules are curved and this develops some difference of pressure and then lift. In the absence of rotation, the lift force is positioned ahead of the center of gravity and this would cause a torque that would disrupt easily the flight. As shown in Fig. 4.12b, the rotation introduces a gyroscope effect so that the “nose up” tendency implies a rotation of the plane of the disc that would make it to bank to the left with respect to direction of flight. The angular momentum of the frisbee is important and for this reason, the rim of the disc is thicker in such a way to maximize the angular momentum when it spins. The motion of the frisbee is quite complex and should be assimilated to the motion of a rigid body in three dimensions. The equations of motion are referred to a system of axes (x1 , y1 , z 1 ) fixed with the Earth and the other (x2 , y2 , z 2 ) attached to the disc (Fig. 4.13b). The two systems are related by a sequence of rotation defined by the

108

4 Aerodynamics and All that (c)

Lift

Flow

y2

bank left

cp

Θ

Φ

Ω

(a)

x2

cg

nose up

x roll, p

pitch, q

y1 z2

yaw, r y

z

x1

Ψ

(b)

z1

Fig. 4.13 In a we show the interaction of the frisbee with the air flow, In the b the torque produced by the lift and the weight coupled with the rotation induces a roll movement to the left and a nose up pitch. In c the references systems of the frisbee (x2 , y2 , z 2 ) and the fixed system (x1 , y1 , z 1 ) are shown

Euler angles φ, θ and ψ. They correspond in this case to the standard aeronautical angles of yaw, pitch and roll. We start by specifying the matrix of the moment of inertia    Ix 0 0     0 Iy 0  (4.84)    0 0 Iz  where Ix = I y and the component of the angular velocity    0 −r q     r 0 − p   −q p 0 

(4.85)

Figure 4.13b shows that with respect the direction of flight the component p corresponds to the roll, q corresponds to the pitch and r to yaw. This is rather counter intuitive but we like to maintain the convention used in the literature. The equation of motions are those for a rigid body in the reference frame of the disc, that is for the center of mass   dv F=m +ω×v (4.86) dt And the momentum M=I

dω + ω × Iω dt

(4.87)

These equations derive from the fact that we go from an inertial system (x1 , y1 , z 1 ) to the one fixed with the disc. Equations (4.86–4.87) do not specify completely the problem because we have to write equations that involves the angles θ , φ and ψ that are function of the components of the angular velocity (4.85) and the momentum. For the detailed treatment the papers by Potts and Crowther (2002) and Hummel and

Appendix

109

Hubbard (2000) are essential. We can illustrate a simple case to derive expression for the disc roll and pitch rate as a function of the applied pitch and rolling moments. We write Eq. (4.87), in the matrix form ⎡

⎤ ⎡ ⎤ ⎡ ⎤ Ix p˙ L qr (Iz − I y ) ⎣ M ⎦ = ⎣ I y q˙ ⎦ + ⎣ r p(Ix − Iz ) ⎦ N Iz r˙ pq(I y − Ix )

(4.88)

This equation simplifies when we remember that Ix = I y L = Ix p˙ + qr (Iz − I y ) M = I y q˙ + r p(Ix − Iz ) N = Iz r˙ In these equations, L is the rolling moment, M the pitching moment and N the yaw. In a horizontal flight path the yawning moment can be neglected and the equations reduce to L = Ix p˙ + qr (Iz − I y ) (4.89) M = I y q˙ + r p(Ix − Iz ) The two moments are given by L=

1 2 ρv cSC L , 2

1 2 ρv cSC M 2

M=

(4.90)

where S is the surface of the disc, C L and C M are the coefficient for pitching and rolling and c is the diameter. We now neglect the angular acceleration that is p˙ = q˙ = r˙ = 0 and assume that the disc is thin and axisymmetric in such a way that Ix = I y = 21 Iz and in that case the Eq. (4.89) reduce to q=

2L ; r Iz

p=

2M r Iz

(4.91)

This result express the fundamental effect of gyroscopic precession on the dynamics of the disc. For a positive spin rate r , a positive rolling moment L causes a positive pitch q, and a positive pitching moment M cause a roll-left p wing down. This is shown in Fig. 4.13b. Since the aerodynamic rolling moment is very small the precessional pitch rate is negligible. To end up with this simplified treatment of the frisbee flight we can consider a very unreal case in which the flight takes place in the z (vertical), x (horizontal) plane. In this case, we need to take into account only the gravity g, the drag force and the lift. The drag force is given by D=

1 C D ρ Sv2 2

(4.92)

110

4 Aerodynamics and All that

Fig. 4.14 The solutions for a very simple two-dimensional flight of the frisbee with different attack angles

3 α = 15°

z(m)

2 α = 10° 1 α = 5° 0

2

4

6 x (m)

8

10

12

where the drag coefficient C D is expressed as C D = C D0 + C Dα (α − α0 )2

(4.93)

where C D0 is called form drag and C Dα is called induced drag and α is the angle of attack while α0 is the angle that produces the minimum lift that is about −4◦ degrees. The lift L is given by 1 (4.94) L = C L ρ Sv2 2 where again C L is the lift coefficient.The equations of motions are then 1 d2x C D ρ Sv2 =− dt 2 2m d2z 1 = −g + C L ρ Sv2 2 dt 2

(4.95)

With the initial conditions d x/dt = v0x = v0 cos(α) and dz/dt = v0z = v0 sin(α). In Fig. 4.14 some results are shown for an initial velocity of 14 m/s and different angles of attack.

Ground Effects When a plane is landing we may have the impression of some floating of the aircraft before touching the ground. This is much more evident for a glider or for the deadstick landing of the space shuttle. The most obvious explanation would be that some kind of “air cushion” develops below the plane which is responsible for the floating. Actually, we will show that the effect is due to a decrease of the induced drag on the wings of the plane. The same effect is responsible for some catastrophic crash at

Appendix

111 Di

V

αi

streamlines

L

V

αeff

α

V αi

local relative

αi w wind

low pressure high pressure

(b)

(a)

(c)

Fig. 4.15 In a the mechanism that produces the tip vortexes (b) it is shown. In c the quantification of the downwash is illustrated

take-off especially for overweight planes. As soon as they take off the reduced drag facilitates the ascent but then as soon as some altitude is gained the ground effect is lost and the plane may crash. First, we need to explain what is the induced drag and then we need to apply that to our problem. We have seen that the lift of the wing is generated by the pressure difference between the underside and the surface of the wing, and we determined that the velocity around the wing may be related to circulation  by the relationship v = /2πr . These considerations imply that vortexes will be generated around the boundary of a finite wing as shown in Fig. 4.15. In particular, the tendency around the wing tip will produce a velocity component toward the wing root of the upper side and the opposite on the underside so that vortexes are generated from the wingtips. The vortexes in large planes (like a 747) may generate enough disturbance of the flow to affect seriously the flight of smaller planes. The vortexes tend to drag the surrounding air producing a small downward component of the velocity that corresponds to the downwash. If we indicate this component with w, we see that combined with the free stream velocity V∞ , it will produce a relative velocity that has a different, smaller inclination than the real angle of attack as shown in Fig. 4.15. In turn, this different inclination will produce a horizontal component of the lift L, and thus what it is called an induced drag. To calculate this induced drag, we may refer to an example reported in the book by Houghton, for a plane of weight W and span 2s flying near the ground at altitude h and speed V . We recur to the method of images. In this case, the wing is assimilated to a line with the associate vortexes on the tip (Fig. 4.16b). The ground plane is simulated by placing a wing “below” the ground which is symmetric with respect the real wing so that there are no vertical component of velocity on the separation horizontal plane. From Fig. 4.16, we can evaluate the change in downwash as w =

0 0 cosθ1 + cosθ2 4πr1 4πr2

(4.96)

112

4 Aerodynamics and All that Γ

Γ

wing

θ1

Γ0

Γ

Γ

Γ0

y h

Γ

ground plane

ground plane

x

-Γ

h

-Γ y

θ2

-Γ

image

-Γ

- Γ0 -Γ

θ2

θ1 s’

- Γ0

s’

Fig. 4.16 On the extreme left the three dimensional images are shown for a schematic horseshoe vortex. At the center the front view is shown and on the extreme left the calculation of the downwash is illustrated

Expressing the cosine as cos θ1 = (s + y)/r1 and cos θ2 = (s − y)/r2 we have w =

0 4π



s + y s − y + 2 r1 r22

 (4.97)

On an elementary span δy the changes in lift δl and in vortex drag dv will be δl = ρV 0 δy

dv =

ρV 0 δy w δl w = V V

(4.98)

The total change in drag Dv is obtained by integrating (4.98) over the wing span

s

− Dv = 2 0

ρ02 4π



s + y s − y + 2 r1 r22

 dy

From the geometry, r12 = 4h 2 + (s + y)2 and r22 = 4h 2 + (s − y)2 we obtain

 s 4h 2 + (s + y)2  ρ02 ln 2 − Dv = 4π 4h + (s − y)2 0   2  s ρ0 ln 1 + = 4 pi h And if we assume an elliptical distribution such that W = ρV 0 π s and s = (π/4)s we get   2W 2 π 2s2 ln 1 + Dv = ρV 2 s 2 π 3 16h 2 Now we can imagine a plane with W = 2.2 × 105 N, h = 15.2 m, s = 13.7 m and V = 45 ms−1 to obtain Dv = 1390 N that is around 0.6% of the weight.

Appendix

113

Notice that today the term ground effect is used also for competition cars but that is another story related to the creation of low pressure below the car using a kind of Venturi tube.

The Great Red Spot The Great Red Spot is the most spectacular feature of the Jovian atmosphere. It is assimilated to a giant storm observed the first time in the seventeenth century but monitored regularly since the mid nineteenth century. It has radius around 16.000 km and rotates in counterclockwise fashion like a low pressure center. Its explanation is the object of speculations and among them (but good for our purpose) we like to talk about the one formulated by Andrew Ingersoll in a 1969 paper on the Journal of the Atmospheric Sciences. The atmosphere in this model is reduced to a layer confined between two rotating planes with an obstacle attached to the bottom plane. The flow it assumed undisturbed away from the obstacle, so that if certain conditions are satisfied a Taylor column is formed above the obstacle following the Taylor-Proudman theorem. We follow a simple approach inspired to a similar problem solved by Kasahara (1966) in a paper published some year earlier. We consider a channel in a x − y plane of width W between the boundaries y = ±W/2. In the absence of an obstacle the solution for the fluid with depth h are the following.

∂ h¯ = 0, ∂x

u = u, ¯ v=0 ¯ ∂h = − f u/g ¯ ∂y

where u¯ is the average velocity and h¯ is the height of the free surface that can be obtained integrating the last equation assuming f = cost u(y) ¯ = h0 −

u¯ fy g

(4.99)

where h 0 is the value of h¯ at y = 0. Now we put a cylinder of thickness H at the bottom of the channel and we invoke the conservation of potential vorticity D Dt



ζ+ f h−H

 =0

(4.100)

Now we assume h = h + h¯ − H where h is the deviation from the average value of h. Equation (4.100) can be rewritten as

114

4 Aerodynamics and All that

h

 ζ+ f D 1 D  ¯ (ζ + f ) − h +h− H =0 2 + h¯ − H Dt (h + h¯ − H ) Dt

We now put (ζ + f )/(h + h¯ − H ) = f 0 / h 0 and obtain (ζ + f ) −

f0 ¯ (h + h − H ) = 0 h0

If ψ is the stream function we have ζ = ∇ 2 ψ so the equation becomes ∇2ψ + f −

f0 ¯ f0 (h + h) = − H h0 h0

The second and third term of the left hand side are roughly the same so we have ∇2ψ = −

f0 H h0

(4.101)

That is the same equation of Kasahara when f = cost. Ingersoll in his paper has a similar equation but it is normalized, The dimensions of (4.101) are s −1 so that we may divide both sides by V /L with V a typical velocity and L a typical length. We have then f0 L H L 2 ∇ ψ =− V V h0 The left hand side of the equation is simply normalize to a new ψ, while on the right hand side we have the Rossby number Ro = V / f 0 L and the normalized thickness of the obstacle H1 = H/ h 0 . The Eq. (4.101) boils down to ∇2ψ = −

H1 Ro

(4.102)

Ingersoll introduces a coordinate system such that r = (x 2 + y 2 )1/2

 = tan−1 (y/x)

(4.103)

Then he assume the velocity be unity at infinity and directed in the negative x direction so that the boundary conditions are ψ y = 1,

ψx = 0, (r → ∞) ∇ψ = 0 (ψ = 0, closed portion)

(4.104)

Its is assumed that obstacle is a cylinder of radius r = 1 so that H1 (r ) = cost = h 1 for r < 1 and H1 (r ) = 0 for r > 1. The solution of (4.102) satisfying the boundary conditions exist and it is given by

Appendix

115 y

H1/R0=1

2

H1/R0=2

H1/R0=4 2

1.5

w

1

x

1

Y

0.5

0

0

-0.5

-1

-1

-1.5

h

-2 -2

-1

H

0

1

2

-2

-1

(a)

0

1

2

-2

(b)

-1

0

1

2

(c)

X

Fig. 4.17 The cylindrical obstacle of height H in a channel at the extreme left. The resulting streamlines when the ratio H1 /R0 = 1, (a). The dashed circle is the top view of the obstacle. The other two cases b and c refer to different boundary conditions (see text) and different ratios H1 /R0 , and the Taylor column is shaded in (c). The separation of the streamlines is 0.5 in (b) and 0.25 in (b) and (c). Notice the similarities with Fig. 4.4

ψ=

⎧ H 1 ⎪ ⎨− 2Ro ln r + r sin  − 1, ⎪ ⎩

if r > 1. (4.105)

H1 − 4R (r 2 o

− 1) + r sin  − 1, if r < 1.

These solutions are very similar to the case of a cylinder in a uniform flow (Eq. (4.52)), and depend essentially on the ratio between h − 1 and the Rossby number. When the height of the obstacle is less than 2Ro , the streamline are distorted over the obstacle but do not cross (Fig. 4.17a). There is clockwise circulation π H1 /Ro and a force in the negative y direction π h 1 . When the ratio H1 > 2Ro the last condition (4.105) is no longer satisfied, and we have to recur to a different solution ⎧ H K r 1 ⎪ if r > 1. ⎨− 2Ro ln r + 2π ln rc + r sin  − 1, ψ= (4.106) ⎪ ⎩ h1 2 K r − 4Ro (r − 1) + 2π ln rc + r sin  − 1, if r < 1, r > rc . where rc is the radius of the Taylor column rc = 2Ro /H1 and the column is centered at x = 0, y = 2Ro /H1 . r is the distance from its center to the point x = r cos , y = r 1sin. The term π H1rc2 /Ro is the additional circulation necessary to satisfy the boundary conditions. Examples of this solutions are plotted in Fig. 4.17b and c. This Taylor column could be the base of the Great Red Spot.

116

4 Aerodynamics and All that

References Textbooks Acheson DJ (2005) Elementary fluid dynamics. Oxford University Press, Oxford Anderson BD (2003) The physics of sailing explained. Sheridan House, New York Anderson J (2012) Fundamentals of aerodynamics. Mc Graw Hill, New York Anderson J (2003) Introduction to flight, Schaum. McGrawHill, New York Eckert M (2003) The dawn of fluid dynamics. Wiley-VCH, New York Houghton EL, Carpenter PW (2003) Aerodynamics for engineering students. Butterworth Heinemann, Oxford Katz J, Plotkin A (2010) Low speed aerodynamics. Cambridge University Press, Cambridge Kundu PK, Cohen IM (2015) Fluid mechanics. Academic Press, New York

Articles Anderson BD (2008) The physics of sailing. Phys Today 61:38 Bloomfield LA (1999) The flight of Frisbee. Sci Am 280:132 Goldenbaum GC (1988) Equilibrium sailing velocities. Am J Phys 56:209 Hubbard H, Hummel SA (2000) Simulation of frisbee flight. In: Cohen G (ed) 5 Conference on mathematics and computers in sport, University of Technology, Sydney, New South Wales, Australia, 14–16 June 2000 Ingersoll A (1969) Inertial Taylor columns and Jupiter’s great red spot. J Atmos Sci 26:744 Kasahara A (1966) The dynamical influence of orography on the large scale motion of the atmosphere. J Atmos Sci 23:259 Kharlamov A, Chara Z, Vlasak P (2007) Magnus and drag forces acting on a golf ball. In: Colloquium fluid dynamics Mehta RD (1985) Aerodynamics of sports balls. Ann Rev Fluid Mech 17:151 Potts JR, Crowther JW (2002) Frisbee aerodynamics. In: 20th AIAA applied aerodynamics Rhadakrishnan V (1997) From square sails to wing sails: the physics of sailing craft. Curr Sci 73:503 Wilson RM. The Physics of Sailing, http://grizzly.colorado.edu/~rmw/files/papers/Physicsof Sailing.pdf

Chapter 5

Waves

Waves may be the most familiar phenomena in the physics of fluid. We will find waves everywhere, from the sea surface to the bathtub up to the atmosphere. To be produced, waves need a restoring force and again the most common may be either pressure or gravity. In the first case, we have sound waves that are quite common in everyday experience, while in the second case, we have gravity waves not to be confused with gravitational waves which are in realm of general relativity. Talking about sound waves, we need to talk about object that moves faster than the speed of sound and this is a good occasion to introduce the Mach Number. Waves transport is characterized by phase and group velocity, and we will discover that the first is not like a vector. Gravity waves have a lot of applications so we study in the appendix “tsunamis” and a variety of waves that populate the atmosphere and the oceans. Then, we will push the envelope a bit and talk about seismic waves and some of their applications.

5.1 Sound Waves We start with the continuity and momentum equations in the Eulerian form ∂ρ + ∇ · (ρu) = 0 ∂t ∂u 1 + u · ∇u = − ∇p ∂t ρ

(5.1) (5.2)

and consider small deviations (ρ1 , p1 , u1 ) from the unperturbed state of the fluid at rest u = 0 with the variables at p0 , ρ0 that is

© Springer Nature Switzerland AG 2020 G. Visconti and P. Ruggieri, Fluid Dynamics, https://doi.org/10.1007/978-3-030-49562-6_5

117

118

5 Waves

p = p0 + p1

(5.3)

ρ = ρ0 + ρ1 u = u1

(5.4) (5.5)

Substituting in Eqs. (5.1)–(5.2) and retaining only the first order terms we get ∂ρ1 + ρ0 ∇ · u1 = 0 ∂t 1 ∂u 1 dp ∇ρ1 = ∇ p1 = − ∂t ρ0 dρ ρ0

(5.6) (5.7)

In this latter equation we have assumed that p1 = (dp/dρ)ρ1 . Now just take the divergence of Eq. (5.7) to obtain ρ0

∂ ∇ · u1 = ∇ 2 p1 ∂t

Then substituting for ∇ · u1 from (5.6) we have ∂ 2 ρ1 = ∇ 2 p1 ∂t 2 With the final substitution p1 = (dp/dρ)ρ1 we have the classical wave equation dp 2 ∂ 2 ρ1 ∇ ρ1 = 2 ∂t dρ

(5.8)

These waves are due to the compressibility of the fluid and they propagate with a speed  (5.9) cs = dp/dρ This result can be demonstrated if one assumes a solution for (5.8) of the form ρ1 = A exp{[i(kx − ωt)]}

(5.10)

Substituting in the Eq. (5.8) we have ω2 /k 2 = dp/dρ. We will find in a while that this is the so-called phase velocity of the wave. It is interesting to note that once we have assumed a dependence like (5.10), this expression can be substituted in (5.6) to obtain iωρ1 = ikρ0 u 1 , and we get a relation between perturbation velocity and perturbation density   ρ1 ω (5.11) u1 = ρ0 k This relation tell us that perturbation velocity is in phase with density perturbation (because their ratio is a real number) and because ρ1 1 that is when h 0 >> λ. In this case (5.49) becomes  (5.52) ω = gk Deep water waves are dispersive waves with the group velocity vg = ∂ω/∂k = √ 1√ g/k that is half the phase velocity v = ω/k = g/k. It is interesting to note φ 2 that for this kind of waves, when solving (5.44) we can neglect the solution ∝ e−kz that would explode for large negative z and then our solution looks like f (z) = f 0 ekz

(5.53)

Following the same method we used above we can find the velocity components using a potential like

128

5 Waves

φ = f (z) exp [iω(t − x/c)]

(5.54)

that implies u=

∂φ = −ik f 0 ekz exp [i(ωt − kx)] , ∂x

w=

∂φ = k f 0 ekz exp [i(ωt − kx)] ∂z

For a fixed depth (z) the amplitude of the two components is the same, that means that the trajectories of motion are circular with the horizontal components lagging by 90◦ the vertical components (i.e. the motion is clockwise as shown in Fig. 5.3).

5.4.2 Shallow Water Waves Again consider (5.49), in the case kh 0 < 1 by expanding tanh(kh 0 ), we have tanh(kh 0 ) = kh 0 − (kh 0 )3 /3 + · · · . Then the dispersion relation becomes ω=



  1 gh 0 1 − k 2 h 20 k 6

(5.55)

On the other hand, in the same approximation the velocities obtained from (5.45) show the horizontal component to be much larger than the vertical component and are essentially independent of depth h 0 . As a matter of fact the semi major axis defined by (5.51) becomes in the case of kh 0 gh 0 and there is only a value of β satisfying (5.79). Then the situation is illustrated in Fig. 5.5 where there is only a strong wave maximum B AC. As a matter of fact, the wave generated when the ship was in the generic position A has travelled to B  and C  . The faster the ship goes the narrower the angle β is. The energy produced by the ship is all concentrated in such a wave because in this case there is no propagation of transverse waves considering √ that they cannot keep up with the ship travelling faster than the critical velocity gh 0 . Now √ we turn to deep water waves, and in such case the phase velocity becomes v p = g/k, so that (5.79) can be written as

y

y

D’

D

B D’

vp t A

β

A Vt

A’

β β

A’

x

x C

C

C’

Fig. 5.5 Oblique plane wave generated on the surface by a moving ship (left) and a shallow wave generated by the same moving ship (right)

134

5 Waves 35°16’

D

V

t

β B

3 2 1

v

g

A

transverse waves

19°28’ β=π/2

β=0

y/ λ0

P

4

divergent waves

C

0 -1 -2

A’

-3

Vt

-4 -2

(a)

0

2

4

6

8

10

12

x/ λ0

(b)

(c)

Fig. 5.6 The formation of an interference maximum in a deep water wake (a). The locus of the interference maximum with the different wave generated (b). The wake of a ship moving to the left (c)

vp = sin β = V



g kV 2

(5.80)

In this case, the critical values are for the wavenumber that exceeds the value g/V 2 . The motion of the ship excite a wide range of wave number and each one propagates according to (5.80). The resulting wake is produced by the interference pattern due to such waves that constitute actually a wave packet propagating with group velocity vg . It is easy to show that vg = dω/dk = (1/2)v p . Consider now a generic interference maximum A P D as shown in Fig. 5.6, with A representing the instantaneous position of the ship. A point P of coordinates (x, y) is part of the wavefront BC originated some time earlier t at point A . Because the interference maxima propagate at the group velocity we have A P = vg t so that referring to the figure, we have x = V t − vg t sin β y = vg t cos β We also have

dy = tan β dx

(5.81)

(5.82)

We can eliminate v p from (5.81) if we consider that v p = V sin β. Equations (5.81) become   x = X 1 − 21 sin2 β (5.83) y = 21 X sin β cos β Where X = V t. The derivative of (5.82) implies dX dβ

  sin β cos β X sin3 β tan β − = − 2 2 cos β 2

(5.84)

5.6 Ship Waves

135

So that

dX = X cot β dβ

(5.85)

This equation can be solved to give X = X 0 sin β

(5.86)

Where X 0 is a constant. The locus of the crest of the waves is given by the parametric curves   x = X 0 sin β 1 − 21 sin2 β (5.87) y = 21 X 0 sin2 β cos β This function is shown in Fig. 5.6 with angle β going from −π/2 to π/2. In the figure A is the instantaneous position of the ship with the curves AD and AB the corresponding crests about θ = 0 that is with respect to the x axis. The cusp in D has coordinates x/ X 0 = (2/3)3/2 and y/ X 0 = 3−3/2 so that the ratio y/x = 2−3/2 which corresponds to an angle of 19◦ 47 . We can now identify tree different waves. For β = π/2 waves propagate in the same direction of the ship and can only exist for y = 0. These are called transverse waves, For β = 0 we have waves propagating at π/2 with respect the ship trajectory and these are called dispersive waves while for √ the maximum of the function that happens for β = sin−1 (1/ 3) ≈ 35◦ 16 we have waves propagating at an angle of ≈35◦ 16 with respect to the ship track. Finally the deep wave approximation implies that k0 h 0 >> 1 and considering that V 2 = g/k0 √ means√ that V N 2 the buoyancy force is not enough to maintain oscillation and the perturbation decays with height. We get the phase and group velocities using ω from (5.144) ω = ±√

Nk k2 + l2

(5.145)

So that the phase velocities c px =

N ω = ±√ , 2 k k + l2

c pz =

Nk ω =± √ l l k2 + l2

(5.146)

cgz =

−N kl ∂ω =±  ∂l l (k 2 + l 2 )3

(5.147)

And the group velocities cgx =

Nl 2 ∂ω = ± , ∂k (k 2 + l 2 )3

We see that upward wave propagation (cgz > 0, minus sign in (5.147)) corresponds to downward phase propagation (c pz < 0). What we have done so far is just a “kinematic” description of a gravity wave, and now we can treat an almost real case.

Appendix

149

Mountain Waves We can treat the most simple example of internal buoyancy waves which are those produced by an isolated sinusoidal mountain. We consider a two dimensional x, z space and write the equation of motion, the continuity equation and the conservation of entropy. The primed variables are perturbations du dt dw dt ∂w ∂u + ∂x ∂z dθ dt

1 ∂ p ρ¯ ∂ x 1 ∂ p θ =− +g ρ¯ ∂z θ =−

(5.148) (5.149)

=0

(5.150)

=0

(5.151)

We now can linearize these equations by assuming that each quantity can be expressed as a basic state (¯) plus a perturbation ( ). We also assume a basic flow in the x direction u 0 . We get ∂u  ∂u  + u0 ∂t ∂x ∂w ∂w + u0 ∂t ∂x  ∂w ∂u + ∂x ∂z ∂θ  ∂θ  + u0 ∂t ∂x

1 ∂ p ρ¯ ∂ x 1 ∂ p θ =− +g ρ¯ ∂z θ0

=−

=0 = −w

dθ0 dz

Derive the first with respect to z and the second with respect to x and subtract we get 

∂ ∂ + u0 ∂t ∂x



∂w ∂u  − ∂x ∂z

 −

g ∂θ  =0 θ0 ∂ x

(5.152)

With the help of the last two equations, u  and θ  can be eliminated from (5.152) to get     2  ∂ 2 ∂ 2 w ∂ ∂ 2 w 2∂ θ + u0 + N + =0 (5.153) ∂t ∂x ∂x2 ∂z 2 ∂x2 We now assume a solution of the form w = w¯ exp{i[(kx + lz − ωt)]} That substituted in (5.153) gives the dispersion relation

(5.154)

150

5 Waves z Θ

u

k

Lv

δs

δz Cp

Cg,a C g,s

l

WEST

EAST L

H

L

LH = 2π/λ

Fig. 5.11 On the left we show the lines of constant phase resulting from a periodic forcing at the bottom. On the right, the relation between phase velocities and group velocity is shown in a similar situation. Notice the alternation of low-pressure zones (where the air is pushed upward) and high-pressure zones

 ω − u 0 k = ±N k/ k 2 + l 2

(5.155)

The minus sign represents westward phase propagation with respect to mean wind while the positive sign is for eastward phase propagation. If k > 0 and l < 0 then lines of constant phase tilt eastward with increasing height as shown in Fig. 5.11 because for constant phase φ = kx + lz if x increase z must also increase. The positive root in (5.155) corresponds to eastward and downward phase propagation. It is easy to show that group velocities are the same as Eqs. (5.146) and (5.147) except that for the x component the velocity. We turn now to a lower boundary in the form of a sinusoidal profile that will produce stationary waves that corresponds to put ω = 0 in (5.154) so that w will depends only on x and z. The partial derivative with respect to time in (5.153) goes to zero and the same equation simplifies to 

∂ 2 w ∂ 2 w + ∂x2 ∂z 2

 +

N2  w =0 u 20

(5.156)

Substituting (5.154) we get the dispersion relation l 2 = N 2 /u 20 − k 2

(5.157)

So that vertical propagation is only possible if N 2 /u 2o > k 2 . If this does not happens m is imaginary and the wave is attenuated with height. In the first case the form of the wave is w = w¯ exp{[i(kx + lz)]} while the attenuated wave should be w = w¯ exp{i(kx)} exp{(−μz)}

Appendix

151

Fig. 5.12 Streamlines for mountain wave forced by a periodic profile (in grey) at the bottom. On the left, the non-attenuated wave is shown with the dashed line representing constant phase. On the right, the same wave it is shown attenuated (see Eq. (5.158))

where μ = |l|. If the bottom topography is of the kind h(x) = h M cos kx Then the lower boundary condition should be w (x, 0) = (dh/dt)z=0 ≈ u 0 ∂h/∂ x = −u 0 kh M sin kx The solutions to (5.156) become  w(x, z) =

−u 0 h M ke−μz sin kx, if u 0 k > N . −u 0 h M k sin (kx + lz), if u 0 k < N .

(5.158)

These two solutions are shown in Fig. 5.12. There are other important waves in the atmosphere and some of them are also present in the ocean so it may be useful to talk about them in the next paragraph.

Waves in the Ocean We have neglected so far the effects of rotation that corresponds on a large-scale at considering the effects of the Coriolis acceleration. This will give rise to a quite number of waves both in the ocean and atmosphere. We will consider this aspect mainly in the ocean.

152

5 Waves

Kelvin Waves We will start from the simplest example of what are called coastal Kelvin waves that are found in coastal water and we will see why. As usual consider a basin of unperturbed depth H whose surface is perturbed by an amount η. We have the geostrophic equilibrium ∂η ∂u − f v = −g , ∂t ∂x (5.159) ∂v ∂η + f u = −g ∂t ∂y A simplified form of the continuity equation can be found considering that the horizontal component of water flux are (H + η)u ≈ H u and (H + η)v ≈ H v so that the rate of change of η is simply ∂η ∂ (H + η) + ∇ · (H u) = +H ∂t ∂t



∂u ∂v + ∂x ∂y

 =0

(5.160)

To simplify further the problem we assume the basin is bounded on one side (x = 0) by a rigid wall so that one of the component for the velocity is zero u = 0. This assumption can be extended to all the basin and Eqs. (5.159) and (5.160) reduce to fv = g

∂η , ∂x

∂v ∂η = −g , ∂t ∂y

∂η ∂v +H =0 ∂t ∂y

(5.161)

Deriving the second with respect to y and the third with respect to time we get ∂ 2v ∂ 2v = c2 2 2 ∂t ∂y where c =

√

(5.162)

g H . This equation will admit solution of the form v = F1 (x, y + ct) + F2 (x, y − ct)

So we get from the previous equations

η=

H (F2 (x, y − ct) − F1 (x, y + ct)) g

(5.163)

To determine the form of F1 and F2 we recur to the simplified Eqs. (5.161) so that f ∂ F1 = − F1 , ∂x c

f ∂ F2 = F2 ∂x c

Appendix

153 Coriolis pressure current current

pressure Coriolis

Propagation direction

Fig. 5.13 Kelvin wave in the northern hemisphere. The orientation is such that the wave propagates in the direction parallel to the upper wall. The currents shown are those required for the geostrophic equilibrium to hold

That give the solutions F1 = F10 (y + ct)e−x/R ,

F2 = F20 (y − ct)e x/R

Where R is the so called radius of deformation and it is defined as √ R=

gH f

(5.164)

this corresponds to the distance travelled by a wave 2π c/ f over the period 2π/ f . This could quite a number for deep basin and a few tens of km for shallow ones. Of the two solution, only the one that decay over x is meaningful so the final result u = 0,  v = g H F(y + ct)e−x/R , η = −H F(y + ct)e

(5.165)

−x/R

The general appearance of a Kelvin wave is given in Fig. Fig. 5.13. In this case the F(y + ct) is a gaussian with the amplitude decaying over a scale R. Actually, this is some what a controversial point because even shallow basin (≈40 m) would give a quite large R at middle latitude (≈200 km) while the observed scale is of the order of 30 m which is the value used in the figure. This show that our theory for such real case is only qualitative.

154

5 Waves

Equatorial Waves Equatorial waves have some peculiar characteristics because at the equator the Coriolis acceleration is zero and in the neighbourhood of it can be assumed to change linearly with the coordinate y like f = βy with β = d f /dy = 2 cos φ/a with a Earth’s, φ latitude and  the rotation rate. With this approximation Eqs. (5.159) become ∂η ∂u − βyv = −g , ∂t ∂x (5.166) ∂v ∂η + βyu = −g ∂t ∂y while continuity stay the same. Now we assume propagation of the wave towards the east (we will see there are reasons for that) along x while the meridional velocity (along y) will be zero. The equations reduce to ∂η ∂u = −g , ∂t ∂x ∂η βyu = −g , ∂y ∂η ∂u +H =0 ∂t ∂x

(5.167)

The procedure now is the same as before and we get the same expressions for u and η  (5.168) u = G(x − ct)F(y), η = H/gG(x − ct)F(y) F(y) is determined by satisfying the second Eq. (5.167) with the result F(y) = ex p(−y 2 β/2c). The final solution is then u = G(x − ct) exp{(−y 2 β/2c)}

(5.169)

This is a wave propagating in the eastward direction that decays along y with a scale √ (2c/β)1/2 . If we assume for c = g H this scale could be quite large considering that β ≈ 2.3 × 10−11 m−1 s−1 . We must consider that the dominant velocity is of the order of 2 − 3 m s−1 so that the meridional scale is of the order of 500 km. It is noteworthy that the Kelvin wave is trapped along the equator and that the current under the crest is in the same direction of the wave propagation while the current under the through is in the opposite direction. This can be understood referring to Fig. 5.14 where a cross section across the crest is shown. The pressure forces tend to flatten the bulge and the only way for the Coriolis forces to balance them is an eastward current. On the other hand the same cross section across a through shows that the tendency to fill out can only be balanced by a westward current.

Appendix

155

pressure

Coriolis wave propagation C H

eastward current SOUTH

NORTH

westward current I A

F B

G E

Coriolis

pressure

Fig. 5.14 The propagation of a gravity wave in the ocean (left). The dashed curve represents the wave some time later with respect the black line. The currents at depth converge at I and diverge at G in such way that current are in the propagation direction under the crest and opposite under the trough. The currents are consistent with geostrophic equilibrium maintained within the wave for the crest (upper right) and a trough (below right)

Similarly to the Kelvin wave along the cost the currents below the wave alternate the direction between crests and troughs while the wave will propagate towards the east. The equator is acting as a waveguide. We turn now our attention to the most general case and to simplify the notation we normalize Eqs. (5.159)–(5.160). The normalization uses the variables √ the system of √ T = (1/cβ), and L = c/β so that the new non dimensional variables are (primed) u = cu  , v = cv ,

x = L X ,

y = L y,

t = T t ,

η = φc2 /g

(5.170)

The non dimensional equations result to be (dropping the prime) ∂φ ∂u − yv + = 0, ∂t ∂x ∂v ∂φ + yu + = 0, ∂t ∂y ∂φ ∂u ∂v + + =0 ∂t ∂x ∂y

(5.171)

We now assume solutions of the form u = u(y) ¯ exp{i[(kx + ωt)]}, v = v¯ (y) exp{i[(kx + ωt)]}, ¯ φ = φ(y) exp{i[(kx + ωt)]}

(5.172)

We can eliminate u by differentiating the first in (5.171) with respect to x and the third with respect to t to obtain

156

5 Waves

∂ 2v ∂ 2u ∂ 2u ∂v + =0 − = y ∂t 2 ∂x2 ∂t ∂ y∂ x

(5.173)

A similar operation can be done with the second and third equation to get ∂ 2v ∂ 2u ∂ 2v ∂u + =0 − 2 = −y 2 ∂t ∂y ∂t ∂ y∂ x

(5.174)

Now we operate with (∂ 2 /∂t 2 − ∂ 2 /∂ x 2 ) to the last equation and integrating one respect to time ∂ 3v ∂v ∂v ∂ 3 v ∂ 3v − 2 + − y2 − 3 =0 (5.175) 2 ∂ x ∂t ∂ y ∂t ∂x ∂t ∂t Assume now a solution of the form v = exp{[i(kx − ωt)]}ψ(y)

(5.176)

Substituting in Eq. (5.176) we have   k ∂ 2ψ 2 2 2 ψ=0 − y + ω − k − ∂ y2 ω The requirement on ψ is that it must go to zero for | y |→ 0 and this can happen only when ω2 − k 2 −

k = 2m + 1, ω

m = 0, 1, . . . , or

k2 +

k − ω2 + 2m + 1 = 0 ω (5.177)

In that case the ψ coincide with the Hermite function d 2 ψm + (2m + 1 − y 2 )ψm dy 2

(5.178)

To find the equivalent of the dispersion relation we solve the second (5.177) for k to find  1 1 ± ω2 + k=− − (2m + 1) (5.179) 2ω 4ω2 For m = 0 we have two roots k =ω−

1 , ω

and,

k = −ω

(5.180)

The corresponding solution is called Yanay wave and for large ω we have k≈ω

(5.181a)

Appendix

157

which correspond to a non dimensional Kelvin wave. For small ω we have k ≈ −1/ω

(5.181b)

Which corresponds to a high zonal wave number for a Rossby wave. The other root (5.180) corresponds to a westward propagating Kelvin wave. Now at large distance from the equator ψ approaches zero so that from (5.174) −yωu = k

∂u ∂u , ⇒ yu = ∂y ∂y

So that the solution become proportional to exp{y 2 /2} and become unbounded for large y so must be discarded as we discussed before. For the low frequency limit in (5.177) the term ω2 can be neglected and we have the dispersion relation ω≈−

k2

k + (2n + 1)

(5.181)

That is a typical dispersion relation for Rossby waves. These waves are travelling westward because the phase velocity ω/k is always negative. For the high frequency limit we can  ω ≈ ± k 2 + (2n + 1) (5.182) That corresponds to Poincar waves that travel in both direction. At this point we remember that the above relations are non dimensional and to make them real the Rossby waves dispersion is given by ω≈−

kc βk ≈− k 2 + (2n + 1)β/c 2n + 1

Where c is the phase speed ω (k of the Kelvin wave). So the first Rossby mode will move at a speed one-third of a Kelvin wave. For < a speed of 2.3 m s−1 it will take 2.5 months to travel a distance of 15,000 km while a Rossby wave will employ about 8 months to travel the same distance. This distance corresponds to the El Nino theatre in the Pacific and 11 months it is just the typical period of its duration. Figure 5.15 shows the dispersion relations for different equatorial waves. Notice that this figure shows the exact solution (5.179) and this can be noticed by the asymmetry in the Poincaré waves with respect to k. As a matter of fact, in the approximation, we have shown the dispersion relation is symmetric with respect to a change of sign for k.

158

5 Waves 3

inertia-gravity n=2

2.5

Frequency ω

n=1 2

1.5

Yanai Kelvin

1

0.5

n=1 Rossby

0 -3

n=2

-2

-1

0

1

2

3

Wavenumber k

Fig. 5.15 The dispersion relation for equatorial wave. The dashed line indicate a Kelvin wave propagating in the westward direction

Seismic Waves The title is rather pretentious because seismic wave is a very complicate matter and in this case we will discuss the simplest results that come from the theory of elasticity. Again if you want to understand something about this just read Chap. 39 of the Feynman’s lectures. We must assume at this point that you know something about elasticity and start from the equation that relates the stress tensor Si j to the strain i j that is    Si j = 2μi j + λ kk δi j (5.183) k

Where λ and μ are the Lamé constants and δi j is the Kronecker delta. The strain on the other hand can be written as   ∂u j 1 ∂u i (5.184) + i j = 2 ∂x j ∂ xi We now consider and elementary volume of dimensions δxδyδz and from the definition of stress we just equate the force per unit volume to the acceleration ∂ 2u ρδxδyδz 2 = ∂t



∂ Sx y ∂ Sx z ∂ Sx x + + ∂x ∂y ∂z

 δxδyδz

(5.185)

Appendix

159

According to (5.183), we substitute for Si j Sx x = (λ + 2μ)x x + λ( yy + zz ),

Sx y = 2μx y ,

Sx z = 2μx z

And for the  according to (5.184) x x

∂u , = ∂x

x y

1 = 2



∂u ∂v + ∂y ∂x

 ,

x z

1 = 2



∂u ∂w + ∂z ∂x



Doing all the substitution we end up with the x component of the acceleration ρ

∂ 2u ∂ = (λ + μ) ∇ · u + μ∇ 2 u ∂t 2 ∂x

Similar equations are obtained for the other components so that in vector terms we have ∂ 2u ρ 2 = (λ + μ)∇(∇ · u) + μ∇ 2 u (5.186) ∂t Now we assume the vector field u decomposed in u = u1 + u2

(5.187)

Such that ∇ · u 1 = 0,

∇ × u2 = 0

(5.188)

Substituting (5.187) into (5.186), we have ρ

∂2 [u1 + u2 ] = (λ + μ)∇(∇ · u2 ) + μ∇ 2 (u1 + u2 ) ∂t 2

(5.189)

This equation can be splitted in two by taking the divergence and using the first of (5.188) ρ

∂2 (∇ · u2 ) = (λ + μ)∇ 2 (∇ · u2 ) + μ∇ · ∇ 2 (u2 ) = ∂t 2   2 ∂ ∇ · ρ 2 u2 − (λ + 2μ)∇u2 ∂t

(5.190)

Since ∇ × u2 = 0 the curl of the quantity in parenthesis is zero as it is the bracket itself and we have ∂2 (5.191) ρ 2 u2 = (λ + 2μ)∇u2 ∂t

160

5 Waves

This is a wave equation which describes waves with no shearing stress (∇ × u 2 0 = 0 √ and that move at a speed α = (λ + 2μ)/ρ. Taking now the curl of Eq. (5.189) we get the wave equation ∂2 (5.192) ρ 2 u1 = μ∇u1 ∂t √ These waves moves at a speed β = μ/ρ and because ∇ · u1 = 0 for these waves there is no change in volume so they correspond to shear waves. These two types of waves are known in seismology as P waves with velocity α and S waves with velocity β that someone interpret also as Primary waves or Secondary waves because α > β. We have assumed this result when we mentioned the Earth hydrostatic models but the application of seismic waves to the sounding is quite important in geophysics. One of the most obvious results is the discovery that part of the Earth’s core is in the liquid state. As we have mentioned S waves do not propagate in liquids that do not support shear stresses. As a consequence waves that travel through the core can be only of the P type. Studying the lack of signals from S waves contributed to the discovery of the Earth Core in 1906. We cannot resist to make a simple example of what is called refractive seismology

Refractive Seismology If we consider two contiguous layers with velocity α1 and α2 (Figure A.5.8a) a seismic ray will obey the Snell’s law α1 sin i = sin r α2

(5.193)

A seismic ray as its parent in optics is the direction of propagation of the seismic wave. If you apply the Snell law to the interior of the Earth α2 < α1 (because the density increases with depth) so that the refracted ray will move away from the vertical. At some point r = π/2 so that we can find a critical angle i c for which we have a total reflection (5.194) i c = sin−1 (α1 /α2 ) Consider now a simple model like the one shown in Figure A.5.8b. Between the source S and the receiver R we can have three different ray paths. A direct wave that travel on the surface, reflected wave that has an incidence angle larger than the critical angle and a refracted wave that after being refracted travels at the interface of the two layers. It is easy to show that for the direct wave the travel time would be t = x/α1

(5.195)

Appendix

x

R Reflected wave

α1

z

TIME

S

161

Head wave slope 1/ α2

α2

Direct Wave slope 1/ α1 Xc

Xcross

DISTANCE

Fig. 5.16 A two-layer model to illustrate refractive seismology (left). In this case, α1 < α2. On the right, the curve x − t which results from the simple model are shown

while for the reflected wave  2 z 2 + x 2 /4, ⇒ α12 t 2 = 4z 12 + x 2 t= α1 1

(5.196)

That is an equation for hyperbola. Finally, the refracted wave will give t=

2z 1  x 1 − (α1 /α2 )2 + α1 α2

(5.197)

 That is a straight line with slope (1/α2 ) and intercept (2z 1 /α1 ) 1 − (α1 /α2 )2 . Figure 5.16 shows the results of such a simple model, In particular, it shows that below the point xc given by 2zα1 (5.198) xc =  (α22 − α12 ) that correspond to the total reflected ray for the critical incidence angle. It is easy to see that a measure of the slope of the signal will give information about the seismic velocities of the layer. Refractive seismology has evolved toward a very sophisticated technique known as seismic tomography that gives 3D sections of the internal structure of the Earth.

162

5 Waves

References Textbooks Clarke AJ (2008) Dynamics of El Nino and Southern oscillation. Academic, New York Cushman-Roisin B, Beckers JN (2011) Introduction to geophysical fluid dynamics: physical and numerical aspects. Academic, New York Drazin PG, Johnson RS (1989) Solitons, an introduction. Cambridge University Press, Cambridge Fowler CRM (2004) The solid Earth: an introduction to global geophysics. Cambridge University Press, Cambridge Lighthill J (1978) Waves in fluids. Cambridge University Press, Cambridge Pedloski J (2010) Waves in the ocean and atmosphere: introduction to wave dynamics. Springer, Berlin Thorne KS, Blandford RD (2017) Modern classical physics: optics, fluids, plasmas, elasticity, relativity, and statistical physics. Princeton University Press, Princeton

Articles Pelinovsky E, Talipova T, Kurkin A, Kharif C (2001) Nonlinear mechanism of tsunami wave generation by atmospheric disturbances. Nat Hazards Earth Syst Sci 1:243 Pelinovsky I, Poplavsky A (1996) Simplified model of tsunami generation by sub-marine landslides. Phys Chem Earth 21 Tinti S, Bortolucci E (2000) Energy of water waves induced by submarine land-slides. Pure Appl Geophys 157:281

Chapter 6

Instabilities

We have already laid some of the elements necessary for this chapter when we dealt with the Benard convection in Chap. 3. These elements have to do with the equilibrium states of the dynamical system. We must consider some more elementary notions of mechanical stability and then go back to study dynamical systems. The approach we use will emphasize those instabilities that are quite relevant and produce visible effects like the Kelvin–Helmholtz or the gravitational instability. Specific applications will be treated in the appendix where we will discover that many more examples of instability can be found in the atmospheric physics and astrophysics. Again, for the time being, we will neglect the plasma instabilities that will be studied in the chapter on the magnetohydrodynamics.

6.1 Stability, Instability and Bifurcations Start with a simple example for the stability of a material point in the potential V (x). We assume that from this potential we can derive a conservative force F(x) such that F(x) = −

dV dx

(6.1)

The equilibrium may correspond to the case where F(x) = 0 that is d V /d x = 0, so that the equilibrium states may correspond to maxima or minima of the function V (x). Let us call x0 the point for the maximum or minimum and we can expand the function F(x) around this point. We have F(x0 + x) = −

   d V  d 2 V  1 d 3 V  − x − x 2 − .... d x x0 d x 2 x0 2 d x 3 x0

© Springer Nature Switzerland AG 2020 G. Visconti and P. Ruggieri, Fluid Dynamics, https://doi.org/10.1007/978-3-030-49562-6_6

(6.2)

163

164

6 Instabilities

The first term on the right hand side at the equilibrium is zero so that we have   d 2 V  1 d 3 V  x 2 − .... F(x0 + x) = − 2  x − d x x0 2 d x 3 x0

(6.3)

If the second derivative is positive (a minimum in the potential) then the force in the neighbours of x0 is negative (restoring) for a positive change x and vice versa. On the other hand if d 2 V /d x 2 < 0 (a maximum in the potential) then the force at x0 + x has the same sign of x so that x0 is a non-equilibrium point. This can easily visualized with a material little ball that rolls over the potential curve. We now ask something more general about the stability of the equilibrium points. We saw that when discussing the Benard convection and we can make now another example with the damped oscillator that obeys to the differential equation x¨ + γ x˙ + ω2 x = 0

(6.4)

This equation can be split in the system x˙ = v

(6.5)

v˙ = −(γ v + ω x) 2

(6.6)

The equilibrium point where x˙ = v˙ = 0 in the space x, v is obviously v = 0 and x = 0 but we want to see if this is a stable point that is if it remains as such for small perturbations around the same point. We assume the perturbation of the form δx = δx0 exp(σ t) and δv = δv0 exp(σ t) that once substituted in (6.5–6.6) will give σ δx0 − δv0 = 0

ω2 δx0 + (σ + γ )δy0 = 0

(6.7)

In order to have solutions such a homogeneous system must have the determinant of the coefficients equal to zero and that brings to the solution for σ σ + γ σ + ω = 0, 2

2

γ σ =− ± 2



γ2 − ω2 4

(6.8)

We have different cases based on the sign of γ and its value with respect to ω. If γ is negative the perturbation will grow with time either exponentially or modulated in case γ /2 < ω so the point will be unstable. In case γ is positive and γ /2 > ω the point will be stable and the perturbation will decay back to the initial point. Finally, if γ /2 < ω the solution will have an imaginary component and the perturbation will decay modulated by the square root term. We have so far reproduced the familiar damped oscillator behaviour and we can pass on to some more sophisticated stuff. Of particular interest is to go back to the Benard convection we have studied in Chap. 3 to introduce the concept of bifurcation. To this end, we consider a toy model for a system that obeys the equation

6.1 Stability, Instability and Bifurcations

165

dx = ax − bx 3 dt

(6.9)

√ This system has three equilibrium points x1 = 0 and x2,3 = ± a/b. To ascertain if these are stable or unstable points, we proceed as before. For example, for x1 = 0 we have for a perturbation like x  = A exp{σ t} σ x  = ax  − bx 3 ≈ ax 

(6.10)

So that, x1 is a stable point for a < 0 and unstable for a > 0. The points x2,3 are treated in the same way dx = ax  − 3bx22 x  (6.11) dt So that they are stable for a > 0 and unstable for a < 0. These results can be summarized as in Fig. 6.1. For a < 0 the only solution is x = 0 while solution x2 and x3 exist only for a > 0 and b > 0. This is called a supercritical bifurcation. These solution can be assimilated to the particle in a well analogy where the unique solution for a < 0 (only a well) while for a > 0 we have two possible equilibria. Again this case is similar to the Bernard convection where we have only the conduction state for r < 1 while we have two possible convective styles for r > 1. We will see that this conclusion is not valid for large values of r . On the other hand, when b < 0 states x2 and x3 can exist only if a < 0.In this case we have a subcritical bifurcation and the only stable state is for a < 0 and x = 0 so that the well analogy is something like Fig. 6.1b. Finally, we examine a higher dimensional version of the bifurcation considering the system

x

x s s

u

u

s

u a

a

s

u

u

u

u s

s

s

s

Fig. 6.1 The stability analysis for the Eq. (6.9). On the left it is shown the case when b > 0 for a > 0 and a < 0. On the right the case for b < 0 is shown. Also shown are is the analogy with the potential wells

166 Fig. 6.2 The stability analysis of the system described by Eq. (6.14). For μ ≤ 0 the stability path is a spiral toward the center. While for μ > 0 the limit cycle is reached either from infinity or the center

6 Instabilities

y stable periodic orbit

stable branch

μ unstable branch x

dx = −y + (μ − x 2 − y 2 )x dt dy = x + (μ − x 2 − y 2 )y dt

(6.12)

The stability problem can be worked out easily if we introduce a change of variables x = ρ cos θ y = ρ sin θ

(6.13)

dθ =1 dt dρ = ρ(μ − ρ 2 ) dt

(6.14)

The system (6.12) becomes

The equilibrium point is clearly x = y = 0 that is ρ = 0. Since θ˙ = 1 the trajectories are anticlockwise about the origin and at the same time if μ = 0 then ρ˙ = −ρ 3 so that the resultant motion is spiralling anticlockwise toward the origin. If μ < 0 then μ − ρ 2 < 0 for all ρ and we have the same spiralling motion. However, if μ > 0 then √ √ ρ˙ > 0 for 0 < ρ < μ and ρ˙ < 0 for μ < ρ < ∞. In the first case we will have √ a divergent flow from the origin to a limit cycle of radius μ while in the second case we have a spiralling motion from infinity to the same limit cycle. This is shown in Fig. 6.2. All these considerations may look a little bit too theoretical but they will be substantiated in due time.

6.2 Gravitational (Jeans) Instability Consider a star like as a massive sphere of gas where the equilibrium is determined by the pressure against the gravitational pull. There is no motion in this case and still we can write the perturbed equation of motion

6.2 Gravitational (Jeans) Instability

167

∂u ∇p =− − ∇  ∂t ρ

(6.15)

This equation is obtained from the Euler equation by perturbing pressure p = p0 + p  , density ρ = ρ0 + p  and the gravitational potential = 0 +  . In the same fashion the continuity equation becomes ∂ρ  + ρ∇ · u = 0 ∂t

(6.16)

The Poisson equation in the perturbed form becomes ∇ 2  = 4π Gρ 

(6.17)

where G is the gravitational constant. These equations assume a constant density and pressure that in the adiabatic case are related by p =

γp  ρ = cs2 ρ  ρ

(6.18)

where we have used the definition for the speed of sound cs . We now use a slight variation of our stability analysis by assuming that all the variables appearing above are in the form exp[i(ωt + k · r)], Then Eqs. (6.15–6.17) become ωu = −kcs2

ρ − k  ρ

ρ +k·u=0 ρ k 2  = 4π Gρ 

ω

(6.19)

Taking the scalar product of the first by k and substituting in the second we have  ρ  2 ω − k 2 cs2 = k 2  ρ Then the third equation can be used to eliminate  and ρ to obtain the dispersion relation (6.20) ω2 = cs2 (k 2 − k 2j ) where kj =

4π Gρ cs2

(6.21)

In the absence of gravity (G = 0) Eq. (6.20) √ gives the sound waves. The presence of gravity would add a frequency ωG = 4π Gρ. If this term becomes larger than

168

6 Instabilities

k 2 cs2 then ω2 < 0 and the instability sets in. Equation (6.20) gives the wavelength that corresponds to this critical point known as Jeans length λ J given by 2π = cs λ J == kJ



π Gρ

We can associate a typical time scale to the frequency ωG that is 2π/ωG = The wavelength λ J implies a Jean mass M J given by MJ =

4 3 πλ ρ 3 J

(6.22) √ π/Gρ.

(6.23)

The physical interpretation given to these quantities is based on the √ fact that given a typical length we can associate to it a free fall time /v √ where v ≈ G M/ , where the mass M ≈ ρ 3 . The resulting free fall time is t ≈ 1/ Gρ while the time it takes for a sound wave to cross the same distance ts ≈ /cs . When these two √ times are the same the resulting length is of the order of the Jeans length ≈ cs / Gρ. It is interesting to evaluate the Jeans mass for some typical condition in the universe. To this end it is convenient to express (6.23) as a function of temperature assuming an isothermal gas. In this case, cs = p/ρ and p/ρ = kT /μm p where m p is the proton mass and μ the number of protons in the molecule. We arrive then to the expression 4 M J = π 5/2 3



kT μGm p ρ 1/3

3/2

 3/2 T ≈ 10 ρ −1/2 μ 22

(6.24)

Using a densities between 10−22 and 10−20 g cm −3 a temperature of 10 K and μ ≈ 2.4 we arrive at masses between 28 and 280 (solar mass 2 × 1030 kg). This may indicates that stars do not form directly from collapse but rather there is an intermediate state of larger mass.

6.3 Instability of Shear Flows We will study the most general class of stability of stream flows when layers of fluid with different densities and velocity are subject to perturbation that may change their status. A drastic simplification can be obtained by using the Squires theorem that is demonstrated in the appendix. This theorem allows to reduce the analysis most of the time to a two-dimensional problem. In particular, we will consider two fluids at rest in a gravitational field with densities ρ1 for the fluid below the interface and ρ2 for the fluid above. The fluids are two-dimensional with the x axis along the interface and the z axis normal to it. The fluids are incompressible and we assume to be in the irrotational regime so that they can be described by potential function ψ such that the velocity u is given by

6.3 Instability of Shear Flows

169

u = −∇ψ. Then we could write the momentum equation in the form of (5.109) −

1 p ∂ψ + u 2 + + φ = F(t) ∂t 2 ρ

(6.25)

where φ is the gravitational potential and F(t) is a time dependent constant. The two fluids as shown in Fig. 6.3 have parallel velocities U1 and U2 so that their potential would be ψ2 = −U2 x + ψ2 (6.26) ψ1 = −U1 x + ψ1 , where we have assumed that each potential is made by a basic state plus a small perturbation ψ1 and ψ2 . These perturbations will produce ripples at the interface ξ(x, t) such that dξ ∂ξ ∂ξ ∂ψ1 = = + U1 − ∂z dt ∂t ∂x (6.27) dξ ∂ξ ∂ξ ∂ψ2 = = + U2 − ∂z dt ∂t ∂x These equations express the vertical component of the velocity and we seek solutions of the form ξ = A exp{[i(kx − ωt)]} (6.28) However, the incompressibility (∇ · u = 0 ) implies ∇ 2 ψ1 = ∇ 2 ψ2 = 0 so that the form of ψ1 and ψ2 will be ψ1 = C1 exp{[i(kx − ωt) + kz]},

ψ2 = C2 exp{[i(kx − ωt) − kz]} (6.29)

Notice that we take z = 0 at the interface so that ψ2 will decay with z above the interface and ψ1 will decay with z below the interface. These quantities can now be substituted in Eq. (6.27) to get i(kU1 − ω)A = −kC1 i(kU2 − ω)A = kC2

(6.30)

The three unknown are A, C1 and C2 , so we need another equation that can be found in the (6.25) from which we obtain the pressure below the interface   ∂ψ  u2 p = −ρ1 − 1 + 1 + gξ + ρ1 F(t) ∂t 2

(6.31)

where the gravitational potential has been written as gξ . The pressure at the interface must be the same so we have     ∂ψ  ∂ψ  u2 u2 ρ1 − 1 + 1 + gξ = ρ2 − 2 + 2 + gξ + ρ F1 (t) − ρ2 F2 (t) (6.32) ∂t 2 ∂t 2

170

6 Instabilities

The time dependent term can be assumed to be constant also in time because the perturbation must vanish at z = ±∞ then we can easily find that ρ1 F1 (t) − ρ2 F2 (t) =

 1 ρ1 U12 − ρ2 U22 2

(6.33)

The perturbation velocities can be determined from (6.26) as u 21 = (U1 − ∇ψ1 )2 = U12 − 2U1

∂ψ1 ∂x

neglecting the square of the perturbed quantity. From these we can get u 21 /2 and u 22 /2 that can be substituted in (6.32) to have     ∂ψ1 ∂ψ2 ∂ψ1 ∂ψ2 − U1 + gξ = ρ2 − − U2 + gξ ρ1 − ∂t ∂x ∂t ∂x

(6.34)

Substituting now from (6.28) and (6.29), we obtain ρ1 [−i(kU1 − ω)C1 + g A] = ρ2 [−i(kU2 − ω)C2 + g A]

(6.35)

We can use now (6.30) to get ω2 (ρ1 + ρ2 ) − 2ω − k(ρ1 U1 + ρ2 U2 ) + k 2 (ρ1 U12 + ρ2 U22 ) + kg(ρ2 − ρ1 ) = 0 (6.36) This is the basic equation that give rise to the two instabilities and we will study the Rayleigh–Taylor and the Kelvin–Helmholtz.

6.3.1 Rayleigh–Taylor Instability If we consider the fluid at rest U1 = U2 = 0, then (6.36) gives the dispersion relation ω2 =

(ρ1 − ρ2 )kg ρ1 + ρ2

(6.37)

It is clear that in the usual situation with the denser fluid at the bottom (ρ1 > ρ2 ) the small perturbation will not grow, while for the opposite situation with the denser fluid on top (ρ1 < ρ2 ) ω is imaginary ad the growth rate is given by  σ =

(ρ1 − ρ2 )gk ρ1 + ρ2

1/2 = (At gk)1/2

(6.38)

6.3 Instability of Shear Flows

ρ2

171

ρ2

U2 ε

ρ1

U1

ρ1 L

Fig. 6.3 The distribution of density and velocity in a two-layer sheared flow (left). On the right, the same fluids are shown to consider the effect of surface tension

where At sometime is called √ the Atwood number, At = (ρ1 − ρ2 )/(ρ1 + ρ1 ). Notice that if ρ2 ρ1 then ω ≈ gk that is we get again the gravity waves. It is quite interesting to note the role that surface tension plays in the Rayleigh– Taylor instability. As shown in Fig. 6.3, we consider the interface as an elastic membrane between the two fluids with the lighter on the bottom. The elementary pressure difference δp ≈ (ρ1 − ρ2 )g where  is the elementary vertical displacement. This pressure difference must be balanced by the effect of surface tension δpt ≈ γ /R where γ is the surface tension. R and the horizontal dimension L can be related by  R ≈ L 2 so that the ratio of the two pressure perturbation is given by ρgL 2 ρg R δp ≈ ≈ δpt γ γ A more detailed theory shows that this expression must be slightly modified as ρgL 2 = 4π 2 γ

(6.39)

That gives for the air-water interface a critical length of the order of 1.7 cm. Applications of the Rayleigh–Taylor instability are frequent in plasma physics and astrophysics. We will see some of them in due course.

172

6 Instabilities

6.3.2 Kelvin–Helmholtz Instability In the Eq. (6.36), we can put ρ1 = ρ2 and solve for ω we get c=

ω = Uave + ±iU k

(6.40)

where Uave = (U1 + U2 )/2 and U = (U1 − U2 )/2 We have then a grow rate for the perturbation given by kωi where the imaginary component it is just U . This means that no matter how small is the velocity difference the instability sets in and the wave is not dispersive with the phase velocity proportional to k. Charru gives a mechanistic interpretation of the Kelvin–Helmholtz instability when he consider that above a perturbation η > 0 of the shear layer the fluid is accelerated owing to the fact that the cross-sectional area perpendicular to the flow is decreased. The resulting velocity perturbation is of the order of u ≈ U kη. This increase in velocity above the crest implies a decrease in pressure of the order of p ≈ −ρU u following the Bernoulli law (increase in velocity—decrease in pressure) that for the horizontal momentum balance implies a growth rate sσ σ =

1 ∂p kp 1 du =− ≈− ≈ kU u dt ρu ∂ x ρu

(6.41)

There is, however, another way to show the intrinsic instability of the shear flow is we consider that the interface at z = 0 corresponds to a vortex line. We can again assume a disturbance of the form (6.28) and calculate across the interface the total vorticity disturbance as 

0+ 0−

  ∂u  ∂ψ1  ∂ψ2  dz = u 1 (z = 0+ ) − u 2 (z = 0− ) = − ∂z ∂ x 0 ∂ x 0

(6.42)

We are talking about the y component of the vorticity that in this case is given by ∂u/∂z − ∂w/∂ x. From Eq. (6.29) we have  ∂ψ1  = ikC1 ei(kx−ωt) = −(ω + kU1 )Aei(kx−ωt) = −(ω − kU1 )ξ ∂ x 0

(6.43)

And similarly  ∂ψ2  = ikC2 ei(kx−ωt) = −(ω + kU2 )Aei(kx−ωt) = (ω − kU2 )ξ ∂ x 0 So the change in vorticity is simply [−2ω + k(U1 + U2 )]ξ

(6.44)

6.3 Instability of Shear Flows

173

We now assume that ρ1 = ρ2 and U1 = −U2 = U > 0 so that (6.40) gives ω = ±ikU . The change in vorticity becomes −2ωξ = ∓2ikU ξ = 2kU A exp[(ikx ∓ iπ/2) ± kU t] where ωt has been substituted with ±ikU t and the i has been taken as exp{iπ/2}. The unstable solution becomes then 

 λ + kU t −2ikU ξ = 2kU A exp ik x − 4 And the vorticity disturbance changes as  

 λ π kU t e 2kU A cos k x − ekU t = 2kU A cos kx − 4 2 While the interface varies as AekU t cos kx The vorticity disturbance then grows most in the point A (Fig. 6.4) where ξ = 0 and ∂ξ/∂ x < 0 while decreases in points like B where ξ = 0 and ∂ξ/∂ x > 0. This behaviour tends to lift the interface crests and suppress the troughs. Beside the disturbance moves with an average velocity   ∂ψ2  ∂ψ1  + = k(U1 − U2 )ξ = 2U ξ ∂ x 0 ∂ x 0 u B

B z A

A -u

vorticity direction y x

Fig. 6.4 The evolution of the Kelvin–Helmholtz instability. In the upper panel, it is shown the initial deformation of the interface with the circles indicating the positive vorticity (clockwise) directed along y and the negative vorticity (anticlockwise). In the bottom panel, it is shown schematically the growth of the perturbation

174

6 Instabilities

That is positive near the crests and negative near the troughs. Therefore, clockwise disturbances accumulate around G while counterclockwise disturbances are swept from the troughs F. The results are the evolution depicted in Fig. 6.4. This analysis, however, is valid only in the limit of linear perturbation.

6.4 Stratified Shear Flows We now consider a parallel shear flow specified by a basic profile u(z). ¯ We consider the plane x − z and write down the equations of velocities zero divergence momentum and mass conservation already in perturbed form ∂u ∂w + =0 ∂x ∂z ∂u ∂ u¯ 1 ∂p ∂u + u(z) ¯ +w =− ∂t ∂x ∂z ρ0 ∂ x ∂w 1 ∂ p ρg ∂w + u(z) ¯ =− − ∂t ∂x ρ0 ∂z ρ0 ∂ρ ∂ρ d ρ¯ + u(z) ¯ +w =0 ∂t ∂x dz

(6.45) (6.46) (6.47) (6.48)

These equations can be reduced if we assume a stream function such that u=−

∂ψ , ∂z

w=

∂ψ ∂x

(6.49)

So that (6.48) is written as ∂ρ ∂ρ ∂ψ d ρ¯ + u(z) ¯ + =0 ∂t ∂x ∂ x dz

(6.50)

While pressure can be eliminated between (6.46) and (6.47) by cross differentiating to obtain   ∂ ∂ g ∂ρ d 2 u¯ ∂ψ + u(z) ¯ ∇2ψ − 2 =− (6.51) ∂t ∂x dz ∂ x ρ0 ∂ x As usual we seek a solution of the form ψ = (z) exp[i(kx − ωt)]

(6.52)

ρ = R(z) exp[i(kx − ωt)]

(6.53)

6.4 Stratified Shear Flows

175

Substituted in the Eqs. (6.50) and (6.51) we have the system  d 2 u¯ g d 2 2 − k  − =− R (u¯ − c) 2 2 dz dz ρ0 d ρ¯ =0 (u¯ − c)R + dz 

(6.54) (6.55)

where c = ω/k. R can be eliminated to obtain the so called Taylor Goldstein equation  (u¯ − c)

   2 d 2 d 2u N 2 =0 − − k  + dz 2 u¯ − c dz 2

(6.56)

where N is the buoyancy frequency we met before. Such an equation is quite difficult to solve so we will make some consideration about the stability.

6.4.1 The Richardson Number We start making the substitution φ=

 u¯ − c

And after some algebra (6.56) can be written as



u 2z − 4N 2 dφ d 1 2 (u¯ − c) − k (u¯ − c) + u zz + φ=0 dz dz 2 4(u¯ − c)

(6.57)

If the flow is contained between z = 0 and z = H the boundary conditions are φ(0) = φ(H ) = 0. This equation has the appearance of a forcing term (the first on the left) and the restoration term on the right. We now multiply (6.57) by the complex conjugate φ ∗ and integrate along the vertical limits 

  ¯ − c) |φz |2 + k|φ|2 dz + 1 (u 2





u¯ 2z − 4N 2 2 |φ| dz = 0 u¯ − c 0 0 0 (6.58) The integral in the middle is always real so that the imaginary parts of the equation reduce to the imaginary parts of the first and the third and this implies H

 ci 0

H





H

u¯ zz |φ|2 dz +

ci (u¯ − c) |φz | + k|φ| dz = 4 2

2

 0

H

1 4

H

u¯ 2z − 4N 2 2 |φ| dz u¯ − c

(6.59)

At this point we could have ci = 0 and in that case ω is real and the wave perturbation is stable. If ci is not zero the two integrals appearing in (6.59) must be equal with

176

6 Instabilities

the first being always positive. This require that the quantity u¯ 2z − 4N 2 be positive at the least in some portion of the domain. So instability is possible only if N2
0, therefore, d 2 u/dz once in the interval 0, H . In this case the flow is unstable but we have not introduced the Richardson number.

6.4 Stratified Shear Flows

177

6.4.2 A Useful Example We will now proceed with an example that could clarify a bit what we have done so far and justify some of the talks about bifurcation we made at the beginning of the chapter. Just consider a different distribution of velocity as shown in Fig. 6.5 where we can consider the region with u¯ constant and z positive (region I). The region where the velocity changes linearly comprised between D/2 and −D/2 (region II) and the region with u¯ constant and z negative (region III). We do consider ρ constant so that Eq. (6.56) reduces to (6.62) that can be derived with respect to x to obtain (remember (6.49))   1 d 2 u¯ d 2w 2 −w k + =0 (6.65) dz 2 u¯ − c dz 2 That can be further simplified if we notice that u¯ changes linearly so the second derivative is zero d 2w − k2w = 0 (6.66) dz 2 The general solution to this equation are the following I : II : III :

w = Ae−K y w = Be−K y + Ce K y w = Fe

(6.67)

Ky

Wee need now to apply boundary conditions to the different regions that requires displacements and pressure to be continuous. The relation between velocity and displacement is   ∂ ∂ d + u¯ δz w = δz = dt ∂t ∂x ¯ exp{ik(x − ct)} so that We assume a δz of the form δz ¯ w = ik(u¯ − c)δz And the continuity of δz demands that w/ik(u¯ − c) to be continuous. This condition is obviously satisfied because u¯ is continuous. As for the pressure we assume a pressure dependence of the form p(z) ¯ exp{ik(x − ct)} that substituted in (6.46) implies that the quantity d u¯ w ik(u¯ − c)u + dz

178

6 Instabilities

be continuous From the continuity equation (6.45) we have ik u¯ = dw/dz so that continuity is required on the quantity (u¯ − c)

dw d u¯ − w dz dz

(6.68)

At this point things get a little bit labour intensive with the first step to impose continuity of w at z = ±D/2. Then from (6.67) we have A = B + Ce K D

(6.69)

F = C + Be K D

At y = D/2 we evaluate the difference between the solution (I) and (II) and then derive to obtain   dw (6.70) = −K e−K D A − B + Ce K D dz So that we have e−K D w(D/2) d u¯ = −A u(D/2) ¯ − c dz u0 − c



2u 0 D

 =

dw dz

where the last equivalence derive from (6.68). We cam eliminate A from (6.70) using (6.68) to obtain  

c −1 B = Ce K D K D 1 − u0 And a similar expression at z = −D/2 C = Be

KD

 

c KD 1+ −1 u0

Eliminating B and C we have the dispersion relation 

DK c uo

2

= (D K − 1)2 − e−2K D

(6.71)

Wee see that c is purely real if (D K − 1)2 ≥ e−2K D with two possible solutions. It is purely imaginary if (D K − 1)2 ≤ e−2K D . The cross over being K D = 1.28. The flux is the unstable for long wavelengths (small K ). In this limit (6.71) would give lim c = ±iu 0

k→0

And the growth rate becomes lim σ = ±i K u 0

k→0

6.4 Stratified Shear Flows

179

z

3 0.4

U

0.2

II

x

σD/U0

ρ -U

2

I

1 KDc/U0

ρ

0.0

0 -1

- 0.2

-2

III - 0.4 0

1

2 KD

3

4

0

1

2

3

4

KD

Fig. 6.5 On the extreme left the distribution of density and velocity is shown. The center figure illustrates the growth rate σ D/U0 as a function of K D. On the extreme right, it is shown the real solution of Eq. (6.71)

So that for small K the growth is linear with K . Figure 6.5 shows the complete solutions to (6.71). We notice the bifurcation at k D = 1.28 for the two modes of the wave while before that we had the modes of growth and decay of the perturbation.

Appendix Supernovae Remnants Supernovae is one of the last stage of the life of a star. When the nuclear fuel of a star is near exhaustion the star grows up to a red giant or supergiant that may explode generating the supernovae. A supernova remnant (SNR) is the structure resulting from the explosion. The supernova remnant is bounded by an expanding shock wave and consists of ejected material expanding from the explosion, and the interstellar material it sweeps up and shocks along the way. Three phases can be distinguished in the evolution of SNR. In the first phase that lasts from a few hundred to thousands of years the mass of ejecta is much larger than the swept-up mass. In the second phase, the swept-up mass dominates and lasts tens of thousands of years. In the third phase, the radiative losses are important and eventually, SNR dissipates. In our galaxy, there are supernovae events every 30–50 years, while in the universe, it is estimated that there is a supernova every second. The SNR may be “visible” up to 100,000 light years primarily in radio and X-ray bands. Supernovae are important because they are thought to be an important source of heavy elements. In the phase when the swept-up mass becomes larger than the ejected mass a shock front is formed as shown in Fig. 6.6 that lies between the shocked gas and the undisturbed circumstellar medium (CSM). This front can reach very high Mach numbers. In the reference frame of this shock front the CSM moves toward the front (Fig. 6.6b) and finally, the Deceleration of ejecta drives a second shock “backwards” heating the ejecta (“reverse shock” in the moving frame), Fig. 6.6c): Between the primary shock front and the reverse shock, a contact discontinuity develops that separates

180

6 Instabilities unshocked gas flowing toward shock front

unshocked gas

shocked gas

gaseous shocked piston stellar ejecta

shocked gas

piston

unshocked gas

shocked CSM

piston shock front

shock front

(a)

(b)

(c)

Fig. 6.6 The different phases for the evolution of gas in a supernovae. In (a) the shock front compress the unshocked gas. In (b) the return shock is depicted while in (c) the shocked interstellar medium is shown

the shocked ejects from the shocked CSM. For typical outflows at early times, this contact discontinuity is unstable and is where the Rayleigh–Taylor instability takes place (Duffell 2018). This problem has been studied since the first paper on SRM in the  70 while today computer programs are quite common. However, we need to maintain some simple style so we will refer to a paper by Velazquez et al. (1998) that follows the classical paper by Plesset and Whipple (1974) and Cowie (1975). We already know that with two fluids with density ρ1 and ρ2 the Rayleigh–Taylor instability sets in with a growth rate given by σ (k) =



kg(α2 − α1 )

(6.72)

where α1,2 = ρ1,2 /(ρ1 + ρ2 ). The solution (6.72) gives very large growth rate for small wavelength and this results can be corrected introducing some viscosity of the fluid. We start by considering a small perturbation of the form η(x, t) = a(t) sin kx

(6.73)

where a(t) obey for stable perturbations to a¨ + ω02 a = 0 √ where dimensional considerations give ω0 = gk. In a fluid with kinematic viscosity ν the simplest form of the damped oscillator could be a¨ + f νk 2 a˙ + ω02 a = 0

(6.74)

Appendix

181

where f is a proportionality factor that must be determined. It is assumed that a(t) is of the form a(t) = a(0) exp{γ t} so that (6.74) implies γ 2 + 2νk 2 γ + ω02 = 0 where f has been set to 2 based on the small damping of surface waves. If νk 2 > ωo2 stability is assured at least for one solution while we have instability for the case γ 2 + 2νk 2 γ − σ 2 = 0

(6.75)

With σ given by (6.72). It is easy to show by derivation of (6.75) that γ has a maximum for 1/3 2π 1g = (α − α ) (6.76) k∗ = 2 1 λ∗ 2 ν2 With the maximum value of γ given by

1/3 1 1 g2 2 γ∗ = (α2 − α1 ) = τ∗ 2 ν

(6.77)

where τ∗ is the growth time. In order to compare these results with the observations hypotheses have to be made on the quantity appearing in (6.76–6.77). The kinematic viscosity can be written as 5/2 Tp (6.78) ν = 3.5 × 107 np Expressed in cm2 s−1 and where T p and n p are the temperature and density of the piston region. As we know the quantity (α2 − α1 ) can be expressed as α. Figure 6.7 gives some result obtained with different values for the temperature. The most notable result is that the maximum growth rate is obtained in the range of k∗ around 2–3 ×10−18 cm−1 that correspond to dimension of the order of 0.9 pc (parsec). The effect of viscosity is then to produce a dimensional range for the instability and reduce the growth rate. A further approximation can be obtained with the introduction of magnetic field effects as it is discussed in the Chapter on magnetohydrodynamics.

Rayleigh–Taylor Instability in the Solid Earth In the beautiful book by Turcotte and Schubert Geodynamics there is an interesting paragraph on the generation of salt domes. These formations are important because many oil and gas fields are found near them. The theory presented there, however, is quite laborious and the interested reader can follow it there. Here we will use a much simpler treatment due to Geoffrey Davies and presented in his book Dynamic Earth. Consider then two fluids like those shown in Fig. 6.8 with the denser fluid above the

182

6 Instabilities 8

6

σ ( 10-10 s-1)

no viscosity 4

2

T= 2 107 K T= 3.5 107 K

0 0

5

10 15 k (10-16 m-1)

20

25

Fig. 6.7 The growth rate of the Rayleigh–Taylor instability in supernovae remnants. The reported case is for different temperatures and absence of viscosity

D w h l Fig. 6.8 The bulge at the base of the mantle that produces the Rayleigh–Taylor instability within the Earth

lighter which present a bulge of height h and width l. This bulge is buoyant relative to the overlying fluid and its buoyancy is approximately (if h is small) B ≈ gρlh This is the force per unit length (in the third dimension perpendicular to the sheet). This buoyancy will make the sheet to grow with a velocity of the order of w = ∂h/∂t with this growth being resisted by the viscous stresses. These stresses will have

Appendix

183

different forms depending on the ratio between l and D the depth of the layer. In the case lD the dominant stress will be proportional to the velocity gradient w/l so that the resisting force will be Rs = μ(w/l)l = μw = μ∂h/∂t where μ is the dynamic viscosity. The balance of buoyancy and resistance will give ∂h gρl = h ∂t μ

(6.79)

That results in an exponential growth for h h = h 0 exp{t/τs } With τs =

μ gρl

(6.80)

(6.81)

The interesting result is that τs decreases with the increasing width. However, there is a limit to this because if l becomes comparable to the depth the top of the layer starts to interfere with the flow and to increase viscous resistance. With the width larger than the depth the horizontal stresses dominates which depend on the horizontal component of the velocity u. Based on continuity we could write u D = wl and the velocity gradient for the shear flow becomes u/D = wl/D 2 . The resisting force in this case (the stress acts across the width l) Rl = μ(u/D)l = μw(l/D)2 Again equating buoyancy and resistance we have gρ D 2 ∂h = h ∂t μl

(6.82)

We have an exponential growth but in this case the time scale is given by τs =

μl gρ D 2

(6.83)

We see that the wider is the bulge the slower is the growth that is the opposite behaviour we found with (6.81). It is easy to argue that the fastest growth is obtained for l = D with the time scale given by τ RT =

μ gρ D

(6.84)

184

6 Instabilities

where RT is for Rayleigh–Taylor instability. The moral of this simple story is that small imperfections in the interface between the two fluids that have a width comparable to the layer depth will grow rapidly and quickly will dominate. There is a very interesting consequence of this analysis because the growing bulge will be smoothed by the presence of the diffusion with a characteristic time τκ = D 2 /κ

(6.85)

where κ is the diffusion coefficient. In order for the bulge to grow τ RT must be smaller than τκ so that μ D 2 /κ gρ D That means

gρ D 3 = Ra ≥ c κμ

(6.86)

We have restated in physical terms the significance of the Rayleigh number. The only trouble is that the constant c is much larger than 1000 rather than 1 as in this case. It is easy to see that Ra is just the ratio τκ /τ RT and in the case of the Earth’s mantle is around 3 × 106 much larger than the critical value 1000.

Instabilities in the Atmosphere In the atmosphere, there are different and very important instabilities that influence our everyday life. We will show that they can be treated with a little extension of our knowledge. This may indicate that meteorology, sometime, maybe just an application of fluid dynamics.

The Centrifugal Instability Consider a cylindrical tank of spinning fluid like in Fig. 6.9 and consider a cylindrical coordinate system with v and u respectively the azimutal and radial velocity. The equations of motion are 1 ∂ p v2 du =− + dt ρ0 ∂r r (6.87) 1 ∂p d(vr ) =− dt ρ0 ∂θ With ρ0 reference density and θ angular coordinate. In terms of the angular momentum M = vr the same equations read

Appendix

185

Ω

Fig. 6.9 The rotating fluid to illustrate the cylindrical instability

r0

r

1 ∂p M2 du =− + 3 dt ρ0 ∂r r dM =0 dt

(6.88)

With the last equation expressing the conservation of angular momentum. Using the symmetry of the problem and the conservation of the angular momentum we displace a ring of fluid at the distance r0 + r and define a mean state as the one which describe the equilibrium of the first of (6.88) 1 ∂ p¯ M¯ 2 − 3 =0 ρ0 ∂r r

(6.89)

Then we define p = p¯ + p  , and M = M¯ + M  and subtract (6.89) from the first of (6.88) to obtain  1 ∂ p 1  du =− + 3 M 2 − M¯ 2 (6.90) dt ρ0 ∂r r We just neglect the effect of the perturbation pressure so that (6.90) reduces to  1  du = 3 M 2 − M¯ 2 dt r

(6.91)

The displaced ring will retain its angular momentum M 2 while the derivative appearing in (6.91) can approximated by du/dt = d(dr/dt)/dt ≈ d 2 r/dt 2 so that (6.91) becomes d 2 r M 2 − M¯ 2 = (6.92) dr 2 (r0 + r )3

186

6 Instabilities

We can use the approximation d M¯ 2 r M¯ 2 = M02 + dr That one substituted in (6.92) will give d 2 r 1 d M¯ 2 r = 0 + dr 2 ro3 dr

(6.93)

This is the equation of the harmonic oscillator resulting in free oscillation (and then equilibrium) if (1/ro )3 (d M 2 /d 2 ) > 0 while and unstable state could result if the same term is negative. Stable vortexes require then that the angular momentum increase with the radius. A situation confirmed in tornadoes and hurricanes.

Inertial Instability We assume a zonal geostrophic flow u g which is regulated by the equations of motion du dy = fv = f dt dt   1 ∂p dv =− − f u = f ug − u dt ρ ∂y

(6.94)

where we have used u g = −(1/ρ)∂ p/∂ y. We now displace a parcel across the stream of a quantity δy from its initial position y0 . The new zonal velocity of the parcel will be obtained by integrating the first of (6.94) u(y0 + δy) = u g (y0 ) + f δy

(6.95)

And the geostrophic wind at y0 + δy can be approximated as u g (y0 + δy) = u g (y0 ) +

∂u g δy ∂y

(6.96)

Now (6.95) and (6.96) can be used to evaluate the second of (6.94) to get dv d 2 δy = =−f dt dt 2



∂u g f − ∂y

 δy = f

∂ Mg δy ∂y

(6.97)

where we have defined the absolute geostrophic angular momentum as Mg = u g − f y and its derivative   ∂ Mg ∂u g − = f − ∂y ∂y

Appendix

187

Equation (6.97) is now of the familiar form and will give stable (oscillatory) solutions for f (∂ Mg /∂ y) < 0 and unstable solution for the positive values. So that in general

−

∂ Mg = ∂y

 f −

∂u g ∂y



⎧ ⎪ ⎨< 0 stable = 0 neutral ⎪ ⎩ > 0 unstable

(6.98)

In the northern hemisphere where f is positive the instability results if the absolute vorticity, that is f − ∂u g /∂ y is negative. Notice that what has been defined as Mg it is just the integration of the geostrophic absolute vorticity. We have      dv = f u g − f y − (u − f y) = f Mg − M dt

(6.99)

So that every time M < Mg the parcel deviates toward north with the opposite happening when M > Mg . In the case of stability f (∂ M/∂ y) > 0 we have a restoring force dv = − f M (6.100) dt With M  deviation from the initial Mg , M  = M − Mg .

Symmetric Instability We are now in a situation with respect to instability very similar to what we found for the potential temperature. In that case, we had an equation for the vertical motion on the kind ∂ 2 δz + N 2 δz = 0 (6.101) ∂t 2 where N2 =

g ∂ θ¯ θ¯ ∂z

Is the Brunt-Vaisala frequency and θ¯ is the average potential temperature. We have an oscillatory behaviour of the particle (stable) when ∂ θ¯ /∂z > 0, which is potential temperature increase with height. Now Eq. (6.101) is quite similar to Eq. (6.97) with the curve of geostrophic momentum surfaces in place of the potential temperature. If a tube of parcels oriented in the y direction is displaced from its equilibrium position (y0 , z o ) a restoring force due to the Coriolis and pressure force returns the parcels to their initial position (Fig. 6.10a). Consider now the situation of the stability of the tube of parcels with respect to both potential temperature and momentum as shown in Fig. 6.10b. The initial position is indicated by A and the final by B. It’s clear

188

6 Instabilities θ + 2Δθ

θ + Δθ

v+2Δv

B A

v+Δv

θ A

B

resulting force

v A

South

North South

South

North

B

a

North

c

b

Fig. 6.10 In a the parcel goes from A to B between different potential temperatures. In b the same displacement is between surfaces of different momentum while in c the more general displacement is illustrated with both the potential temperature and momentum

that in the final position the buoyancy is positive with the temperature lower than the environment while the horizontal restoring force is in the north direction. The resulting acceleration is clearly destabilizing. Our intent now is to evaluate the vertical acceleration av = g

θ − θ¯ θ¯

where it is easy to show that θ − θ¯ = −

∂ θ¯ ∂ θ¯ y − z ∂y ∂z

So hat the vertical acceleration g av = − θ¯



∂ θ¯ ∂ θ¯ y + z ∂y ∂z

 (6.102)

In the same way the horizontal acceleration can be written as  ah = − f (M − Mg ) = f

∂ Mg ∂ Mg y + z ∂y ∂z

 (6.103)

Indicating with s the direction of the resulting motion we have  ah = − f (M − Mg ) = f

∂ Mg ∂ Mg y + z ∂y ∂z

 (6.104)

After labouring a bit with the algebra we obtain a rather cumbersome relation

Appendix

189



     ∂ Mg z z − f cos α+ ∂z y y Mg   

  z z 2 sin α δs cos α − +N y θ¯ y

d 2 δs = dt 2

(6.105)

The sign of d 2 δs/dt 2 will now depends on the quite complicated term between braces in (6.105). In a situation like that in Fig. 6.9a we see a positive acceleration if the slope of the parcels is between the slope of constant M and constant θ . In case the slope of Mg surface exceeds the slope of θ¯ surfaces both terms in square brackets are positive and then d 2 δs/dt 2 < 0 and the parcel is accelerated back to its initial position. On the other hand if the slope of the θ¯ exceeds the slope of the Mg surface with the displacement lying between them d 2 δs/dt 2 > 0. So that a necessary condition for instability is     z z > (6.106) y θ¯ y Mg It is possible now to relate the slope to something we just saw a little earlier, the Richardson number. To do this we just remember the thermal wind relation ∂u g g ∂θ =− ∂z f θ ∂y From this equation we get easily 

z y

 θ¯

=

f θ (∂u g /∂z) ∂θ/∂ y = ∂θ/∂z ∂θ/∂ y

(6.107)

=

f + ζg 1 − ∂u g /∂ y) = ∂u g /∂z) ∂u g /∂z

(6.108)

And in the same way 

z y

 Mg

where ζg is the relative vorticity. Now according to our definition of Richardson number g/θ (∂θ/∂z) f (6.109) < Ri = 2 (∂u g /∂z) ζg + f Considering that at mid latitudes ζg is an order of magnitude smaller than f this condition translate in Ri < 1.

190

6 Instabilities

Barotropic Instability We consider now a particular instability that manifest itself in a barotropic atmosphere one in which there is no horizontal temperature gradient and the zonal wind (parallel to the equator) does not change with height. In such an atmosphere the absolute vorticity is conserved d (6.110) (ζ + f ) = 0 dt We assume a zonal flow u¯ so that (6.110) can be written 

∂ ∂ +u ∂t ∂x

 ζ +v

∂ (ζ + f ) = 0 ∂y

(6.111)

We have assumed a constant f . Now we perturb (6.111) in such a way that u = u¯ + u  ,

v = v ,

ζ = ζ¯ + ζ 

These are substituted in (6.111) and when the second order terms are neglected we have    ∂ ∂ ∂  + u¯ ζ  + v ζ¯ + f = 0 (6.112) ∂t ∂x ∂y where ζ  = ∂v /∂ x − ∂u  /∂ y. Again we introduce a stream function ψ such that u = −

∂ψ , ∂y

v =

∂ψ ∂x

That substituted in (6.113) will give 

∂ ∂ + u¯ ∂t ∂x

 ∇2ψ +

 ∂ψ ∂  ζ¯ + f = 0 ∂x ∂y

(6.113)

We seek a wave solution of the form ψ = ψ0 (y) exp[ik(x − ct)]

(6.114)

where the phase velocity may be complex c = cr + ici

(6.115)

In case Ci > 0 the wave will grow as unstable. We substitute (6.114) in (6.113) to obtain   2 ∂ ∂ ψ0 2 − k ψ0 + ψ0 (ζg + f ) = 0 (6.116) (u¯ − c) 2 ∂y ∂y

Appendix

191

We now substitute ∂/∂ y(ζg + f ) with β − d 2 u¯ g /dy 2 (where β = d f /dy) and divide (6.116) by u¯ − c to get  2 ¯ β − d 2 u/dy ∂ 2 ψ0 2 ψ0 = 0 − k − ∂ y2 u¯ − c

(6.117)

The phase velocity may be complex c = cr + ici so also 1/(u¯ − c) is complex with real and imaginary parts δr =

u − cr , (u − cr )2 + ci2

δi =

ci (u − cr )2 + ci2

(6.118)

Equation (6.117) can be then separated in the real and imaginary parts that look very similar 

   d 2 ψr,i d 2 u¯ d 2 u¯ 2 − k − β − 2 δr ψr,i ∓ β − 2 δi ψi,r (6.119) dy 2 dy dy Multiplying the real part by ψi and the imaginary by ψr and subtracting the latter obtained equation from the former we get ψi

  d 2 ψr d 2 ψi d 2 u¯ δi (ψi2 + ψr2 ) = 0 − ψ − β − r dy 2 dy 2 dy 2

That can also be written       d d 2 ψr d 2 u¯ d 2 ψi = β − 2 δi ψr2 + ψi2 − ψr dy dy 2 dy 2 dy

(6.120)

(6.121)

Integrating (6.121) over y from 0 to W with the boundary conditions ψr = ψi = 0, at y = 0,

and

y=W

The term on the let is zero and we are left with the integral 

W 0



d 2 u¯ β− 2 dy

 δi |ψ0 |2 dy = 0

(6.122)

where |ψ0 |2 = ψr2 + ψi2 . For the instability to exists is necessary that the gradient of the absolute vorticity β−

d 2 u¯ =0 dy 2

(6.123)

192

6 Instabilities

must change sign in the interval 0 − W . This instability could be responsible for the generation of low pressure zones in the tropics. A specific case for the barotropic instability can be made for a flow of the type u = tanh y

(6.124)

So that ∂ 2u ∂ζ = −2sech2 y tanh y =− 2 ∂y ∂y

∂u = −ζ¯ = sech2 y, ∂y

(6.125)

The inflection pint for the profile (6.124) is found at y = 0 and the Eq. (6.116) can be applied with the simplification of β = 0 to obtain  ∂ 2u d 2ψ 2 − k ψ − ψ =0 (u¯ − c) dy 2 ∂ y2 

(6.126)

We can easily find a solution for the stationary case c = 0. Substituting (6.125) in (6.126) we have d 2ψ − (k 2 − 2sech2 y)ψ = 0 (6.127) dy 2 With two independent solutions ψ1 = k cosh ky − sinh ky tanh ky ψ2 = k sinh ky − cosh ky tanh ky

(6.128)

The solution must be bounded for y ± ∞ so we have to choose k = 1 and ψ1 = sechy,

ψ2 = 0

(6.129)

For c = 0 the solution must be found numerically. This particular solution helps to give an interpretation of the barotropic instability. We can use a profile like the one shown in Fig. 6.11 with the relative absolute vorticity ζa . The inflection point is at some point ys . For y > ys , u(y) > u(ys ) and dζa /dy > 0. On the other hand for y < ys , u(y) < u(ys ) and dζa /dy < 0. The resulting phase speed for the Rossby waves will be westward for y > ys and eastward for y < ys . The instability will set in when the meridional velocities of the two waves are in phase (Fig. 6.11).

Baroclinic Instability Baroclinic instability is also an atmospheric instability although a little bit more complicated so it deserves a section of its own. We have mentioned several times the

Appendix

193

y

Westward propagation

u

dζ > 0 dy

-

+

-

+

-

+

ys

-

dζ < 0 dy

-

+

+

-

+ Eastward propagation

Fig. 6.11 The u profile on the left presents an inflection point at ys . In the upper part, the wave propagates westward while propagates eastward in the lower part. The phases of the waves are shown differential heating no circulation

no differential heating ......

......

θ+2Δθ

South

θ+Δθ

θ+Δθ

θ

a

θ+2Δθ

θ+2Δθ

θ+Δθ

North

differential heating and circulation

θ

θ North

South

b

North

South

c

Fig. 6.12 The potential temperature as a function of latitude and height in the absence of circulation and differential heating (a); with differential heating (b) and with differential heating and direct circulation (c)

baroclinic character of the atmosphere. Essentially the atmosphere is characterized by horizontal temperature gradients that lead to the isobar surfaces not being parallels to density surfaces. We may concentrate in particular on the potential temperature θ = T ( p0 / p) R/c p that we have introduced before. In an isothermal Earth’s atmosphere, without taking into account solar or infrared radiation, the surface of constant potential temperature would be parallel as shown in Fig. 6.12a. In a more realistic atmosphere (with the pole colder than the equator) the surfaces would be tilted with respect the horizontal (Fig. 6.12b). That configuration would induce a circulation (Fig. 6.12c) that would reduce the inclination of the surfaces by reducing the temperature gradient between the equator and the pole. This circulation regime is valid for the region where the Coriolis acceleration can be neglected or when the Rossby number R0 = U/ f a is not small. This happens when the zonal velocity u is large and the Coriolis parameter f is negligible. Such conditions we found at the tropics where we have ascending air at the equator and descending air at the tropics. This regime corresponds to the Hadley circulation. When the Rossby number decreases with increasing latitude along the zonal low waves could develop that facilitates the transport of heat across latitude. These waves

194

6 Instabilities

faster than geostrophic

North z θ+Δθ

z West divergence warmer

convergence colder

θ

East

z E

slower than geostrophic

(a)

West

East

(b)

N

(c)

Fig. 6.13 The formation of convergence and divergence zones in Rossby waves (a); the vertical movements associated to a long and short wave (b) and the 3-D picture that illustrates the movement of an air parcel with respect to the potential temperature (c)

must have a sufficient amplitude to do the job and the amplification should come by converting the potential energy created by the latitudinal variation of radiation in kinetic energy of the wave.

More on Rossby Waves When we talk about waves we intend mostly Rossby waves that we have seen before although we need now to expose some more features of them. Consider a wave like that shown in Fig. 6.13. At the inflection point, between a crest and a through, we can associate a vertical velocity. This is because between the crest and the trough there is a decrease in velocity. Remember that around the crest the motion can be assimilated to what happens around a high-pressure zone and there the velocity is higher than the geostrophic value because the Coriolis term must balance the additional centrifugal force. On the other hand, flow around the trough is analogous to the one around a low-pressure zone and here the velocity is less the geostrophic value because now the centrifugal force has the same direction of the Coriolis term. As a result of the convergence between the crest and through a vertical downward velocity develops. The opposite happens between the trough and the crest were we have an upward velocity. There is more. We recall the equation for the Rossby waves that can be obtained from the vorticity conservation (2.67). ∂ζ ∂ζ + u¯ + βv = 0 ∂t ∂x With β = ∂ f /∂ y. And the phase velocity for the same wave obtained with a streamfunction ψ = ψ0 exp[i(kx + ly − ωt)] c = u¯ −

β k2 + l2

Appendix

195

We notice from the last equation that for short waves (large k) the second term can be neglected and the phase velocity has the same direction as u¯ while for long waves (small k) the motion could be retrograde. This can be seen also from the first equation because ∂ζ /∂t < 0 means eastward propagation while ∂ζ /∂t > 0 corresponds to westward propagation. (retrograde) The sign will depends on the advection of relative vorticity u∂ζ ¯ /∂ x and the advection of planetary vorticity βv. We can make an example by taking a stream function of the type ψ = ψ0 − u¯ y + A cos kx sin ly

(6.130)

The it is easy to show that ζ = −(k 2 + l 2 )A sin kx cos ly u = −l A sin kx cos ly v = k A cos kx cos ly

(6.131)

As consequence the advection of planetary vorticity is given by βv = βk A cos kx cos ly

(6.132)

While the advection of relative vorticity −u

∂ζ ∂ζ −v = k u(k ¯ 2 + l 2 )A cos kx cos ly ∂x ∂y

(6.133)

Keeping A constant we see that for short wavelength (large k) advection of relative vorticity will dominate while for long wavelength (small k) planetary vorticity dominate. Remember that planetary vorticity advection will move the wave upwind while advection of relative vorticity will move the wave pattern downwind. The wavelength separating the two effects can be found by equating (6.132) and (6.133) that is k2 + l2 =

β u¯

(6.134)

Taking u ≈ 10 ms−1 , β = 1.6 × 10−11 s−1 m−1 and l ≈ k we get a wavelength of about 6.500 km.

The Physics of the Instability Here we will give a simple interpretation of the baroclinic instability. We have seen that Rossby waves can produce north-south movement (from negative to positive vorticity) to which we associate a downward motion. On the other hand, the northward component of this wave (from positive to negative vorticity) implies an upward

196

6 Instabilities

θ+Δθ

Fig. 6.14 The parcel trajectories (dashed lines) in the y − z plane. The inclination of potential temperature is given by α with respect to the equatorial plane

z

θ θ-Δθ α γ

y

motion. The entity of the upward motion depends on the wavelength: the shorter is the wavelength the larger is the vertical velocity. But there is more because the vertical motion can be interpreted as a thermodynamic effect. Air moving southward is responsible for cold advection so that in the convergence zone the atmosphere is colder and then heavier and may sink. On the other hand, the northward branch of the wave is responsible for heating the atmosphere in the divergence zone and produces ascending air. Shorter wavelength implies a larger temperature gradient with a corresponding larger colder of warmer advection. The resulting situation is shown in Fig. 6.13 where the particle trajectories are shown for short and long wavelengths. If we consider the inclination of the trajectories with respect to the constant potential temperature we see that there is energy gain only when the inclination of the parcels trajectories is less than the inclination of the isothermal. For the southward moving air colder air is moving into a warmer environment and as such will gain kinetic energy. On the other hand, for northward moving air warmer parcels are moving in a colder environment and then will continue to rise gaining potential energy. This is the mechanism of the baroclinic instability and we can give a rough estimate of what is the wavelength for which this happens. We consider a section in the plane y − z like the one shown in Fig. 6.14 where the lines at the same potential temperature are drawn. The inclination of such lines with respect to the horizontal is γ while the inclination of the trajectories is indicated with α. We have already shown that the instability can arise when α < γ and actually simple arguments would give the maximum energy transfer would be for α = γ /2. If w is the vertical velocity and u the horizontal (zonal) velocity we should have α≈

∂θ/∂ y w < u ∂θ/∂z

(6.135)

where the inclination is expressed as a function of the horizontal and vertical temperature gradients. The vertical gradient is related to the change of the zonal wind with height ∂u g ∂θ u =− ≈ (6.136) ∂z f θ ∂z H

Appendix

197

We can obtain a relation for w using the vorticity equation   ∂u ∂v ∂w ∂w d (ζ + f ) = −(ζ + f ) + = (ζ + f ) ≈ f dt ∂x ∂y ∂z ∂z

(6.137)

We introduce a typical horizontal scale L so that (6.137) becomes f

∂w d ≈ (ζ ) ∂z dt

We express ζ ≈ u/L while we assume t ≈ L/u so that w 1  u 2 ≈ H f L

(6.138)

Then using (6.135), (6.136) and (6.138) we get   ∂θ/∂ y w 1  u 2 H < = u f L u ∂θ/∂z This equation gives the length for instability H L> 2 f



g ∂θ θ ∂z

√

1/2 =

gθ f

(6.139)

where the approximation 1/H = (1/θ )(∂θ/∂z) has been used. We reach the conclusion that the critical length coincide with the Rossby derformation ratio that is 3000–4000 km.

The Eady Model To proceed with a quite detailed model of the baroclinic instability we have to introduce the so called quasi geostrophic vorticity equation. We have already found the vorticity equation in its simplest form   ∂u ∂v d (ζ + f ) = −(ζ + f ) + dt ∂x ∂y We assume zero divergence, a constant f and the perturbation values, u = u¯ + u  , v = v and w = w . We obtain considering only first order terms 

∂ ∂ + u¯ ∂t ∂x



ζ − f

∂w =0 ∂z

(6.140)

198

6 Instabilities

We have also neglected in the right hand side ζ with respect to f . We would like to express such equation as a function of the stream function only, To this end we consider the perturbation form of the hydrostatic equation ρ ∂( p¯ + p  ) 1 ∂ p + (ρ¯ + ρ  )g = 0 ⇒ + g=0 ∂z ρ¯ ∂z ρ¯

(6.141)

And in the same way the perturbation form of the thermodynamic equation 

∂ ∂ + u¯ ∂t ∂x



ρ w ∂θ − =0 ρ¯ θ¯ ∂z

(6.142)

Where θ is the potential temperature and we have used ρ  /ρ¯ = −θ  /θ¯ . This equation can be obtained by the constancy of entropy which implies dθ/dt = 0 that is the same as   ∂ w ∂θ ∂ θ + + u¯ =0 ∂t ∂ x θ¯ θ¯ ∂z Notice that N 2 /g = (1/θ¯ )(∂θ/∂z). We derive (6.142) with respect to z and use (6.141) to eliminate ρ we get 

∂ ∂ + u¯ ∂t ∂x

 ζ +

f ∂ 2 p ρ¯ N 2 ∂z 2

 =0

(6.143)

At this point we introduce a stream function ψ= Such that ug = −

∂ψ , ∂y

p f ρ¯ vg =

(6.144) ∂ψ ∂x

(6.145)

And Eq. (6.143) becomes 

∂ ∂ + u¯ ∂t ∂x

  f 2 ∂ 2ψ ∇2ψ + 2 2 = 0 N ∂z

(6.146)

This is known as quasi geostrophic vorticity equation because we use geostrophy in the wind equation (6.145) but keep the divergence term which originates the last term in (6.148). Eric Eady in 1949 developed a model for the baroclinic instability in which he considered a portion of atmosphere comprised between z = ± 21 D, He also considered β = 0 and a basic zonal wind that changes with height U = z with  indicating the gradient. We can apply (6.146) and choose the constant to be zero so we have  2 2 ∂ ψ ∂ 2ψ f ∂ 2ψ + + =0 (6.147) 2 2 ∂x ∂y N ∂z 2

Appendix

199

We look at solutions of the form ψ = Re { (z) exp i [k(x − ct) + ly]}

(6.148)

That substituted in (6.147) would give d 2

− α2 = 0 dz 2 where

  α2 = N 2 k 2 + l 2 / f 2

and we have as a general solution

(z) = A sinh(αz) + B cosh(αz)

(6.149)

We now need to impose the boundary condition that at z = ± 21 D, w = 0. To this end we write the thermodynamic equation in a slight different form 

∂ ∂ +U ∂t ∂x



ρ  + v

∂ ρ¯ ∂ ρ¯ + w =0 ∂y ∂z

(6.150)

In this notation we follow Alan Plumb lectures and exchange density with potential temperature so that θ  /θ¯ ≈ −ρ  /ρ. ¯ Now the perturbation hydrostatic equation gives ¯ y = f ρ¯0 /g, so ρ  = −( f ρ¯0 /g)∂ψ/∂z while the thermal wind equation gives ∂ ρ/∂ imposing the condition w = 0 on the boundary gives on each boundary (U − c)

d

−  = 0 dz

(6.151)

It is convenient to introduce the length scale L = N D/ f that we√ may call internal radius of deformation as opposed to the external radius L e = (g D)/ f so that N κz/ f = κ Lz/D. We now impose that (6.149) satisfy (6.151) at each boundary and after some algebra we have c A − (tanh(α H ) − 1/α H )B = 0 cB − (coth(α H ) − 1/α H )A = 0

(6.152)

This is a homogeneous system where the solutions exists only if the determinant of the coefficient is zero. After some algebra we obtain for the phase velocity c = ±H [(tanh(α H ) − 1/α H )(coth(α H ) − 1/α H )]1/2 The function in the square root is negative when

(6.153)

200

6 Instabilities

tanh(α H ) < α H that is for α H < 1.1997 and c is purely imaginary while for α H > 1.1997 the phase velocity is real. For the latter case we expect a wave propagating while for Im{c} > 0 we expect a growing wave. It is possible to find the maximum for kci (growth rate or minimum time constant) that happens for α H = 0.8031 so that kci = 0.3098( f /N ) We may use the following data to make some realistic estimation: H ≈ 10 km, N ≈ 1.2 × 10−2 s−1 , f ≈ 1. × 10−4 s−1 , and  of the order of 2.5 × 10−3 s−1 . The fastest growing rate is found for k = 1.61 f /N H = 1.34 × 10−6 m−1 which correspond to a wavelength of 4700 km. The growing rate is ≈6.5 × 10−5 s−1 meaning an e folding time of almost 2 days. These are very reasonable numbers. We can also find the ratio A/B from (6.152) to obtain A = B



1/α H − tanh α H coth α H − α H

1/2 (6.154)

Once A and B are obtained it is possible to determine the function ψ  (x, z) using (6.149) sinh αz cosh αz ψ(x, z) = cos kx + sin kx (6.155) sinh α H cosh α H Notice that this it is just the real part of (6.148). From the stream function using (6.149) it is possible to obtain the perturbation temperatures. As a matter of fact we consider ψ to be the geopotential so that we have RT ∂ψ = ∂z H With R the gas constant. We can obtain easily Hα T = R

  cosh αz sinh αz coskx + sin kx sinh α H cosh α H

(6.156)

Finally the vertical velocity can be obtained writing the thermodynamic equation as a function of the stream function ψ 

∂ ∂ +U ∂t ∂x



∂ψ ∂ψ ∂U N2 − +w =0 ∂z ∂ x ∂z f

(6.157)

Through which the vertical velocity results f w=− 2 N



∂ ∂ψ ∂ ∂ψ ∂ψ + z − ∂t ∂z ∂ x ∂z ∂x

 (6.158)

Appendix

201

Figure 6.15 shows the section z − x for the stream function, the vertical velocity and the temperature for the most unstable mode. One of the results of the Eady model is that the phase of the perturbation geopotential tilts westward as it is seen in Fig. 6.15. This feature can be explained quite easily. If we call θ  the temperature perturbation of a moving parcel (like the one in Fig. 6.14) and w the vertical perturbation velocity it is easy to see that in order to gain kinetic energy the correlation between these two deviations must be positive w θ   > 0 that corresponds in terms of density deviations to ρ  θ   > 0. If we refer now to the movement in the z − y plane a similar correlation exists between v and ρ  and it easy to see that southward moving air (v < 0) has θ  < 0 so that the correlation is given by θ  v  > 0. That is the same if we write f v ρ   < 0. Now geostrophic equilibrium requires f v =

∂φ  1 ∂ p = ρ ∂x ∂x

where φ  = gz is the geopotential. Also we have ρ=−

ρ0 ∂φ  g ∂z

stream function

0.5

0.0

H

L

H

-0.5

vertical velocity

z/H

0.5

0.0

-0.5

Perturbation Temperature 0.5

0.0

W

W

C

-0.5

kx Fig. 6.15 The results of the Eady model for the stream function (top), the vertical velocity (middle) and the perturbation temperature. H and L is for high and low while W and C is for warm and cold

202

6 Instabilities

From the perturbation hydrostatic equation. The correlation then becomes   ρ0 ∂φ  ∂φ  f v ρ  = − g ∂ x ∂z  

The slope of the lines at constant φ  is tan α = −

∂φ  /∂ x ∂φ  /∂z

And the correlation term becomes f v ρ   =

ρ0 g



∂φ  ∂x

2

 tan α

The requirement that f v ρ   < 0 implies tan α < 0 and the lines of constant φ  tilts westward (Fig. 6.15 top).

The Squire’s Theorem Consider a fluid of uniform density ρ which obeys to the continuity and momentum equations ∇p ∂u + (u · ∇)u = − ∇ · u = 0, ∂t ρ We now consider a perturbation to the velocity field U + u where U = (U (z), 0, 0) and u = (u, v, w) while for the pressure we have p = p0 + p  . The perturbed equations read ∂v ∂w ∂u + + =0 ∂x ∂y ∂z ∂u ∂U 1 ∂ p ∂u + U (z) +w =− ∂t ∂x ∂z ρ ∂x (6.159) ∂v 1 ∂ p ∂v + U (z) =− ∂t ∂x ρ ∂y ∂w 1 ∂ p ∂w + U (z) =− ∂t ∂x ρ ∂z We assume all quantities of the form ex p[i(ωt − kx − ly)] and for simplicity put ρ = 1. Then (6.159) become

Appendix

203

∂w =0 ∂z i(ω − kU )u + U  w = ikp 

−i(ku + lv) +

i(ω − kU )v = ilp  ∂ p i(ω − kU )w = − ∂z

(6.160)

where we have used U  = ∂U/∂z. We now define a new vector K 2 = k 2 + l 2 and K u˜ = ku + lv. Then the first of Eq. (6.160) can be written as − i K u˜ +

∂w =0 ∂z

(6.161)

Also we multiply the second by k/K and the third by l/K and sum to obtain i(ω − K U )u˜ +

k  U w = i K p K

And defining U˜ = kU/K this becomes i(ω − K U˜ )u˜ + U˜  w = i K p 

(6.162)

The last of Eq. (6.160) becomes 

∂p i(ω − K U˜ )w = − ∂z

(6.163)

Comparing (6.161), (6.162) and (6.163) with Eq. (6.160) we see the substitutions k → K , l → 0, u → u, ˜ v → 0, U → U˜ makes the two systems equivalent. Without loss of generality we can simply put l = 0 and v = 0 and treat the instability as a two dimensional problem in the x − z plane. This corresponds to the Squire’s theorem that we have used implicitly in the previous sections.

References Textbooks Beltrami E (1987) Mathematics for dynamic modeling. Academic Press, New York Chandrasekhar S (1961) Hydrodynamic and hydromagnetic stability. Dover, New York Charru F, De Forcarnd Millard P (2011) Hydrodynamic instabilities. Cambridge University Press, Cambridge Clarke CJ, Carswell RF (2014) Principles of astrophysical fluid dynamics. Cambridge University Press, Cambridge

204

6 Instabilities

Davies G (2009) Dynamic earth. Cambridge University Press, Cambridge Drazin PG (2001) Introduction to hydrodynamic stability. Cambridge University Press, Cambridge Guyon E, Hulin JP, Petit L, Mitescu CD (2001) Physical hydrodynamics. Oxford University Press, Oxford Hess SL (1959) Introduction to theoretical meteorology. Holt, New York Hoskins BJ, James IN (2014) Fluid dynamics of midlatitude atmosphere. Wiley Blackwell, New York Markowski P, Richardson Y (2010) Mesoscale meteorology in midlatitudes. Wiley Blackwell, New York Pringle J, King A (2007) Astrophysical flows. Cambridge University Press, Cambridge Satoh M (2014) Atmospheric circulation dynamics and general circulation models. Springer, Berlin

Articles Cowie LL (1975) Hydrodynamic instabilities and the formation of radio shells in young supernovae remnants. MNRAS 173:429 Duffell PC (2018) A one-dimensional model for Rayleigh-Taylor instability in supernova remnants. Astrophys J 76:821 Plesset MS, Whipple CG (1974) Viscous effects in Rayleigh-Taylor instability. Phys Fluids 17:1 Plumb A. Instability of zonal flows (QG) http://www-eaps.mit.edu/~rap/courses/12810_notes/ instability.pdf Velazquez PF et al (1998) Study of the Rayleigh-Taylor instability in Tycho’s supernova remnant. Astron Astrophys 334:1060

Chapter 7

Non-linearities, Randomness and Chaos

Initially, we planned to write this chapter including both chaos and turbulence but then realized that this choice could be misleading and even erroneous. We have mentioned (when introducing viscosity) the implication of Reynolds number and how that separates some laminar flow from turbulent flow. It is also true that what we now call generically as Chaos theory was born right out from the Benard convection as studied by Edward Lorenz, while many concepts that concur to define it were already there, i.e. bifurcations or Poincaré sections. Although a chaotic fluid may resemble a turbulent fluid in many aspects, we will discover that chaos obeys some precise laws and definitions. Chaos in fluids has to do with non-linearities and we have had an outstanding example of that in the equation of Navier–Stokes, hence we will discuss that first. Another quite intriguing issue is randomness or the stochastic character of some fluids. What we can say now is that a chaotic behaviour has to do mainly with extreme sensitivity to the initial conditions while randomness has more to do with the stochastic forcing of the process. These are quite complex issues over which entire books have been written. Our approach will be as usual on the soft side and our main interest in this chapter will be chaos in dynamical systems.

7.1 Non-linear Behaviour and Navier–Stokes Equation One of the sources of non-linearity in the Navier–Stokes equation is the advection term u · ∇. This can be easily shown if you consider the one-dimensional problem ∂u ∂u +c =0 ∂t ∂x

© Springer Nature Switzerland AG 2020 G. Visconti and P. Ruggieri, Fluid Dynamics, https://doi.org/10.1007/978-3-030-49562-6_7

(7.1)

205

206

7 Non-linearities, Randomness and Chaos

where c is a constant with the dimensions of velocity. Assume a solution of the form u(x, t) = f (t)eikx with the initial condition u(x, 0) = U eikx then the solution is given by u(x, t) = U exp[ik(x − ct)] (7.2) That is a wave that propagates along the x direction and maintains the initial form. This is what we call a linear behaviour. On the other hand, if we consider the nonlinear problem ∂u ∂u +u =0 (7.3) ∂t ∂x This equation assumes the form of a diffusion equation if we consider that u = −δ∂u/∂ x. Then (7.3) becomes ∂ 2u ∂u −δ 2 =0 ∂t ∂x

(7.4)

For the same initial conditions as before, we find that the solution has the form u(x, t) = U e−δk t sin kx 2

(7.5)

This represents a wave that is attenuated with time. Another possible variation is to add to (7.4) an advective linear term like ∂u ∂ 2u ∂u +c −δ 2 =0 ∂t ∂x ∂x

(7.6)

We may assume an initial condition like u(x, 0) = U eikx and a solution of the form u(x, t) = f (t)eikx . It is then easy to show that the solution is given by u(x, t) = U e−δk t sin[k(x − ct)] 2

(7.7)

This is a travelling wave as shown in Fig. 7.1a with decaying amplitude and is clearly distorted. Large wavelength (small k) will decay slower than small wavelength. The most interesting case is the complete Burgers’ equation given by ∂u ∂ 2u ∂u +u −δ 2 =0 ∂t ∂x ∂x

(7.8)

It is possible to find solutions of the form u(x, t) = f (ξ ) = f (x − ct)

(7.9)

7.1 Non-linear Behaviour and Navier–Stokes Equation

t=0.0

t=0.2

t=0.6

207

δ = 1.0 δ = 0.4

t=0.4

δ = 0.16 δ = 0.064

x - ct

x

Fig. 7.1 Travelling wave with decaying amplitude at different times (a). The solution represented by Eq. (7.10) for different diffusion coefficients (b)

We show in the appendix that a solution of this form can be obtained and it is given by   1 1 ( f 1 − f 2 )(x − ct) (7.10) u(x, t) = c − ( f 1 − f 2 ) tanh 2 4δ where f 1 and f 2 are constant related to the phase velocity and the initial condition c=

1 ( f1 + f2 ) 2

(7.11)

Figure 7.1b show this solution as a function of different “diffusion coefficient”, δ. We see that the smaller δ is, the sharper is the transition between the values f 1 and f 2 and for the inviscid limit the solution is a sharp shock wave at x = 0. We see that the advection (non-linearity term) in the Navier–Stokes equation has quite important consequences.

7.2 Stochastic Behaviour In a chapter where we plan to talk about randomness, we could not neglect basic notions about noise and some elementary introduction to stochastic physics. There are several textbooks on the subject from the unsurpassed 1954 bible edited by Nelson Wax (with contribution from Chandrasekhar and Uhlenbeck among others) to the very practical Kurt Jacobs book. We will concentrate here on a few items smoothing the corner on a few occasions. More detailed discussions will be postponed to the Appendix.

208

7 Non-linearities, Randomness and Chaos

7.2.1 White or Gaussian Noise A white noise signal is constituted by a series of samples that are independent and generated from the same probability distribution. For example, you can generate a white noise signal using a random number generator in which all the samples are taken from a given Gaussian distribution. When using a random number generator, you should assign the mean value (μ) and the standard deviation (σ ). We have done that with a MATLAB program and the results are shown in Fig. 7.2. The number of realizations is 1000 with a standard deviation σ = 2 and an average μ = 0. It is possible to see that the normal distribution   (x − μ)2 1 f (x) = √ exp − 2σ 2 σ 2π is already well sampled. Having a continuous signal f (t), we define continuous autocorrelation R(τ ) the quantity

Sample Values

10

5

0

-5 0

100

200

300

400

500

600

700

800

900

1000

Samples 0.3

Arbitrary units

0.25 0.2 0.15 0.1 0.05 0 -6

-4

-2

0

2

4

6

Bins

Fig. 7.2 Sampling of white noise (upper panel) for 1000 realizations, and the Gaussian distribution resulting from the same values

7.2 Stochastic Behaviour

 R(τ ) =

209 ∞ −∞

f (t + τ ) f ∗ (t)dt =



∞ −∞

f (t) f ∗ (t − τ )dt

(7.12)

where the asterisk is indicating the complex conjugate. It is quite easy to show that for a periodic signal like cos(t) the autocorrelation is proportional to the cos(τ ) so it is maximum for the superimposed signals τ = 0 and it is zero for τ = π/2. We would like now to evaluate the autocorrelation function of the white noise, that is, for the stochastic process x(t) we should evaluate g(τ ) = (x(t)x(t + τ )) This is actually a short end writing of the more complex quantity g(τ ) =

1 cov[x(t), x(t + τ )] = var [x(t)] T −τ



T −τ

0

[x(t) − x)][x(t ¯ + τ ) − x]dt ¯ var [x(t)]

In this equation, T is the total length of the time series, x¯ is the mean value and var [x(t)] is the standard deviation. For white noise, the single values are uncorrelated except for τ = 0 so that we could write g(τ ) = σ 2 δ(τ )

(7.13)

Where δ(τ ) is the Dirac delta function Consider now the signal x(t) and its Fourier transform representation ∞ 

x(t) =

Fn en2πiνt

n=−∞

or in integral form

 x(t) =

∞

F(ν)e2πiνt dν

(7.14)

−∞

where ν is the frequency and F is the Fourier transform of x(t) 

T

F(ν) =

x(t)e−2πiνt dt

(7.15)

0

According to the Parseval’s theorem (that we do not demonstrate), the total energy is now  ∞ E[x(t)] =

−∞

|F(ν)|2 dν

so that the average power in the interval 0 < t < T

210

7 Non-linearities, Randomness and Chaos

1 P= T



T /2 −T /2

|F(ν)|2 dν.

(7.16)

If the irregular periodic signal is made up of n waves then the power spectrum is discrete and composed of single frequencies corresponding to different waves. As n → ∞, the spectrum will consist of an infinite number of vertical lines and we will talk about a spectral density function, which is the amount of variance per interval of frequency. The spectral density S(ν) for x(t) is then S(ν) =

1 |F(ν)|2 T

(7.17)

For a white noise, the spectral density oscillates irregularly around zero. Both the correlation function and the spectral density contain the same information except that in one case it is presented in the temporal domain while the other is in the frequency domain. Particularly important is another theorem known as Wiener–Khintchin that states that the spectral density function can be expressed as the Fourier transform of the autocorrelation function  ∞ g(τ )e2πiνt dν g(τ ) = x(t)x(t + τ ) (7.18) S(ν) = −∞

We have now all and we need to make some examples of stochastic processes starting from the most classical one (the Brownian motion) and then see how we can extend that to the simplest stochastic climate model.

7.3 A Couple of Examples of Randomness As promised we have shown that non-linearity plays a quite important role in the Navier–Stokes equation through the advection term and now we are going to illustrate a couple of cases about randomness starting from the most classical example which is Brownian motion.

7.3.1 The Brownian Motion We will follow again Feynman’s lectures. He starts by considering the infamous drunken sailor, who comes out of a pub making his way with random steps of length L. It is easy to show that after the N th step he has walked a distance RN (vector) that can be expressed as RN = RN−1 + L, where L is a vector with random direction and modulus L. The modulus of RN , therefore, is

7.3 A Couple of Examples of Randomness

211

RN · RN = R2N = R2N−1 + 2RN−1 · L + L2

(7.19)

2 2  + L2 because RN−1 · Now, if we take average of many trials we have RN2  = RN−1 L = 0 (L can take any direction). Repeating the process we have

R 2N  = N L 2 = αt

(7.20)

assuming that the number of steps is proportional to the elapsed time. We now consider a particle of mass m subject to a random force F(t) and to a viscous type of drag proportional to the velocity d x/dt so that the equation describing the motion is given by dx d2x = A(t) (7.21) m 2 +μ dt dt where A(t) = A(t)A(t ) = δ(t − t ) and A is the amplitude of the noise. We want to find the mean square distance that the particle drifts in the x direction that would be the same in the y and z directions y. Multiply (7.16) by x and take the average  dx   d2x  = Ax F(t). m x 2 +μ x dt dt

(7.22)

Starting from the right-hand side the average of x F(t) is zero because x can be in any possible value positive or negative. On the other hand, we could write the first term as  2 d2x dx d[x(d x/dt)] mx 2 = m −m dt dt dt Taking the average, the first term will be zero because x can take any value so we are left only with the term mv2 and Eq. (7.17) simplifies to −mv2  +

μ d 2 x  = 0 2 dt

The mean value of kinetic energy is just kT /2 with k the Boltzmann constant, so we are left with dx 2  kT =2 dt μ It can be integrated to give x 2  = 2kT t/μ. We would obtain the same result for the other two directions and the results to be compared with (7.20) is R 2  = 6

kT t μ

(7.23)

212

7 Non-linearities, Randomness and Chaos

That was simple enough and now we will see how the rigorous stochastic treatment can be more complicated. We will follow Jacob approach starting from the same equation as (7.21) m

d2x = −γ mv + gξ(t) = −γ p + gξ(t) dt 2

(7.24)

where we have substituted the momentum p for the velocity and where g is the amplitude of the stochastic forcing ξ(t) with a correlation function. ξ(t)ξ(t + τ ) = δ(t)

(7.25)

The same equation can be written as dp = −γ p + gξ(t) dt In stochastic formalism, it assumes the following form: dp = −γ pdt + gdW

(7.26)

This is also called Ornstein–Uhlenbeck Equation and the difference with a usual differential equation is that we have to apply the rules of stochastic integral and derivative. In this particular case, dW is the noise accumulated in the interval dt. This equation can be integrated and keeping in mind that the noise is not derivable we get a solution that formally is the same as the usual differential equation p(t) = e

−γ t

 p(0) + g

t

e−γ (t−s) dW (s)

(7.27)

0

And the variance is the square of the integral  V [ p(t)] = g 2 0

t

e−2γ (t−s) ds =

 g2

1 − e−2γ t 2γ

(7.28)

We observe now that the average kinetic energy E =  p 2 /2m is also kT /2 with k Boltzmann constant. Based on that, we can determine the value of g. Consider the steady-state situation for t → ∞, we see that the variance reaches the value g 2 /2μ and because for t → ∞, p(t) = 0 it results that  p 2  = V [ p(t)] so that E =

V ( p) g2 g2 p2 = = = 2m 2m 4γ m 24π ηd

(7.29)

where we have used the Stokes formula mγ = 6π ηd with η dynamic viscosity and r radius of the particle. We have then for g

7.3 A Couple of Examples of Randomness

g=

213



12π ηdkT

(7.30)

From the definition of p = mv, we can evaluate v and then x(t) that results  1

g 1 − e−γ t p(0) + x(t) = mγ mγ



t

 1 − e−γ s dW (s)

(7.31)

0

And taking the mean we have x(t) =

 1

1 − e−γ t p(0) mγ

(7.32)

we can also evaluate the variance  V [x(t)] =

g mγ

2 t+

 g 2 −μt − e−2γ t − 3 4e 2 3 2m γ

(7.33)

In practice however γ t 1, so that (7.33) simplifies to 2    3 g 2 kT t− ≈ t V [x(t)] ≈ t= 2γ mγ 3π ηd (7.34) Notice that this formula is the same as (7.20) because γ m = μ = 6π ηr . So we have 

g mγ

2

3g 2 t− = 2m 2 γ 3



g mγ

6kT 6kT kT = = μ γm 3π ηd

(7.35)

Yes we got the same result found in (7.23) but it is rather more complicated and opens the door for the next application.

7.3.2 Climate Variability We can make another example of stochastic physics about the variability of the sea surface temperature (SST). We consider what they call a “slab ocean” that is all the ocean is assimilated to a layer of water of depth h with heat capacity γ O = ρ O c O h. In this case, ρ O is the average ocean density, C O its specific heat and h the depth of the mixed layer that is the first 100 m of ocean. Then it is easy to calculate γ O = 1000 × 4.18 × 103 × 100 = 4.18 × 108 jm−2 K−1 . While for the atmosphere we would have γa = 1.2 · 1000 · 104 = 1.2 × 107 that is about 40 times lower. Considering that the ocean covers 70% of the Earth’s surface the contribution of the atmosphere is negligible. Experimental data show that the ocean has a damping time of about γ O /λ, where λ ≈ 15 wm−2 K−1 that corresponds to a time constant of about 10 months. We could then write a simple equation

214

7 Non-linearities, Randomness and Chaos

λ dT = − T + Q net dt γO

(7.36)

In this case, Q net is the net heat flux between atmosphere and ocean. In Eq. (7.36) if we put Q net = 0, we get a signal which decays exponentially with a time scale of λ/γ O . Now this very long time constant determines the response of the ocean to all possible fluctuations in Q net which may have different time scales from days to months. We can have an idea of such response assuming an input like Q net = Q ω eiωt while the temperature will have a similar response T = Tω eiωt It is easy to show by substitution in (7.36) that the amplitudes are related by 1 Qω γ0 iω +

Tω =

λ γO

with the squares of the amplitude  Tω2 =

Qω γ0

2

1 ω2 + ωc2

(7.37)

where ωC = λ/γ0 . At frequencies larger than ωc , the temperature signal will decrease rapidly while for ω < ωc the temperature response will become independent of frequency. This means that variations with time scales shorter than 300d will be damped out leaving variability on time scales longer than this. A forcing of the ocean climatic system by a stochastic “white noise” will result in “reddening” of the variability spectrum, that is, high frequencies will be damped out while low frequencies will be maintained. This is shown in Fig. 7.3. Notice that what is indicated as atmosphere forcing is the result of the spectrum of the isolated atmosphere that results to be white. This model was invented many years ago by Klaus Hasselmann that formulated the problem using a detailed and elegant formalism that here we have briefly introduced. It is worth exploring the problem a bit further. We start by rewriting Eq. (7.36) in the form dT = −aT − bW (7.36a) dt where T is the temperature perturbation and bW is some stochastic forcing. In the formalism of Stochastic Differential Equation (SDE), this corresponds to = Tk − aTk t − bWk Tk+1

(7.38)

where Wk is the stochastic forcing accumulated in the interval t. Then as shown before this is equivalent to write





T (t) = T (0) − 0

t

aT +

√

DW

7.3 A Couple of Examples of Randomness

215

10 1

atmospheric forcing

spectral power

10 0

10

-1

10 -2 10 -2

10 -1

10 0

10 1

frequency (cycle per year)

Fig. 7.3 The power spectrum resulting from the simple model for the atmosphere–ocean interaction

with D variance of the forcing. In the proper form, this equation reads dT = −aT dt +

√

DdW (t)

(7.39)

It can be integrated to give T (t) = T (0)e−at +

= T (0)e

−at

+

√ √



t



0 t

D

D



e−a(t−t ) dW (t ) (7.40)

dt e−a(t−t ) ξ(t )

0

where ξ(t)ξ(t ) = δ(t − t ). We want to use at this point the Wiener–Kinthchin theorem and calculate the spectral density as the Fourier transform of the correlation function. We can evaluate the correlation function  t+τ  t dt dt e−a(2t+τ −t −t ) ξ(t )ξ(t ) T (t)T (t + τ ) = D 0 0 (7.41)  t  D  −aτ −a(2t+τ −2t ) −a(2t+τ ) e dt e = −e =D 2a 0 The same calculation can be repeated for a negative τ and in the limit of t → ∞ obtain the correlation function

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7 Non-linearities, Randomness and Chaos

R(τ ) =

D −t|τ | e 2a

(7.42)

So that the temperature spectrum will result from the Fourier transform T (ω) = T0

ω2

D + a2

(7.43)

That is the same as (7.37). We see that going by the rules is much more complicated than direct physical approach. However, you need to solve most of these equations by computer and this may help. As a matter of fact, your stochastic forcing on a computer is just a random number generator and your equations need to be integrated with special techniques as we will see in the case of stochastic resonance.

7.4 Chaotic Behaviour In his beautiful book Chaos: from theory to applications, Anastasios Tsonis gives a very interesting definition of chaos when he affirms that “...chaos is randomness generated by deterministic systems, and that chaotic evolutions are irregular, unpredictable evolutions exhibiting spectra practically indistinguishable from spectra of pure random processes”. This is absolutely a good way to connect what we have said about non-linearity and randomness to the last topic of this chapter, that is, chaos.

7.4.1 Generating Chaos A simple way to introduce chaos is through the so-called logistic map and it is based on the recursive relation (7.44) xn+1 = axn (1 − xn ) . This recursive relation is based on a non-linear differential equation, which describes the evolution of population  x x dx (7.45) = 1− dt τ xe where the first term represents births and the second deaths. τ is a characteristic time and xe a representative population. Actually, the situation is much more complicated than that but we will regard (7.44) only as a way to generate a sequence of numbers that may exhibit chaotic behaviour. Consider as a first example x0 = 0.5 and a = 0.8, we get the sequence x1 = 0.2, x2 = 0.128, x3 = 0.0892928 up to x f = 0 when the iteration stops. When we change to a = 2.5, the parabolic curve f (x) = ax(1 − x) intercept the diagonal at two fixed points x f = 0 and x f = 0.6. Starting with x0 =

7.4 Chaotic Behaviour

217

0.1 we obtain x1 = 0.225, x2 = 0.43594, x3 = 0.61474 and the convergence is for x f = 0.6. The same happens with x0 = 0.8 when two iterations are enough to reach the same equilibrium point. The stability of such points can be discussed in general by assuming that the population will repeat after one generation, that is, x f = ax f (1 − x f )

(7.46)

with roots 0 and 1 − 1/a. The stability of the map at these points can be studied by making small perturbations around them = a(x f + xn )[1 − (x f + xn )] x f + xn+1

(7.47)

Neglecting all terms involving higher order, we get = a(x f + xn ) − a(x 2f + 2x f xn ) x f + xn+1

(7.48)

And considering (7.46), we have = axn (1 − 2x f ) xn+1

(7.49)

We now recognize that a(1 − 2x f ) = d f /d x|x f And we have = f (x f )xn xn+1

(7.50)

→ 0 as long as | f (x f )| < 1 that As the number of iterations increases, we get xn+1 translates into or − 1 < a < 3 (7.51) − 1 < a − 2ax f < 1,

Equation (7.51) implies that as long as −1 < a < 3 the population remains constant in time. In the previous cases with a = 0.5 and a = 0.8, we satisfy such criteria. Now see what happens when a = 3 the period of the oscillation doubles between two values x f1 and x f2 as shown in Fig. 7.4. The two values can be found by solving x f2 = ax f1 (1 − x f1 ) x f1 = ax f2 (1 − x f2 )

(7.52)

We obtain that for a = 3.1 x f1 = 0.558 and x f1 = 0.765. Again this regime is maintained for 3 < a < 3.449479 when the period doubles again with period 4 and limit values x f1 = 0.403, x f2 = 0.479, x f3 = 0.835 and x f4 = 0.866. These period doubling points are actually bifurcation and the values of a can be predicted based on the Feigenbaum relation. But the most interesting point is that for some value of a equilibrium is not reached and the recursive relation gives numbers that jump around in a fully chaotic behaviour. Figure 7.4 shows the sequence of bifurcation for the logistic map. Each period doubling actually represents a bifurcation.

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7 Non-linearities, Randomness and Chaos

a = 3.1

a = 0.8

a = 3.5

a = 3.8282

0.5 0.0 -0.5 -1 -1.5 -2 -2.5

3

3.2

3.4

3.6

3.8

a

4

2

2.5

3.0

3.5

4

a

Fig. 7.4 Construction of logistic map for different values of the parameter a. The lowest two panels are the Lyapunov exponent and the bifurcation diagram of the logistic map

7.4.2 Lyapunov Exponents In the case of the logistic map, by changing the parameter a we have encountered a fundamental property of chaos, i.e. intermittence. The dependence upon initial conditions can be measured by the Lyapunov exponents. Suppose the system is allowed

7.4 Chaotic Behaviour

219

to evolve from two slightly different initial states, x and x + , then after n iterations their divergence may be characterized by (n) ≈ eλn

(7.53)

where λ is the Lyapunov exponent that gives a measure of divergence rate. If λ is negative the slightly different trajectories converge and the evolution is not chaotic. On the other hand, we obtain a chaotic behaviour if λ is positive. Consider, for example, the logistic map at its nth iteration, xn+1 = f (xn ), the difference between two initially nearby states after n iteration can be written as f n (x + ) − f n (x) ≈ eλn ,

 =⇒, log

 f n (x + ) − f n (x) ≈ nλ 

(7.54)

and for small  the expression becomes λ≈

 n d f  1  log  n dx 

(7.55)

Using the chain rule, we obtain n−1 1 log | f (xi )| n→∞ n i=0

λ = lim

(7.56)

Figure 7.4 shows the Lyapunov coefficients for the logistic map. This seems a straightforward calculation for the logistic map. For continuous system like those we will see in a while the calculation may be more complex.

7.4.3 Visualization of Chaos: Attractors, Poincarè Sections and Fractals A fundamental tool for the visualization of chaos is the phase space. We have mentioned that several time for example discussing bifurcation and equilibrium in Chap. 6. In particular, we have seen that two-dimensional systems may be represented in a ρ − θ space giving rise to an equilibrium point or a limit cycle. The simplest example that can be made is that of a pendulum, which describes a closed curve in the phase space with coordinate θ (the angle with respect the equilibrium position) and θ˙ the derivative with respect to time. If we add a damping term, then the resulting trajectories in the same phase space are a spiral. These two geometrical objects in the phase space are called attractors. An attractor that assumes the form of a closed curve is representative of a periodic motion and indicates a limit cycle. These attractors are also called “well behaved” and are a particular case of systems that

220

7 Non-linearities, Randomness and Chaos

are predictable. Often they are also called non-chaotic attractors and mathematically speaking are characterized by an integer dimension that belongs to a submanifold of the relative phase space. For example, the attractor for the simple pendulum is a point (zero dimensions) with respect to two-dimensional phase space. The attractor of the damped forced pendulum has one dimension with respect to a two-dimensional or three-dimensional phase space. In order to deal with more complex attractors, we need to introduce two other concepts, the Poincarè section and the fractal dimension. The system of the simple pendulum can be modified with the introduction of a third coordinate φ = ωt so that we have the complete system ˙ dθ/dt = θ,

d θ˙ /dt = (g/l)θ − β θ˙ ,

dφ/dt = ω

(7.57)

where β is the damping. In the three-dimensional phase space θ, θ˙ , φ will have a trajectory very similar to a helix, as shown in Fig. 7.5a. If we make a cross section of this trajectory at a fixed time interval corresponding to a period T = 2π/ω, we will have in the plane θ, θ˙ a point with the same coordinates. However, the system could be such to have a different trajectory in phase space so that the intersection could change in the different sections. In the appendix, we illustrate the case of the non-linear pendulum dθ = θ˙ dt θ˙ d θ˙ (7.58) = − sin θ − + F cos ωt dt q dφ =ω dt 1

P1

ω (radians/second)

. θ

P2

0.5 0 -0.5 -1

t -1.5

θ

-2 -4

-3

-2

-1

0

1

θ (radians)

2

3

4

Fig. 7.5 The left panel shows the construction of the Poincarè map: we need to add a motion along the t axis. On the right, the resulting Poincarè map for the non-linear pendulum is shown

7.4 Chaotic Behaviour

221

where the approximation θ ≈ sin θ has been dropped. As shown in the appendix, such a system does not always give periodic solutions and the parameter that controls mainly the results is the forcing term F. When we calculate the Poincarè section for this non-linear case, we obtain a rather surprising result, shown in Fig. 7.5b. The pattern in the figure is composed of about 10000 intersections of the kind shown in Fig. 7.5a, with the pattern more and more defined as the number of points increases. All these points taken together form the attractor to which we may now give a different interpretation. We actually could think of solving the non-linear system starting from different initial conditions (chosen, e.g. with a random number generator). After a certain time, we would find that the system is actually in one of the points that make up the attractor or in any case in the general area occupied by the attractor. It is quite interesting at this point to pose the question of the dimension for the attractor which in this particular case is not an integer number. The attractor dimension is important because it is somewhat related to the minimum number of variables required to describe the dynamic system. The evaluation of these dimensions can only be made through the introduction of the concept of Fractal dimension. Fractals can be defined as patterns made up of the superposition of similar shapes but on different scales. Typical examples are the dendrites, snowflakes or tree branches. In meteorology, lightning, clouds and turbulence (to name a few) may exhibit a fractal behaviour. The fractal dimension is a measure of the space-filling capability of a particular object. In Euclidean geometry, the dimensions are expressed by integers: a point has 0 dimensions, a line D = 1 and so on. Fractal geometry allows also decimal dimensions. Consider, for example, childish drawings like those shown in Fig. 7.6. In part (a) of this figure, the simple line has an Euclidean dimension D = 1. In part (b), the wiggly line has a fractal dimension between 1 and 2 with the larger dimension corresponding to the case when the line (c) would completely fill the plane. There are different methods to determine the fractal dimension. The simplest one is based on the fact that when we add a spatial dimension the fractal dimension increases by one unit. As a matter of fact, we may imagine the line in Fig. 7.6b as a cross section of a newspaper sheet that has been crinkled. When the sheet is flat on the table, it has dimension D = 2. When it is folded or wadded, its dimension is between

(a)

Fig. 7.6 The fractal dimension and its reduction

(b)

(c)

222

7 Non-linearities, Randomness and Chaos

Fig. 7.7 The determination of fractal dimension of an attractor

. . . . . ... . . . . . . . . .. . . . . . . . . . . . . . . . . .. . . . . . .. . . . . .. .. . . . . . . . . . . . . . .. . . . . . .. . . . . . .. . . .. .. ... . . . . . . . . . . . .. . . .. .. .. .. . ... . . . . .. . . . . . . . .. .. . .. . . . . . .. . ... . .. . . .. .. . . .. . . .. . . .. . . . . . . .. . . . . .. .. . .. . . . . . . . .. . .. . . . . . . . .. . . . .

ε

. . . .. . . . .. . . . . .. . .

2 and 3 and may approach D = 3 when it is well compressed into a perfect ball. If the folded or crinkled newspaper is cut along a plane, the cross section of the cut has a dimension between 1 and 2. For example, if its dimension in three-dimensional space was 2.3, now the cut will have a dimension 2.3 − 1 = 1.3, and its aspect could be that of Fig. 7.6c. If this section is now cut with a line, as in Fig. 7.6b, the intersection points will generate a pattern with dimension 1.3 − 1 = 0.3. Reducing the dimension makes it easier to measure it. As an example, we could take the “cloud” of points shown in Fig. 7.7. In this case, the plane is divided into elementary squares each of size  and the number of squares N () necessary to cover all the points is counted. Now the dimension of the square size is decreased and the count is repeated so that a plot can be made of the number N () as a function of 1/. The slope of the resulting line will be defined as fractal dimension dc (c is for “capacity”), that is, dc = lim

→0

N () 1/

(7.59)

If the attractor has been obtained from Poincarè sections, as those of the pendulum, its fractal dimension must be increased by one unit because we reduced one dimension at the beginning. There are several other methods to determine the fractal dimension, and in the specific case of the pendulum, it can be found to be about 2.3. In threedimensional phase space, the volume of the attractor is zero because it can be shown that trajectories never intersect so they actually constitute surfaces that do not fill out all the space. If the dimension of the attractor is not an integer, we are in the presence of a strange attractor. This is a rather limiting definition because the most important characteristic is that the strange attractors, to repeat Lorenz’s definition, are “an infinite number of curves, surfaces or higher dimension manifolds—generalization of surfaces to multidimensional space—often occurring in parallel set, with a gap between two members of the set”.

7.5 Chaos in a Fluid: The Loop Oscillator The first physical example of chaos in a fluid is the study of the so-called loop oscillator. This device is a torus and it is somewhat inspired by a previous work on convection in a vertical torus as in Fig. 7.8 that was published by Welander in 1967,

7.5 Chaos in a Fluid: The Loop Oscillator

223

Fig. 7.8 The configuration of the loop oscillator

Evaporation Heating

r

R

θ

Precipitation Cooling

that is, 4 years after the seminal work by Lorenz. The results we will illustrate here were published by Dewar and Haung in 1995. Suppose to have a torus of circular section (Fig. 7.8), where the difference between evaporation (q) and precipitation ( p) is given by p − q = E cos θ. (7.60) We use a cylindrical coordinate system with tangential component denoted by θ , so we write the continuity equation as divergence equal to freshwater flux: E cos θ ∂u = 2R . ∂θ r

(7.61)

This equation can be integrated to give u = 2E R sin θ/r + C =⇒ ω = 2E sin θ/r + C

(7.62)

where the angular velocity is ω = u/R so that its average value 1 = 2π



2π

ωdθ = C

(7.63)

0

ω =  + 2E sin θ/r

(7.64)

and we write the continuity equation for the salinity (S) in the same coordinate system ∂ K ∂2S ∂S + (ωS) + 2 2 = 0 ∂t ∂θ R ∂θ

(7.65)

224

7 Non-linearities, Randomness and Chaos

with K diffusivity for the salt. We write now the momentum balance for a section of the tube  ∂P ∂u ∂u =− ρ0 +u − ρg sin θ − ρu (7.66) ∂t ∂l ∂l where ρ0 is the basic density, l the coordinate along the tube, P the pressure. The second term on the right is the component of the gravity and the last term is just the viscous contribution with coefficient . Again integrating the equation along the entire loop we obtain g ∂ = − − ∂t 2π



2π 0

ρ sin θ dθ ρ0

(7.67)

using the dependence of density on salinity ρ = ρ0 (1 + β S) we get ∂ gβ = − − .S sin θ  ∂t R

(7.68)

Now we proceed to make non-dimensional equations (7.65) and (7.67) by putting ¯ S = S/ S,

t = t/T,

ω = ωT

(7.69)

where S¯ is the average salinity and T can be determined from the non-dimensional (7.67) ∂ gβT 2 ¯ = −T  − (7.70) SS sin θ  ∂t R Imposing the coefficient of the second term on the right to be unity we obtain T =  ¯ Dropping the primes, we can write R/gβ S. ∂ = −α − S sin θ  ∂t

(7.71)

where α = T . Similarly, after normalization equation (7.65) becomes ∂S ∂2S ∂ =κ 2 − [( + λ sin θ )S] ∂t ∂θ ∂θ

(7.72)

where λ = 2E T /r and κ = K T /R 2 . The constants we have introduced have an important physical meaning. The fractional density change due to the salinity is β S¯ that give rise to an acceleration ≈ gρ/ρ so that the time it takes to cross a distance R is just T . κ is the ratio between T and the diffusion time R 2 /K while λ is the ratio between T and the filling time r/2E. To find a solution to (7.71–7.72), we assume the salinity written in the form of a highly truncated Fourier series S = 1 + 2a1 (t) sin θ + 2b1 (t) cos θ

(7.73)

7.5 Chaos in a Fluid: The Loop Oscillator

225

And once substituted in Eqs. (7.71)and (7.72) it gives the system ∂ = − α − a1 ∂t ∂a1 = − b1 − κa1 ∂t ∂b1 λ = − a1 − κb1 − ∂t 2

(7.74)

We see this to be a non-linear system considering the terms like b1 or a1 . The system can be made similar to the Lorenz system by putting X = , And (7.74) becomes

Y =−

a1 , α

Z=

b1 λ + α 2κα

∂X =α(Y − X ) ∂t ∂Y =X (r0 − Z ) − κY ∂t ∂Z =X Y − κ Z ∂t

(7.75)

(7.76)

where r0 = λ/2ακ. We will see next that this is not exactly the Lorenz system unless we use κ = 1 and in the Lorenz equations b = 1. Before comparing this system with the quite famous Lorenz system, we want to discuss the stability of the solutions. A trivial solution is X = Y = Z = 0 while for the steady-state solution we need to solve the system α(Y − X ) = 0,

X (r0 − Z ) − κY = 0,

XY − κ Z = 0

(7.77)

with the possible solutions  X = Y = ± κ(r0 − κ) = ±0

Z = r − κ = 20 /κ

(7.78)

Summarizing, we have three different solutions 0, 0 , −0 ,

0,

0 0 ,

20 /κ

−0 ,

20 /κ

(7.79)

The two non-trivial solutions indicate a clockwise or anti-clockwise circulation. In terms of the physical constant, we have for X, Y, Z (see (7.75))

226

7 Non-linearities, Randomness and Chaos

 X = ±0 = ± κ(r0 − κ),

Y =−

a1 , α

Z=

b1 λ + a 2ακ

(7.80)

Based on the expression for the salinity (7.73), we also have the steady-state salinity as a function of θ S = 1 − 2α(0 sin θ + κ cos θ ) (7.81) Now we can study the stability of such solutions following the standard procedure. We start with a perturbation of the (7.76) system that reads ⎡

⎤ −α − σ α 0 A(σ ) = ⎣ r0 − Z e −κ − σ −X e ⎦ Ye Xe κ − σ

(7.82)

where X e , Ye , Z e are the equilibrium values defined by (7.80). This gives rise to a cubic equation for σ σ 3 + σ 2 (α − 2κ) + σ (ακ + κ 2 + 20 ) + 2α20 = 0

(7.83)

We assume a solution that contains a real root (a) and two complex conjugate roots (σ ± ib. The real part c may be negative and results in stable solution for a given sets of parameters. These can be changed and c becomes progressively zero and then positive and so producing instability. The point where c crosses zero corresponds to a Hopf bifurcation and in this case our solutions have the form σ1 = a,

σ2,3 = ±ib

(7.84)

Using these solutions in (7.63), we arrive at a = −(α + 2κ),

b2 = ακ + κ 2 + 20 ,

ab2 = −2α0

(7.85)

And we can determine 20 as 20 =

κ(α + κ)(α + 2κ α − 2κ

(7.86)

And using the convective solution (7.78), we obtain a condition for r r0 =

α(α + 4κ) α − 2κ

(7.87)

From the definition of λ, we have a limiting value λc =

2α 2 κ(α + 4κ) α − 2κ

(7.88)

7.5 Chaos in a Fluid: The Loop Oscillator

227 0.1

0.2 0.15

0.05

0.1

0

0

αy

b1

0.05

-0.05

-0.05

-0.1 -0.15

-0.1

-0.2 -0.25 0

500

1000

1500

2000

Time

2500

3000

-0.15 -0.8

-0.6

-0.4

-0.2

x

0

0.2

0.4

0.6

Fig. 7.9 The behaviour of the coefficient b1 as a function of time (left panel) and the phase space diagram x − aY

Remember that λ is the freshwater flux so that (7.88) is saying that given α and κ there is a limit to the freshwater flux λc beyond which the steady convection will be unstable. Also for α 2κ the steady solution is unconditionally stable. It interesting to note that b is actually the frequency of the limit cycle near the bifurcation so that the period is given by 2π 2π =√ (7.89) TH = b k(r0 + α) The advective time scale is the time it takes for the water to flow around the loop and it is given as 2π TS = √ (7.90) k(r0 − κ) This shows that the period for the limit cycle is shorter than the advective timescale. A problem with this model is that the freshwater flux may become negative which is unrealistic. Figure 7.9 shows some chaotic behaviour of the system that is quite similar to the Lorenz system (see next paragraph). The loop oscillator has been studied as a maximum simplification for the thermohaline circulation of the ocean.

7.6 The Lorenz System We have given the basis of the Benard convection in the Appendix “The Bénard convection”. We discussed there the two equations for the streamfunction ψ and the temperature θ . Now we want to be a little more formal, so write the two equations for the stream function and the temperature

228

7 Non-linearities, Randomness and Chaos



∂ 2 ∂θ ∇ ψ =∇ 4 ψ + ∂t ∂x ∂ψ ∂θ + v · ∇θ = Ra +∇ 2 θ ∂t ∂x 1 Pr

(7.91)

At this point, we take into account the possible non-linear character of the equations and write ψ(x, z, t) =ψ1 (t) cos π z sin q x (7.92) θ (x, z, t) =θ1 (t) cos π z cos q x + θ2 (t) sin(2π z) These two quantities still satisfy the boundary conditions θ = 0 for z = ±1/2 and w = 0 for z = ±1/2. We substitute for ψ and θ in the first of (7.86) to get 1 ∂ψ1 qθ1 − (π 2 + q 2 )ψ1 = 2 Pr ∂t π + q2

(7.93)

The procedure for θ is a little more complex. First we write the advection term for the temperature ∂ψ ∂θ ∂ψ ∂θ ∂θ ∂ψ + − = Ra + ∇2θ (7.94) ∂t ∂ x ∂z ∂z ∂ x ∂x where we have ∂ψ ∂θ ∂ψ ∂θ − = ∂ x ∂z ∂z ∂ x   θ1 πqψ1 − sin (2π z) + θ2 (cos (q x) cos (π z) + cos (3π z)) 2

(7.95)

Then we will equate the coefficients of sin(2π z) and cos(qz) cos(π z) and neglect the cos(3π z) term so we get θ˙1 = −πqψ1 θ2 + q Ra ψ1 − (π 2 + q 2 )θ1 θ˙2 = 21 πqψ1 θ1 − 4π 2 θ2

(7.96)

Equations (7.93) and (7.96) constitute a system of three equations and again it is convenient to change the variables t = (π 2 + q 2 )t, Y =

πq 2 θ1 , 21/2 (π 2 + q 2 )3

X= Z=

πq ψ1 + q 2)

21/2 (π 2

πq 2 θ2 21/2 (π 2 + q 2 )3

(7.97)

7.6 The Lorenz System

229

Also we define the constants r=

(π 2

q2 Ra , + q 2 )3

b=

4π 2 + q2

π2

(7.98)

So that we have the Lorenz system ∂X =Pr (Y − X ) ∂t ∂Y =− XZ +rX −Y ∂t ∂Z =X Y − bZ ∂t

(7.99)

It is quite similar to the one we found for the loop oscillator. It is interesting to discuss now some properties of the system (7.94) with respect to the volume in phase space. To this end, we can take a detour to the damped pendulum which obeys the system θ˙ = ω ω˙ = −γ ω − ω02 θ

(7.100)

where ω02 = g/l and γ the damping coefficient. The energy of the system is the kinetic ml 2 θ˙ 2 /2 plus the potential mlg(1 − cos θ ) ≈ mlgθ 2 /2 for small oscillation. The total energy is then E(θ, θ˙ ) =

1 2 2 ml (θ˙ + ω02 θ 2 ) 2

(7.101)

The change in time of the energy 1 dE ˙ = ml 2 (2θ˙ θ¨ + 2ω02 θθ) dt 2 Substituting for θ¨ from (7.100) and simplifying we get dE = −γ ml 2 θ˙ 2 dt

(7.102)

This show that energy is conserved in the absence of damping (γ = 0) and decreases in time with (γ > 0). This conclusion can be extended to the volume of phase space. Consider a volume δV = δ X δY δ Z that changes in time. The derivative can be expressed as d d d d δV = δY δ Z δ X + δ X δ Z δY + δ X δY δ Z dt dt dt dt

(7.103)

230

7 Non-linearities, Randomness and Chaos

Equivalent to d δV = δY δ Z δ X˙ + δ X δ Z δ Y˙ + δ X δY δ Z˙ dt

(7.104)

Dividing all for δV and going to the limit δ → 0, we have 1 dV ∂ X˙ ∂ Y˙ ∂ Z˙ = + + =∇ ·F V dt ∂X ∂Y ∂Z

(7.105)

where F is the vector of components X˙ , Y˙ , Z˙ . For the pendulum case, we have X = θ, Y = θ˙ so that the divergence ∂ X˙ ∂ Y˙ + =0−γ 1. In the same region, we have the stationary points that depend on r > 1. For the stability near the origin, we have three real eigenvalues for the stability analysis ω1,2 =

1 2

   −(σ + 1) ± (σ − 1)2 + 4σ r ) ,

ω3 = −b

(7.109)

ω2 and ω3 are negative while ω1 is positive. We have −ω2 > ω1 > −ω3 providing that r > 1 + b(σ + 1 + b)/σ . For σ = 10 and b = 8/3 this gives r > 4.644. For the other stationary point, the eigenvalues λ are found by solving the equation

7.6 The Lorenz System

231 unstable limit cycle X

40

r=13.234 30

r=24.74

Z

r=1 20

10

10

0 20

-10 10

0

Y

X -10

-20

Fig. 7.10 Structure of the Lorenz attractor (left) and the bifurcation diagram (right)

λ3 + λ2 (σ + b + 1) + λb(σ + r ) + 2σ b(r − 1)

(7.110)

We assume λ = iω with ω real. When inserted in Eq. (7.110) we obtain for ω ω2 = (r + σ )b =

2bσ (r − 1) σ +b+1

(7.111)

Equating real and imaginary parts from (7.111) we can get r and ω r=

σ (b + σ + 3 , σ −b−1

ω2 = 2σ b

σ −1 σ −b−1

(7.112)

The result is that ω is real if σ > b + 1 and that if r < σ (b + σ + 3)/(σ − b − 1) all the three roots have negative real part. For σ = 10 and b = 8/3 this gives r < 24.74. This means that the stationary points are stable for 1 < r < 24.74 Fig. 7.10 shows on the left the three-dimensional Lorenz attractor with its impressive resemblance to a butterfly. The right figure shows the structure of the bifurcations.

7.7 ENSO: A Very Complex System A very complete application of nonlinearity and chaos is given by the study of ENSO, (El Nino Southern Oscillation). El Nino was a well-known phenomenon to the ancient people living along the Pacific coast of South America. They noticed that the fishing along the coast was not abundant when the sea was unusually warm. If the fishes were scarce, then this would influence the birds that fed on the fishes and deposited the guano at the interior. The warmer sea means also more rain on the

232

7 Non-linearities, Randomness and Chaos El Nino Conditions

Normal Conditions

Convective loop

Increased convection

warm

warm

cold

thermocline thermocline cold water

120°W

80°E

120°W

80°E

Fig. 7.11 The El Nino conditions (left) and the normal conditions (right) for ENSO

coast. At that time people noticed that such warming took place around Christmas time so they call the thing “El Nino” (Holy Child). To explain El Nino, we have to wait the year 1966 when Jacob Bjerknes proposed a theory for the occurrence of El Nino. During normal atmospheric conditions, the trade winds blow over the tropical Pacific Ocean creating an easterly (or westward current) on the sea surface that recall cold, deep water from the South American coast establishing a cold tongue of water that extends towards the Pacific Ocean (see Fig. 7.11). This deep water is rich in nutrients so that fishes can thrive on them maintaining a food chain that was essential to the economy of those regions. The Western Pacific water maintains a warmer temperature and forced by the winds accumulates creating a difference of about 40 cm with the eastern part. The temperature gradient produces an atmospheric circulation (known as Walker circulation) with low pressure over the west and high pressure on the east. In the sea, the surface layer known as thermocline presents a thickening on the western side and a thinning on the eastern side. Occasionally, the trade winds weaken and the first consequence is the appearance of a warm pool of water on the eastern side: it is the beginning of El Nino. As we have seen, there is a complex interaction between the ocean and the atmosphere so the entire stuff has been renamed as El Nino Southern Oscillation (ENSO). As a matter of fact, the cold phase (also called La Nina) and the warm phase alternate with periods going from 4 to 7 years and one of the problems is how to explain this loose periodicity. We will concentrate here only on one of the non-linear aspects of ENSO to show how the stochastic forcing by the winds may have an important role. We will use a simple model proposed in the late 1990s by Fei-Feu Jin. He divided the equatorial Pacific into an eastern half (cold tongue) and western half (warm pool). The temperature of the well-mixed surface layer Te of the eastern basin is written as

7.7 ENSO: A Very Complex System

233

dTe M(w)(Te − Tse ) = −T (Te − Tt ) − dt Hm

(7.113)

The first term of this equation can be easily explained. It is simply a relaxation term proportional to the difference between the temperature of the mixed layer and the equilibrium temperature Tt and T is a damping constant. The second term is more tricky and the reader must remember when we talked about Ekman’s pumping that can be expressed as (the continuity equation) ∂v ∂u w + = Hm ∂x ∂y

(7.114)

The dominant term is ∂u/∂ x and this can be assumed to be dependent on the stress τ of the wind so that finally the vertical velocity is w = −ατ = −α[τ0 − μ(Te − Tt )]

(7.115)

The reason for the two terms is that the first represents the stress due to the Hadley circulation (most properly the trade winds) while the second is due to the Walker circulation which ascends over the warm pool in the western Pacific and descends over the eastern Pacific (see Fig. 7.11). We identify M(w) with w in such a way that  M(w) =

0, if w< 0 w, if w> 0

So actually the second term in (7.103) is the temperature advection due to the upwelling of subsurface water at temperature Tse . Such temperature depends strongly on the thermocline depth h e . Based on the available data such temperature can be parameterized as   Tse = Tt − 0.5(Tt − Tr 0 )1 − tanh (H + h e − z 0 )/ h ∗ 

(7.116)

where Tr 0 is the temperature beneath the thermocline, h e is the departure of the thermocline depth from its reference depth, H , z 0 is the depth where w takes its characteristic value and h ∗ measures the sharpness of the thermocline. As we mentioned before, the thermocline depth during the El Nino condition is larger to the west than to the east and the relation among the two can be written as h e = h w + bLτ

(7.117)

where τ is given by (7.115), b measures the efficiency of the stress to drive the thermocline tilt and L is the basin size. The thermocline depth h w adjusts slowly to the zonally integrated Sverdrup meridional mass transport due to the equatorial Rossby waves and can be described by the equation

234

7 Non-linearities, Randomness and Chaos

dh w r bLτ = −r h w + dt 2

(7.118)

Again the first term is the relaxation term with rate r while the second term is the total Sverdrup meridional mass transport. Equations (7.110–7.118) constitute a system that can be solved with a computer program using the following constants: Tt = 30 ◦ C, Tr 0 = 18 ◦ C, H = 100 m, z 0 = 75 m, h ∗ = 50 m, μα/Hm = 0.0025 ◦ C/day, μbL = 12.5m/◦ C, τ0 /μ = −1 ◦ C, r = 1/300 days and T = 1/150 days. We can find steady state by putting ∂h w /∂t = ∂ Te /∂t = 0 and obtain for the Pacific basin an equilibrium temperature of about 24.5 ◦ C. However, such a state is not stable ad this can be verified by writing the linearized equations for (7.113) and (7.118) that reads dTe = RTe + γ h w dt (7.119) dh w r bLμTe = −r h w − dt 2 where

M(w) ¯  + γ bLμ − 1+ Hw τ0 =− μ(Tt − Te ) M(w) ¯ ∂ T¯w γ = Hm ∂ Te R = −T

(7.120)

We can carry out the stability analysis on the system (7.109) using Tw = T eσ t and h w = heσ t obtaining for the growth rate σ σ = 21 (R − r ) ±

1 2



(R − r )2 + 4r (R − 0.5μbL)

That shows that the system is unstable for R > r while it is stable for R < r . The frequency for the neutral case r = R is given by ω=



0.5μbLr − r 2

The instability translates into an periodic solution as shown in Fig. 7.12, where we notice that both temperature Tw and h w have period of roughly 4 years and are out of phase by roughly 1 year. The same figure shows that the rise of El Nino (warm) phase produces the thinning of the western thermocline depth or, reads in another way, the deepening of the western thermocline leads to the warm phase in the East Pacific. Now what about the stochastic forcing. The role of the surface stress is quite clear now, a decrease in the intensity of the trade winds produces a warm pool in the Eastern Pacific. However the oscillation produced by this model is too regular and do not compare well with the data. A more realistic behaviour can be obtained if a stochastic forcing is added to the zonal component of the stress τ0 . The results are

Appendix

235

30

60 Te

52

hw

26

44

24

36

22

28

20 0

4

8

12

Time (years)

16

20 20

hw(m)

Te (°C)

28

0

4

8

12

16

20

Time (years)

Fig. 7.12 The behaviour of the eastern temperature Te and depth of the thermocline h w (left). On the right the same quantities when a stochastic forcing is added

shown in Fig. 7.12b, where a white noise forcing has been added with an amplitude of about 40% of the steady signal. In this run the relaxation constant r has been halved. At this point, we would like to mention an elaboration of the above model known as the delayed oscillator model of ENSO. The idea is that the initial warming in the Eastern Pacific produces a weakening of the easterly wind in the Central Pacific. The change in the winds excites a downwelling Kelvin wave that propagates to the East (look at the appropriate chapter) and Rossby waves that move in the opposite direction. These are reflected by the western boundary and produces upwelling Kelvin waves that move eastward to counter the downwelling Kelvin waves. The authors of this heuristic models are Eli Tziperman and coworkers that set up a simple equation for the eastern thermocline depth h(t) h(t) =a A{h [t − (L/(2Ck ))]} − b A{h [t − (L/Ck + L/(2C R ))]} + c cos(ωa )

(7.121)

where t is time, L is the basin length and ωa is the seasonal forcing frequency. C K and C R are Kelvin and Rossby wave velocities, respectively. L/Ck is the time it takes for the Kelvin wave to reach the eastern boundary from the middle of the basin while the second term is due to the westward travelling wave excited by the wind at time t − [L/Ck + L/(2C R )] and reflected as a Kelvin wave. A(h) relates the wind stress to the sea surface temperature (SST) and thermocline depth. The slope of A(h) is fixed by a parameter κ that is a measure of the strength of the coupling between ocean and atmosphere. We made a run for the model to show that by increasing κ there is a classical route to chaos as shown in Fig. 7.13. This simply shows the effect of non-linearity in a more complex model.

236

7 Non-linearities, Randomness and Chaos

0.4

0.6 0.4 0.2 0 -0.2 -0.4 -0.6 -0.8 -1 -1.2

0.3 0.2 0.1 0 -0.1 -0.2 -0.3 -0.4

1 0.5 0 -0.5 -1 -1.5

TIME

Fig. 7.13 The transition to chaos for the Tziperman model by changing the coupling between ocean and atmosphere. The quantity shown is the thermocline depth

Appendix Burgers Equation: The Travelling Wave Solution We consider an advection–diffusion equation of the form ∂u ∂ 2u ∂u +u =ν 2 ∂t ∂x ∂x

(7.122)

u(x, t) = f (ξ ) = f (x − vt)

(7.123)

We seek a solution of the form

We have

∂ f ∂ξ ∂u = = −v f ∂t ∂ξ ∂t

And in the same fashion ∂u = f (ξ ) ∂x

∂ 2u = f "(ξ ) ∂x2

Equation (7.122) becomes ν f (ξ ) + f (ξ )[v − f (ξ )] = 0

(7.124)

We notice now that f f = 0.5(∂/∂ξ ) f 2 so that (7.124) can be directly integrated to give 1 (7.125) − v f + f 2 − νf = B 2 It can be written as 2ν

∂f = ( f − f 1 )( f − f 2 ) ∂ξ

(7.126)

Appendix

237

where f1 = v +



v2 + B,

f2 = v −



v2 + B

(7.127)

Because f s are real we have f 1 > f 2 and Eq. (7.126) can be integrated by separating the variables to get f 1 + f 2 exp{[ f 1 − f 2 ]ξ/2ν} f (ξ ) = (7.128) 1 + exp{[ f 1 − f 2 ]ξ/2ν} After some algebraic labouring, we can write in a different form   1 1 ( f 1 − f 2 )(x − vt) f (ξ ) = u(x, t) = v − ( f 1 − f 2 ) tanh 2 4ν

(7.129)

where from (7.127) v = 0.5( f 1 + f 2 ). It is to notice that (7.129) does not depend on time in the reference frame moving with velocity v. From the definition of tanh, we see that f (ξ ) → f 1 for ξ → −∞ while f (ξ ) → f 2 for ξ → ∞. This solution is shown in Fig. 7.1 and goes smoothly from f 1 to f 2 . As a matter of fact, the shape of this solution depends on the value of the viscosity ν becoming sharper as ν decreases as shown again in Fig. 7.1. This shape resembles that of a shock wave. We can have a more precise idea if we multiply both numerator and denominator in (7.118) by exp[−( f 1 − f 2 )ξ/2ν]. We have f (ξ ) =

f 2 + f 1 exp[−( f 1 − f 2 )ξ/2ν] 1 + exp[−( f 1 − f 2 )ξ/2ν]

(7.130)

And if we put f 1 = 2 and f 1 = 0 we get

−1 f (ξ ) = 2e−ξ/ν 1 + e−ξ/ν

(7.131)

This shows the existence of a layer of thickness δ ≈ ν/( f 1 − f 2 ) that tends to zero when ν → 0

Chaotic Pendulum We will consider a damped and forced linear pendulum which obeys the equation of motion d 2θ dθ + mgl sin θ = l F(t) (7.132) ml 2 2 + bl 2 dt dt where l is the length of the pendulum, m its mass, b some damping coefficient and F(t) the forcing. This equation can be made non-dimensional with the substitution dt =

 l/gdτ ⇒ dt 2 = (l/g)dτ 2

(7.133)

238

7 Non-linearities, Randomness and Chaos

Equation (7.132) becomes replacing dτ with dt b dθ d 2θ g F0 + + sin θ = cos ωt 2 dt m dt l ml

(7.134)

where F0 cos ωt is the periodic forcing. We simplify further the equation with the substitutions g F0 b 2β = , ω02 = , γ ω02 = m l ml So that (7.134) becomes d 2θ dθ + ω02 sin θ = γ ω02 cos ωt + 2β dt 2 dt

(7.135)

From this equation, we obtain the system θ˙ =y y˙ =γ ω02 cos ωt − 2βy − ω02 sin θ +

(7.136)

z˙ =ω Before discussing the solution to this system, we can explore a little non-linearity of this pendulum. We use the most simple approximation to sin θ , that is, 1 sin θ ≈ θ − θ 3 6 When substituted in (7.125) it gives  1 3 dθ d 2θ 2 + ω0 θ − θ = γ ω02 cos ωt + 2β dt 2 dt 6

(7.137)

It can be easily found that a solution of θ ∝ exp(iωt) will give rise to a term containing cos 3ω. This shows that as the forcing increase there are new harmonics nω arising and to show this we have to recur to a numerical solution for Eq. (7.135). This has been done using the following parameter ω = 2π , ω0 = 1.5ω and β = ω0 /4. The forcing is raised by changing γ . We see that for small γ like 0.9 we have an apparent oscillatory behaviour as shown in Fig. 7.14 around zero. When forcing is increased beyond 1 we observe a quite different behaviour like that shown in Fig. 7.14. The first case is the result for γ = 1.07 and shows that after an initial adjustment there is almost pure harmonic. A more accurate analysis shows that actually there are two harmonics sign of a first bifurcation. Raising more the forcing to 1.078 we have a regular behaviour but now the harmonics have become 4. This behaviour can be illustrated in the bifurcation diagram shown in Fig. 7.15. We notice the first and second bifurcation and then for same values of the forcing the chaotic behaviour

Appendix

239

γ = 1.06 0.4

20

0.3

15

0.2

10

0.1

5

0 0

-0.1

-5

-0.2

-10

-0.3

-15

-0.4

-20 0

5

10

15

20

25

0

30

5

10

15

20

25

30

γ = 1.073 2 .5

25

2

20

1 .5

15 10

1

θ

5

0 .5

θ

0

0 -5

-0. 5

-10

-1

-15

-1. 5

-20

-2

-25

γ =1.09 2.5

20 15

2

10 1. 5

5

1

0 -5

0. 5

-10 0

-15

-0.5

-20

TIME

d θ /dt

Fig. 7.14 The deviation angle as a function of the quantity γ (left column) and the phase space diagrams on the right column

shows up. In the phase space for small forcing we have a limit cycle as shown in Fig. 7.14 top. Figure 7.14 while the double frequency is evident for γ = 1.073 (Fig. 7.14 middle) and a chaotic behaviour is present for γ = 1.09. The non-linear pendulum is particularly useful to illustrate the Poincaré sections. This is actually a three-dimensional concept so we have to invent a third coordinate z as specified in the system (7.136). We then will study the intersection of the phase

240

7 Non-linearities, Randomness and Chaos

Fig. 7.15 The bifurcation diagram for the non-linear pendulum

space trajectories with the planes z = nωT where T is the period. The result is shown in Fig. 7.5.

The Lorenz–Malkus Water Wheel Lorenz’s seminal paper on deterministic chaos was published in 1963. In 1972, Willem Malkus found a mechanistic analogue of the Lorenz system in a rather strange wheel as shown in Fig. 7.16. The following treatment is taken from Matson’s paper. Consider a wheel on which cups are hanged on its perimeter and are filled with a flux of water. Each cup is allowed to have a leak but not to have overflow. The friction torque is proportional to the angular velocity of the wheel and the leakage is proportional to the mass of the single cup. The wheel is initially at rest and when the cup starts filling up the wheel will move in one of the possible directions. If λ is the leakage constant of the cup and Q is the flux of water the change in the total mass M = K Mk (with M K mass of the single cup) can be written as dM = Q − λM dt

(7.138)

Appendix

241

water input at y = 0, z = R

z

center of mass of water

ω (y,z)

ith cup at (yi,zi)

y R

Fig. 7.16 The water wheel on the left and the coordinate system on the right

With the solution

M(t) = (Q/λ)(1 − eλt )

(7.139)

In the following treatment, it is assumed that the filling operation has already been finished so that the total mass q/λ = M stays constant. For the coordinate system shown in Fig. 7.16b, the centre of mass of all the water in the cups is given by y=

 (Mk yk )  , Mk

z=

 (Mk z k )  Mk

(7.140)

Now we establish a time t0 such that for time t > t0 we consider the total mass of water made of two components one containing the water that entered prior t0 and the other with water added following t0 . We consider the motion of the two mass of water Ma (t) and Mb (t) and rewrite the coordinate of the centre of mass as in (7.140) y=

(Ma ya + Mb yb ) , Ma + M b

z=

(Ma z a + Mb z b ) Ma + M b

The derivatives of such coordinates  1 Ma Mb dya dyb dy = Ma + ya + Mb + yb dt M dt dt dt dt  1 Ma Mb dz a dz b dz = Ma + z a + Mb + zb dt M dt dt dt dt

(7.141)

(7.142)

For t > t0 , the Ma cups do not receive water so that each cup leaks at the rate λM and their leakage does not affect the location of their centre of mass. Referring again

242

7 Non-linearities, Randomness and Chaos

to Fig. 7.16b we define r=



ya2 + z a2 ,

θ = arctan(ya /z a )

(7.143)

We may then write dya = v cos θ = ωr cos θ = ωz a dt dz a = −v sin θ = −ωya dt

(7.144)

Knowing that z a = r cos θ and ya = r sin θ , v = ωr and ω = dθ/dt. For the mass Ma at t = t0 , we have (Ma = M),

d Ma dya dz a = −λM , (ya = y), = ωz , (z a = z), = −ωy dt dt dt

(7.145) For Mb at t = t0 , its cups are empty but the top cup y = 0, z = R is beginning to fill so we have d Mb = λM (7.146) Mb = 0, yb = 0, z b = R, dt We now substitute (7.145) and (7.146) in (7.142) to obtain dy = ωz − λy dt dz = ωy + λ(R − z) dt

(7.147)

Actually, the angular velocity of the wheel is changing because the torques on the wheel are changing. There are three different torques acting on the wheel, Tg due to gravity acting on the water centre mass, Tμ from axle friction and Tw due to bringing the incoming water the stream into the cups I

dω = Tg + Tμ + Tw dt

(7.148)

The torque due to gravity and the one due to friction are very simple to write being Tg = Mgy,

Tμ = −αω

(7.149)

The remaining torque is more complicated. The force on the bucket is given by the amount of water deposited in the unit time (λM) multiplied by the change in the velocity of the same amount of water f or ce = λM × (bucket speed − water speed) = λM(ω R − 0)

(7.150)

Appendix

243

So that the torque Tw = −R × R = −ωλR 2 = −ωλI

(7.151)

where we have made the substitution Iw = M R 2 for the moment of inertia. Substituting the torques in Eq. (7.148), we have Mg dω = y− dt l



α Iw +λ l I

ω

(7.152)

where I is the total moment of inertia I = I0 + Iw with I0 moment of inertia of the wheel. We now define  α Iw Mg , f = +λ (7.153) a= l l I So that the differential equations for the wheel are dω = ay − f ω dt dy = ωz − λy dt dz = −ωy + λ(R − z) dt

(7.154)

These equations can be made dimensionless if we chose two quantities for normalization D for length and τ for time. In this case the primed quantity is dimensionless t = t/τ,

ω = ωτ,

y = y/D,

z = z/D

(7.155)

λ = λτ,

R = R/D

(7.156)

While for the parameters, we have a = a Dτ 2 ,

f = τ f,

An obvious choice is to put R = 1 so that τ = (Ra)−1/2 . The dimensionless equations now become dω = y − fω dt dy (7.157) = ωz − λy dt dz = −ωy + λ(1 − z) dt where all the primes have been dropped. Notice that these equations are simpler than the Lorenz system because they have a constant term in the second equation.

244

7 Non-linearities, Randomness and Chaos

Some Analysis The stationary points of the water wheel are found by setting to zero equations (7.157) y − f ω = 0,

ωz − λy = 0,

λ(1 − z) = ωy

(7.158)

With the solutions z = 1,

y = 0,

And the last two z = λ f,

ω=0

(7.159)

 y = ± z(1 − z)

(7.160)

The first stationary point that we will call C0 correspond to a stopped wheel with all the water in the cups at the top. The other two solutions must have λ f < 0 and this corresponds to a solution with negative y and ω we will call C1 or to a solution with y and ω negative that we will call C2 . When λ f > 0 the wheel comes to a stop whatever be the initials values of (ω, f, z). We can study the stability of solution (7.159) by perturbing Eqs. (7.157) around the solution or finding the eigenvalues of the jacobian that in this case is quite simple ⎡

⎤ −f 1 0 J = ⎣ 0 −λ ω ⎦ 0 −ω −λ

(7.161)

The eigenvalues are obtained by solving the determinant 1σ − J that gives σ 2 + σ (λ + f ) + f λ − 1 = 0,

   ⇒ σ = 0.5 ∗ −(λ + f ) ± (λ + f )2 − 4(λ f − 1)

(7.162) This shows that when λ f < 1 one of the solutions is unstable. With the same method a solution could be found for λ = 0.14 and f = 0.5 with the values ω = 0.51, y = 0.2551 and z = 0.07 which produces all stable solutions. In the work by Matson, there is an ample discussion on the bifurcations that we report only for the conclusions. The first case shown in Fig. 7.17 refers to what is called a preturbulence stage when an initial apparent chaos evolves to a situation of steady rotation. The second case reported in Fig. 7.17 refers to a possible bifurcation that should happen around λ = 0.1536 and f = 0.2. Another possible bifurcation is shown in Fig. 7.17 for λ = 0.1122 and 0.1123 with the chaotic behaviour reported in the same figure (bottom). Here we have some contradiction between the procedure adopted by Matson and a general procedure to find the bifurcation points. For example, Strogatz suggests to solve the jacobian for the eigenvalues. For the water wheel the equation to solve is then (7.163) (σ + f )(σ + λ)2 − yω − z(σ + λ) + ω2 (σ + f ) = 0

Appendix

245

1.5

f = 0.5

λ = 0.14

ω

z 0

-1 0

100

200

300

y

time 1.8

f = 0.2

λ = 0.1537

1

ω

z 0 -1 λ=0.1535

-1.8 0

10

20

30

40

50

time f = 0.2

y

λ = 0.1522

1 0 -1 f=0.2 -2 0

40

80

120

λ=0.1523 160

200

time 1.5

f = 0.4

λ = 0.1122

ω

0

100

200

300

400

500

time

Fig. 7.17 The behaviour of ω as a function of time for different parameters (left column) and the relative phase space diagrams (right column)

246

7 Non-linearities, Randomness and Chaos

Then we substitute the values for y, ω at the equilibrium points to get z = λ f,

yω =

y2 = λ(1 − λ f ), f

ω2 =

λ(1 − λ f f

We obtain the equation σ 3 + σ 2 (2λ + f ) + σ ( f λ + λ/ f ) = 0

(7.164)

It has the solution σ = 0 and 2σ = −(2λ + f ) ±



(2λ + f )2 − 4( f λ + λ/ f )

(7.165)

The quantity within square root has negative values when f = 0.2 only if 0.02 < λ < 4.99, an interval that contains the point λ = 0.1536. Finally, we would like to return to the question of the similarity between the Lorenz system and the waterwheel. Matson starts from the Lorenz equations dx = σ (y − x) dt dy = x(r − z) − y dt dz = x y − bz dt

(7.166)

With the substitutions z → r − z and x → ω that corresponds to a fluid cooled at the top these equations become dω = σ (y − ω) dt dy = ωz − y dt dz = −ωy + b(r − z) dt

(7.167)

This system is then compared with the set proper of the waterwheel dω = ay − f ω dt dy = ωz − λy dt dz = −ωy + λ(R − z) dt

(7.168)

Appendix

247

It is shown that the parameter used in the Lorenz system is related to those of the waterwheel by the relations  f =

σ , r

1 λ= √ σr

(7.169)

Stochastic Resonance The mechanism of stochastic resonance was invented by a group of Italian physicists in the early 1980s. They proposed a theory to explain the 100.000-year cycle of the ice age. This period coincides with a change in the eccentricity of the Earth’s orbit which in principle could modulate the solar radiation reaching the Earth. However, such changes are too small to have any effect and the reasoning was that when this forcing signal was summed to the noise (the weather) on some occasion the glaciation could be triggered. The mechanism is illustrated in Fig. 7.18. A sinusoidal signal is summed with the noise. When this composite signal reaches some threshold a pulse is produced. It is clear that the probability of such events is higher when the sinusoidal

Fig. 7.18 The mechanism of the stochastic resonance, the noise (grey) is superimposed to a periodic signal. When their sum is higher than the threshold a signal is produced threshold

0.15

0. 1

0.05

0

-0.05

-0.1

-0.15 0

5

10

TIME

15

248

7 Non-linearities, Randomness and Chaos

signal has a positive value so that the pulse spectrum should have the frequency of the periodic signal. From a dynamic point of view, we can think of a particle in double potential well described by the function V (x) to which we can associate a force −d V /d x. Then in presence of a viscous force the motion of the mass could be described by dV d2x dx =− −β (7.170) 2 dt dx dt When the acceleration is zero (high friction limit), by adding a stochastic forcing F(t) and a periodic forcing we have β

dx dV =− + F(t) + A sin ωt dt dx

(7.171)

We can obtain a similar equation for the most simple energy balance climate model that we can write as dT = Q(1 − α) − (A + BT ) (7.172) C dt where C is the heat capacity of the system, Q the solar radiation, α the albedo and A + BT the emitted infrared radiation. Now suppose the stationary solution corresponds to T = T0 we can write T = T0 + T that substituted in (7.172) gives dT + BT = Q dt



d H0 dT0



T

(7.173)

where H0 is a function to be determined that satisfies the relation A + BT0 = Q H0 (T0 ). To eliminate B we derive the last equation with respect to T0 and get B = H0

dQ d H0 +Q dT0 dT0

(7.174)

Now this can be used to eliminate B from the (7.173) to get dT = −H0 dt



dT0 dQ

−1

T

(7.175)

We have stability dT /dt that has the opposite sign of T if dT0 /d Q > 0, that is the equilibrium temperature must increase with increasing T0 or vice versa. We can argue based on that the stability criteria is related to some kind of potential that we may write as the function F(T0 ) = AT0 +

1 BT 2 − Q 2 0



T0 0

H0 (T0 )dT0

(7.176)

249

Fig. 7.19 The graphical solution to Eq. (7.178). The straight line represents the left-hand side divided by Q. The solutions are indicated by I, I I, I I I

FRACTION ABSOR: SOLAR RADIATION

Appendix 0.8

I 0.7

Q decrease

II

0.6

Q increase 0.5

0.4

III 0.3 -50

-40

-30

-20

-10

0

10

20

TEMPERATURE (˚C)

The evaluation of the integral depends on the form assumed by H0 (T0 ). For heuristic purposes, we consider a planet such that polar caps disappear completely if the average temperature is above 15 ◦ C s than the sine of latitude of the ice xs is 1, while xs = 0 if the temperature is less than −15◦ . In that case it can be shown that the planetary albedo is given by 1 − α p = H0 [xs (T0 )] = ai + (a f − ai )[xs + S2 (xs − xs3 )/2]

(7.177)

where α p is the planetary albedo, ai is the ice co-albedo (1 − αi ) and a f is the ice free co-albedo. S2 is an appropriate coefficient. According to Eq. (7.169), the equilibrium temperature is given by (7.178) A + BT0 = Q H0 [xs (T0 )] The graphical solution of such an equation is given in Fig. 7.19 and shows three possible solutions. However, according to the criteria (7.172) only the solutions I and I I are stable so that we may assume following (7.173) that the “climate potential” is two well potential of the form V (T ) ≈ −1/2T 2 + 4T 4 . The depth of the well is of the order of ≈5 ◦ C that corresponds to the temperature difference between the glacial and inter-glacial period. In analogy with (7.168), we write √ dT = T − T 3 + A sin ωt + σ η(t) dt

(7.179)

where η(t) is a Gaussian random variable with unit variance η(t)η(t ) = δ(t − t ), σ is the amplitude of the noise. It can be shown that in presence of the noise the transition time between two equilibrium points is given by τ L ≈ τ exp{T 2 /σ τ }

(7.180)

250

7 Non-linearities, Randomness and Chaos

0.2

0.2

0.2

0.15

0.15

0.15

0.1

0.1

0.1

0.05

0.05

0.05

0

0

0

-0.05

-0.05

-0.05

-0.1

-0.1

-0.1

-0.15

-0.15

-0.15

-0.2

0

1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

-0.2

-0.2 0

1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

0

1000 2000 3000 4000 5000 6000 7000 8000 9000 10000

Time

Fig. 7.20 Three different realizations of the solution to the stochastic equation (7.179) with the same initial conditions. They show a possible transition between the two wells of the potential at the forcing frequency

Using τ ≈ 10 years, we have τ L ≈ 50,000 years. The spontaneous transitions are too fast for the 100,000 years cycle so we need to add a forcing term A sin ωt. Figure 7.20 shows some of the results for the simulation. It is clear that such a model is too much simple to deal with very complex case such as the ice ages but has only an heuristic value. However, there is a much more complex problem. In the figure, we report just three different realizations for the stochastic equation (7.179) that were obtained with the same initial conditions. It is clear that the noise makes each solution different from the other and as a matter of fact the proper way to present results for a stochastic equation is the average or the quadratic mean. We suspect that some of the early results published could also depend on the numerical methods used in solving the equations. The only information we can get from these results is that sometimes the resonant mechanism works.

References Textbooks Bergé P, Pomeau Y, Vidal C (1987) Order within Chaos: towards a deterministic approach to turbulence. Wiley, New York Lorenz EN (1993) The essence of chaos. University of Washington Press, Seattle, WA Jacobs K (2010) Stochastic processes for physicists: understanding noisy systems. Cambridge University Press, Cambridge Palmer T, Williams P (2010) Stochastic physics and climate modeling. Cambridge University Press, Cambridge Sparrow C (1982) The Lorenz equations: bifurcations, chaos, and strange attractors. Springer, Berlin Strogatz SH (2014) Non linear dynamics and chaos: with applications to physics, biology, chemistry, and engineering. Westview Press, Boulder Taylor JR (2005) Classical mechanics. University Science Books, Sausalito Tsonis AA (1992) Chaos from theory to applications. Springer, Berlin Wax N (1954) Selected papers on noise and stochastic processes. Dover Publications, New York

References

251

Articles Benzi R (2010) Stochastic resonance: from climate to biology. Nonlinear Processes Geophys 17:431 Benzi R et al (1983) A theory of stochastic resonance in climatic change. SIAM J Appl Math 43:565 Czaja A, Robertson A, Huck T (2003) The role of coupled processes in producing NAO variability. Geophys Monogr 134:147–172 Dewar WK, Huang RX (1995) Fluid flow in loops driven by freshwater and heat fluxes. J Fluid Mech 291:153 Higham DJ (2001) An algorithmic introduction to numerical simulation of stochastic differential equations. SIAM Rev 43:525 Huang RX, Dewar WK (1996) Haline circulation: bifurcation and chaos. J Phys Oceanogr 2093 Jin F-F (1996) Tropical ocean-atmosphere interaction, the Pacific cold tongue, and the El NifnoSouthern oscillation. Science 274:76 Matson LE (2007) The Malkus-Lorenz water wheel revisited. Am J Phys 75:1114 Tziperman E et al (1994) El Nino: overlapping of resonances between the seasonal cycle and the Pacific ocean-atmosphere oscillator. Science 264:72 Welander P (1967) On the oscillatory instability of a differentially heated fluid loop. J Fluid Mech 29:17 Wiesenfeld K, Moss F (1995) Stochastic resonance and the benefits of noise: from ice ages to crayfish and SQUIDs. Nature 373:33

Chapter 8

Turbulence

In their precious book on turbulence, Hendrik Tennekes and John Lumley suggest that there are two closure problems in the study of turbulence. The first problem is fundamental for this topic and that is there are more unknowns than equations. The other is the gap between the “several dozen introductory texts in general fluid dynamics” and the advanced or professional treatment of the subject. Since the publication of the book (1970), the number of the elementary treatments may have increased and possibly this one may be the latest. Paradoxically the mentioned text is the main one that this chapter is inspired by. We are mostly interested in the application of turbulence (for example, to the instrumentation and the turbulent diffusion) so we will go rather fast (but not superficially) through the subject to describe the applications mostly in the appendix. On the other hand, turbulence is a good excuse to return and expand some of the statistical concepts we developed in the previous chapter.

8.1 Some General Matter One of the difficulties in studying turbulence is in its definitions. We know already when a fluid is not turbulent. For example, a laminar flow is a smooth and predictable flow in space and time. On the other hand in a turbulent flow, the velocity u( ¯ x, ¯ t) at some fixed point x¯ changes with time in a somehow random manner or conversely at some time t the velocity varies randomly at different points. In conclusion, we cannot study turbulence in a deterministic way and we must recur to a statistical approach. We may draw a parallelism between the Brownian motion (mentioned in the previous chapter) and the turbulent motion. In the former, the individual gas particles on average travel a distance  (the mean free path) before encountering another particle and exchanging energy and momentum. In turbulent motion, the single particles are replaced by eddies of much larger dimensions that moves for a mixing length before exchanging their properties. We may hence decompose the © Springer Nature Switzerland AG 2020 G. Visconti and P. Ruggieri, Fluid Dynamics, https://doi.org/10.1007/978-3-030-49562-6_8

253

254

8 Turbulence

velocity into a laminar component U(x) and a turbulent component u (x) so that the velocity u(x) is given by (8.1) u(x) = U(x) + u (x) It is rather obvious that the turbulent component should be obtained by subtracting an average to the velocity u = u − u. where the average is defined with respect to time or with respect to space, that is,  1 t+T udt u = T t  1 ud3 x. u = V V

(8.2)

In fully developed turbulence, according to the Taylor hypothesis, (see appendix) the two averages are equivalent. The real measurement of velocity is not continuous but rather taken at discrete instants of time so that the first of (8.2) should be written as u =

N 1  ui N 1

(8.3)

where N is the number of discrete points. Once the turbulent velocity has been isolated we can define isotropic and homogeneous turbulence when the statistical properties of u’ are, respectively, independent of direction and position. If a simple mean is performed on the turbulent component the result will be zero and we can get an idea of the strength of the turbulence by calculating the root mean square velocity urms  urms =

u (t)2 

(8.4)

So that the turbulent intensity can be defined as u r ms /u. At this point, we could start writing the turbulent fluxes starting from the simple case of a concentration of a gas C that can be written again as the sum of the average and the turbulent part C(t) = C + C  (t)

(8.5)

This concentration is subject to transport in the z direction by a turbulent vertical velocity w (8.6) w(t) = w + w (t) We have for the flux Fz = C(t)w(t) =(Cw + Cw + C  w + wc ) =Cw + w C  

(8.7)

8.1 Some General Matter

255

The contribution of the turbulence to the vertical flux is given by the term w C   that is the correlation term between concentration and velocity. The simple arguments used so far indicates that a more appropriate description of turbulence could be done using statistical concepts.

8.2 Statistical Description of Turbulence Here we will follow the classical treatment given by Sebastian Pope starting with the definition of random variable. Given a controlled experiment, if an event A inevitably occurs then such an event is certain or sure. If the same event cannot occur then it is impossible and finally if the event A may occur or may not and in this case we call it random. Turbulent flows are treated as random phenomena. Hence, one can see a contradiction between the fact that the Navier–Stokes equation is deterministic while the turbulence that is a product of that equation is actually a random process. We have seen since the early chapters that the non-linear character of the Navier–Stokes equation makes some solutions extremely sensitive to initial conditions (Lorenz stuff) and this is particularly true for high Reynolds numbers. At the moment, we will introduce some notions about random variables and the title of this paragraph may appear inappropriate. We will find in due time the applications of these concepts. Tennekes and Lumley in their book introduce the probability density function or pdf in a quite original way. Suppose to have a steady signal like the one shown in Fig. 8.1 that could be a fluctuating velocity u(t). We are interested in finding the amount of time that u(t) spends around a certain value. To this end, we imagine to have a gating circuit that generates a signal every time u(t) remains in the window of width u around that value. We may expect that the averaged output of the gating circuit is proportional to the window width u. It is then convenient to define a quantity B(u) such that 1  (t) (8.8) B(u)u = lim T →∞ T The function B(u) is called probability density and is the probability of finding u(t) between u and u + u Because B(u) is a fraction of time is always positive while from its definition it also results  ∞ B(u)du = 1 (8.9) −∞

The most common form of B(u) is Gaussian as shown in Fig. 8.1. We actually already found such a result when we discussed the white Gaussian noise in Chap. 7. As a matter of fact, we can assimilate the signal u(t) to a white noise so that the probability density function becomes a Gaussian as shown in Fig. 7.2. It is possible to evaluate the pd f for periodic waveforms like a square wave and a triangle wave as shown in Fig. 8.2. In the case of a square wave, the pd f is given by two Dirac-delta in

256

8 Turbulence

u B(u)

Δu

time (t) 1 Δt 0 time (t)

Fig. 8.1 The construction of a probability density function from a signal

Amplitude

PDF

Time

PDF

Time

Fig. 8.2 Two examples of PDF for a square wave and a triangular wave

correspondence of the only two values the function may assume. As for the triangle wave, the pd f is constant between the maximum and minimum value it spans. The pd f can be obtained from the so-called cumulative distribution function, cd f . Consider a turbulent velocity field U(x, t) and the probability P that after a measurement of finding a value of the velocity less than Va that is P(A) = P{U < Va }

(8.10)

8.2 Statistical Description of Turbulence

257

We may define the cumulative distribution function as F(x) = P{U < x}

(8.11)

Then the derivative of such a function is just the pd f f (x) =

d F(x) dx

(8.12)

As we stated before, f (x) ≥ 0 and the integral over the entire x space equals unity. With the definitions introduced above, we can derive a number of properties in a simple way. If we have a random variable x with discrete values x1 , x2 , ....xn and relative probabilities p1 , p2 , ..... pn the expected value or mean value is given by x = x1 p1 + x2 p2 + ......xn pn =



xi pi

(8.13)

and in a similar way for a continuous variable  x =

∞ −∞

x f (x)d x

(8.14)

This rule can be generalized to any function of x. Consider the variable U and the fluctuation of U defined as u = U − U . (8.15) The variance σ is defined as the mean square fluctuation  σ = u  = 2

∞ −∞

(U − U )2 f (x)d x

(8.16)

While the standard deviation or root mean square, r.m.s is defined as r.m.s.(u) =

√

σ = u 2 

1/2

(8.17)

As generalization of (8.14) and (8.16), we can define the nth moment of x about its mean as  ∞ σ = u n  =

−∞

(U − U n ) f (x)d x

(8.18)

Sometimes it is convenient to work with standardized random variables which by definition have zero mean and unity variance so that the standardized random variable Uˆ uˆ = (U − u)/σ (8.19) Now we give some example of probability distribution.

258

8 Turbulence

8.2.1 Examples of Probability Distribution The most simple distribution is the uniform distribution. When U is uniformly distributed in the interval a ≤ x < b the pd f is given by  f (x) =

1/(b − a), for a ≤ x < b 0, for x < a and x ≥ b

(8.20)

This distribution is shown in Fig. 8.3. In this case given the interval a − b, the height (value) h of the pd f must satisfy the condition (8.9) so it must be (b − a)h = 1. The probability of an interval of amplitude x will be hx/(b − a)h that is x/(b − a). The exponential distribution is described by the probability density 

λ exp(−λx), for x ≥ 0 0, for x < 0

f (x) =

(8.21)

It can be easily verified that the function is normalized. The mean (or expected value) is given by  ∞

λ exp(−λx)xd x = 1/λ

(8.22)

(x − x)2 λ exp(−λx)xd x = 1/λ2

(8.23)

x = 0

While the variance 

∞

x  = 2

0

So that the variance is just the square of the mean. Figure 8.3 shows both f (x) and F(x) for the exponential distribution. Notice that this corresponds to the Poisson distribution, when addressing the discrete case We have mentioned several times the normal distribution which is described by f (x) = N(x, μ, σ ) =

1 √ exp[− 21 (x − μ)2 /σ 2 ] σ 2π

(8.24)

where μ is the mean and σ the standard deviation. It is interesting to know that the expected value of an event between that lies in the interval [a, b] is given by E(x) = μ − σ 2

f (b) − f (a) F(b) − F(a)

(8.25)

where F(x) is the cumulative distribution function. It is interesting to notice that for the normalized variable xˆ = (x − μ)/σ , we have the standardized Gaussian random variable 1 fˆ(x) = N(x, 0, 1) = √ exp(−x 2 /2) (8.26) 2π

8.2 Statistical Description of Turbulence

259

Probability Density Function

Cumulative Distribution Function

uniform

1/(b-a)

a

b

x

a

b

x

exponential

x

x

log normal σ2 = 5 σ2 = 0.5 σ2 = 0.05

x

x

Fig. 8.3 Pdf (left column) and cdf (right column) are shown for the uniform, exponential and log-normal distributions from top to bottom

260

8 Turbulence

with the corresponding cd f 1 ˆ F(x) =√ 2π



∞ −∞

exp(−ξ 2 )dξ =

1 2

 √  1 + erf(x)/ 2

(8.27)

ˆ Both fˆ(x) and F(x) are illustrated in Fig. 8.4. Finally we would like to discuss the log-normal distribution. Consider a variable x normally distributed with mean μ and variance σ 2 then the variable Y = ex

(8.28)

is log-normally distributed. To find the pd f and cd f of Y , we observe that the cd f FY (y) is given by Fy (y) = P{Y < y} = P{e x < y} = P{x < ln y} = F{ln y}

(8.29)

The pd f is obtained by differentiating with respect to y f y (y) =

 1 d 1 Fy (y) = f (ln y) = √ exp − 21 (ln y − μ)2 /σ 2 dy y yσ 2π

(8.30)

Figure 8.5 shows a few example of log-normal distributions with different variances.

8.3 Smoke Plumes as an Application of Pdf A nice application of the pd f is the study of diffusion and propagation of smoke plumes. Following the idea given by Blackadar we consider a particle which leaves the source at some instant and moves in response to the turbulent motion of the air. In the x direction we indicate with F(x) the probability density that the particle is between x and x + d x and if the matter is conserved we have the known relationship 

∞

−∞

F(x)d x = 1

(8.31)

The conservation of matter implies that the particles or gases in the plume are not subject to chemical transformations. We can assume similar functions in the y direction (G(y)) and in the z direction (H (z)). If we also assume that such probabilities are independent then the probability to find a particle between x and x + d x, y and y + dy and z and z + dz is just the product of the single probabilities so that we have 

∞ −∞



∞

−∞



∞

−∞

F(x)G(y)H (z)d xd ydz = 1.

(8.32)

8.3 Smoke Plumes as an Application of Pdf

261

Fig. 8.4 The plume diffusing from a point source in a constant horizontal wind

udt mean wind x

ut

We indicate with χ the concentration of the pollutant and interpret it as the probability density of the pollutant (for example, number of particles per unit volume), then for a point source of strength Q we have χ (x, y, z) = Q F(x)G(y)H (z) And integrating

 Q=

∞ −∞



∞

−∞



∞

−∞

χ (x, y, z)d xd ydz

(8.33)

(8.34)

Our purpose at the present time is to find the concentration for a source that emits continuously and not a single “puff”. We can do that if we substitute to the source of strength Q an emission rate Q such that the instantaneous strength is given by Qdt. Also, we make the situation simple by assuming a basic wind in the x direction so that the dispersion in this direction is negligible with respect to the transport by the wind u. In that case as shown in Fig. 8.4 with the average velocity u¯ we have the probability density in the x direction to be F(x) =

1 udt ¯

(8.35)

Then we apply (8.33) to get χ (x, y, z) =

Qdt G(y)H (z) Q = G(y)H (z) udt ¯ u¯

(8.36)

This simple equation tells us that the concentration of the pollutant is proportional to the source strength and inversely proportional to the wind velocity. At this point, we need to know something about the functions G(y) and H (z). To this end, we write the continuity equation in presence of a source ∂χ + V · ∇χ = q ∂t

(8.37)

262

8 Turbulence

where q is the net production. The term V · ∇χ can be expressed also in terms of the flux that in this case could be imagined as the product of concentration and velocity as in (8.7) so that the continuity equation becomes when q = 0

 ∂χ  ∂   ∂   ∂   ∂χ  + u¯ =− u χ  + v χ  + w χ  ∂t ∂x ∂x ∂y ∂z

(8.38)

The right-hand side of this equation represents the so-called divergence of eddy flux while the equation presents a classical closure problem that is the average terms on the left are expressed as a function of eddy quantities (the primes). We will discuss this question later and for the moment we solve the problem by expressing the eddy fluxes in terms of average quantities that is u  χ   = −K x

∂χ  , ∂x

v χ   = −K y

∂χ  , ∂y

w χ   = −K z

∂χ  ∂z

(8.39)

where K s are the diffusion coefficients in the three directions. If the mean wind is ¯ so that Eq. (8.38) constant we can make a coordinate transformation x  = x − ut becomes with K s constant ∂χ  ∂ 2 χ  ∂ 2 χ  ∂ 2 χ  + K + K = Kx y z ∂t ∂x2 ∂ y2 ∂z 2

(8.40)

where we have dropped the primes. We now solve the problem for an instantaneous source of strength q emitted at t = 0 at x = y = z = 0 so that the boundary conditions are χ → 0 for t → ∞ and x, y, z → ∞. Also we should require  Q=

∞ −∞



∞

−∞



∞

−∞

χ (x, y, z)d xd ydz

The solution to Eq. (8.40) as shown in the appendix is given by

−x 2 exp χ= 4K x t (2π t)3 K x K y K z Q



−y 2 exp 4K y t



−z 2 exp 4K z t

 (8.41)

Remembering (8.33), we have for the probability densities 

x2 exp − 2 2σx 2π σx   1 y2 G(y) = √ exp − 2 2σ y 2π σ y 

1 z2 H (z) = √ exp − 2 2σz 2π σz F(x) = √

1

(8.42)

8.3 Smoke Plumes as an Application of Pdf

263

where σx2 = 2K x t and so on. Once we have found a solution for the instantaneous point source as already shown we find a solution for the continuous source according to (8.36). We have    y2 z2 Q exp − + . χ= 2π uσ ¯ y σz 2σ y2 2σz2

(8.43)

This result is quite similar to what is observed with a significant difference. According to our definition, σ grows with time according to t n where n = 0.5 while in reality this number is between 0.75 and 1. A simple calculation shows that 95% of the pollutant is contained within 2σ .

8.4 The Atmospheric Boundary Layer: The Mixing Length A very useful example of application of the turbulence notions we have learned so far is the atmospheric boundary layer. This is the layer of the atmosphere in contact with the ground (either solid surface or water) where turbulence is fully developed. The layer has a thickness of roughly 1 km and it is where the atmosphere is dragged around by the solid Earth because of the roughness of the surface. In this layer, the simplest equations of motions should be valid and regulate the large scale circulation. The measurements of the wind velocity at some point show the fluctuations we encountered before but the characteristic decay time is short enough to preserve the large scale motions. We will apply now the familiar decomposition of the velocities V = V + V to the horizontal equations of motion and continuity 1 ∂p ∂u + V · ∇u − f v = − ∂t ρ ∂x 1 ∂p ∂v + V · ∇v + f u = − ∂t ρ ∂y ∂(ρu) ∂(ρv) ∂(ρw) ∂ρ + + + =0 ∂t ∂x ∂y ∂z

(8.44)

Now we substitute for u, v, w the quantities u + u  , v + v , w + w and make the averaging. For the first of (8.44), we get (neglecting the fluctuations in density) ∂u ∂v ∂w ∂u + u + v + w − f v = ∂t ∂x ∂y ∂z 1 ∂ p ∂u  ∂u  ∂u  − u − v − w − ρ ∂x ∂x ∂y ∂z

(8.45)

264

8 Turbulence

We have suspended the average for the terms containing products with primes that we treat as

  ∂u ∂v ∂w ∂(u  u  ) ∂(u  v ) ∂(u  w ) − − + u + + − ∂x ∂y ∂z ∂x ∂y ∂z We assume that the continuity equation holds also for the terms with primes and carrying out the averages Eq. (8.45) becomes ∂u ∂v ∂w ∂u + u + v + w − f v = ∂t ∂x ∂y ∂z

 1 ∂ p ∂   ∂   ∂ − − u u  + u v  + u  w  ρ ∂x ∂x ∂y ∂z

(8.46)

And a similar equation holds for the v component ∂v ∂v ∂v ∂v + u + v + w + f u = ∂t ∂x ∂y ∂z

 1 ∂ p ∂   ∂   ∂   − − u v  + v v  + v w  ρ ∂y ∂x ∂y ∂z

(8.47)

We can now understand a little better the origin of Eq. (8.38) and as in that case we are confronted with a closure problem: the changes in the averaged quantities are a function of the eddy stress terms (those with primes). As we did before we recur to the mixing length theory that assumes eddy stress to be proportional to the gradient of the mean wind. In the planetary boundary layer, the vertical gradient is much larger than the horizontal gradients so that we can limit the discussion to the vertical direction. The mixing length theory is somewhat analogous to the mean free path concept. A parcel of fluid displaced vertically will conserve the horizontal velocity for a length lm before mixing with the environment. Consider then a fluid parcel at time t passing the level z = 0 with an instantaneous velocity u and a moment for unit mass ρu. When this parcel reaches the level z at time t, it has a momentum deficit M = ρ[u(z, t) − u(0, 0)]

(8.48)

And using the decomposition u = u + u  we have M = ρ[u(z) − u(0)] + ρ[u  (z, t) − u  (0, 0)]

(8.49)

In this expression we can neglect the contribution of the turbulent term to the momentum deficit and the difference of the averaged value can be written as z∂u/∂z so that M becomes ∂u (8.50) M = ρz ∂z

8.4 The Atmospheric Boundary Layer: The Mixing Length

265

Now we can write w = dz/dt and the momentum flux in the z direction becomes τx z = Mw = ρz

∂u dz 1 ∂u d 2 = ρ z  ∂z dt 2 ∂z dt

(8.51)

The last term can be also written as   d 2 dz z  = z = 2zw  dt dt

(8.52)

If the fluid at any point did not exchange momentum with the environment then the product zw  would continue to increase with time as z increases. This is not a realistic situation and we expect the correlation between w and z to decrease with time and with the distance travelled. We could assume that such correlation is lost for a distance of the order of lm that we call mixing length. In this approximation, 2zw  ≈ 2 Cwlm and the momentum flux becomes τx z = Cρwlm

∂u ∂z

(8.53)

with C a proportionality coefficient. In analogy with the stress in the viscosity theory we can introduce an eddy viscosity νT = Cwlm . This concept has an immediate consequence for the solution of Eqs. (8.46–8.47) because all the eddy terms appearing in them can be neglected except for the terms u  w  and v w , Those terms can now be written as (see also Appendix “Vorticity and Friction in the Boundary Layer”) − ρu  w  = A z

∂u , ∂z

−ρv w  = A z

∂v ∂z

(8.54)

where A z is the exchange coefficient. Application of such relation will be made in the appendix to explain some features of the planetary boundary layer. A simple application of these concepts is to find the change of wind with altitude in the so-called surface layer (the first few tens of metres above the surface. We can assume that u  nnd w are of the same order w ≈ −u  ≈ lm So that 



u w  =

−lm2

∂u ∂z

∂u ∂z

(8.55)

2 = −u 2∗

(8.56)

The quantity u 2∗ is the friction velocity and may be assumed constant in the surface layer. In this layer lm = kz with k the Karman constant. So we have

266

8 Turbulence

u∗ ∂u = ∂z kz

(8.57)

It can be integrated with the condition U  = 0 for z = z 0 to get U  =

 z u∗ ln k z0

(8.58)

It can be used to evaluate the wind in the surface layer.

8.5 Two-Dimensional Turbulence Most of the geophysical fluids are two dimensional especially when considering large scale flows. The ocean has an average depth of 4 km compared to a horizontal extension of 103 km and the same holds in the atmosphere with a depth of roughly a scale height (8 km) when only the troposphere is considered. A good way to study 2D turbulence is to start from the Navier–Stokes equation ∂v 1 + v · ∇v = − ∇ p + ν∇ 2 v ∂t ρ

(8.59)

where for a two-dimensional fluid v = −∇ψ × k with ψ the streamfunction. Taking the curl of (8.59), we obtain the vorticity equation ∂ 2 ∇ ψ + v · ∇(∇ 2 ψ) = ν∇ 4 ψ ∂t

(8.60)

where the vorticity is ω = ∇ 2 ψ. Now in absence of viscosity ν = 0 the kinetic energy is conserved  (∇ψ)2 d V = const (8.61) V

We can also show that another property related to vorticity is conserved. It is easy to show that  (∇ 2 ψ)2 d V = const (8.62) V

The quantity (∇ 2 ψ)2 = ω2 is called enstrophy A useful relation can be found between the conserved energy and enstrophy using 

 (∇ψ)2 d V =

 ∇ · ψ∇ψd V −

ψ∇ 2 ψd V

8.5 Two-Dimensional Turbulence

267

And because the first member on the right-hand side is zero we have 

 (∇ψ)2 d V = −

ψ∇ 2 ψd V

(8.63)

Which is equivalent to E=

1 2 1 u  = ψω 2 2

(8.64)

The above considerations are valid when ν = 0 while on the contrary we expect both energy and enstrophy to be dissipated for the viscous case. First of all in (8.59), we use the equivalence ∇ 2 u = ∇(∇ · u) − ∇ × (∇ × u) = −∇ × ω So that we have

dv = −ν∇ × ω dt

(8.65)

d  1 2 u = −ν[ω2 + ∇ · (ω × v)] dt 2

(8.66)

Multiplication by v gives

We now average this equation and notice that ensemble average is equivalent to space average. Also the isotropic turbulence implies that all the divergences are integrated to zero. The energy dissipation rate becomes =

d 1 2 dE = u  = −νω2  = −2ν Z dt dt 2

(8.67)

where Z = 21 ω2  is the just defined enstrophy. As for the vorticity, we start with the equation dω = ν∇ 2 ω (8.68) dt Multiplication by ω gives  d  1 2 ω = −ν (∇ω)2 − ∇ · (ω∇ω) 2 dt

(8.69)

That once averaged gives η=

dZ d 1 2 = ω  = −ν(∇ω)2  = −2ν P dt dt 2

(8.70)

268

8 Turbulence

where P = 21 (∇ω)2  is the palinstrophy. Enstrophy was introduced by Cecil Leith in a paper in 1968 and is from the greek word σ τρωφη meaning “turning”. On the other end palinstrophy was introduced three years later by Annick Pouquet as a combination of π αλιν (palin meaning “again”) and σ τρωφη (strofi meaning ‘turning’). For the considerations that follow it is interesting to express some of these quantities in spectral form. Kolmogorov first hypothesis states that the statistical equilibrium of turbulence depends only on the dissipation rate of energy  and the viscosity ν. It is assumed turbulence to be isotropic such that its spectrum can be obtained from the averaged kinetic energy per unit volume 1 V



1 2 u dV = 2



∞

E(k)dk

(8.71)

0

where V is the volume of integration, E(k) is the energy spectrum and k the wave number. The formulation (8.69) is equivalent to assuming a streamfunction with a Fourier transform  ψ(k, t) = dkψ(k, t) exp(ik · x) (8.72) and from the definition of enstrophy we have 

∞

Z=

k 2 E(k)dk.

(8.73)

0

The dimension of the energy dissipation , viscosity, wavenumber and energy spectrum can be easily found [] =

L2 energy = 3, time T

[ν] =

L2 , T

[k] =

1 , L

[E(k)] =

L3 T2

(8.74)

where L is the length unit and T is the time unit. Just from dimensional considerations we have  (8.75) [k] =  1/4 ν −3/4 = kk called the Kolmogorov wave number which corresponds to a length scale lk = (ν 3 /)1/4 . This length represents the size of the smallest eddy present in the fluid. In the case of the Earth atmosphere with ν ≈ 0.1 cm2 s−1 and k ≈ 10 cm2 s−3 so that the dissipational length scale is of the order of 0.1 cm several orders of magnitude larger than the mean free path (10−4 cm). Eddies smaller than that are readily dissipated. The eddies of size k rotates with velocity u k ≈ Z k that is uk ≈

  1/2 ν

k = (ν)1/4

(8.76)

8.5 Two-Dimensional Turbulence

269

Their energy is dissipated in a time τk =

 ν 1/2 k ≈ uk 

(8.77)

In the previous example of the atmosphere this time is about 0.1 s, Kolmogorov did assume that the energy spectrum should have the form E(k) = u 2k k f (k k)

(8.78)

where f is a dimensionless function of the dimensionless quantity k k. The wave number k is comprised between the scale of the volume containing the fluid L and k . In the range L −1 ≤ k ≤ −1 k (called the inertial subrange), there is no dissipation of energy while the eddies break up in smaller eddies and the energy is simply transferred between them. In this range the energy must be independent of scale so it is assumed that the function f is simply a power law f (k k) = α(k k)n

(8.79)

In this range also E(k) must be independent of viscosity because of the negligible dissipation so if we substitute the value of u k and k in (8.76) we have E(k) = ν 1/2  1/2 ν 1/4  −1/4 αν 3n/4  −n/4 k n The requirement of E(k) independent on ν impose n = −5/3 so that the spectrum results (see also Appendix “An Example of Similarity Theories and Dimensional Analysis”) (8.80) E(k) = α 2/3 k −5/3 At this point we may ask if a deterministic theory of turbulence is possible. We can evaluate the number of degrees of freedom (N ), for example, in the case of the atmosphere. In this case we can assume the dimension of the “vessel” to be 1000 km while the smallest eddy has dimension of the order of 1 mm. We have for the N ≈ (L/k )3 ≈ (109 /1)3 = 1027 which is a very large number indeed larger than the Avogadro number. In general if we use the value for k we have N ≈ L3

  3/4 ν3

(8.81)

and approximating  ≈ u 3 /L

N≈

UL ν

9/4 = (Re )9/4 .

(8.82)

270

8 Turbulence

For large atmospheric flows U ≈ 10 ms−1 , L = 106 m, ν = 103 m2 s−1 we have Re = 1012 and N = 1027 . You do not need further proof to abandon the idea of a deterministic theory of turbulence.

8.5.1 Energy and Enstrophy Transfer Until now we did not mention explicitly what makes 2D turbulence so peculiar with respect to the 3D analogue. The main difference is found rewriting the vorticity equation for a 3D flow where Eq. (8.60) becomes ∂ 2 ∇ ψ + v · ∇(∇ 2 ψ) = ∇ 2 ψ · ∇v + ν∇ 4 ψ. ∂t

(8.83)

The extra term of the right-hand side is absent in the two-dimensional case because in this case the vorticity is always normal to the velocities. The net result is an extra source term corresponding to the so-called vortex stretching. The absence of such term in 2D turbulence implies that both energy and enstrophy are conserved. To demonstrate this we calculate the kinetic energy K =

1 2

 (u 2 + v2 )d A = A

and the enstrophy Z=

1 2

 ω2 d A = A

1 2

 (∇ψ)2 d A

(8.84)

A



1 2

(∇ 2 ψ)2 d A

(8.85)

A

where in both cases A is a finite area. When ν = 0 the vorticity is conserved d∇ 2 ψ/dt = 0. In this case the vorticity is conserved on parcels so that the integral of any function of vorticity is zero when integrated over A. So we have d f (ω) = 0, dt

d dt

 f (ω)d A = 0.

(8.86)

A

To understand how energy and enstrophy are transferred, we will follow the example given in Salmon. Suppose that both energy and enstrophy are contained at wavenumber k0 and then transferred to wavenumbers k1 and k2 with k1 = k0 /2 and k2 = 2k0 . The conservation of energy requires E0 = E1 + E1

(8.87)

while for conservation of enstrophy we have

E o k02

= E1

1 2 k 2 0

2 + E 2 (2k0 )2

(8.88)

8.5 Two-Dimensional Turbulence

271

Solving for E 1 and E 2 we have E 1 = 0.8E 0 ,

E 2 = 0.2E 0 ,

(8.89)

so that 80% of energy ends up in the lower wave number and 20% in the higher wave number. It can be easily found that Z 1 = 0.2Z 0

(8.90)

which show the lower wavenumber contains only 20% of the initial enstrophy, i.e. energy and enstrophy move in opposite directions. A more general proof that energy and enstrophy move in opposite direction can be found is if we consider the quantity  I =

(k − ke )2 E(k)dk

where ke is the centroid of the distribution  k E(k)dk ke =  . E(k)dk

(8.91)

(8.92)

The quantity I is an indication of the width of the distribution. Expanding (8.89) we obtain    2 2 E(k)dk. I = k E(k)dk − 2ke k E(k)dk + ke  The second term can be substituted by ke E(k)dk following the definition (8.90) so that we have   2 2 E(k)dk (8.93) I = k E(k)dk − ke 

so that

k 2 E(k)dk − I  E(k)dk

(8.94)

d 2 1 dI ke = − 0. dt

272

8 Turbulence

Expanding we have d dt

 (k − 2

ke2 )2 E(k)dk

    d 2 2 4 = Edk k4 Edk − 2ke k Edk + ke dt (8.97)  d = k 4 Edk dt

From the definition of enstrophy Z (k) = k 2 E(k) so that if (8.94) is positive we have d dt

 2  k Z (k)dk  >0 Z (k)dk

(8.98)

and the conclusion is that the centroid of the enstrophy moves to higher wavenumber and shorter scales. The conclusion we may draw at this point is that in two dimensional turbulence energy is transferred to larger scales while the opposite happens for enstrophy. If the fluid turbulence is excited at some wavenumber ki , the energy propagates to lower wavenumbers and larger scale. The spectrum is represented by (8.78). For the case of enstrophy the cascade rate is given by enstrophy at wavenumber k, k 2 E(k) divided by the eddy turnover time analogous to (8.75) which we now calculate in a different way. If u(l) is the velocity for eddies of size l the energy density E(k) = lu(l)2 so that (8.99) E(k) ≈ lu(l)2 ≈  2/3 l5/3 so that

u(l) ≈  1/3 l1/3 ⇒ τk = k −1 u(l) = k −2/3  −1/3 .

(8.100)

The rate of enstrophy cascade is then η≈ and using τk we have

k 2 E(k) τk

E(k) = βη−2/3 k −3

(8.101)

(8.102)

with β an universal constant similar to the Kolmogorov constant. The turnover time can be found as lk (8.103) tk ≈ vk where from dimensional arguments vk ≈ η1/3 k −1

(8.104)

8.5 Two-Dimensional Turbulence

273

Fig. 8.5 The transfer on energy and the enstrophy in the Kolmogorov spectrum

ε2/3 k-5/3

energy

η2/3 k-3 energy transfer, ε

kν

stirring ε, η

1

(η/ν3)

enstrophy transfer, η

dissipation

wavenumber

with the result that tk ≈ η−1/3 at all scales. The viscous scale can be obtained from t ≈ l2 /ν and using tk we have η1/6 kν ≈ 1/2 . (8.105) ν Now we calculate the dissipation rate for enstrophy d Z =ν dt

 ω∇ 2 ωd A ≈ ν A

vk2 ≈ νkν4 vk2 ≈ η l4k

(8.106)

where we have used (8.102) and (8.103). At this point we can locate three different regions in the Kolmogorov spectrum. If the turbulence is excited at some initial wave number energy is transferred to longer scale (shorter wavenumber) according to (8.97) while enstrophy is transferred to lower scales (longer wavenumber) according to (8.100). This happens up to wavenumber kν when enstrophy is dissipated at a rate independent of the viscosity ν according (8.104). These processes are summarized in Fig. 8.5.

Appendix Analytical Solution to the Diffusion Equation We consider the one-dimensional diffusion equation ∂ 2u ∂u =D 2 ∂t ∂x

(8.107)

and we look for a solution of the form u(x, t) with initial condition u(x, 0) = ϕ(x). A general solution could be found using the Fourier transform

274

8 Turbulence

u(k, ˆ t) = so that

1 2π



∂ 1 u(x, ˆ t) = ∂t 2π

∞ −∞



u(x, t)e−ikx d x =

∞ −∞

(8.108)

∂ u(x, t)e−i(kx d x ∂t

(8.109)

and similarly  ∞ 2 ∂ 1 ∂2 u(k, ˆ t) = u(x, t)e−i(kx d x = 2 ∂x 2π −∞ ∂ x 2  ∞  1 ∂ k2 ∞ u(x, t)(−ik)e−i(kx d x = − u(x, t)e−i(kx d x = −k 2 u(k, ˆ t). 2π −∞ ∂ x 2π −∞ (8.110) We now take the Fourier transform of (8.107) and the boundary condition to get ∂ u(k, ˆ t) + Dk 2 u(k, ˆ t) = 0, ∂t that is equivalent to

ˆ u(k, ˆ 0) = ϕ(k)

(8.111)

 ∂  Dk 2 t e u(k, ˆ t) = 0 ∂t

(8.112)

and can be integrated with respect to t to give u(k, ˆ t) = f (k)e−Dk t , 2

(8.113)

where f (k) is a function of k to be determined. The initial condition gives u(k, ˆ 0) = ϕ(k). ˆ To find the solution u(x, t) we could start from a function S(x, t) with Fourier ˆ t) = e−Dk 2 t . In this case we have transform S(k, 1 S(x, t) = 2π



∞

ˆ t)eikx d x = 1 S(k, 2π −∞



∞

e−Dk

2

t+ikx

−∞

dx = √

1 4π Dt

x2

e− 4Dt

(8.114) 2 Then we use the rule to find the Fourier transform of a product u(k, ˆ t) = e−Dk t ϕ(k) = ˆ t)ϕ(k) we recur to the convolution S(k,  u(x, t) =

∞

−∞

S(x − y)ϕ(y)dy = √

1 4π Dt



∞

−∞

e−

(x−y)2 4Dt

ϕ(y)dy.

(8.115)

This is the solution to the diffusion equation where the initial condition must be specified. Suppose, for example, ours is a point source such that at time zero we have the source in x0 u(x, 0) = ϕ(x) = Qδ(x − x0 ). (8.116)

Appendix

275

The substitution of such condition in Eq. (8.113) gives (x−x0 )2 1 u(x, t) = √ e− 4Dt 4π Dt

(8.117)

which is the familiar form of the diffusion from a point source.

Kolmogorov and Parachutes An unthinkable application of the Kolmogorov theory is reported by Marc Brachet. He considers an object of dimension l and mass m in free fall. In the laminar regime the terminal velocity is determined by the equilibrium between gravity and viscous force. Dimensional analysis gives viscosity as [L 2 T −1 ] and force as [M L T −2 ] and ρ as [M L −3 ] so that the viscous force is FV = νρ lv = mg ⇒ v ≈

mg ρν l

(8.118)

To find the corresponding turbulent force we notice that the power is given by =

v2 v3 ≈ t l

(8.119)

And because power is the product of velocity times force we have  = Ftur b v → Ftur b = ρ l2 v2

(8.120)

where ρ has been added to have consistent dimension. Then we have the terminal velocity for the turbulent case  mg (8.121) vtur b = ρ l2 And the ratio between the forces Ftur b lv ρ l2 v 2 = = Re = Fv νρ lv ν

(8.122)

Notice that (8.118) is quite similar to Stokes terminal velocity except for some numerical factor. However if we calculate this velocity with g = 10 m2 s−1 , m = 100 kg, ρ = 1 kgm−3 , l = 1 m and ν = 10−5 m2 s−1 we obtain v = 108 ms−1 . With the same data we obtain a vtur b of 33 ms−1 with l = 1 m and 3.3 ms−1 after opening the parachute l = 10 m. The Reynolds number is very high confirming that the motion has to be considered turbulent.

276

8 Turbulence

Watching the River Flow We take this example from Falkovich. In principle, the flow of a river is a simple application of the Navier–Stokes equation. Assume that the river bed is an inclined plane as in Fig. 8.6 then the equation of Navier Stokes becomes at steady state ∇p + ν∇ 2 v + g = 0 ρ

(8.123)

It can be projected in the x and z directions to obtain (notice that v has only the x component) dp + ρg cos α = 0 dz (8.124) d 2v ν 2 + g sin α = 0 dz Boundary conditions in this case are important and are for v to be zero at z = 0 while the stress should be normal and balance the pressure τx z (h) = ρν

dv = 0, dz

τzz (h) = − p(h) = − p0

and the solution is then p(z) = p0 + ρg(h − z) cos α,

v(z) =

g z(2h − z) sin α 2ν

(8.125)

Using ν = 10−2 cm2 s−1 and taking a “serious” river with h = 10 m and α = 10−4 (100 m/1000 km) we get v(h) ≈ 100 km s−1 which is an impossible result. To correct this result we need to invoke turbulence. At small Reynolds number the viscous drag is of the order ν∇ 2 v ≈ νv/ h 2 and at high Reynolds number the drag is just v2 / h (see Appendix “The Oseen Correction to the Stokes Drag and the Stokes Paradox”) that must balance the gravity acceleration along the plane g sin α ≈ gα. Thus we have v≈

h



gαh

(8.126)

p0 v river bed

v

river

river

z river bed

α x

Fig. 8.6 The river flowing on a smooth river bed (left) where the velocity is always normal to its gradient. On the right, the river flows over an irregular bed

Appendix

277

that for the previous data gives a more reasonable 10 cms−1 . Another possible way to address this problem is to introduce the eddy or turbulent viscosity that was defined in (8.53). Qualitatively from (8.46–8.47) we see that the acceleration Du/Dt is determined by terms like ∂u  v /∂ y and where according to the mixing length theory (Eq. (8.56)) the correlation term can be expressed as lm2 (∂u/∂ x)2 so that we can assume an equivalent turbulent viscosity given by νT =

∂u 2 l ≈ uh ≈ ν Re ∂y m

(8.127)

where the Reynolds number Re = uh/ν and ν is the fluid viscosity. The drag is then given (as stated earlier) by (8.128) ν∇ 2 u ≈ νu/ h 2 with ν → νT . This simple example on the flow of a river is good to introduce the concept of similarity starting from the definition of Reynolds number Re = u L/ν. However, we know that the kinematic viscosity is just the product of the mean free path (l) by the thermal velocity (vT ) so that the Reynolds number Re = u L/vT l We see that Re can be both large and small depending on the ratio u/vT . The conclusion is that the dimensionless velocity is a function of dimensionless variables v = u f (r/L , Re ). Flows with the same Re can be obtained from one another by simply changing the units of v and r : such flows are called similar. We mentioned this property when we talk about ships or aircraft. Now if we consider the Navier–Stokes equation (v · ∇)v = −∇( p/ρ) + ν∇ 2 v When v is normal to ∇v the non-linear terms cancel out. However if β is the angle between the normal to v and ∇v we have Re (β) ≈ v(h)hβ/ν At steady state, the viscous term and the acceleration balance so that gα ≈ v/ h 2 and so Re (β) ≈ gαβh 3 /ν 2 For a rain puddle with h = 1 mm and α ≈ 10−2 we have Re (β) ≈ 100β while for a river Re (β) ≈ 1012 β. The laminar regimes may be possible for the puddle but not for a river.

278

8 Turbulence

An Example of Similarity Theories and Dimensional Analysis The Kolmogorov spectrum is an example of how dimensional analysis could describe a complex phenomenon like turbulence. We will give here another example taken from the micrometeorology book by Pal Arya. Micrometeorology studies what happens in the planetary boundary layer where as we have seen turbulence plays a quite important role. The idea is to find in the layer just above the surface a relation between a number of atmospheric parameters like: the potential temperature gradient ∂θ/∂z as a function of height, the surface heat flux H0 , what they call the buoyancy parameter g/T0 (whose meaning will see in a minute), and of course the relevant properties of the fluid like density ρ and specific heat at constant pressure c p . The quantity g/T0 appears, for example, when we calculate the buoyancy of a parcel of air with a temperature difference T with respect to the environment at temperature T0 . In this case the acceleration is just gT /T0 . To summarize we would like to have a functional form of the type

f

∂θ , H0 , g/T0 , z, ρ, c p ∂z

 =0

(8.129)

Then assume that the product of the quantities appearing in (8.116) to be equal to a dimensionless quantity D1 such that

D1 =

∂θ ∂z



H0 ρc p

a

g T0

b zc

(8.130)

where we have grouped H0 and ρ, c p in a single variable. Assigning the right basic dimensions (length L, time, T and temperature K ) we have [L 0 T 0 K 0 ] = [K L −1 ][K L T −1 ]a [L T −2 K −1 ]b [L]c

(8.131)

And equating the exponents on both sides we get a linear system of equations −1 + a + b + c = 0 −a − 2b = 0 1+a−b =0

(8.132)

It once solved gives the solutions a = −2/3, b = 1/3, c = 4/3 so that we get from (8.117)

 

 ∂θ H0 −2/3 g −1/3 4/3 z =C (8.133) ∂z ρc p T0 Another way to obtain similarity relation is to formulate a scale (for example, for z) and then scale all the other accordingly. For example, we can use (8.129) to find

Appendix

279

∂θ ∂z

An then



θf = z

=

∂θ ∂z

H0 ρc p



=

2/3

H0 ρc p

g T0

1/3

2/3

g T0

z −4/3

1/3

z −1/3

(8.134)

Similarly for the velocity  H0 g 1/3 = [L T ] = [K L T ][L T K z ρc p T0 (8.135) This approach is particularly useful to find the fluctuations of temperature and vertical velocity that obey in the convective surface layer to u 3f

3

−3

−1

−2

−1

σθ = Cθ , θf

H0 g ][L] = z ⇒ uf = ρc p T0

σw = Cw uf

where the constant can be measured. Then we have σθ = Cθ (H0 /ρc p )2/3 (g/T0 )−1/3 z −1/3 θf  1/3 σw = Cw (H0 /ρc p )(g/T0 )z uf

(8.136)

This similarity theory is quite successful in predicting the fluctuations of temperature and vertical velocity and was formulated by the Russian scientist O.M. Obukhov from which originated also the Monin–Obukhov length.

The Wind Profile Similarity We would like to introduce the heat flux in the surface layer whose horizontal component is different from zero. What is really important is the vertical component that according to what we have seen in the previous paragraph can be written as H = c p ρw T  .

(8.137)

In this form the temperature is not conserved for vertical motion so we recur to a trick using a dry adiabatic vertical profile for the atmospheric temperature Ta = T + d z where d = g/c p is the adiabatic lapse rate and this can be assimilated to the potential temperature θ which is conserved

280

8 Turbulence

θ=T

p0 p

 R/c p (8.138)

and it can be easily verified that ∂θ θ = ∂z T

 ∂T + d . ∂z

(8.139)

Based on the K parameterization of the flux we have H = −c p ρ K h

∂ Ta = −c p ρ K h ∂z

∂T + d ∂z

 = −c p ρ

 ∂ θ¯ T¯ Kh ∂z θ¯

(8.140)

where overbars indicate average values. Our purpose is to find a length that characterizes the vertical motion of the atmosphere. To this end we consider the production of mechanical energy in a layer of fluid of thickness z with the upper surface moving with velocity u 1 with stress τ2 and the lower surface with velocity u 2 and stress τ2 . The work done per unit mass is then W =

1 ∂(uτ ) 1 τ1 u 1 − τ2 u 2 = ρ z ρ ∂z

(8.141)

τ ∂u u ∂τ + . ρ ∂z ρ ∂z

(8.142)

that can be written as W =

Considering that in the first term the stress is proportional to the velocity gradient and it can be written as

2 τ ∂u ∂u M= =ν (8.143) ρ ∂z ∂z and it is always positive. This term represents the dissipation of mechanical energy into heat. The second term depends entirely on the gradient of the viscous stress and thus affects only the kinetic energy of the fluid. We can then assume that (8.142) represents the production of mechanical energy. Using the K parameterization and Eq. (8.57) we have

 ∂u 2 u3 (8.144) ≈ ∗ M = Km ∂z kz where we have used K m = ku ∗ z. Being proportional to 1/z the mechanical production decreases very rapidly with altitude above the surface. The production of buoyant energy can be found starting from the change of potential energy φ ρ

dφ dz = ρg = ρgw. dt dt

(8.145)

Appendix

281

Then assuming ρ = ρ + ρ  and w = w (absence of a steady vertical component) we have for the buoyant energy B=−

g gH g ρ  w  = w T   = ρ T  c p ρT 

(8.146)

where we have used H = c p ρw T  . In presence of a temperature inversion, the production of buoyant energy is negative and turbulence must do work against gravity. Sometime the negative ratio between M and B is also called the flux Richardson number Rf = −

gH B g K h ∂θ /∂z = = M c p ρT τ (∂u/∂z)2 θ  K m (∂u/∂z)2

(8.147)

although this number can be expressed in terms of observable quantities the ratio between K h and K m is known to be approximately unity. Also in this case the boundary for the onset of the turbulent regime is regulated by the Richardson number Ri =

g ∂θ /∂z . ρ (∂u/∂z)2

(8.148)

The transition to turbulence is for this ratio to be around 0.25. Now equating B and M we obtain a length L M c p ρT u 3∗ gH u 3∗ = =⇒ L M = . kz c p ρT  kg H

(8.149)

This measure is known as Monin–Obukhov length and it is used to describe the behaviour of wind and temperature in the first portion of the planetary boundary layer. These profiles are expressed as a function of the dimensionless height ζ = z/L M . The values for L M are in the range −∞ to ∞ with the first limit reached for the flux approaching zero from the negative side (stable condition) and +∞ reached with the flux approaching zero from the positive side (unstable conditions). Usually however is |ζ |  1, that is, in the layer near the surface the buoyancy effects can be neglected while for |ζ |  1 the buoyancy effect dominates with respect to the shear dominated turbulence. In simplified terms this means that, if properly scaled, all mean wind and temperature profiles are the same and we can introduce functions of just ζ which describe both φm (ζ ) = (kz/u ∗ )(∂u/∂z) (8.150) φh (ζ ) = (kz/θ∗ )(∂θ /∂z) where the surface flux is H0 = −ρc p u ∗ θ0 . In principle the temperature and wind appearing in (8.150) can be integrated to give, for example, for the wind u

282

8 Turbulence

Fig. 8.7 The behaviour of φm as a function of the scaled height ζ = z/L

3.5

3

6u φm= kz u 6z *

2.5

2

1.5

1

0.5

0 -2.5

-2

-1.5

-1

-0.5

0

0.5

ξ

 z  z u 1 ln − ψm = u∗ k z0 L where ψm

z L

 =

z/L z 0 /L



dζ . 1 − φm (ζ ) ζ

(8.151)

(8.152)

To determine the temperature profile it is then necessary to determine the function φm (ζ ) that is taken in the form  φm =

(1 − γ1 ζ )−1/4 , (1 + β1 ζ ),

for ζ < 0, (unstable) for ζ ≥ 0, (stable)

(8.153)

and the coefficient to be determined are γ1 and β1 . Figure 8.7 shows a typical result for a function φm .

Measuring Turbulence with SODAR SOund Detection And Ranging (SODAR) is an instrument that using turbulence can obtain measurement of important parameters in the boundary layer. A possible scheme of a sodar is illustrated in Fig. 8.8. There is a transmitter antenna that emits

Fig. 8.8 .

monostatic antenna

scattering volume

w

cΔ τ

S0

bistatic antenna

scattering angle u θ

wind

z

S

S(θ)

ΔΩ

Appendix 283

284

8 Turbulence

pulses of sounds that are scattered at some altitude. The scattered signal can be detected either with the same transmitter antenna (monostatic sodar) or by an antenna located at distance d from the transmitter (bistatic sodar). The power received by a monostatic sodar in this case is given by P(r ) = P0

A exp(−2αr )σ (r ) r2

(8.154)

where P0 is the transmitted power, A is the effective area of the receiver, α is some kind of atmospheric attenuation and sigma the scattering cross section. We may consider that the scattering volume is determined by the pulse length τ while only part of the incident power is scattered at the angle θ . The differential scattering cross section is defined as the portion of the incident scattered in the solid angle d in the direction θ .  π   2π dσ dσ dσ dφ dθ sin θ →σ = d = (8.155) d d d 0 0 A quite complicated theory of the interaction of sound waves with the atmosphere gives for this cross section

T (K ) + cos2 σ (θ ) = 2π k cos θ (2T )2 4

where

2

  θ v (K ) 2 c2

T (K ) = 0.033C T2 K −11/3 V (K ) = 0.061C V2 K −11/3

And

(8.156)

(8.157)

−2/3

C T2 = (T1 − T2 )2 r1,2

−2/3

C V2 = (v1 − v2 )2 r1,2

(8.158)

where the subscripts on the temperature and velocity indicate two different points of measurement at distance r1 and r2 . In the above relations x is the speed of sound, k is the wavenumber of the incident sound k = 2π/λ and K is related to the Kolmogorov length scale lt , K = 2π/lt . For backscattering (θ = 2π ) in (8.155) only the first term survives and the cross section becomes

 4 T (K ) (8.159) σπ = 2π k (2T )2 where now K = 2k so that with the help of (8.157) we have σπ = 7.15 · 10−3 λ−1/3 C T2 /T 2

(8.160)

Appendix

285

This relation is true only if the scale of the irregularities lt = λ/2 is within the inertial range of the Kolmogorov spectrum as shown in Fig. 8.5. This condition is satisfied with sodar with carrier frequency between 1.5 and 6 kHz, that is, λ = 4–20 cm. The scattering cross section can be obtained from (8.154) as σπ =

Pr 2 exp(2αr ) P0 A

(8.161)

This paragraph pretends to be just an introduction to the sodar techniques and on its capabilities.

The Taylor Frozen Turbulence Hypothesis We have mentioned the Taylor frozen turbulence hypothesis in connection with time and space averages of the turbulent velocity. Notice that this is a rather important fact because it simplifies dramatically all the measurements related to turbulence because it is much more simpler to observe temporal variations at some location and relate them to space changes. Consider the local change of temperature at some point that results by summing up the total change to the advection term dT ∂T = − V · ∇T ∂t dt The temperature advection is the change of temperature at the sensor due to the advection of warmer or colder air (V · ∇T ). The total derivative is the change in temperature of an air parcel moving past the sensor. Taylor’s hypothesis says that we can assume that the turbulent eddies (which we can think of as large air parcels) are frozen as they advect past the sensor and thus the change in temperature within each eddy is negligible that is dT /dt = 0. As a consequence we have ∂T = −V · ∇T ∂t

(8.162)

This means that local temperature gradients, which might be present from one side of an eddy to another, are advected across the sensor by the mean wind without the eddy changing, which corresponds to the Taylor frozen field hypothesis. An example made by Stull may clarify the process. Suppose to have an eddy which measures 100 m in length with its front having a temperature 5 K higher than the back that is ∂ T /∂ x = 0.05 K/m. Suppose now that the wind velocity is 10 m/s, then after 10 s the front of the eddy hits the sensor it will measure a temperature of 5 K, assuming that the eddy has not changed as it is advected. The local change in time will then be ∂ T /∂t = −0.5 K/s so that (8.162) is satisfied −0.5 = −10 × 0.05

286

8 Turbulence

This implies that v(t) = v

x V

(8.163)

where v(t) may be the fluctuating velocity over the average wind velocity. This relation shows that time can be substituted with the scaled space coordinate and so justifies the equivalence between the time average and space average. Consider however that in practice such approximation is valid only if v/V  1. Formally the hypothesis can be formulated by stating that in a turbulent flow for which the magnitude of the fluctuations is not too great, it is possible to deduce spatial turbulence quantities from time series measured at a single point in the flow.

References Textbooks Blackadar AK (2013) Turbulence and diffusion in the atmosphere: lectures in environmental sciences. Springer, Berlin Blackadar PA (2004) Turbulence. Cambridge University Press, Cambridge Emeis S (2010) Measurement methods in atmospheric sciences: in situ and remote. Borntraeger Science Publishers, Germany Falkovich G (2011) Fluid mechanics for physicist. Cambridge University Press, Cambridge Pal Arya S (2001) Introduction to micrometeorology. Academic Press, New York Pope SB (2000) Turbulent flows. Cambridge University Press, Cambridge Salmon R (1998) Lectures in geophysical fluid dynamics. Oxford University Press, New York Stull RB (1988) An introduction to boundary layer meteorology. Kluwer, Dordrecht Tennekes H, Lumley JL (1972) A first course in turbulence. MIT Press, Cambridge Vallis JK (2017) Atmospheric and oceanic fluid dynamics. Cambridge University Press, Cambridge Wyngaard JC (2010) Turbulence in the atmosphere. Cambridge University Press, Cambridge

Article Brachet M. A primer in classical turbulence theory. http://www.lps.ens.fr/~brachet/Turbulence_ Lecture_Notes_files/Cours_Chili3.pdf

Chapter 9

Magnetohydrodynamics

Magnetohydrodynamics (or MHD in short) is the study of the fluid dynamics of conducting fluids. The name originates from the fact that conducting fluids may carry currents that generate magnetic fields. We know that in presence of a magnetic field B a charge q moving at velocity v is subject to a force F = qv × B. This relation is equivalent to the force acting on a current j that is j × B and the interesting thing is that, because of this interaction, the magnetic field will change the current that will fed back into B. Among the conducting fluids (with major examples the molten metals), the most important is plasma that is a substance made up of an equal number of positive and negative charges. Magnetohydrodynamics is of a great interest for astrophysics or space physics but also for planetary scientists when they talk about the origin of planetary magnetic fields. The main application of MHD for earthbound scientists is the confinement of hot plasma to produce thermonuclear fusion or the production of energy. At the same end, liquid metals are forced through magnetic fields. So we have so many interests in introducing MHD and discover its many applications.

9.1 Some Preliminaries We will review here some matters related to elementary plasma physics including the interaction of charged particle with electromagnetic fields, plasma oscillations and so on.

© Springer Nature Switzerland AG 2020 G. Visconti and P. Ruggieri, Fluid Dynamics, https://doi.org/10.1007/978-3-030-49562-6_9

287

288

9 Magnetohydrodynamics

9.1.1 Particles Motion We start by considering the motion of an electron subject to a magnetic field B and the equation of the motion is me

dv = −ev × B dt

(9.1)

where m e is the mass of the electron and e its charge. It is evident that if initially the charge has a velocity component parallel to the field it is not affected. On the other hand if perpendicular to the field, an equilibrium is established between the Lorenz force v × B and the centrifugal force such that ev B = m e ω2 r so that the particle rotates around the field direction with angular velocity ωc =

eB me

(9.2)

which is called electron cyclotron frequency. If we take into account the angle α between the magnetic field and the velocity v, the component of the velocity perpendicular to the field is v sin α and we define a Larmor radius given by rL =

v sin α v⊥ = ωe ωe

(9.3)

And the magnetic moment associated to the current 2π e/ωe becomes μ = πr L2

mv⊥ 2 eωe = 2π 2B

(9.4)

9.1.2 Plasma Oscillations We discuss some properties related to the collective behaviour of the particles (ions and electrons) that constitute the plasma. Within the plasma the electrons are moved in some direction so that their density is changed. This will determine a repulsive force (if the density is increased) that will try to restore the initial situation. In an onedimensional model, we consider to move the electrons in the direction x as shown in Fig. 9.1. Then their initial density n 0 will change to n=

  s n0 n 0 x = ≈ n0 1 − x + s 1 + s/x x

(9.5)

In the hypothesis s  x. The positive ions are assumed not to move due to their greater mass and so their density stays the same. The change in charge is then given by

9.1 Some Preliminaries

289

Fig. 9.1 The plasma volume contained between A and B moves to A and B  creating a different charge distribution

A

B

x

Δx A’

B’ s +Δs

s

x+s

ρ = −e(n − n o ) = en 0

Δx+Δs

s ∂s = en 0 x ∂x

(9.6)

The electric field E can be found from the Gauss theorem in differential form ∇ ·E=

en 0 ∂s ρ ∂ Ex = ⇒ 0 ∂x 0 ∂ x

(9.7)

It can be integrated to give with E x = 0 when s = 0 Ex = e

n0 s 0

(9.8)

So that the force is proportional to s and the equation of motion becomes − eE x = m e

d 2s e2 n 0 = − s dt 2 0

(9.9)

It is a harmonic oscillator with frequency ω2p =

e2 n 0 0 m e

(9.10)

9.1.3 Debye Length In a plasma, particles of opposite sign tend to attract one another and repel particles of the same sign. The attraction will then create a cloud of oppositely charged particles around the attracting particle that will create a shield which will tend to mask its

290

9 Magnetohydrodynamics

charge from the outside. Consider a test charge q surrounded by protons and electrons ¯ with respective densities n p and n e while the average density will be denoted by n. The electric potential generated by such distribution obeys the Poisson equation ∇ 2 = −e

n p − ne q − δ(r) 4π 0 4π 0

(9.11)

with δ as the Dirac δ. The electrostatic potential energy of the proton in such a field is e so that the mean density will be altered according to the Boltzmann distribution ex p(−e /k B T ) and analogously for the electrons ex p(e /k B T ). So we have np = exp{−e /kb T } ≈ 1 − e /kb T n¯ ne = exp{+e /kb T } ≈ 1 + e /kb T n¯

(9.12)

where we have used the fact that e  kb T . The first term in Eq. (9.11) becomes −e

n p − ne ¯ e2 n =− 4π 0 2π 0 k B T

And the Poisson equation becomes ∇2 = −

q ¯ e2 n − δ(r) 2π 0 k B T 4π 0

(9.13)

This is now composed of the point charge contribution (the second term) while the first term is the effect of the electrons cloud. The solution of this equation is given by √ q e− 2r/D = (9.14) 4π 0 r 

where D=

0 k B T ne ¯ 2

(9.15)

is the so-called Debye length and gives an idea of the thickness of the sheath surrounding the charge q. This thickness decreases with increasing density n¯ and decreasing temperature. We have now introduced a few concepts and it is interesting to estimate their value in different situations. Table 9.1 is adapted from the beautiful book by Thorne and Blandford summarizes the data for different plasma. These data are important for the study of different environments but also for some practical applications. For the ionosphere, for example, the electromagnetic waves propagate for frequencies larger than ω P and are reflected for smaller ones. It was important some years ago for long-distance communications.

9.2 The MHD Equations

291

Table 9.1 Typical values for electron density (n), temperature (T ), magnetic field (B), Debye length (D), plasma frequency (ω P ), cyclotron frequency (ωC ) and Larmor radius (r L ) for some plasma environments Plasma n (m−3 ) T (K) B (T) D (m) ω P (rad ωc (rad r L (m) s−1 ) s−1 ) Gas discharge Fusion Experiments Ionosphere Magnetosphere Sun’s core Solar wind Interstellar medium Intergalactic medium

1016 1020 1012 107 1032 106 105 1

104 108 103 107 107 105 104 106

– 101 10−5 10−8 – 10−9 10−10 –

10−4 10−4 10−3 102 10−11 10 10 105

1010 1012 1012 105 1018 105 104 102

– 1012 108 103 – 102 10 –

– 10−5 10−1 104 – 104 104 –

9.2 The MHD Equations We are now confronted with a problem related to the fact that we have to apply Maxwell’s equations to a moving fluid. In the fluid reference frame, the fields E and B are related to the inertial frame fields E and B by the Lorenz transformations. The electric field transforms as   v·E  v + γ (E + v × B) (9.16) E = (1 − γ ) v2 with γ = (1 − v2 /c2 )−1/2 and c speed of light. This field determines the current density j through the relation (9.17) j = σ E  where σ is the electric conductivity. Then in the approximation that we can neglect relativistic effects (v  c), the electric field in the fluid is E = E + v × B

(9.18)

That in non-relativistic case means E   E so that E ≈ −v × B

(9.19)

And with similar considerations B ≈ B, Then we can rewrite (9.17) as

j ≈ j

(9.20)

292

9 Magnetohydrodynamics

j = σ (E + v × B)

(9.21)

We can now proceed to rewrite the Maxwell equations for the fluid so that the time evolution for the magnetic field ∂B = −∇ × E ∂t

(9.22)

Assume that the fluid is characterized by a typical length L and time T and the order of magnitude of the above equation is B/T ≈ E/L ⇒ E ≈ L/T B ≈ U B In the equation ∇ × B = μ0 j + μ0 0

∂E ∂t

(9.23)

We compare the order of magnitude of the last term (displacement currents) E/(c2 T ) with the left-hand side B/L. Their ratio is L EU U2B U2 E · ≈ = ≈ 1 c2 T B c2 B c2 B c2 Displacement currents can be neglected in a plasma and so electromagnetic waves as well. The current is then simply μ0 j = ∇ × B

(9.24)

We substitute (9.19) into (9.22) to get ∂B = ∇ × (v × B) ∂t

(9.25)

Finally, the force on the fluid is given by the sum of the electric and Lorenz terms F = ρE + j × B

(9.26)

where ρ is the charge density which produces the electric field ∇ · E = ρ/0 Again the ratio between the electric force and the magnetic force is given by 0 E 2 μ0 L 1 U2 ρE ≈ · · ≈ 2 1 JB L B B c The electric force can then be neglected and the force can be written as F=J×B

(9.27)

9.3 Vorticity and Magnetic Field

293

We now proceed to obtain an equation describing the time evolution of B involving only the fields and the velocity of the fluid. This is done by eliminating E between (9.21)–(9.26). Obtain the electric field from (9.21) E = j/σ − v × B so that E=

∇ ×B j −v×B= −v×B σ μσ

where we have used (9.24). Then we take the rotor of the last equation and substitute in (9.22) to get ∇ × (∇ × B) ∂B = + ∇ × (v × B) (9.28) ∂t μσ Using the vector rules and the conditions ∇ · B = 0 and ∇ · v = 0 we get ∂B = B · ∇v + η∇ 2 B = 0 ∂t

(9.29)

where η = (μσ )−1 is a magnetic diffusion coefficient. This equation shows a source term (the first on the right hand side) and a diffusion dissipative term and will be our main tools to discuss the origin of planetary magnetic fields. The same equations were discussed in Sect. 3.2.1.

9.3 Vorticity and Magnetic Field In Chap. 2, we treated vorticity and discussed the Kelvin theorem. The description can be made from another point of view starting from the equation of motion ∇p ∂v + v · ∇v = − ∇ ∂t ρ Using the vector identity ∇ × v = ∇ ∂v +∇ ∂t



(9.30)

 1 2 v − (v · ∇)v (9.30) becomes 2

 1 2 1 v ) − v × ω + ∇ p = −∇ 2 ρ

(9.31)

where ω = ∇ × v is the vorticity. Taking the curl of (9.30) we have the vorticity equation ∂ω 1 + v · ∇ω − ω · ∇v + ω∇ · v − 2 ∇ρ × ∇ p = 0 ∂t ρ For a barotropic fluid this reduces to ∂ω = ∇ × (v × ω) ∂t

(9.32)

294

9 Magnetohydrodynamics

And using the zero divergence for the velocity 1 ∂ω + v · ∇ω − ω · ∇v − 2 ∇ρ × ∇ p = 0 ∂t ρ Then multiplying by 1/ρ and rearranging we get d dt

  ω ω 1 = · ∇v + 3 ∇ρ × ∇ p = 0 ρ ρ ρ

(9.33)

In the case of η = 0 (that is infinity conductivity), Eq. (9.32) is identical with (9.28) with very interesting implications.

9.4 The Frozen Field Equation (9.32) can be used to demonstrate the Kelvin’s theorem that states that in a barotropic fluid the circulation is conserved. The same can be shown for the flux of the magnetic field as stated by the Alfven’s frozen flux theorem. To this end, we consider a surface S that is moving with the fluid and which is bounded  by a material curve C. We are interested to the flux of the magnetic field B = S B · nd S and how it changes with time  d d b = B · nd S dt dt S The changes to B may come either from changes in B or movement of S. The changes in the surface can be accounted for considering an element of the curve C, δl that during the interval δt sweeps out an area (Fig. 9.2) δS = vδt × δl And the magnetic flux through the area is

Fig. 9.2 The flux of the magnetic field through a surface S that moves to S  remain constant

dS’

S’

vdt

dS S δ C

9.4 The Frozen Field

295

B · δS = B · (vδt × δl) = [(B × v) · δl]δt And the change of flux is then 

 (B × v) · δl = C

∇ × (B × v) · nd S S

where we have used the Stokes theorem. The total rate of change is then d B = dt

 S

∂B · nd S + ∂t

 

 ∇ × (B × v) · nd S = S

S

 ∂B − ∇ × (v × B) · nd S ∂t

In the limit of infinite conductivity η = 0, Eq. (9.28) requires d B /dt = 0 that shows how the magnetic flux through a perfectly conducting fluid is a constant. There is also a more elegant way to obtain the same conclusion and that is to write (9.25) in a slightly different form using ∇ × (v × B) = v(∇ · B) − B(∇ · v) + (B · ∇)v − (v · ∇)B. So that ∂B + (v · ∇)B = (B · ∇)v − B(∇ · v) ∂t where we have taken into account the divergence of B to be zero. The equation of mass conservation dρ + ρ∇ · v = 0 dt And the equation for B becomes d dt

    B B = · ∇v ρ ρ

(9.34)

It has the same form of (9.33) for a barotropic fluid. This equation is the same satisfied by an infinitesimal material line element δr as it is stretched by a velocity gradient d δr = δv = δr · ∇v (9.35) dt The lines of force of the magnetic field are then material lines, that is, they follow the deformations of the plasma. This will be the basis for some theory on the origin of planetary magnetic fields. Remember, however, that this will be strictly valid only for σ → ∞ and all the other cases (9.28) must be taken into account and we can specify two characteristic times: an advection and diffusion time tadv =

L , U

tdi f f =

L2 η

(9.36)

296

9 Magnetohydrodynamics

B0 z

v

x

L

B

B0 Fig. 9.3 The frozen field concept applied to a channel of width L traversed by a conductor with a velocity distribution shown on the left. The unperturbed field is B0

And their ratio is called the magnetic Reynolds number Rm =

tdi f f LU = tadv η

(9.37)

Low values of Rm mean that the advection is not important and the magnetic field will relax to a diffusive state while for Rm  1 advection will dominate. We will return to the frozen field concept but it is worth to make a simple example. Consider a channel like to one shown in Fig. 9.3 with a rectangular section. In the channel, there is a flow of some conducting material and a magnetic field B0 is applied along the y direction while the channel is oriented in the x direction. The frozen field concept gives immediately a field component in the x direction. At the equilibrium (9.28) reads (9.38) ∇ × (v × B) = −η∇ 2 B The cross product (v × B) gives the only component along z of the order of u(y)B0 so that if W is the width of the channel the order of magnitude of (9.38) will be B0

Bx ∂ u(y) ≈ η 2 ∂y W

And calling ζ the shear ∂u/∂ y we have Bx ≈

ζW2 B0 η

In a while, we will examine some interesting applications of these concepts.

(9.39)

9.5 The Planetary Dynamos

297

Table 9.2 The relevant magnetic characteristics of the planets of the solar system. Notice the large tilt for Uranus and Neptune. Adapted from Schubert and Soderlund (2011) Planet Surface B (µT) Dipolarity Axial tilt (deg) Mercury Venus Earth Mars Jupiter Saturn Uranus Neptune

0.30 – 38 0,

⇒

α >1 ηk

As for the omega effect we consider a Cartesian system x − z and a magnetic field given by B = (Bx , 0, Bz ) exp(σ t + iky) and the dynamo equation then becomes σ Bx = −ηk 2 Bx + ωBz σ Bz = −ηk 2 Bz − iαk Bx

(9.64)

We can eliminate the field components and obtain an equation for σ σ 2 + 2ηk 2 σ + (η2 k 4 + iαωk) = 0 

which has the solution σ = −ηk 2 ± (1 − i)

αωk 2

So that we have a positive real σ if α ω >2 ηk ηk 2

(9.65)

In the context of a spherical geometry, the x component can be interpreted as the toroidal field that generates a poloidal field. These conclusions were reached by Eugene Parker in the late 50s and are the base for any dynamo model. Some of these conclusions are shown in Fig. 9.6.

304

9 Magnetohydrodynamics

v B

ω

j

B

B

Fig. 9.6 On the left the α effect is shown when the differential rotation changes a poloidal field in a toroidal component. On the right, the  effect is shown. In this case, an initial straight line of force is deformed by the convection v and then twisted by the rotation ω with a possible final detachment which induces a current j

9.7 Magnetohydrodynamic Waves Plasma is a medium that can propagate waves that have some peculiar properties. As a matter of fact, we have known sound waves where the restoring force is pressure. In the case of plasma, we have to consider the Lorentz force on current as an additional player. The derivation of the magnetohydrodynamics waves is rather involved so that we will describe them in the appendix. Here we will use a much more simple and intuitive approach. However, we need to introduce some notion of electromagnetism starting with the energy of a magnetic field. We consider a solenoid made up of n coil per unit length and cross section A and length l. We know that the energy stored in such a solenoid is given by E=

1 2 LI 2

(9.66)

where I is the current and L is the self-inductance. The current as a function of the magnetic field B is given by B (9.67) I = μ0 n with μ0 magnetic permeability. The self-inductance of the solenoid on the other hand L = μ0 n 2 Al

(9.68)

So that the energy density is given by u=

LI2 B2 = 2 Al 2μ0

(9.69)

9.7 Magnetohydrodynamic Waves

305

We now consider the equation of motion for a plasma which contains the Lorentz force and reads   ∂v ρ + v · ∇v = j × B − ∇ p (9.70) ∂t And at the equilibrium the pressure gradient equilibrates the Lorentz force. The current density is expressed as a function of the field B so that j×B=

∇ ×B ×B = ∇p μ0

(9.71)

In the appendix we show that the term involving the field may have two contributions which depends on ∇(B 2 /2μ0 ) and ∇ · (BB/μ0 ). However, when there are no torsional forces the stress is reduced to B 2 /μ0 . At this point, we need to make a distinction about the propagation direction of the wave. The first case to consider is a wave which propagates in direction parallel to the magnetic field. In this case like sound waves, the velocity of the wave is the square root of the ratio between energy density and mass density. So we have B vA = √ μ0 ρ

(9.72)

Incidentally a sound wave with frequency ω has an energy density ρω2 s 2 /2 with s displacement so that applying the same rule we get the phase velocity ωs. Relation (9.72) was originally found by Hannes Alfven in 1942. He made an example for a region of the Sun with a field of 15 × 10−4 T and density 5 × 10−6 kg m−3 to find a velocity of ≈ 0.6 ms−1 . The other case is when the propagation is in direction perpendicular to the field. In this case we need to recur to the analogy with sound waves which obey to the equation of state pρ −γ = cost with γ the specific heat ratio. For small deviations p1 and ρ1 from the basic state p0 and ρ0 , it is easy to show that p1 / p0 = γρ1 /ρ0

(9.73)

And if we define vs2 = γ p0 /ρ0 as the sound velocity we have p1 = vs2 ρ1

(9.74)

Now in case of the MHD waves we need to add the magnetic pressure to the gas pressure and the equivalent relation to (9.74) is  ∇

p+

B2 2μ0

 2 = vsm ∇ρm

We can eliminate pressure by observing that

(9.75)

306

9 Magnetohydrodynamics

∇p =

γp ∇ρm , ρm

∇ B2 =

2B 2 ∇ρm ρm

And the velocity of the waves is then  vsm =

γp B2 + ρm μ0 ρm

1/2 =



vs2 + v2A

(9.76)

The velocity is a combination of the speed of sound and the Alfven waves. These are also called magnetosonic waves.

9.7.1 Dispersion Relations for MHD Waves In the appendix, we will show that for plane wave solutions of the type v1 (r, t) = v1 (r, t) exp{i(k · r − ωt)} The following dispersion relation is obtained for the MHD waves: −ω2 v1 + (vs2 + v2A )(k · v1 )k +(k · vA )[(k · vA )v1 − (vA · v1 )k − (k · v1 )vA ] = 0

(9.77)

Now we can discuss the different wave modes. For propagation perpendicular to the magnetic field k ⊥ B0 (where B0 is the basic magnetic field) we have k · vA = 0 so that the dispersion relation (9.77) becomes −ω2 v1 + (vs2 + v2A )(k · v1 )k = 0 So that v1 = (vs2 + v2A )(k · v1 )k/ω2 and so k  v1 And we get v1 = (vs2 + v2A )k 2 v1 /ω2 ,

⇒

ω = k



vs2 + v2A

(9.78)

That is we obtain the magnetosonic wave as expected. This case is shown in Fig. 9.7. We can go a little farther if we write Eq. (9.25) for the perturbed field ∂B1 = ∇ × (v1 × B0 ) ∂t Assume now a plane wave solution for B1 so that the ∇ operator becomes k and the last equation reads ωB1 = −k × (v1 × B0 )

(9.79)

Appendix

307 oscillatory motion

oscillatory motion

B parallel to the propagation

B perpendicular to the propagation

Fig. 9.7 The tick straight lines represent the magnetic field while the grey areas are density enhancements of the fluid. On the left the propagation is perpendicular to the field while on the left is parallel

And then make the simplifying assumptions k ⊥ B0 , k  v1 and v1 ⊥ B0 . We have immediately v1 B0 (9.80) B1 = ω/k where the phase velocity ω/k becomes equal to the sound velocity vs if the fluid pressure is greater than the magnetic pressure or to the Alfven velocity v A for the opposite case. For the Alfven wave we can evaluate the ratio between the magnetic energy of the wave and the kinetic energy 1 ρv12 2 B12 /2μ0

From (9.80), we have v12 = v2A (B1 /B0 )2 that once substituted in the above equation gives 1 for the ratio of the two energies. In the Alfven wave, the energy is divided equally between kinetic and magnetic. For propagation parallel to the magnetic field we have k · vA = kv A and the dispersion relation reduces to  (k 2 v2A

− ω )v1 + 2

vs vA

2

 − 1 k 2 (v1 · vA )vA = 0

(9.81)

The first case is when v1  B0  k and (9.81) gives vs = ω/k we have a sound wave. On the other hand, if v1 ⊥ B0  k then v1 · vA = 0 and then ω/k = va .

308

9 Magnetohydrodynamics

Appendix Waves in Plasma We start then from the continuity equation, equation of motion and the thermodynamic equation relating pressure to density. ∂ρ + ∇ · (ρv) = 0 ∂t  ∂v ρ + v · ∇v = j × B − ∇ p ∂t

(9.82)

p = Kρ 5/3 Then we have the Maxwell equations ∂B ∂t ∇ × B = μ0 j

∇ × (v × B) =

(9.83)

∇ ·B=0 At this point, we use the perturbation approach with respect to a situation in which the fluid is at rest with a basic field B0 perturbed by B while the velocity v will be indicated by v. Equation (9.82) becomes ∂ρ + ρ0 ∇ · v = 0 ∂t 1 ∂v ρ0 = −∇ p − B0 × (∇ × B) ∂t μ0 p/ p0 = γρ/ρ0

(9.84)

We can substitute the last equation to the equation of motion to get ρ0

∂v 1 B0 × (∇ × B) = 0 + vs2 ∇ρ + ∂t μ0

(9.85)

with vs2 = γ p0 /ρ0 the sound velocity. We write the first of (9.83) in the perturbed quantities to obtain ∂B (9.86) ∇ × (v × B0 ) = ∂t Take the derivative with respect to t of (9.85)

Appendix

309

  B0 ∂ 2v vs2 ∂ρ ∂B + =0 + ∇ × ∇× ∂t 2 ρ0 ∂t μ0 ρ0 ∂t Substituting ∂ρ/∂t from (9.82) and ∂B/∂t from (9.86) we have ρ0

∂ 2v = vs2 ∇(ρ0 ∇ · v) − B0 × {∇ × [∇ × (v × B0 )]} ∂t 2

(9.87)

We define the vector analogous to (9.72) vA = √

B0 ρ0 μ0

(9.88)

And (9.87) becomes ρ0

∂ 2v = vs2 ∇(ρ0 ∇ · v) − vA × {∇ × [∇ × (v × vA )]} ∂t 2

(9.89)

This rather complicated equation conceals a wave equation and to show that we use a development in Fourier component writing v(r, t) = v(k, ω) exp[i(ωt + k · r]] So that ∂ 2 /∂t 2 → −ω2 and ∇ → ik and (9.89) becomes −ω2 v + vs2 k(k · v) − vA × {k × [k × (v × vA )]} Following the rules of vector calculus [k × (v × vA )] = (k · vA )v − (k · v)vA That is once substituted in the previous equation gives the results (9.77) [ω2 − (k · vA )2 ]v − (vs2 + v2A )(k · v)k + (k · vA )(v · vA )k + (k · vA )(k · v)vA = 0 (9.90) The different modes implied by such relation have been discussed before.

The Chaotic Dynamo Sir Edward Bullard published his paper on the homopolar dynamo in 1955. We have discussed that before and found that the main results are a possible instability of the dynamo when an additional resistance is added to the equivalent circuit. In normal

310

9 Magnetohydrodynamics

conditions, the dynamo gives rise to a steady oscillation but not to a change in the sign of the current and then of the magnetic field. A couple of years later Tsuneji Rikitake published a paper in which he studied the coupling of two disk dynamos that produced a chaotic behaviour. At that time (1957), Lorenz was to yet publish his seminal paper (1963) so that Rikitake reported now inversion of the current and magnetic field but failed to discover Chaos. About 40 years later Raymond Hide showed some inconsistency in the Rikitake disk dymano. Before going to the description of the double dynamo, we need to return to the Bullard dynamo and find some strange behaviour that could be an indication of some peculiar behaviour if we consider an external load of resistance R1 and inductance L 1 the equations for the dynamo become   Mω R dI =I − dt L L Mω d I1 R1 = I− I1 dt L1 L1 G M dω = − I (I + I1 ) dt C C

(9.91)

where the symbols are the same as in (9.46). We now make the following transformations: I =



G/M I  ,

I1 =



G/M I1 ,

ω=



G L/C Mω ,

t=

 C L/G Mt  (9.92)

So that (9.91) becomes     dI R LC 1/2   =I ω − dt  L MG   L   R1 LC 1/2  d I1 = ωI − I1 dt  L1 L1 M G dω = 1 − I  (I  + I1 ) dt 

(9.93)

We know that the dynamo has an unstable behaviour if L/R < L 1 /R1 so that nondimensional parameters appearing in (9.93) which determine the behavior should be R L1 L R1

R1 L1



LC MG

and

L L1

In the Bullard paper, the parameters were chosen in such a way that (9.93) read dI = I  (ω − 4), dt 

d I1 = I  ω − 2I1 , dt 

dω = 1 − I  (I  + I1 ) dt 

Appendix

311

12

7

ω‘ I’1 I’

10

5

6

4

I’1

Amplitude

8

6

4

3

2

2

0

1

-2

0

-4

-1 0

20

40

60

80

100

-1

0

1

Time

2 I’

3

4

5

Fig. 9.8 On the left we show the results for the Bullard dynamo with an additional load. Notice the growing instability and the change of sign. On the right, the phase space diagram is represented for i  and ω and the curves have a counterclockwise sense

The stability of the solution can be found as usual assuming the perturbation changes as ex p( pt) so that the characteristic equation becomes p3 + 2 p2 +

5 p+4=0 3

The roots of this equation are −2.1 and 0.0545 ± 1.3761i. In t  units, this corresponds to having a period for small oscillations of 2π/1.3761 = 4.57. Small disturbances will grow of a factor e in (0.0545 × 4.57)−1 = 4.02 period. Figure 9.8 shows some of these solutions. The additional electrical load produce oscillations that grows in time but again do not show sign of inversion. The double disc dynamo is shown in Fig. 9.9 and the equations of motions are readily written M1 ω1 R1 d I1 = I2 − I1 dt L1 L1 M2 ω2 d I2 R2 = I1 − I2 dt L2 L2 G1 dω1 M1 = − I1 I2 dt C1 C1 G2 dω2 M2 = − I1 I2 dt C2 C2

(9.94)

Where the symbols are the same as before simply refer to the different dynamo. We make them all equals so that L1 = L2 = L ,

R1 = R2 = R,

Then we use the normalization

M1 = M2 = M,

C1 = C2 = C

G! = G2 = G

312

9 Magnetohydrodynamics ω1

Fig. 9.9 The Rikitake dynamo is made up of two coupled Bullard dynamos. The current from dynamo 1 creates the field for the dynamo 2 and vice versa.

ω2

i1

i2

 I1 =

G x, M

 I2 =

G y, M

 ω1 =

 GL GL z, ω2 = (z − a), CM CM   CM  C t= t, v= GL GLM

(9.95)

where a and v are two parameters. We find the system x˙ = −vx + zy y˙ = −vy + (z − a)x z˙ = 1 − x y

(9.96)

It is easy to show that the volume in the phase space decrease with time that is V˙ =

 ∇ · Fd V

where F has components x, ˙ y˙ , z˙ . This means the system is dissipative like the Lorenz system. From (9.95), it is evident that ω1 − ω2 = a and is constant with time. The equilibrium point can be obtained putting to zero the system (9.96) so that vx0 − z 0 Y0 = 0,

vy0 − (z 0 − a)x0 = 0,

x0 y0 − 1 = 0

(9.97)

And putting in the third equation k = x0 with k parameter, we have the solution x0 = ±k,

y0 = ±1/k,

z 0 = vk 2

(9.98)

where v(k 2 − 1/k 2 ) = a. The stability of such points carried out with the usual Jacobian method shows the two points are unstable. As shown in Fig. 9.10, the resulting trajectories in the phase space develop in two planes one parallel to the x − y plane assuming both positive and negative values that is in principle changing the sign of the magnetic field. Hide in a 1995b paper pointed out the effects of introducing mechanical friction in the Rikitake dynamo. This may be done by adding a term like γ ω1 on the right-hand

Appendix

313 60 40 20

x

60

z

0

40

-20

20

-40

0

-60 60

-20 -40

40

-60 60

20

y

40 20

y

0 -20

0

-40

-20 -40 -60

-60

-40

-20

0

20

40

60

-60

1

2

3

4

5

6

7

8

9

10

60

x

40 20

z

0 -20 -40 -60

TIME

Fig. 9.10 The solution of the Rikitake dynamo with the parameters v = a = 0.1. On the left, the phase space diagram is shown with the characteristic trajectories lying in two different planes. On the right, the behaviour of the single variables as a function of time. Notice the slight decrease of the amplitudes with time

side of third Eq. (9.94) and a similar term on the last equation. So we subtract the resulting equations d (ω2 − ω1 ) = −γ (ω2 − ω1 ) (9.99) dt with the solution ω2 (t) − ω1 (t) = [ω2 (0) − ω1 (0)] exp(−γ t)

(9.100)

The difference between angular velocity is shown to be no longer constant. The reduction with time of this difference precludes any chaotic behaviour of the dynamo.

The Bow Shock of the Magnetosphere A nice application of what we have learned so far is the study of the interaction of the solar wind with the magnetic dipole field of the Earth. The solar wind is a flow of plasma originating from the Sun. Near the Earth 95% of is made up of ionized hydrogen, and 4% is doubly ionized helium. The speed of the particle is between 200 and 800 km s−1 and the density between 3 and 10 cm−3 . It is then possible to define two Mach numbers referred to the sonic waves Ms = V /vs and Alfvenic waves

314

9 Magnetohydrodynamics interplanetary magnetic field lines

magnetosheat

ψ

downstream region

bow shock magnetotail

upstream region

Fig. 9.11 A very simplified scheme of the interaction of the solar wind with the Earth magnetic field

M A = V /V A and these numbers are very large of the order of 10 so that the solar wind flow is supersonic. The interaction between the solar wind and the dipole field is determined by the balance between the kinetic energy of the wind and the magnetic pressure. It can be shown that this balance occurs at a distance  d=a

Be2 2π nm p V 2

1/6 (9.101)

where a is the equatorial radius of the Earth, Be the magnetic field on the equatorial plane, n the density of the solar wind, m p the proton mass and V the velocity. This gives a ratio d/a ≈ 10. Because of the supersonic flow a bow shock is formed upstream that has some peculiar characteristics. For example, it does not move like a shock wave produced by an airplane. Also because of the low density and the very long mean free path this shock is collisionless. After crossing the bow shock, the flow becomes slower, denser and hotter and its kinetic pressure compresses the magnetic field like shown in Fig. 9.11. The boundary between the bow shock and the internal cavity formed is called magnetopause. If we call ψ the angle between the normal to the discontinuity and the direction of the flow and consider a cylinder of length V and unitary section, its projection in the incoming direction is V cos ψ and the flux F = ρV cos ψ is the same upstream and downstream. The tangential momentum carried by the flux is then F V sin ψ = ρV 2 sin ψ cos ψ. The component of the tangential velocity V sin ψ is constant across the discontinuity.

Appendix

315

We now denote with suffixes u and d the upstream and downstream quantities so that vu = V sin ψ and (9.102) F = ρu vu = ρd vd If we call Pu and Pd the pressures upstream and downstream we should also have ρu vu2 + Pu = ρd vd2 + Pd

(9.103)

To get an energy equation, we observe that the energy carried out by the flux is given by F( 21 V 2 + ) where  is the internal energy per unit mass. We must consider that when the cylinder of length V is pushed through the discontinuity it performs a work pV = F P/ρ. The balance is then 1 2 Pu 1 Pd vu + u + = vd2 + d + 2 ρu 2 ρd

(9.104)

We have now F 2 /ρu = ρu vu2 and F 2 /ρd = ρd vd2 and from (9.103) we get 

1 1 F = (Pd − Pu ) − ρu ρd

−1

2

(9.105)

And similarly Eq. (9.104) gives   1 1 1 −1 u − d + (Pu + Pd ) − =0 2 ρu ρd

(9.106)

This relation simply says that the change in internal energy is given by the work done by the average pressure between the upstream and downstream values. We now rewrite the Mach number as  ρu Mu = vu γ Pu with γ the ratio of the specific heat. Based on the above relation in the approximation of large Mach numbers (M1  1), it is possible to find the ratios between densities, pressure and temperature on both sides of the bow shock. We have γ +1 ρd , = ρu γ −1

Pd 2γ Mu2 , = Pu γ +1

Td 2γ (γ − 1)Mu2 = Tu (γ + 1)2

(9.107)

We notice that the downstream density is larger by a factor of 4 and independent of the Mach number for γ = 5/3 while both compression ratio and temperature grow with the square of the Mach number according to 1.25Mu2 and 0.3225Mu2 , respectively. The relation between the Mach numbers on the other hand is

316

9 Magnetohydrodynamics

Md2 =

2 + (γ − 1)Mu2 2γ Mu2 − γ + 1

(9.108)

γ −1 2γ

(9.109)

For large Mach numbers it becomes Md2 =

In all this treatment taken from the very useful book by Bertotti and Farinella we have neglected the presence of the electromagnetic fields. Because of the zero divergence of the magnetic field and the irrotational nature of the electric field both the fields are continuous across the shock.

The Rayleigh Taylor Instability with Magnetic Field We treated the Rayleigh Taylor instability in paragraph 6.3 and we now ask what happens if we add a magnetic field. This is a problem familiar to the astrophysicists so we recur to its treatment in the book by Clarke and Carswell on astrophysical fluid dynamics. We assume two fluids of density ρ1 and ρ2 stratified parallel to the x direction and subject to a gravitational field in the z direction (see Fig. 9.11). The equations for conservation of mass and for the motions are ∂ρ + ∇ · (ρv) =0 ∂t  ∂v ρ + v · ∇v =j × B − ∇ p + ρg ∂t

(9.110)

We also assume g to have components only along z and B to have the unperturbed component (B0 ) along x. A perturbation to the displacement of the fluid is performed in the form η = η(x) exp[i(kx − ωt)] (9.111) So that v=

∂η = −iωη ∂t

(9.112)

The second Eq. (9.110) can be written in its components −ρω2 ηx =ikp1 ∂p − ρg + (j × B0 ) −ρω2 ηz = − ∂z

(9.113)

Appendix

317

Fig. 9.12 The scheme for the Rayleigh Taylor instability in the presence of a magnetic field B. The two fluids are initially stratified in the direction of B. The sine wave is the perturbation

z

ρ1

B x

ρ2

where p1 is the perturbation pressure. We now assume that the magnetic field is given by the basic field (B − 0) plus a perturbation B = B0 + B1 where the perturbation has the form (Fig. 9.12) B1 = B1 (x) exp[i(kx − ωt)] And using (9.93) we have − iωB1 = ∇ × (−iωη × B0 ),

⇒ B1 = ∇ × (η × B0 )

(9.114)

Consider now that the product in parenthesis has the only component in the direction normal to the plane x − z so that the rotor will have the components B1z = B0 ikηz ,

B1x = −B0

∂ηz ∂z

(9.115)

From Eq. (9.93), the only current component will be along the y direction (normal to the x − z plane) and is given by  jx = B0

d 2 ηz − k 2 ηz dz 2

 (9.116)

The terms corresponding to pressure gradient and the density we consider first of all the first of (9.113) so that p1 = −

ω 2 dη ρω2 z ηx = ρ ik k dz

(9.117)

where we have used (9.112). We have for the gradient

  d ω 2 dηz dp1 = ρ dz dz k dz

(9.118)

The term involving the density perturbation can be written as ρ1 = −ηz

dρ0 dx

(9.119)

318

9 Magnetohydrodynamics

The second Eq. (9.113) becomes  2     

k dηz dρ0 1 k B0 2 d 2 ηz 2 g − k η =0 ρ0 − k 2 ρ0 ηz − ηz − z dz ω dx μ0 ω dz 2 (9.120) This equation must be integrated along the interface and must satisfy the conditions to be finite for z → ±∞ and for ηz to be continuous across the interface. These conditions produce the dispersion relation d dx



ρ2 − ρ1 ω = kg ρ2 + ρ1



2

2 + μ0



k 2 B02 ρ2 + ρ1

 (9.121)

Compared with the expression (6.37) we see that the effect of the magnetic field is of stabilization. As a matter of fact, the additional term on the right is always positive. However, we have to keep in mind that this case refers to a field parallel to the interface. In general (9.121) is substituted by 

ρ2 − ρ1 ω = kg ρ2 + ρ1 2

 +

2 (k · B0 )2 μ0 ρ2 + ρ1

We see that the stabilization is still effective if the second term is larger than the first one that is for a wavelength λ=

2B02 2π < cos θ 2 k gμ0 ρ

where ρ = ρ1 − ρ2 and θ is the angle between k and B0 . One of the effects of the presence of the magnetic field is the modification of the convection and a lower gas pressure necessary to achieve balance in the outer region of solar photosphere. This could account for the dark appearance of the sunspot. Chandrasekhar notes that the additional term has the same effect of an additional surface tension as we saw in Chap. 6. This tension could arise from the magnetic pressure term B 2 /2μ0 . Finally, in the case of homogeneous plasma ρ1 = ρ2 = ρ the frequency (9.121) is the same as the Alfven wave.

Plasma Confinement One of the most promising energy sources for the future could be nuclear fusion. One of the techniques is to produce very-high-temperature plasma where fusion can take place according to the reaction deuterium + tritium ⇒ α + neutr on + 22.4 MeV

(9.122)

Appendix

319

where the fast neutron can be absorbed in a lithium blanket and the heat drives a generator. A crucial point is the confinement of such a hot plasma than cannot be done in a vessel and the idea is to use magnetic fields instead. To understand how it works we need to reconsider the equation of motion, for example, (9.84) where the magnetic force is given by Fm = j × B =

(∇ × B) × B B2 (B · ∇)B =− + μ0 2μ0 μ0

(9.123)

And the equation of motion becomes ∂v = −∇ ρm ∂t



B2 p+ 2μ0

 +

(B · ∇)B μ0

(9.124)

We now remember that according to the frozen field hypothesis for a fluid of infinite conductivity the field lines move with the fluid so that if the magnetic field is static then also the fluid will not move, that is it will be confined. In this case according to Eq. (9.124) we have the equilibrium  ∇

p+

B2 2μ0

 =

(B · ∇)B μ0

(9.125)

This relation in principle gives the value of B able to confine a conductive fluid at pressure p. The equilibrium given by (9.125) is equivalent to j × B = ∇p so that the pressure gradient is normal to both the current density and the magnetic field so we have B · ∇ p = 0, j·∇p = 0 (9.126) The first of these equations state that the pressure must be constant along the magnetic field lines that is the field lines must lie along surface at constant pressure. The same is true for the second so that surfaces with p = const (isobars) normal to ∇ p are both magnetic surfaces and current surfaces. Also from the second relation (9.83) ∇ × B = μ0 j it follows (9.127) ∇ · (∇ × B) = μ0 ∇ · j = 0 So that ∇ ·j=0

(9.128)

This means that both the magnetic field and the current have null divergence so that both the field and current lines are closed. It can be shown that if the variations of the variables B, j, p are smooth and if the magnetic and isobaric surfaces are contained in a finite volume then the surfaces assume a toroidal shape. We now examine two cases of confinement starting with the linear geometry as shown in Fig. 9.13a. The geometry corresponds to a solenoid of infinite length that produces an axial field B. In this case according to (9.125), the maximum pressure

320

9 Magnetohydrodynamics (a) Induced currents

jθ B z

Bθ

r

z

θ jz

(b)

Fig. 9.13 The scheme for θ -pinch confinement in (a) and z-pinch in (b). In a, a field is generated along the axis of the cylinder while in b an axial current is generated by the azimuthal field Bθ

that can be confined is given by Pmax = B 2 /2μ0 We can add some detail to this method (known also as θ − pinch) by considering a cylindrical coordinate system so that we have (j × B)r = (∇ p)r (∇ × B)θ = (μ0 j)θ which translates into jθ Bz =

∂ p ∂r

∂ Bz = −μ0 jθ ∂r

(9.129)

(9.130)

So that we can eliminate j to obtain ∂ ∂r So that



 B2 Bz2 + p = 0, ⇒ z + p = const 2μ0 2μ0 Bz2 B2 + p = ext 2μ0 2μ0

(9.131)

Appendix

321

where Bext is the external applied field. In practice the axial field produces an azimuthal current (called diamagnetic) that develops a pressure gradient according to j × B = ∇p that counteracts the gas pressure. Another method of confinement (known as Z-pinch) utilizes a current along the axis of an infinite cylinder of radius a (Fig. 9.13b). This current generates an azimuthal magnetic field Bθ so that the force j × B is again in the radial direction. In this case Eq. (9.125) reduces to in a cylindrical coordinate system d dp 1 =− Bθ (r Bθ ) dr μ0 r dr

(9.132)

Multiplying both sides by r 2 and integrating between 0 and a we have 

a

r2 0



1 dp dr = − dr μ0

a

r Bθ 0

d (r Bθ )dr dr

(9.133)

Integrating by parts we obtain 

a

2

1 [a Bθ (a)]2 2μ0

r pdr =

0

(9.134)

If the plasma can be considered a perfect gas the pressure is given by p = nkT with k Boltzmann constant and n number density. So that  2 0

a

 r pdr = 2

a

r nkT dr =

0

kT π



a

2πr ndr =

0

kNT π

(9.135)

where N is the number of particles per unit length 

a

N=

2πr ndr

(9.136)

0

From the circulation theorem we have a Bθ =

μ0 I 2π

And using (9.134) we obtain the a relation between temperature and current kT N =

μ0 I 2 8π

(9.137)

The plasma temperature is then proportional to the square of the current and decreases with increasing N . The Z-pinch method is unstable and an additional axial magnetic field must be added to increase the stability. However, the Z-pinch is important because on it is based the most used Tokamak technique.

322

9 Magnetohydrodynamics

To show the requirements for fusion we can make a numerical example out of the (9.137) equation (known also as Bennet equation). We use an equivalent temperature of 10 keV corresponding to a temperature 104 × 1.6 × 10−19 /1.38 × 10−23 = 1.15 × 108 K and a current of 1.3 × 106 A. The density N can be calculated from (9.137) 10−7 (1.3 × 106 )2 μ0 I 2 = N= ≈ 5 × 1019 m −1 8π kT 2 104 × 1.6 × 10−19 As a didactic application of the Z-pinch effect, we can mention the so-called can crusher. It consists of putting a soft drink can in a solenoid where a strong current is produced by a discharge of a large capacitor. The field developed by the current produces a radial force (according to Eq. (9.123)) that crushes the can. Typical values are 900 V with a 7–8 ×10−3 F storing about 3000 J of energy. When discharged this may produce a magnetic field of the order of 10 T with a power of about 20 M watt. A natural Z-pinch can be observed in the lighting bolts.

References Textbooks Bertotti B, Farinella P (1990) Physics of the earth and the solar system. Kluwer, Dordrecht Chandrasekhar S (1961) Hydrodynamic and hydromagnetic stability. Dover, New York Clarke CJ, Carswell RF (2014) Principles of astrophysical fluid dynamics. Cambridge University Press, Cambridge Ghil M, Childress S (1987) Topics in geophysical fluid dynamics: atmospheric dynamics, dynamo theory and climate dynamics. Springer, Berlin Pringle J, King A (2007) Astrophysical flows. Cambridge University Press, Cambridge Reid WH (ed) (1978) Mathematical problems in the geophysical sciences, 2. Inverse problems, dynamo theory and tides. American Mathematical Society, Providence

Articles Bullard E (1955) The stability of a homopolar dynamo. Math Proc Camb Philos Soc 51:744 Busse FH, Simitev RD (2015) Planetary dynamos. Treatise Geophys 10:239 Hide R (1995) Structural stability of the Rikitake, dynamo. Geophys Res Lett 22:1057 Ito K (1980) Chaos in the Rikitake two-disc dynamo system. Earth Planet Sci Lett 51:451 Rikitake T (1958) Oscillations of a system of disk dynamo. Math Proc Camb Philos Soc 54:89 Schubert G, Soderlund KM (2011) Planetary magnetic fields: observations and models. Phys Earth Planet Int 187:92 Stevenson DJ (2013) Planetary magnetic fields. Earth Planet Sci Lett 208:1

Index

A Adams Williamson equation, 7 Adiabatic gradient, 146 motion, 3 Advection, 52, 68, 71, 195, 205, 228, 236 warm, cold, 21 Aerodynamics, 85 Angular momentum, 33, 44, 186 Archimedes’ law, 4 Atmospheric pressure, 2

B Baroclinic fluid, 37 Barotropic, 2, 34, 294 instability, 190, 192 Bénard convection, 73 Bernoulli theorem, 23 Bifurcation, 163, 164, 217, 226, 230, 238, 244 subcritical, 165 supercritical, 165 Blasius theorem, 56 Blausius theorem, 92 Borda mouthpiece, 26 Boundary layer, 56, 76 Brownian motion, 210 Brunt Väisälä, 147 Burgers equation, 236

C Cascade rate, 272 Centrifugal acceleration, 8, 10 Chaos, 216 Circulation, 31 © Springer Nature Switzerland AG 2020 G. Visconti and P. Ruggieri, Fluid Dynamics, https://doi.org/10.1007/978-3-030-49562-6

Climate potential, 249 Climate variability, 213 Closure problem, 253, 262, 264 Complex variables, 94, 97 Compressible fluid, 24, 49 Conformal mapping, 99 Continuity equation, 19 Coriolis acceleration, 39, 77, 151, 187, 193, 301 Cowling theorem, 297 Cumulative distribution function, 256

D D’Alembert paradox, 91 Debye length, 290 Density, 7, 13, 16 Derivative, substantial, Lagrangian, 20, 22 Diffusion, 51 Diffusion equation, 51, 206 analytical solution, 273 Dimensional analysis, 52, 275 Distribution exponential, 258 lognormal, 260 normal, 208, 258 uniform, 258 Divergence, 20, 21, 32, 125, 196 Drag coefficient, 54, 64, 106 Dynamo, 297 alpha and omega effects, 304 Bullard, 310 chaotic, 309 Faraday, 298 Rikitake, 310 turbulent, 302 323

324 E Eady model, 197 Earthquakes, 7 Earth’s interior, 6, 160 Eddies, 253 stress, 264 viscosity, 265 Efflux coefficient, 26 Ekman pumping, suction, 81 Ekman pumping, 83 ENSO, El Nino, 231 Enstrophy, 266 cascade, 272 transfer, 270 Entropy, 3, 142, 146 Equations of motion, 22 Equatorial radius, 8, 314 Euler equation, 36, 141, 167 Exchange coefficient, 265 Expected value, 257

F Faraday Dynamo, 298 self-excited Dynamo, 298 Feigenbaum relation, 217 Flattening, 9 Flow from a tank, 25 Flow past a sphere, 60 Fractals, 219 Frisbee, flight of, 107 Froude number, 54 Frozen field, 294

G Gaussian noise, 208, 255 Geostrophic approximation, 39, 186 equilibrium, 76, 152 motion, 38, 40, 42 Geostrophic approximation, 44 Golf balls, 106 Gravitational potential, 2, 9 Great Red Spot, 113 Ground effects, 110 Gulf current, 79

H Hydrostatic adjustment, 9

Index approximation, 1 earth interior, 6 equation, 1, 13 equilibrium, 2, 4 I Incompressible, fluid, 25, 42, 49 Inertia, moment of, 9, 33 Instabilities, 163 baroclinic, 192 barotropic, 190 centrifugal, 184 gravitational (Jeans), 166 inertial, 186 Kelvin-Helmholtz, 172 Rayleigh Taylor, 170, 181, 316 shear flows, 168 symmetric, 187 Irrotational flow, 85 Isostasy, 5, 13 Isothermal, 14 J Jupiter, 16, 43 K Karman constant, 265 Kelvin circulation theorem, 34 Kinetic energy, 55, 123, 131, 194, 211, 266, 314 Kolmogorov and parachutes, 275 constant, 272 length scale, 284 spectrum, 273 wave number, 268 Kutta-Zhukhovsky lift theorem, 92, 101 L Laminar flow, 205, 253 Lane–Emden equation, 13 Laplace equation, 86 Lift deflection of air stream, 95 Lifting flow, 91 Line vortex, 87 Logistic map, 216 Loop oscillator, 222 Lorenz-Malkus water wheel, 240 Lorenz system, 225 Lyapunov exponents, 218

Index

325

M Mach number, 56 Magnetic diffusion coefficient, 293 Magnetic field, 293 Rayleigh Taylor instability, 316 toroidal, poloidal, 298 Magnetohydrodynamics, 287 equations, 288 waves, 304 Magnetosphere, bow shock, 313 Magnus effect, 91 Mean free path, 253, 264, 277 Mixing length, 253, 263, 277 Monin-Obukhov length, 279

Q Quantum mechanics, 35

N Navier Stokes, 50 Navier Stokes equation, 50, 53, 58, 205, 255, 266, 276 Non lifting flow, cylinder, 89 Non-linearities, in Navier–Stokes equation, 205

S Sailing, physics of, 103 Saturn, 16 Scallop theorem, 65 Sea breeze, 37 Shape of the Earth, 8 Ship waves, 133 Similarity theory, wind profile, 279 Sodar, 282 Solitons, 137 bump, 140 single, 140 Solitons, two soliton solutions, 140 Stability, 3, 146, 163, 217 Stagnation point, 90 Stellar structure, 13 Stochastic resonance, 247 Stokes drag, 63 Stokes formula, 212 Stokes paradox, 68 Stratified fluid, 3, 174 Stratified shear flows, 174 Stream function, 40 Stream tube, 24 Supernovae remnants, 179 Supersonic flow, 119, 314

O Oblateness, 9 Oseen correction, 68

P Palinstrophy, 268 Pendulum, 220 chaotic, 237 non linear, 220 Pinch-effect, 320 Pipe flow, 58 Pitot tube, 27 Plane source, doublet, 87, 88 Planetary dynamos, 297 Plasma confinement, 318 Plasma frequency, 291 Plasma oscillations, 287, 288 Poincarè maps, 220 sections, 219 Poiseuille flow, 58 Polytrope, 14 Potential flow, 86 Potential temperature, 146, 187, 278 Probability density function, 255 Propeller, thrust, 27

R Rayleigh friction, 82 Rayleigh Taylor Earth, 181 Refractive seismology, 160 Reynolds number, 53, 60 life at low, 64 Richardson number, flux, 175 River flow, 276 Rossby number, 39, 114 waves, 157, 192, 233

T Taylor Proudman, theorem, 42 Tea leaves, 77 Thermocline, 232 Thermohaline circulation, 227 Theta-pinch, 320 Thrust, propeller, 27 Tokamak, 321 Tsunami, 117, 143

326 Turbulence, 56, 253 fluxes, 254 homogeneous, 254 isotropic, 254, 268 statistical description, 255 two dimensional, 266

V Valley wind, 70 Velocity potential, 85 Vena contracta, 26, 30 Viscosity, 47 coefficient, 49 Viscous stress, 57, 280 Vortex, 87 Vortex line, 32, 87 Vorticity, 31 absolute, 44 and magnetic field, 293 equation, 44, 79, 197, 266, 293 geostrophic, 44

Index planetary, 44 potential, 113 relative, 44 Vorticity equation, 43

W Wave drag, 135 Waves, 117 atmospheric, 146 buoyancy, 147 capillary, 128 deep water, 127 equatorial, 154 Kelvin, 152 mountain, 149 ocean, 151 seismic, 6, 158 shallow water, 128 sound, 117 Western boundary current intensification, 79 Wind generator, 28