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marshall baker green's functions

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A book-length text by M. Baker and S. Sutlief, Version 1, revised December 19, 2003, found among downloaded physics books. It starts with the vibrating string, boundary conditions and Green's identities, then covers eigenfunction expansions, steady-state and dynamic problems, surface waves and membranes, N-dimensional problems, the method of images, cylindrical and spherical symmetry, and heat conduction.

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Extracted text (machine-read; may contain errors)

Green’s Functions in Physics Version 1 M. Baker, S. Sutlief Revision: December 19, 2003 Contents 1 The Vibrating String 1 1.1 The String . . . . . . . . . . . . . . . . . . . . . . . . . . 2 1.1.1 Forces on the String . . . . . . . . . . . . . . . . 2 1.1.2 Equations of Motion for a Massless String . . . . 3 1.1.3 Equations of Motion for a Massive String . . . . . 4 1.2 The Linear Operator Form . . . . . . . . . . . . . . . . . 5 1.3 Boundary Conditions . . . . . . . . . . . . . . . . . . . . 5 1.3.1 Case 1: A Closed String . . . . . . . . . . . . . . 6 1.3.2 Case 2: An Open String . . . . . . . . . . . . . . 6 1.3.3 Limiting Cases . . . . . . . . . . . . . . . . . . . 7 1.3.4 Initial Conditions . . . . . . . . . . . . . . . . . . 8 1.4 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . 8 1.4.1 No Tension at Boundary . . . . . . . . . . . . . . 9 1.4.2 Semi-inﬁnite String . . . . . . . . . . . . . . . . . 9 1.4.3 Oscillatory External Force . . . . . . . . . . . . . 9 1.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 10 1.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . 11 2 Green’s Identities 13 2.1 Green’s 1st and 2nd Identities . . . . . . . . . . . . . . . 14 2.2 Using G.I. #2 to Satisfy R.B.C. . . . . . . . . . . . . . . 15 2.2.1 The Closed String . . . . . . . . . . . . . . . . . . 15 2.2.2 The Open String . . . . . . . . . . . . . . . . . . 16 2.2.3 A Note on Hermitian Operators . . . . . . . . . . 17 2.3 Another Boundary Condition . . . . . . . . . . . . . . . 17 2.4 Physical Interpretations of the G.I.s . . . . . . . . . . . . 18 2.4.1 The Physics of Green’s 2nd Identity . . . . . . . . 18 i ii CONTENTS 2.4.2 A Note on Potential Energy . . . . . . . . . . . . 18 2.4.3 The Physics of Green’s 1st Identity . . . . . . . . 19 2.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 20 2.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . 21 3 Green’s Functions 23 3.1 The Principle of Superposition . . . . . . . . . . . . . . . 23 3.2 The Dirac Delta Function . . . . . . . . . . . . . . . . . 24 3.3 Two Conditions . . . . . . . . . . . . . . . . . . . . . . . 28 3.3.1 Condition 1 . . . . . . . . . . . . . . . . . . . . . 28 3.3.2 Condition 2 . . . . . . . . . . . . . . . . . . . . . 28 3.3.3 Application . . . . . . . . . . . . . . . . . . . . . 28 3.4 Open String . . . . . . . . . . . . . . . . . . . . . . . . . 29 3.5 The Forced Oscillation Problem . . . . . . . . . . . . . . 31 3.6 Free Oscillation . . . . . . . . . . . . . . . . . . . . . . . 32 3.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 32 3.8 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . 34 4 Properties of Eigen States 35 4.1 Eigen Functions and Natural Modes . . . . . . . . . . . . 37 4.1.1 A Closed String Problem . . . . . . . . . . . . . . 37 4.1.2 The Continuum Limit . . . . . . . . . . . . . . . 38 4.1.3 Schr¨ odinger’s Equation . . . . . . . . . . . . . . . 39 4.2 Natural Frequencies and the Green’s Function . . . . . . 40 4.3 GF behavior near λ=λn. . . . . . . . . . . . . . . . . . 41 4.4 Relation between GF & Eig. Fn. . . . . . . . . . . . . . . 42 4.4.1 Case 1: λNondegenerate . . . . . . . . . . . . . . 43 4.4.2 Case 2: λnDouble Degenerate . . . . . . . . . . . 44 4.5 Solution for a Fixed String . . . . . . . . . . . . . . . . . 45 4.5.1 A Non-analytic Solution . . . . . . . . . . . . . . 45 4.5.2 The Branch Cut . . . . . . . . . . . . . . . . . . . 46 4.5.3 Analytic Fundamental Solutions and GF . . . . . 46 4.5.4 Analytic GF for Fixed String . . . . . . . . . . . 47 4.5.5 GF Properties . . . . . . . . . . . . . . . . . . . . 49 4.5.6 The GF Near an Eigenvalue . . . . . . . . . . . . 50 4.6 Derivation of GF form near E.Val. . . . . . . . . . . . . . 51 4.6.1 Reconsider the Gen. Self-Adjoint Problem . . . . 51 CONTENTS iii 4.6.2 Summary, Interp. & Asymptotics . . . . . . . . . 52 4.7 General Solution form of GF . . . . . . . . . . . . . . . . 53 4.7.1δ-fn Representations & Completeness . . . . . . . 57 4.8 Extension to Continuous Eigenvalues . . . . . . . . . . . 58 4.9 Orthogonality for Continuum . . . . . . . . . . . . . . . 59 4.10 Example: Inﬁnite String . . . . . . . . . . . . . . . . . . 62 4.10.1 The Green’s Function . . . . . . . . . . . . . . . . 62 4.10.2 Uniqueness . . . . . . . . . . . . . . . . . . . . . 64 4.10.3 Look at the Wronskian . . . . . . . . . . . . . . . 64 4.10.4 Solution . . . . . . . . . . . . . . . . . . . . . . . 65 4.10.5 Motivation, Origin of Problem . . . . . . . . . . . 65 4.11 Summary of the Inﬁnite String . . . . . . . . . . . . . . . 67 4.12 The Eigen Function Problem Revisited . . . . . . . . . . 68 4.13 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 69 4.14 References . . . . . . . . . . . . . . . . . . . . . . . . . . 71 5 Steady State Problems 73 5.1 Oscillating Point Source . . . . . . . . . . . . . . . . . . 73 5.2 The Klein-Gordon Equation . . . . . . . . . . . . . . . . 74 5.2.1 Continuous Completeness . . . . . . . . . . . . . 76 5.3 The Semi-inﬁnite Problem . . . . . . . . . . . . . . . . . 78 5.3.1 A Check on the Solution . . . . . . . . . . . . . . 80 5.4 Steady State Semi-inﬁnite Problem . . . . . . . . . . . . 80 5.4.1 The Fourier-Bessel Transform . . . . . . . . . . . 82 5.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 83 5.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . 84 6 Dynamic Problems 85 6.1 Advanced and Retarded GF’s . . . . . . . . . . . . . . . 86 6.2 Physics of a Blow . . . . . . . . . . . . . . . . . . . . . . 87 6.3 Solution using Fourier Transform . . . . . . . . . . . . . 88 6.4 Inverting the Fourier Transform . . . . . . . . . . . . . . 90 6.4.1 Summary of the General IVP . . . . . . . . . . . 92 6.5 Analyticity and Causality . . . . . . . . . . . . . . . . . 92 6.6 The Inﬁnite String Problem . . . . . . . . . . . . . . . . 93 6.6.1 Derivation of Green’s Function . . . . . . . . . . 93 6.6.2 Physical Derivation . . . . . . . . . . . . . . . . . 96 iv CONTENTS 6.7 Semi-Inﬁnite String with Fixed End . . . . . . . . . . . . 97 6.8 Semi-Inﬁnite String with Free End . . . . . . . . . . . . 97 6.9 Elastically Bound Semi-Inﬁnite String . . . . . . . . . . . 99 6.10 Relation to the Eigen Fn Problem . . . . . . . . . . . . . 99 6.10.1 Alternative form of the GRProblem . . . . . . . 101 6.11 Comments on Green’s Function . . . . . . . . . . . . . . 102 6.11.1 Continuous Spectra . . . . . . . . . . . . . . . . . 102 6.11.2 Neumann BC . . . . . . . . . . . . . . . . . . . . 102 6.11.3 Zero Net Force . . . . . . . . . . . . . . . . . . . 104 6.12 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 104 6.13 References . . . . . . . . . . . . . . . . . . . . . . . . . . 105 7 Surface Waves and Membranes 107 7.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 107 7.2 One Dimensional Surface Waves on Fluids . . . . . . . . 108 7.2.1 The Physical Situation . . . . . . . . . . . . . . . 108 7.2.2 Shallow Water Case . . . . . . . . . . . . . . . . . 108 7.3 Two Dimensional Problems . . . . . . . . . . . . . . . . 109 7.3.1 Boundary Conditions . . . . . . . . . . . . . . . . 111 7.4 Example: 2D Surface Waves . . . . . . . . . . . . . . . . 112 7.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 113 7.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . 113 8 Extension to N-dimensions 115 8.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 115 8.2 Regions of Interest . . . . . . . . . . . . . . . . . . . . . 116 8.3 Examples of N-dimensional Problems . . . . . . . . . . . 117 8.3.1 General Response . . . . . . . . . . . . . . . . . . 117 8.3.2 Normal Mode Problem . . . . . . . . . . . . . . . 117 8.3.3 Forced Oscillation Problem . . . . . . . . . . . . . 118 8.4 Green’s Identities . . . . . . . . . . . . . . . . . . . . . . 118 8.4.1 Green’s First Identity . . . . . . . . . . . . . . . . 119 8.4.2 Green’s Second Identity . . . . . . . . . . . . . . 119 8.4.3 Criterion for Hermitian L0. . . . . . . . . . . . . 119 8.5 The Retarded Problem . . . . . . . . . . . . . . . . . . . 119 8.5.1 General Solution of Retarded Problem . . . . . . 119 8.5.2 The Retarded Green’s Function in N-Dim. . . . . 120 CONTENTS v 8.5.3 Reduction to Eigenvalue Problem . . . . . . . . . 121 8.6 Region R. . . . . . . . . . . . . . . . . . . . . . . . . . 122 8.6.1 Interior . . . . . . . . . . . . . . . . . . . . . . . 122 8.6.2 Exterior . . . . . . . . . . . . . . . . . . . . . . . 122 8.7 The Method of Images . . . . . . . . . . . . . . . . . . . 122 8.7.1 Eigenfunction Method . . . . . . . . . . . . . . . 123 8.7.2 Method of Images . . . . . . . . . . . . . . . . . . 123 8.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 124 8.9 References . . . . . . . . . . . . . . . . . . . . . . . . . . 125 9 Cylindrical Problems 127 9.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 127 9.1.1 Coordinates . . . . . . . . . . . . . . . . . . . . . 128 9.1.2 Delta Function . . . . . . . . . . . . . . . . . . . 129 9.2 GF Problem for Cylindrical Sym. . . . . . . . . . . . . . 130 9.3 Expansion in Terms of Eigenfunctions . . . . . . . . . . . 131 9.3.1 Partial Expansion . . . . . . . . . . . . . . . . . . 131 9.3.2 Summary of GF for Cyl. Sym. . . . . . . . . . . . 132 9.4 Eigen Value Problem for L0. . . . . . . . . . . . . . . . 133 9.5 Uses of the GF Gm(r,r/prime;λ) . . . . . . . . . . . . . . . . . 134 9.5.1 Eigenfunction Problem . . . . . . . . . . . . . . . 134 9.5.2 Normal Modes/Normal Frequencies . . . . . . . . 134 9.5.3 The Steady State Problem . . . . . . . . . . . . . 135 9.5.4 Full Time Dependence . . . . . . . . . . . . . . . 136 9.6 The Wedge Problem . . . . . . . . . . . . . . . . . . . . 136 9.6.1 General Case . . . . . . . . . . . . . . . . . . . . 137 9.6.2 Special Case: Fixed Sides . . . . . . . . . . . . . 138 9.7 The Homogeneous Membrane . . . . . . . . . . . . . . . 138 9.7.1 The Radial Eigenvalues . . . . . . . . . . . . . . . 140 9.7.2 The Physics . . . . . . . . . . . . . . . . . . . . . 141 9.8 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 141 9.9 Reference . . . . . . . . . . . . . . . . . . . . . . . . . . 142 10 Heat Conduction 143 10.1 Introduction . . . . . . . . . . . . . . . . . . . . . . . . . 143 10.1.1 Conservation of Energy . . . . . . . . . . . . . . . 143 10.1.2 Boundary Conditions . . . . . . . . . . . . . . . . 145 vi CONTENTS 10.2 The Standard form of the Heat Eq. . . . . . . . . . . . . 146 10.2.1 Correspondence with the Wave Equation . . . . . 146 10.2.2 Green’s Function Problem . . . . . . . . . . . . . 146 10.2.3 Laplace Transform . . . . . . . . . . . . . . . . . 147 10.2.4 Eigen Function Expansions . . . . . . . . . . . . . 148 10.3 Explicit One Dimensional Calculation . . . . . . . . . . . 150 10.3.1 Application of Transform Method . . . . . . . . . 151 10.3.2 Solution of the Transform Integral . . . . . . . . . 151 10.3.3 The Physics of the Fundamental Solution . . . . . 154 10.3.4 Solution of the General IVP . . . . . . . . . . . . 154 10.3.5 Special Cases . . . . . . . . . . . . . . . . . . . . 155 10.4 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 156 10.5 References . . . . . . . . . . . . . . . . . . . . . . . . . . 157 11 Spherical Symmetry 159 11.1 Spherical Coordinates . . . . . . . . . . . . . . . . . . . . 160 11.2 Discussion of Lθϕ. . . . . . . . . . . . . . . . . . . . . . 162 11.3 Spherical Eigenfunctions . . . . . . . . . . . . . . . . . . 164 11.3.1 Reduced Eigenvalue Equation . . . . . . . . . . . 164 11.3.2 Determination of um l(x) . . . . . . . . . . . . . . 165 11.3.3 Orthogonality and Completeness of um l(x) . . . . 169 11.4 Spherical Harmonics . . . . . . . . . . . . . . . . . . . . 170 11.4.1 Othonormality and Completeness of Ym l. . . . . 171 11.5 GF’s for Spherical Symmetry . . . . . . . . . . . . . . . 172 11.5.1 GF Diﬀerential Equation . . . . . . . . . . . . . . 172 11.5.2 Boundary Conditions . . . . . . . . . . . . . . . . 173 11.5.3 GF for the Exterior Problem . . . . . . . . . . . . 174 11.6 Example: Constant Parameters . . . . . . . . . . . . . . 177 11.6.1 Exterior Problem . . . . . . . . . . . . . . . . . . 177 11.6.2 Free Space Problem . . . . . . . . . . . . . . . . . 178 11.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 180 11.8 References . . . . . . . . . . . . . . . . . . . . . . . . . . 181 12 Steady State Scattering 183 12.1 Spherical Waves . . . . . . . . . . . . . . . . . . . . . . . 183 12.2 Plane Waves . . . . . . . . . . . . . . . . . . . . . . . . . 185 12.3 Relation to Potential Theory . . . . . . . . . . . . . . . . 186 CONTENTS vii 12.4 Scattering from a Cylinder . . . . . . . . . . . . . . . . . 189 12.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 190 12.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . 190 13 Kirchhoﬀ’s Formula 191 13.1 References . . . . . . . . . . . . . . . . . . . . . . . . . . 194 14 Quantum Mechanics 195 14.1 Quantum Mechanical Scattering . . . . . . . . . . . . . . 197 14.2 Plane Wave Approximation . . . . . . . . . . . . . . . . 199 14.3 Quantum Mechanics . . . . . . . . . . . . . . . . . . . . 200 14.4 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . 201 14.5 Spherical Symmetry Degeneracy . . . . . . . . . . . . . . 202 14.6 Comparison of Classical and Quantum . . . . . . . . . . 202 14.7 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 204 14.8 References . . . . . . . . . . . . . . . . . . . . . . . . . . 204 15 Scattering in 3-Dim 205 15.1 Angular Momentum . . . . . . . . . . . . . . . . . . . . 207 15.2 Far-Field Limit . . . . . . . . . . . . . . . . . . . . . . . 208 15.3 Relation to the General Propagation Problem . . . . . . 210 15.4 Simpliﬁcation of Scattering Problem . . . . . . . . . . . 210 15.5 Scattering Amplitude . . . . . . . . . . . . . . . . . . . . 211 15.6 Kinematics of Scattered Waves . . . . . . . . . . . . . . 212 15.7 Plane Wave Scattering . . . . . . . . . . . . . . . . . . . 213 15.8 Special Cases . . . . . . . . . . . . . . . . . . . . . . . . 214 15.8.1 Homogeneous Source; Inhomogeneous Observer . 214 15.8.2 Homogeneous Observer; Inhomogeneous Source . 215 15.8.3 Homogeneous Source; Homogeneous Observer . . 216 15.8.4 Both Points in Interior Region . . . . . . . . . . . 217 15.8.5 Summary . . . . . . . . . . . . . . . . . . . . . . 218 15.8.6 Far Field Observation . . . . . . . . . . . . . . . 218 15.8.7 Distant Source: r/prime→ ∞ . . . . . . . . . . . . . . 219 15.9 The Physical signiﬁcance of Xl. . . . . . . . . . . . . . . 219 15.9.1 Calculating δl(k) . . . . . . . . . . . . . . . . . . 222 15.10Scattering from a Sphere . . . . . . . . . . . . . . . . . . 223 15.10.1 A Related Problem . . . . . . . . . . . . . . . . . 224 viii CONTENTS 15.11Calculation of Phase for a Hard Sphere . . . . . . . . . . 225 15.12Experimental Measurement . . . . . . . . . . . . . . . . 226 15.12.1 Cross Section . . . . . . . . . . . . . . . . . . . . 227 15.12.2 Notes on Cross Section . . . . . . . . . . . . . . . 229 15.12.3 Geometrical Limit . . . . . . . . . . . . . . . . . 230 15.13Optical Theorem . . . . . . . . . . . . . . . . . . . . . . 231 15.14Conservation of Probability Interpretation: . . . . . . . . 231 15.14.1 Hard Sphere . . . . . . . . . . . . . . . . . . . . . 231 15.15Radiation of Sound Waves . . . . . . . . . . . . . . . . . 232 15.15.1 Steady State Solution . . . . . . . . . . . . . . . . 234 15.15.2 Far Field Behavior . . . . . . . . . . . . . . . . . 235 15.15.3 Special Case . . . . . . . . . . . . . . . . . . . . . 236 15.15.4 Energy Flux . . . . . . . . . . . . . . . . . . . . . 237 15.15.5 Scattering From Plane Waves . . . . . . . . . . . 240 15.15.6 Spherical Symmetry . . . . . . . . . . . . . . . . 241 15.16Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 242 15.17References . . . . . . . . . . . . . . . . . . . . . . . . . . 243 16 Heat Conduction in 3D 245 16.1 General Boundary Value Problem . . . . . . . . . . . . . 245 16.2 Time Dependent Problem . . . . . . . . . . . . . . . . . 247 16.3 Evaluation of the Integrals . . . . . . . . . . . . . . . . . 248 16.4 Physics of the Heat Problem . . . . . . . . . . . . . . . . 251 16.4.1 The Parameter Θ . . . . . . . . . . . . . . . . . . 251 16.5 Example: Sphere . . . . . . . . . . . . . . . . . . . . . . 252 16.5.1 Long Times . . . . . . . . . . . . . . . . . . . . . 253 16.5.2 Interior Case . . . . . . . . . . . . . . . . . . . . 254 16.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 255 16.7 References . . . . . . . . . . . . . . . . . . . . . . . . . . 256 17 The Wave Equation 257 17.1 introduction . . . . . . . . . . . . . . . . . . . . . . . . . 257 17.2 Dimensionality . . . . . . . . . . . . . . . . . . . . . . . 259 17.2.1 Odd Dimensions . . . . . . . . . . . . . . . . . . 259 17.2.2 Even Dimensions . . . . . . . . . . . . . . . . . . 260 17.3 Physics . . . . . . . . . . . . . . . . . . . . . . . . . . . . 260 17.3.1 Odd Dimensions . . . . . . . . . . . . . . . . . . 260 CONTENTS ix 17.3.2 Even Dimensions . . . . . . . . . . . . . . . . . . 260 17.3.3 Connection between GF’s in 2 & 3-dim . . . . . . 261 17.4 Evaluation of G2. . . . . . . . . . . . . . . . . . . . . . 263 17.5 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 264 17.6 References . . . . . . . . . . . . . . . . . . . . . . . . . . 264 18 The Method of Steepest Descent 265 18.1 Review of Complex Variables . . . . . . . . . . . . . . . 266 18.2 Speciﬁcation of Steepest Descent . . . . . . . . . . . . . 269 18.3 Inverting a Series . . . . . . . . . . . . . . . . . . . . . . 270 18.4 Example 1: Expansion of Γ–function . . . . . . . . . . . 273 18.4.1 Transforming the Integral . . . . . . . . . . . . . 273 18.4.2 The Curve of Steepest Descent . . . . . . . . . . 274 18.5 Example 2: Asymptotic Hankel Function . . . . . . . . . 276 18.6 Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 280 18.7 References . . . . . . . . . . . . . . . . . . . . . . . . . . 280 19 High Energy Scattering 281 19.1 Fundamental Integral Equation of Scattering . . . . . . . 283 19.2 Formal Scattering Theory . . . . . . . . . . . . . . . . . 285 19.2.1 A short digression on operators . . . . . . . . . . 287 19.3 Summary of Operator Method . . . . . . . . . . . . . . . 288 19.3.1 Derivation of G= (E−H)−1. . . . . . . . . . . 289 19.3.2 Born Approximation . . . . . . . . . . . . . . . . 289 19.4 Physical Interest . . . . . . . . . . . . . . . . . . . . . . 290 19.4.1 Satisfying the Scattering Condition . . . . . . . . 291 19.5 Physical Interpretation . . . . . . . . . . . . . . . . . . . 292 19.6 Probability Amplitude . . . . . . . . . . . . . . . . . . . 292 19.7 Review . . . . . . . . . . . . . . . . . . . . . . . . . . . . 293 19.8 The Born Approximation . . . . . . . . . . . . . . . . . . 294 19.8.1 Geometry . . . . . . . . . . . . . . . . . . . . . . 296 19.8.2 Spherically Symmetric Case . . . . . . . . . . . . 296 19.8.3 Coulomb Case . . . . . . . . . . . . . . . . . . . . 297 19.9 Scattering Approximation . . . . . . . . . . . . . . . . . 298 19.10Perturbation Expansion . . . . . . . . . . . . . . . . . . 299 19.10.1 Perturbation Expansion . . . . . . . . . . . . . . 300 19.10.2 Use of the T-Matrix . . . . . . . . . . . . . . . . 301 x CONTENTS 19.11Summary . . . . . . . . . . . . . . . . . . . . . . . . . . 302 19.12References . . . . . . . . . . . . . . . . . . . . . . . . . . 302 A Symbols Used 303 List of Figures 1.1 A string with mass points attached to springs. . . . . . . 2 1.2 A closed string, where aandbare connected. . . . . . . 6 1.3 An open string, where the endpoints aandbare free. . . 7 3.1 The pointed string . . . . . . . . . . . . . . . . . . . . . 27 4.1 The closed string with discrete mass points. . . . . . . . 37 4.2 Negative energy levels . . . . . . . . . . . . . . . . . . . 40 4.3 Theθ-convention . . . . . . . . . . . . . . . . . . . . . . 46 4.4 The contour of integration . . . . . . . . . . . . . . . . . 54 4.5 Circle around a singularity. . . . . . . . . . . . . . . . . . 55 4.6 Division of contour. . . . . . . . . . . . . . . . . . . . . . 56 4.7λnear the branch cut. . . . . . . . . . . . . . . . . . . . 61 4.8θspeciﬁcation. . . . . . . . . . . . . . . . . . . . . . . . 63 4.9 Geometry in λ-plane . . . . . . . . . . . . . . . . . . . . 69 6.1 The contour Lin theλ-plane. . . . . . . . . . . . . . . . 92 6.2 Contour LC1=L+LUHP closed in UH λ-plane. . . . . . 93 6.3 Contour closed in the lower half λ-plane. . . . . . . . . . 95 6.4 An illustration of the retarded Green’s Function. . . . . . 96 6.5GRatt1=t/prime+1 2x/prime/cand att2=t/prime+3 2x/prime/c. . . . . . . . 98 7.1 Water waves moving in channels. . . . . . . . . . . . . . 108 7.2 The rectangular membrane. . . . . . . . . . . . . . . . . 111 9.1 The region Ras a circle with radius a. . . . . . . . . . . 130 9.2 The wedge. . . . . . . . . . . . . . . . . . . . . . . . . . 137 10.1 Rotation of contour in complex plane. . . . . . . . . . . . 148 xi xii LIST OF FIGURES 10.2 Contour closed in left half s-plane. . . . . . . . . . . . . 149 10.3 A contour with Branch cut. . . . . . . . . . . . . . . . . 152 11.1 Spherical Coordinates. . . . . . . . . . . . . . . . . . . . 160 11.2 The general boundary for spherical symmetry. . . . . . . 174 12.1 Waves scattering from an obstacle. . . . . . . . . . . . . 184 12.2 Deﬁnition of γandθ.. . . . . . . . . . . . . . . . . . . . 186 13.1 A screen with a hole in it. . . . . . . . . . . . . . . . . . 192 13.2 The source and image source. . . . . . . . . . . . . . . . 193 13.3 Conﬁgurations for the G’s. . . . . . . . . . . . . . . . . . 194 14.1 An attractive potential. . . . . . . . . . . . . . . . . . . . 196 14.2 The complex energy plane. . . . . . . . . . . . . . . . . . 197 15.1 The schematic representation of a scattering experiment. 208 15.2 The geometry deﬁning γandθ. . . . . . . . . . . . . . . 212 15.3 Phase shift due to potential. . . . . . . . . . . . . . . . . 221 15.4 A repulsive potential. . . . . . . . . . . . . . . . . . . . . 223 15.5 The potential VandVeﬀfor a particular example. . . . . 225 15.6 An inﬁnite potential wall. . . . . . . . . . . . . . . . . . 227 15.7 Scattering with a strong forward peak. . . . . . . . . . . 232 16.1 Closed contour around branch cut. . . . . . . . . . . . . 250 17.1 Radial part of the 2-dimensional Green’s function. . . . . 261 17.2 A line source in 3-dimensions. . . . . . . . . . . . . . . . 263 18.1 Contour C& deformation C0with point z0. . . . . . . . 266 18.2 Gradients of uandv. . . . . . . . . . . . . . . . . . . . . 267 18.3f(z) near a saddle-point. . . . . . . . . . . . . . . . . . . 268 18.4 Deﬁning Contour for the Hankel function. . . . . . . . . 277 18.5 Deformed contour for the Hankel function. . . . . . . . . 278 18.6 Hankel function contours. . . . . . . . . . . . . . . . . . 280 19.1 Geometry of the scattered wave vectors. . . . . . . . . . 296 Preface This manuscript is based on lectures given by Marshall Baker for a class on Mathematical Methods in Physics at the University of Washington in 1988. The subject of the lectures was Green’s function techniques in Physics. All the members of the class had completed the equivalent of the ﬁrst three and a half years of the undergraduate physics program, although some had signiﬁcantly more background. The class was a preparation for graduate study in physics. These notes develop Green’s function techiques for both single and multiple dimension problems, and then apply these techniques to solv- ing the wave equation, the heat equation, and the scattering problem. Many other mathematical techniques are also discussed. To read this manuscript it is best to have Arfken’s book handy for the mathematics details and Fetter and Walecka’s book handy for the physics details. There are other good books on Green’s functions available, but none of them are geared for same background as assumed here. The two volume set by Stakgold is particularly useful. For a strictly mathematical discussion, the book by Dennery is good. Here are some notes and warnings about this revision: •Text This text is an ampliﬁcation of lecture notes taken of the Physics 425-426 sequence. Some sections are still a bit rough. Be alert for errors and omissions. •List of Symbols A listing of mostly all the variables used is in- cluded. Be warned that many symbols are created ad hoc, and thus are only used in a particular section. •Bibliography The bibliography includes those books which have been useful to Steve Sutlief in creating this manuscript, and were xiii xiv LIST OF FIGURES not necessarily used for the development of the original lectures. Books marked with an asterisk are are more supplemental. Com- ments on the books listed are given above. •Index The index was composed by skimming through the text and picking out places where ideas were introduced or elaborated upon. No attempt was made to locate all relevant discussions for each idea. A Note About Copying: These notes are in a state of rapid transition and are provided so as to be of beneﬁt to those who have recently taken the class. Therefore, please do not photocopy these notes. Contacting the Authors: A list of phone numbers and email addresses will be maintained of those who wish to be notiﬁed when revisions become available. If you would like to be on this list, please send email to [email protected] before 1996. Otherwise, call Marshall Baker at 206-543-2898. Acknowledgements: This manuscript beneﬁts greatly from the excellent set of notes taken by Steve Griﬃes. Richard Horn contributed many corrections and suggestions. Special thanks go to the students of Physics 425-426 at the University of Washington during 1988 and 1993. This ﬁrst revision contains corrections only. No additional material has been added since Version 0. Steve Sutlief Seattle, Washington 16 June, 1993 4 January, 1994 Chapter 1 The Vibrating String 4 Jan p1 p1prv.yr. Chapter Goals: •Construct the wave equation for a string by identi- fying forces and using Newton’s second law. •Determine boundary conditions appropriate for a closed string, an open string, and an elastically bound string. •Determine the wave equation for a string subject to an external force with harmonic time dependence. The central topic under consideration is the branch of diﬀerential equa- tion theory containing boundary value problems. First we look at an pr:bvp1 example of the application of Newton’s second law to small vibrations: transverse vibrations on a string. Physical problems such as this and those involving sound, surface waves, heat conduction, electromagnetic waves, and gravitational waves, for example, can be solved using the mathematical theory of boundary value problems. Consider the problem of a string embedded in a medium with a pr:string1 restoring force V(x) and an external force F(x,t). This problem covers pr:V1 pr:F1most of the physical interpretations of small vibrations. In this chapter we will investigate the mathematics of this problem by determining the equations of motion. 1 2 CHAPTER 1. THE VIBRATING STRING PPPPPPPPPP XXXXXXXXXXXXXXXXXXXXXX XXXXXXXXXXXXXXXXXXXXXXXXXXXXXX uuθu ###""""""!!!!!! ui+1 ui ui−1 xi−1xixi+1mi−1mimi+1 a aFτi iyFτi+1 iy ki−1kiki+1 Figure 1.1: A string with mass points attached to springs. 1.1 The String We consider a massless string with equidistant mass points attached. In the case of a string, we shall see (in chapter 3) that the Green’s function corresponds to an impulsive force and is represented by a complete set of functions. Consider Nmass points of mass miattached to a massless pr:N1 pr:mi1 string, which has a tension τbetween mass points. An elastic force at pr:tau1each mass point is represented by a spring. This problem is illustrated in ﬁgure 1.1 We want to ﬁnd the equations of motion for transverseﬁg1.1 pr:eom1vibrations of the string. 1.1.1 Forces on the String For the massless vibrating string, there are three forces which are in- cluded in the equation of motion. These forces are the tension force, elastic force, and external force. Tension Force4 Jan p2 For each mass point there are two force contributions due to the tension pr:tension1 on the string. We call τithe tension on the segment between mi−1 andmi,uithe vertical displacement of the ith mass point, and athe pr:ui1 pr:a1 horizontal displacement between mass points. Since we are considering transverse vibrations (in the u-direction) , we want to know the tensionpr:transvib1 1.1. THE STRING 3 force in the u-direction, which is τi+1sinθ. From the ﬁgure we see that pr:theta1 θ≈(ui+1−ui)/afor small angles and we can thus write Fτi+1 iy=τi+1(ui+1−ui) a and pr:Fiyt1 Fτi iy=−τi(ui−ui−1) a. Note that the equations agree with dimensional analysis: Grif’s uses Taylor exp pr:m1 pr:l1 pr:t1Fτi iy= dim(m·l/t2), τ i= dim(m·l/t2), ui= dim(l), anda= dim(l). Elastic Forcepr:elastic1 We add an elastic force with spring constant ki: pr:ki1 Felastic i =−kiui, where dim( ki) = (m/t2). This situation can be visualized by imagining pr:Fel1 vertical springs attached to each mass point, as depicted in ﬁgure 1.1. A small value of kicorresponds to an elastic spring, while a large value ofkicorresponds to a rigid spring. External Force We add the external force Fext i. This force depends on the nature of pr:ExtForce1 pr:Fext1 the physical problem under consideration. For example, it may be a transverse force at the end points. 1.1.2 Equations of Motion for a Massless String The problem thus far has concerned a massless string with mass points attached. By summing the above forces and applying Newton’s second law, we have pr:Newton1 pr:t2 Ftot=τi+1(ui+1−ui) a−τi(ui−ui−1) a−kiui+Fext i=mid2 dt2ui.(1.1) This gives us Ncoupled inhomogeneous linear ordinary diﬀerential eq1force equations where each uiis a function of time. In the case that Fext ipr:diﬀeq1 is zero we have free vibration, otherwise we have forced vibration.pr:FreeVib1 pr:ForcedVib1 4 CHAPTER 1. THE VIBRATING STRING 1.1.3 Equations of Motion for a Massive String 4 Jan p3 For a string with continuous mass density, the equidistant mass points on the string are replaced by a continuum. First we take a, the sep- aration distance between mass points, to be small and redeﬁne it as a= ∆x. We correspondingly write ui−ui−1= ∆u. This allows us to pr:deltax1 pr:deltau1 write (ui−ui−1) a=/parenleftbigg∆u ∆x/parenrightbigg i. (1.2) The equations of motion become (after dividing both sides by ∆ x) 1 ∆x/bracketleftBigg τi+1/parenleftbigg∆u ∆x/parenrightbigg i+1−τi/parenleftbigg∆u ∆x/parenrightbigg i/bracketrightBigg −ki ∆xui+Fext i ∆x=mi ∆xd2ui dt2.(1.3) In the limit we take a→0,N→ ∞ , and deﬁne their product to be eq1deltf lim a→0 N→∞Na≡L. (1.4) The limiting case allows us to redeﬁne the terms of the equations of motion as follows: pr:sigmax1 mi→0mi ∆x→σ(xi)≡mass length= mass density; ki→0ki ∆x→V(xi) = coeﬃcient of elasticity of the media; Fext i→0Fext ∆x= (mi ∆x·Fext mi)→σ(xi)f(xi) (1.5) where f(xi) =Fext mi=external force mass. (1.6) Since xi=x xi−1=x−∆x xi+1=x+ ∆x we have pr:x1/parenleftbigg∆u ∆x/parenrightbigg i=ui−ui−1 xi−xi−1→∂u(x,t) ∂x(1.7) 1.2. THE LINEAR OPERATOR FORM 5 so that 1 ∆x/bracketleftBigg τi+1/parenleftbigg∆u ∆x/parenrightbigg i+1−τi/parenleftbigg∆u ∆x/parenrightbigg i/bracketrightBigg =1 ∆x/bracketleftBigg τ(x+ ∆x)∂u(x+ ∆x) ∂x−τ(x)∂u(x) ∂x/bracketrightBigg =∂ ∂x/bracketleftBigg τ(x)∂u ∂x/bracketrightBigg . (1.8) This allows us to write 1.3 as 4 Jan p4 ∂ ∂x/bracketleftBigg τ(x)∂u ∂x/bracketrightBigg −V(x)u+σ(x)f(x,t) =σ(x)∂2u ∂t2. (1.9) This is a partial diﬀerential equation. We will look at this problem in eq1diﬀ pr:pde1 detail in the following chapters. Note that the ﬁrst term is net tension force overdx. 1.2 The Linear Operator Form We deﬁne the linear operator L0by the equation pr:LinOp1 L0≡ −∂ ∂x/parenleftBigg τ(x)∂ ∂x/parenrightBigg +V(x). (1.10) We can now write equation (1.9) as eq1LinOp /bracketleftBigg L0+σ(x)∂2 ∂t2/bracketrightBigg u(x,t) =σ(x)f(x,t) ona<x<b. (1.11) This is an inhomogeneous equation with an external force term. Note eq1waveone that each term in this equation has units of m/t2. Integrating this equation over the length of the string gives the total force on the string. 1.3 Boundary Conditions pr:bc1 To obtain a unique solution for the diﬀerential equation, we must place restrictive conditions on it. In this case we place conditions on the ends of the string. Either the string is tied together (i.e. closed), or its ends are left apart (open). 6 CHAPTER 1. THE VIBRATING STRING r r' &$ %a b Figure 1.2: A closed string, where aandbare connected. 1.3.1 Case 1: A Closed String A closed string has its endpoints aandbconnected. This case is illus- pr:ClStr1 pr:a2 trated in ﬁgure 2. This is the periodic boundary condition for a closed ﬁg1loop pr:pbc1string. A closed string must satisfy the following equations: u(a,t) =u(b,t) (1.12) which is the condition that the ends meet, and eq1pbc1 ∂u(x,t) ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a=∂u(x,t) ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=b(1.13) which is the condition that the ends have the same declination (i.e., eq1pbc2 the string must be smooth across the end points). 1.3.2 Case 2: An Open String sec1-c2 4 Jan p5 For an elastically bound open string we have the boundary condition pr:ebc1 pr:OpStr1that the total force must vanish at the end points. Thus, by multiplying equation 1.3 by ∆ xand setting the right hand side equal to zero, we have the equation τa∂u(x,t) ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a−kau(a,t) +Fa(t) = 0. The homogeneous terms of this equation are τa∂u ∂x|x=aandkau(a,t), and the inhomogeneous term is Fa(t). The term kau(a) describes how the string is bound. We now deﬁne pr:ha1 ha(t)≡Fa τaandκa≡ka τa. 1.3. BOUNDARY CONDITIONS 7 r r- ˆn ˆna b Figure 1.3: An open string, where the endpoints aandbare free. The termha(t) is the eﬀective force and κais the eﬀective spring con- pr:EﬀFrc1 stant. pr:esc1 −∂u ∂x+κau(x) =ha(t) forx=a. (1.14) We also deﬁne the outward normal, ˆ n, as shown in ﬁgure 1.3. This eq1bound pr:OutNorm1 ﬁg1.2allows us to write 1.14 as ˆn· ∇u(x) +κau(x) =ha(t) forx=a. The boundary condition at bcan be similarly deﬁned: ∂u ∂x+κbu(x) =hb(t) forx=b, where hb(t)≡Fb τbandκb≡kb τb. For a more compact notation, consider points aandbto be elements of the “surface” of the one dimensional string, S={a,b}. This gives pr:S1 us ˆnS∇u(x) +κSu(x) =hS(t) forxonS, for allt. (1.15) In this case ˆ na=−/vectorlxand ˆnb=/vectorlx. eq1osbc pr:lhat1 1.3.3 Limiting Cases 6 Jan p2.1 It is also worthwhile to consider the limiting cases for an elastically bound string. These cases may be arrived at by varying κaandκb. The pr:ebc2 termsκaandκbsignify how rigidly the string’s endpoints are bound. The two limiting cases of equation 1.14 are as follows: pr:ga1 8 CHAPTER 1. THE VIBRATING STRING κa→0−∂u ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a=ha(t) (1.16) κa→ ∞u(x,t)|x=a=ha/κa=Fa/ka. (1.17) The boundary condition κa→0 corresponds to an elastic media, and pr:ElMed1 is called the Neumann boundary condition. The case κa→ ∞ corre- pr:nbc1 sponds to a rigid medium, and is called the Dirichlet boundary condi- tion. pr:dbc1 IfhS(t) = 0 in equation 1.15, so that pr:hS1 [ˆnS· ∇+κS]u(x,t) =hS(t) = 0 for xonS, (1.18) then the boundary conditions are called regular boundary conditions . eq1RBC pr:rbc1 Regular boundary conditions are either see Stakgold p2691.u(a,t) =u(b,t),d dxu(a,t) =d dxu(b,t) (periodic), or 2. [ˆnS· ∇+κS]u(x,t) = 0 for xonS. Thus regular boundary conditions correspond to the case in which there is no external force on the end points. 1.3.4 Initial Conditionspr:ic1 6 Jan p2 The complete description of the problem also requires information about the string at some reference point in time:pr:u0.1 u(x,t)|t=0=u0(x) fora<x<b (1.19) and∂ ∂tu(x,t)|t=0=u1(x) fora<x<b. (1.20) Here we claim that it is suﬃcient to know the position and velocity of the string at some point in time. 1.4 Special Cases This material was originally in chapter 3 8 Jan p3.3We now consider two singular boundary conditions and a boundary pr:sbc1condition leading to the Helmholtz equation. The conditions ﬁrst two cases will ensure that the right-hand side of Green’s second identity (introduced in chapter 2) vanishes. This is necessary for a physical system. 1.4. SPECIAL CASES 9 1.4.1 No Tension at Boundary For the case in which τ(a) = 0 and the regular boundary conditions hold, the condition that u(a) be ﬁnite is necessary. This is enough to ensure that the right hand side of Green’s second identity is zero. 1.4.2 Semi-inﬁnite String In the case that a→ −∞ , we require that u(x) have a ﬁnite limit as x→ −∞ . Similarly, if b→ ∞ , we require that u(x) have a ﬁnite limit asx→ ∞ . If botha→ −∞ andb→ ∞ , we require that u(x) have ﬁnite limits as either x→ −∞ orx→ ∞ . 1.4.3 Oscillatory External Force sec1helm In the case in which there are no forces at the boundary we have ha=hb= 0. (1.21) The termsha,hbare extra forces on the boundaries. Thus the condition of no forces on the boundary does not imply that the internal forces are zero. We now treat the case where the interior force is oscillatory and write pr:omega1 f(x,t) =f(x)e−iωt. (1.22) In this case the physical solution will be Ref(x,t) =f(x) cosωt. (1.23) We look for steady state solutions of the form pr:sss1 u(x,t) =e−iωtu(x) for all t. (1.24) This gives us the equation /bracketleftBigg L0+σ(x)∂2 ∂t2/bracketrightBigg e−iωtu(x) =σ(x)f(x)e−iωt. (1.25) Ifu(x,ω) satisﬁes the equation [L0−ω2σ(x)]u(x) =σ(x)f(x) with R.B.C. on u(x) (1.26) (the Helmholtz equation), then a solution exists. We will solve this eq1helm pr:Helm1 equation in chapter 3. 10 CHAPTER 1. THE VIBRATING STRING 1.5 Summary In this chapter the equations of motion have been derived for the small oscillation problem. Appropriate forms of the boundary conditions and initial conditions have been given. The general string problem with external forces is mathematically the same as the small oscillation (vibration) problem, which uses vectors and matrices. Let ui=u(xi) be the amplitude of the string at the point xi. For the discrete case we have Ncomponent vectors ui=u(xi), and for the continuum case we have a continuous function u(x). These considerations outline the most general problem. The main results for this chapter are: 1. The equation of motion for a string is /bracketleftBigg L0+σ(x)∂2 ∂t2/bracketrightBigg u(x,t) =σ(x)f(x,t) ona<x<b where L0u=/bracketleftBigg −∂ ∂x/parenleftBigg τ(x)∂ ∂x/parenrightBigg +V(x)/bracketrightBigg u. 2.Regular boundary conditions refer to the boundary conditions for either (a) a closed string: u(a,t) =u(b,t) (continuous) ∂u(a,t) ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a=∂u(b,t) ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=b(no bends) or (b) an open string: [ˆnS· ∇+κS]u(x,t) =hS(t) = 0xonS, allt. 3. The initial conditions are given by the equations u(x,t)|t=0=u0(x) fora<x<b (1.27) 1.6. REFERENCES 11 and ∂ ∂tu(x,t)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle t=0=u1(x) fora<x<b. (1.28) 4. The Helmholtz equation is [L0−ω2σ(x)]u(x) =σ(x)f(x). 1.6 References See any book which derives the wave equation, such as [Fetter80, p120ﬀ], [Griﬃths81, p297], [Halliday78, pA5]. A more thorough deﬁnition of regular boundary conditions may be found in [Stakgold67a, p268ﬀ]. 12 CHAPTER 1. THE VIBRATING STRING Chapter 2 Green’s Identities Chapter Goals: •Derive Green’s ﬁrst and second identities. •Show that for regular boundary conditions, the lin- ear operator is hermitian. In this chapter, appropriate tools and relations are developed to solve the equation of motion for a string developed in the previous chapter. In order to solve the equations, we will want the function u(x) to take on complex values. We also need the notion of an inner product. The note pr:InProd1 inner product of Sanduis deﬁned as pr:S2 /angbracketleftS,u/angbracketright=/braceleftBigg/summationtextn i=1S∗ iui for the discrete case/integraltextb adxS∗(x)u(x) for the continuous case.(2.1) In the uses of the inner product which will be encountered here, for the eq2.2 continuum case, one of the variables Soruwill be a length (amplitude of the string), and the other will be a force per unit length. Thus the inner product will have units of force times length, which is work. 13 14 CHAPTER 2. GREEN’S IDENTITIES 2.1 Green’s 1st and 2nd Identities 6 Jan p2.4 In the deﬁnition of the inner product we make the substitution of L0u foru, where L0u(x)≡/bracketleftBigg −d dx/parenleftBigg τ(x)d dx/parenrightBigg +V(x)/bracketrightBigg u(x). (2.2) This substitution gives us eq2.3 /angbracketleftS,L 0u/angbracketright=/integraldisplayb adxS∗(x)/bracketleftBigg −d dx/parenleftBigg τ(x)d dx/parenrightBigg +V(x)/bracketrightBigg u(x) =−/integraldisplayb adxS∗(x)/parenleftBigg −d dx/parenleftBigg τ(x)d dxu/parenrightBigg/parenrightBigg +/integraldisplayb adxS∗(x)V(x)u(x). We now integrate twice by parts (/integraltext¯ud¯v= ¯u¯v−/integraltext¯vd¯u), letting ¯u=S∗(x) =⇒d¯u=dS∗(x) =dxdS∗(x) dx and d¯v=dxd dx/parenleftBigg τ(x)d dxu/parenrightBigg =d/parenleftBigg τ(x)d dxu/parenrightBigg =⇒¯v=τ(x)d dxu so that /angbracketleftS,L 0u/angbracketright=−/vextendsingle/vextendsingle/vextendsingle/vextendsingleb aS∗(x)τ(x)/parenleftBiggd dxu(x)/parenrightBigg +/integraldisplayb adxdS∗ dxτ(x)/parenleftBiggd dxu(x)/parenrightBigg +/integraldisplayb adxS∗(x)V(x)u(x) =−/vextendsingle/vextendsingle/vextendsingle/vextendsingleb aS∗(x)τ(x)d dxu(x) +/integraldisplayb adx/bracketleftBigg/parenleftBiggd dxS∗/parenrightBigg τ(x)d dxu(x) +S∗(x)V(x)u(x)/bracketrightBigg . Note that the ﬁnal integrand is symmetric in terms of S∗(x) andu(x). This is Green’s First Identity : pr:G1Id1 2.2. USING G.I. #2 TO SATISFY R.B.C. 15 /angbracketleftS,L 0u/angbracketright=−/vextendsingle/vextendsingle/vextendsingle/vextendsingleb aS∗(x)τ(x)d dxu(x) (2.3) +/integraldisplayb adx/bracketleftBigg/parenleftBiggd dxS∗/parenrightBigg τ(x)d dxu(x) +S∗(x)V(x)u(x)/bracketrightBigg . Now interchange S∗anduto get eq2G1Id /angbracketleftu,L 0S/angbracketright∗=/angbracketleftL0S,u/angbracketright =−/vextendsingle/vextendsingle/vextendsingle/vextendsingleb au(x)τ(x)d dxS∗(x) (2.4) +/integraldisplayb adx/bracketleftBigg/parenleftBiggd dxu/parenrightBigg τ(x)d dxS∗(x) +u(x)V(x)S∗(x)/bracketrightBigg . When the diﬀerence of equations 2.3 and 2.4 is taken, the symmetric eq2preG2Id terms cancel. This is Green’s Second Identity : pr:G2Id1 /angbracketleftS,L 0u/angbracketright − /angbracketleftL0S,u/angbracketright=/vextendsingle/vextendsingle/vextendsingle/vextendsingleb aτ(x)/bracketleftBigg u(x)d dxS∗(x)−S∗(x)d dxu(x)/bracketrightBigg .(2.5) In the literature, the expressions for the Green’s identities take τ=−1eq2G2Id andV= 0 in the operator L0. Furthermore, the expressions here are for one dimension, while the multidimensional generalization is given in section 8.4.1. 2.2 Using G.I. #2 to Satisfy R.B.C. 6 Jan p2.5 The regular boundary conditions for a string (either equations 1.12 and 1.13 or equation 1.18) can simplify Green’s 2nd Identity. If Sandu correspond to physical quantities, they must satisfy RBC. We will verify this statement for two special cases: the closed string and the open string. 2.2.1 The Closed String For a closed string we have (from equations 1.12 and 1.13) u(a,t) =u(b,t), S∗(a,t) =S∗(b,t), 16 CHAPTER 2. GREEN’S IDENTITIES τ(a) =τ(b),d dxS∗/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a=d dxS∗/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=b,d dxu/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a=d dxu/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=b. By plugging these equalities into Green’s second identity, we ﬁnd that /angbracketleftS,L 0u/angbracketright=/angbracketleftL0S,u/angbracketright. (2.6) eq2twox 2.2.2 The Open String For an open string we have −∂u ∂x+Kau= 0 for x=a, −∂S∗ ∂x+KaS∗= 0 for x=a, ∂u ∂x+Kbu= 0 for x=b, ∂S∗ ∂x+KbS∗= 0 for x=b. (2.7) These are the conditions for RBC from equation 1.14. Plugging these eq21osbc expressions into Green’s second identity gives /vextendsingle/vextendsingle/vextendsingle/vextendsingle aτ(x)/bracketleftBigg udS∗ dx−S∗du dx/bracketrightBigg =τ(a)[uKaS∗−S∗Kau] = 0 and/vextendsingle/vextendsingle/vextendsingle/vextendsingleb τ(x)/bracketleftBigg udS∗ dx−S∗du dx/bracketrightBigg =τ(b)[uKbS∗−S∗Kbu] = 0. Thus from equation 2.5 we ﬁnd that /angbracketleftS,L 0u/angbracketright=/angbracketleftL0S,u/angbracketright, (2.8) just as in equation 2.6 for a closed string. eq2twox2 2.3. ANOTHER BOUNDARY CONDITION 17 2.2.3 A Note on Hermitian Operators The equation /angbracketleftS,L 0u/angbracketright=/angbracketleftL0S,u/angbracketright, which we have found to hold for both a closed string and an open string, is the criterion for L0to be a Hermitian operator . By using the deﬁnition 2.1, this expression can be pr:HermOp1 rewritten as /angbracketleftS,L 0u/angbracketright=/angbracketleftu,L 0S/angbracketright∗. (2.9) Hermitian operators are generally generated by nondissipative phys- ical problems. Thus Hermitian operators with Regular Boundary Con- ditions are generated by nondissipative mechanical systems. In a dis- sipative system, the acceleration cannot be completely speciﬁed by the position and velocity, because of additional factors such as heat, fric- tion, and/or other phenomena. 2.3 Another Boundary Condition 6 Jan p2.6 If the ends of an open string are free of horizontal forces, the tension at the end points must be zero. Since lim x→a,bτ(x) = 0 we have lim x→a,bτ(x)u(x)∂ ∂xS∗(x) = 0 and lim x→a,bτ(x)S∗(x)∂ ∂xu(x) = 0. In the preceding equations, the abbreviated notation lim x→a,bis introduced to represent either the limit as xapproaches the endpoint aor the limit asxapproaches the endpoint b. These equations allow us to rewrite Green’s second identity (equation 2.5) as /angbracketleftS,L 0u/angbracketright=/angbracketleftL0S,u/angbracketright (2.10) for the case of zero tension on the end points This is another way of eq2G2Id getting at the result in equation 2.8 for the special case of free ends. 18 CHAPTER 2. GREEN’S IDENTITIES 2.4 Physical Interpretations of the G.I.s sec2.4 Certain qualities of the Green’s Identities correspond to physical situ- ations and constraints. 2.4.1 The Physics of Green’s 2nd Identity 6 Jan p2.6 The right hand side of Green’s 2nd Identity will always vanish for phys- ically realizable systems. Thus L0is Hermitian for any physically real- izable system. We could extend the deﬁnition of regular boundary conditions by letting them be those in which the right-hand side of Green’s second identity vanishes. This would allow us to include a wider class of prob- lems, including singular boundary conditions, domains, and operators. This will be necessary to treat Bessel’s equation. For now, however, we only consider problems whose boundary conditions are periodic or of the form of equation 1.18. 2.4.2 A Note on Potential Energy The potential energy of an element dxof the string has two contribu- tions. One is the “spring” potential energy1 2V(x)(u(x))2(c.f.,1 2kx2in U=−/integraltextFdx =−/integraltext(−kx)dx=1 2kx2[Halliday76, p141]). The other is the “tension” potential energy, which comes from the tension force in section 1.1.3, dF=∂ ∂x[τ(x)∂ ∂xu(x)]dx, and thusUtension is U=−/integraldisplay∂ ∂x/parenleftBigg τ∂u ∂x/parenrightBigg dx, so dU dt=−d dt/integraldisplay∂ ∂x/parenleftBigg τ∂u ∂x/parenrightBigg dx=−/integraldisplay x∂ ∂x/parenleftBigg τ∂u ∂x/parenrightBigg/parenleftBigg∂u ∂t/parenrightBigg dx, and so the change in potential energy in a time interval dtis Udt =−/integraldisplayb a∂ ∂x/parenleftBigg τ∂u ∂x/parenrightBigg/parenleftBigg∂u ∂t/parenrightBigg dtdx parts=/integraldisplayb a/parenleftBigg τ∂u ∂x/parenrightBigg∂ ∂t∂u ∂xdtdx−/bracketleftBigg/parenleftBigg τ∂u ∂x/parenrightBigg∂u ∂tdt/bracketrightBiggb a 2.4. PHYSICAL INTERPRETATIONS OF THE G.I.S 19 =/integraldisplayb a/parenleftBigg τ∂u ∂x/parenrightBigg∂ ∂t∂u ∂xdtdx = ∂ ∂t/integraldisplayb a1 2τ/parenleftBigg∂u ∂x/parenrightBigg2 dx t+dt t. The second term in the second equality vanishes. We may now sum the diﬀerentials of Uin time to obtain the potential energy: U=/integraldisplayt t/prime=0Udt = /integraldisplayb a1 2τ/parenleftBigg∂u ∂x/parenrightBigg2 dx t 0=/integraldisplayb a1 2τ/parenleftBigg∂u ∂x/parenrightBigg2 dx. 2.4.3 The Physics of Green’s 1st Identity sec2.4.2 6 Jan p2LetS=u. Then 2.3 becomes /angbracketleftu,L 0u/angbracketright=−/vextendsingle/vextendsingle/vextendsingle/vextendsingleb au∗(x)τ(x)d dxu(x) (2.11) +/integraldisplayb adx/bracketleftBigg/parenleftBiggd dxu∗/parenrightBigg τ(x)d dxu(x) +u∗(x)V(x)u(x)/bracketrightBigg . For a closed string we have /angbracketleftu,L 0u/angbracketright=/integraldisplayb adx τ(x)/parenleftBiggdu dx/parenrightBigg2 +V(x)(u(x))2 = 2U (2.12) since each quantity is the same at aandb. For an open string we found eq2x 8 Jan p3.2 (equation 1.15) du dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a=Kau (2.13) anddu dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=b=−Kbu (2.14) so that /angbracketleftu,L 0u/angbracketright=τ(a)Ka|u(a)|2+τ(b)Kb|u(b)|2 +/integraldisplayb adx τ(x)/parenleftBiggdu dx/parenrightBigg2 +V(x)(u(x))2  = 2U, 20 CHAPTER 2. GREEN’S IDENTITIES twice the potential energy. The term1 2τ(a)Ka|u(a)|2+1 2τ(b)Kb|u(b)|2see FW p207, expl. p109 p126 eq2y pr:pe1is the potential energy due to two discrete “springs” at the end points, and is simply the spring constant times the displacement squared. The termτ(x) (du/dx )2is the tension potential energy. Since du/dx represents the string stretching in the transverse direction, τ(x) (du/dx )2 is a potential due to the stretching of the string. V(x)(u(x))2is the elastic potential energy. For the case of the closed string, equation 2.12, and the open string, equation 2.15, the right hand side is equal to twice the potential energy. IfKa,Kb,τandVare positive for the open string, the potential energy Uis also positive. Thus /angbracketleftu,L 0u/angbracketright>0, which implies that L0is a positive deﬁnite operator. pr:pdo1 2.5 Summary 1. The Green’s identities are: (a) Green’s ﬁrst identity: /angbracketleftS,L 0u/angbracketright=−/vextendsingle/vextendsingle/vextendsingle/vextendsingleb aS∗(x)τ(x)d dxu(x) +/integraldisplayb adx/bracketleftbigg/parenleftBiggd dxS∗/parenrightBigg τ(x)d dxu(x) +S∗(x)V(x)u(x)/bracketrightbigg , (b) Green’s second identity: /angbracketleftS,L 0u/angbracketright−/angbracketleftL0S,u/angbracketright=/vextendsingle/vextendsingle/vextendsingle/vextendsingleb aτ(x)/bracketleftBigg u(x)d dxS∗(x)−S∗(x)d dxu(x)/bracketrightBigg . 2. For a closed string and an open string (i.e., RBC) the linear op- eratorL0is Hermitian: /angbracketleftS,L 0u/angbracketright=/angbracketleftu,L 0S/angbracketright∗. 2.6. REFERENCES 21 2.6 References Green’s formula is described in [Stakgold67, p70] and [Stakgold79, p167]. The derivation of the potential energy of a string was inspired by [Simon71,p390]. 22 CHAPTER 2. GREEN’S IDENTITIES Chapter 3 Green’s Functions Chapter Goals: •Show that an external force can be written as a sum ofδ-functions. •Find the Green’s function for an open string with no external force on the endpoints. In this chapter we want to solve the Helmholtz equation, which was obtained in section 1.4.3. First we will develop some mathematical principles which will facilitate the derivation. 8 Jan p3.4 Lagrangian stuﬀ com- mented out 3.1 The Principle of Superposition Suppose that pr:a1.1 f(x) =a1f1(x) +a2f2(x). (3.1) Ifu1andu2are solutions to the equations (c.f., 1.26) [L0−ω2σ(x)]u1(x) =σ(x)f1(x) (3.2) [L0−ω2σ(x)]u2(x) =σ(x)f2(x) (3.3) with RBC and such that (see equation 1.15) eq3q (ˆnS· ∇+κS)u1= 0 (ˆnS· ∇+κS)u2= 0/bracerightBigg forxonS 23 24 CHAPTER 3. GREEN’S FUNCTIONS then their weighted sum satisﬁes the same equation of motion [L0−ω2σ(x)](a1u1(x) +a2u2(x)) =a1[L0−ω2σ(x)]u1(x)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright σ(x)f1(x)+a2[L0−ω2σ(x)]u2(x)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright σ(x)f2(x) =σ(x)f(x). and boundary condition [ˆnS· ∇+κS][a1u1(x) +a2u2(x)] =a1[ˆnS· ∇+κS]u1+a2[ˆnS· ∇+κS]u2 =a1(0) +a2(0) = 0. We have thus shown that L0[a1u1+a2u2] =a1L0u1+a2L0u2. (3.4) This is called the principle of superposition , and it is the deﬁning prop- pr:pos1 erty of a linear operator . 3.2 The Dirac Delta Function 11 Jan p4.1 We now develop a tool to solve the Helmholtz equation (which is also called the steady state equation), equation 1.26: [L0−ω2σ(x)]u(x) =σ(x)f(x). The delta function is deﬁned by the equation pr:DeltaFn1 pr:Fcd Fcd=/integraldisplayd cdxδ(x−xk) =/braceleftBigg 1 ifc<x k<d 0 otherwise .(3.5) whereFcdrepresents the total force over the interval [ c,d]. Thus we see eq3deltdef pr:Fcd1 that the appearance of the delta function is equivalent to the application of a unit force at xk. The Dirac delta function has units of force/length. On the right-hand side of equation 1.26 make the substitution σ(x)f(x) =δ(x−xk). (3.6) 3.2. THE DIRAC DELTA FUNCTION 25 Integration gives us/integraldisplayd cσ(x)f(x)dx=Fcd, (3.7) which is the total force applied over the domain. This allows us to write eq3fdc [L0−σω2]u(x,ω) =δ(x−xk)a<x<b, RBC (3.8) where we have written RBC to indicate that the solution of this equa- tion must also satisfy regular boundary conditions. We may now use 11 Jan p2 the principle of superposition to get an arbitrary force. We deﬁne an element of such an arbitrary force as Fk=/integraldisplayxk+∆x xkdxσ(x)f(x) (3.9) = the force on the interval ∆ x. (3.10) eq3Fsubk We now prove that σ(x)f(x) =N/summationdisplay k=1Fkδ(x−x/prime k) (3.11) wherexk<x/prime k<x k+ ∆x. We ﬁrst integrate both sides to get eq3sfd x/primereplaces x soδ-fn isn’t on boundary/integraldisplayd cdxσ(x)f(x) =/integraldisplayd cdxN/summationdisplay k=1Fkδ(x−x/prime k). (3.12) By deﬁnition (equation 3.7), the left-hand side is the total force applied over the domain, Fcd. The right-hand side is /integraldisplayd cN/summationdisplay k=1Fkδ(x−x/prime k)dx =N/summationdisplay k=1/integraldisplayd cdxF kδ(x−x/prime k) (3.13) =/summationdisplay c<xk<dFk (3.14) =/summationdisplay c<xk<d/integraldisplayxk+∆x xkdxσ(x)f(x) (3.15) N→∞−→/integraldisplayd cdxσ(x)f(x) (3.16) =Fcd. (3.17) 26 CHAPTER 3. GREEN’S FUNCTIONS In the ﬁrst equality, 3.13, switching the sum and integration holds for eq3sum1-5 all well behaved Fk. Equality 3.14 follows from the deﬁnition of the delta function in equation 3.5. Equality 3.15 follows from equation 3.9. By taking the continuum limit, equality 3.16 completes the proof. The Helmholtz equation 3.2 can now be rewritten (using 3.11) as pr:Helm2 [L0−σ(x)ω2]u(x,ω) =N/summationdisplay k=1Fkδ(x−xk). (3.18) By the principle of superposition we can write u(x) =N/summationdisplay k=1Fkuk(x) (3.19) whereuk(x) is the solution of [ L0−σ(x)ω2]uk(x,ω2) =δ(x−xk). Thus, if we know the response of the system to a localized force, we can ﬁnd the response of the system to a general force as the sum of responses to localized forces. 11 Jan p3 We now introduce the following notation uk(x)≡G(x,x k;ω2) (3.20) whereGis the Green’s function ,xksigniﬁes the location of the distur- pr:Gxxo1 bance, and ωcorresponds to frequency. This allows us to write u(x) =N/summationdisplay k=1Fkuk(x) =N/summationdisplay k=1/integraldisplayxk+∆x xkdx/primeσ(x/prime)f(x/prime)G(x,x k;ω2) N→∞−→/integraldisplayb adx/primeG(x,x/prime;ω2)σ(x/prime)f(x/prime). We have deﬁned the Green’s function by [L0−σ(x)ω2]G(x,x/prime;ω2) =δ(x−x/prime)a<x,x/prime<b,RBC.(3.21) The solution will explode for ω2whenωis a natural frequency of the 11 Jan p4 pr:NatFreq1 system, as will be seen later. where will nat freq be deﬁned 3.2. THE DIRAC DELTA FUNCTION 27 eee llQQHHPPXX``%%% ,, ee ee ee%% %% %%d dxG|x=x/prime−εd dxG|x=x/prime+εG(x,x/prime;ω2) x/prime−ε x/primex/prime+ε a bA A AA U Figure 3.1: The pointed string Letλ=ω2be an arbitrary complex number. Since the squared pr:lambda1frequencyω2cannot be complex, we relabel it λ. So now we want to ﬁx thissolve /bracketleftBigg −d dx/parenleftBigg τ(x)d dx/parenrightBigg +V(x)−σ(x)λ/bracketrightBigg G(x,x/prime;ω2) =δ(x−x/prime) (3.22) a<x,x/prime<b,RBC Note thatGwill have singularities when λis a natural frequency. To eq3.19a obtain a condition which connects solutions on either side of the sin- gularity, we integrate equation 3.22. Consider ﬁgure 3.1. In this case ﬁg3.1/integraldisplayx/prime+/epsilon1 x/prime−/epsilon1dx/bracketleftBigg −d dx/parenleftBigg τ(x)d dx/parenrightBigg +V(x)−σ(x)λ/bracketrightBigg G(x,x/prime;λ) =/integraldisplayx/prime+/epsilon1 x/prime−/epsilon1δ(x−x/prime)dx which becomes pr:epsilon1 −τ(x)d dxG(x,x/prime;λ)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglex/prime+/epsilon1 x/prime−/epsilon1= 1 (3.23) since the integrals over V(x) andσ(x) vanish as ε→0. Note that in this last expression “1” has units of force. 28 CHAPTER 3. GREEN’S FUNCTIONS 3.3 Two Conditions 11 Jan p5 3.3.1 Condition 1 The previous equation can be written as d dxG/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x/prime+/epsilon1−d dxG/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x/prime−/epsilon1=−1 τ. (3.24) This makes sense after considering that a larger tension implies a smaller eq3other kink (discontinuity of ﬁrst derivative) in the string. eq3b 3.3.2 Condition 2 We also require that the string doesn’t break: G(x,x/prime)|x=x/prime+/epsilon1=G(x,x/prime)|x=x/prime−/epsilon1. (3.25) This is called the continuity condition . eq3d pr:ContCond1 3.3.3 Application To ﬁnd the Green’s function for equation 3.22 away from the point x/prime, we study the homogeneous equation pr:homog1 [L0−σ(x)λ]u(x,λ) = 0x/negationslash=x/prime,RBC. (3.26) This is called the eigen function problem. Once we specify G(x0,x/prime;λ) pr:efp1 andd dxG(x0,x/prime;λ), we may use this equation to get all higher derivatives and thus determine G(x,x/prime;λ). We know, from diﬀerential equation theory, that two fundamental solutions must exist. Let u1andu2be the solutions to [L0−σ(x)λ]u1,2(x,λ) = 0 (3.27) whereu1,2denotes either solution. Thus pr:AABB G(x,x/prime;λ) =A1u1(x,λ) +A2u2(x,λ) forx<x/prime, (3.28) and eq3ab1 3.4. OPEN STRING 29 G(x,x/prime;λ) =B1u1(x,λ) +B2u2(x,λ) forx>x/prime. (3.29) We have now deﬁned the Green’s function in terms of four constants. eq3ab2 11 Jan p6 We have two matching conditions and two R.B.C.s which determine these four constants. 3.4 Open String 13 Jan p2 where is 13 Jan p1We will solve for an open string with no external force h(x), which was ﬁrst discussed in section 1.3.2. G(x,x/prime;λ) must satisfy the boundary condition 1.18. Choose u1such that it satisﬁes the boundary condition for the left end −∂u1 ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a+Kau1(a) = 0. (3.30) This determines u1up to an arbitrary constant. Choose u2such that it satisﬁes the right end boundary condition ∂u2 ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=b+Kbu2(b) = 0. (3.31) We ﬁnd that in equations 3.28 and 3.29, A2=B1= 0. Thus we have two remaining conditions to satisfy. 13 Jan p1 We now have G(x,x/prime;λ) =A1(x/prime)u1(x,λ) forx<x/prime. (3.32) Note that only the boundary condition at aapplies since the behavior ofu1(x) does not matter at b(sinceb>x/prime). This gives Gdetermined up to an arbitrary constant. We can also write G(x,x/prime;λ) =B2(x/prime)u2(x,λ) forx>x/prime. (3.33) We also note that AandBare constants determined by x/primeonly. Thus we can write the previous expressions in a more symmetric form: pr:CD G(x,x/prime;λ) =Cu1(x,λ)u2(x/prime,λ) forx<x/prime, (3.34) G(x,x/prime;λ) =Du 1(x/prime,λ)u2(x,λ) forx>x/prime. (3.35) 30 CHAPTER 3. GREEN’S FUNCTIONS In one of the problem sets we prove that G(x,x/prime;λ) =G(x/prime,x;λ). This can also be stated as Green’s Reciprocity Principle : ‘The ampli- pr:grp1 tude of the string at xsubject to a localized force applied at x/primeis equivalent to the amplitude of the string at x/primesubject to a localized force applied at x.’ We now apply the continuity condition. Equation 3.25 implies that C=D. 13 Jan p3 Now we have a function symmetric in xandx/prime, which veriﬁes the Green’s Reciprocity Principle. By imposing the condition in equation 3.24 we will be able to determine C: dG dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x/prime+/epsilon1=Cdu1 dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x/primeu2(x/prime) (3.36) eq3trionedG dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x/prime−/epsilon1=Cu1(x/prime)du2 dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle x/prime(3.37) Combining equations (3.24), (3.36), and (3.37) gives us eq3tritwo C/bracketleftBigg u1du2 dx−du1 dxu2/bracketrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=x/prime=−1 τ(x/prime). (3.38) The Wronskian is deﬁned as pr:wronsk1 W(u1,u2)≡u1du2 dx−u2du1 dx. (3.39) This allows us to write C=1 −τ(x/prime)W(u1(x/prime,λ),u2(x/prime,λ)). (3.40) Thus 13 Jan p4 G(x,x/prime;λ) =u1(x<,λ)u2(x>,λ) −τ(x/prime)W(u1(x/prime,λ),u2(x/prime,λ)), (3.41) where we deﬁne eq3.39 pr:xless1 u(x<)≡/braceleftBigg u(x) ifx<x/prime u(x/prime) ifx/prime<x and u(x>)≡/braceleftBigg u(x) ifx>x/prime u(x/prime) ifx/prime>x. 3.5. THE FORCED OSCILLATION PROBLEM 31 Theu’s are two diﬀerent solutions to the diﬀerential equation: [L0−σ(x)λ]u1= 0 [L0−σ(x)λ]u2= 0. (3.42) Multiply the ﬁrst equation by u2and the second by u1. Subtract one FW p249 equation from the other to get −u2(τu/prime 1)/prime+u1(τu/prime 2)/prime= 0 (where we have used equation 1.10, L0=−∂ ∂x(τ∂ ∂x) +V). Rewriting this as a total derivative gives d dx[τ(x)W(u1,u2)] = 0. (3.43) This implies that the expression τ(x)W(u1(x,λ),u2(x,λ)) is indepen- but isn’t W/prime(x) = 0 also true?dent ofx. ThusGis symmetric in xandx/prime. The case in which the Wronskian is zero implies that u1=αu2, since then 0 = u1u/prime 2−u2u/prime 1, oru/prime 2/u2=u/prime 1/u1, which is only valid for all xifu1is proportional to u2. Thus ifu1andu2are linearly independent, pr:LinIndep1 the Wronskian is non-zero. 3.5 The Forced Oscillation Problem 13 Jan p5 The general forced harmonic oscillation problem can be expanded into pr:fhop1 equations having forces internally and on the boundary which are sim- ple time harmonic functions. Consider the eﬀect of a harmonic forcing term /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg u(x,t) =σ(x)f(x)e−iωt. (3.44) We apply the following boundary conditions: eq3ss −∂u(x,t) ∂x+κau(x,t) =hae−iωtforx=a, (3.45) and ∂u(x,t) ∂x+κbu(x,t) =hbe−iωt. forx=b. (3.46) We want to ﬁnd the steady state solution. First, we assume a steady state solution form, the time dependence of the solution being u(x,t) =e−iωtu(x). (3.47) 32 CHAPTER 3. GREEN’S FUNCTIONS After making the substitution we get an ordinary diﬀerential equation inx. Next determine G(x,x/prime;λ=ω2) to obtain the general steady state solution. In the second problem set we use Green’s Second Identity to solve this inhomogeneous boundary value problem. All the physics of the exciting system is given by the Green’s function. 3.6 Free Oscillation Another kind of problem is the free oscillation problem. In this case pr:fop1 f(x,t) = 0 andha=hb= 0. The object of this problem is to ﬁnd the natural frequencies and normal modes. This problem is characterized pr:NatFreq2 by the equation:/bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg u(x,t) = 0 (3.48) with the Regular Boundary Conditions : eq3fo •uis periodic. ( Closed string ) •[ˆn· ∇+KS]u= 0 for xinS. (Open string ) The goal is to ﬁnd normal mode solutions u(x,t) =e−iωntun(x). The natural frequencies are the ωnand the natural modes are the un(x). pr:NatMode1 We want to solve the eigenvalue equation [L0−σω2 n]un(x) = 0 with R.B.C. (3.49) The variable ω2 nis called the eigenvalue of L0. The variable un(x) is eq3.48 called the eigenvector (or eigenfunction) of the operator L0. pr:EVect1 3.7 Summary 1. The Principle of superposition is L0[a1u1+a2u2] =a1L0u1+a2L0u2, whereL0is a linear operator, u1andu2are functions, and a1and a2are constants. 3.7. SUMMARY 33 2. The Dirac Delta Function is deﬁned as /integraldisplayd cdxδ(x−xk) =/braceleftBigg 1 ifc<x k<d 0 otherwise . 3. Force contributions can be constructed by superposition. σ(x)f(x) =N/summationdisplay k=1Fkδ(x−x/prime k). 4. The Green’s Function is the solution to to an equation whose inhomogeneous term is a δ-function. For the Helmholtz equation, the Green’s function satisﬁes: [L0−σ(x)ω2]G(x,x/prime;ω2) =δ(x−x/prime)a<x,x/prime<b,RBC. 5. At the source point x/prime, the Green’s function satisﬁesd dxG|x=x/prime+/epsilon1− d dxG|x=x/prime−/epsilon1=−1 τandG(x,x/prime)|x=x/prime+/epsilon1=G(x,x/prime)|x=x/prime−/epsilon1. 6. Green’s Reciprocity Principle is ‘The amplitude of the string at xsubject to a localized force applied at x/primeis equivalent to the amplitude of the string at x/primesubject to a localized force applied atx.’ 7. The Green’s function for the 1-dimensional wave equation is given by G(x,x/prime;λ) =u1(x<,λ)u2(x>,λ) −τ(x/prime)W(u1(x/prime,λ),u2(x/prime,λ)). 8. The forced oscillation problem is /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg u(x,t) =σ(x)f(x)e−iωt, with periodic boundary conditions or the elastic boundary condi- tions with harmonic forcing. 9. The free oscillation problem is /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg u(x,t) = 0. 34 CHAPTER 3. GREEN’S FUNCTIONS 3.8 Reference See [Fetter81, p249] for the derivation at the end of section 3.4. A more complete understanding of the delta function requires knowl- edge of the theory of distributions, which is described in [Stakgold67a, p28ﬀ] and [Stakgold79, p86ﬀ]. The Green’s function for a string is derived in [Stakgold67a, p64ﬀ]. Chapter 4 Properties of Eigen States 13 Jan p7 Chapter Goals: •Show that for the Helmholtz equation, ω2 n>0,ω2 n is real, and the eigen functions are orthogonal. •Derive the dispersion relation for a closed massless string with discrete mass points. •Show that the Green’s function obeys Hermitian analyticity. •Derive the form of the Green’s function for λnear an eigen value λn. •Derive the Green’s function for the ﬁxed string problem. By deﬁnition 2.1 ω2>0 /angbracketleftS,u/angbracketright=/integraldisplayb adxS∗(x)u(x). (4.1) In section 2.4 we saw (using Green’s ﬁrst identity) for L0as deﬁned in equation 2.2, and for all uwhich satisfy equation 1.26, that V > 0 implies /angbracketleftu,L 0u/angbracketright>0 . We choose u=unand use equation 3.49 so that 0</angbracketleftun,L0un/angbracketright=/angbracketleftun,σu n/angbracketrightω2 n. (4.2) 35 36 CHAPTER 4. PROPERTIES OF EIGEN STATES Remember that σsigniﬁes the mass density, and thus σ > 0. So we conclude ω2 n=/angbracketleftun,L0un/angbracketright /angbracketleftun,σu n/angbracketright>0. (4.3) This all came from Green’s ﬁrst identity. 13 Jan p8 Next we apply Green’s second identity 2.5,ω2real /angbracketleftS,L 0u/angbracketright=/angbracketleftL0S,u/angbracketright forS,usatisfying RBC (4.4) LetS=u=un. This gives us ω2 n/angbracketleftun,σu n/angbracketright=/angbracketleftun,L0un/angbracketright (4.5) =/angbracketleftL0un,un/angbracketright (4.6) = (ω2 n)∗/angbracketleftun,σu n/angbracketright. (4.7) We used equation 3.49 in the ﬁrst equality, 2.5 in the second equality, and both in the third equality. From this we can conclude that ω2 nis real. Now let us choose u=unandS=um. This gives us orthogonality /angbracketleftum,L0un/angbracketright=/angbracketleftL0um,un/angbracketright. (4.8) Extracting ω2 ngives (note that σ(x) is real) ω2 n/angbracketleftum,σu n/angbracketright=ω2 m/angbracketleftσum,un/angbracketright=ω2 m/angbracketleftum,σu n/angbracketright. (4.9) So (ω2 n−ω2 m)/angbracketleftum,σu n/angbracketright= 0. (4.10) Thus ifω2 n/negationslash=ω2 mthen/angbracketleftum,σu n/angbracketright= 0: /integraldisplayb adxu∗ m(x)σ(x)un(x) = 0 if ω2 n/negationslash=ω2 m. (4.11) That is, two eigen vectors umandunofL0corresponding to diﬀerent eq4.11 13 Jan p8 eigenvalues are orthogonal with respect to the weight function σ. If pr:ortho1the eigen vectors umandunare normalized, then the orthonormality pr:orthon1condition is /integraldisplayb adxu∗ m(x)σ(x)un(x) =δmn ifω2 n/negationslash=ω2 m, (4.12) where the Kronecker delta function is 1 if m=nand 0 otherwise. eq4.11p 4.1. EIGEN FUNCTIONS AND NATURAL MODES 37 u u1 2 Na LLuZZZ uPPPuhhu u u u ((uLLu ZZZu PPPu hhuu uu((u Figure 4.1: The closed string with discrete mass points. 4.1 Eigen Functions and Natural Modes 15 Jan p1 We now examine the natural mode problem given by equation 3.49. To ﬁnd the natural modes we must know the natural frequencies ωnand pr:NatMode2 the normal modes un. This is equivalent to the problem pr:lambdan1 L0un(x) =σ(x)λnun(x), RBC. (4.13) To illustrate this problem we look at a discrete problem. eq4A 4.1.1 A Closed String Problem pr:dcs1 15 Jan p2This problem is illustrated in ﬁgure 4.11. In this problem the mass ﬁg4wdensityσand the tension τare constant, and the potential Vis zero. The termu(xi) represents the perpendicular displacement of the ith mass point. The string density is given by σ=m/a wheremis the mass of each mass point and ais a unit of length. We also make the deﬁnitionc=/radicalBig τ/σ. Under these conditions equation 1.1 becomes m¨ui=Ftot=τ a(ui+1+ui−1−2ui). Substituting the solution form ui=eikxieiωtinto this equation gives mω2= 2τ a/parenleftBigg −eika+e−ika 2+ 1/parenrightBigg = 2τ a(1−coska) = 4τ asin2ka 2. But continuity implies u(x) =u(x+Na), soeikNa= 1, orkNa = 2πn, so that k=/parenleftbigg2π Na/parenrightbigg n, n = 1,...,N. 1See FW p115. 38 CHAPTER 4. PROPERTIES OF EIGEN STATES The natural frequencies for this system are then ω2 n=c2sin2(kna/2) a2/4(4.14) wherekn=2π Lnandncan take on the values 0 ,±1,...,±N−1 2for odd eq4r N, and 0,±1,...,±N 2−1,+N 2for evenN. The constant is c2=aτ/m . Equation 4.14 is called the dispersion relation . Ifnis too large, the un pr:DispRel1 take on duplicate values. The physical reason that we are restricted to a ﬁnite number of natural modes is because we cannot have a wavelength λ<a . The corresponding normal modes are given by φn(xi,t) =e−iωntun(xi) (4.15) =e−i[ωnt−knxi](4.16) =e−i[ωnt−2π Lnxi]. (4.17) The normal modes correspond to traveling waves. Note that ωnis eq4giver pr:travel1 doubly degenerate in equation 4.14. Solutions φn(x) fornwhich are larger than allowed give the same displacement of the mass points, but with some nonphysical wavelength. Thus we are restricted to Nmodes and a cutoﬀ frequency. 4.1.2 The Continuum Limit We now let abecome increasingly small so that Nbecomes large for L ﬁxed. This gives us ∆ kn= 2π/L forL=Na. In the continuum limit, the number of normal modes becomes inﬁnite. Shorter and shorter wavelengths become physically relevant and there is no cutoﬀ frequency. pr:cutoﬀ1 Lettingaapproach zero while Lremains ﬁxed gives frequency expression ω2 n=c2sin2kna 2 a2 4Na=L a→0−→c2k2 n (4.18) and so ωn=c|kn| (4.19) ∆ωn=c|∆kn|=c/vextendsingle/vextendsingle/vextendsingle/vextendsingle2π L/vextendsingle/vextendsingle/vextendsingle/vextendsingle (4.20) 4.1. EIGEN FUNCTIONS AND NATURAL MODES 39 ωn=c2π Ln. (4.21) Equation (4.17) gives us the un’s for alln. We have found characteristics •For a closed string, the two eigenvectors for every eigenvalue (called degeneracy ) correspond to the two directions in which a pr:degen1 wave can move. The eigenvalues are ω2 n. •The natural frequencies ωnare always discrete, with a separation distance proportional to 1 /L. •For open strings there is no degeneracy. This is because the po- sition and slope of the Green’s function at the ends is ﬁxed by the open string boundary conditions, whereas the closed string boundary conditions do not determine the Green’s function at any particular point. For the discrete closed case, the ωn’s are discrete with double degeneracy, 15 Jan p4 givingu±. We also ﬁnd the correspondence ∆ ωn∼c/Lwherec∼/radicalBig τ/σ. We also found that there is no degeneracy for the open discrete case. 4.1.3 Schr¨ odinger’s Equation pr:Schro1 Consider again equation 4.13 L0un(x) =σ(x)λnun(x) RBC (4.22) where L0=−d dxτ(x)d dx+V(x). (4.23) We now consider the case in which τ(x) = ¯h2/2mandσ= 1, both quantities being numerical constants. The linear operator now becomes L0=−¯h2 2md2 dx2+V(x). (4.24) 40 CHAPTER 4. PROPERTIES OF EIGEN STATES xV(x) E E EE D D D C CC BB AAQQ Figure 4.2: Negative energy levels This is the linear operator for the Schr¨ odinger equation for a particle of massmin a potential V: /bracketleftBigg−¯h2 2md2 dx2+V(x)/bracketrightBigg un(x) =λnun(x) + RBC (4.25) In this case λgives the allowed energy values. The potential V(x) can be either positive or negative. It needs to be positive for L0to be positive deﬁnite, in which case λ0>0. For V < 0 we can have a ﬁnite number of of eigenvalues λless that zero. On a physical string the condition that V > 0 is necessary. verify this paragraph 15 Jan p5One can prove that negative energy levels are discrete and bounded from below. The bound depends on the nature of V(x) (Rayleigh quo- tient idea). Suppose that Vhas a minimum, as shown in ﬁgure 4.2 forpr:RayQuo1 example. By Green’s ﬁrst identity the quantity L0−Vmingives a newﬁg4negoperator which is positive deﬁnite. 4.2 Natural Frequencies and the Green’s Function We now look at a Green’s function problem in the second problem set. Under consideration is the Fredholm equation pr:Fred1 [L0−σ(x)ω2 n]u(x) =σ(x)f(x), RBC. (4.26) 4.3. GF BEHAVIOR NEAR λ=λN 41 In problem 2.2 one shows that the solution un(x) for this equation only exists if /integraldisplay dxu∗ n(x)σ(x)f(x) = 0. (4.27) This is the condition that the eigenvectors un(x) are orthogonal to eq4.26 the function f(x). We apply this to the Green’s function. We choose λ=λn=ω2 nand evaluate the Green’s function at this point. Thus for Ask Baker Isn’t this just hitting a sta- tionary point?the equation [L0−σ(x)λn]G(x,x/prime;λ) =δ(x−x/prime) (4.28) there will be a solution G(x,x/prime;λ) only if (using 4.27)eq4.26p clarify this. G(x,x;λ) =/integraldisplay dx/primeG(x,x/prime;λ)δ(x−x/prime) = 0. (4.29) The resultG(x,x;λ) = 0 implies that u∗ n(x/prime) = 0 (using equation 3.41). eq4.26b 15 Jan p6 There will be no solution unless x/primeis a node. In physical terms, this means that a natural frequency can only be excited at a node. 4.3 GF behavior near λ=λn 15 Jan p6 From the result of the previous section, we expect that if the driving frequencyωis not a natural frequency, everything will be well behaved. So we show that G(x,x/prime;λ) is good everywhere (in the ﬁnite interval [a,b]) except for a ﬁnite number of points. The value of G(x,x/prime;λ) becomes inﬁnite near λn, that is, as λ→λn. Forλnearλnwe can write the Green’s function as pr:gnxx1 G(x,x/prime;λ)∼1 λn−λgn(x,x/prime) + ﬁnite λ→λn (4.30) where ﬁnite is a value always of ﬁnite magnitude. We want to ﬁnd gn, so we put [ L0−λσ(x)] in front of each side of the equation and then add and subtract λnσ(x)G(x,x/prime;λ) on the the right-hand side. This gives us (using 4.28) δ(x−x/prime) = [L0−λnσ(x)]/bracketleftbigg1 λn−λgn(x,x/prime) + ﬁnite/bracketrightbigg + (λn−λ)σ(x)/bracketleftbigg1 λn−λgn(x,x/prime) + ﬁnite/bracketrightbigg λ→λn = [L0−λnσ(x)]/bracketleftbigg1 λn−λgn(x,x/prime)/bracketrightbigg + ﬁnite λ→λn. 42 CHAPTER 4. PROPERTIES OF EIGEN STATES The left-hand side is also ﬁnite if we exclude x=x/prime. This can only 15 Jan p7 occur if [L0−λnσ(x)]gn(x,x/prime) = 0x/negationslash=x/prime. (4.31) From this we can conclude that gnhas the form gn(x,x/prime) =un(x)f(x/prime), x /negationslash=x/prime, (4.32) wheref(x/prime) is a ﬁnite term and un(x) satisﬁes [L0−λnσ(x)]un(x) = 0 eq4guf pr:fxprime1 with RBC. Note that here the eigen functions un(x) are not yet nor- malized. This is the relation between the natural frequency and thepr:NatFreq3 Green’s function. 4.4 Relation between GF & Eig. Fn. 20 Jan p2 We continue developing the relation between the Green’s function and spectral theory. So far we have discussed the one dimensional problem. pr:SpecThy1 This problem was formulated as [L0−σ(x)λ]G(x,x/prime;λ) =δ(x−x/prime), RBC. (4.33) To solve this problem we ﬁrst solved the corresponding homogeneous eq4.31m problem [L0−σ(x)λn]un(x) = 0, RBC. (4.34) The eigenvalue λnis called degenerate if there is more than one unper λn. We note the following properties in the Green’s function: 1.G∗(x,x/prime;λ∗) =G(x,x/prime;λ). Recall that σ,L0, and the boundary condition terms are real. First we take the complex conjugate of equation 4.33, [L0−σ(x)λ∗]G∗(x,x/prime;λ) =δ(x−x/prime), RBC (4.35) and then we take the complex conjugate of λto get [L0−σ(x)λ]G∗(x,x/prime;λ∗) =δ(x−x/prime), RBC (4.36) which gives us G∗(x,x/prime;λ∗) =G(x,x/prime;λ). (4.37) 4.4. RELATION BETWEEN GF & EIG. FN. 43 2.Gis symmetric. In the second problem set it was seen that G(x,x/prime;λ) =G(x/prime,x;λ). (4.38) 3. The Green’s function Ghas the property of Hermitian analyticity. 18 Jan p3 pr:HermAn1 By combining the results of 1 and 2 we get G∗(x,x/prime;λ) =G(x/prime,x;λ∗). (4.39) This may be called the property of Hermitian analyticity . eq4ha In the last section we saw that G(x,x/prime;λ)λ→λn−→gn(x,x/prime) λn−λ(4.40) forgsuch that eq4cA [L0−σ(x)λn]gn(x,x/prime) = 0. (4.41) For the open string there is no degeneracy and for the closed string there is double degeneracy. (There is also degeneracy for the 2- and give ref 3-dimensional cases.) 4.4.1 Case 1: λNondegenerate Assume that λnis non-degenerate. In this case we can write (using equation 4.32) gn(x,x/prime) =un(x)fn(x/prime). (4.42) Hermitian analyticity and the complex conjugate of equation 4.40 give eq4cB (note thatλn∈R) g∗ n(x,x/prime) λn−λ∗=gn(x/prime,x) λn−λ∗(4.43) asλ→λn. This implies that g∗ n(x,x/prime) =gn(x/prime,x). (4.44) eq4cC So now 4.42 becomes 20 Jan p4 44 CHAPTER 4. PROPERTIES OF EIGEN STATES u∗ n(x)f∗ n(x/prime) =fn(x)un(x/prime) (4.45) so that (since xandx/primeare independent) if un(x) is normalized (accord- ing to equation 4.12) fn(x) =u∗ n(x). (4.46) In the non-degenerate case we have (from 4.40 and 4.42) G(x,x/prime;λ)λ→λn−→un(x)u∗ n(x/prime) λn−λ. (4.47) whereun(x) are normalized eigen functions. eq4.47 pr:normal1 4.4.2 Case 2: λnDouble Degenerate In the second case, the eigenvalue λnhas double degeneracy like the closed string. The homogeneous closed string equation is is it +? (L0+λnσ)u(±) n(x) = 0. The eigenfunctions corresponding to λnareu+ n(x) andu− n(x). By using the same reasoning that lead to equation 4.47 we can write why? G(x,x/prime;λ)→1 λn−λ[u(+) n(x)u(+)∗ n(x) +u(−) n(x)u(−)∗ n(x)] (4.48) for the equation [L0−σ(x)λn]u(±) n(x) = 0, RBC. (4.49) The eigenfunction unmay be written as un=A+u++A−u−. Double How does this ﬁt in? degeneracy is the maximum possible degeneracy in one dimension. In the general case of α-fold degeneracy 20 Jan p5 G(x,x/prime;λ)λ→λn−→1 λn−λ/summationdisplay α[uα n(x)uα∗ n(x/prime)] (4.50) whereuα n(x) solves the equation [L0−σ(x)λn]uα n(x) = 0, RBC. (4.51) The mathematical relation between the Green’s function and the eigen functions is the following: The eigenvalues λnare the poles of G. pr:poles1 The sum of bilinear products/summationtext αuα n(x)uα∗ n(x/prime) is the residue of the pole λ=λn. 4.5. SOLUTION FOR A FIXED STRING 45 4.5 Solution for a Fixed String We want to solve equation 3.22. Further, we take V= 0,σandτ constant, and a= 0,b=L. /bracketleftBigg −τd2 dx2−λσ/bracketrightBigg G(x,x/prime;λ) =δ(x−x/prime) for 0 <x,x/prime<L (4.52) for the case of a ﬁxed end string. Our boundary conditions are G(x,x/prime;λ) = 0 for x=a,b. 4.5.1 A Non-analytic Solution We know from equation 3.41 the solution is 20 Jan p6 G(x,x/prime;λ) =u1(x<,λ)u2(x>,λ) −τW(u1,u2). (4.53) This solution only applies to the one dimensional case. This is because eq4.54 the solution was obtained using the theory of ordinary diﬀerential equa- tions. The corresponding homogeneous equations are given by pr:homog2 /bracketleftBiggd2 dx2+λ c2/bracketrightBigg u1,2(x,λ) = 0. (4.54) In this equation we have used the deﬁnition 1 /c2≡σ/τ. The variables eq4lcu pr:c1 u1andu2also satisfy the conditions u1(0,λ) = 0 and u2(L,λ) = 0. (4.55) The solution to this homogeneous problem can be found to be u1= sin/radicalBigg λ c2x andu2= sin/radicalBigg λ c2(L−x). (4.56) In these solutions√ λappears. Since λcan be complex, we must deﬁne a branch cut2. pr:branch1 46 CHAPTER 4. PROPERTIES OF EIGEN STATES u ReλImλ θλ Figure 4.3: The θ-convention 4.5.2 The Branch Cut SinceG∼1 λn−λ(see 4.40) it follows (from λn>0) thatGnis analytic pr:analytic1 for Re (λ)<0. As a convention, we choose θsuch that 0 < θ < 2π. This is illustrated in ﬁgure 4.3. ﬁg4one Using this convention,√ λcan be represented by √ λ=/radicalBig |λ|eiθ/2(4.57) =/radicalBig |λ|/bracketleftBigg cosθ 2+isinθ 2/bracketrightBigg . (4.58) Note that√ λhas a discontinuity along the positive real axis. The eq4.58-59 20 Jan p7 functionGis analytic in the complex plane if the positive real axis is removed. This can be expressed mathematically as the condition that Im√ λ>0. (4.59) 4.5.3 Analytic Fundamental Solutions and GF We now look back at the ﬁxed string problem. We found that pr:u1.1 u1(x,λ) = sin/radicalBigg λ c2x so that du1 dx=/radicalBigg λ c2cos/radicalBigg λ c2x. This gives us the boundary value but only if x= a= 0.2See also FW p485. 4.5. SOLUTION FOR A FIXED STRING 47 du1 dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a=/radicalBigg λ c2. (4.60) Because of the√ λ, this is not analytic over the positive real axis. We eq4fs choose instead the solution pr:baru1 u1→u1 du1 dx|x=a=1/radicalBig λ c2sin/radicalBigg λ c2x≡u1. (4.61) The function u1has the properties u1(a) = 0 anddu1 dx/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a= 1. This satisﬁes the diﬀerential equation. 18 Jan p8 /bracketleftBiggd2 dx2+λ c2/bracketrightBigg u1(x,λ) = 0. (4.62) Sou1is analytic for λwith no branch cut. Similarly, for the substitution justify this u2≡u2/(du2 dx|x=b) we obtain u2= (λ/c2)−1/2sin/radicalBig λ/c2(L−x). One can always ﬁnd u1andu2as analytic functions of λwith no branch cut. Is this always true? 4.5.4 Analytic GF for Fixed String We have been considering the Green’s function equation [L0−λσ(x)]G(x,x/prime;λ) =δ(x−x/prime) for 0<x,x/prime<b (4.63) with the open string RBC (1.18) [ˆnS· ∇+κS]G(x,x/prime;λ) = 0 for xonS (4.64) and linear operator L0deﬁned as L0=−d dxτ(x)d dx+V(x). (4.65) 48 CHAPTER 4. PROPERTIES OF EIGEN STATES We found that the solution to this equation can be written (3.41) G(x,x/prime;λ) =u1(x<,λ)u2(x>,λ) −τ(x)W(u1,u2)(4.66) whereu1andu2are solutions to the homogeneous equation with the eq4cD same boundary conditions as (1.18) [L0−λσ(x)]u1,2(x,λ) = 0 for a<x<b (4.67) −∂ ∂xu1(x,λ) +kau1(x,λ) = 0 for x=a (4.68) +∂ ∂xu2(x,λ) +kbu2(x,λ) = 0 for x=b. (4.69) We have been calculating the Green’s function for a string with ﬁxed tension ( τ= constant) and ﬁxed string density ( σ= constant) in the absence of a potential ﬁeld ( V= 0) and ﬁxed end points. This last condition implies that the Green’s function is restricted to the boundary condition that G= 0 atx=a= 0 andx=b=L. We saw that u1= sin/radicalBigg λ c2x (4.70) and u2= sin/radicalBigg λ c2(L−x). (4.71) We also assigned the convention that 22 Jan p2 √ λ=/radicalBig |λ|eiθ/2. (4.72) This is shown in ﬁgure 4.3. The Wronskian in equation 4.53 for this problem can be simpliﬁed as W(u1,u2) =u1∂u2 ∂x−u2∂u1 ∂x =/radicalBigg λ c2 −sin/radicalBigg λ c2xcos/radicalBigg λ c2(L−x) 4.5. SOLUTION FOR A FIXED STRING 49 −sin/radicalBigg λ c2(L−x) cos/radicalBigg λ c2x  =−/radicalBigg λ c2sin/radicalBigg λ c2L. (4.73) Thus we can write the full solution for the ﬁxed string problem as eq4.75b G(x,x/prime;λ) =sin/radicalBig λ c2x<sin/radicalBig λ c2(L−x>) τ/radicalBig λ c2sin/radicalBig λ c2L. (4.74) eq4ss 4.5.5 GF Properties We may now summarize the properties of Gin the complex λplane. •Branch Cut :Ghas no branch cut. It is analytic except at isolated simple poles. The λ1/2branch vanishes if the eigen functions are properly chosen. This is a general result for discrete spectrum. pr:DiscSpec1 Justify•Asymptotic limit :Ggoes to zero as λgoes to inﬁnity. If λispr:asymp1real and we do not go through the poles, this result can be seen immediately from equation 4.74. For complex λ, we let |λ| → ∞ and use the deﬁnition stated in equation 4.57 which is valid for 0<θ< 2π. This deﬁnition gives Im√ λ>0. We can then write sin/radicalBig λ/c2x =ei√ λ/c2x−e−i√ λ/c2x 2i(4.75) |λ|→∞−→e−i√ λ/c2x 2i(4.76) forθ>0. Thus from equation 3.44 we get 23 Jan p3 G(x,x/prime;λ)|λ|→∞−→ −1 2ie−i√ λ/c2x<e−i√ λ/c2(L−x>) τ/radicalBig λ/c2e−i√ λ/c2L(4.77) =−1 2ie+i√ λ/c2(x>−x<) τ/radicalBig λ/c2. (4.78) 50 CHAPTER 4. PROPERTIES OF EIGEN STATES By convention x>−x<>0, and thus we conclude G(x,x/prime;λ)|λ|→∞−→0. (4.79) eq4.81a •Poles : The Green’s function can have poles. The Green’s function is a ratio of analytic functions. Thus the poles occur at the zeros of the denominator. We now look at sin/radicalBig λn c2L= 0 from the denominator of equation 3.44. The poles are at λ=λn. We can write/radicalBig λn/c2L=nπor λn=/parenleftbiggcnπ L/parenrightbigg2 forn= 1,2,... (4.80) We delete the case n= 0 since we have a removable singularity at eq4et λ= 0. Equation 4.80 occurs when the Wronskian vanishes. This happens when u1= constant ×u2(not linearly independent). Both u1andu2satisfy the boundary conditions at both boundaries and are therefore eigenfunctions. Thus the un’s are eigenfunctions and the λn’s are the eigenvalues. So 23 Jan p4 [L0−λnσ]un(x) = 0 RBC (4.81) is satisﬁed for λnbyun. eq4star 4.5.6 The GF Near an Eigenvalue We now look at equation 4.74 near an eigenvalue. First we expand the denominator in a power series about λ=λn: sin/radicalBigg λ c2L=(λ−λn)L ccos√λnL c 2√λn+O(λ−λn)2. (4.82) So forλnear an eigenvalue we have eq4.85 τ/radicalBigg λ c2sin/radicalBigg λ c2Lλ→λn−→τL 2c2(λ−λn) cosnπ. (4.83) 4.6. DERIVATION OF GF FORM NEAR E.VAL. 51 eq4.86 Now we look at the numerator of 4.74. We can rewrite sin/radicalBig λ/c2(L−x>) =−sinnπx >cos/radicalBig λ/c2L. (4.84) Note thatf(x<)f(x>) =f(x)f(x/prime). So, with σ=τ/c2, and substitut- eq4.87 ing 4.83 and 4.84, equation 4.74 becomes G(x,x/prime;λ)λ→λn−→2 σLsinnπx Lsinnπx/prime L λn−λ. (4.85) So we conclude that Gλ→λn−→un(x)un(x/prime) λn−λ(4.86) as in 4.47 where the eigenfunction is eq4cC2 22 Jan p5 un(x) =/radicalBigg 2 σLsinnπx L, (4.87) which satisﬁes the completeness relation/integraltextL 0um(x)u∗ n(x)σdx=δmn. eq4.91 pr:CompRel1 4.6 Derivation of GF form near E.Val. 4.6.1 Reconsider the Gen. Self-Adjoint Problem We now give an indirect proof of equation 4.86 based on the speciﬁc Green’s function deﬁned in equation 4.53, G(x,x/prime;λ) =u1(x<,λ)u2(x>,λ) −τ(x)W(u1,u2). (4.88) The boundary conditions are (see 1.18) eq4nt −∂u1 ∂x+kau1= 0 for x=a= 0 (4.89) +∂u2 ∂x+kbu2= 0 for x=b=L. (4.90) The function u1(respectively u2) may be any solution which is an analytic function of λ, and independent of λatx=a(respectively 52 CHAPTER 4. PROPERTIES OF EIGEN STATES x=b). Thus both the numerator and the denominator of equation 4.88 are analytic functions of λ, so there is no branch cut. Note that 22 Jan p6 G(x,x/prime;λ) may only have poles when W(u1(x,λ)u2(x,λ)) = 0, which only occurs when u1(x,λ n) =dnu2(x,λ n). (4.91) where thednare constants. Look at the Green’s function near λ=λn. Finding the residue will give the correct normalization. We have (using 4.73 and 4.82) τ(x)W(u1,u2)λ→λn−→(λ−λn)cn, (4.92) wherecnis some normalization constant. In this limit equation 4.88 becomes G(x,x/prime;λ)λ→λn−→1 dn¯un(x<)¯un(x>) (λ−λn)cn(4.93) wherednis some constant. So un≡¯un/radicalBig 1 cndnis the normalized eigen- function. Equation 4.88 then implies G(x,x/prime;λ)λ→λn−→un(x)u∗ n(x/prime) (λ−λn)(4.94) whereunsatisﬁes equation 4.81. 4.6.2 Summary, Interp. & Asymptotics 25 Jan p1 In the previous sections we looked at the eigenvalue problem [L0−λnσ(x)]un(x) = 0 for a<x<b , RBC (4.95) and the Green’s function problem [L0−λnσ(x)]G(x,x/prime;λ) =δ(x−x/prime) fora<x,x/prime<b, RBC (4.96) where eq4fooo L0=−d dx/parenleftBigg τ(x)d dx/parenrightBigg +V(x), (4.97) which is a formally self-adjoint operator. The general problem requires pr:selfadj1 4.7. GENERAL SOLUTION FORM OF GF 53 ﬁnding an explicit expression for the Green’s function for a force local- ized atx/prime. Forλ→λnwe found G(x,x/prime;λ)→un(x)u∗ n(x/prime) λn−λ. (4.98) This equation shows the contribution of the nth eigenfunction. We saw eq4.102 thatGis an analytic function of λ(with poles at λn) whereG→0 as |λ| → ∞ . We can think of Gas the inverse operator of L0−λnσ: G=1 L0−λnσ. (4.99) Thus the poles are at L0=λnσ. For largeλthe behavior of Gis determined by thed2 dx2Gterm since it brings down the highest power of λ. Thus for our simple example of τconstant, L0≈ −τd2 dx2(4.100) and forλlarge G∼exp(i/radicalBig λ/c2x), G/prime∼√ λG, G/prime/prime∼(√ λ)2G (4.101) where the derivatives are taken with respect to x. 4.7 General Solution form of GF 25 Jan p2 ﬁg4ﬀ In this section we obtain a general form (equation 4.108) for the Green’s function which is constructed using the solutions to the corresponding eigen value equation. This is done by evaluating a particular complex integral. We have seen that G(x,x/prime;λ) is analytic in the complex λ- plane except for poles on the real axis at the eigen values λn. We consider the following complex integral /contintegraldisplay c1+c2dλ/primeG(x,x/prime;λ/prime) λ/prime−λ≡/contintegraldisplay c1+c2dλ/primeF(λ/prime) (4.102) where we have deﬁned F(λ/prime)≡G(x,x/prime;λ/prime)/(λ/prime−λ). Let the contour of integration be the contour illustrated in ﬁgure 4.4. This equation has 54 CHAPTER 4. PROPERTIES OF EIGEN STATES λ/prime-plane Reλ/primeImλ/prime C1- PPPQQQJJJBBB PPPQQQ J JJ B BBC2QQ k 3 q λS Figure 4.4: The contour of integration a singularity only at λ. Thus we need only integrate on the contour pr:singular1 aroundλ. This is accomplished by deforming the contour C1+C2to the contour S(following Cauchy’s theorem) pr:Cauchy1 /contintegraldisplay C1+C2dλ/primeF(λ/prime) =/contintegraldisplay Sdλ/primeF(λ/prime). (4.103) See ﬁgure 4.5. Note that although F(λ/prime) blows up as λ/primeapproaches λ, G(x,x/prime;λ/prime)→G(x,x/prime;λ) in this limit. Thus the integration about the small circle around λcan be written /contintegraldisplay Sdλ/primeF(λ/prime)λ/prime→λ−→G(x,x/prime;λ)/contintegraldisplay Sdλ/prime λ/prime−λ. Now make the substitution λ/prime−λ=/epsilon1eiα(4.104) dλ/prime=i/epsilon1eiαdα (4.105) dλ/prime λ/prime−λ=idα. (4.106) This allows us to write /contintegraldisplay Sdλ/prime λ/prime−λ=ilim /epsilon1→0/integraldisplay2π 0dα= 2πi. 4.7. GENERAL SOLUTION FORM OF GF 55 "!# qε λ Figure 4.5: Circle around a singularity. We conclude /contintegraldisplay Sdλ/primeF(λ/prime) = 2πiG(x,x/prime;λ) and thus 2πiG(x,x/prime;λ) =/contintegraldisplay C1+C2dλ/primeG(x,x/prime;λ/prime) λ/prime−λ. (4.107) eq4tpig We now assume that G(x,x/prime;λ)→0 asλ→ ∞ . We must check this for each example we consider. An intuitive reason for this limit is the following. The Green’s function is like the inverse of the diﬀerential operator:G∼1/(L0−λσ). Thus as λbecomes large, Gmust vanish. This assumption allows us to evaluate the integral around the large circleC2. We parameterize λ/primealong this contour as λ/prime−λ=Reiα, dλ/prime→Reiαidα asR→ ∞. So lim R→∞/contintegraldisplay C2G(x,x/prime;λ/prime) λ/prime−λdλ/prime= lim R→∞/integraldisplay2π 0Reiαdα ReiαG(x,x/prime;Reiθ)→0 sinceG(x,x/prime;Reiθ)→0 asR→ ∞ . ﬁg4fof We now only need to evaluate the integral for the contour C1. /contintegraldisplay C1dλ/primeG(x,x/prime;λ/prime) λ/prime−λ=/summationdisplay n/contintegraldisplay cndλ/primeG(x,x/prime;λ/prime) λ/prime−λ. In this equation we replaced the contour C1by a sum of contours around the poles, as shown in ﬁgure 4.6. Recall that ﬁg4.6 G(x,x/prime;λ/prime)λ/prime→λn−→/summationtext αuα n(x)uα∗ n(x/prime) λn−λ/prime. 56 CHAPTER 4. PROPERTIES OF EIGEN STATES C1- qλ1qλ2qλ3qλ4... = q λ1c1 q λ2c2 q λ3c3 q λ4c4 ... Figure 4.6: Division of contour. We note that /contintegraldisplay cndλ/prime (λ/prime−λ)(λn−λ/prime)=1 λn−λ/contintegraldisplay cndλ/prime λn−λ/prime =1 λn−λ/contintegraldisplaydλ/prime λ/prime−λn =1 λn−λ2πi. The ﬁrst equality is valid since 1 /(λ/prime−λ) is well behaved as λ/prime→λn. The last equality follows from the same change of variables performed above. The integral along the small circle containing λnis thus /contintegraldisplay cndλ/prime λ/prime−λG(x,x/prime;λ/prime) =2πi λn−λ/summationdisplay αuα n(x)uα∗ n(x/prime). The integral along the contour C1(and thus the closed contour C1+C2) is then /contintegraldisplay C1dλ/prime λ/prime−λG(x,x/prime;λ/prime) = 2πi/summationdisplay n/summationtext αuα n(x)uα∗ n(x/prime) λn−λ. Substituting equation 4.107 gives the result G(x,x/prime;λ) =/summationdisplay n1 λn−λ/parenleftBigg/summationdisplay αuα λn(x)uα∗ λn(x/prime)/parenrightBigg , (4.108) where the indices λnsumn= 1,2,...andαsums over the degeneracy. eq4.124 4.7. GENERAL SOLUTION FORM OF GF 57 4.7.1δ-fn Representations & Completeness Using the above result (equation 4.108) we can write δ(x−x/prime) = [L0−λσ(x)]G(x,x/prime;λ) (4.109) = [L0−λσ(x)]∞/summationdisplay n=1un(x)u∗ n(x/prime) λn−λ(4.110) =∞/summationdisplay n=1[(L0−λnσ) + (λn−λ)σ]un(x)u∗ n(x/prime) λn−λ(4.111) =∞/summationdisplay n=1σun(x)u∗ n(x/prime). (4.112) In the last equality we used ∞/summationdisplay n=1(L0−λnσ)un(x)u∗ n(x/prime) λn−λ=∞/summationdisplay n=1un(x/prime) λn−λ(L0−λnσ)un(x) = 0 sinceL0is a diﬀerential operator in terms of x, From this we get the completeness relation pr:CompRel2 δ(x−x/prime) =∞/summationdisplay n=1σ(x)un(x)u∗ n(x/prime) (4.113) or δ(x−x/prime) σ(x)=∞/summationdisplay n=0un(x)u∗ n(x/prime). (4.114) This is called the completeness relation because it is only true if the eq4.128b unare a complete orthonormal set of eigenfunctions, which means that anyf(x) can be written as a sum of the un’s weighted by the projection off(x) onto them. This notion is expressed by the expansion theorem (4.117 and 4.118). We now derive the expansion theorem. Consider pr:ExpThm1 25 Jan p6 f(x) =/integraldisplayb adx/primef(x/prime)δ(x−x/prime) fora<x<b (4.115) =/integraldisplayb adx/primef(x/prime)σ(x/prime)∞/summationdisplay n=1un(x)u∗ n(x/prime) (4.116) So eq4.129 58 CHAPTER 4. PROPERTIES OF EIGEN STATES f(x) =∞/summationdisplay n=1un(x)fn, (4.117) where eq4fo32 fn=/integraldisplayb adx/primeu/prime(x/prime)σ(x/prime)f(x/prime) (4.118) is the generalized nth Fourier coeﬃcient for f(x). Equation 4.117 rep- eq4.132 pr:FourCoef1 resents the projection of f(x) onto theun(x) normal modes. This was obtained using the completeness relation. Now we check normalization. The Green’s function Gis normalized pr:normal2 because the u’s are normalized. We check the normalization of the u’s by looking at the completeness relation δ(x−x/prime) =σ(x/prime)/summationdisplay nun(x)u∗ n(x/prime). (4.119) Integrate both sides by/integraltextdx/primeum(x/prime). On the left hand side we immedi- ately obtain/integraltextdx/primeum(x/prime)δ(x−x/prime) =um(x). On the right hand side /integraldisplay dx/primeum(x/prime)σ(x/prime)/summationdisplay nun(x)u∗ n(x/prime) =/summationdisplay nun(x)/integraldisplay dx/primeu∗ n(x/prime)σ(x/prime)um(x/prime) where we used 4.11 in the equality. But um(x) =/summationdisplay nun(x)δn,m. Thus we conclude that normalized eigen functions are used in the com- pleteness relation: /integraldisplayb adxu∗ m(x)σ(x)un(x) =δn,m. (4.120) This is the condition for orthonormality . pr:orthon2 4.8 Extension to Continuous Eigenvalues 27 Jan p1 AsL(the length of the string) becomes large, the eigen values become27 Jan p2closer together. The normalized eigen functions un(x) and eigen values λnfor the ﬁxed string problem, equation 4.54, can be written un=1√ σLe±i√ λn/c2x(4.121) 4.9. ORTHOGONALITY FOR CONTINUUM 59 and factor of 2? λn= (ckn)2kn=2πn Ln= 0,1,2,.... (4.122) The separation between the eigen values is then ∆λn=λn−λn−1∼c2/parenleftbigg2π L/parenrightbigg2/parenleftBig n2−(n−1)2/parenrightBigL→∞−→0. (4.123) We now consider the case of continuous eigen values. Let λbe complex (as before) and let ∆ λn→0. This limit exists as long as λis not on the positive real axis, which means that the denominator will not blow up as ∆ λn→0. In the continuum case equation 4.108 becomes lim ∆λn→0G(x,x/prime;λ) =/integraldisplaydλn λn−λ/summationdisplay αuα λn(x)uα∗ λn(x). (4.124) eq4.124ab The completeness relation, equation 4.114, becomes δ(x−x/prime) σ(x/prime)=/integraldisplay dλn/summationdisplay αuλn(x)u∗ λn(x/prime). (4.125) Now we take any function f(x) and express it as a superposition using theδ-function representation (in direct analogy with equations 4.115 and 4.118 in the discrete case) f(x) =/integraldisplay dλn/summationdisplay αfα λnuα λn(x). (4.126) This is the generalized Fourier integral, with generalized Fourier coef- eq4.175 pr:GenFourInt1 ﬁcients fα λn=/integraldisplay dxuα∗ λn(x)σ(x)f(x). (4.127) The coeﬃcients fα λnmay be interpreted as the projection of f(x) with eq4.176 respect toσ(x) onto the eigenfunction uα λn(x). 4.9 Orthogonality for Continuum 27 Jan p3 We now give the derivation of the orthogonality of the eigenfunctions for the continuum case. The method of derivation is the same as we 60 CHAPTER 4. PROPERTIES OF EIGEN STATES used in the discrete spectrum case. First we choose f(x) =uα λm(x) for f(x) in equation 4.126. Equation 4.127 then becomes fα/prime∗ λ/primen=/integraldisplay dxuα/prime λ/primen(x)σuα λm(x). (4.128) The form of equation 4.126 corresponding to this is uα λm(x) =/summationdisplay α/integraldisplay dλ/prime mfα/prime λ/primenuα/prime λ/primen(x). (4.129) This equation can only be true if fα/prime λ/primen=δαα/primeδ(λ/prime n−λm). (4.130) So we conclude that /integraldisplay dxuα/prime λ/primen(x)σuα λm(x) =fα/prime λ/primen=δαα/primeδ(λ/prime n−λm). (4.131) This is the statement of orthogonality, analogous to equation 4.114. All of these results come from manipulations on equation 4.108. We now have both a Fourier sum theorem and a Fourier integral theorem. We now investigate equation 4.124 in more detail G(x,x/prime;λ) =/integraldisplaydλn λn−λ/parenleftBigg/summationdisplay αuα λn(x)uα λn(x/prime)/parenrightBigg . (Generally, the integration is over the interval from zero to inﬁnity.) Where are the singularities? Consider λapproaching the positive real axis. It can’t ever get there. The value approaching the negative side may be diﬀerent from the value approaching the positive side. There- fore this line must be a branch cut corresponding to a continuous spec- trum. Note that generally there is only a positive continuous spectrum, although we may have a few negative bound states. So G(x,x/prime;λ) is analytic on the entire complex cut λ-plane. It is in the region of non- analyticity that all the physics occurs. The singular diﬀerence of a 27 Jan p4 branch cut is the diﬀerence in the value of Gabove and below. See ﬁgure 4.7. ﬁg4555 We now examine the branch cut in more detail. Using equation pr:branch2 4.9. ORTHOGONALITY FOR CONTINUUM 61 rλ/primerλ/prime+iε rλ/prime−iε Figure 4.7: λnear the branch cut. 4.108 we can write lim /epsilon1→0G(x,x/prime;λ/prime+i/epsilon1)−G(x,x/prime;λ/prime−i/epsilon1) 2πi =1 2πi/integraldisplay∞ 0dλn/summationdisplay αuα λn(x)uα∗ λn(x/prime)/parenleftbigg1 λn−λ/prime−i/epsilon1−1 λn−λ/prime+i/epsilon1/parenrightbigg where 1 2πi/parenleftbigg1 λn−λ/prime−i/epsilon1−1 λn−λ/prime+i/epsilon1/parenrightbigg =1 2πi2i/epsilon1 (λn−λ/prime)2+/epsilon12 =/epsilon1 π1 (λn−λ/prime)2+/epsilon12. In the ﬁrst problem set we found that lim /epsilon1→0/epsilon1 π1 (λ−λ/prime)2+/epsilon12=δ(λ/prime−λ). (4.132) So lim /epsilon1→0G(x,x/prime;λ/prime+i/epsilon1)−G(x,x/prime;λ/prime−i/epsilon1) 2πi =/integraldisplay dλn/summationdisplay αuα λn(x)uα∗ λn(x/prime)δ(λ/prime−λn) =/summationdisplay αuα λ/prime(x)uα λ/prime(x/prime). Therefore the discontinuity gives the product of the eigenfunctions. 27 Jan p5 We derived in the second problem set the property G∗(x,x/prime;λ) =G(x,x/prime;λ∗). (4.133) 62 CHAPTER 4. PROPERTIES OF EIGEN STATES Now takeλ=λ/prime+i/epsilon1. This allows us to write G∗(x,x/prime;λ/prime+i/epsilon1) =G(x,x/prime;λ/prime−i/epsilon1) (4.134) and G(x,x/prime;λ/prime+i/epsilon1)−G(x,x/prime;λ/prime−i/epsilon1) 2πi(4.135) =G(x,x/prime;λ/prime+i/epsilon1)−G∗(x,x/prime;λ/prime+i/epsilon1) 2πi(4.136) =1 πImG(x,x/prime;λ/prime+i/epsilon1) (4.137) =/summationdisplay αuα λ/prime(x)uα∗ λ/prime(x/prime). (4.138) So we can say that the sum over degeneracy of the bilinear product eq4.151 of the eigen function uα λis proportional to the imaginary part of the Green’s function. 4.10 Example: Inﬁnite String 29 Jan p1 Consider the case of an inﬁnite string. In this case we take the endpr:InfStr1pointsa→ −∞ ,b→ ∞ and the density σ, tensionτ, and potential V as constants. The term Vis the elastic constant of media. 4.10.1 The Green’s Function In this case the Green’s function is deﬁned as the solution to the equa- tion /bracketleftBigg −τd2 dx2+V−λσ/bracketrightBigg G(x,x/prime;λ) =δ(x−x/prime) for −∞<x,x/prime<∞. (4.139) To get the solution we must take λ(=ω2) to be imaginary. The solution for the Green’s function can be written in terms of the normal modes (3.41) G(x,x/prime;λ) =u1(x<)u2(x>) −τW(u1,u2)(4.140) 4.10. EXAMPLE: INFINITE STRING 63 u uθ c2k2ReλImλλ Figure 4.8: θspeciﬁcation. whereu1andu2satisfy the equation /bracketleftBigg −τd2 dx2+V−λσ/bracketrightBigg u1,2= 0. (4.141) The boundary conditions are that u1is bounded and converges as x→eq4.139 −∞ and thatu2is bounded and converges as x→ ∞ . Divide both ‘and converges’ - R. Horn sides of equation 4.141 by −τand substitute the deﬁnitions σ/τ= 1/c2 andV/τ=k2. This gives us the equation pr:k2.1 /bracketleftBiggd2 dx2+λ−c2k2 c2/bracketrightBigg u(x) = 0. (4.142) The general solution to this equation can be written as eq4.132a u(x) =Aei√ λ−c2k2x/c+Be−i√ λ−c2k2x/c. (4.143) We specify the root by the angle θextending around the point c2k2oneq4.132 the real axis, as shown in ﬁgure 4.8. This is valid for 0 <θ< 2π. The ﬁg4.6a correspondence of θis as follows: λ−c2k2=|λ−c2k2|eiθ. (4.144) Thus √ λ−c2k2=/radicalBig |λ−c2k2|eiθ/2(4.145) =/radicalBig |λ−c2k2|/parenleftBigg cosθ 2+isinθ 2/parenrightBigg . (4.146) 64 CHAPTER 4. PROPERTIES OF EIGEN STATES So that θ→0⇐⇒/radicalBig λ−c2|k2| (4.147) θ→π⇐⇒i/radicalBig λ−c2|k2| (4.148) θ→2π⇐⇒ −/radicalBig λ−c2|k2|. (4.149) This is good everywhere except for values on the real line greater than c2k2. 4.10.2 Uniqueness 29 Jan p2 The Green’s function is unique since it was found using the theory of ordinary diﬀerential equations. We identify the fundamental solutions u1,u2by looking at the large xbehavior of 4.143. ei√ λ−c2k2x/cx→∞−→e−∞ande−i√ λ−c2k2x/cx→∞−→e+∞(4.150) ei√ λ−c2k2x/cx→−∞−→e+∞ande−i√ λ−c2k2x/cx→−∞−→e−∞(4.151) so u1(x) =e−i√ λ−c2k2x c (4.152) u2(x) =ei√ λ−c2k2x c. (4.153) The boundary condition has an explicit dependence on λso the solution is not analytic. Notice that this time there is no way to get rid of the branch cut. The branch cut that comes in the solution is unavoidable because satisfaction of the boundary condition depends on the value of λ. 4.10.3 Look at the Wronskian In the problem we are considering we have W(u1,u2) =u1u/prime 2−u2u/prime 1 (4.154) = +i c√ λ−c2k2−/parenleftbigg −i c√ λ−c2k2/parenrightbigg (4.155) =2i c√ λ−c2k2. (4.156) 4.10. EXAMPLE: INFINITE STRING 65 4.10.4 Solution This gives the Greens’ function G(x,x/prime;λ) =ice−i√ λ−c2k2x</cei√ λ−c2k2x>/c 2τ√ λ−c2k2(4.157) =ic 2τei√ λ−c2k2|x−x/prime|/c √ λ−c2k2. (4.158) We now have a branch cut for Re ( λ)≥c2k2with branch point at λ=c2k2. Outside of this, Gis analytic with no poles. This is analytic forλin the cutλ-planeλ>c2k2. Consider the special case of large λ29 Jan p3 Gλ→∞−→ic 2τei√ λ|x−x/prime|/c √ λ→0. (4.159) Notice that this is the same asymptotic form we obtained in the discrete case when we looked at Gasλ→ ∞ (see equation 4.79). Now consider the caseλ < c2k2on the real axis. This corresponds to the case that θ=π. So G(x,x/prime;λ) =c 2τe−√ c2k−λ|x−x/prime| √ c2k2−λ. (4.160) This function is real and exponentially decreasing. For λ > c2k2this eq4.159 function oscillates. If θ= 0, it oscillates one way, and if θ= 2πit oscillates the other way, so the solution lacks uniqueness. The solutions correspond to diﬀerent directions of traveling waves. 4.10.5 Motivation, Origin of Problem We want to understand the degeneracy in the Green’s function for an inﬁnite string. So we take a look at the physics behind the problem. How did the problem arise? It came from the time dependent problem with forced oscillation imposed by an impulsive force at x/prime: /bracketleftBigg −τd2 dx2+V+σ∂2 ∂t2/bracketrightBigg u(x,t) =δ(x−x/prime)e−iωtwhereω2<c2k2. (4.161) 66 CHAPTER 4. PROPERTIES OF EIGEN STATES We wanted the steady state solution: u(x,t) =e−iωtG(x,x/prime;λ). (4.162) By substituting 4.160 this can be rewritten as u(x,t) =c 2τe−iωte−√ c2k2−ω2|x−x/prime| −√ c2k2−ω2. (4.163) We consider the cases ω2<c2k2andω2>c2k2separately. eq4.163ab Ifω2< c2k2, then there is a unique solution and the exponential dies oﬀ. This implies that there is no wave propagation. This agrees pr:WaveProp1 with what the physics tells us intuitively: k2c2> ω2implies large k, which corresponds to a large elastic constant V, which in turn means there will not be any waves. Forω>c2k2there is propagation of waves. For ω>c2k2we denote 29 Jan p4 the two solutions: u±(x,t) =e−iωtG(x,x/prime;λ=ω2±i/epsilon1) asε→0. (4.164) Asωincreases, it approaches the branch point. We deﬁne the cutoﬀ pr:cutoﬀ2 frequency as being at the branch point, ω2 c=c2k2. We have seen that forω < ω cthere is no wave propagation, and for ω > ω cthere is propagation, but we don’t know the direction of propagation, so there is no unique solution. The natural appearance of a branch cut with two solutions means that all the physics has not yet been given. We may rewrite equation 4.163 as: u±(x,t) =e−iωt±i(√ ω2−c2k2/c)|x−x/prime| 2τ c√ ω2−c2k2. (4.165) These solutions to the steady state problem can be interpreted as follows. The solution u+represents a wave traveling to the right for points to the right of (i.e. on the positive side of) the source and a wave traveling to the left for points to the left of (i.e. on the negative side of) the source. Mathematically this means u+=/braceleftBigg ∼e−iω(t−x√ ω2−c2k2/cω)forx>x/prime ∼e−iω(t+x√ ω2−c2k2/cω)forx<x/prime. 4.11. SUMMARY OF THE INFINITE STRING 67 Similarly, the solution u−represents a wave traveling to the left for points to the right of the source and a wave traveling to the right for points to the left of the source. Mathematically this means u−=/braceleftBigg ∼e−iω(t+x√ ω2−c2k2/cω)forx>x/prime ∼e−iω(t−x√ ω2−c2k2/cω)forx<x/prime. These results can be rephrased by saying that u+is a steady state solution having only waves going out from the source, and u−is a steady state solution having only waves going inward from the outside absorbed by the point. So the equation describes two situations, and the branch cut corresponds to the ambiguity in the situation. 4.11 Summary of the Inﬁnite String 2 Feb p1 We have considered the equation (L0−λσ)G=δ(x−x/prime) (4.166) where L0=−τd2 dx2+V. (4.167) We found that isV/negationslash= 0? G(x,x/prime;λ) = i 2τ/radicalBig λ c2−k2 ei/radicalBig λ c2−k2|x−x/prime|(4.168) where we have introduced the substitutions V/τ≡k2andσ/τ≡1/c2. The time dependent response is u±(x,t) =1 2τ/radicalBig λ c2−k2e−iωt−i/radicalBig k2−λ c2|x−x/prime|. (4.169) Ifω2<c2k2≡ω2 cthere is exponential decay, in which case there is no singularity of Gatλ=ω2. The other case is that u±(x,t) =e−iωtG(x,x/prime;λ=ω2±i/epsilon1). (4.170) 68 CHAPTER 4. PROPERTIES OF EIGEN STATES This case occurs when ω2>ω2 c, for which there is a branch cut across the real axis. In this case we have traveling waves. Note that in the case that k= 0 we always have traveling waves. The relevance of the equation k2=V/τ is that the resistance of the 1 Feb p2 medium to propagation determines whether waves are produced. When k= 0 there is no static solution — wave propagation always occurs. If ω2<ω2 c, then the period of the external force is small with respect to the response of the system, so that the media has no time to respond — the system doesn’t know which way to go, so it exponentially decays. Recall that the solution for the ﬁnite string (open or closed) allowed incoming and outgoing waves corresponding to reﬂections (for the open string) or diﬀerent directions ( for the closed string). Recall the periodic boundary condition problem: u(x,t) =e−iωtG(x,x/prime;λ) (4.171) =e−iωtc 2ωcos[ω c(L 2− |x−x/prime|)] sin[ω cl 2]. (4.172) This is equal to the combination of incoming and outgoing waves, which can be seen by expanding the cosine. We need the superposition to satisfy the boundary conditions and physically correspond to reﬂections at the boundaries. The sum of the two waves superimpose to satisfy the boundary conditions. 4.12 The Eigen Function Problem Revis- ited 1 Feb p3 We now return to the the connection with the eigen function problem. We have seen before that the expression pr:efp2 G(λ=λ/prime+i/epsilon1)−G(λ=λ/prime−i/epsilon1) 2πi(4.173) vanishes ifλ/prime<c2k2. In the case that λ/prime>c2k2we have G(λ=λ/prime+i/epsilon1)−G(λ=λ/prime−i/epsilon1) 2πi=/summationdisplay αuα λ/prime(x)(uα λ/prime(x/prime))∗. (4.174) 4.13. SUMMARY 69 q c2k2 /radicalBig λ/c2−k2→+|λ/c2−k2|1/2 BBM/radicalBig λ/c2−k2→ −|λ/c2−k2|1/2 Figure 4.9: Geometry in λ-plane The geometry of this on the λ-plane is shown in ﬁgure 4.9. This gives eq4777 us G(λ=λ/prime+i/epsilon1)−G(λ=λ/prime−i/epsilon1) 2πi =1 2πi1 2τ|λ c2−k2|1/2(ei|λ c2−k2|1/2|x−x/prime|−e−i|λ c2−k2|1/2|x−x/prime|) =1 π1 2τ|λ c2−k2|1/2sin /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleλ c2−k2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle1/2 |x−x/prime|  =1 πImG(λ=λ/prime+i/epsilon1). Forλ/prime>c2k2we can write (using equation 4.138) u± λ/prime(x) =1/radicalbigg 4πτ/radicalBig λ/prime c2−k2e±i√ λ/prime−c2k2x/c(4.175) forλ/prime> c2k2. We now see that no eigen functions exist for λ/prime< c2k2eq4.188 since it exponentially increases as λ/prime→ ∞ and we must kill both terms. The Green’s function is all right since λ∈C. 4.13 Summary 1. For the Helmholtz equation, ω2 n>0,ω2 nis real, and the eigen functions are real. 2. The dispersion relation for a closed massless string with discrete mass points is ω2 n=c2sin2(kna/2) a2/4. 70 CHAPTER 4. PROPERTIES OF EIGEN STATES 3. The Green’s function obeys Hermitian analyticity: G∗(x,x/prime;λ) =G(x/prime,x;λ∗). 4. The form of the Green’s function for λnear an eigen value λnis G(x,x/prime;λ)λ→λn−→un(x)u∗ n(x/prime) λn−λ. 5. The Green’s function for the ﬁxed string problem is G(x,x/primeλ) =sin/radicalBig λ c2x<sin/radicalBig λ c2(L−x>) τ/radicalBig λ c2sin/radicalBig λ c2L. 6. The completeness relation is δ(x−x/prime) =∞/summationdisplay n=1σ(x)un(x)u∗ n(x). 7. The expansion theorem is f(x) =∞/summationdisplay n=1un(x)fn where fn=/integraldisplayb adx/primeu∗ n(x/prime)σ(x/prime)f(x/prime). 8. The Green’s function near the branch cut is related to the eigen functions by 1 πImG(x,x/prime;λ/prime+i/epsilon1) =/summationdisplay αuα λ/prime(x)uα λ/prime(x/prime). 9. The Green’s function solution for an inﬁnite string is G(x,x/prime;λ) =ic 2τei√ λ−c2k2|x−x/prime|/c √ λ−c2k2. 4.14. REFERENCES 71 4.14 References The Rayleigh quotient is described in [Stakgold67a, p226ﬀ] and [Stak- gold79, p339ﬀ]. For other ideas in this chapter, see Fetter and Stakgold. A discussion of the discrete closed string is given in [Fetter80, p115]. The material in this chapter is also in [Fetter81, p245ﬀ]. 72 CHAPTER 4. PROPERTIES OF EIGEN STATES Chapter 5 Steady State Problems Chapter Goals: •Interpret the eﬀect of an oscillating point source on an inﬁnite string. •Construct the Klein-Gordon equation and interpret its steady state solutions. •Write the completeness relation for a continuous eigenvalue spectrum and apply it to the Klein- Gordon problem. •Show that the solutions for the string problem with σ=x,τ=x, andV=m/x2on the interval 0<x< ∞are Bessel functions. •Construct the Green’s function for this problem. •Construct and interpret the steady state solutions for this problem with a source point. •Derive the Fourier-Bessel transform.moved from few pages later 5.1 Oscillating Point Source pr:ops1 We now look at the problem with an oscillating point source. In the notation of the previous chapter this is 3 Feb p1 73 74 CHAPTER 5. STEADY STATE PROBLEMS /bracketleftBigg L0+σ(x)∂2 ∂t2/bracketrightBigg u(x,t) =δ(x−x/prime)e−iωt−∞<x,x/prime<∞+ R.B.C. (5.1) and can be written in terms of the Green function G(x,x/prime;λ) which satisﬁes [L0−σ(x)λ]G(x,x/prime;λ) =δ(x−x/prime)−∞<x,x/prime<∞+ R.B.C. (5.2) The steady state solution corresponding to energy radiated outward to inﬁnity is u(x,t) =e−iωtG(x,x/prime,λ=ω2+i/epsilon1). (5.3) The solution for energy radiated inward from inﬁnity is the same equa- tion withλ=ω2−iε, but this is generally not a physical solution. This contrasts with the case of a ﬁnite region. In that case there are no branch cuts and there is no radiation. 1 Feb p5 5.2 The Klein-Gordon Equation pr:kge1 We now apply the results of the previous chapter to another physical problem. Consider the equations of relativistic quantum mechanics. In the theory of relativity we have the energy relation E2=m2c4+p2c2(5.4) In the theory of quantum mechanics we treat momentum and energy as operators p→ −i¯h∇ (5.5) E→i¯h∂ ∂t(5.6) where dim[¯h] = Action (5.7) We want to derive the appropriate wave equation, so we start with E2−(m2c4+p2c2) = 0 (5.8) 5.2. THE KLEIN-GORDON EQUATION 75 Now substitute the operators into the above equation to get  /parenleftBigg i¯h∂ ∂t/parenrightBigg2 − m2c4+/parenleftBigg¯h i∇/parenrightBigg2 c2  Φ = 0 (5.9) This is the Klein–Gordon equation, which is a relativistic form of the pr:phi1 Schr¨ odinger Equation. Note that |Φ|2still has a probability interpre- tation, as it does in non-relativistic quantum mechanics. Now specialize this equation to one dimension. 1 Feb p6 /bracketleftBigg −¯h2c2d2 dx2+m2c4+ ¯h2∂2 ∂t2/bracketrightBigg Φ(x,t) = 0 (5.10) This is like the equation of a string (c.f., 1.11). In the case of the eq5.energy string the parameters were tension τ= dim[E/t], coeﬃcient of elasticity V= dim[E/l3], and mass density σ= [t2E/l3], whereEis energy,t is time, and lis length. The overall equation has units of force over length (since it is the derivative of Newton’s second law). By comparing equation 1.11 with equation 5.10 we note the correspondence V→m2c4σ→¯h2τ→¯h2c2andf(x,t)→0.(5.11) Note that/radicalBig V/τ=mc/¯his a fundamental length known as the Comp- ton wavelength, λc. It represents the intrinsic size of the particle. The expression expression σ/τ = 1/c2shows that particle inertia corresponds to elasticity, which prevents the particle from responding quickly. Equation 5.10 has dimensions of energy squared (since it came ? from an energy equation). We look again for steady state solutions Φ(x,t) =e−iE/primet/¯hΦE/prime(x) (5.12) to get the eigen value problem /bracketleftBigg −¯h2c2d2 dx2+m2c4−E/prime2/bracketrightBigg ΦE/prime(x) = 0 (5.13) Thus we quote the previous result (equation 4.175 which solves equation 4.142) Φ± E/prime(x) =u±(x) (5.14) 76 CHAPTER 5. STEADY STATE PROBLEMS where we let k2→(mc ¯h)2andλ/prime→E/prime2/¯h2. As a notational shorthand, letp/prime=√ E/prime2−m2c4/c. Thus we write Φ± E/prime(x) =e±i(√ E/prime2−m2c4/¯hc)x /radicalBig 4π¯hc√ E2−m2c4 =e±p/prime/¯h √4π¯hp/primec2. The cut-oﬀ energy is mc2. We have the usual condition on the solution thatλ/prime>c2k2. This eigen value condition implies E2>m2c4. In relativistic quantum mechanics, if E <mc2, then no free particle is emitted at large distances. In the case that E >mc2there is radia- tion. Also note that mbecomes inertia. For the case that m→0 there pr:rad1 is always radiation. This corresponds to V→0 in the elastic string analogy. The potential Vacts as an elastic resistance. The Green’s function has the form G(x,x/prime;E)∼exp{(√ m2c4−E2/¯hc)|x−x/prime|}. SoG(x,x/prime;E) has a characteristic half-width of |x−x/prime| ∼¯hc/√ m2c4−E2= ¯h/p. This is a manifestation of the uncertainty principle. As x→ ∞ , for E < mc2, the Green’s function vanishes and no particle is radiated, while forE >mc2the Green’s function remains ﬁnite at large distances which corresponds to the radiation of a particle of mass m. 5.2.1 Continuous Completeness pr:CompRel3 3 Feb p2Recall that the completeness condition in the discrete case is δ(x−x/prime) σ(x)=/summationdisplay λ/prime,αuα λ/prime(x)uα∗ λ/prime(x/prime). (5.15) The corresponding equation for the case of continuous eigenvalues is R. Horn says noλ/primein sum.δ(x−x/prime) σ(x)=/summationdisplay λ/prime,α=±/integraldisplay∞ ω2cdλ/primeuα λ/prime(x)uα∗ λ/prime(x/prime). (5.16) 5.2. THE KLEIN-GORDON EQUATION 77 Recall that in this case the condition for an eigen function to exist is λ/prime>ω2 c. In this case u± λ/prime(x) =e±i/radicalBig λ/prime c2−k2x /radicalBig 4πτ(λ/prime c2−k2)1/2. (5.17) Substituting the u±into the continuous completeness relation gives δ(x−x/prime) =σ(x)c 4πτ/integraldisplay∞ ω2cdλ/prime (λ/prime c2−k2)1/2/bracketleftbigg ei√ λ/prime−ω2c(x−x/prime c)+e−i√ λ/prime−ω2c(x−x/prime c)/bracketrightbigg (5.18) whereω2 c=k2c2. We now make a change of variables. We deﬁne the wave number as k=/radicalBig λ/prime−ω2 c c(5.19) It follows that the diﬀerential of the wave number is given by dk=1 2cdλ/prime /radicalBig λ/prime−ω2 c. (5.20) With this deﬁnition we can write 3 Feb p3 δ(x−x/prime) =2σc2 4πτ/integraldisplay∞ 0dk/bracketleftBig eik(x−x/prime)+e−ik(x−x/prime)/bracketrightBig . (5.21) Note the symmetry of the transformation k→ −k. This property allows us to write δ(x−x/prime) =1 2π/integraldisplay∞ −∞dkeik(x−x/prime). (5.22) This is a Fourier integral. In our problem this has a wave interpretation. We now apply this to the quantum problem just studied. In this caseλ/prime= (E/¯h)2andωc2=m2c4/¯h2. With these substitutions we can write k=(E2 ¯h2−m2c4 ¯h2)1/2 c(5.23) =1 ¯h/radicalBigg E2 c2−m2c2=p ¯h(5.24) 78 CHAPTER 5. STEADY STATE PROBLEMS so δ(x−x/prime) =1 2π/integraldisplay∞ −∞dp/primeei(x−x/prime)p/prime/¯h. (5.25) This Fourier integral has a particle interpretation. pr:FI1 5.3 The Semi-inﬁnite Problem Consider the following linear operator in the semi-inﬁnite region 3 Feb p4 L0=−d dx/parenleftBigg xd dx/parenrightBigg +m2 x0<x< ∞. (5.26) Here we have let the string tension be τ=xand the potential be V= m2/x. We will let the density be σ(x) =x. This is like a centrifugal potential. The region of consideration is 0 <x< ∞. This gives us the following Green’s function equation (from 3.22) /bracketleftBigg −d dx/parenleftBigg xd dx/parenrightBigg +m2 x−λx/bracketrightBigg G(x,x/prime;λ) =δ(x−x/prime) (5.27) deﬁned on the interval 0 <x,x/prime<∞. We now discuss the boundary conditions appropriate for the semi- inﬁnite problem. We require that the solution be bounded at inﬁnity, as was required in the inﬁnite string problem. Note that in the above equationτ= 0 atx= 0. Then at that point the right hand side of Green’s second identity vanishes, as long as the amplitude at x= 0 is ﬁnite. Physically, τ→0 atx= 0 means that the string has a free end. So there is a solution which becomes inﬁnite at x= 0. This non-physical solution is eliminated by the boundary condition that the amplitude of that end is ﬁnite. Under these boundary conditions L0is hermitian for 0 ≤x<∞. The solution can be written in the form (using 3.41) G(x,x/prime;λ) =−u1(x<,λ)u2(x>,λ) xW(u1,u2)(5.28) The function u1andu2are the solutions to the equation eq5.29 5.3. THE SEMI-INFINITE PROBLEM 79 (L0−λx)u1,2= 0 (5.29) where we restrict u1to be that function which is regular at x= 0, and u2to be that function which is bounded at inﬁnity. We note that the equations are Bessel’s equations of order m: pr:Bessel1 /bracketleftBigg y2d2 dy2+yd dy+ (y2−m2)/bracketrightBigg Xm(y) = 0 (5.30) wherey=x√ λandXm(y) is any solution. The solutions are then u1(x) =Jm(x√ λ) (5.31) and u2(x) =H(1) m(x√ λ). (5.32) The function H(1) m(x√ λ) is known as the Hankel function. For large x3 Feb p5 it may be approximated as H(1) m(x√ λ)x→∞∼/radicalBigg 2 πx√ λei(x√ λ−mπ 2−π 4)(5.33) and since Im(√ λ)>0, we have decay as well as out going waves. Now we get the Wronskian: Justify this W(u1,u2) =W(Jm(x√ λ),H(1) m(x√ λ)) (5.34) =iW(Jm(x√ λ),Nm(x√ λ)) (5.35) =i√ λ/parenleftBigg2 πx√ λ/parenrightBigg =2i πx. (5.36) In the second equality we used the deﬁnition H(1) m(x) =Jm(X)+iNm(x), where Nm(x)≡Jm(x) cosmπ−J−m(x) sinmπ. The third equality is veriﬁed in the problem set. So τWis independent ofx, as expected. Therefore (using equation 5.28) G(x,x/prime;λ) = −1 xπx 2iJm(x<√ λ)H(1) m(x>√ λ) (5.37) =iπ 2Jm(x<√ λ)H(1) m(x>√ λ) (5.38) 80 CHAPTER 5. STEADY STATE PROBLEMS 5.3.1 A Check on the Solution Suppose that λ<0. In this case we should have Greal, since there is no branch cut. We have √ λ=i|λ|1/2atθ=π. (5.39) We can use the deﬁnitions pr:Bessel2 Im(x)≡e−imπ/ 2Jm(ix), K m(x)≡(πi/2)eimπ/ 2H(1) m(ix) to write G(x,x/prime;λ) =iπ 2Jm(i|λ|1/2x<)H(1) m(i|λ|1/2x>) (5.40) =iπ 2imIm(x<|λ|1/2)/parenleftbigg i−m2 πiKm(x>|λ|1/2/parenrightbigg (5.41) =Im(x<|λ|1/2)Km(x>|λ|1/2)∈R (5.42) Note also that G→0 asx→ ∞ , so we have decay, and therefore no propagation. This is because asymptotically Im(z)≈(2zπ)−1/2ez|argz|<1/2,|z| → ∞ Km(z)≈(2z/π)−1/2e−z|argz|<3π/2,|z| → ∞. 5.4 Steady State Semi-inﬁnite Problem 5 Feb p1 For the equation /bracketleftBigg −d dx/parenleftBigg xd dx/parenrightBigg +m2 x−λx/bracketrightBigg G(x,x/prime;λ) =δ(x−x/prime) for 0<x,x/prime<∞ (5.43) the solution we obtained was G(x,x/prime;λ) =iπ 2Jm(√ λx<)H(1) m(√ λx>). (5.44) We now look at the steady state solution for the wave equation, /bracketleftBigg L0+x∂2 ∂t2/bracketrightBigg u(x,t) =δ(x−x/prime)e−iωt. (5.45) 5.4. STEADY STATE SEMI-INFINITE PROBLEM 81 We only consider outgoing radiation ( ω2→ω2+iε). We take ε >0, pr:rad2 sinceε <0 corresponds to incoming radiation. So δ(x−x/prime) acts as a point source, but not as a sink. u(x,t) =e−iωtG(x,x/prime;λ=ω2+iε) (5.46) =e−iωtiπ 2Jm(ωx<)H(1) m(ωx>) (5.47) =e−iωtiπ 2Jm(ωx/prime)H(1) m(ωx>) forx>x/prime.(5.48) Next letωx/greatermuch1, so u(x,t) =iπ 2Jm(ωx/prime)e−iω(t−x) √ωxforωx/greatermuch1. The condition ωx/greatermuch1 allows us to use the asymptotic form of the 5 Feb p2 Hankel function. In this case we have outgoing (right moving) waves. These waves are composed of radiation reﬂected from the boundary x= 0 and from direct radiation. If in addition to ωx/greatermuch1 we takeωx/prime→0, then we Explain why? have u(x,t) =iπ 2(ωx/prime)me−iω(t−x) √ωx(5.49) We now look at the case x<x/primewithωlarge. In this case we have u(x,t) =e−iωtiπ 2H(1) m(ωx/prime)Jm(ωx) =e−iωtiπ 2H(1) m(ωx/prime)1 2[H(1) m(ωx) +H(2) m(ωx)] ∼e−iωtH(1) m(ωx/prime)/bracketleftBiggeiωx √ωx+e−iωx √ωx/bracketrightBigg ωx/greatermuch1 ∼H(1) m(ωx/prime)[e−iω(t−x)+e−iω(t+x)]. We now look at the Green’s function as a complete set of eigen functions. First we consider 5 Feb p3 1 2πi[G(x,x/prime;λ/prime+iε)−G(x,x/prime;λ/prime−iε)] 82 CHAPTER 5. STEADY STATE PROBLEMS =1 2πi[Jm(√ λ/primex<)H(1) m(√ λ/primex>)−Jm(−√ λ/primex<)H(1) m(−√ λ/primex>)] =1 4[Jm(√ λ/primex>)][H(1) n(√ λ/primex) +H(2) n(√ λ/primex)] =1 2Jm(√ λ/primex)Jm(√ λ/primex/prime) =1 πImG(x,x/prime;λ/prime+iε). 5.4.1 The Fourier-Bessel Transform The eigen functions uλ/primesatisfy /bracketleftBigg−d dx/parenleftBigg xd dx/parenrightBigg +m2 x−λx/bracketrightBigg uλ/prime= 0 for 0 <x< ∞. (5.50) In this case since there is a boundary at the origin, waves move only to 5 Feb p4 the right. There is no degeneracy, just one eigen function: uλ/prime=/radicalBigg 1 2Jm(√ λ/primex). (5.51) We know from the general theory that if there is no degeneracy, then 1 σ(x)δ(x−x/prime) =/integraldisplay∞ 0dλ/primeuλ/prime(x)u∗ λ/prime(x/prime) (5.52) so 1 xδ(x−x/prime) =1 2/integraldisplay∞ ∞dλ/primeJm(√ λ/primex)Jm(√ λ/primex/prime) (5.53) =/integraldisplay∞ 0ω/primedω/primeJm(ω/primex)Jm(ω/primex/prime). (5.54) This is valid for 0 <x,x/prime<∞. Thus forf(x) on 0<x< ∞we have f(x) =/integraldisplay∞ 0dx/primef(x/prime)δ(x−x/prime) (5.55) =/integraldisplay∞ 0dx/primef(x/prime)x/prime/integraldisplay∞ 0ω/primedω/primeJm(ω/primex)Jm(ω/primex/prime) (5.56) =/integraldisplay∞ 0ωdω/primeJm(ω/primex)/integraldisplay∞ 0dx/primex/primef(x/prime)Jm(ω/primex/prime).(5.57) 5.5. SUMMARY 83 Thus for a given f(x) on 0<x< ∞we can write 5 Feb p5 f(x) =/integraldisplay∞ 0ω/primedω/primeJm(ω/primex)Fm(ω/prime). (5.58) This is the inversion theorem. Fm(ω) =/integraldisplay∞ 0x/primedx/primef(x/prime)Jm(ωx/prime). (5.59) This is the Fourier-Bessel transform of order m. Explain why this is useful? pr:FBt15.5 Summary 1. The string equation of an oscillating point source on an inﬁnite string has solutions corresponding to energy radiated in from or out to inﬁnity. 2. The Klein-Gordon equation is /bracketleftBigg −¯h2c2d2 dx2+m2c4+ ¯h2∂2 ∂t2/bracketrightBigg Φ(x,t) = 0. Steady-state solutions for a point source with |E|> mc2corre- spond to a mass mparticle radiated to ( ±) inﬁnity, where as solutions with |E|< mc2die oﬀ with a characteristic range of x∼¯h/p. pr:chRan1 3. The string problem with σ=x,τ=x, andV=m/x2on the interval 0<x< ∞corresponds the Bessel’s equation /bracketleftBigg y2d2 dy2+yd dy+ (y2−m2)/bracketrightBigg Xm(y) = 0 wherey=x√ λ. The linearly independent pairs of solutions to this equation are the various Bessel functions: (i) Jm(y) and Nm(y), and (ii)H(1) m(y) andH(2) m(y). 4. The Green’s function for this problem is G(x,x/prime;λ) =iπ 2Jm(√ λx<)H(1) m(√ λx>). 84 CHAPTER 5. STEADY STATE PROBLEMS 5. The steady state solutions for this problem with point source are u(x,t) =e−iωtiπ 2Jm(ωx<)H(1) m(ωx>). The outgoing solutions consist of direct radiation and radiation reﬂected from the x= 0 boundary. 6. The Fourier-Bessel transform is Fm(ω/prime) =/integraldisplay∞ 0x/primedx/primef(x/prime)Jm(ωx/prime). The inversion theorem for this transform is f(x) =/integraldisplay∞ 0ω/primedω/primeJm(ω/primex)Fm(ω/prime). 5.6 References The Green’s function related to Bessel’s equation is given in [Stak- gold67a, p75]. Chapter 6 Dynamic Problems 8 Feb p1 4 Jan p1 p1prv.yr.Chapter Goals: •State the problem which the retarded Green’s func- tionGRsolves, and the problem which the ad- vanced Green’s function GAsolves. Give a physical interpretation for GRandGA. •Show how the retarded Green’s function can be written in terms of the Green’s function which solves the steady state problem. •Find the retarded Green’s function for an inﬁnite string with σandτconstant, and V= 0. •Find the retarded Green’s function for a semi- inﬁnite string with a ﬁxed end, σandτconstant, andV= 0. •Find the retarded Green’s function for a semi- inﬁnite string with a free end, σandτconstant, andV= 0. 85 86 CHAPTER 6. DYNAMIC PROBLEMS •Explain how to ﬁnd the retarded Green’s function for an elastically bound semi-inﬁnite string with σ andτconstant, and V= 0. •Find an expression for the retarded Green’s func- tion in terms of the eigen functions. •Show how the retarded boundary value problem can be restated as an initial value problem. 6.1 Advanced and Retarded GF’s Consider an impulsive force, a force applied at a point in space along pr:impf1 the string at an instant in time. This force is represented by σ(x)f(x,t) =δ(x−x/prime)δ(t−t/prime). (6.1) As with the steady state problem we considered in chapter 5, we apply no external forces on the boundary. If we can solve this problem, then we can solve the problem for a general time dependent force density f(x,t). We now examine this initial value problem (in contrast to the steady state problems considered in the previous chapter). We begin with the string at rest. Then we apply a blow at the point x/primeat the time t/prime. For this physical situation we want to ﬁnd the solution u(x,t) =GR(x,t;x/prime,t/prime) (6.2) whereGRstands for the retarded Green’s function: pr:GR1 /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg GR(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime) fora<x,x/prime<b; allt,t/prime. Now we look at the form of the two possible Regular Boundary Conditions. These two sets of conditions correspond to the case of an open and closed string. In the case of an open string the boundary condition is characterized by the equation [ˆns∇+κs]GR(x,t;x/prime,t/prime) = 0x∈S,a<x/prime<b;∀t,t/prime(6.3) 6.2. PHYSICS OF A BLOW 87 whereSis the set of end points {a,b}. In the case of a closed string eq6RBC the boundary condition is characterized by the equations GR(x,t;x/prime,t/prime)|x=a=GR(x,t;x/prime,t/prime)|x=b fora<x/prime<b,∀t,t/prime, (6.4) ∂ ∂xGR(x,t;x/prime,t/prime)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=a=∂ ∂xGR(x,t;x/prime,t/prime)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=bfora<x/prime<b,∀t,t/prime. (6.5) We now apply the condition that the string begins at rest: 8 Feb p2 GR(x,t;x/prime,t/prime) = 0 for t<t/prime. (6.6) This is called the retarded Green’s function since the motionless string becomes excited as a result of the impulse. This cause–eﬀect relation- ship is called causality . The RBC’s are the same as in previous chapters pr:caus1 but now apply to all times. Another Green’s function is GA, which satisﬁes the same diﬀerential equation as GRwith RBC with the deﬁnition GA(x,t;x/prime,t/prime) = 0 for t>t/prime(6.7) This is called the advanced Green’s function since the string is in an pr:AGF1 excited state until the impulse is applied, after which it is at rest. In what follows we will usually be concerned with the retarded Green function, and thus write GforGR(suppressing the R) except when contrasting the advanced and retarded Green functions. 6.2 Physics of a Blow We now look at the physics of a blow. Consider a string which satisﬁes the inhomogeneous wave equation with arbitrary force σ(x)f(x). The momentum applied to the string, over time ∆ t, is then pr:momch1 ∆p=p(t2)−p(t1) =/integraldisplayt2 t1dtdp dt =/integraldisplayt2 t1dt/integraldisplayx2 x1dxσ(x)f(x,t). 88 CHAPTER 6. DYNAMIC PROBLEMS The third equality holds because dp/dt is the force, which in this case is/integraltextx2 x1dxσ(x)f(x,t). We now look at the special case where the force is theδfunction. In this case 8 Feb p3 ∆p=/integraldisplayt2 t1dt/integraldisplayx2 x1δ(t−t/prime)δ(x−x/prime)dx= 1 (6.8) forx1< x/prime< x 2,t < t/prime< t 2. Thus a delta force imparts one unit of momentum. Therefore we ﬁnd that GRis the response of our system to a localized blow at x=x/prime,t=t/primewhich imparts a unit impulse of momentum to the string. 6.3 Solution using Fourier Transform We consider the Green’s function which solves the following problem given by a diﬀerential equation and an initial condition: /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg G(x,t;x/prime,t/prime) =δ(t−t/prime)δ(x−x/prime) + RBC ,(6.9) eq6de1 G(x,t;x/prime,t/prime) = 0 for t<t/prime. For ﬁxedxwe note that G(x,t;x/prime,t/prime) is a function of t−t/primeand nott andt/primeseparately, since only ∂2/∂t2andt−t/primeappear in the equation. Thus the transformation t→t+aandt/prime→t/prime+adoes not change anything. This implies that the Green’s function can be written G(x,t;x/prime,t/prime) =G(x,x/prime;t−t/prime) =G(x,x/prime;τ), where we deﬁne τ≡t−t/prime. By this deﬁnition, G= 0 forτ <0. This problem can be solved by taking the complex Fourier trans- form: pr:FTrans1 8 Feb p4˜G(x,x/prime;ω) =/integraldisplay∞ −∞dτeiωτG(x,x/prime;τ). (6.10) Note that ˜G(x,x/prime;ω) is convergent everywhere in the upper half ω- eq6Ftran plane, which we now show. The complex frequency can be written as ω=ωR+iωI. Thus eiωτ=eiωRτe−ωIτ. 6.3. SOLUTION USING FOURIER TRANSFORM 89 For˜G(x,x/prime;ω) to exist, the integral must converge. Thus we require eiωτ→0 asτ→ ∞ , which means we must have e−ωIτ→0 asτ→ ∞ . This is only true when ωis in the upper half plane, ωI>0. Thus ˜G exists for all ωsuch that Im ω > 0. Note that for ˜GAeverything is reversed and ωis deﬁned in the lower half plane. By taking the derivative of both sides of ˜G(x,x/prime;ω) in the Fourier transform, equation 6.10, we have d dω˜G(x,x/prime;ω) =/integraldisplay∞ −∞dτd dωeiωτG(x,x/prime;τ) =i/integraldisplay∞ 0dττG (x,x/prime;τ)e−iωτ. which is ﬁnite. Therefore the derivative exists everywhere in the upper Ask Baker but how do we know −i/integraltextdττG (τ) converges?halfω-plane. Thus ˜Gis analytic in the upper half ω-plane. We have thus seen that the causality condition allows us to use the Fourier trans- form to show analyticity and pick the correct solution. The condition thatG= 0 forτ <0 (causality) was only needed to show analyticity; it is not needed anymore. 8 Feb p5 We now Fourier transform the boundary condition of an open string 6.3: 0 =/integraldisplay∞ −∞dτeiωτ((ˆnS· ∇+κS)G) = (ˆnS· ∇+κS)/integraldisplay∞ −∞dτeiωtG = (ˆnS· ∇+κS)˜G. Similarly, in the periodic case we regain the periodic boundary condi- tions of continuity ˜Ga=˜Gband smoothness ˜G/prime a=˜G/prime b. So ˜Gsatisﬁes the same boundary conditions as Gsince the boundary conditions do not involve any time derivatives. Consider equation 6.9 rewritten as /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg G(x,x/prime;τ) =δ(x−x/prime)δ(τ). The Fourier transform of this equation is L0˜G(x,x/prime;ω) +σ(x)/integraldisplay∞ −∞dτeiωτ∂2 ∂τ2G(x,x/prime;τ) =δ(x−x/prime).(6.11) 90 CHAPTER 6. DYNAMIC PROBLEMS Using the product rule for diﬀerentiation, we can “pull out a divergence term”: eiωτ∂2 ∂τ2G=/parenleftBigg∂2 ∂τ2eiωτ/parenrightBigg G+∂ ∂τ/parenleftBigg eiωτ∂ ∂τG−G∂ ∂τeiωτ/parenrightBigg . Thus our equation 6.11 becomes δ(x−x/prime) =L0˜G(x,x/prime;ω) +σ(x)(−ω2˜G(x,x/prime;ω)) +σ(x)/bracketleftBigg eiωτ∂ ∂tG−G∂ ∂teiωτ/bracketrightBiggt=+∞ t=−∞. Now we evaluate the surface term. Note that G= 0 forτ <0 implies ∂G/∂t = 0, and thus |−∞= 0. Similarly, as τ→ ∞ ,eiωτ→0 since Imω>0, and thus |∞= 0. So we can drop the boundary term. We thus ﬁnd that GRsatisﬁes the diﬀerential equation 8 Feb p6 [L0−ω2σ(x)]˜G(x,x/prime;ω) =δ(x−x/prime) RBC, (6.12) with Imω>0. We now recognize that the Green function must be the eq6stst same as in the steady state case: ˜G(x,x/prime;ω) =G(x,x/prime;λ=ω2). Recall that from our study of the steady state problem we know that the function G(x,x/prime;λ) is analytic in the cut λ-plane. Thus by analytic continuation we know that ˜G(x,x/prime;ω) is analytic in the whole cut plane. The convention ω=√ λcompresses the region of interest to the upper this is unclear half plane, where λsatisﬁes [L0−λσ(x)]G(x,x/prime;λ) =δ(x−x/prime) + RBC, (6.13) All that is left is to invert the Fourier transform. eq6ststb 6.4 Inverting the Fourier Transform In the previous section we showed that for the Green’s function Gwe have the Fourier Transform ˜G(x,x/prime;ω) =/integraldisplay∞ −∞dτe−iωτG(x,x/prime;τ) 6.4. INVERTING THE FOURIER TRANSFORM 91 whereτ=t−t/prime, and we also found ˜G(x,x/prime;ω) =G(x,x/prime;λ=ω2) whereω=ωR+iωIwithωI>0, andG(x,x/prime;λ=ω2) is the solution of the steady state problem. Now we only need to invert the Fourier but didn’t we use analytic continuation to ge the whole λ plane.Transform to get the retarded Green’s function. We write eiωτ=eiωRτe−ωIτ so that ˜G(x,x/prime;ωR+iωI)/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright ˜F(ωR)=/integraldisplay∞ −∞dτeiωRτ[e−ωIτGR(x,x/prime;τ)]/bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright F(τ). This is a real Fourier Transform in terms of F(τ). We now apply the Fourier Inversion Theorem: pr:FIT1 F(τ) =e−ωIτGR(x,x/prime;τ) =1 2π/integraldisplay∞ −∞dωRe−iωRτ˜G(x,x/prime;ωR+iωI) so GR(x,x/prime;τ) =1 2π/integraldisplay∞ −∞dωRe−iωRτ˜G(x,x/prime;ω)eωIτ(6.14) ﬁxωI=εand integrate over ωR (6.15) =1 2π/integraldisplay Ldωe−iωτ˜G(x,x/prime;ω) (6.16) where the contour Lis a line in the upper half plane parallel to the ωR10 Feb p2 axis, as shown in ﬁgure 6.1.a. The contour is oﬀ the real axis because of ﬁg6Lcont the branch cut. We note that any line in the upper half plane parallel to the real axis may be used as the contour of integration. This can be seen by considering the rectangular integral shown in ﬁgure 6.1.b. Because ˜G= 0 asωR→ ∞ , we know that the sides LS1andLS2vanish. And sinceeiωτand ˜Gare analytic in the upper half plane, Cauchy’s theorem tells use that the integral over the closed contour is zero. Thus the integrals over path L1and pathL2must be equal. 92 CHAPTER 6. DYNAMIC PROBLEMS - ?6ωI ωRε L 1 (a) The line L1- ? 6L2 L1LS1 LS2 (b) Closed contour with L1 Figure 6.1: The contour Lin theλ-plane. 6.4.1 Summary of the General IVP pr:IVP1 We have considered the problem of a string hit with a blow of unit momentum. This situation was described by the equation /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg GR(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime) + RBC (6.17) with the condition that GR(x,t;x/prime,t/prime) = 0 fort < t/prime. The Green’s eq6GFxt function which satisﬁes this equation was found to be GR(x,t;x/primet/prime) =/integraldisplay Ldω 2πe−iω(t−t/prime)˜G(x,x/prime;ω). (6.18) where ˜G(x,x/prime;ω) satisﬁes the steady state Green’s function problem. eq6GFT 6.5 Analyticity and Causality To satisfy the physical constraints on the problem, we need to have GR= 0 fort<t/prime. This condition is referred to as causality. This con- dition is obtained due to the fact that the product e−iω(t−t/prime)˜G(x,x/prime;ω2) appearing in equation 6.17 is analytic. In this way we see that the ana- Need to show somewhere that ˜G|ω|→∞−→0.lyticity of the solution allows it to satisfy the causality condition. As a check, for the case t<t/primewe writee−iω(t−t/prime)=e−iωR(t−t/prime)eωI(t−t/prime)and close the contour as shown in ﬁgure 6.2. The quantity e−iω(t−t/prime)˜G(x,x/prime;ω2) ﬁg6Luhp 6.6. THE INFINITE STRING PROBLEM 93 λ/prime-plane ωRωI L - PPPQQQ J JJ B BBLUHPQQ k Figure 6.2: Contour LC1=L+LUHP closed in UH λ-plane. vanishes on the contour LC1=L+LUHP sincee−iωI(t−t/prime)→0 as ωI→ ∞ and|e−iωR(t−t/prime)|= 1, while we required |˜G(x,x/prime;ω2)| → 0 as|ω| → ∞ . 6.6 The Inﬁnite String Problem pr:ISP1 10 Feb p3We now consider an inﬁnite string where we take σandτto be a constant, and V= 0. Thus our linear operator (cf 1.10) is given by L0=−τd2 dx2. 6.6.1 Derivation of Green’s Function We want to solve the equation /bracketleftBigg −τ∂2 ∂x2+σ∂2 ∂t2/bracketrightBigg GR(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime) for −∞<x,x/prime<∞ (6.19) with the initial condition GR(x,t;x/primet/prime) = 0 for t<t/prime. 94 CHAPTER 6. DYNAMIC PROBLEMS We write the Fourier transform of the Green’s function in terms of λ: ˜G(x,x/prime;ω) =G(x,x/prime;λ=ω2). From 6.13, we know that G(x,x/prime;λ) satisﬁes /bracketleftBigg −τd2 dx2−σλ/bracketrightBigg G(x,x/prime;λ) =δ(x−x/prime) for −∞<x,x/prime<∞. For the case of λlargewe found the solution (4.159) Actually since V= 0 G=i 2√ λc τei√ λ/c2|x−x/prime| or (λ→ω2) ˜G(x,x/prime;ω) =i 2ωc τeiω|x−x/prime|/c. This gives us the retarded Green’s function GR(x,t;x/prime,t/prime) =1 2π/integraldisplay Ldωe−iω(t−t/prime)i 2ωc τeiω c|x−x/prime|. (6.20) Now consider the term eq6GRInfStr e−iω/bracketleftBig (t−t/prime)−|x−x/prime| c/bracketrightBig . We treat this term in two cases: •t−t/prime<|x−x/prime| c. In this case e−iω/bracketleftBig (t−t/prime)−|x−x/prime| c/bracketrightBig →0 asωI→ ∞. But the term ( i/2ω)→+∞asωR→0 in equation 6.20. Thus the integral vanishes along the contour LUHP shown in ﬁgure 6.2 so (using Cauchy’s theorem) equation 6.20 becomes GR(x,t;x/prime,t/prime) =/contintegraldisplay L=/contintegraldisplay L+LUHP= 0. 6.6. THE INFINITE STRING PROBLEM 95 λ/prime-plane ωRωI L - B BB J JJQQQPPP LLHP + Figure 6.3: Contour closed in the lower half λ-plane. •t−t/prime>|x−x/prime| c. In this case 10 Feb p4 e−iω/bracketleftBig (t−t/prime)−|x−x/prime| c/bracketrightBig →0 asωI→ −∞ so we close the contour below as shown in ﬁgure 6.3. Since the ﬁg6Llhp integral vanishes along LLHP, we have GR=/integraltext L=/integraltext L+LLHP. Cauchy’s theorem says that the integral around the closed con- tour is −2πitimes the sum of the residues of the enclosed poles. The only pole is at ω= 0 and its residue is1 2πic 2τ. For this case we obtain GR(x,t;x/prime,t/prime) =−2πi/parenleftbigg1 2πic 2τ/parenrightbigg =c 2τ, which is constant. From these two cases we conclude that GR(x,t;x/prime,t/prime) =c 2τθ/parenleftBigg t−t/prime−|x−x/prime| c/parenrightBigg . (6.21) The function θis deﬁned by the equation θ(u) =/braceleftBigg 0 foru<0 1 foru>0. 96 CHAPTER 6. DYNAMIC PROBLEMS G= 0 G= 0 x/prime−c(t−t/prime) x/prime+c(t−t/prime) x/primeGR(x,t;x/prime,t/prime) c 2τ Figure 6.4: An illustration of the retarded Green’s Function. The situation is illustrated in ﬁgure 6.4. This solution displays some ﬁg6retGF interesting physical properties. •The function is zero for x<x/prime−c(t−t/prime) and forx>x/prime+c(t−t/prime), so it represents an expanding pulse. •The amplitude of the string is c/2τ, which makes sense since for a smaller string tension τwe expect a larger transverse amplitude. •The traveling pulse does not damp out since V= 0. is this right? 6.6.2 Physical Derivation 10 Feb p5 We now explain how to get the solution from purely physical grounds. Consider an impulse ∆ papplied at position x/primeand timet/prime. Applying symmetry, at the ﬁrst instant ∆ py= 1/2 for movement to the left and ∆py= 1/2 for movement to the right. We may also write the velocity ∆vy= ∆py/∆m=1 2/σdx since ∆py= 1/2 and ∆m=σdx. By substi- tutingdx=cdt, we ﬁnd that in the time dta velocity ∆ vy= 1/2cσdt is imparted to the string. This ∆ vyis the velocity of the string portion atdx. By conservation of momentum, the previous string portion must now be stationary. In time dtthe disturbance moves in the ydirection an amount ∆ y=vydt=1 2σc=c 2τ. In these equalities we have used the identity 1 /c2=σ/τ. Thus momentum is continually transferred from point to point (which satisﬁes the condition of conservation of momentum). 12 Feb p1 6.7. SEMI-INFINITE STRING WITH FIXED END 97 6.7 Semi-Inﬁnite String with Fixed End We now consider the problem of an inﬁnite string with one end ﬁxed. We will get the same form of Green’s function. The deﬁning equation is (c.f. 6.19) /bracketleftBigg −τ∂2 ∂x2+σ∂2 ∂t2/bracketrightBigg GR(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime) (6.22) for−∞<t,t/prime<∞; 0<x,x/prime<∞ with the further condition eq6qdblst GR(x,t;x/prime,t/prime) = 0 for x= 0. (6.23) This is called the Dirichlet boundary condition. We could use transform eq6qdblstbc methods to solve this problem, but it is easier to use the method of images and the solution 6.21 to the inﬁnite string problem. To solve this problem we consider an inﬁnite string with sources at 12 Feb p2 x/primeand−x/prime. This gives us a combined force σf(x,t) = [δ(x−x/prime)−δ(x+x/prime)]δ(t−t/prime) and the principle of superposition allows us to write the solution of the problem as the sum of the solutions for the forces separately: GR(x,t;x/prime,t/prime) =c 2τ/bracketleftBigg θ/parenleftBigg t−t/prime−|x−x/prime| c/parenrightBigg −θ/parenleftBigg t−t/prime−|x+x/prime| c/parenrightBigg/bracketrightBigg (6.24) whereuis the solution (c.f. 6.21) of the inﬁnite string with sources at eq6qsst xandx/prime. This solution is shown in ﬁgure 6.5. Since usatisﬁes 6.22 ﬁg6LandG and 6.23, we can identify GR=uforx≥0. The case of a ﬁnite string leads to an inﬁnite number of images to solve (c.f., section 8.7). 6.8 Semi-Inﬁnite String with Free End We now consider a new problem, that of a string with two free ends. The free end Green’s function is GR=c 2τ/bracketleftBigg θ/parenleftBigg t−t/prime−|x−x/prime| c/parenrightBigg +θ/parenleftBigg t−t/prime−|x+x/prime| c/parenrightBigg/bracketrightBigg . 98 CHAPTER 6. DYNAMIC PROBLEMS --GR x −x/prime x/prime (a)GRat timet--GR x−x/prime x/prime (a)GRat timet Figure 6.5: GRatt1=t/prime+1 2x/prime/cand att2=t/prime+3 2x/prime/c. This satisﬁes the equation /bracketleftBigg −τ∂2 ∂x2+σ∂2 ∂t2/bracketrightBigg GR(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime) (6.25) for−∞<t,t/prime<∞;a<x,x/prime<b with the boundary condition eq6qdblst2 d dxGR(x,t;x/prime,t/prime)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=0= 0 which corresponds to κa= 0 andha= 0 is equation 6.3 (c.f., section 1.3.3). The derivative of GR(x= 0) is always zero. Note that 12 Feb p3 /integraldisplayb ad dxθ(x) =θ(b)−θ(a) =/braceleftBigg 1 fora<0<b 0 otherwise .(6.26) which implies d dxθ(x) =δ(x). But for any ﬁxed t−t/primewe can chose an /epsilon1such that the interval [0 ,/epsilon1] is is ﬂat. Therefore ask Bakerd dxGfree end R = 0. 6.9. ELASTICALLY BOUND SEMI-INFINITE STRING 99 Notes about the physics: For a string with a free end, the force on the end point is Fy=τdG dx= 0 atx= 0 which impliesdG dx= 0 atx= 0 if the tension τdoes not vanish. If the tension does vanish at x= 0, then we have a singular point at the origin and do not restrictdG dx= 0 atx= 0. 6.9 Elastically Bound Semi-Inﬁnite String We now consider the problem with boundary condition /bracketleftBigg −d dx+κ/bracketrightBigg GR= 0 for x= 0. The solution can be found using the standard transform method. Do an inverse Fourier transform of the Green’s function in eq. 6.13 for the related problem [ −d/dx +κ]˜G= 0. The frequency space part of this problem is done in problem 4.3. 6.10 Relation to the Eigen Fn Problem We now look at the relation between the general problem and the eigen function problem (normal modes and natural frequencies). The normal mode problem is used in solving [L0−λσ]G(x,x/prime;λ) =δ(x−x/prime) +RBC for whichG(x,x/prime;λ) has poles at the eigen values of L0. We found in chapter 4 that G(x,x/prime;λ) can be written as a bilinear summation G(x,x/prime;λ) =/summationdisplay nun(x)u∗ n(x/prime) λn−λ(6.27) where theun(x) solve the normal mode problem: eq6qA [L0−λnσ]un(x) = 0 + RBC . (6.28) Here we made the identiﬁcation λn=ω2 nwhere theωn’s are the natural eq6qE 100 CHAPTER 6. DYNAMIC PROBLEMS frequencies and the un’s are the normal modes. Recall also that the steady state solution for the force δ(x−x/prime)e−iωt isu(x,t) =G(x,x/prime;λ=ω2+iε)e−iωt.The non-steady state response is GR(x,t;x/prime,t/prime) which is given by GR(x,t;x/prime,t/prime) =/integraldisplaydω 2πe−iω(t−t/prime)˜G(x,x/prime;ω) where ˜G(x,x/prime;ω) =G(x,x/prime;λ=ω2). PlugG(x,x/prime;λ) (from equation 6.27) into the Fourier transformed expression (equation 6.18). This 12 Feb p4 gives GR(x,t;x/prime,t/prime) =1 2π/integraldisplay Ldωe−iω(t−t/prime)/summationdisplay nun(x)u∗ n(x/prime) λn−ω2(6.29) In this equation ωcan be arbitrarily complex. (This equation is very eq6qC diﬀerent (c.f. section 4.6) from the steady state problem G(x,x/prime;λ= ω2+iε)e−iωtwhereωwas real.) Note that we are only interested in t>t/prime, since we have shown already that GR= 0 fort<t/prime. Now we add the lower contour since e−iω(t−t/prime)is small for t/prime>tand This is back- wards ωI<0. This contour is shown in ﬁgure 6.3. The integral vanishes over the curved path, so we can use Cauchy’s theorem to solve 6.29. The poles are at λn=ω2, orω=±√λn. We now perform an evaluation of the integral for one of the terms of the summation in equation 6.29. 12 Feb p5 /integraldisplay L+LLHPdω 2πe−iω(t−t/prime) λn−ω2=−/integraldisplay L+LLHPdω 2πe−iω(t−t/prime) (ω−√λn)(ω+√λn)(6.30) =2πi 2π/bracketleftBigge−i√λn(t−t/prime) 2√λn−ei√λn(t−t/prime) 2√λn/bracketrightBigg =sin√λn(t−t/prime)√λn. By equation 6.29 we get eq6qFa GR(x,t;x/prime,t/prime) =/summationdisplay nun(x)u∗ n(x/prime)sin√λn(t−t/prime)√λn(6.31) where√λn=ωnandt>t/prime. This general solution gives the relationship eq6qF between the retarded Green’s function problem (equation 6.17) and the eigen function problem (eq. 6.28). 6.10. RELATION TO THE EIGEN FN PROBLEM 101 6.10.1 Alternative form of the GRProblem Fort−t/primesmall, eq. 6.31 becomes GR(x,t;x/prime,t/prime)∼/summationdisplay nun(x)u∗ n(x/prime)√λn/radicalBig λn(t−t/prime) = (t−t/prime)/summationdisplay nun(x)u∗ n(x/prime) = (t−t/prime)δ(x−x/prime) σ(x). Where we used the completeness relation 4.113. Thus for t−t/primesmall, 12 Feb 6 theGRhas the form GR(x,t;x/prime,t/prime)|t→t/prime∼(t−t/prime)δ(x−x/prime) σ(x). Diﬀerentiating, this equation gives ∂ ∂tGR(x,t;x/prime,t/prime)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle t→t/prime+=δ(x−x/prime) σ(x). Also, astapproaches t/primefrom the right hand side GR(x,t;x/prime,t/prime) = 0 for t→t/prime−. These results allow us to formulate an alternative statement of the GR problem in terms of an initial value problem. The GRis speciﬁed by the following three equations: /bracketleftBigg L0+σ(x)∂2 ∂t2/bracketrightBigg GR(x,t;x/prime,t/prime) = 0 for t>t/prime+ RBC GR(x,t;x/prime,t/prime) = 0 for t=t/prime σ(x)∂ ∂tGR=δ(x−x/prime) fort=t/prime Hereσ(t)∂ ∂tGR(x,t;x/prime,t/prime) represents a localized unit of impulse at x/prime,t/prime (like ∆p= 1). Thus we have the solution to the initial value problem for which the string is at rest and given a unit of momentum at t/prime. 102 CHAPTER 6. DYNAMIC PROBLEMS We have now cast the statement of the GRproblem in two forms, as a retarded boundary value problem (RBVP) and as an initial value problem (IVP): RBVP =/braceleftBigg/bracketleftBig L0+σ∂2 ∂t2/bracketrightBig GR(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime) + RBC , GR(x,t;x/prime,t/prime) = 0 for t<t/prime IVP =  /bracketleftBig L0+σ(x)∂2 ∂t2/bracketrightBig GR(x,t;x/prime,t/prime) = 0 for t>t/prime,RBC, GR(x,t;x/prime,t/prime) = 0 for t=t/prime, σ(x)∂ ∂tGR=δ(x−x/prime) fort=t/prime. 6.11 Comments on Green’s Function 17 Feb p1 6.11.1 Continuous Spectra In the previous section we obtained the spectral expansion for discrete eigenvalues: GR(x,t;x/prime,t/prime) =∞/summationdisplay n=1un(x)u∗ n(x/prime)√λnsin/radicalBig λn(t−t/prime) (6.32) This gives us an expansion of the Green’s function in terms of the eq6rst natural frequencies. For continuous spectra the sum is replaced by an integral GR=/integraldisplay dλn/summationdisplay αuα λnu∗α λnsin√λn(t−t/prime)√λn where we have included a sum over degeneracy index α(c.f. 4.108). Note that this result follows directly because the derivation in the pre- vious section did not refer to whether we had a discrete or continuous spectrum. 6.11.2 Neumann BC Recall that the RBC for an open string, equation 1.18, is /bracketleftBigg −d dx+κa/bracketrightBigg GR= 0 for x=a. 6.11. COMMENTS ON GREEN’S FUNCTION 103 Ifκa→0 (Neumann boundary condition) then the boundary condition for the normal mode problem will be ( d/dx )u(x) = 0 which will have a constant solution, i.e., λ1= 0. We cannot substitute λ1= 0 into Ask why equation 6.32, but instead must take the limit as λ1approaches zero. Physically, this corresponds to taking the elasticity κaas a small quan- tity, and then letting it go to zero. In this case we can write equation 6.32 with the λ1eigen value separated out: GR(x,t;x/prime,t/prime)λ1→0−→u1(x)u∗ 1(x/prime)(t−t/prime) (6.33) +/summationdisplay nun(x)u∗ n(x/prime)√λnsin/radicalBig λn(t−t/prime) t→∞−→u1(x)u∗ 1(x/prime)(t−t/prime). The last limit is true because the sum oscillates in t. The Green’s function represents the response to a unit momentum, but κa= 0 which means there is no restoring force. Thus a change in momentum ∆p= 1 is completely imparted to the string, which causes the string to acquire a constant velocity, so its amplitude increases linearly with time. Note that equation 6.33 would still be valid if we had taken λ1= 0 in our derivation of the Green’s function as a bilinear sum. In this case equation 6.30 would have a double pole for λ1= 0, so the residue would involve the derivative of the numerator, which would give the linear factor of t−t/prime. Consider a string subject to an arbitrary force σ(x)f(x,t). Remember 17 Feb p2 thatσ(x)f(x,t) =δ(t−t/prime)δ(x−x/prime) givesGR(x,t;x/prime,t/prime). A general force σ(x)f(x,t) gives a response u(x,t) which is a superposition of Green’s pr:GenResp1 functions: u(x,t) =/integraldisplayt 0dt/prime/integraldisplayb adx/primeG(x,t;x/prime,t/prime)σ(x/prime)f(x/prime,t/prime) with no boundary terms ( u(x,0) = 0 =d dtu(x,0)). Now plug in the Green’s function expansion 6.33 to get u(x,t) =u1(x)/integraldisplayt 0dt/prime(t−t/prime)/integraldisplayb adx/primeu∗ 1(x/prime)σ(x/prime)f(x/prime,t/prime) +∞/summationdisplay n=2un(x)√λn/integraldisplayt 0dt/primesin/radicalBig λn(t−t/prime)/integraldisplayb adx/primeu∗ n(x/prime)σ(x/prime)f(x/prime,t/prime). 104 CHAPTER 6. DYNAMIC PROBLEMS Note that again the summation terms oscillate with frequency ωn. The spatial dependence is given by the un(x). The coeﬃcients give the projection of σ(x)f(x,t) ontou∗ n(x). In theλ1= 0 case the u1(x) term is constant. 6.11.3 Zero Net Force Now let F(t)≡(const.)/integraldisplay dt/primeσ(x/prime)f(x/prime,t/prime) whereF(t/prime) represents the total applied force at time t/prime. IfF(t/prime) = 0, then there are no terms which are linearly increasing in time contribut- ing to the response u(x,t). This is a meaningful situation, correspond- ing to a disturbance which sums to zero. The response is purely oscil- latory; there is no growth or decay. ask baker about last part not included here 6.12 Summary 1. The retarded Green’s function GRsolves /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg GR(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime) + RBC fora<x,x/prime<b; allt,t/prime. with the condition GR(x,t;x/prime,t/prime) = 0 for t<t/prime. The advanced Green’s function GAsolves the same equation, but with the condition GA(x,t;x/prime,t/prime) = 0 for t>t/prime. The retarded Green’s function gives the response of the string (initially at rest) to a unit of momentum applied to the string at a point in time t/primeat a pointx/primealong the string. The advanced Green’s function gives the initial motion of the string such that a unit of momentum applied at x/prime,t/primecauses it to come to rest. 6.13. REFERENCES 105 2. The retarded Green’s function can be written in terms of the steady state Green’s function: GR(x,t;x/primet/prime) =/integraldisplay Ldω 2πe−iω(t−t/prime)˜G(x,x/prime;ω). 3. The retarded Green’s function for an inﬁnite string with σandτ constant, and V= 0 is GR(x,t;x/prime,t/prime) =c 2τθ/parenleftBigg t−t/prime−|x−x/prime| c/parenrightBigg . 4. The retarded Green’s function for a semi-inﬁnite string with a ﬁxed end,σandτconstant, and V= 0 is GR(x,t;x/prime,t/prime) =c 2τ/bracketleftBigg θ/parenleftBigg t−t/prime−|x−x/prime| c/parenrightBigg −θ/parenleftBigg t−t/prime−|x+x/prime| c/parenrightBigg/bracketrightBigg . 5. The retarded Green’s function for a semi-inﬁnite string with a free end,σandτconstant, and V= 0 is GR=c 2τ/bracketleftBigg θ/parenleftBigg t−t/prime−|x−x/prime| c/parenrightBigg +θ/parenleftBigg t−t/prime−|x+x/prime| c/parenrightBigg/bracketrightBigg . 6. The retarded Green’s function can be written in terms of the eigen functions as GR(x,t;x/prime,t/prime) =/summationdisplay nun(x)u∗ n(x/prime)sin√λn(t−t/prime)√λn. 6.13 References A good reference is [Stakgold67b, p246ﬀ]. This material is developed in three dimensions in [Fetter80, p311ﬀ]. 106 CHAPTER 6. DYNAMIC PROBLEMS Chapter 7 Surface Waves and Membranes Chapter Goals: •Show how the equation describing shallow water surface waves is related to our most general diﬀer- ential equation. •Derive the equation of motion for a 2-dimensional membrane and state the corresponding regular boundary conditions. 7.1 Introduction In this chapter we formulate physical problems which correspond to equations involving more than one dimension. This serves to moti- vate the mathematical study of N-dimensional equations in the next chapter. 107 108 CHAPTER 7. SURFACE WAVES AND MEMBRANES x b (x)h(x) Figure 7.1: Water waves moving in channels. 7.2 One Dimensional Surface Waves on Fluids 7.2.1 The Physical Situation 17 Feb p3 Consider the physical situation of a surface wave moving in a channel1. This situation is represented in ﬁgure 7.1. The height of equilibrium pr:surf1 ﬁg:7.1 ish(x) and the width of the channel is b(x). The height of the wave pr:hww1z(x,t) can then be written as z(x,t) =h(x) +u(x,t) whereu(x,t) is the deviation from equilibrium. We now assume the shallow wave case u(x,t)/lessmuchh(x). This will allow us to linearize the Navier–Stokes equation. 7.2.2 Shallow Water Casepr:shal1 This is the case in which the height satisﬁes the condition h(x)/lessmuch λwhereλis the wavelength. In this case the motion of the water pr:lambda2 is approximately horizontal. Let S(x) =h(x)b(x). The equation of continuity and Newton’s law (i.e., the Navier–Stokes equation) then give −∂ ∂x/parenleftBigg gS(x)∂ ∂x/parenrightBigg u+b(x)∂2 ∂t2u(x,t) = 0, 1This material corresponds to FW p. 357–363. 7.3. TWO DIMENSIONAL PROBLEMS 109 which is equivalent to the 1-dimensional string, where σ(x)⇒b(x) and 17 Feb p4 τ(x)⇒gS(x). Consider the case in which b(x) is independent of x: −∂ ∂x/parenleftBigg gh(x)∂ ∂x/parenrightBigg u+∂2 ∂t2u(x,t) = 0. (7.1) This corresponds to σ= 1 andτ=gh(x). eq7shallow Propagation of shallow water waves looks identical to waves on a string. For example, in problem 3.5, h(x) =xgives the Bessel’s equa- tion, with the identiﬁcation τ(x) =xandV(x) =m2/x. As another example, take h(x) to be constant. This gives us wave propagation with c=/radicalBig τ/σ,τ=gh, andσ= 1. So the velocity of a water wave is c=√gh. The deeper the channel, the faster the velocity. This partially explains wave breaking: The crest sees more depth than the trough. 7.3 Two Dimensional Problems 17 Feb p5 We now look at the 2-dimensional problem, that of an elastic membrane2. We denote the region of the membrane by Rand the perimeter (1- pr:elmem1 dimensional “surface”) by S. The potential energy diﬀerential for an element of a 1-dimensional string is dU=1 2τ(x)/parenleftBiggdu dx/parenrightBigg2 dx. In the case of a 2-dimensional membrane we replace u(x) withu(x,y) = u(x). In this case the potential energy diﬀerence is (see section 2.4.2) dU =1 2τ(x,y) /parenleftBiggdu dx/parenrightBigg2 +/parenleftBiggdu dy/parenrightBigg2 dxdy (7.2) =1 2τ(x)(∇u)2dx (7.3) 2This is discussed on p. 271–288 of FW. 110 CHAPTER 7. SURFACE WAVES AND MEMBRANES whereτis the tension of the membrane. Note that there is no mixed termd dxd dyusince the medium is homogeneous. The total potential energy is given by the equation U=/integraldisplay Rdx1 2τ(x)(∇u)2, whereτ(x) is the surface tension. In this equation dx=dxdy and (∇u)2= (∇u)·(∇u). The total kinetic energy is T=1 2mv2=/integraldisplay Rdx1 2σ(x)/parenleftBigg∂u ∂t/parenrightBigg2 . Now think of the membrane as inserted in an elastic media. We then get an addition to the U(x) energy due to elasticity,1 2V(x)u(x,t)2. We also add an additional force which will add to the potential energy: f(x,t)σ(x)dxu(x,t) =/parenleftBiggforce mass/parenrightBigg/parenleftBiggmass length/parenrightBigg (length) (displacement) . The Lagrangian is thus pr:lagr1 L=1 2/integraldisplay Rdx×  σx/parenleftBigg∂ ∂tu/parenrightBigg2 −τ(x)(∇u)2−V(x)u(x)2−f(x,t)σ(x)dxu(x,t) . Notice the resemblance of this Lagrangian to the one for a one dimen- sional string (see section 2.4.2). 19 Feb p1 We apply Hamiltonian Dynamics to get the equation of motion: 19 Feb p2 /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg u(x,t) =σ(x)f(x,t) whereL0=−∇(τ(x)∇)−V(x). This is identical to the equation of motion for a string except now in two dimensions. It is valid everywhere forxinside the region R. 7.3. TWO DIMENSIONAL PROBLEMS 111 r r (0,0) ( a,0)(a,b) (0 ,b) (0,y) ( a,y) Figure 7.2: The rectangular membrane. 7.3.1 Boundary Conditions pr:bc2 Elastically Bound Surface The most general statement of the boundary condition for an elastically bound surface is [ˆn· ∇+κ(x)]u(x,t) =h(x,t) forxonS. (7.4) In this equation the “surface” Sis the perimeter of the membrane, ˆ nissone the outward normal for a point on the perimeter, κ(x) =k(x)/τ(x) is the eﬀective spring constant at a point on the boundary, and h(x,t) = f(x,t)/τ(x) is an external force acting on the boundary S. Periodic Boundary Conditions 19 Feb p3 pr:pbc2We now consider the case of a rectangular membrane, illustrated in ﬁgure 7.2, with periodic boundary conditions:ﬁg:7.2 u(0,y) =u(a,y) for 0 ≤y≤b, (7.5) u(x,0) =u(x,b) for 0 ≤x≤a, and ∂2 ∂x2u(x,y)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=0=∂2 ∂x2u(x,y)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=afor 0≤y≤b, 112 CHAPTER 7. SURFACE WAVES AND MEMBRANES ∂2 ∂x2u(x,y)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle y=0=∂2 ∂x2u(x,y)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle y=bfor 0≤x≤a. In this case we can consider the region Rto be a torus. 7.4 Example: 2D Surface Waves We now give one last 2-dimensional example3. We consider a tank of 19 Feb p4 water whose bottom has arbitrary height h(x) and look at the surface waves. This example connects the 1-dimensional surface wave problem and the 2-dimensional membrane problem. For this problem the vertical displacement is given by z(x,t) =h(x) +u(x,t), withλ/greatermuchh(x) for the shallow water case and u/lessmuchh(x). Thus (using 7.1) our equation of motion is /bracketleftBigg −∇ · (gh(x)∇) +∂2 ∂t2/bracketrightBigg u(x,t) =f(x,t). Note that in this equation we have σ= 1. For this problem we take the Neumann natural boundary condition: ˆn· ∇u(x,t) = 0 and∂ ∂tu⊥=−g∇⊥u|S forxonS. This is the case of rigid walls. The latter equation just means that there is no perpendicular velocity at the surface. The case of membranes for a small displacement is the same as for surface waves. We took σ= 1 andτ(x) =gh(x). The formula for all these problems is just 19 Feb p5 /bracketleftBigg L0+σ(x)∂2 ∂t2/bracketrightBigg u(x,t) =σ(x)f(x,t) for xinR whereL0=−∇(τ(x)∇) +V(x). (Note that τ(x) is not necessarily tension.) We let x= (x1,...,x n) and ∇= (∂/∂x 1,...,∂/∂x n). The boundary conditions can be elastic or periodic. 3This one comes from FWp. 363–366. 7.5. SUMMARY 113 7.5 Summary 1. The equation for shallow water surface waves and the equation for string motion are the same if we identify gravity times the cross-sectional area with “tension”, and the width of the channel with “mass density”. 2. The two-dimensional membrane problem is characterized by the wave equation /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg u(x,t) =σ(x)f(x,t) whereL0=−∇(τ(x)∇)−V(x), subject to either an elastic boundary condition, [ˆn· ∇+κ(x)]u(x,t) =h(x,t) forxonS, or a periodic boundary condition. 7.6 References The material on surface waves is covered in greater depth in [Fetter80, p357ﬀ], while the material on membranes can be found in [Fetter80, p271]. 114 CHAPTER 7. SURFACE WAVES AND MEMBRANES Chapter 8 Extension to N-dimensions Chapter Goals: •Describe the diﬀerent sorts of boundaries and boundary conditions which can occur for the N- dimensional problem. •Derive the Green’s identities for the N-dimensional case. •Write the solution for the N-dimensional problem in terms of the Green’s function. •Describe the method of images.22 Feb p1 8.1 Introduction In the previous chapter we obtained the general equation in two dimen- sions: /bracketleftBigg L0+σ(x)∂2 ∂t2/bracketrightBigg u(x,t) =σ(x)f(x,t) for xinR (8.1) where eq8usf L0=−∇ ·τ(x)∇+V(x). This is immediately generalizable to N-dimensions. We simply let x= (x1,...,x n) and ∇= (∂/∂x 1,...,∂/∂x n). and we introduce the 115 116 CHAPTER 8. EXTENSION TO N-DIMENSIONS notation ˆn· ∇u≡∂u/∂n .Ris now a region in N-dimensional space andSis the (N−1)-dimensional surface of R. The boundary conditions can either be elastic or periodic: pr:bc3 1.Elastic : The equation for this boundary condition is [ˆn· ∇+K(x)]u(x,t) =h(x,t) for xonS. (8.2) The termK(x) is like a spring constant which determines the eq8.1b properties of the medium on the surface, and h(x,t) is an exter- nal force on the boundary. These terms determine the outward gradient of u(x,t). 2.Periodic : For the two dimensional case the region looks like 7.2 pr:pbc3 and the periodic boundary conditions are 7.5 and following. In theN-dimensional case the region Ris anN-cube. Connecting matching periodic boundaries of Syields anN-torus in (N+ 1)- dimensional space. To uniquely specify the time dependence of u(x,t) we must specify the initial conditions u(x,t)|t=0=u0(x) for xinR (8.3) ∂ ∂t(x,t)|t=0=u1(x) for xinR. (8.4) For the 1-dimensional case we solved this problem using Green’s Identi- ties. In section 8.3 we will derive the Green’s Identities for N-dimensions. 22 Feb p2 8.2 Regions of Interest There are three types of regions of interest: the Interior problem, the Exterior problem, and the All-space problem. 1.Interior problem : HereRis enclosed in a ﬁnite region bounded by pr:intprob1 S. In this case we expect a discrete spectrum of eigenvalues, like one would expect for a quantum mechanical bound state problem or for pressure modes in a cavity. 8.3. EXAMPLES OF N-DIMENSIONAL PROBLEMS 117 2.Exterior problem : HereRextend to inﬁnity in all directions but is pr:extprob1 excluded by a ﬁnite region bounded by S. In this case we expect a continuum spectrum if V > 0 and a mixed spectrum is V < 0. This is similar to what one would expect for quantum mechanical scattering. 3.All-space problem : HereRextends to inﬁnity in all directions pr:allprob1 and is not excluded from any region. This can be considered a degenerate case of the Exterior problem. 8.3 Examples of N-dimensional Problems 8.3.1 General Response pr:GenResp2 In the following sections we will show that the N-dimensional general response problem can be solved using the Green’s function solution to the steady state problem. The steps are identical to those for the single dimension case covered in chapter 6. G(x,x/prime;λ=ω2) = ˜G(x,x/prime;λ) →GR(x,t;x/prime,t/prime) retarded →u(x,t) General Response . 8.3.2 Normal Mode Problempr:NormMode3 The normal mode problem is given by the homogeneous diﬀerential equation/parenleftBigg L0+σ(x)∂2 ∂t2/parenrightBigg u(x,t) = 0. Look for solutions of the form u(x,t) =e−iωntun(x). The natural frequencies are ωn=√λn, whereλnis an eigen value. The normal modes are eigen functions of L0. Note: we need RBC to ensure thatL0is Hermitian. 118 CHAPTER 8. EXTENSION TO N-DIMENSIONS 8.3.3 Forced Oscillation Problem24 Feb p2 pr:fhop2 The basic problem of steady state oscillation is given by the equation /parenleftBigg L0+σ∂2 ∂t2/parenrightBigg u(x,t) =e−iωtσ(x)f(x) for x∈R and (for example) the elastic boundary condition (ˆn· ∇+K)u=h(x)e−iωtforx∈S. We look for steady state solutions of the form u(x,t) =e−iωtu(x,ω). The value of ωis chosen, so this is not an eigen value problem. We assert u(x,ω) =/integraldisplay x/prime∈Rdx/primeG(x,x/prime;λ=ω+iε)σ(x/prime)f(x/prime) +/integraldisplay x/prime∈Sdx/primeτ(x/prime)G(x,x/prime;λ=ω+iε)σ(x)h(x/prime). The ﬁrst term gives the contribution due to forces on the volume and the second term gives the contribution due to forces on the surface. In the special case that σ(x)f(x) =δ(x−x/prime), we have u(x,t) =e−iωtG(x,x/prime;λ=ω2). 8.4 Green’s Identities In this section we will derive Green’s 1st and 2nd identities for the N-dimensional case. We will use the general linear operator for N- dimensions L0=−∇ · (τ(x)∇) +V(x) and the inner product for N-dimensions /angbracketleftS,L 0u/angbracketright=/integraldisplay dxS∗(x)L0u(x). (8.5) eq8tst 8.5. THE RETARDED PROBLEM 119 8.4.1 Green’s First Identity pr:G1Id2 The derivation here generalizes the derivation given in section 2.1. /angbracketleftS,L 0u/angbracketright=/integraldisplay RdxS∗(x)[−∇ · (τ(x)∇) +V(x)]u(x) (8.6) integrate 1st term by parts =/integraldisplay Rdx[−∇ · (S∗τ(x)∇u) + (∇S∗)τ(x)∇u+S∗Vu] integrate 1st term using Gauss’ Theorem =−/integraldisplay SdSˆn·(S∗τ(x)∇u) +/integraldisplay R[S∗Vu+ (∇ ·S∗τ(x)∇u)]dx This is Green’s First Identity generalized to N-dimensions: /angbracketleftS,L 0u/angbracketright=−/integraldisplay SdSˆn·(S∗τ(x)∇u)+/integraldisplay R[S∗Vu+(∇·S∗τ(x)∇u)]dx.(8.7) Compare this with 2.3. 8.4.2 Green’s Second Identity pr:G2Id2 22 Feb p3We now interchange Sandu. In the quantity /angbracketleftS,L 0u/angbracketright − /angbracketleftL0S,u/angbracketrightthe symmetric terms will drop out, i.e., the second integral in 8.7 is can- celled. We are left with /angbracketleftS,L 0u/angbracketright − /angbracketleftL0S,u/angbracketright=/integraldisplay SdSˆn·[−S∗τ(x)∇u+uτ(x)∇S∗]. 8.4.3 Criterion for Hermitian L0pr:HermOp2 Ifu,S∗satisfy the RBC, then the surface integral in Green’s second identity vanishes. This leaves /angbracketleftS,L 0u/angbracketright−/angbracketleftL0S,u/angbracketright, which means that L0 is a hermitian (or self-adjoint) operator: L=L†. 8.5 The Retarded Problem 8.5.1 General Solution of Retarded Problem We now reduce 8.1 to a simpler problem. If this is an initial value problem, then u(x,t) is completely determined by equations 8.1, 8.2, pr:IVP2 120 CHAPTER 8. EXTENSION TO N-DIMENSIONS 8.3, and 8.4. We look again at GRwhich is the response of a system to a unit force:/bracketleftBigg L0+σ(x)∂2 ∂t2/bracketrightBigg GR(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime) for x,x/primeinR. (8.8) We also require the retarded Green’s function to satisfy RBC and the eq8tB initial condition GR= 0 fort<t/prime. We now use the result from problem 22 Feb p4 4.2: u(x,t) =/integraldisplay x/prime∈Rdx/prime/integraldisplayt 0dt/primeGR(x,t;x/prime,t/prime)σ(x/prime)f(x/prime,t/prime) +/integraldisplay x/prime∈Sdx/primeτ(x/prime)/integraldisplayt 0dt/primeGR(x,t;x/prime,t/prime)h(x/prime,t/prime) +/integraldisplay x/prime∈Rdx/primeσ(x/prime)/bracketleftbigg GR(x,t;x/prime,0)u1(x/prime) −∂ ∂t/primeGR(x,t;x/prime,t/prime)u0(x/prime)/bracketrightbigg (8.9) The ﬁrst line gives the volume sources, the second gives the surface eq8ugen R Horn says eval ˙GRatx ’=0sources, and the third and fourth gives the contribution from the initial conditions. Since the deﬁning equations for GRare linear in volume, surface, and initial condition terms, we were able to write down the solutionu(x,t) as a linear superposition of the GR.ask Baker about limits Note that we recover the initial value of u(x,t) in the limit t→t/prime. This is true since in the above equation we can substitute lim t→t/primeGR(x,t;x/prime,t/prime) = 0 and lim t→t/primeGR(x,t;x/prime,t/prime) =δ(x−x/prime) σ(x/prime). 8.5.2 The Retarded Green’s Function in N-Dim.pr:GR2 By using 8.9 we need only solve 8.8 to solve 8.1. In section 6.3 we found thatGRcould by determined by using a Fourier Transform. Here we follow the same procedure generalized to N-dimensions. The Fourier transform of GRinN-dimensions is GR(x,t;x/prime,t/prime) =/integraldisplay Ldω 2πe−iω(t−t/prime)˜G(x,x/prime;ω) 8.5. THE RETARDED PROBLEM 121 whereLis a line in the upper half plane parallel to the real axis (c.f., 22 Feb p5 section 6.4), since ˜Gis analytic in the upper half plane (which is due to the criterion GR= 0 fort<t/prime). Now the problem is simply to evaluate the Fourier Transform. By the same reasoning in section 6.3 the Fourier transform of GRis iden- tical to the Green’s function for the steady state case: ˜G(x,x/prime;ω) =G(x,x;λ=ω2). Recall that the Green’s function for the steady state problem satisﬁes [L0−λσ]G(x,x/prime;λ) =δ(x−x/prime) for x,x/prime∈R,RBC (8.10) We have reduced the general problem in Ndimensions (equation8.1) to the steady state Green’s function problem in N-dimensions. 22 Feb p6 8.5.3 Reduction to Eigenvalue Problem pr:efp3 The eigenvalue problem (i.e., the homogeneous equation) in N-dimensions is L0un(x) =λnσun(x)x∈R,RBC (8.11) Theλn’s are the eigen values of L0. SinceL0is hermitian, the λn’sLots of work are real. The un(x)’s are the corresponding eigenfunctions of L0. We can also prove orthonormality (using the same method as in the single dimension case) /integraldisplay Rdx/summationdisplay αuα∗ n(x)uα m(x) = 0 ifλn/negationslash=λm. This follows from the hermiticity of L0. Note that because we are now inN-dimensions, the degeneracy may now be inﬁnite. 22 Feb p7 By using the same procedure as in chapter 4, we can write a solution of eq. 8.8 expanded in terms of a solution of 8.11: G(x,x/prime;λ) =/summationdisplay nun(x)u∗ n(x/prime) λn−λ. Note that the sum would become an integral for a continuous spectrum. The methods of chapter 4 also allow us to construct the δ-function 22 Feb p8 122 CHAPTER 8. EXTENSION TO N-DIMENSIONS representation δ(x−x/prime) =σ(x/prime)/summationdisplay nun(x)u∗ n(x/prime), which is also called the completeness relation. All we have left is to discuss the physical interpretation of G. 24 Feb p1 8.6 Region R 24 Feb p3 8.6.1 Interiorpr:intprob2 In the interior problem the Green’s function can be written as a discrete spectrum of eigenvalues. In the case of a discrete spectrum we have G(x,x/prime;λ=ω2) =/summationdisplay nun(x)u∗ n(x/prime) λn−ω2 8.6.2 Exteriorpr:extprob2 For the exterior problem the sums become integrals and we have a continuous spectrum: G(x,x/prime;λ) =/integraldisplay λndλn/summationtext αun(x)u∗ n(x/prime) λn−λ. In this case we take λ=ω2+iε.Gnow has a branch cut for all real λ’s, which means that there will be two linearly independent solutions which correspond to whether we approach the real λaxis from above or below. We choose ε>0 to correspond to the physical out going wave solution. ask Baker about omitted material 24 Feb p4 8.7 The Method of Images 24 Feb p5 pr:MethIm1We now present an alternative method for solving N-dimensional prob- lem which is sometimes useful when the problem exhibits suﬃcient 8.7. THE METHOD OF IMAGES 123 symmetry. It is called the Method of Images . For simplicity we con- sider a one dimensional problem. Consider the GRproblem for periodic boundary conditions with constant coeﬃcients. /parenleftBigg L0−λσ∂2 ∂t2/parenrightBigg GR=δ(x−x/prime)δ(t−t/prime) 0 ≤x≤l. 8.7.1 Eigenfunction Method We have previously solved this problem by using an eigen function expansion solution (equation 6.31) GR=/summationdisplay nun(x)u∗ n(x/prime)√λnsin/radicalBig λn(t−t/prime). For this problem the eigen functions and eigen values are un(x) =/parenleftbigg1 l/parenrightbigg1/2 e2πinx/ln= 0,±1,±2,... λn=/parenleftbigg2πin l/parenrightbigg2 n= 0,±1,±2,... 8.7.2 Method of Images The method of images solution uses the uniqueness theorem. Put im- ages over −∞ to∞in region of length λ. Φ =/parenleftbiggc 2τ/parenrightbigg θ/parenleftBigg t−t/prime−|x−x/prime| c/parenrightBigg . This is not periodic over 0 to l. Rather, it is over all space. Our GRis GR=c 2τ∞/summationdisplay n=−∞θ/parenleftBigg t−t/prime−|x−x/prime−nl| c/parenrightBigg . Notice that this solution satisﬁes /parenleftBigg L0+σ∂2 ∂t2/parenrightBigg GR=∞/summationdisplay n=−∞δ(x−x/prime−nl)− ∞<x,x/prime<∞. 124 CHAPTER 8. EXTENSION TO N-DIMENSIONS However, we only care about 0 <x<l /parenleftBigg L0+σ∂2 ∂t2/parenrightBigg GR=∞/summationdisplay n=−∞δ(x−x/prime−nl) 0<x,x/prime<l. Since the other sources are outside the region of interest they do not aﬀect this equation. Our Green’s function is obviously periodic. The relation between these solution forms is a Fourier series. 8.8 Summary 1. For the exterior problem, the region is outside the boundary and extends to the boundary. The the interior problem, the boundary is inside the boundary and has ﬁnite extent. For the all-space problem, there is no boundary. The boundary conditions can be either elastic or periodic, or in the case that there is no boundary, the function must be regular at large and/or small values of its parameter. 2. The Green’s identities for the N-dimensional case are /angbracketleftS,L 0u/angbracketright=−/integraldisplay SdSˆn·(S∗τ(x)∇u)+/integraldisplay R[S∗Vu+(∇·S∗τ(x)∇u)]dx, /angbracketleftS,L 0u/angbracketright − /angbracketleftL0S,u/angbracketright=/integraldisplay sdSˆn·[−S∗τ(x)∇u+uτ(x)∇S∗]. 3. The solution for the N-dimensional problem in terms of the Green’s function is u(x,t) =/integraldisplay x/prime∈Rdx/prime/integraldisplayt 0dt/primeGR(x,t;x/prime,t/prime)σ(x/prime)f(x/prime,t/prime) +/integraldisplay x/prime∈Sdx/primeτ(x/prime)/integraldisplayt 0dt/primeGR(x,t;x/prime,t/prime)h(x/prime,t/prime) +/integraldisplay x/prime∈Rdx/primeσ(x/prime)/bracketleftbigg GR(x,t;x/prime,0)u1(x/prime) −∂ ∂t/primeGR(x,t;x/prime,t/prime)u0(x/prime)/bracketrightbigg . 8.9. REFERENCES 125 4. The method of images is applicable if the original problem ex- hibits enough symmetry. The method is to replace the original problem, which has a boundary limiting region of the solution, with a new problem in which the boundary is taken away and sources are placed in the region which was excluded by the bound- ary such that the solution will satisfy the boundary conditions of the original problem. 8.9 References The method of images is covered in most electromagnetism books, for example [Jackson75, p54ﬀ], [Griﬃths81, p106ﬀ]; a Green’s function application is given in [Fetter80, p317]. The other material in this chapter is a generalization of the results from the previous chapters. 126 CHAPTER 8. EXTENSION TO N-DIMENSIONS Chapter 9 Cylindrical Problems Chapter Goals: •Deﬁne the coordinates for cylindrical symmetry and obtain the appropriate δ-function. •Write down the Green’s function equation for the case of circular symmetry. •Use a partial expansion for the Green’s function to obtain the radial Green’s function equation for the case of cylindrical symmetry. •Find the Green’s function for the case of a circular wedge and for a circular membrane.29 Feb p1 9.1 Introduction In the previous chapter we considered the Green’s function equation (L0−λσ(x))G(x,x/prime;λ) =δ(x−x/prime) for x,x/prime∈R where L0=−∇ · (τ(x)∇) +V(x) subject to RBC, which are either for the elastic case (ˆn· ∇+κ(S))G(x,x/prime;λ) = 0 for x∈S (9.1) 127 128 CHAPTER 9. CYLINDRICAL PROBLEMS or the periodic case. In this chapter we want to systematically solve this problem for 2-dimensional cases which exhibit cylindrical symmetry. 9.1.1 Coordinates A point in space can be represented in cartesian coordinates as x=ˆix+ˆjy. Instead of the coordinate pair ( x,y) we may choose polar coordinates pr:CartCoord1 (r,φ). The transformation to cartesian coordinates is x=rcosφ y =rsinφ while the transformation to polar coordinates is (for tan φdeﬁned on the interval −π/2<φ<π/ 2) pr:r1 r=/radicalBig x2+y2 φ=  tan−1(y/x) for x>0,y> 0 tan−1(y/x) +πforx<0 tan−1(y/x) + 2πforx>0,y< 0 A diﬀerential of area for polar coordinates is related by that for carte- sian coordinates by a Jacobian (see Boas, p220): pr:jak1 dxdy =dA =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleJ/parenleftBiggx,y r,φ/parenrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingledrdφ =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle/parenleftBigg∂(x,y) ∂(r,φ)/parenrightBigg/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingledrdφ =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle(∂x/∂r ) (∂x/∂φ ) (∂y/∂r ) (∂y/∂φ )/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingledrdφ =/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglecosφ−rsinφ sinφ r cosφ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingledrdφ =rdrdφ 9.1. INTRODUCTION 129 By expanding dxanddyin terms of dranddφ, we can write the diﬀerential of arc length in polar coordinates (see Boas, p224) ds=/radicalBig dx2+dy2=/radicalBig dr2+r2dφ2. The diﬀerential operator becomes (see Boas, p252,431) pr:grad1 gradient ∇u= ˆr∂u ∂r+ˆφ1 r∂u ∂φ, divergence ∇ ·B=1 r∂ ∂r(rBr) +1 r∂ ∂φ(Bφ). LetB=τ(x)∇x, then ∇ ·(τ(x)∇u(x)) =1 r∂ ∂r/parenleftBigg rτ(x)∂u ∂r/parenrightBigg +1 r∂ ∂φ/parenleftBigg τ(x)∂u ∂φ/parenrightBigg . (9.2) 9.1.2 Delta Function29 Feb p2 TheN-dimensional δ-function is deﬁned by the property pr:DeltaFn2 f(x) =/integraldisplay d2xf(x)δ(x−x/prime). In polar form we have (since dxdy =rdrdφ ) f(x) =f(r,φ) =/integraldisplay r/primedr/primedφ/primef(r/prime,φ/prime)δ(x−x/prime). By comparing this with f(r,φ) =/integraldisplay dr/primedφ/primef(r/prime,φ/prime)δ(r−r/prime)δ(φ−φ/prime) we identify that the delta function can be written in polar coordinates in the form pr:delta1 δ(x−x/prime) =δ(r−r/prime) rδ(φ−φ/prime). 29 February 1988 130 CHAPTER 9. CYLINDRICAL PROBLEMS "!# aJJ ] 3S Rˆnˆφˆrφ Figure 9.1: The region Ras a circle with radius a. 9.2 GF Problem for Cylindrical Sym. The analysis in the previous chapters may be carried into cylindrical coordinates. For simplicity we consider cylindrical symmetry: τ(x) = τ(r),σ(x) =σ(r), andV(x) =V(r). Thus 9.2 becomes ∇(τ(r)∇u(x)) =1 r∂ ∂r/parenleftBigg rτ(r)∂u ∂r/parenrightBigg +τ(r) r2∂2u ∂φ2. The equation for the Green’s function (L0−λσ)G(r,φ;r/prime,φ/prime) =1 rδ(r−r/prime)δ(φ−φ/prime) becomes (for r,φ∈R) /bracketleftBigg −1 r∂ ∂r/parenleftBigg rτ(r)∂ ∂r/parenrightBigg −τ(r) r2∂2 ∂φ2+V(r)−λσ(r)/bracketrightBigg G=1 rδ(r−r/prime)δ(φ−φ/prime). (9.3) HereRmay be the interior or the exterior of a circle. It could also be a wedge of a circle, or an annulus, or anything else with circular symmetry. For deﬁniteness, take the region Rto be the interior of a circle of radiusa(see ﬁgure 9.1) and apply the elastic boundary condition 9.1. We now deﬁne the elasticity on the boundary S,κ(S) =κ(φ). We ﬁg10a must further specify κ(φ) =κ, a constant, since if κ=κ(φ), then we might not have cylindrical symmetry. Cylindrical symmetry implies ˆn· ∇=∂/∂r so that the boundary condition (ˆn· ∇+κ(S))G(r,φ;r/prime,φ/prime) = 0 for r=a is now /parenleftBigg∂ ∂r+κ/parenrightBigg G(r,φ;r/prime,φ/prime) = 0 for r=a. 9.3. EXPANSION IN TERMS OF EIGENFUNCTIONS 131 We also need to have Gperiodic under φ→φ+ 2π. So 29 Feb p3 G(r,0;r/prime,φ/prime) =G(r,2π;r/prime,φ/prime) and ∂G ∂φ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle φ=0=∂G ∂φ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle φ=2π. We have now completely respeciﬁed the Green’s function for the case of cylindrical symmetry. 9.3 Expansion in Terms of Eigenfunctions Since the Green’s function is periodic in φand sinceφonly appears in the operator as ∂2/∂φ2, we use an eigenfunction expansion to separate out theφ-dependence. Thus we look for a complete set of eigenfunctions pr:efexp1 which solve −∂2 ∂φ2um(φ) =µmum(φ) (9.4) forum(φ) periodic. The solutions of this equation are um(φ) =1√ 2πeimφform= 0,±1,±2,... and the eigenvalues are µm=m2form= 0,±1±2,.... (Other types of regions would give diﬀerent eigenvalues µm). Since pr:CompRel4 this set of eigenfunctions is complete it satisﬁes the expansion δ(φ−φ/prime) =/summationdisplay mum(φ)u∗ m(φ/prime). 9.3.1 Partial Expansion We now want to ﬁnd Gm(r,r/prime;λ) which satisﬁes the partial expansion (using the principle of superposition) pr:partExp1 G(r,φ;r/prime,φ/prime) =/summationdisplay mum(φ)Gm(r,r/prime;λ)u∗ m(φ/prime). (9.5) 132 CHAPTER 9. CYLINDRICAL PROBLEMS We plug this and 9.4 into the partial diﬀerential equation 9.3: ask Baker where this comes from 29 Feb p4/summationdisplay mum(φ)/bracketleftBigg −1 rd dr/parenleftBigg r2d dr/parenrightBigg +µmτ r2+V(r)−λσ(r)/bracketrightBigg Gmu∗ m(φ/prime) =1 rδ(r−r/prime)/summationdisplay mum(φ)u∗ m(φ/prime) We now deﬁne the reduced linear operator pr:rlo1 Lµm 0≡rL0=−d dr/parenleftBigg rτ(r)d dr/parenrightBigg +r/bracketleftBiggµmτ(r) r2+V(r)/bracketrightBigg (9.6) soGm(r,r/prime;λ) must satisfy eq9rLo (Lµm 0−λrσ(r))Gm(r,r/prime;λ) =δ(r−r/prime), for 0<r,r/prime<a and the boundary condition /parenleftBigg∂ ∂r+κ/parenrightBigg Gm(r,r/prime;λ) = 0 for r=a,0<r/prime<a. Comments on the eigenvalues µm: The RBC will always lead to µn>0. Ifµm<0 then the term µmτ(r)/r2in 9.6 would act like an attractive sink and there would be no stable solution. Since µm>0, this term instead looks like a centrifugal barrier at the origin. 29 Feb p5 Note that the eﬀective “tension” in this case is rτ(r), sor= 0 is a singular point. Thus we must impose regularity at r= 0:|G(r= 0,r/prime;λ)|<∞. 9.3.2 Summary of GF for Cyl. Sym. We have reduced the Green’s function for cylindrical symmetry to the 1-dimensional problem: (Lµm 0−λrσ(r))Gm(r,r/prime;λ) =δ(r−r/prime), for 0<r,r/prime<a, /parenleftBigg∂ ∂r+κ/parenrightBigg Gm(r,r/prime;λ) = 0 for r=a,0<r/prime<a, |G(0,r/prime;λ)|<∞. 9.4. EIGEN VALUE PROBLEM FOR L0 133 9.4 Eigen Value Problem for L0 To solve the reduced Green’s function problem which we have just ob- tained, we must solve the reduced eigen value problem pr:efp4 Lµm 0u(m) n(r) =λ(m) nrσ(r)u(m) n(r) for 0<r<a, du(m) n dr+κu(m) n(r) = 0 for r=a, |u(m) n(r)|<∞ atr= 0. In these equation λ(m) nis thenth eigenvalue of the reduced operator L(µm) 0andu(m) n(r) is thenth eigenfunction of L(µm) 0. From the general theory of 1-dimensional problems (c.f., chapter 4) we know that Gm(r,r/prime;λ) =/summationdisplay nu(m) n(r)u∗(m) n(r/prime) λ(m) n−λform= 0,±1,±2,... It follows that (using 9.5) 29 Feb p6 G(r,φ,r/prime,φ/prime;λ) =/summationdisplay mum(φ)/parenleftBigg/summationdisplay nu(m) n(r)u∗(m) n(r/prime) λ(m) n−λ/parenrightBigg u∗ m(φ/prime) =/summationdisplay n,mu(m) n(r,φ)u∗(m) n(r/prime,φ/prime) λ(m) n−λ whereu(m) n(r,φ) =um(φ)u(m) n(r). Recall that Gsatisﬁes (L0−λσ)G= δ(x−x/prime). with RBC. Thus we can conclude L0u(m) n(r,φ) =λ(m) nσ(r)u(m) n(r,φ) RBC. Theseu(m) n(r,φ) also satisfy a completeness relation ﬁgure this part out 29 Feb p7/summationdisplay m,nu(m) n(r,φ)u∗(m) n(r/prime,φ/prime) =δ(x−x/prime) σ(x) =δ(r−r/prime)δ(φ−φ/prime) r/primeσ(r/prime) 2 Mar p1The radial part of the Green’s function, Gm, may also be constructed directly if solutions satisfying the homogeneous equation are known, 134 CHAPTER 9. CYLINDRICAL PROBLEMS where one of them also satisﬁes the r= 0 boundary condition and the other also satisﬁes the r=aboundary condition. The method from chapter 3 (which is valid for 1-dimensional problems) gives Gm(r,r/prime;λ) =−u1(r<)u2(r>) rτ(r)W(u1,u2) where τ(r)? (Lµm 0−λσr)u1,2= 0 |u1|<∞ atr= 0 ∂u2 ∂r+κu2= 0 for r=a. The eﬀective mass density is rσ(τ), the eﬀective tension is rτ(r), and the eﬀective potential is r(µmτ/r2+V(r)). 9.5 Uses of the GF Gm(r,r/prime;λ) 2 Mar p3 9.5.1 Eigenfunction Problem pr:efp5 OnceGm(r,r/prime;λ) is known, the eigenvalues and normalized eigenfunc- tions can be found using the relation Gm(r,r/prime;λ)λ→λ(m) n∼u(m) n(r)u(m) n(r/prime) λ(m) n−λ. The eigen values come from the poles, the eigen functions come from the residues. 9.5.2 Normal Modes/Normal Frequencies pr:NormMode4 In the general problem with no external forces the equation of motion is homogeneous /parenleftBigg L0+σ∂2 ∂t2/parenrightBigg u(x,t) = 0 + RBC . We look for natural mode solutions: this section is still rough 9.5. USES OF THE GF GM(R,R/prime;λ) 135 u(x,t) =e−iω(m) ntu(m) n(r,φ). The natural frequencies are given by ω(m) n=/radicalBig λ(m) n. The eigen functions (natural modes) are (cf section 9.4) u(m) n(r,φ) =u(m) n(r)um(φ). The normal modes are u(m) n(x,t) =e−iω(m) ntu(m) n(r,φ). The normalization of the factored eigenfunctions u(m) n(r) andum(φ) is /integraldisplaya 0drrσ(r)u(m) n(r)u∗(m) n/prime(r/prime) =δn,n/prime forn= 1,2,... /integraldisplay2π 0dφu m(φ)u∗ m/prime(φ) =δm,m/prime. The overall normalization of the ( r,φ) eigen functions is pr:normal3 2 Mar p4 /integraldisplay2π 0dφ/integraldisplaya 0dr(rσ(r))u(m) n(r,φ)u∗(m/prime) n/prime(r,φ) =δn,n/primeδm,m/prime or /integraldisplay2π 0/integraldisplaya 0rdrdφσ (r)u(m) n(r,φ)u∗(m/prime) n/prime(r,φ) =/integraldisplay Rd3xσ(x)u∗(x)u(x) =δn,n/primeδm,m/prime. 9.5.3 The Steady State Problem pr:sss2 This is the case of a periodic driving force: /parenleftBigg L0+σ∂2 ∂t2/parenrightBigg u(r,φ,t ) =σf(r,φ)e−iωt /parenleftBigg∂ ∂r+κ/parenrightBigg u(r,φ,t ) =h(φ)e−iωt 136 CHAPTER 9. CYLINDRICAL PROBLEMS Note: As long as the normal mode solution has circular symmetry, we may perturb it with forces f(r,φ) andh(φ). It is not necessary to have circularly symmetric forces. The solution is (using 9.5) u(r,φ) =/summationdisplay mum(φ) ×/parenleftbigg/integraldisplaya 0r/primeσ(r/prime)dr/primeGm(r,r/prime;λ=ω2+iε)/integraldisplay2π 0dφ/primeu∗ m(φ/prime)f(r/prime,φ/prime) +Gm(r,a;λ=ω2+iε)/integraldisplay2π 0adφ/primeτ(a)u∗ m(φ/prime)f(r/prime,φ/prime)/parenrightbigg . In this equation/integraltext2π 0dφ/primeu∗ m(φ/prime)f(r/prime,φ/prime) is themth Fourier coeﬃcient of the interior force f(r/prime,φ/prime) and/integraltext2π 0adφ/primeτ(a)u∗ m(φ/prime)f(r/prime,φ/prime) is themth Fourier coeﬃcient of the surface force τ(a)h(φ/prime). 9.5.4 Full Time Dependence 2 Mar p5 For the retarded Green’s function we havepr:GR3 GR(r,φ,t ;r/prime,φ/prime,t/prime) =/summationdisplay mum(φ)GmR(r,t,r/prime,t/prime)u∗ m(φ/prime) where GmR(r,t;r/prime,t/prime) =/integraldisplay Ldω 2πe−iω(t−t/prime)Gm(r,r/prime;λ=ω2) Gm(r,r/prime;λ=ω2) =−1 rτ(r)u1(r<)u2(r>) W(u1,u2). Note: In the exterior case, the poles coalesce to a branch cut. All space has circular symmetry. All the normal limits (/summationtext→/integraltext,δn,n/prime→δ(n−n/prime), etc.,) hold. 9.6 The Wedge Problem pr:wedge1 We now consider the case of a wedge. The equations are similar for the internal and external region problems. We consider the internal region problem. The region Ris now 0<r <a , 0<φ<γ and its boundary is formed by φ= 0,φ−2π, andr=a. 9.6. THE WEDGE PROBLEM 137 AA B BBγ r Figure 9.2: The wedge. 9.6.1 General Case See the ﬁgure 9.2. The angular eigenfunction equation is again 9.4: ﬁg10.3 −∂2 ∂φ2um(φ) =µmum(φ) RBC. Note that the operator ∂2/∂φ2is positive deﬁnite by Green’s 1st iden- tity. The angular eigenvalues are completely determined by the angular boundary conditions. For RBC it is always true the µm>0. This is physically important since if it were negative, the solutions to 9.4 would be real exponentials, which would not satisfy the case of periodic bound- ary conditions. The boundary condition is now (ˆn· ∇+κ)G= 0 x∈S. This is satisﬁed if we choose κ1(r) =κ1/r,κ2(r) =κ2/r, andκ3(φ) = κ3, withκ1,κ2≥0. The boundary condition (ˆ n·∇+κ)G= 0 becomes /parenleftBigg −∂ ∂φ+κ1/parenrightBigg G= 0 for φ= 0 /parenleftBigg∂ ∂φ+κ2/parenrightBigg G= 0 for φ=γ /parenleftBigg∂ ∂r+κ3/parenrightBigg G= 0 for r=a,0<φ<γ. We now choose the um(φ) to satisfy the ﬁrst two boundary conditions. The rest of the problem is the same, except that Lµm 0gives diﬀerent µm eigenvalues. 138 CHAPTER 9. CYLINDRICAL PROBLEMS 9.6.2 Special Case: Fixed Sides 2 Mar p6 The caseκ1→ ∞ andκ2→ ∞ corresponds to ﬁxed sides. We thus haveG= 0 forφ= 0 andφ=γ. So theumeigenvalues must satisfy −∂2 ∂φ2um=µmum and um= 0 for φ= 0,γ. The solution to this problem is um(φ) =/radicalBigg 2 γsinmπφ γ with µm=/parenleftBiggmπ γ/parenrightBigg2 m= 1,2,.... The casem= 0 is excluded because its eigenfunction is trivial. As γ→2πwe recover the full circle case. ask Baker why not γ→ 2π? 2 Mar p7 9.7 The Homogeneous Membrane pr:membrane1Recall the general Green’s function problem for circular symmetry. By4 Mar p1substituting the completeness relation for um(φ), our diﬀerential equa- tion becomes (L0−λσ)G(x,x/prime;λ) =δ(x−x/prime) =δ(r−r/prime) r/summationdisplay mum(φ)u∗ m(φ/prime) where L0um(φ) =1 rLµm 0um(φ), Lµm 0=−d dr/parenleftBigg rτ(r)d dr/parenrightBigg +r/parenleftBiggµmτ(r) r2+V(r)/parenrightBigg . We now consider the problem of a complete circle and a wedge. 4 Mar p2 9.7. THE HOMOGENEOUS MEMBRANE 139 We look at the case of a circular membrane or wedge with V= 0,σ= constant,τ= constant. This corresponds to a homogeneous membrane. We separate the problem into radial and angular parts. First we consider the radial part. To ﬁnd Gm(r,r/prime;λ), we want to solve the problem /bracketleftBigg −d dr/parenleftBigg rd dr/parenrightBigg +µm r−λr c2/bracketrightBigg Gm(r,r/prime;λ) =1 τδ(r−r/prime) withG= 0 andr=a, which corresponds to ﬁxed ends. This problem was solved in problem set 3: Gm=π 2τJ√µm(r</radicalBig λ/c2) J√µm(a/radicalBig λ/c2)/parenleftbigg J√µm(r>/radicalBig λ/c2)N√µm(a/radicalBig λ/c2) −J√µm(aλ/c2)N√µm(r>/radicalBig λ/c2)/parenrightbigg . (9.7) Using 9.5, this provides an explicit solution of the full Green’s function problem. Now we consider the angular part, where we have γ= 2π, so 4 Mar p3 that√µm=±mwhich means the angular eigenfunctions are the same as for the circular membrane problem considered before: um=1√ 2πeimφforµm=m2,m= 0,±1,±2,.... The total answer is thus a sum over both positive and negative m G(r,φ,r/prime,φ/prime;λ) =∞/summationdisplay m=−∞um(φ)Gm(r,r/prime;λ)u∗ m(φ/prime). We now redo this with κ→ ∞ and arbitrary γ. This implies that the eigen functions are the same as the wedge problem considered before um(φ) =/radicalBigg 2 γsin/parenleftBiggmπφ γ/parenrightBigg , µm=/parenleftBiggmπ γ/parenrightBigg2 . 140 CHAPTER 9. CYLINDRICAL PROBLEMS We now get Jmπ/γ(r/radicalBig λ/c2) andNmπ/γ(r/radicalBig λ/c2). We also get the orig- Show why (new Bessel op. inal expansion for G: G(r,φ;r/prime,φ/prime;λ) =∞/summationdisplay m=1um(φ)Gm(r,r/prime;λ)u∗ m(φ/prime) 9.7.1 The Radial Eigenvalues pr:efp6 The poles of 9.7 occur when J√µm(a/radicalBig λ/c2) = 0. We denote the nth zero ofJ√µmbyx√µmnThis gives us λmn=/parenleftbiggx√µmnc a/parenrightbigg2 forn= 1,2,... whereJ√µm(x√µm,n) = 0 is the nth root of the µmBessel function. To ﬁnd the normalized eigenfunctions, we look at the residues of Gmλ→λn−→u(m) n(r)u(m) n(r/prime) λ(m) n−λ. We ﬁnd 4 Mar p4 um n(r) =/radicalBigg 2 σa2J√µm(x√µmnr a) J/prime√µm(x√µmn). Thus the normalized eigen functions of the overall operator L0u(m) n(r,φ) =σλ(m) nun(r,φ) are u(m) n(r,φ) =u(m) n(r)um(φ) where the the form of um(φ) depends on whether we are considering a wedge or circular membrane. 9.8. SUMMARY 141 9.7.2 The Physics The normal mode frequencies are given by the radial eigenvalues ωm,n=/radicalBig λ(m) n=c ax√µ,n. The eigen values increase in two ways: as nincreases and as mincreases. For smallx(i.e.,x/lessmuch1),J√µm∼(x)√µmwhich implies that for larger µmthe rise is slower. Asmincreases,µmincreases, so the ﬁrst root occurs at larger x. As we increase m, we also increase the number of angular nodes in eimφor sin(mnφ/γ ). This also increases the centrifugal potential. Thus ωm,2is this true? increases with m. The more angular modes that are present, the more angular kinetic energy contributes to the potential barrier in the radial equation. Now consider behavior with varying γfor a ﬁxedm.µmincreases 4 Mar p5 as we decrease γ, so thatωm,nincreases. Thus the smaller the wedge, the larger the ﬁrst frequency. The case γ→0 means the angular eigen- functions oscillate very quickly and this angular energy gets thrown into the radial operator and adds to the centrifugal barrier. 9.8 Summary 1. Whereas cartesian coordinates measure the perpendicular dis- tance from two lines, cylindrical coordinates measure the length of a line from some reference point in its angle from some reference line. 2. Theδ-function for circular coordinates is δ(x−x/prime) =δ(r−r/prime) rδ(φ−φ/prime). 3. The Green’s function equation for circular coordinates is /bracketleftBigg −1 r∂ ∂r/parenleftBigg rτ(r)∂ ∂r/parenrightBigg −τ(r) r2∂2u ∂φ2+V(r)−λσ(r)/bracketrightBigg G=δ(x−x/prime). 142 CHAPTER 9. CYLINDRICAL PROBLEMS 4. The partial expansion of the Green’s function for the circular problem is G(r,φ;r/prime,φ/prime) =/summationdisplay mum(φ)Gm(r,r/prime;λ)u∗ m(φ/prime). 5. The radial Green’s function for circular coordinates satisﬁes (Lµm 0−λrσ(r))Gm(r,r/prime;λ) =δ(r−r/prime), for 0<r,r/prime<a, where the reduced linear operator is Lµm 0≡rL0=−d dr/parenleftBigg rτ(r)d dr/parenrightBigg +r/bracketleftBiggµmτ(r) r2+V(r)/bracketrightBigg , and the boundary condition /parenleftBigg∂ ∂r+κ/parenrightBigg Gm(r,r/prime;λ) = 0 for r=a,0<r/prime<a. 9.9 Reference The material in this chapter can be also found in various parts of [Fet- ter80] and [Stakgold67]. The preferred special functions reference for physicists seems to be [Jackson75]. Chapter 10 Heat Conduction Chapter Goals: •Derive the conservation law and boundary condi- tions appropriate for heat conduction. •Construct the heat equation and the Green’s func- tion equation for heat conduction. •Solve the heat equation and interpret the solution.7 Mar p1 10.1 Introduction pr:heat1 We now turn to the problem of heat conduction.1The following physi- cal parameters will be used: mass density ρ, speciﬁc heat per unit mass cp, temperature T, and energy E. Again we consider a region Rwith boundarySand outward normal ˆ n. 10.1.1 Conservation of Energy The speciﬁc heat, cp, gives the additional amount of thermal energy which is stored in a unit of mass of a particular material when it’s temperature is raised by one unit: ∆ E=cp∆T. Thus the total energy can be expressed as Etotal=E0+/integraldisplay Rd3xρcpT. 1The corresponding material in FW begins on page 408 143 144 CHAPTER 10. HEAT CONDUCTION Diﬀerentiating with respect to time gives dE dt=/integraldisplay Rd3xρcp/parenleftBigg∂T ∂t/parenrightBigg . There are two types of energy ﬂow: from across the boundary S and from sources/sinks in R. 1. Energy ﬂow into RacrossS. This gives /parenleftBiggdE dt/parenrightBigg boundary=−/integraldisplay ˆn·jndS=−/integraldisplay Rdx∇ ·jn where the heat current is deﬁned pr:heatcur1 jn=−kT∇T. kTis the thermal conductivity. Note that since ∇Tpoints toward the hot regions, the minus sign in the equation deﬁning heat ﬂow indicates that heat ﬂows from hot to cold regions. 2. Energy production in Rdue to sources or sinks, /parenleftBiggdE dt/parenrightBigg sources=−/integraldisplay Rd3xρ˙q where ˙qis the rate of energy production per unit mass by sources insideS. Thus the total energy is given by /integraldisplay Rd3xρcp∂T ∂t=dE dt =/parenleftBiggdE dt/parenrightBigg boundary+/parenleftBiggdE dt/parenrightBigg sources =/integraldisplay Rd3x(ρ˙q− ∇ · jn). By taking an arbitrary volume, we get the relation ρcp∂T ∂t=∇ ·(kT∇T) +ρ˙q. (10.1) 10.1. INTRODUCTION 145 10.1.2 Boundary Conditions pr:bc4 7 Mar p2There are three types of boundary conditions which we will encounter: 1.Tgiven onS. This is the case of a region surrounded by a heat bath. 2. ˆn· ∇Tforx∈Sgiven. This means that the heat current normal to the boundary, ˆ n·jn, is speciﬁed. In particular, if the boundary is insulated, then ˆ n· ∇T= 0. 3.−kT(x)ˆn· ∇T=α(T−Texternal) forx∈S. In the ﬁrst case the temperature is speciﬁed on the boundary. In the second case the temperature ﬂux is speciﬁed on the boundary. The third case is a radiation condition, which is a generalization of the ﬁrst two cases. The limiting value of αgive α/greatermuch1 =⇒T≈Texternal→#1 α/lessmuch1 =⇒ˆn· ∇T≈0→#2 with insulated boundary . We now rewrite the general boundary condition (3) as [ˆn· ∇T+θ(S)]T(x,t) =h(x,t) for x∈S (10.2) whereθ(S) =α/k T(S) andh(S,t) = (α/k T(S))Texternal.In the limit θ/greatermuch1,Tis given. In this case we recover boundary condition #1. The radiation is essentially perfect, which says that the temperature of the surface is equal to the temperature of the environment, which corresponds to α→ ∞ . In the other limit, for θ/lessmuch1, ˆn· ∇Tis given. Thus we recover boundary condition # 2 which corresponds to α→0. By comparing the general boundary conditions for the heat equation with the general N-dimensional elastic boundary condition, [ˆn· ∇+κ(x)]u(x,t) =h(x,t) we identify u(x,t)→T(x,t) andκ(x)→θ(x). 146 CHAPTER 10. HEAT CONDUCTION 10.2 The Standard form of the Heat Eq. 10.2.1 Correspondence with the Wave Equation We can make the conservation of energy equation 10.1 look more fa- miliar by writing it in our standard diﬀerential equation form pr:heateq1 /parenleftBigg L0+ρcp∂ ∂t/parenrightBigg T=ρ˙q(x,t) for x∈G (10.3) where the linear operator is eq10DE L0=−∇ · (kT(x)∇). The correspondence with the wave equation is as follows: Wave Equation Heat Equation τ(x)kT(x) σ(x)ρ(x)cp σ(x)f(x,t)ρ(x) ˙q(x,t) V(x) no potential For the initial condition, we only need T(x,0) to fully specify the solu- tion for all time. 7 Mar p3 10.2.2 Green’s Function Problem We know that because equation 10.3 is linear, it is suﬃcient to con- sider only the Green’s function problem (which is related to the above problem by p˙q(x,t) =δ(x−x/prime)δ(t−t/prime) andh(S,t) = 0): /parenleftBigg L0+ρcp∂ ∂t/parenrightBigg G(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime), [ˆn· ∇+θ(S)]G(x,t;x/prime,t/prime) = 0 for x∈S, G(x,t;x/prime,t/prime) = 0 for t<t/prime. 10.2. THE STANDARD FORM OF THE HEAT EQ. 147 We lose symmetry in time since only the ﬁrst time derivative appears. We evaluate the retarded Green’s functions by applying the standard Fourier transform technique from chapter 6: G(x,t;x/prime,t/prime) =/integraldisplay Ldω 2πe−iω(t−t/prime)˜G(x,x/prime;ω). We know by G= 0 fort<t/primethat ˜Gis analytic in the Im ω>0 plane. Thus we take Lto be a line parallel to the real ω-axis in the upper half plane. The Fourier Transform of the Green’s function is the solution of the problem (L0−ρcpiω)˜G(x,x/prime;ω) =δ(x−x/prime), (ˆn· ∇+θ(x))˜G(x,x/prime;ω) = 0 for x∈S, which is obtained by Fourier transforming the above Green’s function problem. 10.2.3 Laplace Transform pr:LapTrans1 We note that this problem is identical to the forced oscillation Green’s function problem with the substitutions σ→ρcpandτ→kT. Thus we identify ˜G(x,x/prime;ω) =G(x,x/prime;λ=iω). The single time derivative causes the eigenvalues to be λ=iω. To evaluate this problem we thus make the substitution s=−iω. This substitution results in the Laplace Transformation . Under this trans- 7 Mar p4 formation the Green’s function in transform space is related by ˜G(x,x/prime;ω=is) =G(x,x/prime;λ=−s). ˜Gis now analytic in the right hand side plane: Re ( s)>0. This variable substitution is depicted in ﬁgure 10.1. The transformed contour is labeledL/prime. The Laplace transform of the Green’s function satisﬁes ﬁg10a the relation G(x,t;x/prime,t/prime) =i 2π/integraldisplay L/prime↓ds˜G(x,x/prime;λ=−s/prime)es(t−t/prime) 148 CHAPTER 10. HEAT CONDUCTION - ?ω sL L/prime=⇒ Figure 10.1: Rotation of contour in complex plane. or, by changing the direction of the path, we have G(x,t;x/prime,t/prime) =1 2πi/integraldisplay L/prime↑ds˜G(x,x/prime;λ=s/prime)es(t−t/prime). (10.4) In the following we will denote L/prime↑asL. The inversion formula ˜G(ω) =/integraldisplay∞ 0dτeiωτG(x,x/prime,τ=t−t/prime) is also rotated to become ˜G(s) =/integraldisplay∞ 0dτe−sτG(x,x/prime;τ). ˜G(s) is analytic for all Re ( s)>0. Note that the retarded condition allows us start the lower limit at τ= 0 rather than τ=−∞. 10.2.4 Eigen Function Expansions 7 Mar p5 We now solve the Green’s function by writing it as a bilinear sum of eigenfunctions: G(x,x/prime;λ) =/summationdisplay nun(x)u∗ n(x/prime) λn−λ. (10.5) The eigenfunctions un(x) solve the problem L0un(x) =λnρcpun(x) for x∈R 10.2. THE STANDARD FORM OF THE HEAT EQ. 149 s-plane ResIms B BB J JJQQQPPP C + QQ s6 q λ1C1 q λ2C2 q λ3 Figure 10.2: Contour closed in left half s-plane. whereL0=−∇ · (kT(x)∇) with the elastic boundary condition (ˆn· ∇+θ(s))un= 0 for x∈S. Because of the identiﬁcation ˜G(x,x/prime;s) =G(x,x/prime;λ=−s) we can substitute 10.5 into the transform integral 10.4 to get G(x,t;x/prime,t/prime) =/integraldisplay Lds 2πi/summationdisplay nun(x)u∗ n(x/prime) λn+ses(t−t/prime) =/summationdisplay nun(x)u∗ n(x)1 2πi/contintegraldisplay Lds λn+ses(t−t/prime). This vanishes for t < t/prime. Close the contour in the left half s-plane fort−t/prime>0, as shown in ﬁgure 10.2. This integral consists of ﬁg10a1 contributions from the residues of the poles at −λn, wheren= 1,2,.... So1 2πi/contintegraldisplay Cnds λn+ses(t−t/prime)=e−λn(t−t/prime). Thus G(x,t;x/prime,t) =/summationdisplay nun(x)u∗ n(x/prime)e−λn(t−t/prime). (10.6) 150 CHAPTER 10. HEAT CONDUCTION We now consider the two limiting cases for t. Suppose that t→t/prime. Then 10.6 becomes Gt→t/prime−→/summationdisplay nun(x)u∗ n(x/prime) =δ(x−x/prime) ρcp. Thus we see that another interpretation of Gis as the solution of an 7 Mar p6 initial value problem with the initial temperature T(x,0) =δ(x−x/prime) ρcp and no forcing term. why is this? Now suppose we have the other case, t−t/prime/greatermuch1. We know λn>0 for allnsinceL0is positive deﬁnite (physically, entropy requires k>0 so that heat ﬂows from hot to cold). Thus the dominant term is the one for the lowest eigenvalue: G∼u1(x)u∗ 1(x/prime)e−λ1(t−t/prime)(t−t/prime)/greatermuch1. In particular, this formula is valid when ( t−t/prime)>1/λ2. We may thus interpret 1/λn=τnas the lifetime of these states. After ( t−t/prime)/greatermuchτN, all contributions to Gfrom eigen values with n≥Nare exponentially small. This is the physical meaning for the eigen values. The reason that the lowest eigen function contribution is the only one that contributes fort−t/prime/greatermuch1 is because for higher Nthere are more nodes in the eigenfunction, so it has a larger spatial second derivative. This means (using the heat equation) that the time derivative of temperature is large, so the temperature is able to equilize quickly. This smoothing or diﬀusing process is due to the term with a ﬁrst derivative in time, which gives the non-reversible nature of the problem. 10.3 Explicit One Dimensional Calculation 9 Mar p1 We now consider the heat equation in one dimension. 10.3. EXPLICIT ONE DIMENSIONAL CALCULATION 151 10.3.1 Application of Transform Method pr:fsp1 Recall that the 1-dimensional Green’s function for the free space wave equation is deﬁned by 9 Mar p2 (L0−σλ)G=δ(x−x/prime) for −∞<x< ∞. We found that the solution for this wave equation is G(x,x/prime;λ) =1 2√ λei√ λ|x−x/prime|/c σc. Transferring from the wave equation to the heat equation as discussed above, we substitute τ→kT,σ→ρ,c=/radicalBig τ/σ→√κwhereκ= KT/ρcpis the thermal diﬀusivity, and√ λ→i√swhich means Im λ> pr:kappa1 kT→KTand c→cp?0 becomes Re s>0. The substitutions yield ˜G(x,x/prime;s) =/parenleftBigg1 2√sρcp√κ/parenrightBigg e−√s/parenleftBig |x−x/prime|√κ/parenrightBig or ρcp˜G(x,x/prime;s) =1 2√κse−√ s/κ|x−x/prime|. We see that√κplays the role of a velocity. Now invert the transform 9 Mar p3 to obtain the free space Green’s function for the heat equation: ρcpG(x,t;x/prime,t/prime) =/integraldisplay Lds 2πies(t−t/prime)ρcp˜G(x,x/prime;s) =/integraldisplay Lds 2πies(t−t/prime)−√ s/κ|x−x/prime| 2√sκ. 10.3.2 Solution of the Transform Integral Our result has a branch on√s. We parameterize the s-plane: s=|s|eiθfor−π<θ<π This gives us Re√s=|s|1/2cos(θ/2)>0.We choose the contour of integration based on t. 9 Mar p5 152 CHAPTER 10. HEAT CONDUCTION s-plane ResIms L2L3 L4- B BB J JJQQQPPP L1 L5 + QQ s6 - -√s=i/radicalBig |s| √s=−i/radicalBig |s| Figure 10.3: A contour with Branch cut. Fort<t/primewe have the condition G= 0. Thus we close the contour in the right half plane so that exp/bracketleftBig s(t−t/prime)−√s|x−x|/√κ/bracketrightBigs→∞−→0 since both terms are increasingly negative. Since the contour encloses no poles, we recover G= 0 as required. Fort−t/prime>0, close contour in the left half plane. See ﬁgure 10.3. We know by Cauchy’s theorem that the integral around the closed contour L+L1+L2+L3+L4+L5vanishes. We perform the usual Branch cut evaluation, by treating the diﬀerent segments separately. For L3it is convenient to use the parameterization s=εeiθfor−π < θ < π as shown in ﬁgure 10.3. In this case the integral becomes 1 2π1 2√κ/integraldisplayπ −πdθ|√ε|[1 +O(ε(t−t/prime)) +O(ε1/2|x−x/prime|/√κ)]ε→0−→0. In this equation we assert that it is permissible to take the limit ε→0 before the other quantities are taken arbitrarily large. For the contour L2above the branch cut we have√s=i/radicalBig |s|, and for the contour L4below the branch cut we have√s=−i/radicalBig |s| 10.3. EXPLICIT ONE DIMENSIONAL CALCULATION 153 Combining the integrals for these two cases gives lim ε→0/integraldisplay−ε −∞ds 2πes(t−t/prime) 2√κ2 cos/radicalBig |s|/κ|x−x/prime| /radicalBig |s|. ForL1andL5the integral vanishes. By letting s=Reiθwhere −π<θ<π we have /vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleexp [−√s|x−x/prime|/√κ]√s/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle≤exp/bracketleftBig −|x−x/prime|√κR2cos1 2φ/bracketrightBig √ RR→∞−→0. Our ﬁnal result is ρcpG(x,t;x/prime,t/prime) =1 2π√κ/integraldisplay∞ 0ds√se−s(t−t/prime)cos/radicalBig s/κ(x−x/prime). Substituting s=u2gives ρcpG(x,t;x/prime,t/prime) =/integraldisplay∞ 02udu 2πu√κcosu√κ|x−x/prime|e−u2(t−t/prime) or ρcpG(x,t;x/prime,t/prime) =1√κI(t−t/prime,|x−x/prime|/√κ) where (since the integrand is even) I(t,y) =1 2/integraldisplay∞ −∞du πe−u2t+iuy. This can be made into a simple Gaussian by completing the square: pr:gaus1 I(t,y) =e−y2/(4t)/integraldisplay∞ −∞du 2πe±(u−iy 2t)2. By shifting u→u+iy/2t, the result is 9 Mar p6 I(t,y) =e−y2/(4t) √ 4πt. The free space Green function in 1-dimension is thus ρcpG(x,t;x/prime,t/prime) = 1/radicalBig 4πκ|t−t/prime| e−(x−x/prime)2 4κ(t−t/prime). (10.7) 154 CHAPTER 10. HEAT CONDUCTION 10.3.3 The Physics of the Fundamental Solution This solution corresponds to a pure initial value problem where, if x/prime= t/prime= 0, we have ρcpG(x,t) =e−x2/4κt √ 4πκt. At the initial time we have ρcpGt→0−→δ(x−x/prime) =δ(x). 1. Forx2>4κt, the amplitude is very small. Since Gis small for x≥√ 4κt, diﬀusion proceeds at rate proportional to√ t, nott as in wave equation. The average propagation is proportional to t1/2. This is indicative of a statistical process (Random walk). It 9 Mar p7 is non-dynamical in that it does not come from Newton’s laws. Rather it comes from the dissipative–conduction nature of ther- modynamics. 2. For any t>0 we have a non-zero eﬀect for all space. This corre- sponds to propagation with inﬁnite velocity. Again, this indicates the non-dynamical nature of the problem. This is quite diﬀerent from the case of wave propagation, where an event at the origin does not aﬀect the position xuntil timex/c. 3. Another non-dynamical aspect of this problem is that it smoothes the singularity in the initial distribution, whereas the wave equa- tion propagates all singularities in the initial distribution forward in time. 4.κis a fundamental parameter whose role for the heat equation is analogous to the role of cfor the wave equation. It deter- mines the rate of diﬀusion. κ(=kT/ρcp) has the dimensions of (distance)2/time, whereas chas the dimensions of distance/time.11 Mar p1 10.3.4 Solution of the General IVP11 Mar p2 We now use the Green’s function to solve the initial value problem: /parenleftBigg −kT∂2 ∂x2+ρcpd dt/parenrightBigg T(x,t) = 0 for −∞<x,x/prime,∞ 10.3. EXPLICIT ONE DIMENSIONAL CALCULATION 155 T(x,0) =T0(x) T→0 for |x| → ∞ The method of the solution is to use superposition and 10.8: T(x,t) =/integraldisplay∞ −∞dx/primeT0(x/prime)ρcpG(x/prime,0;x,t) =1√ 4πκt/integraldisplay∞ −∞dx/primee−(x−x/prime)2/(4κt)T0(x/prime) (10.8) 10.3.5 Special Cases Initialδ-function SupposeT0(x) =δ(x−x/prime). Then we have T(x,t) =ρcpG(x/prime/prime,0;x,t). Thus we see that Gis the solution to the IVP with the δ-function as the initial condition and no forcing term. Initial Gaussian Functionpr:gaus2 11 Mar p3 We now consider the special case of an initial Gaussian temperature distribution. Let T0(x) = (a/π)1/2e−ax2. The width of the initial dis- tribution is (∆ x)0= 1/√a. Plugging this form of T0(x) into 10.8 gives T(x,t) =1 π√ 4κta/integraldisplay∞ −∞dx/primee−(x−x/prime)2/(4κt)−ax/prime2 T(x,t) =1√π1/radicalBig (1/a) + 4κte−x2 (1/a)+4κt =1√π/parenleftbigg1 ∆x/parenrightbigg e(x/∆x)2 where ∆x=/radicalBig (∆x0)2+ 4κt. The packet is spreading as (∆ x)2= 4κt+ (∆x0)2. Again, ∆ x∼t1/2like a random walk, (again non- dynamical). Suppose t/greatermuchτ≡(∆x0)/4κ. This is the simplest quantity with dimensions of time, so τis the characteristic time of the system. We rewrite ∆ x= (∆x0)/radicalBig 1 +t/τ. Thus fort/greatermuchτ, ∆x∼/radicalBig t/τ(∆x0). 156 CHAPTER 10. HEAT CONDUCTION τ= (∆x0)2/4κis a fundamental unit of time in the problem. Since ask Baker about omitted the region is inﬁnite, there does not exist any characteristic distance for the problem. 11 Mar p4 11 Mar p5 10.4 Summary 1. Conservation of energy for heat conduction is given by the equa- tion ρcp∂T ∂t=∇ ·(kT∇T) +ρ˙q, whereρis the mass density, cpis the speciﬁc heat, Tis the temper- ature,kTis the thermal conductivity, and ˙ qis the rate of energy production per unit mass by sources inside the region. 2. The general boundary condition for heat conduction is [ˆn· ∇T+θ(S)]T(x,t) =h(x,t) for x∈S. 3. The heat equation is /parenleftBigg L0+ρcp∂ ∂t/parenrightBigg T=ρ˙q(x,t) for x∈G, where the linear operator is L0=−∇ · (kT(x)∇). 4. The Green’s function equation for the heat conduction problem is/parenleftBigg L0+ρcp∂ ∂t/parenrightBigg G(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime). 5. The solution of the heat equation for the initial value problem in one dimension is T(x,t) =1√ 4πκt/integraldisplay∞ −∞dx/primee−(x−x/prime)2/(4κt)T0(x/prime), which is a weighted integration over point sources which individ- ually diﬀuse with a gaussian shape. 10.5. REFERENCES 157 10.5 References A similar treatment (though more thorough) is given in [Stakgold67b, p194ﬀ]. See also [Fetter80, p406ﬀ]. The deﬁnitive reference on heat conduction is [Carslaw86]. 158 CHAPTER 10. HEAT CONDUCTION Chapter 11 Spherical Symmetry Chapter Goals: •Derive the form of the linear operator in spherical coordinates. •Show that the angular part of the linear operator Lθφis hermitian. •Write the eigenvalue equations for Ym l. •Write the partial wave expansion for the Green’s function. •Find the Green’s function for the free space prob- lem.28 Mar p1 (17) Our object of study is the Green’s function for the problem [L0−λσ(x)]G(x,x/prime;λ) =δ(x−x/prime) (11.1) with the regular boundary condition (RBC) eq11.1 [ˆn· ∇+K(S)]G(x,x/prime;λ) = 0 forx/primein a region Randxin the regions boundary S. The term xis a ﬁeld point, and x/primeis a source point. The unit vector ˆ nis the outward normal of the surface S. The operator L0is deﬁned by the equation L0=−∇ · (τ(x)∇) +V(x). 159 160 CHAPTER 11. SPHERICAL SYMMETRY 7 lllll rθ ϕ xyz (r,θ,ϕ ) Figure 11.1: Spherical Coordinates. We have solved this problem for the one and two dimensional cases in which there was a certain degree of symmetry. 11.1 Spherical Coordinates 28 Mar p2 We now treat the problem in three dimensions. For this we use spherical coordinates (since we will later assume angular independence). A point pr:spher1 in spherical coordinates is denoted ( r,θ,ϕ ), where the range of each variable is 0≤r<∞, 0< θ≤π, 0≤ϕ<2π. We use the following transformation of coordinate systems: z=rcosθ, x=rsinθcosϕ, y=rsinθsinϕ. This relationship is illustrated in ﬁgure 11.1 For an arbitrary volume ﬁg11.1 11.1. SPHERICAL COORDINATES 161 element we have d3x= (dr)(rdθ)(rsinθdϕ) =r2dΩdr where Ω is the solid angle, and an inﬁnitesimal of solid angle is dΩ = pr:Omega1 sinθdθdϕ . We further deﬁne the delta function f(r,θ,ϕ ) =f(x) =/integraldisplay d3x/primef(x/prime)δ(x−x/prime) =/integraldisplay dr/primer/prime2sinθ/primedθ/primedϕ/primef(r/prime,θ/prime,ϕ/prime)δ(x−x/prime). From this we can extract the form of the δ-function for spherical coor- dinates: δ(x−x/prime) =1 r2sinθδ(r−r/prime)δ(θ−θ/prime)δ(ϕ−ϕ/prime) =δ(r−r/prime) r2δ(Ω−Ω/prime) where the solid angle δ-function is 28 Mar p3 pr:delta2 δ(Ω−Ω/prime) =δ(θ−θ/prime)δ(ϕ−ϕ/prime) sinθ. We want to rewrite equation 11.1 in spherical coordinates. First we deﬁne the gradient pr:grad2 ∇= ˆr∂ ∂r+ˆθ r∂ ∂θ+ˆϕ rsinθ∂ ∂ϕ. See [Boas] for derivations of identities involving ∇. The divergence is 87’ notes have Gauss’ law, p18∇ ·A=1 r2∂ ∂r(r2Ar) +1 rsinθ∂ ∂θ(sinθAθ) +1 rsinθ∂ ∂ϕAϕ. When we apply this to the case A=τ(x)∇. 162 CHAPTER 11. SPHERICAL SYMMETRY the result is ∇ ·(τ∇) =1 r2∂ ∂r/parenleftBigg r2τ∂ ∂r/parenrightBigg +1 rsinθ∂ ∂θ/parenleftBigg sinθτ r∂ ∂θ/parenrightBigg +1 rsinθ∂ ∂ϕ/parenleftBiggτ rsinθ∂ ∂ϕ/parenrightBigg (11.2) whereτ=τ(x) =τ(r,θ,ϕ ). eq11.2 28 Mar p4 Now we can write L0. We assume that τ,σ, andVare spherically symmetric, i.e., they are only a function of r:τ(x) =τ(r),σ(x) =σ(r), V(x) =V(r). In this case the linear operator is L0=−1 r2∂ ∂r/parenleftBigg r2τ(r)∂ ∂r/parenrightBigg +τ(r) r2Lθϕ+V(r) (11.3) where eq11.2b Lθϕ=−1 sinθ∂ ∂θ/parenleftBigg sinθ∂ ∂θ/parenrightBigg −1 sin2θ∂2 ∂2ϕ, which is the centrifugal term from equation 11.2. In the next few sec- pr:Lthph1 tions we will study the properties of L0. 11.2 Discussion of Lθϕ Note thatLθϕis a hermitian operator on the surface of the sphere, as shown by the following argument. In an earlier chapter we derived the Green’s Identity /integraldisplay d3xS∗(x)L0u(x) =/integraldisplay d3x(u∗(x)L0S(x))∗(11.4) whereuandSsatisfy RBC. We use this fact to show the hermiticity eq11.3 ofLθϕ. Consider the functions 28 Mar p5 S(x) =S(r)S(θ,ϕ) and u(x) =u(r)u(θ,ϕ) whereuandSsatisfy RBC. Such functions are a subset of the functions which satisfy equation 11.4. Choose u(θ,ϕ) andS(θ,ϕ) to be periodic in the azimuthal angle ϕ: u(θ,ϕ) =u(θ,ϕ+ 2π), S (θ,ϕ) =S(θ,ϕ+ 2π). 11.3. SPHERICAL EIGENFUNCTIONS 163 Now substitute d3x=r2drdΩ andL0(as deﬁned in equation 11.3) into equation 11.4. The term −1 r2∂ ∂r/parenleftBigg r2τ(r)∂ ∂r/parenrightBigg +V(r) inL0is hermitian so it cancels out in 11.4. All that is left is /integraldisplay r2drS∗(r)τ(r) r2u(r)/integraldisplay dΩS∗(θ,ϕ)Lθϕu(θ,ϕ) = /integraldisplay r2drS∗(r)τ(r) r2u(r)/integraldisplay dΩ (u∗(θ,ϕ)LθϕS(θ,ϕ))∗ This can be rewritten as 28 Mar p6 /integraldisplay r2drS∗(r)τ(r) r2u(r)dΩ/bracketleftbigg/integraldisplay S∗(θ,ϕ)Lθϕu(θ,ϕ) −/integraldisplay dΩ (u∗(θ,ϕ)LθϕS(θ,ϕ))∗/bracketrightbigg = 0. The bracket must then be zero. So /integraldisplay dΩ(S∗(θ,ϕ)Lθϕu(θ,ϕ) =/integraldisplay dΩ (u∗(θ,ϕ)LθϕS(θ,ϕ))∗. (11.5) This is the same as equation 11.4 with d3x→dΩ andL0→Lθϕ. Thus eq11.4 Lθϕis hermitian. If the region did not include the whole sphere, we just integrate the region of physical interest and apply the appropriate boundary conditions. Equation 11.5 can also be obtained directly from 28 Mar p7 the form of Lθϕby applying integration by parts on Lθϕtwice, but usingL0=L∗ 0is much more elegant. Note that the operators∂2 ∂ϕ2andLθϕcommute: /bracketleftBigg∂2 ∂ϕ2,Lθϕ/bracketrightBigg = 0. Thus we can reduce equation 11.1 to a one dimensional case and expand the Green’s function Gin terms of a single set of eigenfunctions which are valid for both −∂2/∂ϕ2andLθϕ. We know that L0andLθϕare hermitian operators, and thus the eigenfunctions form a complete set. For this reason this method is valid. 164 CHAPTER 11. SPHERICAL SYMMETRY 11.3 Spherical Eigenfunctions 28 Mar p8 We want to ﬁnd a common set of eigenfunctions valid for both Lθϕand −∂2/∂ϕ2. Note that −∂2 ∂ϕ2eimϕ √ 2π=m2eimϕ √ 2πm= 0,±1,±2,.... So (2π)−1/2eimϕare normalized eigen functions of −∂2/∂ϕ2. We deﬁne the functions Ym l(θ,ϕ) as the set of solutions to the equation pr:sphHarm1 LθϕYm l(θ,ϕ) =l(l+ 1)Ym l(θ,ϕ) (11.6) for eigen values l(l+ 1) and periodic boundary conditions, and the eq11.5 equation −∂2 ∂ϕ2Ym l(θ,ϕ) =m2Ym l(θ,ϕ)m= 0,±1,±2,.... (11.7) We can immediately write down the orthogonality condition (due to eq11.6 pr:orthon3 the hermiticity of the operator Lθϕ): is this true?/integraldisplay dΩYm l∗(θ,ϕ)Ym/prime l/prime(θ,ϕ) =δl,l/primeδm,m/prime. (If there are degeneracies, we may use Gram–Schmidt techniques to arrive at this result). We can choose the normalization coeﬃcient to be one. There is also a completeness relation which will be given later. 11.3.1 Reduced Eigenvalue Equation 28 Mar p9 We now separate the eigenfunction into the product of a ϕ-part and a θ-part: pr:ulm1 Ym l(θ,ϕ) =eimϕ √ 2πum l(cosθ) (11.8) (which explicitly solves equation 11.7, the diﬀerential equation involv- eq11.6b ingϕ) so that we may write equation 11.6 as /bracketleftBigg −1 sinθ∂ ∂θ/parenleftBigg sinθ∂ ∂θ/parenrightBigg +m2 sin2θ/bracketrightBigg um l(cosθ) =l(l+ 1)um l(cosθ). 11.3. SPHERICAL EIGENFUNCTIONS 165 All we have left to do is solve this eigenvalue equation. The original region was the surface of the sphere because the solid angle represents area on the surface. We make a change of variables: x= cosθ. The derivative operator becomes d dθ=dx dθd dx=−sinθd dx so −1 sinθd dθ=d dx. The eigen value equation for ubecomes /bracketleftBigg −d dx/parenleftBigg (1−x2)d dx/parenrightBigg +m2 1−x2/bracketrightBigg um l(x) =l(l+ 1)um l(x) (11.9) deﬁned on the interval −1<x< 1. Thusx= 1 corresponds to θ= 0, eq11.7 andx=−1 corresponds to θ=π. Note that τ, which represents the eﬀective tension, is proportional to 1 −x2, so both end points are singular points. On account of this we get both regular and irregular 28 Mar p10 solutions. A solution occurs only if the eigen value ltakes on a special value. Requiring regularity at x=±1 implieslis an integer. Note also Verify this that equation 11.9 represents an inﬁnite number of one dimensional eigenvalue problems (indexed by m), which makes sense because we started with a partial diﬀerential equation eigenvalue problem. The way to solve the equation near a singular point is to look for solutions of the form xp·[power series], as in the solutions to Bessel’s change this equation. 30 Mar p1 11.3.2 Determination of um l(x) 30 Mar p2 We now determine the function um l(x) which is regular at x=±1. Suppose that it is of the form um l(x) = (1 −x2)β(power series) . (11.10) We want to determine the power term β. First we compute eq11.8 166 CHAPTER 11. SPHERICAL SYMMETRY (1−x2)d dx(1−x2)β=β(−2x)(1−x2)β, and then d dx/bracketleftBigg (1−x2)d dx(1−x2)β/bracketrightBigg =β2(1−x2)β−1(−2x)2 +β(−2x)(1−x2)β+.... For the case x→1 we can drop all but the leading term: d dx/bracketleftBigg (1−x2)d dx(1−x2)β/bracketrightBigg ≈4x2β2(1−x2)β−1x→1.(11.11) Plugging equations 11.10 and 11.11 into equation 11.9 gives eq11.9 (1−x2)β−1A[−4β2+m2]≈(l+ 1)lA(1−x2)β, x →1. whereAis the leading constant from the power series. Note however that (1 −x2)βapproaches zero faster that (1 −x2)β−1asx→1. So we getm2= 4β2, or β=±m 2. We thus look for a solution of the form um l(x) = (1 −x2)m/2Cm(x), (11.12) whereCm(x) is a power series in xwith implicit ldependence. We eq11.9a expect regular and irregular solutions for Cm(x). We plug this equation 30 Mar p3 into equation 11.9 to get an equation for Cm(x). The result is −(1−x2)C/prime/prime m+ 2x(m+ 1)C/prime m−(l−m)(l+m+ 1)Cm(x) = 0.(11.13) We still have the boundary condition that um l(x) is ﬁnite. In the case eq11.9b thatm= 0 we have −(1−x2)C/prime/prime 0+ 2xC/prime 0−l(l+ 1)C0= 0. (11.14) This is called Legendre’s equation. We want to ﬁnd the solution of eq11.10 pr:LegEq1 this equation which is regular at x= 1. We deﬁne Pl(x) to be such 11.3. SPHERICAL EIGENFUNCTIONS 167 a solution. The irregular solution at x= 1, called Q(x), is of interest if the region Rin our problem excludes x= 1 (which corresponds to cosθ= 1 orθ= 0). Note that we consider lto be an arbitrary complex number. (From the “general theory”, however, we know that the eigen values are real.) By convention we normalize: Pl(1) = 1. We know C0(x) =Pl(x), because we deﬁned C0(x) to be regular at x= 1. But x=−1 is also a singular point. We deﬁne ˜Rl(x) to be the regular solution and ˜Il(x) to be the irregular solution at x=−1 We can write C0(x) =Pl(x) =A(l)˜Rl(x) +B(l)˜Il(x). (11.15) But if we further require that Plmust be ﬁnite (regular) at x=−1, we then haveB(l) = 0 forl= 0,1,2,.... 30 Mar p4 We now take Legendre’s equation, 11.14, with C0(x) replaced by Pl(x) (the regular solution), and diﬀerentiate it mtimes using Leibnitz formula pr:LeibForm1 dm dxm(f(x)g(x)) =m/summationdisplay i=0/parenleftbiggm i/parenrightbiggdif dxidm−ig dxm−i. This yields −(1−x2)d2 dx2/parenleftBiggdm dxmPl(x)/parenrightBigg + 2(m+ 1)xd dx/parenleftBiggdm dxmPl(x)/parenrightBigg −(l−m)(l+m+ 1)dm dxmPl(x) = 0. (11.16) Thus (dm/dxm)Pl(x) is also a solution of equation 11.13. We thus see eq11.11 that Cm(x) =αdm dxmPl(x), (11.17) whereαstill needs to be determined. Once we ﬁnd out how to chose leq11.11b soPl(x) is 0 atl,Pl/primeforl/prime/negationslash=lwill also be zero. So we see that once we determine constraints on lsuch that the Pl(x) which solves the m= 0 equation is zero at zero at x=±1, we can generate a solution for the casem/negationslash= 0. We now calculate a recurrence relation for Pl(x). Setx= 1 in pr:recrel1 equation 11.16. This gives 2(m+1)/bracketleftBiggdm+1 dxm+1Pl(x)/bracketrightBigg x=1= (l−m)(l+m+1)/bracketleftBiggdm dxmPl(x)/bracketrightBigg x=1.(11.18) 168 CHAPTER 11. SPHERICAL SYMMETRY This tells us the ( m+ 1)thderivative of Pl(x) in terms of the mth eq11.12 derivative. We can diﬀerentiate ltimes iflis an integer. Take lto be an integer. The case m=lyields /bracketleftBiggdl+1 dxl+1Pl(x)/bracketrightBigg x=1= 0. (11.19) So all derivatives are zero for m > l atx= 1. This means that P(x) ask Baker isn’t this valid only atx= 1is anlth order polynomial, since all of its Taylor coeﬃcients vanish for m > l . SincePlis regular at x= 1 and is a polynomial of degree l, 30 Mar p5 it must be regular at x=−1 also. Iflwere not an integer, we would obtain a series which diverges at x=±1. Thus we conclude that lmust be an integer. Note that even for the solution which is not regular at x=±1, for which lis not an integer, equation 11.18 is still valid for calculating the series. For a general m, we substitute equation 11.17 into equation 11.12 to get um l(x) =α(1−x2)m/2dm dxmPl(x). (11.20) This equation holds for m= 0,1,2,.... Furthermore this equation solves equation 11.13. Note that becausedm dxmPl(x) is regular at x= 1 andx=−1,um l(x) is also. 30 Mar p6 We now compute the derivative. Using equation 11.18, for m= 0 we get 2d dxPl(x)|x=1=l(l+ 1)Pl(1) =l(l+ 1). By repeating this process for m= 1,2,...and using induction, we ﬁnd that the following polynomial satisﬁes equation 11.18 yet to be veri- ﬁed Pl(x) =1 2ll!dl dxl(x2−1)l. (11.21) This is called Rodrigues formula for the Legrendre function. eq11.13 pr:rodform1 We deﬁne the associated Legendre polynomial Pm l(x) = ( −1)m(1−x2)m/2dm dxmPl(x)m≥0 =(1−x2)m/2 2ll!dl+m dxl+m(x2−1)l, m ≥0.(11.22) 11.3. SPHERICAL EIGENFUNCTIONS 169 Form>l ,Pm l(x) = 0. So the allowed range of mis−l≤m≤l. Thus eq11.13b the value of maﬀects what the lowest eigenvalue, l(l+ 1), can be. 11.3.3 Orthogonality and Completeness of um l(x) We want to choose um l(x) to be normalized. We deﬁne the normalized pr:normal4 eigen functions as the set of eigen functions which satisﬁes the condition (withσ= 1)/integraldisplay1 −1dx(um l(x))∗um l/prime(x) = 1. (11.23) We need to evaluate/integraltext1 −1dx|Pm l(x)|2. Using integration by parts and eq11.13c equation 11.22 we get (a short exercise) /integraldisplay1 −1dx|Pm l(x)|2=/integraldisplay1 −1dx(Pm l(x))∗Pm l(x) =2 2l+ 1(l+m)! (l−m)!,(11.24) so the normalized eigenfunctions are 30 Mar p7 um l(x) =/radicaltp/radicalvertex/radicalvertex/radicalbt2 2l+ 1(l+m)! (l−m)!Pm l(x). (11.25) The condition for orthonormality is eq11.14 pr:orthon4 /angbracketleftum l(x),um l/prime(x)/angbracketright=/integraldisplay1 −1dx(um l(x))∗um l/prime(x) =δll/prime, (11.26) The corresponding completeness relation is (as usual, with σ= 1) ∞/summationdisplay l=mum l(x)um l(x/prime) =δ(x−x/prime). (11.27) The problem we wanted to solve was equation 11.6, so we substitute back inx= cosθinto the completeness relation, which gives ∞/summationdisplay l=mum l(cosθ)um l(cosθ/prime) =δ(cosθ−cosθ/prime) =δ((θ−θ/prime)(−sinθ)) =δ(θ−θ/prime) sinθ. 170 CHAPTER 11. SPHERICAL SYMMETRY In the second equality we used a Taylor expansion for θnearθ/prime, which yields cosθ−cosθ/prime=−(θ−θ/prime) sinθ. In the third equality we used the δ-function property δ(ax) =|a|−1δ(x). The completeness condition for um l(cosθ) is thus ∞/summationdisplay l=mum∗ l(cosθ)um l(cosθ/prime) =δ(θ−θ/prime) sinθ. (11.28) Similarly, the orthogonality condition becomes (since/integraltext1 −1d(cosθ) = 30 Mar p8/integraltextπ 0sinθdθ)/integraldisplayπ 0dθsinθum l/prime(cosθ)um l(cosθ) =δll/prime. (11.29) 11.4 Spherical Harmonics pr:spherH1 1 Apr p1aWe want to determine the properties of the functions Ym l, such as completeness and orthogonality, and to determine their explicit form. We postulated that the solution of equations 11.6 and 11.7 has the form (c.f., equation 11.8) Ym l(θ,ϕ) =eimϕ √ 2πum l(cosθ) for integer l=m,m + 1,m+ 2,...andm≥0, where we have found (equation 11.25) um l(cosθ) =/radicaltp/radicalvertex/radicalvertex/radicalbt(l−m)! (lm)!2l+ 1 2(sinθ)m/parenleftBiggd dcosθ/parenrightBiggm Pl(cosθ). We deﬁne the Y−m l(θ,ϕ), form> 0, as Y−m l(θ,ϕ)≡(−1)mYm l(θ,ϕ)∗ = (−1)me−imϕ √ 2πum l(cosθ). The term ( −1)mis a phase convention ande−imϕ√ 2πis an eigen function. This is often called the Condon-Shortley phase convention. pr:consho1 11.4. SPHERICAL HARMONICS 171 11.4.1 Othonormality and Completeness of Ym l We saw that the functions um lsatisfy the following completeness con- dition: ∞/summationdisplay l=mum l(cosθ)um l(cosθ/prime) =δ(θ−θ/prime) sinθfor allm wheremis ﬁxed and positive. We know that ∞/summationdisplay m=−∞eimϕ/prime √ 2πe−imϕ √ 2π=δ(ϕ−ϕ/prime). (11.30) Multiply1 sinθδ(θ−θ/prime) into equation 11.30, so that eq11.15 1 GApr 1b δ(ϕ−ϕ/prime)δ(θ−θ/prime) sinθ=∞/summationdisplay m=−∞eimϕ √ 2π ∞/summationdisplay l≥|m|u|m| l(cosθ)u|m| l∗(cosθ) e−imϕ/prime √ 2π =∞/summationdisplay m=−∞∞/summationdisplay l≥|m|Ym l(θ,ϕ)Ym l∗(θ,ϕ/prime), since Ym l= (−1)meimϕ √ 2πu|m| l(cosθ) form< 0, and Ym l∗= (−1)me−imϕ √ 2πu|m| l(cosθ) form< 0. Thus we have the completeness relation δ(Ω−Ω/prime) =∞/summationdisplay l=0l/summationdisplay m=−lYm l(θ,ϕ)Ym l∗(θ/prime,ϕ/prime). (11.31) We also note that Lθϕhas (2l+ 1)–fold degenerate eigenvalues l(l+ 1) eq11.16 in LθϕYm l(θ,ϕ) =l(l+ 1)Ym l(θ,ϕ). Thusmis like a degeneracy index in this equation. Next we look at the orthogonality of the spherical harmonics. The pr:orthon5 172 CHAPTER 11. SPHERICAL SYMMETRY orthogonality relation becomes /integraldisplay dΩYm l/prime(θ,ϕ)Ym l(θ,ϕ) =/integraldisplay2π 0dϕ 2πδmm/prime/integraldisplay1 −1dcosθum l(cosθ)um/prime l/prime(cosθ) =δll/primeδmm/prime, wheredΩ =dϕdθ sinθon the right hand side, Because the u’s are orthogonal and the e−imϕ’s are orthogonal, the right hand side is zero whenl/negationslash=l/primeorm/negationslash=m/prime. 11.5 GF’s for Spherical Symmetry 1 Apr 2a We now want to solve the Green’s function problem for spherical sym- metry. 11.5.1 GF Diﬀerential Equation The ﬁrst step is to convert the diﬀerential equation into spherical co- ordinates. The equation we are considering is [L0−λσ(x)]G(x,x/prime;λ) =δ(x−x/prime). (11.32) By substituting the L0for spherically symmetric problems, which we eq11.16a found in equation 11.3, we have I don’t know how to ﬁx this. /bracketleftBigg −1 r2d dr/parenleftBigg r2τ(r)d dr/parenrightBigg +τ(r) r2Lθϕ+V(r)−λσ/bracketrightBigg Glm(x,x/prime;λ/prime) =δ(r−r/prime) r2δ(Ω−Ω/prime) (11.33) =δ(r−r/prime) r2∞/summationdisplay l=0l/summationdisplay m=lYm l(θ,ϕ)Ym∗ l(θ/prime,ϕ/prime). where the second equality follows from the completeness relation, equa- eq11.17 tion 11.31. Thus we try the solution form pr:ExpThm2 G(x,x/prime;λ) =∞/summationdisplay l=0l/summationdisplay m=lYm l(θ,ϕ)Glm(r,r/prime;λ/prime)Ym l∗(θ,ϕ). (11.34) 11.5. GF’S FOR SPHERICAL SYMMETRY 173 Note that the symmetry of θ,φandθ/prime,φ/primein this solution form means eq11.17b that Green’s reciprocity principle is satisﬁed, as required. Substituting this into equation 11.33 and using equation 11.6 results in Lθφbeing replaced by the eigenvalue of Ym l, which isl(l+ 1). Superposition says that we can look at just one term in the series. Since the linear operator no longer involves θ,φ, we may divide out the Ym l’s from both sides to get the following radial equation /bracketleftBigg −1 r2d dr/parenleftBigg r2τ(r)d dr/parenrightBigg +τ(r) r2l(l+ 1) +V(r)/bracketrightBigg Glm(r,r/prime;λ/prime) =δ(r−r/prime) r2. (11.35) The linear operator for this equation has no mdependence. That is, meq11.18 is just a degeneracy index for the 2 l+ 1 diﬀerent solutions of the Lθφ equation for ﬁxed l. Thus we can rewrite our radial Green’s function GlmasGland deﬁne the radial operator as L(l) 0=−d dr/parenleftBigg r2τ(r)d dr/parenrightBigg +r2/bracketleftBiggτ(r) r2l(l+ 1) +V(r)/bracketrightBigg . (11.36) We have reduced the three dimensional Green’s function to the standard eq11.18b single dimensional case with eﬀective tension r2τ(r), eﬀective potential energyV(r), and a centripetal kinetic energy term ( τ(r)/r2)l(l+ 1). 11.5.2 Boundary Conditions pr:bc5 What about the boundary conditions? If the boundary conditions are1 Apr 2bnot spherically symmetric, we need to take account of the angles, i.e., [ˆn· ∇+κ(S)]G(x x/prime,λ) = 0, forxonSandx/primeinR. Consider a spherical region as shown in ﬁgure 11.2. For spherically ﬁg11b symmetric boundary conditions, we can set κint(S) =kaandκext(S) = kb. Thus/bracketleftBigg −∂ ∂r+ka/bracketrightBigg G= 0 forr=a, /bracketleftBigg∂ ∂r+kb/bracketrightBigg G= 0 forr=b. 174 CHAPTER 11. SPHERICAL SYMMETRY bb ' &$ %S2-R S1- a b- Figure 11.2: The general boundary for spherical symmetry. If we insert Gfrom equation 11.34 into these conditions, we ﬁnd the following conditions on how Glbehaves: /bracketleftBigg −∂ ∂r+ka/bracketrightBigg Gl= 0 forr=a, /bracketleftBigg∂ ∂r+kb/bracketrightBigg Gl= 0 forr=b, for alll. These equations, together with equation 11.35, uniquely de- termineGl. With these deﬁnitions we can examine three interesting cases: (1) the internal problem a→ ∞ , (2) the external problem b→0, and (3) the all space problem a→0 andb→ ∞ . pr:spcProb1 These cases correspond to bound state, scattering, and free space prob- lems respectively. 11.5.3 GF for the Exterior Problem4 Apr p1 We will now look at how to determine the radial part of the Green’s function for the exterior problem. essential idea is that we have taken a single partial diﬀerential equation and broken it into several ordinary diﬀerential equations. For the spherical exterior problem the region R of interest is the region outside a sphere of radius a, and the boundary S is the surface of the sphere. The physical parameters are all spherically 11.5. GF’S FOR SPHERICAL SYMMETRY 175 symmetric: τ(r),σ(r),V(r), andκ(S) =ka. Our boundary condition is /bracketleftBigg −∂ ∂r+ka/bracketrightBigg G(x,x/prime;λ) = 0, wherer=afor allθ,ϕ(that is, |x/prime|>|x|=a). This implies 4 Apr p2,3 /bracketleftBigg −∂ ∂r+ka/bracketrightBigg Gl(r,r/prime;λ) = 0 forr/prime>r=a. The other boundary condition is that Glis bounded as r→ ∞ . We now want to solve Gl(r,r/prime;λ). Recall that we have seen two ways of expressing the Green’s function in terms of solutions of the homogeneous equation. One way is to write the Green’s functions as a product of the solution satisfying the upper boundary condition and the solution satisfying the lower boundary condition, and then divide by the Wronskian to ensure continuity. Thus we write Gl(r,r/prime;λ) =−1 r2τ(r)ul 1(r<,λ)ul 2(r>,λ) W(ul 1,ul 2), (11.37) whereul 1andul 2satisfy the equations eq11.18c [Ll 0−λσ(r)r2]ul 1(r,λ) = 0, [Ll 0−λσ(r)r2]ul 2(r,λ) = 0, and the boundary conditions /bracketleftBigg −∂ ∂r+ka/bracketrightBigg ul 1= 0 forr=a, ul 2(r,λ)<∞whenr→ ∞. The other way of expressing the Green’s function is to look at how it behaves near its poles (or branch cut) and consider it as a sum of residues. This analysis was performed in chapter 4 where we obtained the following bilinear sum of eigenfunctions: 4 April p4 Gl(r,r/prime,λ) =/summationdisplay nu(l) n(r)u(l) n(r/prime) λ(l) n−λ. (11.38) 176 CHAPTER 11. SPHERICAL SYMMETRY Note that what is meant here is really a generic sum which can mean eq11.19 either a sum or an integral depending on the spectrum of eigenvalues. For the external problem we are considering, the spectrum is a pure continuum and sums over nshould be replaced by integrals over λn. Theu(l) n(r) solve the corresponding eigen value problem L(l) 0u(l) n(r) =λ(l) nr2σ(r)u(l) n(r) with the boundary conditions that /bracketleftBigg −∂ ∂r+ka/bracketrightBigg u(l) n= 0 forr=a, andu(l) nis ﬁnite as r→ ∞ . The interior problem, with ul nﬁnite as r→0, has a discrete spectrum. The normalization of the u(l) n(r) is given by the completeness relation /summationdisplay nu(l) n(r)u(l) n(r/prime) =δ(r−r/prime) r2σ(r). We insert equation 11.38 into 11.34 to get 4 Apr p5 G(x,x/prime;λ) =/summationdisplay nu(l,m) n(x)u(l,m) n(x/prime) λ(l) n−λ(11.39) where eq11.20 u(l,m) n(x) =Ym l(θ,ϕ)u(l) n(r). andλ(l) nis the position of the nth pole ofGl. So the eigenvalues λnare theλ(l) ndetermined from the r-space eigenvalue problem. The corre- sponding eigenfunctions u(l,m) n(x) satisfy L(l) 0u(l,m) n(x) =λ(l) nr2σ(r)u(l,m) n(x). The completeness relation for u(l,m) n(x) is found by substituting equation 11.39 into 11.32 and performing the same analysis as in chapter 4. The result is /summationdisplay nu(l) n(r)u(l) n(r/prime) =δ(x−x/prime) σ(x). 11.6. EXAMPLE: CONSTANT PARAMETERS 177 11.6 Example: Constant Parameters 4 Apr p6 We now look at a problem from the homework. We apply the aboveOld HW#4analysis to the case where V= 0 andτandσare constant. Our operator for Lbecomes (c.f., equation 11.36) L(l) 0=τ/bracketleftBigg −d dr/parenleftBigg r2d dr/parenrightBigg +l(l+ 1)/bracketrightBigg . The equation for the Green’s function becomes (after dividing by τ): /bracketleftBigg −d dr/parenleftBigg r2d dr/parenrightBigg +l(l+ 1)−k2r2/bracketrightBigg Gl(r,r/prime;λ) =1 τδ(r−r/prime), wherek2=λσ/τ =λ/c2. 11.6.1 Exterior Problem We again have the boundary conditions /bracketleftBigg −∂ ∂r+ka/bracketrightBigg Gl(r,r/prime;λ) = 0, (11.40) wherer=a,r/prime>a, andGlbounded as r→ ∞ . As usual, we assume (eq11.21d the solution form Gl(r,r/prime,λ) =−1 r2τ(r)u(l) 1(r<,λ)u(l) 2(r>,λ) W(u(l) 1,u(l) 2. (11.41) We solve for u1andu2: eq11.21a 4 Apr p7 /bracketleftBigg −d dr/parenleftBigg r2d dr/parenrightBigg +l(l+ 1)−k2r2/bracketrightBigg u(l) 1,2= 0, whereu1andu2satisfy the boundary conditions as r=aandr→ ∞ respectively and k≡ω/c=/radicalBig λ/c2. This is the spherical Bessel equa- tion from the third assignment of last quarter. We found the solution u(r) =R(r)√r, 178 CHAPTER 11. SPHERICAL SYMMETRY whereR(r) satisﬁes the regular Bessel equation /bracketleftBigg −d dr/parenleftBigg rd dr/parenrightBigg +(l+1 2)2 r−k2r/bracketrightBigg R(r) = 0. The solutions for this equation are: R∼Jl+1 2,Nl+1 2. By deﬁnition pr:sphBes jl(x) =/radicalbiggπ 2xJl+1 2(x), nl(x) =/radicalbiggπ 2xNl+1 2(x), h(1) l(x) =/radicalbiggπ 2xHl+1 2(x), wherejl(kr) is a spherical Bessel function, nl(kr) is a spherical Neu- mann function, and h(1) l(kr) is a spherical Hankel function. So we can write u1=Ajl(kr<) +Bnl(kr<), u2=h1 l(kr>), Note that the u2solution is valid because it is bounded for large r: Is this correct? limx→∞h(1) l(x) =−i xeix(−i)l. 11.6.2 Free Space Problem 4 Apr p8 We now take the special case where a= 0. The boundary condition becomes the regularity condition at r= 0, which kills the nl(kr<) solution. The solutions u(1) landu(2) lare then u(1) l=jl(kr), u(2) l=h(1) l(kr). Letλbe an arbitrary complex number. Note that since h(1) l(x) = 4 Apr p9 11.6. EXAMPLE: CONSTANT PARAMETERS 179 jl(x) +inl(x),we have W(jl(x),h(1) l(x)) =iW(jl(x),nl(x)) =i x2. The last equality follows immediately if we evaluate the Wronskian for largerusing jl(x)≈cos(x−(l+ 1)π/2)x→ ∞, nl(x)≈sin(x−(l+ 1)π/2)x→ ∞, and recall that τWis a constant (for the general theory, c.f., problem 1, set 3) where “ τ” isr2τfor this problem. In particular we have W(jl,h(1) l) =i (kr)2. We then get from equation 11.41 Gl=1 r2τjl(kr<)h(1) l(kr>) ki (kr)2 =ik τjl(kr<)h(1) l(kr>). (11.42) (eq11.21c We have thus found the solution of /bracketleftBigg −∇2−λ c2/bracketrightBigg G=δ(x−x/prime) τ, (11.43) which is the fundamental three-space Green’s function. We found (eq11.22 4 Apr p10 G(x,x/prime,λ) =ik τ/summationdisplay lmYm l(θ,ϕ)jl(kr<)hl(kr>)Ym l∗, (11.44) which is a simple combination of equation 11.34 and 11.42. In the ﬁrst eq11.22a homework assignment we solve this by a diﬀerent method to ﬁnd an explicit form for equation 11.43. Here we show something related. In equation 11.43 we have G=G(|x−x/prime|) sinceV= 0 andσandτ 180 CHAPTER 11. SPHERICAL SYMMETRY constant. This gives translational and rotational invariance, which cor- responds to isotropy and homogeneity of space. We solve by choosing x= 0 so that jl(kr<) =j0(kr) + 0/primes asr/prime→0. Only the l= 0 term survives, since as r/prime→0 we have 4 Apr p11 l= 0 −→jl(0) = 1, l/negationslash= 0 −→jl(0) = 0. Thus we have G(x,x/prime;λ) =ik τ|Y0 0|2h0(kr). Sincel= 0 implies m= 0, we get Y0 0= const.= 1/4πsince it satisﬁes the normalization/integraltextdΩ|Y|2= 1. We also know that h(1) 0(x) =−i xeix. This gives us G(x x/prime;λ) =1 τ4πreikr. We may thus conclude that G(|x−x/prime|) =eik|x−x/prime| 4π|x−x/prime|τ. (11.45) eq11.25 stuﬀ omitted 11.7 Summary 1. The form of the linear operator in spherical coordinates is L0=−1 r2∂ ∂r/parenleftBigg r2τ(r)∂ ∂r/parenrightBigg +τ(r) r2Lθϕ+V(r), where Lθϕ=−1 sinθ∂ ∂θ/parenleftBigg sinθ∂ ∂θ/parenrightBigg −1 sin2θ∂2 ∂2ϕ. 11.8. REFERENCES 181 2.Lθφis hermitian. 3. The eigenvalue equations for Ym lare LθϕYm l(θ,ϕ) =l(l+ 1)Ym l(θ,ϕ), −∂2 ∂ϕ2Ym l(θ,ϕ) =m2Ym l(θ,ϕ)m= 0,±1,±2,.... 4. The partial wave expansion for the Green’s function is G(x,x/prime;λ) =∞/summationdisplay l=0l/summationdisplay m=lYm l(θ,ϕ)Glm(r,r/prime;λ/prime)Ym l∗(θ,ϕ). 5. The Green’s function for the free spce problem is G(|x−x/prime|) =eik|x−x/prime| 4π|x−x/prime|τ. 11.8 References The preferred special functions reference for physicists seems to be [Jackson75]. Another good source is [Arfken85]. This material is developed by example in [Fetter81]. 182 CHAPTER 11. SPHERICAL SYMMETRY Chapter 12 Steady State Scattering Chapter Goals: •Find the free space Green’s function outside a circle of radiusadue to point source. •Find the free space Green’s function in one, two, and three dimensions. •Describe scattering from a cylinder. 12.1 Spherical Waves 6 Apr p1 We now look at the important problem of steady state scattering. Consider a point source at x/primewith sinusoidal time dependence pr:sssc1 f(x/prime,t) =δ(x−x/prime)e−iωt, whose radiated wave encounters an obstacle, as shown in ﬁgure 12.1 ﬁg11.3 We saw in chapter 1 that the steady state response for free space with a point source at x/primesatisﬁes /bracketleftBigg L0−σ∂2 ∂t2/bracketrightBigg u0(x,t) =δ(x−x/prime)e−iωt, and was solved in terms of the Green’s function, u0(x,ω) =G0(x,x/prime,λ=ω2+i/epsilon1)e−iωt(12.1) 183 184 CHAPTER 12. STEADY STATE SCATTERING sx/prime-sx $ !% $ #' " & ! %AA AA @@@ Figure 12.1: Waves scattering from an obstacle. =eik|x−x/prime|−iwt 4πτ|x−x/prime|(12.2) where the Green’s function G0satisﬁes the equation 11.43 and the eq11.25b pr:uoxo1 second equality follows from 11.45. The equation for G0can be written [−∇2−k2]G0(x,x/prime;λ) =1 τδ(x−x/prime) (12.3) with the deﬁnition k=/radicalBig λ/c2=ω/c(Remember that λ=ω2+iε). eq11.25c 6 Apr p3 We combine these observations to get u0(x,x/prime;ω) =1 4πτ|x−x/prime|e−iω(t−(x−x/prime)/c). If there is an obstacle (i.e., interaction), then we have a new steady state response u(x,x/prime;ω) =G(x,x/prime;λ=ω2+iε)e−iωt, where [L0−λσ]G(x,x/prime;λ) =δ(x−x/prime) RBC. (12.4) This is the steady state solution for all time. We used u0andG0for the free space problem, and uandGfor the case with a boundary. Note that equation 12.4 reduces to 12.2 if there is no interaction. The scattered part of the wave is GSe−iωt= (G−G0)e−iωt. pr:GS1 6 Apr p4 12.2. PLANE WAVES 185 12.2 Plane Waves We now look at the special case of an incident plane wave instead of an incident spherical wave. This case is more common. Note that once we solve the point source problem, we can also solve the plane wave problem, since plane waves may be decomposed into spherical waves. An incident plane wave has the form Φ0=ei(wt−k·x) where k= (ω/c)ˆn. This is a solution of the homogeneous wave equation pr:Phi0 /bracketleftBigg −∇2+1 c2∂2 ∂t2/bracketrightBigg Φ0= 0. So Φ 0is the solution to the equation without scattering. Let Φ be the wave when an obstacle is present, /bracketleftBigg −∇2+1 c2∂2 ∂t2/bracketrightBigg Φ = 0 which solves the homogeneous wave equation and the regular boundary condition at the surface of the obstacle. To obtain this plane wave problem, we let x/primego to −∞. We now describe this process. To obtain the situation of a plane wave approach- ing the origin from −∞ˆz, we let x/prime=−rˆz, asr/prime→ ∞ . We want to ﬁnd out what eﬀect this limit has on the plane wave solution we obtained in equation 11.45, G0=eik|x−x/prime| 4πτ|x−x/prime|. We deﬁne the angles γandθas shown in ﬁgure 12.2. From the ﬁgure pr:gamma1 ﬁg12a we see that |x−x/prime|=r/prime+rcosθ=r/prime−rcosγ, |x/prime| → ∞. We further recall that the dot product of unit vectors is equal to the cosine of the separation angle, cosγ=x·x/prime rr/prime=−cosθ. 186 CHAPTER 12. STEADY STATE SCATTERING 1 - x/primex x−x/prime r r/primeγ rcosθθˆzorigin Figure 12.2: Deﬁnition of γandθ.. So the solution is G0(x,x/prime;λ) =eik|x−x/prime| 4πτ|x−x/prime| =1 4πτr/primeeik(r−rcosγ) =1 4πτr/primeeikr/primeeikx·(−ˆx/prime) =eikr/prime 4πτr/primeeik·x where k=w c(−ˆx/prime). Note that we have used the ﬁrst two terms of the approximation in the exponent, but only the ﬁrst term in the denomi- nator. Thus we see that in this limit the spherical wave u0in free space due to a point source is lim |x/prime|→∞u0=G0e−iωt =eikr/prime 4πr/primeτΦ0 where Φ0=e−i(ωt−k·x). 12.3 Relation to Potential Theory pr:PotThy1 8 Apr p1Consider the problem of ﬁnding the steady state response due to a point source with frequency ωlocated at x/primeoutside a circular region of radiusa. The steady state response must satisfy the regular boundary condition /bracketleftBigg −∂ ∂rG+κaG/bracketrightBigg = 0 forr=a. (12.5) 12.3. RELATION TO POTENTIAL THEORY 187 In particular we want to ﬁnd this free space Green’s function outside a eq11A1 circle of radius a, whereV= 0 andσandτare constant. The Green’s function Gsatisﬁes the inhomogeneous wave equation [−∇2−k2]G(x,x/prime) =δ(x−x/prime) τ wherek2=λσ/τ =λ/c2. The solution was found to be (from problem 1 of the ﬁnal exam of last quarter) G=1 4πτ∞/summationdisplay m=−∞eim(ϕ−ϕ/prime)[Jm(kr<) +XmH(1) m(kr<)]H(1) m(kr>).(12.6) with eq11A3 pr:Xm1 Xm=−[kJ/prime m(ka)−KaJm(ka)] [kH(1) m(ka)−KaH(1) m(ka)]. (12.7) Note thatH(1) m(kr>) comes from taking Im√ λ > 0. If we consider (Eq.BE) Im√ λ < 0, we would have H(2) minstead. All the physics is in the functionsXm. Note that from this solution we can obtain the solution 8 Apr p2 to the free space problem (having no boundary circle). Our boundary condition is then then Gmust be regular at |x|= 0: Gregular,|x|= 0. (12.8) (That is, equation 12.5 becomes 12.8.) How do we get this full space eq11A2 solution? From our solution, let ago to zero. So Xmin equation 12.6 goes the zero as agoes to zero, and by deﬁnition G→G0. The free space Green’s function is then G0=i 4πτ∞/summationdisplay m=−∞eim(ϕ−ϕ/prime)Jm(kr<)H(1) m(kr>). (12.9) This is the 2-dimensional analog of what we did in three dimensions. eq11A4 We use this to derive the plane wave expression in 3-dimensions. We may now obtain an alternative expression for the free space Green’s function in two dimensions by shifting the origin. In particular, we place the origin at x/prime. This gives us r/prime= 0, for which Jm(kr/prime)H(1) m(kr)|r/prime=0=/braceleftBigg H(1) 0(kr)m= 0 0 else. 188 CHAPTER 12. STEADY STATE SCATTERING Thus equation 12.9 reduces to G0(r) =i 4πτH(1) 0(kr), which can also be written as G0(|x−x/prime|) =i 4πτH(1) 0(k|x/prime−x|). (12.10) This is the expression for the two dimensional free space Green’s func- eq11A5 tion. In the process of obtaining it, we have proven the Hankel function addition formula H(1) 0(k|x/prime−x|) =∞/summationdisplay m=−∞eim(ϕ−ϕ/prime)Jm(kr<)H(1) m(kr>). We now have the free space Green’s functions for one, two, and three dimensions: G1D 0(|x−x/prime|) =i 2kτeik|x−x/prime|, G2D 0(|x−x/prime|) =i 4πτH(1) 0(k|x/prime−x|), G3D 0(|x−x/prime|) =eik|x−x/prime| 4πτ|x−x/prime|. (12.11) We can interpret these free space Green’s functions physically as fol- eq11A6 pr:fsp2 lows. The one dimensional Green’s function is the response due to a 8 Apr 5plane source, for which waves go oﬀ in both directions. The two dimen- sional Green’s function is the cylindrical wave from a line source. The three dimensional Green’s functions is the spherical wave from a point source. Note that if we let k→0 in each case, we have eik|x−x/prime|= 1 +ik|x−x/prime|, and thus we recover the correct potential respectively for a sheet of charge, a line charge, and a point charge. 12.4. SCATTERING FROM A CYLINDER 189 12.4 Scattering from a Cylinder We consider again the Green’s function for scattering from a cylinder, equation 12.6 G=1 4πτ∞/summationdisplay m=−∞eim(ϕ−ϕ/prime)[Jm(kr<) +XmH(1) m(kr<)]H(1) m(kr>).(12.12) What is the physical meaning of [ Jm(kr<)+XmH(1) m(kr<)] in this equa- eq11A6b tion? This is gives the ﬁeld due to a point source exterior to the cylin- der: u=G(x,x/prime,λ=ω2+iε)e−iωt =G0e−iωt /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright u0+ (G−G0)e−iωt /bracehtipupleft/bracehtipdownright/bracehtipdownleft/bracehtipupright us(12.13) Note thatXmcontains the physics of the boundary condition, eq11A6c /bracketleftBigg −∂G ∂r+κaG/bracketrightBigg = 0 forr=a. From equations 12.12 and 12.13 we identify the scattered part of the solution,us, as us=e−iωt 4πτ∞/summationdisplay m=0eim(φ−φ/prime)XmH(1) m(kr)H(1) m(kr/prime). (12.14) So we have expanded the total scattered wave in terms of H(1) m, where eq11A7 Xmgives themth amplitude. Why are the r>andr<in equation 12.11 but not in equation 12.14? Because there is a singular point at x=x/prime in equation 12.11, but uswill never have a singularity at x=x/prime. (Eq.j) Now consider the more general case of spherical symmetry: V(r), τ(r),σ(r). If these parameters are constant at large distances, V(r) = 0, τ(r) =τ= constant, σ(r) =σ= constant, 190 CHAPTER 12. STEADY STATE SCATTERING then at large distances Gmust have the form of equation 12.11, G0(|x−x/prime|) =eik|x−x/prime| 4πτ|x−x/prime|. We shall see that this formula is basic solution form of quantum me- chanical scattering. 12.5 Summary 1. The free space Green’s function outside a circle of radius adue to point source is G=1 4πτ∞/summationdisplay m=−∞eim(ϕ−ϕ/prime)[Jm(kr<) +XmH(1) m(kr<)]H(1) m(kr>). with Xm=−[kJ/prime m(ka)−KaJm(ka)] [kH(1) m(ka)−KaH(1) m(ka)]. 2. The free space Green’s function in one, two, and three dimensions is G1D 0(|x−x/prime|) =i 2kτeik|x−x/prime|, G2D 0(|x−x/prime|) =i 4πτH(1) 0(k|x/prime−x|), G3D 0(|x−x/prime|) =eik|x−x/prime| 4πτ|x−x/prime|. 3. For the problem of scattering from a cylinder, the total response uis easily decomposed into an incident part a scattered part us, whereuscontains the coeﬃcient Xl. 12.6 References See any oldnuclear of high energy physics text, such as [Perkins87]. Chapter 13 Kirchhoﬀ’s Formula As a further application of Green’s functions to steady state problems, let us derive Kirchhoﬀ’s formula for diﬀraction through an aperture. pr:Kirch1 Suppose we have a point source of sound waves of frequency ωat some point x0in the left half plane and at x= 0 we have a plane with a hole. We want to ﬁnd the diﬀracted wave at a point xin the right-half plane. ﬁg19a If we look at the screen directly, we see the aperture is the yz-plane atx= 0 with a hole of shape σ/primeas shown in the ﬁgure. The solution G(x,x0;ω) satisﬁes the equations −[∇2+k2]G(x,x0;ω) =δ(x−x0), ∂G ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglex=0 x/negationslash∈σ/prime= 0. We will reformulate this problem as an integral equation. This integral equation will have as a kernel the solution in the absence of the hole due to a point source at x/primein the R.H.P. This kernal is the free space Green’s function, G0(x,x/prime;ω), which satisﬁes the equations −[∇2+k2]G0(x,x/prime;ω) =δ(x−x/prime), ∂G0 ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglex=0 ally,z= 0. This we solve by the method of images. The boundary condition is pr:MethIm2 191 192 CHAPTER 13. KIRCHHOFF’S FORMULA S R σ/primexx0 vv Figure 13.1: A screen with a hole in it. satisﬁed by adding an image source at x/prime∗, as shown in ﬁgure 13.2 Thus the Green’s function for this boundary value problem is G(x,x/prime) =1 4π/bracketleftBiggeik|x−x/prime| |x−x/prime|+e−ik|x−x/prime| |x−x/prime|/bracketrightBigg Now by taking L0=−(∇2+k2) we may apply Green’s second identity /integraldisplay x∈R(S∗L0u−uL0S∗) =/integraldisplay x∈SdSˆn·[u∇S∗−S∗∇u] withu=G(x,x0) andS∗=G0(x,x/prime) whereRis the region x>0 and Sis theyz-plane. This gives us L0u= 0 for x>0, ∂u ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsinglex=0 x/negationslash∈σ= 0, and L0S∗=δ(x−x/prime), ∂S∗ ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=0= 0. 193 x= 0x/prime∗ vx/prime vxv * x0vx/prime v Figure 13.2: The source and image source. These identities allow us to rewrite Green’s second identity as −/integraldisplay dxG(x,x0)δ(x−x/prime) =/integraldisplay dydzG 0(x,x/prime)/parenleftBigg −∂ ∂xG(x,x0)/parenrightBigg x=0 and therefore G(x/prime,x0) =−/integraldisplay dydzG 0(x,x/prime)∂ ∂xG(x,x0)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=0. (13.1) Thus the knowledge of the disturbance, i.e., the normal component of eq19a the velocity at the aperture, determines the disturbance at an arbitrary point xin the right half plane. We have then only to know∂ ∂xGat the aperture to know Geverywhere. Furthermore, if x0approaches the aperture, equation 13.1 becomes an integral equation for Gfor which we can develop approximation methods. G(x/prime,x0) =−/integraldisplay x∈σ/primedydzG 0(x,x/prime)∂ ∂xG(x,x0). (13.2) The conﬁgurations for the diﬀerent G’s are shown in ﬁgure 13.3. Note eq19b ﬁg19b G0(x,x/prime)|x/prime=0=1 2πeik|x−x/prime| |x−x/prime|, and equation 13.2 becomes G(x/prime,x0) =−1 2π/integraldisplay x∈σ/primedydzeik|x−x/prime| |x−x/prime|∂ ∂xG(x,x0)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=0. 194 CHAPTER 13. KIRCHHOFF’S FORMULA G0(x,x/prime)xvx/prime v G(x0,x/prime)x0vx/prime v G(x,x0)x0vxv Figure 13.3: Conﬁgurations for the G’s. Now suppose that the size aof the aperture is much larger than the wavelength λ= 2π/k of the disturbance which determines the distance scale. In this case we expect that the wave in the aperture does not diﬀer much from the undisturbed wave except for within a few wavelengths near the aperture. Thus for ka/lessmuch1 we can write G(x,x0)≈ −1 2π/integraldisplay x∈σ/primedydzeik|x−x/prime| |x−x/prime|∂ ∂xG(x,x0)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=0∂ ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleeik|x−x/prime| 4π|x−x/prime|/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=0, where we have used the substitution ∂ ∂xG(x,x0)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=0=∂ ∂x/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleeik|x−x/prime| 4π|x−x/prime|/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle x=0. This equation then gives us an explicit expression for G(x/prime,x0) in terms of propagation from the source at xto the ﬁeld point x/primeof the velocity disturbance at xof the velocity distribution∂ ∂xeik|x−x/prime| 4π|x−x/prime|produced by free propagation to xfrom the point x0of the disturbance. This yields Huy- ﬁg19c gen’s principle and other results of physical optics (Babenet’s principle, pr:Huyg1 etc.). 13.1 References See [Fetter80, pp327–332] for a discussion of these results. Chapter 14 Quantum Mechanics Chapter Goals: •State the Green’s function equation for the inho- mogeneous Schr¨ odinger equation. •State the Green’s function for a bound-state spec- tra in terms of eigen wave functions. •State the correspondence between classical wave theory and quantum particle theory.8 Apr p8 pr:QM1 The Schr¨ odinger equation is /bracketleftBigg H−i¯h∂ ∂t/bracketrightBigg ψ(x,t) = 0 where the Hamiltonian His given by H=−¯h2 2m∇2+V(x). This is identical to our original equation, with the substitutions τ= ¯h2/2m,L0=H, and in the steady state case λσ=E. For the free space problem, we require that the wave function ψbe a regular function. The expression |ψ(x,t)|2is the probability given by the probability amplitudeψ(x,t). For the time dependent Schr¨ odinger equation, we have the same form as the heat equation, with ρcp→i¯h. In making the 195 196 CHAPTER 14. QUANTUM MECHANICS xV(x) E E EE D D D C CC BB AAQQ x E < 0E > 0 Figure 14.1: An attractive potential. transition from classical mechanics to quantum mechanics, we use H= p2/2m+V(x) with the substitution p→(¯h/i)∇; this correspondence for momentum means that the better we know the position, and thus the more sharply the wave function falls oﬀ, the worse we know the subsequent position. This is the essence of the uncertainty principle. We now look at the steady state form. Steady state solutions will pr:sss3 be of the form ψ(x,t) =e−iωtψω(x), whereψω(x) satisﬁes the equation [H−¯hω]ψω(x) = 0. The allowed energy levels for ¯ hωare the eigen values EofH: pr:enLev1 8 Apr p9 Hψ=Eψ. (14.1) So the allowed frequencies are ω=E/¯h, for energy eigenvalues E. The eq13.1 energy spectrum can be either discrete or continuous. Consider the potential shown in ﬁgure 14.1. For E=En< V(0), the energy levels ﬁg13.1 are discrete and there are a ﬁnite number of such levels; for E >V (0), the energy spectrum is continuous: any energy above V(0) is allowed. A plot of the complex energy plane for this potential is shown in ﬁgure 14.2. The important features are that the discrete energies appear as poles on the negative real axis, and the continuous energies appear as a 14.1. QUANTUM MECHANICAL SCATTERING 197 u u u u u ReEImEE-plane Figure 14.2: The complex energy plane. branch cut on the positive real axis. Note that where as in this problem there are a ﬁnite number of discrete levels, for the coulomb potential there are instead an inﬁnite number of discrete levels. If, on the other hand, we had a repulsive potential, then there would be no discrete spectrum. The Green’s function solves the Schr¨ odinger equation with an inho- mogeneous δ-function term: 8 Apr p10 (H−E)G(x,x/prime;E) =δ(x−x/prime) (14.2) whereEis a complex variable. The boundary condition of the Green’s eq13.2 function for the free space problem is that it be a regular solution. Un- like previously considered problems, in the quantum mechanical prob- lems, the eﬀect of a boundary, (e.g., the surface of a hard sphere) is enforced by an appropriate choice of the potential (e.g., V=∞for r>a ). Once we have obtained the Green’s function, we can look at its energy spectrum to obtain the ψ’s andEn’s, using the formula obtained in chapter 4: G(x,x;E) =/summationdisplay nψn(x)ψn(x/prime) En−E. This formula relates the solution of equation 14.1 to the solution of equation 14.2. ﬁg13.2 11 Apr p1 11 Apr p2 11 Apr p314.1 Quantum Mechanical Scattering pr:QMS1 We now look at the continuum case. This corresponds to the prob- lem of scattering. We use the Green’s function to solve the problem of 198 CHAPTER 14. QUANTUM MECHANICS quantum mechanical scattering. In the process of doing this, we will see that the quantum mechanics is mathematically equivalent to the classical mechanics of waves. Both situations involve scattering. The solutionufor the classical wave problem is interpreted as a velocity potential, whereas the solution ψfor the quantum mechanical problem is interpreted as the probability amplitude. For the classical wave prob- lem,u2is interpreted as intensity, while for quantum mechanics, |ψ|2is interpreted as probability density. Thus, although the mathematics for these problems is similar, the general diﬀerence is in the interpretation. The case of quantum mechanical scattering is similar to classical scattering, so we consider classical scattering ﬁrst. We use the Green’s function in classical wave theory, [L0−λσ]G(x,x;λ) =δ(x−x/prime), withλ=ω2+iεfor causality, to obtain the steady state response due to a point source, u=e−iωtG(x,x;λ=ω2+iε). This steady state solution solves the time dependent classical wave equation,/bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg u(x,t) =δ(x−x/prime). (14.3) We may decompose the solution uinto two parts, eq13.3 u=u0+uscat, where u0=e−iωtG0=e−i(ωt−kR) 4πτR whereR=|x−x/prime|, and uscat=e−iωt[G−G0]. Note thatu0is the steady state solution for a point source at x=x/prime, pr:uscat1 solution for a point source at x,G0is the solution for the free case, andusis the solution for outgoing scattered waves. Note also that the outgoing scattered waves have no singularity at x=x/prime. 14.2. PLANE WAVE APPROXIMATION 199 14.2 Plane Wave Approximation pr:PlWv1 11 Apr p4If one solves the problem of the scattering of the spherical wave from a point source, that is, for the Green’s function problem, then we also have the solution for scattering from a plane wave, merely by letting |x/prime| → ∞ . We now deﬁne Φ( x,t) as the solution of equation 14.3 for the special case in which x/prime→ −∞ ˆz, /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg Φ(x,t) = 0. (14.4) eq13.4 11 Apr p5 For the steady state solution Φ(x,t) =e−iωtΦ(x,ω), we get the equation [L0−σω2]Φ(x,ω) = 0. How is this equation solved for positive ω? There isn’t a unique solution for this, just like there wasn’t a unique solution for the Green’s function. We have already found a solution to this equation by considering the Green’s function with the source point going to inﬁnity in the above method, but it does not satisfy the boundary condition appropriate for 11 Apr p6 scattering. In order to determine the unique scattering solution, we must in- troduce a new boundary condition appropriate for scattering. This is necessary because although our previous equation [L0−σλn]un= 0 (14.5) with RBC gave the eigenfunctions, it does not give unique physical eq13abc solutions for the case of scattering. To ﬁnd the un’s in equation 14.5 we would extract them from the Green’s function using, for the discrete case, G(x,x/prime;λ) =/summationdisplay nunu∗ n λn−λ, 200 CHAPTER 14. QUANTUM MECHANICS or for the continuum case, un=1 2πi[G(x,x/prime;λ=λn+iε)−G(x,x/prime;λ=λn−iε)] =1 πIm [G(x,x/prime;λ=λn+iε)]. But theseun’s are not solutions corresponding to the scattering bound- ary condition, since they contain both incoming and outgoing waves. 14.3 Quantum Mechanics 11 Apr p7 We now apply what we have said for particles to the case of quantum mechanics. In this case the steady state solutions are of the form ψ=e−i(E/¯h)tψ0. We want the total wave to be a superposition of an incident plane wave and a scattered wave. ψ=eik·x−iωt+ψs whereω=E/¯h. Note that eik·xcorresponds to an incident plane wave, andψscorresponds to outgoing waves. To get this form, we use the 11 Apr p9 Green’s function for the free space problem (no potential) G0=eikR 4πRτ=m 2π¯h2eikR R, where k2=λσ τ=E ¯h2/2m=2mE ¯h2=p2 ¯h2. Take the limit |x/prime| → ∞ , the free space Green’s function becomes e−i(E/¯h)tG0(x,x/prime;E) =ψ0m 2π¯h2eikr r, whereψ0=e−ik·x−i(E/¯h)t. 11 Apr p9 14.4. REVIEW 201 14.4 Review We have been considering the steady state response problem 13 Apr p1 /bracketleftBigg L0+σ∂2 ∂t2/bracketrightBigg u(x,t) =δ(x−x/prime)e−iωt. The steady state response for outgoing waves (i.e., that which satisﬁes the boundary condition for scattering) is u(x,t) =e−iωtG(x,x/prime,λ=ω2+iε) (14.6) whereGsolves eq14.g [L0−σλ]G(x,x/prime;λ) =δ(x−x/prime). We want to get the response Φ( x,t) for scattering from a plane wave. We only need to let |x/prime|go to inﬁnity: lim |x/prime|→∞u(x/prime,t) =eikr/prime 4πr/primeΦ(x,t). This gives scattering from a plane wave. Φ is the solution of [L0+σ∂2 ∂t2]Φ(x,t) = 0 which satisﬁes the boundary condition of scattering. For the case of steady state response we can write Φ(x,t) =e−iωtΦ(x,ω) where Φ( x,ω) solves the equation [L0−σω2]Φ(x,ω) = 0. This equation satisﬁes the boundary condition of scattering: 13 Apr p2 Φ(x,ω) =eik·x+ Φ s(x), where Φ s(x) has only outgoing waves. Note that k2=ω2σ τ=ω2 c2andk=kˆn, whereσ= lim r→∞σ(r) andτ= lim r→∞τ(r). 202 CHAPTER 14. QUANTUM MECHANICS 14.5 Spherical Symmetry Degeneracy We now compare the mathematics of the plane wave solution Φ with that of the eigen function u. The eigenvalue equation can be written [L0−σω2]uα(x,ω2) = 0. In this equation uαis a positive frequency eigen function with degener- acy number αand eigen value ω2. The eigen functions can be obtained directly from the Green’s function by using 1 πImG(x,x/prime;λ=ω2+i/epsilon1) =/summationdisplay αuα(x,ω2)u∗ α(x/prime,ω2). For the case of spherical symmetry we have uα(x,ω2) =ulm(x,ω2) =Ylm(θ,ϕ)ul(r,ω2) wherel= 0,1,...andm=−l,..., 0,...,l . The eigenvalues ω2are 13 Apr p3 continuous: 0 <ω2<∞. Because of the degeneracy, a solution of the diﬀerential equation may be any linear combination of the degenerate eigen functions: Φ(x,ω) =/summationdisplay cαuα, where thecα’s are arbitrary coeﬃcients. This relates the eigen function to the plane wave scattering solution. In the next chapter we will see how thecα’s are to be chosen. 14.6 Comparison of Classical and Quan- tum pr:ClasMech1 Mathematically we have seen that classical mechanics and quantum mechanics are similar. Here we summarize the correspondences between their interpretations. classical wave theory quantum particle theory u= wave amplitude ψ= probability amplitude 14.6. COMPARISON OF CLASSICAL AND QUANTUM 203 u2= energy density |ψ|2= probability density ωn= natural frequencies En= energy eigenvalues Normal modes: Stationary states un(x)e−iωnt=un(x,t) ψn(x)e−i(En/¯h)t=ψn(x,t) L0un=σω2 nun Hψ n(x) =Enψn(x) L0is positive deﬁnite: H+=H: ω2 n>0 En∈R 13 Apr p4 The scattering problem is like the eigen value problem but we look at the region of continuous spectrum. For scattering, we require that En>0. In this continuum case, look at the eigen values from: (H−E)ψα(x,E) = 0, (14.7) whereαlabels wave functions with degenerate eigenvalues. Rather eq13.25 than a wave we have a beam of particles characterized by some energy E. The substitution from classical mechanics to quantum mechanics is as follows: ω2σ→E, τ →¯h2 2m, k2→E ¯h2/2m=2mE ¯h2=/parenleftbiggp ¯h/parenrightbigg2 . This last equation is the De Broglie relation. pr:DeBr1 13 Apr p5 Now we want to look at the solution for quantum mechanical scat- tering using Green’s functions. Suppose we have a beam of particles coming in. This incident free wave has the form eik·x−iEt/ ¯hwhich solve the free space hamiltonian H0=¯h2 2m∇2. We want the solution to equation 14.7 which corresponds to scattering. That is, we want the solution for (H−E)Φ(x,E) = 0 which is of the form Φ(x,E) =e−k·x+ Φ s where Ψ shas only outgoing waves. 204 CHAPTER 14. QUANTUM MECHANICS To solve this we look at the Green’s function. (H−E)G(x,x/prime;E) =δ(x−x/prime) for ImE > 0, with the appropriate boundary conditions. We make the substitution ψ(x,t) =e−i(E/¯h)tG(x,x/prime;E+iε). This corresponds not to a beam of particles but rather to a source of particles. 13 Apr p6 Stuﬀ omitted 14.7 Summary 1. The Green’s function equation for the inhomogeneous Schr¨ odinger equation is (H−E)G(x,x/prime;E) =δ(x−x/prime), where H=−¯h2 2m∇2+V(x). 2. The Green’s function for a bound-state spectra in terms of eigen wave functions is G(x,x;E) =/summationdisplay nψn(x)ψn(x/prime) En−E. 3. There is a close connection between classical and quantum me- chanics which is discussed in section 14.6. 14.8 References See your favorite quantum mechanics text. Chapter 15 Scattering in 3-Dim Chapter Goals: •State the asymptotic form of the response function due to scattering from a localized potential. •Derive the scattering amplitude for a far-ﬁeld ob- server due to an incident plane wave. •Derive the far-ﬁeld form of the scattering ampli- tude. •Deﬁne the diﬀerential cross section and write it in terms of the scattering amplitude. •Derive and interpret the optical theorem. •Derive the total cross section for scattering from a hard sphere in the high energy limit. •Describe the scattering of sound waves from an os- cillating sphere.15 Apr p1 We have seen that the steady state case reduces the Green’s function problem to the equation [L0−λσ]G(x,x/prime;λ) =δ(x−x/prime) 205 206 CHAPTER 15. SCATTERING IN 3-DIM with RBC, where the linear operator is given by L0=−∇ ·τ(x)∇+V(x). In the spherically symmetric case we have V(r),σ(r), andτ(r). In chapter 11 we saw that the Green’s function can be written as an ex- pansion in terms of spherical harmonics, G(x,x/prime;λ) =/summationdisplay lmYm l(θ,ϕ)Gl(r,r/prime;λ)Ym∗ l(θ/prime,ϕ/prime). In the last chapter we saw how to solve for scattering from a point source and scattering from a plane wave. We did this for a particular case in the problem set. This all had nothing to due with spherical symmetry. Only partly true. Now consider the case of spherical symmetry. From chapter 3 we know that the radial Green’s function can be written Gl(r,r/prime;λ) =−ul 1(r<,λ)ul 2(r>,λ) r2τ(r)W(ul 1,ul 2). (15.1) Theu’s solve the same radial eigenvalue equation eq14.0 /bracketleftBigg −1 r2d dr/parenleftBigg r2τ(r)d dr/parenrightBigg +τ(r)l(l+ 1) r2+V(r)−λσ(r)/bracketrightBigg ul 1,2= 0,(15.2) but diﬀerent boundary conditions. The eigenfunction u1satisﬁes the eq14.1 15 Apr p2 boundary condition ∂ ∂rul 1−κul 1= 0 forr=a and asa→0 we replace this with the boundary condition ul 1(r) ﬁnite atr= 0. The eigenfunction u2satisﬁes the boundary condition u2 ﬁnite asr→ ∞ . By comparing equation 15.2 with previous one-dimensional equa- tions we have encountered, we identify the second and third terms as an eﬀective potential, Veﬀ=τ(r)l(l+ 1) r2+V(r). In quantum mechanics we have pr:Veﬀ1 Veﬀ=¯h2l(l+ 1) 2mr2+V(r). 15.1. ANGULAR MOMENTUM 207 15.1 Angular Momentum The above spherical harmonic expansion for the Green’s function was obtained by solving the corresponding eigenfunction equation for the angular part, ¯h2LθϕYm l= ¯h2l(l+ 1)Ym l. The diﬀerential operator Lθϕcan be related to angular momentum by recalling that the square of the angular momentum operator satisﬁes the equation L2 opYm l= ¯h2l(l+ 1)Ym l. Thus we identify ¯ h2LθϕasL2 op, the square of the angular momentum operator. ¯h2Lθϕ≡L2 op. In the central potential problem of classical mechanics it was found that Veﬀ=L2 2mr2+V(r), where pr:bfL1 L=x×p and L2=L·L. In quantum mechanics the momentum operator is p= (¯h/i)∇, so that L=x×p→Lop=¯h ix× ∇ and thus L2 op=/parenleftBigg¯h ix× ∇/parenrightBigg ·/parenleftBigg¯h ix× ∇/parenrightBigg = ¯h2Lθϕ. This gives the relation between angular momentum in classical mechan- ics and quantum mechanics. 208 CHAPTER 15. SCATTERING IN 3-DIM sx/prime-sx $ !% $ #' " & ! %AA AA @@@ Figure 15.1: The schematic representation of a scattering experiment. 15.2 Far-Field Limit We now take the far ﬁeld limit, in which r→ ∞ , meaning the ﬁeld is measured far from the obstacle. This situation is accurate for ex- perimental scattering measurements and is shown in ﬁgure 15.1. We pr:ExpScat1 ﬁg14a assume that in this r→ ∞ limit, we have σ(r)→σ,τ(r)→τ, and rV(r)→0. If instead the potential went as V(r) =γ/r, e.g., the Coulomb potential, then our analysis would change somewhat. We will also use the wave number k=/radicalBig σλ/τ , whereλ=ω2+i/epsilon1classically, andλ=E+i/epsilon1for the quantum case. In classical mechanics we then havek=ω/cand in quantum mechanics we have k=p/¯h. 15 Apr p3 Our incident wave is from a point source, but by taking the source- to-target separation r/primebig, we have a plane wave approximation. After making these approximations, equation 15.2 becomes /bracketleftBigg −1 r2d dr/parenleftBigg r2d dr/parenrightBigg +l(l+ 1) r2−k2/bracketrightBigg ul 1,2= 0. (15.3) If we neglect the term l(l+ 1)/r2, compared to k2we would need kr/greatermuch eq14.2 l, which we don’t want. Instead we keep this term in order to keep conventional solutions. In fact, it will prove easier to keep it, even though it may vanish faster than V(r) asr→ ∞ , and since we also have to consider llarge, we don’t want to kill it. In this limit the Green’s function is proportional to a product of the u’s, limr→∞Gl(r,r/prime;λ) =Aul 1(r/prime,λ)ul 2(r,λ). (15.4) 15.2. FAR-FIELD LIMIT 209 We assume the the point source is not in the region where things are eq14.3 really happening, but far away. In this case V(r/prime) = 0,σ(r/prime) =σ, and τ(r/prime) =τ, for larger/prime. Thus we are looking at the far ﬁeld solution where the point source is outside the region of interaction. 15 Apr p4 We already know the explicit asymptotic solution to the radial equa- tion: ul 2(r>)≈h(1) l(kr>) forkr>/greatermuch1, (15.5) ul 1(r<)≈j(1) l(kr<) +Xlh(1) l(kr<) forkr</greatermuch1.(15.6) Xlcontains all the physics, which arises due to the boundary condition. eq14.5 In general,Xlmust be evaluated numerically. For speciﬁc cases such as in the problem set, V(r) = 0 so equation 15.3 is valid everywhere, and thus we may obtain Xlexplicitly. For our present situation we have assumed the far ﬁeld approximation and an interaction-free source, for which the asymptotic form of the Green’s function may be written in terms of equation 15.5 and 15.6 as limr→∞Gl(r,r/prime;λ) =A[jl(kr/prime) +Xlh(1) l(kr/prime)]h(1) l(kr). We have not yet speciﬁed r>r/prime, only that r/greatermuch1 andr/prime/greatermuch1. To obtain the scattered wave at large r, we look at G−G0. In particular we will evaluate Gl−Gl0. So we look at Gl 0(r,r/prime;λ) for r/greatermuchr/prime. Recall that Gl0has the form 15 Apr p5 Gl0(r,r/prime;λ) =ik τjl(kr<)h(1) l(kr>). This is for the free problem; it solves all the way to the origin. Now take the diﬀerence. Gl−Gl0=/parenleftBigg A−ik τ/parenrightBigg jl(kr<)h(1) l(kr>) +AX lh(1) l(kr<)h(1) l(kr>). This equation assumes only that we are out of the range of of interac- tion. The term h(1) l(kr>) gives the discontinuity on dG/dr atr=r/prime. We now assert that Amust equal ik/τ because the scattering wave What does this mean? 15 Apr p6Gl−Gl0has to be nonsingular. Now look at the case r>r/prime: pr:scatWv1 210 CHAPTER 15. SCATTERING IN 3-DIM Gl−Gl0=ik τXlh(1) l(kr/prime)h(1) l(kr). All that is left to see is what the scattered wave looks like. We take kr to be large, as in the problem set. To solve this equation not using Green’s function, we ﬁrst look for a solution of the homogeneous problem which at large distances gives scattered plus incident waves. Φ =∞/summationdisplay m=−∞cmum(r,ϕ). At large distances we have 15 Apr p7 Φ→eixt+ outgoing waves . 18 Apr p1 18 Apr p2 15.3 Relation to the General Propagation Problem We could instead consider the general problem of propagation, but at this time we are just considering the case of scattering, for which the source lies in a homogeneous region where V(r) = 0 andσandτare constant. The propagation problem is more general because it allows the source to be anywhere. 15.4 Simpliﬁcation of Scattering Problem For the scattering problem, we are considering a beam of particles from a distant (r/prime/greatermuch1) point source in a homogeneous medium incident on a target, which scatter and are detected by detectors far away ( r/greatermuch1). This latter condition is called the far ﬁeld condition. In this case we have seen that the problem can be simpliﬁed, and that we may explicitly calculate the scattered Green’s function GlS(r,r/prime;λ): pr:scGF1 GlS=Gl(r,r/prime,λ)−Gl0(r,r/prime,λ) =ik cXlh1 l(kr)h1 l(kr/prime), 15.5. SCATTERING AMPLITUDE 211 withk=/radicalBig λσ/τ =√ λ/c, wherecis the speed. The value of Xl depends on kand is obtained from the behavior of u1at larger=|x|. We already know that ul 1(r) must be of the form 18 Apr p3 limr→∞ul 1(r) =jl(kr) +Xl(k)h1 l(kr), since this is the asymptotic form of the solution of the diﬀerential equa- What does this mean Physi- cally? Isn’t Ul 1 for distances less than rsource ? Don’t we need r < r/prime?tion. Thus scattering reduces to this form. All we need is Xl, which is obtained from the behavior of ul 1(r). In particular, we don’t need to know anything about ul 2(r) if we are only interested in the scattering problem, because at large distances it cancels out. The large distance behavior of the function which satisﬁes the bound- ary condition at small distances is what determines the scattering so- lution. In this case randr/primeare both large enough that we are in essentially a homogeneous region. 15.5 Scattering Amplitude pr:scAmp1 Consider the special problem where V= 0,σ= const., and τ= const., with the boundary condition ∂u1 ∂r+ku1= 0 forr=a. In the problem set we found Xlby satisfying this condition. The result was Xl=−[kj/prime l(ka)−κjl(ka)] [kh1 l/prime(ka)−κh1 l(ka)]. This equation is valid for r>a . 18 Apr p4 GivenXlwe can calculate the diﬀerence, Gl−Gl0so we can calculate GlS. Thus we can determine the scattered wave, which we now do. We calculate the scattered piece by recalling the expansion in terms of spherical harmonics, GS=G−G0=/summationdisplay l,mYm l(θ,ϕ)[Gl(r,r/prime;λ)−Gl0(r,r/prime;λ)]Ym∗ l(θ/prime,ϕ/prime). (15.7) We substitute into this the radial part of the the scattered Green’s eq14.6 212 CHAPTER 15. SCATTERING IN 3-DIM 1 - x/primex x−x/prime r r/primeγ rcosθθˆzorigin Figure 15.2: The geometry deﬁning γandθ. function, G(r,r/prime;λ)−G0(r,r/prime;λ) =∞/summationdisplay l=0ik cXl(k)h(1) l(kr)h(1) l(kr/prime), (15.8) and the spherical harmonics addition formula eq14.7 18 Apr p5 pr:addForm1l/summationdisplay m=−lYm l(θ,ϕ)Ym∗ l(θ/prime,ϕ/prime) =2l+ 1 4πPl(cosγ) =(−1)l 4π(2l+ 1)Pl(cosθ) (15.9) where cosγ= ˆx·ˆx/prime. The geometry is shown in ﬁgure 15.2. The result eq14.8 ﬁg14b of plugging equations 15.8 and 15.9 into equation 15.7 is GS=G−G0=∞/summationdisplay l=0ik τXl(k)h1 l(kr)h1 l(kr/prime)(−1)l 4πPl(cosθ)(2l+ 1). 15.6 Kinematics of Scattered Waves We take the limit kr→ ∞ to get the far ﬁeld behavior. In the asymp- totic limit, the spherical Hankel function becomes h(1)(x)x→∞−→ −i x(−i)leix. Thus in the far ﬁeld limit the scattered Green’s function becomes G−G0→ik τeikr kr(−i) 4π∞/summationdisplay l=0Xkh1 l(kr/prime)(i)l(2l+ 1)Pl(cosθ). This in the case for a detector very far away. We can write also this as 18 Apr p6 15.7. PLANE WAVE SCATTERING 213 G−G0=eikr r˜f(θ,r/prime,k), where ˜f(θ,r/prime,k) =1 4πτ∞/summationdisplay l=0(2l+ 1)(i)lPl(cosθ)Xl(k)h1 l(kr/prime). (15.10) This is an independent proof that the scattered Green’s function, G−eq14.9 G0, is precisely an outgoing wave with amplitude ˜f. The scattered part of the solution for the steady state problem is given by us=e−iωt(G−G0). The energy scattered per unit time per unit solid angle will be propor- tional to the energy per unit area, which is the energy ﬂux u2. This in turn is proportional to the scattering amplitude ˜f. Thus dE dtdΩ∼ |f|2. 18 Apr p7 We know the radial diﬀerential ds=r2drof the volume dV=dsdΩ for a spherical shell, so that we get dE dtdΩ=dE dtdsds dr Note that dimensionally we havedE dtds=1 r2andds dr=r2, so thatdE dtdΩis dimensionless. Thus it is the 1 /rterm in the scattered spherical wave which assures conservation of energy. pr:ConsE1 15.7 Plane Wave Scattering We now look at scattering from a plane wave. Let r/prime=|x/prime|go to inﬁnity. This gives us h(1) l(kr/prime)r/prime→∞−→(−i)l+1e|kr/prime| kr/prime. In this limit equation 15.10 becomes ˜f(θ,r/prime,k)→e|kr/prime| 4πτr/primef(θ,k) 214 CHAPTER 15. SCATTERING IN 3-DIM where 18 Apr p8 ekr/prime r/primef(θ,k) =−i k∞/summationdisplay l=0(2l+ 1)Pl(cosθ)Xl. (15.11) f(θ,k) is called the scattering amplitude for a ﬁeld observer from an eq14fth pr:ftrk1 incident plane wave. We can now compute the total wave for the far ﬁeld limit with incident plane wave. It is u=e−iωtG=e−iωt[G−G0+G0] =e−iωt/parenleftBigg −eikr/prime 4πτr/prime/parenrightBigg/parenleftBigg eik·x+eikr rf/parenrightBigg . In this equation the term eik·xcorresponds to a plane wave and the termeikr rfcorresponds to an outgoing scattered wave. So |f2| ∼dE dtdΩ This is a problem in the problem set. 15.8 Special Cases 20 Apr p1 So far we have considered the case in which all the physics occurs within some region of space, outside of which we have essentially free space. We thus require that in the area exterior to the region, V0= 0,τ= constant, and σ= constant. The source emits waves at x/prime, and we want to ﬁnd the wave amplitude at x. Note that for the Coulomb potential, we have no free space, but we may instead establish a distance after which we may ignore the potential. 20 Apr p3 15.8.1 Homogeneous Source; Inhomogeneous Ob- server In this case x/primeis in a region where V(r)≈0, andσandτare constant. We deﬁneu0to be the steady state solution to the point source problem without a scatterer present, i.e., u0is the free space solution. u0=e−iωtG0, 15.8. SPECIAL CASES 215 where G0=eik|x−x/prime| 4πτ|x−x/prime|. Further we deﬁne the scattered solution us≡u−u0. To ﬁnduswe use equation 15.1 to get the spherical wave expansion G0l=ik τjl(kr<)h(1) l(kr>). Thus us=e−iωt(G−G0). For the case of a homogeneous source and an inhomogeneous observer r>=r/prime,r<=r= 0. We take u(l) 2(r/prime) =h(1) l(kr/prime). Remember that r/primeis outside the region of scattering, so ul 2solves the 20 Apr p4 free space equation, 15.3, /bracketleftBigg −1 r2d dr/parenleftBigg r2d dr/parenrightBigg +l(l+ 1) r2−k2/bracketrightBigg u2= 0, (15.12) wherek2=λσ/τ with the condition ul 2(r) ﬁnite asr→ ∞ . The general eq14.10 solution to equation 15.12 is u(l) 2(r/prime) =h(1) l(kr/prime). We still need to solve the full problem for u1with the total eﬀective potentialV(r)/negationslash= 0. 15.8.2 Homogeneous Observer; Inhomogeneous Source In this case the source point is in the interior region. We want to ﬁnd 20 Apr p5 uforxinside the medium, but we cannot use the usmethod as we did in case 1. The reason why it is not reasonable to separate u0anduSin this case is because the source is still inside the scattering region. 216 CHAPTER 15. SCATTERING IN 3-DIM We replace r>→randr<→r/primeso that u(l) 2(r)→h(1) l(kr) andu(l) 1(r) satisﬁes the full potential problem So once again we only need to solve for u(l) 1(r). The physics looks the same in case 1 and case 2, and the solutions in these two cases are reciprocal. This is a manifestation of Green’s reciprocity principle. The case of a ﬁeld inside due to a source outside looks like the case of a ﬁeld outside due to a source inside. 15.8.3 Homogeneous Source; Homogeneous Observer For this case both points are in exterior region. By explicitly taking |x|>|x/prime|we make this a special case of the previous case. Thus we haver>→randr<→r/prime. Now both u1andu2satisfy the reduced ordinary diﬀerential equation /bracketleftBigg −1 r2d dr/parenleftBigg r2d dr/parenrightBigg +l(l+ 1) r2−k2/bracketrightBigg u1,2= 0, (15.13) whereu1satisﬁes the lower boundary condition and u2satisﬁes the eq14rad upper boundary condition. As we have seen, the asymptotic solutions to this equation are u(l) 1(r/prime)→jl(kr/prime) +Xlh(1) l(kr/prime). (15.14) u(l) 2(r)→h(1) l(kr). (15.15) To obtainXlwe must solve eq14.11,12 20 Apr p6/bracketleftBigg −1 r2d dr/parenleftBigg r2τ(r)d dr/parenrightBigg +/parenleftBiggl(l+ 1) r2+V(r)/parenrightBigg τ(r)−λσ/bracketrightBigg u1= 0, and then take r/greatermuch1. We can then get Xl(k) simply by comparing equations 15.14 and 15.15. The scattered wave is then us=e−iωt(G−G0), where Gl−G0 l→ik τe−iωtXlh(1) l(kr)h(1) l(kr/prime). We see that the ﬁeld at xis due to source waves u0and scattered waves uS. 15.8. SPECIAL CASES 217 Homogeneous Source and Observer, Far Field For this case the source and the ﬁeld point are out of the region of interaction. We take r>r/primeandr/prime→ ∞ . For these values of randr/primewe haveV= 0 andτandσconstant. In this case e−iωtG→e−iωtG0=eik|x−x/prime|−iωt 4πτ|x−x/prime|(15.16) (Eq.f) G0=∞/summationdisplay l=0(2l+ 1) 4π(−1)lPl(cosθ)G0 l (15.17) where (Eq.g) G0 l=ik τjl(kr<)h(1) l(kr>) (15.18) (Eq.h) u=e−iωtG=e−iωtG0+us (15.19) where (Eq.i) 22 Apr p3 us=e−iωt(G−G0) =ik τe−iωt∞/summationdisplay l=0Xl(2l+ 1)(−1)l 4πPl(cosθ)h(1) l(kr)h(1) l(kr/prime) where for large r, u1→jl(kr) +Xlh(1) l(kr) (15.20) This is the large rbehavior of the solution satisfying the small rbound- (Eq.k) ary condition. 15.8.4 Both Points in Interior Region We put xvery far away, next to a detector. The assumption that xlies in the vicinity of a detector implies kr/greatermuch1. This allows us to make the following simpliﬁcation from case 2: h(1) l(kr)→(−i)l(−i) kreikr. (15.21) Thus we can rewrite u1. We have (Eq.m) 20 Apr p7 218 CHAPTER 15. SCATTERING IN 3-DIM us=e−iωt(G−G0) (15.22) and the simpliﬁcation (Eq.n) Gl 0→jl(kr/prime)h(1) l(kr). (15.23) (Eq.0) 22 Apr p1 15.8.5 Summary Here is a summary of the cases we have looked at case 4 need to know u1,u2everywhere cases 1, 2 need to know u1everywhere case 3 need to know u1at largeronly We now look at two more special cases. 22 Apr p4 15.8.6 Far Field Observation Make a large rexpansion ( r→ ∞ ): h(1) l(kr)→(−i)l(−i kr)e−ikr(15.24) (EQ.l) u=e−i(ωt−k|x−x/prime|) 4πτ|x−x/prime|+e−i(ωt−kr) r˜f(θ,r/prime,k). (15.25) The terme−i(ωt−kr) r˜f(θ,r/prime,k) is explicitly just the outgoing wave. We (Eq.m) found ˜f(θ,r/prime,k) =1 4πτ∞/summationdisplay l=0(2l+ 1)(i)lPl(cosθ)Xlh(1) l(kr/prime) (15.26) The term ˜f(θ,r/prime,k) is called the scattering amplitude for a point source (Eq.n) 22 Apr p5 atr/prime. The ﬂux of energy is proportional to ˜f2. 15.9. THE PHYSICAL SIGNIFICANCE OF XL 219 15.8.7 Distant Source: r/prime→ ∞ Let the distance of thee source go to inﬁnity. Deﬁne k=k(−ˆx/prime) (15.27) and in ˜f, letr/prime→ ∞ . This gives us (Eq.o) u→eikr/prime 4πr/primeτ/bracketleftBigg e−i(ωt−k·x)+e−i(ωt−kr) rf(θ,k)/bracketrightBigg (15.28) We can then get (Eq.p) 22 Apr p6 f=−i k∞/summationdisplay l=0(2l+ 1)Pl(cosθ)Xl (15.29) This equation is seen in quantum mechanics. ˜fis called the scattering (Eq.q) amplitude at angle θ, and does not depend on ϕdue to symmetry. The basic idea is that plane waves come in, and a scattered wave goes out. The wave number kcomes from the incident plane wave. 22 Apr p7 15.9 The Physical signiﬁcance of Xl Recall that Xlis determined by the large distance behavior of the solu- pr:Xl1 tion which satisﬁes the short distance boundary condition. Xlis deﬁned by u(1) l(kr)→jl(kr) +Xl(k)h(1) l(kr). (15.30) This equation holds for large rwithV= 0 andσ,τconstant. By using eq14.20 the identity jl(kr) =1 2/parenleftBig h(1) l(kr) +h(2) l(kr)/parenrightBig , we can rewrite equation 15.30 as u(1) l(kr)→1 2/bracketleftBig h(2) l(kr) + (1 + 2Xl)h(1) l(kr)/bracketrightBig . (15.31) We now deﬁne δl(k) by eq14.21 1 + 2Xl=e2iδl(k), 220 CHAPTER 15. SCATTERING IN 3-DIM We will prove that δl(k) is real. This deﬁnition allows us to rewrite equation 15.31 as u(1) l(kr)→1 2/bracketleftBig h(2) l(kr) +e2iδl(k)h(1) l(kr)/bracketrightBig , or u(1) l(kr) =1 2eiδl/bracketleftBig e−iδlh(2) l+eiδlh(1) l/bracketrightBig . (15.32) The solution ul 1satisﬁes a real diﬀerential equation. The boundary eq14.22 condition at r→0 gives real coeﬃcients. Thus ul 1is real up to an overall constant factor. This implies δlreal. Another way of seeing this is to note that by the deﬁnition of h(1) landh(2) lwe have h(2) l(kr) =/bracketleftBig h(1) l(kr)/bracketrightBig∗. Thus the bracketed expression in equation 15.32 is an element plus its complex conjugate, which is therefore real. If ul 1(kr)∈R, then δl(kr)∈R. Ask Baker We now look at the second term in equation 15.32 for far ﬁelds, 22 Apr p8 eiδlh(1) l(kr)r→∞−→eiδl(k)−i kr(−i)leikr. Note that (−i)l=e−iπl/2. This gives eiδlh(1) l(kr) =−i krei(kr−πl/2+δl) So ul 1(kr)∼1 krsin(kr−πl/2 +δl(k))r→ ∞. (15.33) Thusδl(k) is the phase shift of the lth partial wave at wave number k. eq14.23 In the case that V= 0 we have ul 1(kr)→ul 1,0(kr). If there is no potential, then we have Xl(k)→0, 15.9. THE PHYSICAL SIGNIFICANCE OF XL 221 u(l) 1(r)u(l) 1,0(r)u r R nR(0) n Figure 15.3: Phase shift due to potential. and by using the asymptotic expansion of j, we see that equation 15.30 22 Apr p9 becomes ul 1,0(r)∼1 krsin/parenleftBigg kr−πl 2/parenrightBigg . (15.34) Thus the phase shift δl(k) is zero if the potential is zero. eq14.24 25 Apr p1 25 Apr p2 25 Apr p3Consider the values of rfor which the waves u1andu1,0are zero in the far ﬁeld limit. For equation 15.33 and equation 15.34 respectively, the zeros occur when kRn−πl 2+δl=nπ, and kR0 n−πl 2=nπ. By taking the diﬀerence of these equations we have k(Rn−R0 n) =−δl(k). (15.35) Thusδl(k) gives the large distance diﬀerence of phase between solutions (eq14.25 with interaction and without interaction. This situation is shown in ﬁg- ure 15.3. For the case shown in the ﬁgure, we have R0 n> R n, which ﬁg14c meansδl>0. Note that turning on the interaction “pulls in” the scat- tered wave. Thus we identify two situations. δl>0 corresponds to an attractive potential, which pulls in the wave, while δl<0 corresponds to a repulsive potential, which pushes out the wave. 222 CHAPTER 15. SCATTERING IN 3-DIM We now verify this behavior by looking at the diﬀerential equation for the quantum mechanical case. We now turn to the quantum me- 25 Apr p4 chanical case. In this case we set τ(r) = ¯h2/2mandk2(x) =2mE ¯h2in the equation /bracketleftBigg −1 r2d dr/parenleftBigg r2d dr/parenrightBigg +Vl eﬀ(r) τ(r)−λσ(r) τ(r)/bracketrightBigg ul 1(r) = 0. So for the radial equation with no interaction potential we have λσ/τ = 2mE/ ¯h2, while for the radial equation with an interaction potential we haveλσ/τ = 2m(E−V)/¯h2. Thus the eﬀect of the interaction is to change the wave number from k2 0=2mE ¯h2, to an eﬀective wave number k2(r) =2m(E−V) ¯h2. (15.36) eq14.26 Suppose we have an attractive potential, V(r)>0. Then from equation 15.36 we see k2(r)>k2 0, which means momentum is increasing. Also, since k2(r)>0, increasing k2increases the curvature of u, which means the wavelength λ(r) decreases and the kinetic energy increases. Thus the case k2(r)>l2 0corresponds to an attractive potential pulling in a wave, which means δl(k)>0. The phase shift δl(k)>0 is a measure of how much the wave is pulled in. Note that this situation is essentially that of a wave equation for a wave moving through a region of variable index of refraction. 25 Apr p5 Now consider a repulsive potential with l= 0, as shown in ﬁgure 15.4. We have V(r) =Eforr=r0, andV(r)> E forr > r 0. In ﬁg14d this latter case equation 15.36 indicates that k2(r)>0, which means the wave will be attenuated. Thus, as the wave penetrates the barrier, there will be exponential decay rather than propagation. 15.9.1 Calculating δl(k) From Griﬃes, 28 April, p2b It is possible to calculate δl(k) directly from ul 1without calculating Xl(k) as an intermediate step. To do this, let r→ ∞ and then compare 15.10. SCATTERING FROM A SPHERE 223 V(r) r r 0E Figure 15.4: A repulsive potential. thisul 1with the general asymptotic form from equation 15.33, u∼ sin(kr−lπ/2 +δl(k))/r. A diﬀerent method for calculating δl(k) is presented in a later section. 15.10 Scattering from a Sphere pr:ScSph1 We now look at the example of scattering from a sphere, which was already solved in the homework. We have the boundary conditions V= 0 atr=a ∂ ∂rul 1+κu1= 0 atr=a. We found in problem set 2, that Xlfor this problem is 25 Apr p7 Xl=[kj/prime l(ka)−κjl(ka)] [kh(1)/prime l(ka)−κh(1) l(ka)](15.37) by solving the radial equation. Stuﬀ missing Now look at the long wavelength limit, which is also the low energy limit. In this case ka/lessmuch1 wherek= 2π/λ. We know asymptotically that jl(ka)∼(ka)l, 224 CHAPTER 15. SCATTERING IN 3-DIM and h(1) l(ka)∼1 (ka)l+1. Thus we have Xl(k)ka/lessmuch1−→(ka)2l+1/parenleftBiggl−κa l+κa/parenrightBigg /lessmuch1 since (ka)2l+1=(ka)l (ka)−l−1. Again,k=/radicalBig 2mE/ ¯h2. We now look at the phase shift for low energy 25 Apr p8 scattering. We use the fact Xl(k)∼(ka)2l+1 to write 1 + 2Xl(k) =ei2δ(k) = 1 + 2iδl+···. Thus we have δl(k)∼(ka)2l+1. 15.10.1 A Related Problem25 Apr p9 We now turn to a related problem. Take an arbitrary potential, for example V=V0e−r/a. In this case the shape of Veﬀis similar, except that is has a potential barrier for low values of r.VandVeﬀfor this example are shown in ﬁgure 15.5. The centrifugal barrier increases as lincreases, that is, it ﬁg14e gets steeper. Thus, as lincreases, the scattering phase shift gets smaller and smaller since the centrifugal barrier gets steeper. Recall that arepresents the range of the potential and 2 l+ 1 rep- resents the eﬀect of a potential barrier. We assert that in the long 15.11. CALCULATION OF PHASE FOR A HARD SPHERE 225 V(r) Veﬀ(r) r r r0a Figure 15.5: The potential VandVeﬀfor a particular example. wave length limit, that is, low energy scattering, the phases shift goes generally as δl(k)∼(ka)2l+1forka/greatermuch1. This is a great simpliﬁcation for low energy scattering. It means that as long aska/lessmuch1, we need only consider the ﬁrst few lin the inﬁnite series for the scattering amplitude f(θ). In particular, the dominant contribution will usually come from the l= 0 term. For the case l= 0, the radial equation is easier to solve, and X0(k) is easier to obtain. Thus the partial wave expansion is very useful in the long wavelength, or low energy, limit. This limit is the opposite of the geometrical or physical optics limit. The low energy limit is useful, for example, in the study of the nuclear force, where the range of the potential is a∼10−13cm, which giveska/lessmuch1. Note that in the geometrical optics limit, ka/greatermuch1, it is also possible to sum the series accurately. The summation is diﬃcult in the middle region, ka∼1. In this case many terms of the series must be retained. 27 Apr p1 27 Apr p2 27 Apr p315.11 Calculation of Phase for a Hard Sphere We use the “special case” from above. Take κ→ ∞ (very high elastic constant, very rigid media, a hard sphere). In this case u→0 when 226 CHAPTER 15. SCATTERING IN 3-DIM r=a. Thus we get from equation 15.37 X0(k) =−sinka ka −ieika ka =−ieika−e−ika 2ieika =−1 2[1−e−2ika]. So e2iδ0= 1−[1−e−2ika] =e−2ika, and thus δ0(k) =−ka. (15.38) In terms of quantum mechanics, this is like having eq14do 27 Apr p4V(r) =∞forr<a, V(r) = 0 forr>a. Outside we get the asymptotic solution form given in equation 15.33. Forl= 0 and substituting equation 15.38, this becomes u(0) 1=1 rsin(kr−ka). By substituting this into equation 15.13 it is easy to verify that this is an exact solution for ul=0 1. This is exactly what we would expect: a free space spherical wave which satisﬁes the boundary condition at r=a. The wave is pushed out by an amount ka. We thus see that δl(k) is determined by the boundary condition. This situation is shown in ﬁgure 15.6. ﬁg14f 15.12 Experimental Measurement pr:ExpMeas1 We now look at the experimental consequences. Assume that we have solved foru1and knowXl(k) and thus know δl(k). By writing the scattering amplitude from equation 15.11 in terms of the phase shift δl(k), we have 27 Apr p5 15.12. EXPERIMENTAL MEASUREMENT 227 V(r) r aul=0 1(r) r a Figure 15.6: An inﬁnite potential wall. f(θ) =1 k∞/summationdisplay l=0(2l+ 1)Pl(cosθ)e2iδl−1 2i. To getδlfor the solution for u1, we look at large r. Note that e2iδl−1 2i=eiδl[eiδl−e−iδl] 2i =eiδlsinδl. So f(θ) =1 k∞/summationdisplay l=0(2l+ 1)eiδlsinδlPl(cosθ). (15.39) eq14.55 15.12.1 Cross Sectionpr:CrSec1 This scattering amplitude is the quantity from which we determine the energy or probability of the scattered wave. However, the scattering amplitude is not a directly measurable experimental quantity. Recall our original conﬁguration of a source, an obstacle, and a detector. The detector measures the number of particles intercepted per unit time, dN/dt . (It may also distinguish energy of the intercepted pr:N2 particle.) This number will be proportional to the solid angle covered by the detector and the incident ﬂux of particles. If we denote the proportionality factor as σ(θ,φ), then this relationship says that the rate at which particles are scattered into an element of solid angle is is dN/dt =jincdσ=jinc(dσ/d Ω)dΩ. Note that an element of solid angle pr:jinc1 is related to an element of area by r2dΩ =dA. The scattered current through area dAis thendN/dt =jinc(σ(θ,φ)/dΩ)dA/r2. From this we 228 CHAPTER 15. SCATTERING IN 3-DIM identify the scattered current density jscat=jincσ(θ,φ) dΩ1 r2ˆr. (15.40) Now the quantum mechanical current density jis deﬁned in terms of eq14cs1 the wave function: j(r) = Re/bracketleftBigg ψ†¯h im∇ψ/bracketrightBigg , where in the far-ﬁeld limit the boundary condition of scattering tells us that the wave function goes as ψscatr→∞−→N/parenleftBigg eikz+eikr rf(θ,φ)/parenrightBigg . The wave function has an incident plane wave part and a scattered spherical wave part. The current density for the incident wave is then jinc=|N|2¯hk mˆz=jincˆz, and the current density for the scattered wave is jscat=|N|2¯hk m|f(θ,φ)|2 r2ˆr+O(r−3)≈jinc|f(θ,φ)|2 r2ˆr. (15.41) By comparing equations 15.40 and 15.41, we identify the diﬀerential eq14cs2 cross section asdσ dΩ≡ |f(θ,k)|2. (15.42) eq14.57 This relationship between cross section and scattering amplitude agrees with dimensional analysis. Note that the only dimensionful quantity appear in equation 15.39 for fisk: dim(f(θ)) = dim(k−1) = dim(length) . On the other hand, the dimension of the diﬀerential cross section is dim[dσ/d Ω] = dim[l2/1] and dim[ |f(θ,k)|2] = dim[l2]. Thus equation 15.42 is dimensionally valid. Note also that, because the diﬀerential cross section is an area per solid angle, it must be real and positive, which also agrees with |f|2. The total cross section is σ(k) =/integraldisplay dΩdσ dΩ=/integraldisplay dΩ|f(θ,k)|2. 15.12. EXPERIMENTAL MEASUREMENT 229 15.12.2 Notes on Cross Section By using equation 15.39 we can calculate the diﬀerential cross section: dσ dΩ=1 k2∞/summationdisplay l,l/prime=0(2l+ 1)(2l/prime+ 1)eiδlsinδle−iδl/primesinδl/primePl(cosθ)Pl/prime(cosθ) (15.43) In this equation we get interference terms (cross terms). These interfer- eq14.60 27 Apr p6 ence terms prevent us from being able to think of the diﬀerential cross section as a sum of contributions from each partial wave individually. If we are measuring just σ, we can integrate equation 15.43 to get σ=/integraldisplay dΩdσ dΩ(15.44) =/integraldisplay dΩ1 k2∞/summationdisplay ll/prime=0(2l+ 1)(2l/prime+ 1)eiδlsinδle−iδl/primesinδl/primePl(cosθ)Pl/prime(cosθ) We can simplify this by using the orthogonality of the Legendre poly- eq14.61 nomials: /integraldisplay dΩPl(cosθ)Pl/prime(cosθ) =δll/prime4π 2l+ 1. In equation 15.44 the terms eiδlcancel. So we now have σ(k)≡/integraldisplay dΩ|f(θ,k)|2 =4π k2∞/summationdisplay l=0(2l+ 1) sin2δl. (15.45) From this we can conclude that eq14.64 σ=∞/summationdisplay l=0σl, where σl=4π k2(2l+ 1) sin2δl. (15.46) Note thatσlis the contribution of the total cross section of scattering eq14.65 from the 2l+ 1 partial waves which have angular momentum l. There are no interference eﬀects, which is because of spherical symmetry. 27 Apr p7 230 CHAPTER 15. SCATTERING IN 3-DIM Another way to think about this point is that the measuring appa- ratus has introduced an asymmetry in the ﬁeld, and the we have inter- ference eﬀects in dσ/d Ω. On the other hand, in the whole measurement ofσ, there is still spherical symmetry, and thus no interference eﬀects. A measurement of σ(k) is much more crude than a measurement of dσ/d Ω. Because sine is bounded by one, the total cross section of the partial waves are also bounded: σl≤σmax l=4π k2(2l+ 1), or, by using λ= 2π/k, σmax l= 4π/parenleftBiggλ 2π/parenrightBigg2 (2l+ 1). (15.47) Note that the reality of δl(k) puts a maximum value on the contribution eq14.67 σl(k) of thelth partial wave on the total cross section. 15.12.3 Geometrical Limitpr:GeoLim1 In the geometrical limit we have ka/greatermuch1, which is the long wavelength limit,λ/greatermucha. Recall that in this limit δl∼(ka)2l+1. Thus the dominant contribution to the cross section will come from σ0=4π k2sin2δ0=4π k2(ka)2= 4πa2. (15.48) From equation 15.47 we have σmax 0∼4π/parenleftBiggλ 2π/parenrightBigg2 , and from equation 15.48 we have 27 Apr p8 σ0= 4πλ2/parenleftbigga λ/parenrightbigg2 . Comparing these gives us σ0/σmax 0/lessmuch1 fora/λ/lessmuch1, which is the fraction of the incident beam seen by an observer. 15.13. OPTICAL THEOREM 231 15.13 Optical Theorem pr:OptThm1 We now take the imaginary part of equation 15.46: Imf(θ) =1 k∞/summationdisplay l=0(2l+ 1) sin2δlPl(cosθ). In the case that θ= 0 we get Imf(0) =1 k∞/summationdisplay l=0(2l+ 1) sin2δl. By comparing this with equation 15.45 we obtain Imf(0) =σk 4π. This is called the optical theorem. The meaning of this is that the imaginary part of the energy taken out of the forward beam goes into scattering. This principle is called unitarity or conservation of momen- tum. The quantity Im f(θ)|θ=0represents the radiation of the intensity in the incident beam due to interference with the forward scattered beam. This is just conservation of energy: energy removed from the incident beam goes into the scattered wave. 29 Apr p1 15.14 Conservation of Probability Inter- pretation: 29 Apr p2 σk/4π as a proportional- ity factor15.14.1 Hard Sphere pr:HardSph1For the case of a hard sphere of radius awe found that δ0=−ka. In the case of ( ka)/lessmuch1, only the lower terms of equation 15.39 matter. Exact scattering amplitude from a hard sphere k= 0? So for a sphere of radiusa, we have δl∼(ka)2l+1. 232 CHAPTER 15. SCATTERING IN 3-DIM πa2πa2 strong forward peak Figure 15.7: Scattering with a strong forward peak. In the case that k→0, we get (noting that eiδl→0): f(θ) =i keiδlsinδ0(k)P0(cosθ) =i keiδl(−ka) =i k(−ka) =−ia, ask→0. Thus 29 Apr p3dσ dΩ=a2, ask→0. Note that the hard sphere diﬀerential cross section is spherically sym- metric at low energy (that is, when ka/lessmuch1). In this case the total cross section is σ/integraldisplaydσ dΩ= 4πa2. For the geometrical optics limit, ka/greatermuch1, corresponding to short wavelength and high energy, we would expect σ∼πa2since the sphere looks like a circle, but instead we get σ∼2(πa2). The factor of two comes from contributions from all partial waves and has a strong forward peak. The situation has the geometry shown in ﬁgure 15.7. The ﬁgure is composed of a spherically symmetric part and ﬁg14g a forward peak, which each contribute πa2to the total cross section σ. 29 Apr p4 15.15 Radiation of Sound Waves pr:soundWv1 We consider a non-viscous medium characterized by a sound velocity v. In this medium is a hard sphere oscillating about the origin along 15.15. RADIATION OF SOUND WAVES 233 thez-axis. The motion of the center of the sphere is given by xc=εae−iωtˆz, whereε/lessmuch1, and the velocity of the center of the sphere is then given by vc=−iωae−iωtˆz. Note that the normal component of the velocity at the surface of the sphere is ˆn·vsphere =−iεωae−iωtcosθ. (15.49) where we have used ˆ n·ˆz= cosθwithθmeasures from the ˆ z-axis. The eq14.88 minus sign appears because we choose nto point into the sphere. For the velocity of the ﬂuid outside of the hard sphere we have vﬂuid=∇Φ, where Φ is the velocity potential. Thus near the surface of the sphere we have, up to ﬁrst order in ε, ˆn·v|r=a=∂Φ ∂r/vextendsingle/vextendsingle/vextendsingle/vextendsingle r=a. We want to ﬁnd the velocity potential where the velocity potential satisﬁes the equation /bracketleftBigg ∇2+1 c2∂2 ∂t2/bracketrightBigg Φ(x,t) = 0, r>a, with the hard sphere boundary condition that the ﬂuid and the sphere move at the same radial velocity near the surface of the sphere, ˆn·vsphere = ˆn·vﬂuid. The velocity of the ﬂuid is then given by (using equation 15.49) −∂Φ ∂r|r=a=−iεωaeiωtcosθ. (15.50) eq14.93 234 CHAPTER 15. SCATTERING IN 3-DIM 15.15.1 Steady State Solution pr:sss4 The steady state solution is of the form Φ(x,t) =e−iωtΦ(x,ω), where Φ( x,ω) satisﬁes [∇2+l2]Φ(x,ω), r>a, with the outgoing wave boundary condition (from equation 15.50) 29 Apr p5 ∂Φ ∂r=−iεωa cos(θ), r =a. Our boundary condition is of the form ∂Φ ∂r/vextendsingle/vextendsingle/vextendsingle/vextendsingle r=a=g(θ,ϕ), where for our speciﬁc case g(θ,ϕ) =−iεωa cosθ. (15.51) We want to solve the steady state equation subject to the boundary eq14.98 condition. A more general form of the boundary condition is ∂Φ ∂r+κΦ =g(θ,ϕ). (15.52) We write eq14.99 [−∇2−k2]G(x,x/prime;λ) =δ(x/prime−x)/c2, wherek2=λ/c2. This is the standard form of the Green’s function in the case that τ=c2,L0=τ∇2, and |x/prime||x|>0. The solution of this equation, which we found previously, is c2G(x,x/prime;λ) =ik/summationdisplay lmYlm(θ,ϕ)Y∗ lm(θ/prime,ϕ/prime)zl(kr<)h(1) l(kr>) κh(1) l(ka)−kh(1) l/prime(ka). (15.53) The general solution is given by a superposition of the Green’s function (eq14.101 solution for source points on the surface of the sphere, 29 Apr p6 15.15. RADIATION OF SOUND WAVES 235 Φ(x) =c2/integraldisplay x/prime∈SG(x,x/prime;λ=ω2+iε)g(θ,ϕ)a2dΩ/prime. (15.54) By setting λ=ω2+iε, we have automatically incorporated the out- eq14.102 going wave condition. Physically, g(θ,ϕ)a2dΩ/primeis the strength of the disturbance. We use equation 15.53 with r>= 0 andr<=a. Thuszlis zl(kr) = [κh(1) l(ka)−kh(1) l/prime(ka)]jl(kr) −[κj/prime l(ka)−j/prime l(ka)]h(1) l(kr) =−kW(jl(ka)h(1) l(ka)). Recall that we have evaluated this Wronskian before, and plugging in the result gives zl(ka) =−ki (ka)2=−i ka2(15.55) By combining equations 15.55 and 15.53 into equation 15.54, we obtain (eq14.106 29 Apr p7 Φ(x) = −a2(i ka2)(ik)/summationdisplay lmYlm(θ,ϕ)h(1) l(kr) κh(1) l(ka)−kh(1) l(ka)/integraldisplay dΩ/primeY∗ lm(θ/prime,ϕ/prime)g(θ/prime,ϕ/prime) =/summationdisplay lmYlm(θ,ϕ)h(1) l(kr) κh(1) l(ka)−kh(1) l(ka)glm. (15.56) This is called the multipole expansion, where we also have deﬁned eq14.108 glm≡/integraldisplay dΩ/primeY∗ lm(θ/prime,ϕ/prime)g(θ/prime,ϕ/prime). (15.57) glmis the (l,m)th multipole moment of g(θ,ϕ). eq14.109 2 May p2 15.15.2 Far Field Behaviorpr:farFld1 At distances far from the origin ( r→ ∞ ) the spherical Hankel functions can be approximated by h(1) l(k,r) =−i kr(−i)leilr, r /greatermuch1. 236 CHAPTER 15. SCATTERING IN 3-DIM In this limit the velocity potential can be written Φ(x) =eikr rf(θ,ϕ), r → ∞, where the amplitude factor fis given by f(θ,ϕ) =−i k/summationdisplay l,mgl,mYm l(θ,ϕ)(−i)l κh(1) l(k,a)−kh(1)/prime l(k,a). This amplitude may be further decomposed into components of partic- ularlandm: f=/summationdisplay l,mfl,mYm l(θ,ϕ), where fl,m=(−i)l+1 kgl,m κh(1) l(k,a)−kh(1)/prime l(k,a). The interpretation of the l’s is l= 1 dipole radiation m= 0,±1 l= 2 quadrapole radiation m= 0,±1,±2 l= 3 octopole radiation m= 0,±1,±2,±3 For the case l= 1, the 3 possible m’s correspond to diﬀerent polariza- tions. 2 May p3 15.15.3 Special Case We now return to the speciﬁc case of the general boundary condition, equation 15.52, which applies to a hard sphere executing small oscilla- tions. In this case the hard surface implies κ= 0 and the oscillatory motion implies that gis given by equation 15.51, which can be rewritten in terms of the spherical harmonic Y0 1: g(θ,ϕ) =−iωaε/radicalBigg 3 4πY0 l(θϕ). 15.15. RADIATION OF SOUND WAVES 237 By plugging this into equation 15.57, we have obtained the ( l,m) com- ponents ofg(θ,φ), gl,m=/integraldisplay dΩYm l(θ,φ)g(θ,φ) =/integraldisplay dΩ(−iωaε)Ym l(θ,φ)/radicalBigg 3 4πY0 1(θ,ϕ) =−iωaε/radicalBigg 3 4πδm0δl1. This shows that the oscillating sphere only excites the Y0 lmode. f=−1 k−iωat −kh(1)/prime l(ka)/radicalBigg 3 4πY0 l(θ,ϕ) =−iaεω cosθ k2h(1)/prime l(ka). Thus we have pure dipole radiation for this type of oscillation. This ﬁnal equation gives the radiation and shows the dependence on k. omitted qm stuﬀ 2 May p415.15.4 Energy Flux Consider a sound wave with velocity v=−∇Φ(x,t), where in the far ﬁeld limit the velocity potential is Φ(x,t)→eikr rf(θ,ϕ)e−iωt, x→ ∞. We now obtain the rate dE/dt at which energy ﬂows through a surface. This is given by the energy ﬂux through the surface, dE dt≡/integraldisplay ds·jE, where jEis the energy ﬂux vector. For sound waves the energy ﬂux vector can be expressed as a product of velocity and pressure, jE=vp. (15.58) 238 CHAPTER 15. SCATTERING IN 3-DIM This can be intuited as follows. The ﬁrst law of thermodynamics says eq14jvp FW p299 that for an ideal ﬂuid undergoing a reversible isentropic process, the change in internal energy dEmatches the work done on the element, −pdV. The total energy ﬂowing outof through the surface Sis /integraldisplay Sds·jE=dEs dt=/integraldisplay SdsdE dt=/integraldisplay SpdV =/integraldisplay Sds·/parenleftBigg pdr dt/parenrightBigg . By comparing integrands we obtain equation 15.58, as desired. Note thatjE=vphas the correct dimensions for ﬂux — that is, velocity times pressure gives the correct dimensions for energy. 2 May p5 The velocity and pressure are deﬁned in terms of the velocity po- tential and density, x=−∇Φ, p =ρ∂Φ ∂t. We now look at the real parts of the velocity and the pressure for the steady state solution, Rev=1 2ve−iωt+v∗eiωt, Rep=1 2(pe−iωt+p∗e−iωt). The ﬂux is then ask Baker about this. j= RevRRepR =1 4(vp∗+v∗p) +e−2iωtvp+e2iωtv∗p∗. The time averaged ﬂux is then /angbracketleftj/angbracketright=1 4(vp∗+v∗p) =1 2Re (xp∗), where we have used /angbracketlefte−2iωt/angbracketright+/angbracketlefte+2iωt/angbracketright= 0. 15.15. RADIATION OF SOUND WAVES 239 The angled brackets represents the average over time. Note that xand p∗are still complex, but with their time dependence factored out. To obtainpwe use p(t) =e−iωtp=−ρωΦe−iωt, from which we obtain p=−ρωΦ. Thus the time averaged ﬂux is /angbracketleftjE/angbracketright=1 2Re(−∇Φ)(+iωe)Φ∗. (15.59) The radial derivative of Φ is eq14.149 −ˆr∇Φ =−∂Φ ∂r=−ikΦ. Thus in this case the time averaged energy rate is /angbracketleftBiggdE dt/angbracketrightBigg =r2dΩˆr· /angbracketleftjE/angbracketright=r2dΩ|f|2 r2ωρk. Therefore /angbracketleftBiggdE dt/angbracketrightBigg =1 2ρkω|f|2, r /greatermuch1. (15.60) eq14.152 Plane Wave Approximation pr:PlWv2 Now suppose that instead of a spherical wave, we have a plane wave, Φ(x,t) =eik·x−iωt. In this special case the velocity and pressure are given by v=−∇Φ =ikΦr/greatermuch1, p=ρ(−iω)Φ. Using equation 15.59, the energy ﬂux is j=1 2kω=1 2kωρ. 240 CHAPTER 15. SCATTERING IN 3-DIM The power radiated through the area element dAis then 2 May p8 dEA dt=/integraldisplay dAds·j=ρkω 2dA, and we have1 dAdEA dt≡Incident ﬂux =ρkω 2. (15.61) (eq14.158 15.15.5 Scattering From Plane Waves The far ﬁeld response to scattering from an incident plane wave is Φ(x) =eik·x+feikr r. Note that in the limit r→ ∞ , the scattered wave Φ Sisfeikr/r. So dEs dt dΩ=1 2ρkω|f|2. By deﬁnition, the diﬀerential cross section is given by the amount of 2 May p5 energy per unit solid angle per unit time divided by the incident energy ﬂux, dσ dΩ≡dEs dt dΩ Incident ﬂux =1 2ρkω|f|2 1 2ρkω=|f|2. In the second equality we have used equation 15.60 and 15.61. This duplicates our earlier result, equation 15.42. If we are just interested in the radiated wave and not the incident ﬂux, the angular distribution of power is dP dΩ=dE dtdΩ=1 2kρω|f|2. Now expand fin terms of spherical harmonics, f=/summationdisplay l,mYl,mfl,m. 15.15. RADIATION OF SOUND WAVES 241 The radiated diﬀerential power can then be written dP dΩ=1 2kρω|f|2=1 2kρω /summationdisplay l,mYm lfl,m  /summationdisplay l/prime,m/primeYm/prime∗ l/primef∗ l/prime,m/prime , where we have interference terms. The total power is 2 May p6 P=/integraldisplay dΩdP dΩ=/summationdisplay l,m|fl,m|2. In this case there is no interference. This is the analogue for sound wave of the diﬀerential cross section we studied earlier. For the case of a sphere fl,0/negationslash= 0. 4 May p1 15.15.6 Spherical Symmetry We now consider the situation where the properties of the medium surrounding the ﬂuid exhibit spherical symmetry. In this case the scat- tering amplitude can be expanded in terms of spherical harmonics, f(θ,ϕ) =/summationdisplay l,mfl,mYm l(θ,ϕ). This is called the multipole expansion. The term fl,mcorresponds to the mode of angular momentum radiation. Spherical symmetry here means that the dynamic terms are spherically symmetric: σ(r),τ(r), andV(r). However, any initial condition or disturbance, such as g, may have asymmetry. We now look at the external distance problem. gl,m=/integraldisplay dΩYm∗ l(θ,ϕ)g(θ,ϕ). For the general boundary condition the scattering amplitude is related togby fl,m=(−i)l+1 kgl,m κh(1) l(ka)−kh(1)/prime l(ka). 242 CHAPTER 15. SCATTERING IN 3-DIM For our case of small oscillations of a hard sphere, we have κ= 0 and 4 May p2 gl,m=−δl,1δm,0/radicalBigg 3 4πiεaω. In this case the scattering amplitude becomes f(θ,ϕ) =/summationdisplay l,mFlmYm l(θϕ) =−ia/epsilon1ω k2h(1) l=1(ka)cosθ=−ia/epsilon1ccosθ kh(1)/prime l(ka). Thus the diﬀerential power radiated is dP dΩ=1 2ρk2/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingleaεc kcosθ h(1)/prime 1(ka)/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle =1 2ρa2ε2c2 /vextendsingle/vextendsingle/vextendsingleh(1)/prime 1(ka)/vextendsingle/vextendsingle/vextendsingle2cos2θ. Notice that this cos2θdependence is opposite that of dipole radiation, which goes like sin2θ. The total power radiated is in general given by P=/integraldisplay dΩdP dΩ=1 2ρck2/summationdisplay l,m|fl,m|2. Note that there are no interference terms. It is simply a sum of power 4 May p3 from each partial wave. 15.16 Summary 1. The asymptotic form of the response function is limr→∞ul 1(r) =jl(kr) +Xl(k)h1 l(kr). 2. The scattering amplitude for a far-ﬁeld observer due to an inci- dent plane wave is f(θ,k) =−i k∞/summationdisplay l=0(2l+ 1)Pl(cosθ)Xl. 15.17. REFERENCES 243 3. The phase shift δl(k) is deﬁned by the relation 1 + 2Xl=e2iδl(k), which results in a scattered wave solution of the form ul 1(kr)∼1 krsin(kr−πl/2 +δl(k))r→ ∞, whereδl(k) appears as a simple shift in the phase of the sine wave. 4. The scattering amplitude is given by f(θ) =1 k∞/summationdisplay l=0(2l+ 1)eiδlsinδlPl(cosθ). 5. The diﬀerential cross section represents the eﬀective area of the scatterer for those particle which are deﬂected into the solid angle dΩ, and can be written in terms of the scattering amplitude as dσ dΩ≡ |f(θ,k)|2. 6. The optical theorem is Imf(θ)|θ=0=σk 4π. It relates forward wave to the scattered wave. 7. The total cross section for scattering from a hard sphere in the high energy limit is σ∼2(πa2). 15.17 References See any oldnuclear or high energy physics text, such as [Perkins87]. 244 CHAPTER 15. SCATTERING IN 3-DIM Chapter 16 Heat Conduction in 3D Chapter Goals: •State the general response to the time-dependent inhomogeneous heat equation. •Describe the physical signiﬁcance of the boundary condition. •Derive the temperature exterior to a ﬁxed temper- ature circle. 16.1 General Boundary Value Problem We saw in an earlier chapter that the heat equation is /bracketleftBigg L0+ρcp(x)∂ ∂t/bracketrightBigg T(x,t) =ρq(x,t) forxinR, with the linear operator L0=−∇κT(x)∇. For the time dependent problem need both an initial condition and a boundary condition to determine a unique solution. The initial condi- tion is T(x,t) =T0(x) fort= 0. 245 246 CHAPTER 16. HEAT CONDUCTION IN 3D For our boundary condition we take the radiation condition, κTˆn· ∇T=α[Text(s,t)−T(x,t)] for xons. Recall the the radiation condition came from the equilibrium condition for radiation conduction balance. As an example of this sort of problem, consider the boundary to be the surface of the earth. In the evening time the temperature of the surface is determined by radiation. This is a faster method of transfer than heat conduction. The above radiation condition says that there exists a radiation conduction balance. Note that when we consider convection, we must keep the velocity dependent term x· ∇and the problem becomes non-linear. In this context vis the motion of the medium due to convection. The solution in terms of the Green’s function is given by the prin- ciple of superposition pr:GenSolHeat1 T(x,t) =/integraldisplayt 0dt/prime/integraldisplay Rdx/primeG(x,t;x/prime,t/prime)ρ(x/prime) ˙q(x/prime,t/prime) +/integraldisplayt 0dt/prime/integraldisplay x∈Sds/primeG(x,t;x/prime,t/prime)αText(s/prime,t/prime) +/integraldisplay RG(x,t;x/prime,0)ρ(x/prime)cp(x/prime)T0(x/prime). The integral containing ρ(x/prime) ˙q(x/prime,t/prime) represents contributions due to eq15.0 4 May p4 volume sources; the integral containing αText(s/prime,t/prime) represents contribu- tions due to surface sources; and the integral containing ρ(x/prime)cp(x/prime)T0(x/prime) represents contributions due to the initial conditions. The integrations over time and space can be done in either order, which ever is easiest. The Green’s function is given by G(x,t;x/prime,t/prime) =/integraldisplay Lds 2πies(t−t/prime)G(x,x/prime;λ=−s), (16.1) whereLis the upward directed line along any constant Re s>0. This eq15.1 choice of contour is necessary since L0is positive deﬁnite, which means that all the singularities of G(x,x/prime;λ=−s) lie on the negative real saxis. This integral, which gives the inverse Laplace transform, is sometimes called the Bromwich integral. The Laplace space Green’s pr:Brom1 function satisﬁes the diﬀerential equation [L0−λρcp]G(x,x/prime;λ) =δ(x−x/prime)x,x/prime∈R, 16.2. TIME DEPENDENT PROBLEM 247 and the boundary condition [κTˆn· ∇+α]G(x,x/prime,λ) = 0 x∈R,x/prime∈S. If the dynamical variables cp(x),ρ(x), andκT(x) are spherically sym- metric, then the Green’s function can be written as bilinear product of spherical harmonics, G(x,x/prime;λ) =/summationdisplay Ym l(θ,ϕ)Gl(r,r/prime;λ)Ym∗ l(θ/prime,ϕ/prime). By plugging this into equation 16.1, we obtain G(x,t,x/prime,t/prime) =/summationdisplay l,mYm l(θ,ϕ)Gl(r,t;r/prime,t/prime)Ym∗ l(θ/prime,ϕ/prime) where G(r,t;r/prime,t/prime) =/integraldisplay Lds 2πies(t−t/prime)Gl(r,r/prime;λ=−s). 7 May p1 16.2 Time Dependent Problem We now consider the case in which the temperature is initially zero, and the volume and surface sources undergo harmonic time dependence: T0(x,t) = 0 ρ˙q(x,t) =ρ˙q(x)e−iωt αText(s/prime,t) =αText(s/prime)e−iωt. We want to ﬁnd T(x,t) fort >0. Note that if T0(x)/negationslash= 0 instead, then in the following analysis we would also evaluate the third integral in equation 16.1. For the conditions stated above, the temperature 7 May p2 response is T(x,t) =/integraldisplayt 0dt/prime/integraldisplay Rdx/primeG(x,t;x/prime,t/prime)ρ(x/prime) ˙q(x/prime)e−iωt/prime +/integraldisplayt 0dt/prime/integraldisplay x∈Sds/primeG(x,t;x/prime,t/prime)αText(s/prime)e−iωt/prime. 248 CHAPTER 16. HEAT CONDUCTION IN 3D We are looking for the complete time response of the temperature rather than the steady state response. The time integration is of the form /integraldisplayt 0dtG(x,t;x/prime,t/prime)e−iωt/prime=/integraldisplay Lds 2πiestG(x,x/prime;λ=−s)/integraldisplayt 0e−st/prime−iωt/primedt/prime =/integraldisplay Lds 2πiestG(x,x/prime;λ=−s)1−e−(s+iω)t s+iω =/integraldisplay Lds 2πiG(x,x/prime;λ=−s) s+iω[est−e−iωt].(16.2) The contour of integration, L, is any upward-directed line parallel to eq15.10 the imaginary axis in the left half plane. We got the ﬁrst equality by substituting in equation 16.1 and interchanging the sandtintegrations. The second equality we got by noting /integraldisplayt 0e−st/prime−iωt/primedt=/integraldisplayt 0e−(s+iω)t/primedt/prime =1 s+iω/parenleftBig 1−e−(s+iω)t/parenrightBig . If we allow T0(x)/negationslash= 0, then in evaluating the third integral of equation 16.1 we would also need to calculate the free space Green’s function, as was done in chapter 10. G(x,t;x/prime,0) =/integraldisplayds 2πiestG(x,x/prime;λ=−s) =e−(x−x/prime)2/4κt √ 4πκt. (16.3) This applies to the special case of radiation in the inﬁnite one-dimensional eq15.15 7 May p3 plane. 16.3 Evaluation of the Integrals Recall that the Green’s function can also be written as a bilinear ex- pansion of the eigenfunctions. The general form of solution for equation 16.3 is G(x,t;x/prime,0) =/braceleftBigg/summationtext ne−λntun(x)u∗ n(x/prime) interior/integraltext∞ 0dλ/primee−λ/primet1 πImG(x,x/prime,λ/prime+iε) exterior.(16.4) 16.3. EVALUATION OF THE INTEGRALS 249 In the case when there is explicit time dependence, it may prove useful eq15.16 7 May p4 to integrate over tﬁrst, and then integrate over s. The expressions in equation 16.4 are particularly useful for large times. In this limit only a small range of λnmust be used in the evaluation. In contrast, for short times, an expression like equation 16.3 is more useful. If the functions Text,T0,qandρare spherically symmetric, then we only need the spherically symmetric part of the Green’s function, G0. This was done in the second problem set. In contrast, for the problem presently being considered, the boundary conditions are arbitrary, but the sources are oscillating in time. Ask Baker about rotating. Now we will simplify the integral expression in equation 16.2. 7 May p5 To evaluate equation 16.2, we will use the fact from chapter 10 that G(x,x/prime;s) has the form G(x,x/prime,s)∝e−√s √s. Note that the second term in equation 16.2 is /integraldisplayds 2πiG(x,x/prime;λ=−s) s+iωe−iωt= 0, because the integrand decays in the right-hand plane as e−√s. Thus the fact that we have oscillating sources merely amounts to a change in denominator, /integraldisplayds 2πiG(x,x/prime;λ=−s)e−stosc.−→/integraldisplayds 2πiG(x,x/prime;λ=−s) s+iωe−st.(16.5) eq15osc We thus need to evaluate the ﬁrst term in equation 16.2. We close the contour in the left-hand s-plane, omitting the branch along the negative real axis, as shown in ﬁgure 16.1. By Cauchy’s theorem, the ﬁg15a closed contour gives zero: /integraldisplay Lds 2πiG(x,x/prime;λ=−s) s+iωe−st= 0. The integrand vanishes exponentially along L1,L5. Over the small circle around the origin we have /integraldisplayds 2πiG(x,x/prime;λ=−s) s+iωe−st=e−iωtG(x,x/prime,λ=iω). 250 CHAPTER 16. HEAT CONDUCTION IN 3D s-plane ResIms L2L3 L4- B BB J JJQQQPPP L1 L5 + QQ s6 - -√s=i/radicalBig |s| √s=−i/radicalBig |s| Figure 16.1: Closed contour around branch cut. This looks like a steady state piece. We can use equation 16.4 to write 7 May p6 7 May p7/integraldisplayds 2πiG(x,x/prime,λ=−s) s+iωest=/integraltext=/summationtext∞ n=0e−λmtun(x)u∗ nx/prime) (ω−λn)discrete, /integraltext∞ 0dλ/primee−λ/primet1 πImG(x,x/prime,λ/prime−iε) iω−λ/prime continuum. /integraldisplayds 2πiG(x,x/prime,λ=−s)est s+iω=1 2πi/bracketleftBigg/integraldisplay0 −∞ds/prime s+iωG(x,x/prime;λ=s/prime+iε) +/integraldisplay−∞ 0ds/prime s/prime+iωG(x,x/prime;λ=s/prime−iε)/bracketrightBigg . Change variables 7 May p8 λ/prime=−s/prime, to obtain /integraldisplayds 2πiG(x,x/prime,λ=−s) s+iωest=/integraldisplay∞ 0dλ/prime iω−λ/prime1 2πi[G(x,x/prime,λ+iε)−G(x,x/prime,λ/prime−iε)] =/integraldisplay∞ 0dλ/primee−λ/primet1 πImG(x,x/prime,λ/prime−iε) iω−λ/prime. 9 May p1 16.4. PHYSICS OF THE HEAT PROBLEM 251 16.4 Physics of the Heat Problem We have been looking at how to evaluate the general solution of the heat equation, T(x,t) =/integraldisplayt 0dt/prime/integraldisplay Rdx/primeG(x,t,x/prime,t)ρ(x/prime) ˙q(x/prime,t/prime) +/integraldisplayt 0dt/prime/integraldisplay xinRds/primeG(x,t,x/prime,t/prime)αText(s/prime,t/prime) +/integraldisplay xinRG(x,t,x/prime,0)ρ(x/prime),cp(x/prime)T0(x/prime). We now look at the physics. We require the solution to satisfy the initial condition T(x,t) =T0(x) = 0 for t= 0, and the general regular boundary condition [κT(x)ˆn· ∇+α]T(x,t) =αText(x,t)x∈S. This boundary condition represents the balance between conduction and radiation. 16.4.1 The Parameter Θpr:Theta1 We can rewrite the regular boundary condition as ˆn· ∇T|xonS=α κTh[Text(s,t)−T(x,t)]xonS = Θ [Text(s,t)−T(x,t)]xonS, where Θ = α/κ Th. The expression on the left had side is the conduction in the body, while the expression on the right hand side is the radiation into the body. Thus, this equation is a statement of energy balance. The dynamic characteristic parameter in this equation is Θ, which has the dimensions of inverse distance: Θ =α κTh∼1 distance. We now consider large and small values of Θ. 9 May p2 252 CHAPTER 16. HEAT CONDUCTION IN 3D Case 1: radiation important In this region we have Θ =α κth/greatermuch1. For this case we have radiation at large sand conduction is small, which means T(x,t)≈Text(s,t) forx∈S. Case 2: heat ﬂux occurs This applies to the case Θ =α κth/lessmuch1. Thus we take lim Tint→∞ α→0αTint≡F(s,t), whereF(s,t) is some particular heat ﬂux at position sand timet. The boundary condition then becomes a ﬁxed ﬂux condition, κthˆn· ∇T(x,t)|xons=F(s,t). 9 May p3 HW comments omitted16.5 Example: Sphere The region is the exterior region to a sphere with Text(θ,ϕ;t) =Text(t). So we can write G(x,t,x/prime,t/prime) =1 4πG0(r,t,r/prime,t/prime). So we just need to evaluate equation 16.1. We will get the typical functions of the theory of the heat equation. 9 May p4 We take the temperature Texton the surface of the sphere to be uniform in space and constant in time: Tt(θ,ϕ,t ) =Text. 16.5. EXAMPLE: SPHERE 253 In this case plugging equation 16.3 into 16.1 yields T(r,t) =/integraldisplay dt/integraldisplay dxG(x,t;x/prime,0)αText =/integraldisplay dt/integraldisplay dxe−(x−x/prime)2/4κt √ 4πκtαText =Texta rerfc/parenleftBiggr−a√ 4κt/parenrightBigg , where we deﬁne ask Baker erfx=2√π/integraldisplayx 0dze−z2, and erfcx= 1−erfx=2√π/integraldisplay0 xdze−z2. (16.6) For short times, xis large, and for small times xis small and ( Eq.14) eq15.32 is easy to evaluate. For large xwe use integration by parts, Mysterious equation omit- ted erfcx=2√π/integraldisplay∞ xzdze−z21 z =2√π/bracketleftbigg −1 2e−z21 z−/integraldisplay∞ xdz/parenleftbigg −1 2e−z2/parenrightbigg/parenleftbigg −1 z2/parenrightbigg/bracketrightbigg =2√π/parenleftBigge−x2 2x−1 2/integraldisplay∞ xdz z2e−z2/parenrightBigg =2e−x2 √π/parenleftbigg1 2x−1 4x3+···/parenrightbigg . Thus we have a rapidly converging expansion for large x. Forx/lessmuch1,9 May p5 we can directly place the Taylor series of e−z2inside the integral. 16.5.1 Long Times We have standard diﬀusion phenomena. As t→ ∞ , the solution goes T(r,t)t→∞−→a rText. This is the steady state solution. It satisﬁes the con- ditions ∇2T(x,t) = 0,|x|>a, T(x,t) =Text,|x|=a. 254 CHAPTER 16. HEAT CONDUCTION IN 3D If x2=(r−a)2 4κt/greatermuch1 then we satisfy the steady state condition, and we can deﬁne τby (r−a)2 4κτ= 1 and so τ=(r−a)2 4κt. The variable τis the characteristic time which determines the rate of diﬀusion. So for t/greatermuchτ, the temperature Tis if the form of the steady state solution. 16.5.2 Interior Case Having considered the region exterior to the sphere, we now consider the problem for the interior of the sphere. In particular, we take the surface source Textto have harmonic time dependence and arbitrary spatial independence: Text(t,s) =eiωtText(s). We further assume that there are no volume source and that the internal 9 May p6 temperature is initially zero: ρ˙q(x,t) = 0,T(x,t= 0) = 0. In this case equation 16.2 reduces to T(x,t) =/integraldisplay xonsds/prime/integraldisplayt 0dt/primeG(x,t;x/prime,t/prime)αeiωtText(t,s), or T(x,t) =/integraldisplay xonsds/prime/bracketleftbigg αText(s/prime)/integraldisplay∞ 0dt/primeG(x,t,x/prime,t/prime)e−iωt/prime/bracketrightbigg . This equation was computed previously for an external region. The solution was T(x,t) =/integraldisplay x/primeds/primeαText(s/prime)/bracketleftbigg e−iωtG(x,x/prime;λ=iω) +/summationdisplay me−λntun(x)u∗ n(x/prime) iω−λn/bracketrightbigg . 16.6. SUMMARY 255 This holds for the discrete case, which occurs when the region is the interior of a sphere. For the continuous case, which is valid for the external problem, we have T(x,t) =/integraldisplay x/primeds/primeαText(s/prime)/bracketleftbigg e−iωtG(x,x/prime;λ=iω) +1 π/integraldisplay∞ 0dλ/primee−λ/primetImG(x,x/prime;λ=λ/prime+iω) iω−λ/prime/bracketrightbigg . The ﬁrst term in the bracketed expression is the steady state part of 9 May p7 the response. The second term is the transient part of the response. These transient terms do not always vanish, as is the case in the ﬁxed ﬂux problem, in which there is a zero eigenvalue. Many HW comments omitted 11 May p1 11 May p2 11 May p3 11 May p4 11 May p516.6 Summary 1. The general response to the time-dependent inhomogeneous heat equation is T(x,t) =/integraldisplayt 0dt/prime/integraldisplay Rdx/primeG(x,t;x/prime,t/prime)ρ(x/prime) ˙q(x/prime,t/prime) +/integraldisplayt 0dt/prime/integraldisplay x∈Sds/primeG(x,t;x/prime,t/prime)αText(s/prime,t/prime) +/integraldisplay RG(x,t;x/prime,0)ρ(x/prime)cp(x/prime)T0(x/prime). 2. The boundary condition for the heat equation can be written ˆn· ∇T|xonS= Θ [Text(s,t)−T(x,t)]xonS, where Θ = α/κ Th. If Θ /greatermuch1, then radiation is dominant, other- wise if Θ /lessmuch1, then heat ﬂux is dominant. 3. The temperature exterior to a ﬁxed temperature circle is T(r,t) =Texta rerfc/parenleftBiggr−a√ 4κt/parenrightBigg , where erfcx= 1−erfx=2√π/integraldisplay0 xdze−z2. 256 CHAPTER 16. HEAT CONDUCTION IN 3D 16.7 References See the references of chapter 10. Chapter 17 The Wave Equation Chapter Goals: •State the free space Green’s function in n- dimensions. •Describe the connection between the even– and odd-dimensional Green’s functions. 17.1 introduction The Retarded Green’s function for the wave equation satisﬁes /bracketleftBigg −τ∇2+σ∂2 ∂t2/bracketrightBigg G(x,t;x/prime,t/prime) =δ(x−x/prime)δ(t−t/prime) with the retarded boundary condition that GR= 0 fort < t/prime. The solution to this equation is GR(x,t;x/prime,t/prime) =/integraldisplay Ldω 2πe−iω(t−t/prime)G(x,x/prime;λ=ω2), (17.1) where the integration path Lis any line in the upper half plane parallel eq16ft1 to the real axis and R=|x−x/prime|and whereG(x,x/prime;λ) satisﬁes [−τ∇2−σλ]G(x,x/prime;λ) =δ(x−x/prime). (17.2) We denote the solution of equation 17.2 in n-dimensions as Gn. We eq16B 257 258 CHAPTER 17. THE WAVE EQUATION then have G1(x,x/prime;λ) =iei√ λ/c2R 2τ/radicalBig λ/c2(17.3) G2(x,x/prime;λ) =i 4τH(1) 0(k,R) (17.4) G3(x,x/prime;λ) =ei√ λ/c2R 4πτR, (17.5) wherek=/radicalBig λ/c2. It is readily veriﬁed that these three equations can be written in the more general form Gn(R;λ) =i 4τ/parenleftBiggk 2πR/parenrightBiggn 2−1 H(1) n 2−1(k,R). The Fourier transform, equation 17.1, for the 3-dimensional case can be reduced to the Fourier transform for the one dimensional case, which we have already solved. The trick to do this is to rewrite the integral as a derivative with respect to the constant parameter R, and then pull the diﬀerential outside the integral. GR(x,t,x/prime,t/prime) =/integraldisplay Ldω 2πe−iω(t−t/prime)G3(x,x/prime;λ=ω2) =/integraldisplay Ldω 2πe−iω(t−t/prime)e(iω/c)R 4πτR =/integraldisplay Ldω 2πe−iω(t−t/prime)/parenleftbigg −1 2πR/parenrightbigg∂ ∂Ri 2eiω cR 2τω c =−1 2πR∂ ∂R/integraldisplay Ldω 2πeiω cR 2τω cie−iω(t−t/prime) =1 2πR∂ ∂RG1(x,t;x/prime,t/prime) =1 2πR∂ ∂R/bracketleftbiggc 2τθ(c(t−t/prime)−R)/bracketrightbigg where theθ-function satisﬁes dθ(x/dx=δ(x). Note that f(ax) =1 |a|δ(x). 17.2. DIMENSIONALITY 259 Thus we can write 11 May p6 G3(x,t;x/prime,t/prime) =1 2πR∂ ∂R/bracketleftbiggc 2τθ(c(t−t/prime)−R)/bracketrightbigg =1 4πRτδ(c(t−t/prime)−R/c) =1 4πRτδ(t−t/prime−R/c). Our result is then G3(R,t−t/prime) =1 τπR∂ ∂RG1(R,t−t/prime) =δ(t−t/prime−R/c) 4πRτ. (17.6) eq16.4 13 May p1 17.2 Dimensionality 17.2.1 Odd Dimensionspr:oddDim1 Note thatH(1) n 2−1(k,R) is a trigonometric function for any odd integer. Thus fornodd, we get 13 May p2 Gn(R,t−t/prime) =/parenleftBigg −1 2πR∂ ∂R/parenrightBiggn−1 2 G1(R,t−t/prime) and Gn(R,λ) =/parenleftBigg −1 2πR∂ ∂R/parenrightBiggn−1 2 G1(R,λ). Thus Gn(R,λ) =/parenleftbigg −i 2πR/parenrightbiggn−1 2 G1(R,λ) fornodd. We also have Gn(R,t−t/prime) =/parenleftBigg −1 2πR∂ ∂R/parenrightBiggn−3 2 G3(R,t−t/prime) (17.7) =/parenleftBigg −1 2πR∂ ∂R/parenrightBiggn−3 2δ(t−t/prime−R c) 4πτR. (17.8) 260 CHAPTER 17. THE WAVE EQUATION 17.2.2 Even Dimensionspr:evDim1 Recall that the steady state Green’s function for 2-dimensions is G2(R,λ) =i 4τH(1) 0(k,R). If we insert this into equation 17.1 we obtain the retarded Green’s function, G2(R,t−t/prime) =c 2πτθ(c(t−t/prime)−R)/radicalBig c2(t−t/prime)2−R2. 13 May p3 17.3 Physics There are two ways to deﬁne electrostatics. The ﬁrst is by Guass’s law and the second is by it’s solution, Coulomb’s law. The same relationship is true here. 17.3.1 Odd Dimensions We consider the n= 3-dimensional case, G3(R,t−t/prime) =δ(t−t/prime−R/c) 4πRτ. At timetthe disturbance is zero everywhere except at the radius R= c(t−t/prime) from x/prime. We only see a disturbance on the spherical shell. 17.3.2 Even Dimensions In two dimensions the disturbance is felt at locations other than the surface of the expanding spherical shell. In two dimensions we have 13 May p3 G2=c 2τθ[c(t−t/prime)−R]/radicalBig c2(t−t/prime)2−R2=/braceleftBigg = 0R>c (t−t/prime), /negationslash= 0R<c (t−t/prime).(17.9) The caseG= 0 forR>c (t−t/prime) makes sense since the disturbance has not yet had time to reach the observer. We also have G2=c 2τθ[c(t−t/prime)−R]/radicalBig c2(t−t/prime)2−R2→ ∞ asR→c(t−t/prime). 17.3. PHYSICS 261 G2 R c (t−t/prime) Figure 17.1: Radial part of the 2-dimensional Green’s function. Thus the maximum disturbance occurs at R→c(t−t/prime). Finally,G/negationslash= 0 forR < c (t−t/prime). Thus we have propagation at speed c, as well as all smaller velocities. This is called a wake. The disturbance is shown in ﬁgure 17.1. We have not yet given a motivation for why G2/negationslash= 0 for ﬁg16a R>c (t−t/prime). This will be done in the next section, where we will also give an alternative derivation of this result. 17.3.3 Connection between GF’s in 2 & 3-dim We now calculate the Green’s function in 2-dimensions using the Green’s function in 3-dimensions. This will help us to understand the diﬀerence between even and odd dimensions. Consider the general inhomogeneous wave equation in three dimensions, /bracketleftBigg −τ∇2 3+σ∂2 ∂t2/bracketrightBigg u(x,t) =σf(x,t). (17.10) From our general theory we know that the solution of this equation can eq16.1 be written in terms of the Green’s function as u(x,t) =/integraldisplayt 0dt/prime/integraldisplay dx/primeG3(x,t;x/prime,t/prime)σf(x/prime,t/prime). (17.11) We now consider a particular source, eq16.2 σf(x/prime,t/prime) =δ(x/prime)δ(y/prime)δ(t−t0). This corresponds to a line source along the z-axis acting at time t=t0. What equation does usatisfy for this case? The solution will be 13 May p6 completely independent of z:u(x,t) =u(x,y,t ) =u(x2+y2,t) =u(ρ,t) 262 CHAPTER 17. THE WAVE EQUATION whereρ=x2+y2, and the second equality follows from rotational invariance. For this case equation 17.10 becomes /bracketleftBigg −∇2 2+σ τ∂ ∂t2/bracketrightBigg u(x,y,t ) =1 τδ(x)δ(y)δ(t−t0). Thus u(x,y,t ) =G2(ρ,t−t0), whereG2was given in equation 17.9. We should be able to get the same result by plugging the expression for G3, equation 17.6, in equation 17.11. Thus we have u(x,t) =/integraldisplayt 0dt/prime/integraldisplay dx/primedy/primedz/prime1 4πRδ(t−t/prime−R/c)δ(x/prime)δ(y/prime)δ(t/prime−t0) (17.12) Now letu(x,t) =u(x,y,0,t) =G2(ρ,t−t/prime) on the left hand side of eq16.6 13 May p7 equation 17.12 and partially evaluate the right hand side to get G2(ρ,t−t0) =/integraldisplayt 0dt/prime/integraldisplay∞ −∞dz/prime1 4πτδ(t−t0−R/c)√ρ2+z/prime2. (17.13) The disturbance at time tat the ﬁeld point will be due to contributions eq16.7 atz= 0 fromρ=c(t−t/prime). We also have disturbances at farther distances which were emanated at an earlier time. This is shown in ﬁgure 17.2. ﬁg16b 16 May p1 Note that only the terms at z/primecontribute, where z/prime2+ρ2=c2(t−t0)2. So we deﬁne z/prime=z±=±/radicalBig c2(t−t0)2−ρ2 We now consider the value of G2using equation 17.13 for three diﬀerent regions. •G2= 0 ifρ>c (t−t0) for a signal emitted at z. This is true since a signal emitted at any zwill not have time to arrive at ρsince in travels at velocity c. •Ifρ=c(t−t0), then the signal emitted from the point z= 0 at timet0arrives atρat timet. Thusz±= 0. 16 May p2 •Finally ifρ < c (t−t0), then the signals emitted at time t=t0 from the points z=z±arrive at time t. This is the origin of the wake. 17.4. EVALUATION OF G2 263 ρ xy line source onz-axisAAAAA Kρ z− zz+z= 0z/prime√ρ2+z/prime2ﬁeld pointHHH j Figure 17.2: A line source in 3-dimensions. 17.4 Evaluation of G2 We make a change of variables in equation 17.9, R/prime=/radicalBig ρ2+z/prime2 and thus dR/prime=zdz/prime R/prime, so dz/prime R/prime=dR/prime √R/prime2−ρ2. The Green’s function G−2 is then G2=1 4πτ(2)/integraldisplay∞ ρdR/prime/c√R/prime2−ρ2δ(t−t0−R/prime/c) so the answer is G2=c 2πτθ(c(t−t0)−ρ)/radicalBig c2(t−t0)2−ρ2. We would get the same result if we took the inverse Fourier transform 16 May p3 ofH(1) 0. For heat equation, the character of the Green’s function is independent of dimension; it is always Gaussian. 264 CHAPTER 17. THE WAVE EQUATION 17.5 Summary 1. The free space Green’s function in n-dimensions is Gn(R;λ) =i 4τ/parenleftBiggk 2πR/parenrightBiggn 2−1 H(1) n 2−1(k,R). 2. The connection between the fact that the 3-dimensional Green’s function response propagates on the surface of a sphere and the fact the the 2-dimensional Green’s function response propagates inside of a cylinder is illustrated. 17.6 References See [Fetter80] and [Stakgold67]. This chapter is mostly just an explo- ration of how the number of dimensions aﬀects the solution form. Chapter 18 The Method of Steepest Descent pr:StDesc1 Chapter Goals: •Find the solution to the integral I(ω) =/integraltext Cdzeωf(z)g(z). •Find the asymptotic form of the Gamma function. •Find the asymptotic behavior of the Hankel func- tion. Suppose that we integrate over a contour Csuch as that shown in ﬁgure 18.1: I(ω) =/integraldisplay Cdzeωf(z)g(z). (18.1) We want to ﬁnd an expression for I(ω) for large ω. Without loss of eq17.a pr:Iint1 generality, we take ωto be real and positive. This simply reﬂects the choice of what we call f(z). The ﬁrst step will be to take the indeﬁnite integral. The second step will will then be to deform the contour C into a contour C0such that df dz/vextendsingle/vextendsingle/vextendsingle/vextendsingle z=z0= 0 wherez0lies on the contour C0. In order to perform these operations we will ﬁrst digress to a review 265 266 CHAPTER 18. THE METHOD OF STEEPEST DESCENT z0 CC0 Figure 18.1: Contour C& deformation C0with point z0. of the methods of complex analysis which are needed to compute this integral. Then we shall explicitly solve the integral. ﬁg17.1 18.1 Review of Complex Variables Letz=x+iyandf(x,y) =u(x,y) +iv(x,y) wheref(z) is analytic on the region which we are considering. In general a function fof pr:anal2 the complex variable zisanalytic (orholomorphic ) at a point z0if its derivative exists not only at z0but also at each point zin some neighborhood of z0, and a function fis said to be analytic in a region Rif it is analytic at each point in R. In this case we have: df dz=d dz(u+iv) =du dz+idv dz. Since the function is analytic, its derivative is independent of the path of approach. If we diﬀerentiate with respect to an inﬁnitesimal change dz=dx, we get df dz=du dx+idv dx, (18.2) and if we diﬀerentiate with respect to an inﬁnitesimal change dz=idy, eq17.1 we get df dz=du d(iy)+idv d(iy)=−idu dy+dv dy. (18.3) By comparing equations 18.2 and 18.3 and separating the resulting eq17.2 equation into real and imaginary parts we get the Cauchy-Riemann equations: pr:CReq1 18.1. REVIEW OF COMPLEX VARIABLES 267 Imz Rez∇v ∇u vconstant uconstant Figure 18.2: Gradients of uandv. du dx=dv dyanddv dx=−du dy. These facts allow us to make the following four observations about diﬀerentiation on the complex plane: Observation 1. The gradient of a complex valued function is de- picted in ﬁgure 18.2 for an integral curve of an analytic function. ﬁg17.2 The product of gradients is given by the equation ∇u· ∇v=du dxdv dx+du dydv dy= 0. The last equality follows from the Cauchy-Riemann equations. This means that the lines for which uis constant are perpendicular (i.e., orthogonal) to the lines for which vis constant. Observation 2. For the second derivatives we have the following relations: d2u dx2=d dx/parenleftBiggdu dx/parenrightBigg =d dxdv dy(18.4) and eq17.3 d2u dy2=d dy/parenleftBiggdu dy/parenrightBigg =d dy/parenleftBigg −dv dx/parenrightBigg . (18.5) The diﬀerentials commute, so by combining equations 18.4 and 18.5 eq17.4 we get d2u dx2+d2u dy2= 0, 268 CHAPTER 18. THE METHOD OF STEEPEST DESCENT u= Ref(z) x= Rez y= Imy Figure 18.3: f(z) near a saddle-point. and similarly d2v dx2+d2v dy2= 0. This means that analytic functions satisfy Laplace’s equation. pr:LapEq1 Observation 3. From Observation 2 we ﬁnd that: Ifd2u dx2>0 thend2u dy2<0. (18.6) Thus we cannot have a maximum or a minimum of both uandvoccur eq17.5 anywhere in the complex plane. The point z0=x0+iy0for which du dx|z0= 0 anddu dy|z0= 0 is called a saddle point. The Cauchy–Riemann equations and equation 18.6 imply that if df/dz = 0 atz0, thenz0is a saddle point of both u(x,y) andv(x,y). This is illustrated in ﬁgure 18.3. ﬁg17.3 Observation 4. For an analytic function f=u+ivand a diﬀerential dlwe have df=dl· ∇f =dl· ∇u+idl· ∇v. Note that |df/dz|is independent of the direction of dldue to analyticity. Suppose that we chose dlto be perpendicular to ∇u. In this case dl· ∇v= 0, so df=dl· ∇u fordl/bardbl∇u. 18.2. SPECIFICATION OF STEEPEST DESCENT 269 As the magnitude of fchanges, the change dl· ∇uis purely real, since u(x,y) is real. Thus the real part of fhas maximum change in the direction where dl· ∇v= 0, since |df/dz|is independent of direction. Thereforedl·∇v= 0 gives the path of either steepest descent or steepest ascent. The information given so far is insuﬃcient to determine which. 18.2 Speciﬁcation of Steepest Descent We want to evaluate the integral from equation 18.1, I(ω) =/integraldisplay Cdzeωf(z)g(z), forωlarge. We take ωto be real and positive. In the previous section we wrotef(z) =u(z)+iv(z). Thus we want to know Re ( f(z)) in order to determine the leading order behavior of I(ω) forω/greatermuch1. To solve for I(ω) we deform C→C0such that most of the contribu- tion of the integral when ω/greatermuch1 comes from a small region on C0. Thus we need to make an optimal choice of contour. We want df/dz = 0 at some point z=z0on the deformed contour C0. We parameterize C0 with the line z(τ) =x(τ) +iy(τ). We want the region of the curve where u(τ= Re (f(τ)) to be as lo- calized as possible. Thus we want the contour to run in the direction whereu(τ) has maximal change. As we saw at the end of the previ- ous section, this occurs when v(z(τ)) =v(τ) remains constant. So our deformed contour C0has the property that v(τ) = a constant on C0. (18.7) This will uniquely determine the contour. eq17.6 Note that we assume there is only one point where df/dz = 0. If there were more than one such point, then we would merely repeat this process at the new point and add its contribution. Equation 18.7 is equivalent to the condition Im[f(z(τ))−f(z0)] = 0. 270 CHAPTER 18. THE METHOD OF STEEPEST DESCENT The path for which this condition is satisﬁed is also the one for which Re[f(z)−f(z0)] =u(z)−u(z0) changes most rapidly. We want to evaluate I(ω) forωlarge. Recalling the condition for a local maximum or minimum that df/dz = 0, we note that it is useful to rewrite the integral deﬁned in equation 18.1 as I(ω) =/integraldisplay Cdzeωf(z)g(z) =eωf(z0)/integraldisplay C0dzeω(f(z)−f(z0))g(z). Note that since f(z) is an analytic function, the integral over C0is equal to the integral over C. The main contribution is at the maximum of the diﬀerencef(z)−f(z0). We want to ﬁnd the curve with the maximum change, which has a local maximum at z0, which means we want the quantityf(z)−f(z0) to be negative. Thus we want the curve along which Re[f(z)−f(z0)] =u(z)−u(z0) changes most rapidly and is negative. This is called the curve of steepest descent. This condition speciﬁes which of the two curves speciﬁed by Im[f(z(τ))−f(z0)] = 0 we choose: we choose the path of steepest descent. 18.3 Inverting a Series pr:invSer1 We choose the parameterization f(z)−f(z0)≡ −τ2 so we get z(τ= 0) =z0. Note thatτis real since ∆ v(τ) = 0 along the curve and f(z)<f(u0). We need to invert the integral. Expand f(z)−f(z0) in a power series about z0: f(z)−f(z0) =f/prime/prime(z0) 2!(z−z0)2+f/prime/prime/prime(z0) 3!(z−z0)3+...=−τ2.(18.8) 18.3. INVERTING A SERIES 271 We also get eq17.7 z−z0=∞/summationdisplay n=1anτn=a1τ+a2τ2+a1τ3+.... (18.9) Note that there is no constant term in this series. This is because τ= 0 eq17.8 impliesz−z0= 0. Thus, if we had an n= 0 term, we couldn’t satisfy this stipulation. Plug equation 18.9 into 18.8: −τ2=f/prime/prime(z0) 2!/parenleftBigg∞/summationdisplay n=1anτn/parenrightBigg2 +f/prime/prime/prime(z0) 3!/parenleftBigg∞/summationdisplay n=1anτn/parenrightBigg3 . To calculate a1, forget the terms ( f/prime/prime/prime(z0)/3!)(z−z0)3on. The calcu- lation ofa2includes this term and the calculation of a3includes the following term. Thus −τ2=f/prime/prime(z0) 2!a2 1τ2+O(τ3). (18.10) Now let eq17.9 f/prime/prime(z0) 2!≡Re+iθ. Plugging this into equation 18.10 and canceling τ2yields −1 =a2 1Reiθ so a2 1=eiπ−iθ R where −1 =eiπ. So a1=1√ Rei(−θ 2±π 2). (18.11) The calculation of the a/prime isis the only messy part involved in ﬁnding eq17.10 subsequent terms of the inverted series. For our purposes, it is suﬃcient to have calculated a1. The ±in equation 18.11 gives us two curves for the ﬁrst term: z−z0≈a1τ=τ√ Rei(−θ±π)/2. 272 CHAPTER 18. THE METHOD OF STEEPEST DESCENT We now assume that we have calculated the whole series, and use the series to rewrite I(ω). We now take our integral I(ω) =eωf(z0)/integraldisplay C0dzeω(f(z)−f(z0))g(z), and make a variable substitution dz=dz dτdτ to obtain I(ω) =eωf(z0)/integraldisplayτ+ τ−dτe−ωτ2dz dτg(z(τ)), whereτ+andτ−are on the curve C0on opposite sides of τ= 0. We expand the z(τ) in the function g(z(τ)) as z=a1τ+a2τ2+... and thus dz dτg(z(τ)) =∞/summationdisplay n=0cnτn, (18.12) where thecncan be determined from the anandg(z(τ)). Thus we can eq17.11 write I(ω) =eωf(z0)/summationdisplay n/integraldisplayτ+ τ−dτe−ωτ2cnτn. Thus, with no approximations being made so far, we can assert I(ω) =eωf(z0)/summationdisplay ncn/integraldisplayτ+ τ−dτe−ωτ2τn. Now letτ−→ ∞ andτ+→ ∞ . Our integral becomes I(w) =ewf(z0)∞/summationdisplay n=0cn/integraldisplay∞ −∞dτe−wτ2τn. This is an elementary integral. We know /integraldisplay∞ −∞dτe−wτ2=/radicalbiggπ w, 18.4. EXAMPLE 1: EXPANSION OF Γ–FUNCTION 273 /integraldisplay∞ −∞dττ2e−wτ2=d dω/integraldisplay∞ −∞dτe−wτ2=d dω/radicalbiggπ w=√π 2ω3/2, (this is called diﬀerentiating with respect to a parameter), and similarly /integraldisplay∞ −∞dττ2me−wτ2=/parenleftBiggd dω/parenrightBiggn/integraldisplay∞ −∞dτe−wτ2=√π1·3·5···(2m−1) 2mω(2m+1)/2. Since oddngives zero by symmetry, we have I(w) =ewf(z0)∞/summationdisplay n=0,2,4,...cn/integraldisplay∞ −∞dτe−wτ2τn. All this gives I(w) =ewf(z0)/bracketleftBigg c0/radicalbiggπ w+c2 2√π w3/2+√π∞/summationdisplay m=2c2m1·3·5···(2m−1) 2mw(2m+1)/2/bracketrightBigg . The termc0/radicalBig π/ωcorresponds to Sterling’s formula and the termc2 2√π w3/2 is the ﬁrst correction to Sterling’s formula. The only computation re- maining is the dz/dτ in equation 18.12. 18.4 Example 1: Expansion of Γ–function pr:Gamma1 We want to evaluate the integral I(w) =/integraldisplay∞ 0e−ttwdt. 18.4.1 Transforming the Integral We want to get this equation into the standard form. We make an elementary transformation to get it into the form /integraldisplay dzewf(z)g(z). We substitute t=zwto get I(w) =/integraldisplay∞ 0dze−wz(zw)w(18.13) =ww+1/integraldisplay∞ 0dzew[logz−z]. (18.14) 274 CHAPTER 18. THE METHOD OF STEEPEST DESCENT Forg= 1 we have f(z) = logz−z and we havedf dz=1 z−1 = 0 atz= 1. This is a saddle point. We chose to deﬁne ϕon the interval −π<ϕ<π so that z=reiϕ, logz= logr+iϕ. This is analytic everywhere except the negative real axis, which we don’t need. 18.4.2 The Curve of Steepest Descent Since we know the saddle point, we can write f(z)−f(z0) = logz−z+ 1. So we just need to calculate 0 = Im[f(z)−f(z0)] (18.15) =ϕ−rsinϕ. (18.16) We expect the lines of steepest assent and descent passing through z0 to be perpendicular to each other. The two solutions of this equation correspond to these curves. The solution ϕ= 0 gives a line on the positive real axis. The other solution is r=ϕ sinϕ(18.17) ≈1 +ϕ2 6forϕ/lessmuch1. (18.18) We haven’t yet formally shown which one is the line of steepest ascent and descent. This is determined by looking at the behavior of f(z) on each line. 18.4. EXAMPLE 1: EXPANSION OF Γ–FUNCTION 275 By looking at log z−z+ 1 we see that f(z)−f(z0) can be written f(z)−f(z0) = logz−(z−1) (18.19) =−τ2 =/summationdisplay ncn(z−1)n. We have to invert this in order to get the asymptotic expansion of the gamma function. We expand equation 18.19 in a power series, after noting that log z= log[(z−1) + 1]: −τ2=−1 2(z−1)2+1 3(z−1)3−1 4(z−1)4+.... We plug in z(τ)−1 =Aτ+Bτ2+Cτ3+O(τ4) −τ2=−1 2(Aτ+Bτ2+Cτ3)2+1 3(Aτ+Bτ2)3−1 4(Aτ)4+O(τ5) =−1 2A2τ2−A/parenleftBigg B−A2 3/parenrightBigg τ3−/parenleftBiggB2 2+AC−A2B+A4 4/parenrightBigg τ4+O(τ5). Comparing coeﬃcients on the left and right hand side, we get τ2:A2= 2 τ3:A(B−A2/3) = 0 τ4:B2 2+AC−A2B+A4 4= 0. This method is called inverting the power series. We ﬁnd A=√ 2, C=√ 2/8. The positive roots were chosen for convenience. Now we calculate what thecn’s are from dz dτg(z(τ)) =∞/summationdisplay n=0cnτn. 276 CHAPTER 18. THE METHOD OF STEEPEST DESCENT Sinceg(z) = 1 and dz dτ=A+ 2Bτ+ 3Cτ2+..., we know that C0=√ 2, C2=√ 2 6. Finally, we plug in these values: /integraldisplay∞ 0e−ttwdt=I(w) =ewf(z0)/bracketleftBigg c0/radicalbiggπ w+c2 2√π w3/2+.../bracketrightBigg =e−w /radicalBigg 2π w+√ 2π 12w3/2+... , which agrees with Abramowitz & Stegun, formula 6.1.37. 18.5 Example 2: Asymptotic Hankel Func- tion pr:Hankel1 We want to ﬁnd the asymptotic form of the Hankel function, starting with the integral representation H(1) ν(z) =1 πi/integraldisplay∞+πi −∞ezsinhw−νwdw The contour of integration is the ﬁgure 18.4. The high index and argu- ﬁg17.4 ment behavior of the Hankel function H(1) ν(z) are important in high energy scattering. The index νis related to the eﬀect of an angular momentum barrier, and zto an energy barrier. In this equation νis an arbitrary complex number and zis an arbitrary complex number in a certain strip of the plane. 18.5. EXAMPLE 2: ASYMPTOTIC HANKEL FUNCTION 277 RewImw π Figure 18.4: Deﬁning Contour for the Hankel function. We now relabel the Hankel function as H(1) p(z), where p z= cosω0. Note thatp/zis real and 0 <p/z < 1, which implies 0 <ω 0<π/ 2. So Hp(z) =1 πi/integraldisplay Cdzezf(w)g(w) whereg(w) = 1, with f(w) = sinhw−pw/z = sinhw−wcosω0. To examine asymptotic values |z| /greatermuch1 withω0ﬁxed, we want to deform the contour so that it goes through a saddle point. Using the usual method, we have df(w) dw= coshw−cosω0= 0. We deﬁne w0=iω0 so that coshw0coshiω0= cosω0. Thus f(w=iω) = sinhiω0−iω0cosω0=i[sinω0−ω0cosω0] so that f(w)−f(w0) = sinhw−wcosω0−i[sinω0−ω0cosω0]. 278 CHAPTER 18. THE METHOD OF STEEPEST DESCENT Ascent Descent RewImw π/4 Figure 18.5: Deformed contour for the Hankel function. We want to ﬁnd out what the curves are. Note that d2f(w) dw2/vextendsingle/vextendsingle/vextendsingle/vextendsingle w=iω0= sinhw/vextendsingle/vextendsingle/vextendsingle/vextendsingle w=iω0=isinω0 so we can write −τ2=f(w)−f(iω0) =1 2i(sinω0)(w−iω0)2+.... To invert this series, we write w−iω0=Aτ+Bτ2+Cτ3+.... For now, we are just interested in the leading order term. So −τ2=1 2A2τ2i(sinω0) which implies A2=2i sinω0. Recall that we are looking for the tangent of C0atw0. Thus we have A=±eiπ/4/radicalBigg 2 sinω0, so w−iω0=±/radicalBigg 2 sinω0eiπ/4τ. The deformed curve C0has the form shown in ﬁgure 18.5. The picture ﬁg17.5 neglects to take into account higher order terms. We choose the plus 18.5. EXAMPLE 2: ASYMPTOTIC HANKEL FUNCTION 279 sign to get the direction correct. Note that the curve of steepest ascent is obtained by a rotation of π/2 of the tangent, not the whole curve. In the power series g(z(τ))dz dτ=∞/summationdisplay n=0cnτn whereg(z(τ)) = 1, we have dw dτ=A+ higher order terms, so c0=A=eiπ/4/radicalBigg 2 sinω0. Note that sinω0=/radicalBigg 1−p2 z2=1 z/radicalBig z2−p2. Usually we consider z/greatermuchp, so that 1 z/radicalBig z2−p2= 1. The equation for C0comes from Im [f(w)−f(w0)] = 0 wherew=u+iv. So in the equation Im [f(w)−f(w0)] = coshusinv−vcosω0−(sinω0−ω0cosω0) = 0. Thus, u→+∞ implies cosh u→+∞ sov= 0,π, u→ −∞ implies cosh u→ −∞ sov= 0,π, This gives the line of steepest assent and descent. The orientation of the curves of ascent and descent are shown in ﬁgure 18.6 ﬁg17.6 280 CHAPTER 18. THE METHOD OF STEEPEST DESCENT Descent RewImw Figure 18.6: Hankel function contours. 18.6 Summary 1. The asymptotic solution of the integral I(ω) =/integraldisplay Cdzeωf(z)g(z) is I(w) =ewf(z0)/bracketleftBigg c0/radicalbiggπ w+c2 2√π w3/2+√π∞/summationdisplay m=2c2m1·3·5···(2m−1) 2mw(2m+1)/2/bracketrightBigg . 2. The asymptotic expansion for the Gamma function is /integraldisplay∞ 0e−ttwdt=e−w /radicalBigg 2π w+√ 2π 12w3/2+... . 3. The asymptotic behavior of the Hankel function is discussed in section 17.4. 18.7 References See [Dennery], as well as [Arfken85]. Chapter 19 High Energy Scattering Chapter Goals: •Derive the fundamental integral equation of scat- tering. •Derive the Born approximation. •Derive the integral equation for the transition op- erator.25 May p1 The study of scattering involves the same equation (Schr¨ odinger’s) as before, but subject to speciﬁc boundary conditions. We want solutions for the Schr¨ odinger equation, i¯h∂ ∂tΨ(x,t) =HΨ(x,t), (19.1) where the Hamiltonian is eq18.1 H=−¯h2 2m∇2+V(x) =H0+V. We look for steady state solutions of the form Ψ(x,t) =e−i(E/¯h)tΨE(x), (19.2) using the association E= ¯hω. In particular we want E > 0 solutions, eq18.3 since the solutions for E < 0 are bound states. By substituting equation pr:bound1 281 282 CHAPTER 19. HIGH ENERGY SCATTERING 19.2 into 19.1 we ﬁnd that Ψ Esatisﬁes (E−H)ΨE(x) = 0. (19.3) The boundary condition of scattering requires that the wave function eq18.4 be of the form ΨE(x) =eiki·x+ Ψ S(x) (19.4) where the incident wave number kiis eq18.5 ki=/radicalBigg 2mE ¯h2ˆez. We interpret equation 19.4 as meaning that the total wave function is the sum of an incident plane wave eiki·xwith wavelength λ= 2π/ki, and a wave function due to scattering. This solution is illustrated by the following picture: ψE(x) =- - - eik·x+z ψS(x) 25 May p2 At distances far from the scatterer ( r/greatermuch1), the scattered wave function becomes1 Ψs(x) =eikr rf(ki,kf;E) forr/greatermuch1 (19.5) where the ﬁnal wave number kfis eq18.7 kf=pf ¯h=ˆx/radicalBigg 2mE ¯h2. The unit vector ˆ xsimply indicates some arbitrary direction of interest. Equation 19.5 is the correct equation for the scattered wave function. The angular function f(ki,kf;E) is called the form factor and contains the physical information of the interaction. 1Again, see most any quantum mechanics text. 19.1. FUNDAMENTAL INTEGRAL EQUATION OF SCATTERING 283 For the case of spherical symmetry f(kf,ki;E) =f(kf·ki;E). where kf·ki=kfkicosθ. We now would like to formulate the scattering problem for an arbi- trary interaction. Thus we look at the relation of the above formulation to Green’s functions. The form of equation 19.3 appropriate for Green’s functions is (E−H)G(x,x/prime;E) =δ(x−x/prime). Note the minus sign (used by convention) on the left hand side of this G 5/25/88 equation. We solved equation 19.3 by writing (for the asymptotic limit |x/prime s| → ∞ ) G→ −m 2π¯h2eikr/prime r/primeΨE(x) =−m 2π¯h2eikr/prime r/prime[eik·x+ Ψ s(x)] where Ψ E(x) satisﬁes (E−H)ΨE(x) = 0. The Green’s function holds asymptotically since δ(x−x/prime)→0 as|x/prime| → ∞. This is the solution of the Schr¨ odinger equation which has the This needs ﬁxin’ boundary condition of scattering. pr:bcos1 19.1 Fundamental Integral Equation of Scat- tering G 5/25/88 ¡25 May p3The equation for a general Green’s function is (E−H)G(x,x/prime;E) =δ(x−x/prime). (19.6) SinceH=H0+V, whereH0=−¯h2∇2/2m, the free space Green’s eq18.13 function satisﬁes (E−H0)G0(x,x/prime;E) =δ(x−x/prime). As we have seen, the solution to this equation is See also Jack- son, p.224. 284 CHAPTER 19. HIGH ENERGY SCATTERING G0=−m 2π¯h2eikR R, (19.7) whereR=|x−x/prime|. We now convert equation 19.6 into an integral eq18.15 equation. The general Green’s function equation can be written (E−H0)G=δ(x−x/prime) +V(x)G. (19.8) We can now use Green’s second identity. We deﬁne an operator eq18.16 L0≡E−H0. The operator L0is hermitian, since both EandH0are. Recall that Green’s second identity is 25 May p4 /integraldisplay (S∗L0u) =/integraldisplay (uL0S), where we now choose S∗=G(x,x/prime;E), u=G0(x,x/prime/prime;E), with theL0from above. We now have (using equation 19.8) /integraldisplay dxG(x,x/prime;E)δ(x/prime−x/prime/prime) = /integraldisplay dxG 0(x,x/prime/prime;E)[δ(x−x/prime) +V(x)G(x,x/prime/prime;E)], so G(x/prime/prime,x/prime) =G0(x/prime,x/prime/prime) +/integraldisplay dxG 0(x,x/prime/prime)V(x)G(x,x/prime). We can use the fact that Gis symmetric (see equation 19.7) to write G(x/prime/prime,x/prime;E) =G0(x/prime/prime,x/prime;E) +/integraldisplay dxG 0(x/prime/prime,x;E)V(x)G(x,x/prime;E). So, for x/prime/prime→xandx→x1, we have G(x,x/prime;E) =G0(x,x/prime;E) +/integraldisplay dx1G0(x,x1;E)V(x1)G(x1,x/prime;E). (19.9) 19.2. FORMAL SCATTERING THEORY 285 This is called the fundamental integral equation of scattering. This eq18.23 pr:FundInt1 integral is completely equivalent to equation 19.6. We now describe a way to write equation 19.9 diagrammatically. We establish the following correspondences. G(x,x/prime;E) = x G x/prime G0(x,x/prime;E) = x x/prime /integraldisplay dx1G0(x,x1;E)V(x1)G(x1,x/prime;E) = xz x1 G x/prime Thus a line indicates a free Green’s function. A dot indicates a poten- tial, and aGin a circle represents the Green’s function in the presence of the potential. The point x1represents the position of the last inter- action. Thus equation 19.9 can be written 25 May p5 x G x/prime= x x/prime+ xz x1 G x/prime The arrowheads indicate the line of causality. This helps us to remember the ordering of x/prime,x1andx. 19.2 Formal Scattering Theory Now we want to derive this equation again more formally. We will use the operator formalism, which we now introduce. The free Green’s pr:OpForm1 function equation is /bracketleftBigg E+¯h2 2m∇2/bracketrightBigg G0(x,x/prime;E) =δ(x−x/prime), 286 CHAPTER 19. HIGH ENERGY SCATTERING where the right hand side is just the identity matrix, /angbracketleftx|1|x/prime/angbracketright=δ(x−x/prime), and (1) ij=δij. We also write G0(x,x/prime;E)/bracketleftBigg E+¯h2 2m∇/prime2/bracketrightBigg =δ(x−x/prime), where the operator ∇/prime2operates to the right. From these equations we can write symbolically [E−H0]G0= 1 (19.10) and eq18.25 G0= 1[E−H0]. (19.11) This uses the symmetry of G0operating on xorx/prime. Thus our manipula- eq18.26 tions are essentially based on hermiticity. Because Gis also symmetric, we may also write [E−H]G= 1, and G[E−H] = 1. We now want to rederive equation 19.9. We write 25 May p6 [E−H]G= 1 and [E−H0−V]G= 1. We now multiply on the left by G0to get G0[E−H0−V]G=G0·1 =G0. With the aid of equation 19.11 this becomes G−G0VG=G0, or G=G0+G0VG, (19.12) 19.2. FORMAL SCATTERING THEORY 287 where the term G0VGsymbolizes matrix multiplication, which is thus eq18.31 as integral. This is equivalent to equation 19.9, which is what we wanted to derive. Note that /angbracketleftx|G|x/prime/angbracketright=G(x,x/prime) and /angbracketleftx|V|x/prime/angbracketright=V(x)δ(x−x/prime). 19.2.1 A short digression on operators If an integral of the form C(x1,x2) =/integraldisplay dx/primeA(x1,x/prime)Bx/prime,x2) were written as a discrete sum, we would let x1→i,x2→j, and x/prime→k. We could then express it as Cij=/summationdisplay kAikBkj. But nowA,B, andCare just matrices, so we can express Cas a matrix product C=AB. This can also be viewed as an operator equation. Quantum mechanically, this can be represented as a product of expectation values, either for a discrete spectrum, /angbracketlefti|C|j/angbracketright=/summationdisplay k/angbracketlefti|A|k/angbracketright/angbracketleftk|B|j/angbracketright, or for a continuous spectrum, /angbracketleftx1|C|x2/angbracketright=/integraldisplay dx/prime/angbracketleftx1|A|x/prime/angbracketright/angbracketleftx/prime|B|x2/angbracketright. We now show that the form of the fundamental integral of scattering expressed in equation 19.12 is equivalent to that in equation 19.9. If we reexpress equation 19.12 in terms of expectation values, we have /angbracketleftx|G|x/prime/angbracketright=/angbracketleftx|G0|x/prime/angbracketright+/angbracketleftx|G0VG|x/prime/angbracketright. 288 CHAPTER 19. HIGH ENERGY SCATTERING By comparing with equation 19.10, we see that the last term can be written as /angbracketleftx|G0VG|x/prime/angbracketright=/integraldisplay dx1dx2/angbracketleftx|G0|x1/angbracketright/angbracketleftx1|V|x2/angbracketright/angbracketleftx2|G|x/prime/angbracketright =/integraldisplay dx1dx2/angbracketleftx|G0|x1/angbracketrightV(x1)δ(x1−x2)/angbracketleftx2|G|x/prime/angbracketright =/integraldisplay dx1/angbracketleftx|G0|x1/angbracketrightV(x1)/angbracketleftx1|G|x/prime/angbracketright. So by identifying /angbracketleftx|G|x/prime/angbracketright=G(x,x/prime) /angbracketleftx|G0|x/prime/angbracketright=G0(x,x/prime) we have our ﬁnal result, identical to equation 19.9, G(x,x/prime) =G0(x,x/prime) +/integraldisplay dx1G0(x,x1)V(x1)G(x1,x/prime) which we obtained using equation 19.12. 27 May p1 19.3 Summary of Operator Method We started with H=H0+V and the algebraic formulas (E−H)G= 1, (E0−H)G0= 1. We then found that Gsatisﬁes the integral equation G+G0+G0VG. By noting that G(E−H) = 1, we also got G=G0+GVG 0. 27 May p2 19.3. SUMMARY OF OPERATOR METHOD 289 19.3.1 Derivation of G= (E−H)−1 The trick is to multiply by G0. Thus G(E−H0−V) = 1·G0, and so G−GVG 0=G0. Operators are nothing more than matrices. By inverting the equa- tion, we get G=1 E−H. SoGis the inverse operator of [ E−H].In this context it is useful to deﬁneGin terms of its matrix elements: G(x,x/prime;E)≡ /angbracketleftx|G(E)|x/prime/angbracketright where ImE= 0. This arithmetic summarizes the arithmetic of Greens Second Identity. We also found that the Green’s function solves the 27 May p3 following integral equation G(x,x/prime;E) =G0(x,x/prime;E) +/integraldisplay dxG 0(x,x1;E)V(x1)G(x1,x/prime;E). We were able to express this graphically as well. The other equation gives G(x,x/prime;E) =G0(x,x/prime;E) +/integraldisplay dxG(x,x1;E)V(x1)G0(x1,x/prime;E). 19.3.2 Born Approximation Suppose the V(x) is small. Then in the ﬁrst approximation G∼G0. We originally used this to calculate the scattering amplitude f. We 27 May p4 now use perturbation methods to obtain a power series in V. 290 CHAPTER 19. HIGH ENERGY SCATTERING 19.4 Physical Interest We now want to look at E=E+iε.We place the source point at x/prime→ −r/primeˆz, wherer/prime→ ∞ . We can then write the free space Green’s function as G0(x,x/prime;E) =−m 2π¯h2eikr/prime r/primeeikx where k=/radicalBigg 2mE ¯h2. In this limit the full Green’s function becomes, for the fundamental integral equation of scattering, equation 19.9, lim x→r/primeˆlz r/prime→∞G(x,x/prime;E) =−m 2π¯h2ekr/prime r/prime/bracketleftbigg eikx+/integraldisplay dx1G(x,x1;E)V(x1)eik·x/bracketrightbigg . (19.13) Note that since eq18.13a H=−¯h2 2m∇2 we have [E−H0]eikx= 0. We deﬁne 27 May p5 Ψ(+) k(x)≡eikx+/integraldisplay dx1G(x,x1;E)V(x1)eik·x1. (19.14) Then what we have shown is twsedblst lim r/prime→∞G(x,x/prime;E) =−m 2π¯h2eikr/prime r/primeΨ(+) l(x) where [E−H]Ψ+ k= 0, and Ψ+ ksatisﬁes outgoing wave boundary condi- tion for scattering. We can get Ψ(+) kto any order in perturbation since we have an explicit expression for it and G. Now consider the case in which r/prime→ ∞ with 1.G=G0+G0VG 2.G=G0+GVG 0 19.4. PHYSICAL INTEREST 291 Case 2 implies that G=−m 2π¯h2eikr/prime r/primeΨ(+) n(x) asr/prime→ ∞ . (19.15) By inserting Eq. (19.15) into 1., we get eq:twoseva −m 2π¯h2eikr/prime r/primeΨ(+) N(x) =−m 2π¯h2eikr/prime r/prime/bracketleftbigg eik·x+/integraldisplay G0(x,x1;E)V(x1)Ψ(+) k(x1)/bracketrightbigg . This implies that Ψ(+) k(x) satisﬁes 27 May p6 Ψ(+) k(x) =eik·x+/integraldisplay G0(x,x1;E)V(x1)Ψ(+) k(x1). (19.16) This is the fundamental integral expression for Ψ. Compare with Eq. twosevst (19.14). We now prove that Ψ(+) ksatisﬁes scattering equation with the scat- tering condition. We use the form of Ψ(+)in Eq. (19.14). (E−H)Ψ(+) k= (E−H0−V)eik·x +/integraldisplay dx1(E−H0−V)xG(x,x1;E)V(x1)eik·x1 =−V(x)eik·x+/integraldisplay dx1δ(x−x/prime)V(x1)eik·x1 =−V(x)eik·x+V(x)eik·x = 0. 27 May p7 19.4.1 Satisfying the Scattering Condition We use the form of equation (19.16) to prove that it does satisfy the scattering condition. Let x= ˆerrwherer→ ∞ . We use the result lim x=rˆer r→∞G0(x,x1;E) =−m 2π¯h2eikr re−ikf·x1 292 CHAPTER 19. HIGH ENERGY SCATTERING where kf=kˆer. There is a minus in the exponent since we are taking the limit as xapproaches ˆ er∞rather than −ˆer∞as in equation (19.15). Thus the limiting case is Ψ(+) k(x) =eik·x+eikr r/bracketleftbigg −m 2π¯h2/integraldisplay dx1e−ikf·x1V(x1)Ψ(+) k(x1)/bracketrightbigg . We further deﬁne lim x=rˆer r→∞Ψ(+) k(x)≡eik·x+eikr rf(k,kf;E) wherefis the scattering amplitude to scatter a particle of incident wave kto outgoing kfwith energy E. This is the outgoing wave boundary condition. 27 May p8 19.5 Physical Interpretation We deﬁned the wave function Ψ(+) k(x) using equation 19.13 whose com- ponents have the following interpretation. Ψ(+) k(x) =eik·x+/integraldisplay dx1G0(x,x1;E)V(x1)Ψ(+) k(x1) ≡Ψincident (x) + Ψ scattered (x) where lim |x|=r→∞Ψs(x) =eikr rf(k,kf;E). The physical interpretation of this is shown graphically as follows. Ψ(+) k(x) =- - - Ψincident (x)+z Ψscattered (x) 19.6 Probability Amplitude 27 May p9 The diﬀerential cross section is given bypr:diﬀcrsec1 19.7. REVIEW 293 dσ dΩ(k→kf) =|f(k,kf;E)|2 wherefis the scattering amplitude for scattering with initial momen- tump= ¯hkand ﬁnal amplitude pf= ¯hkffrom a potential V(x). The scattering amplitude fis sometimes written f(k,kf;E) =−m 2π¯h2/angbracketleftkf|V|Ψ(+) k/angbracketright. 19.7 Review 1 Jun p1 We have obtained the integral equation for the Green’s function, G(x,x/prime;E) =G0(x,x/prime;E) +/integraldisplay dxG 0(x,x1;E)V(x1)G(x1,x/prime;E). For the case of a distant source we have seen lim x/prime→−∞ ˆzG(x,x/prime;E) =−m 2π¯h2eikr/prime r/primeΨ(+) E(x), where Ψ(+) E(x) =eik·x+/integraldisplay dx1G0(x,x/prime;E)V(x1)Ψ(+) E(x1). The ﬁrst term is a plane wave. The integral represents a distorted wave. Note that Ψ(+) E(x) automatically satisﬁes the outgoing wave boundary condition. (This is the advantage of the integral equation approach over the diﬀerential equation approach.) To verify this, we took the limitx→ ∞.We also obtained Ψ(+) E(x) =eik·x+/integraldisplay dx1G(x1,x/prime;E)V(x1)eik·x1. We letE→E+i/epsilon1to get a scattering solution, G0(x,x/prime;E+i/epsilon1) =eik·|x−x/prime| |x−x/prime|/parenleftbigg −m 2π¯h2/parenrightbigg , where |k|=/radicalBig 2mE/ ¯h2. We also have shown that Ψ(+) E(x) satisﬁes [E−H]Ψ(+) E(x) = 0. 294 CHAPTER 19. HIGH ENERGY SCATTERING The wave function Ψ(+) E(x) can also be written in the form Ψ(+) E(x)x→∞ ˆn−→eik·x+f(k,k1;E)eikr r, where we have obtained the following unique expression for f, f(k,k1;E) =−m 2π¯h2/integraldisplay dxe−ikf·xV(x)Ψ(+) E(x), where the integral represents a distorted wave. In particular, the term e−ikf·xis a free wave with the ﬁnal momentum, V(x) is the interaction potential, and Ψ(+) E(x) is the distorted wave. So the integral expression forfis the overlap of Ψ and Vwith the outgoing ﬁnal wave. Note that we have made no use of spherical symmetry. All xcontribute, so we still need short distance behavior even for far distance results. The diﬀerential cross section can be written in terms of fas dσ dΩ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle k→kf=|f|2. 19.8 The Born Approximation pr:BornAp1 We now study a particular approximation technique to evaluate Ψ(+) E(x) in Ψ(+) E(x) =eik·x+/integraldisplay dx1G0(x,x/prime;E)V(x1)Ψ(+) E(x1). We assume that the potential is weak so that the distortion, as repre- sented by the integral, is small. The condition that the distortion is small is small distortion ⇐⇒ | Ψ(+) E(x)−eik·x| /lessmuch1. In this case the potential must be suﬃciently small, such that /integraldisplay dx1G0(x,x/prime;E)V(x1)Ψ(+) E(x1)/lessmuch1. We now introduce the short hand of representing this integral by VB, the Born parameter: VB≡/integraldisplay dx1G0(x,x/prime;E)V(x1)Ψ(+) E(x1)/lessmuch1. 19.8. THE BORN APPROXIMATION 295 ForVB/lessmuch1 we may let Ψ(+) E(x1) be replaced by eik·xinf(k,kf;E). In this casefbecomesfBorn(k,kf;E), deﬁned by f(k,kf;E) =−m 2π¯h2/integraldisplay dxe−ix·(k−xf)V(x). In this approximation the cross section becomes dσ dΩBorn−→dσB dΩ=|fB|2. This is called the ﬁrst Born approximation. This approximation is valid in certain high energy physics domains. We now introduce the matrix notation/integraldisplay dxe−ix·kfV(x)Ψ(+) k(x)≡ /angbracketleftxf|V|Ψ(+) k/angbracketright. So in terms of this matrix element the diﬀerential cross section is dσ dΩ=|f(k,kf;E)|2, where the scattering amplitude is given by f(k,kf;E) =−m 2π¯h2/angbracketleftxf|V|Ψ(+) k/angbracketright. We also deﬁne the “wave number” transfer q, q=kf−k= (pf−pi)/¯h. Thus qis the same as (momentum transfer) /¯h. This allows us to write fB=−m 2π¯h2˜V(q), where the fourier transformed potential, ˜V(q), is given by ˜V(q) =/integraldisplay dxe−iq·xV(x). So in the ﬁrst Born approximation, fBdepends only on q. Suppose that q→0. In this case the potential simpliﬁes to ˜V(q)q→0−→/integraldisplay dxV(x). So the ﬁrst Born approximation just gives us the fdependence on the average of the potential. Notice that the ﬁrst Born approximation looses the imaginary part of f(k,kf;E) forfBinR, the real numbers. 296 CHAPTER 19. HIGH ENERGY SCATTERING - > 1 - −kqkf θ Figure 19.1: Geometry of the scattered wave vectors. 19.8.1 Geometry The relationship between k,kf,qandθis shown in ﬁgure 19.1. For ﬁg18a the special case of elastic scattering we have k2 f=k2= 2mE/ ¯h2(elastic scattering) . In this case q2is given by q2= (kf−k)·(kf−k) = 2k2−2k2cosθ = 2k2(1−cosθ) = 4k2sin2(θ/2). Thus we have q= 2πsin(θ/2). We thus know that qwill be small for eitherk→0 (the low energy limit) or sin( θ/2)→0 (forward scatter- ing). 19.8.2 Spherically Symmetric Case In this case the potential V(x) is replaced by V(r). We choose the z-axis along qand use spherical coordinates. The fourier transform of the potential then becomes ˜V(q) =/integraldisplay r2drdφd (cosθ)e−iqrcosθV(r) = 2π/integraldisplay∞ 0r2drV(r)/integraldisplay1 −1d(cosθ)e−iqrcosθ =4π q/integraldisplay∞ 0rdrV (r) sinqr. This is a 1-dimensional fourier sine transform. 19.8. THE BORN APPROXIMATION 297 19.8.3 Coulomb Case We now choose a speciﬁc V(r) so that we can do the integral. We choose the shielded Coulomb potential, V(r) =V0 re−r/a. In the problem set set we use αinstead ofV0. The parameter αchar- acterizes the charge. The fourier sine transform of this potential is ˜(q) =4πα q/integraldisplay∞ 0drsinrqe−r/a, and the diﬀerential cross section is then dσ dΩ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle Born=|fB|2 =/vextendsingle/vextendsingle/vextendsingle/vextendsingle−m 2π¯h2˜V(q)/vextendsingle/vextendsingle/vextendsingle/vextendsingle2 = 4a2/parenleftbiggαma ¯h2/parenrightbigg2/parenleftBigg1 q2a2+ 1/parenrightBigg2 . This is the shielded Coulomb scattering diﬀerential cros section in the ﬁrst Born approximation, where q= 2ksin(θ/2). Notice that as α→ ∞ this reduces to Rutherford scattering, which is a lucky accident. We now look at characteristics of the diﬀerential cross section we have obtained. Most of the cross section contribution comes from qa= 2ksin(θ/2)/lessmuch1. Now ifka/greatermuch1, then we must require θ/lessmuch1, which means that we can use the small angle approximation. In this case out dominant cross section condition becomes qa≈2ka(θ/2)/lessmuch1, orθ/lessmuch1/ka. This gives a quantitative estimation of how strongly for- ward peaked the scattered wave is. The condition ka/greatermuch1 corresponds to the small λ, or high energy, limit. In this case the wavelength is much smaller that the particle, which means that most of the scatter- ing will be in the forward direction. We can see how good the ﬁrst Born approximation is by evaluating the Born parameter in this limit. We ﬁnd VB=V0ma2 k21 ka/lessmuch1. 298 CHAPTER 19. HIGH ENERGY SCATTERING In this equation V0is the strength of the potential and ais the range of the potential. Notice that ka/greatermuch1 can make VB/lessmuch1 even ifV0is large. Thus we have a dimensionless measure of the strength of the potential. 19.9 Scattering Approximation We now want to look at the perturbation expansion for the diﬀerential cross section, dσ dΩ/vextendsingle/vextendsingle/vextendsingle/vextendsingle/vextendsingle k→kf=|f(k,kf)|2, where the scattering amplitude f(k,kf) is f(k,k1;E) =−m 2π¯h2/integraldisplay dxe−ikf·xV(x)Ψ(+) E(x), (19.17) where the incident wave function is eq18b1 Ψ(+) E(x) =eik·x+/integraldisplay dx1G(x1,x/prime;E)V(x1)eik·x1. (19.18) Ψ(+) E(x) satisﬁes the outgoing wave condition. By combining equation eq18b2 19.17 into 19.18 be obtain f(k,k1;E) = −m 2π¯h2/braceleftbigg/integraldisplay dxe−kf·xV(x)e−k·x /integraldisplay dxdx/primee−kf·xV(x)G(x,x/prime;E)V(x/prime)ek·x/bracerightbigg . The ﬁrst integral represents a single interaction, while the second inte- gral represents two or more interactions. By introducing the transition operator, we can simplify the expression for the scattering amplitude, f(k,k1;E) =−m 2π¯h2/integraldisplay dxdx/primee−kf·xG(x,x/prime;E)ek·x. We now deﬁne the transition operator T. In function notation it is pr:transOp1 T(x,x/prime;E)≡V(x)δ(x−x/prime) +V(x)G(x,x/prime;E)V(x/prime). In operator notation, we can rewrite this equation as T=V+VGV. 19.10. PERTURBATION EXPANSION 299 Thus in matrix notation, our old equation for f, f(k,kf;E) =−m 2π¯h2/angbracketleftxf|V|Ψ(+) k/angbracketright, is replaced by f(k,k/prime) =−m 2π¯h2/angbracketleftxf|T|k/angbracketright, Thus we now have two equivalent forms for expressing f. In the ﬁrst Born approximation we approximate T=V.Tplays the role in the exact theory what Vplays in the ﬁrst Born approximation. 19.10 Perturbation Expansion pr:pertExp1 We now look at how the transition operator Tcan be used in dia- gramatic perturbation theory. We make the following correspondences between terms in the formulas and the graphical counterparts (these are the “Feynman rules”): pr:FeynRul1 An incoming line:HH jHHHxk representseik·x. An outgoing line: *kfxrepresentse−ikf·x. A vertex point: uxrepresentsV(x). A free propagator: -x1x2representsG0(x2,x1). A circledG: -x/prime mGx-representsG(x,x/prime). Thus we can write the transition operator matrix element as /angbracketleftxf|T|k/angbracketright=HH jHHHxk u *kf+HH jHHHk ux/prime -mG-ux *kf The ﬁrst diagram represents the ﬁrst Born approximation, which cor- responds to a single scatterer. The second diagram represents two or 300 CHAPTER 19. HIGH ENERGY SCATTERING more scatterer, where the propagation occurs via any number of inter- actions through G. The integral equation for the full Green’s function, G(x,x/prime;E) =G0(x,x/prime;E) +/integraldisplay dx1G(x,x1;E)V(x1)G(x1,x/prime;E), has the following symbolic representation: -x/prime mGx-=-x/primex+-xux1-mGx- where x/primeis the source point, xis the ﬁeld point, and x1is one of the interaction points. 19.10.1 Perturbation Expansion In matrix language the integral equation for the full Green’s function is G=G0+GVG 0, which implies G=G0(1−VG 0)−1. Thus the following geometric series gives the solution to the integral equation, G=G0(1 +VG 0+ (VG 0)(VG 0) + (VG 0)(VG 0)(VG 0) +···). In symbolic notation, this expansion corresponds to -x/prime mGx-=-x/primex+-x/primex1u-x+-x/primex1u-x2u-x +-x/primex1u-x2u-x3u-x+···. We could also write the series expansion in integral notation. In this case the third order in Vterm, (VG 0)(VG 0)(VG 0), is (writing right to left) /integraldisplay dx1dx2dx3G0(x,x3)V(x3)G0(x3,x2)V(x2)G0(x2,x1)V(x1)G0(x1,x/prime), 19.10. PERTURBATION EXPANSION 301 where, for example, G0(x3,x2) =−m 2π¯h2eik·|x2−x3| |x2−x3|. Think of these terms as multiply scattered terms. Now we can use this series to get a perturbation expansion for the scattering amplitude f, that is, for the matrix element /angbracketleftxf|T|k/angbracketright. In symbolic language it is is this correct? /angbracketleftxf|T|k/angbracketright=HH jHHHxk u *kf+HH jHHHk ux/prime -mG-ux *kf =HH jHHHxk u *kf+HH jHHHx/primek u-ux *kf+HH jHHHx/primek u-ux1-ux *kf +HH jHHHx/primek u-ux1-ux2-ux *kf+···. To convert this to integral language we note that, for example, the fourth Born approximation term is HH jHHHx/primek u-ux1-ux2-ux *kf In integral notation this is expressed as /integraldisplay dxdx/primedx1dx2/bracketleftBig e−ikf·xV(x)G0(x,x2)V(x2)G0(x2,x1)V(x1)G0(x1,x/prime)ek·x/prime/bracketrightBig . We must integrate over all space since each of the interaction points may occur at any place. 19.10.2 Use of the T-Matrix An alternative approach is to eliminate all direct reference to Gwith- out perturbation theory. We then obtain an integral equation for the transition matrix. By using G=G0(1−VG 0)−1, we have VG=VG 0(1−VG 0)−1, 302 CHAPTER 19. HIGH ENERGY SCATTERING so we can write the transition matrix as T=V+VG 0(1−VG 0)−1V = [1 +VG 0(1−VG 0)−1]V = [(1 −VG 0) +VG 0](1−VG 0)−1V = (1 −VG 0)−1V. This provides us with a new solution for T: T= (1−VG 0)−1V. We can write this as an integral equation, which would have the oper- ator form (1−VG 0)T=V, or T=V+VG 0T. This gives us another Lippman/Schwinger equation. Notice that T= (1−VG 0)−1Vmay be expanded in a power series in Vjust as was the previous expression for G. 19.11 Summary 1. The fundamental integral equation of scattering is G(x,x/prime;E) =G0(x,x/prime;E) +/integraldisplay dx1G0(x,x1;E)V(x1)G(x1,x/prime;E). 19.12 References See [Neyfeh, p360ﬀ] for perturbation theory. Appendix A Symbols Used /angbracketleftS,u/angbracketrightthe brackets denote an inner product, 13. ∗as a superscript, represents complex conjugation, 13. ∇nabla, the diﬀerential operator in an arbitrary number of dimen- sions, 7. A1,A2constants used in determining the Green’s function, 28. athe horizontal displacement between mass points on a string; an ar- bitrary position on the string, 2, the left endpoint of a string, 6. a1,a2constants used in discussion of superposition, 23. B1,B2constants used in determining the Green’s function, 28. bthe right endpoint of a string 6. b(x) width of a water channel, 108. Ca constant used in determining the Green’s function, 29. cleft endpoint used in the discussion of the δ-function, 24; constant characterizing velocity, 39, 45. Da constant used in determining the Green’s function, 29. 303 304 APPENDIX A. SYMBOLS USED dthe diﬀerential operator; right endpoint used in the discussion of the δ-function, 24. ∆pchange in momentum, 87. ∆uithe transverse distance between adjacent points ( ui−ui−1) on a discrete string, 4. ∆xthe longitudinal distance between adjacent points on a discrete string, 4. δ(x−x/prime) the delta function, 24, 129, 161. δmnthe Kroneker delta function, 36. Eenergy, 74, 143. e= 2.71···. /epsilon1a small distance along the string, 27. F(x,t) the external force on a continuous string, 1. Fcdthe force over the interval [ c,d], used in the discussion of the δ- function, 24. Felastic i the elastic force on the ith mass point of a discrete string, 3. Fext ithe external force on the ith mass point of a discrete string, 3. Fτi iythe transverse force at the ith mass point on a string due to tension, 3. Ftotthe total force on the ith mass point of the string. f(x) is the external force density divided by the mass density at posi- tionx, 4. f(x/prime) a ﬁnite term used in discussion of asymptotic Green’s function, 42. f1,f2force terms used in discussion of superposition, 23. 305 f(θ,k) the scattering amplitude for a ﬁeld observer from an incident plane wave, 214. ˜f(θ,r/prime,k) scattering amplitude, 214. G(x,x k;ω2) the Green’s function for the Helmholtz equation, 26. GAthe advances Green’s function, 87. GSthe scattered part of the steady state Green’s function, 184. GRthe retarded Green’s function, 86. Gm(r,r/prime;λ) reduced Green’s function, 132. ˜Gthe Fourier transform of the Green’s function, 88, the Laplace trans- form of the Green’s function, 147. gn(x,x/prime) asymptotic coeﬃcient for Green’s function near an eigen value, 41. γangular diﬀerence between xandx/primeused in scattering discussion, 185. Hthe Hamiltonian, 195 H(1) m(x),H(2) m(x) the ﬁrst and second Hankel functions, 79. h(x) equilibrium height of a surface wave, 108. hl(x) the spherical Hankel function, 178. ha(t) the eﬀective force exerted by the string: Fa/τa, 6. hS(t) same asha(t), generalized for both endpoints, 8. ¯hthe reduced Plank’s constant, 74. I(ω) a general integral used in discussion of method of steepest descent, 265. Imanother Bessel function, 80. 306 APPENDIX A. SYMBOLS USED ithe index of mass points on a string, 2. ˆiunit vector in the x-direction, 128. JJacobian function, 128. j(r) the quantum mechanical current density, 227. jincthe incident ﬂux, 227. jl(x) the spherical Bessel function, 178. jnheat current, 144. ˆjunit vector in the y-direction, 128. Kmanother Bessel function, 80. kthe wave number, 38. k2a short hand for V/τused in inﬁnite string problem, 63. kithe spring constant at the ith mass point, 3. κthe thermal diﬀusity, 151. κathe eﬀective spring constant exerted by the string at endpoint a: ka/τa, 6. L0linear operator, 5. Lθϕcentrifugal linear operator, 162. Lthe angular momentum vector, 207. ldimension of length, 3. ˆlthe direction along the string in the positive xdirection, 7. λan arbitrary complex number representing the square of the fre- quency continued into the complex plane, 27; wavelength of sur- face waves, 108. 307 λnnth eigen value for the normal mode problem, 37. λ(m) nthenth eigenvalue of the reduced operator L(µm) 0, 133. mdimension of mass, 3. mithe mass of the particle at point ion the discrete string, 2. µmeigenvalues for circular eigenfunctions, 131. Nthe number of mass particles on the discrete string, 2; the number of particles intercepted in a scattering experiment, 227. nl(x) the spherical Neumann function, 178. ˆnthe outward normal, 7. Ω solid angle, 161. ωangular frequency, 9. ωnthe natural frequency of the nth normal mode, 32. pmomentum, 74, 207. Φ solution of the Klein Gordon equation, 75; total response due to a plane wave scattering on an obstacle, 185. Φ0incident plane wave used in scattering discussion, 185. φangular coordinate, 128, 160. φn(xi,t) the normal modes, 38. ψquantum mechanical wave function, 195 R(r) function used to obtain Bessel’s equation, 178. Re take real value of whatever term imediately follows. rradial coordinate, 128. 308 APPENDIX A. SYMBOLS USED Sthe “surface” (i.e., endpoints) of a one dimensional string, 7; an arbitrary function used in the derivation of the Green’s identities, 13. S(x) cross sectional area of a surface wave, 108. σthe cross section, 227. σ(x) the mass density of the string at position x4. T(x,x/prime;E) transition operator, 298. ttime, dimension 3, variable, 3. τithe tension on the segment between the ( i−1)th andith mass points on a string, 2. Θ parameter in RBC for the heat equation, 251. θthe angle of the string between mass points on a discrete string, 3; angle in parameterization of complex plane, 63. u(x,t) transverse displacement of string, 5; displacement of a surface wave from equilibrium height, 108. u0(x) an arbitrary function used in the derivation of the Green’s iden- tities, 13. u0(x) value of the transverse amplitude at t= 0, 8. u0(x,ω) steady state in free space due to a point source, 184. u1(x) value of the derivative of the transverse amplitude at t= 0, 8. u1,u2functions used in discussion of superposition, 23. u1solution of the homogeneous ﬁxed string problem, 46. uithe vertical displacement of the ith mass particle on a string, 2. u(m) n(r) thenth eigenfunction of L(µm) 0, 133. um l(x) the normalized θ-part of the spherical harmonic, 164. 309 uscatthe scattered part of the steady state response, 198. u1modiﬁed solution of the homogeneous ﬁxed string problem, 47. V(x) the coeﬃcient of elasticity of the string at position x, 1. Veﬀthe eﬀective potential, 206. W(u1,u2) the Wronskian, 30. Xlcoeﬃcient of the scattered part of the wave relative to the incident part, 187, 219. xcontinuous position variable, 4. x<the lower of the position point and source point, 30. x>the higher of the position point and source point, 30. x/primethe location of the δ-function disturbance, 24. xidiscrete position variable, 4. xkthe location of the δ-function disturbance, 26. Ym l(θ,ϕ) the spherical harmonics, 164. z(x) hight of a surface wave, 108. 310 APPENDIX A. SYMBOLS USED Bibliography [1] Arfken, George B. Mathematical Methods for Physicists. Academic Press, 1985. [2] Barton, Gabriel, Elements of Green’s Functions and Propagation: potentials, diﬀusion, and waves. Oxford, 1989, 1991. [3] *Boas, Mary, Mathematical Methods in the Physical Sciences . [4] Carslaw, Horatio Scott & Jaeger, J. C. Conduction of heat in solids. Oxford, 1959, 1986. [5] Dennery, Phillippi and Andr´ e Krzywicki, Mathematics for Physi- cists. [6] Fetter, Alexander L. & Walecka, John Dirk. Theoretical Mechanics of Particles and Continua , Chapters 9–13. McGraw-Hill, 1980. [7] *Griﬃths, David, Indroduction to Electrodynamics , Prentice-Hall, 1981. [8] *Halliday, David and Robert Resnick, Physics , John Wiley, 1978. [9] *Jackson, David, Classical Electrodynamics , John Wiley, 1975. [10] Morse, Philip M. & Feshbach, Herman. Methods of Theoretical Physics. McGraw-Hill, 1953. [11] Neyfeh, Perturbation Methods . [12] Stakgold, Ivar. Boundary Value Problems of Mathematical Physics. Macmillan, 1967. 311 312 BIBLIOGRAPHY [13] Stakgold, Ivar. Green’s Functions and Boundary Value Problems. John Wiley, 1979. [14] *Symon, Keith R., Mechanics , Addison-Wesley, 1971. Index addition formula 212 advanced Green’s function 87 all-spce problem 117, 174 analytic 46, 266 angular momentum 207 associated Legendre polynomial 168 asymptotic limit 49 Babenet’s principle 194 Bessel’s equation 79 Born approximation 294 bound states 281 boundary conditions 5, 111, 116, 145, 173; of scattereing 283 boundary value problem 1 branch cut 45, 60 Bromwich integral 246 Cartesian coordinates 128 Cauchy’s theorem 54 Cauchy-Riemann equations 266 causality 87 characteristic range 83 classical mechanics (vs. quantum mechanics) 202, 207 closed string 6, discrete 37 coeﬃcient of elasticity 4 completeness relation 51, 57, 76, 131, 169Condon-Shortley phase conven- tion 170 conservation of energy 144, 213 continuity condition 28 Coulomb potential 208 cross section 227 cutoﬀ frequency 38, 66 De Broglie relation 203 degeneracy 39 delta function 24, 129 diﬀerential cross section 292 diﬀerential equation 3 diﬀraction 191 Dirichlet boundary conditions 8 discrete spectrum 49 dispersion relation 38 divergence 129, 161 eﬀective force 7 eﬀective spring constant 7 eigen function 32 eigenfunction expansion 131 eigen value problem 28, 68, 121, 133, 134, 140 eigen vector 32 elastic boundary conditions 6, 116 limiting cases, 7 elastic force 3 elastic media 8 elastic membrane 109 313 314 INDEX energy 74, 143 energy levels 196 even dimensions, Green’s func- tion in, 260 equations of motion 2 expansion theorem 57, 172 experimental scattering 208, 226 exterior problem 117, 122, 174 external force 3 far-ﬁeld limit 208, 235 Feynman rules 299 forced oscillation problem 31, 73, 118 forced vibration 3 Fourier coeﬃcient 58 Fourier integral 78, see also ex- pansion theorem Fourier Inversion Theorem see in- verse Fourier transform Fourier-Bessel transform 83 Fourier transform 88 Fredholm equation 40 free oscillation problem 32 free space problem 151, 188 free vibration 3 fundamental integral equation of scattering 285 Gamma function 273 Gaussian 153, 155 gradient 129, 161 general response problem 103, 117, 119 general solution, heat equation, 246 Generalized Fourier Integral 59 geometrical limit of scattering 230 Green’s ﬁrst identity 14, 119Green’s function for the Helmholtz equation, 26 Green’s reciprocity principle 30, Green’s second identity 15, 119 Hamiltonian 195 Hankel function 79; asymptotic form, 276 hard sphere, scattering from a, 231 heat conduction 143 heat current 144 heat equation 146 Helmholtz equation 9, 26 Hermitian analyticity 43 Hermitian operator 17, 119 holomorphic see analytic homogeneous equation 28, 45 Huygen’s principle 194 impulsive force 86 inﬁnite string 62 initial conditions 8 initial value problem 92, 119 inner product 13 inverting a series 270 interior problem 116, 122, 174 inverse Fourier transform 91 Jacobian 128 Kirchhoﬀ’s formula 191 Klein Gordon equation 74 Lagrangian 110 Laplace transform 147 Laplace’s equation 268 Legendre’s equation 166 Legendre polynomial 168 Leibnitz formula 167 linear operator 5, 24 linearly independent 31 INDEX 315 mass density 4 membrane problem 138 method of images 122, 191 momentum operator 207 natural frequency 26, 32, 37, 42 natural modes 32, 37 Neumann boundary conditions 8 Newton’s Second Law 3 normal modes 32, 37, 117, 134 normalization 44, 58, 135, 169 odd dimensions, Green’s function in, 259 open string 6 operator formalism 285 optical theorem 231 orthogonal 36 orthonormality 36, 58, 164, 169, 171 oscillating point source see forced oscillation problem outward normal 7 partial expansion 131 partial diﬀerential equation 5 periodic boundary conditions 6, 111, 116 perturbation expansion 299 plane wave 199, 213, 239 polar coordinates 128 poles 44 positive deﬁnite operator 20 potential energy 20 potential theory 186 principle of superposition 24, 131 quantum mechanical scattering 197 quantum mechanics 195radiation 81 Rayleigh quotient 40 recurrence relation 167 reduced linear operator 132 regular boundary conditions 8 residues 44 retarded Green’s function 86, 120, 136 Rodrigues formula 168 scattered Green’s function 210 scattering Amplitude 211 scattering from a sphere 223 scattering wave 209 Schr¨ odinger equation 39, 195 self-adjoint operator 52, 119 singular boundary conditions 8 shallow water condition 108 singularity 54 sound waves, radiation of, 232 speciﬁc heat 143 spectral theory 42 spherical coordinates 160 spherical harmonics 170 steady state scattering 183 steady state solution 9, 135, 196, 234 steepest descent, method of, 265 string 1 superposition see principle of surface waves 108 temperature 143 tension 2 transition operator 298 transverse vibrations 2 travelling wave 38 wave propagation 66 wedge problem 136 316 INDEX Wronskian 30