Mathematical Tripos: Part II Waves

Contents
0 Waves i
0.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i
0.2 Examples of Wave Motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i
0.3 Outline of Course . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . i
0.4 Appropriate Books . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii
0.5 Extras . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . ii

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
1 Acoustics/Sound Waves 1
1.1 Examples of Applications . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2 The Equations of Motion for Compressible Fluids . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2.1 Solids, liquids and gases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 1
1.2.2 Notation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2.3 Conservation of mass . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.2.4 Conservation of momentum . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2
1.3 Thermodynamics (on a need-to-know basis) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3.1 Isentropic flows and a closed system . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3
1.3.2 Homentropic flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 4
1.4 Linear Acoustic Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4.1 Linearised equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 5
1.4.2 Wave equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.4.3 The acoustic velocity potential . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 6
1.4.4 Wave–energy density and wave–energy flux . . . . . . . . . . . . . . . . . . . . . . . . . . . 7
1.4.5 * Energy density and energy flux: respectable method (unlectured) * . . . . . . . . . . . . . 8
1.5 Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 8
1.5.1 Properties of plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 9
1.5.2 Harmonic plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 10
1.5.3 Energies of harmonic plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 11
1.6 Reflection and Transmission (or Why Bass Notes Ignore Walls) . . . . . . . . . . . . . . . . . . . . 11
1.7 Waves in Three Dimensions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.7.1 Acoustic waves in a rectangular duct . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 12
1.7.2 Spherically symmetric waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 15
1.7.3 Solution for a pulsating sphere . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 16
1.7.4 Loud speakers: why tall and thin is good . . . . . . . . . . . . . . . . . . . . . . . . . . . . 17

Mathematical Tripos: Part II Waves I © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Appendices a
1.A * Thermodynamics [for Dummies] (unlectured) * . . . . . . . . . . . . . . . . . . . . . . . . . . . . a
1.A.1 0th Law – there exists temperature . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . a
st
1.A.2 1 Law – energy is conserved if all forms are counted . . . . . . . . . . . . . . . . . . . . . a
1.A.3 Adiabatic processes, isolated systems and reversible changes . . . . . . . . . . . . . . . . . . a
1.A.4 2nd Law – chaos increases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . b
1.B * Perfect Gases (unlectured) * . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . c
1.B.1 Pressure . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . c
1.B.2 Internal energy and specific heats . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . c
1.B.3 Entropy of a perfect gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . d

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
2 Finite Amplitude 1D Waves 19
2.1 Nonlinear Acoustics in 1D: Governing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.2 Riemann Form of Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 19
2.2.1 Example: perfect gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.3 Solution by Characteristics . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.3.1 The Cauchy initial-value problem (IVP) . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.3.2 General form of the solution . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 21
2.3.3 * Numerical method * . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.4 Simple Waves and Shock Formation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.4.1 Simple waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 22
2.4.2 Generic form . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 24
2.4.3 Wave steepening . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.4.4 Shock formation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.4.5 Alternative estimate of shock formation time . . . . . . . . . . . . . . . . . . . . . . . . . . 25
2.4.6 Example . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.5 Piston Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 26
2.5.1 Piston accelerates away from the gas . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 27
2.5.2 Piston accelerates towards the gas (at some point) . . . . . . . . . . . . . . . . . . . . . . . 28
2.6 Shock Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 29
2.6.1 Rankine-Hugoniot relations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.6.2 The Hugoniot adiabatic . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 30
2.6.3 The entropy jump . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 31
2.6.4 Upstream/downstream flow is supersonic/subsonic . . . . . . . . . . . . . . . . . . . . . . . 32
2.6.5 Upstream/downstream flow is supersonic/subsonic – perfect gas (unlectured) . . . . . . . . 32
2.6.6 Moving shocks . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 33
2.6.7 Shock propagating into a perfect gas at rest. . . . . . . . . . . . . . . . . . . . . . . . . . . 34
2.7 Nonlinear Shallow Water Waves in 1D . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.7.1 Governing Equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 35
2.7.2 * Hydraulic jumps * . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 36

Mathematical Tripos: Part II Waves II © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Appendices e
2.A * Shock Formation in Trombones * . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . e

3 Linear Elastic Waves 38

3.1 Governing Equations of Linear Elasticity . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.1.1 Deformation of a solid . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.1.2 The stress tensor . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 38
3.1.3 Momentum, angular momentum and energy . . . . . . . . . . . . . . . . . . . . . . . . . . . 39
3.1.4 Fluid and solid continua . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40
3.1.5 Strain . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 40

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
3.1.6 Constitutive equation for a linear elastic solid . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.1.7 Simple deformations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 41
3.1.8 Equations of motion . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 42
3.1.9 Energy equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 43
3.2 Compressional Waves and Shear Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.2.1 Waves of dilatation and rotation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 44
3.2.2 Plane waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 45
3.2.3 Harmonic waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 46
3.2.4 Interface boundary conditions . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.2.5 Rayleigh waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 47
3.3 Reflection and Refraction of Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 50
3.3.1 Reflection and refraction of SH waves at an interface . . . . . . . . . . . . . . . . . . . . . . 50
3.3.2 Reflection of P waves at a rigid interface . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 52
3.3.3 Seismic tomography . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 53

4 Linear Dispersive Waves 55

4.1 Geometric Dispersion in Wave Guides . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.1.1 Acoustic waves in a rectangular duct . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.1.2 Love waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 55
4.2 Initial Value (Cauchy) Problems . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.2.1 General solution of the beam equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 57
4.2.2 Exact solution of the beam equation for Gaussian modulation . . . . . . . . . . . . . . . . . 59
4.2.3 Modulation of periodic wavetrains . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 60
4.2.4 Large time solution — Kelvin’s method of stationary phase . . . . . . . . . . . . . . . . . . 60
4.2.5 Example: beam equation . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 63
4.2.6 Example: dispersion in a waveguide (unlectured) . . . . . . . . . . . . . . . . . . . . . . . . 64
4.3 Capillary-Gravity Waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 64
4.3.1 Limiting Cases . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 66
4.4 Internal Gravity Waves in a Stratified Incompressible Fluid . . . . . . . . . . . . . . . . . . . . . . 69

Mathematical Tripos: Part II Waves III © S.J.Cowley@maths.cam.ac.uk, Lent 2019

5 Ray Theory 72
5.1 Ray Tracing . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
5.1.1 Multiple scales . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 72
5.1.2 Phase and wavecrests . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 73
5.1.3 Ray-tracing equations . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 74
5.1.4 Hamilton’s equations of classical mechanics . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
5.1.5 * Wave action * . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 75
5.1.6 Example: gravity waves approaching a beach (surfing) . . . . . . . . . . . . . . . . . . . . . 76
5.2 Fermat’s Principle . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
5.2.1 Time independent case . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 77
5.2.2 Snell’s law of refraction . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 78
5.2.3 Example: sound waves in inhomogeneous media . . . . . . . . . . . . . . . . . . . . . . . . . 78
5.3 Media with Mean Flows . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 79
5.3.1 The Doppler effect . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 80
5.3.2 Stationary capillary-gravity waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 81
5.3.3 Mach cone . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
5.3.4 Ship/duck waves . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 82
5.3.5 Internal gravity waves in a horizontal wind . . . . . . . . . . . . . . . . . . . . . . . . . . . 85

Mathematical Tripos: Part II Waves IV © S.J.Cowley@maths.cam.ac.uk, Lent 2019

0 Waves
0.1 Introduction
• A wave is one of those things that if you see it you know what it is!
• Intuition: A wave is something that goes up and down – or just up – or just down.
• Dictionary: A wave is a disturbance propagating in a medium at a finite speed . . . but what about
standing waves?

Wave motion is a mechanism for energy communication over large distances

without any necessary bulk motion of the medium.

• Waves are not necessarily oscillatory, e.g. one-signed pulse on a string.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
• Waves do not necessarily have small amplitudes, e.g. shocks, breaking waves.
• There is no implication of single-speed propagation, i.e. dispersion is possible.

0.2 Examples of Wave Motion

• Waves with elastic restoring forces, e.g. sound waves (aircraft noise, ultrasound), seismic waves
(earthquakes, nuclear explosions, prospecting), your pulse.
• Waves arising from gravity and density inhomogeneities, e.g. water waves (including duck-generated
waves), internal waves.
• Waves arising from Coriolis forces and other rotation effects, e.g. inertial waves, Rossby waves.
• Waves associated with a magnetic field ‘frozen’ in a fluid, e.g. Alfven waves.
• Electromagnetic waves such as light (see also Electrodynamics).
• Gravitational waves (see also GR).
• The wave-function in quantum mechanics (see QM courses).

0.3 Outline of Course

Uses IB Methods and little bits of IB Complex Methods and IB Fluid Dynamics (but will recap).

Sound waves. Elementary properties of linear acoustic waves, i.e. linear compression waves in simple flu-
ids (no dissipation, no dispersion) – or ‘Why can you hear me?’. Nonlinear finite-amplitude effects in
one dimensional gas dynamics – cumulative nonlinear effects can lead to shock waves (sonic booms);
relation to hydraulic jumps, traffic flow, etc. ‘Weak’, i.e. discontinuous, solutions of nonlinear PDEs.
See also Differential Equations, Methods, Further Complex Methods and Electrodynamics.
Formally:
Sound waves. Equations of motion of an inviscid compressible fluid (without discussion
of thermodynamics). Mach number. Linear acoustic waves; wave equation; wave-energy
equation; plane waves; spherically symmetric waves. [4]
Non-linear waves. One-dimensional unsteady flow of a perfect gas. Water waves. Riemann
invariants; development of shocks; rarefaction waves; ‘piston’ problems. Rankine-Hugoniot
relations for a steady shock. Shallow-water equations *and hydraulic jumps*. [5]
Linear elastic waves. Similarities and differences between solids and fluids. Longitudinal (P) and trans-
verse (S) [seismic] waves, and their interaction with interfaces.
Formally:
Elastic waves. Momentum balance; stress and infinitesimal strain tensors and hypothesis
of a linear relation between them for an isotropic solid. Wave equations for dilatation
and rotation. Compressional and shear plane waves; simple problems of reflection and
transmission; Rayleigh waves. [5]

Mathematical Tripos: Part II Waves i © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Dispersive waves, i.e. waves whose speed and direction of propagation vary with wavelength, e.g. acous-
tic waveguides, Love waves, surface gravity waves, waves in a density-stratified medium (see also
Electrodynamics). Solution by Fourier analysis. Approximate solutions by stationary phase (see also
Asymptotic Methods). The difference between the directions of crest and energy propagation.
Formally:
Dispersive waves. Rectangular acoustic wave guide; Love waves; cut-off frequency. Rep-
resentation of a localised initial disturbance by a Fourier integral (one-dimensional case
only); modulated wave trains; stationary phase. Group velocity as energy propagation
velocity; dispersing wave trains. Water waves; internal gravity waves. [6]
Modulated wavetrains and ray theory. Rays, Fermat’s Principle (see also Variational Principles),
and the duck’s wake.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Formally:
Ray theory. Group velocity from wave-crest kinematics; ray tracing equations. Doppler
effect; ship wave pattern. Cases where Fermat’s Principle and Snell’s law apply. [4]

0.4 Appropriate Books

Fluids
• H. Ockendon & J.R. Ockendon, Waves and Compressible Flow, Springer. Recommended by a su-
pervisor.
• J. Billingham & A.C. King, Wave Motion: Theory and Application. Accessible.
• G.B. Whitham, Linear and Nonlinear Waves, Wiley-Interscience. Very good for the nonlinear part
of the course. Rather expensive.
• A.R. Paterson, A First Course in Fluid Dynamics, CUP. Recommended by a student (many years
ago) for characteristics, etc.
• L.D. Landau & E.M. Lifshitz, Fluid Mechanics, Pagan Press. This is where I learnt ‘sound’ from in
1976/77, but some students do not like the dated style.
• M.J. Lighthill, Waves in Fluids, CUP. This is the book of a former version of the Waves course (the
version I took). Rather too wordy for my liking.
• A.P. Dowling & J.E. Ffowcs Williams, Sound and Sources of Sound, Ellis Horwood. Even more
wordy book written for engineers.
• M. Van Dyke, An Album of Fluid Motion, Parabolic. An excellent book for your coffee-table.

Solids

• J. Billingham & A.C. King, Wave Motion: Theory and Application. Accessible.
• J.D. Achenbach, Wave Propagation in Elastic Solids, North Holland. Lots of detail (i.e. written for
Americans).
• D.R. Bland, Wave Theory and Applications, Clarendon Press.
• J.A. Hudson, The Excitation and Propagation of Elastic Waves, CUP. Compact, but the key points
are there.

0.5 Extras
Classic Film. Aerodynamic Generation of Sound staring Lighthill (former Lucasian Professor) and
Ffowcs-Williams (former Master of Emmanuel): http://www.youtube.com/watch?v=8BmESsMroRM.
Extra Credit. Remember that you can obtain extra credit by doing the CATAM Project on theWaves
course, i.e. Project 2.10 on Phase and Group Velocity.

Mathematical Tripos: Part II Waves ii © S.J.Cowley@maths.cam.ac.uk, Lent 2019

1 Acoustics/Sound Waves
The theory of compression/rarefaction waves in compressible fluids (e.g. air, water).

1.1 Examples of Applications

• The noise generated by machinery, e.g. jet engines, for which an understanding of the power radiated
is important. Note that:
(i) the acoustic power of a large jet aircraft ≡ all the world shouting at once;
(ii) the acoustic power generated by a Wembley crowd during a match ≡ the power to fry one egg!

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
• Sonar – both navigation and detection.
• Ultrasonics – both medical (e.g. imaging of tissue, destruction of kidney stones) and industrial (e.g.
testing for cracks in pressure vessels).

• Design of concert halls, ‘hi-fi’.

• Explanation of how musical instruments work.
• Snoring.

1.2 The Equations of Motion for Compressible Fluids

1.2.1 Solids, liquids and gases

Solids: A material behaves as a solid if it suffers only a finite deformation on application of a finite force.
The solid is said to be elastic if the deformation tends to zero when the applied force is released.
Fluids: A material is said to behave as a fluid if its deformation increases indefinitely when a deforming
force, i.e. not a purely compressive force, is applied.

Certain materials can behave sometimes like a fluid and sometimes like a solid.
In solids there are strong intermolecular forces. Fluids can be divided into liquids and gases. In liquids
the intermolecular forces are still fairly strong. The intermolecular distance is similar to that in solids,
but the molecules can move relative to each other. In gases the intermolecular forces are much weaker,
and the intermolecular spacing is much greater. There is a random arrangement of molecules. Gases are
much more compressible than liquids or solids.
We begin by studying fluids. In particular we consider fluids that are:

(a) homogeneous, i.e. the same at all points,

(b) dissipationless, i.e. we ignore viscous effects, etc., on the basis that dissipation is small over many
characteristic wavelengths (c.f. waves in water and treacle); e.g. middle C has a wavelength of ap-
proximately 4 feet while dissipation occurs over hundreds of feet.

Suitable revision/introductory books include:

• Acheson, D.J., Elementary Fluid Dynamics

• Batchelor, G.K., An Introduction to Fluid Dynamics

• Paterson, A.R., A First Course in Fluid Dynamics

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1.2.2 Notation

We make the continuum assumption. Since it is not possible to describe the motion of each molecule
separately we assume that fluids are continuous, and that fluid properties, e.g. the density ρ, can be
defined at a point x. This is usually possible as long as the molecular spacing is much less than the
macroscopic lengthscale (e.g. the wavelength of a sound wave). Hence we define

ρ(x, t) : The density of the fluid at x, t 
p(x, t) : The pressure in the fluid at x, t 5 variables in three dimensions
u(x, t) : The velocity of the fluid at x, t


Remark. We need a closed system of governing equations for these five variables, i.e. we need five equations.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
1.2.3 Conservation of mass

We consider an arbitrary fixed volume V with surface S (a pretty standard approach if we are going to
apply a conservation argument). Then
Rate of increase of mass in V = Rate of flow of mass across S into V
+ Source strength inside V
i.e. Z Z Z
d
ρ dV = − ρ u.dS + QdV ,
dt
where Q(x, t) is the rate of creation of mass per unit volume at x, t.

Since the volume V is fixed, we can differentiate under

the integral sign. Then using the divergence theorem we
conclude that
Z Z Z
∂ρ
dV = − ∇.(ρu) dV + QdV .
∂t
Since the volume is arbitrary it follows that
∂ρ
+ ∇.(ρu) = Q(x, t) , (1.1a)
∂t
or equivalently
Dρ ∂ρ
≡ + u.∇ρ = −ρ∇.u + Q . (1.1b)
Dt ∂t

Remarks

(i) This is the first equation that we require.

(ii) Note that since the fluid is not incompressible, ∇.u is not necessarily zero (cf. IB Fluid Dynamics).

1.2.4 Conservation of momentum

Apply conservation of momentum to the same fixed volume V. Then (remember that we are neglecting
any small viscous effects)
Z Z Z Z Z
d
ρu dV = − (ρu)u.dS − p dS + Qu dV + ρF dV ,
dt
where F(x, t) is the body force per unit mass (e.g. gravity). Since the volume is fixed, and again using
the divergence theorem, it follows that
Z Z  
∂ ∂ ∂p
(ρui ) dV = − (ρui uj ) − + Qui + ρFi dV .
∂t ∂xj ∂xi

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 Because the volume is arbitrary
∂ ∂ ∂p
(ρui ) + (ρui uj ) = − + Qui + ρFi . (1.2a)
∂t ∂xj ∂xi
An alternative form of this equation, namely Euler’s Equation, can be obtained by means of the
manipulation [(1.2a) − (1.1b)ui ]:
∂u
ρ + ρ(u.∇)u = −∇p + ρF . (1.2b)
∂t

Kinematic boundary condition. At a stationary rigid boundary, the boundary condition u · n = 0 is sat-
isfied. Alternatively, if the boundary is moving with speed U, then we require that u · n = U · n.
Since we are assuming that the fluid is inviscid there are no conditions on u × n.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Dynamic boundary condition. At a free boundary, e.g. the atmosphere, p = const., or wlog, p = 0. At
+
an interface between fluids, assuming that there is no surface tension, [p]− = 0. If there is surface
+
tension, σ, then [p]− = σκ, where κ is the curvature.
Remark. Euler’s equation provides three more of the equations that we require. Together, (1.1b) and
(1.2b) constitute four scalar equations for the five scalar variables ρ, p and u. To complete the
system of governing equations we need another relation between ρ, p and u. We obtain this relation
by considering the thermodynamics of the flow (but the Schedules state that I am to do this
‘without discussion of thermodynamics’ §).

1.3 Thermodynamics (on a need-to-know basis)

From school physics we know that Boyle’s law for a perfect gas states that
pv = nR̄T , i.e. p = ρRT , (1.3)
where v is the volume, T is the temperature, ρ is the density, n is the number of moles of gas, R̄ is the
n
universal gas constant and R = ρv R̄ is the specific gas constant. If T is known to be constant, i.e. if the
flow is isothermal, then we have the required fifth equation relating ρ, p and u. Unfortunately, T usually
varies.
01/16 Remark. We are no further forward: we now have five scalar equations for six scalar variables.
01/17
01/18
01/19 1.3.1 Isentropic flows and a closed system

In general, as in (1.3), p ≡ p(ρ, T ). Hence to close the system we need another equation for T , e.g. by
considering heat transport by conduction/advection and/or thermodynamics.
We will take an easy way out and assume that
timescale of motion  timescale of thermal diffusion ,
so that the heat content of each fluid parcel/particle is constant (this is not the same at the temperature
of each fluid parcel/particle being constant). This is a good approximation for sound in air where typically
f = 1kHz, period=10−3 s, λ = 30cm, so that
λ2
tdiffusion = ≈ 104 s  10−3 s = tperiod ,
k/ρcP
where k is the thermal conductivity, cP is the specific heat capacity at constant pressure, and k/ρcP is
the thermal diffusivity.
Technically, we will assume reversible adiabatic motion, also referred to as isentropic motion. For those
that are interested, the thermodynamic theory behind this assumption is expounded in Appendix 1.A.
In short, a motion is isentropic if the entropy, S, of a fluid parcel/particle is constant, i.e. if
DS
= 0. (1.4a)
Dt

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The Second Law of Thermodynamics states that the entropy, S, exists, and, like the temperature T ,
pressure p and density ρ, is a parameter of state. Each of these quantities can be viewed as a function of
two of the others, e.g. p ≡ p(ρ, T ), or more pertinently,
p ≡ p(ρ, S) . (1.4b)

A closed system. The conservation of mass equation (1.1b), Euler’s equation (1.2b), the isentropic motion
equation (1.4a) and the equation of state (1.4b) together constitute six equations for the six variables
p, ρ, S and u. Hence we now have a closed system.

1.3.2 Homentropic flows

In fact, other than in shocks, we make the further assumption that the entropy takes a uniform constant

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
value everywhere, i.e. that S is a constant. The flow is then said to be homentropic, and equations (1.1b),
(1.2b), and (1.4b) constitute a closed system of five equations for the five effective variables p, ρ and u.

Perfect gas. If the flow of a perfect gas is homentropic, i.e. reversible, adiabatic and uniform, then it
can be shown, e.g. see Appendix 1.B, that
 γ
p ρ
= , (1.5a)
p0 ρ0
where p0 and ρ0 are initial/reference values, γ = cP /cV is the constant ratio of specific heats, and
cV is the specific heat capacity at constant volume. This equation, together with the conservation
of mass equation (1.1b) and Euler’s equation (1.2b), constitute a closed system of five equations for
the five unknowns p, ρ, and u.
The speed of sound. Subsequently we shall see that the square of the speed of sound, c2 , is given by
 
∂p p
c2 ≡ = γ = γRT. (1.5b)
∂ρ S ρ
Newton made a mistake of assuming an isothermal change (∂p/∂ρ)T = RT , so missing the factor
of γ (which is 1.4 for dry air).
Internal energy and enthalpy. For an isentropic flow (of which a homentropic flow is a special case),
it follows from (1.A.5a) that
p
de = −pdV = 2 dρ , (1.5c)
ρ
where e is internal energy per unit mass. Alternatively view this equation as ‘God given’ based on
conservation of [mechanical] energy for an adiabatic reversible process. Thence from (1.5a)
 γ
de p p0 ρ
= 2 = 2 , (1.5d)
dρ ρ ρ ρ0
and
1 p
e(p, ρ) = , (1.5e)
γ−1ρ
where the constant of integration is normalised to be zero. As shown in Appendix 1.B, the internal
energy of a perfect gas, when viewed as a function of T and ρ, can be expressed as
e(T, ρ) = cV T . (1.5f)
The combination h = e + pV , is known as enthalpy. For a perfect gas it is given by
p γ p
h(p, ρ) ≡ e + = . (1.5g)
ρ γ−1ρ

Finally, when studying shocks we will need the entropy of a perfect gas. This can be calculated as,
see (1.B.6d),   γ 
p ρ0
S(p, ρ) = S0 + cV ln . (1.5h)
p0 ρ

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1.4 Linear Acoustic Waves

1.4.1 Linearised equations

Conservation of mass from (1.1b):

∂ρ
+ ∇ · (ρu) = Q(x, t) . (1.1b)
∂t

Conservation of momentum from (1.2a):

∂ ∂ ∂p
(ρui ) + (ρui uj ) = − + Qui + ρFi . (1.2a)
∂t ∂xj ∂xi

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
∂ ∂
∂t (1.1b) − ∂xi (1.2a) :

∂2ρ ∂2 ∂Q ∂ ∂2
2
− p= − (ρFi + Qui ) + (ρui uj ) . (1.6)
∂t ∂xi ∂xi ∂t ∂xi ∂xi ∂xj

An exact solution is a uniform state of rest:

ρ = ρ0 , p = p0 , u = 0, Q = 0, F = 0. (1.7)

Consider small perturbations to this state:1

ρ = ρ0 + ρ̃, |ρ̃|  ρ0 (1.8a)

p = p0 + p̃, |p̃|  p0 (1.8b)
0
u, Q, F ‘small

Then

ρFi =ρ0 Fi + ρ̃Fi ≈ ρ0 Fi

↑ ↑
small doubly small

Qui and ρui uj are also doubly small.

Hence at leading order (1.6) linearises to

∂ 2 ρ̃ ∂Q
− ∇2 p̃ = − ρ0 ∇  F . (1.9)
∂t2 ∂t
From thermodynamics, e.g. (1.4b),
p ≡ p(ρ, S) . (1.10a)
Assume that the state of rest is homentropic, i.e. S = S0 , and remains so under perturbation. Then

p0 + p̃ = p(ρ0 + ρ̃, S0 )
∂p
= p(ρ0 , S0 ) + ρ̃ (ρ0 , S0 ) + ... .
∂ρ
Hence
p̃ = c20 ρ̃ , (1.10b)
where
∂p
c20 = (ρ0 , S0 ) . (1.10c)
∂ρ
1 Some students do not like approximations, however we make them all the time. For instance, we ignore relativistic

effects when modelling the flight of a cricket ball (although not in the design of GPS systems).

Mathematical Tripos: Part II Waves 5 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 1.4.2 Wave equation

Substitute (1.10b) into (1.9), then

1 ∂2
 
∂Q
∇2 − p̃ = − + ρ0 ∇  F . (1.11)
c20 ∂t2 ∂t

Assume no mass sources (Q = 0) and no body forces (F = 0). Then p̃, and ρ̃ (from (1.10b)), satisfy the
3D wave equation with wave speed c0 , while the linearised versions of (1.1b) and (1.2a) are
∂ ρ̃
+ ρ0 ∇  u = 0 , (1.12a)
∂t
∂u
ρ0 = −∇p̃ . (1.12b)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
∂t
∂ ∂
From ∂xi (1.11), ∂t (1.11), (1.10b), (1.12a) and (1.12b), it follows that
 
∂u
2 =0 and 2 (∇  u) = 0 , (1.13)
∂t
where
1 ∂2
 
2
2
 = ∇ − 2 2 . (1.14)
c0 ∂t
From integrating the first equation of (1.13), and then using the second equation of (1.13),

2 u = α(x) where ∇  α = 0. (1.15)

Assume the motion starts from rest, then α = 0 and we conclude that

2 (p̃, ρ̃, u) = 0 . (1.16)

02/04

1.4.3 The acoustic velocity potential

Curl the linearised momentum equation (1.12b), then

∂ω
=0 where ω = curl u = vorticity.
∂t
Hence within linear theory
u = ∇ϕ(x, t) + urot (x) . (1.17a)
For motions with no vorticity at t = 0, urot ≡ 0. Then using (1.17a) and (1.10b) to integrate (1.12b) we
obtain  
p̃ ρ0 ∂ϕ
ρ̃ = 2 = − 2 (x, t) + β(t) . (1.17b)
c0 c0 ∂t
R
Wlog we may assume that β ≡ 0 (the transformation ϕ → ϕ − β dt does not change ∇ϕ), so

∂ϕ ρ0 ∂ϕ
p̃ = −ρ0 , ρ̃ = − , u = ∇ϕ. (1.17c)
∂t c20 ∂t

Substitute (1.17b) into (1.12a) to conclude that

2 ϕ = 0 . (1.18a)

Hence ‘everything’ satisfies the wave equation with wave speed

 1/2
∂p
c0 = . (1.18b)
∂ρ S

Sound speed.

Mathematical Tripos: Part II Waves 6 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 • For dry air at 293o K, c0 = 340 ms−1 , i.e. at 20o C, c0 = 760 mph. (Homework: repeat Newton’s
experiment in Nevile’s Court.)
• For water c0 ≈ 1500 m s−1 .

Linear equations. The equations for p̃, ρ̃, ϕ and u are linear, so we can superimpose solutions, e.g. if ϕ1
and ϕ2 are solutions then so is λϕ1 + µϕ2 for constant λ and µ.
Linearisation: can we do it? When linearising the momentum equations we neglected ρ0 (u.∇)u and kept
ρ0 ∂u
∂t . Suppose that |u| has a typical magnitude U , and varies on a lengthscale L. Then in an order
of magnitude sense (and using the wave equation)

1 1 ∂
∇∼ ∼ .
L c0 ∂t

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Hence
ρ0 (u.∇)u ρ0 U 2 /L U
∂u
∼ = ≡M, (1.18c)
ρ0 ∂t ρ0 c0 U/L c0
where M is the Mach number. Hence for linearisation we need

02/16 M 1 i.e. fluid speed  wave speed.

02/17
02/18
02/19 1.4.4 Wave–energy density and wave–energy flux

There are difficulties in defining energy density and energy flux for many media, especially non-uniformly
moving media where second-order terms in u, etc. need to be retained in the equations.
For perturbations to a state of rest we can ‘cheat’. Form
p̃
u  Momentum Eqn (1.12b) + Mass Eqn (1.12a) ,
ρ

and use p̃ = c20 ρ̃ to obtain the Wave-Energy Equation

∂
(Ek + Ep ) + ∇  I = 0 , (1.19)
∂t
where
Ek = 21 ρ0 u2

: K.E. density 
2 2
1 c0 ρ̃
(1.20a)
Ep = 2 ρ : P.E. density due to compression 
0

I = p̃u : Wave-energy flux or ‘acoustic intensity’. (1.20b)

Integrate over a fixed volume V to obtain
Z Z Z
d
(Ek + Ep ) dV = − ∇  I dV = − p̃ u  dS
dt V V S

Rate of increase of Rate of working of

acoustic energy density pressure over S

Loudness. Define the intensity or loudness in decibels as

loudness ≡ 120 + 10 log10 |I| , with I = |I| measured in Wm−2 .

10−12 Wm−2 → 0 dB

: which can just be heard at 3000Hz;
I=
1 Wm−2 → 120 dB : the threshold of pain.
The acoustic power of a voice in conversation ≈ 10−5 W.
Remark. The intensity depends on distance like 1/r2 (see later).

Mathematical Tripos: Part II Waves 7 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

1.4.5 * Energy density and energy flux: respectable method (unlectured) *

The total energy of the system is given by

 Z ρ 
1 2 p(ρ̂, S)
E = ρ 2u + dρ̂ , (1.21a)
ρ̂2

where the second term on the RHS is the integral of (−pdV ) at constant entropy (where V is the volume
per unit mass and ρ = V −1 ).
Using conservation of mass (1.1b) with Q = 0, and Euler’s equation (1.2b) with F = 0, it follows that
∂E ∂u p ∂ρ E ∂ρ E
+ ∇. (uE + pu) =ρu + + + E∇. u + (u. ∇)ρ
∂t ∂t ρ ∂t ρ ∂t ρ

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
p
+ ρ(u. ∇)( 21 u2 ) + (u. ∇)ρ + (u. ∇)p + p∇. u
ρ
=0 , (1.21b)

or, in integral form for a fixed volume V, we have the exact result
Z Z Z
d
E dV = − Eu·n dS − pu·n dS
dt V S S
Change of energy Advection of Rate of working
in V = energy into V + by p at surface S

The following derivation for small-amplitude waves is slightly tricky because of the need to include
quadratic terms. So write ρ = ρ0 + ρ̃, p = p0 + p̃, etc., and expand (1.21a) including quadratic terms:
Rρ  
E = 12 ρ0 u2 + ρ 0 ρp2 dρ + ρ ρp02 ρ̃ + 12 ρ ∂ρ ∂ p
ρ2 ρ̃2 + . . .
0 0
R ρ0 2
1 2 p p0 1 c0 2
= 2 ρ0 u + ρ ρ2 dρ + ρ0 ρ̃ + 2 ρ0 ρ̃ + ...

= Ek + E0 + Er + Ep + ... ,

where the quadratic terms Ek and Ep are as in (1.20a), and

R ρ0 R 
p ρ0 p p0
E0 = ρ0 ρ2 dρ and Er = ρ2 dρ + ρ0 ρ̃ .
Internal energy per unit Change in energy per unit (1.21c)
volume when fluid at rest volume due to change in mass

From the conservation of mass equation (1.1b) it follows, after some very careful algebra, that (E0 + Er )
exactly satisfies
∂
(E0 + Er ) + ∇. (u(E0 + Er ) + p0 u) = 0 , (1.21d)
∂t
Hence from (1.21b) and (1.21d), it follows that consistent to second order we can write, as in (1.19),
∂
(Ek + Ep ) + ∇(p̃u) ≈ 0 .
∂t

1.5 Plane Waves

A solution to (1.18a), i.e.

1 ∂2
 
∇2 − 2 2 ϕ = 0 , (1.18a)
c0 ∂t
is
ϕ = f (lx + my + nz − ωt) (1.22a)
for any function f if
(l2 + m2 + n2 )c20 = ω 2 . (1.22b)

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The name plane waves follows from the observation that at any fixed time, ϕ is constant on the planes

lx + my + nz = constant .

Take axes OXYZ with OX k to k = (l, m, n). Then

ϕ = f (kX − ωt) = f (k.x − ωt) , (1.23a)

where
k 2 = |k|2 = l2 + m2 + n2 , and ω = ±kc0 . (1.23b)

Definition and remark. The combination θ = (k.x − ωt) is the phase of the wave. ϕ is constant on lines of
constant θ, and hence solution (1.23a) describes a wave propagating in the k direction with speed c0 ,

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
whatever the choice of f .

By superposition another solution to (1.18a) is

ϕ = f (k(X − c0 t)) + g(k(X + c0 t)) , (1.24a)

i.e. two plane waves propagating in opposite directions with sound speeds ±c0 , whatever the choice of f
and g.

Characteristics. Recall from Part IB that D’Alembert’s general solution to ϕtt − c20 ϕXX = 0 is

ϕ = F (X − c0 t) + G(X + c0 t) (1.24b)

for any functions F and G. The natural coordinates here are the characteristics

ξ = X − c0 t and η = X + c0 t . (1.24c)

Remarks
(i) The characteristics are straight, but are a non-
orthogonal co-ordinate system (unless c0 = 1).
(ii) The characteristics ξ and η represent right and left
travelling waves respectively.

(iii) In the case of a single ‘simple’ linear wave, crests and

troughs, and indeed any points of constant phase,
propagate along the relevant characteristic.

1.5.1 Properties of plane waves

Longitudinal waves. From (1.22a) and (1.24a)

u = ∇ϕ = k (f 0 (k.x − kc0 t) + g 0 (k.x + kc0 t)) . (1.25a)

Hence sound consists of longitudinal waves since u is k to the propagation vector k.

Acoustic Impedance. From (1.17c)

p̃ = −ρ0 ϕt = ρ0 kc0 (f 0 (k.x − kc0 t) − g 0 (k.x + kc0 t)) . (1.25b)

Hence, for a single wave with u = k̂u,

)
p̃ = ρo c0 u (g = 0)
(1.25c)
p̃ = −ρ0 c0 u (f = 0)

Hence p̃ and u are in phase. ρ0 c0 is referred to as the acoustic impedance of the medium.

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 Equi-partition of energy. For a single wave (e.g. g = 0) it follows from (1.10b), (1.20a), (1.25a), (1.25b)
and (1.25c) that )
Ek = 21 ρ0 k 2 f 02
(1.26a)
Ep = 12 ρ0 k 2 f 02 = Ek
Hence there is instantaneous equi-partition of energy between kinetic and elastic modes (equi-
partition of energy is often only true after time averaging). Further, from (1.20b), (1.25a), (1.25b)
and (1.25c)
I = ρ0 c0 k 2 f 02 k̂ = (Ek + Ep )c0 k̂ . (1.26b)
Hence I is generated by the transport of acoustic energy at the energy propagation velocity
cg = c0 k̂. (1.26c)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
1.5.2 Harmonic plane waves

Harmonic plane waves correspond to the choice:

f (α) = Aeiα , (1.27a)
i.e.
ϕ = A exp(ik. x − iωt) , (1.27b)
where A is the complex amplitude, and it is understood that ϕ is actually equal to the real part. Equation
(1.27b) is a solution to (1.18a) if the dispersion relation (1.23b) is satisfied, i.e. if
ω = ±|k|c0 . (1.27c)
Remarks

(i) Harmonic waves play an important role because, by Fourier decomposition in k, a general distur-
bance can be written as the sum/integral of harmonic waves. Note that since there are two roots,
ω = ±|k|c0 , it is necessary to specify both ϕ and ϕt in an IVP.
(ii) The dispersion relation (1.27c) is isotropic, i.e. ω does not depend on the direction of k.

Key notation
Wavenumber: |k| ≡ k
|ω| b
Wavevector: k= k
c0
(
λ = 2πc
|ω|
0
k k
b
Wavelength: 3cm-30m for audible sound
λ=∞ ⊥r kb
Radian frequency: |ω| cycles per radian of time
1 |ω|
Frequency: F = = cycles per unit of time
T 2π
2π
Period: T = 10−4 -10−1 s for audible sound
|ω|
Phase: θ(x, t) = k.x − ωt crests given by θ(x, t) = constant
∇θ k
Normal to crests: n= = =k
b ∇θ = k
|∇θ| |k|
 
dx ω ωb
Phase velocity: c= .n n = k= k velocity of crests ⊥ to crests
dt θ |k|2 k
Remark. The phase velocity of sound waves,
ωb
c≡ k = ±c0 k
b = ±c0 n , (1.27d)
k
is independent of frequency and has constant magnitude.
This has important consequences, since it results in sound
03/04 waves being non-dispersive (see later).
03/17
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03/19
 1.5.3 Energies of harmonic plane waves

Energies are quadratic; hence you must take the real part before multiplication. However, there
is an exception, and an energy saving device, when you are interested in the average over a period. For in-
stance, assume that θ and ψ are harmonic with period 2π
ω , and that they have the complex representations
(recall, with real part assumed)

θ = Θ exp(ik. x − iωt) , ψ = Ψ exp(ik. x − iωt) .

Then
Z 2π Z 2π
ω ω ω ω
< θψ > ≡ θ(x, t)ψ(x, t)dt = (Θr c − Θi s)(Ψr c − Ψi s)dt
2π 0 2π 0

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
= 12 (Θr Ψr + Θi Ψi )
= 12 Re (ΘΨ∗ ) = 12 Re (θψ ∗ ) , (1.28)

where c = cos(k. x − ωt), s = sin(k. x − ωt), Θr = Re (Θ), Θi = Im (Θ), Ψr = Re (Ψ), Ψi = Im (Ψ), and
∗
03/16 indicates a complex conjugate.
As an example, we note from (1.17c), (1.20a), etc., that for ϕ given by (1.27b):

< Ek > = 12 Re ( 12 ρ0 u.u∗ )

= 14 ρ0 Re (∇ϕ.∇ϕ∗ )
= 14 ρ0 Re (ikA exp(ik. x − iωt).(−ik)A∗ exp(−(ik. x − iωt)))
= 41 ρ0 k 2 |A|2 . (1.29a)

Similarly
< Ep > = 14 ρ0 k 2 |A|2 , (1.29b)
and
< I > = ± 12 ρ0 k 2 |A|2 c0 k
b. (1.29c)
Since this is a special case, it is consistent with (1.26a) and (1.26b).

1.6 Reflection and Transmission (or Why Bass Notes Ignore Walls)

We will do a relatively simple example here, but return to similar problems when we get to elastic waves.
So, consider sound bouncing off a thin ( λ) wall between two rooms assuming normal incidence. Express
all quantities in terms of eiωt , or alternatively e−iωt , implicitly taking the real part. Assume that the wall

• has a mass m per unit area;

• only moves a small amount so that it can be assumed to be at x = X(t) ≈ 0, i.e. linearise the
evaluation of quantities at the wall to x = 0 so that

p̃(X(t), t) = p̃(0, t) + X(t)p̃x (0, t) + . . . ≈ p̃(0, t) .

x<0 p̃ = Aeiω(x−c0 t)/c0 + Re−iω(x+c0 t)/c0 .

given incident wave reflected wave

Hence from (1.25c)

1  iω(x−c0 t)/c0 
u= Ae − Re−iω(x+c0 t)/c0 .
ρ0 c0

x>0 p̃ =T eiω(x−c0 t)/c0 ,

1
u= T eiω(x−c0 t)/c0 .
ρ0 c0

Mathematical Tripos: Part II Waves 11 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Remark. We assume that there are no waves of the form e−iω(x+c0 t)/c0 in x > 0, i.e. there are no waves
bringing energy in from +∞ (e.g. there is only one person playing their sound system in x < 0);
this is a radiation condition.
Boundary conditions. These are linearised onto x = 0:
Kinematic:
Ẋ(t) = u(0−, t) = u(0+, t) ,
Dynamic:
+
mẌ(t) = p̃(0−, t) − p̃(0+, t) = − [p̃]− .

Solution. Put X = X0 e−iωt and solve, then

1 1

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
−iωX0 = (A − R) = T,
ρ 0 c0 ρ 0 c0
−mω 2 X0 = A + R − T .

Introduce the dimensionless group

ρ0 c0 ρ0 /k mass of gas in wavelength
α= = = .
ωm m mass of wall
Then
1 2iα
R= A, T = A.
1 + 2iα 1 + 2iα

Checks

• m = 0 =⇒ α = ∞, R = 0, T = A (perfect transmission).
• m = ∞ =⇒ α = 0, R = A, T = 0 (total reflection).

Remarks

• The wall blocks high frequencies (α  1) better than low frequencies (α  1).
• After a lot of work (at least for x < 0) one can show that

 2ρ10 c0 |A|2 − |R|2


in x < 0 ,
< Ix >=< p̃u >=
 1 2
2ρ0 c0 |T | in x > 0 .

Note that the energy flux is independent of x.

1.7 Waves in Three Dimensions

1.7.1 Acoustic waves in a rectangular duct

Consider the model problem of acoustic waves in a duct

(a crude model of a musical instrument). For simplicity
we take the duct to be rectangular (a very crude model).
The wave equation, and the boundary conditions of zero
normal velocity at the walls, are then

1 ∂2
 
2 ϕ = ∇ 2 − 2 2 ϕ = 0 ,
c0 ∂t

ϕy = 0 on y = 0, h
∀x.
ϕz = 0 on z = 0, b

Mathematical Tripos: Part II Waves 12 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Assume the waves propagate in the x−direction, and seek a separable harmonic wave solution of the form

ϕ = eikx−iωt f (y, z) . (1.30a)

This satisfies the wave equation and the boundary conditions if
 2 
ω 2
fyy +fzz + − k f =0, (1.30b)
c20
fy = 0 on y = 0, h , (1.30c)
fz = 0 on z = 0, b . (1.30d)

Equations (1.30a)-(1.30d) define an eigenvalue problem

(cf. Schrödinger bound states), i.e. for given k seek eigen-

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
values ω∗ , and associated eigenfunctions (modes) f∗ .
Try a separable harmonic solution (in terms of sin and
cos) that satisfies the boundary conditions:
mπy nπz
fmn = Amn cos cos , m, n ∈ Z , (1.31)
h b
where, wlog, m, n ∈ N. Then the wave equation is satis-
fied when ω is given by the dispersion relation
  mπ 2  nπ 2 
2 2 2 2
ω = ωmn (k) ≡ c0 k + + . (1.32)
h b

The phase velocity. For a given mode (m, n), the phase velocity is given by

ω
  mπ 2  nπ 2  12
c = = ± c0 1 + + . (1.33)
k kh kb

Hence, unless the waves are plane waves (i.e. m = n = 0), the phase speed, c, varies with wavenum-
ber; the waves are then said to be dispersive.
The ‘cut-off ’ frequency. A mode of a given frequency ω propagates only if k is real. Hence to excite the
(m, n) mode we require
 1
mπ 2  nπ 2 2
|ω| > c0 + . (1.34)
h b
This minimum frequency is called the cut-off frequency. Attempts to excite the (m, n) mode (m 6= 0,
n 6= 0) below this frequency yield exponentially decaying evanescent modes with imaginary k.
Hence for given ω, there exist only a finite number of propagating waves: the plane wave and a
04/17 limited number of ‘cross-modes’. The other modes are ‘cut-off’ and do not propagate.
04/18
The velocity of energy propagation. From (1.30a) and (1.31)
 mπy nπz mπ mπy nπz nπ mπy nπz  ikx−iωt
u = ∇ϕ =Amn ik cos cos ,− sin cos ,− cos sin e .
h b h h b b h b
Averaging both in time and over a cross-section, and using (1.28) and the dispersion relation (1.32),
yields
Z b Z h
< Ek >= dz dy 21 Re[ 12 ρ0 u  u∗ ]
0 0
m2 π 2 n2 π 2
 
1 2 2 bh
= 4 ρ0 |Amn | k + 2
+ 2
h b χmn
ρ0 |Amn |2 bhω 2
= ,
4χmn c20

Mathematical Tripos: Part II Waves 13 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 where χmn = (2 − δm0 )(2 − δn0 ) accounts for the difference between averaging (cos 0)2 compared
with (cos πy 2 2
h ) , etc. Similarly, using definition (1.20a), and p̃ = c0 ρ̃ = iωρ0 ϕ from (1.17c),
Z b Z h  ∗

1 1 2 ρ̃ρ̃
< Ep >= dz dy 2 Re 2 c0 ρ
0 0 0
ρ0 |Amn | bhω 2 2
= =< Ek > .
4χmn c20

04/19 Remark. We again have equi-partition of energy.

Further, from (1.17c) and (1.20b),
Z b Z h
< Ix >= dz dy 21 Re [p̃ϕ∗x ]

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
0 0
Z b Z h
mπy nπz
= dz dy 12 ωρ0 k|Amn |2 cos2 cos2
0 0 h b
ωρ0 k|Amn |2 bh
= .
2χmn
Define the mean energy propagation velocity to be

< Ix > kc2

U (k) = = o, (1.35a)
< Ek + Ep > ω

and define the group velocity to be

∂ω
cg (k) = . (1.35b)
∂k
Then from (1.32)
c0 kc20
cg (k) = ±  1 = . (1.35c)
2

nπ 2 2 ω
1 + mπ



kh + kb

Hence for this example

∂ω
U (k) = cg (k) =
. (1.35d)
∂k
This is an important result, and holds much more generally (e.g. there exists a vector equivalent).
Remarks
(i) Note that

cg (k) 6= c(k)
i.e.
Energy propagation velocity 6= Wave crest/trough velocity

(ii) Other than for plane waves (m = n = 0), it follows from (1.33) and (1.35c) that
(a) |cg | < c0 , while |c| > c0 .
(b) Both |cg | → c0 and |c| → c0 as k → ∞ (short
waves).
(c) |cg | → 0 while |c| → ∞ as k → 0 (long waves),
so the phase speed can, in principle, be larger
than the speed of light.

Mathematical Tripos: Part II Waves 14 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 (iii) It is possible to interpret the waveguide as a superposition of reflecting plane waves.

For instance with m 6= 0, n = 0, let sin ψ = mπc0 /ωh. Then

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
h iω iω
i
ϕ ∝ e c0 (x cos ψ+y sin ψ−c0 t) + e c0 (x cos ψ−y sin ψ−c0 t) .

The apparent phase speed in the x-direction is

c0
c= > c0 ,
cos ψ
while the component of the plane wave speed in the x-direction is c0 cos ψ = cg < c0 .

1.7.2 Spherically symmetric waves

In spherical polar co-ordinates

1 ∂2
∇2 p̃ = (rp̃) ,
r ∂r2
1 ∂ ∂
= 2 r2 p̃ .
r ∂r ∂r

So for spherically symmetric waves the wave equation (1.16), 2 p̃ = 0, becomes (after multiplying by r)

∂2 ∂2
2
(rp̃) − c20 2 (rp̃) = 0 .
∂t ∂r
This is the 1D wave equation for rp̃, so

f (r − c0 t) g(r + c0 t)
p̃(r, t) = + . (1.36)
r r
outgoing wave incoming wave

Remark. Often one has a radiation condition of no incoming waves from ∞: then g ≡ 0.
It is conventional to write the outgoing solution in the form

q̇(t − r/c0 )
p̃(r, t) = . (1.37a)
4πr
Then the density perturbation is given by

1 q̇(t − r/c0 )
ρ̃ = 2 p̃ = , (1.37b)
c0 4πc20 r

and from (1.17c), i.e. p̃ = −ρ0 ∂ϕ

∂t , the potential is given by

q(t − r/c0 )
ϕ=− . (1.37c)
4πρ0 r

Mathematical Tripos: Part II Waves 15 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

The velocity is given by
 
∂ϕ 1 q̇(t − r/c0 ) q(t − r/c0 )
u = ∇ϕ = r= + r. (1.37d)
r2
b b
∂r 4πρ0 c0 r

dominates dominates
for r  1 for r  1
Hence
Mass flux out of a
= 4πr2 ρ0 u(r, t)
sphere of radius r
r
= q̇(t − r/c0 ) + q(t − r/c0 )
c0

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
→ q(t) as r → 0 .

q(t) is the mass flux from a point source at the origin.

1.7.3 Solution for a pulsating sphere

Consider a sphere performing small radial pulsations

r = a + εeiωt with ε  a , (1.38a)

where is it assumed that the real part of the equation is taken.

Then on the surface of the sphere, r = a + εeiωt , the velocity
of the sphere is
∂ϕ
u= = iωεeiωt . (1.38b)
∂r
But
u(a + εeiωt , t) ≈ u(a, t) + εeiωt ur (a, t) + . . . .
| {z }
doubly small

Hence apply the boundary condition (b.c.) on r = a. Based on the form of the forcing, seek a solution
with q(t) ∝ eiωt , e.g. from (1.37c)
r−a
eiω(t− c0 )
ϕ=A .
4πρ0 r
Then from (1.37d)  
A 1 iω r−a
u=− + eiω(t− c0 )
.
4πρ0 r2 c0 r
Evaluate u on r = a to deduce from (1.38b) that
 
A iωa
− 1+ = iωε ,
4πρ0 a2 c0
i.e.
4πiρ0 a2 ωε
A=− ,
(1 + iωa/c0 )
and hence
εa2 ω 2 ρ0 eiω(t−(r−a)/c0 )
p̃ = −iωρ0 ϕ = − .
(1 + iωa/c0 ) r
Remarks
2πc0
(i) The sound has wavelength λ = ω , hence we define
2πc0 ωa
a ‘large’ sphere to be one such that a , i.e. = ka  1 ,
ω c0
2πc0 ωa
a ‘small’ sphere to be one such that a , i.e. = ka  1 ,
ω c0

Mathematical Tripos: Part II Waves 16 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 (ii) A small sphere is said to be ‘compact’. One can show that compact sources are inefficient sound
sources because p̃ and u are out of phase (see question 5 on Example Sheet 1).

1.7.4 Loud speakers: why tall and thin is good

A small/compact vibrating disk is an inefficient sound

source, because the flow is locally incompressible; it acts
as a dipole (the strength of which can be shown to be
proportional to the force exerted to move the disk).
One way forward to improve efficiency is to use an ex-
ponential horn (megaphone) longer than a wavelength, so
that a quasi-2D flow is set up (see Example Sheet 1).

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
04/16
Slower next
year
Alternatively, effectively change the ‘dipole’ to a monopole
by inhibiting the reverse flow, e.g. with a baffle. The
boundary conditions then become, say at x = 0+,

U (y, z, t) on moving part,
un =
0 on stationary baffle.

Using the method of images, this is equivalent to having

no baffle, and an additional image boundary condition of
un = −U at x = 0−. For a disk of area dy dz, this is
equivalent to a source of strength q = 2ρ0 U (y, z, t)dy dz.

To find the solution for a loud speaker, we add up lots of little moving parts (see also Tripos question
2009/1/38). Then, from the solution for a point source, i.e. (1.37a),
2ρ0 U̇ (Y, Z, t − R/c0 )
ZZ
p̃(x, y, z, t) = dY dZ ,
4πR
where R is the distance between the field point (x, y, z) and the source point (0, Y, Z):

1/2
R = x2 + (y − Y )2 + (z − Z)2 .
04/04
05/17 When far from the loudspeaker
05/18
05/19 
2(yY + zZ)
1/2
R = (x2 + y 2 + z 2 )1/2 1 − 2 + . . .
x + y2 + z2
yY + zZ
∼ r − + ... .
r
O(r) O(1)
For simplicity, consider a single frequency ω so that
U = U (Y, Z)eiωt ,
then in far field
ZZ
iωρ0 iω(t−r/c0 )
p̃ ∼ e U (Y, Z)ei(yY +zZ)ω/rc0 dY dZ .
2πr
| {z }
Fourier transform of U

Mathematical Tripos: Part II Waves 17 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Note that it is OK to use the leading-order approx to R in denominator but not in exponent (since
cos(2019)  cos(2020)).
Suppose further that

U0 |Y | 6 a, |Z| 6 b
U= ,
0 otherwise.

and that
ωa ωb
 1, & 1,
c0 c0
then
iωρ0 eiω(t−r/c0 )
 
sin(ωbz/rc0 )
p̃ ∼ 4abU0 .

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
2πr ωbz/rc0

For this solution most power is radiated in directions with

ωbz
. π,
rc0
i.e. for
z πc0
sin θ = . .
r ωb
If b is large enough, then it is possible to ensure that little
sound hits the roof of a theatre, etc.

Mathematical Tripos: Part II Waves 18 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Appendix

1.A * Thermodynamics [for Dummies] (unlectured) *

1.A.1 0th Law – there exists temperature

After several collisions with other molecules, the velocity of a molecule becomes randomised and the gas
attains thermal equilibrium characterised by a temperature T ; this time for air is 10−9 s and for water
10−12 s. A reversible change is a change to the system that is slow compared with the above times, so
that the system always remains near to thermal equilibrium.
More generally we say that a fluid is in thermodynamic equilibrium if all past history is forgotten, and all
macroscopic quantities do not change with time.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Parameters of State. Parameters of state (also sometimes called functions of state) are macroscopic
quantities which, in any change in thermodynamic equilibrium, vary by an amount independent of
the path taken between the initial and final states. Examples are the pressure, p, the density, ρ, and
the [absolute] temperature, T .

Equations of State. An equation of state is a functional relation between the parameters of state for a
system in thermodynamic equilibrium, e.g.

f (p, ρ, T ) = 0 . (1.A.1a)

E.g. for a perfect gas

p = ρRT . (1.A.1b)

1.A.2 1st Law – energy is conserved if all forms are counted

Consider a unit mass of gas, then the specific volume (volume per unit mass) is V = 1/ρ. The First Law
of Thermodynamics states that this volume has a (specific) internal energy e(ρ, T ) that arises from the
kinetic energy of thermal motion of molecules and possibly from the potential energy of excited vibrational
modes. The internal energy depends strongly on the current temperature and density, but not on past
values.
More generally the First Law of Thermodynamics states that every thermodynamic system possesses a
parameter of state, the internal energy, e. This parameter can be increased by adding heat energy, δQ,
to the system, or by performing mechanical work, δW , on the system:

de = δQ + δW . (1.A.2)

Q and W are not parameters of state.

1.A.3 Adiabatic processes, isolated systems and reversible changes

An adiabatic process2 is a process in which there is no heat change, i.e.

δQ = 0 . (1.A.3a)

• A system is isolated when it does not interact with its surroundings, i.e. when no heat can be
transferred to/from it, and no work can be performed on it.

2 Some authors, e.g. Pippard, call this an adiathermal process.

Mathematical Tripos: Part II Waves a © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 • A reversible change3 from one equilibrium state to another is an ‘infinitely’ slow quasi-static process
during which the thermodynamic system remains infinitesimally close to thermodynamic equilib-
rium. At each stage of a reversible process, the parameters of state have a well-defined meaning,
hence we can write
de = dQ − pdV , (1.A.3b)
since the mechanical work done on a simple gas against pressure is −p dV (note that positive work
is done under compression, i.e. dV < 0).

Specific Heats. The specific heat of a material is defined as dQ/dt in a reversible change (i.e. it is
effectively the heat input to increase the temperature of a unit mass by 1 K in a reversible change).
The specific heats at constant volume and constant pressure are therefore respectively equal to
   
dQ ∂e

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
cV = = , (1.A.3c)
dT V ∂T V or ρ
     
dQ ∂e ∂V
cp = = +p . (1.A.3d)
01/04
dT p ∂T p ∂T p

1.A.4 2nd Law – chaos increases

The Second Law of Thermodynamics states that all thermodynamic systems possess a parameter of state
called the entropy, S, such that in a reversible change
T dS = dQ . (1.A.4)
Further, the entropy of an isolated system can only increase, i.e. for an irreversible adiabatic change
dS > 0.

Remark. Roughly, S is proportional to the number of arrangements of the molecules at a given volume V
and temperature T . Thence increasing entropy means increasing chaos, e.g. heat flows from hot to
cold.
An Exact Differential For Entropy. For a reversible process, (1.A.3b) and (1.A.4) imply that
T dS = de + pdV . (1.A.5a)
This equation involves only parameters of state, and hence must be valid for any infinitesimal
transition, whether reversible or not; thus S ≡ S(e, V ). However if the transition is irreversible (e.g.
because of work done against friction) then
δQ 6= T dS , δW 6= −pdV . (1.A.5b)

The Maxwell Relations. It is very important to display what is a function of what, and so what is
being held constant during partial differentiation. One is permitted to take any two parameters of
state as independent variables, e.g. e & V , or ρ & T , or ρ & p, or S & ρ.
Consider de = T dS − p dV , and view all variables (e, T , ρ) as functions of S & V . Then
   
∂e ∂e
T = and p = − .
∂S V ∂V S
But by cross differentiation
∂2e
   
∂T ∂p
= =− .
∂V S ∂S∂V ∂S V
Another Maxwell relation is obtained by the following trick d(e + pV ) = T dS + V dp, so
   
∂T ∂V
= .
∂p S ∂S p
The combination h = e + pV , the enthalpy, is important in fluid mechanics.

3 Beware: some authors, e.g. Pippard, call a reversible adiathermal process an adiabatic process!

Mathematical Tripos: Part II Waves b © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Ideal Fluids. Ideal fluids are fluids in which all molecular diffusivities are zero. No heat can therefore
be transferred between fluid particles, and the fluid must always be in thermodynamic equilibrium.
Hence for a dissipationless fluid
DS
=0, (1.A.5c)
Dt
i.e. the flow is isentropic.
If the entropy takes a uniform constant value everywhere, the flow is homentropic. For future
reference note that molecular diffusivities are not negligible in shocks.
A Closed System. Equation (1.A.5c), together with the equation of state relating S to p and ρ, close
the system of governing equations.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
1.B * Perfect Gases (unlectured) *

Consider dilute gases, e.g. air for which ρair = 10−3 ρliquid air .

1.B.1 Pressure

The pressure force on a surface is the momentum exchanged from collisions in unit time, i.e. p = N mhvx2 i,
where

• N = nNA /v is the number of molecules per unit volume (n is the number of moles of gas,
NA = 6.02 1023 is Avogadro’s number, and v is volume), and
• it follows from statistical physics that mhvx2 i = kB T (kB = 1.38 10−23 J K−1 is Boltzmann’s con-
stant).

Hence
pv = nR̄T ,
where the universal gas constant R̄ = NA kB .

1.B.2 Internal energy and specific heats

In a perfect gas, the internal energy e varies with temperature T but not with density ρ — because
‘dilute’ means the size of the molecules is irrelevant. If we further assume that the specific heats do not
vary with temperature (OK for 100-2000 K when vibrational modes are not excited), then
   
∂e ∂e
cV = = constant and = 0.
∂T V ∂V T
So
e = cV T , (1.B.6a)
setting constant of integration to zero (so that there is no energy at 0o K).
From kinetic theory, each molecule has 21 kB T internal energy per degree of freedom of internal motion.
Hence cV = 32 N kB = 32 R for monatomic gases He and Ar, and cV = 52 N kB = 52 R for diatomic gases H2 ,
N2 , O2 , CO (since the molecules can have rotational KE represented by, say, two Euler angles, but no
vibrational energy).
The other specific heat is    
∂e ∂V
cp = +p = cV + R ,
∂T p ∂T p

from e = cV T and V = 1/ρ = RT /p. Thence the ratio of specific heats

cp cV + R
γ= = ,
cV cV

Mathematical Tripos: Part II Waves c © S.J.Cowley@maths.cam.ac.uk, Lent 2019

is equal to 53 for monatomic gases and 7
5 for diatomic gases. A useful alternative expression for the internal
energy is, cf. (1.5e),
cV 1 p
e = cV T = pV = . (1.B.6b)
R γ−1 ρ
Hence the enthalpy of a perfect gas is given by, cf. (1.5g),
p γ p
h≡e+ = . (1.B.6c)
ρ γ−1 ρ

1.B.3 Entropy of a perfect gas

From (1.A.5a), (1.B.6a) and (1.A.1b), T dS = de + p dV , e = cV T and p = ρRT = RT /V . Hence

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
dT R
dS = cV + dV .
T V
By integrating we obtain
T V
S − S0 =cV ln + R ln
T V
"0   0 #
R/cV
T V
=cV ln
T0 V0
"  γ−1 #
pV V
=cV ln .
p0 V0 V0

So   γ 
p ρ0
S = S0 + cV ln . (1.B.6d)
p0 ρ

Mathematical Tripos: Part II Waves d © S.J.Cowley@maths.cam.ac.uk, Lent 2019

2 Finite Amplitude 1D Waves
Our aim is to examine the effects of nonlinearity on (plane) sound waves. In doing so we will learn why
waves ‘break’ and shocks form. The theory is also applicable to why you can feel your pulse and why
traffic jams develop.

2.1 Nonlinear Acoustics in 1D: Governing Equations

Assume

(a) 1D motion;

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
(b) homentropic flow, i.e. from (1.10a)
p ≡ p(ρ, S0 ) . (2.1)

The exact 1D version of conservation of mass (1.1b) is, with ∂y = ∂z = 0, u = (u, 0, 0), and Q = F = 0,
∂ρ ∂
+ (ρu) = 0 . (2.2a)
∂t ∂x
Similarly, the exact 1D version of Euler’s equation, (1.2b), is
∂u ∂u 1 ∂p
+u =− , (2.2b)
∂t ∂x ρ ∂x
while from (1.10a)
∂p dp ∂ρ ∂ρ
= = c2 , (2.2c)
∂x dρ ∂x ∂x
where
dp
c2 (ρ) = (ρ) . (2.2d)
dρ

2.2 Riemann Form of Equations

We can simplify the system of equations (2.2a)-(2.2d) for u(x, t) and ρ(x, t) by a change of coordinates.4
First form λ.(2.2a) + (2.2b), where λ(x, t) is to be determined, to obtain

c2
     
∂ ∂ ∂ ∂
+ (u + λρ) u+λ + u+ ρ = 0.
∂t ∂x ∂t λρ ∂x
We fix λ by choosing
c2 c
u + λρ = u + , i.e. λ=± ,
λρ ρ
so that    
∂ ∂ c ∂ ∂
+ (u ± c) u± + (u ± c) ρ = 0.
∂t ∂x ρ ∂t ∂x
Define Z ρ
c(ρ̂)
Q(ρ) = dρ̂ , (2.3)
ρ0 ρ̂
then
c c
Qt = ρt , Qx = ρx ,
ρ ρ
and hence  
∂ ∂
+ (u ± c) (u ± Q) = 0 . (2.4)
∂t ∂x

4 A transcript of Riemann’s article can be downloaded from http://www.emis.de/classics/Riemann/Welle.pdf. There

is also an English translation, but not for download: http://www.kendrickpress.com/Riemann.htm.

Mathematical Tripos: Part II Waves 19 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Define the Riemann invariants, R± (x, t) to be
R± = u ± Q , (2.5)
then we obtain the Riemann equations
 
∂ ∂
+ (u ± c) R± = 0 . (2.6)
∂t ∂x
We can integrate these equations by changing variables from (x, t) to (ξ, η) where the lines ξ =const. are
defined by
dx
= u(x, t) + c(x, t) , (the C+ characteristics, ξ =const.) , (2.7a)
dt
where we have supposed that u and c are in principle known. Similarly the lines η =const. are defined by

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
dx
= u(x, t) − c(x, t) , (the C− characteristics, η =const.) . (2.7b)
dt
These two intersecting families of lines are referred to as the C± characteristics respectively.
On the C+ characteristics, i.e. on lines ξ =const., it follows from
the first Riemann equation (2.6) and the first characteristic equation
(2.7a) that

dR+ ∂R+ dx ∂R+ ∂R+ ∂R+

= + = + (u + c) = 0; (2.8a)
dt ∂t dt ∂x ∂t ∂x
hence
R+ ≡ R+ (ξ) . (2.8b)
Similarly
R− ≡ R− (η) . (2.8c)

Remark. The physical interpretation of these equations is that waves carrying constant values of R±
propagate at speeds ±c relative to the local flow speed u.
Check (unlectured). We have that
∂t ∂t ∂x ∂x
dt = dη + dξ , dx = dη + dξ .
∂η ξ ∂ξ η ∂η ξ ∂ξ η

Hence on ξ =const.,
∂x dx ∂t ∂t
dξ = 0 and = = (u + c) .
∂η ξ dt ∂η ξ ∂η ξ

It follows from (2.6), consistent with (2.8b), that

∂R+ ∂R+ ∂t ∂R+ ∂x
= +
∂η ξ ∂t x ∂η ξ ∂x t ∂η ξ
 
∂R+ ∂R+ ∂t
= + (u + c)
∂t x ∂x t ∂η ξ
=0.

Relationship to D’Alembert’s solution. From (2.5),

u = 21 (R+ (ξ) + R− (η)) , (2.9a)
Q = 21 (R+ (ξ) − R− (η)) . (2.9b)
This can be viewed as a generalisation of D’Alembert’s linear solution, cf. (1.24b):
u = F (x − c0 t) + G(x + c0 t) ,
as may be seen by solving (2.7a) and (2.7b) with u = 0 and c = c0 .

Mathematical Tripos: Part II Waves 20 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 2.2.1 Example: perfect gas

From (1.5a), (1.5b) and (2.2d)

 γ   γ−1
p ρ 2 dp γp c ρ 2
= , c = = and = . (2.10a)
p0 ρ0 dρ ρ c0 ρ0

Hence from the definitions of Q, see (2.3), and R± , see (2.5),

2 2(c − c0 )
Q= (c − c0 ) , R± = u ± , (2.10b)
γ−1 γ−1
and hence

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
u = 12 (R+ (ξ) + R− (η)) , c = c0 + 41 (γ − 1)(R+ (ξ) − R− (η)) . (2.10c)
05/04
06/17

2.3 Solution by Characteristics

2.3.1 The Cauchy initial-value problem (IVP)

Suppose that initial conditions are known at t = 0, e.g.

u(x, 0) = U (x), ρ(x, 0) = P(x) , (2.11a)

or since c ≡ c(ρ), equivalently

u(x, 0) = U (x), c(x, 0) = C(x) . (2.11b)
For simplicity assume ‘compact support’, i.e.

U = 0 , C = c0 and P = ρ0 for x < xB and x > xF . (2.11c)

2.3.2 General form of the solution

05/16
06/18
06/19 Regions I, II and III. At t = 0 on characteristics that pass through x < xB or x > xF :

u(x, 0) = 0 , c(x, 0) = c0 , Q(x, 0) = 0 . (2.12a)

Hence
(i) on ξ < ξB and ξ > ξF , R+ = 0;
(ii) on η < ηB and η > ηF , R− = 0.
It follows that in regions I, II and III,

R+ = R− = 0 =⇒ u = 0, c = c0 =⇒ undisturbed flow. (2.12b)

Mathematical Tripos: Part II Waves 21 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Regions IV and V. In region IV
R− = u − Q = 0 =⇒ u=Q (2.13)
On C+ , from (2.8b),
u + Q = R+ = const.
Hence from (2.3) and (2.13), on C+

u = const., Q = const., ρ = const., c = const.,

and
dx
= u + c = const. .
dt
Hence the C+ are straight lines in region IV, corresponding to a right-running simple wave (see
below).

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Similarly, in region V the C− characteristics are straight, and there is a left-running simple wave.
Region VI. In region VI there is compound flow, where neither C+ nor C− are straight. At t = tc the
compound flow disentangles to give two simple waves with undisturbed regions ahead, between and
behind them.

2.3.3 * Numerical method *

Aim. Find the solution at x = X for δt  1.

Method. Draw the characteristics through (X, δt) back to t = 0. Since δt  1, the C+ characteristic can
be approximated, using a Taylor expansion for 0 6 t 6 δt, as

x(t) = X + (u(X, δt) + c(X, δt))(t − δt) + O(δt2 )

| {z }
slope

= X + (u(X, 0) + c(X, 0)) (t − δt) + O(δt2 ) . (2.14a)

Note that x = X at t = δt. Hence at t = 0 it follows from using the initial conditions that

x(0) = X − (U (X) + C(X)) δt + O(δt2 ) . (2.14b)

Since R+ =const. on C+ this implies

R+ (X, δt) = R+ (X − (U (X) + C(X))δt , 0) + O(δt2 ) . (2.15a)

Similarly
R− (X, δt) = R− (X − (U (X) − C(X))δt , 0) + O(δt2 ) . (2.15b)
Numerical Recipe.
(a) At t = t0 assume that u and c are known.
(b) Calculate R+ and R− at t = t0 , e.g. using (2.10b).
(c) Calculate R+ and R− at t = t0 + δt using (2.15a) and (2.15b).
(d) Calculate u and c at t = t0 + δt, e.g. using (2.10c).
06/04 (e) Iterate after t0 → t0 + δt (remembering that interpolation is your friend).

2.4 Simple Waves and Shock Formation

2.4.1 Simple waves

Definition. A simple wave is a wave where one of R+ or R− is uniformly constant (wlog zero).

As an example, consider a right-running simple wave (as in region IV) with

R− = 0 . (2.16a)

Hence from (2.9a), (2.9b) and (2.5) we infer that

Mathematical Tripos: Part II Waves 22 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 u = Q and R+ = 2u = 2Q . (2.16b)
However, R+ ≡ R+ (ξ) from (2.8b), i.e. on each C+ characteristic
R+ = const. . (2.17a)
Hence from (2.16b), (2.3) and (2.2d)
u ≡ u(ξ), Q ≡ Q(ξ), ρ ≡ ρ(ξ), c ≡ c(ξ) , (2.17b)
and so these quantities are also constant on each C+ characteristic. Further, from the definition (2.7a) of
a C+ characteristic,
dx
= u + c ≡ u(ξ) + c(ξ) , (2.17c)
dt
it follows that for a right-running simple wave the C+ characteristics are straight lines.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Solution. Suppose that at t = 0 consistent initial conditions are

u(x, 0) = U (x) and c(x, 0) = C(x) . (2.18a)

Then we deduce from (2.17b) and (2.17c) that

u(x, t) = U (ξ) and c(x, t) = C(ξ) , (2.18b)

and that the equations of the C+ characteristics are, from

imposing x = ξ at t = 0,

x(t; ξ) = ξ + (U (ξ) + C(ξ))t . (2.18c)

Thus, for given (x, t), one can in principle solve for ξ, and
thence obtain u(x, t) and c(x, t) from (2.18b).
Perfect gas. In the case of a perfect gas it follows from (2.16b) and (2.10b) that
4
R+ = 2u = (c − c0 ) = const. , (2.19a)
γ−1
c = c0 + 21 (γ − 1)u . (2.19b)
Hence on a C+ characteristic,
dx
= c0 + 21 (γ + 1)u , (2.20a)
dt
where, as above, the C+ are straight. Hence the problem for right-running simple waves in a perfect
gas can be posed either as
du dx
= 0 on = c0 + 12 (γ + 1)u , (2.20b)
dt dt
or, from the Riemann equations (2.6) with R− = 0 and u + c = c0 + 12 (γ + 1)u, as
 
∂
∂
+ c0 + 12 (γ + 1)u u = 0. (2.20c)
∂t ∂x
Further, from (2.18c), or from (2.20b) and imposing x = ξ at t = 0, the equation for the C+
characteristics is
x(t; ξ) = ξ + c0 + 21 (γ + 1)U (ξ) t .


(2.20d)
Thus, for given (x, t), one can in principle solve for ξ, and thence obtain
u from (2.18b) ,

c = c0 + 12 (γ − 1)u from (2.19b) ,

 2
 γ−1
ρ (γ − 1)u
= 1+ from (2.10a) .
ρ0 2c0

Mathematical Tripos: Part II Waves 23 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 2.4.2 Generic form

Equation (2.20c) can be written in a more suggestive form. Let

X = x − c0 t , (2.21a)

and introduce the excess wavespeed

 
γ+1
v= u = u + c − c0 . (2.21b)
2

Then (2.20c) becomes

∂v ∂v
+v =0. (2.22)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
∂t ∂X
This is a special case of the Kinematic Wave Equation. The equation has C+ characteristics

X(t; ξ) = ξ + v(ξ, 0)t , (2.23a)

on which v is a constant, i.e.
v(X, t) = v(ξ, 0) . (2.23b)

Hence from (2.23a)

v(X, t) = v(X − v(ξ, 0)t, 0) , (2.23c)

and thence
ρ(X, t) = ρ(X − v(ξ, 0)t, 0) , (2.23d)

i.e. the density ρ propagates at speed v (in a frame moving with velocity c0 ).

Each element of the wave form propagates at speed v. The deformation is pure distortion – i.e. there are
06/16 no new values of v (or ρ).
07/17
07/18
07/19

Mathematical Tripos: Part II Waves 24 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

2.4.3 Wave steepening

Calculate the slope of the wave:

∂v ∂v(ξ, 0) ∂ξ
= . (2.24a)
∂X t ∂ξ ∂X t
From (2.23a)  
∂ξ ∂v(ξ, 0)
1= 1+ t , (2.24b)
∂X t ∂ξ
and hence
∂v vξ (ξ, 0)
= . (2.24c)
∂X t 1 + vξ (ξ, 0) t

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Hence parts of the profile with vξ (ξ, 0) < 0 get steeper, while parts of the profile with vξ (ξ, 0) > 0 get
flatter. If vξ (ξ, 0) < 0 somewhere, then a triple-valued waveform will form at some finite time. This is
unphysical. The moment at which the slope becomes infinite is referred to as the shock formation time.

2.4.4 Shock formation

The profile is on the verge of becoming multi-valued at the first time that
∂v
=∞.
∂X t

From (2.24c), vX is infinite when

1
t=− . (2.25)
vξ (ξ, 0)

The shock formation time is thus

   
min 1 2 1
ts = − = max . (2.26a)
ξ vξ (ξ, 0) γ+1 ξ (−uξ (ξ, 0))

Suppose that [−uξ (ξ, 0)] is maximised at ξ = ξm , then the position at which the shock forms is

Xs = ξm + v(ξm , 0)ts , (2.26b)

i.e. the position of shock formation in the original co-ordinates is given by

xs = Xs + c0 ts
= ξm + c0 + 12 (γ + 1)u(ξm , 0) ts .


(2.26c)

2.4.5 Alternative estimate of shock formation time

The shock forms when neighbouring C+ first touch, i.e. when contradicting information from different
C+ characteristics first propagate to the same point. From (2.20d)

x = ξ + (c0 + 12 (γ + 1)u(ξ, 0))t .

The C+ touch when

∂x
=0,
∂ξ t
i.e. the C+ touch when
2
t=− .
(γ + 1)uξ (ξ, 0)
Hence, in agreement with (2.26a),
 
2 min 1
ts = − . (2.27)
(γ + 1) ξ uξ (ξ, 0)

Mathematical Tripos: Part II Waves 25 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

2.4.6 Example

Assume

u = u0 sin kx at t=0, u0 > 0 ,

so that
uξ (ξ, 0) = ku0 cos kξ .
max(−uξ ) occurs at kξ = π, 3π, . . ., so that it follows

2
ts = ,
(γ + 1)ku0
(2n + 1)π

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
xs = + c0 ts .
k

Remarks.

(i) Note that, using ω = kc,

ts 2 c0 k 1 c0 1
= = ∝ ,
tperiod (γ + 1)ku0 2π π(γ + 1) u0 M

where, as defined in (1.18c), the Mach number, M , is

u0
M= .
c0

(ii) For an incredibly large sound of 160 dB,

u0
≈ 10−3 ,
c0
which implies that a wave will propagate for 1000 periods before a shock forms.
(iii) For normal speech of about 60 dB,
u0
≈ 10−13 ,
c0
which implies that a shock never forms before viscous effects dissipate the wave.
(iv) For t  ts and x  xs we can use linear theory for small amplitude waves. We conclude that for
many practical applications of acoustics, linear theory is sufficient.

2.5 Piston Problems

• At t = 0 assume a perfect gas at rest in x > 0; so u = 0, c = c0 in x > 0.

• Piston moves with path x = Xp (t) where Xp (0) = 0.

What happens in the gas? Partly a BVP since need to take into account information from the piston.

Mathematical Tripos: Part II Waves 26 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 2.5.1 Piston accelerates away from the gas

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Assume all C− characteristics originate from t = 0, x > 0, so that

R− = 0 everywhere.

Hence the flow is a simple wave, and the C+ are straight. Specifically, from the C+ characteristic equation
(2.20a),

dx
C+ : = c0 + 21 (γ + 1)u, u = const.
dt

Region I. If a C+ meets t = 0 at x > 0, then

R+ = 0 , u = 0, c = c0 .
07/04
Region II. Assume that the piston is moving slow enough that the C+ intersect the piston (see below),
and that the C+ meet the piston path at

t=τ, x = Xp (τ ) where u = Ẋp (τ ) . (2.28a)

| {z }
boundary condition

The flow is then a simple wave (u=Q), with the C+ satisfying the equation (cf. (2.20d))
 
x − Xp (τ ) = c0 + 21 (γ + 1)Ẋp (τ ) (t − τ ) , (2.28b)

where τ plays the role of ξ here. Given (x, t) in region II, this is an implicit equation for τ (x, t);
solve for τ , then from (2.28b)
u(x, t) = Ẋp (τ ) , (2.28c)
and from (2.19b) and (2.10a)
2
! γ−1
γ − 1 Ẋp (τ )
ρ = ρ0 1+ . (2.28d)
2 c0
07/16
08/17
08/18

Mathematical Tripos: Part II Waves 27 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Characteristics ‘diverge’. While the value of Ẋp decreases
with τ , the slope dx
dt decreases with τ (and the diagram
dt
slope dx increases); in this case the characteristics ‘di-
2c0
verge’. A characteristic with Ẋp = − γ+1 is ‘vertical’.
Escape speed. Note that
2c0
ρ=0 if Ẋp = − .
γ−1
2c0
γ−1 is the maximum speed of the piston for contact
with gas to be maintained. For speeds greater than
this, the C+ characteristics no longer reach the piston,
and there is a vacuum between the limiting C− char-

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
2c0
acteristic and the piston. Hence, γ−1 is the speed of
expansion of a gas into a vacuum: the ‘escape’ speed
(M ∼ 5; M & 5 sometimes referred to as hypersonic).
Expansion fan. Consider rapid acceleration up to a constant
speed −V . The slopes of C+ are uniform in each of
regions I and IIB, and range over intermediate values
in region IIA.
Now take the limit of an impulsive start. Region IIA
reduces to a wedge, called an expansion fan with many
C+ coming out of the origin. The slopes of C+ in region
IIA are constant and must be t/x in order to go through
08/19
the origin.

2.5.2 Piston accelerates towards the gas (at some point)

A shock will form if the piston moves towards the

gas whatever the form of Ẋ(t) > 0, since the C+
characteristics have slope c0 in region I, and slope
c0 + 21 (γ +1)Ẋ(τ ) > c0 in region II, and hence the char-
acteristics must eventually cross. A similar argument
for successive C+ characteristics shows that a shock
also forms if Ẍ(t) > 0 for any t.
As for a receding piston, the equations of the C+ characteristics are, see (2.28b),
 
x(t; τ ) = Xp (τ ) + c0 + 12 (γ + 1)Ẋp (τ ) (t − τ ) .

The characteristics touch when

∂x
= 0,
∂τ t

i.e. when
1
2 (γ + 1)Ẍp t = 21 (γ + 1)Ẍp τ + c0 + 21 (γ − 1)Ẋp .
Hence a shock forms at !
2c0 + (γ − 1)Ẋp
t = ts ≡ min τ + ,
τ >0 (γ + 1)Ẍp
on the characteristic corresponding to the minimising τ = τm , i.e. at x = xs ≡ x(ts ; τm ), subject to the
requirement that xs > Xp (ts ) and no vacuum has formed.

Example. Uniform positive acceleration: Xp (τ ) = 21 f τ 2 , with f > 0. Two adjacent characteristics touch
when
1
(γ + 1)f τ + c0 + 21 (γ − 1)f τ 2(γf τ + c0 )
t=b t(τ ) ≡ 2 1 = for τ > 0 .
2 (γ + 1)f
(γ + 1)f

Mathematical Tripos: Part II Waves 28 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 The minimum overlap time, i.e. the shock formation time, is thus:
2c0
ts = min b
t(τ ) = .
τ >0 (γ + 1)f
This corresponds to the characteristic τ = 0 through x = t = 0, and hence (since γ + 1 > 0)
2c20
xs = c0 ts = > Xp (ts ) .
(γ + 1)f

2.6 Shock Waves

A triple–valued region is unphysical. In practice it is prevented by mechanisms/physics so far neglected,

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
e.g. dissipation. When the wave form is steep, viscous and heat conduction terms which were small become
large.
For instance, the simple wave equation (2.20c) can be modified to
ut + (c0 + 12 (γ + 1)u)ux = νuxx , (2.29)
where the extra term is proportional to the kinematic viscosity, ν (10−5 m2 s−1 for air). An analytic solution
to (2.29) exists (e.g. see Part III). The triple–valued region is avoided by a region of rapid change, the
thickness of which is proportional to ν.

Outside the so–called ‘shock’ region, we can use the characteristic solution of the simple wave equation.
For ‘real’ problems we cannot find exact solutions to equations like (2.29). However, if ν is very small,
shocks are very thin, and across them p, ρ, u and T jump almost discontinuously in value. Hence, in
such circumstances, might we treat the shock as a discontinuity. Further, without knowing the internal
details of the shock:

• Might we be able to impose ‘jump conditions’ across the shock?

• Might we be able to identify where the shock is, and how fast it moves?

For example, we know that mass is conserved, i.e.

Z ∞
ρ dx = const. .
−∞

Hence, in a plot of ρ against x, a shock should be inserted to ensure that it cuts off lobes of equal area
either side of the shock.

Mathematical Tripos: Part II Waves 29 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

2.6.1 Rankine-Hugoniot relations

Consider a steady shock separating two uniform regions (where by uniform we mean uniform on the
scale of the shock width). Take coordinates in shock:

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Mass conservation. Equate the mass entering and leaving the control volume:
(δtδyδz) ρ1 u1 = ρ2 u2 (δtδyδz) ,
i.e.
ρ1 u1 = ρ2 u2 (2.30)

Momentum conservation. Similarly for momentum and the rate of forcing:

p1 + (ρ1 u1 )u1 = p2 + (ρ2 u2 )u2 . (2.31)

Energy conservation. For the energy budget we need to include e, the internal energy per unit mass, cf.
(1.5e); thence
p1 u1 + 21 ρ1 u21 + ρ1 e1 u1 = p2 u2 + 21 ρ2 u22 + ρ2 e2 u2 .



(2.32a)
It follows from dividing through by ρ1 u1 and, using (2.30), that
1 2 p1 1 2 p2
2 u1 + e1 + ρ = 2 u2 + e2 + ρ , (2.32b)
1 2

which is a statement of Bernoulli’s principle. From (1.5g), e + p/ρ is the enthalpy of the gas.
Rankine-Hugoniot relations. (2.30), (2.31) and (2.32b) are known as the Rankine-Hugoniot relations.
There are three equations relating the six variables ρ1 , p1 , u1 , ρ2 , p2 and u2 .
Oblique shocks. In the case of oblique shocks, the velocity parallel to the shock is unchanged across the
shock.

2.6.2 The Hugoniot adiabatic

From (2.30) and (2.31)

p2 − p1 (ρ2 u2 )2 p2 − p1 ρ1 u21
u21 = + = + ,
ρ1 ρ1 ρ2 ρ1 ρ2
so
ρ2 (p2 − p1 )
u21 = . (2.33)
ρ1 (ρ2 − ρ1 )
Hence
either p2 > p1 & ρ2 > ρ1 ,
or p2 < p1 & ρ2 < ρ1 .
Substitute for u21 and u22 into (2.32b) to obtain
 
1 1
e2 − e1 = 21 (p1 + p2 ) − . (2.34)
ρ1 ρ2
This is the Hugoniot adiabatic: it relates thermodynamic quantities across the shock.

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 Remark. Physical intuition might suggest that it is ‘obvious’ that ρ1 < ρ2 if p1 < p2 ; however, this
08/04
09/17 intuition is based on constant entropy (or constant temperature) thinking.
09/18
2.6.3 The entropy jump

Let
1 1
V1 = and V2 = .
ρ1 ρ2
Then for a given mass of gas, plot p versus V
at constant entropy, S, assuming that

d2 p
> 0.
dV 2 S

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Suppose that ρ2 > ρ1 ; hence V2 < V1 , and from
(2.33) p2 > p1 .

09/19

• If entropy were to be conserved, say S = S0 , then p2 and p1 would be given by A and B in the
diagram (where we do not [yet] assume that ρ2 , etc. are necessarily ‘downstream’). Since we are
assuming that dS = 0, it follows from (1.5c), or (1.A.5a), that
Z V2
e(V2 , S0 ) − e(V1 , S0 ) = − pdV = area under curve AB.
V1

• However, from the Hugoniot adiabatic, (2.34), we know that we require

e(V2 , p2 ) − e(V1 , p1 ) = 21 (p1 + p2 )(V1 − V2 ) (2.35)
= area under chord AB
> e(V2 , S0 ) − e(V1 , S0 ) .
2 2
d p
Hence if dV 2 > 0 (or equivalently ddpV2 > 0), a constant entropy shock is impossible. This is
s s
not surprising given that inside the shock there is viscous dissipation, heat generation, thermal
conduction, etc., and hence the flow is not adiabatic.
• If we keep V1 and V2 fixed, then the value of the RHS of (2.35), i.e. the area under the chord, is
∂e
unchanged if we ‘rotate’ AB about the mid-point to, say, CD. Moreover, if ∂p > 0, then as we
V
rotate clockwise, e(V2 , S2 ) − e(V1 , S1 ) increases, until (2.35) is satisfied.
• The Second Law of Thermodynamics implies, e.g. see (1.A.5a),
de = T dS − pdV .
Hence
∂e ∂S
=T ,
∂p V ∂p V
and consequently, since T > 0,
the entropy at C is greater than at A,
the entropy at D is less than at B.
So if ρ2 > ρ1 , then
S2 > S1 . (2.36)
But the entropy of a fixed mass of gas cannot decrease. Hence

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 S2 is downstream of S1 .

Indeed, the full equations for the shock’s internal structure have no solution unless this is the case.
We conclude that the transition is from a low density to a high density; we say that we have a
compressive shock.
08/16

2.6.4 Upstream/downstream flow is supersonic/subsonic

Plot p versus V for S = S1 and S = S2 .

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Then from (2.33), with V = 1/ρ:

(p2 − p1 )
u21 = V12 = −V12 × slope of CD
V1 − V2
> −V12 × slope of tangent at D
∂p
= −V12 (V1 ) = c21 ,
∂V S

∂p ∂p
since c2 = ∂ρ = −V 2 ∂V . So the flow is supersonic upstream since
S S

u1
M1 = > 1. (2.37a)
c1
It is also possible to show (but it is messy, e.g. see Landau and Lifschitz), that the flow is subsonic
downstream, i.e.
u2
M2 = < 1. (2.37b)
c2

Exercise: find an easy proof.

2.6.5 Upstream/downstream flow is supersonic/subsonic – perfect gas (unlectured)

For a perfect gas (see (1.5f), (1.5e) and (1.5b))

γp p
c2 = , e = cV T = . (2.38)
ρ (γ − 1)ρ

Define
p2 − p1
Shock Strength = β = , (2.39a)
p1
so that p2 = (1 + β)p1 . Then from (2.34) and (2.38)

ρ2 2γ + (γ + 1)β
= . (2.39b)
ρ1 2γ + (γ − 1)β

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Then from (2.33), (2.38), (2.39a) and (2.39b), and assuming that β > 0 and γ > 0 (in practice γ > 1),
 2  
u1 γ+1
=1+ β>1, (2.39c)
c1 2γ
 2    
u2 1 γ−1
= 1+ β <1. (2.39d)
c2 1+β 2γ
This confirms that, at least for a perfect gas,

shocks are transitions from supersonic to subsonic flow.

Further, from (1.5h)  

p

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
S = cV log + const .
ργ
It follows from (2.39a) and (2.39b) that, if β > 0 and γ > 1,
 
S2 − S1 2γ + (γ + 1)β
= log(1 + β) − γ log (2.39e)
cV 2γ + (γ − 1)β
> 0.

Further, if the shock is weak, i.e. β  1, then

S2 − S1 (γ 2 − 1) 3
= β + ... . (2.39f)
cV 12γ 2
Hence for weak shocks, the entropy jump is much smaller than the jumps in pressure, density, velocity,
etc.

∂2V
Remark. For a general gas, the coefficient in (2.39f) is proportional to ∂p2 , consistent with our earlier
S
2
∂ V
assumption that ∂p2 >0.
S

2.6.6 Moving shocks

Consider a shock moving at the velocity V , with velocities v1 and v2 on either side of the shock. Switch
to the frame in which the shock is at rest, corresponding to velocities
u1 = v1 − V, and u2 = v2 − V respectively.
Then from (2.30), (2.31) and (2.34)
ρ1 (v1 − V ) = ρ2 (v2 − V ) , (2.40a)
2 2
p1 + ρ1 (v1 − V ) = p2 + ρ2 (v2 − V ) , (2.40b)
 
1 1
e2 − e1 = 21 (p1 + p2 ) − . (2.40c)
ρ1 ρ2
Note that if V > v2 , then p2 , ρ2 , u2 become ‘upstream’; it is then necessary to be careful with signs of
square roots in (2.39c), (2.39d) etc.

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 2.6.7 Shock propagating into a perfect gas at rest.
For example, as generated by a steadily moving piston or an explosion.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
There are three independent equations, and six variables: ρ0 , p0 , ρ2 , p2 , v2 and V . However, we know
ρ0 and p0 , and one of ρ2 , p2 and v2 . For a piston problem it might be natural to specify v2 , but we will
specify p2 . It is then convenient to define
p2 − p0
Shock Strength = β = , (2.41)
p0

so that p2 = (1 + β)p0 .

Remark. A word of warning: since the algebra can be ‘detailed’, consider what a question asks, before
embarking on messy algebra to eliminate variables. In this case we first aim for ρ2 (ρ0 , p0 , p2 ).

For a perfect gas (see (1.5f), (1.5e) and (1.5b))

γp p
c2 = , e = cV T = . (2.42)
ρ (γ − 1)ρ

Then from (2.40c), (2.41) and (2.42)

ρ2 (γ − 1)p0 + (γ + 1)p2 2γ + (γ + 1)β

= = . (2.43a)
ρ0 (γ + 1)p0 + (γ − 1)p2 2γ + (γ − 1)β

As before, from conservation of mass and momentum,

ρ0 (−V ) = ρ2 (v2 − V ) , (2.43b)

2 2
p0 + ρ0 V = p2 + ρ2 (v2 − V ) . (2.43c)

Thence from (2.41) and (2.43a),

 
ρ2 (p2 − p0 ) γ+1
V2 = = 1+ β c20 . (2.43d)
ρ0 (ρ2 − ρ0 ) 2γ

Moreover, from (2.43a), (2.43b), (2.43d) and some algebra,

βc20
v2 = . (2.43e)
γV

09/04 Hence, if V > v2 > 0, then β > 0, and from (2.43d) the shock propagates supersonically into the gas at
rest.

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 2.7 Nonlinear Shallow Water Waves in 1D

Shallow water means that the wavelength  water depth. This means that we can assume that the flow
is almost horizontal and independent of depth (cf. hydraulic flows in IB).

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
2.7.1 Governing Equations

Consider a thin control volume of width δx. Then from conservation of mass
∂
(ρw hδx) = (ρw hu)(x) − (ρw hu)(x + δx) ,
∂t
thence by Taylor series, and assuming that the density of water, ρw , is a constant,
∂h ∂(hu)
+ = 0. (2.44a)
∂t ∂x
Similarly, from conservation of momentum
∂
(ρw huδx) = (ρw hu2 )(x) + P (x) − (ρw hu2 )(x + δx) − P (x + δx) ,
∂t
where P is the pressure force integrated over the depth; thence by Taylor series
∂(hu) ∂(hu2 ) 1 ∂P
=− − .
∂t ∂x ρw ∂x
Since we are assuming the the flow is primarily in the x-direction, it is consistent to assume that the
vertical accelerations are negligible, and that the pressure is hydro-static to leading order, i.e.
p = ρw g(h − z) ,
where p = 0 has been assumed at the free surface. Hence
Z h
P = pdz = 21 ρw gh2 , (2.44b)
0

and so the momentum equation becomes

∂(hu) ∂(hu2 ) ∂h
+ = −gh . (2.44c)
10/19 ∂t ∂x ∂x
The shallow water equations are thus, after simplification of (2.44c) using (2.44a),
∂h ∂h ∂u
+u +h = 0, (2.45a)
∂t ∂x ∂x
∂u ∂u ∂h
+u +g = 0. (2.45b)
∂t ∂x ∂x
These are the same as the 1D gas√equations (2.2a), (2.2b) and (2.2c) if we identify the gas density, ρ,
with h, and the wavespeed c with gh, i.e.
√ p
ρ = h and c = gρ = gh . (2.45c)
γ−1
For a perfect gas, c = c0 (ρ/ρ0 ) 2 ; hence the shallow water equations are equivalent to 1D gas dynamics
with γ = 2. As above, these equations can therefore be solved using C± characteristics, with the solutions
10/17 portraying wave steepening, shock formation, etc.
10/18

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 2.7.2 * Hydraulic jumps *

The ‘shocks’ that form from the steepening of shallow water waves are termed hydraulic jumps or bores.
The extra physics that controls a hydraulic jump is different from that in a gas shock. In a gas shock,
p, ρ and u jump so that mass. momentum and energy are conserved across the shock; however, across a
09/16 hydraulic jump only h and u can jump, and energy is not conserved.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Work in the frame of the jump.

Mass conservation. Divide through by ρw to obtain

h1 u1 = h2 u2 . (2.46a)

Momentum conservation.

P1 + ρw h1 u21 = P2 + ρw h2 u22 ,
where from (2.44b) the force due to the integrated hydrostatic pressure is given by
P1 = 21 ρw gh21 , P2 = 12 ρw gh22 .
Hence
2
1
2 gh1 + h1 u21 = 21 gh22 + h2 u22 . (2.46b)

Analogously to a gas shock, these can be manipulated to obtain

u21 u2 (h1 + h2 )h2
F12 ≡ 2 = 1 = = (1 + 21 β)(1 + β) , (2.47a)
c1 gh1 2h21
u22 u2 (h1 + h2 )h1 (1 + 21 β)
F22 ≡ 2 = 2 = 2 = , (2.47b)
c2 gh2 2h2 (1 + β)2
where F1 and F2 are Froude numbers (the ratio of fluid speed to wavespeed, cf. Mach number), and
h2 = (1 + β)h1 , where β is the shock strength.

Energy loss. The energy flux is

1 2 1 2
+ 12 ρw gh uh



I= 2 ρw gh u + 2 ρw u
| {z } | {z }
Work done by pressure Volume flux of (KE+GPE)
1 2


= ρw uh 2u + gh .
1 2


While ρw uh is continuous across the jump, 2u + gh is not, because ‘Bernoulli’ does not apply.
The change in energy flux is given by
2
1
− u22 ) + g(h1 − h2 )


I1 − I2 = ρw u1 h1 2 (u1 (2.48a)
ρw gu1 h21 β 3
= . (2.48b)
4(1 + β)
The energy loss has to be positive, from which we conclude that if the height upstream is h1 , then

β > 0, h2 > h1 , F1 > 1 and F2 < 1 .

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 Hence the flow upstream is ‘supercritical’, i.e. the fluid moves faster than the long-wave speed (cf.
supersonic), and the flow downstream is ‘subcritical’, i.e. the fluid moves slower than the long-wave
speed (cf. subsonic).

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
There are two possibilities for energy loss.

β & 0.5: this generally corresponds to a turbulent bore, where energy is lost to small scale motion
and viscous dissipation.
β . 0.5: this generally corresponds to an undular bore where [semi-long] waves behind the jump
carry energy away.
*Counting characteristics* (unlectured). In the frame in which the jump is at rest, both the C+ and C−
characteristics point towards the shock in the supercritical region, while only the C+ characteristic
points away from the shock in the subsonic region. There is thus too much information at the jump
(two routes in and one route out), and something has to ‘give’ (in this case energy, while in the case
of shocks it was entropy).

The number of characteristics approaching the shock corresponds to the total number of parameters
that can be specified on both sides of the shock.

Example. Consider a jump propagating with speed V into quiescent fluid of depth h0 . On the assumption
that the ‘upstream’ height h2 = (1 + β)h0 is known, find V . Note: there are three characteristics
approaching the jump, and we specify u0 = 0, h0 and h2 .
In the frame in which the jump is at rest, ‘upstream’
is ‘downstream’. It follows from conservation of mass
and momentum, cf. (2.46a) and (2.46b),

h0 (−V ) = h2 (u2 − V ) , (2.49a)

1 2 2 1 2 2
2 gh0 + h0 V = 2 gh2 + h2 (u2 − V ) , (2.49b)

and thence, cf. (2.47a),

V2 (h0 + h2 )h2
=
gh0 2h20
= (1 + 12 β)(1 + β) > 1 . (2.49c)

Hence the jump propagates at a supercritical veloc-

ity.

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Appendix

2.A * Shock Formation in Trombones *

Model the trombone as a semi-infinite straight tube x > 0 with fluid at rest for t < 0. For t > 0 consider
pressure perturbations at x = 0. In the quiescent fluid (from which the C− characteristics emanate),
u = 0 and c = c0 , so R− = 0. Let u(t), c(t) and p(t) denote the velocity, sound speed and pressure at
x = 0 when t > 0, and assume that p(t) is known.
Proceeding as in § 2.5.1, the flow is again a simple wave. Hence the C+ are straight with characteristic
equation, see (2.20a),

dx
= c0 + 21 (γ + 1)u,

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
C+ : u = const. (2.A.1a)
dt
For characteristics emanating from x = 0 for t > 0, the C+ satisfy the equation, cf. (2.28b),

x = c0 + 21 (γ + 1)u(τ ) (t − τ ) ,


(2.A.1b)

where τ is used to label the characteristics. From (2.10a), (2.10b) and (2.10c), u(τ ) is given in terms of
the pressure by
    γ−1 !
2c0 p(τ ) 2γ
u(τ ) = −1 . (2.A.1c)
γ−1 p0

Hence, given (x, t), (2.A.1b) and (2.A.1c) are an implicit system of equations for τ (x, t) in terms of the
known pressure at x = 0.
Shock waves form when the characteristics intersect, i.e. when dx
dτ t = 0. Hence the time of shock formation
is given by
c0 + 12 (γ + 1)u(τ )
 
ts = min τ + 1 . (2.A.2a)
2 (γ + 1)u̇(τ )
τ >0

If the RHS of (2.A.2a) is minimised for τ = τm , then the position of shock formation is given by

2
c0 + 12 (γ + 1)u(τm )
xs = 1 . (2.A.2b)
2 (γ + 1)u̇(τm )

For simplicity, assume that the pressure perturbations are small and that |u|  c0 , then the above
expression can be linearised to obtain
 
2γp0 2γc0 p0
ts ∼ min τ + and xs ∼ . (2.A.2c)
τ >0 (γ + 1)ṗ(τ ) (γ + 1)ṗ(τm )

Thus for a shock to form downstream of the mouthpiece, we require ṗ(τm ) > 0. Moreover, the shock will
first form on the characteristic, or at least close to the characteristic, where ṗ(τ ) is largest.
For instance, if for τ > 0
p(τ ) = A(1 − e−ατ ) , so ṗ(τ ) = Aαe−ατ , (2.A.3a)
then
2γp0 eατ
 
2γp0
ts ∼ min τ + = (τm = 0) , (2.A.3b)
τ >0 (γ + 1)Aα (γ + 1)Aα
2γc0 p0
xs ∼ . (2.A.3c)
(γ + 1)Aα

Typical values of max(ṗ) at the mouthpiece of a trombone range from around 5, 000 kPa s−1 when played
at piano (soft) level to around 20, 000 kPa s−1 when played at fortissimo (loud) for low frequencies. Using
values of γ ≈ 1.4, p0 = 101 kPa and c0 = 340 m s−1 , this gives a shock formation distance of around

Mathematical Tripos: Part II Waves e © S.J.Cowley@maths.cam.ac.uk, Lent 2019

6.8 m (piano) to around 1.7 m (fortissimo). At higher frequencies this distance is approximately in the
order of tens of metres (piano) to around 1 m (fortissimo). With the typical length of a trombone’s tubing
around 2-3 metres, one would expect shocks to form at higher dynamic playing levels. Indeed, this is what
is observed by Hirschberg et al. (1996)5 when measuring the horn exit pressure.
For a video see

http://www.bbc.co.uk/news/science-environment-13574197
and/or https://www.youtube.com/watch?v=TOhxr643YuA.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.

5 Hirschberg, A., Gilbert, J., Msallam, R. & Wijnands, A.P.J, Shock Waves in Trombones. Journal of the Acoustical

Society of America, Acoustical Society of America, 1995, 99(3), 1754-1758.

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x

3 Linear Elastic Waves

3.1 Governing Equations of Linear Elasticity

3.1.1 Deformation of a solid

As in fluid dynamics, we make the continuum approximation (averaging over volumes large enough to
contain many molecules but much smaller than the scales in interest) in order to define fields of density
ρ(x, t), etc.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
When a body is deformed, a particle moves from its original ‘reference’ position ξ to a new position x(ξ, t);
since the position must be invertible, ξ ≡ ξ(x, t). A key feature of an elastic solid is that it remembers
the original reference configuration ξ. It is conventional in solid mechanics to denote the

particle displacement by u(x, t) = x(ξ, t) − ξ , (3.1a)

du
particle velocity by v(x, t) = , (3.1b)
dt
dv
particle acceleration by a(x, t) = , (3.1c)
dt
d ∂
where = is the rate of change moving with the particle, written more usually in fluid dynamics
dt ∂t ξ
as
D ∂
≡ + v·∇ .
Dt ∂t x
Warning: in solid mechanics u denotes displacement not velocity.
It is convenient to write all fields as functions of the current (Eulerian) position x, rather than the original
(Lagrangian) position ξ.

3.1.2 The stress tensor

As in fluid dynamics, the forces acting the material can be divided into

(a) surface forces, or tractions/stresses, acting on and proportional to the surface area, like pressure;
(b) body forces acting on and proportional to the volume, like gravity, where we denote the body force
by F, or ρf .

Consider a small area element n dS at position x within a solid body. The force exerted by the outside
on the solid inside of the element (with ‘outside’ defined by the outward normal n) is assumed to
be a surface force τ (x, t; n) dS; by Newton three, τ (x, t; −n) = −τ (x, t; n).

The traction or stress, τ , acting on the area element n dS depends on the orientation n. (The units of
traction are force/area, and forces are obtained by integrating tractions over an area.) In an inviscid fluid
τ = −pn. In general, τ (x, t; n) is linearly related to n by a second-rank tensor σ. To see this we proceed
as follows.

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 Consider a small material tetrahedron with three faces aligned parallel to the Cartesian coordinate planes
and a fourth sloping face with area ε2 and normal n. By geometry, the areas of the other faces are ε2 ni
and the normals are −ei , where ei is the ith basis vector.

The surface forces on the tetrahedron are ∝ ε2 , whereas both the body forces and the inertia of the

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
tetrahedron are ∝ ε3 .It follows that the surface forces must balance each other as ε → 0, i.e.

ε2 τ (x, t; n) − τ (x, t; e1 )n1 − τ (x, t; e2 )n2 − τ (x, t; e3 )n3 = O(ε3 ) ,



or equivalently in the limit ε → 0,

τi (x, t; n) = τi (x, t; ej )nj . (3.2a)

Hence the traction τ exerted by the outside on the inside of a surface with outward normal n is given by

τ = σ·n , (3.2b)

where σ is the second-rank tensor, the (Cauchy) stress tensor, given by

σij (x, t) = τi (x, t; ej ) . (3.2c)

Warning: ‘stress’ is also often used to refer to the vector τ as well as the tensor σ.

3.1.3 Momentum, angular momentum and energy

Momentum. Consider an arbitrary material control volume V(t) with surface S(t) and normal n. Integrate
‘F = ma’ over all the particles in the control volume, to yield
Z Z Z
ρ(x, t)a(x, t) dV = F(x, t) dV + σ(x, t)·n dS . (3.3a)
V V S

Using the divergence theorem and the arbitrariness of V, we obtain the momentum equation
∂σij
ρai = Fi + . (3.3b)
∂xj

Angular Momentum. Provided that long-range forces exert no body couple on the material (a counter-
example is provided by a solid containing magnetic particles in an external magnetic field), we can
show that the stress tensor is symmetric from the angular momentum balance. Taking moments
about 0,
Z Z Z
ρx × a dV = x × F dV + x × σ·n dS . (3.4a)
V V S

and hence
∂ ∂σij
pqi xq (ρai − Fi ) = (pqi xq σij ) = pqi xq + pji σij . (3.4b)
∂xj ∂xj

Subtracting pqi xq × (3.3b) gives

pji σij = 0 .
Hence σ is symmetric:
σij = σji . (3.4c)
11/17

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 Energy. We obtain the energy equation by multiplying the momentum equation (3.3b) by the velocity vi .
As a preliminary note that from the definition of acceleration, conservation of mass, and the product
rule,
d 1 2
vi ai = ( v ), (3.5a)
dt 2
d
(ρ dV ) = 0 for a material volume element , (3.5b)
dt
∂σij ∂(vi σij ) ∂vi
vi = − σij . (3.5c)
∂xj ∂xj ∂xj
Hence, using the symmetry of the stress tension, (3.4c), we obtain by integrating over a material
volume

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Z Z Z Z  
d 1 2 1 ∂vi ∂vj
ρv dV = v · F dV + v·σ·n dS − + σij dV . (3.5d)
dt V(t) 2 V S V
2 ∂xj ∂xi
| {z } | {z } | {z } | {z }
Change in KE Rate of work Rate of work Rate of release
by body forces by surface forces of internal energy

R final term; in the meantime note that in the simple case σij = −pδij the final
We will return to the
11/18 term reduces to + p ∇·v dV , the result for the reversible adiabatic compression of a gas.
11/19
3.1.4 Fluid and solid continua

All of the arguments above apply to any continuum – fluid or solid, whether viscous, inviscid, elastic,
plastic, viscoelastic or whatever. They are presented for a viscous fluid in Part II Fluid Dynamics. The
main difference between these various sorts of continua lies in the constitutive relationship between the
stress σ and the deformation or rate of deformation. (The notation also varies from one application to
another e.g. u or v for velocity, D/Dt or d/dt for material derivative, e or ė for strain rate.)
10/16 In a linear elastic solid the stress is linearly related to the strain.

3.1.5 Strain

Consider the elastic stresses that result from the change in separa-
tion of two neighbouring material elements:

dxi − dξi = ui (x + dx, t) − ui (x, t)

∂ui
= dxj + h.o.t. . (3.6a)
∂xj

Assume further that the deformation is small, i.e.

|dx − dξ|  |dx| , (3.6b)

so that
∂ui
 1, (3.6c)
∂xj

as is appropriate for hard solids like metal and rock (but not rub-
ber), where a small deformation produces a large restoring force.
Write
   
∂ui 1 ∂ui ∂uj 1 ∂ui ∂uj
= + + − (3.7a)
∂xj 2 ∂xj ∂xi 2 ∂xj ∂xi
= eij + ωij , (3.7b)
∂ui
where eij and ωij are the symmetric and antisymmetric parts of the tensor ∂xj . Also write

ωij = − 12 ijk ωk , where ω = curl u ,

then

Mathematical Tripos: Part II Waves 40 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 1
ωij δxj = 2 (ω × δx)i .

With small deformations this term gives only a rigid rotation, and hence induces no elastic stresses. curl u
is referred to as the rotation; so there is plenty of potential for confusion between solids mechanics and
fluid mechanics (in the latter case u, ω and e are the velocity, vorticity and the strain rate).
If e is zero then to first order there is no deformation. e is called the (Cauchy, or infinitesimal) strain
tensor.
Exercise. Show that with small deformations, that e is related to changes in length by

|δx|2 − |δξ|2 = 2 δξ · e · δξ .

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
3.1.6 Constitutive equation for a linear elastic solid

The constitutive equation is the relationship between stress and strain. For a elastic body subject to small
strains about the reference state e = 0:

(i) We will assume that the constitutive equation is instantaneous and local so that σ ≡ σ(e), i.e. there
is no dependence on
∂eij ∂eij
or or . . . .
∂t ∂xk
(ii) We will assume that the constitutive equation is linear, i.e.

σij (x, t) = cijk` ek` (x, t) , (3.8a)

where c is a fourth-order tensor that is a property of the material (cf. Hooke’s law for which force
∝ extension, with σ =force/area ∝ e =extension/original-length).
Since σij = σji , and ek` = e`k ,
cijk` = cjik` = cij`k . (3.8b)
Hence there are up to 36 parameters in a general anisotropic material.
(iii) We will also assume that the material is isotropic; then c is an isotropic tensor

cijk` = λ δij δk` + µ δik δj` + µ̄ δi` δjk . (3.8c)

Moreover, (3.8b), or equivalently the symmetry of σ, implies that µ̄ = µ; hence

 
∂uk ∂ui ∂uj
σij = λ δij ekk + 2µ eij = λ δij +µ + . (3.8d)
∂xk ∂xj ∂xi
λ and µ are the Lamé constants/moduli.
Since σkk = (3λ + 2µ)ekk , it follows from (3.8d) that
 
1 λ
eij = σij − δij σkk . (3.8e)
2µ 3λ + 2µ

3.1.7 Simple deformations

Dilatation.
Consider a small change of material volume
Z Z
∆V = u · n dS = ∇ · u dV .

We define the [local] dilatation to be

∂uk
θ ≡∇·u= = ekk . (3.9)
∂xk

Mathematical Tripos: Part II Waves 41 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Hydrostatic pressure. In the case of hydrostatic pressure, i.e. when

σij = −p δij and τ (x, t; n) = −pn , (3.10a)

then from (3.8d),

 
2µ
p = − 13 σ`` = − λ + ekk = −κθ , (3.10b)
3

where κ = (λ + 32 µ) is the bulk modulus or modulus of incompressibility. On physical grounds

we expect that
κ > 0. (3.10c)
Otherwise. When not in a state of hydrostatic pressure, we define

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
p = − 31 σkk , (3.11a)
and write
σij = −p δij + σ ij , (3.11b)

where σ ij is the deviatoric stress.

Simple shear.
Consider the simple shear u = (γy, 0, 0). Then

0 12
   
0 0 1 0
e = γ  12 0 0 , σ = γµ 1 0 0 .
0 0 0 0 0 0

µ is referred to as the shear modulus or modulus of rigidity. In a

fluid µ = 0; more generally we expect on physical grounds that

µ > 0. (3.12)
Uniaxial extension.
Consider a uniaxial extension induced by a stress
 
1 0 0
F 
σ= 0 0 0 .
A
0 0 0

Then from (3.8e)

   
2(λ + µ) 0 0 1 0 0
F F
e=  0 −λ 0 = 0 −ν 0 ,
2A(3λ + 2µ)µ EA
0 0 −λ 0 0 −ν

where
(3λ + 2µ)µ 3κµ
E= = > 0, Young’s modulus , (3.13a)
λ+µ λ+µ
λ
ν= , Poisson’s ratio . (3.13b)
2(λ + µ)
10/04 Remark. ν is not the kinematic viscosity, and can be negative (in auxetic materials).

3.1.8 Equations of motion

From substituting the isotropic constitutive relation (3.8d) into the momentum equation (3.3b),
   
∂ ∂uk ∂ ∂ui ∂uj
ρaj = ρFj + λ + µ + . (3.14)
∂xj ∂xk ∂xi ∂xj ∂xi

Mathematical Tripos: Part II Waves 42 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Next, recall that the velocity v(x, t) of a material point is given by
∂ d ∂ui ∂ui dxj ∂ui ∂ui
vi = xi (ξ, t) = (ui (x(ξ, t), t)) = + = + vj .
∂t ξ dt ∂t ∂xj dt ∂t ∂xj

∂ui
However, since we are assuming that the deformation is small, i.e. ∂x j
 1 from (3.6c), it is consistent
to neglect the convective derivative in the material derivative, so that
∂ui
vi ≈ . (3.15a)
∂t
Similarly
∂vi ∂vi ∂vi ∂ 2 ui
ai = + vj ≈ ≈ . (3.15b)
∂t ∂xj ∂t ∂t2

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Further, for small deformations the dilatation is small, and hence the density can be taken to be that in
the undeformed state (relieving the need to consider the conservation of mass equation henceforth):

ρ(x, t) = ρ(ξ + u, t) ≈ ρ(ξ, 0) . (3.15c)

From substituting the above linearisations into the momentum equation (3.3b), and using (3.8d), it follows
that
∂ 2 uj
    
∂ ∂uk ∂ ∂ui ∂uj
ρ 2 = ρFj + λ + µ + . (3.16a)
∂t ∂xj ∂xk ∂xi ∂xj ∂xi
Henceforth we will assume that the material is homogeneous, i.e. that ρ, λ and µ are constants throughout
the material. Then
∂2u
ρ = ρF + (λ + µ) ∇(∇.u) + µ∇2 u (3.16b)
∂t2
12/17 = ρF + (λ + 2µ) ∇(∇.u) − µ∇ × (∇ × u) . (3.16c)
12/19
Boundary conditions
Rigid/clamped boundary.

u = 0. (3.17a)

Free surface. At a constant pressure free surface

τ = σ·n = −p0 n ( =⇒ n × σ·n = 0) . (3.17b)

Solid/solid interface. No relative motion and Newton three

imply the six conditions

[u]+ +
− = [σ·n]− = 0 . (3.17c)

Solid/inviscid-fluid interface. No relative normal motion, and continutity of stress (an inviscid fluid
only has a pressure traction/stress), imply the four conditions

12/18 [u·n]+
− = 0, σ·n = −p0 n . (3.17d)

3.1.9 Energy equation

Recall the energy equation (3.5d)

Z Z Z Z  
d 1 1 ∂vi ∂vj
ρv · v dV = v · F dV + v·σ·n dS − + σij dV .
dt V 2 V S V
2 ∂xj ∂xi
Assuming small deformations, so that v ≈ u̇, this becomes
Z Z Z Z
d 1
ρ u̇ · u̇ dV = u̇ · F dV + u̇·σ·n dS − ėij σij dV . (3.18a)
dt V 2 V S V

Mathematical Tripos: Part II Waves 43 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Further, for small displacements and an isotropic material
∂ 1
= 12 cijk` (ek` ėij + ėk` eij )


∂t 2 σij eij from(3.8a)
=cijk` ek` ėij cijk` = ck`ij from(3.8b)
=σij ėij . (3.18b)

Hence, using the fact that the density is approximately constant,

Z Z Z
d 1
(ρu̇ · u̇ + σij eij ) dV − u̇·σ·n dS = u̇ · F dV , (3.18c)
dt V 2 S V

i.e.
∂
(Ek + Ep ) + ∇·I = u̇ · F , (3.18d)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
∂t
where, assuming no strain energy when the material is undeformed,
Kinetic Energy = Ek = 21 ρu̇2 , (3.18e)
1
Elastic Potential Energy = Ep = ρE = 2 σij eij , (3.18f)
Energy Flux Vector = I = −u̇·σ . (3.18g)

Here E is the strain energy density per unit mass. Further, from (3.8d),

ρE = 21 (λekk ejj + 2µeij eij )

= 12 λ + 32 µ (ekk ejj ) + µ eij − 13 ekk δij eij − 13 e`` δij .

Hence E positive definite iff

κ = λ + 32 µ > 0 , µ > 0. (3.19)
11/16 These are the same restrictions as obtained in (3.10c) and (3.12) by physical reasoning.

3.2 Compressional Waves and Shear Waves

3.2.1 Waves of dilatation and rotation

Recall from (3.16b) that linear disturbances to an isotropic homogeneous material are governed by,
assuming no body force,
∂2u
ρ = (λ + µ)∇(∇.u) + µ∇2 u . (3.20a)
∂t2
= (λ + 2µ) ∇(∇.u) − µ∇ × (∇ × u) (3.20b)

Primary/compressional/pressure waves. Take the divergence of (3.20a) to obtain

∂2θ
− c2P ∇2 θ = 0 , (3.21a)
∂t2
where θ = ∇ · u and
κ + 43 µ
 
λ + 2µ
c2P = = > 0. (3.21b)
ρ ρ
cP is the dilatational (or compressional) wave speed.
Secondary/shear waves. Next take the curl of (3.20a) to obtain
∂2ω
− c2S ∇2 ω = 0 (3.22a)
∂t2
where ω = curl u and
µ
0< = c2S < c2P . (3.22b)
ρ
cS is the shear wave speed.

Mathematical Tripos: Part II Waves 44 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Hence a general disturbance, e.g. as generated by an earthquake, propagates [at least] at two speeds. The
first to arrive, the primary P wave, is the dilatational part. The second to arrive, the secondary S wave,
is the shear wave.

Elastic fluids. If µ = 0 (i.e. an elastic fluid) then cS = 0 and only compressional, i.e. acoustic/sound,
waves are supported with
κ dp
c2P = = , (3.23)
ρ dρ
since for an elastic fluid

p = − 13 σkk = −(λ + 23 µ)ekk = −κθ ,

and (with some slight abiguity in the use of ρ = ρ0 + ρ̃)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
∂ ρ̃ ∂θ κ dp κ
≈ − ρ0 ∇.u̇ ≈ −ρ0 =⇒ ρ̃ ≈ −ρ0 θ =⇒ p≈ ρ̃ =⇒ ≈ .
∂t ∂t ρ0 dρ ρ

11/04 Example. For steel, cP ∼ 6 . 103 ms−1 , cS ∼ 3 . 103 ms−1 .

3.2.2 Plane waves

Look for solutions to (3.20b) of the form

 
b − ct ,
u = p k·x (3.24a)

where p is the polarisation and k

b is a unit vector. Then we require
 
c2 p00 = c2P k( b 00 ) − c2 k
b k·p
S
b× kb × p00 . (3.24b)

It follows that
b 00 = 0 ,
(c2 − c2P ) k·p (3.24c)
b × p00 = 0 .
(c2 − c2S ) k (3.24d)

Since cP 6= cS there are two possibilities.

b × p00 = 0, and we let 6

Longitudinal plane waves. If c = cP , then k
   
u=k b − cP t , θ = ∇ · u = f 0 k·x
b f k·x b − cP t and ω = curl u = 0 . (3.25)

This is a plane wave of arbitrary shape with u parallel to k,

b i.e. the displacement is parallel to the
direction of travel of the wave. This is a longitudinal wave.
b · p00 = 0, , and we let
Transverse plane waves. If c = cS , then k
   
u=k b − cS t , θ = ∇ · u = 0 and ω = curl u = −g0 k·x
b × g k·x b − cS t , (3.26)

where, without loss of generality, we can assume that k b · g = 0. This is a plane wave of arbitrary
shape with u·k = 0, i.e. the displacement is perpendicular to the direction of travel of the wave.
b
b × g is fixed, then the wave is said to be polarised.
This is a transverse wave. If the direction of k
Plane waves: stress tensor. From (3.8d)
 
 λδij + 2µb kj f 0
ki b : P wave,
σij =  
µ i`m b 0
kj + j`m b
ki bk` gm : S wave.

6 We do not consider the cases of constant uniform displacement or uniform translation of the solid.

Mathematical Tripos: Part II Waves 45 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Plane waves: energy density. From (3.18e) and (3.18f)
(
1
ρc2 f 02 : P wave,
Ek = 12 ρu̇·u̇ = 12 P 2 b 0 b × g0 ) : S wave,
2 ρcS (k × g )·(k
(
1 02
Ep = 12 λekk ejj + µeij eij = 2 (λ + 2µ) f : P wave,
1 0 0
2 µ(k × g )·(k × g ) : S wave,
b b

Hence, since ρc2P = λ + 2µ from (3.21b), and ρc2S = µ from (3.22b), for both P and S waves there
is equi-partition of energy:
Ek = Ep .

Plane waves: energy flux. From (3.18g)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
(
(λ + 2µ)cP f 02 k
b : P wave,
I = −σ·u̇ = 0 b × g0 ) k
b × g )·(k
µcS (k b : S wave.

Plane waves: energy propagation. Hence, for either type of plane wave separately,

I = (Ek + Ep ) c k
b,

where c is the corresponding phase speed (i.e. cP or cS ). Hence energy is propagated at the wavespeed
in the direction of travel of the waves (because the waves are non-dispersive). However, remember
that while linear displacements can be added, quadratic energies cannot in general.
13/17
13/19 3.2.3 Harmonic waves

P waves. Proceeding as in §1.5.2, it follows from (3.25) that a harmonic P-wave solution is

u = Ak ei(k·x−ωt) , (3.27)

where k = |k|, ω = kcP and A is a constant.

S waves and polarisation. From (3.26) a harmonic S-wave solution is

u = k × B ei(k·x−ωt) , (3.28a)

where ω = kcS and B is a constant. For an S wave, the displacement is perpendicular to k, with
12/16 k × B being the direction of polarisation.
To fix ideas, suppose that we have a plane boundary, say
y = 0, defining the ‘horizontal’ (as in the earth). W.l.o.g. we
need only consider two-dimensional waves, e.g. by rotating
axes so that k lies in the (x, y) plane, say k = k(sin θ, cos θ, 0).
Recalling that k·b z = 0, we can then decompose k × B into

(i) a part that is horizontally polarised, i.e. parallel to b

z,
called a SH-wave

u = (k × B·b z ei(k·x−ωt) ;
z) b (3.28b)

(ii) a part that is ‘vertically’ polarised, i.e. which lies in the

same plane as k and the vertical y b and is parallel to
k×b z, called a SV-wave

u = (B·b z ei(k·x−ωt) .
z) k × b (3.28c)

Note that despite the name, displacements are not

purely “vertical” in SV waves.
13/18 Remark. A general disturbance contains P, SV and SH waves.

Mathematical Tripos: Part II Waves 46 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Attenuated/evanescent waves (near a boundary). If k is allowed to be complex, say k = kr + iki , then,
taking P waves as an example,

u = Ak exp(ikr . x − ki . x − iωt) ,
where
ω 2 = c2P (kr + iki )·(kr + iki )
= c2P (kr2 − ki2 ) + 2ic2P kr ·ki . (3.29a)

Hence
• there is propagation in the kr direction;
• there is attenuation/evanescence in the ki direction;

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
• if ω is real, then we need kr2 > ki2 and kr ·ki = 0, in which case
1
frequency: ω = ±cP (kr2 − ki2 ) 2 , (3.29b)
ω 1
phase speed: = ±cP (1 − ki2 /kr2 ) 2 . (3.29c)
kr

Hence attenuated waves move slower than unattenuated waves, i.e. propagation along a boundary
is slower than through the interior (evanescent waves do not occur in the interior of a solid, but
may occur near an interface).

3.2.4 Interface boundary conditions

Suppose there is a plane interface at y = 0 when undisturbed.

Let the disturbed position be

y = η(x, z, t). (3.30)

If the solid material is not to fracture, then we require the dis-

placement to be continuous at the interface, i.e., as in (3.17c),
+
[u(x, η, z, t)]− = 0. (3.31a)
If the displacements are small, i.e. |η|  1, we can linearise
this to
+
[u]− = 0, on y = 0. (3.31b)

Throughout our study of solid dynamics we have built into our analysis an assumption that, by Newton
three, the traction on the material on one side of an element of surface, due to the material on the other
side of the element of surface, is equal and opposite. Hence we also have the boundary condition (again
see (3.17c))
+
[σ · n]− = 0 on y = η, where n = (−ηx , 1, −ηz ) . (3.32a)
On linearising this becomes
+ + +
[σxy ]− = [σyy ]− = [σzy ]− = 0 on y = 0, where n = (0, 1, 0) . (3.32b)

3.2.5 Rayleigh waves

It is not possible to have a self-sustained SH wave at a boundary, but is is possible to have a self-contained
combination of P and SV waves, called a Rayleigh wave.

Mathematical Tripos: Part II Waves 47 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Suppose that an elastic material is bounded by a stress-free
boundary, initially at y = 0. Suppose that the surface is
slightly displaced to y = η(x, z, t). If |η|  1, then the lin-
earised stress-free boundary condition becomes

σ·n = 0 on y = 0 , where n = (0, 1, 0) . (3.33a)

The is no restriction on u since the surface is free to move.

Seek 2D solutions; then we have two boundary conditions,

σxy = 0 = σyy on y = 0 , (3.33b)

and hence we need both P and SV waves.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Seek attenuated wave solutions

u = A kP ei(kx−ωt)+kαy + B kS × b
z ei(kx−ωt)+kβy , (3.34a)
= A ekαy − iβB ekβy , −iαA ekαy − B ekβy k ei(kx−ωt) ,


(3.34b)

where
x − ikαb
kP = kb y, x − ikβb
kS = kb y, (3.34c)
and we have taken both the P wave and the SV wave to have the same frequency and x-wavenumber
so that the boundary conditions can be satisfied (also see below). Further, the trick in many of these
calculations is to keep the algebra under control; to this end we have chosen a notation such that k is the
x-wavenumber, and −kα and −kβ are the complex y-wavenumbers (and k 6= |kj |).

If k > ω/cS (> ω/cP ), decaying solutions exist as y → −∞ if

we choose (see the dispersion relation (3.29b) and the analo-
gous result for a shear wave)

21
21
α = + 1 − c2 /c2P , β = + 1 − c2 /c2S , (3.35)

where c = ω/k is the phase velocity of the wave.

12/04
Now apply the boundary conditions at y = 0:
 
∂u1 ∂u2
0 = σxy = µ +
∂y ∂x
= µ 2αA − i 1 + β 2 B k 2 ei(kx−ωt)

c2 
  
= µ 2αA − i 2 − 2 B k 2 ei(kx−ωt) ,
cS
∂u1 ∂u2
0 = σyy = λ + (λ + 2µ)
∂x ∂y
= −i((λ + 2µ)α − λ) A − 2µβB k 2 ei(kx−ωt)
2

c2
   
= µ −i 2 − 2 A − 2βB k 2 ei(kx−ωt) ,
cS

using (λ + 2µ)c2S = µc2P . For a non-trivial solution for A, B we require that the dispersion relation
2 1  1
c2 c2 2 c2 2
 
2− 2 = 4αβ = 4 1 − 2 1− 2 , (3.36a)
cS cP cS

is satisfied. This equation for the unknown wave speed c is Rayleigh’s equation (which is one of many).
It can be rewritten, given c 6= 0, as

2c2 c2
   
f (ξ) = ξ 3 − 8ξ 2 + 8 3 − 2S ξ − 16 1 − 2S = 0, (3.36b)
cP cP

Mathematical Tripos: Part II Waves 48 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 c2
where ξ = c2S
. Note that

c2S
 
f (0) = −16 1 − 2 < 0, and f (1) = 1 > 0 ;
cP

hence ∃ at least one real root with 0 < c2 < c2S (in fact ∃ only one root). This is the Rayleigh wave.
Remarks.

(i) Rayleigh waves are non-dispersive, i.e. c is independent of k.

(ii) The displacements are given by

c2
   
ekαy − 1 − 2 ekβy ei(kx−ωt) ,

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
u1 = kA
2cS
−1 !
c2

kαy kβy
u2 = −ikαA e − 1− 2 e ei(kx−ωt) .
2cS

• Hence at y = 0, u1 is − π2 out of phase with u2 .

• Also, since α > β, u1 changes sign for some y, while u2 is one signed for all y.
(iii) Rayleigh waves form the principal part of the seismic signal since P and S waves spread through a
3D volume, while Rayleigh waves spread out in a 2D region.
(iv) In earthquakes:
• P waves arrive first, then
• S waves arrive, and then
• Rayleigh waves arrive (and knock your buildings down).

Vertical-component seismogram showing the arrival of the P(rimary), (S)econdary, Love and Rayleigh
waves.7 The Rayleigh waves have the largest amplitude; for Love waves see § 4.1.2.
13/16
14/17
7 Gabi Laske, http://igppweb.ucsd.edu/˜gabi/sio15/sio15-13/lectures/Lecture07.html.

Mathematical Tripos: Part II Waves 49 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 3.3 Reflection and Refraction of Plane Waves

If an elastic wave is incident on an interface where the properties of the medium change discontinuously,
reflection and refraction can occur.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
In general, an incident wave of one type generates reflected and refracted waves of other types.

• For instance, SV and P waves can excite each other at an interface (cf. Rayleigh waves).
• However, SH waves can only excite other SH waves (since they are the only types of waves with
displacements in the spanwise z-direction).
14/18
14/19

3.3.1 Reflection and refraction of SH waves at an interface

The waves on either side of the interface must satisfy the relevant dispersion relation. Hence

Incident SH wave:
ωI
uI = (0, 0, 1) eikI ·x−iωI t where kI = (sin θ, cos θ, 0) , (3.37a)
cS
Reflected SH wave:
ωR
uR = (0, 0, R) eikR ·x−iωR t where kR = (sin Θ, − cos Θ, 0) , (3.37b)
cS

Transmitted SH wave:
ωT  
uT = (0, 0, T ) eikT ·x−iωT t where kT = sinθ, cosθ, 0 . (3.37c)
cS

The appropriate boundary conditions on y = 0 are (since σxy = σyy = 0)

uI + uR = uT , (3.38a)
(σyz )I + (σyz )R = (σyz )T . (3.38b)

Mathematical Tripos: Part II Waves 50 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 These can only be satisfied ∀x and t if the frequencies are the same, i.e. if

ωI = ωR = ωT ≡ ω , (3.39a)

and the wavelengths in the x

b direction are equal, i.e. if

sin θ sin Θ sinθ

= = . (3.39b)
cS cS cS
Hence

sin θ sinθ
θ| =
{z Θ} , = . (3.39c)
cS cS
Angles of incidence | {z }

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
and reflection equal Snell’s law
of refraction

Now
∂(uI )z µ cos θ ikI ·x−iωt
(σyz )I = µ = iω e , etc.,
∂y cS
hence from applying the boundary conditions (3.38a) and (3.38b)

1+R= T , (3.40a)
µ cos θ µ cosθ
(1 − R) = T . (3.40b)
cS cS
Solve for R and T to obtain
Z −Z 2Z
R= , T = , (3.41a)
Z +Z Z +Z
where Z and Z are impedances defined by
cS cS
Z = , Z = . (3.41b)
13/04 µ cos θ µ cosθ

Energy flux. From (3.18g), the flux of energy is given by

∂u
I = −σ. , (3.42)
∂t
and so
µω 2
hIy iI = − 21 Re [σyz (u̇)∗z ]I = cos θ ,
2cS
µω 2 |R|2
hIy iR = − cos θ ,
2cS
µω 2 |T |2
hIy iT = cosθ .
2cS
Thus
µω 2
cos θ 1 − |R|2


hIy iI + hIy iR =
2cS
2ω 2Z ω2
=
2 = |T |2 = hIy iT . (3.43)
Z +Z 2Z

It follows that the mean incident energy flux is either reflected or transmitted.

Remarks
(i) There is total transmission if the impedances match, i.e. if Z = Z.

Mathematical Tripos: Part II Waves 51 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 (ii) If cS < cS then from (3.39c)

cS
sinθ = sin θ < sin θ .
cS
Hence, whatever the value of θ, there exists a maximum
value of θ given by
 
−1 cS
θmax = sin . (3.44)
cS

There can be no transmitted wave with θ > θmax .

π
(iii) If cS > cS then θ > θ, and there exists an incident angle for which θ = 2,namely

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
 
−1 cS
θ = θmax = sin . (3.45a)
cS
For incident angles greater than this critical angle, i.e. for
θ > θmax , the solutions in y > 0 are evanescent waves of
the form (from using (3.37c) and (3.39c) )
 
ω ω
uT = (0, 0, T ) exp i sin θ x − βy − iωt , (3.45b)
cS cS

where (cf. (3.29b))

1
c2 2

β= sin2 θ − 2S . (3.45c)
cS

uT is an evanescent wave, and there is no energy flux

away from the interface.
The only modification necessary to (3.41a), (3.41b), etc. is
cS
i cosθ → − β.
cS
It follows from (3.41a) that
1
c2 2

iµ cos θ + µ sin2 θ − 2S
cS
R=  1 , (3.46)
c2 2
iµ cos θ − µ sin2 θ − 2S
cS
and hence |R| = 1. So for θ > θmax all incident energy is reflected and there is total internal
reflection.

3.3.2 Reflection of P waves at a rigid interface

Consider reflection at a rigid barrier so that u = 0

on y = 0 (alternatively one might consider a free
surface). As before
(a) all waves must have the same frequency, ω;

(b) the angles of incidence and reflection of the P

wave must be equal, i.e. Θ = θ.

Incident P wave:
ω
uI = (sin θ, cos θ, 0)eikI .x−iωt where kI = (sin θ, cos θ, 0) , (3.47a)
cP

Mathematical Tripos: Part II Waves 52 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Reflected P wave:
ω
uR = R(sin θ, − cos θ, 0)eikR .x−iωt where kR = (sin θ, − cos θ, 0) , (3.47b)
cP

Reflected SV wave:
ω
u = R(cosθ, sinθ, 0)eik R .x−iωt where k R = (sinθ, − cosθ, 0) . (3.47c)
R cS

From matching the phases

 
sin θ sinθ cS
= , i.e. θ = sin−1 sin θ , (3.48a)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
cP cS cP

and hence
θ < θ. (3.48b)

Applying the rigid boundary condition, u = 0 on y = 0:

(1 + R) sin θ + R cosθ = 0 ,
(1 − R) cos θ + R sinθ = 0 .

Hence
cos(θ + θ) sin 2θ
R= , R=− . (3.49)
cos(θ − θ) cos(θ − θ)

Remarks

(i) The SV waves reflect at a smaller angle than the P waves.

(ii) Both R and R are real, so the waves are in phase at y = 0.
(iii) For normal incidence, θ = 0, there is no reflected SV wave (since R = 0).
π cP
14/16 (iv) If θ + θ = 2, i.e. tan θ = cS , then a pure SV wave is reflected (‘mode conversion’).
15/17
3.3.3 Seismic tomography

Travel-time measurements of P and S waves generated by earthquakes, or man-made explosions, can be

used to make inferences about the material the waves have passed through, since the wavespeed, etc.
varies with material properties. This is an example of an inverse problem where a set of observations are
used to deduce the causal factors that produced them; in inverse problems you start with the results and
then calculate the causes (while in forward problems you start with the causes and then calculate the
results). There are many similar problems, e.g. PET/brain scans.
An important early discovery using such methods was that the outer core of the earth is liquid (since
µ = 0 in fluids, S waves cannot propagate through the liquid core).

Mathematical Tripos: Part II Waves 53 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Typical ray paths of reflected and refracted P and S waves in the interior of the earth.8 For the reason
why changes in material properties mean that the paths are curved, see Ray Theory in §5.

15/18
15/19 Ray paths of reflected and refracted P and S waves.9 P wave paths illustrating shadow zones.10

8 United States Geological Survey, http://commons.wikimedia.org/wiki/File:Earthquake wave paths.svg.

9 An Introduction to Seismology, Earthquakes, and Earth Structure, Stein, S. & Wysession, W., Blackwell Publishing
(2003); see also http://levee.wustl.edu/seismology/book/.
10 Steven Dutch, http://www.uwgb.edu/dutchs/earthsc202notes/quakes.htm.

Mathematical Tripos: Part II Waves 54 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 4 Linear Dispersive Waves
In the plane-wave, eikx−iωt , systems studied so far (with the exception of the waveguide in § 1.7.1), the
[signed] phase speed,
ω
c= , (4.1)
|k|
has been independent of k (because there has been nothing in the geometry or material parameters to
give a lengthscale, or direction, to measure k against). However, for most systems the phase speed c
depends on k (e.g. water waves). If the phase speed depends on the wavenumber, k, then the different
Fourier components of an initial local disturbance travel at different speeds and ‘disperse’. This has
consequences for the speed of energy propagation, which now differs from the velocity of propagation of
b 11
the crests/troughs, c = c k.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
4.1 Geometric Dispersion in Wave Guides
4.1.1 Acoustic waves in a rectangular duct

In § 1.7.1, we have already seen that for acoustic waves in a rectangular waveguide, the phase speed
15/16 depends on the wavenumber, k.

4.1.2 Love waves

A Rayleigh wave is a mixed interfacial P /SV wave. The question might be posed whether it is possible
to have an interfacial SH waves. The answer is yes, but only within a layered half space.

Love waves are trapped in a finite layer (cf. bound states in a finite well in QM ) by

(i) reflection at a free surface;

(ii) total internal reflection at the lower interface . . . hence we require c̄S > cS (i.e. the ‘crust’ is a low
velocity layer).

For SH waves, where u is the displacement,

(
c2S ∇2 u − utt = 0 , 0 < y < h ,
u = (0, 0, u) ,
c̄2S ∇2 u − utt = 0 , y < 0.
Again insist that waves in both layers have the same x-wavenumber and frequency. Seek solutions of the
form (that satisfy the dispersion relation)
0<y<h: u = (A1 sin `y + A2 cos `y)ei(kx−ωt) ,
 2  12  2  12
ω c
`=k − 1 = k − 1 > 0,
k 2 c2S c2S
ω
where we have assumed cS < c ≡ k for a propagating mode,12 and

11 The solution to cg ≡ ∇k ω = ωk
b ≡ c, i.e. group velocity equals phase velocity, is ω = const.
k k
12 Some prefer the solution (A1 sin `(h − y) + A2 cos `(h − y))ei(kx−ωt) , but it is swings and roundabouts.

Mathematical Tripos: Part II Waves 55 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 y<0: u = αeβy ei(kx−ωt) ,
1
c2 2

β =k 1− 2 > 0,
c̄S
15/04 where we have assumed c < c̄S for an evanescent mode.
est.
The boundary conditions are
y=h: σyz = 0 A1 cos `h − A2 sin `h = 0 ,
y=0: [σyz ]+
− =0 µA1 ` = µ̄αβ ,
y=0: [u]+− =0 A2 = α .
Hence

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
µ̄β = µ` tan `h ,
which, on substitution, yields the dispersion relation relating c(k) to k:
  21
"  21 # c2
c2 µ̄  1 − c̄2S 
tan −1 kh = . (4.2)
c2S µ cc22 − 1
S

This equation can be solved graphically by plotting the LHS and RHS for cS < c < c̄S , noting that over
 2  12  2  21
c̄
this range, φ = cc2 − 1 kh increases from 0 to cS2 − 1 kh.
S S

Remarks
(i) From the graph we see that there is at least one solution with cS < c < c̄S for all k. Alternatively
this can also be deduced by considering
  12
"  12 # c2
c2 µ̄  1 − c̄2S 
f (c, k) = tan −1 kh − . (4.3)
c2S µ cc22 − 1
S

The dispersion relation is f (c, k) = 0, while

f (c̄S , k) > 0 , and f (c, k) → −∞ as c → cS + .
Thus there exists at least one root for all k.
(ii) There are n solutions if
 12
c̄2S

(n − 1)π 6 −1 kh < nπ .
c2S
(iii) As k and φ increase, the tan curves ‘squash’ to the left, which implies that each new mode appears
with c = c̄S and the phase speed then decreases towards cS . The nth mode has a minimum ‘cut-off’
frequency (using the fact that ω ≡ kc = kc̄S at cut-off)
− 12
nπc̄S c̄2S

ω= −1 .
h c2S

Mathematical Tripos: Part II Waves 56 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

(iv) Since c depends on k, i.e. c ≡ c(k), the waves are dispersive.
(v) If the lower layer is not present, i.e. µ̄ = 0, then the dispersion relation (4.2) reduces to
"  21 #
c2
tan −1 kh = 0 , (4.4a)
c2S
i.e.   mπ 2 
2 2 2
ω = cS k + . (4.4b)
h
This is the dispersion relation for the ‘classical’ 2D waveguide (cf. (1.32)).
(vi) Mean energy propagation velocity (unlectured). If sufficiently determined/perverse, one can confirm
that the mean energy propagation velocity is equal to the group velocity.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Z 0 Z h
1 ∗ ∗
1  1
1 
< Ek >= 2 Re 2 ρ̄ u̇u̇ dy + 2 Re 2 ρu̇u̇ dy
−∞ 0
c2S
 
µ̄A2 ω 2 1− c̄2S
2
2β h 
= 22 
c2
+ .
8βcS c2S
−1 ` sin 2`h
Z h
1 ∗
< Ep >= 4 Re [σij eij ]dy
−∞
Z 0 Z h
1
exz e∗xz + ezx e∗zx + eyz e∗yz + ezy e∗zy dy + µRe exz e∗xz + eyz e∗yz dy
   
= 2 µ̄Re
−∞ 0
c2S
 
µ̄A22 ω 2  1 − 2β 2 h 
c̄2S
= 2 c2 +
8βcS c2
− 1 ` sin 2`h
S

= < Ek >
Z h
< Ix >= − 12 Re [σxz u̇∗ ]
−∞
c2S
   
2
µ̄A22 ωk  c 1 − c̄2S 2β 2 h 
=  +
4β 2
c2 c2 − 1 ` sin 2`h
S cS
 
c2S
−1
c2S
 
2
∂ω c2S  c 1 − 2β 2 h   1 − c̄2S
c̄2S 2β 2 h 
cg ≡ =  + c2
+ .
∂k c c2 c22 − 1 ` sin 2`h c2S
− 1 ` sin 2`h
S c S

Hence
< Ix >
Mean energy propagation velocity = U (k) = = cg 6= c .
< Ek + Ep >

4.2 Initial Value (Cauchy) Problems

4.2.1 General solution of the beam equation

For clarity reduce from 2D/3D to 1D and consider the simpler beam equation (imagine a twanged ruler):

ϕtt + γ 2 ϕxxxx = 0 , (4.5a)

B
where γ 2 = m > 0, and m and B are the mass and bending moment per unit length of the beam. Since
the equation is second order in time, two initial conditions are required; suppose that at t = 0
ϕ(x, 0) = ϕ0 (x) and ϕt (x, 0) = v0 (x) . (4.5b)
Seek a solution by Fourier transform:
Z ∞
ϕ(x, t) = b t) eikx dk .
ϕ(k, (4.6a)
−∞

Mathematical Tripos: Part II Waves 57 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Then we require that
btt + γ 2 k 4 ϕ
ϕ b = 0, (4.6b)
with solution
b = A(k) e−iωt ,
ϕ where − ω2 + γ 2 k4 = 0 . (4.6c)
i(kx−ωt)
Hence ω satisfies the dispersion relation for plane waves, e ,
ω = ±γk 2 = ±Ω(k) where Ω(k) = γk 2 . (4.6d)

The general solution to (4.5a) is thus a linear superposition of plane waves ei(kx±Ωt) :
Z ∞ Z ∞
ϕ(x, t) = A(k)eikx−iΩ(k)t dk + B(k)eikx+iΩ(k)t dk . (4.7)
−∞ −∞

If the initial conditions (4.5b) are to be satisifed then

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Z ∞
ϕ0 (x) = [A(k) + B(k)]eikx dk , (4.8a)
−∞
Z ∞
v0 (x) = − i Ω(k)[A(k) − B(k)]eikx dk . (4.8b)
−∞

Invert these Fourier transforms to obtain

Z ∞
1
A+B = ϕ0 (x)e−ikx dx , (4.9a)
2π −∞
Z ∞
1
2
−iγk (A − B) = v0 (x)e−ikx dx . (4.9b)
2π −∞
For definiteness, suppose further that v0 = 0, so that A(k) = B(k). Then to obtain a solution one needs
to be able to evaluate Z ∞
ϕ± (x, t) = A(k)eikx±iΩ(k)t dk . (4.10)
−∞
16/17
Remarks

(i) All information that (4.7) is a solution to the beam equation (4.5a), rather than another equation,
is contained in the solutions to the dispersion relation, i.e. from (4.6d), ω = ±Ω(k).
(ii) Solutions to many equations have the generic form (4.7), where ω = ±Ω(k), or similar, are solutions
to the dispersion relation. For instance, suppose that
L(∂t , ∂x )ϕ = 0 , (4.11a)
where L is another linear, constant coefficient, operator. Then the dispersion relation is
L(−iω, ik) = 0 . (4.11b)
Solve this equation for ω = ωj (k), where the number of roots equals the order, say n, of the equation
in time. The general solution then consists of the sum of n Fourier integrals.
Example. Suppose
∂ϕ p ∂ϕ p ∂3ϕ
+ gh + h2 gh 3 = 0 ,
∂t ∂x ∂x
then
p p
L(∂t , ∂x ) = ∂t + gh ∂x + h2 gh ∂x3 ,
and the dispersion relation is given by
p p
L(−iω, ik) = −iω + gh ik − ih2 gh k 3 = 0 ,
i.e.
p p
ω= gh k − h2 gh k 3 .
This is a first-order equation in time, so one initial condition is needed, there is one root of the
16/04 dispersion relation, and there will be one Fourier integral in the solution.
16/16
16/18
16/19 Mathematical Tripos: Part II Waves 58 © S.J.Cowley@maths.cam.ac.uk, Lent 2019
4.2.2 Exact solution of the beam equation for Gaussian modulation

Consider the initial conditions

   
x 2
ϕ0 (x) = exp − exp(iαx) , v0 = 0 , (4.12a)
λ
where α and λ are positive constants, and the real part is understood. This represents a ‘carrier’ harmonic
mode with wavenumber α, modulated by a Gaussian of width λ. Then
Z ∞
1 2
A(k) = e−(x/λ) e−i(k−α)x dx
4π −∞
 2
λ (k − α)2

λ
= √ exp − . (4.12b)
4 π 4

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
From (4.6d) and (4.10)
∞  2
λ (k − α)2
Z 
λ
ϕ± = √ exp − + ikx ± iγk 2 t dk
−∞ 4 π 4
(x ± 2γαt)2
 
λ 2
= 1 exp iαx ± iγα t − . (4.13)
2 λ2 ∓ 4iγt
2 (λ ∓ 4iγt) 2

Slow modulation. Suppose that the carrier wave is slowly modulated, i.e. αλ  1, and consider times
t  λ2 /γ. Since λ is large, the latter restriction still includes ‘moderate’ times when t ∼ λ/(αγ).
Then by Taylor expansion of (4.13)
 2 !
x ± 2γαt
ϕ± ∼ 12 exp (iα(x ± γαt)) exp −
λ
 2 !
1 x ± cg t
∼ 2 exp (iα(x ± ct)) exp − , (4.14)
| {z } λ
carrier wave | {z }
modulation envelope
where, in agreement with (4.6d), the phase speed is given by c(α) = Ω(α)/α = γα, and the group
velocity by cg (α) = ∂Ω
∂k (α) = 2γα.

Mathematical Tripos: Part II Waves 59 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Remarks
(i) The factor of 12 arises because half of the initial disturbance propagates to the right and half
to the left.
(ii) The slowly varying modulation envelope, i.e. the ‘group’ of waves occupying many wavelengths,
propagates with the group velocity, cg , of the carrier wave.
(iii) Within the group, individual phase fronts of the carrier wave propagate with the phase speed, c,
which maybe faster or slower than the group velocity; here c = 12 cg .
(iv) Waves must be continually created/destroyed at the front/back of the packet, or vice versa.
(v) To make the two scales explicit, once can introduce ‘slow’ variables
x t
X= and τ = . (4.15a)
λ λ

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Then (4.14) becomes
 
2
ϕ± ∼ 12 exp (iα(x ± ct)) exp − (X ± cg τ ) . (4.15b)

(vi) We consider larger times below, i.e. times when the wavepacket has travelled many times its
initial width (plus a bit).

4.2.3 Modulation of periodic wavetrains

Consider the superposition of two waves of equal amplitude and nearly equal wavenumbers and frequen-
cies, say
ϕ = cos(k1 x − ω1 t) + cos(k2 x − ω2 t)
1
+ k2 )x − 12 (ω1 + ω2 )t cos 1
− k2 )x − 12 (ω1 − ω2 )t



= 2 cos 2 (k1 2 (k1
= 2 cos (kx − ωt) cos (ε(κx − Ωt)) (4.16a)

where
k = 21 (k1 + k2 ) , ω = 12 (ω1 + ω2 ) , εκ = 21 (k1 − k2 ) , εΩ = 12 (ω1 − ω2 ) . (4.16b)

This is a modulated wave where the individual wavecrests

move at the phase speed
ω ω1 + ω2
c=
= , (4.17a)
k k1 + k2
but the envelope moves at the group velocity
εΩ ω1 − ω2 dω
cg = = ≈ . (4.17b)
εκ k1 − k2 dk

4.2.4 Large time solution — Kelvin’s method of stationary phase

Dispersive waves disperse!

For an initially compact disturbance, at large times we might expect to find disturbances with wavenum-
ber α a distance cg (α)t away. Hence if cg is not constant, the initial disturbance will spread out, and we
expect the wavenumber and amplitude to be slowly varying functions of x at large times.

Mathematical Tripos: Part II Waves 60 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Seek the large time behaviour of
Z ∞
Φ(x, t) = A(k)eikx−iω(k)t dk . (4.18a)
−∞

For large times t  1 we expect x to be large (since x = O(cg t)); hence write

x=Vt . (4.18b)

The integral (4.18a) can then be rewritten as

Z ∞
Φ= A(k)eitψ(k) dk , (4.18c)
−∞

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
where the phase ψ is given by
ψ(k) = kV − ω(k) . (4.18d)

In order to find a large-time approximation of

(4.18c) we first note that ‘rapid oscillations lead
to cancellation’.

The ‘tails’ of Fourier transforms. For example, consider

Z ∞
J(t) = A(k)eitk dk ,
−∞

2
where A is C ∞ , and A(n) (k) → 0 as |k| → ∞ (e.g. A(k) = e−k ). Then, from integration by parts,

1 ∞ 0
Z
J(t) = − A (k)eitk dk
it −∞
 2 Z ∞
i
= A00 (k)eitk dk ,
t −∞
 n Z ∞
i
= A(n) (k)eitk dk .
t −∞
R∞
If −∞ A(n) (k) dk is well behaved, it follows that as t → ∞, J(t) → 0 faster than any algebraic
power of t. Colloquially J(t) is said to be exponentially small as t → ∞.
17/04
17/17
It is possible to prove similar exponential decay for Φ(x, t) as t → ∞ if ψ(k) is a monotonic function
of k. However, if this is not the case, i.e. if there exists k = α such that

ψ 0 = V − ω 0 (α) = V − cg (α) = 0 , (4.19a)

then the cancellation is mitigated for wavenumbers k close to α. To see this, Taylor expand ψ(k) and
1 1
eitψ(k) for |k − α|  1 (or, to be more precise, for (k − α) = O( |ω 00 (α)|− 2 t− 2 ) to obtain

ψ(k) ∼ ψ(α) + 12 ψ 00 (α) (k − α)2 + . . . ,

eitψ(k) ∼ eitψ(α) cos β 2 + iσ sin β 2 + . . . ,

where
 21
t |ψ 00 (α)|

β= (k − α) , (4.19b)
2
σ = sgn [ψ 00 (α)] = −sgn [ω 00 (α)] . (4.19c)

The cancellation due to the rapid change in phase is least, and hence the contributions to Φ greatest, in
17/16 the neighbourhood of wavenumbers, α, where the phase ψ is stationary (Kelvin).

Mathematical Tripos: Part II Waves 61 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 1 1

0.5 0.5

cos(β 2 )
sin(β 2 )

0 0

-0.5 -0.5

-1 -1
-8 -6 -4 -2 0 2 4 6 8 -8 -6 -4 -2 0 2 4 6 8
β β

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Rescale for t  1. Suppose there exists a single stationary point, and let
  12
2
= 1. (4.20)
t |ψ 00 (α)|

Then using (4.19b), i.e. k = α + β, rewrite (4.10) in terms of the scaled wavenumber β, and Taylor
expand:
Z ∞
A(α + β) exp itψ(α) + iσβ 2 + O(β 3 )  dβ


Φ=
−∞
Z ∞
2
= A(α) eitψ(α) eiσβ (1 + O()) dβ .
−∞

From contour integration or otherwise,

Z ∞
2 1
e±iu du = π 2 e±iπ/4 , (4.21)
−∞

thus from (4.19b)

  12
2π  1

Φ= A(α) eiψ(α)t+iσπ/4 1 + O(t− 2 ) , (4.22)
|ψ 00 (α)| t
where we have assumed that

(i) A(α) 6= 0;
(ii) ψ 00 (α) = −ω 00 (α) 6= 0.
17/18
17/19
Summary. For t  1,
  21
2π
Φ∼ A(α) eiαx−iω(α)t+iσπ/4 , (4.23a)
|ω 00 (α)| t
where σ = −sgn[ω 00 (α)], and the wavenumber α satisfies, see (4.18d) and (4.19a),

ψ 0 (α) = 0 ,

i.e.
x = ω 0 (α) t = cg (α) t . (4.23b)
Hence to find the solution for given large x and t, solve (4.23b) for the wavenumber α, and substitute
back into (4.23a).
Interpretation. Harmonic waves of a given wavenumber α, and corresponding frequency ω(α), are ob-
served in the neighbourhood of points propagating with the group velocity. The amplitude of the
1
waves decays like t− 2 .

Mathematical Tripos: Part II Waves 62 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Radiation condition. Energy must travel outwards from a source, i.e. cg must be directed away from a
source. The direction of the phase velocity is immaterial.
Further, suppose that t = 0 oscillations of frequency ω0 and wavenumber k0 , where ω(k0 ) = ω0 , are
generated at x = 0. Then there will be no disturbance at x = x0 for
x0
t< .
cg (k0 )
Additional remarks.
(i) Only waves with k’s for which A(k) is significant will be observed.
(ii) Originally localised waves are dispersed, and hence the amplitude decreases. The decay like
t−1/2 follows from
R conservation of energy; the extent1 of the disturbance is proportional to t,
while energy ∝ |ϕ|2 dx, hence ϕ must decay like t− 2 .

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
(iii) There may be more than one point of stationary phase, i.e. more than one α may satisfy
(4.23b); e.g. if ω(k) is odd, then cg (k) is even, and so ±α satisfy (4.23b). The contributions
from each point of stationary phase need to be added.
(iv) If ω(k) is even, then waves with wavenumber (−α) will be found where
x
= −ω 0 (α) .
t
(v) If ω 00 (α) = 0, then it is necessary to include higher-order terms in the Taylor expansion.

4.2.5 Example: beam equation

Consider, from (4.7) and (4.10),

Z ∞ Z ∞
i(kx−Ω(k)t)
ϕ(x, t) = ϕ− (x, t) + ϕ+ (x, t) = A(k)e dk + A(k)ei(kx+Ω(k)t) dk . (4.24a)
−∞ −∞

where Ω(k) = γk 2 . From (4.18d)

ω

ψ± (k) = V k − ω(k) = V k ∓ Ω(k) .

Hence the stationary points satisfy
x
V = = ±Ω0 (k) ,
t k
i.e. the stationary points are at
x
k=± = ±α ,
2γt
waveguide
beam

where α = x/2γt. Note that, compared

with our earlier notation, the stationary
All branches of the dispersion relations for the waveguide,
points are at +α and −α, so there is a
(1.32) & (4.25a), and the beam equation, (4.6d); that for the
need to keep careful track of signs. waveguide has a bounded slope (⇒ bounded cg ).

Next, substitute into (4.23a), first with ω = Ω(k) = γk 2 and then with ω = −Ω(k) = −γk 2 , noting that
at a stationary point ω 00 (k) = ±2γ, and σ = −sgn [ω 00 (k)] = ∓1. It follows that
  12 
π 2 2

ϕ∼ A(α)eiαx−iγα t−iπ/4 + A(−α)e−iαx+iγα t+iπ/4 . (4.24b)
γt
Since ϕ0 is real and A(k) = B(k) (because we have assumed v0 = 0), we conclude from (4.9a) that
A(−k) = A∗ (k) ,
where ∗ indicates the complex conjugate (c.c). Hence ϕ is reassuringly real:
  12 
π 2

ϕ∼ A(α)eiαx−iγα t−iπ/4 + c.c. (4.24c)
γt
  12 
π 2

= A(α)ei(x /γt−π)/4 + c.c. . (4.24d)
γt

Mathematical Tripos: Part II Waves 63 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 4.2.6 Example: dispersion in a waveguide (unlectured)

From the dispersion relation for the waveguide, (1.32),

12
ω = ± ω02 + k 2 c20 = ±Ω(k) (4.25a)

m2 π 2 n2 π 2
where ω02 = h2 + b2 for some fixed choice of m and n. Then

21
ψ± = V k ∓ ω02 + k 2 c20 , (4.25b)
0
and ψ± = 0, when

kc20

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
V =± 1 .
(ω02 + k 2 c20 ) 2
0
For each −c0 < V = x/t < c0 there is one root on each branch; specifically ψ± = 0 at

ω0 V
k=± 1 = ±α . (4.25c)
c0 (c20 − V 2 ) 2

From (4.5a), (4.6d) and A(k) = B(k),

Z ∞
A(k) eiψ+ t + eiψ− t dk ,


ϕ(x, t) =
−∞

and
00 ω02 c20 ω02 c20
ψ± (k) = ∓ 3 =∓ .
(ω02 + k 2 c20 ) 2 Ω3 (k)

So from (4.22), (4.23a) and (4.23b)

 12 
2πΩ3 (α)
 
iπ iπ
ϕ(x, t) ≈ A(α) eiαx−iΩ(α)t− 4 + A(−α) e−iαx+iΩ(−α)t+ 4 .
ω02 c20 t

As in the previous example, if ϕ0 is real, then it follows from A(k) = B(k) and (4.8a) that

A(−k) = A∗ (k) .
18/04
Hence, since Ω(−k) = Ω(k),
 12 
2πΩ(α)3
 
iπ
ϕ(x, t) ≈ A(α) eiαx−iΩ(α)t− 4 + c.c. , (4.26a)
ω 2 c2 t
| 0 0 {z }
real

where
ω0 V ω0 x
α= 1 = 1 . (4.26b)
c0 (c20 −V 2) 2 c0 (c20 t2 − x2 ) 2

Remark. There is no stationary phase point if |V | > c0 , i.e. |x| > c0 t. ψ is then monotonic, and the
solution is exponentially small.

4.3 Capillary-Gravity Waves

Assume inviscid, irrotational, incompressible flow. Hence

u = ∇ϕ , ∇.u = 0 and ∇2 ϕ = 0 . (4.27)

Mathematical Tripos: Part II Waves 64 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Bottom boundary condition. The boundary condition at the rigid bottom is

ϕy = 0 on y = −h . (4.28)

Kinematic boundary condition at surface.

D
(y − η(x, z, t)) = 0 on y=η ,
Dt
i.e.
v = ηt + uηx + wηz on y=η . (4.29a)
Linearise for small amplitude waves such that

|η|  min(h, λ) and |u|  ωλ , (4.29b)

where λ and ω are the wavelength and wave frequency, respectively. Then, since

v(x, η, z, t) ≈ v(x, 0, z, t) + ηvy (x, 0, z, t) + . . . ,

we have
ϕy = v = ηt on y=0. (4.29c)

Dynamic boundary condition at surface. Assume that the free surface is at constant pressure and, unlike
Part IB, allow for surface tension. Then

p − patm = −T (κ1 + κ2 ) = T ∇ · n on y = η , (4.30a)

where T is a the coefficient of surface tension, κ1 and κ2 are the principal curvatures (positive if
concave upward), and n is the unit normal.13 From Bernoulli’s equation for unsteady potential flow
with gravity:
ρ(ϕt + 21 |∇ϕ|2 + gy) + p − patm = F (t) . (4.30b)
R
W.l.o.g we can assume F (t) = 0, by ϕ → ϕ + F dt. Substitute into (4.30a) to obtain

ρ(ϕt + 12 |∇ϕ|2 + gη) + T ∇ · n = 0 on y=η . (4.30c)

Linearise to obtain
ρϕt + ρgη − T (ηxx + ηzz ) = 0 on y=0. (4.30d)
19/04
18/16
18/17 13 (κ1 + κ2 ) is twice the mean curvature, and is given by (see Vector Calculus for 2D),
 
∇(y − η)
κ1 + κ2 = −∇ · n = −∇ ·
|∇(y − η)|
"  2 ! 2  ! #  2 !−3/2
∂η ∂η ∂ 2 η ∂η 2 ∂ 2 η ∂η 2
  
∂η ∂ η ∂η
= 1+ − 2 + 1 + 1 + + .
∂z ∂x2 ∂x ∂z ∂x∂z ∂x ∂z 2 ∂x ∂z

Mathematical Tripos: Part II Waves 65 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Seek Fourier-mode solution. Try
ϕ = f (y)eik1 x+ik3 z−iωt , (4.31a)
η = A eik1 x+ik3 z−iωt . (4.31b)
2
From (4.27), ∇ ϕ = 0, which implies
1
f 00 − k 2 f = 0 , where k = ±(k12 + k32 ) 2 .
The bottom boundary condition, (4.28), implies that f 0 (−h) = 0, and hence
f = B cosh k(y + h) . (4.32a)
Then from the kinematic boundary condition, (4.29c), it follows that the amplitude, A, of the wave
is given by

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
ikB
A= sinh kh . (4.32b)
ω
Finally, the dynamic boundary condition, (4.30d), yields the dispersion relation for gravity waves
with surface tension, i.e. capillary-gravity waves:
T k2
 
ω 2 = gk 1 + tanh kh . (4.33a)
ρg
18/18
18/19 Group velocity. The group velocity is given by
1 + 3βk 2
 
∂ω ω 2kh
cg = = 2
+ , (4.33b)
∂k 2k 1 + βk sinh 2kh
where the surface tension parameter, β, is defined by
T
β= . (4.33c)
ρg
As in previous examples, it is possible to show that the group velocity, cg , is again the mean
wave-energy propagation speed.

4.3.1 Limiting Cases

(i) Gravity waves: T = 0, i.e. β = 0.

ω 2 = gk tanh kh , (4.34a)
 
ω 2kh ω
cg = 1+ < = cp , (4.34b)
2k sinh 2kh k
where, to fix signs, we assume ωk > 0. ‘New’ waves appear at the back of the group.
(ii) Shallow-water waves: kh → 0 by letting the depth, h, become small.
ω 2 ≈ ghk 2 1 + βk 2 ,


(4.35a)
ω 1 + 2βk 2
 
ω
cg ≈ > = cp , (4.35b)
k 1 + βk 2 k
where again we assume ωk > 0. ‘New’ waves appear at the front of the group. For surface tension
to be important need h ≈ 1 mm, at which point it is debatable as to whether viscous effects are
negligible.
(iii) Long gravity waves (or ocean waves): kh → 0, βk 2 → 0 by letting the wavenumber, k, become small.

ω 2 ≈ ghk 2 , (4.36a)
i.e.
p
ω ≈ ±ck where c= gh . (4.36b)
Non-dispersive waves: cg ≈ cp . Relevant, for instance, to tides, tsunamis (λ = 100km, h = 1km, ⇒
c = 100 ms−1 ).

Mathematical Tripos: Part II Waves 66 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 (iv) Deep-water, capillary-gravity waves: |k|h → ∞ by letting the depth, h, become large.
 
T
ω 2 = g|k| 1 + k 2 = g|k| 1 + βk 2 ,


(4.37a)
ρg
1
ω 1 + 3βk 2 as k → 0,
 
cp
cg = → 32 (4.37b)
2k 1 + βk 2 c
2 p as k → ∞.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Remarks (for definiteness, concentrate on the first quadrant of the dispersion relation).
(a) Minimum phase speed. There is a wavenumber,
1
k2 = β − 2 , (4.37c)
at which the phase velocity has a minimum. Since
∂cp ∂ ω 1
= = (cg − cp ) , (4.37d)
∂k ∂k k k
it follows that at the minimum, cp (k2 ) = cg (k2 ). Further
cp >
< cg according as k<
> k2 . (4.37e)

(b) Ocean waves. If βk 2  1 (wavelength & 2 cm), then we have ‘ocean’ waves, with cp > cg .
(c) Capillary waves. If βk 2  1 (wavelength . 2 cm), then we have capillary waves/ripples, with
cp < cg .
(d) Degenerate stationary point. There is a wavenumber, say k1 , at which the group velocity has a
minimum, and the stationary point is degenerate, i.e.
 
0 00 2 1 2
cg (k1 ) = ω (k1 ) = 0 for k1 = √ −1 . (4.37f)
β 3
Since   12   12
1 1
amplitude ∝ ∝ ,
ψ 00 (k) ω 00 (k)
the amplitude increases without bound as k → k1 . In fact the stationary phase approximation
fails for x ≈ cg (k1 ) t, with the ‘infinite’ amplitude problem being resolved by Airy functions.
(e) Example. Suppose
A(k) = 0 for k < k0 and k > k3 ,
where
0 < k0 < k1 < k2 < k3 and cg (k0 ) > cg (k3 ) .
Then at large times and x  1,
• for x < cg (k1 ) t, x > cg (k0 ) t, there will be ‘no waves’,
• for cg (k3 ) t < x < cg (k0 ) t, there will be one wave,
20/04 • for cg (k1 ) t < x < cg (k3 ) t, there will be two waves.

Mathematical Tripos: Part II Waves 67 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Mathematical Tripos: Part II Waves
68
© S.J.Cowley@maths.cam.ac.uk, Lent 2019
This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
 4.4 Internal Gravity Waves in a Stratified Incompressible Fluid
Change convention and put gravity in the z direction. Then,
consider a fluid medium in which the mean pressure, p0 (z),
and the density, ρ0 (z), are in hydrostatic balance when
there is no motion, i.e.
dp0
= −ρ0 g . (4.38)
dz
Assume that the vertical lengthscale for O(1) changes in ρ0
19/17
is L, e.g. L is the lengthscale over which the density halves.
Assume also that the fluid is incompressible, i.e.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Dρ
= ρt + uρx + vρy + wρz = 0 , (4.39a)
Dt
and
∇.u = ux + vy + wz = 0 , (4.39b)

where u = (u, v, w). For an inviscid fluid, the motion is governed by momentum equation

Du
ρ = −∇p − ρgb
z. (4.39c)
Dt
Finally, assume that there are small perturbations to the mean state, i.e.

p = p0 (z) + p̃(x, t) , ρ = ρ0 (z) + ρ̃(x, t) , etc.

Linearise (4.39a), (4.39b) and (4.39c), and use the hydrostatic balance equation (4.38), to obtain

ρ̃t + wρ00 (z) = 0 , (4.40a)

ux + vy + wz = 0 , (4.40b)
ρ0 ut = −p̃x , (4.40c)
ρ0 vt = −p̃y , (4.40d)
ρ0 wt = −p̃z − ρ̃g . (4.40e)
19/16
19/18 Then

(4.40a) and (4.40e) ⇒ ρ0 wtt = −p̃zt + ρ00 gw , (4.41a)

(4.40b), (4.40c) and (4.40d) ⇒ ρ0 wzt = p̃xx + p̃yy . (4.41b)

Hence

(ρ0 wzz + ρ00 wz )tt − (∂xx

2 2
+ ∂yy )(gρ00 − ρ0 ∂tt
2
)w = 0 , (4.42a)
19/19 or equivalently
ρ00
 
2
∂tt ∂xx w + ∂yy w + ∂zz w + ∂z w = −N 2 (z)(∂xx
2 2 2 2 2
w + ∂yy w) , (4.42b)
ρ0

where the Brunt–Väisälä frequency, N , is given by

gρ00 (z)
N 2 (z) = − > 0, (4.42c)
ρ0 (z)

and it has been assumed that the stratification is stable with the density decreasing with increasing
height. Hence w satisfies a linear equation of the form

L(∂t , ∂x , ∂y , ∂z ; z)w = 0 . (4.42d)

Mathematical Tripos: Part II Waves 69 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Slowly varying assumption. We now make a slowly varying assumption, in that we assume that a plane
harmonic wave has a relatively short vertical lengthscale, i.e.
2π ρ0
 L where = O(L) . (4.43a)
k3 ρ00

Then
ρ00
 
wk3
wzz = O wk32 ,


wz = O , (4.43b)
ρ0 L
and so
ρ00 wz  ρ0 wzz . (4.43c)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Under this assumption, (4.42b) simplifies to
2 2
) N 2 (z) + ∂tt
2


wzztt + (∂xx + ∂yy w = 0. (4.44)

Suppose, for the time being, that N 2 is a constant. From (4.42c), this is the case if

ρ0 = ρ̄ exp(−z/L) , (4.45)

where ρ̄ is a constant and L = g/N 2 . Then a harmonic plane wave is an exact solution of (4.44),
and from (4.11b) the dispersion relation is given by

(−k32 )(−ω 2 ) − (k12 + k22 ) N 2 − ω 2 = 0 ,

or equivalently
N 2 (k12 + k22 )
ω 2 = Ω2 (k) ≡ . (4.46a)
k12 + k22 + k32

Remarks.
(i) If as assumed N 2 > 0, wave solutions to (4.46a) are possible for frequencies ω 2 6 N 2 .
(ii) If N (z) is a slowly-varying function of z, i.e. if the vertical wavelength is much shorter than
the length over which N varies by an order one amount, then for a given value of z, ‘local’
harmonic-wave solutions to (4.44) can be sought (as we shall see in § 5 is the case in ray theory).
Under this assumption the [leading-order] dispersion relation becomes a ‘slow’ function of the
vertical co-ordinate, i.e.
N 2 (z)(k12 + k22 )
ω 2 = Ω2 (k; z) ≡ . (4.46b)
k12 + k22 + k32
(iii) (4.46a), or more generally (4.46b), are anisotropic dispersion relations.
(iv) The phase and group velocities are given by
ωb
c= k, (4.46c)
k
and
N 2 k3
 
∂ω ∂ω ∂ω
k1 k3 , k2 k3 , −(k12 + k22 ) ,


cg ≡ , , = (4.46d)
∂k1 ∂k2 ∂k3 ωk 4

respectively. Hence
N2
(k12 + k22 )k32 − (k12 + k22 )k32 = 0 ,


c · cg = 6
(4.46e)
k
i.e. the phase velocity and the group velocity are perpendicular.

Mathematical Tripos: Part II Waves 70 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Experiment. Consider an experiment where waves are generated by a two–dimensional oscillatory cylinder
with ω < N . So that the waves travel in straight lines,14 assume that N is constant. Then with

k = k (sin φ, 0, − cos φ) , k > 0, −π < φ 6 π ,

(4.46b) and (4.46d) imply that

ω N cos φ
= ± sin φ , cg = ± (cos φ, 0, sin φ) and c = ±N sin φ (sin φ, 0, − cos φ) .
N k
Here the signs need to be chosen so that, according to the radiation condition, the group velocity,
cg , is directed away from the cylinder, i.e. chosen so that ± cos φ > 0.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Downward propagating crests imply upward propagating energy.

14 Once ray theory is at our disposal, we can deduce, from the ray tracing equations (5.10b) and (5.10a), that ω, k and
1
k2 are constant on rays since
∂Ω ∂Ω ∂Ω
= = =0. (4.47a)
∂t ∂x ∂y
Also, from (5.10b),
dk3 ∂Ω Ω
=− = − N0 . (4.47b)
dt ∂z N
Hence if (4.45) is satisfied, k3 is also constant on rays, with the consequence that all rays are straight.

Mathematical Tripos: Part II Waves 71 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

5 Ray Theory
Ray theory (geometric optics) is the generalisation to PDEs of the WKB[JLG] method for ODEs (but
do not worry if you have not done Asymptotic Methods; we will be starting from scratch). It is an
asymptotic approximation for long-distance propagation through ‘slowly varying’ (s.v.) media, where by
slowly varying we mean that the variations must be slow on the scale of the wavelength or period (as in
(4.43a) for internal waves), i.e.
media lengthscales  k −1 and/or media timescales  ω −1 . (5.1)

Ray theory is motivated by the fact that waves often transport energy, etc. over distances  k −1 and
times  ω −1 into regions with different properties, e.g.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
• seismic waves through the earth;
• sound waves, or internal gravity waves, propagating though the lower/upper atmosphere;
• water waves propagating towards a [gently] sloping beach.

The aim is to use local solutions (on the scale of the waves) to solve on the much larger scale of the
medium. In what follows we will neglect

(i) dissipation,
(ii) nonlinear effects,
(iii) rapid variations (e.g. effects near boundaries, at interfaces or at focusing points),
(iv) diffraction (i.e. spreading effects).

5.1 Ray Tracing

5.1.1 Multiple scales

We will assume that the medium varies on an O(1) lengthscale/timescale, while the wavelength/period
of the waves is O(ε) where ε  1. Hence seek slowly-varying solutions to
L(∂t , ∂x , ∂y , ∂z ; x, t) ϕ = 0 , (5.2a)
where the inhomogeneity in the medium is represented by the x dependence, and we have allowed the
medium to be time dependent (e.g. arising from diurnal variations). The dispersion relation for internal
gravity waves is of this form (see (4.42b) and (4.42d)), as is the equation for sound waves in a stratified
atmosphere:
∂ 2 p̃
 
2 ∂ 1 ∂ p̃
Lp̃ ≡ 2 − ρ0 (z)c0 (z) = 0. (5.2b)
∂t ∂z ρ0 (z) ∂z
Because the wavevector/frequency may vary across the medium, a simple local harmonic wave, ei(k.x−ωt) ,
with phase (k.x − ωt), is unlikely to work on the medium’s scale. The key idea is to focus on the phase
of the wave by seeking a solution of the form
i
ϕ = A(x, t; ε)e ε θ(x,t) . (5.3)

Mathematical Tripos: Part II Waves 72 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 The factor of 1ε ensures that variations in phase, θ/ε, are rapid, while variations in amplitude, A, are
slow[er]. This has consequences for derivatives, e.g.
∂ϕ i ∂θ 1 ∂A i ∂θ
= ϕ+ ϕ∼ ϕ. (5.4a)
∂t |ε{z
∂t} |A{z∂t} ε ∂t
large O(1)

Similarly
∂ϕ i ∂θ
∼ ϕ, etc. (5.4b)
∂x ε ∂x
For a harmonic wave, ϕ = ei(k.x−ωt) ,
∂ϕ
= −iωϕ , ∇ϕ = ikϕ . (5.4c)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
∂t
This is somewhat suggestive, so define the local frequency, ω, and local wavevector, k, over O(ε) scales,
by
1 1
ω = − θt , k = ∇θ . (5.4d)
ε ε
Substitute into (5.2a), then we obtain, at leading order, the local dispersion relation
 
iθt iθx iθy iθz
L , , , ; x, t = L(−iω, ik1 , ik2 , ik3 ; x, t) = 0 . (5.5a)
ε ε ε ε
or, on solving,
ω = Ω(k; x, t) , (5.5b)
where the dependence on

• k represents dispersion;
• x represents the [slow] spatial inhomogeneity of the medium;
20/16 • t represents the [slow] time dependence of the medium.
20/17
20/18
20/19 5.1.2 Phase and wavecrests

Assuming that partial derivatives of the phase commute, we note using (5.4d) that
∂2θ ∂2θ ∂ki ∂kj
= ⇒ = , (5.6a)
∂xj ∂xi ∂xi ∂xj ∂xj ∂xi
 
∂ ∂θ ∂k
(∇θ) = ∇ ⇒ = −∇x ω . (5.6b)
∂t ∂t ∂t

(5.6a) is equivalent to ∇ × k = 0. Hence, by Stokes’ theorem,

Z x2 Z x2
θ(x2 ) − θ(x1 ) = ∇θ · dx = ε k · dx , (5.6c)
x1 x1

is independent of path; it follows that the number of crests

between x1 and x2 does not vary by path..
(5.6b) is the continuity of phase equation. It is a conservation
equation stating that the rate of change of phase density, k, is
balanced by the flux, or phase flow rate, ω. In integral form

d x2
Z
k · dx = ω(x1 ) − ω(x2 ) . (5.6d)
dt x1

Since wavecrests are points of constant θ, we can interpret this

as the rate of change in the number of wavecrests in [x1 , x2 ]
equals the wavecrests entering minus those leaving.

Mathematical Tripos: Part II Waves 73 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

5.1.3 Ray-tracing equations

As a wavepacket moves through the medium, the wavevector, k, and frequency, ω, must vary in order to
satisfy the local dispersion relation (5.5b):

ω(x, t) = Ω(k(x, t); x, t) . (5.7)

From the chain rule

∂ω ∂Ω ∂ki ∂Ω
= + , (5.8a)
∂t x ∂ki x,t ∂t x ∂t k,x

and hence from (5.6b)

∂ω ∂Ω

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
+ (cg · ∇x ) ω = , (5.8b)
∂t ∂t
where the local group velocity is given by
cg = ∇k Ω . (5.8c)

Similarly,
∂ω ∂Ω ∂ki ∂Ω
= + , (5.9a)
∂xj t ∂ki x,t ∂xj t ∂xj k,t

and hence from (5.6a) and (5.6b)

∂k
+ (cg · ∇x ) k = −∇x Ω . (5.9b)
∂t

Equations (5.8b) and (5.9b) can be written in character-

istic form as
dω ∂Ω
= , (5.10a)
dt cg ∂t
dk
= −∇x Ω , (5.10b)
dt cg
where
d ∂
= + cg · ∇x , (5.10c)
dt cg ∂t

and the characteristics are the paths, x = x(t), satisfying

dx
= cg (k(x, t); x, t) . (5.10d)
dt

Definition. A ray is a path satisfying (5.10d).

Remarks.
(i) In general rays are curved.
(ii) Equation (5.10c) represents the time derivative of an observer moving with velocity cg along
a ray.
(iii) From (5.4d), the evolution of the phase is given by
 
1 dθ 1 ∂θ
= + cg · ∇x θ = −ω + cg .k . (5.10e)
ε dt cg ε ∂t

Solution. Given Ω, and initial conditions k0 , ω0 and θ0 at a position x0 , it is possible to integrate the ray
tracing equations (5.10a), (5.10b), (5.10d) and (5.10e) to solve for k, ω and θ at x. It is necessary
to do this for each ray (with lots of initial positions and wavevectors) to find k and ω everywhere.

Mathematical Tripos: Part II Waves 74 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Evolution of frequency. From (5.10a), the frequency, ω, of a wavepacket moving with velocity cg changes
iff the medium varies with t. If the medium is time independent, i.e.
∂Ω
= 0, (5.11a)
∂t
then from (5.10a)
dω
= 0, (5.11b)
dt
i.e. the frequency, ω, is constant along rays.
Evolution of wavenumber. Analogously, from (5.10b), the wavevector, k, of a wavepacket moving with
velocity cg changes iff there is spatial inhomogeneity. This is referred to as refraction. If the medium
is homogeneous, i.e.
∇x Ω = 0 , (5.11c)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
then from (5.10b)
dk
= 0, (5.11d)
dt
i.e. the wavevector, k, is constant along rays.
Homogeneous time-independent media. If both (5.11a) and (5.11c) are satisfied, i.e. if Ω ≡ Ω(k), then
cg ≡ cg (k), and hence, from the fact that k is constant along a ray, the rays/characteristics are
straight.

5.1.4 Hamilton’s equations of classical mechanics

From (5.10b) and (5.10d)

dkj ∂Ω dxj ∂Ω
=− . = . (5.12a)
dt ∂xj dt ∂kj
These are Hamilton’s equations with
x = q, k = p, Ω(k; x, t) = H(q, p, t) . (5.12b)
Hence waves are like particles travelling with the group velocity. Further, let S = θ/ε, then from (5.4d)
and (5.5b)
∂S
+ Ω (∇S; x, t) = 0 . (5.12c)
∂t
This is the Hamilton-Jacobi equation, with the phase as the action.

Remark. The Hamiltonian is a first integral and, thus, so is the dispersion relation. Since a first integral
of the ray equations exists, the number of ray equations that need to be solved is reduced by one.

5.1.5 * Wave action *

The local energy density, E, is not necessarily conserved for a slowly-varying wavetrain, but the wave
action, I, is (statement with no proof):
∂I
+ ∇.(cg I) = 0 , (5.13a)
∂t
where
E
I= . (5.13b)
ω
Remarks.
(i) This is the equivalent of an adiabatic invariant in classical mechanics.
(ii) If the medium is time independent, so that the frequency is constant along a ray, then energy is
conserved.
(iii) Equations (5.13a) and (5.13b) can be used to calculate the amplitude of waves in slowly-varying
21/04 media (e.g. see Question 6 on Example Sheet 4).

Mathematical Tripos: Part II Waves 75 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 5.1.6 Example: gravity waves approaching a beach (surfing)

Suppose that there is a sloping beach with a shore

line at x = 0. Let the depth of the water be

h ≡ h(x, z) . (5.14)

Assume
(i) that h → ∞ as x → −∞, and

(ii) that k = (k1 , k3 ) → k∞ (cos φ∞ , sin φ∞ ) as

x → −∞ , i.e. that far from the beach we have
uniform plane waves.

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
From (4.34a) the dispersion relation is

ω 2 = Ω2 (k; x) = gk tanh kh , (5.15a)

where surface tension has been neglected, and
k 2 = k12 + k32 . (5.15b)

From (5.11a) and (5.11b), ω is a constant on each ray. Further, all rays originate from x = −∞, and
hence
ω 2 = gk∞ (5.16a)
everywhere. Thus
k∞
tanh kh = , (5.16b)
k
and so k ≡ k(x, z), and
21/16 {measuring k} ⇔ {measuring h} .
Further, water waves are an example of isotropic dispersion, i.e.
Ω ≡ Ω(k; x) , where k = |k| . (5.17a)
In such cases
∂Ω b
cg = k, (5.17b)
∂k
21/17 i.e. the rays are parallel to k. It follows from the characteristic equation (5.10d) that the rays are specified
21/18 by the equation
dx
dx dt k1
= dz = . (5.17c)
dz dt
k3

To solve for the wavevector, one more equation is required; from the ray-tracing equations (5.10b), on a
ray
dk3
dk3 dt Ωz
= dk = . (5.17d)
21/18
dk1 dt
1 Ωx

Wavecrests. As noted both in § 1.5.2 and above, wavecrests are lines (or in 3D, surfaces) of constant θ.
The normal to these lines (or surfaces) is in the direction of ∇θ. Hence, as concluded in § 1.5.2, or
from (5.4d), the unit normal n is given by
n=k
b. (5.18a)
It follows from (5.17b) that, for isotropic dispersion relations, the wavecrests are perpendicular to
the rays (in contrast with internal gravity waves). Further, for our 2D example, the equation for
the wavecrests is
θx dx + θz dz = 0 ,
or, from (5.4d),
dx k3
=− . (5.18b)
dz k1
Check. Consistent with (5.17b) and (5.18a), the rays (5.17c) and wavecrests (5.18b) are orthogonal.

Mathematical Tripos: Part II Waves 76 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Uniform beach. Suppose h ≡ h(x). Then
Ω ≡ Ω(k; x) ,
and so
dk3 dk3
= 0, or equivalently = 0.
dk1 dt

Thus, since all characteristics originate from −∞,

k3 = k∞ sin φ∞

everywhere. Further, from (5.16b), as h → 0

r
k∞ p
k∼ → ∞ , and kh ∼ k∞ h → 0 .

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
h
But k3 is fixed, so k1 → ∞ as h → 0. Hence at the
shoreline, k is perpendicular to the beach, while
the wavecrests are parallel to the beach. Good
news for surfing. ©

5.2 Fermat’s Principle

Claim. The ray-tracing equations (5.10d) and (5.10b) are equivalent to the variational principle
Z t2
δ Φ(x, ẋ, k, t)dt = 0, (5.19a)
t1

where
Φ = k.ẋ − Ω(k; x, t), (5.19b)
t1 , t2 are fixed, and δx and δk are independently varied s.t.
δx(t1 ) = δx(t2 ) = 0 (5.19c)

Proof. (5.19a) can be rewritten

Z t2
{(ẋ − ∇k Ω) · δk + k · δ ẋ − ∇x Ω · δx} dt = 0 . (5.20a)
t1

Integrate by parts and use (5.19c). Since δk and δx can be independently varied, their coefficients
must be zero; thus
ẋ = ∇k Ω , k̇ = −∇x Ω . (5.20b)
Hence from (5.10d) and (5.10b), the stationary path is a ray.

5.2.1 Time independent case

Suppose that the medium is time independent, i.e. ∂Ω∂t = 0, and that the frequency, ω = Ω, is a uniform
constant. We then need to to consider a restricted class of variations δk, δx such that
Ω(k + δk; x + δx) = Ω(k; x) , (5.21a)
i.e., such that
Z t2
δ Ω dt = 0 . (5.21b)
t1

(5.19a) and (5.19b) then become

Z t2
δ k · ẋ dt = 0 , (5.21c)
t1
or equivalently

Mathematical Tripos: Part II Waves 77 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Z x2
δ k · dx = 0 . (5.21d)
x1

Isotropic case. When the dispersion relation is isotropic, then the wavevector, k, and the group velocity,
cg , are parallel, e.g. see (5.17b). Since a ray, i.e. stationary path, is parallel to cg = ∇k Ω, it follows
that on a ray k is parallel to dx; hence
Z x2 Z x2 Z x2
ds
k.dx = kds = ω , (5.22a)
x1 x1 x1 c
where  ω   ω  ω2
k2 = k · k , ds2 = dx · dx , c2 =
b ·
k k
b = . (5.22b)
k k k2
Since ω is a constant, (5.21d) can be written in the classical form

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Z x2
ds
δ = 0. (5.22c)
x1 c

5.2.2 Snell’s law of refraction

Suppose that the dispersion relation is isotropic and depends

only on x3 , i.e.
Ω ≡ Ω(k; x3 ) . (5.23a)
Then from (5.11d) and (5.11b), k1 , k2 and ω are constant on
rays, and the rays are parallel to k. Suppose that k subtends
an angle α with the x3 direction. Then
1 1
(k12 + k22 ) 2 (k 2 + k22 ) 2
sin α = = 1 c,
k ω
and hence on a ray
sin α sin α0
= = const . (5.23b)
c c0
This is Snell’s law of refraction.

5.2.3 Example: sound waves in inhomogeneous media

Consider a sound ray in the (x, z) plane, and suppose that

∂Ω ∂Ω ∂Ω
= = = 0. (5.24a)
∂x ∂y ∂t
Then, since sound waves are non-dispersive,

ω = Ω(k; z) = kc(z) . (5.24b)

Seek a ray joining (±a, 0, 0); then from (5.22c)

Z a
ds
δ = 0, (5.25a)
−a c
or
Z a  dz 2  12 dx
δ 1+ =0. (5.25b)
−a dx c(z)
Standard calculus of variations, and manipulation, show that on a ray
 12
c20

dz
=± 2 −1 , (5.26a)
dx c (z) sin2 α0

Mathematical Tripos: Part II Waves 78 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 where c0 / sin α0 is a constant.15
Alternatively, from Snell’s law
sin α
= constant on a ray.
c
For an isotropic dispersion relation k is parallel to a ray, hence
dx c
1 = sin α0 ,
(dx2 + dz 2 ) 2 c0

or, as in (5.26a),
 21
c20

dz

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
=± − 1 . (5.26b)
dx c2 (z) sin2 α0

Exercise. Recover (5.26a), or (5.26b), from the dispersion relation (5.24b) and the ray-tracing equations
(5.10d), (5.10b) and (5.10a).
∂c dz
Remark. Take the + sign, and suppose that c decreases as z increases ( ∂z < 0); then dx increases with
dz
height. Hence dx > 0 for all z, and thus a ray cannot return to original level.
∂c
Alternatively suppose that c increases as z increases ( ∂z > 0), and that there exits (x0 , z0 ) such
that
dz
=0.
dx
Then the ray has a maximum at this point.
22/16
Example. As an example suppose that

c = c0 (z + 1) .

Then from solving (5.26a), we find that a ray passing

through (0, 0) has the equation

(x − cot α0 )2 + (z + 1)2 = cosec2 α0 ,

i.e. the rays are arcs of circles in x–z plane.

Remark. On still summer nights there can be an inversion in the atmosphere so that close to the ground
the temperature, and thence the sound speed, increase with height. On such nights sound can travel
22/04 much further.
22/17
22/19
5.3 Media with Mean Flows
So far we have assumed that the medium that the
waves are propagating through has no mean flow.
This is not always the case. As an example where
mean flow effects are important, consider a source
emitting waves with a frequency ωs and moving with
22/18 a [uniform] velocity U through a fluid medium.

15 Since the integrand of the functional is independent of x, the Euler-Lagrange equation

1 ! 1 !!
∂ (1 + z 02 ) 2 d ∂ (1 + z 02 ) 2
− = 0.
∂z c(z) dx ∂z 0 c(z)

has the first integral

1 1 !
(1 + z 02 ) 2 ∂ (1 + z 02 ) 2 sin α0
− z0 0 = .
c(z) ∂z c(z) c0
On expansion and rearrangement, this yields (5.26a).

Mathematical Tripos: Part II Waves 79 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Frame in which the fluid is at rest. In the (x, t) frame in which the fluid is at rest, suppose that the
dispersion relation for plane harmonic waves of frequency ωr is known:
ωr = Ωr (k; x, t) . (5.27)

Frame in which the source is at rest.

Transform to a (X, t) frame moving with the source, in
which the fluid has velocity −U. It is in this frame that
the source ‘emits’ waves with frequency ωs . The frames
are related by a Galilean transformation:

X = x − Ut . (5.28a)

Thus, the transformation formula for the time derivative

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
is given by
∂ ∂ ∂
→ − Uj . (5.28b)
∂t x ∂t X ∂Xj
It follows that for plane harmonic waves the frequencies in the ‘at rest’ and ‘source/moving’ frames
are related by
−ı ωr = −ı ωs − ı k · U ,
i.e.
ωs = ωr − k · U , (5.28c)
with corresponding group velocity, from (5.27),
cgs = ∇k (Ωr (k) − k · U) = cgr − U . (5.28d)
Remarks.
(i) In general this is an anisotropic dispersion relation, even if Ωr is isotropic (i.e. Ωr ≡ Ωr (k)),
since U introduces a direction.
(ii) Group velocities obey Galilean transformations, whereas phase velocities do not — because
phase velocities are not ‘true’ velocities.

5.3.1 The Doppler effect

Consider a source moving with velocity U emitting sound

waves with constant frequency ωs . We know that in a
medium which is at rest, the dispersion relation for sound
waves is
ωr = c0 k , (5.29a)
where c0 is the constant sound speed. Let φ be the angle
between the direction of motion and the ray, or equiva-
lently the wavevector, k.

Then from (5.29a)

U ωr cos φ
k · U = U k cos φ = ,
c0
and hence, from (5.28c), the frequency heard by a stationary observer is
ωs
ωr = U
. (5.29b)
1 − c0 cos φ
If U < c0 , then

(i) if cos φ > 0, i.e. if the observer is in front of the moving source, then ωr > ωs and the frequency
heard is larger than generated.
(ii) if cos φ < 0, i.e. if the observer is behind the moving source, then ωr < ωs and the frequency heard
is smaller than generated.

Exercise. What happens if U > c0 ?

Mathematical Tripos: Part II Waves 80 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 5.3.2 Stationary capillary-gravity waves

In general waves are unsteady, because of the e−ıωt

factor. However, they can be steady in the frame of
a moving source.
Suppose that there is a cm size object moving with
velocity (U, 0, 0) in deep water (e.g. a trailing finger
from a punt). If U > 0 is large enough, then this
steady force can generate a stationary wave pattern
in the frame of the object.
Work in 1D for simplicity. Then, in the frame where
the forcing is at rest, and the fluid has a uniform

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
mean velocity (−U, 0, 0), the dispersion relation is,
from (5.27) and (5.28c),

ωs = Ωr (k) − U k , (5.30a)

where, see (4.37a),

1
21
Ωr (k) = ±|gk| 2 1 + βk 2 , (5.30b)

is the dispersion relation for deep-water waves in the

absence of a mean flow.
For a wave pattern generated by a steadily moving
force we require that ωs = 0, i.e.
1 1
U k = ±|gk| 2 (1 + βk 2 ) 2 . (5.30c)
1 1
• If 0 < U < (2g/k2 ) 2 = (4g 2 β) 4 , which is about 0.25 ms−1 in water, then there is no non-zero (i.e.
finite-wavelength) solution to (5.30c), and there can be no steady wave pattern. Here, as in (4.37c),
1
k2 = β − 2 is the [positive] wavenumber of the wave with the minimum phase speed when there is
no mean flow
1
• If U > (4g 2 β) 4 , then there are four non-zero solutions to (5.30c). The radiation condition necessi-
tates that the waves should propagate outwards, and from (5.28d), cgs = cgr − U. Hence

|k| < k2 , cgr (k) < U , cgs (k) < 0. |k| > k2 , cgr (k) > U , cgs (k) > 0.

Long waves occur downstream of, i.e. behind, Short waves occur upstream, i.e. ahead, of the
the steady disturbance steady disturbance

23/04 Combined

Mathematical Tripos: Part II Waves 81 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 5.3.3 Mach cone

Consider an aircraft moving supersonically with velocity U = (M c0 , 0, 0), where c0 is the speed of sound
and M > 1; suppose that linear theory has something to say.
In the frame of the aircraft, from (5.28c) and (5.29a),

ωs = c0 k − U · k = c0 k(1 − M cos φ) . (5.31a)

Steady waves are possible if φ = cos−1 (1/M ), which

implies
 1 
b = (cos φ, sin φ, 0) = 1 , 1 − 1 2 , 0 . (5.31b)

k
M M2

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
Remark. Note that the angle is unique (up to a sign);
hence waves of a variety of wavenumbers will
propagate at this angle.
For these waves, from (5.28d),

b−U
cgs = c0 k (5.31c)
1
2
= c0 (M − 1) (− sin φ, cos φ) ,
2 (5.31d)

where we note that cgs · U < 0 and cgs · k b = 0. Hence

steady waves develop on a Mach cone behind the air-
craft at a semi-angle of α = sin−1 (cos φ) = sin−1 (1/M ).
Further, since (5.31c) also holds for unsteady waves,
the locus of unsteady sound waves is a sphere within
23/16 the Mach cone.
23/17
5.3.4 Ship/duck waves

Consider the wave pattern generated by a ship, or duck,

travelling with a uniform velocity U = (U, 0, 0). In the
frame in which the fluid is at rest, the dispersion rela-
tion for deep water gravity waves with no surface ten-
sion is (see (4.37a))
p
Ωr = gk , (5.32a)
where we take the wavevector, k, to be given by
23/18 k = k(cos φ, 0, sin φ) . (5.32b)
In the frame moving with the duck, the dispersion relation for deep-water gravity waves is

(ωs + k · U)2 = gk . (5.32c)

Henceforth drop the subscript ‘s’, and solve for ω to obtain

p
ω = Ω(k1 , k3 ) = ± gk − U k cos φ . (5.32d)

Again this is an anisotropic dispersion relation since Ω 6≡ Ω(k). As expected the group velocity is not
parallel to k: r
gb
cg = ± 21 k − U. (5.32e)
23/19 k
Seek a steady wave pattern with
ω=0.

Mathematical Tripos: Part II Waves 82 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

From (5.32d) we need the ± signs according as cos φ >
< 0; so
r
g
U cos φ = ± , (5.33a)
k
and
b − U = 1 U (cos2 φ − 2, 0, sin φ cos φ) .
cg = 1 U cos φ k (5.33b)
2 2

Thus cg · x
b < 0, and the waves appear behind the duck.
Moreover, the rays are straight since

Ωt = 0 and ∇x Ω = 0 . (5.34a)

On such rays let

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
dz
= − tan ψ . (5.34b)
dx
Then from (5.10d)
dz ∂ω
(cg )z tan φ
dt
tan ψ = − dx = − ∂k3
=− = , (5.34c)
dt
∂ω
∂k1
(cg )x 1 + 2 tan2 φ
and
cg = (− cos ψ, 0, sin ψ) .
b (5.34d)

Hence for given φ,

g
solve (5.33a) for k: k= , (5.35a)
U 2 cos2 φ
 
tan φ
solve (5.34c) for ψ: ψ = arctan . (5.35b)
1 + 2 tan2 φ
We can restrict attention to
π π
− <φ< , (5.36a)
2 2
or equivalently cos φ > 0, since this leads to no less choice for ψ in (5.35b). However, in order to have
waves behind the duck, we need
π
|ψ| < . (5.36b)
2
Further, there exists a maximum value of |ψ|. In particular

∂ tan ψ
=0, (5.37a)
∂ tan φ
when
1
tan2 φ = 1
2 , tan |ψ| = √ . (5.37b)
2 2
◦
The maximum value of |ψ| is approximately 19 12 , i.e. the
◦
waves are all confined to a wedge of semi–angle 19 12 behind
the duck.

Geometric interpretation. Rewrite (5.33b) as

cg = 14 U (cos 2φ, 0, sin 2φ) − 43 U (1, 0, 0) . (5.38)

Hence the locus of cg lies on a circle confined within a

wedge of semi-angle ψmax = sin−1 13 = tan−1 2√ 1
2
.

While the duck travels a distance U τ , the distance trav-

elled by waves with wavenumber k = g(U cos φ)−2 is
1


cgr τ = 2 U cos φ τ .

Mathematical Tripos: Part II Waves 83 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

Wavecrests. The phase can be calculated by writing
Z
θ= ∇θ · dx
any path
=k·x choose a ray, on which k is constant, as the path
= kr cos(π − ψ − φ)
= −kr cos ψ cos φ(1 − tan ψ tan φ)
 
cos ψ 1
= −kr from (5.34c) . (5.39a)
cos φ 1 + 2 tan2 φ

Hence
cos φ(1 + 2 tan2 φ)
x = −r cos ψ = −θ . (5.39b)

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
k
Thus from (5.35a) we obtain a parametric equation for the crest shape:

U 2θ
x=− cos3 φ (1 + 2 tan2 φ) , (5.40a)
g
U 2θ
z = −x tan ψ = cos2 φ sin φ . (5.40b)
g

Pattern of wavecrests behind a steadily moving object.

Remarks.
(i) Unsteady waves overlie this pattern, especially in the wake.
(ii) Stationary phase fails in the immediate neighbourhood of the cusps.

Credit: Daderot. Cloud pattern in the wake of the Île Amsterdam

(NASA, http://tinyurl.com/Cloud-Wake).

Mathematical Tripos: Part II Waves 84 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

5.3.5 Internal gravity waves in a horizontal wind

Assume the all motion is two-dimensional and confined

to the x–z plane. Suppose that the fluid medium has a
background wind:

U = (U (z), 0, 0) . (5.41)

Assume that the lengthscale L over which the wind

changes velocity is long compared with the wavelength
of any wave, i.e. assume that the wind is slowly varying.
An observer fixed on the ground sees a moving medium with velocity +U (cf. the start of §5.3 where
in the frame is which the source is stationary, the fluid was assumed to have velocity −U). Hence from

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
(5.28c), with a change of sign,
ω ≡ ωs = ωr + k1 U , (5.42a)
where ωr is the frequency in a frame moving at the wind speed. Thus from (4.46b) with k2 = 0:

N (z)k1
ω = Ω(k; z) ≡ + k1 U (z) . (5.42b)
k
From the ray-tracing equations (5.10b) and (5.10a), ω and k1 are constant on rays since
∂Ω ∂Ω
= = 0. (5.43a)
∂t ∂x
In addition
N0
 
dk3 ∂Ω
=− = −k1 + U0 , (5.43b)
dt ∂z k
while from (5.10d)
dx ∂Ω N k2
= = 33 + U , (5.43c)
dt ∂k1 k
dz ∂Ω N k1 k3
= =− . (5.43d)
dt ∂k3 k3
For definiteness assume that:
(i) N = constant; i.e. from (4.42c) assume that

ρ0 = ρ̄ exp(−zN 2 /g) , where ρ̄ is a constant. (5.44)

(ii) k1 > 0 .
(iii) k3 < 0 , so that the waves are propagating upwards.

Then from (5.43b) and (5.43d)

dk3 U 0 k3
= , (5.45a)
dz N k3
and thus since k1 is a constant
U (z) − U (0) 1 1
= 1 − 1 . (5.45b)
N 2 2
(k1 + k3 (0)) 2
(k1 + k32 ) 2
2

As z increases, and U (z) increases from U (0), k3 must increase. If U (z) increases sufficiently that
N
U (zc ) − U (0) = 1 (5.46a)
(k12 + k32 (0)) 2
then
k3 (zc ) = −∞ . (5.46b)

Mathematical Tripos: Part II Waves 85 © S.J.Cowley@maths.cam.ac.uk, Lent 2019

 Exercise (see also Question 9 on Example Sheet 4). Show that as
z → zc −,
1
k3 ∼ − → −∞ , (5.47a)
U 0 (zc )(zc − z)
U (zc ) 1
x∼ 2 →∞, (5.47b)
N k1 (U 0 (z
c )) zc−z
1 1
t∼ 2 z − z → ∞. (5.47c)
0
N k1 (U (zc )) c

Thus a wave packet takes an infinite time to approach

the critical level, z = zc , along a ray. Show also that
U (zc ) = ω/k1 , i.e. that at the critical level the fluid veloc-

This is a specific individual’s copy of the notes. It is not to be copied and/or redistributed.
ity equals the apparent phase velocity in the same direction.

* Conservation of wave action (unlectured) *. For the current system, conservation of wave action can be
expressed as    
∂ Er Er
+∇· ∇k Ω = 0 , (5.48)
∂t ωr ωr
where Er is the wave energy density in a frame moving with the wind, and ωr and Ω are as earlier.
For the normalisation, ı 
ρ0 w = A(x, t; ) exp θ(x, t) , (5.49a)

it follows (after some manipulation) that

|A|2 k 2
Er = . (5.49b)
2ρ0 k12

Steady wave pattern. First consider a steady wave pattern. Then the upward component of wave-
action flux is constant, i.e.
Er ∂Ω
= const . (5.50a)
ωr ∂k3
As a critical level is approached:
N k1 N k1
ωr = ω − k1 U = ≈ , (5.50b)
k k3
∂Ω N k1 k3 N k1
=− ≈− 2 . (5.50c)
∂k3 k3 k3

Hence as the critical level is approached

Er ≈ const.(−k3 ) → ∞ ,
const.
|A| ≈ 1 → 0 . (5.50d)
(−k3 ) 2

Wave packet. Alternatively, consider a wave packet. Then (5.48) implies that the total wave action
of the wavepacket will be conserved. As the wavepacket approaches the critical level, then from
(5.50b),
ωr → 0 . (5.51)
Hence the total wave energy of the packet must also go to zero. There is no dissipation, so
where does the energy go? The answer is that is transferred to the mean flow (see the Part III
24/16 Geophysical Fluid Dynamics course or similar).
24/17
24/18
24/19

Mathematical Tripos: Part II Waves 86 © S.J.Cowley@maths.cam.ac.uk, Lent 2019