Theory of Vortex Sound (Cambridge Texts in Applied Mathematics) by M. S. Howe

http://www.cambridge.org/9780521812818

This page intentionally left blank

Theory of Vortex Sound

Theory of Vortex Sound is an introduction to the theory of sound generated byhydrodynamic flows. Starting with a review of elementary theoretical acoustics,the book proceeds to a unified treatment of low Mach number vortex-surfaceinteraction noise in terms of the compact Green’s function. Problems are pro-vided at the end of each chapter, many of which can be used for extended studentprojects, and a whole chapter is devoted to worked examples.

It is designed for a one-semester introductory course at the advanced un-dergraduate or graduate levels. Great care is taken to explain underlying fluidmechanical and acoustic concepts, and to describe as fully as possible the stepsin a complicated derivation.

M.S. Howe has been Professor in the Department of Aerospace and MechanicalEngineering at Boston University since 1992. He is a Fellow of the Institute ofAcoustics (U.K.) and of the Acoustical Society of America.

Cambridge Texts in Applied Mathematics

Maximum and Minimum PrinciplesM. J. Sewell

SolitonsP. G. Drazin and R. S. Johnson

The Kinematics of MixingJ. M. Ottino

Introduction to Numerical Linear Algebra and OptimisationPhilippe G. Ciarlet

Integral EquationsDavid Porter and David S. G. Stirling

Perturbation MethodsE. J. Hinch

The Thermomechanics of Plasticity and FractureGerard A. Maugin

Boundary Integral and Singularity Methods for Linearized Viscous FlowC. Pozrikidis

Nonlinear Wave Processes in AcousticsK. Naugolnykh and L. Ostrovsky

Nonlinear SystemsP. G. Drazin

Stability, Instability, and ChaosPaul Glendinning

Applied Analysis of the Navier–Stokes EquationsC. R. Doering and J. D. Gibbon

Viscous FlowH. Ockendon and J. R. Ockendon

Scaling, Self-Similarity, and Intermediate AsymptoticsG. I. Barenblatt

A First Course in the Numerical Analysis of Differential EquationsArieh Iserles

Complex Variables: Introduction and ApplicationsMark J. Ablowitz and Athanassios S. Fokas

Mathematical Models in the Applied SciencesA. C. Fowler

Thinking About Ordinary Differential EquationsRobert E. O’Malley

A Modern Introduction to the Mathematical Theory of Water WavesR. S. Johnson

Rarefied Gas DynamicsCarlo Cercignani

Symmetry Methods for Differential EquationsPeter E. Hydon

High Speed FlowC. J. Chapman

Wave MotionJ. Billingham and A. C. King

An Introduction to MagnetohydrodynamicsP. A. Davidson

Linear Elastic WavesJohn G. Harris

Vorticity and Incompressible FlowAndrew J. Majda and Andrea L. Bertozzi

Infinite Dimensional Dynamical SystemsJames C. Robinson

An Introducion to Symmetry AnalysisBrian J. Cantwell

Backlund and Darboux TransformationsC. Rogers and W. K. Schief

Finite-Volume Methods for Hyperbolic ProblemsRandall J. LeVeque

Introduction to Hydrodynamic StabilityP. G. Drazin

Theory of Vortex SoundM. S. Howe

Theory of Vortex Sound

M. S. HOWEBoston University

Cambridge, New York, Melbourne, Madrid, Cape Town, Singapore, São Paulo

Cambridge University PressThe Edinburgh Building, Cambridge , United Kingdom

First published in print format

- ----

- ----

- ----

© Cambridge University Press 2003

2002

Information on this title: www.cambridge.org/9780521812818

This book is in copyright. Subject to statutory exception and to the provision ofrelevant collective licensing agreements, no reproduction of any part may take placewithout the written permission of Cambridge University Press.

- ---

- ---

- ---

Cambridge University Press has no responsibility for the persistence or accuracy ofs for external or third-party internet websites referred to in this book, and does notguarantee that any content on such websites is, or will remain, accurate or appropriate.

Published in the United States of America by Cambridge University Press, New York

www.cambridge.org

hardback

paperbackpaperback

eBook (NetLibrary)eBook (NetLibrary)

hardback

http://www.cambridge.org

http://www.cambridge.org/9780521812818

To Shon Ffowcs Williams

Contents

Preface page xiii

1 Introduction 11.1 What is Vortex Sound? 11.2 Equations of Motion of a Fluid 21.3 Equation of Linear Acoustics 41.4 The Special Case of an Incompressible Fluid 71.5 Sound Produced by an Impulsive Point Source 101.6 Free-Space Green’s Function 121.7 Monopoles, Dipoles, and Quadrupoles 131.8 Acoustic Energy Flux 181.9 Calculation of the Acoustic Far Field 20Problems 1 23

2 Lighthill’s Theory 252.1 The Acoustic Analogy 252.2 Lighthill’s v8 Law 292.3 Curle’s Theory 322.4 Sound Produced by Turbulence Near a Compact Rigid Body 362.5 Radiation from a Noncompact Surface 37Problems 2 39

3 The Compact Green’s Function 413.1 The Influence of Solid Boundaries 413.2 The Helmholtz Equation 443.3 The Reciprocal Theorem 463.4 Time-Harmonic Compact Green’s Function 49

ix

x Contents

3.5 Compact Green’s Function for a Rigid Sphere 533.6 Compact Green’s Function for Cylindrical Bodies 583.7 Symmetric Compact Green’s Function 633.8 Low-Frequency Radiation from a Vibrating Body 653.9 Compact Green’s Function Summary and Special Cases 70Problems 3 79

4 Vorticity 824.1 Vorticity and the Kinetic Energy of Incompressible Flow 824.2 The Vorticity Equation 844.3 The Biot–Savart Law 884.4 Surface Force in Incompressible Flow Expressed in Terms

of Vorticity 934.5 The Complex Potential 1004.6 Motion of a Line Vortex 106Problems 4 112

5 Vortex Sound 1145.1 The Role of Vorticity in Lighthill’s Theory 1145.2 The Equation of Vortex Sound 1165.3 Vortex–Surface Interaction Noise 1245.4 Radiation from an Acoustically Compact Body 1285.5 Radiation from Cylindrical Bodies of Compact Cross Section 1305.6 Impulse Theory of Vortex Sound 131Problems 5 132

6 Vortex–Surface Interaction Noise in Two Dimensions 1366.1 Compact Green’s Function in Two Dimensions 1366.2 Sound Generated by a Line Vortex Interacting with

a Cylindrical Body 1396.3 Influence of Vortex Shedding 1456.4 Blade–Vortex Interaction Noise in Two Dimensions 150Problems 6 154

7 Problems in Three Dimensions 1567.1 Linear Theory of Vortex–Airfoil Interaction Noise 1567.2 Blade–Vortex Interactions in Three Dimensions 1587.3 Sound Produced by Vortex Motion near a Sphere 1627.4 Compression Wave Generated When a Train Enters a Tunnel 166Problems 7 172

Contents xi

8 Further Worked Examples 1758.1 Blade–Vortex Interactions in Two Dimensions 1758.2 Parallel Blade–Vortex Interactions in Three Dimensions 1868.3 Vortex Passing over a Spoiler 1918.4 Bluff Body Interactions: The Circular Cylinder 1948.5 Vortex Ring and Sphere 1998.6 Vortex Pair Incident on a Wall Aperture 204

Bibliography 209Index 213

Preface

Vortex sound is the branch of fluid mechanics concerned with the conversionof hydrodynamic (rotational) kinetic energy into the longitudinal disturbanceswe call sound. The subject is itself a subsection of the theory of aerodynamicsound, which encompasses a much wider range of problems also involving, forexample, combustion and ‘entropy’ sources of sound. The book is based on anintroductory one-semester graduate level course given on several occasions atBoston University. Most students at this level possess an insufficient grasp ofbasic principles to appreciate the subtle coupling of the hydrodynamic andacoustic fields, and many are ill-equipped to deal with the novel analyticaltechniques that have been developed to investigate the coupling. Great carehas therefore been taken to discuss underlying fluid mechanical and acousticconcepts, and to explain as fully as possible the steps in a complicated derivation.

A considerable number of practical problems occur at low Mach numbers(say, less than about 0.4). It seems reasonable, therefore, to confine an intro-ductory discussion specifically to low Mach number flows. It is then possibleto investigate a number of idealized hydrodynamic flows involving elementarydistributions of vorticity adjacent to solid boundaries, and to analyze in detailthe sound produced by these vortex–surface interactions. For a broad range ofsuch problems, and a corresponding broad range of noise problems encounteredin industrial applications, the effective acoustic sources turn out to be localizedto one or more regions that are small compared to the acoustic wavelength. Thispermits the development of a unified theory of sound production by vortex–surface interactions in terms of the compact Green’s function, culminating in aroutine procedure for estimating the sound, and providing, at the same time, aneasy identification of those parts of a structure that are likely to be importantsources of sound. Many examples of this type are discussed, and they are simpleenough for the student to acquire an intuitive understanding of the method of

xiii

xiv Preface

solution and the underlying physics. By these means the reader is encouraged toinvestigate both the hydrodynamics and the sound generated by a simple flow.Experience has shown that the successful completion of this kind of project,involving the implementation of a widely applicable yet standard procedure forthe prediction of sound generation at low Mach numbers, motivates a studentto understand the ostensibly difficult parts of the theory. One or more of theproblems appended to some of the later chapters can form the basis of a project.The final chapter contains a set of worked examples that have been investi-gated by students at Boston University. I wish to thank my former studentsH. Abou-Hussein, A. DeBenedictis, N. Harrison, M. Kim, M. A. Rodrigues,and F. Zagadou for their considerable help in preparing that chapter.

The mathematical ability assumed of the reader is roughly equivalent tothat taught in an advanced undergraduate course on Engineering Mathematics.In particular, the reader should be familiar with basic vector differential and in-tegral calculus and with the repeated suffix summation convention of Cartesiantensors (but a detailed knowledge of tensor calculus is not required). An el-ementary understanding of the properties of the Dirac δ function is desir-able (Lighthill, 1958), including its interpretation as the formal limit of anε-sequence, such as

δ(x) = ε

π (x2 + ε2), ε → +0.

Much use is made of the formula

δ( f (x)) =∑

n

δ(x − xn)

| f ′(xn)| ,

where the summation is over real simple roots of f (x) = 0.M. S. Howe

1Introduction

1.1 What is Vortex Sound?

Vortex sound is the sound produced as a by-product of unsteady fluid mo-tions (Fig. 1.1.1). It is part of the more general subject of aerodynamic sound.The modern theory of aerodynamic sound was pioneered by James Lighthillin the early 1950s. Lighthill (1952) wanted to understand the mechanisms ofnoise generation by the jet engines of new passenger jet aircraft that werethen about to enter service. However, it is now widely recognized that anymechanism that produces sound can actually be formulated as a problem ofaerodynamic sound. Thus, apart from the high speed turbulent jet – which maybe regarded as a distribution of intense turbulence velocity fluctuations that gen-erate sound by converting a tiny fraction of the jet rotational kinetic energy intothe longitudinal waves that constitute sound – colliding solid bodies, aeroenginerotor blades, vibrating surfaces, complex fluid–structure interactions in the lar-ynx (responsible for speech), musical instruments, conventional loudspeakers,crackling paper, explosions, combustion and combustion instabilities in rock-ets, and so forth all fall within the theory of aerodynamic sound in its broadestsense.

In this book we shall consider principally the production of sound by un-steady motions of a fluid. Any fluid that possesses intrinsic kinetic energy, thatis, energy not directly attributable to a moving boundary (which is largely with-drawn from the fluid when the boundary motion ceases), must possess vorticity.We shall see that in a certain sense and for a vast number of flows vorticity maybe regarded as the ultimate source of the sound generated by the flow. Ourobjective, therefore, is to simplify the general aerodynamic sound problem toobtain a thorough understanding of how this happens, and of how the soundcan be estimated quantitatively.

1

2 1 Introduction

Fig. 1.1.1. Typical vortex sound problems.

1.2 Equations of Motion of a Fluid

At time t and position x = (x1, x2, x3), the state of a fluid is defined when thevelocity v and any two thermodynamic variables are specified. Five scalarequations are therefore required to determine the motion. These equations arestatements of the conservation of mass, momentum, and energy.

1.2.1 Equation of Continuity

Conservation of mass requires the rate of increase of the fluid mass within afixed region of space V to be equal to the net influx due to convection across theboundaries of V. The velocity v and the fluid density ρ must therefore satisfy

1.2 Equations of Motion of a Fluid 3

the equation of continuity, which has the following equivalent forms

∂ρ

∂t+ div(ρv) = 0,

1

ρ

Dρ

Dt+ div v = 0,

div v = ρD

Dt

(1

ρ

)

, (1.2.1)

where

D

Dt= ∂

∂t+ v · ∇ ≡ ∂

∂t+ v j

∂

∂x j(1.2.2)

is the material derivative; the repeated suffix j implies summation over j =1, 2, 3. The last of Equations (1.2.1) states that div v is equal to the rate ofchange of fluid volume per unit volume following the motion of the fluid. Foran incompressible fluid this is zero, i.e., div v = 0.

1.2.2 Momentum Equation

The momentum equation is also called the Navier–Stokes equation; it expressesthe rate of change of momentum of a fluid particle in terms of the pressure p,the viscous or frictional force, and body forces F per unit volume. We consideronly Stokesian fluids (most liquids and monatomic gases, but also a good ap-proximation in air for calculating the frictional drag at a solid boundary) forwhich the principal frictional forces are expressed in terms of the shear coeffi-cient of viscosity η, which we shall invariably assume to be constant. Then themomentum equation is

ρDvDt

= −∇ p + η

(∇2v + 1

3∇(div v)

)+ F. (1.2.3)

Values of ρ, η and ν = η/ρ (the ‘kinematic’ viscosity) for air and water at10 C and one atmosphere pressure are given in the Table 1.2.1:

Table 1.2.1. Density and viscosity

ρ, kg/m3 η, kg/ms ν,m2/s

Air 1.23 1.764 × 10−5 1.433 × 10−5

Water 1000 1.284 × 10−3 1.284 × 10−6

4 1 Introduction

1.2.3 Energy Equation

This equation must be used in its full generality in problems where energy istransferred by heat conduction, where frictional dissipation of sound is occur-ring, when shock waves are formed by highly nonlinear events, or when soundis being generated by combustion and other heat sources. For our purposes itwill usually be sufficient to suppose the flow to be homentropic; namely, thespecific entropy s of the fluid is uniform and constant throughout the fluid, sothat the energy equation becomes

s = constant. (1.2.4)

We may then assume that the pressure and density are related by an equationof the form

p = p(ρ, s), s = constant. (1.2.5)

This equation will be satisfied by both the mean (undisturbed) and unsteadycomponents of the flow. Thus, for an ideal gas

p = constant × ργ , γ = ratio of specific heats. (1.2.6)

1.3 Equation of Linear Acoustics

The intensity of a sound pressure p in air is usually measured on a decibel scaleby the quantity

20 × log10

( |p|pref

),

where the reference pressure pref = 2 × 10−5 N/m2. Thus, p = p0 ≡ 1 atmo-sphere ( = 105 N/m2) is equivalent to 194 dB. A very loud sound ∼120 dBcorresponds to

p

p0≈ 2 × 10−5

105× 10( 120

20 ) = 2 × 10−4 1.

Similarly, for a ‘deafening’ sound of 160 dB, p/p0 ∼ 0.02. This correspondsto a pressure of about 0.3 lbs/in2 and is loud enough for nonlinear effects tobegin to be important.

The passage of a sound wave in the form of a pressure fluctuation is, ofcourse, accompanied by a back-and-forth motion of the fluid at the acoustic

1.3 Equation of Linear Acoustics 5

particle velocity v, say. We shall see later that

acoustic particle velocity ≈ acoustic pressure

mean density × speed of sound.

In air the speed of sound is about 340 m/sec. Thus, at 120 dB v ∼ 5 cm/sec; at160 dB v ∼ 5 m/sec.

In most applications the acoustic amplitude is very small relative to the meanpressure p0, and sound propagation may be studied by linearizing the equations.To do this we shall first consider sound propagating in a stationary inviscid fluidof mean pressure p0 and densityρ0; let the departures of the pressure and densityfrom these mean values be denoted by p′, ρ ′, where p′/p0 1, ρ ′/ρ0 1. Thelinearized momentum equation (1.2.3) becomes

ρ0∂v∂t

+ ∇ p′ = F. (1.3.1)

Before linearizing the continuity equation (1.2.1), we introduce an artificialgeneralization by inserting a volume source distribution q(x, t) on the right-hand side

1

ρ

Dρ

Dt+ div v = q, (1.3.2)

where q is the rate of increase of fluid volume per unit volume of the fluid, andmight represent, for example, the effect of volume pulsations of a small bodyin the fluid. The linearized equation is then

1

ρ0

∂ρ ′

∂t+ div v = q. (1.3.3)

Now eliminate v between (1.3.1) and (1.3.3):

∂2ρ ′

∂t2− ∇2 p′ = ρ0

∂q

∂t− div F. (1.3.4)

An equation determining the pressure p′ alone in terms of q and F is obtainedby invoking the homentropic relation (1.2.5). In the undisturbed and disturbedstates we have

p0 = p(ρ0, s), p0 + p′ = p(ρ0 + ρ ′, s) ≈ p(ρ0, s) +(∂p

∂ρ(ρ, s)

)0

ρ ′,

s = constant. (1.3.5)

6 1 Introduction

The derivative is evaluated at the undisturbed values of the pressure and density(p0, ρ0). It has the dimensions of velocity2, and its square root defines the speedof sound

c0 =√(

∂p

∂ρ

)s

, (1.3.6)

where the derivative is taken with the entropy s held fixed at its value in theundisturbed fluid. The implication is that losses due to heat transfer betweenneighboring fluid particles by viscous and thermal diffusion are neglectedduring the passage of a sound wave (i.e., that the motion of a fluid particleis adiabatic).

From (1.3.5): ρ ′ = p′/c20. Hence, substituting for ρ ′ in (1.3.4), we obtain(

1

c20

∂2

∂t2− ∇2

)p = ρ0

∂q

∂t− div F, (1.3.7)

where the prime (′) on the acoustic pressure has been discarded. This equationgoverns the production of sound waves by the volume source q and the forceF. When these terms are absent the equation describes sound propagationfrom sources on the boundaries of the fluid, such as the vibrating cone of aloudspeaker.

The volume source q and the body force F would never appear in a completedescription of sound generation within a fluid. They are introduced only whenwe think we understand how to model the real sources of sound in terms ofvolume sources and forces. In general this can be a dangerous procedure be-cause, as we shall see, small errors in specifying the sources of sound in a fluidcan lead to very large errors in the predicted sound. This is because only a tinyfraction of the available energy of a vibrating fluid or structure actually radiatesaway as sound.

When F = 0, Equation (1.3.1) implies the existence of a velocity potentialϕ such that v = ∇ϕ, in terms of which the perturbation pressure is given by

p = −ρ0∂ϕ

∂t. (1.3.8)

It follows from this and (1.3.7) (with F = 0) that the velocity potential is thesolution of (

1

c20

∂2

∂t2− ∇2

)ϕ = −q(x, t). (1.3.9)

This is the wave equation of classical acoustics.

1.4 The Special Case of an Incompressible Fluid 7

Table 1.3.1. Speed of sound and acoustic wavelength

c0 λ at 1 kHz

m/s ft/s km/h mi/h m ft

Air 340 1100 1225 750 0.3 1Water 1500 5000 5400 3400 1.5 5

For future reference, Table 1.3.1 lists the approximate speeds of sound in airand in water, and the corresponding acoustic wavelength λ at a frequency of1 kHz (sound of frequency f has wavelength λ = c0/ f ).

1.4 The Special Case of an Incompressible Fluid

Small (adiabatic) pressure and density perturbations δp and δρ satisfy

δp

δρ≈ c2

0.

In an incompressible fluid the pressure can change by the action of externalforces (moving boundaries, etc.), but the density must remain fixed. Thus,c0 = ∞, and Equation (1.3.9) reduces to

∇2ϕ = q(x, t). (1.4.1)

1.4.1 Pulsating Sphere

Consider the motion produced by small amplitude radial pulsations of a sphereof mean radius a. Let the center of the sphere be at the origin, and let its normalvelocity be vn(t). There are no sources within the fluid, so that q ≡ 0. Therefore,

∇2ϕ = 0, r > a,

∂ϕ/∂r = vn(t), r = a

where r = |x|.

The motion is obviously radially symmetric, so that

∇2ϕ = 1

r2

∂

∂r

(r2 ∂

∂r

)ϕ = 0, r > a.

Hence,

ϕ = A

r+ B,

8 1 Introduction

where A ≡ A(t) and B ≡ B(t) are functions of t . B(t) can be discarded be-cause the pressure fluctuations (∼ −ρ0∂ϕ/∂t) must vanish as r → ∞. Apply-ing the condition ∂ϕ/∂r = vn at r = a, we then find

ϕ = −a2vn(t)

r, r > a. (1.4.2)

Thus, the pressure

p = −ρ0∂ϕ

∂t= ρ0

a2

r

dvn

dt(t)

decays as 1/r with distance from the sphere, and exhibits the unphysical char-acteristic of changing instantaneously everywhere when dvn/dt changes itsvalue. For any time t , the volume flux q(t) of fluid is the same across any closedsurface enclosing the sphere. Evaluating it for any sphere S of radius r > a, asshown in Fig. 1.4.1, we find

q(t) =∮

S∇ϕ · dS = 4πa2vn(t),

and we may also write

ϕ = −q(t)

4πr, r > a. (1.4.3)

Fig. 1.4.1.

1.4 The Special Case of an Incompressible Fluid 9

1.4.2 Point Source

The incompressible motion generated by a volume point source of strength q(t)at the origin is the solution of

∇2ϕ = q(t)δ(x), where δ(x) = δ(x1)δ(x2)δ(x3). (1.4.4)

The solution must be radially symmetric and given by

ϕ = A

rfor r > 0. (1.4.5)

To find A, we integrate (1.4.4) over the interior of a sphere of radius r = R > 0,and use the divergence theorem

∫r<R ∇2ϕ d3x = ∮

S ∇ϕ · dS, where S is thesurface of the sphere. Then

∮S∇ϕ · dS ≡

(−A

R2

)× (4π R2) = q(t).

Hence, A = −q(t)/4π and ϕ = −q(t)/4πr , which agrees with the solution(1.4.3) for the sphere with the same volume outflow in the region r > a = radiusof the sphere. This indicates that when we are interested in modelling the effectof a pulsating sphere at large distances r a, it is permissible to replace thesphere by a point source (a monopole) of the same strength q(t) = rate ofchange of the volume of the sphere. This conclusion is valid for any pulsatingbody, not just a sphere. However, it is not necessarily a good model (especiallywhen we come to examine the production of sound by a pulsating body) in thepresence of a mean fluid flow past the sphere.

The Solution (1.4.5) for the point source is strictly valid only for r > 0,where it satisfies ∇2ϕ = 0. What happens as r → 0, where its value is actuallyundefined? To answer this question, we write the solution in the form

ϕ = limε→0

−q(t)

4π (r2 + ε2)12

, ε > 0, in which case ∇2ϕ = limε→0

3ε2q(t)

4π (r2 + ε2)52

.

The last limit is just equal to q(t)δ(x). Indeed when ε is small 3ε2/4π (r2 + ε2)52

is also small except close to r = 0, where it attains a large maximum ∼3/4πε3.Therefore, for any smoothly varying test function f (x) and any volume V

10 1 Introduction

enclosing the origin

limε→0

∫V

3ε2 f (x) d3x

4π (r2 + ε2)52

= f (0) limε→0

∫ ∞

−∞

3ε2 d3x

4π (r2 + ε2)52

= f (0)∫ ∞

0

3ε2r2 dr

(r2 + ε2)52

= f (0),

where the value of the last integral is independent of ε. This is the definingproperty of the three-dimensional δ function.

Thus, the correct interpretation of the solution

ϕ = −1

4πrof ∇2ϕ = δ(x) (1.4.6)

for a unit point source (q = 1) is

−1

4πr= lim

ε→0

−1

4π (r2 + ε2)12

, r ≥ 0, (1.4.7)

where

∇2

( −1

4πr

)= lim

ε→0∇2

( −1

4π (r2 + ε2)12

)= lim

ε→0

3ε2

4π (r2 + ε2)52

= δ(x). (1.4.8)

1.5 Sound Produced by an Impulsive Point Source

The sound generated by the unit, impulsive point source δ(x)δ(t) is the solutionof (

1

c20

∂2

∂t2− ∇2

)ϕ = δ(x)δ(t). (1.5.1)

The source exists only for an infinitesimal instant of time at t = 0; therefore atearlier times ϕ(x, t) = 0 everywhere.

It is evident that the solution is radially symmetric, and that for r = |x|> 0we have to solve

1

c20

∂2ϕ

∂t2− 1

r2

∂

∂r

(r2 ∂

∂r

)ϕ = 0, r > 0. (1.5.2)

The identity

1

r2

∂

∂r

(r2 ∂

∂r

)ϕ ≡ 1

r

∂2

∂r2(rϕ) (1.5.3)

1.5 Sound Produced by an Impulsive Point Source 11

permits us to write Equation (1.5.2) in the form of the one-dimensional waveequation for rϕ:

1

c20

∂2

∂t2(rϕ) − ∂2

∂r2(rϕ) = 0, r > 0. (1.5.4)

This has the general solution rϕ = (t − r/c0) + (t + r/c0), where and are arbitrary functions. Hence, the general solution of (1.5.2) is

ϕ = (t − r

c0

)r

+ (t + r

c0

)r

, r > 0. (1.5.5)

The first term on the right represents a spherically symmetric disturbance thatpropagates in the direction of increasing values of r at the speed of sound c0 as tincreases, whereas the second represents an incoming wave converging towardx = 0. We must therefore set = 0, since it represents sound waves generatedat r = ∞ that approach the source rather than sound waves generated by thesource and radiating away from the source. This is a causality or radiationcondition, that (in the absence of boundaries) sound produced by a sourcemust radiate away from the source. It is also consistent with the Second Lawof Thermodynamics, which requires natural systems to change in the moreprobable direction. An event in which sound waves converge on a point fromall directions at infinity is so unlikely as to be impossible in practice; it would bethe acoustic analogue of the far-scattered pieces of a broken cup spontaneouslyreassembling.

To complete the solution it remains to determine the function . We do thisby extending the solution down to the source at r = 0 by writing (c.f., (1.4.7))

ϕ = (t − r

c0

)r

= limε→0

(t − r

c0

)(r2 + ε2)

12

, r ≥ 0. (1.5.6)

Let us substitute this into Equation (1.5.1) and examine what happens as ε → 0.By direct calculation we find

∇2ϕ = 1

r

∂2

∂r2(rϕ) = −3ε2(t − r/c0)

(r2 + ε2)52

− 2ε2′(t − r/c0)

c0r (r2 + ε2)32

+ ′′(t − r/c0)

c20(r2 + ε2)

12

1

c20

∂2ϕ

∂t2= ′′(t − r/c0)

c20(r2 + ε2)

12

12 1 Introduction

Therefore,

1

c20

∂2ϕ

∂t2− ∇2ϕ = 3ε2(t − r/c0)

(r2 + ε2)52

+ 2ε2′(t − r/c0)

c0r (r2 + ε2)32

→ 4π(t)δ(x) + 0 as ε → 0, (1.5.7)

where the δ function in the last line follows from (1.4.8), and the ‘+ 0’ isobtained by noting that for any smoothly varying test function f (x) and anyvolume V enclosing the origin∫

V

2ε2 f (x) d3x

r (r2 + ε2)32

≈ f (0)∫ ∞

−∞

2ε2 d3x

r (r2 + ε2)32

= f (0)∫ ∞

0

8πε2r dr

(r2 + ε2)32

= 8πε f (0) → 0 as ε → 0.

Hence, comparing (1.5.7) with the inhomogeneous wave equation (1.5.1),we find

(t) = 1

4πδ(t),

and the Solution (1.5.6) becomes

ϕ(x, t) = 1

4πrδ

(t − r

c0

)≡ 1

4π |x|δ(

t − |x|c0

). (1.5.8)

This represents a spherical pulse that is nonzero only on the surface of thesphere r = c0t > 0, whose radius increases at the speed of sound c0; it vanisheseverywhere for t < 0, before the impulsive source is triggered.

1.6 Free-Space Green’s Function

The free-space Green’s function G(x, y, t −τ ) is the causal solution of the waveequation generated by the impulsive point source δ(x − y)δ(t − τ ), located atthe point x = y at time t = τ . The formula for G is obtained from the solution(1.5.8) for a source at x = 0 at t = 0 simply by replacing x by x − y and t byt − τ . In other words, if(

1

c20

∂2

∂t2− ∇2

)G = δ(x − y)δ(t − τ ), where G = 0 for t <τ, (1.6.1)

then

G(x, y, t − τ ) = 1

4π |x − y| δ(

t − τ − |x − y|c0

). (1.6.2)

1.7 Monopoles, Dipoles, and Quadrupoles 13

This represents an impulsive, spherically symmetric wave expanding from thesource at y at the speed of sound. The wave amplitude decreases inversely withdistance |x − y| from the source point y.

Green’s function is the fundamental building block for forming solutions ofthe inhomogeneous wave equation (1.3.7) of linear acoustics. Let us write thisequation in the form

(1

c20

∂2

∂t2− ∇2

)p = F(x, t), (1.6.3)

where the generalized source F(x, t) is assumed to be generating waves thatpropagate away from the source region, in accordance with the radiationcondition.

This source distribution can be regarded as a distribution of impulsive pointsources of the type on the right of Equation (1.6.1), because

F(x, t) =∫∫ ∞

−∞F(y, τ ) δ(x − y) δ(t − τ ) d3y dτ.

The outgoing wave solution for each constituent source of strength

F(y, τ )δ(x − y)δ(t − τ ) d3y dτ is F(y, τ )G(x, y, t − τ ) d3y dτ,

so that by adding up these individual contributions we obtain

p(x, t) =∫∫ ∞

−∞F(y, τ )G(x, y, t − τ ) d3y dτ (1.6.4)

= 1

4π

∫∫ ∞

−∞

F(y, τ )

|x − y| δ

(t − τ − |x − y|

c0

)d3y dτ (1.6.5)

i.e., p(x, t) = 1

4π

∫ ∞

−∞

F(y, t − |x−y|c0

)|x − y| d3y. (1.6.6)

The integral formula (1.6.6) is called a retarded potential; it represents thepressure at position x and time t as a linear superposition of contributions fromsources at positions y, which radiated at the earlier times t−|x−y|/c0, |x−y|/c0

being the time of travel of sound waves from y to x.

1.7 Monopoles, Dipoles, and Quadrupoles

A volume point source q(t)δ(x) of the type considered in Section 1.4 as a modelfor a pulsating sphere is also called a monopole point source. For a compressible

14 1 Introduction

medium the corresponding velocity potential it produces is the solution of theequation (

1

c20

∂2

∂t2− ∇2

)ϕ = −q(t)δ(x). (1.7.1)

The solution can be written down by analogy with the Solution (1.6.6) ofEquation (1.6.3) for the acoustic pressure. Replace p by ϕ in (1.6.6) and setF(y, τ ) = −q(τ )δ(y). Then,

ϕ(x, t) = −q(t − |x|

c0

)4π |x| ≡ −q

(t − r

c0

)4πr

. (1.7.2)

This differs from the corresponding solution (1.4.3) for a pulsating sphereor volume point source in an incompressible fluid by the dependence on theretarded time t − r

c0. This is physically more realistic; any effects associated

with changes in the motion of the sphere (i.e., in the value of the volume out-flow rate q(t)) are now communicated to a fluid element at distance r afteran appropriate time delay r/c0 required for sound to travel outward from thesource.

1.7.1 The Point Dipole

Let f = f(t) be a time-dependent vector. Then a source on the right of theacoustic pressure equation (1.6.3) of the form

F(x, t) = div(f(t)δ(x)) ≡ ∂

∂x j( f j (t)δ(x)) (1.7.3)

is called a point dipole (located at the origin). As explained in the Preface, arepeated italic subscript, such as j in this equation, implies a summation overj = 1, 2, 3. Equation (1.3.7) shows that the point dipole is equivalent to a forcedistribution F(t) = −f(t)δ(x) per unit volume applied to the fluid at the origin.

The sound produced by the dipole can be calculated from (1.6.6), but it iseasier to use (1.6.5):

p(x, t) = 1

4π

∫∫ ∞

−∞

∂

∂y j( f j (τ )δ(y))

δ(t − τ − |x−y|

c0

)|x − y| d3y dτ.

Integrate by parts with respect to each y j (recalling that δ(y) = 0 at y j = ±∞),and note that

∂

∂y j

δ(t − τ − |x−y|

c0

)|x − y| = − ∂

∂x j

δ(t − τ − |x−y|

c0

)|x − y| .


Then,

p(x, t) = 1

4π

∫∫ ∞

−∞f j (τ )δ(y)

∂

∂x j

(δ(t − τ − |x−y|

c0

)|x − y|

)d3y dτ

= 1

4π

∂

∂x j

∫∫ ∞

−∞f j (τ )δ(y)

(δ(t − τ − |x−y|

c0

)|x − y|

)d3y dτ.

Thus,

p(x, t) = ∂

∂x j

(f j(t − |x|

c0

)4π |x|

). (1.7.4)

The same procedure shows that for a distributed dipole source of the typeF(x, t) = div f(x, t) on the right of Equation (1.6.3), the acoustic pressurebecomes

p(x, t) = 1

4π

∂

∂x j

∫ ∞

−∞

f j(y, t − |x−y|

c0

)|x − y| d3y. (1.7.5)

A point dipole at the origin orientated in the direction of a unit vector nis entirely equivalent to two point monopoles of equal but opposite strengthsplaced a short distance apart (much smaller than the acoustic wavelength) onopposite sides of the origin on a line through the origin parallel to n. Forexample, if n is parallel to the x axis, and the sources are distance ε apart, thetwo monopoles would be

q(t)δ(

x − ε

2

)δ(y)δ(z) − q(t)δ

(x + ε

2

)δ(y)δ(z)

≈ −εq(t)δ′(x)δ(y)δ(z) ≡ − ∂

∂x(εq(t)δ(x)).

This is a fluid volume dipole. The relation p = −ρ0∂ϕ/∂t implies that theequivalent dipole source in the pressure equation (1.3.7) or (1.6.3) is

−ρ0∂

∂x(εq(t)δ(x)),

where the dot denotes differentiation with respect to time.

1.7.2 Quadrupoles

A source distribution involving two space derivatives is equivalent to a combi-nation of four monopole sources (whose net volume source strength is zero),

16 1 Introduction

and is called a quadrupole. A general quadrupole is a source of the form

F(x, t) = ∂2Ti j

∂xi∂x j(x, t) (1.7.6)

in Equation (1.6.3). The argument above leading to Expression (1.7.5) can beapplied twice to show that the corresponding acoustic pressure is given by

p(x, t) = 1

4π

∂2

∂xi∂x j

∫ ∞

−∞

Ti j (y, t − |x − y|/c0)

|x − y| d3y. (1.7.7)

1.7.3 Vibrating Sphere

Let a rigid sphere of radius a execute small amplitude oscillations at speed U (t)in the x1 direction (Fig. 1.7.1a). Take the coordinate origin at the mean positionof the center. In Section 3.5, we shall prove that the motion induced in an idealfluid when the sphere is small is equivalent to that produced by a point volumedipole of strength 2πa3U (t) at its center directed along the x1 axis, determined

Fig. 1.7.1.


by the solution of (1

c20

∂2

∂t2− ∇2

)ϕ = ∂

∂x1(2πa3U (t)δ(x)). (1.7.8)

By analogy with (1.7.3) and (1.7.4), we have

ϕ(x, t) = ∂

∂x1

(2πa3U

(t − |x|

c0

)4π |x|

). (1.7.9)

Now,

∂

∂x j|x| = x j

|x| . (1.7.10)

Applying this formula for j = 1, we find (putting r = |x| and x1 = r cos θ )

ϕ = −a3 cos θ

2r2U

(t − r

c0

)− a3 cos θ

2c0r

∂U

∂t

(t − r

c0

).

near field far field

The near-field term is dominant at sufficiently small distances r from the originsuch that

1

r 1

c0

1

U

∂U

∂t∼ f

c0,

where f is the characteristic frequency of the oscillations of the sphere. But,sound of frequency f travels a distance

c0/ f = λ≡ one acoustic wavelength

in one period of oscillation 1/ f . Hence, the near-field term is dominant when

r λ.

The motion becomes incompressible when c0 → ∞. In this limit the solutionreduces entirely to the near-field term, which is also called the hydrodynamicnear field; it decreases in amplitude like 1/r2 as r → ∞.

The far field is the acoustic region that only exists when the fluid is com-pressible. It consists of propagating sound waves, carrying energy away fromthe sphere, and takes over from the near field when r λ. There is an interme-diate zone where r ∼ λ in which the solution is in a state of transition from thenear to the far field. The analytical model (1.7.8), in which the sphere is replacedby a point dipole at its center, involves the implicit assumption that the motion

18 1 Introduction

close to the sphere is essentially the same as if the fluid is incompressible. It fol-lows from what we have just said that a λ, that is, the diameter of the sphereis much smaller than the acoustic wavelength. In general, a body is said to beacoustically compact when its characteristic dimension is small compared tothe wavelengths of the sound waves it is producing or with which it interacts.

The intensity of the sound generated by the sphere in the far field is propor-tional to ϕ2:

ϕ2 → a6

4c20r2

(∂U

∂t

(t − r

c0

))2

cos2 θ.

The dependence on θ determines the directivity of the sound. For the dipole ithas the figure of eight pattern illustrated in Fig. 1.7.1b, with peaks in directionsparallel to the dipole axis (θ = 0, π ); there are radiation nulls at θ = π

2 (thecurve should be imagined to be rotated about the x1 axis).

1.8 Acoustic Energy Flux

At large distances r from a source region we generally have

p(x, t) ∼ ρ0(θ, φ, t − r

c0

)r

, r → ∞, (1.8.1)

where the function depends on the nature of the source distribution, and θ

and φ are polar angles determining the directivity of the sound. From the radialcomponent of the linearized momentum equation

∂vr

∂t= − 1

ρ0

∂p

∂r

≡ 1

r2

(θ, φ, t − r

c0

)+ 1

c0r

∂

∂t

(θ, φ, t − r

c0

). (1.8.2)

The first term in the second line can be neglected when r → ∞, and therefore

vr ∼ 1

c0r

(θ, φ, t − r

c0

)≡ p

ρ0c0. (1.8.3)

By considering the θ andφ components of the momentum equation we can showthat the corresponding velocity components vθ , vφ , say, decrease faster than 1/ras r → ∞. We therefore conclude from this and (1.8.3) that the acoustic particlevelocity is normal to the acoustic wavefronts (the spherical surfaces r = c0t).In other words, sound consists of longitudinal waves in which the fluid particlesoscillate backwards and forwards along the local direction of propagation ofthe sound.

1.8 Acoustic Energy Flux 19

The acoustic power radiated by a source distribution can be computedfrom the formula

=∮

Spvr d S =

∮S

p2

ρ0c0d S, (1.8.4)

where the integration is over the surface S of a large sphere of radius r centeredon the source region. Because the surface area = 4πr2, we only need to knowthe pressure and velocity correct to order 1/r on S in order to evaluate theintegral. Smaller contributions (such as that determined by the first term inthe second line of (1.8.2)) decrease too fast as r increases to supply a finitecontribution to the integral as r → ∞.

In acoustic problems we are therefore usually satisfied if we can calculatethe pressure and velocity in the acoustic far field correct to order 1/r ; this willalways enable us to determine the radiated sound power. The formula vr =p/ρ0c0 is applicable at large distances from the sources, where the wavefrontscan be regarded as locally plane, but it is true identically for plane sound waves.In the latter case, and for spherical waves on the surface of the large sphere ofFig. 1.8.1, the quantity

I = pvr = p2

ρ0c0(1.8.5)

is called the acoustic intensity. It is the rate of transmission of acoustic energyper unit area of wavefront.

Fig. 1.8.1.

20 1 Introduction

Fig. 1.9.1.

1.9 Calculation of the Acoustic Far Field

We now discuss the approximations necessary to evaluate the sound in the farfield from the retarded potential representation:

p(x, t) = 1

4π

∫ ∞

−∞

F(y, t − |x−y|c0

)|x − y| d3y. (1.9.1)

We assume that F(x, t) = 0 only within a finite source region (Fig. 1.9.1), andtake the coordinate origin O within the region.

When |x| → ∞ and y lies within the source region (so that |x| |y|)

|x − y| ≡ (|x|2 − 2x · y + |y|2)12 = |x|

1 − 2x · y

|x|2 + |y|2|x|2

12

≈ |x|

1 − x · y|x|2 + O

( |y|2|x|2

)

Then,

|x − y| ≈ |x| − x · y|x| when

|y||x| 1. (1.9.2)

Also,

1

|x − y| ≈ 1(|x| − x · y|x|) ≈ 1

|x|(

1 + x · y|x|2

)

Therefore,1

|x − y| ≈ 1

|x| + x · y|x|3 when

|y||x| 1. (1.9.3)

The Approximation (1.9.3) shows that, in order to obtain the far-field approx-imation of the Solution (1.9.1) that behaves like 1/r = 1/|x| as |x| → ∞, it issufficient to replace |x−y| in the denominator of the integrand by |x|. However,in the argument of the source strength F it is important to retain possible phase

1.9 Calculation of the Acoustic Far Field 21

differences between the sound waves generated by components of the sourcedistribution at different locations y; we therefore replace |x − y| in the retardedtime by the right-hand side of (1.9.2). Hence,

p(x, t) ≈ 1

4π |x|∫ ∞

−∞F(

y, t − |x|c0

+ x · yc0|x|

)d3y, |x| → ∞. (1.9.4)

This is called the Fraunhofer approximation.The source region may extend over many characteristic acoustic wavelengths

of the sound. By retaining the contribution x · y/c0|x| to the retarded time weensure that any interference between waves generated at different positionswithin the source region is correctly described by the far-field approximation.In Fig. 1.9.1 the acoustic travel time from a source point y to the far field pointx is equal to that from the point labelled A to x when |x| → ∞. The travel timeover the distance O A is just x · y/c0|x|, so that |x|/c0 − x · y/c0|x| gives thecorrect value of the retarded time when |x| → ∞.

1.9.1 Dipole Source Distributions

By applying the far-field formula (1.9.4) to a dipole source F(x, t) = div f(x, t)we obtain (from (1.7.5))

p(x, t) ≈ 1

4π

∂

∂x j

[1

|x|∫ ∞

−∞f j

(y, t − |x|

c0+ x · y

c0|x|)

d3y]

≈ 1

4π |x|∂

∂x j

∫ ∞

−∞f j

(y, t − |x|

c0+ x · y

c0|x|)

d3y, |x| → ∞, (1.9.5)

because the differential operator ∂/∂x j need not be applied to 1/|x| as this wouldgive a contribution decreasing like 1/r2 at large distances from the dipole.

However, it is useful to make a further transformation that replaces ∂/∂x j bythe time derivative ∂/∂t , which is usually more easily estimated in applications.To do this, we observe that

∂ f j

∂x j

(y, t − |x|

c0+ x · y

c0|x|)

= ∂ f j

∂t

(y, t − |x|

c0+ x · y

c0|x|)

∂

∂x j

(t − |x|

c0+ x · y

c0|x|)

= ∂ f j

∂t

(y, t − |x|

c0+ x · y

c0|x|)(

− x j

c0|x| + y j

c0|x| − (x · y)x j

c0|x|3)

≈ − x j

c0|x|∂ f j

∂t

(y, t − |x|

c0+ x · y

c0|x|)

as |x| → ∞.

22 1 Introduction

Hence, the far field of a distribution of dipolesF(x, t) = div f(x, t) is given by

p(x, t) = −x j

4πc0|x|2∂

∂t

∫ ∞

−∞f j

(y, t − |x|

c0+ x · y

c0|x|)

d3y. (1.9.6)

Note that

x j

|x|2 = x j

|x|1

|x| ,

where x j/|x| is the j th component of the unit vector x/|x|. Thus, the additionalfactor of x j/|x| in (1.9.6) does not change the rate of decay of the sound withdistance from the source (which is still like 1/r ), but it does have an influenceon the acoustic directivity.

A comparison of (1.9.5) and (1.9.6) leads to the following rule for inter-changing space and time derivatives in the acoustic far field:

∂

∂x j←→ − 1

c0

x j

|x|∂

∂t. (1.9.7)

1.9.2 Quadrupole Source Distributions

For the Quadrupole (1.7.6)

F(x, t) = ∂2Ti j

∂xi∂x j(x, t),

and

p(x, t) = 1

4π

∂2

∂xi∂x j

∫ ∞

−∞

Ti j (y, t − |x − y|/c0)

|x − y| d3y.

By applying (1.9.4) and the rule (1.9.7), we find that the acoustic far field isgiven by

p(x, t) ≈ xi x j

4πc20|x|3

∂2

∂t2

∫ ∞

−∞Ti j

(y, t − |x|

c0+ x · y

c0|x|)

d3y, |x| → ∞.

(1.9.8)

1.9.3 Example

For the (1, 2) point quadrupole

F(x, t) = ∂2

∂x1∂x2(T (t)δ(x))

Problems 1 23

Fig. 1.9.2.

Equation (1.9.8) shows that in the acoustic far field

p(x, t) ≈ x1x2

4πc20|x|3

∂2T

∂t2

(t − |x|

c0

), |x| → ∞.

If we use spherical polar coordinates, such that

x1 = r cos θ, x2 = r sin θ cosφ, x3 = r sin θ sinφ,

we can write the pressure in the form

p(x, t) ≈ sin 2θ cosφ

8πc20|x|

∂2T

∂t2

(t − |x|

c0

), |x| → ∞.

The directivity of the sound (∝ p2) is therefore represented by sin2 2θ cos2 φ.Its shape is plotted in Fig. 1.9.2 for radiation in the x1, x2 plane (φ = 0, π ). Thefour-lobe cloverleaf pattern is characteristic of a quadrupole Ti j for which i = j .

Problems 1

1. A plane sound wave propagating parallel to the x axis satisfies the equation

(1

c20

∂2

∂t2− ∂2

∂x2

)ϕ = 0,

24 1 Introduction

with general solution

ϕ =

(t − x

c0

)+

(t + x

c0

),

where and are arbitrary functions respectively representing waves pro-pagating in the positive and negative x directions.

Show that for a wave propagating in the positive x direction in an ideal gas

v = p

ρ0c0, ρ = p

c20

, T = p

ρ0cp,

where v is the acoustic particle velocity; p, ρ, and T are respectively theacoustic pressure, density, and temperature variations, and cp is the specificheat at constant pressure.

2. Calculate the acoustic power (1.8.4) radiated by an acoustically compactsphere of radius R executing small amplitude translational oscillations offrequency ω and velocity U (t) = U0 cos(ωt), where U0 = constant.

3. As for Problem 2, when the sphere executes small amplitude radial oscilla-tions at normal velocity vn = U0 cos(ωt), U0 = constant.

4. A volume point source of strength q0(t) translates at constant, subsonic ve-locity U. The velocity potential ϕ(x, t) of the radiated sound is determinedby the solution of (

1

c20

∂2

∂t2− ∇2

)ϕ = −q0(t)δ(x − Ut).

Show that

ϕ(x, t) = −q0(t − R/c0)

4π R(1 − M cos), M = U

c0,

where R is the distance of the reception point x from the source position atthe time of emission of the sound received at x at time t , and is the anglebetween U and the direction of propagation of this sound.

2Lighthill’s Theory

2.1 The Acoustic Analogy

The sound generated by turbulence in an unbounded fluid is usually calledaerodynamic sound. Most unsteady flows of technological interest are of highReynolds number and turbulent, and the acoustic radiation is a very small by-product of the motion. The turbulence is usually produced by fluid motionover a solid boundary or by flow instability. Lighthill (1952) transformed theNavier–Stokes and continuity equations to form an exact, inhomogeneous waveequation whose source terms are important only within the turbulent region.He argued that sound is a very small component of the whole motion and that,once generated, its back-reaction on the main flow can usually be ignored. Theproperties of the unsteady flow in the source region may then be determinedby neglecting the production and propagation of the sound, a reasonable ap-proximation if the Mach number M is small, and there are many importantflows where the hypothesis is obviously correct, and where the theory leads tounambiguous predictions of the sound.

Lighthill was initially interested in solving the problem, illustrated inFig. 2.1.1a, of the sound produced by a turbulent nozzle flow. However, hisoriginal theory actually applies to the simpler situation shown in Fig. 2.1.1b, inwhich the sound is imagined to be generated by a finite region of rotational flowin an unbounded fluid. This avoids complications caused by the presence of thenozzle. The fluid is assumed to be at rest at infinity, where the mean pressure,density, and sound speed are respectively equal to p0, ρ0, c0. Lighthill com-pared the equations for the production of acoustic density fluctuations in thereal flow with those in an ideal linear acoustic medium that coincides with thereal fluid at large distances from the sources.

25

26 2 Lighthill’s Theory

Fig. 2.1.1.

To do this, body forces are neglected, and the i th component of the momentumequation (1.2.3) is cast in the form

ρ∂vi

∂t+ ρv j

∂vi

∂x j= − ∂p

∂xi+ ∂σi j

∂x j≡ − ∂

∂x j(pδi j − σi j ). (2.1.1)

δi j is the Kronecker delta (= 1 for i = j, and 0 for i = j), and σi j is theviscous stress tensor defined (for a Stokesian fluid) by

σi j = 2η(ei j − 1

3 ekkδi j), (2.1.2)

where

ei j = 1

2

(∂vi

∂x j+ ∂v j

∂xi

)(2.1.3)

2.1 The Acoustic Analogy 27

is the rate of strain tensor. Next multiply the continuity equation (1.2.1) by vi :

vi∂ρ

∂t+ vi

∂(ρv j )

∂x j= 0.

By adding this to Equation (2.1.1), we obtain the Reynolds form of the momen-tum equation

∂(ρvi )

∂t= −∂πi j

∂x j, (2.1.4)

where

πi j = ρviv j + (p − p0)δi j − σi j , (2.1.5)

is called the momentum flux tensor, and the constant pressure p0 is insertedfor convenience.

In an ideal, linear acoustic medium, the momentum flux tensor contains onlythe pressure

πi j → π0i j = (p − p0)δi j ≡ c2

0(ρ − ρ0)δi j , (2.1.6)

and the momentum equation then reduces to

∂(ρvi )

∂t+ ∂

∂xi

[c2

0(ρ − ρ0)] = 0. (2.1.7)

If the continuity equation (1.2.1) is written in the slightly modified form

∂

∂t(ρ − ρ0) + ∂(ρvi )

∂xi= 0, (2.1.8)

we can eliminate the momentum density ρvi between (2.1.7) and (2.1.8) toobtain the equation of linear acoustics satisfied by the perturbation densityρ − ρ0 (

1

c20

∂2

∂t2− ∇2

)[c2

0(ρ − ρ0)] = 0. (2.1.9)

Because the turbulence is neglected in this approximation, and there are noexternally applied forces or moving boundaries, the unique solution of thisequation that satisfies the radiation condition of outgoing wave behavior issimply ρ − ρ0 = 0.

It can now be asserted that the sound generated by the turbulence in thereal fluid is exactly equivalent to that produced in the ideal, stationary acoustic


medium (which is governed by (2.1.9) in turbulence-free regions) forced by thestress distribution

Ti j = πi j − π0i j

= ρviv j + ((p − p0) − c2

0(ρ − ρ0))δi j − σi j , (2.1.10)

where Ti j is called the Lighthill stress tensor. This is Lighthill’s acousticanalogy.

Indeed, we can rewrite (2.1.4) as the momentum equation for an ideal, sta-tionary acoustic medium of mean density ρ0 and sound speed c0 subject to theexternally applied stress Ti j

∂(ρvi )

∂t+ ∂π0

i j

∂x j= − ∂

∂x j

(πi j − π0

i j

),

or

∂(ρvi )

∂t+ ∂

∂xi

[c2

0(ρ − ρ0)] = −∂Ti j

∂x j. (2.1.11)

By eliminating the momentum density ρvi between this and the continuityequation (2.1.8) (the same procedure used above for the linear problem), weobtain Lighthill’s equation(

1

c20

∂2

∂t2− ∇2

)[c2

0(ρ − ρ0)] = ∂2Ti j

∂xi∂x j. (2.1.12)

This is the exact, nonlinear counterpart of (2.1.9). The problem of calculatingthe turbulence generated sound is therefore equivalent to solving this equationfor the radiation into a stationary, ideal fluid produced by a distribution ofquadrupole sources whose strength per unit volume is the Lighthill stress tensorTi j . The quadrupole character of the turbulence sources is one of the mostimportant conclusions of Lighthill’s theory; it implies (see Section 2.2) thatfree-field turbulence is an extremely weak sound source, and that in a typicallow Mach number flow only a tiny fraction of the available flow energy isconverted into sound.

In the definition (2.1.10) of Ti j , the term ρviv j is called the Reynolds stress.For the simplified problem of Fig. 2.1.1b it is a nonlinear quantity that can beneglected except where the motion is turbulent. The second term representsthe excess of momentum transfer by the pressure over that in the ideal (linear)fluid of density ρ0 and sound speed c0. This is produced by wave amplitudenonlinearity, and by mean density variations in the source flow. The viscous

2.2 Lighthill’s v8 Law 29

stress tensor σi j is linear in the perturbation quantities, and properly accountsfor the attenuation of the sound; in most applications the Reynolds numberin the source region is very large, and σi j can be neglected, and the viscousattenuation of the radiating sound is usually ignored.

2.2 Lighthill’s v8 Law

The formal solution of Lighthill’s equation (2.1.12) with outgoing wave be-havior is given by (1.7.7) with p(x, t) replaced by c2

0(ρ − ρ0)

c20(ρ − ρ0)(x, t) = 1

4π

∂2

∂xi∂x j

∫ ∞

−∞

Ti j (y, t − |x − y|/c0)

|x − y| d3y. (2.2.1)

This is strictly an alternative, integral equation representation of Equation(2.1.12); it provides a useful prediction of the sound only when Ti j is knownor has been determined by some other means. This is because the terms in thedefinition (2.1.10) of Ti j not only account for the generation of sound, but alsogovern acoustic self-modulation caused by acoustic nonlinearity, the convec-tion of sound waves by the turbulent velocity, refraction caused by sound speedvariations, and attenuation due to thermal and viscous actions. The influenceof acoustic nonlinearity and of thermoviscous dissipation is usually sufficientlyweak to be neglected within the source region, although they may affect prop-agation to a distant observer. Convection and refraction of sound within andnear the source flow can be important, for example in the presence of a meanshear layer (when the Reynolds stress will include terms like ρUi u j , where Uand u respectively denote the mean and fluctuating components of v), or whenthere are large variations in the mean thermodynamic properties of the mediumwithin the source region; such effects are described by the presence of unsteadylinear terms in Ti j (Ffowcs Williams, 1974).

Thus, to predict the radiated sound from Lighthill’s equation (2.2.1) it isusually necessary to suppose that all of these acoustic effects in the source flow(which really depend on fluid compressibility) are in some sense negligible.This means that in practice it must be possible to derive a good approximationfor Ti j by taking the source flow to be effectively incompressible. This is oftenpossible when the characteristic Mach number M ∼ v/c0 is small (specifically,when M2 1), and when the wavelength of the sound is much larger than thesize of the source region.

Consider the particular but important case in which the mean density andsound speed are uniform throughout the fluid. The variations in the densityρ within a low Mach number, high Reynolds number source flow are then of


order ρ0 M2 (Batchelor, 1967). Thus, ρviv j = ρ0(1 + O(M2))viv j ≈ ρ0viv j .Similarly, if c(x, t) is the local speed of sound in the source region, it may alsobe shown that c2

0/c2 = 1 + O(M2), so that

p − p0 − c20(ρ − ρ0) ≈ (p − p0)

(1 − c2

0

/c2) ∼ O(ρ0v

2 M2).

Hence, if viscous dissipation is neglected we make the approximation

Ti j ≈ ρ0viv j , provided that M2 1. (2.2.2)

In the acoustic region outside the source flow c20(ρ − ρ0) = p − p0. If the

irrelevant constant pressure p0 is suppressed, the Solution (2.2.1) of Lighthill’sequation therefore becomes

p(x, t) ≈ ∂2

∂xi∂x j

∫ρ0viv j (y, t − |x − y|/c0)

4π |x − y| d3y

≈ xi x j

4πc20|x|3

∂2

∂t2

∫ρ0viv j

(y, t − |x|

c0+ x · y

c0|x|)

d3y, |x| → ∞,

(2.2.3)

where in the second line we have used the formula (1.9.8) for the far fieldof a quadrupole distribution. Quantitative predictions can be made from thisformula provided the behavior of the Reynolds stress ρ0viv j is known.

To determine the order of magnitude of p, we introduce a characteristicvelocity v and length scale (of the energy-containing eddies) of the turbulencesources. The value of depends on the mechanism responsible for turbulenceproduction, such as the width of a jet mixing layer. Fluctuations inviv j occurringin different regions of the turbulent flow separated by distances >O() will tendto be statistically independent, and the sound may be considered to be generatedby a collection of V0/

3 independent eddies, where V0 is the volume occupiedby the turbulence (Fig. 2.2.1). The characteristic frequency of the turbulentfluctuations f ∼ v/, so that the wavelength (c0/ f ) of the sound ∼/M

(because M = v/c0 1). Hence, we arrive at the important conclusion thatthe turbulence eddies are each acoustically compact. This means that whenthe integral in (2.2.3) is confined to a single eddy, the retarded time variationsx · y/c0|x| across that eddy can be neglected; that is, if the coordinate originis temporarily placed at O within the eddy, we can set x · y/c0|x| = 0 in theintegration over that eddy. The value of the integral over the eddy then may beestimated to be of order ρ0v

23.

2.2 Lighthill’s v8 Law 31

Fig. 2.2.1.

The order of magnitude of the time derivative for changes in the source regionis

∂

∂t∼ v

.

Therefore, it follows from (2.2.3) that, for one eddy, the far-field acoustic pres-sure satisfies

p ∼

|x|ρ0v

4

c20

=

|x|ρ0v2 M2. (2.2.4)

The acoustic power radiated by the eddy is determined by the surface inte-gral (1.8.4) taken over a large sphere centered on the eddy. Thus, in order ofmagnitude,

acoustic power radiated by one eddy ∼4π |x|2 p2

ρ0c0∼ 2ρ0v

8

c50

= 2ρ0v3 M5.

(2.2.5)This is Lighthill’s ‘eighth power’ law.

The total power radiated from the whole of the turbulent region of volumeV0, containing V0/

3 independent eddies, is

q ≈ V0

3× (2ρ0v

3 M5) = v

ρ0v

2 M5V0.


Dimensional arguments and experiment indicate that the rate 0, say, at whichenergy must be supplied by the action of external forces to maintain the kineticenergy of a statistically steady turbulent field occupying a volume V0 is givenin order of magnitude by

0 ∼ v

ρ0v

2V0.

Therefore, the mechanical efficiency with which turbulence kinetic energy isconverted into sound is

q

0∼ M5. (2.2.6)

This is smaller than about 0.01 for Mach numbers M < 0.4, confirmingLighthill’s hypothesis that the flow generated sound is an infinitesimal by-product of the turbulent motion.

2.3 Curle’s Theory

In most applications of Lighthill’s theory it is necessary to generalize the so-lution (2.2.1) to account for the presence of solid bodies in the flow. Indeed,turbulence is frequently generated in the boundary layers and wakes of flowpast such bodies (airfoils, flow control surfaces, etc.), and the unsteady surfaceforces (dipoles) that arise are likely to make a significant contribution to theproduction of sound. The procedure in such cases is to introduce a system ofmathematical control surfaces that can be deformed to coincide with the sur-faces of the different moving or stationary bodies, although for the momentwe shall discuss only cases involving stationary bodies. Before doing this weestablish an integral transformation formula that is used repeatedly in problemsof this kind.

2.3.1 Volume and Surface Integrals

Let V be the fluid outside a closed control surface S (Fig. 2.3.1) defined by theequation

f (x) = 0, where

f (x) > 0 for x in V,

f (x) < 0 for x within S,(2.3.1)

and consider the Heaviside unit function

H ( f ) =

1 for x in V,

0 for x within S.

2.3 Curle’s Theory 33

Fig. 2.3.1.

Then, for an arbitrary function (x) defined in V and on S,

∫ ∞

−∞(x)∇ H d3x =

∮S(x) n d S ≡

∮S(x) dS, (2.3.2)

or∫ ∞

−∞(x)

∂H

∂x jd3x =

∮S(x) n j d S ≡

∮S(x) d Sj , (2.3.3)

where H ≡ H ( f ) and n is the unit normal on S directed into V.

Proof

∇ H ( f ) ≡ δ( f ) ∇ f (2.3.4)

is nonzero only on S, where ∇ f is in the direction of n. The volume integralis therefore confined to the region between the inner and outer faces of a shellof infinitesimal thickness (between the broken line surfaces in Fig. 2.3.1) thatjust encloses S, and in which the volume element is

d3x = ds⊥d S,

where s⊥ = 0 on S and s⊥ is measured parallel to n. Because f = 0 on S wecan write, for small values of s⊥,

f =(

∂ f

∂s⊥

)S

s⊥,

where (∂ f /∂s⊥)S ≡ |∇ f | > 0 is evaluated on S.


Therefore,

δ( f ) = δ(|∇ f |s⊥) ≡ δ(s⊥)

|∇ f | .

Hence,

∫ ∞

−∞(x)∇ H d3x ≡

∫ ∞

−∞(x)∇ f δ( f ) d3x =

∫ ∞

−∞(x)

∇ f

|∇ f |δ(s⊥) ds⊥ d S

=∮

S(x)n dS, because n = ∇ f

|∇ f | .

2.3.2 Curle’s Equation

Curle (1955) has derived a formal solution (called Curle’s equation) ofLighthill’s equation (2.1.12) for the sound produced by turbulence in the vicin-ity of an arbitrary, fixed surface S, defined as above by an equation f (x) = 0(Fig. 2.3.2). This surface may either enclose a solid body, or merely constitutea control surface used to isolate a fixed region of space containing both solidbodies and fluid or just fluid.

To derive Curle’s equation, multiply the momentum equation (2.1.11) byH ≡ H ( f ), and use the definition (2.1.10) of Ti j to obtain

∂

∂t(ρvi H ) + ∂

∂xi

(Hc2

0(ρ − ρ0)) = − ∂

∂x j(H Ti j ) + (ρviv j + p′

i j )∂H

∂x j,

(2.3.5)

Fig. 2.3.2.

2.3 Curle’s Theory 35

where

p′i j = (p − p0)δi j − σi j (2.3.6)

is the compressive stress tensor. Repeat this operation for the continuity equation(2.1.8):

∂

∂t(H (ρ − ρ0)) + ∂

∂xi(Hρvi ) = (ρvi )

∂H

∂xi. (2.3.7)

The Formula (2.3.4), ∇ H = ∇ f δ( f ), shows that Equations (2.3.5) and (2.3.7)formally determine the momentum density ρvi and the density fluctuation(ρ − ρ0) in the exterior region V (where H ( f ) ≡ 1) in terms of the Lighthillstresses Ti j in V and sources distributed over the control surface.

An analog of Lighthill’s equation (2.1.12) can now be obtained by eliminat-ing Hρvi between (2.3.5) and (2.3.7). This is the differential form of Curle’sequation(

1

c20

∂2

∂t2− ∇2

)[Hc2

0(ρ − ρ0)]

= ∂2(H Ti j )

∂xi∂x j− ∂

∂xi

((ρviv j + p′

i j )∂H

∂x j

)+ ∂

∂t

(ρv j

∂H

∂x j

). (2.3.8)

The equation is valid throughout all space, including the region enclosed byS where H ( f ) vanishes. The second and third terms on the right-hand siderespectively represent dipole and monopole sources distributed over S. Theyhave the following interpretations:

1. If S is merely an artificial control surface it will enclose fluid, possibly alsosolid bodies, and may or may not contain turbulence; the surface dipole andmonopole sources then represent the influence of this region on the soundradiated in V ; in other words the aggregate effect of the dipole and monopolesources accounts for the presence of solid bodies and turbulence within S(when Ti j = 0 in S) and also for the interaction of sound generated outsideS with the fluid and solid bodies in S.

2. If S is the boundary of a solid body, the surface dipole represents the pro-duction of sound by the unsteady surface force that the body exerts on theexterior fluid, whereas the monopole is responsible for the sound producedby volume pulsations (if any) of the body.

Because Curle’s form of Lighthill’s equation is valid throughout all space, theoutgoing wave solution is found from the general solution (1.6.6) of the wave


equation (1.6.3), as before, using the special form (1.7.5) for dipole sources.When account is taken of the transformation formula (2.3.3) this yields Curle’sequation

Hc20(ρ − ρ0) = ∂2

∂xi∂x j

∫V

[Ti j ]d3y

4π |x − y| − ∂

∂xi

∮S[ρviv j + p′

i j ]d Sj (y)

4π |x − y|

+ ∂

∂t

∮S[ρv j ]

d Sj (y)

4π |x − y| , (2.3.9)

where the square bracket notation such as [Ti j ] ≡ Ti j (y, t −|x − y|/c0) impliesevaluation at the retarded time. Note that, because H ( f ) ≡ 0 inside S, the sumof the three integrals on the right-hand side must also vanish when the fieldpoint x is within S.

2.4 Sound Produced by Turbulence Near a Compact Rigid Body

When the surface S (in Fig. 2.3.2) is rigid, Curle’s equation (2.3.9) reduces to

Hc20(ρ − ρ0) = ∂2

∂xi∂x j

∫V

[Ti j ]d3y

4π |x − y| − ∂

∂xi

∮S[p′

i j ]d Sj (y)

4π |x − y| . (2.4.1)

We now use this solution to determine the order of magnitude of the sound gen-erated by an acoustically compact body within a turbulent flow. Compactnessusually requires the Mach number M ∼ v/c0 1, and we shall assume this tobe the case in the following.

The contribution from the quadrupole integral in (2.4.1) is estimated as inSection 2.2. To deal with the surface dipole, note first that for turbulence ofvelocity v and correlation scale , the orders of magnitude of the pressure andviscous components of the compressive stress tensor

p′i j = (p − p0)δi j − σi j

are

(p − p0) ∼ ρ0v2, σ ∼ η

v

;

that is,

(p − p0)

σ∼ ρ0v

η= v

ν,

where ν = η/ρ0 is the kinematic viscosity. The dimensionless ratio Re = v/ν

is the Reynolds number and is always very large (∼104 or more) in turbulentflow. This means that viscous contributions to the surface force can be neglected.

2.5 Radiation from a Noncompact Surface 37

In the far field the pressure p(x, t) = c20(ρ − ρ0)(x, t), and H ( f ) = 1. Thus,

applying the far-field dipole approximation (1.9.6), and neglecting retarded timevariations x · y/c0|x| because S is compact, the dipole sound pressure pd, say,can be written

pd ≈ xi

4πc0|x|2∂

∂t

∮S(p − p0)

(y, t − |x|

c0

)d Si = xi

4πc0|x|2d Fi

dt

(t − |x|

c0

),

|x| → ∞, (2.4.2)

where F(t) is the unsteady force exerted on the fluid by the body. The contri-bution to pd from a surface element of diameter within which the turbulencesurface pressure fluctuations are correlated is evidently of order

1

c0|x|v

× (ρ0v

22) =

|x|ρ0v2 M,

which exceeds by an order of magnitude (1/M 1) the sound pressure (2.2.4)produced by a quadrupole in V of length scale . If A is the total surface areawetted by the turbulent flow, there are A/2 independently radiating surfaceelements, and the total power radiated by the dipoles is

d ∼ 4π |x|2 ×(

p2d

ρ0c0

)∼ Aρ0v

3 M3. (2.4.3)

The direct power radiated by quadrupoles occupying a volume V0 is q ∼(V0/)ρ0v

3 M5, the same as in the absence of the body (see Section 2.2). Thesound produced by the turbulence near S is therefore dominated by the dipolewhen M is small, and as M → 0 the acoustic power exceeds the quadrupolepower by a factor ∼1/M2 1. Precisely how small M should be for this tobe true depends on the details of the flow, which determine the appropriatevalues of A and V0/.

This increase in acoustic efficiency brought about by surface dipoles on anacoustically compact body occurs also for arbitrary, noncompact bodies whenturbulence interacts with compact structural elements, such as edges, corners,and protuberances.

2.5 Radiation from a Noncompact Surface

Consider a compact turbulent eddy in x2 > 0 adjacent to an infinite, plane rigidwall at x2 = 0 (Fig. 2.5.1). Let us apply the rigid surface form (2.4.1) of Curle’sequation to calculate the radiation. At high Reynolds numbers and at x in the


Fig. 2.5.1.

acoustic far field (where p(x, t) ≡ c20(ρ −ρ0) and H ( f ) ≡ H (x2) = 1) we find

p(x, t) ≈ xi x j

4πc20|x|3

∂2

∂t2

∫Ti j

(y, t − |x|

c0

)d3y

+ x2

4πc0|x|2∂

∂t

∮y2=0

(p − p0)

(y, t − |x|

c0+ x1 y1 + x3 y3

c0|x|)

dy1 dy3,

|x| → ∞. (2.5.1)

Retarded time variations have been neglected in the integral over the volumeof the compact turbulent eddy. We have not done this in the surface pressureintegral, because this tends to extend over a larger region than the Reynoldsstress fluctuations responsible for it (indeed, the acoustic component of thepressure, as opposed to the near field hydrodynamic pressure, extends out toinfinity on the wall, decaying only very slowly like 1/|x|).

The value of the surface integral cannot be estimated by a naive order-of-magnitude calculation of the kind performed in Section 2.4 for a compact body,because for an infinite plane wall the domain of integration includes the acousticregion, and therefore involves an unknown and possibly important contributionfrom the acoustic pressure that we are trying to calculate! The difficulty wasresolved by Powell (1960) by the ingenious device of applying Curle’s solution

Problems 2 39

(2.4.1) at the image x = (x1,−x2, x3) in the wall of the far field observationpoint x. At the image point H ( f ) = H (x2) ≡ 0, and therefore

0 ≈ x i x j

4πc20|x|3

∂2

∂t2

∫Ti j

(y, t − |x|

c0

)d3y

− x2

4πc0|x|2∂

∂t

∮y2=0

(p − p0)

(y, t − |x|

c0+ x1 y1 + x3 y3

c0|x|)

dy1 dy3,

|x| → ∞. (2.5.2)

The surface integral term in this formula is equal but opposite in sign to thatin the original solution (2.5.1), which is now seen to exactly represent thequadrupole sound generated by a system of image quadrupoles in the wall!Adding (2.5.1) and (2.5.2), we find

p(x, t) ≈ (xi x j + x i x j )

4πc20|x|3

∂2

∂t2

∫Ti j

(y, t − |x|

c0

)d3y

≈ (xi x j + x i x j )

4πc20|x|3

∂2

∂t2

∫ρ0viv j

(y, t − |x|

c0

)d3y, |x| → ∞. (2.5.3)

Therefore, the apparently strong contribution from the surface pressure di-poles actually integrates to a term of quadrupole strength. This is a consequenceof the Kraichnan–Phillips theorem (see Howe, 1998a), according to which thenet unsteady component of the normal force between an infinite plane wall andan incompressible fluid must vanish identically∮

y2=0(p − p0)(y, t) dy1 dy3 ≡ 0. (2.5.4)

Thus, extreme care must be exercised when using Curle’s equation to estimatethe sound produced by turbulence interacting with large surfaces. As a generalrule, the surface contribution will be comparable to that from the turbulencequadrupoles whenever the characteristic wavelength of the sound is smallerthan the radius of curvature of the surface.

Problems 2

1. Show that the acoustic efficiency of a compact sphere of radius R executingsmall amplitude translational oscillations at velocity U = U0 sin(ωt) is

a

0∼(ωR

c0

)3

,


where

a = πωR3

6ρ0U 2

0

(ωR

c0

)3

, 0 = 2ωR3

3ρ0U 2

0

are respectively the average acoustic and hydrodynamic powers fed into thefluid during the quarter cycle 0 < ωt < π/2.Explain the significance of averaging only over 0 < ωt < π/2.

2. What is the efficiency in Problem 1 when the sphere pulsates with smallamplitude normal velocity vn = U0 sin(ωt)?

3. The wake behind a bluff body fixed in a nominally steady, low Mach numberflow at speed U produces a drag force equal to CD A 1

2ρ0U 2, where CD

is the drag coefficient (which may be regarded as constant), and A is theprojected cross-sectional area of the body in the flow direction. Derive anapproximate formula for the far-field acoustic pressure radiated by the bodywhen U contains a small amplitude, time-harmonic component such thatU = U0 + u cos(ωt), where U0 and u are constant and u U0.

4. Show that Powell’s solution (2.5.3) for the sound generated by turbulenceadjacent to a rigid plane wall is identical with the solution of Curle’s differ-ential equation (2.3.8) determined by the modified Green’s function

G(x, y, t − τ ) = 1

4π |x − y|δ(

t − τ − |x − y|c0

)

+ 1

4π |x − y|δ(

t − τ − |x − y|c0

),

where x = (x1,−x2, x3). G(x, y, t − τ ) is the solution of what problem oflinear acoustics?

3The Compact Green’s Function

3.1 The Influence of Solid Boundaries

Let us return to the general problem of linear acoustics. To fix ideas, we shallframe the present discussion in terms of Equation (1.3.9)(

1

c20

∂2

∂t2− ∇2

)ϕ = −q(x, t) (3.1.1)

for the velocity potential, but our conclusions will be applicable quite generally.This equation determines ϕ in terms of a specified source distribution q(x, t). Inthe absence of solid boundaries (in free space) the results of Section 1.6 enableus to represent ϕ in the form

ϕ(x, t) =∫∫ ∞

−∞−q(y, τ )G(x, y, t − τ ) d3y dτ, (3.1.2)

where G(x, y, t − τ ) is the free space Green’s function

G(x, y, t − τ ) = 1

4π |x − y|δ(

t − τ − |x − y|c0

), (3.1.3)

that is, G(x, y, t − τ ) is the outgoing wave solution of

(1

c20

∂2

∂t2− ∇2

)G = δ(x−y)δ(t −τ ), where G = 0 for t < τ. (3.1.4)

In our discussion of Curle’s extension of Lighthill’s theory in Chapter 2, itwas found that the presence of a solid boundary S in the vicinity of the tur-bulence quadrupole sources Ti j resulted in the appearance of additional dipoleand monopole sources distributed over S, represented by the second and thirdterms on the right-hand side of Equation (2.3.8). Curle’s solution (2.3.9) of this

41

42 3 The Compact Green’s Function

Fig. 3.1.1.

equation was derived by using the retarded potential formula (1.6.6), obtainedby use of the free space Green’s function (3.1.3). A similar representation in-volving surface distributions of dipoles and monopoles is obtained for ϕ whenwe attempt to solve Equation (3.1.1) using Green’s function (3.1.3) in the sit-uation illustrated in Fig. 3.1.1, where the source distribution q(x, t) is adjacentto a rigid boundary S on which the normal derivative ∂ϕ/∂xn = 0.

It would be very convenient if we could modify the functional form ofG(x, y, t − τ ) so that it automatically takes account of the contributions fromthe dipole and monopole sources on S, inasmuch that no surface integrals occurin the final solution. To do this, we must find a solution of Green’s functionequation (3.1.4) that satisfies appropriate boundary conditions on S. The so-lution ϕ of (3.1.1) is then once again given by Formula (3.1.2) in terms ofthe modified Green’s function, there being no additional surface integrals toevaluate.

The main practical difficulty is the calculation of the modified Green’s func-tion. Although it is always possible in principle, exact analytical representationsare known only for solid bodies of very simple shapes (such as spheres, circularcylinders, and half-planes). However, it turns out that a relatively simple andgeneral approximate formula can be found for the modified Green’s functionfor those problems where it is known that the typical wavelength of the soundproduced by the source distribution q(x, t) is large compared to one or moreprincipal dimensions of the solid body S. This is called the compact Green’sfunction.

3.1 The Influence of Solid Boundaries 43

To simplify the calculation of the compact Green’s function, we use theFourier integral formula

δ(t − τ ) = 1

2π

∫ ∞

−∞e−iω(t−τ ) dω, (3.1.5)

which expresses the δ function as a linear combination of time-harmonic os-cillations of frequency ω. The formula is proved by observing that no realsystem can oscillate at infinitely large frequencies, and therefore in all practicalproblems e−iω(t−τ ) can be replaced by e−iω(t−τ )−ε|ω| for arbitrarily small ε > 0.Then,

1

2π

∫ ∞

−∞e−iω(t−τ )dω ≡ lim

ε→+0

1

2π

∫ ∞

−∞e−iω(t−τ )−ε|ω| dω

= limε→+0

ε

π [ε2 + (t − τ )2].

The final term on the right is the usual definition of δ(t − τ ) as the limit of an‘ε-sequence’ (Lighthill, 1958).

If we now put

G(x, y, t − τ ) = −1

2π

∫ ∞

−∞G(x, y, ω)e−iω(t−τ ) dω, (3.1.6)

then the substitution of this and (3.1.5) into the Green’s function equation (3.1.4)shows that, for each frequency ω, G(x, y, ω) is the solution of

(∇2 + κ20

)G(x, y, ω) = δ(x − y), (3.1.7)

where κ0 = ω/c0 is called the acoustic wave number. Sound of frequency ω

has wavelength

λ = 2π

κ0.

Thus, a solid body of characteristic dimension is compact for waves offrequency ω provided that

λ= κ0

2π 1. (3.1.8)

This condition will be used below in Section 3.4 to calculate the compactGreen’s function.


3.2 The Helmholtz Equation

The equations (∇2 + κ20

)ϕ = 0,

(∇2 + κ20

)ϕ = q(x, ω) (3.2.1)

are known respectively as the Helmholtz equation and the inhomogeneousHelmholtz equation. The source term q(x, ω) represents one frequency com-ponent of the source q(x, t) of Equation (3.1.1), so that

q(x, t) =∫ ∞

−∞q(x, ω)e−iωt dω. (3.2.2)

Therefore, because (differentiating under the integral sign)

1

c20

∂2

∂t2

∫ ∞

−∞ϕ(x, ω)e−iωt dω= 1

c20

∫ ∞

−∞−ω2ϕ(x, ω)e−iωt dω

≡ −∫ ∞

−∞κ2

0 ϕ(x, ω)e−iωt dω,

the solution ϕ(x, ω) of the inhomogeneous equation is related to the solution of(3.1.1) by

ϕ(x, t) =∫ ∞

−∞ϕ(x, ω)e−iωt dω. (3.2.3)

3.2.1 The Point Source

Equation (3.1.7) determines Green’s function G(x, y, ω) for the inhomoge-neous Helmholtz equation. The free space Green’s function can be found bythe method used in Section 1.5 for the wave equation. If we temporarily sety = 0, we have to find the radially symmetric solution of(∇2 + κ2

0

)G = δ(x). (3.2.4)

In the usual way (see Section 1.5), we have

∂2

∂r2(r G) + κ2

0 (r G) = 0 for r = |x| > 0,

and therefore

G = Aeiκ0r

r+ Be−iκ0r

r, (3.2.5)

where A, B remain to be determined.

3.2 The Helmholtz Equation 45

To do this recall that our solution represents one component of a time-dependent acoustic problem of frequency ω. Since the time factor is e−iωt ,the two terms on the right-hand side of (3.2.5) correspond to propagating soundwaves of the form

Ae−iω(

t− rc0

)r

+ Be−iω(t+ rc0

)

r,

the second of which represents waves converging on the source from infinity,and must therefore be rejected because of the radiation condition. Hence, B = 0.The value of the remaining constant A is found by extending the solution toinclude the region occupied by the source at r = 0 by writing

G = limε→0

Aeiκ0r

(r2 + ε2)12

.

By substituting the solution into (3.2.4) and using the Formula (1.4.8) we findA = −1/4π .

The free space Green’s function for the inhomogeneous Helmholtz equation(the solution of (3.1.7)) is now obtained by replacing r = |x| by |x − y|

G(x, y, ω) = −eiκ0|x−y|

4π |x − y| . (3.2.6)

Because the source q(x, ω) in the second of Equations (3.2.1) can be ex-pressed as a superposition of point sources by means of

q(x, ω) =∫ ∞

−∞q(y, ω)δ(x − y) d3y,

the solution of the inhomogeneous Helmholtz equation in an unbounded me-dium can be written

ϕ(x, ω) =∫ ∞

−∞G(x, y, ω)q(y, ω) d3y ≡ −1

4π

∫ ∞

−∞

q(y, ω)eiκ0|x−y|

|x − y| d3y.

(3.2.7)

3.2.2 Dipole and Quadrupole Sources

The method of integration by parts described in Section 1.7 can be used to showthat the corresponding solutions for the dipole and quadrupole sources

q(x, ω) = ∂ f j

∂x j(x, ω) and q(x, ω) = ∂2Ti j

∂xi∂x j(x, ω) (3.2.8)


are respectively

ϕ(x, ω) = −1

4π

∂

∂x j

∫ ∞

−∞

f j (y, ω)eiκ0|x−y|

|x − y| d3y,

and

ϕ(x, ω) = −1

4π

∂2

∂xi∂x j

∫ ∞

−∞

Ti j (y, ω)eiκ0|x−y|

|x − y| d3y. (3.2.9)

Example For a point dipole at the origin orientated in the x1 direction

q(x, ω) = ∂

∂x1( f1δ(x)).

Therefore,

ϕ(x, ω) = −1

4π

∂

∂x1

∫ ∞

−∞

f1δ(y)eiκ0|x−y|

|x − y| d3y ≈ −iκ0x1 f1eiκ0|x|

4π |x|2 , |x| → ∞.

(3.2.10)

3.2.3 Green’s Function for the Wave Equation

Let us verify the general relation (3.1.6) between the Green’s functions G(x, y,t−τ ) and G(x, y, ω), respectively, for the wave equation and the inhomogeneousHelmholtz equation in the special case in which there are no solid boundaries.According to this formula we find, using the expression (3.2.6) for the freespace Green’s function G,

G(x, y, t − τ ) = −1

2π

∫ ∞

−∞G(x, y, ω)e−iω(t−τ ) dω

= 1

8π2|x − y|∫ ∞

−∞e−iω(t−τ− |x−y|

c0)dω

= 1

4π |x − y|δ(

t − τ − |x − y|c0

),

which is precisely Equation (3.1.3).

3.3 The Reciprocal Theorem

The calculation of the compact Green’s function is greatly simplified by appli-cation of the reciprocal theorem. We need to consider only a special case of

3.3 The Reciprocal Theorem 47

Fig. 3.3.1.

this very general theorem of mechanics, which was first used with great effectin acoustics by Lord Rayleigh (1945).

Consider the two acoustic problems indicated in Fig. 3.3.1, in which soundof frequency ω is generated by two unit point sources at x = xA and x = xB inthe presence of a solid body S. We denote the functional forms of the respectivevelocity potentials generated by these sources by G(x, xA, ω) and G(x, xB, ω),where

(∇2 + κ20

)G(x, xA, ω) = δ(x − xA), (3.3.1)(∇2 + κ2

0

)G(x, xB, ω) = δ(x − xB). (3.3.2)

In addition G(x, xA, ω) and G(x, xB, ω) must satisfy appropriate mechani-cal boundary conditions on S. We take these to have the same general linearform

∂G

∂xn(x, xA, ω) = G(x, xA, ω)

Z(x, ω),

∂G

∂xn(x, xB, ω) = G(x, xB, ω)

Z(x, ω),

for x on S, (3.3.3)


where xn is measured in the normal direction from S into the fluid and Z(x, ω)is the surface impedance. For a rigid surface, Z(x, ω) = ∞.

At large distances from S, in the acoustic far field, both solutions are assumedto exhibit the characteristics of outgoing sound waves, such that (with implicittime dependence e−iωt )

G(x, xA, ω) ∼ f A(θ, φ)eiκ0r

r, G(x, xB, ω) ∼ fB(θ, φ)eiκ0r

r,

r ≡ |x| → ∞, (3.3.4)

where it may be supposed that the coordinate origin is in the neighborhood ofS. The angular dependencies of the far-field radiations from the two sourcesare determined by the factors f A(θ, φ) and fB(θ, φ), which are functions of thepolar angles θ, φ defining the orientation of the far field point x, and generallydepend strongly on the details of the interaction of the volume flows from eachsource with S.

The reciprocal theorem states that

G(xA, xB, ω) = G(xB, xA, ω). (3.3.5)

That is, the potential at xA produced by the point source at xB is equal to thepotential at xB produced by an equal point source at xA.

Proof. Multiply Equation (3.3.1) by G(x, xB, ω) and Equation (3.3.2) byG(x, xA, ω), subtract the resulting equations and integrate over the volumebounded by the surface S and by a large surface in the acoustic far field.Green’s identity

G(x, xB, ω)∇2G(x, xA, ω) − G(x, xA, ω)∇2G(x, xB, ω)

= div(G(x, xB, ω)∇G(x, xA, ω) − G(x, xA, ω)∇G(x, xB, ω)),

and the divergence theorem permit the volume integral of the term obtainedfrom the left-hand sides to be expressed as surface integrals over S and ,whereas the integrals involving the δ functions can be evaluated explicitly. Thisprocedure gives

∮S+

(G(x, xA, ω)

∂G

∂xn(x, xB, ω) − G(x, xB, ω)

∂G

∂xn(x, xA, ω)

)dS

= G(xB, xA, ω) − G(xA, xB, ω).

3.4 Time-Harmonic Compact Green’s Function 49

The surface integral over S vanishes because of the impedance conditions(3.3.3). The surface integral over vanishes because of conditions (3.3.4) andbecause ∂θ/∂xn and ∂φ/∂xn are each of order 1/r as r → ∞, and therefore

∂G

∂xn(x, xA, ω) ∼ f A(θ, φ)

iκ0eiκ0r

r

∂r

∂xn,

∂G

∂xn(x, xB, ω) ∼ fB(θ, φ)

iκ0eiκ0r

r

∂r

∂xn, as r → ∞.

This proves the theorem.The result is usually expressed as the simple reciprocal relation

G(x, y, ω) = G(y, x, ω). (3.3.6)

3.4 Time-Harmonic Compact Green’s Function

We are now ready to derive the compact Green’s function G(x, y, ω) for theproblem depicted in Fig. 3.4.1. We have to solve

(∇2 + κ20

)G(x, y, ω) = δ(x − y),

∂G

∂xn= 0 on S, (3.4.1)

where the rigid body S is assumed to be acoustically compact. The influence of

Fig. 3.4.1.


a solid body on the production of sound by neighboring sources is equivalent toan additional distribution of monopoles and dipoles on S. The compact Green’sfunction includes a first approximation for the net effect of these monopole anddipole distributions, obviating the need to evaluate surface integrals.

In practice, we are interested primarily in calculating the sound in the farfield of the body. Let denote the characteristic diameter of the body, and takethe coordinate origin at O within S. The source point y is assumed to be closeto S (so that |y| ∼ ) and the observer at x is taken to be in the acoustic farfield. The compactness condition (3.1.8) therefore implies that

κ0 1 and κ0|y| 1.

In these circumstances the compact approximation for G(x, y, ω) can befound very easily from the solution of the reciprocal problem:

(∂2

∂y21

+ ∂2

∂y22

+ ∂2

∂y23

+ κ20

)G(y, x, ω) = δ(y − x),

∂G

∂yn= 0 on S,

(3.4.2)

where the source is at the far-field point x, and G(y, x, ω) is determined as afunction of y close to S. The solution of (3.4.1) is then given by the reciprocaltheorem (Section 3.3) G(x, y, ω) = G(y, x, ω) (the potential G(x, y, ω) at thefar-field point x produced by the point source at y is exactly equal to the potentialG(y, x, ω) produced at the near-field point y by an equal point source at thefar-field point x).

To solve (3.4.2), we put

G(y, x, ω) = G0(y, x, ω) + G ′(y, x, ω)

≡ −eiκ0|x−y|

4π |x − y| + G ′(y, x, ω)

where G0(y, x, ω) is the spherically spreading wave generated by the pointsource at x when the presence of the solid is ignored. The term G ′(y, x, ω)is the velocity potential of the motion produced in the fluid when this waveimpinges on S.

When |x| → ∞, the approximations

|x − y| ≈ |x| − x · y|x| ≡ |x| − x j y j

|x| and1

|x − y| ≈ 1

|x| + x · y|x|3 ≈ 1

|x|

3.4 Time-Harmonic Compact Green’s Function 51

and the condition κ0 y j ∼ κ0 1 imply that

G0(y, x, ω) ≡ −eiκ0|x−y|

4π |x − y| ≈ −eiκ0|x|

4π |x| × e− iκ0 x j y j|x|

≈ −eiκ0|x|

4π |x|(

1 − iκ0x j y j

|x| + O(κ0)2

)(3.4.3)

The linear dependence on y j in the second line of this formula represents the firstapproximation (of order κ0) in a power series expansion of rapidly decreasingterms that describes the variation of the incident spherical wave close to thebody. Thus, regarded as a function of y, the terms shown explicitly in

G0(y, x, ω) = −eiκ0|x|

4π |x| + eiκ0|x|

4π |x|iκ0x j y j

|x| + · · · ≡ constant + U j y j + · · · ,

where U j = eiκ0|x|

4π|x|iκ0x j

|x| , (3.4.4)

can be regarded as the velocity potential of a uniform flow at velocity U j im-pinging on the solid.

At distances |y| from S, the distortion of this flow produced by the bodymust be small. Let it be represented by the velocity potential

G ′(y, x, ω) = −U jϕ∗j (y) + O(κ0)2, where ϕ∗

j (y) → 0 when |y| .

The function ϕ∗j has the dimensions of length and ∼ in order of magnitude

(Batchelor 1967). Then,

G(y, x, ω) = G0(y, x, ω) + G ′(y, x, ω) = −eiκ0|x|

4π |x| + U j (y j − ϕ∗j (y)) + · · · ,

(3.4.5)

where the terms shown explicitly represent a potential flow past the body.Near the body G(y, x, ω) satisfies (3.4.2) with the right-hand side replaced

by zero (because the source is in the far field). Hence,

U j∇2(y j − ϕ∗j (y)) + O(κ0)2 = 0.

But U j (y j − ϕ∗j (y)) = O(κ0), and therefore, correct to the neglect of small

terms of order O(κ0)2,

∇2(y j − ϕ∗j (y)) = 0, i.e. ∇2ϕ∗

j (y) = 0,


where the rigid surface condition requires

∂

∂yn(y j − ϕ∗

j (y)) = 0 on S. (3.4.6)

Summarizing our conclusions from Equations (3.4.4)–(3.4.6): When x is inthe acoustic far field, and y is close to the body

G(x, y, ω) = −eiκ0|x|

4π |x|(

1 − iκ0x j

|x| (y j − ϕ∗j (y)) + O(κ0)2

),

y ∼ O(), |x| → ∞. (3.4.7)

The first term in the large brackets represents the contribution from the sphericalwave G0(x, y, ω) evaluated at y = 0. The next term is O(κ0) and includes a com-ponent −iκ0x j y j/|x| from the incident wave plus a correction iκ0x jϕ

∗j (y)/|x|

produced by S.The vector field

Y(y) ≡ y − ϕ∗(y) (3.4.8)

is called the Kirchhoff vector for the body; the j th component

Y j (y) ≡ y j − ϕ∗j (y)

satisfies Laplace’s equation ∇2Y j = 0 with ∂Y j/∂yn = 0 on S, and can beinterpreted as the velocity potential of an incompressible flow past S that hasunit speed in the j direction at large distances from S. The functionϕ∗

j (y) decayswith distance from S, and satisfies

∂ϕ∗j

∂yn(y) = n j on S, (3.4.9)

because ∂y j/∂yn ≡ ni∂y j/∂yi = niδi j = n j . Hence, ϕ∗j (y) is just the instanta-

neous velocity potential of the motion that would be produced by translationalmotion of S as a rigid body at unit speed in the j direction.

Definition

G(x, y, ω) = −eiκ0|x|

4π |x|(

1 − iκ0x j

|x| (y j − ϕ∗j (y))

), y ∼ O(), |x| → ∞,

(3.4.10)

is called the compact Green’s function for source points y near the body andobserver positions x in the acoustic far field.

3.5 Compact Green’s Function for a Rigid Sphere 53

In Section 3.7, we shall introduce a very much more elegant representationof the compact Green’s function that greatly expands its utility.

3.5 Compact Green’s Function for a Rigid Sphere

Let the sphere have radius a and take the coordinate origin O at its center, asillustrated in Fig. 3.5.1. We have to determine the Kirchhoff vector whose j thcomponent

Y j (y) = y j − ϕ∗j (y) for j = 1, 2, 3

is equal to the velocity potential of incompressible flow past the sphere havingunit speed in the j direction at large distances from the sphere.

Consider the case j = 1 shown in the figure. The flow is evidently symmetricabout the y1 axis. Take spherical polar coordinates (r, ϑ, φ) with ϑ measuredfrom the positive y1 axis. Then, y1 = r cosϑ and the condition (3.4.9) to besatisfied on the sphere is

∂ϕ∗1

∂r= cosϑ at r = a. (3.5.1)

The axisymmetry of the problem suggests that we look for a solution ofLaplace’s equation in the form

ϕ∗1 = (r ) cosϑ,

which satisfies the axisymmetric form of Laplace equation(1

r2

∂

∂r

(r2 ∂

∂r

)+ 1

r2 sinϑ

∂

∂ϑ

(sinϑ

∂

∂ϑ

))(r ) cosϑ = 0,

Fig. 3.5.1.


provided that

r2 d2

dr2+ 2r

d

dr− 2 = 0.

The solutions of this equation are proportional to rn where n is a root of thequadratic equation

n2 + n − 2 = 0, i.e., n = −2, 1.

Hence,

Y1 ≡ y1 − ϕ∗1 = r cosϑ −

(Ar + B

r2

)cosϑ,

where A and B are constants. The condition that ϕ∗1 → 0 as r → ∞ implies

that A = 0, and condition (3.5.1) supplies B = −a3/2. Therefore,

Y1 = r cosϑ + a3

2r2cosϑ ≡ y1

(1 + a3

2r3

).

Because of the symmetry of the sphere it is clear that we also have

Y2 = y2

(1 + a3

2r3

), Y3 = y3

(1 + a3

2r3

), r = |y|.

Thus, the compact Green’s function (3.4.10) for the sphere is

G(x, y, ω) = −eiκ0|x|

4π |x|

1 − iκ0x j y j

|x|(

1 + a3

2|y|3)

, y ∼ O(a),

|x| → ∞. (3.5.2)

This represents the far-field acoustic potential produced by a point source at yclose to the sphere. Because κ0|y| is small the second term in the brace bracketsis always small compared to 1. This appears to suggest that, after all, the spherehas a relatively small effect on the production of sound! This is certainly truefor monopole sources, but most sources of interest in applications are dipoles orquadrupoles, and in these circumstances we shall see that it is the small, secondterm that determines the leading order approximation for the far-field sound.

3.5.1 Radiation from a Dipole Adjacent to a Compact Sphere

Let us apply the compact Green’s function (3.5.2) to determine the far-fieldsound generated by a dipole source close to a sphere of radius a λ = acoustic


Fig. 3.5.2.

wavelength. With the origin at the center of the sphere, we consider the outgoingwave solution of

(∇2 + κ20

)ϕ = f1

∂

∂x1δ(x1 − L)δ(x2)δ(x3), where

∂ϕ

∂xn= 0 on |x| = a.

The dipole is orientated in the x1 direction and lies on the x1 axis at (L , 0, 0),as in Fig. 3.5.2. The solution is given by

ϕ(x, ω) =∫

f1∂

∂y1δ(y1 − L)δ(y2)δ(y3) G(x, y, ω) d3y,

where the integration is over the fluid, and ∂G/∂xn = 0 on the sphere. Thesource term is zero everywhere except at (L , 0, 0). To evaluate the integral wewrite

ϕ(x, ω) = f1

∫∂

∂y1G(x, y, ω)δ(y1 − L)δ(y2)δ(y3) d3y

− f1

∫δ(y1 − L)δ(y2)δ(y3)

∂G

∂y1(x, y, ω) d3y.

The first integral is zero because δ(y1 − L) = 0 on the boundaries of the regionof integration, so that

ϕ(x, ω) = −f1

(∂G

∂y1(x, y, ω)

)y=(L ,0,0)

. (3.5.3)

Thus far the calculation is exact. To determine the solution in the far fieldgiven that the sphere is acoustically compact we use the compact approxima-tion (3.5.2) for G(x, y, ω). We see immediately that the differentiation withrespect to y1 will be applied only to the small second term in the braces


of (3.5.2), giving

ϕ(x, ω) = −iκ0 f1x j eiκ0|x|

4π |x|2

∂

∂y1

[y j

(1 + a3

2|y|3)]

y=(L ,0,0)

= −iκ0 f1x1eiκ0|x|

4π |x|2(

1 − a3

L3

)

= −iκ0 f1 cos θ eiκ0|x|

4π |x|(

1 − a3

L3

), |x| → ∞, κ0a 1,

where θ is the angle between the x1 axis and the x direction (so that x1 =|x| cos θ).

By setting a = 0 in this formula, we recover the far field (3.2.10) of a dipolesource in the absence of the sphere. The presence of the sphere accordinglyreduces the amplitude of the sound relative to that produced by a free-fielddipole. The amplitude is zero when L → a, because in this limit the surfaceof the sphere is effectively plane in the vicinity of the dipole and an equal andopposite image dipole is formed in the sphere. The net radiation is thereforeequivalent to that produced by a quadrupole source, and to calculate the soundin this case it would be necessary to use a more accurate approximation toG(x, y, ω). This conclusion applies only to dipoles orientated radially withrespect to the sphere (see Problem 1), but it is also true for any compact rigidsurface when a dipole orientated in the direction of the local surface normalapproaches the surface.

3.5.2 Sound Produced by a Vibrating Sphere

Let the surface S of a fixed body execute small amplitude vibrations with normalvelocity vn(x, ω) (Fig. 3.5.3). The fluid motion is the same as that generated bya distribution of monopoles of strength vn(x, ω) per unit area of S when S is

Fig. 3.5.3.


assumed to be stationary (rigid). The corresponding source strength q(x, ω) inthe inhomogeneous Helmholtz equation (3.2.1) is

q(x, ω) = vn(x, ω)δ(s⊥ − ε), (ε → +0),

where s⊥ is distance measured in the normal direction from S into the fluid,and ε > 0 places the sources just within the fluid adjacent to S. The velocitypotential ϕ(x, ω) is therefore

ϕ(x, ω) =∫

fluidvn(y, ω)δ(s⊥ − ε)G(x, y, ω) d3y (ε → +0)

=∮

Svn(y, ω)G(x, y, ω) d S(y) where

∂G

∂xn(x, y, ω) = 0 on S.

(3.5.4)

Consider the sound produced when the sphere of Fig. 3.5.1 vibrates withsmall amplitude about its undisturbed position centred at the origin with velocityU (ω)e−iωt along the x1 axis. Then,

vn(y, ω) = U (ω) cosϑ.

If the vibrations are at sufficiently low frequency, the sphere will be compact,and when the observer at x is in the acoustic far field the integral in (3.5.4) canbe evaluated using the compact approximation (3.5.2) for G(x, y, ω):

ϕ(x, ω)

≈ −eiκ0|x|

4π |x|[∮

Svn(y, ω) d S(y) − iκ0x j

|x|∮

Sy j

(1 + a3

2|y|3)vn(y, ω) d S(y)

].

The first integral represents the net volume flux through S and vanishes identi-cally for rigid body translational motion. The second integral is nonzero onlyfor j = 1, when y1 = a cosϑ and |y| = a on S, and we can take d S = 2πa2

sinϑ dϑ (so that the surface integral becomes 3πa3U (ω)∫ π

0 cos2 ϑ sinϑ dϑ =2πa3U (ω)). Hence,

ϕ(x, ω) ≈ iκ0U (ω)a3x1eiκ0|x|

2|x|2 ≡ iωU (ω)a3 cos θeiκ0|x|

2c0|x| , |x| → ∞,

where θ is the angle between the x1 axis and the radiation direction x (seeFig. 1.7.1).

The solution for a sphere oscillating at an arbitrary time dependent velocityU (t) can be derived from this result provided the sphere remains compact. This


means that if we write

U (t) =∫ ∞

−∞U (ω)e−iωt dω,

then U (ω) = 0 only for κ0a 1. If this condition is satisfied we can use theFormulae (3.2.2) and (3.2.3) to obtain the time-dependent velocity potential inthe form

ϕ(x, t) ≈ a3 cos θ

2c0|x|∫ ∞

−∞iωU (ω)e−iω(t−|x|/c0) dω

= −a3 cos θ

2c0|x|∂

∂t

∫ ∞

−∞U (ω)e−iω(t−|x|/c0) dω

= −a3 cos θ

2c0|x|∂U

∂t(t − |x|/c0), |x| → ∞.

This agrees with the far-field result obtained in Section 1.7, and confirms themodel used there in which the vibrating sphere was replaced by a point dipoleof strength 2πa3U (t) at its center.

3.6 Compact Green’s Function for Cylindrical Bodies

The reciprocal calculation of the Green’s function described in Section 3.4 forthe compact body in Fig. 3.4.1 can be immediately extended to the case of acylindrical body of compact cross section.

Figure 3.6.1 illustrates the situation for an infinite circular cylinder of radiusa whose axis lies along the y3 axis, and whose diameter 2a ∼ is acousticallycompact. The source point y is adjacent to the cylinder and for the moment(see Section 3.7) is assumed to be within an axial distance |y3| λ fromthe coordinate origin O. In this region the Expansion (3.4.7) remains validwith ϕ∗

3 (y) ≡ 0, because the impinging flow described by the velocity potential(3.4.4) for j = 3 is unaffected by the cylinder. Hence, we can take

G(x, y, ω) = −eiκ0|x|

4π |x|(

1 − iκ0x j Y j

|x|), y ∼ O(), |x| → ∞, (3.6.1)

where the Kirchhoff vector Y has the components

Y1 = y1 − ϕ∗1 (y), Y2 = y2 − ϕ∗

2 (y), Y3 = y3. (3.6.2)

3.6.1 Circular Cylinder

The potentials ϕ∗1 (y), ϕ∗

2 (y) for the circular cylinder of radius a can be foundby the method of Section 3.5.

3.6 Compact Green’s Function for Cylindrical Bodies 59

Fig. 3.6.1.

For j = 1 the flow is symmetric about the y1 axis and is independent ofthe spanwise coordinate y3 (Fig. 3.6.2). Using polar coordinates (y1, y2) =r (cosϑ, sinϑ), the condition (3.4.9) to be satisfied on the cylinder is

∂ϕ∗1

∂r= cosϑ at r = a. (3.6.3)

Fig. 3.6.2.


As in the case of the sphere, we try a solution of the form

ϕ∗1 = (r ) cosϑ,

which satisfies the polar form of Laplace’s equation(1

r

∂

∂r

(r∂

∂r

)+ 1

r2

∂2

∂ϑ2

)(r ) cosϑ = 0,

provided that

r2 d2

dr2+ r

d

dr− = 0.

The general solution is = Ar + B/r . The component Ar must be rejectedbecause it does not decay as r → ∞. Therefore,

Y1 ≡ y1 − ϕ∗1 = r cosϑ − B

rcosϑ,

and condition (3.6.3) yields B = −a2. Therefore,

Y1 = r cosϑ + a2

rcosϑ ≡ y1

(1 + a2

r2

).

Similarly,

Y2 = y2

(1 + a2

r2

).

Hence, the compact Green’s function for a circular cylinder, with sourcenear the origin, is

G(x, y, ω) = −eiκ0|x|

4π |x|(

1 − iκ0x j Y j

|x|), y ∼ O(), |x| → ∞, (3.6.4)

where

Y j = y j

(1 + a2

y21 + y2

2

), j = 1, 2; Y3 = y3. (3.6.5)

3.6.2 Rigid Strip

The rigid strip of chord 2a and infinite span provides a simple model of asharp-edged airfoil. In Fig. 3.6.3 the airfoil occupies −a < y1 < a, y2 = 0,−∞ < y3 < ∞. The airfoil has no influence on a uniform mean flow in the y1-direction, nor on one in the y3-direction, so that potential functions ϕ∗

1 (y) ≡ 0and ϕ∗

3 (y) ≡ 0.

3.6 Compact Green’s Function for Cylindrical Bodies 61

Fig. 3.6.3.

The potential ϕ∗2 (y) ≡ ϕ∗

2 (y1, y2) can be determined by the method of con-formal transformation. (Readers unfamiliar with this procedure should consultSection 4.5.) If z = y1 + iy2, the cross section of the airfoil in the z plane ismapped onto the circular cylinder |Z | = a in the Z plane by the transformation

Z = z +√

z2 − a2.

Because Z ∼ 2z as |z| → ∞ a uniform flow at unit speed in the y2 directionin the z plane at large distances from the airfoil corresponds to a uniformflow at speed 1

2 in the direction of the imaginary Z axis at large distancesfrom the cylinder. This flow can be found by the method discussed abovefor the circular cylinder (or see Example 3 of Section 4.5), and determinesY2 = y2 − ϕ∗

2 (y1, y2) = Re[w(z)], where w is the complex potential

w(z) = − i

2

(Z − a2

Z

)

= − i

2

(z +

√z2 − a2 − a2

z + √z2 − a2

)

= −i√

z2 − a2.

Thus, the compact Green’s function for a strip, with source near the origin,is given by

G(x, y, ω) = −eiκ0|x|

4π |x|(

1 − iκ0x j Y j

|x|), y ∼ O(a), |x| → ∞, (3.6.6)

where the components of the Kirchhoff vector are

Y1 = y1, Y2 = Re(−i√

z2 − a2), Y3 = y3, z = y1 + iy2. (3.6.7)


Fig. 3.6.4.

Figure 3.6.4 depicts the streamline pattern of the flow past the strip definedby the velocity potential Y2(y). The streamlines crowd together and change veryrapidly near the sharp edges. This is an indication that edges can be importantsources of noise when located in the near field of a dipole or quadrupole (orany higher order multipole) source, because ∇Y2 becomes very large there.

Example Calculate the far-field velocity potential when

(∇2 + κ20

)ϕ = f2

∂

∂x2[δ(x1 − L)δ(x2)δ(x3)] , L > a, κ0L 1,

where∂ϕ

∂x2= 0 on the airfoil −a < x1 < a, x2 = 0, −∞ < x3 < ∞.

The dipole source is orientated in the x2 direction and is positioned just to theright of the edge at y1 = a in Fig. 3.6.3. The solution is given by the followingform of Equation (3.5.3)

ϕ(x, ω) = − f2

(∂G

∂y2(x, y, ω)

)y=(L ,0,0)

≈ − i f2κ0x2eiκ0|x|

4π |x|2(∂Y2

∂y2

)y=(L ,0,0)

where, from (3.6.7),

∂Y2

∂y2= Re

(−i

∂

∂y2

√z2 − a2

), z = y1 + iy2

= Re

(z√

z2 − a2

),

3.7 Symmetric Compact Green’s Function 63

Therefore,

ϕ(x, ω) ≈ − i f2κ0x2eiκ0|x|

4π |x|2L√

L2 − a2

= − i f2κ0L cos eiκ0|x|

4π |x|√L2 − a2, |x| → ∞,

where = cos−1(x2/|x|) is the angle between the normal to the strip and theradiation direction (x/|x|) indicated in Fig. 3.6.3.

The amplitude of the sound is increased by a factor L/√

L2 − a2 relative tothat produced by the same dipole in free space, and is unbounded as L → a,when the dipole approaches the edge.

3.7 Symmetric Compact Green’s Function

The definition (3.4.10) of the compact Green’s function can be recast to exhibitthe reciprocal nature of the source and observer positions y and x. To do thiswe first observe that, for a body of characteristic diameter ,

Y j (y) = y j − ϕ∗j (y) ∼ O(),

and, therefore, that κ0Y j 1. Hence,

G(x, y, ω) ≈ −eiκ0|x|

4π |x|(

1 − iκ0x j Y j

|x|)

= −eiκ0|x|

4π |x|(

1 − iκ0x · Y|x|

)

≈ −1

4π |x|eiκ0|x|− iκ0x·Y|x| ,

≈ −eiκ0|x−Y|

4π |x| , Y ∼ O(), |x| → ∞, (3.7.1)

where on the last line we have used the usual far-field approximation (1.9.2)

|x − Y| ≈ |x| − x · Y|x| , |x| → ∞.

Now let X(x) denote the Kirchhoff vector for the body expressed in terms of x,i.e., let

X j (x) = x j − ϕ∗j (x). (3.7.2)


Then, because ϕ∗j (x) → 0 as |x| → ∞, we also have |X| ∼ |x| as |x| → ∞,

and, therefore, from (1.9.2) and (1.9.3)

|X − Y| ≈ |x| − x · Y|x|

1

|X − Y| ≈ 1

|x| + x · Y|x|3 ≈ 1

|x|

when |x| → ∞. (3.7.3)

Thus, to the same approximation, (3.7.1) can be written

G(x, y, ω) ≈ −eiκ0|X−Y|

4π |X − Y| , Y ∼ O(), |x| → ∞.

This result is the basis of our revised definition of the

Compact Green’s Function for the Inhomogeneous Helmholtz Equation

G(x, y, ω) = −eiκ0|X−Y|

4π |X − Y| , (3.7.4)

where X = x − ϕ∗(x), Y = y − ϕ∗(y) are the Kirchhoff vectors for the bodyexpressed respectively in terms of x and y. The components X j and Y j are thevelocity potential of incompressible flow past the body having unit speed inthe j direction at large distances from the body; ϕ∗

j is the velocity potential ofthe incompressible flow that would be produced by rigid body motion of S atunit speed in the j direction.

Our generalized definition clearly satisfies the reciprocal theorem. Also, be-cause of the symmetrical way in which x and y enter this formula we may nowremove any restriction on the position of the coordinate origin. The approxima-tion is valid for arbitrary source and observer locations provided that at leastone of them lies in the far field of the body. When both x and y are in the farfield (so that X ∼ x and Y ∼ y) predictions made with the compact Green’sfunction will be the same as when the body is absent. This is because for dis-tant sources the amplitude of the sound scattered by a compact rigid object isO((κ0)2) smaller than the incident sound, that is, is of quadrupole intensity(Lighthill 1978; Howe 1998a). When x is close to the body the source must bein the far field; G(x, y, ω) then determines the modification by the body of lowfrequency sound received by an observer near the body.

The definition (3.7.4) is easily recalled because it is an obvious generaliza-tion of the free space Green’s function (3.2.6). In applications it is necessaryto remember also that it is valid for determining only the leading order ap-proximation to the surface monopole and dipole sources induced on the body

3.8 Low-Frequency Radiation from a Vibrating Body 65

by neighboring sources in the fluid. In practice this means that when used incalculations G(x, y, ω) will normally be expanded only to first order in theKirchhoff source vector Y(y).

We can go further and use the formula (3.1.6) relating Green’s functions forthe wave equation and the Helmholtz equation to derive the compact approxi-mation for Green’s function of the wave equation:

G(x, y, t − τ ) = −1

2π

∫ ∞

−∞G(x, y, ω)e−iω(t−τ ) dω

≈ 1

8π2|X − Y|∫ ∞

−∞e−iω(t−τ− |X−Y|

c0)dω

= 1

4π |X − Y|δ(

t − τ − |X − Y|c0

).

This remarkable result is formally identical with the classical free space Green’sfunction (1.6.2) with X, Y substituted for x, y. However, its use is subjectto the same restrictions as (3.7.4), and it will be valid only when applied totime-dependent source terms producing sound whose wavelength is large com-pared to the characteristic body dimension . With this understanding we candefine the

Compact Green’s Function for the Wave Equation

G(x, y, t − τ ) = 1

4π |X − Y|δ(

t − τ − |X − Y|c0

), (3.7.5)

where X = x −ϕ∗(x), Y = y −ϕ∗(y) are Kirchhoff vectors for the body. Thecomponents X j and Y j are the velocity potential of incompressible flow pastthe body having unit speed in the j direction at large distances from the body;ϕ∗

j is the velocity potential of the incompressible flow that would be producedby rigid body motion of S at unit speed in the j direction.

3.8 Low-Frequency Radiation from a Vibrating Body

The compact Green’s function (3.7.5) for the wave equation will now be usedto give a complete theory of the low-frequency sound produced by a vibratingbody. The maximum frequency of the vibrations must be small enough toensure that the body (or its cross section, in the case of vibrating a cylinder)is acoustically compact. The argument follows closely the discussion of thevibrating sphere in Section 3.5, except that we now work directly with timedependent quantities.


Fig. 3.8.1.

Let the closed surface S (Fig. 3.8.1) vibrate with normal velocity vn(x, t). Asbefore, the velocity potential in the fluid (governed by Equation (3.1.1)) is thesame as that generated by the distribution of volume sources

q(x, t) = vn(x, t)δ(s⊥ − ε) (ε → +0) (3.8.1)

distributed over S regarded as a rigid, stationary surface, where s⊥ is distancemeasured in the normal direction from S into the fluid. The velocity potentialϕ(x, t) is given exactly by

ϕ(x, t) = −∫ ∞

−∞

∫fluid

vn(y, τ )δ(s⊥ − ε)G(x, y, t − τ ) d3y dτ (ε → +0)

= −∫ ∞

−∞

∮Svn(y, τ )G(x, y, t − τ ) d S(y) dτ,

where∂G

∂xn(x, y, t − τ ) = 0 on S. (3.8.2)

At low frequencies the first approximation to the far-field sound is obtainedby replacing G(x, y, t − τ ) in (3.8.2) by its compact approximation (3.7.5).The details are given below; they illustrate the general procedure that should beadopted when using the compact Green’s function (in particular, the techniqueof expanding to first order in Y):

ϕ(x, t) ≈ −∫ ∞

−∞

∫fluid

vn(y, τ )δ(s⊥ − ε)

4π |X − Y| δ

(t − τ − |X − Y|

c0

)d3y dτ

(ε → +0), |x| → ∞

= −∫ ∞

−∞

∮S

vn(y, τ )

4π |X − Y|δ(

t − τ − |X − Y|c0

)d S(y) dτ,

= − 1

4π |x|∫ ∞

−∞

∮Svn(y, τ )δ

(t − τ − |x|

c0+ x · Y

c0|x|)

d S(y) dτ

(X ∼ x as |x| → ∞)


= − 1

4π |x|∫ ∞

−∞

∮Svn(y, τ )

δ

(t − τ − |x|

c0

)

+ δ′(

t − τ − |x|c0

)x j Y j

c0|x|

d S(y) dτ,

where the prime denotes differentiation with respect to time. Performing theintegration with respect to τ :

ϕ(x, t) ≈ − 1

4π |x|∮

Svn

(y, t − |x|

c0

)d S(y)

− x j

4πc0|x|2∂

∂t

∮Svn

(y, t − |x|

c0

)Y j (y) d S(y).

The first integral represents an omnidirectional monopole sound wave, and isnonzero only if the volume enclosed by S changes with time (i.e., only for apulsating body). It is then the most important component of the far field sound –the second integral is smaller by a factor ∼O(ω/c0) 1 (because ∂/∂t ∼ ω

and Y j ∼ ).The monopole term vanishes for a rigid body executing small amplitude

translational oscillations at velocity U(t), say. Then,

vn(y, τ ) = n(y) · U(τ ) = ni (y)Ui (τ ),

where n(y) is the surface normal directed into the fluid. Making the substitutionY j = y j − ϕ∗

j (y) in the second integral we obtain an acoustic field of dipoletype, given by

ϕ(x, t) ≈ − x j

4πc0|x|2∂Ui

∂t

(t − |x|

c0

)∮S

ni (y)Y j (y) dS(y) (3.8.3)

= − x j

4πc0|x|2∂Ui

∂t

(t − |x|

c0

)∮S[ni y j − niϕ

∗j ] dS, |x| → ∞.

(3.8.4)

Example: The vibrating sphere Consider a rigid sphere of radius a centred atthe origin and oscillating in the x1 direction at velocity U (t) (Fig. 3.8.2). ThenU = (U, 0, 0), and it is only necessary to take i = 1 in (3.8.3) or (3.8.4).

In terms of spherical polar coordinates (r, ϑ, φ) we have

y = r (cosϑ, sinϑ cosφ, sinϑ sinφ)


Fig. 3.8.2.

Therefore,

Y ≡ y(

1 + a3

2|y|3)

= 3a

2(cosϑ, sinϑ cosφ, sinϑ sinφ)

on the sphere and n1 = cosϑ

Hence,∮S

n1Y j d S = 3a3

2

∮S(cosϑ, sinϑ cosφ, sinϑ sinφ) cosϑ sinϑ dϑ dφ

=

2πa3, j = 1,0, j = 2, 3

and, therefore, (3.8.3) becomes

ϕ(x, t) ≈ −a3 cos θ

2c0|x|∂U

∂t(t − |x|/c0), |x| → ∞, and

x1

|x| = cos θ,

(3.8.5)

which is the result already obtained in Section 3.5 using the solution derivedfrom the Helmholtz equation.

3.8.1 Far Field Pressure Produced by a Vibrating Body

A more general and illuminating discussion of the low-frequency sound pro-duced by a vibrating rigid body can be given in terms of the added mass tensorMi j (Batchelor 1967), defined by the surface integral

Mi j = −ρ0

∮S

niϕ∗j d S.


The Condition (3.4.9) satisfied by ϕ∗j on S implies that Mi j = M ji , because ni

can be replaced in the integrand by ∂ϕ∗i /∂yn, and

Mi j = −ρ0

∮S

niϕ∗j d S = −ρ0

∮S

∂ϕ∗i

∂ynϕ∗

j d S ≡ −ρ0

∮Sϕ∗

i

∂ϕ∗j

∂ynd S = M ji .

(3.8.6)

The final integral is deduced from the second by referring to Fig. 3.3.1, recallingthat ∇2ϕ∗

j = ∇2ϕ∗i = 0, and applying the divergence theorem as follows:

∮S+

(∂ϕ∗

i

∂ynϕ∗

j − ϕ∗i

∂ϕ∗j

∂yn

)d S =

∫fluid

(ϕ∗i ∇2ϕ∗

j − ϕ∗j ∇2ϕ∗

i ) d3y ≡ 0.

The integration over vanishes as the surface recedes to infinity (Batchelor1967) because

ϕ∗i, j (y) ∼ O

(1

|y|2)

as |y| → ∞.

By evaluating the net force on S produced by the unsteady surface pressure(or by the method described below in Section 4.4) it can be verified that whenthe body translates at velocity U(t) without rotation in an ideal, incompressiblefluid, it exerts a force on the fluid in the i direction given by

Fi (t) = Mi jdU j

dt. (3.8.7)

For a body of mass m, this means that when an external force Fi acts throughits centre of mass, the equation of motion of the body can be written

(mδi j + Mi j )dU j

dt= Fi .

The added mass tensor determines the effective mass of fluid dragged along bythe body in its accelerated motion. The inertia of this fluid, in addition to thatof the body, must also be overcome by the force F when the body accelerates.In general, however, a couple must also be applied to the translating body tocounter a rotational torque also exerted on the body by the fluid (see Batchelor(1967) for further discussion).

Let us now apply these concepts to determine from (3.8.4) the sound pres-sure produced by a rigid compact body executing small amplitude translationaloscillations at velocity U(t). The acoustic pressure is given in the far field byp = −ρ0∂ϕ/∂t (see Section 1.3), and therefore

p(x, t) = x j

4πc0|x|2∂2Ui

∂t2

(t − |x|

c0

)ρ0

∮S

ni y j d S − ρ0

∮S

niϕ∗j d S

,

|x| → ∞. (3.8.8)


The first integral is evaluated by applying the divergence theorem, which trans-forms it into an integral over the volume Vs of the body:

ρ0

∮S

ni y j dS = ρ0

∫Vs

∂y j

∂yid3y = ρ0Vsδi j ≡ m0δi j , (3.8.9)

where m0 is the mass of the fluid displaced by the body. The second term in thebrace brackets of (3.8.8) is just the added mass tensor Mi j .

Thus, the acoustic pressure can be expressed in either of the forms

p(x, t) ≈ xi

4πc0|x|2 (m0δi j + Mi j )∂2U j

∂t2

(t − |x|

c0

)

= xi

4πc0|x|2(

m0∂2Ui

∂t2+ ∂Fi

∂t

)(t − |x|

c0

), |x| → ∞, (3.8.10)

where the second line follows from (3.8.7), where Fi is the force exerted by thebody on the fluid in the i direction.

For a sphere of radius a oscillating at speed U (t) in the x1 direction

m0 = 43πa3ρ0 and Mi j = 1

2 m0δi j

Therefore,

p(x, t) ≈ xi

4πc0|x|2(

m0δi j + 1

2m0δi j

)∂2U j

∂t2

(t − |x|

c0

), |x| → ∞.

= ρ0a3 cos θ

2c0|x|∂2U

∂t2

(t − |x|

c0

),

which is equivalent to (3.8.5).

3.9 Compact Green’s Function Summary and Special Cases

3.9.1 Compact Bodies and Cylindrical Bodies of Compact Cross Section

General Form

G(x, y, t − τ ) = 1

4π |X − Y|δ(

t − τ − |X − Y|c0

)(3.9.1)

X = x − ϕ∗(x)

Kirchhoff vectors for the body.Y = y − ϕ∗(y)

The vector components X j (x) and Y j (y) are the velocity potentials of

3.9 Compact Green’s Function Summary and Special Cases 71

Table 3.9.1. Standard Special Cases

Body X1 X2 X3

Sphere of radius a, withcentre at origin

x1

(1 + a3

2|x|3)

x2

(1 + a3

2|x|3)

x3

(1 + a3

2|x|3)

Circular cylinder of radius acoaxial with the x3-axis

x1

(1 + a2

x21 +x2

2

)x2

(1 + a2

x21 +x2

2

)x3

Strip airfoil−a < x1 < a, x2 = 0,−∞ < x3 < ∞

x1 x3Re(−i

√z2 − a2)

z = x1 + i x2

incompressible flow past the body having unit speed in the j direction at largedistances from the body (special cases are listed in Table 3.9.1); ϕ∗

j is the veloc-ity potential of the incompressible flow that would be produced by rigid bodymotion of S at unit speed in the j direction. For a cylindrical body of compactcross section parallel to the x3 direction, we take

X3 = x3, Y3 = y3.

3.9.2 Airfoil of Variable Chord

The compact Green’s function defined by (3.6.6) and (3.6.7) for a rigid stripcan be generalized to include the finite span, variable chord airfoil illustrated inFig. 3.9.1. The coordinate axes are orientated as in Fig. 3.6.3 for the strip airfoil,with y2 normal to the plane of the airfoil and y3 in the spanwise direction. Theairfoil span is assumed to be large, and the chord 2a ≡ 2a(y3) is a slowly varyingfunction of y3. The potential Y2 of flow past the airfoil in the y2 direction maythen be approximated locally by the formula for an airfoil of uniform chord2a(y3).

Fig. 3.9.1.


Therefore, a first approximation to the compact Green’s function (3.9.1) for anairfoil of span L occupying the interval − 1

2 L < y3 < 12 L is obtained by taking

Y1 = y1, Y2 =

Re(− i

√z2 − a(y3)2

), |y3| < 1

2 Ly2, |y3| > 1

2 L,

Y3 = y3, z = y1 + iy2. (3.9.2)

This model has been found to give predictions within a few percent of thosebased on the exact value of Y2(y) in the case of an airfoil of elliptic planformwhose

aspect ratio = airfoil span

midchord> 5.

3.9.3 Projection or Cavity on a Plane Wall

Let the plane wall be rigid and coincide with x2 = 0 (Fig. 3.9.2). When the pro-jection or cavity is absent the Green’s function with vanishing normal derivativeon the wall is

G0(x, y, t − τ ) = 1

4π |x − y|δ(

t − τ − |x − y|c0

)

+ 1

4π |x − y|δ(

t − τ − |x − y|c0

),

Fig. 3.9.2.


where x = (x1,−x2, x3) is the image of the observer position x in the planewall.

The figure illustrates the case for a projection, but the following discussionapplies without change to compact (but nonresonant) wall cavities. Assume firstthat the origin is close to the projection. Let |x| → ∞ (noting that |x| = |x|) andexpand G0 near the projection to first order in y (i.e., correct to dipole order)

G0(x, y, t − τ )

≈ 1

4π |x|δ

(t − τ − |x|

c0+ x · y

c0|x|)

+ δ

(t − τ − |x|

c0+ x · y

c0|x|)

≈ 1

4π |x|

2δ

(t − τ − |x|

c0

)+ 2(x1 y1 + x3 y3)

c0|x| δ′(

t − τ − |x|c0

).

We require a corrected expression that has vanishing normal derivative (as afunction of y) on the wall and on the projection. By inspection, this is obtainedsimply by replacing the factor

2(x1 y1 + x3 y3)

c0|x| by2(x1Y1 + x3Y3)

c0|x| ,

where Y1 = y1 −ϕ∗1 (y), Y3 = y3 −ϕ∗

3 (y) are the velocity potentials of hori-zontal flows past the projection that are parallel to the wall and have unit speedsrespectively in the y1 and y3-directions as |y| → ∞.

It may now be verified that (in the usual notation) the required compactGreen’s function is

G(x, y, t − τ ) = 1

4π |X − Y|δ(

t − τ − |X − Y|c0

)

+ 1

4π |X − Y|δ(

t − τ − |X − Y|c0

), (3.9.3)

where

Y1 = y1 − ϕ∗1 (y), Y2 = y2, Y3 = y3 − ϕ∗

3 (y)

X1 = x1 − ϕ∗1 (x), X2 = x2, X3 = x3 − ϕ∗

3 (x)

, (3.9.4)

and X = (X1,−X2, X3).These formulae can be used also for a two-dimensional projection or cavity

that is uniform, say, in the x3 direction simply by setting Y3 = y3, X3 = x3.To complete this discussion of compact Green’s function, we now give with-

out proofs a selection of useful examples.


Fig. 3.9.3.

3.9.4 Green’s Function for a Half-Plane (Howe, 1975a)

Analytical representations of the exact Green’s function G(x, y, ω) are knownfor a rigid half-plane x1 < 0, x2 = 0 (which is infinite in the x3 direction,Fig. 3.9.3) but are of limited use in applications. However, we can define acompact Green’s function for a source at y whose distance from the edge issmall compared to the acoustic wavelength, that is, for κ0(y2

1 + y22 )

12 1. To

do this, we introduce cylindrical polar coordinates

x = (r cos θ, r sin θ, x3), y = (r0 cos θ0, r0 sin θ0, y3).

Then, if i3 is a unit vector in the x3 direction (parallel to the edge),

G(x, y, ω) = G0(x, y, ω) + G1(x, y, ω) + · · · , (3.9.5)

where, for |x − y3i3| → ∞ and κ0

√y2

1 + y22 1,

G0(x, y, ω) = −1

4π |x − y3i3|eiκ0|x−y3i3|,

(3.9.6)G1(x, y, ω) = −1

π√

2π i

√κ0ϕ

∗(x)ϕ∗(y)

|x − y3i3|3/2eiκ0|x−y3i3|,

and

ϕ∗(x) = √r sin(θ/2), ϕ∗(y) = √

r0 sin(θ0/2). (3.9.7)


ϕ∗(x) is a velocity potential of incompressible flow around the edge of the half-plane expressed in terms of polar coordinates (x1, x2) = r (cos θ, sin θ ) (andsimilarly for ϕ∗(y)). The component G0 of G represents the radiation from apoint source at y when scattering is neglected; G1 is the first correction due topresence of the half-plane.

3.9.5 Two-Dimensional Green’s Function for a Half-Plane(Howe, 1975a)

The Green’s function for the wave equation in two dimensions, where conditionsare uniform in the x3 direction, satisfies(

1

c20

∂2

∂t2− ∇2

)G = δ(x1 − y1)δ(x2 − y2)δ(t − τ ),

where G = 0 for t < τ. (3.9.8)

When a line source at y = (y1, y2) is close to the edge of the half-plane inFig. 3.9.3 the corresponding compact Green’s function is obtained by integrat-ing (3.9.6) over −∞ < y3 < ∞, using the method of stationary phase for

κ0

√x2

1 + x22 → ∞ (see Example 2 of Section 5.2), and then using the Formula

(3.1.6) to calculate G(x, y, t − τ ) ≈ G0(x, t − τ ) + G1(x, y, t − τ ) + · · ·. Inparticular G1(x, y, t − τ ) is the first term in the expansion that involves y, andis found to be

G1(x, y, t − τ ) ≈ ϕ∗(x)ϕ∗(y)

π |x| δ(t − τ − |x|/c0), |x| → ∞, (3.9.9)

where x = (x1, x2), y = (y1, y2) and ϕ∗ is defined as in (3.9.7).

3.9.6 Two-Dimensional Green’s Function for a Plane with an Aperture

A rigid plane x1 = 0 is pierced by a two-dimensional aperture occupying−a < x2 < a (Fig. 3.9.4). The two-dimensional compact Green’s function (thesolution of (3.9.8)) is applicable for a source at y = (y1, y2) well within anacoustic wavelength of the aperture on either side of the plane. The observer atx = (x1, x2) is in the acoustic far field. The y-dependent part of the compactGreen’s function is

G(x, y, t − τ ) ≈ −√

c0 sgn(x1)

π√

2π |x|χ(t − τ − |x|/c0)√

t − τ − |x|/c0Re

ln

(Z

a+√

Z2

a2− 1

),

Z = y2 + iy1, (3.9.10)


Fig. 3.9.4.

where

χ (t) = H (t)∫ ∞

0

ln(aξ 2/4c0t)e−ξ 2dξ

[ln(aξ 2/4c0t)]2 + π2,

and = 1.781072. . . . Note the definition Z = y2 + iy1.

3.9.7 Green’s Function for Long Waves in a Rigid Walled Duct(Howe, 1975b)

Only plane waves can propagate in a cylindrical duct of cross-sectional areaA when the characteristic wavelength of sound is large compared with thediameter ∼√

A, even if the source region is highly three dimensional. When thiscondition is satisfied the corresponding compact Green’s function satisfies theone-dimensional wave equation, provided the cross-sectional area is uniform.Taking the x1 direction along the axis of the duct (Fig. 3.9.5a), we have

G(x, y, t − τ ) = c0

2AH

(t − τ − |x1 − y1|

c0

), |x − y|

√A, (3.9.11)

where H is the Heaviside step function. For a uniform duct with an acoustically


Fig. 3.9.5.

compact section of variable cross section, such as the neck in Fig. 3.9.5b, thecompact Green’s function becomes

G(x, y, t − τ ) = c0

2AH

(t − τ − |X1 − Y1|

c0

), |x − y|

√A, (3.9.12)

where X1(x) and Y1(y) are the velocity potential of incompressible flow in theduct having unit speed at large distances from the neck.

3.9.8 Compact Green’s Function for a Duct Entrance (Howe, 1998a,b)

The typical geometry is illustrated in Fig. 3.9.6a. Within the duct, severaldiameters from the entrance, the cross-sectional area is uniform and equal toA. However, the geometry of the entrance can be arbitrary, and not necessarilythat of the uniform cylinder shown in the figure. Take the coordinate origin inthe entrance plane of the duct, with the negative x1 axis lying along the axis ofthe duct. Then, there are two cases:

(i) Propagation within the Duct

This is applicable for the case shown in Fig. 3.9.6a, involving a source at y nearthe duct entrance and an observer at x within the duct (or vice versa), when the


Fig. 3.9.6.

characteristic acoustic wavelength is large compared to the duct diameter.

G(x, y, t − τ ) ≈ c0

2A

H

[t − τ − |ϕ∗(x) − ϕ∗(y)|

c0

]

−H

[t − τ + ϕ∗(x) + ϕ∗(y)

c0

], (3.9.13)

where the velocity potential ϕ∗(y) describes incompressible flow from the duct,and satisfies

ϕ∗(x) ≈ x1 − ′ when |x1| √A within the duct,

≈ −A/4π |x| when |x| √A outside the duct, (3.9.14)

in which ′ is the end correction (Rayleigh, 1945) of the duct opening (≈ 0.61Rfor an unflanged circular cylinder of radius R = √A/π ).

(ii) Propagation in Free Space (Fig. 3.9.6b)

When either the source or observer is located at a large distance from the ductentrance in free space

G(x, y, t − τ ) = 1

4π |X − Y|δ(

t − τ − (|X(x) − Y(y)| − [ϕ∗(x) + ϕ∗(y)])

c0

),

(3.9.15)

where ϕ∗ is defined as in (3.9.14), and X(x),Y(y) denote the Kirchhoff vectorwhose i component is the velocity potential of flow past the stationary surfaceformed by the duct entrance having unit speed in the i direction at large dis-tances from the entrance outside the duct (they become exponentially small

Problems 3 79

with distance |x1| or |y1| into the duct). The terms ϕ∗(x), ϕ∗(y) account for theadditional, weak monopole sound generated by a source near the duct entrance;the source compresses the fluid in the duct mouth producing a sound wave in theduct whose reaction on the mouth causes a volume flux equal to the monopolesource strength. The amplitude of this monopole is of the same order as theusual dipole sound determined by the compact Green’s function.

For a uniform, thin-walled cylindrical duct we can take

X(x) ≡ (x1 − ϕ∗(x), X2(x), X3(x)), Y(y) ≡ (y1 − ϕ∗(y), Y2(y), Y3(y)).

If the source coordinate y1 → −∞ within the duct,

G(x, y, t − τ ) = 1

4π |x|δ(t − τ − (|x| − y1)/c0), |x| → ∞. (3.9.16)

This represents a monopole wave centered on the duct entrance. This limitingform of G can be used to calculate the low-frequency free space radiationgenerated by internal sources far from the entrance.

Problems 3

1. Use the compact Green’s function to solve

(∇2 + κ20

)ϕ = f2

∂

∂x2[δ(x1 − L)δ(x2)δ(x3)] ,

where

∂ϕ

∂xn= 0 on |x| = a.

for the sound radiated by an azimuthally orientated dipole adjacent to acompact, rigid sphere.

2. Use the compact Green’s function to solve

(∇2 + κ20

)ϕ = f1

∂

∂x1[δ(x1 − L)δ(x2)δ(x3)],

where

∂ϕ

∂xn= 0 on

(x2

1 + x22

) 12 = a.

for the sound radiated by a radially orientated dipole adjacent to a rigidcircular cylinder of compact cross section.


3. Repeat Question 2 for the dipoles

f2∂

∂x2[δ(x1 − L)δ(x2)δ(x3)] , f3

∂

∂x3[δ(x1 − L)δ(x2)δ(x3)].

4. The constant strength dipole

q(x, t) = f2∂

∂x2(δ(x1 − Ut)δ(x2 − h)δ(x3)), f2 = constant

translates at constant velocity U past a fixed rigid cylinder of radius a < hwhose axis coincides with the x3 axis. Show that when M = U/c0 1, thefar-field acoustic potential determined by Equation (3.1.1) is given by

ϕ ≈ f2 Ma2

2π |x|(h2 + U 2[t]2)3

x1h

|x| (3U 2[t]2 − h2) + x2U [t]

|x| (3h2 − U 2[t]2)

,

where

[t] = t − |x|c0

.

5. The volume source

q(x, t) = q0δ(x1 − Ut)δ(x2 − h)δ(x3), q0 = constant

translates at constant velocity U past a fixed rigid sphere of radius a < hwhose center is at the origin. Determine from Equation (3.1.1) the far fieldacoustic pressure p = −ρ0∂ϕ/∂t given that M = U/c0 1.

6. The point source q(x, t) = q0δ(x1−Ut)δ(x2)δ(x3), (q0 = constant) convectsalong the axis of symmetry of the necked duct shown in Fig. 3.9.5 at constant,low Mach number speed U . If the cross-sectional area of the duct is denotedby S(x1), where S(x1) → A, x1 → ±∞, use the approximations

X1 = A∫ x1

0

dξ

S(ξ ), Y1 = A

∫ y1

0

dξ

S(ξ ),

to calculate the acoustic pressure radiated from the neck during the passageof the source.

7. In incompressible flow the velocity potential generated by a distribution ofsources q(x, t) near a rigid body is determined by the solution of

∇2ϕ = q(x, t).

Problems 3 81

Show that the monopole and dipole components of the solution at largedistances from the body (in the hydrodynamic far field) can be calculatedusing the following incompressible limit of the compact Green’s function

G(x, y, t − τ ) = −δ(t − τ )

4π |X − Y| .

8. A rigid body translates without rotation in the j direction at velocity U j (t)in an ideal, incompressible fluid at rest at infinity. Show that the velocitypotential of the fluid motion is

= U jϕ∗j .

In a fixed reference frame, the pressure can be calculated from Bernoulli’sequation:

∂

∂t+ p

ρ0+ 1

2(∇)2 = 0.

Use these results to prove formula (3.8.7) for the force exerted on the fluidby the body.

9. A compact rigid disc of radius a executes small amplitude vibrations atvelocity U (t) normal to itself. In the undisturbed state it lies in the plane

x1 = 0 with its center at the origin. If ϕ∗1 (x) = ∓(2/π )

√a2 − x2

2 − x23 on the

faces x1 = ±0,√

x22 + x2

3 < a of the disc, show that the acoustic pressuregenerated by the motion is given by

p(x, t) ≈ 2ρ0a3 cos θ

3πc0|x|∂2U

∂t2(t − |x|/c0), |x| → ∞, cos θ = x1

|x| .

4Vorticity

4.1 Vorticity and the Kinetic Energy of Incompressible Flow

4.1.1 Kelvin’s (1867) Definition

Kelvin was responsible for much of the pioneering work on the mechanics ofincompressible flow. He gave the following definition of a vortex in a homo-geneous incompressible fluid,

. . . a portion of fluid having any motion that it could not acquire by fluid pressuretransmitted from its boundary.

To understand this consider the ideal (i.e., inviscid) incompressible flowproduced by arbitrary motion of a solid body with surface S (Fig. 4.1.1). Themotion generated from rest by ‘fluid pressure transmitted from its boundary’can be described by a velocity potential ϕ such that

v(x, t) = ∇ϕ,∂ϕ

∂xn= Un on S,

where Un is the normal component of velocity on S.There are no sources within the instantaneous region V occupied by the fluid,

where ∇2ϕ = 0. Therefore, the kinetic energy T0 of the flow is

T0 = 1

2ρ0

∫V

(∇ϕ)2 d3x = 1

2ρ0

∫V

(div(ϕ∇ϕ) − ϕ∇2ϕ) d3x

= −1

2ρ0

∮Sϕ

∂ϕ

∂xnd S ≡ −1

2ρ0

∮SϕUn d S, (4.1.1)

where the divergence theorem has been used to obtain the second line (thereis no contribution from the surface at infinity in Fig. 4.1.1, where ϕ ∼O(1/|x|2)). This formula implies that if S is suddenly brought to rest (Un → 0)the motion everywhere in the fluid ceases instantaneously, because

∫V (∇ϕ)2 d3x

82

4.1 Vorticity and the Kinetic Energy of Incompressible Flow 83

Fig. 4.1.1.

can vanish only if ∇ϕ ≡ 0. This unphysical behavior is never observed in areal fluid because (i) no fluid is perfectly incompressible, and signals generatedby changes in the boundary conditions propagate at the finite speed of sound,and (ii) diffusion of vorticity from the boundary supplies irrecoverable kineticenergy to the fluid.

For an incompressible, real fluid we write

v = ∇ϕ + u,

and define the vorticity ω by

ω = curl u ≡ curl v.

If ϕ is taken to be defined as above (for ideal flow) then, because div u = 0 andthe normal component un = n · u = 0 on S, the kinetic energy becomes

T = 1

2ρ0

∫V

(∇ϕ + u)2 d3x = 1

2ρ0

∫V

((∇ϕ)2 + 2∇ϕ · u + u2) d3x

= 1

2ρ0

∫V

((∇ϕ)2 + u2) d3x + ρ0

∫V

div(ϕu) d3x

= 1

2ρ0

∫V

((∇ϕ)2 + u2) d3x − ρ0

∮Sϕun d S

= −1

2ρ0

∮SϕUn d S + 1

2ρ0

∫V

u2 d3x ≡ T0 + 1

2ρ0

∫V

u2 d3x. (4.1.2)

84 4 Vorticity

When the surface motion is arrested, the flow described by the velocity poten-tial ϕ stops instantaneously, but that associated with the rotational velocity upersists. The crucial difference between rotational and irrotational flows is that,once established, vortical motions proceed irrespective of whether or not thefluid continues to be driven by moving boundaries or other external agencies.

Equation (4.1.2) also establishes Kelvin’s theorem that T ≥ T0: the kineticenergy of the real flow (for which u = 0) always exceeds that of the correspond-ing ideal, irrotational flow. In other words, the irrotational motion representsthe least possible disturbance that can be produced in the fluid by the movingboundary.

4.2 The Vorticity Equation

Let vA denote the fluid velocity at a point A at x. The velocity vB at a neigh-bouring point B at x + δx can then be written (Goldstein, 1960)

vB ≈ vA + (δx · ∇)v

= vA + 12ω ∧ δx + 1

2∇(ei jδxiδx j ),

where ei j is the rate-of-strain tensor (2.1.3) and the gradient in the second lineis taken with respect to δx. The first two terms on the second line representmotion of A and B as a rigid body, consisting of a translation at velocity vA

together with a rotation at angular velocity 12ω; the last term, being a gradient,

represents an irrotational distortion of the fluid in the neighborhood of A.If we consider a small spherical fluid particle with center at A, the distortion

corresponds to a deformation into an ellipsoid whose principal axes correspondto the principal axes of ei j . If a spherical fluid particle is suddenly solidifiedwithout change of angular momentum, it will rotate at angular velocity ω/2, sothat ω may be defined as twice the initial angular velocity of the solid spherewhen an infinitesimally small sphere of fluid with center at A is suddenly solid-ified without change of angular momentum (but this is not true for arbitrarilyshaped volume elements). The vorticity may therefore be regarded as a measureof the angular momentum of a fluid particle. This is consistent with our conclu-sion above regarding kinetic energy, inasmuch as the conservation of angularmomentum suggests that vorticity is associated the intrinsic kinetic energy ofthe flow, and determines the motion that persists in an incompressible fluidwhen the boundaries are brought to rest.

A vortex line is tangential to the vorticity vector at all points along its length.Vortex lines that pass through every point of a simple closed curve define theboundary of a vortex tube. For a tube of small cross-sectional area δS the product

4.2 The Vorticity Equation 85

ω δS is called the tube strength, and is constant because

divω = div(curl v) ≡ 0,

and the divergence theorem therefore implies that∮ω · dS = 0 for any closed

surface, and in particular for the surface formed by two cross sections of thetube and the tube wall separating them, on the latter of which ω · dS = 0. Itfollows that vortex tubes and lines cannot begin or end within the fluid. Theno-slip condition (Batchelor 1967) requires the velocity at a boundary to be thesame as that of the boundary. A vortex line must therefore form a closed loop,or end on a rotating surface S at which

n ·ω = 2n ·Ω, (4.2.1)

where Ω is the angular velocity of S.We shall show that vorticity is transported by convection and molecular

diffusion. Therefore an initially confined region of vortex loops can frequentlybe assumed to remain within a bounded region. In the absence of body forces F,we first use the identity curl curl A = grad div A−∇2A to write the momentumequation (1.2.3) for homentropic flow in the form

∂v∂t

+ (v · ∇)v + ∇(∫

dp

ρ

)= −ν

(curlω − 4

3∇(div v)

).

By using the vector identity

(v · ∇)v = ω ∧ v + ∇ (12v

2), (4.2.2)

the momentum equation can be cast into Crocco’s form

∂v∂t

+ ω ∧ v + ∇ B = −ν

(curlω − 4

3∇(div v)

), (4.2.3)

where

B =∫

dp

ρ+ 1

2v2 (4.2.4)

is the total enthalpy in homentropic flow. The vector ω ∧ v is sometimescalled the Lamb vector. When the fluid is incompressible (or when the termin div v representing the small effect of compressibility on viscous dissipationis neglected) Crocco’s equation reduces to

∂v∂t

+ ω ∧ v + ∇ B = −ν curlω, (4.2.5)

in which case dissipation occurs only where ω = 0.

86 4 Vorticity

The curl of (4.2.3) and the relation curl curlω ≡ −∇2ω yield the vorticityequation for a Stokesian fluid of constant shear viscosity:

∂ω

∂t+ curl(ω ∧ v) = ν∇2ω. (4.2.6)

For an incompressible fluid divω = div v = 0, and

curl(ω∧ v) ≡ (v · ∇)ω + ω div v − (ω · ∇)v − v divω = (v · ∇)ω − (ω · ∇)v,

therefore the vorticity equation can also be written

Dω

Dt= (ω · ∇)v + ν∇2ω. (4.2.7)

The terms on the right represent the mechanisms that change the vorticity of amoving fluid particle in incompressible flow:

(i) (ω · ∇)v.

Consider a fluid particle at A in Fig. 4.2.1 with velocity v at time t . Let thevorticity at A be ω = ωn, where n is a unit vector, and consider a neighboringparticle at B a small distance s from A in the direction n; that is, sn is the positionof B relative to A. At time t the points A and B lie on the vortex line throughA, and the velocity at B is v + s(n · ∇)v. After a short time δt , A has moved avector distance v δt to A′ and B has moved to B ′ whose position relative to A′

is s(n+(n · ∇)v δt). During this time, the term (ω · ∇)v in the vorticity equationcauses the vorticity of the fluid particle initially at A to change from ωn at A toω(n + (n · ∇)v δt) at A′. Thus, the vortex line through A′ lies along the relativevector s(n + (n · ∇)v δt) from A′ to B ′. Therefore, the fluid particles and thevortex line through A and B have deformed and convected in the flow in the

Fig. 4.2.1.

4.2 The Vorticity Equation 87

same way; in their new positions A and B continue to lie on the same vortexline. In the absence of viscosity (when ν∇2ω does not appear on the right of(4.2.7)) vortex lines are therefore said to move with the fluid; they are rotatedand stretched in a manner determined entirely by the relative motions of A andB. The magnitude of ω increases in direct proportion to the stretching of vortexlines. When a vortex tube is stretched, the cross-sectional area δS decreases andtherefore ω must increase to preserve the strength of the tube.

(ii) ν∇2ω: Molecular Diffusion of Vorticity

This term is important only in regions of high shear, in particular near solidboundaries. Very close to a stationary wall the velocity becomes small and non-linear terms in the vorticity equation (4.2.7) can be neglected. The equation thenreduces to the classical diffusion equation

∂ω

∂t= ν∇2ω.

Vorticity is generated at solid boundaries, and viscosity is responsible for itsdiffusion into the body of the fluid, where it can subsequently be convected bythe flow.

It should be understood that viscosity merely serves to diffuse the vorticityinto the fluid from the surface, and does not generate the vorticity. In an idealfluid the slipping of the flow over the surface creates a singular layer of vorticityat the surface called a vortex sheet whose strength is determined by the tan-gential velocity difference between the surface and the ideal exterior flow. Thisvorticity stays on the surface; it would start to diffuse into the fluid if the fluidwere suddenly endowed with viscosity. The rate of diffusion would depend onthe value of ν, but the amount of the vorticity available for diffusion from thesurface is independent of ν.

The circulation with respect to a closed material contour C is defined by

=∮

Cv · dx =

∫S′

curl v · dS ≡∫

S′ω · dS,

where S′ is any two-sided surface bounded by C . When ν = 0 the motion inhomentropic flow evolves in such a way that the circulation around the movingcontour remains constant:

D

Dt= D

Dt

∮C

v · dx =∮

C∇(

−∫

dp

ρ+ 1

2v2

)· dx ≡ 0. (4.2.8)

This is Kelvin’s circulation theorem. It follows that vorticity can neither becreated nor destroyed in a body of inviscid and homentropic fluid.

88 4 Vorticity

Fig. 4.2.2.

4.2.1 Vortex Sheets

A vortex sheet is a useful model of a thin layer of vorticity when viscous diffu-sion can be neglected. Imagine a thin shear layer (Fig. 4.2.2) across which thevelocity changes rapidly from v− to v+. We approximate the layer by a surfacef (x, t) = 0 with unit normal n across which the normal components of veloc-ity are equal (n · v− = n · v+), but the tangential components are discontinuous(n ∧ v− = n ∧ v+). Let f >< 0 respectively on the ± sides of the surface. Nearthe sheet, on either side, it can be assumed that curl v± = 0, and we can set

v = H ( f )v+ + H (− f )v−

Hence,

ω = ∇ H ∧ (v+ − v−) = n ∧ (v+ − v−)δ(s⊥), (4.2.9)

where ∇ H ≡ ∇ H ( f ) = −∇ H (− f ) = nδ(s⊥), and s⊥ is distance measured inthe normal direction from the sheet.

In a real fluid the vorticity would diffuse out from the sheet and it couldnot therefore persist indefinitely. In an ideal fluid the sheet is subject only toconvection and stretching by the flow at the local mean velocity, which is

v = 12 (v+ + v−), (4.2.10)

where v± are evaluated just above and below the sheet.

4.3 The Biot–Savart Law

In an unbounded fluid the velocity v can always be expressed in terms of scalarand vector potentials ϕ and A such that

v = ∇ϕ + curl A, where div A = 0.

The equations determining ϕ and A are found by taking in turn the divergence

4.3 The Biot–Savart Law 89

and curl (using the formula curl curl A = grad div A − ∇2A):

∇2ϕ = div v, ∇2A = −curl v ≡ −ω.

We can take ϕ = 0 for incompressible, unbounded flow which is at rest at in-finity. To find A we use the Green’s function for Laplace’s equation determinedby (1.4.6) (i.e., by the incompressible limit of (3.2.6), when κ0 = 0) to obtain

A =∫

ω(y, t) d3y4π |x − y| .

The velocity is then given by the Biot–Savart formula

v(x, t) = curl∫

ω(y, t) d3y4π |x − y| . (4.3.1)

This is a purely kinematic relation between a vector v that vanishes at infinityand ω = curl v.

Because vorticity is transported by convection and diffusion, an initiallyconfined region of vorticity will tend to remain within a bounded domain, sothat it may be assumed that ω → 0 as |x| → ∞. The divergence theorem thenshows that ∫

ωi (y, t) d3y = −∮

yiω j (y, t)n j d S(y) ≡ 0,

where the surface (with inward normal n) is large enough to contain all thevorticity. By using this result and the expansion (1.9.3) for |x| → ∞ we derivefrom (4.3.1) the following approximation in the hydrodynamic far field:

v(x, t) ≈ curl

(x j

4π |x|3∫

y jω(y, t) d3y)

∼ O

(1

|x|3), |x| → ∞. (4.3.2)

Furthermore, the divergence theorem also implies that∫div(yi y jω(y, t)) d3y = 0,

and therefore that∫

(yiω j (y, t) + y jωi (y, t)) d3y = 0.

This can be used to express (4.3.2) in either of the following equivalent forms

v(x, t) ≈ curl curl

(I

4π |x|)

= grad div

(I

4π |x|), |x| → ∞,

where I = 1

2

∫y ∧ ω(y, t) d3y, (4.3.3)

90 4 Vorticity

(see Question 2 of Problems 4). The vector I is called the impulse of the vortexsystem, and is an absolute constant in an unbounded flow (see Section 4.4).These formulae supply alternative representations of v in the hydrodynamic farfield (where the motion is entirely irrotational) in terms of either the vector po-tential A = curl (I/4π |x|) or the scalar potential ϕ = div (I/4π |x|). (Batchelor(1967) denotes I by P/ρ0; Lighthill (1978, 1986) uses G.)

4.3.1 Kinetic Energy

Using the Biot–Savart formula it can be verified that the kinetic energy of anunbounded (three-dimensional) incompressible flow is given in terms of thevorticity by

T = ρ0

8π

∫∫ω(x, t) ·ω(y, t)

|x − y| d3x d3y. (4.3.4)

The following representation can also be derived (using (4.2.2))

T = ρ0

∫x · (ω ∧ v)(x, t) d3x. (4.3.5)

4.3.2 Incompressible Flow with an Internal Boundary

Let the rigid body in Fig. 4.1.1 have volume and move in an incompressiblefluid at rest at infinity with velocity

U = U0 + Ω ∧ (x − x0(t)), (4.3.6)

where U0 = dx0/dt is the velocity of its center of volume x0(t), and Ω(t) is itsangular velocity.

In the usual way let f (x, t) vanish on S, with f > 0 in the fluid. Then H ( f )v +H (− f )U is the velocity everywhere, in both the fluid and solid (where it equalsU(x, t)). But

H ( f )v + H (− f )U = ∇ϕ + curl A, (div A = 0).

The body has constant volume (div U = 0), but curl U = 2Ω. Now, the no-slipcondition on S implies that

div(H ( f )v + H (− f )U) = ∇ H ( f ) · (v − U) ≡ 0

curl(H ( f )v + H (− f )U) = H ( f )ω + H (− f )2Ω.

4.3 The Biot–Savart Law 91

Hence, ϕ ≡ 0, and the velocity everywhere is given by the following modifica-tion of the Biot–Savart formula (4.3.1):

v(x, t) = curl∫

V

ω(y, t) d3y4π |x − y| + curl

∫

2Ω(t) d3y4π |x − y| , (4.3.7)

where V is the volume occupied by the fluid. This formula predicts that v =U when x lies in the region occupied by the body. Vortex lines may beimagined to continue into the solid. As for an unbounded flow, the identity∫

curl (H ( f )v + H (− f )U) d3x = 0 implies that v ∼ O(1/|x|3) as |x| → ∞.Similarly, the asymptotic representations (4.3.3) remain valid provided the inte-gration includes the region occupied by the body (whereω = 2Ω). The formulais also applicable in inviscid flow, but the contribution from the bound vorticityin the vortex sheet on the surface of the body must be included in the integrals.

4.3.3 Blowing Out a Candle (Lighthill 1963)

An amusing illustration of the significance of vorticity is depicted in Fig. 4.3.1,where a puff of air is ejected from the tube and directed at the flame of a candle.Suppose the tube has radius R, and that the air is forced out at constant speedV by impulsive movement of the piston over an axial distance L . In an idealfluid the motion outside is irrotational and resembles at large distances fromthe exit a radially symmetric source flow. This flow persists only while the pistonis moving, during which time the velocity potential at a large distance r from the

Fig. 4.3.1.

92 4 Vorticity

exit resembles that produced by a monopole of strength q = π R2V :

ϕ ∼ − V R2

4r,

so that the air blows against the flame at distance at speed

Vϕ ∼ V

4

R2

2.

In reality, vorticity is generated at the tube wall. The air leaves the tube in theform of a jet, and the exiting fluid is initially contained within a cylindrical slugof air of length L , whose displacement from the tube forces the potential flowϕ. In addition, however, vorticity leaves the tube within a circular cylindricalvortex sheet at the periphery of the slug, across which the axial velocity changesfrom V within the jet to 0 outside. The sheet may be pictured as a successionof vortex rings of radius R and infinitesimal core radii. The circulation of theserings per unit length of the jet is V , and they translate at the local mean airvelocity on the sheet equal to 1

2 V . The total circulation ejected from the tubeduring the time L/V in which the piston moves is therefore = 1

2 LV .Shortly after leaving the tube the cylindrical vortex rolls up to form a vortex

ring of circulation which translates by self-induction (as determined by theBiot–Savart law (4.3.1)) at speed estimated by Kelvin to equal

Vt ∼

4π R0

[ln

(8R0

σ

)− 1

4

]≈ V L

8π R0

[ln

(8R0

σ

)− 1

4

],

where R0 ∼ 1.2R is the radius of the vortex ring and σ ∼ 0.2R0 is the radius ofits core (assumed to be of circular cross section). The ring arrives at the flameafter a time t ∼ /Vt.

The air on the axis of the vortex ring at its center forms a localized jet withvelocity on the centerline equal to

Vj ∼ V L

4R0.

If the flame is extinguished it is because the vortex jet blows away the hotcombusting gases from newly vaporized wax.

According to this sequence of events, the candle is only blown out because ofthe presence of the vortex. In its absence the flame would barely flicker under theinfluence of blowing by the potential velocity field Vϕ . The following numeri-cal estimates confirm this conclusion. Take V = 10 m/s, L = 1 cm, R = 0.5 cm,and= 0.3 m. Then, R0 ≈ 0.6 cm, σ ≈ 0.12 cm, Vϕ ≈ 0.0007 m/s, Vt ≈ 2.4 m/s,Vj ≈ 4.2 m/s, and t ≈ 0.13 s.

4.4 Surface Force in Incompressible Flow 93

4.4 Surface Force in Incompressible Flow Expressedin Terms of Vorticity

We now derive the following formula for the force F exerted on an incompress-ible fluid by a rigid body with surface S whose centre of volume has velocityU0:

F + m0dU0

dt= ρ0

dIdt

≡ ρ0

2

d

dt

∫x ∧ ω(x, t) d3x, (4.4.1)

where I is the impulse defined by the integral in (4.3.3), including any contri-butions from bound vorticity within S, and m0 is the mass of fluid displaced bythe body.

With reference to Fig. 4.1.1, let

F = external force applied to the body to maintain its motion,m = mass of the body.

The fluid is assumed to be at rest at infinity. Let V denote the fluid between alarge closed surface containing all of the vorticity and the surface S of thebody, and let V+ denote the interior of including the volume of the body.The center of volume of the body is assumed to be in motion at velocity U0(t),but in general the body may also be rotating at some time dependent angularvelocity Ω. The global equations of motion are

mdUdt

+ ρ0d

dt

∫V

v(x, t) d3x = F +∮

p(x, t) dS

mdUdt

= F − F

where U is the velocity of the centre of mass of the body. Subtracting theseequations and extending the volume integral to include the volume occupiedby the rigid body (where

∫

v d3x = U0), we can also write

F + m0dU0

dt= ρ0

d

dt

∫V+

v(x, t) d3x −∮

p(x, t) dS. (4.4.2)

Now let v(x, t) = curl A, where the vector potential A(x, t) is defined by theBiot–Savart integral (4.3.1) taken over V+, which includes the region occupiedby the body. Then, (4.3.3) implies that A = curl(I(t)/4π |x|) on , so that

∫V+

v(x, t) d3x =∫

V+curl A d3x = −

∫

n ∧ A d S → −∫

n ∧ curl

(I(t)

4π |x|)

d S.

94 4 Vorticity

Similarly, p → −ρ0∂ϕ/∂t on, whereϕ = div(I(t)/4π |x|) (see (4.3.3)). Hence,

−∮

p(x, t) dS → ρ0d

dt

∮

ϕn d S = ρ0d

dt

∮

n div

(I(t)

4π |x|)

d S.

The right-hand side of (4.4.2) can therefore be written

ρ0d

dt

∮

−n ∧ curl

(I(t)

4π |x|)

+ n div

(I(t)

4π |x|)

d S.

By the divergence theorem, the integral in this expression over the large, butarbitrary surface can be replaced by an integration over the surface of a largesphere |x| = R, because

curl curl − ∇div(I(t)/4π |x|) = −∇2(I(t)/4π |x|) ≡ 0 for |x|> 0.

On the sphere n = −x/|x|, and the integrand equals I(t)/4π R2; the integral istherefore just equal to I(t). Thus, (4.4.2) reduces to the desired representation(4.4.1).

4.4.1 Bound Vorticity and the Added Mass

Consider the particular case of a rigid body accelerating without rotation atvelocity U(t) in an otherwise unbounded, ideal incompressible fluid in theabsence of vorticity (Fig. 4.4.1). We have seen previously (Section 3.8) thatforce exerted on the fluid can be written

Fi = Mi jdU j

dt,

Fig. 4.4.1.


where Mi j is the added mass tensor (3.8.6). This result will now be derivedfrom the integral formula (4.4.1).

There is no vorticity in the fluid, but the slipping of the ideal flow over Sgenerates a singular distribution (a vortex sheet) of bound vorticity that mustbe used to evaluate the integral. To calculate the bound vorticity we need anexpression for the velocity everywhere in space, including the region occupiedby the body, where v ≡ U.

In the fluid, we can take

v = U j∇ϕ∗j .

Therefore, by introducing a control surface f (x, t) = 0 that coincides with thesurface S of the body, with f > 0 in the fluid and f < 0 within S, the requiredformula for the velocity is

v = U j H ( f )∇ϕ∗j + U j H (− f )∇x j ,

and the vorticity is

ω = curlU j H ( f )∇ϕ∗j + U j H (− f )∇x j

= −U j∇ H ∧ ∇(x j − ϕ∗j ) (where ∇ H = ∇ H ( f ) = −∇ H (− f ))

≡ curl U j (x j − ϕ∗j )∇ H.

Equation (4.4.1) accordingly gives the force in the form

F = −m0dUdt

+ ρ0

2

d

dt

∫x ∧ curlU j (x j − ϕ∗

j )∇ H d3x.

Now the vector A = U j (x j −ϕ∗j )∇ H vanishes except on the surface S of the

body. The identities

x ∧ curl A = 2A + ∇(x · A) − ∂

∂x j(x j A),

∫(·)∇ H d3x =

∮S(·) dS

(4.4.3)

therefore imply that

Fi = −m0dUi

dt+ ρ0

d

dt

∫U j (x j − ϕ∗

j )∂H

∂xid3x

= −m0dUi

dt+ ρ0

d

dt

∮S

U j (x j − ϕ∗j )ni d S.

But

ρ0

∮S

x j ni d S = ρ0 δi j ≡ m0δi j ,

96 4 Vorticity

where = volume contained by S, and

ρ0

∮S

−ϕ∗j ni d S = Mi j ,

where Mi j is the added mass tensor (3.8.8). Therefore, force on fluid in irrota-tional flow ≡ Fi = Mi j

dU j

dt .

4.4.2 Force Exerted on an Incompressible Fluid by a Moving Body

The integral in (4.4.1) defining the value of dI/dt can be transformed to removethe strong dependence of the integrand on the bound vorticity on S. This vorticityis produced both by motion of S and by relative motion between S and the fluidinduced by free vorticity in the flow. Thus, any attempt to recast dI/dt mustbe strongly influenced by both the shape and motion of S. We consider onlythe important special case of a body in translational motion without rotation atvelocity U(t), and show that the i th component of the force F exerted on thefluid can also be written

Fi = Mi jdU j

dt− ρ0

∫V

∇ Xi ·ω∧ vrel d3x − η

∮S∇ Xi ·ω∧ dS, vrel = v − U,

(4.4.4)

where vrel is the fluid velocity relative to the translational velocity of S, Xi =xi −ϕ∗

i (x, t) is the Kirchhoff vector already encountered in the definition of thecompact Green’s function (Section 3.4), and Mi j = M ji = −ρ0

∮S n jϕ

∗i d S is

the added mass tensor. Xi represents the velocity potential of an ideal flow pastS that has unit speed in the i direction at large distances from S (it depends ont because a fixed coordinate system is being used).

The first term on the right of (4.4.4) represents the inviscid component of theforce, associated with the added mass. The contribution from free vorticity isfurnished by the volume integral; the final term arises from frictional effects onS, which are relatively small at large Reynolds numbers. Now vrel = 0 on S,and therefore the contribution to the volume integral from vorticity close to andon S is negligible; indeed, even in the inviscid limit there is no contribution tothe integral from the surface vortex sheet forming the bound vorticity, because∇ Xi and the relative Lamb vector ω ∧ vrel are orthogonal on S.

To derive this formula from (4.4.1) we introduce the usual control surfacef (x, t) = 0 enclosing S, with f > 0 in the outer fluid region, multiply Crocco’shomentropic momentum equation (4.2.5) by H ≡ H ( f ), and take the curl ofthe resulting equation. Using the formula

DH

Dt≡ ∂H

∂t+ v · ∇ H = 0,


and the no-slip condition on S, we find

∂

∂t(Hω) = − ∂

∂t(∇ H ∧ U) − curl((∇ H · U)U) − ∇ H ∧ ∇ B

− curl(Hω ∧ v) − ν curl(H curlω).

Then, because ω = 0 within S,

d

dt

∫x ∧ ω(x, t) d3x = d

dt

∫x ∧ (Hω) d3x =

∫x ∧ ∂

∂t(Hω) d3x

= −∫

x ∧ ∂

∂t(∇ H ∧ U) d3x −

∫x ∧ curl ((∇ H · U)U) d3x

−∫

x ∧ (∇ H ∧ ∇ B) d3x −∫

x ∧ curl (Hω ∧ v) d3x

− ν

∫x ∧ curl (H curlω) d3x

= 2dUdt

+ 0 + 2∮

SB dS − 2

∫Vω ∧ v d3x − 2ν

∮Sω ∧ dS,

where the last line follows by use of the identities (4.4.3). Thus, adopting suffixnotation,

d Ii

dt≡ 1

2

d

dt

∫(x ∧ ω)i d3x =

dUi

dt+∮

SBni d S −

∫V

∇xi · (ω ∧ v) d3x

− ν

∮S∇xi ·ω∧ dS. (4.4.5)

The surface integral∮

S Bni d S can be eliminated by recalling that

∇2ϕ∗i = 0,

∂ϕ∗i

∂xn≡ n j

∂ϕ∗i

∂x j= ni on S.

Then, because ∇ϕ∗i ∼ O(1/|x|3) as |x| → ∞, the divergence theorem shows

that∮

S Bni d S = − ∫V div(∇ϕ∗

i B) d3x ≡ − ∫V ∇ϕ∗

i · ∇ B d3x. Hence, usingCrocco’s equation (4.2.5)

∮S

Bni d S =∫

Vdiv

(ϕ∗

i

∂v∂t

)d3x +

∫V

∇ϕ∗i ·ω ∧ v d3x

− ν

∫V

div(∇ϕ∗i ∧ ω) d3x.

The first and last integrals on the right are transformed further by the divergence

98 4 Vorticity

theorem, for the first ∫V

div

(ϕ∗

i

∂v∂t

)d3x = Mi j

ρ0

dU j

dt,

where Mi j = M ji = −ρ0∮

S n jϕ∗i d S is the added mass coefficient of (3.8.6).

For the last

−ν

∫V

div(∇ϕ∗i ∧ ω) d3x = ν

∮S∇ϕ∗

i ·ω ∧ dS.

Thus, substituting for∮

S Bni d S in (4.4.5), we find

d Ii

dt=

dUi

dt+ Mi j

ρ0

dU j

dt−∫

V∇ Xi ·ω∧v d3x− ν

∮S∇ Xi ·ω∧dS. (4.4.6)

But the identity

∇ Xi ·ω ∧ U = div(U(v · ∇ Xi ) − v(U · ∇ Xi ) − (v · U)∇ Xi )

implies that∫

V ∇ Xi ·ω ∧ U d3x = 0, so that (4.4.6) can also be written

d Ii

dt=

dUi

dt+ Mi j

ρ0

dU j

dt−∫

V∇ Xi ·ω∧vrel d3x−ν

∮S∇ Xi ·ω∧dS (4.4.7)

where vrel = v − U.Equation (4.4.4) is now obtained by substituting from (4.4.7) into (4.4.1)

(recalling that m0 = ρ0).

4.4.3 Stokes Drag on a Sphere

The first term on the right-hand side of (4.4.4) is the force necessary to acceleratethe added mass of the body. The i th component of the viscous skin friction is−η

∮S(ω ∧ dS)i ≡ −η

∮S ∇xi ·ω ∧ dS. Thus, (because Xi = xi − ϕ∗

i ) the netcontribution of the normal pressure forces on S is represented in (4.4.4) by theterms

−ρ0

∫V

∇ Xi ·ω ∧ vrel d3x + η

∮S∇ϕ∗

i ·ω ∧ dS.

The second, viscous component is comparable in magnitude to the skin friction,and is produced by the pressure field established by the surface shear stress.

The necessity for such a term is vividly illustrated by the Stokes drag on asphere. Let the sphere have radius a and translate at constant velocity U = (U, 0,0),U > 0, along the x1 axis. At very small Reynolds numbers Re = aU/ν 1


the inertial terms ω ∧ v and ∇( 12v

2) can be discarded from Crocco’s equation(4.2.5), which (for incompressible flow) reduces to the creeping flow equation

∇ p = −η curlω (4.4.8)

in a reference frame moving with the sphere. Both the pressure and the vorticitytherefore satisfy Laplace’s equation ∇2 p = 0,∇2ω = 0. By symmetry p mustvary linearly with η and U · x, and the condition that p should vanish at largedistances from the sphere supplies the dipole solution

p = CηU · x|x|3 ≡ −Cη div

(U|x|), |x|> a,

where C is a constant.Similarly, ω must be a linear function of U ∧ x: the identity curl curl(U/

|x|) = grad div(U/|x|) and Equation (4.4.8) imply that ω = C curl (U/|x|) ≡C(U ∧ x)/|x|3 (|x|> a). The value of C is most easily found by substitutingthis expression for ω into the Biot–Savart formula (4.3.1) and evaluating theright-hand side at the centre x = 0 of the sphere, where v ≡ U. This yieldsC = 3a/2, and therefore

ω = curl

(3aU2|x|

), |x|> a. (4.4.9)

The net force F1 on the fluid is in the x1 direction, and is given by the finalintegral on the right of (4.4.4). It is equal in magnitude to Ds + Dp, whereDs and Dp are the respective components of the Stokes drag on the sphereproduced by the skin friction and the viscous surface pressure. For the sphereϕ∗

1 = −a3x1/2|x|3, and we readily calculate F1 = 6πηUa, and

Ds = η

∮S(ω ∧ dS)1 = −4πηUa, Dp = −η

∮S∇ϕ∗

1 ·ω∧ dS = −2πηUa.

The pressure drag is therefore equal to half the skin-friction drag.This interpretation of Dp as the component of drag attributable to the normal

pressure forces on the sphere can be confirmed directly using the creeping flowapproximation (4.4.8). Because n1 = n · ∇ϕ∗

1 on S we find, using the divergencetheorem,

Dp = −∮

Spn1 d S ≡ −

∮S

p∇ϕ∗1 · dS =

∫V

∇ p · ∇ϕ∗1 d3x

= −η

∫V

curlω · ∇ϕ∗1 d3x = −η

∮S∇ϕ∗

1 ·ω ∧ dS,

100 4 Vorticity

where (4.4.8) and the identity div(A ∧ B) = curl A · B − A · curl B have beenused on the second line.

4.5 The Complex Potential

The remainder of this chapter is devoted to a brief outline of the complexpotential representation of two-dimensional, incompressible flows and its ap-plication to determine the equation of motion of a line vortex in such flows. Theresults will be applied in later chapters to investigate simple models of soundproduction by vortices interacting with surfaces.

4.5.1 Laplace’s Equation in Two Dimensions

Suppose that

w(z) = ϕ(x, y) + iψ(x, y), z = x + iy

is regular (analytic) in a region D of the z-plane. The real and imaginary partsϕ(x, y) and ψ(x, y) are solutions of Laplace’s equation.

∂2ϕ

∂x2+ ∂2ϕ

∂y2= 0,

∂2ψ

∂x2+ ∂2ψ

∂y2= 0 in D.

Let f (z) be regular in D, and define a (conformal) transformation Z = f (z)of D into a region D′ in the plane of Z = X + iY . Let W(Z ) be regular in D′

with real and imaginary parts (X, Y ), (X, Y ). Then,

∂2

∂X2+ ∂2

∂Y 2= 0,

∂2

∂X2+ ∂2

∂Y 2= 0 in D′.

The transformation Z = f (z) permits us to define a corresponding functionw(z) ≡ ϕ(x, y) + iψ(x, y) = W( f (z)), which is regular in D, with derivativew′(z) = f ′(z)W ′( f (z)). For corresponding points in D and D′ we have

ϕ(x, y) = (X (x, y), Y (x, y)), ψ(x, y) = (X (x, y), Y (x, y)).

In other words, the solutions and of Laplace’s equation in D′ are alsosolutions of Laplace’s equation in D.

These results have the following significance. The solution of Laplace’s equa-tion within a given two-dimensional bounded region D is equivalent to the so-lution of Laplace’s equation within the transformed regionD′. If it is possible tosolve the latter problem, the solution to the original problem in D can be found

4.5 The Complex Potential 101

by transforming back to the z plane. Difficulties may arise at isolated pointswhere f ′(z) = 0 and at points where f (z) ceases to be regular, but these can usu-ally be dealt with by careful examination of the behavior of the transformationnear such points.

4.5.2 Hydrodynamics in Two Dimensions

Irrotational motion of an ideal, incompressible fluid in planes parallel to the xyplane can be investigated by introducing the complex potentialw(z) =ϕ(x, y) +iψ(x, y). The velocity

v = ∇ϕ = (∂ϕ/∂x, ∂ϕ/∂y).

The function ψ is called the stream function. For steady motion the velocity at(x, y) does not change with time, and the fluid particles travel along a fixed sys-tem of streamlines each of which is a member of the family of curves ψ(x, y) =constant.

Both ϕ(x, y) and ψ(x, y) are solutions of Laplace’s equation and satisfy theCauchy–Riemann equations:

∂ϕ

∂x= ∂ψ

∂y,

∂ϕ

∂y= −∂ψ

∂x,

which imply that ∇ϕ · ∇ψ = 0, i.e., that the streamlines intersect the equipo-tentials ϕ = constant at right angles. If v = (u, v), then the complex velocity

w′(z) = ∂ϕ

∂x− i

∂ϕ

∂y≡ u − iv

is also regular.The fact that w(z) is a regular function of z can greatly simplify the solution

of many problems. This will be illustrated by consideration of two methodsbased on the theory of complex variables.

Method 1 The real and imaginary parts of every regular function w(z) deter-mine the velocity potential and stream function of a possible flow. A catalogof flows can therefore be constructed by studying the properties of arbitrarilyselected w(z).

Example 1 w = U z,U = real constant:

ϕ = U x, ψ = U y. Thus, v = (U, 0).

102 4 Vorticity

The motion is uniform at speed U along streamlines parallel to the x direction.

Example 2

w = U

(z + a2

z

), U = real constant, a > 0, |z|> a. (4.5.1)

At large distances from the origin w → U z, and the motion becomes uniformat speed U parallel to the x axis. In terms of the polar form z = reiθ ,

w = U

(reiθ + a2

re−iθ

). Thus, ϕ = U cos θ

(r + a2

r

).

The radial component of velocity

∂ϕ

∂r= U cos θ

(1 − a2

r2

)

vanishes at r = a. The motion therefore represents steady flow in the x direc-tion past a rigid cylinder of radius a with centre at the origin (Fig. 4.5.1; c.f.,Section 3.6).

Example 3

w = −iU

(z − a2

z

), U = real constant, |z|> a > 0,

(ϕ = U sin θ

(r + a2

r

)), (4.5.2)

describes potential flow in the y direction past a rigid cylinder of radius a withcenter at the origin.

Fig. 4.5.1.


Example 4 The function

w = 1

2πln z,

(ϕ = 1

2πln r, ψ = θ

2π, z = reiθ

),

is regular except at z = 0. The flow is radially outward from the origin alongstreamlines θ = constant, at speed ∂ϕ/∂r = 1/2πr . The origin is a singularityof the flow where fluid is created at a rate equal to

∮C ∇ϕ · n ds, where C is any

simple closed curve enclosing the origin with outward normal n, and ds is theelement of arc length on C . In particular, taking C to be a circle of radius r ,∮

C∇ϕ · n ds =

∫ 2π

0

∂ϕ

∂rr dθ = 1.

The origin is therefore a simple source of unit strength. When the source issituated at z0 = x0 + iy0

w = 1

2πln(z−z0),

(ϕ = 1

2πln |z − z0| = 1

2πln√

(x − x0)2 + (y − y0)2

).


w = −i

2πln z,

(ϕ = θ

2π,ψ = −

2πln r, z = reiθ

),

is regular except at z = 0, and describes the irrotational flow outside a linevortex of strength concentrated at z = 0. The streamlines are circles centeredat z = 0, and the flow speed is ∂ϕ/r∂θ =/2πr in the anticlockwise direction(for > 0). The circulation

∫C v · dx =, where C is any contour encircling

the vortex once, and the contour is traversed in the positive direction (with theinterior on the left). When the vortex is at z0 = x0 + iy0

w = −i

2πln(z − z0).


w = 1

2π(ln(z − z0) + ln(z − z∗

0)),

(ϕ = 1

2π(ln r1 + ln r2)

),

represents the flow produced by two unit point sources located at z0 = x0 + iy0

and z∗0 = x0 − iy0 (Fig. 4.5.2). The motion is symmetric with respect to the x

axis, and ∂ϕ/∂y = 0 on y = 0. Therefore, in the region y > 0 the potential alsodescribes the flow produced by a point source at z0 adjacent to a rigid wall aty = 0 (the presence of the wall is said to be accounted for by an image source).

104 4 Vorticity

Fig. 4.5.2.


w = −i

2πln(z − z0) + i

2πln(z − z∗

0),

represents the flow produced by two line vortices of circulations± respectivelyat z0 = x0 + iy0, z∗

0 = x0 − iy0 (Fig. 4.5.3). The stream functionψ = Imw = 0on the x axis, which is therefore a streamline of the flow, on which ∂ϕ/∂y = 0. Inthe region y > 0 the potential describes the flow produced by a vortex of strength at z0 adjacent to a rigid wall at y = 0 (which is accounted for by an equal andopposite image vortex). Each vortex translates parallel to the wall at speed u =/4πy0 determined by the velocity potential of its image. The mean value of thelocal rotational flow produced by the self-potential of each vortex (Example 5)vanishes on the vortex axis, and cannot therefore cause it to translate.

Method 2 The flow past a system of rigid boundaries in the z plane is repre-sented by means of a conformal transformation Z = f (z) by an equivalent flowin the Z plane. The transformation is usually chosen to simplify the boundary

Fig. 4.5.3.


conditions, thereby permitting the solution in the Z plane to be found in a rela-tively straightforward manner. Point source and vortex singularities of the floware preserved under the transformation. Indeed, if Z = Z0 is the image of avortex of strength at z = z0, the complex potential in the neighborhood ofZ0 (where Z − Z0 ≈ f ′(z0)(z − z0)) is determined by

W(Z ) = w(z) = −i

2πln(z − z0) + terms finite at z0

= −i

2πln

(Z − Z0

f ′(z0)

)+ terms finite at Z0

= −i

2πln(Z − Z0) + terms finite at Z0.

The vortex in the z plane therefore maps into an equal vortex at the image pointin the Z plane.

Example 8 Derive the following formula for the velocity potential of irrota-tional flow around the edge of the rigid half-plane x < 0, y = 0 in terms ofpolar coordinates (r, θ ):

ϕ = α√

r sinθ

2, α = a real constant,

and plot the streamlines.The transformation Z = i

√z maps the z plane cut along the negative real

axis (so that −π < arg z <π ) onto the upper half of the Z plane. The complexpotential of flow in the positive X direction parallel to the boundary Y = 0 in theZ plane corresponds to flow around the edge of the half-plane in the clockwisesense, and has the general representation W = U Z , where U is real. In the zplane this becomes

w = iU√

z ≡ −U√

r sin

(θ

2

)+ iU

√r cos

(θ

2

), − π <θ <π.

The polar representation of the velocity is therefore

v = (vr , vθ ) =(∂ϕ

∂r,

1

r

∂ϕ

∂θ

)= −U

2√

r

(sin

θ

2, cos

θ

2

).

This satisfies the rigid wall condition on the half-plane because the componentof velocity normal to the wall is vθ , which vanishes at θ = ±π . The streamlinesof the flow are the parabolas

√r cos

(θ

2

)= constant, i.e., y = ±2β

√1 − x

β,

106 4 Vorticity

Fig. 4.5.4.

where x <β, β being a positive constant, as shown in Fig. 4.5.4. When U > 0fluid particles travel along the parabolic streamlines around the edge in theclockwise direction. The streamline for β = 0 corresponds to the upper andlower surfaces of the half-plane, which maps into the streamline Y = 0 on thesurface of the wall in the Z plane. The flow velocity becomes infinite like 1/

√r

as r → 0 at the sharp edge.

4.6 Motion of a Line Vortex

In two-dimensional incompressible, inviscid flow in planes parallel to x3 = 0the vortex lines are all parallel to the x3 direction, and the vorticity equation(4.2.7) reduces to

Dω3

Dt= 0.

A line vortex is therefore convected without change at the local velocity at its

4.6 Motion of a Line Vortex 107

core. For a vortex of strength at z = z0(t) in the plane of z = x1 + i x2,the velocity becomes infinite as the core is approached because of the singularvelocity induced by its self-potential

− i

2πln(z − z0).

But the rotational flow around the core induced by the vortex cannot inducemotion in itself, and this potential must be removed from the complex potentialw(z) before calculating the convection velocity of the vortex.

In applications the complex potential w(z) usually arises in the form

w(z) = − i

2πln(ζ (z) − ζ (z0)) + F(z), (4.6.1)

where ζ (z), F(z) are regular functions of z in the neighborhood of the vortexcore at z = z0. In particular, when |z − z0| is small we have

ζ (z) = ζ (z0) + (z − z0)ζ ′(z0) + (z − z0)2

2ζ ′′(z0) + · · · ,

where the primes denote differentiation with respect to z. Thus, subtracting theself-potential from w(z) we find, near the vortex,

W (z) = w(z) + i

2πln(z − z0)

= − i

2πln(ζ (z) − ζ (z0)) + i

2πln(z − z0) + F(z)

≈ − i

2πln

[ζ ′(z0) + 1

2ζ ′′(z0)(z − z0)

]+ F(z). (4.6.2)

The complex velocity of the vortex is W ′(z0) ≡ W ′(z)z=z0 , i.e.,

dz∗0

dt≡ dx01

dt− i

dx02

dt= W ′(z0).

Using (4.6.2) this becomes

dx01

dt− i

dx02

dt= − iζ ′′(z0)

4πζ ′(z0)+ F ′(z0). (4.6.3)

The real and imaginary parts of this equation supply two nonlinear first-orderordinary differential equations for the position (x01(t), x02(t)) of the vortex attime t .

108 4 Vorticity

4.6.1 Numerical Integration of the Vortex Path Equation

In most cases it is necessary to integrate equation (4.6.3) numerically. Thetime and space variables should first be nondimensionalized with respect toconvenient time and length scales defined by the problem (several examples arediscussed in Chapter 8). The integration is started from a prescribed point onthe trajectory through which the vortex is required to pass.

Let us consider integration by means of a fourth-order Runge–Kutta algo-rithm. Write the equation of motion (4.6.3) in the form

dz0

dt= f ∗(z0), where f (z0) = − iζ ′′(z0)

4πζ ′(z0)+ F ′(z0),

and let h be a suitably small integration time step (which need not be constant).Assume that at time tn the vortex is at z0(tn) = zn

0. To determine the complexposition zn+1

0 at time tn+1 = tn + h, we evaluate

k1 = h f ∗(zn0

), k2 = h f ∗

(zn

0 + 1

2k1

), k3 = h f ∗

(zn

0 + 1

2k2

),

k4 = h f ∗(zn0 + k3

),

and then find

zn+10 = zn

0 + 16 (k1 + 2k2 + 2k3 + k4).

Example 1 Calculate the trajectory of a line vortex of strength> 0 adjacent toa rigid half-plane lying along the negative real axis (x1 < 0, x2 = 0; Fig. 4.6.1a).

The transformation

ζ = i√

z, z = x1 + i x2, −π < arg z <π, (4.6.4)

maps the fluid region −π < arg z <π into the upper half Im ζ > 0 of the ζ plane(Fig. 4.6.1b). Let the vortex at z0(t) map into a vortex at ζ = ζ0(t). The velocitypotential w(ζ ) of the motion in the ζ plane is found by introducing an imagevortex of strength − at ζ = ζ ∗

0 (t), as described in Example 7 of Section 4.5,in which case

w = −i

2πln(ζ − ζ0) + i

2πln(ζ − ζ ∗

0 ).

In the z plane this becomes

w(z) = − i

2πln(ζ (z) − ζ (z0)) + i

2πln(ζ (z) − ζ ∗(z0)),


Fig. 4.6.1.

which is of the form (4.6.1). Hence, the equation of motion (4.6.3) becomes

dx01

dt− i

dx02

dt= i

8π z0+ i

4π√

z0[√

z0 + (√

z0)∗].

This can be integrated in closed form. Let z0 = reiθ . Then the real and imag-inary parts of the equation are

dx01

dt≡ cos θ

dr

dt− r sin θ

dθ

dt=

8πr

(sin θ + tan

θ

2

),

dx02

dt≡ sin θ

dr

dt+ r cos θ

dθ

dt= −

8πr(cos θ + 1).

Therefore,

dr

dt= −

8πrtan

θ

2,

dθ

dt= −

4πr2(4.6.5)

that is,

rdθ

dr= 2 cot

1

2θ

110 4 Vorticity

Fig. 4.6.2.

Thus,

r = sec1

2θ, = constant. (4.6.6)

This is the polar equation of the trajectory plotted in Fig. 4.6.2. The constantlength is equal to the distance of closest approach of the vortex to the edgeof the half-plane, which occurs at θ = 0. Substituting for r in the second ofequations (4.6.5), we find

sec2

(1

2θ

)dθ

dt= −

4π2. Thus, θ = 2 tan−1

(− t

8π2

),

where time is measured from the instant at which θ = 0. The dependence of ron t is now obtained by substituting into (4.6.6).

Collecting together these results we have

r =

√1 +

(Ut

)2

, θ = 2 tan−1

(−Ut

);

x01

= 1 − (Ut/)2√

1 + (Ut/)2,

x02

= −2Ut/√

1 + (Ut/)2, (4.6.7)

where U = 8π

. Thus, (for > 0) the vortex starts above the half-plane att = −∞ at x01 = −∞, x02 = 2 and translates towards the edge, initially atspeed U parallel to the plane. It crosses the x1 axis at t = 0 at x01 = , andproceeds along a symmetrical path below the half-plane.


Fig. 4.6.3.

Example 2: Vortex motion outside a cylinder A vortex is located at z0 = reiθ

outside a rigid cylinder of radius a(< r ) with center at the origin (Fig. 4.6.3).There is no net circulation around the cylinder. The complex potential is ob-tained by placing an image vortex − at the inverse point z = a2/z∗

0 togetherwith a vortex + at the center of the cylinder. The two interior vortices ensurethat the total circulation around the cylinder vanishes. Then

w(z) = − i

2πln(z − z0) + i

2πln

(z − a2

z∗0

)− i

2πln z.

The first term on the right is the self-potential of the vortex (in the notation of(4.6.1) ζ ≡ z), so that the equation of motion of the vortex is

dz∗0

dt= iz∗

0

2π (r2 − a2)− i

2π z0≡ ia2

2π z0(r2 − a2).

By multiplying by z0 and adding the complex conjugate equation we see thatr = constant, and

rd

dte−iθ ≡ −ire−iθ dθ

dt= ia2e−iθ

2πr (r2 − a2).

Therefore,

dθ

dt= −a2

2πr2(r2 − a2),

and (for > 0) the vortex trajectory is a circle traversed in the clockwise direc-tion at speed

v0 = a2

2πr (r2 − a2). (4.6.8)

112 4 Vorticity

Problems 4

1. Show that in inviscid, homentropic flow (where div v = 0) the vorticityequation (4.2.7) takes the form

D

Dt

(ω

ρ

)=(ω

ρ· ∇)

v.

2. Use the relation∫

(yiω j (y, t) + y jωi (y, t)) d3y = 0 to show that

1

4π |x|3∫

(x · y)ω(y, t) d3y = ∇(

1

4π |x|)

∧[

1

2

∫y ∧ ω(y, t) d3y

].

Deduce the formulae (4.3.3).3. Calculate the added mass coefficients Mi j for an infinite, rigid strip of width

2a.4. Calculate the added mass coefficients Mi j for an infinite, rigid cylinder of

radius a.5. Calculate the unsteady lift and drag exerted on a rigid circular cylinder of

radius a produced by a parallel line vortex of circulation in the presenceof a uniform mean flow normal to the cylinder. Assume the motion is idealand that the net circulation around the cylinder vanishes.

6. Repeat Question 5 under the assumption that the vortex is convected solelyby the mean flow (i.e., when the induced component of the motion of

produced by image vortices in the cylinder is neglected).7. A rigid sphere of radius a translates at constant velocity U = (U, 0, 0),

U > 0, along the x1 axis. Use the creeping flow approximation ω= curl(3aU/2|x|), where the coordinate origin is taken at the center of the sphere,to deduce the Stokes drag formula D = 6πηUa.

8. A gas bubble in water is set into translational motion at velocity U(t) bysound whose wavelength greatly exceeds the bubble radius. If the acousticparticle velocity near the bubble would equal V(t) in the absence of thebubble, show that U = 3V when the mass of the air within the bubble isneglected.

9. Calculate the path of a line vortex of strength that is parallel to a rigid stripoccupying −a < x1 < a, x2 = 0,−∞< x3 <∞. Assume the fluid is at restat infinity and that there is no net circulation around the strip. Determinethe unsteady force on the strip.

10. Calculate the path of a line vortex of strength that is parallel to a rigidelliptic cylinder of semi-major and minor axes respectively equal to a andb. Assume the fluid is at rest at infinity and that there is no net circulationaround the cylinder.

Problems 4 113

11. A line vortex of strength is adjacent to a rigid right-angle corner whosesides lie along the positive x1 and x2 axes, the vortex being parallel tothe edge of the corner. Show that the vortex traverses a path with polarrepresentation r sin 2θ = constant.

12. Calculate the trajectories of a vortex pair consisting of two parallel linevortices of strengths ± moving under their mutual induction towards arigid plane parallel to the line of centers of the vortices.

13. Calculate the trajectory of a line vortex of strength adjacent to the rigidhalf-plane x1 < 0, x2 = 0 in the presence of a uniform mean flow at speedU in the positive x1 direction.

14. Show that the transformation ζ =√

z2/a2 + 1, a > 0 maps the upper z planecut by a thin rigid barrier along the imaginary axis between z = 0 and z = iaonto the upper ζ plane. Deduce that a line vortex at z = z0(t) traverses apath determined by the equation

dz∗0

dt= − i

4π

a2

z0(z2

0 + a2) − 2z0[

z20 + a2 − ∣∣z2

0 + a2∣∣],

provided the fluid is at rest at infinity.

5Vortex Sound

5.1 The Role of Vorticity in Lighthill’s Theory

At low Mach numbers in unbounded, homentropic flow the value of Lighthill’squadrupole source (2.2.2) can be approximated by means of the Biot–Savartinduction formula (4.3.1):

Ti j ≈ ρ0ui u j , u(x, t) = curl∫

ω(y, t) d3y4π |x − y| . (5.1.1)

To examine this in more detail, consider the acoustically compact eddy ofFig. 2.2.1 consisting of vorticity of characteristic length , and take the coordi-nate origin within the eddy. Put

v = u + ∇ϕ;

u involves the whole incompressible component of velocity, and u ∼ O(1/|x|3)as |x| → ∞ (see Section 4.3). Because div u = 0, the scalar potentialϕ describescompressible motions, and the continuity equation becomes

∇2ϕ + 1

ρ

Dρ

Dt= 0.

But p − p0 ∼ ρ0u2 in the eddy, where the characteristic frequency ∼u/. Thus,

Dp

Dt∼ ρ0u3

. Hence,

1

ρ

Dρ

Dt= 1

ρc2

Dp

Dt∼ u

M2, M = u

c0.

Hence, in order of magnitude

∇ϕ = O(uM2) within the eddy, where |x| ∼ . (5.1.2)

114

5.1 The Role of Vorticity in Lighthill’s Theory 115

Now write

∂2(ui u j )

∂xi∂x j= div(ω ∧ u) + ∇2

(1

2u2

)(5.1.3)

and express the solution p(x, t) = c20(ρ − ρ0) of Lighthill’s equation given by

(2.2.1) in the form

p(x, t) = p1(x, t) + p2(x, t),

where, using (1.9.6) and (1.9.8) as |x| → ∞,

p1(x, t) = −ρ0xi

4πc0|x|2∂

∂t

∫(ω ∧ u)i

(y, t − |x|

c0+ x · y

c0|x|)

d3y, (5.1.4)

p2(x, t) = ρ0

4πc20|x|

∂2

∂t2

∫1

2u2

(y, t − |x|

c0+ x · y

c0|x|)

d3y. (5.1.5)

When retarded time variations x · y/c0|x| within the eddy are neglected theidentity (5.1.3) and the divergence theorem imply that

∫ω∧ u d3y ≡ 0, because

u ∼ O(1/|y|3) as |y| → ∞. To estimate the value of the integral in (5.1.4) it istherefore necessary to expand the integrand to the next higher approximationin the retarded time:

(ω ∧ u)

(y, t − |x|

c0+ x · y

c0|x|)

= (ω ∧ u)

(y, t − |x|

c0

)

+ x · yc0|x|

∂

∂t

(ω ∧ u)

(y, t − |x|

c0

)+ · · · .

We now find

p1(x, t) ≈ −ρ0xi x j

4πc20|x|3

∂2

∂t2

∫yi (ω ∧ u) j

(y, t − |x|

c0

)d3y ∼

|x|ρ0u2 M2,

|x| → ∞. (5.1.6)

The order of magnitude of p2(x, t) is estimated by using the momentumequation (4.2.3). Because div v ∼ O(M2) within the source region, we can write

∂u∂t

+ ∇(∫

dp

ρ+ 1

2v2 + ∂ϕ

∂t

)= −ω ∧ u − ω ∧ ∇ϕ − ν curlω.

Take the scalar product with u

∂

∂t

(1

2u2

)+ div

u(∫

dp

ρ+ 1

2v2 + ∂ϕ

∂t

)= −u ·ω ∧ ∇ϕ − νu · curlω

= −u ·ω ∧ ∇ϕ + ν(div (u ∧ ω) − ω2)

116 5 Vortex Sound

and integrate over the whole of space. The contributions from the divergenceterms vanish because u(

∫dp/ρ + 1

2v2 + ∂ϕ/∂t) tends to zero at least as fast

as 1/|y|3 as |y| → ∞, where also ω = 0. Hence, using the estimate (5.1.2)

∂

∂t

∫1

2u2(y, t) d3y = −

∫(u ·ω ∧ ∇ϕ + νω2)(y, t) d3y ∼ 2u3 M2 + 2u3

Re,

(5.1.7)

where Re = u/ν typically exceeds 104 in turbulent flow. The two terms onthe right-hand side nominally represent the dissipation of the turbulent motionsrespectively by acoustic radiation and by viscous damping. (We have alreadyseen, however, in Chapter 2, Equation (2.2.5) that a more accurate estimate ofthe radiation damping is 2u3 M5.)

Thus, when retarded time variations are neglected in (5.1.5), we find

p2(x, t) ∼

|x|ρ0u2 M4 +

|x|ρ0u2 M2

Re,

and therefore that p2 p1 in turbulent flow where M 1 and Re 1.We conclude that the component

div(ρ0ω ∧ v) of the Lighthill quadrupole∂2(ρ0viv j )

∂xi∂x j

is principal source of sound at low Mach numbers.

5.2 The Equation of Vortex Sound

Lighthill’s equation (2.1.12) can be recast in a form that emphasizes the promi-nent role of vorticity in the production of sound by taking the total enthalpy

B =∫

dp

ρ+ 1

2v2

as the independent acoustic variable, in place of Lighthill’s c20(ρ−ρ0). The total

enthalpy occurs naturally in Crocco’s form (4.2.3) of the momentum equation.In the following we shall actually use the Approximation (4.2.5) of this equation,in which the viscous term 4

3ν∇(div v) is neglected. Indeed, the principal effect ofthis term is to attenuate the sound once it has been generated and is propagatingto a distant observer in the source-free region of the flow. This attenuation can besignificant in applications, but is of no particular interest when studying soundgeneration mechanisms. All viscous stresses can be ignored in a high Reynoldsnumber source flow except possibly within surface boundary layers on bodies

5.2 The Equation of Vortex Sound 117

immersed in the flow. But surface friction is dominated by the vorticity term−ν curlω, which is retained in (4.2.5).

In irrotational flow Crocco’s equation (4.2.5) reduces to

∂v∂t

= −∇ B.

In other words,

B = −∂ϕ

∂tin regions where ω = 0, (5.2.1)

where ϕ(x, t) is the velocity potential that determines the whole motion in the ir-rotational regions of the fluid. B is therefore constant in steady irrotational flow,and at large distances from the acoustic sources perturbations in B representacoustic waves.

If the mean flow is at rest in the far field, the acoustic pressure is given by

p = ρ0 B ≡ −ρ0∂ϕ

∂t. (5.2.2)

To calculate the pressure in terms of B elsewhere in the flow, we use thedefinition ∫

dp

ρ= B − 1

2v2.

Differentiating with respect to time and using Crocco’s equation (4.2.5), we have

1

ρ

∂p

∂t= ∂B

∂t− v · ∂v

∂t

= ∂B

∂t− v · (−∇ B − ω ∧ v − ν curlω)

= DB

Dt+ νv · curlω.

The small viscous correction can be ignored in high Reynolds number sourceflows, where p and B can be taken to be related by

1

ρ

∂p

∂t= DB

Dt. (5.2.3)

5.2.1 Reformulation of Lighthill’s Equation

Multiply Crocco’s equation (4.2.5) by the density ρ and take the divergence

div

(ρ∂v∂t

)+ ∇ · (ρ∇ B) = −div(ρω ∧ v). (5.2.4)

118 5 Vortex Sound

The first term on the left is expressed in terms of B by using the continuityequation in the form

div v = − 1

ρ

Dρ

Dt,

and writing

div

(ρ∂v∂t

)= ∇ρ · ∂v

∂t+ ρ

∂

∂tdiv v

= ∇ρ · ∂v∂t

− ρ∂

∂t

(1

ρ

Dρ

Dt

)

= ∇ρ · ∂v∂t

− ρ∂

∂t

(1

ρ

∂ρ

∂t

)− ∂v

∂t· ∇ρ − ρv · ∇

(1

ρ

∂ρ

∂t

)

= −ρD

Dt

(1

ρ

∂ρ

∂t

)

= −ρD

Dt

(1

ρc2

∂p

∂t

)

= −ρD

Dt

(1

c2

DB

Dt

),

where Equation (5.2.3) has been used on the last line. Substituting into (5.2.4)and dividing by ρ, we obtain the desired vortex sound equation for homen-tropic flow (

D

Dt

(1

c2

D

Dt

)− 1

ρ∇ · (ρ∇)

)B = 1

ρdiv(ρω ∧ v). (5.2.5)

The vortex source on the right-hand side vanishes in irrotational regions; ifω = 0 everywhere, and if there are no moving boundaries, the total enthalpy Bis constant, and there are no sound waves propagating in the fluid. If acousticwaves cannot enter from infinity, it follows that the (homentropic) flow cangenerate sound only if moving vorticity is present, and the right-hand sideof (5.2.5) may be identified as the analytical representation of the acousticsources. The differential operator on the left describes propagation of the soundthrough the nonuniform flow; as in the case of Lighthill’s equation, when thesource region is very extensive it will not normally be permissible to neglectthe interaction of the sound with the vorticity through which it propagates.

5.2.2 Sound Waves in Irrotational Mean Flow

Let an irrotational mean flow be defined by the velocity potential ϕ0(x), withmean velocity U = ∇ϕ0. In an unbounded fluid U = constant; the mean velocity


can vary with position only if the fluid is bounded, either internally by an airfoil,say, or externally by the walls of a duct of variable cross section (ϕ0(x) can bemultiple-valued if the boundaries are multiply connected, but the mean velocityis always single-valued).

Consider an irrotational disturbance ϕ′(x, t), and set

ϕ(x, t) = ϕ0(x) + ϕ′(x, t).

It can be shown that the general, nonlinear equation satisfied by ϕ is

1

c2

∂2ϕ

∂t2+ 1

c2

D

Dt

(1

2(∇ϕ)2

)+ 1

c2

∂

∂t

(1

2(∇ϕ)2

)− ∇2ϕ = 0.

The linearized version of this equation describes the propagation of small ampli-tude sound waves determined by ϕ′(x, t). However, when ω = 0 the linearizedequation for B = −∂ϕ′/∂t ≡ −ϕ is more easily derived from (5.2.5), whichbecomes (

D

Dt

(1

c2

D

Dt

)− 1

ρ∇ · (ρ∇)

)ϕ = 0. (5.2.6)

The coefficients of the differential operators in this equation are functions ofboth mean and perturbation quantities, but the linearized equation is obtainedmerely by replacing these coefficients by their values in the absence of thesound. In homentropic flow the mean density and sound speed can be expressedin terms of the variable mean velocity U(x), and

D

Dt≈ ∂

∂t+ U · ∇ .

Furthermore, because the mean flow does not depend on time, we can take theperturbation potential ϕ′, rather than ϕ, as the acoustic variable. The linearizedequation then becomes(

∂

∂t+ U · ∇

)[1

c2

(∂

∂t+ U · ∇

)]− 1

ρ∇ · (ρ∇)

ϕ′ = 0,

where c ≡ c(x) and ρ ≡ ρ(x) are the local sound speed and density in the steadyflow.

5.2.3 Vortex Sound at Low Mach Numbers

When the characteristic Mach number M is small the local mean values of thedensity and sound speed are related to their uniform respective values ρ0 and

120 5 Vortex Sound

c0 at infinity by relations of the form

c

c0∼ 1 + O(M2),

ρ

ρ0∼ 1 + O(M2).

The vortex sound equation (5.2.5) can therefore be simplified by (a) takingc = c0, and ρ = ρ0, and (b) by neglecting nonlinear effects of propagation andthe scattering of sound by the vorticity. The production of sound is then governedby the simpler equation(

1

c20

∂2

∂t2− ∇2

)B = div(ω ∧ v), (5.2.7)

and in the far field the acoustic pressure is given by the linearized approximation

p(x, t) ≈ ρ0 B(x, t). (5.2.8)

5.2.4 Example 1 (Powell 1963): Sound Generationby a Spinning Vortex Pair

Two parallel vortex filaments each of circulation and distance 2 apart rotateabout the x3 axis midway between them (Fig. 5.2.1) at angular velocity ! =/4π2, provided the Mach number is small enough for the motion to beregarded as incompressible. Their positions at time t are

x = (x1, x2) = ±s ≡ ±(s1(t), s2(t)) = ±(cos!t, sin!t).

Fig. 5.2.1.


The vorticity distribution is

ω = k(δ(x − s) + δ(x + s)),

where k is a unit vector in the x3 direction, parallel to the vortices. The vortexconvection velocities are

v = ±!k ∧ s(t) at x = ±s(t), where U = ! c0.

Hence, because k ∧ (k ∧ s) = −s,

ω ∧ v = −!s(t)[δ(x − s) − δ(x + s)].

If this is expanded in powers of the radius of the circular orbit (and it can beverified that this is equivalent to expanding the acoustic pressure in powers ofM = U/c0 1) the i th component of the first nonzero term is

(ω ∧ v)i ≈ ∂

∂ x j(2!si (t)s j (t)δ(x)),

so that the vortex sound source is equivalent to the quadrupole

div(ω ∧ v) ≈ ∂2

∂ x i∂ x j(2!si (t)s j (t)δ(x)).

The solution of the vortex sound equation (5.2.7) for this quadrupole sourceis (c.f., (2.2.1))

B = 1

4π

∂2

∂ x i∂ x j

∫2! (si s j )

(t − |x − y|

c0

)δ(y) d3y|x − y| , y = (y1, y2).

In the acoustic far field, we use (1.9.7):

∂

∂ x j≈ −x j

c0(r2 + (x3 − y3)2)12

∂

∂t,

where r = (x21 + x2

2 )12 is the perpendicular distance from the centroid of the

vortices (the x3 axis). Thus, by setting ξ = y3 − x3, we can write in the acousticfar field, where B ≈ p/ρ0,

p ≈ ρ0!x i x j

2πc20

∂2

∂t2

∫ ∞

−∞(si s j )

(t − (r2 + ξ 2)

12

c0

)dξ

(r2 + ξ 2)32

, r → ∞.

(5.2.9)

122 5 Vortex Sound

In this formula

(si s j )(t) = 2

2

(1 + cos 2!t sin 2!t

sin 2!t 1 − cos 2!t

),

but the constant terms in the matrix can be omitted because of the time deriva-tives in (5.2.9). The integration in (5.2.9) can now be performed by the approx-imate method described below in Example 2 in the limit that !r/c0 → ∞,i.e., in the limit in which the radial distance r greatly exceeds the acousticwavelength

∫ ∞

−∞(si s j )

(t − (r2 + ξ 2)

12

c0

)dξ

(r2 + ξ 2)32

≈ 2

2r2

(πc0

!r

) 12

(cos

(2![t] − π

4

)sin

(2![t] − π

4

)sin

(2![t] − π

4

) −cos(2![t] − π

4

)),

where [t] = t − r/c0. Hence, introducing polar coordinates x = r (cos θ, sin θ )we find

x i x j2

2r2

(πc0

!r

) 12

(cos

(2![t] − π

4

)sin

(2![t] − π

4

)sin

(2![t] − π

4

) −cos(2![t] − π

4

))

i j

= 2

2

(πc0

!r

) 12

(cos θ, sin θ )

(cos

(2![t] − π

4

)sin(2![t] − π

4

)sin(2![t] − π

4

) −cos(2![t] − π

4

))(

cos θsin θ

)

= 2

2

(πc0

!r

) 12

cos(

2θ − 2![t] + π

4

),

and, therefore, (5.2.9) becomes

p ≈ −ρ0!32

πc20

(πc0

!r

) 12

cos

[2θ − 2!

(t − r

c0

)+ π

4

]

= −4

√π

rρ0U 2 M3/2 cos

[2θ − 2!

(t − r

c0

)+ π

4

],

!r

c0→ ∞.

(5.2.10)

The amplitude of the sound decreases like 1/√

r (instead of 1/r ) because thewaves are spreading cylindrically in two dimensions. The sound power mustnow be calculated by considering the integral∫

p2

ρ0c0d S

over the surface of a large circular cylinder r = constant. Taking the time


average, we can show that the acoustic power per unit length of the vortices∼ρ0U 3 M4. This differs by a factor of the Mach number M from the powerradiated by a compact body of three-dimensional turbulence, and is character-istic of the acoustic power produced by two-dimensional regions of turbulencein an unbounded fluid.

5.2.5 Example 2

Show that∫ ∞

−∞f

(ξ

r

)eiκ0

√r2+ξ 2

dξ ≈ r f (0)

(2π

κ0r

) 12

ei(κ0r+ π4 ), κ0r → ∞. (5.2.11)

Put ξ =µr , then

I ≡∫ ∞

−∞f

(ξ

r

)eiκ0

√r2+ξ 2

dξ = r∫ ∞

−∞f (µ)eiκ0r

√1+µ2

dµ.

As κ0r → ∞ the exponential factor oscillates increasingly rapidly, and the maincontribution to the integral is from the neighborhood of that value of µ wherethe oscillations are stationary. This occurs at µ = 0. The integrand is thereforeexpanded about this point. In the first approximation, f (µ) can be replaced byf (0), and

eiκ0r√

1+µ2 ≈ eiκ0r+iκ0rµ2/2.

Thus,

I = r∫ ∞

−∞f (µ)eiκ0r

√1+µ2

dµ ≈ r f (0)eiκ0r∫ ∞

−∞eiκ0rµ2/2 dµ

= r f (0)eiκ0r

(2π

κ0r

) 12

eiπ4 ,

which yields (5.2.11).In particular,

∫ ∞

−∞

cos 2!

(t − (r2 + ξ 2)

12

c0

)− i sin 2!

(t − (r2 + ξ 2)

12

c0

)dξ

(r2 + ξ 2)32

= e−2i!t∫ ∞

−∞ei 2!

c0

√r2+ξ 2 dξ

(r2 + ξ 2)32

≈ 1

r2

(πc0

!r

) 12

e−i[2!(t − r

c0) − π

4 ],

!r

c0→ ∞.

124 5 Vortex Sound

Hence,

∫ ∞

−∞cos 2!

(t − (r2 + ξ 2)

12

c0

)dξ

(r2 + ξ 2)32

≈ 1

r2

(πc0

!r

) 12

cos

[2!

(t − r

c0

)− π

4

]∫ ∞

−∞sin 2!

(t − (r2 + ξ 2)

12

c0

)dξ

(r2 + ξ 2)32

≈ 1

r2

(πc0

!r

) 12

sin

[2!

(t − r

c0

)− π

4

].

5.3 Vortex–Surface Interaction Noise

The small Mach number vortex sound equation (5.2.7) is now applied to deter-mine the sound generated by vorticity in the neighborhood of a fixed body whosesurface S may be vibrating at small amplitude (Fig. 5.3.1). The developmenthere is analogous to the derivation in Section 2.3 of Curle’s equation.

Introduce a stationary, closed control surface S+ on which f (x) = 0, suchthat f (x) >< 0 according as x lies without or within S+. The body is assumedto be within S+, and S+ will subsequently be allowed to shrink down to co-incide with the body surface S. Multiply equation (5.2.7) by H ≡ H ( f ) and

Fig. 5.3.1.

5.3 Vortex–Surface Interaction Noise 125

form the inhomogeneous wave equation for the new variable H B. We use thetransformations

H∇2 B ≡ H div (∇ B) = div(H∇ B) − ∇ H · ∇ B

= ∇2(H B) − div(B∇ H ) − ∇ H · ∇ B, (5.3.1)

and H div(ω ∧ v) = div(Hω ∧ v) − ∇ H ·ω ∧ v. (5.3.2)

Then, (5.2.7) becomes(1

c20

∂2

∂t2− ∇2

)(H B) = −div(B ∇ H ) − ∇ H · (∇ B +ω∧ v) + div(Hω∧ v).

This equation is formally valid everywhere, including the region within S+where H B ≡ 0. The source terms involving ∇ H are concentrated on the controlsurface. When x lies in the exterior region these surface terms take account ofthe presence of the solid body inside S+; if the body is absent (so that S+ isfilled with fluid), the surface sources constitute a representation ‘to the outsideworld’ in f > 0 of the various hydrodynamic or acoustic processes that may beoccurring within S+.

Using Crocco’s equation (4.2.5), we can make the substitution

∇ H · (∇ B +ω ∧ v) = −∇ H ·(∂v∂t

+ ν curlω

)

≡ −∇ H · ∂v∂t

+ ν div(∇ H ∧ω).

Hence the vortex sound equation becomes(1

c20

∂2

∂t2− ∇2

)(H B) = −div(B∇ H ) + ∇ H · ∂v

∂t

+ div(Hω ∧ v) − ν div(∇ H ∧ ω). (5.3.3)

The sources on the right of this equation are either concentrated on the controlsurface S+ or lie in the fluid outside S+; they completely determine B outsidethis control surface. The solution in this region can therefore be found by usingany Green’s function G(x, y, t − τ ) that satisfies(

1

c20

∂2

∂t2− ∇2

)G = δ(x − y)δ(t − τ ), where G = 0 for t <τ

for x and y anywhere within the fluid. In Fig. 5.3.1 the fluid occupies the regionV outside the surface S of the solid body; the control surface S+ ( f (x) = 0)therefore lies within V .

126 5 Vortex Sound

Thus, for points x within the fluid the solution of (5.3.3) is

H B(x, t) =∫ ∞

−∞

∫V

G(x, y, t − τ )

−div(B∇ H ) + ∇ H · ∂v

∂τ+ div(Hω∧ v)

− ν div(∇ H ∧ ω)

d3y dτ,

where all of the source terms within the brace brackets are functions of y and τ .Those involving ∇ H vanish except on the control surface S+. The divergenceterms are removed by application of the divergence theorem, and then using theformula ∫

V(·)∇ H d3y =

∮S+

(·) dS

(see Section 2.3). Let us illustrate the procedure for the first term in the bracebrackets of the integrand∫

VG−div(B∇ H ) d3y = −

∫Vdiv(G B∇ H ) − B∇G · ∇ H d3y

=∮

S+

G B∇ H · dS +∫

VB∇G · ∇ H d3y

= 0 +∮

S+B∇G · dS

≡∮

S+B(y, τ )

∂G

∂yn(x, y, t − τ ) d S(y),

where all vector operators are with respect to the y dependence, and is alarge, closed ‘surface at infinity’ where ω = 0. There are no contributions fromS and because ∇ H = 0 everywhere except on S+.

The general solution in the region f > 0 outside S+ accordingly becomes

B(x, t) =∮

S+

(B(y, τ )

∂G

∂yn(x, y, t − τ ) + G(x, y, t − τ )

∂vn

∂τ(y, τ )

)dS(y) dτ

−∫

VH ( f (y))(ω ∧ v)(y, τ ) · ∂G

∂y(x, y, t − τ ) d3y dτ

+ ν

∮S+ω(y, τ ) ∧ ∂G

∂y(x, y, t − τ ) · dS(y) dτ, (5.3.4)

where for brevity we have omitted the integration sign for τ , which is understoodto vary over the range −∞<τ <∞.

5.3 Vortex–Surface Interaction Noise 127

We now choose G to have vanishing normal derivative on the surface S ofthe body. When this is done the control surface S+ is allowed to shrink downonto S (whereupon the first term in the first integral of (5.3.4) vanishes), andthe general solution in the fluid becomes

B(x, t) = −∫

V(ω ∧ v)(y, τ ) · ∂G

∂y(x, y, t − τ ) d3y dτ + ν

∮Sω(y, τ )

∧ ∂G

∂y(x, y, t − τ ) · dS(y) dτ +

∮S

G(x, y, t − τ )∂vn

∂τ(y, τ ) dS(y) dτ,

where∂G

∂yn(x, y, t − τ ) = 0 on S. (5.3.5)

In the acoustic far field (|x| → ∞) we can replace B(x, t) by p(x, t)/ρ0.The first integral represents the production of sound by vortex sources dis-

tributed within the fluid. Green’s function takes full account of the influenceof the body on the efficiency with which these sources generate sound. Thesecond, surface integral involving the surface vorticity is the contribution fromfrictional forces on S. To interpret the final term, recall that the control surfaceS+ was taken to be fixed in space. This means that fluid can flow through thesurface. When S+ shrinks down to S the implication is that S is also fixed inspace. However, the normal velocity vn can still be nonzero if the surface of thebody is vibrating at small amplitude, and this term in the solution is actuallyidentical with that given previously in (3.8.2) for a vibrating body in the absenceof vortex sources.

In connection with this, it should also be noted that the reciprocal theoremimplies that the normal derivative conditions

∂G

∂yn(x, y, t − τ ) = 0,

∂G

∂xn(x, y, t − τ ) = 0 respectively for y, x on S,

are always satisfied simultaneously.The contribution to the sound from surface friction (the first surface integral

on the right of (5.3.5)) is nominally of order

1

Re 1, Re = v

ν,

relative to the contribution from the volume vorticity (the first integral), where is the characteristic length scale of the turbulence or body and v is a typicalvelocity. At high Reynolds numbers the surface term can therefore be discarded,and in the important case in which the body does not vibrate the acoustic far

128 5 Vortex Sound

field is then given by

p

ρ0(x, t) = −

∫V

(ω ∧ v)(y, τ ) · ∂G

∂y(x, y, t − τ ) d3y dτ. (5.3.6)

5.4 Radiation from an Acoustically Compact Body

When the surface S is acoustically compact the compact Green’s function (3.9.1)can be used to evaluate the general solution (5.3.5) in the far field. When|x| → ∞ and the origin is within or close to S, we proceed as already describedin Section 3.8 by expanding Green’s function to first order in the retarded timeacross S:

G(x, y, t − τ ) = 1

4π |X − Y|δ(

t − τ − |X − Y|c0

)

≈ 1

4π |x|δ(

t − τ − |x|c0

)+ x j Y j

4πc0|x|2 δ′(

t − τ − |x|c0

),

|x| → ∞,

where the prime denotes differentiation with respect to t . The first term in thisapproximation, which is independent of y, clearly makes no contribution to thefirst two integrals in (5.3.5). It makes a contribution to the final surface integralonly if the volume of the body is pulsating. When this happens the resultingmonopole radiation from the body is usually large compared to all other sources.We shall, therefore, ignore this possibility, and consider only surface vibrationsfor which the volume of S is constant; in particular we shall assume that Svibrates as a rigid body. In this case, therefore, the first approximation in theGreen’s function expansion can again be discarded.

Substituting into (5.3.5) and performing the integrations with respect to τ ,we find in the acoustic far field (where B = p/ρ0)

p(x, t) ≈ −ρ0x j

4πc0|x|2∂

∂t

[∫(ω ∧ v) · ∇Y j d3y − ν

∮Sω ∧ ∇Y j · dS(y)

−∮

S

∂Un

∂tY j d S(y)

], (5.4.1)

where the large square brackets ([ ]) denote that the enclosed quantity is to beevaluated at the retarded time t − |x|/c0, and Un is the normal component ofvelocity of vibration of S.

5.4 Radiation from an Acoustically Compact Body 129

When the body executes translational oscillations at velocity U(t) we have

Un = Ui ni ,

where ni is the i th component of the surface normal directed into the fluid.Then

ρ0

∮S

∂Un

∂tY j d S = ρ0

dUi

dt

∮S(ni y j − niϕ

∗j ) d S

= (m0δi j + Mi j )dUi

dt,

where if the body has volume , then m0 = ρ0 is the mass of fluid displacedby the body, and Mi j is the added mass tensor of the body (see (3.8.6)).

The solution (5.4.1) can therefore be written

p(x, t) ≈ −x j

4πc0|x|2∂

∂t

[ρ0

∫(ω ∧ v) · ∇Y j d3y − η

∮Sω ∧ ∇Y j · dS(y)

− (m0δi j + Mi j )dUi

dt

]. (5.4.2)

Reference to Equation (4.4.4) shows that this can also be written

p(x, t) ≈ x j

4πc0|x|2∂Fj

∂t

(t − |x|

c0

)+ m0x j

4πc0|x|2∂2U j

∂t2

(t − |x|

c0

), |x| → ∞,

(5.4.3)

where F(t) is the unsteady force exerted on the fluid by the body. This is justour earlier conclusion (2.4.2) derived from Curle’s equation, with the additionof the fluid-displacement effect of the vibrating body, and is equivalent to thesolution (3.8.10) obtained in the absence of vorticity.

The relative contributions from the volume and surface distributions of vor-ticity in (5.4.2) (respectively the first and second integrals) for turbulence oflength scale and velocity v are estimated respectively by

ρ0v2 M

|x| and ρ0v2 M

|x|1

Re, Re = v

ν.

Thus, in high Reynolds number turbulent flows the surface frictional contribu-tion to the dipole force F can usually be neglected. For a nonvibrating compactbody the principal component of the acoustic pressure in the far field is therefore

130 5 Vortex Sound

given by

p(x, t) ≈ −ρ0x j

4πc0|x|2∂

∂t

∫(ω ∧ v)

(y, t − |x|

c0

)· ∇Y j (y) d3y, |x| → ∞.

(5.4.4)

5.5 Radiation from Cylindrical Bodies of Compact Cross Section

An important special case occurs when vorticity interacts with a cylindrical (orapproximately cylindrical) surface S of compact cross section (such as the stripairfoil of Fig. 3.6.3). If the vorticity extends over an extensive spanwise sectionof the body it may be important to account for differences in the retarded timesof the sound produced at different spanwise positions.

To do this we first write the Kirchhoff vector in the form

Y = Y⊥ + ky3, Y⊥ = (Y1(y), Y2(y), 0), (5.5.1)

where k is a unit vector in the x3 direction. Then, because variations in thespanwise source position are not necessarily small compared to the acousticwavelength, the compact Green’s function in (5.3.6) is expanded as follows:

G(x, y, t − τ ) = 1

4π |X − Y|δ(

t − τ − |X − Y|c0

)

≈ 1

4π |x − ky3|δ(

t − τ − |x − ky3|c0

)

+ x j Y⊥ j

4πc0|x − ky3|2 δ′(

t − τ − |x − ky3|c0

)

≈ 1

4π |x|δ(

t − τ − |x − ky3|c0

)

+ x j Y⊥ j

4πc0|x|2 δ′(

t − τ − |x − ky3|c0

), |x| → ∞. (5.5.2)

For example, when a stationary cylindrical body interacts with high Reynoldsnumber flow at low Mach number, the monopole term in (5.5.2) can be discardedas before, and the acoustic pressure given by (5.3.6) becomes

p(x, t) ≈ −ρ0x j

4πc0|x|2∂

∂t

∫(ω∧ v)

(y, t − |x − ky3|

c0

)· ∇Y⊥ j (y) d3y, |x| → ∞.

(5.5.3)

The sound is produced by dipole sources orientated in the lift and drag directions

5.6 Impulse Theory of Vortex Sound 131

only ( j = 1, 2). This approximation is applicable also to a thin airfoil of largebut finite span, and in cases where the chord of the airfoil is a slowly varyingfunction of y3 (such as the elliptic airfoil of Fig. 3.9.1).

5.6 Impulse Theory of Vortex Sound

An interesting formula for the sound generated by vorticity near a compact bodycan be derived directly from the representation (4.3.3) of the velocity v(x, t) inthe hydrodynamic far field in terms of the impulse I(t). Indeed,

v(x, t) ≈ ∇ϕ when |x| ,

where is the length scale of the interaction region (the body), and

ϕ(x, t) = div

(I(t)

4π |x|)

where I(t) = 1

2

∫y ∧ ω(y, t) d3y. (5.6.1)

This expression for ϕ(x, t) defines the incompressible motion in the irrota-tional region far from the body. It is the velocity potential of a hydrodynamicdipole that will be recognized as the acoustic near field of an outgoing acousticdipole representing sound production by the flow (see Equations (1.7.4) and(1.7.9)). The acoustic dipole is found simply by replacing I(t) in (5.6.1) byI(t − |x|/c0). At large distances from the body (where the undisturbed fluid isstationary) the pressure p(x, t) = −ρ0∂ϕ/∂t , and this procedure therefore leadsto the following formula for the sound in terms of the vorticity:

p(x, t) ≈ −ρ0∂

∂x j

(1

4π |x|∂ I j

∂t(t − |x|/c0)

)

≈ ρ0x j

4πc0|x|2∂2 I j

∂t2

(t − |x|

c0

), |x| → ∞

= ρ0x j

8πc0|x|2∂2

∂t2

∫(y ∧ ω) j

(y, t − |x|

c0

)d3y. (5.6.2)

Equation (4.4.1) shows that this is equivalent to the compact approximation(5.4.3). Note, however, that ω in (5.6.2) is the generalized vorticity, includingthe bound vorticity on S. This should be contrasted with the representation(5.4.2) involving the Kirchhoff vector, where bound vorticity occurs only in thesurface integral of the frictional contribution to the sound. But, (5.4.2) is validonly for a body in translational motion whereas (5.6.2) is applicable for a bodyexecuting any combination of translations and rotations (Fig. 4.1.1).

132 5 Vortex Sound

The impulse I is constant for vorticity in an unbounded fluid (when com-pressibility is ignored). We know that the sound is now generated by quadrupolesources and that it can be represented in terms of the vorticity as in (5.1.6).Mohring (1978) has shown that it is also possible to express the quadrupolesound as a third-order time derivative of a second-order moment of the vor-ticity, analogous to the first order moment in (5.6.2) (see Problems 5),namely

p(x, t) ≈ ρ0xi x j

12πc20|x|3

∂3

∂t3

∫yi (y ∧ω) j (y, t −|x|/c0) d3y, |x| → ∞. (5.6.3)

Problems 5

1. Kirchhoff’s spinning vortex: A columnar vortex parallel to the x3 axishas elliptic cross section defined by the polar equation r = a1+ε cos(2θ −!t/2), where ε 1 and ! is the uniform vorticity in the core. The ellipserotates at angular velocity 1

4!, and the velocity distribution within the coreis given by

v = (v1, v2) = −1

2!r (sin θ + ε sin(θ −!t/2),− cos θ + ε cos(θ −!t/2)).

Show that to first order in ε the vortex is equivalent to the two-dimensionalquadrupole

div(ω ∧ v) ≈ ∂2

∂xi∂x j(Ti jδ(x1)δ(x2)), i, j = 1, 2

Problems 5 133

where

Ti j = επ!2a4

8

(cos(!t/2) sin(!t/2)

sin(!t/2) −cos(!t/2)

)

and that the acoustic pressure is

p ≈ −ε

8

√2πa

rρ0U 2 M3/2 cos

[2θ − !

2

(t − r

c0

)+ π

4

],

!r

c0→ ∞,

where U = 12 a! and M = U/c0.

2. Coaxial vortex rings: Use Equation (5.2.7) to calculate the sound producedby the unsteady motions of an acoustically compact system of N vortexrings coaxial with the x1 axis. Take the vorticity of the nth vortex to beωn = nδ(x1 − Xn(t))δ(r − Rn(t))iθ , where (r, θ, x1) are cylindrical polarcoordinates, Rn(t) being the vortex ring radius, Xn(t) its location in the x1

direction, and iθ is a unit vector in the azimuthal direction. Show that

p ≈ ρ0

8c20|x| (3 cos2 − 1)

∂2

∂t2

[∑n

n Xnd R2

n

dt

], |x| → ∞,

where is the angle between the observer direction and the positive x1 axisand the term in square braces is evaluated at τ = t − |x|/c0.

In an ideal, incompressible fluid conservation of energy and momentumimplies that∑

n

n Rn

(Rn

d Xn

dt− Xn

d Rn

dt

)= constant,

∑n

n R2n = constant.

Use these equations to show that

p ≈ ρ0

12c20|x| (3 cos2 − 1)

[d3S

dt3

], |x| → ∞,

where S(t) = ∑n n R2

n Xn .

134 5 Vortex Sound

3. Calculate the sound produced by a vortex ring of total circulation , coaxialwith the x1 axis, whose core has elliptic cross section of major and minoraxes 2a, 2b R, where R is the mean radius of the ring. Assume that thex1 coordinate X (t) of the vorticity centroid satisfies

d X

dt=

4π R

[ln

(16R

a + b

)− 1

4+ 3(a − b)

2(a + b)cos 2!t

], ! =

π (a + b)2.

In the notation of Question 2, show that

p ≈ ρ0R2(3 cos2 − 1)

12c20|x|

[d3 X

dt3

]t− |x|

c0

, |x| → ∞

= ρ0U 2 M2

8π3

R(3 cos2 − 1)

|x|(

a − b

a + b

)cos

(2

π (a + b)2

[t − |x|

c0

]),

where U =

a + b, M = U

c0.

4. Derive Mohring’s formula (5.6.3) for the acoustic pressure generated by acompact region of vorticity in an unbounded fluid. Take the cross product of ywith the high Reynolds number vorticity equation ∂ω/∂t + curl (ω∧ v) = 0(expressed in terms of y and t as independent variables), multiply by yi , anduse the identity

y ∧ curl A = 2A + ∇(y · A) − ∂

∂y j(y j A)

to deduce that the integral in (5.1.6) can be written∫yi (ω ∧ v) j d3y = −1

3

∂

∂t

∫yi (y ∧ ω) j d3y + 1

3δi j

∫1

2v2 d3y.

The result now follows by noting that the estimate (5.1.7) permits the secondintegral on the right to be discarded.

5. The free space Green’s function in two dimensions – the solution of

(1

c20

∂2

∂t2− ∇2

)G = δ(x1 − y1)δ(x2 − y2)δ(t −τ ), where G = 0 for t <τ,

– can be derived by integrating the three-dimensional Green’s function

1

4π |x − y|δ(

t − τ − |x − y|c0

)over −∞< y3 <∞.

Problems 5 135

Deduce that

G(x, y, t − τ ) = H (t − τ − |x − y|/c0)

2π√

(t − τ )2 − (x − y)2/

c20

, x = (x1, x2), y = (y1, y2),

and that near the wavefront, where |x − y| ≈ c0(t − τ )

G(x, y, t − τ ) ≈ H (t − τ − |x − y|/c0)

2π√

2|x − y|/c0√

(t − τ ) − |x − y|/c0.

6. Consider sound production by a compact distribution of vorticity in an un-bounded two-dimensional flow (independent of x3), where the vorticity ω

is parallel to the k direction (the x3 axis). Show that∫yi (ω ∧ v) j dy1 dy2 ≈ −1

2

∂

∂t

∫yi (y ∧ ω) j dy1 dy2, where i, j = 1, 2.

Deduce Mohring’s (1980) two-dimensional representation

p(x, t) ≈ ρ0xi x j

4πc20|x|2

∂3

∂t3

∫dy1 dy2

∫ t−|x|/c0

−∞

yi (y ∧ ω) j (y, τ ) dτ√(t − τ )2 − |x|2/c2

0

≈ ρ0xi x j

4πc20|x|2

√c0

2|x|∂3

∂t3

∫dy1 dy2

∫ t−|x|/c0

−∞

yi (y ∧ ω) j (y, τ ) dτ√t − τ − |x|/c0

,

|x| → ∞.

7. Use the result of Problem 6 to derive the Solution (5.2.10) for the soundproduced by a spinning vortex pair.

8. Use the result of Problem 6 to solve Problem 1.

6Vortex–Surface Interaction Noise

in Two Dimensions

6.1 Compact Green’s Function in Two Dimensions

In this chapter, we apply the general high Reynolds number solution (5.3.6)to determine sound produced by two-dimensional interactions of rectilinearvortices with a stationary solid boundary. Conditions are assumed to be uniformin the x3 direction, parallel to the vorticity. We shall derive the two-dimensionalanalogue of the general solution

p

ρ0(x, t) = −

∫V

(ω ∧ v)(y, τ ) · ∂G

∂y(x, y, t − τ ) d3y dτ (6.1.1)

by first determining a suitable representation of G in two dimensions. Both thevorticity convection velocity v and the Lamb vectorω∧ v are parallel to the x1x2

plane, so that only the y1 and y2 components of the gradient ∂G/∂y contributeto the integral. Also, because (ω ∧ v)(y, τ ) depends only on y1, y2 and τ , theintegration with respect to the spanwise coordinate y3 involves only the Green’sfunction, and may be performed prior to any further calculations of the sound.

Let

G2 =∫ ∞

−∞G(x, y, t − τ ) dy3. (6.1.2)

For two-dimensional problems G is a function of y3 − x3, and the function G2

will therefore satisfy the Green’s function equation(1

c20

∂2

∂t2− ∇2

)G2 = δ(x1 − y1)δ(x2 − y2)δ(t − τ ),

where G2 = 0 for t <τ, (6.1.3)

obtained by integrating the three-dimensional Equation (3.1.4) over −∞< y3 <

∞. In two dimensions, G2 represents the field generated by a uniform line sourceparallel to the x3 axis extending along the whole of the line x1 = y1, x2 = y2.

136

6.1 Compact Green’s Function in Two Dimensions 137

Similarly, the compact Green’s function for a cylindrical body (with gener-ators parallel to x3)

G(x, y, t − τ ) = 1

4π |X − Y|δ(

t − τ − |X − Y|c0

),

X1,2 = x1,2 − ϕ∗1,2(x), Y1,2 = y1,2 − ϕ∗

1,2(y), X3 = x3, Y3 = y3,

is a function of y3 – x3, and the corresponding two-dimensional compact Green’sfunction can be found by integrating over −∞< y3 <∞.

Set ξ = y3 − x3, x = (x1, x2) and let |x| = (x21 + x2

2 )12 → ∞. Taking the origin

of coordinates within the cylindrical body, we have

G2 ≈ 1

4π

∫ ∞

−∞δ

(t − τ −

√|x|2 + ξ 2

c0+ x · Y

c0

√|x|2 + ξ 2

)dξ√

|x|2 + ξ 2

as |x| → ∞,

where x · Y = x1Y1 + x2Y2 is independent of ξ . To use this to evaluate (6.1.1)the δ function must be expanded to first order in Y

G2 ≈ 1

4π

∫ ∞

−∞δ

(t − τ −

√|x|2 + ξ 2

c0

)dξ√

|x|2 + ξ 2

+ x · Y4πc0

∫ ∞

−∞δ′(

t − τ −√

|x|2 + ξ 2

c0

)dξ

(|x|2 + ξ 2).

Only the second term on the right depends on y and therefore contributes to theradiation integral (6.1.1). We can therefore take

G2 ≈ x · Y4πc0

∫ ∞

−∞δ′(

t − τ −√

|x|2 + ξ 2

c0

)dξ

(|x|2 + ξ 2)

= x · Y2πc0

∂

∂t

∫ ∞

0δ

(t − τ −

√|x|2 + ξ 2

c0

)dξ

(|x|2 + ξ 2)

= x · Y2πc0

∂

∂t

H (t − τ − |x|/c0)

(|x|2 + ξ 2)∣∣ ∂∂ξ

√|x|2+ξ 2

c0

∣∣

ξ=√

c20(t−τ )2−|x|2

= x · Y2πc0

∂

∂t

H (t − τ − |x|/c0)

(t − τ )√

c20(t − τ )2 − |x|2

.

138 6 Vortex–Surface Interaction Noise in Two Dimensions

We shall henceforth in this chapter regard all space vectors as two-dimen-sional, such as x = (x1, x2), y = (y1, y2), and drop the overbar on x and thesubscript 2 on G2 and soforth. The dipole component of the two-dimensionalcompact Green’s function then becomes

G(x, y, t − τ ) ≈ x · Y2πc0

∂

∂t

H (t − τ − |x|/c0)

(t − τ )√

c20(t − τ )2 − |x|2

, |x| → ∞.

(6.1.4)

In contrast to the three-dimensional Green’s function, which is nonzero onlyon a spherically expanding wavefront, G has an infinite peak at the wavefrontwhere t − τ − |x|/c0 = 0 followed by a slowly decaying tail. At any point xin the far field the first sound arrives from the nearest point on the line sourceof (6.1.3), after propagating along a ray perpendicular to the source over adistance equal to |x|, and therefore after a time delay |x|/c0; but the observer at xreceives sound continuously after the passage of this wavefront, generated atmore distant sections of the infinitely long line source; the wavefront arrives attime τ + |x|/c0, and at a later time t sound is received from those source pointswhose distance from x is equal to c0(t − τ ).

The far-field representation (6.1.4) can be approximated further by expandingabout the wavefront (where t −τ = |x|/c0), which contains most of the acousticenergy. Just to the rear of the wavefront

(t − τ )√

c20(t − τ )2 − |x|2 ≡ (t − τ )

√c0(t − τ ) + |x|

√c0(t − τ ) − |x|

≈ |x|c0

√2|x|

√c0(t − τ ) − |x|, for t − τ ∼ |x|

c0.

Therefore, (6.1.4) becomes

G(x, y, t−τ ) ≈ x · Y

2π√

2c0|x| 32

∂

∂t

H (t − τ − |x|/c0)√

t − τ − |x|/c0

, |x| → ∞, (6.1.5)

(see Question 1 of Problems 6).The special case of a cylindrical body adjacent to a plane, rigid wall at

x2 = 0, or of a cylindrical wall cavity or projection from a wall (see Fig. 3.9.2)is handled by the two-dimensional version of the compact Green’s function(3.9.3). The procedure described above yields the following expression for thedipole component of the two-dimensional compact Green’s function

G(x, y, t − τ ) ≈ x1Y1

π√

2c0|x| 32

∂

∂t

H (t − τ − |x|/c0)√

t − τ − |x|/c0

, |x| → ∞, (6.1.6)

6.2 Sound Generated by a Line Vortex 139

where Y1 ≡ Y1(y1, y2) is the velocity potential of incompressible flow past thecylinder in a direction parallel to the wall (with unit speed at large distancesfrom the body). G represents the field of a dipole orientated parallel to thewall and perpendicular to the cylinder axis, and the effect of the wall is togenerate an equal image dipole that just doubles the magnitude of the soundrelative to the corresponding dipole (6.1.5) of the cylinder in the absence of thewall.

6.2 Sound Generated by a Line Vortex Interactingwith a Cylindrical Body

The sound produced by two-dimensional vortex motion at low Mach numberM ∼ v/c0 near a stationary, rigid cylindrical body of diameter has charac-teristic wavelength ∼/M . The acoustic pressure is determined by thetwo-dimensional version of (6.1.1) using the compact Green’s function (6.1.5),viz,

p(x, t) ≈ −ρ0x j

2π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

dτ√t − τ − |x|/c0

×∫

(ω ∧ v · ∇Y j )(y, τ ) dy1 dy2, |x| → ∞. (6.2.1)

According to the inviscid form of the Formula (4.4.4) applied to a stationarybody, this can also be written

p(x, t) ≈ x j

2π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

Fj (τ ) dτ√t − τ − |x|/c0

, |x| → ∞, (6.2.2)

where Fj is the force per unit length of the cylinder exerted on the fluid in thej direction.

In these two-dimensional problems the vorticity ω is directed along the x3

axis, out of the plane of the paper in Fig. 6.2.1 and parallel to the generators ofthe cylinder. Let k be a unit vector in this direction and consider a line vortexof strength whose position and translational velocity are

x = x0(t), v = dx0

dt(t).

If the motion elsewhere is irrotational, we have

ω = kδ(x − x0(t))


Fig. 6.2.1.

so that

ω ∧ v = k ∧ vδ(x − x0(t)) ≡ k ∧ dx0

dt(t)δ(x − x0(t)), (6.2.3)

and (6.2.1) yields the following general formula for the acoustic pressure as|x| → ∞:

p(x, t)

≈ −ρ0x j

2π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

(k ∧ dx0

dτ(τ ) · ∇Y j (x0(τ ))

)dτ√

t − τ − |x|/c0

= −ρ0x j

2π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

dx01

dτ

∂Y j

∂y2− dx02

dτ

∂Y j

∂y1

x0(τ )

dτ√t − τ − |x|/c0

.

(6.2.4)

6.2.1 Example 1: Sound Produced by Vortex Motion neara Circular Cylinder

Assume there is no mean flow, and let the vortex strength be sufficientlysmall for the local motion to be considered incompressible. Then, the vortexwill traverse the circular orbit discussed in Section 4.6 (Example 2).

Let the cylinder have radius a, the vortex path have radius r0, and take thecoordinate origin at the centre of the cylinder. If > 0 the vortex moves in the


Fig. 6.2.2.

clockwise direction in Fig. 6.2.2 at the speed given by (4.6.8) when r is replacedby r0. Then,

v = dx0

dt(t) = !k ∧ x0(t),

! = −a2

2πr20

(r2

0 − a2) ,

x0 = r0(cos!t, sin!t),

and

k ∧ dx0

dt= !k ∧ (k ∧ x0) = −!x0.

Therefore, (6.2.4) becomes

p(x, t) ≈ ρ0!x jr0

2π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

∂Y j

∂r(x0(τ ))

dτ√t − τ − |x|/c0

, (6.2.5)

where r =√

y21 + y2

2 is the radial distance from the cylinder axis.Table 3.9.1 supplies the components of the two-dimensional Kirchhoff vector

for the cylinder:

Y1 = cosϑ

(r + a2

r

), Y2 = sinϑ

(r + a2

r

),


where (y1, y2) = r (cosϑ, sinϑ). Thus, by introducing polar coordinates for x

x = |x|(cos θ, sin θ )

we find x j∂Y j

∂r= |x|

(1 − a2

r2

)cos(θ − ϑ)

so that

x j∂Y j

∂r(x0(τ )) = |x|

(1 − a2

r20

)cos(θ − !τ ),

and (6.2.5) becomes

p(x, t) ≈ ρ0!r0

2π (2c0|x|) 12

(1 − a2

r20

)∂

∂t

∫ t−|x|/c0

−∞

cos(θ − !τ ) dτ√t − τ − |x|/c0

.

Hence, using the formula

∫ t−|x|/c0

−∞

cos(θ − !τ ) dτ√t − τ − |x|/c0

=(

π

|!|) 1

2

cos

θ − !

[t − |x|

c0

]− π

4

the pressure becomes

p(x, t) ≈ ρ0|!| 32 r0

2(2πc0|x|) 12

(1 − a2

r20

)sin

θ − !

[t − |x|

c0

]− π

4

= ρ0U 2√

M

√πr0

2|x|(r0

a

)2(

1 − a2

r20

)2

sin

θ − !

[t − |x|

c0

]− π

4

,

|x| → ∞, (6.2.6)

where U = |!|r0 is the vortex speed.The acoustic waves decay like 1/

√|x| with distance, which is appropriatefor energy spreading two dimensionally in cylindrically diverging waves, andhave the characteristic dipole amplitude proportional to ρ0U 2

√M . In three

dimensions the pressure would be proportional to ρ0U 2 M , which is smaller bya factor

√M when M 1. The increased amplitude in two dimensions is a

consequence of the infinite extent of the vortex source parallel to the cylinder.At any particular retarded time t −|x|/c0, the acoustic amplitude has the double-lobed directivity pattern illustrated in Fig. 1.7.1b for a dipole. Because !< 0,the peaks of these lobes at a fixed distance |x| from the cylinder rotate in theclockwise direction at angular velocity |!| following the orbiting vortex, butwith a phase lag of π/4 radians. The reader can confirm that the instantaneousforce exerted on the fluid by the cylinder lies in the direction of the vector x0(t)


joining the center of the cylinder to the vortex. The radiation peak thereforealso lags by π/4 the peak in the retarded surface force.

6.2.2 Example 2: Sound Produced by Vortex Motion near a Half-Plane(Crighton 1972)

The trajectory of a line vortex of strength interacting with a rigid half-plane isshown in Fig. 4.6.2 for ideal motion at low Mach number. The sound generatedby the vortex is calculated using the two-dimensional compact Green’s function(3.9.9), which can be written

G1(x, y, t − τ ) ≈ sin(θ/2)ϕ∗(y)

π√|x| δ

(t − τ − |x|

c0

), |x| → ∞,

ϕ∗(y) = √r0 sin(θ0/2), (6.2.7)

where x = |x|(cos θ, sin θ ), the coordinates being defined as in Fig. 4.6.2. Thisis applicable when the distance r0 from the edge of the source at (y1, y2) =r0(cos θ0, sin θ0) is much smaller than the acoustic wavelength. For a line vortexat (r0, θ0) the characteristic frequency ∼v/r0, where v is the vortex translationalvelocity. The wavelength is therefore of order

r0

v× c0 = r0

M r0 for M = v

c0 1,

so that low Mach number motion is sufficient to ensure that the wavelength ofthe sound is much larger than the vortex distance from the edge.

Thus, adopting the notation of (6.2.3) and applying (6.1.1) in two dimensions(Green’s function being given by (6.2.7)), we find

p(x, t) ≈ −ρ0 sin(θ/2)

π√|x|

∫k ∧ dx0

dτ(τ ) · ∇ϕ∗(y)δ(y − x0(τ ))

× δ

(t − τ − |x|

c0

)dy1 dy2 dτ

= −ρ0 sin(θ/2)

π√|x|

[k ∧ dx0

dt· ∇ϕ∗

], |x| → ∞, (6.2.8)

where on the second line the term in the square brackets is evaluated at theretarded position x0(t − |x|/c0) of the vortex.

Now ϕ∗ is the velocity potential of an ideal flow around the edge of thehalf-plane in the anticlockwise sense (with streamlines as in Fig. 6.2.3). It is


Fig. 6.2.3.

the real part of the complex potential

w = ϕ∗ + iψ∗ = −i√

z, z = y1 + iy2,

where ϕ∗ and the stream function ψ∗ satisfy the Cauchy–Riemann equations(see Section 4.5)

∂ϕ∗

∂y1= ∂ψ∗

∂y2,

∂ϕ∗

∂y2= −∂ψ∗

∂y1.

A simple calculation shows that

k ∧ ∇ϕ∗ = ∇ψ∗,

and therefore that

k ∧ dx0

dt· ∇ϕ∗ = −dx0

dt∧ k · ∇ϕ∗ = −dx0

dt· k ∧ ∇ϕ∗ = −dx0

dt· ∇ψ∗.

The acoustic pressure can therefore be put in the form

p(x, t) ≈ ρ0 sin(θ/2)

π√|x|

[dx0

dt· ∇ψ∗

]≡ ρ0 sin(θ/2)

π√|x|

[Dψ∗

Dt

], |x| → ∞,

(6.2.9)

where [Dψ∗/Dt] is evaluated at the retarded position of the vortex.

6.3 Influence of Vortex Shedding 145

Fig. 6.2.4.

The stream function ψ∗ is constant on each of the parabolic streamlines ofthe ideal flow around the edge defined by ϕ∗ (Fig. 6.2.3). A vortex that trans-lated along one of these streamlines would be silent; the actual edge-generatedsound depends on the rate at which the trajectory of the vortex cuts across thestreamlines of this ideal edge flow. This is found as follows

ψ∗ = −√r0 cos

(θ0

2

).

Therefore,

Dψ∗

Dt= − 1

2√

r0

dr0

dtcos

(θ0

2

)+

√r0

2sin

(θ0

2

)dθ0

dt,

where r0, θ0 are given by (4.6.7) (in which r is replaced by r0 and θ by θ0).Performing the calculations we find (see Fig. 6.2.4)

p(x, t) ≈ ρ02

(4π)2

(

|x|) 1

2

sin

(θ

2

)[t/8π2

[1 + (t/8π2)2]5/4

]t−|x|/c0

, |x| → ∞,

where is the distance of closest approach of the vortex to the edge (where itcrosses the x1 axis at time t = 0 in Fig. 4.6.2).

6.3 Influence of Vortex Shedding

We can use Equation (6.2.9) to form a qualitative picture of the influence ofvortex shedding on sound generation. Let the circulation of the vortex in


Fig. 6.2.3 be in the indicated anticlockwise sense, so that fluid near the edgeis induced to flow in a clockwise direction around the edge, as implied bythe dashed curve in the figure. When the Reynolds number is large, inertialforces actually cause the flow to separate from the edge, resulting in the releaseof vorticity of opposite sign from the edge into the wake. Let us assume forsimplicity that this shed vorticity rolls up into a concentrated core of strengths. Equation (6.2.9) then supplies the following prediction for the net acousticpressure

p(x, t) ≈ ρ0 sin(θ/2)

π√|x|

(

[Dψ∗

Dt

]

+ s

[Dψ∗

Dt

]s

), |x| → ∞,

where the derivatives are evaluated at the retarded positions of and s respec-tively. Both vortices translate across the curves ψ∗ = constant in the directionof decreasing ψ∗, and the derivatives therefore have the same sign. Hence, be-cause and s have opposite signs, sound produced by the shed vortex willtend to cancel the edge-generated sound attributable to the incident vortex

alone.This conclusion is applicable to a wide range of fluid–structure interactions. A

typical interaction involves a bounded region of vorticity, called a ‘gust,’ sweptalong in a nominally steady mean flow. The localized velocity field of the gustis determined by the Biot–Savart formula (4.3.1). At high mean flow speedsit is sometimes permissible to neglect changes in the relative configurationof the vorticity distribution during its convection past an observation point,the vorticity is then said to be frozen (at least temporarily) and the inducedvelocity is steady in a frame translating with the gust. If the mean flow carriesthe gust past the surface S of a stationary body, the free field induced velocitydetermined by the Biot–Savart integral is said to produce an upwash velocity onS; the actual velocity consists of the upwash velocity augmented by the velocityfield required to satisfy the no-slip condition on S. (In an ideal fluid only thenormal component of the upwash velocity is cancelled on S.)

When a gust convects past a stationary airfoil the high Reynolds numbersurface force (responsible for the sound) is given by (see (4.4.4))

Fi = −ρ0

∫V

∇Yi (y) · (ω ∧ v)(y, t) d3y, Yi = yi − ϕ∗i (y). (6.3.1)

The vector ∇Yi is the velocity of an ideal flow past the airfoil that has unitspeed in the i direction at large distances from the airfoil. It is singular (or verylarge) at the edges of the airfoil. These singularities have the following signifi-cance, when the vorticity length scale is small compared to the airfoil chord the


principal contribution to the integral is from vorticity in the neighborhoods ofthe singularities. For example, for the strip airfoil of Fig. 3.6.3

Y2 = Re(−i√

z2 − a2), z = y1 + iy2,

and ∇Y2 becomes infinite at the leading and trailing edges z = ∓a. An incident,small-scale gust convecting in the y1 direction in a mean flow at speed U wouldin practice induce shedding from the trailing edge at z = a. When this sheddingis ignored the force calculated from (6.3.1) has two principal components,respectively from gust elements near the leading and trailing edges. To calculatethe overall force, however, it is necessary to include the contribution from theshed vorticity, which affects the motion only near the trailing edge when thelength scale of the wake vorticity is small. In the linearized treatment of thiscase (discussed in more detail below), when both the gust and wake vorticityare taken to convect at the same mean velocity U , it is known from unsteadyaerodynamics that the force component produced by the wake is equal andopposite to that generated by the gust at the trailing edge (Sears 1941).

The effect of this cancellation can be approximated without calculatingany details of the shed vorticity. This is accomplished by formally deletingthe trailing edge singularity from ∇Y2, and then ignoring the contribution to theintegral (6.3.1) from the shed vorticity. To understand this observe that, becausethe value of the integral is dominated by vorticity near the edges, it is only thebehaviors of Y2 near these edges that must be retained in the integrand, and Y2

can therefore be replaced by the leading order terms in its expansions about theedges. For the strip airfoil of Fig. 3.6.3, we would write

Y2 = Re(−i√

z − a√

z + a) ∼ Re(√

2a√

z + a) + Re(−i√

2a√

z − a).(6.3.2)

The last term is singular at the trailing edge; it is deleted and the followingapproximation is used in (6.3.1) with the wake vorticity ignored:

Y2 ∼ Re(√

2a√

z + a), (6.3.3)

where the branch cut for√

z + a is taken along the real axis from z = −a toz = +∞.

6.3.1 Example: Surface Force Produced by a Periodic Gust

To illustrate the procedure consider incompressible flow parallel to the airfoilof Fig. 6.3.1 at speed U in the x1 direction, in which a time harmonic vortex


Fig. 6.3.1.

sheet of vorticity

ωI = γkδ(x2 − h)e−iω(t−x1/U ), h > 0, ω> 0,

is convected past the airfoil at perpendicular distance h, where γ is the cir-culation per unit length of the sheet. The vortex sheet can be regarded as anelementary model of a periodic wake behind a small diameter circular cylin-der upstream of the airfoil. The gust upwash velocity induces the shedding ofvorticity ωS from the trailing edge of the airfoil. When the reduced frequencyωa/U is large the hydrodynamic wavelength 2πU/ω of the gust and wake ismuch smaller than the airfoil chord, and the surface force is produced primarilyby the gust interaction with the leading edge at x1 = −a.

The net force F2 (per unit span) on the fluid in the x2 direction can thereforebe calculated from (6.3.1) by setting ω = ωI, where

ωI ∧ v = γU jδ(x2 − h)e−iω(t−x1/U ),

(j being a unit vector in the x2 direction) and by replacing Y2 by the right-handside of (6.3.3):

F2 = −ρ0γU∫ ∞

−∞

∂Y2

∂y2(y)δ(y2 − h)e−iω(t−y1/U ) dy1 dy2

≈ −ρ0γU√

2a∫ ∞

−∞Re

(i

2√

y1 + ih + a

)e−iω(t−y1/U ) dy1

= −ρ0γUa

(πU

2iωa

) 12

e−ωh/U−iω(t+a/U ). (6.3.4)

The force can also be calculated exactly from linearized thin airfoil theorywith proper account taken of vortex shedding. This is the gust loading problem


of classical aerodynamics (Sears 1941). The linear theory wake is treated asa vortex sheet downstream of the edge, whose elements convect at the meanstream velocity U. The strength of the vortex sheet is determined by imposingthe Kutta condition that the pressure (and velocity) should be finite at the edge(Crighton 1985). For arbitrary values of the reduced frequency ωa/U it is foundthat

F2 = π iρ0γUaS(ωa

U

)e−ωh/U−iωt , (6.3.5)

where S(x) is the Sears function, which is defined in terms of the Hankelfunctions H (1)

0 and H (1)1 by

S(x) = 2

πx[H (1)

0 (x) + i H (1)1 (x)

] . (6.3.6)

In the limit of high reduced frequency

S(ωa

U

)∼(

iU

2πωa

) 12

e−iωa/U ,

and the substitution of this into (6.3.5) yields the prediction (6.3.4) determinedby the leading edge singularity of Y2. The plots in Fig. 6.3.2 of the real andimaginary parts of S(ωa/U ) and its asymptotic limit show that the approx-imation (6.3.4) and the exact value (6.3.5) of the surface force agree whenωa/U > 1.

Fig. 6.3.2.


Linear theory does not permit the corresponding force component at theleading edge to be removed by vorticity production at the edge, because it re-quires vorticity shed there to be swept over the rigid surface of the airfoil onwhich it cannot influence the force because ω ∧ v · ∇Y2 ≡ 0.

6.4 Blade–Vortex Interaction Noise in Two Dimensions

A rigid, two-dimensional airfoil of chord 2a occupies −a < x1 < a, x2 = 0 inthe presence of a uniform mean stream at speed U in the positive x1 direction.A line vortex of strength parallel to the airfoil span convects at the meanflow velocity at a constant distance h above the airfoil (Fig. 6.4.1). This is theapproximation of linearized thin airfoil theory, which is applicable when

h U,

that is, when the influence on the vortex trajectory of the induced velocity(∼/h) of image vortices in the body of the airfoil is negligible. As the vortexpasses the airfoil new vorticity is shed from the trailing edge into the airfoilwake, which is assumed to consist of a vortex sheet stretching along the x1 axisfrom x1 = a to x1 = +∞.

Let us first consider the potential flow interaction of the vortex and airfoil,when no account is taken of vortex shedding. Suppose the vortex passes abovethe midchord of the airfoil at time t = 0, then

ω = kδ(x1 − Ut)δ(x2 − h), v = U i.

Hence,

ω ∧ v = U jδ(x1 − Ut)δ(x2 − h), (6.4.1)

where i and j are unit vectors in the x1 and x2 directions. The acoustic pressure

Fig. 6.4.1.

6.4 Blade–Vortex Interaction Noise in Two Dimensions 151

generated when the wake is ignored is given by (6.2.1) with

Y1 = y1, Y2 = Re (−i√

z2 − a2), z = y1 + iy2.

Thus, ∇Y1 = i and ω ∧ v · ∇Y1 ≡ 0, and (6.2.1) reduces to

p(x, t) ≈ −ρ0U x2

2π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

∂Y2

∂y2(Uτ, h)

dτ√t − τ − |x|/c0

,

= −ρ0U cos

2π√

2c0|x| 12

∂

∂t

∫ t−|x|/c0

−∞

∂Y2

∂y2(Uτ, h)

dτ√t − τ − |x|/c0

,

|x| → ∞, (6.4.2)

where = cos−1(x2/|x|) is the angle between the radiation direction x andthe normal to the airfoil (the x2 axis). The sound can be attributed to a dipolesource orientated in the x2 direction.

The radiation produced when a vortex passes very close to the airfoil (so thath a) is likely to be particularly intense. The dominant interactions occur asthe vortex passes the edges, where the time scales of the motions ∼h/U . Thus,the characteristic frequency

ω ∼ U

hand the reduced frequency

ωa

U∼ a

h 1.

We may therefore regard the leading and trailing edges as independent sourcesof sound, and calculate their individual contributions by using the local approx-imation (6.3.2). For the acoustic pressure pLE, say, produced at the leading edge(x1 = −a) we take

Y2 ∼ Re (√

2a√

z + a),

so that (6.4.2) becomes

pLE ≈ −ρ0U cos

2π√

2c0|x| 12

∂

∂t

∫ t−|x|/c0

−∞Re

(i√

2a

2(Uτ + ih + a)12

)

× dτ√t − τ − |x|/c0

, |x| → ∞.

To evaluate the integral, make the substitution µ= √t − τ − |x|/c0 and


perform the differentiation with respect to time. Then,

pLE ≈ −ρ0U cos

2π√

c0

(a

|x|) 1

2 ∂

∂t

∫ ∞

0Re

[i

(U [t] + ih + a − Uµ2)12

]dµ

= ρ0U 2 cos

4π√

c0

(a

|x|) 1

2∫ ∞

0Re

[i

(U [t] + ih + a − Uµ2)32

]dµ,

where [t] = t − |x|c0

.

The additional substitution µ= 1/ξ transforms the integrand into an exact dif-ferential, leading finally to

pLE ≈ ρ0U 2 cos

4π√

c0

(a

|x|) 1

2∫ ∞

0Re

[iξ

[(U [t] + ih + a)ξ 2 − U ]32

]dξ

= ρ0U√

M cos

4πa

(a

|x|) 1

2(U [t]

a + 1)

(U [t]a + 1

)2 + (ha

)2, |x| → ∞, (6.4.3)

where M = U/c0.The corresponding nondimensional pressure signature

pLE

/ρ0U

√M cos

4πa

(a

|x|) 1

2

is plotted in Fig. 6.4.2 as the solid curve when h/a = 0.2. The pressure field

Fig. 6.4.2.

6.4 Blade–Vortex Interaction Noise in Two Dimensions 153

is generated predominantly as the vortex passes the leading edge of the air-foil at the retarded time [t] = − a/U , with characteristic frequency ω∼ U/h.According to Section 6.3, at high reduced frequencies the sound pressure pTE,say, generated by the potential flow interaction of the vortex with the trailingedge is cancelled by that produced by the wake vorticity; the solid curve inFig. 6.4.2 is therefore representative of the whole of the radiation produced bythe blade–vortex interaction.

To determine pTE the calculation described above for pLE is repeated aftersetting

Y2 = Re(−i√

2a√

z − a)

in (6.4.2), leading to

pTE ≈ −ρ0U cos

4π√

c0

(a

|x|) 1

2 ∂

∂t

∫ t−|x|/c0

−∞Re

(1

(Uτ + ih − a)12

)

× dτ√t − τ − |x|/c0

= −ρ0U√

M cos

4πa

(a

|x|) 1

2(

ha

)(U [t]

a − 1)2 + (

ha

)2, |x| → ∞. (6.4.4)

This is large at the retarded times during which the vortex is close to the trailingedge. Thus, when the contribution from the wake is ignored (which is equaland opposite to pTE) the overall acoustic pressure signature is given nondimen-sionally by

(pLE + pTE)

/ρ0U

√M cos

4πa

(a

|x|) 1

2

,

which is plotted as the broken line curve in Fig. 6.4.2.The leading and trailing edge generated components pLE and pTE have dif-

ferent waveforms, even though they are produced by the vortex interacting withgeometrically identical airfoil edges. This is because the integral in (6.4.2) de-termines the acoustic pressure at the retarded time [t] in terms of interactionsbetween the vortex and the airfoil at all earlier retarded times; it is a furtherconsequence of the two-dimensional character of the acoustic sources, accord-ing to which, after the first arrival of sound from the nearest point on the source,additional contributions to the sound continue to be received indefinitely in timefrom more distant parts of the source.


Problems 6

1. Starting from the formula

G ≈ x · Y2πc0

∂

∂t

∫ ∞

0δ

(t − τ −

√|x|2 + ξ 2

c0

)dξ

(|x|2 + ξ 2),

derive the far-field approximation (6.1.5) for the dipole component of thetwo-dimensional compact Green’s function by writing

δ

(t − τ −

√|x|2 + ξ 2

c0

)= 1

2π

∫ ∞

−∞e−iω(t−τ−

√|x|2+ξ2

c0) dω,

and applying Formula (5.2.11).2. Investigate the production of sound by the low Mach number motion of a

line vortex of strength that is parallel to a rigid circular cylinder of radius awhose axis coincides with the x3 coordinate axis (c.f., Section 4.6, Example2). Assume that there is no net circulation around the cylinder, and that thereis a mean flow past the cylinder which has speed U in the x1 direction when|x1| a. If the vortex is initially far upstream of the cylinder at a distanceh from the x1 axis, examine the production of sound for different values ofthe nondimensional parameter /Uh.

3. A line vortex of strength is parallel to a rigid airfoil occupying−a < x1 < a,x2 = 0,−∞< x3 <∞, in the presence of a mean flow at speed U in thex1 direction. The vortex is initially far upstream of the airfoil at a verticalstand-off distance h above the plane of the airfoil. There is no net circulationaround the airfoil. If the motion occurs at very small Mach number, calculatethe sound produced when the vortex passes the airfoil for different valuesof the nondimensional velocity ratio /Uh. When U /h, estimate theinfluence on the sound of vortex shedding from the trailing edge of the airfoil.

4. A line vortex of strength is parallel to a rigid airfoil occupying

−a < x1 < a, x2 = 0, −∞< x3 <∞

in fluid at rest at infinity. The vortex is in periodic motion around the airfoilunder the influence of image vortices in the absence of a mean circulationaround the airfoil. Calculate the sound produced when the motion occurs ata very small Mach number, and show that it can be attributed to two dipolesources orientated in the x1 and x2 directions. Explain the significance ofthese sources in terms of the corresponding components of the unsteadyforce between the fluid and airfoil.

Problems 6 155

5. A line vortex of strength traverses a path of the kind illustrated in thefigure past a two-dimensional, thin rigid barrier of length d at right anglesto a plane wall at x2 = 0. There is a low Mach number mean potential flowover the barrier that has speed U parallel to the wall at large distances fromthe barrier. If the distance of the vortex from the wall is h when the vortexis far upstream of the barrier, calculate the sound produced as the vortexpasses the barrier for different values of /Uh. Explain what happens when/h U . Discuss the forces exerted on the barrier by the flow, and howthey contribute to the radiation.

7Problems in Three Dimensions

7.1 Linear Theory of Vortex–Airfoil Interaction Noise

Consider an inhomogeneous field of vorticity, a gust, convected in high Reynoldsnumber, homentropic flow past a stationary rigid airfoil (Fig. 7.1.1). The undis-turbed flow has speed U in the x1 direction, where the origin is at a convenientpoint within the airfoil, with x3 along the span and x2 vertically upward. TheMach number M = U/c0 is sufficiently small that convection of sound by theflow can be ignored, and the airfoil chord can be assumed to be acousticallycompact.

The vortex sound source div(ω∧v) includes vorticity in the gust together withany vorticity shed from the airfoil, either in response to excitation by the gust,or as tip vortices responsible for the mean lift. The problem can be linearized byassuming that u U , where curl u = ω, that is, by requiring the gust-inducedvelocity, and the perturbation velocities caused by airfoil thickness, twist, cam-ber, and the angle of attack, to be small. When div(ω ∧ v) is expanded aboutthe undisturbed mean flow, only the gust vorticity and additional vorticity shedwhen the gust encounters the airfoil contribute to the acoustic radiation to firstorder. In other words, thickness, twist, camber, and angle of attack may ignored,and the airfoil regarded as a rigid lamina in the plane x2 = 0. In this approx-imation, quadrupoles are neglected and vorticity convects as a frozen patternof vortex filaments at the undisturbed mean stream velocity U = (U, 0, 0). Inparticular, the wake vorticity is confined to a vortex sheet downstream of thetrailing edge.

When convection of sound by the flow is neglected, the linearized form ofthe vortex sound equation (5.2.5) becomes

(1

c20

∂2

∂t2− ∇2

)B = div(ω ∧ U), (7.1.1)

156

7.1 Linear Theory of Vortex–Airfoil Interaction Noise 157

Fig. 7.1.1.

with solution

B(x, t) = p(x, t)

ρ0≈ −

∫(ω ∧ U)(y, τ ) · ∂G

∂y(x, y, t − τ ) d3y dτ , |x| → ∞,

(7.1.2)

where ∂G/∂y2 = 0 on both sides (y2 = ±0) of the projection of the airfoil plan-form onto the y1, y3 plane. At sufficiently small Mach numbers G may beapproximated by Green’s function for an airfoil of compact chord.

The sound produced when a localized, high Reynolds number frozen gustω(x−Ut) ≡ω(x1 −Ut, x2, x3) is swept past the airfoil of Fig. 7.1.1 is thereforegiven by (5.4.4) with the convection velocity v replaced by U, and the Kirchhoffvector Y by

Y1 = y1, Y2 = y2 − ϕ∗2 (y), Y3 = y3.

Thus, because∫ω d3y ≡ 0, Equation (5.4.4) reduces to

p(x, t) ≈ −ρ0U cos

4πc0|x|∂

∂t

∫ [ω3

∂Y2

∂y2− ω2

∂Y2

∂y3

]t−|x|/c0

d3y, |x| → ∞,

(7.1.3)

where = cos−1(x2/|x|) is the angle between the radiation direction and thenormal to the airfoil, and the origin is taken in the airfoil within the interactionregion.

Let the interaction occur at an inboard location where the chord may beregarded as constant, with both the leading and trailing edges at right angles tothe mean flow (so that ∂Y2/∂y3 ∂Y2/∂y2). The planform in the interaction

158 7 Problems in Three Dimensions

Fig. 7.2.1.

region is then locally the same as that of the two-dimensional airfoil of Fig. 7.2.1,and Y2 can be approximated as in (3.9.2) for constant a ≡ a(y3), such that 2ais equal to the local chord of the airfoil. Then, (7.1.3) becomes


4πc0|x|∂

∂t

∫ [ω3

∂Y2

∂y2

]t−|x|/c0

d3y, |x| → ∞, (7.1.4)

which reveals that only the spanwise component of vorticity contributes to theproduction of sound.

According to Equation (5.4.3) (in which dU j/dt = 0 for a stationary airfoil),this result can also be expressed in the form

p(x, t) ≈ cos

4πc0|x|∂F2

∂t

(t − |x|

c0

), |x| → ∞,

F2(t) = −ρ0U∫

ω3(y, t)∂Y2

∂y2(y) d3y, (7.2.5)

where −F2 is the unsteady airfoil lift during the interaction when the motionis regarded as incompressible, and the vorticity ω3 includes contributions fromthe impinging gust together with any shed into the vortex sheet wake.

7.2 Blade–Vortex Interactions in Three Dimensions

The calculations can be performed explicitly for a gust in the form of a rectilinearline vortex. Let the vortex have circulation and be orientated with its axis inthe direction of the unit vector n, as indicated in Fig. 7.2.1. The mean flow speedis sufficiently large that the vortex can be assumed to maintain its rectilinearform after being cut by the leading edge of the airfoil. Choose the origin on theairfoil midchord such that the axis of the vortex passes through the origin attime t = 0. Any point x on the vortex can then be represented in the parametric

7.2 Blade–Vortex Interactions in Three Dimensions 159

form

x = (Ut, 0, 0) + sn, −∞< s <∞, (7.2.1)

where s is distance measured along the vortex from its point of intersectionwith the plane of the airfoil. Then, if s⊥ denotes vector distance measured inthe normal direction from the vortex axis,

ω = nδ(s⊥),

where the polar angles θ, φ in Figure 7.2.1 define the orientation of the unitvector

n = (sin θ cosφ, sin θ sinφ, cos θ), 0<θ <π, 0<φ < 2π.

Because of vortex shedding from the trailing edge, most of the sound isgenerated when the vortex is cut by the leading edge. The influence of the shedvorticity can be formally included by the procedure described in Sections 6.3and 6.4 by expanding Y2 about its singularity at the leading edge, that is, bysetting

Y2 ∼ Re(√

2a√

z + a), z = y1 + iy2,

with the branch cut for the square root taken along the z axis from z = −a toz = +∞. Thus, by recalling the Relation (7.2.1), Equation (7.1.4) becomes

p(x, t) ≈ −n3ρ0U cos

4πc0|x|∂

∂tRe

∫ [δ(s⊥)

(i√

2a

2√

y1+ iy2 + a

)]t−|x|/c0

d2s⊥ ds

= −n3ρ0U cos

4√

2πc0|x|∂

∂tRe

∫ ∞

−∞

(i√

a√U [t] + s(n1 + in2) + a

)ds

= n3ρ0U 2 cos

8√

2πc0|x| Re∫ ∞

−∞

(i√

a

(U [t] + s(n1 + in2) + a)32

)ds

= −n3ρ0U 2 cos

4√

2πc0|x| Re

[i√

a

(U [t] + s(n1 + in2) + a)12 (n1 + in2)

]∞

−∞,

(7.2.2)

where [t] = t − |x|/c0 is the retarded time.By referring to Fig. 7.2.2 it will be seen that the last line of (7.2.2) is zero

when

U [t] + a < 0,


Fig. 7.2.2.

before the vortex is cut by the leading edge of the airfoil. At later times theintegration along the vortex axis over the infinite range −∞< s <∞ must besplit, as indicated in the figure, into the two parts −∞< s <−0,+0< s <∞,because the square root in the last line of (7.2.2) is discontinuous across theairfoil; for example, when n2 > 0 the square root is real and positive on theupper surface (s = +0) and real and negative at s = −0. Hence, the acousticpressure becomes

p(x, t) ≈ ρ0U M cos

232 π |x|

|sinφ|tan θ

H(U [t]

a + 1)

√U [t]

a + 1. (7.2.3)

The pressure pulse accordingly begins with a singular peak at the instant atwhich the vortex is severed by the leading edge of the airfoil at the retarded timeU [t]/a = −1. The waveform is illustrated by the dotted curve in Fig. 7.2.3,which is a plot of

p(x, t)

(ρ0U M cos/|x|) = |sinφ|2

32 π tan θ

H(U [t]

a + 1)

√U [t]

a + 1for θ = 85, φ = 90.

The infinite singularity in the pressure is absent for a vortex of nonzerocore radius R, say. If, for example, the vorticity is assumed to be distributedaccording to the Gaussian formula

ω(x) = ne−(s⊥/R)2

π R2,

as a function of distance s⊥ from the vortex axis, the acoustic pressure is found

7.2 Blade–Vortex Interactions in Three Dimensions 161

Fig. 7.2.3.

to be given by (Howe 1998a)

p(x, t)

(ρ0U M cos/|x|) ≈ | sinφ| 32

8 tan θ

( a

π R

) 12 (α), |x| → ∞, for R a,

(7.2.4)

where

(α) = |α| 12

I− 1

4

(α2

8

)+ sgn(α)I 1

4

(α2

8

)e−α2/8, α = 2a| sinφ|

R(U [t]

a + 1) ,

and I± 14

are modified Bessel functions of the first kind. The pressure signaturepredicted by (7.2.4) for

R = 0.1a, θ = 85, φ = 90

is plotted as the solid curve in Fig. 7.2.3, and differs negligibly from the linevortex prediction when U [t]/a >−1.

The broken-line curve in Fig. 7.2.3 represents the pressure signature producedby the potential flow interaction between the finite core vortex and airfoil (i.e.,when vortex shedding is ignored). It is an odd function of the retarded time[t]. The large negative peak produced as the vortex crosses the trailing edge (atU [t]/a = 1) is cancelled by an equal and opposite contribution generated by thewake. The reader can easily show that, when the finite size of the vortex core is


ignored, the potential flow, trailing edge generated pressure pulse is given by

p(x, t) ≈ −ρ0U M cos

232 π |x|

|sinφ|tan θ

H(1 − U [t]

a

)√

1 − U [t]a

.

7.3 Sound Produced by Vortex Motion near a Sphere

The sound generated when a nominally rectilinear vortex is swept past a com-pact rigid body can also be treated in a linearized fashion, by assuming thateach element of the vortex core is convected along a streamline of the steadyundisturbed mean flow at the local mean velocity. It is not generally possible,however, to include the influence of vortex shedding in a satisfactory manner,except perhaps for streamlined body shapes that are amenable to treatment bythe strip theory of unsteady aerodynamics.

To illustrate the procedure, consider a rigid sphere of radius a with center atthe coordinate origin in the presence of a low Mach number irrotational meanflow which is in the x1 direction at speed U for |x| a. The mean velocity atx can therefore be written

U = U∇ X1(x), (7.3.1)

where X1(x) is the x1 component of the Kirchhoff vector for the sphere, whichhas the general form (Table 3.9.1)

Xi = xi

(1 + a3

2|x|3). (7.3.2)

Suppose a line vortex of strength is initially far upstream of the sphereand parallel to the x3 axis at a distance h above the plane x2 = 0. The vortex isconvected toward the sphere by the mean flow. The part of the vortex that passesclose to the sphere must evidently be deformed to pass around the sphere; moredistant parts of the vortex (at |x3| a) are unaffected and remain parallel to thex3 direction during the whole of the interaction. The shape of the distorted vortexwill be symmetric with respect to the mid-plane x3 = 0; the vortex elementinitially on x3 = 0 remains on this plane of symmetry as it convects past thesphere along a mean streamline, as illustrated in Fig. 7.3.1, which shows themotion in the plane x3 = 0.

The shape of the vortex at time t is determined by the solution of the equations

dx1

dt= U

∂X1

∂x1(x),

dx2

dt= U

∂X1

∂x2(x),

dx3

dt= U

∂X1

∂x3(x),

7.3 Sound Produced by Vortex Motion near a Sphere 163

Fig. 7.3.1.

for each element of the vortex. If the undistorted parts of the vortex (at |x3| a)are assumed to convect across the plane x1 = 0 at time t = 0, these equationsare to be integrated subject to the initial conditions

x1 = Ut, x2 = h, x3 = x03 t → −∞,

where x03 is the initial spanwise location of the vortex element.

In terms of the nondimensional variables

T = Ut

a, x = x

a,

the equations of motion of a point on the vortex are found to be

dx1

dT= 1 + x2

2 + x23 − 2x2

1

2(x2

1 + x22 + x2

3

) 52

,dx2

dT= −3x1 x2

2(x2

1 + x22 + x2

3

) 52

,

dx3

dT= −3x1 x3

2(x2

1 + x22 + x2

3

) 52

.

These are solved (for example, by the Runge–Kutta method described inSection 4.6) by starting the integration at T = −10, say. It can be safely assumedthat the sphere has no perceptible influence on vortex elements initially locatedat |x3|> 10. Because the motion is symmetric about x3 = 0 the solutions arerequired only for the N +1 vortex elements with respective the initial positions

x1 = T , x2 = h

a, x3 = x n

3 at T = −10,

where x n3 = 10n/N , 0 ≤ n ≤ N , and N is a suitably large integer.


Fig. 7.3.2.

Figure 7.3.2 illustrates successive calculated positions of the vortex with in-creasing values of the time T = Ut/a for the case h/a = 0.2. The distortion ofthe vortex first becomes evident at about T = −2. The hairpin loop is formedbecause the translation velocities of vortex elements close to the sphere are smallin the neighborhood of the stagnation points just in front and just to the rear of thesphere; the accelerated motion over the upper surface of the sphere is insufficientto counteract the formation of the loop. In reality, of course, the motion wouldbe strongly influenced by large self-induced velocities, by image vorticity inthe sphere, and by viscous diffusion of vorticity from the vortex and from thesurface of the sphere, none of which is accounted for in the present calculation.

The sound generated during this potential flow interaction can be calculatedusing the Formula (5.4.4) when the sphere is compact:

p(x, t) ≈ −ρ0x j

4πc0|x|2∂

∂t

∫(ω ∧ v)

(y, t − |x|

c0

)· ∇Y j (y) d3y, |x| → ∞,

(7.3.3)

where

v = U∇Y1(y).

7.3 Sound Produced by Vortex Motion near a Sphere 165

There is no unsteady drag contribution to (7.3.3) from j = 1 because ω∧∇Y1 ·∇Y1 ≡ 0. Similarly, there can be no net side-force on the sphere because ofthe symmetric form of the vortex, and therefore there will be no contributionfrom j = 3. The sound is accordingly produced by a dipole source orientatedin the x2 direction; that is, the interaction produces an unsteady lift force in thisdirection which is responsible for the sound, which has the representation


4πc0|x|∂

∂t

∫(ω · ∇Y1 ∧ ∇Y2)

(y, t − |x|

c0

)d3y, |x| → ∞,

(7.3.4)

where = cos−1(x2/|x|) is the angle between the observer direction x and thex2 axis.

To evaluate the integral, write

ω = δ(s⊥)s, d3y = d2s⊥ ds,

where s is a unit vector locally parallel to ω, s⊥ is the vector distance measuredin the normal direction from the local axis of the vortex, and s is distancemeasured along the vortex in the direction of ω. The integral (7.3.4) can thenbe cast in the following nondimensional form, suitable for numerical evaluation,

p(x, t)

ρ0U M cos/4π |x| = − ∂

∂T

∫ ∞

−∞[s · ∇Y1 ∧ ∇Y2] ds, s = s

a, M = U

c0,

(7.3.5)

where the integrand is evaluated at the retarded position of the distorted vortex.Now the integral in (7.3.5) is divergent, because s · ∇Y1 ∧ ∇Y2 → 1 as

s → ±∞. The divergence is not real, however, but a consequence of the formaloperations used in the application of the compact Green’s function. The infinitecontributions to the integral from large values of s are equal at successiveretarded locations of the vortex, and disappear on differentiation with respectto T . The integral can therefore be evaluated numerically by restricting the rangeof integration to a finite interval, say, −10< s < 10, because the contributionsat larger values of s are the same for all retarded times, and give no contributionto the sound when differentiated.

Typical plots of the calculated nondimensional pressure (7.3.5) are shown inFig. 7.3.3 for two values of h/a; they illustrate how the sound level decreasesas the initial standoff distance h of the vortex increases relative to the radius aof the sphere.


Fig. 7.3.3.

7.4 Compression Wave Generated When a Train Enters a Tunnel

A train entering a tunnel pushes aside the stationary air, most of which flowsover the train and out of the tunnel portal, but the build-up of pressure just aheadof the train propagates into the tunnel as a compression wave at the speed ofsound. In a long tunnel the compression wavefront can experience nonlinearsteepening that is ultimately manifested as a loud, impulsive bang or ‘crack’(called a micro-pressure wave) radiating out of the distant tunnel exit. In additioninaudible low-frequency pressure fluctuations called infrasound (at frequencies∼10–20 Hz) are radiated from the tunnel portal into the open air when the trainenters and leaves the tunnel. All of these waves are indicated schematically inFig. 7.4.1. Their effects become pronounced when the train speed U exceeds

Fig. 7.4.1.

7.4 Compression Wave Generated When a Train Enters a Tunnel 167

Fig. 7.4.2.

about 200 km/h (125 mi/h), when in particular the micro-pressure wave and theinfrasound can cause vibrations and annoying structural rattles in neighboringbuildings.

The formation of the compression wave can be studied in terms of the simplerproblem involving an axisymmetric train entering axisymmetrically a semi-infinite, circular cylindrical duct of radius R and cross-sectional area A = π R2

(Fig. 7.4.2). Let the train be travelling at constant speed U in the negative x1

direction, where the origin O is at the center of the tunnel entrance plane, so thatthe x1 axis coincides with the axis of the tunnel. Denote the pressure, density,and speed of sound in the air respectively by p, ρ and c. They vary with positionand time within the tunnel, and their corresponding undisturbed values arep0, ρ0, c0.

The cross section of the train is assumed to become uniform with constantarea A0 =πh2 at a distance L from the nose of the train, where h is the uniformmaximum train radius. The aspect ratio h/L of the nose is taken to be sufficientlysmall, and the train profile sufficiently streamlined, to ensure that flow separationdoes not occur. In practice the Mach number M = U/c0 does not exceed 0.4,and the blockageA0/A≤ 0.2. If heat transfer and frictional losses are neglectedduring the initial stages of wave formation, the air flow may be regarded ashomentropic, and the compression wave can be calculated using the vortexsound equation (5.2.5)(

D

Dt

(1

c2

D

Dt

)− 1

ρ∇· (ρ∇)

)B = 1

ρdiv(ρω ∧ v). (7.4.1)

The air in the compression wave region ahead of the train may be regarded aslinearly perturbed from its mean state, with B ≈ p/ρ0 ≡ ( p − p0)/ρ0. In this


simple model the vorticity ω vanishes everywhere except within the outer shearlayer of the exit flow of the air displaced when the train enters the tunnel (seeFig. 7.4.2).

Let f ≡ f (x1 + Ut, x2, x3) = 0 be a control surface S contained within thefluid that just encloses the moving train, with f < 0 inside S (in the region oc-cupied by the train) and f > 0 outside. The surface is fixed relative to the train,and the influence of the train on its surroundings can be represented in termsof monopole and dipole sources on S. In the usual way, multiply (7.4.1) byH ≡ H ( f ) and rearrange (noting that DH/Dt = 0) to obtain

(D

Dt

(1

c2

D

Dt

)− 1

ρ∇ · (ρ∇)

)(H B)

= 1

ρdiv(Hρω ∧ v) − (∇ B + ω ∧ v) · ∇ H − 1

ρdiv(ρB∇ H ). (7.4.2)

This is a generalization of Equation (5.3.3). The two terms on the right-hand sideinvolving ∇ H respectively represent monopole and dipole sources distributedover the moving surface f (x1 + Ut, x2, x3) = 0.

When frictional losses are neglected Crocco’s equation (4.2.5) reduces to∂v/∂t = −∇ B − ω ∧ v, so that the source terms can be written

∂

∂t(U · ∇ H ) − v · ∇ ∂H

∂t− 1

ρdiv (ρB∇ H ) + 1

ρdiv(Hρω ∧ v),

where U = (−U, 0, 0). The compressibility of the air adjacent to S and withinthe very low Mach number exterior flow from the tunnel portal can be neglectedwhen M(A0/A)2 1 (Howe et al. 2000), and the source approximated furtherby

∂

∂t(U · ∇ H ) + div (v U · ∇ H ) − div

(p

ρ0+ 1

2v2

)∇ H

+ div(Hω ∧ v),

where the relation ∂H ( f )/∂t = −U · ∇ H ( f ) has been used.Thus, if the nonlinear terms on the left of (7.4.2) (which affect the propagation

of the compression wave) are also ignored, the equation finally reduces to

(1

c20

∂2

∂t2− ∇2

)(H B) = ∂

∂t(U · ∇ H ) + div (v U · ∇ H )

− div

(p

ρ0+ 1

2v2

)∇ H

+ div(Hω ∧ v). (7.4.3)


Fig. 7.4.3.

7.4.1 Linear Theory

When the blockageA0/A is small it is sufficient to retain only the first monopolesource on the right-hand side of Equation (7.4.3). This source can be simplifiedwhen the aspect ratio h/L 1 by introducing a slender body approximation.To do this, consider a system of cylindrical coordinates in which r =

√x2

2 + x23

is the perpendicular distance from the axis of the train. The control surfaceequation

f (x1 + Ut, x2, x3) = 0 can be written r = f (x1 + Ut),

where AT(s) = π f 2(s) is the cross-sectional area of the train at distance s fromthe nose, and the nose is assumed to cross the tunnel entrance plane (x1 = 0)at t = 0. As the train moves (to the left in Fig. 7.4.3) the rate at which air isdisplaced by a section of the train of length dx1 is

U2π f (x1 + Ut) d f (x1 + Ut) = U2π f (x1 + Ut)∂ f

∂x1(x1 + Ut) dx1

≡ U∂AT

∂x1(x1 + Ut) dx1.

Thus,

U∂AT

∂x1(x1 + Ut) = monopole source strength per unit length of the train.

When h/L 1, we can collapse the monopole source distribution over thesurface of the train into a line source concentrated on its axis, and approximatethe monopole on the right-hand side of Equation (7.4.3) by

∂

∂t(U · ∇ H ) (x, t) ≈ ∂

∂t

(U

∂AT

∂x1(x1 + Ut)δ(x2)δ(x3)

).


The source strength is proportional to the rate at which the train cross sectionchanges with distance along the train, and is nonzero only in the vicinity of thetrain nose (and also the tail).

The corresponding approximation of Equation (7.4.3) is therefore

(1

c20

∂2

∂t2− ∇2

)(H B) = ∂

∂t

(U

∂AT

∂x1(x1 + Ut)δ(x2)δ(x3)

), (7.4.4)

where B → p/ρ0 in the linear acoustic region ahead of the train.The monopole in this equation does not depend on time when viewed in a

reference frame moving at the uniform subsonic speed U of the train. The sourcetherefore creates only a nonacoustic near field when travelling within the tunnelor in free space far from the tunnel entrance (c.f., Question 4 of Problems 1 whenq0(t) = constant). The compression wave is produced when the near field of thesource interacts with the tunnel portal, as the train nose enters the tunnel. Thisoccurs over a time ∼R/U , so that the characteristic thickness of the wavefront∼R/M R. Equation (7.4.4) can therefore be solved by using the compactGreen’s function (3.9.13) for a duct entrance

G(x, y; t − τ ) ≈ c0

2A

H

(t − τ − |ϕ∗(x) − ϕ∗(y)|

c0

)

− H

(t − τ + ϕ∗(x) + ϕ∗(y)

c0

),

where ϕ∗(x) is the velocity potential of a uniform incompressible flow out ofthe tunnel portal that has unit speed far inside the tunnel (see (3.9.14)).

Thus, at x within the tunnel, ahead of the train where B = p/ρ0, we have

p ≡ p

(t + x1

c0

)≈ ρ0

∂

∂t

∫∫ ∞

−∞U

∂AT

∂y1(y1 + Uτ )G(x, y1, 0, 0; t − τ ) dy1 dτ

= ρ0Uc0

2A∫ ∞

−∞A′

T(y1 − Mϕ∗(y1, 0, 0) + U [t])

− A′T(y1 + Mϕ∗(y1, 0, 0) + U [t]) dy1, (7.4.5)

where the prime on AT denotes differentiation with respect to the argument,and [t] = t + (x1 − ′)/c0 is the effective retarded time. Because nonlinearpropagation terms have been ignored, this approximation determines the initialform of the compression wave profile, before the onset of nonlinear steepening.It is therefore applicable within the region several tunnel diameters ahead ofthe train, during and just after tunnel entry.


The main contributions to the Integral (7.4.5) are from the vicinities of thenose and tail of the train, where the cross-sectional area AT is changing. Thecompression wave is generated as the nose enters the tunnel, and may be cal-culated by temporarily considering a train of semi-infinite length. During theformation of the wave, and in the particular case in which the Mach number issmall enough that terms ∼O(M2) are negligible, the term Mϕ∗ in the argumentsof A′

T in (7.4.5) is small, and we then find, by expanding to first order in Mϕ∗

and integrating by parts, that

p ≈ ρ0U 2

A∫ ∞

−∞

∂AT

∂y1(y1 +U [t])

∂ϕ∗

∂y1(y1, 0, 0) dy1, M2 1. (7.4.6)

After the nose has passed into the tunnel, ∂ϕ∗/∂y1 = 1 in the region occupiedby the nose, and (7.4.6) predicts the overall (linear theory) pressure rise to bep ≈ ρ0U 2A0/A. But the linear theory, asymptotic pressure rise can also becalculated exactly, with no restriction on Mach number, to be

p = ρ0U 2A0

A(1 − M2),

because this is attained when ϕ∗(y1, 0, 0) ≈ y1 − ′ in (7.4.5) (see Equation(3.9.14)). This implies that the Approximation (7.4.6) can be extrapolated tofinite Mach numbers by writing

p ≈ ρ0U 2

A(1 − M2)

∫ ∞

−∞

∂AT

∂y1(y1 + U [t])

∂ϕ∗

∂y1(y1, 0, 0) dy1. (7.4.7)

This extrapolation of the linear theory to finite values of M turns out to beapplicable for M < 0.4 (Howe et al. 2000).

Figure 7.4.4 illustrates schematically an experimental arrangement used byMaeda et al. (1993) to investigate the compression wave. Wire-guided, axisym-metric model trains are projected into and along the axis of a tunnel consisting ofa 7-m long circular cylinder of internal diameter 0.147 m. The nose aspect ratioh/L = 0.2, the blockage A0/A = 0.116, and the projection speed U ≈ 230km/h (M ≈ 0.188). The train nose profiles include the cone, and the paraboloidand ellipsoid of revolution, with respective cross-sectional areas given by

AT(s)

A0=

s2

L2 ,sL ,

sL

(2 − s

L

), 0< s < L ,

1, s ≥ L .

The data points in the figure are measurements (made at the point labelled T)of the pressure gradient dp/dt 1 m from the entrance for these three different


Fig. 7.4.4.

nose profiles. The solid curves are predictions of Equation (7.4.7), evaluatedusing the following formulae for ∂ϕ∗/∂y1 (Howe 1998b):

∂ϕ∗

∂y1(y) = 1

2− 1

2π

∫ ∞

0I0

(ξr

R

)(2K1(ξ )

I1(ξ )

) 12

sinξ( y1

R+ Z(ξ )

)dξ,

Z(ξ ) = 1

π

∫ ∞

0ln

(K1(µ)I1(µ)

K1(ξ )I1(ξ )

)dµ

µ2 − ξ 2,

where r =√

y22 + y2

3 < R and I0, I1, and K1 are modified Bessel functions.The linear theory underpredicts the maximum observed pressure gradients

by about 8%. The agreement with experiment can be greatly improved by inc-luding contributions from the surface dipoles in Equation (7.4.3) (which in afirst approximation are determined by the drag force exerted on the nose of thetrain by the linear theory pressure rise) and, to a lesser extent, by including thevortex sound generated by the tunnel exit-flow vorticity (the final source termon the right of (7.4.3)).

Problems 7

1. The term ω2∂Y2/∂y3 in the Representation (7.1.3) of the sound produced bya gust interacting with a thin airfoil accounts for the influence of changes in

Problems 7 173

the airfoil chord 2a(y3) over the interaction region. Show that when

da

dy3(y3) 1

the Formula (7.2.3) for the sound produced by a line vortex is given in a firstapproximation by

p(x, t) ≈ ρ0U M cos

232 π |x|

|sinφ|sin θ

×

cos θ − sin θ cosφ

(da

dy3

)0

H(U [t]

a + 1)

√U [t]

a + 1,

where da/dy3 is evaluated at y3 = 0, where the vortex is cut by the airfoil.Show that this result is identical with that given by (7.1.4) provided that

in (7.1.4) ω3 is interpreted as the component of the vorticity parallel to thelocal leading edge of the airfoil and the convection velocity U is replaced byits component normal to the local leading edge.

2. A vortex ring orientated with its axis parallel to the +x1 axis is convected ina low Mach number mean flow at speed U in the x1 direction past the edgeof the rigid half-plane x1 < 0, x2 = 0,−∞< x3 <∞. Use the compactGreen’s function

G(x, y, t − τ ) = 1

2π2√

2π ic0

ϕ∗(x)ϕ∗(y)

|x − y3i3|3/2

∫ ∞

−∞

√ωe−iω(t−|x−y3i3|/c0) dω,

where ϕ∗ is defined as in (3.9.6), to calculate the sound produced as thevortex passes the edge when the influence of the half-plane on the motionof the ring is ignored.

3. Calculate the sound produced within and outside a semi-finite circular cylin-drical rigid pipe when a vortex ring exhausts axisymmetrically from the openend. Neglect the influence of the pipe walls on the motion of the vortex andignore any change in the vortex radius at the exit.

4. Determine the (quadrupole) sound produced by the head-on collision oftwo equal ring vortices. Estimate the sound generated when a ring vortex isincident normally on a plane wall.


5. Use the Green’s function (3.9.15) and Equation (7.4.4) to determine theinfrasound generated by a train entering a tunnel modeled by the unflanged,circular cylindrical duct in Fig. 3.9.6b. Assume that the train travels alongthe axis of the duct and show that the acoustic pressure at the far field pointx outside the tunnel is given approximately by

p(x, t) ≈ ρ0U 2 M

4π |x|(

1 − x1

|x|)

×∫ ∞

−∞

∂AT

∂y1(y1 + U [t])

∂2ϕ∗

∂y21

(y1, 0, 0) dy1, |x| → ∞,

where [t] = t − |x|/c0.

8Further Worked Examples

8.1 Blade–Vortex Interactions in Two Dimensions

The linear theory of the low Mach number, two-dimensional interaction of a linevortex with an airfoil was discussed in Section 6.4. The interaction will now beexamined in more detail, including also the influence of image vortices on themotion. The general problem to be considered is depicted in Fig. 8.1.1, whichshows a vortex of strength moving in the neighbourhood of a rigid airfoil ofchord 2a occupying −a < x1 < a, x2 = 0. There is no mean circulation aboutthe airfoil. We shall consider cases with and without a mean flow in the x1

direction and examine the influence of vortex shedding from the trailing edge.

8.1.1 Equation of Motion of the Vortex

At time t let the vortex be at

x ≡ (x1, x2) = x0(t), and translate at velocity v0 = dx0

dt(t).

If we set z = x1 + i x2, z0 = x01 + i x02, the transformation

ζ = z

a+√

z2

a2− 1 (8.1.1)

maps the fluid region in the z plane of the airfoil into the region |ζ |> 1 inthe ζ plane. The upper and lower faces of the airfoil (x2 = ± 0) respectivelytransform into the upper and lower halves of the unit circular cylinder |ζ | = 1,and the vortex maps into an equal vortex at ζ = ζ0 (Fig. 8.1.2). In the absenceof mean flow (U = 0), and when there is no mean circulation about the cylin-der (and therefore about the airfoil), the complex potential of the motion is ob-tained by placing an image vortex − at the inverse point ζ = 1/ζ ∗

0 together with

175

176 8 Further Worked Examples

Fig. 8.1.1.

a vortex + at the center of the cylinder. The two interior vortices ensure thatthe total circulation around the cylinder vanishes. Then,

w(ζ ) = − i

2πln(ζ − ζ0) + i

2πln

(ζ − 1

ζ ∗0

)− i

2πln ζ.

The velocity potential of the motion in the z plane is given by setting ζ = ζ (z).Because a mean flow in the x1 direction is unaffected by the airfoil, we caninclude its contribution by adding the complex potential U z. Then,

w(z) = − i

2πln(ζ (z) − ζ (z0)) + F(z),

where F(z) = i

2πln

(ζ (z) − 1

ζ (z0)∗

)− i

2πln ζ (z) + U z.

This is of the form given in (4.6.1), so that the corresponding equation of motion

Fig. 8.1.2.

8.1 Blade–Vortex Interactions in Two Dimensions 177

of the vortex at z0 is found from (4.6.3) to be

dz∗0

dt≡ dx01

dt− i

dx02

dt= − iζ ′′(z0)

4πζ ′(z0)+ F ′(z0),

that isdz∗

0

dt= i

4πa√

Z2 − 1

Z√

Z2 − 1− 1 + 2

|ζ0|2 − 1

+ U, (8.1.2)

where Z = z0

a, ζ0 = Z +

√Z2 − 1.

Equation (8.1.2) takes no account of the influence of vortex shedding. In alinearized calculation (in which image effects in the airfoil are neglected) thiscould be done by assuming shed vorticity to lie in a thin vortex sheet downstreamof the trailing edge at x1 = a, as in Section 6.3. This would lead to a solution interms of the Sears function (6.3.6), but we shall not do this, because it limits thediscussion to linearized motions. Instead, we shall apply the method discussedin Section 6.4, where the effects of vortex shedding are modelled by deletingsingularities from the compact Green’s function.

8.1.2 Formula for the Acoustic Pressure

The sound produced by the low Mach number motion of the vortex is calculatedfrom (6.2.4):

p(x, t) ≈ −ρ0x j

2π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

dx01

dτ

∂Y j

∂y2− dx02

dτ

∂Y j

∂y1

x0(τ )

× dτ√t − τ − |x|/c0

, (8.1.3)

where the Kirchhoff vector for the strip airfoil has the components (Table 3.9.1)

Y1 = y1, Y2 = Re(−i√

z2 − a2), z = y1 + iy2. (8.1.4)

By defining the radiation angle for an observer at x in the far field as inFig. 8.1.1, we can write

p(x, t) ≈ ρ0 sin

2π√

2c0|x|∂

∂t

∫ t−|x|/c0

−∞

dx02

dτ(τ )

dτ√t − τ − |x|/c0

− ρ0 cos

2π√

2c0|x|

× ∂

∂t

∫ t−|x|/c0

−∞

(dx01

dτ

∂Y2

∂y2− dx02

dτ

∂Y2

∂y1

)x0(τ )

dτ√t − τ − |x|/c0

.

The two integrals in this formula represent the acoustic fields of dipole sources.According to Curle’s theory (Section 2.3) and the Formula (6.2.2), the strengths


of these dipoles are determined by the unsteady force (F1, F2) exerted on thefluid (per unit span) by the airfoil. The first is aligned with the airfoil chord(the mean flow direction) and represents the influence of suction forces at theleading and trailing edges (Batchelor 1967); the second component F2 is equaland opposite to the unsteady lift experienced by the airfoil during the interaction.

In general the integrals must be evaluated numerically using the solution ofthe equation of motion (8.1.2). Introduce the shorthand notation

W = d

dz(−i

√z2 − a2) = −i Z√

Z2 − 1(8.1.5)

evaluated at the vortex. Then,(dx01

dτ

∂Y2

∂y2− dx02

dτ

∂Y2

∂y1

)x0(τ )

≡ −a Im

(W(Z )

d Z

dτ

),

and the acoustic pressure becomes

p(x, t) ≈ ρ0a

2π√

2c0|x|∂

∂t

sin

∫ [t]

−∞Im

(d Z

dτ

)dτ√

[t] − τ

+ cos∫ [t]

−∞Im

(W(Z )

d Z

dτ

)dτ√

[t] − τ

, (8.1.6)

where [t] = t − |x|/c0 is the retarded time, and it is understood that Z = Z (τ ).Equations (8.1.2) and (8.1.6) for the vortex motion and the acoustic pressure

will now be applied to several different special cases.

8.1.3 Linear Theory

In the linearized approximation the vortex is swept past the airfoil along atrajectory parallel to the x1 direction at precisely the uniform mean flow speedU . This is the case illustrated in Fig. 6.4.1. When the standoff distance h a itwas argued in Section 6.4 that the influence of vortex shedding from the trailingedge could be estimated by deleting the singularity that occurs at the edge fromthe Green’s function and ignoring the shed vorticity. Only the second integralin (8.1.6) contributes to the sound (because d Z/dτ = U/a is real and F1 ≡ 0),and the trailing edge singularity corresponds to the singularity of W(Z ) atZ = 1.W(Z ) is singular at both the leading and trailing edges (Z = ±1), which,other things being equal, are therefore the most significant sources of soundat high frequencies, because the second integral in (8.1.6) is dominated bycontributions from the neighbourhoods of the singularities. By deleting the


contribution from the trailing edge we are asserting that all of the sound isproduced by the interaction of the vortex with the leading edge. Near this edge

W(Z ) ≈ 1√2√

Z + 1, (8.1.7)

where the branch cut for√

Z + 1 runs along the real axis from Z = −1 to Z =+∞. Making this substitution in (8.1.6), measuring time from the instant that thevortex crosses the midchord x1 = 0 of the airfoil, so that Z = Uτ/a + ih/a, andchanging the integration variable to ξ = 1/

√t − τ − |x|/c0, we then recover

the result (6.4.3), which can be written,

p(x, t)

ρ0U√

M cos(a/|x|) 12/

4πa≈

(U [t]a + 1

)(U [t]

a + 1)2 + (

ha

)2, |x| → ∞. (8.1.8)

The nondimensional acoustic pressure signature (the right-hand side of(8.1.8)) is plotted as the solid curves in Fig. 8.1.3 for h/a = 0.2, 0.5, 1.0.

Fig. 8.1.3.


It was pointed out in Section 6.3 that the linearized problem of determining theunsteady force F2 exerted on the fluid when an incompressible, sinusoidal gustconvects past the airfoil can be solved exactly with full account taken of vortexshedding, in terms of the Sears function (6.3.6). This force also determinesthe low Mach number acoustic radiation by Equation (6.2.2) (because F1 ≡ 0),which in the present case can be shown to predict that

p(x, t)

ρ0U√

M cos(a/|x|) 12/

4πa≈

√2πRe

∫ ∞

0(iλ)

12 S(λ)e−λh/a+iU [t]/a dλ,

|x| → ∞. (8.1.9)

The corresponding pressure signatures are plotted as the dotted curves inFig. 8.1.3. The agreement with the approximate theory of Equation (8.1.8) isremarkably good even when h/a is as large as unity, when the characteristicreduced frequency λ=ωa/U of the motion is relatively small, and might beexpected to lie outside the range for which (8.1.8) is valid.

8.1.4 Nonlinear Theory

When account is taken of image vortices in the airfoil the trajectory of thevortex in the neighbourhood of the airfoil is no longer parallel to the mean flowdirection, and must be determined by numerical integration of Equation (8.1.2).To do this it is convenient to introduce a dimensionless velocity ratio ε and timeT defined by

ε =

4πaU, T = Ut

a,

in terms of which (8.1.2) becomes

d Z∗

dT= iε√

Z2 − 1

Z√

Z2 − 1− 1 + 2

|ζ0|2 − 1

+ 1, (8.1.10)

which can be solved for Z by Runge–Kutta integration (Section 4.6).If the initial standoff distance is h at x1 = −∞, the integration is started at a

large distance L upstream of the airfoil midchord, say L = 10a, by prescribingthe initial position of the vortex to be Z = −L/a + ih/a. The upper part ofFig. 8.1.4 shows a calculated trajectory for

ε = 0.2,h

a= 0.2,


Fig. 8.1.4.

where time is measured from the instant that the vortex passes the midchordof the airfoil. The nonlinear influence of the image vorticity is to shift theinitially rectilinear trajectory of the vortex away from the airfoil in the directionof the vortex force ω ∧ U (U = U i). The vortex is closest to the airfoil atUt/a = 0, where x02 ∼ 0.28a, and where convection by the images increasesthe translation speed of the vortex from U to approximately

U +

4πx02= U + εU

x02/a∼ U

(1 + 0.2

0.28

)= 1.71U.

The sound generated as the vortex passes the airfoil is given by (8.1.6). Theinfluence of vortex shedding into the wake is included by using the approxi-mation (8.1.7) for W(Z ). The integrals must be evaluated numerically, and thisis done by defining a dimensionless vortex convection velocity (u(T ), v(T ))by

d Z

dT= u(T ) + iv(T ), where T = Uτ

a.


Then,

p(x, t)

ρ0U√

M(a/|x|) 12/

4πa

≈ 212

d

dT

sin

∫ [T ]

−∞

v(T ) dT√[T ] − T

+ cos∫ [T ]

−∞

Im(W(Z )(u + iv))(T ) dT√[T ] − T

= 232 sin

d

dT

∫ ∞

0v([T ] − λ2) dλ

+ 232 cos

d

dT

∫ ∞

0Im(W(Z )(u + iv))([T ] − λ2) dλ

≈ p1(x, t)

ρ0U√

M(a/|x|) 12/

4πa+ p2(x, t)

ρ0U√

M(a/|x|) 12/

4πa, (8.1.11)

where T = Ut/a, [T ] = U [t]/a, and the integration variable T has been re-placed by λ =

√[T ] − T . The final integrals are easily evaluated numerically

when the path of the vortex has been determined. The upper limit of integra-tion is actually finite, because the source terms must be set to equal zero assoon as [T ] −λ2 reduces to the nondimensional time at which the computationof the vortex path begins (where the vortex is sufficiently far upstream that iteffectively produces no sound by interaction with the airfoil).

The components p1(x, t), p2(x, t) of (8.1.11) correspond respectively to thedipole sound produced by the unsteady suction and lift forces; their nondimen-sional forms

p1(x, t)

ρ0U√

M sin(a/|x|) 12/

4πaand

p2(x, t)

ρ0U√

M cos(a/|x|) 12/

4πa

are plotted in Fig. 8.1.4. Vortex shedding should smooth out the pressure signa-tures at the retarded times when the vortex is close to the trailing edge. But thecalculated pressures exhibit blips shown as dotted curves in the figure. Thesearise because, although our calculation has accounted for vortex shedding inevaluating the dipole source strengths (by means of the approximation (8.1.7)),the effect of shedding was not included in the calculation of the vortex tra-jectory. However, the smoothing influence of shedding at a sharp edge actsto remove the blips, and the pressure signatures have profiles similar to thosedepicted by the solid curves in the figure, obtained by interpolating smoothlybetween the calculated pressures on either side of the blips.

An interesting nonlinear interaction occurs when the initial standoff distanceof the vortex h = 0 (Fig. 8.1.5). In the linearized approximation, the vortexwould strike the leading edge of the airfoil at U [t]/a = −1, at which time the


Fig. 8.1.5.

linear theory acoustic pressure (8.1.8) is infinite. This singular event does notoccur because the vortex trajectory is deflected around the airfoil by the imagevorticity (for a rounded nose the possibility of additional vortex shedding fromthe leading edge may be ignored). The upper part of Fig. 8.1.5 illustrates thisfor the same value of the velocity ratio ε = /4πaU = 0.2 considered above.The maximum convection velocity of the vortex (at Ut/a = 0) is now morethan twice the mean stream velocity:

U +

4π (0.15a)= U

(1 + 0.2

0.15

)= 2.3U.

The corresponding suction- and lift-dipole acoustic pressures p1 and p2 shownin the figure are also greatly increased.

8.1.5 Periodic Vortex Motion

When there is no mean flow (U = 0) the characteristic velocity and dimension-less time become

V =

4πa, T = V t

a,


Fig. 8.1.6.

and the vortex equation of motion (8.1.2) reduces to

d Z∗

dT= i√

Z2 − 1

Z√

Z2 − 1− 1 + 2

|ζ0|2 − 1

.

The solutions are closed trajectories orbiting the airfoil periodically. A typicalorbit is plotted in the upper half of Fig. 8.1.6 for > 0, for the case where thetrajectory passes through the point labelled 0, where x01 = −2a, x02 = 0. Thecalculated period is T0 ≡ V t0/a ≈ 35.84.

An orbiting vortex motion of this kind cannot be realized in practice (becauseof diffusion from the vortex core and the continual shedding of additionalvorticity from the airfoil), but it is still instructive to calculate the sound producedby the motion. By writing

T = V τ

aand

d Z

dT= u(T ) + iv(T )

in the general formula (8.1.6) for the acoustic pressure, the nondimensional


suction and lift acoustic pressures are found to be given by the following mod-ified form of (8.1.11)

p1(x, t)

ρ0V 2√

M(a/|x|) 12

+ p2(x, t)

ρ0V 2√

M(a/|x|) 12

≈ 212

d

dT

sin

∫ [T ]

−∞

v(T ) dT√[T ] − T

+ cos∫ [T ]

−∞

Im(W(Z )(u + iv))(T ) dT√[T ] − T

,

(8.1.12)

where M = V/c0.The function W(Z ) in the integrand is given by (8.1.5) in the absence of

vortex shedding. It follows by inspection and from the numerical solution, thatwhen T is measured from Z = −2, as indicated in Fig. 8.1.6, the suction anddipole source strengths have period T0, and possess Fourier series expansionsof the form

v(T ) =∞∑

n=1

an cos

(2πnT

T0

), Im(W(Z )(u+iv))(T ) =

∞∑n=1

bn sin

(2πnT

T0

),

where the coefficients an, bn can be calculated by using the numerical solutionfor the orbit to evaluate

an = 2

T0

∫ T0

0v(T ) cos

(2πnT

T0

)dT,

bn = 2

T0

∫ T0

0Im(W(Z )(u + iv))(T ) sin

(2πnT

T0

)dT .

By making the change of integration variable λ =√

[T ] − T /√

T0, the right-hand side of (8.1.12) now becomes

4√

2π√T0

∞∑n=1

−ann sin

∫ ∞

0sin

[2πn

([T ]

T0− λ2

)]dλ

+ bnn cos∫ ∞

0cos

[2πn

([T ]

T0− λ2

)]dλ

.

The integrals are evaluated from the real and imaginary parts of

∫ ∞

0e2π in[T ]/T0−λ2 dλ = 1

2√

2ne2n[T ]/T0− 1

4 π i .


Hence, the suction and lift force dipole fields are given respectively by

p1(x, t)

ρ0V 2√

M sin(a/|x|) 12

≈ − 2π√T0

∞∑n=1

an√

n sin

[2nπ t

t0− π

4

]

p2(x, t)

ρ0V 2√

M cos(a/|x|) 12

≈ 2π√T0

∞∑n=1

bn√

n cos

[2nπ t

t0− π

4

], |x| → ∞,

where [ ] denotes evaluation at the retarded time t − |x|/c0. The correspondingnondimensional pressures are plotted in Fig. 8.1.6 (taking the first 26 termsin the series); both have similar orders of magnitude, and exhibit rapid varia-tions at the retarded times at which the vortex is directly above and below theairfoil.

8.2 Parallel Blade–Vortex Interactions in Three Dimensions

We now examine to what extent the simple two-dimensional methods of theprevious section can be adapted to wings of finite span and variable chord forproblems of the kind shown in Fig. 8.2.1. The general representation of thesound produced by vortex–airfoil interactions is discussed in Section 7.1, when

Fig. 8.2.1.

8.2 Parallel Blade–Vortex Interactions in Three Dimensions 187

the airfoil chord can be regarded as compact. The general solution is applicableto airfoils of arbitrary span, but we shall consider only the case where the spanis compact; predictions for a noncompact span will be intermediate betweenthose discussed here and those in Section 8.1.

Consider a planar airfoil of either rectangular or elliptic planform, orientatedas illustrated in Fig. 8.2.1 at zero angle of attack to a mean flow at speed U inthe x1 direction. A spanwise line vortex of strength is swept past the airfoilat an initial standoff distance h above the airfoil, as indicated in the side viewof Fig. 8.2.1b. When h = 0 it will be necessary to take account of nonlinearinteractions with the airfoil.

For an airfoil of compact chord and span the acoustic pressure produced bythe interaction is given by Equation (5.4.4):

p(x, t) ≈ −ρ0x j

4πc0|x|2∂

∂t

∫(ω ∧ v)

(y, t − |x|

c0

)· ∇Y j (y) d3y, |x| → ∞.

(8.2.1)

It is assumed that the section of the line vortex that interacts with the airfoilremains rectilinear, with the representation

ω = kδ(x1 − x01(t))δ(x2 − x02(t)), where x0 = (x01, x02, 0).

For an elliptic airfoil of span L (between − 12 L < x3 <

12 L), the Kirchhoff vector

Y has the components

Y1 = y1,

Y2 =

Re(−i√

z2 − a(y3)2), |y3| < 12 L

y2, |y3| > 12 L

, Y3 = y3, z = y1 + iy2,

where 2a(y3) is the airfoil chord at the spanwise location y3. For the rectangularairfoil a(y3) ≡ a = constant; for the elliptic airfoil a(y3) assumes a maximumvalue of a at y3 = 0, and we shall write

a(y3)

a=√

1 − 4y23

L2, |y3| < 1

2L . (8.2.2)

Vorticity is shed into the wake of the airfoil in accordance with the Kuttacondition of unsteady aerodynamics. This smooths out conditions at the trail-ing edge, so that sound is generated primarily as the vortex passes over theleading edge of the airfoil. As before, this can be dealt with in a first approxima-tion by ignoring the shed vorticity and deleting the trailing edge singularity of


Green’s function, by using the following modification of the x2 componentof Y:

Y2 = Re(√

2a(y3)√

z + a(y3)), |y3| < 1

2L .

Then, (8.2.1) becomes

p(x, t) ≡ p1(x, t) + p2(x, t)

≈ ρ0 cos

4πc0|x|∂

∂t

∫ L2

− L2

[dx02

dt

]dy3 + ρ0 cos

4√

2πc0|x|

× ∂

∂t

∫ L2

− L2

[Im

(dz0

dt

√a(y3)√

z0(t) + a(y3)

)]dy3 |x| → ∞, (8.2.3)

where , are respectively the angles shown in Fig. 8.2.1a between the x1 andx2 directions and the radiation direction, z0(t) = x01(t)+ i x02(t), and quantitiesin square braces are evaluated at the retarded time [t] = t − |x|/c0.

The first term on the right is the suction force dipole, aligned with the airfoilchord, whose strength is determined by the x2 component of the vortex convec-tion velocity. It is assumed to be nonzero only over the section − 1

2 L < y3 < 12 L

of the vortex, where dx02/dt = 0 because of nonlinear interactions with the air-foil. The second term is the conventional lift dipole radiation. Note that ‘infinite’contributions to the integrals from |y3| > L/2 are constant because ω ∧ v isconstant for |y3| > L/2, and have been discarded (c.f., Section 7.3).

8.2.1 Linear Theory

When there is no back-reaction of the airfoil on the vortex the convectionvelocity of the vortex is equal to the mean stream velocity

dx01

dt= U ,

dx02

dt= 0.

The radiation is produced entirely by the lift dipole, and if the vortex crossesthe midchord of the airfoil at time t = 0 (8.2.3) reduces to

p2(x, t)

ρ0U M cos(L/|x|)/4πa≈ − 1

232

∫ 12

− 12

Im

√aa(U [t]

a + aa + i h

a

) 32

d y3,

(8.2.4)

8.2 Parallel Blade–Vortex Interactions in Three Dimensions 189

Fig. 8.2.2.

where y3 = y3/L; a/a = 1 for the rectangular airfoil, and is given by (8.2.2)for the elliptic airfoil. For a rectangular airfoil the integral evaluates to

Im

(U [t]

a+ 1 + i

h

a

)− 32

.

The acoustic pressure signatures (the left-hand side of (8.2.4)) for the rectan-gular and elliptic airfoils are plotted in Fig. 8.2.2 for a vortex standoff distanceh = 0.2a. The profiles are qualitatively similar to the corresponding plot inFig. 8.1.3 for an airfoil of infinite span, although in three dimensions the am-plitude decreases much more rapidly with increasing retarded distance of thevortex from the leading edge. The peak amplitude is larger and the acousticpulse is of smaller duration ∼h/U for the rectangular airfoil because differentsections of the vortex interact with the leading edge of the elliptic airfoil atdifferent times during a total interaction time ∼ a/U > h/U .

8.2.2 Nonlinear Theory

When the standoff distance h = 0, image vorticity in the airfoil prevents thevortex from impinging on the leading edge, and causes the trajectory to belocally deflected above the airfoil (for > 0). If the leading edge of the airfoilis suitably rounded (so that no additional vortex shedding occurs) this casecan be treated for a rectangular airfoil by assuming that only the section of thevortex within the span − 1

2 L < x3 < 12 L of the airfoil is affected in this way, and

that the distorted path can be approximated by that for locally two-dimensionalflow.


Let

ε =

4πaU, T = Ut

a, Z = z0

a,

d Z

dT= u(T ) + iv(T ),

W(Z ) = 1√2√

Z + 1.

Then, the motion of the section of the vortex within the airfoil span (− 12 L <

x3 <12 L) is governed by Equation (8.1.10) (where ζ0 is defined in terms of z0

as in (8.1.1)), and the suction and lift dipole radiation pressures are given by

p1(x, t)

ρ0U M cos(L/|x|)/4πa≈[

dv

dT

]p2(x, t)

ρ0U M cos(L/|x|)/4πa≈ ∂

∂T[Im (W(Z )(u + iv))]

|x| → ∞,

where [ ] denotes evaluation at the retarded time t − |x|/c0.These nondimensional pressures are plotted in Fig. 8.2.3 for a velocity ratio

ε = 0.2 when the vortex is released upstream with h = 0. The upper part of the

Fig. 8.2.3.

8.3 Vortex Passing over a Spoiler 191

figure shows the path followed by those sections of the vortex inboard of theairfoil tips; it is the same as that depicted in Fig. 8.1.5 for the infinite spanairfoil. As in that case, small acoustic blips spuriously predicted during thepassage of the vortex past the trailing edge have been removed (c.f., Fig. 8.1.4).The three-dimensional acoustic pulses are narrower than those predicted in twodimensions.

8.3 Vortex Passing over a Spoiler

The sound produced when vorticity interacts at low Mach number with surfaceirregularities on a nominally plane, rigid wall is produced by dipoles orientatedin the plane of the wall, that is, by the unsteady wall drag. A simple canonicalinteraction (Kasoev, 1976) involving a line vortex near a thin vertical spoiler isillustrated in Fig. 8.3.1. The wall coincides with the plane x2 = 0, and the spoilerextends along the x2 axis from x2 = 0 to x2 = a > 0 for −∞ < x3 < ∞. Thevortex

ω = kδ(x − x0(t)), where x0 = (x01, x02, 0),

is parallel to the spoiler, and is assumed to convect over it in a low Mach number,irrotational mean stream having uniform speed U in the x1 direction.

Define z = x1 + i x2, z0 = x01 + i x02. The transformation

ζ =√

z2

a2+ 1 (8.3.1)

maps the fluid region onto the upper half Im ζ > 0 of the ζ -plane. The left andright faces of the spoiler (x1 = ∓ 0) transform respectively into the intervals−1<ζ < 0 and 0<ζ < 1 of the real ζ axis, and the vortex maps into an equalvortex at ζ = ζ0. The mean flow is parallel to the real axis in the ζ plane,

Fig. 8.3.1.


with complex potential Uaζ . The complex potential of the whole flow in theζ plane is obtained by introducing an image vortex of strength − at thecomplex conjugate point ζ = ζ ∗

0 , and is given by

w(ζ ) = − i

2πln(ζ − ζ0) + i

2πln(ζ − ζ ∗

0 ) + Uaζ.

Hence, setting ζ = ζ (z) the motion in the z plane is defined by

w(z) = − i

2πln(ζ (z) − ζ (z0)) + F(z),

where F(z) = i

2πln(ζ (z) − ζ (z0)∗) + Uaζ (z),

which is in the form (4.6.1). The equation of motion of the vortex at z0 istherefore (see (4.6.3))

dz∗0

dt≡ dx01

dt− i

dx02

dt= − iζ ′′(z0)

4πζ ′(z0)+ F ′(z0),

that is,

d Z∗

dT= −i

1

Z (Z2 + 1)− 2Z

Z2 + 1 − |Z2 + 1|

+ εZ√Z2 + 1

, (8.3.2)

where Z = z0

a, T = V t

a, V =

4πa, ε = U

V.

The compact Green’s function for this problem (applicable when the acousticwavelength ∂) is given by (6.1.6), so that the analogue of Equation (6.2.4)for the far-field acoustic pressure becomes

p(x, t)

≈ −ρ0x1

π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

(k ∧ dx0

dτ(τ ) · ∇Y1(x0(τ ))

)dτ√

t − τ − |x|/c0

= −ρ0x1

π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

dx01

dτ

∂Y1

∂y2− dx02

dτ

∂Y1

∂y1

x0(τ )

dτ√t − τ − |x|/c0

,

(8.3.3)

where the Kirchhoff vector

Y1 = Re(aζ ) = aRe(√

Z2 + 1) at z = z0. (8.3.4)

8.3 Vortex Passing over a Spoiler 193

The radiation is produced by the unsteady drag force F1 exerted on the fluidby the spoiler, given (per unit span) by

F1 = −ρ0

∫ω ∧ v · ∇Y1 dy1 dy2 = −ρ0k ∧ dx0

dt· ∇Y1(x0).

This force vanishes, and therefore no sound is generated, in the linearizedapproximation in which the vortex is assumed to translate at the local meanstream velocity, because in that case

dx0

dt= U∇Y1(x0) and k ∧ ∇Y1 · ∇Y1 ≡ 0.

Following the procedure of Section 8.1, introduce the notations

d Z

dT= u(T ) + iv(T ), W = d

dz(√

z2 + a2 ) = Z√Z2 + 1

(8.3.5)

evaluated at the vortex, and make the substitution T = V τ/a in (8.3.3) to obtainthe acoustic pressure in the form

p(x, t) ≈ ρ0V√

M sin

aπ√

2

(a

|x|) 1

2 ∂

∂T

∫ [T ]

−∞Im

(W(Z )

d Z

dT

)dT√

[T ] − T,

that is,

p(x, t)

ρ0V 2√

M sin(a/|x|) 12

≈ 252

∂

∂T

∫ ∞

0Im(W(Z )(u + iv))([T ] − λ2) dλ,

(8.3.6)

where [T ] = V [t]/a is the nondimensional retarded time and M = V/c0.The vortex path equation (8.3.2) and the acoustic pressure integral (8.3.6)

are evaluated numerically, taking the initial position of the vortex to be severalspoiler heights a upstream, where its motion is unaffected by the spoiler. Theupper part of Fig. 8.3.2 shows the vortex trajectories when the initial distance ofthe vortex from the wall is h = 0.75a for the two cases (i) of no mean flow, U = 0,and (ii) U = V ≡/4πa; the corresponding nondimensional acoustic pressures(8.3.6) are plotted in the lower part of the figure, where time is measured fromthe instant that the vortex passes the spoiler. The effect of mean flow is to drawthe trajectory marginally closer to the spoiler as it passes the tip of the spoilerwhere the interaction is strongest. The convection velocity at this point is alsoincreased from about 1.98V when U = 0 to 3.95V when U = V , and this isresponsible for more than doubling the amplitude and the effective frequencyof the sound.


Fig. 8.3.2.

8.4 Bluff Body Interactions: The Circular Cylinder

Low Mach number, two-dimensional interactions of a line vortex with a circularcylinder provide an interesting contrast to the sharp-edge problems discussedabove. Let the cylinder have radius a and be coaxial with the x3 axis, and letthere be an irrotational mean flow at speed U past the cylinder in the x1 direction,with no mean circulation about the cylinder.

Set z = x1 + i x2 and let the vortex of strength have the complex positionz0 = x01 + i x02 at time t . The complex potential w(z) is found by placing animage vortex − at the inverse point z = a/z∗

0 within the cylinder, a vortex+ at the centre to make the circulation vanish, and by adding the potential forthe uniform mean flow past the cylinder (c.f., Section 8.1):

w(z) = − i

2πln(z−z0)+ i

2πln

(z − a2

z∗0

)− i

2πln z + U

(z + a2

z

). (8.4.1)

The velocity potential governing the motion of the vortex at z0 is obtainedby deleting the self-potential

− i

2πln(z − z0).

8.4 Bluff Body Interactions: The Circular Cylinder 195

Hence we arrive at the equation of motion

d Z∗

dT= i

Z (|Z |2 − 1)+ ε

(1 − 1

Z2

), (8.4.2)

where Z = z0

a, V =

2πa, T = V t

a, ε = U

V,

anddz0

dt= V

d Z

dT≡ V (u + iv).

8.4.1 The Acoustic Pressure

The far-field sound produced by the vortex is calculated from (6.2.4):

p(x, t) ≈ −ρ0x j

2π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

dx01

dτ

∂Y j

∂y2− dx02

dτ

∂Y j

∂y1

x0(τ )

× dτ√t − τ − |x|/c0

, (8.4.3)

where the components of the Kirchhoff vector can be written (see Section 4.5)

Y1 = Re

(z + a2

z

), Y2 = Re

−i

(z − a2

z

), z = y1 + iy2. (8.4.4)

By defining

W1 = d

dz

(z + a2

z

)≡ 1 − 1

Z2, W2 = d

dz

−i

(z − a2

z

)≡ −i

(1 + 1

Z2

)

evaluated at z0, and making the change of integration variable T = V τ/a,Equation (8.4.3) can be written

p(x, t) ≈ ρ0V√

Mx j

2π√

2a|x| 32

∂

∂T

∫ [T ]

−∞Im(W j (u + iv))(T )

dT√[T ] − T

,

where M = V

c0, [T ] = V

a

(t − |x|

c0

).

The subscripts j = 1, 2 in this formula respectively correspond to the acous-tic pressures p1, p2, say, produced by drag and lift dipoles, whose strengthsare determined by the force (F1, F2) exerted on the fluid (per unit span) bythe cylinder. The integrals must be evaluated numerically using the numerical


Fig. 8.4.1.

solution of Equation (8.4.2) for the vortex path. This is done by making thefurther change of integration variable λ =

√[T ] − T , in which case

p1(x, t)

ρ0V 2√

M sin(a/|x|) 12

≈ 212

∂

∂T

∫ ∞

0Im(W1(u + iv))([T ] − λ2) dλ,

p2(x, t)

ρ0V 2√

M cos(a/|x|) 12

≈ 212

∂

∂T

∫ ∞

0Im(W2(u + iv))([T ] − λ2) dλ.

The calculation begins at time T ′, say, by taking the initial position of the vortexto be far upstream of the cylinder at z0 = −L + ih, where L a is sufficientlylarge that the source strengths are negligible for T < T ′ (Fig. 8.4.1). The upperlimits of integration are then finite because the source terms vanish as soon as[T ] − λ2 < T ′.

Figure 8.4.1 illustrates the typical nondimensional waveforms produced whenV ≡ /2πa = 2U and for h/a = ±0.7, time being measured from the instantthat the vortex crosses x1 = 0. The amplitude of the sound decreases rapidlywith increasing distance of closest approach of the vortex to the cylinder; nearthe cylinder the translational velocity of the vortex is increased because the

8.4 Bluff Body Interactions: The Circular Cylinder 197

mean flow velocity is larger, and also because of the increased influence of theimage vorticity. In cases where U V , the lift dipole will tend to predomi-nate because convection by the image vortices can then be neglected in a firstapproximation, and the drag ∼ (ω ∧ U∇Y1) · ∇Y1 ≡ 0.

8.4.2 Wall-Mounted Cylinder

The case of ideal motion of a vortex translating past a cylindrical, semicircularprojection on a rigid wall (Fig. 8.4.2) can be treated by the method used forthe spoiler in Section 8.3. The problem is equivalent to that in which a vortexpair, consisting of a vortex at z0 accompanied by an image of strength − atz∗

0, is incident symmetrically on a circular cylinder. In this case, the lift dipolevanishes identically.

The velocity potential of the unsteady motion is given by augmenting thecomplex potential (8.4.1) by the terms

i

2πln(z − z∗

0) − i

2πln

(z − a2

z0

)+ i

2πln z,

Fig. 8.4.2.


which correspond to the net potential produced by the image. Then, the equationof motion becomes

d Z∗

dT= i

1

Z − Z∗ + Z − Z∗

(|Z |2 − 1)(Z2 − 1)

+ ε

(1 − 1

Z2

), (8.4.5)

where Z = z0

a, V =

2πa, T = V t

a, ε = U

V.

The compact Green’s function is given by (6.1.6), and the far-field acousticpressure by

p(x, t)

≈ −ρ0x1

π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

(k ∧ dx0

dτ(τ ) · ∇Y1(x0(τ ))

)dτ√

t − τ − |x|/c0

= −ρ0x1

π√

2c0|x| 32

∂

∂t

∫ t−|x|/c0

−∞

dx01

dτ

∂Y1

∂y2− dx02

dτ

∂Y1

∂y1

x0(τ )

dτ√t − τ − |x|/c0

,

(8.4.6)

where the Kirchhoff vector

Y1 = Re

(z + a2

z

). (8.4.7)

The radiation is produced by the unsteady drag force exerted on the fluid bythe cylinder, which vanishes in the linearized approximation, when the vortexis assumed to convect passively at the local velocity of the undisturbed meanstream. As before, set

d Z

dT= u(T ) + iv(T ), W1 = d

dz

(z + a2

z

)= 1 − 1

Z2(8.4.8)

evaluated at the vortex. Then,

p(x, t)

ρ0V 2√

M sin(a/|x|) 12

≈ 232

∂

∂T

∫ ∞

0Im(W1(u+iv))([T ]−λ2) dλ, (8.4.9)

where the angle is defined as in Fig. 8.4.1, [T ] = V [t]/a is the nondimensionalretarded time (T = 0 when the vortex is at x1 = 0), and M = V/c0.

The vortex path equation (8.4.5) and the acoustic pressure integral (8.4.9)must be evaluated numerically, taking the initial position of the vortex tobe several cylinder radii a upstream where its motion is unaffected by thecylinder. The upper part of Fig. 8.4.2 shows the vortex trajectories when the

8.5 Vortex Ring and Sphere 199

initial standoff distance of the vortex from the wall h = 0.5a for the two cases(i) of no mean flow, U = 0, and (ii) U = V . The vortex convection velocityat y1 = 0 is increased from about 1.23V when U = 0 to 3.07V when U = V ;this is responsible for the increased acoustic amplitude and for more than dou-bling the effective frequency. The waveforms and these general conclusionsare qualitatively similar to those discussed in Section 8.3 for the sharp-edgedspoiler.

8.5 Vortex Ring and Sphere

Perhaps the simplest low Mach number, inviscid, three-dimensional vortex–surface interaction amenable to analysis is the axisymmetric motion of a ringvortex over a sphere. Let a sphere of radius a be placed with its centre at theorigin in the presence of a uniform mean flow at speed U in the x1 direction.A vortex ring of radius r0(t) and circulation is coaxial with the x1 axis andtranslates in the positive x1 direction under the influence of the mean flow,self-induction and image vorticity in the sphere (Fig. 8.5.1). We shall assumethe vortex core is circular (and remains circular throughout the interaction)with radius σ (t) r0. The self-induced velocity of the ring (in inviscid flow)is parallel to the x1-axis at speed u given approximately by Kelvin’s formula(Batchelor, 1967)

u(r0, σ ) =

4πr0

ln

(8r0

σ

)− 1

4

. (8.5.1)

Fig. 8.5.1.


The image vorticity consists of a coaxial ring vortex whose circulation ′,radius r ′

0, and axial location x ′01 are given by

′ = −(r2

0 + x201

) 12

a, r ′

0 = a2r0

r20 + x2

01

, x ′01 = a2x01

r20 + x2

01

, (8.5.2)

where the planes of symmetry of the ring vortex and its image cut the x1

axis respectively at x01(t), x ′01(t). The motion of the vortex produced by the

combined induction by the image and the mean flow can be expressed in termsof the Stokes stream function ψ( ·r , x1) (Batchelor, 1967; Ting and Klein, 1991)

ψ(r, x1) = Ur2

2

1 − a3(

r2 + x21

) 32

+ ′

2π(+ + −)K (") − E("),

(8.5.3)

where ± =√

(r ∓ r ′0)2 + (x1 − x ′

01)2, " = − − +− + +

,

K (") =∫ π

2

0

dµ√1 − "2 sin2 µ

, E(") =∫ π

2

0

√1 − "2 sin2 µ dµ,

where r denotes perpendicular distance from the x1 axis, and K ("), E(") arerespectively complete elliptic integrals of the first and second kinds.

The radius r0(t) and axial position x01(t) of the ring vortex are then determinedby the equations of motion

dr0

dt= − 1

r0

∂ψ

∂x1(r0, x01),

dx01

dt= u(r0, σ ) + 1

r0

∂ψ

∂r(r0, x01). (8.5.4)

The core radius σ decreases when r0 increases, because the vortex lines movewith the fluid particles. If r0 = h and σ = σ0 are the initial values when thevortex ring is far from the sphere, then at any time t

(2πr0)πσ 2 = (2πh)πσ 20 , i.e., σ (t) = σ0

√h

r0(t)

so that the self-induced velocity (8.5.1) becomes

u =

4πr0

ln

(8h

σ0

[r0(t)

h

] 32

)− 1

4

. (8.5.5)


The equations of motion of the vortex are cast in nondimensional terms bydefining

X = x01

a, R = r0

a, V =

2πa, T = V t

a, ε = U

V.

Then,

d R

dT= − 1

R

∂

∂X(R, X ),

d X

dT= 1

R

∂

∂R(R, X ) + 1

2R

ln

(8h

σ0

[a R

h

] 32

)− 1

4

,

(8.5.6)

where = εR2

2

(1 − 1

(R2 + X2)32

)− (R2 + X2)

12

× (+ + −)K (") − E("),

± =√

(R ∓ R′)2 + (X − X ′)2, " = − − +− + +

,

R′ = R

R2 + X2, X ′ = X

R2 + X2.

Figure 8.5.2 illustrates the sections in the vertical plane of symmetry ofthe sphere of two typical vortex trajectories predicted by Equations (8.5.6). Inboth cases the integration is started five sphere diameters upstream with thefollowing initial values for the vortex ring radius and core radius,

h = 0.8a, σ0 = 0.05h. (8.5.7)

The solid and broken-line curves in the figure correspond respectively to ε = 0

Fig. 8.5.2.


(no mean flow) and ε = 3. The latter value is chosen to make the mean streamvelocity U approximately the same as the self-induced velocity u at largedistances from the sphere. This has a relatively small effect on the trajectory,although the convection speed of the ring past the sphere is greatly increased.

8.5.1 Acoustic Pressure

When the sphere is acoustically compact, equation (5.4.4) gives

p(x, t) ≈ −ρ0x j

4πc0|x|2∂

∂t

∫(ω ∧ v)

(y, t − |x|

c0

)· ∇Y j (y) d3y, |x| → ∞.

(8.5.8)

For the purpose of evaluating the integral we may neglect the finite core size ofthe vortex, and set

ω = θδ(r − r0(t))δ(x1 − x01(t)), (8.5.9)

where θ is a unit azimuthal vector, locally tangential to the vorticity ω and ori-entated in the clockwise direction when the vortex ring is viewed from upstream,as indicated in Fig. 8.5.1.

The force on the sphere is in the mean flow direction – the effective acousticsource is the unsteady drag – and only the component

Y1 = y1

(1 + a3

2|y|3)

≡ y1

1 + a3

2(r2 + y2

1

) 32

(from Table 3.9.1)

of the Kirchhoff vector makes a nontrivial contribution to (8.5.8). The produc-tion of sound is therefore a nonlinear event – the source explicitly involvesonly the self-induced velocity and the velocity induced by the image vortex;the mean flow velocity U = U∇Y1 is absent because

(ω ∧ U∇Y1)

(y, t − |x|

c0

)· ∇Y1(y) ≡ 0.

However, the amplitude and characteristic frequency of the sound both increasewith U .

Substituting (8.5.9) into (8.5.8) and evaluating the integral, we find

p(x, t) ≈ −ρ0 cos

2c0|x|∂

∂t

[r0

(dx01

dt

∂Y1

∂r− dr0

dt

∂Y1

∂y1

)(r0,x01)

], |x| → ∞,


Fig. 8.5.3.

where the quantity in the square braces is evaluated at the retarded positionof the vortex ring, and is the angle between the radiation direction and thex1 axis illustrated in the upper part of Fig. 8.5.3. Expressing this result innondimensional form, we have

p(x, t)

ρ0V 2 M cos(a/|x|) ≈ π∂

∂T

[3R2 X

2(R2 + X2)52

d X

dT

+(

1 + R2 − 2X2

2(R2 + X2)52

)R

d R

dT

],

where M = V/c0, and R and X are the solutions of the vortex equations ofmotion (8.5.6).

The nondimensional acoustic pressure signatures plotted in Fig. 8.5.3 arefor the same the initial conditions (8.5.7) considered above for the vortex ringtrajectories in Fig. 8.5.2. The thick solid curve is the pressure profile in the


absence of mean flow (U = 0). The positions of the vortex ring in this case atseveral different retarded times V [t]/a are marked on the thick curve in theupper part of Fig. 8.5.3 (time being measured from the instant that the ringcrosses the centre of the sphere). Similarly, the thin-line curves in the figuregive the pressure signature and retarded positions for U = 3V , when the self-induction velocity u ≈ U at large distances from sphere. Both the amplitudeand frequency of the sound are increased because of the increased convectionvelocity of the vortex past the sphere.

8.6 Vortex Pair Incident on a Wall Aperture

Hydrodynamic motion in the vicinity of an aperture in a large thin wall generallyproduces an unsteady volume flux through the aperture, which is acousticallyequivalent to a monopole source when the aperture is compact. The upper partof Fig. 8.6.1 depicts a simple model of such a source. The rigid wall coincideswith the plane x1 = 0, and is pierced by a two-dimensional slit aperture ofwidth 2a whose centerline extends along the x3 axis. A vortex pair aligned withthe x3 axis, consisting of vortices of strengths ± at the respective complexpositions

z0 = x01 + i x02 and z∗0 = x01 − i x02

is incident on the aperture from the left (x1 = −∞).The motion is evidently symmetric with respect to the x1 axis, and the trans-

formation

ζ = z√z2 + a2

(z = x1 + i x2)

maps the region Im z > 0 cut along the upper section x2 > a of the wall ontothe upper half of the ζ plane. By the usual method, we accordingly obtain theequation of motion of the vortex pair in the form (Karweit, 1975)

d Z∗

dT= 3i Z

Z2 + 1+ 2i

(Z2 + 1)32 Z/

√Z2 + 1 − (Z/

√Z2 + 1)∗

, (8.6.1)

where Z = z0

a, T = V t

a, V =

4πa,

anddz0

dt= V

d Z

dT≡ V (u + iv).

Let the initial separation of the vortices at x1 = − ∞ be 2h. To integrate theequation, we can set z0 = − L + ih at a convenient initial (but arbitrary) time

8.6 Vortex Pair Incident on a Wall Aperture 205

Fig. 8.6.1.

T = T ′, where L a. When h/a is smaller than 2/332 ≈ 0.385 the vortex pair

passes through the aperture in the manner indicated in Fig. 8.6.1 for h/a = 0.35.For larger values of h/a the trajectories of the two vortices separate; the vorticestravel along symmetric paths parallel to the wall on either side of the aperture,as illustrated for h/a = 0.6.

The acoustic pressure in the far field is given by

p(x, t) ≈ −ρ0

∫ω ∧ v · ∂G

∂yd2y dτ, (8.6.2)

where G is the compact Green’s function (3.9.10) for the wall aperture

G(x, y, t − τ )≈ −√

c0 sgn(x1)

π√

2π |x|χ (t − τ − |x|/c0)√

t − τ − |x|/c0Re

ln

(z

a+√

z2

a2− 1

),

z = y2 + iy1, (8.6.3)


and

χ (t) = H (t)∫ ∞

0

ln(aξ 2/4c0t)e−ξ 2dξ

[ln(aξ 2/4c0t)]2 + π2, = 1.781072.

The dependence on source position y in (8.6.3) is contained entirely in thelogarithmic term, which represents the velocity potential of the ideal flow thatwould be produced through the aperture (from left to right) by a uniform pressuredrop across the wall.

When vortex shedding from the aperture edges is ignored,

ω = kδ(y1 − x01)δ(y2 − x02) − kδ(y1 − x01)δ(y2 + x02),

where k is a unit vector in the x3 direction (out of the plane of the paper inFig. 8.6.1).

If we define

W(Z ) =(

1√Z2 − 1

)Z=i Z∗

, χ (T ) = χ (t), M = V

c0,

and put

T = V τ

ain the integral (8.6.2),

we find

p(x, t) ≈ 252 ρ0V 2 sgn(x1)√

πM

(a

|x|) 1

2∫ [T ]

−∞Re(W∗(Z )(u + iv))(T )

× χ ([T ] − T ) dT√[T ] − T

, [T ] = V [t]

a.

Therefore, by setting λ =√

[T ] − T we can write

p(x, t)

ρ0V 2 sgn(x1)(a/|x|) 12

≈ 272√

πM

∫ ∞

0Re(W∗(Z )(u + iv))([T ] − λ2)χ (λ2) dλ,

|x| → ∞, (8.6.4)

where

χ (λ2) =∫ ∞

0

ln( Mξ 2/4λ2)e−ξ 2dξ

[ln( Mξ 2/4λ2)]2 + π2.

8.6 Vortex Pair Incident on a Wall Aperture 207

As before, the upper limit of integration in (8.6.4) is actually λ = √[T ] − T ′,

where T ′ is the nondimensional initial time from which the motion of the vortexpair is calculated.

The value of the integral depends weakly on the characteristic Mach number

M = V

c0≡

4πac0.

This is just the self-induced convection Mach number of the vortex pair whenseparated by a distance 2a. We have taken M = 0.03 for the far-field acousticpressure signatures plotted in Fig. 8.6.1; in air this would imply that V ∼10 m/sec.

The flow induced by the vortex pair approaching the wall forms a localizedtwo-dimensional jet between the vortices, directed toward the wall. The resis-tance of the wall to this flow causes the pressure just to the left of the wallaperture to rise, forcing fluid through the aperture into the region x1 > 0. Theradiation therefore has the characteristics of an acoustic monopole source forx1 > 0 and a sink for x1 < 0. Numerical results are illustrated in the figure forh/a = 0.35, 0.6. In each case, the time origin has been adjusted to correspondapproximately with the peak in the radiated acoustic pressure, which occurswhen the vortices pass close to the edges of the aperture. When h/a = 0.6 thevortices do not penetrate the aperture but are deflected by the wall; this pro-duces a relatively larger pressure rise than for h/a = 0.35, where the vorticespass through the aperture. The maximum acoustic pressure amplitude is foundto occur when h/a just exceeds the critical value (∼0.385), when the vortextrajectories pass very close to the aperture edges. Further increases of h/a be-yond 0.6 result in a gradual reduction in the amplitude of the sound, and acorresponding increase in the width of the acoustic pulse (i.e., a decrease in thecharacteristic frequency of the sound).

Bibliography

Batchelor, G.K. 1967. An Introduction to Fluid Dynamics. Cambridge: UniversityPress.

Cannell, P. and Ffowcs Williams, J.E. 1973. Radiation from line vortex filamentsexhausting from a two-dimensional semi-infinite duct. Journal of FluidMechanics 58: 65–80.

Crighton, D.G. 1972. Radiation from vortex filament motion near a half plane. Journalof Fluid Mechanics 51: 357–362.

Crighton, D.G. 1975. Basic principles of aerodynamic noise generation. Progress inAerospace Sciences 16: 31–96.

Crighton, D.G. 1985. The Kutta condition in unsteady flow. Annual Reviews of FluidMechanics 17: 411–445.

Crighton, D.G., Dowling, A.P., Ffowcs Williams, J.E., Heckl, M., and Leppington,F.G. 1992. Modern Methods in Analytical Acoustics (Lecture Notes). London:Springer-Verlag.

Curle, N. 1955. The influence of solid boundaries upon aerodynamic sound.Proceedings of the Royal Society of London A231: 505–514.

Dhanak, M.R. 1981. Interaction between a vortex filament and an approaching rigidsphere. Journal of Fluid Mechanics 110: 129–147.

Dowling, A.P. and Ffowcs Williams, J.E. 1983. Sound and Sources of Sound.Chichester: Ellis Horwood.

Ffowcs Williams, J.E. 1963. The noise from turbulence convected at high speed.Philosophical Transactions of the Royal Society of London A255: 469–503.

Ffowcs Williams, J.E. and Hawkings, D.L. 1969. Sound generation by turbulence andsurfaces in arbitrary motion. Philosophical Transactions of the Royal Society ofLondon A264: 321–342.

Ffowcs Williams, J.E. and Hall, L.H. 1970. Aerodynamic sound generation byturbulent flow in the vicinity of a scattering half-plane. Journal of FluidMechanics 40: 657–670.

Ffowcs Williams, J.E. 1974. Sound production at the edge of a steady flow. Journal ofFluid Mechanics 66: 791–816.

Goldstein, M.E. 1976. Aeroacoustics. New York: McGraw-Hill.Goldstein, S. 1960. Lectures on Fluid Mechanics. New York: Interscience.Howe, M.S. 1975a. Contributions to the theory of aerodynamic sound with

209

210 Bibliography

application to excess jet noise and the theory of the flute. Journal of FluidMechanics 71: 625–673.

Howe, M.S. 1975b. The generation of sound by aerodynamic sources in aninhomogeneous steady flow. Journal of Fluid Mechanics 67: 579–610.

Howe, M.S. 1989. On unsteady surface forces, and sound produced by the normalchopping of a rectilinear vortex. Journal of Fluid Mechanics 206: 131–153.

Howe, M.S. 1998a. Acoustics of Fluid–Structure Interactions. Cambridge: CambridgeUniversity Press.

Howe, M.S. 1998b. The compression wave produced by a high-speed train entering atunnel. Proceedings of the Royal Society A454: 1523–1534.

Howe, M.S. 2001. Vorticity and the theory of aerodynamic sound. Journal ofEngineering Mathematics 41: 367–400.

Howe, M.S., Iida, M., Fukuda, T., and Maeda, T. 2000. Theoretical and experimentalinvestigation of the compression wave generated by a train entering a tunnel witha flared portal. Journal of Fluid Mechanics 425: 111–132.

Karweit, M. 1975. Motion of a vortex pair approaching an opening in a boundary.Physics of Fluids 18: 1604–1606.

Kasoev, S.G. 1976. Sound radiation from a linear vortex over a plane with a projectingedge. Soviet Physics Acoustics 22: 71–72.

Kelvin, Lord. 1867. On vortex motion. Transactions of the Royal Society of Edinburgh25: 217–260.

Knio, O.M., Ting, L., and Klein, R. 1998. Interaction of a slender vortex with a rigidsphere: Dynamics and far field sound. Journal of the Acoustical Society ofAmerica 103: 83–98.

Lamb, Horace. 1932. Hydrodynamics. 6th ed. Cambridge: Cambridge University Press.Landau, L.D. and Lifshitz, E.M. 1987. Fluid Mechanics. 2nd ed. Oxford: Pergamon.Lighthill, M.J. 1952. On sound generated aerodynamically. Part I: General theory.

Proceedings of the Royal Society of London A211: 564–587.Lighthill, M.J. 1956. The image system of a vortex element in a rigid sphere.

Proceedings of the Cambridge Philosophical Society 52: 317–321.Lighthill, M.J. 1958. An Introduction to Fourier Analysis and Generalised Functions.

Cambridge: Cambridge University Press.Lighthill, M.J. 1963. Laminar Boundary Layers. Edited by L. Rosenhead. Chs. 1, 2.

Oxford: University Press.Lighthill, James. 1978. Waves in Fluids. Cambridge: Cambridge University Press.Lighthill, J. 1986. An Informal Introduction to Theoretical Fluid Mechanics. Oxford:

Clarendon Press.Maeda, T., Matsumura, T., Iida, M., Nakatani, K., and Uchida, K. 1993. Effect of

shape of train nose on compression wave generated by train entering tunnel.Proceedings of the International Conference on Speedup Technology for Railwayand Maglev Vehicles. Japan Society of Mechanical Engineers (Yokohama, Japan22–26 November) pp. 315–319.

Milne-Thomson, L.M. 1968. Theoretical Hydrodynamics. 5th ed. London: Macmillan.Moore, D.W. 1980. The velocity of a vortex ring with a thin core of elliptical

cross-section. Proceedings of the Royal Society of London A370: 407–415.Mohring, W. 1978. On vortex sound at low Mach number. Journal of Fluid Mechanics

85: 685–691.Mohring, W. 1980. Modelling low Mach number noise. Mechanics of Sound

Generation in Flows. Edited by E.-A. Muller, pp. 85–96. Berlin: Springer-Verlag.Powell, A. 1960. Aerodynamic noise and the plane boundary. Journal of the Acoustical

Society of America 32: 962–990.

Bibliography 211

Powell, A. 1963. Mechanisms of Aerodynamic Sound Production. AGARD Report No.466.

Powell, A. 1964. Theory of vortex sound. Journal of the Acoustical Society of America36: 177–195.

Rayleigh, Lord. 1945. Theory of Sound. 2 vols. New York: Dover.Saffman, P.G. 1993. Vortex Dynamics. Cambridge: University Press.Sears, W.R. 1941. Some aspects of non-stationary airfoil theory and its practical

applications. Journal of the Aeronautical Sciences 8: 104–108.Ting, L. and Klein, R. 1991. Viscous vortical flows. Lecture Notes in Physics, Vol. 374.

New York: Springer-Verlag.

Index

acoustic wave number, 43added mass tensor, 68

contribution to surface force, 69, 94relation to bound vorticity, 94sphere, 70

aerodynamic sound, definition, 1acoustic analogy, 25

airfoil theory, linearized, 148airfoil, elliptic, 187

rectangular, 187, 190analogy, acoustic, 25

Biot-Savart, application to creeping flow, 99formula, 88used to calculate sound source, 114with bound vorticity, 91

blade-vortex interactions, two-dimensions150, 175

linear theory, 178, 188nonlinear, 180, 189three-dimensions, 158, 186

blockage, 167bluff body, 194bound vorticity, 91boundary, solid, 41bubble, in sound field, 112

candle, blowing out, 91causality, 11circulation, definition, 87

Kelvin’s theorem, 87per unit length of wake, 148

compact, acoustically, 18body, 36, 128turbulence eddies, 30, 38

compact Green’s function, preliminarydefinition, 49, 52

airfoil of variable chord, 71

circular cylinder, 58cylindrical bodies, 58definition for Helmholtz equation, 64definition for wave equation, 65duct (tunnel) entrance, 77, 170duct with neck, 77, 80for incompressible flow, 81general form, 70half-plane, 74rigid strip, 60sphere, 53symmetric, 63two-dimensional, 136, 154uniform duct, 76wall aperture, 75wall cavity, 72wall projection, 72

complex potential, 100circular cylinder, 102half-plane, 105source, 103vortex, 103

complex velocity, 101compression wave, generated by train, 166compressive stress, 35conformal transformation, 100

half-plane, 105strip 61,

continuity equation, 2linearized, 5with source, 5

control surface, definition, 32creeping flow, 99Crighton, 143, 149Crocco’s equation, 85Curle, theory, 32

differential equation, 35integral equation, 36

213

214 Index

decibel, 4delta function

Fourier integral for, 43in three dimensions, 9

density, 2values for air and water, 3

diffusion, of vorticity, 87dipole, definition, 14

directivity, 16drag, 195far field form, 21fluid volume, 15in Curle’s equation, 35in Helmholtz equation, 45lift, 188, 190, 195near compact sphere, 54near edge, 62

drag, 191, 195

efficiency, acoustic, 32of quadrupole radiation, 32of surface dipoles, 37

end correction, 78energy, equation, 4

flux, 18kinetic, 82, 90

enthalpy, see total enthalpyentropy, 4

far field, definition, 17acoustic, 19calculation of, 20hydrodynamic, 89in terms of impulse, 89

Ffowcs Williams, 29Force, impulse formula, 93

added mass contribution, 94formula for translating body, 96lift, 182, 185produced by gust, 147role of bound vorticity, 94Sears’ formula, 149

Fraunhofer approximation, 21

Green’s function, see also compactGreen’s function

definition, 12for Helmholtz equation, 45free-space, 12, 46

gust, time harmonic, 148vortex, 146, 156

Heaviside unit function, definition, 32Helmholtz, equation, 44

Green’s function, 45inhomogeneous equation, 44, 57

homentropic, 4

hydrodynamic, pressure, 38far field, 89wavelength, 148

ideal acoustic medium, 25, 27impedance, 48impulse, definition, 90

theory of vortex sound, 131incompressible fluid, 7, 90, 93, 96infrasound, 166, 174integrals, transformation using Heaviside

function, 32intensity, acoustic, 19irrotational flow, sound waves in, 118

Karweit, 204Kasoev, 191Kelvin, circulation theorem, 87

definition of vorticity, 82formula for vortex ring speed, 92, 199theorem on kinetic energy, 84

kinetic energy, 82in terms of vorticity, 90Kelvin’s theorem on, 84

Kirchhoff spinning vortex, 132Kirchhoff vector, definition, 52

circular cylinder, 60rigid strip, 61singularities of, 146, 159, 179special cases, 71sphere, 54

Kraichnan-Phillips theorem, 39Kutta condition, 149

see also singularities

Lamb vector, 85, 96, 136Laplace equation

axisymmetric, 53polar from, 60two dimensions, 100

leading edge, Kutta condition, 150suction, 178, 182, 185

lift, 158lift dipole, 188, 190, 195Lighthill, 1

acoustic analogy, 25eighth power law, 29equation, 28equation reformulated in terms of vorticity,

117stress tensor, 28

linear, acoustics, 4Lighthill’s equation, 27momentum equation, 5

Mach number, 29Maeda, 171

Index 215

micro-pressure wave, 166Mohring, vortex sound formula, 132, 135momentum equation, 3

Crocco’s form, 85linearized, 5Reynolds’ form, 27

momentum flux tensor, 27monopole, 9, 13

in Curle’s equation, 35

Navier-Stokes equation, 3near field, hydrodynamic, 17no slip condition, 85nonlinear steepening in tunnel, 166

plane wave, 19, 23point source, definition, 9

impulsive, 10, 12in Helmholtz equation, 44monopole, 9

potential, velocity, 6vector, 88

potential flow interaction, 150, 161, 164Powell, 38power, acoustic, 19

quadrupole, definition, 15far field form, 22image in plane wall, 39in Helmholtz equation, 45Lighthill’s, 28

radiation condition, 11, 13rate of strain tensor, 26, 84Rayleigh, 47

end correction, 78reciprocal, theorem, 46, 127

problem for compact Green’s function, 50reduced frequency, definition, 148retarded potential, definition, 13retarded time, definition, 14Reynolds number, definition, 36Reynolds stress, definition, 28

in Lighthill tensor, 29linear, 29

Runge-Kutta integration, 108, 163

Sears, 147Sears function, 149self potential, vortex, 107singularities, deleting from Green’s function,

146, 159, 179, 187specific heats, ratio, 4speed of sound, definition, 6

values for air and water, 7sphere, added mass, 68

interacting with a vortex, 162, 199

modelled by point source, 9pulsating, 7, 67vibrating, 16, 57, 67

slender body approximation, 169spinning vortex pair, 120stationary phase, 75Stokes, drag, 98

stream function, 200Stokesian fluid, 3stress tensor

Lighthill’s, 28viscous, 26

suction force, 178, 182, 185dipole, 188, 190

surface, compact, 36impedance, 48noncompact, 37

test function, 9thermodynamics, Second Law, 11total enthalpy, definition for homentropic flow,

85acoustic variable, 116in terms of velocity potential, 117relation to pressure, 117

tunnel, 166, 174turbulent nozzle flow, 25

velocity, acoustic particle, 5, 18in terms of impulse, 89potential, 6

vibrating body, 56low frequency radiation from, 65, 129

vibrating sphere, 16, 57, 67viscosity, 3

kinematic, 3values for air and water, 3

volume source, 5dipole, 15

vortex-airfoil interaction, linear theory, 156,178, 188

nonlinear, 180, 189periodic, 183three-dimensional, 158, 186two-dimensional, 150

vortex, coaxial rings, 133complex potential, 103equation of motion, 106, 175, 180, 192, 195,

198, 204finite core, 160line, 84, 158motion near half-plane, 108outside cylinder, 111ring, 92, 173, 199self potential, 107shedding, 145, 159tube, 84

216 Index

vortex pair, 204spinning, 120

vortex sheet, 87, 88, 91wake, 150

vortex sound, at low Mach numbers,119

for cylindrical bodies, 130, 194for nonvibrating body, 130from spinning vortices, 120from spoiler, 191from vortex interacting with cylinder,

139from vortex near half-plane, 143, 173from vortex near sphere, 162, 199from wall aperture, 204in terms of surface force, 129

vortex sound equation, 116, 118linearized, 156low Mach number approximation, 120

vortex-surface interaction noise, 124general formula, 127high Reynolds number form, 128influence of vortex shedding, 143vortex near sphere, 162

vortex near spoiler, 191wall mounted cylinder, 197

vorticity, bound, 91equation, 84equation for Stokesian fluid, 86in force formulae, 93, 96Kelvin’s definition, 82molecular diffusion, 87source of sound, 114

wake, cylinder, 148vortex sheet, 150

wall drag, 191wave equation

classical acoustics, 6for pressure, 6in irrotational mean flow, 118inhomogeneous, 13radially symmetric, 10

wavelengthacoustic, 7, 17, 43at 1 kHz, 7hydrodynamic, 148

wave number, 43

Documents

Theory of Vortex Sound (Cambridge Texts in Applied Mathematics) by M. S. Howe