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List of Figures 1.1 A string with mass points attached to springs. . . . . . . 2 1.2 A closed string, where a and b are connected. . . . . . . 6 1.3 An open string, where the endpoints a and b are 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 The θ-convention . . . . . . . . . . . . . . . . . . . . . . 46 4.4 The contour of integration . . . . . . . . . . . . . . . . . 54 4.5 Circle around a singularity. . . . . . . . . . . . . . . . . . 55 4.6 Division of contour. . . . . . . . . . . . . . . . . . . . . . 56 4.7 λ near the branch cut. . . . . . . . . . . . . . . . . . . . 61 4.8 θ speciﬁcation. . . . . . . . . . . . . . . . . . . . . . . . 63 4.9 Geometry in λ-plane . . . . . . . . . . . . . . . . . . . . 69 6.1 The contour L in the λ-plane. . . . . . . . . . . . . . . . 92 6.2 Contour LC1 = L + LU HP 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 1 6.5 GR at t1 = t + 2 x /c and at t2 = t + 3 x /c. . . 2 . . . . . 98 7.1 Water waves moving in channels. . . . . . . . . . . . . . 108 7.2 The rectangular membrane. . . . . . . . . . . . . . . . . 111 9.1 The region R as a circle with radius a. . . . . . . . . . . 130 9.2 The wedge. . . . . . . . . . . . . . . . . . . . . . . . . . 137 10.1 Rotation of contour in complex plane. . . . . . . . . . . . 148 xi
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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 γ and θ.. . . . . . . . . . . . . . . . . . . . 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 γ and θ. . . . . . . . . . . . . . . 212 15.3 Phase shift due to potential. . . . . . . . . . . . . . . . . 221 15.4 A repulsive potential. . . . . . . . . . . . . . . . . . . . . 223 15.5 The potential V and Veﬀ for 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 C0 with point z0 . . . . . . . . 266 18.2 Gradients of u and v. . . . . . . . . . . . . . . . . . . . . 267 18.3 f (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
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PrefaceThis manuscript is based on lectures given by Marshall Baker for a classon Mathematical Methods in Physics at the University of Washingtonin 1988. The subject of the lectures was Green’s function techniques inPhysics. All the members of the class had completed the equivalent ofthe ﬁrst three and a half years of the undergraduate physics program,although some had signiﬁcantly more background. The class was apreparation for graduate study in physics. These notes develop Green’s function techiques for both single andmultiple 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 handyfor the mathematics details and Fetter and Walecka’s book handy forthe physics details. There are other good books on Green’s functionsavailable, but none of them are geared for same background as assumedhere. The two volume set by Stakgold is particularly useful. For astrictly 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
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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 asto 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 ofthose who wish to be notiﬁed when revisions become available. If youwould like to be on this list, please send email to sutlief@u.washington.edubefore 1996. Otherwise, call Marshall Baker at 206-543-2898.Acknowledgements: This manuscript beneﬁts greatly from the excellent set of notestaken by Steve Griﬃes. Richard Horn contributed many correctionsand suggestions. Special thanks go to the students of Physics 425-426at the University of Washington during 1988 and 1993. This ﬁrst revision contains corrections only. No additional materialhas been added since Version 0. Steve Sutlief Seattle, Washington 16 June, 1993 4 January, 1994
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Chapter 1The Vibrating String 4 Jan p1 Chapter Goals: p1prv.yr. • 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:bvp1example of the application of Newton’s second law to small vibrations:transverse vibrations on a string. Physical problems such as this andthose involving sound, surface waves, heat conduction, electromagneticwaves, and gravitational waves, for example, can be solved using themathematical theory of boundary value problems. Consider the problem of a string embedded in a medium with a pr:string1restoring force V (x) and an external force F (x, t). This problem covers pr:V1most of the physical interpretations of small vibrations. In this chapter pr:F1we will investigate the mathematics of this problem by determining theequations of motion. 1
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2 CHAPTER 1. THE VIBRATING STRING mi+1 τi+1 ui+1 u Fiy 322 2 mi 33 $ $ $ $ 33θ Fiyi4 u $ $ τ ui 3 $ $ 4 $ $ $ $ $ $ $ $ mi−1 44 $ $ $ $ $ $ $ $ u 4 $ $ $ $ ui−1 4 $ $ $ $ $ $ $ $ $ $ 5 $ $ $ $ $ $ 5 $ 5 ki−1 $ $ $ ki $ $ $ $ ki+1 $ $ $ $ $ $ $ $ $ $ a a xi−1 xi xi+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 setpr:N1 of functions. Consider N mass points of mass mi attached to a masslesspr:mi1 string, which has a tension τ between mass points. An elastic force atpr:tau1 each mass point is represented by a spring. This problem is illustratedﬁg1.1 in ﬁgure 1.1 We want to ﬁnd the equations of motion for transverse vibrations of the string.pr:eom1 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 p2pr:tension1 For each mass point there are two force contributions due to the tension on the string. We call τi the tension on the segment between mi−1pr:ui1 and mi , ui the vertical displacement of the ith mass point, and a thepr:a1 horizontal displacement between mass points. Since we are consideringpr:transvib1 transverse vibrations (in the u-direction) , we want to know the tension
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1.1. THE STRING 3force in the u-direction, which is τi+1 sin θ. From the ﬁgure we see that pr:theta1θ ≈ (ui+1 − ui )/a for small angles and we can thus write τ (ui+1 − ui ) Fiyi+1 = τi+1 aand pr:Fiyt1 τ (ui − ui−1 ) Fiyi = −τi . aNote that the equations agree with dimensional analysis: Grif’s uses τ Taylor exp Fiyi = dim(m · l/t2 ), τi = dim(m · l/t2 ), pr:m1 ui = dim(l), and a = dim(l). pr:l1 pr:t1Elastic Force pr:elastic1We add an elastic force with spring constant ki : pr:ki1 Fielastic = −ki ui ,where dim(ki ) = (m/t2 ). This situation can be visualized by imagining pr:Fel1vertical springs attached to each mass point, as depicted in ﬁgure 1.1.A small value of ki corresponds to an elastic spring, while a large valueof ki corresponds to a rigid spring.External ForceWe add the external force Fiext . This force depends on the nature of pr:ExtForce1the physical problem under consideration. For example, it may be a pr:Fext1transverse force at the end points.1.1.2 Equations of Motion for a Massless StringThe problem thus far has concerned a massless string with mass pointsattached. By summing the above forces and applying Newton’s secondlaw, we have pr:Newton1 (ui+1 − ui ) (ui − ui−1 ) d2 pr:t2 Ftot = τi+1 − τi − ki ui + Fiext = mi 2 ui . (1.1) a a dtThis gives us N coupled inhomogeneous linear ordinary diﬀerential eq1forceequations where each ui is a function of time. In the case that Fiext pr:diﬀeq1is zero we have free vibration, otherwise we have forced vibration. pr:FreeVib1 pr:ForcedVib1
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4 CHAPTER 1. THE VIBRATING STRING 1.1.3 Equations of Motion for a Massive String4 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 aspr:deltax1 a = ∆x. We correspondingly write ui − ui−1 = ∆u. This allows us topr:deltau1 write (ui − ui−1 ) ∆u = . (1.2) a ∆x i The equations of motion become (after dividing both sides by ∆x) 1 ∆u ∆u ki F ext mi d2 ui τi+1 − τi − ui + i = . (1.3) ∆x ∆x i+1 ∆x i ∆x ∆x ∆x dt2eq1deltf In the limit we take a → 0, N → ∞, and deﬁne their product to be lim N a ≡ L. (1.4) a→0 N →∞ The limiting case allows us to redeﬁne the terms of the equations ofpr:sigmax1 motion as follows: mi mass mi → 0 ∆x → σ(xi ) ≡ length = mass density; ki ki → 0 ∆x → V (xi ) = coeﬃcient of elasticity of the media; F ext mi F ext Fiext → 0 ∆x = ( ∆x · mi ) → σ(xi )f (xi ) (1.5) where F ext external force f (xi ) = = . (1.6) mi mass Since xi = x xi−1 = x − ∆x xi+1 = x + ∆xpr:x1 we have ∆u ui − ui−1 ∂u(x, t) = → (1.7) ∆x i xi − xi−1 ∂x
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1.2. THE LINEAR OPERATOR FORM 5so that 1 ∆u ∆u τi+1 − τi ∆x ∆x i+1 ∆x i 1 ∂u(x + ∆x) ∂u(x) = τ (x + ∆x) − τ (x) ∆x ∂x ∂x ∂ ∂u = τ (x) . (1.8) ∂x ∂xThis allows us to write 1.3 as 4 Jan p4 ∂ ∂u ∂2u τ (x) − V (x)u + σ(x)f (x, t) = σ(x) 2 . (1.9) ∂x ∂x ∂tThis is a partial diﬀerential equation. We will look at this problem in eq1diﬀdetail in the following chapters. Note that the ﬁrst term is net tension pr:pde1force over dx.1.2 The Linear Operator FormWe deﬁne the linear operator L0 by the equation pr:LinOp1 ∂ ∂ L0 ≡ − τ (x) + V (x). (1.10) ∂x ∂xWe can now write equation (1.9) as eq1LinOp ∂2 L0 + σ(x) u(x, t) = σ(x)f (x, t) on a < x < b. (1.11) ∂t2This is an inhomogeneous equation with an external force term. Note eq1waveonethat each term in this equation has units of m/t2 . Integrating thisequation over the length of the string gives the total force on the string.1.3 Boundary Conditions pr:bc1To obtain a unique solution for the diﬀerential equation, we must placerestrictive conditions on it. In this case we place conditions on the endsof the string. Either the string is tied together (i.e. closed), or its endsare left apart (open).
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6 CHAPTER 1. THE VIBRATING STRING $ ar br & % Figure 1.2: A closed string, where a and b are connected. 1.3.1 Case 1: A Closed Stringpr:ClStr1 A closed string has its endpoints a and b connected. This case is illus-pr:a2 trated in ﬁgure 2. This is the periodic boundary condition for a closedﬁg1loop string. A closed string must satisfy the following equations:pr:pbc1 u(a, t) = u(b, t) (1.12)eq1pbc1 which is the condition that the ends meet, and ∂u(x, t) ∂u(x, t) = (1.13) ∂x x=a ∂x x=beq1pbc2 which is the condition that the ends have the same declination (i.e., the string must be smooth across the end points). 1.3.2 Case 2: An Open Stringsec1-c24 Jan p5 For an elastically bound open string we have the boundary conditionpr:ebc1 that the total force must vanish at the end points. Thus, by multiplying equation 1.3 by ∆x and setting the right hand side equal to zero, wepr:OpStr1 have the equation ∂u(x, t) τa − ka u(a, t) + Fa (t) = 0. ∂x x=a The homogeneous terms of this equation are τa ∂u |x=a and ka u(a, t), and ∂x the inhomogeneous term is Fa (t). The term ka u(a) describes how thepr:ha1 string is bound. We now deﬁne Fa ka ha (t) ≡ and κa ≡ . τa τa
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1.3. BOUNDARY CONDITIONS 7 b a E n r ˆ r n ˆ Figure 1.3: An open string, where the endpoints a and b are free.pr:EﬀFrc1 The term ha (t) is the eﬀective force and κa is the eﬀective spring con-pr:esc1 stant. ∂u − + κa u(x) = ha (t) for x = a. (1.14) ∂x We also deﬁne the outward normal, n, as shown in ﬁgure 1.3. This eq1bound ˆ allows us to write 1.14 as pr:OutNorm1 ﬁg1.2 n· ˆ u(x) + κa u(x) = ha (t) for x = a. The boundary condition at b can be similarly deﬁned: ∂u + κb u(x) = hb (t) for x = b, ∂x where Fb kb hb (t) ≡ and κb ≡ . τb τb For a more compact notation, consider points a and b to be elements of the “surface” of the one dimensional string, S = {a, b}. This gives pr:S1 us nS u(x) + κS u(x) = hS (t) ˆ for x on S, for all t. (1.15) In this case na = −lx and nb = lx . ˆ ˆ 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 κa and κb . The pr:ebc2 terms κa and κb signify how rigidly the string’s endpoints are bound. The two limiting cases of equation 1.14 are as follows: pr:ga1
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8 CHAPTER 1. THE VIBRATING STRING ∂u κa → 0 − = ha (t) (1.16) ∂x x=a κa → ∞ u(x, t)|x=a = ha /κa = Fa /ka . (1.17) The boundary condition κa → 0 corresponds to an elastic media, and pr:ElMed1pr:nbc1 is called the Neumann boundary condition. The case κa → ∞ corre- sponds to a rigid medium, and is called the Dirichlet boundary condi-pr:dbc1 tion.pr:hS1 If hS (t) = 0 in equation 1.15, so that [ˆ S · n + κS ]u(x, t) = hS (t) = 0 for x on S, (1.18)eq1RBC then the boundary conditions are called regular boundary conditions.pr:rbc1 Regular boundary conditions are either d dsee Stakgold 1. u(a, t) = u(b, t), dx u(a, t) = dx u(b, t) (periodic), orp269 2. [ˆ S · n + κS ]u(x, t) = 0 for x on S. Thus regular boundary conditions correspond to the case in which there is no external force on the end points. 1.3.4 Initial Conditionspr:ic16 Jan p2 The complete description of the problem also requires information aboutpr:u0.1 the string at some reference point in time: u(x, t)|t=0 = u0 (x) for a < x < b (1.19) and ∂ u(x, t)|t=0 = u1 (x) for a < x < b. (1.20) ∂t Here we claim that it is suﬃcient to know the position and velocity of the string at some point in time. 1.4 Special CasesThis materialwas originally We now consider two singular boundary conditions and a boundaryin chapter 3 condition leading to the Helmholtz equation. The conditions ﬁrst two8 Jan p3.3 cases will ensure that the right-hand side of Green’s second identitypr:sbc1 (introduced in chapter 2) vanishes. This is necessary for a physical system.
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1.4. SPECIAL CASES 91.4.1 No Tension at BoundaryFor the case in which τ (a) = 0 and the regular boundary conditionshold, the condition that u(a) be ﬁnite is necessary. This is enough toensure that the right hand side of Green’s second identity is zero.1.4.2 Semi-inﬁnite StringIn the case that a → −∞, we require that u(x) have a ﬁnite limit asx → −∞. Similarly, if b → ∞, we require that u(x) have a ﬁnite limitas x → ∞. If both a → −∞ and b → ∞, we require that u(x) haveﬁnite limits as either x → −∞ or x → ∞.1.4.3 Oscillatory External Force sec1helmIn the case in which there are no forces at the boundary we have ha = hb = 0. (1.21)The terms ha , hb are extra forces on the boundaries. Thus the conditionof no forces on the boundary does not imply that the internal forcesare zero. We now treat the case where the interior force is oscillatoryand write pr:omega1 −iωt f (x, t) = f (x)e . (1.22)In this case the physical solution will be Re f (x, t) = f (x) cos ωt. (1.23)We look for steady state solutions of the form pr:sss1 u(x, t) = e−iωt u(x) for all t. (1.24)This gives us the equation ∂ 2 −iωt L0 + σ(x) 2 e u(x) = σ(x)f (x)e−iωt . (1.25) ∂tIf u(x, ω) satisﬁes the equation [L0 − ω 2 σ(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 eq1helmequation in chapter 3. pr:Helm1
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10 CHAPTER 1. THE VIBRATING STRING1.5 SummaryIn this chapter the equations of motion have been derived for the smalloscillation problem. Appropriate forms of the boundary conditions andinitial conditions have been given. The general string problem with external forces is mathematicallythe same as the small oscillation (vibration) problem, which uses vectorsand matrices. Let ui = u(xi ) be the amplitude of the string at the pointxi . For the discrete case we have N component vectors ui = u(xi ), andfor the continuum case we have a continuous function u(x). Theseconsiderations outline the most general problem. The main results for this chapter are: 1. The equation of motion for a string is ∂2 L0 + σ(x) u(x, t) = σ(x)f (x, t) on a < x < b ∂t2 where ∂ ∂ L0 u = − τ (x) + V (x) u. ∂x ∂x 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) ∂u(b, t) = (no bends) ∂x x=a ∂x x=b or (b) an open string: [ˆ S · n + κS ]u(x, t) = hS (t) = 0 x on S, all t. 3. The initial conditions are given by the equations u(x, t)|t=0 = u0 (x) for a < x < b (1.27)
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1.6. REFERENCES 11 and ∂ u(x, t) = u1 (x) for a < x < b. (1.28) ∂t t=0 4. The Helmholtz equation is [L0 − ω 2 σ(x)]u(x) = σ(x)f (x).1.6 ReferencesSee any book which derives the wave equation, such as [Fetter80, p120ﬀ],[Griﬃths81, p297], [Halliday78, pA5]. A more thorough deﬁnition of regular boundary conditions may befound in [Stakgold67a, p268ﬀ].
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Chapter 2Green’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 solvethe 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 takeon complex values. We also need the notion of an inner product. The noteinner product of S and u is deﬁned as pr:InProd1 pr:S2 n Si∗ ui i=1 for the discrete case S, u = b ∗ (2.1) a dxS (x)u(x) for the continuous case.In the uses of the inner product which will be encountered here, for the eq2.2continuum case, one of the variables S or u will be a length (amplitudeof the string), and the other will be a force per unit length. Thus theinner product will have units of force times length, which is work. 13
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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 L0 u for u, where d d L0 u(x) ≡ − τ (x) + V (x) u(x). (2.2) dx dxeq2.3 This substitution gives us b d d S, L0 u = dxS ∗ (x) − τ (x) + V (x) u(x) a dx dx b d d = − dxS ∗ (x) − τ (x) u a dx dx b + dxS ∗ (x)V (x)u(x). a We now integrate twice by parts ( ud¯ = uv − v d¯), letting ¯ v ¯¯ ¯ u dS ∗ (x) u = S ∗ (x) =⇒ d¯ = dS ∗ (x) = dx ¯ u dx and d d d d d¯ = dx v τ (x) u = d τ (x) u =⇒ v = τ (x) u ¯ dx dx dx dx so that b d S, L0 u = − S ∗ (x)τ (x) u(x) a dx b dS ∗ d b + dx τ (x) u(x) + dxS ∗ (x)V (x)u(x) a dx dx a b d = − S ∗ (x)τ (x) u(x) a dx b d ∗ d + dx S τ (x) u(x) + S ∗ (x)V (x)u(x) . a dx dx Note that the ﬁnal integrand is symmetric in terms of S ∗ (x) and u(x).pr:G1Id1 This is Green’s First Identity:
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2.2. USING G.I. #2 TO SATISFY R.B.C. 15 b d S, L0 u = − S ∗ (x)τ (x) u(x) (2.3) a dx b d ∗ d + dx S τ (x) u(x) + S ∗ (x)V (x)u(x) . a dx dxNow interchange S ∗ and u to get eq2G1Id ∗ u, L0 S = L0 S, u b d ∗ = − u(x)τ (x) S (x) (2.4) a dx b d d + dx u τ (x) S ∗ (x) + u(x)V (x)S ∗ (x) . a dx dxWhen the diﬀerence of equations 2.3 and 2.4 is taken, the symmetric eq2preG2Idterms cancel. This is Green’s Second Identity: pr:G2Id1 b d ∗ d S, L0 u − L0 S, u = τ (x) u(x) S (x) − S ∗ (x) u(x) . (2.5) a dx dxIn the literature, the expressions for the Green’s identities take τ = −1 eq2G2Idand V = 0 in the operator L0 . Furthermore, the expressions here arefor one dimension, while the multidimensional generalization is givenin section 8.4.1.2.2 Using G.I. #2 to Satisfy R.B.C. 6 Jan p2.5The regular boundary conditions for a string (either equations 1.12 and1.13 or equation 1.18) can simplify Green’s 2nd Identity. If S and ucorrespond to physical quantities, they must satisfy RBC. We willverify this statement for two special cases: the closed string and theopen string.2.2.1 The Closed StringFor a closed string we have (from equations 1.12 and 1.13) u(a, t) = u(b, t), S ∗ (a, t) = S ∗ (b, t),
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16 CHAPTER 2. GREEN’S IDENTITIES d ∗ d ∗ d d τ (a) = τ (b), S = S , u = u . dx x=a dx x=b dx x=a dx x=b By plugging these equalities into Green’s second identity, we ﬁnd that S, L0 u = L0 S, u . (2.6)eq2twox 2.2.2 The Open String For an open string we have ∂u − + Ka u = 0 for x = a, ∂x ∂S ∗ − + Ka S ∗ = 0 for x = a, ∂x ∂u + Kb u = 0 for x = b, ∂x ∂S ∗ + Kb S ∗ = 0 for x = b. (2.7) ∂xeq21osbc These are the conditions for RBC from equation 1.14. Plugging these expressions into Green’s second identity gives dS ∗ du τ (x) u − S∗ = τ (a)[uKa S ∗ − S ∗ Ka u] = 0 a dx dx and b dS ∗ du τ (x) u − S∗ = τ (b)[uKb S ∗ − S ∗ Kb u] = 0. dx dx Thus from equation 2.5 we ﬁnd that S, L0 u = L0 S, u , (2.8)eq2twox2 just as in equation 2.6 for a closed string.
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2.3. ANOTHER BOUNDARY CONDITION 17 2.2.3 A Note on Hermitian Operators The equation S, L0 u = L0 S, u , which we have found to hold for both a closed string and an open string, is the criterion for L0 to be apr:HermOp1 Hermitian operator. By using the deﬁnition 2.1, this expression can be rewritten as S, L0 u = u, L0 S ∗ . (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 τ (x) = 0 x→a,b we have ∂ ∗ lim τ (x)u(x) S (x) = 0 x→a,b ∂x and ∂ lim τ (x)S ∗ (x) u(x) = 0. x→a,b ∂x In the preceding equations, the abbreviated notation lim is introduced x→a,b to represent either the limit as x approaches the endpoint a or the limit as x approaches the endpoint b. These equations allow us to rewrite Green’s second identity (equation 2.5) as S, L0 u = L0 S, u (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.
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18 CHAPTER 2. GREEN’S IDENTITIES 2.4 Physical Interpretations of the G.I.ssec2.4 Certain qualities of the Green’s Identities correspond to physical situ- ations and constraints. 2.4.1 The Physics of Green’s 2nd Identity6 Jan p2.6 The right hand side of Green’s 2nd Identity will always vanish for phys- ically realizable systems. Thus L0 is 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 dx of the string has two contribu- tions. One is the “spring” potential energy 1 V (x)(u(x))2 (c.f., 1 kx2 in 2 2 U = − F dx = − (−kx)dx = 1 kx2 [Halliday76, p141]). The other is 2 the “tension” potential energy, which comes from the tension force in ∂ ∂ section 1.1.3, dF = ∂x [τ (x) ∂x u(x)]dx, and thus Utension is ∂ ∂u U =− τ dx, ∂x ∂x so dU d ∂ ∂u ∂ ∂u ∂u =− τ dx = − τ dx, dt dt ∂x ∂x x ∂x ∂x ∂t and so the change in potential energy in a time interval dt is b ∂ ∂u ∂u U dt = − τ dt dx a ∂x ∂x ∂t b b parts ∂u ∂ ∂u ∂u ∂u = τ dt dx − τ dt a ∂x ∂t ∂x ∂x ∂t a
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2.4. PHYSICAL INTERPRETATIONS OF THE G.I.S 19 b ∂u ∂ ∂u = τ dt dx a ∂x ∂t ∂x t+dt b 2 ∂ 1 ∂u = τ dx . ∂t a 2 ∂x t The second term in the second equality vanishes. We may now sumthe diﬀerentials of U in time to obtain the potential energy: t t b 2 b 2 1 ∂u 1 ∂u U= U dt = τ dx = τ dx. t =0 a 2 ∂x a 2 ∂x 02.4.3 The Physics of Green’s 1st Identity sec2.4.2Let S = u. Then 2.3 becomes 6 Jan p2 b d u, L0 u = − u∗ (x)τ (x) u(x) (2.11) a dx b d ∗ d + dx u τ (x) u(x) + u∗ (x)V (x)u(x) . a dx dxFor a closed string we have b 2 du u, L0 u = dx τ (x) + V (x)(u(x))2 = 2U (2.12) a dxsince each quantity is the same at a and b. For an open string we found eq2x(equation 1.15) 8 Jan p3.2 du = Ka u (2.13) dx x=aand du = −Kb u (2.14) dx x=bso that u, L0 u = τ (a)Ka |u(a)|2 + τ (b)Kb |u(b)|2 b 2 du + dx τ (x) + V (x)(u(x))2 a dx = 2U,
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20 CHAPTER 2. GREEN’S IDENTITIES twice the potential energy. The term 1 τ (a)Ka |u(a)|2 + 1 τ (b)Kb |u(b)|2 2 2 see FW p207 is the potential energy due to two discrete “springs” at the end points, expl. p10 and is simply the spring constant times the displacement squared. p126 The term τ (x) (du/dx)2 is the tension potential energy. Since du/dx eq2y represents the string stretching in the transverse direction, τ (x) (du/dx)2 pr:pe1 is a potential due to the stretching of the string. V (x)(u(x))2 is 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. If Ka , Kb , τ and V are positive for the open string, the potential energy U is also positive. Thus u, L0 u > 0, which implies that L0 is a positivepr:pdo1 deﬁnite operator. 2.5 Summary 1. The Green’s identities are: (a) Green’s ﬁrst identity: b d S, L0 u = − S ∗ (x)τ (x) u(x) a dx b d ∗ d + dx S τ (x) u(x) a dx dx + S ∗ (x)V (x)u(x) , (b) Green’s second identity: b d ∗ d S, L0 u − L0 S, u = τ (x) u(x) S (x) − S ∗ (x) u(x) . a dx dx 2. For a closed string and an open string (i.e., RBC) the linear op- erator L0 is Hermitian: S, L0 u = u, L0 S ∗ .
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2.6. REFERENCES 212.6 ReferencesGreen’s formula is described in [Stakgold67, p70] and [Stakgold79,p167]. The derivation of the potential energy of a string was inspired by[Simon71,p390].
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Chapter 3Green’s Functions Chapter Goals: • Show that an external force can be written as a sum of δ-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 wasobtained in section 1.4.3. First we will develop some mathematicalprinciples which will facilitate the derivation. 8 Jan p3.4 Lagrangian stuﬀ com-3.1 The Principle of Superposition mented outSuppose that pr:a1.1 f (x) = a1 f1 (x) + a2 f2 (x). (3.1)If u1 and u2 are solutions to the equations (c.f., 1.26) [L0 − ω 2 σ(x)]u1 (x) = σ(x)f1 (x) (3.2) [L0 − ω 2 σ(x)]u2 (x) = σ(x)f2 (x) (3.3)with RBC and such that (see equation 1.15) eq3q (ˆ S · n + κS )u1 = 0 for x on S (ˆ S · n + κS )u2 = 0 23
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24 CHAPTER 3. GREEN’S FUNCTIONS then their weighted sum satisﬁes the same equation of motion [L0 − ω 2 σ(x)](a1 u1 (x) + a2 u2 (x)) = a1 [L0 − ω 2 σ(x)]u1 (x) +a2 [L0 − ω 2 σ(x)]u2 (x) σ(x)f1 (x) σ(x)f2 (x) = σ(x)f (x). and boundary condition [ˆ S · n + κS ][a1 u1 (x) + a2 u2 (x)] = a1 [ˆ S · + κS ]u1 + a2 [ˆ S · n n + κS ]u2 = a1 (0) + a2 (0) = 0. We have thus shown that L0 [a1 u1 + a2 u2 ] = a1 L0 u1 + a2 L0 u2 . (3.4)pr:pos1 This is called the principle of superposition, and it is the deﬁning prop- erty of a linear operator. 3.2 The Dirac Delta Function11 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 − ω 2 σ(x)]u(x) = σ(x)f (x).pr:DeltaFn1 The delta function is deﬁned by the equationpr:Fcd d 1 if c < xk < d Fcd = dxδ(x − xk ) = (3.5) c 0 otherwise.eq3deltdef where Fcd represents the total force over the interval [c, d]. Thus we seepr: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)
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3.2. THE DIRAC DELTA FUNCTION 25 Integration gives us d σ(x)f (x)dx = Fcd , (3.7) ceq3fdc which is the total force applied over the domain. This allows us to write [L0 − σω 2 ]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 xk +∆x Fk = dxσ(x)f (x) (3.9) xk = the force on the interval ∆x. (3.10) eq3Fsubk We now prove that N σ(x)f (x) = Fk δ(x − xk ) (3.11) k=1 where xk < xk < xk + ∆x. We ﬁrst integrate both sides to get eq3sfd d d N x replaces x dxσ(x)f (x) = dx Fk δ(x − xk ). (3.12) so δ-fn isn’t on c c k=1 boundary By deﬁnition (equation 3.7), the left-hand side is the total force applied over the domain, Fcd . The right-hand side is d N N d Fk δ(x − xk )dx = dxFk δ(x − xk ) (3.13) c k=1 k=1 c = Fk (3.14) c<xk <d xk +∆x = dxσ(x)f (x) (3.15) c<xk <d xk d N →∞ −→ dxσ(x)f (x) (3.16) c = Fcd . (3.17)
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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.pr:Helm2 The Helmholtz equation 3.2 can now be rewritten (using 3.11) as N 2 [L0 − σ(x)ω ]u(x, ω) = Fk δ(x − xk ). (3.18) k=1 By the principle of superposition we can write N u(x) = Fk uk (x) (3.19) k=1 where uk (x) is the solution of [L0 − σ(x)ω 2 ]uk (x, ω 2 ) = 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 responses11 Jan p3 to localized forces. We now introduce the following notation uk (x) ≡ G(x, xk ; ω 2 ) (3.20)pr:Gxxo1 where G is the Green’s function, xk signiﬁes the location of the distur- bance, and ω corresponds to frequency. This allows us to write N u(x) = Fk uk (x) k=1 N xk +∆x = dx σ(x )f (x )G(x, xk ; ω 2 ) k=1 xk b N →∞ −→ dx G(x, x ; ω 2 )σ(x )f (x ). a We have deﬁned the Green’s function by [L0 − σ(x)ω 2 ]G(x, x ; ω 2 ) = δ(x − x ) a < x, x < b, RBC. (3.21)11 Jan p4 The solution will explode for ω 2 when ω is a natural frequency of thepr:NatFreq1 system, as will be seen later.where will natfreq be deﬁned
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3.2. THE DIRAC DELTA FUNCTION 27 G(x, x ; ω 2 ) d d dx G|x=x −ε dx G|x=x +ε e ¡ e 7e ¡ e D7 e e l ¡¡ D7 l e 7 e ¨ 7 e r ¨ 7 e r 2 $ 2 $ 7 e 7 e a x −ε x x +ε b Figure 3.1: The pointed string Let λ = ω 2 be an arbitrary complex number. Since the squared frequency ω 2 cannot be complex, we relabel it λ. So now we want topr:lambda1 solveﬁx this d d − τ (x) + V (x) − σ(x)λ G(x, x ; ω 2 ) = δ(x − x ) (3.22) dx dx a < x, x < b, RBC Note that G will have singularities when λ is 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 x+ d d dx − τ (x) + V (x) − σ(x)λ G(x, x ; λ) x− dx dx x+ = δ(x − x )dx x− which becomes pr:epsilon1 x+ d −τ (x) G(x, x ; λ) =1 (3.23) dx x− since the integrals over V (x) and σ(x) vanish as ε → 0. Note that in this last expression “1” has units of force.
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28 CHAPTER 3. GREEN’S FUNCTIONS 3.3 Two Conditions11 Jan p5 3.3.1 Condition 1 The previous equation can be written as d d 1 G − G =− . (3.24) dx x=x + dx x=x − τeq3other This makes sense after considering that a larger tension implies a smallereq3b kink (discontinuity of ﬁrst derivative) in the string. 3.3.2 Condition 2 We also require that the string doesn’t break: G(x, x )|x=x + = G(x, x )|x=x − . (3.25)eq3d This is called the continuity condition.pr:ContCond1 3.3.3 Application To ﬁnd the Green’s function for equation 3.22 away from the point x ,pr:homog1 we study the homogeneous equation [L0 − σ(x)λ]u(x, λ) = 0 x = x , RBC. (3.26)pr:efp1 This is called the eigen function problem. Once we specify G(x0 , x ; λ) d and dx G(x0 , x ; λ), we may use this equation to get all higher derivatives and thus determine G(x, x ; λ). We know, from diﬀerential equation theory, that two fundamental solutions must exist. Let u1 and u2 be the solutions to [L0 − σ(x)λ]u1,2 (x, λ) = 0 (3.27)pr:AABB where u1,2 denotes either solution. Thus G(x, x ; λ) = A1 u1 (x, λ) + A2 u2 (x, λ) for x < x , (3.28)eq3ab1 and
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3.4. OPEN STRING 29 G(x, x ; λ) = B1 u1 (x, λ) + B2 u2 (x, λ) for x > x . (3.29)eq3ab2 We have now deﬁned the Green’s function in terms of four constants.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 We will solve for an open string with no external force h(x), which was where is 13 Jan ﬁrst discussed in section 1.3.2. G(x, x ; λ) must satisfy the boundary p1 condition 1.18. Choose u1 such that it satisﬁes the boundary condition for the left end ∂u1 − + Ka u1 (a) = 0. (3.30) ∂x x=a This determines u1 up to an arbitrary constant. Choose u2 such that it satisﬁes the right end boundary condition ∂u2 + Kb u2 (b) = 0. (3.31) ∂x x=b 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 ; λ) = A1 (x )u1 (x, λ) for x < x . (3.32) Note that only the boundary condition at a applies since the behavior of u1 (x) does not matter at b (since b > x ). This gives G determined up to an arbitrary constant. We can also write G(x, x ; λ) = B2 (x )u2 (x, λ) for x > x . (3.33) We also note that A and B are constants determined by x only. Thus we can write the previous expressions in a more symmetric form: pr:CD G(x, x ; λ) = Cu1 (x, λ)u2 (x , λ) for x < x , (3.34) G(x, x ; λ) = Du1 (x , λ)u2 (x, λ) for x > x . (3.35)
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30 CHAPTER 3. GREEN’S FUNCTIONS In one of the problem sets we prove that G(x, x ; λ) = G(x , x; λ). This can also be stated as Green’s Reciprocity Principle: ‘The ampli- pr:grp1 tude of the string at x subject to a localized force applied at x is equivalent to the amplitude of the string at x subject to a localized force applied at x.’ We now apply the continuity condition. Equation 3.25 implies that13 Jan p3 C = D. Now we have a function symmetric in x and x , which veriﬁes the Green’s Reciprocity Principle. By imposing the condition in equation 3.24 we will be able to determine C: dG du1 =C u2 (x ) (3.36) dx x=x + dx xeq3trione dG du2 = Cu1 (x ) (3.37) dx x=x − dx xeq3tritwo Combining equations (3.24), (3.36), and (3.37) gives us du2 du1 −1 C u1 − u2 = . (3.38) dx dx x=x τ (x )pr:wronsk1 The Wronskian is deﬁned as du2 du1 W (u1 , u2 ) ≡ u1 − u2 . (3.39) dx dx This allows us to write 1 C= . (3.40) −τ (x )W (u1 (x , λ), u2 (x , λ))13 Jan p4 Thus u1 (x< , λ)u2 (x> , λ) G(x, x ; λ) = , (3.41) −τ (x )W (u1 (x , λ), u2 (x , λ))eq3.39 where we deﬁnepr:xless1 u(x) if x < x u(x< ) ≡ u(x ) if x < x and u(x) if x > x u(x> ) ≡ u(x ) if x > x.
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3.5. THE FORCED OSCILLATION PROBLEM 31The u’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 u2 and the second by u1 . Subtract one FW p249equation from the other to get −u2 (τ u1 ) + u1 (τ u2 ) = 0 (where we ∂ ∂have used equation 1.10, L0 = − ∂x (τ ∂x ) + V ). Rewriting this as atotal derivative gives d [τ (x)W (u1 , u2 )] = 0. (3.43) dxThis implies that the expression τ (x)W (u1 (x, λ), u2 (x, λ)) is indepen- but isn’tdent of x. Thus G is symmetric in x and x . W (x) = 0 also The case in which the Wronskian is zero implies that u1 = αu2 , true?since then 0 = u1 u2 − u2 u1 , or u2 /u2 = u1 /u1 , which is only valid for allx if u1 is proportional to u2 . Thus if u1 and u2 are linearly independent, pr:LinIndep1the Wronskian is non-zero.3.5 The Forced Oscillation Problem 13 Jan p5The general forced harmonic oscillation problem can be expanded into pr:fhop1equations having forces internally and on the boundary which are sim-ple time harmonic functions. Consider the eﬀect of a harmonic forcingterm ∂2 L0 + σ 2 u(x, t) = σ(x)f (x)e−iωt . (3.44) ∂tWe apply the following boundary conditions: eq3ss −∂u(x, t) + κa u(x, t) = ha e−iωt for x = a, (3.45) ∂xand ∂u(x, t) + κb u(x, t) = hb e−iωt . for x = b. (3.46) ∂x We want to ﬁnd the steady state solution. First, we assume a steadystate solution form, the time dependence of the solution being u(x, t) = e−iωt u(x). (3.47)
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32 CHAPTER 3. GREEN’S FUNCTIONS After making the substitution we get an ordinary diﬀerential equation in x. Next determine G(x, x ; λ = ω 2 ) 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 Oscillationpr:fop1 Another kind of problem is the free oscillation problem. In this case f (x, t) = 0 and ha = hb = 0. The object of this problem is to ﬁnd thepr:NatFreq2 natural frequencies and normal modes. This problem is characterized by the equation: ∂2 L0 + σ 2 u(x, t) = 0 (3.48) ∂teq3fo with the Regular Boundary Conditions: • u is periodic. (Closed string) • [ˆ · n + KS ]u = 0 for x in S. (Open string) The goal is to ﬁnd normal mode solutions u(x, t) = e−iωn t un (x). Thepr:NatMode1 natural frequencies are the ωn and the natural modes are the un (x). We want to solve the eigenvalue equation 2 [L0 − σωn ]un (x) = 0 with R.B.C. (3.49) 2eq3.48 The variable ωn is called the eigenvalue of L0 . The variable un (x) ispr:EVect1 called the eigenvector (or eigenfunction) of the operator L0 . 3.7 Summary 1. The Principle of superposition is L0 [a1 u1 + a2 u2 ] = a1 L0 u1 + a2 L0 u2 , where L0 is a linear operator, u1 and u2 are functions, and a1 and a2 are constants.
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3.7. SUMMARY 33 2. The Dirac Delta Function is deﬁned as d 1 if c < xk < d dxδ(x − xk ) = c 0 otherwise. 3. Force contributions can be constructed by superposition. N σ(x)f (x) = Fk δ(x − xk ). k=1 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)ω 2 ]G(x, x ; ω 2 ) = δ(x − x ) a < x, x < b, RBC. d 5. At the source point x , the Green’s function satisﬁes dx G|x=x + − d 1 dx G|x=x − = − τ and G(x, x )|x=x + = G(x, x )|x=x − . 6. Green’s Reciprocity Principle is ‘The amplitude of the string at x subject to a localized force applied at x is equivalent to the amplitude of the string at x subject to a localized force applied at x.’ 7. The Green’s function for the 1-dimensional wave equation is given by u1 (x< , λ)u2 (x> , λ) G(x, x ; λ) = . −τ (x )W (u1 (x , λ), u2 (x , λ)) 8. The forced oscillation problem is ∂2 L0 + σ u(x, t) = σ(x)f (x)e−iωt , ∂t2 with periodic boundary conditions or the elastic boundary condi- tions with harmonic forcing. 9. The free oscillation problem is ∂2 L0 + σ 2 u(x, t) = 0. ∂t
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34 CHAPTER 3. GREEN’S FUNCTIONS3.8 ReferenceSee [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].
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Chapter 4Properties of Eigen States 13 Jan p7 Chapter Goals: 2 2 • Show that for the Helmholtz equation, ωn > 0, ωn 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 λ near an eigen value λn . • Derive the Green’s function for the ﬁxed string problem.By deﬁnition 2.1 ω2 > 0 b S, u = dxS ∗ (x)u(x). (4.1) aIn section 2.4 we saw (using Green’s ﬁrst identity) for L0 as deﬁnedin equation 2.2, and for all u which satisfy equation 1.26, that V > 0implies u, L0 u > 0 . We choose u = un and use equation 3.49 so that 2 0 < un , L0 un = un , σun ωn . (4.2) 35
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36 CHAPTER 4. PROPERTIES OF EIGEN STATES Remember that σ signiﬁes the mass density, and thus σ > 0. So we conclude 2 un , L0 un ωn = > 0. (4.3) un , σun13 Jan p8 This all came from Green’s ﬁrst identity.ω 2 real Next we apply Green’s second identity 2.5, S, L0 u = L0 S, u for S, u satisfying RBC (4.4) Let S = u = un . This gives us 2 ωn un , σun = un , L0 un (4.5) = L0 un , un (4.6) = (ωn )∗ un , σun . 2 (4.7) We used equation 3.49 in the ﬁrst equality, 2.5 in the second equality, 2 and both in the third equality. From this we can conclude that ωn is real.orthogonality Now let us choose u = un and S = um . This gives us um , L0 un = L0 um , un . (4.8) 2 Extracting ωn gives (note that σ(x) is real) 2 2 2 ωn um , σun = ωm σum , un = ωm um , σun . (4.9) So 2 2 (ωn − ωm ) um , σun = 0. (4.10) 2 2 Thus if ωn = ωm then um , σun = 0: b dxu∗ (x)σ(x)un (x) = 0 m 2 2 if ωn = ωm . (4.11) aeq4.11 That is, two eigen vectors um and un of L0 corresponding to diﬀerent13 Jan p8 eigenvalues are orthogonal with respect to the weight function σ. Ifpr:ortho1 the eigen vectors um and un are normalized, then the orthonormality condition ispr:orthon1 b dxu∗ (x)σ(x)un (x) = δmn m 2 2 if ωn = ωm , (4.12) aeq4.11p where the Kronecker delta function is 1 if m = n and 0 otherwise.
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4.1. EIGEN FUNCTIONS AND NATURAL MODES 37 u@ u h u @ h u u 1 a 2 & N u u & & u vu v vu v u & u u & & u h h u @ u @ Figure 4.1: The closed string with discrete mass points.4.1 Eigen Functions and Natural Modes 15 Jan p1We now examine the natural mode problem given by equation 3.49. Toﬁnd the natural modes we must know the natural frequencies ωn and pr:NatMode2the normal modes un . This is equivalent to the problem pr:lambdan1 L0 un (x) = σ(x)λn un (x), RBC. (4.13)To illustrate this problem we look at a discrete problem. eq4A4.1.1 A Closed String Problem pr:dcs1This problem is illustrated in ﬁgure 4.11 . In this problem the mass 15 Jan p2density σ and the tension τ are constant, and the potential V is zero. ﬁg4wThe term u(xi ) represents the perpendicular displacement of the ithmass point. The string density is given by σ = m/a where m is themass of each mass point and a is a unit of length. We also make thedeﬁnition c = τ /σ. Under these conditions equation 1.1 becomes τ m¨i = Ftot = u (ui+1 + ui−1 − 2ui ). aSubstituting the solution form ui = eikxi eiωt into this equation gives τ eika + e−ika τ τ ka mω 2 = 2 − + 1 = 2 (1 − cos ka) = 4 sin2 . a 2 a a 2But continuity implies u(x) = u(x + N a), so eikN a = 1, or kN a = 2πn,so that 2π k= n, n = 1, . . . , N. Na 1 See FW p115.
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38 CHAPTER 4. PROPERTIES OF EIGEN STATES The natural frequencies for this system are then 2 c2 sin2 (kn a/2) ωn = (4.14) a2 /4 −1eq4r where kn = 2π n and n can take on the values 0, ±1, . . . , ± N 2 for odd L N , and 0, ±1, . . . , ± N − 1, + N for even N . The constant is c2 = aτ /m. 2 2pr:DispRel1 Equation 4.14 is called the dispersion relation. If n is too large, the un 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−iωn t un (xi ) (4.15) = e−i[ωn t−kn xi ] (4.16) 2π = e−i[ωn t− L nxi ] . (4.17)eq4giver The normal modes correspond to traveling waves. Note that ωn ispr:travel1 doubly degenerate in equation 4.14. Solutions φn (x) for n which are larger than allowed give the same displacement of the mass points, but with some nonphysical wavelength. Thus we are restricted to N modes and a cutoﬀ frequency. 4.1.2 The Continuum Limit We now let a become increasingly small so that N becomes large for L ﬁxed. This gives us ∆kn = 2π/L for L = N a. In the continuum limit, the number of normal modes becomes inﬁnite. Shorter and shorterpr:cutoﬀ1 wavelengths become physically relevant and there is no cutoﬀ frequency.frequency Letting a approach zero while L remains ﬁxed givesexpression 2 c2 sin2 kn a N a=L ωn = a2 2 −→ c2 kn a→0 2 (4.18) 4 and so ωn = c|kn | (4.19) 2π ∆ωn = c|∆kn | = c (4.20) L
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4.1. EIGEN FUNCTIONS AND NATURAL MODES 39 2π ωn = c n. (4.21) LEquation (4.17) gives us the un ’s for all n. 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 2 wave can move. The eigenvalues are ωn . • The natural frequencies ωn are 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 p4giving u± . We also ﬁnd the correspondence ∆ωn ∼ c/L where c ∼ τ /σ. We also found that there is no degeneracy for the open discretecase.4.1.3 Schr¨dinger’s Equation o pr:Schro1Consider again equation 4.13 L0 un (x) = σ(x)λn un (x) RBC (4.22)where d d L0 = − τ (x) + V (x). (4.23) dx dxWe now consider the case in which τ (x) = h2 /2m and σ = 1, both ¯quantities being numerical constants. The linear operator now becomes −¯ 2 d2 h L0 = + V (x). (4.24) 2m dx2
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40 CHAPTER 4. PROPERTIES OF EIGEN STATES V (x) i i ii 2 $ 2 $2 h $ h h g gg ¡ f ¢ f ¢ e ¡ x Figure 4.2: Negative energy levels This is the linear operator for the Schr¨dinger equation for a particle o of mass m in a potential V : −¯ 2 d2 h + V (x) un (x) = λn un (x) + RBC (4.25) 2m dx2 In this case λ gives the allowed energy values. The potential V (x) can be either positive or negative. It needs to be positive for L0 to be positive deﬁnite, in which case λ0 > 0. For V < 0 we can have a ﬁnite number of of eigenvalues λ less that zero.verify this On a physical string the condition that V > 0 is necessary.paragraph One can prove that negative energy levels are discrete and bounded15 Jan p5 from below. The bound depends on the nature of V (x) (Rayleigh quo-pr:RayQuo1 tient idea). Suppose that V has a minimum, as shown in ﬁgure 4.2 forﬁg4neg example. By Green’s ﬁrst identity the quantity L0 − Vmin gives a new operator 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.pr:Fred1 Under consideration is the Fredholm equation 2 [L0 − σ(x)ωn ]u(x) = σ(x)f (x), RBC. (4.26)
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4.3. GF BEHAVIOR NEAR λ = λN 41In problem 2.2 one shows that the solution un (x) for this equation onlyexists if dxu∗ (x)σ(x)f (x) = 0. n (4.27)This is the condition that the eigenvectors un (x) are orthogonal to eq4.26the function f (x). We apply this to the Green’s function. We choose 2λ = λn = ωn and evaluate the Green’s function at this point. Thus for Ask Bakerthe equation Isn’t this just [L0 − σ(x)λn ]G(x, x ; λ) = δ(x − x ) (4.28) hitting a sta- tionary point?there will be a solution G(x, x ; λ) only if (using 4.27) eq4.26p G(x, x; λ) = dx G(x, x ; λ)δ(x − x ) = 0. (4.29) clarify this.The result G(x, x; λ) = 0 implies that u∗ (x ) = 0 (using equation 3.41). eq4.26b nThere will be no solution unless x is a node. In physical terms, this 15 Jan p6means that a natural frequency can only be excited at a node.4.3 GF behavior near λ = λn 15 Jan p6From the result of the previous section, we expect that if the drivingfrequency ω is not a natural frequency, everything will be well behaved.So we show that G(x, x ; λ) is good everywhere (in the ﬁnite interval[a, b]) except for a ﬁnite number of points. The value of G(x, x ; λ)becomes inﬁnite near λn , that is, as λ → λn . For λ near λn we canwrite the Green’s function as pr:gnxx1 1 G(x, x ; λ) ∼ gn (x, x ) + ﬁnite λ → λn (4.30) λn − λwhere 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 thenadd and subtract λn σ(x)G(x, x ; λ) on the the right-hand side. Thisgives us (using 4.28) 1 δ(x − x ) = [L0 − λn σ(x)] gn (x, x ) + ﬁnite λn − λ 1 + (λn − λ)σ(x) gn (x, x ) + ﬁnite λ → λn λn − λ 1 = [L0 − λn σ(x)] gn (x, x ) + ﬁnite λ → λn . λn − λ
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42 CHAPTER 4. PROPERTIES OF EIGEN STATES15 Jan p7 The left-hand side is also ﬁnite if we exclude x = x . This can only occur if [L0 − λn σ(x)]gn (x, x ) = 0 x=x. (4.31) From this we can conclude that gn has the form gn (x, x ) = un (x)f (x ), x=x, (4.32)eq4guf where f (x ) is a ﬁnite term and un (x) satisﬁes [L0 − λn σ(x)]un (x) = 0pr:fxprime1 with RBC. Note that here the eigen functions un (x) are not yet nor-pr:NatFreq3 malized. This is the relation between the natural frequency and the Green’s function. 4.4 Relation between GF & Eig. Fn.20 Jan p2 We continue developing the relation between the Green’s function andpr:SpecThy1 spectral theory. So far we have discussed the one dimensional problem. This problem was formulated as [L0 − σ(x)λ]G(x, x ; λ) = δ(x − x ), RBC. (4.33)eq4.31m To solve this problem we ﬁrst solved the corresponding homogeneous problem [L0 − σ(x)λn ]un (x) = 0, RBC. (4.34) The eigenvalue λn is called degenerate if there is more than one un per λn . We note the following properties in the Green’s function: 1. G∗ (x, x ; λ∗ ) = G(x, x ; 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 ; λ) = δ(x − x ), RBC (4.35) and then we take the complex conjugate of λ to get [L0 − σ(x)λ]G∗ (x, x ; λ∗ ) = δ(x − x ), RBC (4.36) which gives us G∗ (x, x ; λ∗ ) = G(x, x ; λ). (4.37)
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4.4. RELATION BETWEEN GF & EIG. FN. 43 2. G is symmetric. In the second problem set it was seen that G(x, x ; λ) = G(x , x; λ). (4.38) 3. The Green’s function G has the property of Hermitian analyticity. 18 Jan p3 By combining the results of 1 and 2 we get pr:HermAn1 G∗ (x, x ; λ) = G(x , x; λ∗ ). (4.39) This may be called the property of Hermitian analyticity. eq4ha In the last section we saw that λ→λ n gn (x, x ) G(x, x ; λ) −→ (4.40) λn − λfor g such that eq4cA [L0 − σ(x)λn ]gn (x, x ) = 0. (4.41)For the open string there is no degeneracy and for the closed stringthere is double degeneracy. (There is also degeneracy for the 2- and give ref3-dimensional cases.)4.4.1 Case 1: λ NondegenerateAssume that λn is non-degenerate. In this case we can write (usingequation 4.32) gn (x, x ) = un (x)fn (x ). (4.42)Hermitian analyticity and the complex conjugate of equation 4.40 give eq4cB(note that λn ∈ R) ∗ gn (x, x ) gn (x , x) ∗ = (4.43) λn − λ λn − λ ∗as λ → λn . This implies that ∗ gn (x, x ) = gn (x , x). (4.44) eq4cC So now 4.42 becomes 20 Jan p4
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44 CHAPTER 4. PROPERTIES OF EIGEN STATES u∗ (x)fn (x ) = fn (x)un (x ) n ∗ (4.45) so that (since x and x are independent) if un (x) is normalized (accord- ing to equation 4.12) fn (x) = u∗ (x). n (4.46) In the non-degenerate case we have (from 4.40 and 4.42) λ→λ n un (x)u∗ (x ) n G(x, x ; λ) −→ . (4.47) λn − λeq4.47 where un (x) are normalized eigen functions.pr:normal1 4.4.2 Case 2: λn Double Degenerate In the second case, the eigenvalue λn has double degeneracy like theis it +? closed string. The homogeneous closed string equation is (L0 + λn σ)u(±) (x) = 0. n The eigenfunctions corresponding to λn are u+ (x) and u− (x). By using n nwhy? the same reasoning that lead to equation 4.47 we can write 1 G(x, x ; λ) → [u(+) (x)u(+)∗ (x) + u(−) (x)u(−)∗ (x)] (4.48) λn − λ n n n n for the equation [L0 − σ(x)λn ]u(±) (x) = 0, n RBC. (4.49)How does this The eigenfunction un may be written as un = A+ u+ + A− u− . Doubleﬁt in? degeneracy is the maximum possible degeneracy in one dimension.20 Jan p5 In the general case of α-fold degeneracy λ→λ 1 G(x, x ; λ) −→n [uα (x)uα∗ (x )] n n (4.50) λn − λ α where uα (x) solves the equation n [L0 − σ(x)λn ]uα (x) = 0, n RBC. (4.51) The mathematical relation between the Green’s function and thepr:poles1 eigen functions is the following: The eigenvalues λn are the poles of G. The sum of bilinear products α uα (x)uα∗ (x ) is the residue of the pole n n λ = λn .
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4.5. SOLUTION FOR A FIXED STRING 454.5 Solution for a Fixed StringWe want to solve equation 3.22. Further, we take V = 0, σ and τconstant, and a = 0, b = L. d2 −τ 2 − λσ G(x, x ; λ) = δ(x − x ) for 0 < x, x < L (4.52) dxfor the case of a ﬁxed end string. Our boundary conditions are G(x, x ; λ) = 0 for x = a, b.4.5.1 A Non-analytic SolutionWe know from equation 3.41 the solution is 20 Jan p6 u1 (x< , λ)u2 (x> , λ) G(x, x ; λ) = . (4.53) −τ W (u1 , u2 )This solution only applies to the one dimensional case. This is because eq4.54the solution was obtained using the theory of ordinary diﬀerential equa-tions. The corresponding homogeneous equations are given by pr:homog2 d2 λ 2 + 2 u1,2 (x, λ) = 0. (4.54) dx cIn this equation we have used the deﬁnition 1/c2 ≡ σ/τ . The variables eq4lcuu1 and u2 also satisfy the conditions pr:c1 u1 (0, λ) = 0 and u2 (L, λ) = 0. (4.55)The solution to this homogeneous problem can be found to be λ λ u1 = sin 2 x and u2 = sin (L − x). (4.56) c c2 √In these solutions λ appears. Since λ can be complex, we must deﬁnea branch cut2 . pr:branch1
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46 CHAPTER 4. PROPERTIES OF EIGEN STATES Im λ λ u θ Re λ Figure 4.3: The θ-convention 4.5.2 The Branch Cut Since G ∼ λn1 (see 4.40) it follows (from λn > 0) that Gn is analytic pr:analytic1 −λ for Re (λ) < 0. As a convention, we choose θ such that 0 < θ < 2π.ﬁg4one This is illustrated in ﬁgure√4.3. Using this convention, λ can be represented by √ λ = |λ|eiθ/2 (4.57) θ θ = |λ| cos + i sin . (4.58) 2 2 √eq4.58-59 Note that λ has a discontinuity along the positive real axis. The20 Jan p7 function G is 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 GFpr:u1.1 We now look back at the ﬁxed string problem. We found that λ u1 (x, λ) = sin x c2 so that du1 λ λ = cos x. dx c2 c2but only if x = This gives us the boundary valuea = 0. 2 See also FW p485.
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4.5. SOLUTION FOR A FIXED STRING 47 du1 λ = . (4.60) dx x=a c2 √Because of the λ, this is not analytic over the positive real axis. We eq4fschoose instead the solution pr:baru1 u1 1 λ u1 → du1 = sin x ≡ u1 . (4.61) | dx x=a λ c2 c2The function u1 has the properties du1 u1 (a) = 0 and = 1. dx x=aThis satisﬁes the diﬀerential equation. 18 Jan p8 d2 λ + 2 u1 (x, λ) = 0. (4.62) dx2 cSo u1 is analytic for λ with no branch cut. Similarly, for the substitution justify thisu2 ≡ u2 /( du2 |x=b ) we obtain dx u2 = (λ/c2 )−1/2 sin λ/c2 (L − x).One can always ﬁnd u1 and u2 as analytic functions of λ with no branchcut. Is this always true?4.5.4 Analytic GF for Fixed StringWe have been considering the Green’s function equation [L0 − λσ(x)]G(x, x ; λ) = δ(x − x ) for 0 < x, x < b (4.63)with the open string RBC (1.18) [ˆ S · n + κS ]G(x, x ; λ) = 0 for x on S (4.64)and linear operator L0 deﬁned as d d L0 = − τ (x) + V (x). (4.65) dx dx
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48 CHAPTER 4. PROPERTIES OF EIGEN STATES We found that the solution to this equation can be written (3.41) u1 (x< , λ)u2 (x> , λ) G(x, x ; λ) = (4.66) −τ (x)W (u1 , u2 )eq4cD where u1 and u2 are solutions to the homogeneous equation with the same boundary conditions as (1.18) [L0 − λσ(x)]u1,2 (x, λ) = 0 for a < x < b (4.67) ∂ − u1 (x, λ) + ka u1 (x, λ) = 0 for x = a (4.68) ∂x ∂ + u2 (x, λ) + kb u2 (x, λ) = 0 for x = b. (4.69) ∂x 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 at x = a = 0 and x = b = L. We saw that λ u1 = sin x (4.70) c2 and λ u2 = sin (L − x). (4.71) c222 Jan p2 We also assigned the convention that √ λ = |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 ∂u2 ∂u1 W (u1 , u2 ) = u1 − u2 ∂x ∂x λ λ λ = 2 − sin x cos (L − x) c c2 c2
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4.5. SOLUTION FOR A FIXED STRING 49 λ λ − sin (L − x) cos x c2 c2 λ λ = − sin L. (4.73) c2 c2Thus we can write the full solution for the ﬁxed string problem as eq4.75b λ λ sin x c2 < sin c2 (L − x> ) G(x, x ; λ) = . (4.74) λ λ τ c2 sin c2 L eq4ss4.5.5 GF PropertiesWe may now summarize the properties of G in the complex λ plane. • Branch Cut: G has no branch cut. It is analytic except at isolated simple poles. The λ1/2 branch vanishes if the eigen functions are properly chosen. This is a general result for discrete spectrum. pr:DiscSpec1 Justify • Asymptotic limit: G goes to zero as λ goes to inﬁnity. If λ is pr:asymp1 real 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 √ 2 √ 2 ei λ/c x − e−i λ/c x sin λ/c2 x = (4.75) √ 2i −i λ/c2 x |λ|→∞ e −→ (4.76) 2i for θ > 0. Thus from equation 3.44 we get 23 Jan p3 √ √ −i λ/c2 x< −i λ/c2 (L−x> ) |λ|→∞ 1e e G(x, x ; λ) −→ − √ (4.77) 2i τ λ/c 2 e−i λ/c2 L √ 2 1 e+i λ/c (x> −x< ) = − . (4.78) 2i τ λ/c2
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50 CHAPTER 4. PROPERTIES OF EIGEN STATES By convention x> − x< > 0, and thus we conclude |λ|→∞ G(x, x ; 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. λn We now look at sin c2 L = 0 from the denominator of equation 3.44. The poles are at λ = λn . We can write λn /c2 L = nπ or cnπ 2 λn = for n = 1, 2, . . . (4.80) Leq4et We delete the case n = 0 since we have a removable singularity at λ = 0. Equation 4.80 occurs when the Wronskian vanishes. This happens when u1 = constant × u2 (not linearly independent). Both u1 and u2 satisfy the boundary conditions at both boundaries and are therefore eigenfunctions. Thus the un ’s are eigenfunctions and the λn ’s23 Jan p4 are the eigenvalues. So [L0 − λn σ]un (x) = 0 RBC (4.81)eq4star is satisﬁed for λn by un . 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 : √ λ (λ − λn ) L cos λn L sin L= √c c + O(λ − λn )2 . (4.82) c2 2 λneq4.85 So for λ near an eigenvalue we have λ λ λ→λn τ L τ sin L −→ 2 (λ − λn ) cos nπ. (4.83) c2 c2 2c
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4.6. DERIVATION OF GF FORM NEAR E.VAL. 51eq4.86 Now we look at the numerator of 4.74. We can rewrite sin λ/c2 (L − x> ) = − sin nπx> cos λ/c2 L. (4.84)eq4.87 Note that f (x< )f (x> ) = f (x)f (x ). So, with σ = τ /c2 , and substitut- ing 4.83 and 4.84, equation 4.74 becomes λ→λ n 2 sin nπx sin nπx L L G(x, x ; λ) −→ . (4.85) σL λn − λ So we conclude that un (x)un (x ) λ→λ n G −→ (4.86) λn − λ as in 4.47 where the eigenfunction is eq4cC2 22 Jan p5 2 nπx un (x) = sin , (4.87) σL L L which satisﬁes the completeness relation 0 um (x)u∗ (x)σdx = δmn . n 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, u1 (x< , λ)u2 (x> , λ) G(x, x ; λ) = . (4.88) −τ (x)W (u1 , u2 ) The boundary conditions are (see 1.18) eq4nt ∂u1 − + ka u1 = 0 for x = a = 0 (4.89) ∂x ∂u2 + + kb u2 = 0 for x = b = L. (4.90) ∂x The function u1 (respectively u2 ) may be any solution which is an analytic function of λ, and independent of λ at x = a (respectively
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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 ; λ) may only have poles when W (u1 (x, λ)u2 (x, λ)) = 0, which only occurs when u1 (x, λn ) = dn u2 (x, λn ). (4.91) where the dn are 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) λ→λ n τ (x)W (u1 , u2 ) −→ (λ − λn )cn , (4.92) where cn is some normalization constant. In this limit equation 4.88 becomes 1 λ→λn dn un (x< )¯n (x> ) ¯ u G(x, x ; λ) −→ (4.93) (λ − λn )cn 1 where dn is some constant. So un ≡ un¯ cn dn is the normalized eigen- function. Equation 4.88 then implies ∗ λ→λn un (x)un (x ) G(x, x ; λ) −→ (4.94) (λ − λn ) where un satisﬁes equation 4.81. 4.6.2 Summary, Interp. & Asymptotics25 Jan p1 In the previous sections we looked at the eigenvalue problem [L0 − λn σ(x)]un (x) = 0 for a < x < b, RBC (4.95) and the Green’s function problem [L0 − λn σ(x)]G(x, x ; λ) = δ(x − x ) for a < x, x < b, RBC (4.96)eq4fooo where d d L0 = − τ (x) + V (x), (4.97) dx dxpr:selfadj1 which is a formally self-adjoint operator. The general problem requires
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4.7. GENERAL SOLUTION FORM OF GF 53ﬁnding an explicit expression for the Green’s function for a force local-ized at x . For λ → λn we found un (x)u∗ (x ) n G(x, x ; λ) → . (4.98) λn − λThis equation shows the contribution of the nth eigenfunction. We saw eq4.102that G is an analytic function of λ (with poles at λn ) where G → 0 as|λ| → ∞. We can think of G as the inverse operator of L0 − λn σ: 1 G= . (4.99) L 0 − λn σThus the poles are at L0 = λn σ. d2 For large λ the behavior of G is determined by the dx2 G term sinceit brings down the highest power of λ. Thus for our simple example ofτ constant, d2 L0 ≈ −τ 2 (4.100) dxand for λ large √ √ G ∼ exp(i λ/c2 x), G ∼ λG, G ∼ ( λ)2 G (4.101)where the derivatives are taken with respect to x.4.7 General Solution form of GF 25 Jan p2In this section we obtain a general form (equation 4.108) for the Green’s ﬁg4ﬀfunction which is constructed using the solutions to the correspondingeigen value equation. This is done by evaluating a particular complexintegral. We have seen that G(x, x ; 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 dλ G(x, x ; λ ) ≡ dλ F (λ ) (4.102) c1 +c2 λ −λ c1 +c2where we have deﬁned F (λ ) ≡ G(x, x ; λ )/(λ − λ). Let the contour ofintegration be the contour illustrated in ﬁgure 4.4. This equation has
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54 CHAPTER 4. PROPERTIES OF EIGEN STATES λ -plane Im λ C St 2 k λq t t ¢ f ¢ C1 f Re λ f ¢ E f ¢ f ¢ f ¢ t t t Q Figure 4.4: The contour of integration a singularity only at λ. Thus we need only integrate on the contour pr:singular1 around λ. This is accomplished by deforming the contour C1 + C2 topr:Cauchy1 the contour S (following Cauchy’s theorem) dλ F (λ ) = dλ F (λ ). (4.103) C1 +C2 S See ﬁgure 4.5. Note that although F (λ ) blows up as λ approaches λ, G(x, x ; λ ) → G(x, x ; λ) in this limit. Thus the integration about the small circle around λ can be written λ →λ dλ dλ F (λ ) −→G(x, x ; λ) . S S λ −λ Now make the substitution λ − λ = eiα (4.104) dλ = i eiα dα (4.105) dλ = idα. (4.106) λ −λ This allows us to write dλ 2π = i lim dα = 2πi. S λ −λ →0 0
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4.7. GENERAL SOLUTION FORM OF GF 55 # λ q ε "! Figure 4.5: Circle around a singularity.We conclude dλ F (λ ) = 2πiG(x, x ; λ) Sand thus G(x, x ; λ ) 2πiG(x, x ; λ) = dλ . (4.107) C1 +C2 λ −λ eq4tpig We now assume that G(x, x ; λ) → 0 as λ → ∞. We must checkthis for each example we consider. An intuitive reason for this limit isthe following. The Green’s function is like the inverse of the diﬀerentialoperator: G ∼ 1/(L0 − λσ). Thus as λ becomes large, G must vanish.This assumption allows us to evaluate the integral around the largecircle C2 . We parameterize λ along this contour as λ − λ = Reiα , dλ → Reiα idα as R → ∞.So G(x, x ; λ ) 2π Reiα dα lim dλ = lim G(x, x ; Reiθ ) → 0 R→∞ C2 λ −λ R→∞ 0 Reiαsince G(x, x ; Reiθ ) → 0 as R → ∞. ﬁg4fof We now only need to evaluate the integral for the contour C1 . dλ G(x, x ; λ ) dλ G(x, x ; λ ) = . C1 λ −λ n cn λ −λIn this equation we replaced the contour C1 by a sum of contours aroundthe poles, as shown in ﬁgure 4.6. Recall that ﬁg4.6 λ →λ α uα (x)uα∗ (x ) n n G(x, x ; λ ) −→n . λn − λ
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56 CHAPTER 4. PROPERTIES OF EIGEN STATES EC1 c3 c1 c 2 4 c qλ1 qλ2 qλ3 qλ4 ... = qλ1 qλ2 qλ3 qλ4 ... Figure 4.6: Division of contour. We note that dλ 1 dλ = cn (λ − λ)(λn − λ ) λn − λ cn λn − λ 1 dλ = λn − λ λ − λn 1 = 2πi. λn − λ The ﬁrst equality is valid since 1/(λ − λ) is well behaved as λ → λn . The last equality follows from the same change of variables performed above. The integral along the small circle containing λn is thus dλ 2πi G(x, x ; λ ) = uα (x)uα∗ (x ). n n cn λ −λ λn − λ α The integral along the contour C1 (and thus the closed contour C1 +C2 ) is then dλ α uα (x)uα∗ (x ) n n G(x, x ; λ ) = 2πi . C1 λ −λ n λn − λ Substituting equation 4.107 gives the result 1 G(x, x ; λ) = uαn (x)uα∗ (x ) , λ λn (4.108) n λn − λ αeq4.124 where the indices λn sum n = 1, 2, . . . and α sums over the degeneracy.
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4.7. GENERAL SOLUTION FORM OF GF 574.7.1 δ-fn Representations & CompletenessUsing the above result (equation 4.108) we can write δ(x − x ) = [L0 − λσ(x)]G(x, x ; λ) (4.109) ∞ un (x)u∗ (x ) n = [L0 − λσ(x)] (4.110) n=1 λn − λ ∞ un (x)u∗ (x ) n = [(L0 − λn σ) + (λn − λ)σ] (4.111) n=1 λn − λ ∞ = σun (x)u∗ (x ). n (4.112) n=1In the last equality we used ∞ ∞ un (x)u∗ (x ) n un (x ) (L0 − λn σ) = (L0 − λn σ)un (x) = 0 n=1 λn − λ n=1 λn − λsince L0 is a diﬀerential operator in terms of x, From this we get thecompleteness relation pr:CompRel2 ∞ δ(x − x ) = σ(x)un (x)u∗ (x ) n (4.113) n=1or ∞ δ(x − x ) = un (x)u∗ (x ). n (4.114) σ(x) n=0This is called the completeness relation because it is only true if the eq4.128bun are a complete orthonormal set of eigenfunctions, which means thatany f (x) can be written as a sum of the un ’s weighted by the projectionof f (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 b 25 Jan p6 f (x) = dx f (x )δ(x − x ) for a < x < b (4.115) a b ∞ = dx f (x )σ(x ) un (x)u∗ (x ) n (4.116) a n=1So eq4.129
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58 CHAPTER 4. PROPERTIES OF EIGEN STATES ∞ f (x) = un (x)fn , (4.117) n=1 where eq4fo32 b fn = dx u (x )σ(x )f (x ) (4.118) aeq4.132 is the generalized nth Fourier coeﬃcient for f (x). Equation 4.117 rep-pr:FourCoef1 resents the projection of f (x) onto the un (x) normal modes. This was obtained using the completeness relation.pr:normal2 Now we check normalization. The Green’s function G is normalized because the u’s are normalized. We check the normalization of the u’s by looking at the completeness relation δ(x − x ) = σ(x ) un (x)u∗ (x ). n (4.119) n Integrate both sides by dx um (x ). On the left hand side we immedi- ately obtain dx um (x )δ(x − x ) = um (x). On the right hand side dx um (x )σ(x ) un (x)u∗ (x ) = n un (x) dx u∗ (x )σ(x )um (x ) n n n where we used 4.11 in the equality. But um (x) = un (x)δn,m . n Thus we conclude that normalized eigen functions are used in the com- pleteness relation: b dxu∗ (x)σ(x)un (x) = δn,m . m (4.120) apr:orthon2 This is the condition for orthonormality. 4.8 Extension to Continuous Eigenvalues27 Jan p127 Jan p2 As L (the length of the string) becomes large, the eigen values become closer together. The normalized eigen functions un (x) and eigen values λn for the ﬁxed string problem, equation 4.54, can be written 1 √ 2 un = √ e±i λn /c x (4.121) σL
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4.9. ORTHOGONALITY FOR CONTINUUM 59factor of 2? and 2πn λn = (ckn )2 kn = n = 0, 1, 2, . . . . (4.122) L The separation between the eigen values is then 2 2π L→∞ ∆λn = λn − λn−1 ∼ c2 n2 − (n − 1)2 −→ 0. (4.123) L We now consider the case of continuous eigen values. Let λ be complex (as before) and let ∆λn → 0. This limit exists as long as λ is 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 dλn lim G(x, x ; λ) = uαn (x)uα∗ (x). λ λn (4.124) ∆λn →0 λn − λ α eq4.124ab The completeness relation, equation 4.114, becomes δ(x − x ) = dλn uλn (x)u∗ n (x ). λ (4.125) σ(x ) α Now we take any function f (x) and express it as a superposition using the δ-function representation (in direct analogy with equations 4.115 and 4.118 in the discrete case) f (x) = dλn fλn uαn (x). α λ (4.126) α This is the generalized Fourier integral, with generalized Fourier coef- eq4.175 ﬁcients pr:GenFourInt1 fλn = dxuα∗ (x)σ(x)f (x). α λn (4.127) α The coeﬃcients fλn may be interpreted as the projection of f (x) with eq4.176 respect to σ(x) onto the eigenfunction uαn (x). λ 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
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60 CHAPTER 4. PROPERTIES OF EIGEN STATES used in the discrete spectrum case. First we choose f (x) = uαm (x) for λ f (x) in equation 4.126. Equation 4.127 then becomes fλn∗ = α dxuαn (x)σuαm (x). λ λ (4.128) The form of equation 4.126 corresponding to this is uαm (x) = λ dλm fλn uαn (x). α λ (4.129) α This equation can only be true if α fλn = δαα δ(λn − λm ). (4.130) So we conclude that dxuαn (x)σuαm (x) = fλn = δαα δ(λn − λm ). λ λ α (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 dλn G(x, x ; λ) = uαn (x)uαn (x ) . λ λ λn − λ α (Generally, the integration is over the interval from zero to inﬁnity.) Where are the singularities? Consider λ approaching 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 ; λ) is analytic on the entire complex cut λ-plane. It is in the region of non-27 Jan p4 analyticity that all the physics occurs. The singular diﬀerence of a branch cut is the diﬀerence in the value of G above and below. Seeﬁg4555 ﬁgure 4.7.pr:branch2 We now examine the branch cut in more detail. Using equation
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4.9. ORTHOGONALITY FOR CONTINUUM 61 rλ + iε λr rλ − iε Figure 4.7: λ near the branch cut.4.108 we can write G(x, x ; λ + i ) − G(x, x ; λ − i ) lim →0 2πi 1 ∞ 1 1 = dλn uαn (x)uα∗ (x ) λ λn − 2πi 0 α λn − λ − i λn − λ + iwhere 1 1 1 1 2i − = 2πi λn − λ − i λn − λ + i 2πi (λn − λ )2 + 2 1 = . π (λn − λ )2 + 2 In the ﬁrst problem set we found that 1 lim = δ(λ − λ). (4.132) →0 π (λ − λ )2 + 2So G(x, x ; λ + i ) − G(x, x ; λ − i ) lim →0 2πi = dλn uαn (x)uα∗ (x )δ(λ − λn ) λ λn α = uα λ (x)uα (x ). λ αTherefore the discontinuity gives the product of the eigenfunctions. 27 Jan p5 We derived in the second problem set the property G∗ (x, x ; λ) = G(x, x ; λ∗ ). (4.133)
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62 CHAPTER 4. PROPERTIES OF EIGEN STATES Now take λ = λ + i . This allows us to write G∗ (x, x ; λ + i ) = G(x, x ; λ − i ) (4.134) and G(x, x ; λ + i ) − G(x, x ; λ − i ) (4.135) 2πi G(x, x ; λ + i ) − G∗ (x, x ; λ + i ) = (4.136) 2πi 1 = ImG(x, x ; λ + i ) (4.137) π = uα (x)uα∗ (x ). λ λ (4.138) αeq4.151 So we can say that the sum over degeneracy of the bilinear product of the eigen function uα is proportional to the imaginary part of the λ Green’s function. 4.10 Example: Inﬁnite String29 Jan p1pr:InfStr1 Consider the case of an inﬁnite string. In this case we take the end points a → −∞, b → ∞ and the density σ, tension τ , and potential V as constants. The term V is 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 d2 −τ + V − λσ G(x, x ; λ) = δ(x − x ) for −∞ < x, x < ∞. dx2 (4.139) To get the solution we must take λ (= ω 2 ) to be imaginary. The solution for the Green’s function can be written in terms of the normal modes (3.41) u1 (x< )u2 (x> ) G(x, x ; λ) = (4.140) −τ W (u1 , u2 )
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4.10. EXAMPLE: INFINITE STRING 63 Im λ λ u θ u Re λ 2 2 ck Figure 4.8: θ speciﬁcation.where u1 and u2 satisfy the equation d2 −τ + V − λσ u1,2 = 0. (4.141) dx2The boundary conditions are that u1 is bounded and converges as x → eq4.139−∞ and that u2 is bounded and converges as x → ∞. Divide both ‘and converges’sides of equation 4.141 by −τ and substitute the deﬁnitions σ/τ = 1/c2 - R. Hornand V /τ = k 2 . This gives us the equation pr:k2.1 d2 λ − c2 k 2 + u(x) = 0. (4.142) dx2 c2The general solution to this equation can be written as eq4.132a √ √ λ−c2 k2 x/c λ−c2 k2 x/c u(x) = Aei + Be−i . (4.143)We specify the root by the angle θ extending around the point c2 k 2 on eq4.132the real axis, as shown in ﬁgure 4.8. This is valid for 0 < θ < 2π. The ﬁg4.6acorrespondence of θ is as follows: λ − c2 k 2 = |λ − c2 k 2 |eiθ . (4.144)Thus √ λ − c2 k 2 = |λ − c2 k 2 |eiθ/2 (4.145) θ θ = |λ − c2 k 2 | cos + i sin . (4.146) 2 2
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64 CHAPTER 4. PROPERTIES OF EIGEN STATES So that θ → 0 ⇐⇒ λ − c2 |k 2 | (4.147) θ → π ⇐⇒ i λ − c2 |k 2 | (4.148) θ → 2π ⇐⇒ − λ − c2 |k 2 |. (4.149) This is good everywhere except for values on the real line greater than c2 k 2 . 4.10.2 Uniqueness29 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 , u2 by looking at the large x behavior of 4.143. √ √ λ−c2 k2 x/c x→∞ −∞ λ−c2 k2 x/c x→∞ +∞ ei −→ e and e−i −→ e (4.150) √ √ i λ−c2 k2 x/c x→−∞ +∞ −i λ−c2 k2 x/c x→−∞ −∞ e −→ e and e −→ e (4.151) so √ λ−c2 k2 x u1 (x) = e−i c (4.152) √ i λ−c2 k2 x u2 (x) = e c . (4.153) The boundary condition has an explicit dependence on λ so 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 ) = u1 u2 − u2 u1 (4.154) i√ i√ = + λ − c2 k 2 − − λ − c2 k 2 (4.155) c c 2i √ = λ − c2 k 2 . (4.156) c
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4.10. EXAMPLE: INFINITE STRING 654.10.4 SolutionThis gives the Greens’ function √ √ λ−c2 k2 x< /c i λ−c2 k2 x> /c ice−i e G(x, x ; λ) = √ (4.157) 2τ λ − c2 k 2 √ 2 2 ic ei λ−c k |x−x |/c = √ . (4.158) 2τ λ − c2 k 2We now have a branch cut for Re (λ) ≥ c2 k 2 with branch point atλ = c2 k 2 . Outside of this, G is analytic with no poles. This is analyticfor λ in the cut λ-plane λ > c2 k 2 . Consider the special case of large λ 29 Jan p3 √ λ→∞ic ei λ|x−x |/c G −→ √ → 0. (4.159) 2τ λNotice that this is the same asymptotic form we obtained in the discretecase when we looked at G as λ → ∞ (see equation 4.79). Now considerthe case λ < c2 k 2 on the real axis. This corresponds to the case thatθ = π. So √ 2 c e− c k−λ|x−x | G(x, x ; λ) = √ . (4.160) 2τ c2 k 2 − λThis function is real and exponentially decreasing. For λ > c2 k 2 this eq4.159function oscillates. If θ = 0, it oscillates one way, and if θ = 2π itoscillates the other way, so the solution lacks uniqueness. The solutionscorrespond to diﬀerent directions of traveling waves.4.10.5 Motivation, Origin of ProblemWe want to understand the degeneracy in the Green’s function for aninﬁnite string. So we take a look at the physics behind the problem.How did the problem arise? It came from the time dependent problemwith forced oscillation imposed by an impulsive force at x : d2 ∂2 −τ + V + σ 2 u(x, t) = δ(x − x )e−iωt where ω 2 < c2 k 2 . dx2 ∂t (4.161)
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66 CHAPTER 4. PROPERTIES OF EIGEN STATES We wanted the steady state solution: u(x, t) = e−iωt G(x, x ; λ). (4.162) By substituting 4.160 this can be rewritten as √ 2 2 2 c e−iωt e− c k −ω |x−x | u(x, t) = √ . (4.163) 2τ − c2 k 2 − ω 2eq4.163ab We consider the cases ω 2 < c2 k 2 and ω 2 > c2 k 2 separately. If ω 2 < c2 k 2 , then there is a unique solution and the exponentialpr:WaveProp1 dies oﬀ. This implies that there is no wave propagation. This agrees with what the physics tells us intuitively: k 2 c2 > ω 2 implies large k, which corresponds to a large elastic constant V , which in turn means there will not be any waves.29 Jan p4 For ω > c2 k 2 there is propagation of waves. For ω > c2 k 2 we denote the two solutions: u± (x, t) = e−iωt G(x, x ; λ = ω 2 ± i ) as ε → 0. (4.164)pr:cutoﬀ2 As ω increases, it approaches the branch point. We deﬁne the cutoﬀ frequency as being at the branch point, ωc = c2 k 2 . We have seen that 2 for ω < ωc there is no wave propagation, and for ω > ωc there 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: √ 2 2 2 e−iωt±i( ω −c k /c)|x−x | u± (x, t) = 2τ √ . (4.165) c ω 2 − c2 k 2 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 √ 2 2 2 ∼ e−iω(t−x√ω −c k /cω) for x > x u+ = 2 2 2 ∼ e−iω(t+x ω −c k /cω) for x < x .
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4.11. SUMMARY OF THE INFINITE STRING 67Similarly, the solution u− represents a wave traveling to the left forpoints to the right of the source and a wave traveling to the right forpoints to the left of the source. Mathematically this means √ 2 2 2 ∼ e−iω(t+x√ω −c k /cω) for x > x u− = 2 2 2 ∼ e−iω(t−x ω −c k /cω) for x < x .These results can be rephrased by saying that u+ is a steady statesolution having only waves going out from the source, and u− is asteady state solution having only waves going inward from the outsideabsorbed by the point. So the equation describes two situations, andthe branch cut corresponds to the ambiguity in the situation.4.11 Summary of the Inﬁnite String 2 Feb p1We have considered the equation (L0 − λσ)G = δ(x − x ) (4.166)where d2 L0 = −τ + V. (4.167) dx2We found that is V = 0? i i λ c2 −k2 |x−x | G(x, x ; λ) = e (4.168) λ 2τ c2 − k2where we have introduced the substitutions V /τ ≡ k 2 and σ/τ ≡ 1/c2 .The time dependent response is 1 −iωt−i k2 − λ c2 |x−x | u± (x, t) = e . (4.169) λ 2τ c2 − k2If ω 2 < c2 k 2 ≡ ωc there is exponential decay, in which case there is no 2singularity of G at λ = ω 2 . The other case is that u± (x, t) = e−iωt G(x, x ; λ = ω 2 ± i ). (4.170)
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68 CHAPTER 4. PROPERTIES OF EIGEN STATES This case occurs when ω 2 > ωc , for which there is a branch cut across 2 the real axis. In this case we have traveling waves. Note that in the case that k = 0 we always have traveling waves.1 Feb p2 The relevance of the equation k 2 = V /τ is that the resistance of the medium to propagation determines whether waves are produced. When k = 0 there is no static solution — wave propagation always occurs. If ω 2 < ωc , then the period of the external force is small with respect to 2 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−iωt G(x, x ; λ) (4.171) c cos[ ω ( L − |x − x |)] = e−iωt c 2 . (4.172) 2ω sin[ ω 2 ] c l 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- ited1 Feb p3 We now return to the the connection with the eigen function problem.pr:efp2 We have seen before that the expression G(λ = λ + i ) − G(λ = λ − i ) (4.173) 2πi vanishes if λ < c2 k 2 . In the case that λ > c2 k 2 we have G(λ = λ + i ) − G(λ = λ − i ) = uα (x)(uα (x ))∗ . λ λ (4.174) 2πi α
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4.13. SUMMARY 69 λ/c2 − k 2 → +|λ/c2 − k 2 |1/2 q ¢ ¢ c2 k 2 w f f λ/c2 − k 2 → −|λ/c2 − k 2 |1/2 Figure 4.9: Geometry in λ-planeeq4777 The geometry of this on the λ-plane is shown in ﬁgure 4.9. This gives us G(λ = λ + i ) − G(λ = λ − i ) 2πi 1 1 λ 2 1/2 λ 2 1/2 = λ 2 |1/2 (ei| c2 −k | |x−x | − e−i| c2 −k | |x−x | ) 2πi 2τ | c2 − k 1/2 1 1 λ = λ 2 |1/2 sin 2 − k 2 |x − x | π 2τ | c2 − k c 1 = ImG(λ = λ + i ). π For λ > c2 k 2 we can write (using equation 4.138) 1 √ λ −c2 k2 x/c u± (x) = λ e±i (4.175) λ 4πτ c2 − k2 for λ > c2 k 2 . We now see that no eigen functions exist for λ < c2 k 2 eq4.188 since it exponentially increases as λ → ∞ and we must kill both terms. The Green’s function is all right since λ ∈ C. 4.13 Summary 2 2 1. For the Helmholtz equation, ωn > 0, ωn is real, and the eigen functions are real. 2. The dispersion relation for a closed massless string with discrete mass points is 2 c2 sin2 (kn a/2) ωn = . a2 /4
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70 CHAPTER 4. PROPERTIES OF EIGEN STATES 3. The Green’s function obeys Hermitian analyticity: G∗ (x, x ; λ) = G(x , x; λ∗ ). 4. The form of the Green’s function for λ near an eigen value λn is nλ→λ un (x)u∗ (x ) n G(x, x ; λ) −→ . λn − λ 5. The Green’s function for the ﬁxed string problem is λ λ sin x c2 < sin c2 (L − x> ) G(x, x λ) = . λ λ τ c2 sin c2 L 6. The completeness relation is ∞ δ(x − x ) = σ(x)un (x)u∗ (x). n n=1 7. The expansion theorem is ∞ f (x) = un (x)fn n=1 where b fn = dx u∗ (x )σ(x )f (x ). n a 8. The Green’s function near the branch cut is related to the eigen functions by 1 ImG(x, x ; λ + i ) = uα (x)uα (x ). λ λ π α 9. The Green’s function solution for an inﬁnite string is √ 2 2 ic ei λ−c k |x−x |/c G(x, x ; λ) = √ . 2τ λ − c2 k 2
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4.14. REFERENCES 714.14 ReferencesThe 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].
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Chapter 5Steady 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. moved from few pages later • Show that the solutions for the string problem with σ = x, τ = x, and V = m/x2 on 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.5.1 Oscillating Point Source pr:ops1We now look at the problem with an oscillating point source. In thenotation of the previous chapter this is 3 Feb p1 73
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74 CHAPTER 5. STEADY STATE PROBLEMS ∂2 L0 + σ(x) u(x, t) = δ(x − x )e−iωt −∞ < x, x < ∞+ R.B.C. ∂t2 (5.1) and can be written in terms of the Green function G(x, x ; λ) which satisﬁes [L0 −σ(x)λ]G(x, x ; λ) = δ(x−x ) −∞ < x, x < ∞+ R.B.C. (5.2) The steady state solution corresponding to energy radiated outward to inﬁnity is u(x, t) = e−iωt G(x, x , λ = ω 2 + i ). (5.3) The solution for energy radiated inward from inﬁnity is the same equa- tion with λ = ω 2 − iε, but this is generally not a physical solution. This contrasts with the case of a ﬁnite region. In that case there are1 Feb p5 no branch cuts and there is no radiation. 5.2 The Klein-Gordon Equationpr: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 E 2 = m2 c4 + p2 c2 (5.4) In the theory of quantum mechanics we treat momentum and energy as operators p → −i¯h (5.5) ∂ E → i¯ h (5.6) ∂t where dim[¯ ] = Action h (5.7) We want to derive the appropriate wave equation, so we start with E 2 − (m2 c4 + p2 c2 ) = 0 (5.8)
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5.2. THE KLEIN-GORDON EQUATION 75Now substitute the operators into the above equation to get 2 2 ∂ h ¯ i¯ h − m2 c4 + c 2 Φ = 0 (5.9) ∂t iThis is the Klein–Gordon equation, which is a relativistic form of the pr:phi1Schr¨dinger Equation. Note that |Φ|2 still has a probability interpre- otation, as it does in non-relativistic quantum mechanics. Now specialize this equation to one dimension. 1 Feb p6 d2 ∂2 −¯ 2 c2 h + m2 c4 + h2 2 Φ(x, t) = 0 ¯ (5.10) dx2 ∂tThis is like the equation of a string (c.f., 1.11). In the case of the eq5.energystring the parameters were tension τ = dim[E/t], coeﬃcient of elasticityV = dim[E/l3 ], and mass density σ = [t2 E/l3 ], where E is energy, tis time, and l is length. The overall equation has units of force overlength (since it is the derivative of Newton’s second law). By comparingequation 1.11 with equation 5.10 we note the correspondence V → m2 c4 σ → h2 ¯ τ → h2 c2 ¯ and f (x, t) → 0. (5.11)Note that V /τ = mc/¯ is a fundamental length known as the Comp- hton wavelength, λc . It represents the intrinsic size of the particle.The expression expression σ/τ = 1/c2 shows that particle inertiacorresponds to elasticity, which prevents the particle from respondingquickly. 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 t/¯ ΦE (x) h (5.12)to get the eigen value problem d2 2 2 −¯ c h 2 + m2 c4 − E 2 ΦE (x) = 0 (5.13) dxThus we quote the previous result (equation 4.175 which solves equation4.142) Φ± (x) = u± (x) E (5.14)
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76 CHAPTER 5. STEADY STATE PROBLEMS where we let k 2 → ( mc )2 and λ → E 2 /¯ 2 . As a notational shorthand, h √ h ¯ let p = E 2 − m2 c4 /c. Thus we write √ 2 2 4 h e±i( E −m c /¯ c)x Φ± (x) = E √ 4π¯ c E 2 − m2 c4 h e±p /¯h = √ . 4π¯ p c2 h The cut-oﬀ energy is mc2 . We have the usual condition on the solution that λ > c2 k 2 . This eigen value condition implies E 2 > m2 c4 . In relativistic quantum mechanics, if E < mc2 , then no free particle is emitted at large distances. In the case that E > mc2 there is radia-pr:rad1 tion. Also note that m becomes inertia. For the case that m → 0 there is always radiation. This corresponds to V → 0 in the elastic string analogy. The potential V acts as an elastic resistance. The Green’s function has the form √ G(x, x ; E) ∼ exp{( m2 c4 − E 2 /¯ c)|x − x |}. h So G(x, x ; E) has a characteristic half-width of √ |x − x | ∼ hc/ m2 c4 − E 2 = 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 for E > mc2 the Green’s function remains ﬁnite at large distances which corresponds to the radiation of a particle of mass m. 5.2.1 Continuous Completenesspr:CompRel33 Feb p2 Recall that the completeness condition in the discrete case is δ(x − x ) = uα (x)uα∗ (x ). λ λ (5.15) σ(x) λ ,αR. Horn says The corresponding equation for the case of continuous eigenvalues isno λ in sum. δ(x − x ) ∞ = dλ uα (x)uα∗ (x ). λ λ (5.16) σ(x) λ ,α=± 2 ωc
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5.2. THE KLEIN-GORDON EQUATION 77Recall that in this case the condition for an eigen function to exist is 2λ > ωc . In this case λ ±i −k2 x c2 e u± (x) λ = . (5.17) λ 4πτ ( c2 − k 2 )1/2Substituting the u± into the continuous completeness relation gives c ∞ dλ √ √ i λ −ωc ( x−x ) 2 −i λ −ωc ( x−x ) 2δ(x − x ) = σ(x) e c +e c 2 ( λ − k 2 )1/2 4πτ ωc c2 (5.18) 2 2 2where ωc = k c . We now make a change of variables. We deﬁne thewave number as λ − ωc 2 k= (5.19) cIt follows that the diﬀerential of the wave number is given by 1 dλ dk = . (5.20) 2c 2 λ − ωcWith this deﬁnition we can write 3 Feb p3 2σc2 ∞ δ(x − x ) = dk eik(x−x ) + e−ik(x−x ) . (5.21) 4πτ 0Note the symmetry of the transformation k → −k. This propertyallows us to write 1 ∞ δ(x − x ) = dkeik(x−x ) . (5.22) 2π −∞This is a Fourier integral. In our problem this has a wave interpretation. We now apply this to the quantum problem just studied. In thiscase λ = (E/¯ )2 and ωc2 = m2 c4 /¯ 2 . With these substitutions we can h hwrite 2 m2 c4 1/2 ( E2 − h ¯ ¯2 h ) k = (5.23) c 1 E2 p = 2 − m2 c2 = (5.24) h ¯ c h ¯
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78 CHAPTER 5. STEADY STATE PROBLEMS so 1 ∞ δ(x − x ) = dp ei(x−x )p /¯ . h (5.25) 2π −∞pr:FI1 This Fourier integral has a particle interpretation. 5.3 The Semi-inﬁnite Problem3 Feb p4 Consider the following linear operator in the semi-inﬁnite region d d m2 L0 = − x + 0 < x < ∞. (5.26) dx dx x Here we have let the string tension be τ = x and 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) d d m2 − x + − λx G(x, x ; λ) = δ(x − x ) (5.27) dx dx x deﬁned on the interval 0 < x, x < ∞. 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 equation τ = 0 at x = 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 at x = 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 L0 is hermitian for 0 ≤ x < ∞. The solution can be written in the form (using 3.41) u1 (x< , λ)u2 (x> , λ) G(x, x ; λ) = − (5.28) xW (u1 , u2 )eq5.29 The function u1 and u2 are the solutions to the equation
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5.3. THE SEMI-INFINITE PROBLEM 79 (L0 − λx)u1,2 = 0 (5.29)where we restrict u1 to be that function which is regular at x = 0, andu2 to be that function which is bounded at inﬁnity. We note that the equations are Bessel’s equations of order m: pr:Bessel1 d2 d y2 + y + (y 2 − m2 ) Xm (y) = 0 (5.30) dy 2 dy √where y = x λ and Xm (y) is any solution. The solutions are then √ u1 (x) = Jm (x λ) (5.31)and √ (1) u2 (x) = Hm (x λ). (5.32) (1) √The function Hm (x λ) is known as the Hankel function. For large x 3 Feb p5it may be approximated as √ x→∞ 2 √ mπ π (1) Hm (x λ) ∼ √ ei(x λ− 2 − 4 ) (5.33) πx λ √and since Im( λ) > 0, we have decay as well as out going waves. Now we get the Wronskian: Justify this √ (1) √ W (u1 , u2 ) = W (Jm (x λ), Hm (x λ)) (5.34) √ √ = iW (Jm (x λ), Nm (x λ)) (5.35) √ 2 2i = i λ √ = . (5.36) πx λ πx (1)In the second equality we used the deﬁnition Hm (x) = Jm (X)+iNm (x),where Jm (x) cos mπ − J−m (x) Nm (x) ≡ . sin mπThe third equality is veriﬁed in the problem set. So τ W is independentof x, as expected. Therefore (using equation 5.28) 1 πx √ (1) √ G(x, x ; λ) = − Jm (x< λ)Hm (x> λ) (5.37) x 2i iπ √ (1) √ = Jm (x< λ)Hm (x> λ) (5.38) 2
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80 CHAPTER 5. STEADY STATE PROBLEMS 5.3.1 A Check on the Solution Suppose that λ < 0. In this case we should have G real, since there is no branch cut. We have √ λ = i|λ|1/2 at θ = π. (5.39)pr:Bessel2 We can use the deﬁnitions Im (x) ≡ e−imπ/2 Jm (ix), Km (x) ≡ (πi/2)eimπ/2 Hm (ix) (1) to write π G(x, x ; λ) = i Jm (i|λ|1/2 x< )Hm (i|λ|1/2 x> ) (1) (5.40) 2 iπ m 2 = i Im (x< |λ|1/2 ) i−m Km (x> |λ|1/2 (5.41) 2 πi 1/2 1/2 = Im (x< |λ| )Km (x> |λ| ) ∈ R (5.42) Note also that G → 0 as x → ∞, so we have decay, and therefore no propagation. This is because asymptotically Im (z) ≈ (2zπ)−1/2 ez | arg z| < 1/2, |z| → ∞ Km (z) ≈ (2z/π)−1/2 e−z | arg z| < 3π/2, |z| → ∞. 5.4 Steady State Semi-inﬁnite Problem5 Feb p1 For the equation d d m2 − x + − λx G(x, x ; λ) = δ(x − x ) for 0 < x, x < ∞ dx dx x (5.43) the solution we obtained was iπ √ (1) √ G(x, x ; λ) = Jm ( λx< )Hm ( λx> ). (5.44) 2 We now look at the steady state solution for the wave equation, ∂2 L0 + x 2 u(x, t) = δ(x − x )e−iωt . (5.45) ∂t
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5.4. STEADY STATE SEMI-INFINITE PROBLEM 81pr:rad2 We only consider outgoing radiation (ω 2 → ω 2 + iε). We take ε > 0, since ε < 0 corresponds to incoming radiation. So δ(x − x ) acts as a point source, but not as a sink. u(x, t) = e−iωt G(x, x ; λ = ω 2 + iε) (5.46) iπ = e−iωt Jm (ωx< )Hm (ωx> ) (1) (5.47) 2 iπ = e−iωt Jm (ωx )Hm (ωx> ) (1) for x > x . (5.48) 2 Next let ωx 1, so iπ e−iω(t−x) u(x, t) = Jm (ωx ) √ for ωx 1. 2 ωx The condition ωx 1 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 1 we take ωx → 0, then we Explain why? have iπ e−iω(t−x) u(x, t) = (ωx )m √ (5.49) 2 ωx We now look at the case x < x with ω large. In this case we have iπ (1) u(x, t) = e−iωt H (ωx )Jm (ωx) 2 m iπ (1) 1 (1) = e−iωt Hm (ωx ) [Hm (ωx) + Hm (ωx)] (2) 2 2 eiωx e−iωx ∼ e−iωt Hm (ωx ) √ (1) +√ ωx 1 ωx ωx ∼ Hm (ωx )[e−iω(t−x) + e−iω(t+x) ]. (1) We now look at the Green’s function as a complete set of eigen functions. First we consider 5 Feb p3 1 [G(x, x ; λ + iε) − G(x, x ; λ − iε)] 2πi
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82 CHAPTER 5. STEADY STATE PROBLEMS 1 √ (1) √ √ (1) √ = [Jm ( λ x< )Hm ( λ x> ) − Jm (− λ x< )Hm (− λ x> )] 2πi 1 √ (1) √ (2) √ = [Jm ( λ x> )][Hn ( λ x) + Hn ( λ x)] 4 1 √ √ = Jm ( λ x)Jm ( λ x ) 2 1 = Im G(x, x ; λ + iε). π 5.4.1 The Fourier-Bessel Transform The eigen functions uλ satisfy −d d m2 x + − λx uλ = 0 for 0 < x < ∞. (5.50) dx dx x5 Feb p4 In this case since there is a boundary at the origin, waves move only to the right. There is no degeneracy, just one eigen function: 1 √ uλ = Jm ( λ x). (5.51) 2 We know from the general theory that if there is no degeneracy, then 1 ∞ δ(x − x ) = dλ uλ (x)u∗ (x ) λ (5.52) σ(x) 0 so 1 1 ∞ √ √ δ(x − x ) = dλ Jm ( λ x)Jm ( λ x ) (5.53) x 2 ∞ ∞ = ω dω Jm (ω x)Jm (ω x ). (5.54) 0 This is valid for 0 < x, x < ∞. Thus for f (x) on 0 < x < ∞ we have ∞ f (x) = dx f (x )δ(x − x ) (5.55) 0 ∞ ∞ = dx f (x )x ω dω Jm (ω x)Jm (ω x ) (5.56) 0 0 ∞ ∞ = ωdω Jm (ω x) dx x f (x )Jm (ω x ). (5.57) 0 0
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5.5. SUMMARY 835 Feb p5 Thus for a given f (x) on 0 < x < ∞ we can write ∞ f (x) = ω dω Jm (ω x)Fm (ω ). (5.58) 0 This is the inversion theorem. ∞ Fm (ω) = x dx f (x )Jm (ωx ). (5.59) 0 This is the Fourier-Bessel transform of order m. Explain why this is useful? pr:FBt1 5.5 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 d2 ∂2 −¯ 2 c2 h + m2 c4 + h2 2 Φ(x, t) = 0. ¯ dx2 ∂t Steady-state solutions for a point source with |E| > mc2 corre- spond to a mass m particle radiated to (±) inﬁnity, where as solutions with |E| < mc2 die oﬀ with a characteristic range of x ∼ h/p. ¯ pr:chRan1 3. The string problem with σ = x, τ = x, and V = m/x2 on the interval 0 < x < ∞ corresponds the Bessel’s equation d2 d y2 2 + y + (y 2 − m2 ) Xm (y) = 0 dy dy √ where y = x λ. The linearly independent pairs of solutions to this equation are the various Bessel functions: (i) Jm (y) and (1) (2) Nm (y), and (ii) Hm (y) and Hm (y). 4. The Green’s function for this problem is iπ √ (1) √ G(x, x ; λ) = Jm ( λx< )Hm ( λx> ). 2
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84 CHAPTER 5. STEADY STATE PROBLEMS 5. The steady state solutions for this problem with point source are iπ u(x, t) = e−iωt (1) Jm (ωx< )Hm (ωx> ). 2 The outgoing solutions consist of direct radiation and radiation reﬂected from the x = 0 boundary. 6. The Fourier-Bessel transform is ∞ Fm (ω ) = x dx f (x )Jm (ωx ). 0 The inversion theorem for this transform is ∞ f (x) = ω dω Jm (ω x)Fm (ω ). 05.6 ReferencesThe Green’s function related to Bessel’s equation is given in [Stak-gold67a, p75].
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Chapter 6Dynamic Problems 8 Feb p1 Chapter Goals: 4 Jan p1 • State the problem which the retarded Green’s func- p1prv.yr. tion GR solves, and the problem which the ad- vanced Green’s function GA solves. Give a physical interpretation for GR and GA . • 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 σ and τ constant, and V = 0. • Find the retarded Green’s function for a semi- inﬁnite string with a ﬁxed end, σ and τ constant, and V = 0. • Find the retarded Green’s function for a semi- inﬁnite string with a free end, σ and τ constant, and V = 0. 85
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86 CHAPTER 6. DYNAMIC PROBLEMS • Explain how to ﬁnd the retarded Green’s function for an elastically bound semi-inﬁnite string with σ and τ constant, 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’spr:impf1 Consider an impulsive force, a force applied at a point in space along the string at an instant in time. This force is represented by σ(x)f (x, t) = δ(x − x )δ(t − t ). (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 at the time t . For this physical situation we want to ﬁnd the solution u(x, t) = GR (x, t; x , t ) (6.2)pr:GR1 where GR stands for the retarded Green’s function: ∂2 L0 + σ GR (x, t; x , t ) = δ(x − x )δ(t − t ) ∂t2 for a < x, x < b; all t, t . 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 [ˆ s n + κs ]GR (x, t; x , t ) = 0 x ∈ S, a < x < b; ∀t, t (6.3)
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6.2. PHYSICS OF A BLOW 87eq6RBC where S is the set of end points {a, b}. In the case of a closed string the boundary condition is characterized by the equations GR (x, t; x , t )|x=a = GR (x, t; x , t )|x=b for a < x < b, ∀t, t , (6.4) ∂ ∂ GR (x, t; x , t ) = GR (x, t; x , t ) for a < x < b, ∀t, t . ∂x x=a ∂x x=b (6.5) We now apply the condition that the string begins at rest: 8 Feb p2 GR (x, t; x , t ) = 0 for t < t . (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 GR with RBC with the deﬁnition GA (x, t; x , t ) = 0 for t > t (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 G for GR (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 ) t2 dp = dt t1 dt t2 x2 = dt dxσ(x)f (x, t). t1 x1
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88 CHAPTER 6. DYNAMIC PROBLEMS The third equality holds because dp/dt is the force, which in this case x is x12 dxσ(x)f (x, t). We now look at the special case where the force is8 Feb p3 the δ function. In this case t2 x2 ∆p = dt δ(t − t )δ(x − x )dx = 1 (6.8) t1 x1 for x1 < x < x2 , t < t < t2 . Thus a delta force imparts one unit of momentum. Therefore we ﬁnd that GR is the response of our system to a localized blow at x = x , t = t which 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: ∂2 L0 + σ G(x, t; x , t ) = δ(t − t )δ(x − x ) + RBC, (6.9) ∂t2eq6de1 G(x, t; x , t ) = 0 for t < t . For ﬁxed x we note that G(x, t; x , t ) is a function of t − t and not t and t separately, since only ∂ 2 /∂t2 and t − t appear in the equation. Thus the transformation t → t + a and t → t + a does not change anything. This implies that the Green’s function can be written G(x, t; x , t ) = G(x, x ; t − t ) = G(x, x ; τ ), where we deﬁne τ ≡ t − t . By this deﬁnition, G = 0 for τ < 0. This problem can be solved by taking the complex Fourier trans-pr:FTrans1 form: ∞8 Feb p4 ˜ G(x, x ; ω) = dτ eiωτ G(x, x ; τ ). (6.10) −∞eq6Ftran ˜ Note that G(x, x ; ω) is convergent everywhere in the upper half ω- plane, which we now show. The complex frequency can be written as ω = ωR + iωI . Thus eiωτ = eiωR τ e−ωI τ .
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6.3. SOLUTION USING FOURIER TRANSFORM 89 ˜For G(x, x ; ω) to exist, the integral must converge. Thus we requiree → 0 as τ → ∞, which means we must have e−ωI τ → 0 as τ → ∞. iωτThis is only true when ω is in the upper half plane, ωI > 0. Thus G ˜ ˜exists for all ω such that Im ω > 0. Note that for GA everything isreversed and ω is deﬁned in the lower half plane. ˜ By taking the derivative of both sides of G(x, x ; ω) in the Fouriertransform, equation 6.10, we have d ˜ ∞ d iωτ ∞ G(x, x ; ω) = dτ e G(x, x ; τ ) = i dτ τ G(x, x ; τ )e−iωτ . dω −∞ dω 0which is ﬁnite. Therefore the derivative exists everywhere in the upper Ask Baker ˜half ω-plane. Thus G is analytic in the upper half ω-plane. We have but how do wethus seen that the causality condition allows us to use the Fourier trans- knowform to show analyticity and pick the correct solution. The condition −i dτ τ G(τ )that G = 0 for τ < 0 (causality) was only needed to show analyticity; converges?it is not needed anymore. 8 Feb p5 We now Fourier transform the boundary condition of an open string6.3: ∞ 0 = dτ eiωτ ((ˆ S · n + κS )G) −∞ ∞ = (ˆ S · n + κS ) dτ eiωt G −∞ = (ˆ S · n ˜ + κS )G.Similarly, in the periodic case we regain the periodic boundary condi- ˜ ˜ ˜ ˜ ˜tions of continuity Ga = Gb and smoothness Ga = Gb . So G satisﬁesthe same boundary conditions as G since the boundary conditions donot involve any time derivatives. Consider equation 6.9 rewritten as ∂2 L0 + σ 2 G(x, x ; τ ) = δ(x − x )δ(τ ). ∂tThe Fourier transform of this equation is ˜ ∞ ∂2 L0 G(x, x ; ω) + σ(x) dτ eiωτ G(x, x ; τ ) = δ(x − x ). (6.11) −∞ ∂τ 2
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90 CHAPTER 6. DYNAMIC PROBLEMS Using the product rule for diﬀerentiation, we can “pull out a divergence term”: ∂2 ∂ 2 iωτ ∂ ∂ ∂ eiωτ G= e G+ eiωτ G − G eiωτ . ∂τ 2 ∂τ 2 ∂τ ∂τ ∂τ Thus our equation 6.11 becomes ˜ ˜ δ(x − x ) = L0 G(x, x ; ω) + σ(x)(−ω 2 G(x, x ; ω)) t=+∞ ∂ ∂ + σ(x) eiωτ G − G eiωτ . ∂t ∂t t=−∞ Now we evaluate the surface term. Note that G = 0 for τ < 0 implies ∂G/∂t = 0, and thus |−∞ = 0. Similarly, as τ → ∞, eiωτ → 0 since Im ω > 0, and thus |∞ = 0. So we can drop the boundary term.8 Feb p6 We thus ﬁnd that GR satisﬁes the diﬀerential equation ˜ [L0 − ω 2 σ(x)]G(x, x ; ω) = δ(x − x ) RBC, (6.12)eq6stst with Im ω > 0. We now recognize that the Green function must be the same as in the steady state case: ˜ G(x, x ; ω) = G(x, x ; λ = ω 2 ). Recall that from our study of the steady state problem we know that the function G(x, x ; λ) is analytic in the cut λ-plane. Thus by analytic ˜ continuation we know that G(x, x ; ω) is analytic in the whole cut plane. √this is unclear The convention ω = λ compresses the region of interest to the upper half plane, where λ satisﬁes [L0 − λσ(x)]G(x, x ; λ) = δ(x − x ) + RBC, (6.13)eq6ststb All that is left is to invert the Fourier transform. 6.4 Inverting the Fourier Transform In the previous section we showed that for the Green’s function G we have the Fourier Transform ∞ ˜ G(x, x ; ω) = dτ e−iωτ G(x, x ; τ ) −∞
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6.4. INVERTING THE FOURIER TRANSFORM 91where τ = t − t , and we also found ˜ G(x, x ; ω) = G(x, x ; λ = ω 2 )where ω = ωR + iωI with ωI > 0, and G(x, x ; λ = ω 2 ) is the solutionof the steady state problem. Now we only need to invert the Fourier but didn’tTransform to get the retarded Green’s function. We write we use analytic continuation to iωτ iωR τ −ωI τ ge the whole λ e =e e plane.so that ∞ ˜ G(x, x ; ωR + iωI ) = dτ eiωR τ [e−ωI τ GR (x, x ; τ )] . −∞ ˜ F (ωR ) F (τ )This is a real Fourier Transform in terms of F (τ ). We now apply theFourier Inversion Theorem: pr:FIT1 1 ∞ F (τ ) = e−ωI τ GR (x, x ; τ ) = ˜ dωR e−iωR τ G(x, x ; ωR + iωI ) 2π −∞so 1 ∞ ˜ GR (x, x ; τ ) = dωR e−iωR τ G(x, x ; ω)eωI τ (6.14) 2π −∞ ﬁx ωI = ε and integrate over ωR (6.15) 1 ˜ = dωe−iωτ G(x, x ; ω) (6.16) 2π Lwhere the contour L is a line in the upper half plane parallel to the ωR 10 Feb p2axis, as shown in ﬁgure 6.1.a. The contour is oﬀ the real axis because of ﬁg6Lcontthe branch cut. We note that any line in the upper half plane parallelto the real axis may be used as the contour of integration. This canbe seen by considering the rectangular integral shown in ﬁgure 6.1.b. ˜Because G = 0 as ωR → ∞, we know that the sides LS1 and LS2 vanish. ˜And since eiωτ and G are analytic in the upper half plane, Cauchy’stheorem tells use that the integral over the closed contour is zero. Thusthe integrals over path L1 and path L2 must be equal.
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92 CHAPTER 6. DYNAMIC PROBLEMS ωI L2 E LS1 T E L c S2 ε T c L1 ωR L1 (a) The line L1 (b) Closed contour with L1 Figure 6.1: The contour L in the λ-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 ∂2 L0 + σ GR (x, t; x , t ) = δ(x − x )δ(t − t ) + RBC (6.17) ∂t2eq6GFxt with the condition that GR (x, t; x , t ) = 0 for t < t . The Green’s function which satisﬁes this equation was found to be dω −iω(t−t ) ˜ GR (x, t; x t ) = e G(x, x ; ω). (6.18) L 2πeq6GFT ˜ where G(x, x ; ω) satisﬁes the steady state Green’s function problem. 6.5 Analyticity and Causality To satisfy the physical constraints on the problem, we need to have GR = 0 for t < t . This condition is referred to as causality. This con- ˜ dition is obtained due to the fact that the product e−iω(t−t ) G(x, x ; ω 2 )Need to show appearing in equation 6.17 is analytic. In this way we see that the ana-somewhere lyticity of the solution allows it to satisfy the causality condition. As a ˜ |ω|→∞0. check, for the case t < t we write e−iω(t−t ) = e−iωR (t−t ) eωI (t−t ) and closethat G −→ﬁg6Luhp ˜ the contour as shown in ﬁgure 6.2. The quantity e−iω(t−t ) G(x, x ; ω 2 )
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6.6. THE INFINITE STRING PROBLEM 93 λ -plane ωI L k U HP t t t ¢ f ¢ L f ωR f ¢ E Figure 6.2: Contour LC1 = L + LU HP closed in UH λ-plane.vanishes on the contour LC1 = L + LU HP since e−iωI (t−t ) → 0 as ˜ωI → ∞ and |e−iωR (t−t ) | = 1, while we required |G(x, x ; ω 2 )| → 0as |ω| → ∞.6.6 The Inﬁnite String Problem pr:ISP1We now consider an inﬁnite string where we take σ and τ to be a 10 Feb p3constant, and V = 0. Thus our linear operator (cf 1.10) is given by d2 L0 = −τ 2 . dx6.6.1 Derivation of Green’s FunctionWe want to solve the equation ∂2 ∂2 −τ 2 + σ 2 GR (x, t; x , t ) = δ(x−x )δ(t−t ) for −∞ < x, x < ∞ ∂x ∂t (6.19)with the initial condition GR (x, t; x t ) = 0 for t < t .
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94 CHAPTER 6. DYNAMIC PROBLEMS We write the Fourier transform of the Green’s function in terms of λ: ˜ G(x, x ; ω) = G(x, x ; λ = ω 2 ). From 6.13, we know that G(x, x ; λ) satisﬁes d2 −τ − σλ G(x, x ; λ) = δ(x − x ) for −∞ < x, x < ∞. dx2Actually since For the case of λ largewe found the solution (4.159)V =0 i c √ 2 G = √ ei λ/c |x−x | 2 λτ or (λ → ω 2 ) ˜ i c iω|x−x |/c G(x, x ; ω) = e . 2ω τ This gives us the retarded Green’s function 1 i c iω |x−x | GR (x, t; x , t ) = dωe−iω(t−t ) ec . (6.20) 2π L 2ω τeq6GRInfStr Now consider the term |x−x | −iω (t−t )− c e . We treat this term in two cases: |x−x | • t−t < c . In this case |x−x | −iω (t−t )− c e →0 as ωI → ∞. But the term (i/2ω) → +∞ as ωR → 0 in equation 6.20. Thus the integral vanishes along the contour LU HP shown in ﬁgure 6.2 so (using Cauchy’s theorem) equation 6.20 becomes GR (x, t; x , t ) = = = 0. L L+LU HP
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6.6. THE INFINITE STRING PROBLEM 95 λ -plane ωI L E ωR f ¢ f ¢ f ¢ t LLHP t t C Figure 6.3: Contour closed in the lower half λ-plane. |x−x |10 Feb p4 • t−t > c . In this case |x−x | −iω (t−t )− c e →0 as ωI → −∞ﬁg6Llhp so we close the contour below as shown in ﬁgure 6.3. Since the integral vanishes along LLHP , we have GR = L = L+LLHP . Cauchy’s theorem says that the integral around the closed con- tour is −2πi times the sum of the residues of the enclosed poles. 1 ic The only pole is at ω = 0 and its residue is 2π 2τ . For this case we obtain 1 ic c GR (x, t; x , t ) = −2πi = , 2π 2τ 2τ which is constant. From these two cases we conclude that c |x − x | GR (x, t; x , t ) = θ t−t − . (6.21) 2τ c The function θ is deﬁned by the equation 0 for u < 0 θ(u) = 1 for u > 0.
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96 CHAPTER 6. DYNAMIC PROBLEMS GR (x, t; x , t ) c 2 τ G=0 G=0 x − c(t − t ) x x + c(t − t ) Figure 6.4: An illustration of the retarded Green’s Function.ﬁg6retGF The situation is illustrated in ﬁgure 6.4. This solution displays some interesting physical properties. • The function is zero for x < x − c(t − t ) and for x > x + c(t − t ), so it represents an expanding pulse. • The amplitude of the string is c/2τ , which makes sense since for a smaller string tension τ we expect a larger transverse amplitude.is this right? • The traveling pulse does not damp out since V = 0. 6.6.2 Physical Derivation10 Feb p5 We now explain how to get the solution from purely physical grounds. Consider an impulse ∆p applied at position x and time t . 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 1 ∆vy = ∆py /∆m = 2 /σdx since ∆py = 1/2 and ∆m = σdx. By substi- tuting dx = cdt, we ﬁnd that in the time dt a velocity ∆vy = 1/2cσdt is imparted to the string. This ∆vy is the velocity of the string portion at dx. By conservation of momentum, the previous string portion must now be stationary. In time dt the disturbance moves in the y direction 1 c an amount ∆y = vy dt = 2σc = 2τ . In these equalities we have used 2 the identity 1/c = σ/τ . Thus momentum is continually transferred from point to point (which satisﬁes the condition of conservation of12 Feb p1 momentum).
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6.7. SEMI-INFINITE STRING WITH FIXED END 976.7 Semi-Inﬁnite String with Fixed EndWe now consider the problem of an inﬁnite string with one end ﬁxed.We will get the same form of Green’s function. The deﬁning equationis (c.f. 6.19) ∂2 ∂2 −τ + σ 2 GR (x, t; x , t ) = δ(x − x )δ(t − t ) (6.22) ∂x2 ∂t for −∞ < t, t < ∞; 0 < x, x < ∞with the further condition eq6qdblst GR (x, t; x , t ) = 0 for x = 0. (6.23)This is called the Dirichlet boundary condition. We could use transform eq6qdblstbcmethods to solve this problem, but it is easier to use the method ofimages and the solution 6.21 to the inﬁnite string problem. To solve this problem we consider an inﬁnite string with sources at 12 Feb p2x and −x . This gives us a combined force σf (x, t) = [δ(x − x ) − δ(x + x )]δ(t − t )and the principle of superposition allows us to write the solution of theproblem as the sum of the solutions for the forces separately: c |x − x | |x + x | GR (x, t; x , t ) = θ t−t − −θ t−t − 2τ c c (6.24)where u is the solution (c.f. 6.21) of the inﬁnite string with sources at eq6qsstx and x . This solution is shown in ﬁgure 6.5. Since u satisﬁes 6.22 ﬁg6LandGand 6.23, we can identify GR = u for x ≥ 0. The case of a ﬁnite stringleads to an inﬁnite number of images to solve (c.f., section 8.7).6.8 Semi-Inﬁnite String with Free EndWe now consider a new problem, that of a string with two free ends.The free end Green’s function is c |x − x | |x + x | GR = θ t−t − +θ t−t − . 2τ c c
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98 CHAPTER 6. DYNAMIC PROBLEMS GR GR x −x E −x E x E E x x (a) GR at time t (a) GR at time t Figure 6.5: GR at t1 = t + 1 x /c and at t2 = t + 3 x /c. 2 2 This satisﬁes the equation ∂2 ∂2 −τ + σ 2 GR (x, t; x , t ) = δ(x − x )δ(t − t ) (6.25) ∂x2 ∂t for −∞ < t, t < ∞; a < x, x < beq6qdblst2 with the boundary condition d GR (x, t; x , t ) =0 dx x=0 which corresponds to κa = 0 and ha = 0 is equation 6.3 (c.f., section 1.3.3).12 Feb p3 The derivative of GR (x = 0) is always zero. Note that b d 1 for a < 0 < b θ(x) = θ(b) − θ(a) = (6.26) a dx 0 otherwise. which implies d θ(x) = δ(x). dx But for any ﬁxed t − t we can chose an such that the interval [0, ] isask Baker is ﬂat. Therefore d free end G = 0. dx R
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6.9. ELASTICALLY BOUND SEMI-INFINITE STRING 99Notes about the physics: For a string with a free end, the force on theend point is Fy = τ dG = 0 at x = 0 which implies dG = 0 at x = 0 dx dxif the tension τ does not vanish. If the tension does vanish at x = 0,then we have a singular point at the origin and do not restrict dG = 0 dxat x = 0.6.9 Elastically Bound Semi-Inﬁnite StringWe now consider the problem with boundary condition d − + κ GR = 0 for x = 0. dxThe solution can be found using the standard transform method. Doan 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 thisproblem is done in problem 4.3.6.10 Relation to the Eigen Fn ProblemWe now look at the relation between the general problem and the eigenfunction problem (normal modes and natural frequencies). The normalmode problem is used in solving [L0 − λσ]G(x, x ; λ) = δ(x − x ) + RBCfor which G(x, x ; λ) has poles at the eigen values of L0 . We found inchapter 4 that G(x, x ; λ) can be written as a bilinear summation un (x)u∗ (x ) n G(x, x ; λ) = (6.27) n λn − λwhere the un (x) solve the normal mode problem: eq6qA [L0 − λn σ]un (x) = 0 + RBC. (6.28) 2Here we made the identiﬁcation λn = ωn where the ωn ’s are the natural eq6qE
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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 )e−iωt is u(x, t) = G(x, x ; λ = ω 2 + iε)e−iωt . The non-steady state response is GR (x, t; x , t ) which is given by dω −iω(t−t ) ˜ GR (x, t; x , t ) = e G(x, x ; ω) 2π ˜ where G(x, x ; ω) = G(x, x ; λ = ω 2 ). Plug G(x, x ; λ) (from equation12 Feb p4 6.27) into the Fourier transformed expression (equation 6.18). This gives 1 −iω(t−t ) un (x)u∗ (x ) n GR (x, t; x , t ) = dωe (6.29) 2π L n λn − ω 2eq6qC In this equation ω can be arbitrarily complex. (This equation is very diﬀerent (c.f. section 4.6) from the steady state problem G(x, x ; λ = ω 2 + iε)e−iωt where ω was real.) Note that we are only interested in t > t , since we have shown already that GR = 0 for t < t .This is back- Now we add the lower contour since e−iω(t−t ) is small for t > t andwards 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 = ω 2 , or ω = ± λn . We now perform an evaluation of the integral for one of the terms12 Feb p5 of the summation in equation 6.29. dω e−iω(t−t ) dω e−iω(t−t ) = − √ √ (6.30) L+LLHP 2π λn − ω 2 L+LLHP 2π (ω − λn )(ω + λn ) √ √ −i λn (t−t ) 2πi e ei λn (t−t ) = √ − √ 2π 2 λn 2 λn √ sin λn (t − t ) = √ . λneq6qFa By equation 6.29 we get √ sin λn (t − t ) GR (x, t; x , t ) = un (x)u∗ (x n ) √ (6.31) n λn √eq6qF where λn = ωn and t > t . This general solution gives the relationship between the retarded Green’s function problem (equation 6.17) and the eigen function problem (eq. 6.28).
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6.10. RELATION TO THE EIGEN FN PROBLEM 1016.10.1 Alternative form of the GR ProblemFor t − t small, eq. 6.31 becomes un (x)u∗ (x ) GR (x, t; x , t ) ∼ √ n λn (t − t ) n λn = (t − t ) un (x)u∗ (x ) n n δ(x − x ) = (t − t ) . σ(x)Where we used the completeness relation 4.113. Thus for t − t small, 12 Feb 6the GR has the form δ(x − x ) GR (x, t; x , t )|t→t ∼ (t − t ) . σ(x)Diﬀerentiating, this equation gives ∂ δ(x − x ) GR (x, t; x , t ) = . ∂t t→t + σ(x)Also, as t approaches t from the right hand side GR (x, t; x , t ) = 0 for t → t − .These results allow us to formulate an alternative statement of the GRproblem in terms of an initial value problem. The GR is speciﬁed bythe following three equations: ∂2 L0 + σ(x) GR (x, t; x , t ) = 0 for t > t + RBC ∂t2 GR (x, t; x , t ) = 0 for t = t ∂ σ(x) GR = δ(x − x ) for t = t ∂t ∂Here σ(t) ∂t GR (x, t; x , t ) represents a localized unit of impulse at x , t(like ∆p = 1). Thus we have the solution to the initial value problemfor which the string is at rest and given a unit of momentum at t .
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102 CHAPTER 6. DYNAMIC PROBLEMS We have now cast the statement of the GR problem in two forms, as a retarded boundary value problem (RBVP) and as an initial value problem (IVP): ∂ 2 L0 + σ ∂t2 GR (x, t; x , t ) = δ(x − x )δ(t − t ) + RBC, RBVP = GR (x, t; x , t ) = 0 for t < t L + σ(x) ∂ 2 G (x, t; x , t ) = 0 for t > t , RBC, 0 ∂t2 R IVP = GR (x, t; x , t ) = 0 for t = t , σ(x) ∂ G = δ(x − x ) for t = t . ∂t R 6.11 Comments on Green’s Function17 Feb p1 6.11.1 Continuous Spectra In the previous section we obtained the spectral expansion for discrete eigenvalues: ∞ un (x)u∗ (x ) GR (x, t; x , t ) = √ n sin λn (t − t ) (6.32) n=1 λneq6rst This gives us an expansion of the Green’s function in terms of the natural frequencies. For continuous spectra the sum is replaced by an integral √ α ∗α sin λn (t − t ) GR = dλn uλn uλn √ α 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 d − + κa GR = 0 for x = a. dx
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6.11. COMMENTS ON GREEN’S FUNCTION 103If κa → 0 (Neumann boundary condition) then the boundary conditionfor the normal mode problem will be (d/dx)u(x) = 0 which will havea constant solution, i.e., λ1 = 0. We cannot substitute λ1 = 0 into Ask whyequation 6.32, but instead must take the limit as λ1 approaches zero.Physically, this corresponds to taking the elasticity κa as a small quan-tity, and then letting it go to zero. In this case we can write equation6.32 with the λ1 eigen value separated out: λ →0 GR (x, t; x , t ) −→ u1 (x)u∗ (x )(t − t ) 1 1 (6.33) un (x)u∗ (x ) + √ n sin λn (t − t ) n λn t→∞ −→ u1 (x)u∗ (x )(t − t ). 1The last limit is true because the sum oscillates in t. The Green’sfunction represents the response to a unit momentum, but κa = 0which means there is no restoring force. Thus a change in momentum∆p = 1 is completely imparted to the string, which causes the stringto acquire a constant velocity, so its amplitude increases linearly withtime. Note that equation 6.33 would still be valid if we had taken λ1 = 0in our derivation of the Green’s function as a bilinear sum. In thiscase equation 6.30 would have a double pole for λ1 = 0, so the residuewould involve the derivative of the numerator, which would give thelinear factor of t − t . Consider a string subject to an arbitrary force σ(x)f (x, t). Remember 17 Feb p2that σ(x)f (x, t) = δ(t − t )δ(x − x ) gives GR (x, t; x , t ). A general forceσ(x)f (x, t) gives a response u(x, t) which is a superposition of Green’s pr:GenResp1functions: t b u(x, t) = dt dx G(x, t; x , t )σ(x )f (x , t ) 0 a dwith no boundary terms (u(x, 0) = 0 = dt u(x, 0)). Now plug in theGreen’s function expansion 6.33 to get t bu(x, t) = u1 (x) dt (t − t ) dx u∗ (x )σ(x )f (x , t ) 1 0 a ∞ t b un (x) + √ dt sin λn (t − t ) dx u∗ (x )σ(x )f (x , t ). n n=2 λn 0 a
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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) onto u∗ (x). In the λ1 = 0 case the u1 (x) term n is constant. 6.11.3 Zero Net Force Now let F (t) ≡ (const.) dt σ(x )f (x , t ) where F (t ) represents the total applied force at time t . If F (t ) = 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- ask baker latory; there is no growth or decay.about last partnot includedhere 6.12 Summary 1. The retarded Green’s function GR solves ∂2 L0 + σ GR (x, t; x , t ) = δ(x − x )δ(t − t ) + RBC ∂t2 for a < x, x < b; all t, t . with the condition GR (x, t; x , t ) = 0 for t < t . The advanced Green’s function GA solves the same equation, but with the condition GA (x, t; x , t ) = 0 for t > t . 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 at a point x along the string. The advanced Green’s function gives the initial motion of the string such that a unit of momentum applied at x , t causes it to come to rest.
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6.13. REFERENCES 105 2. The retarded Green’s function can be written in terms of the steady state Green’s function: dω −iω(t−t ) ˜ GR (x, t; x t ) = e G(x, x ; ω). L 2π 3. The retarded Green’s function for an inﬁnite string with σ and τ constant, and V = 0 is c |x − x | GR (x, t; x , t ) = θ t−t − . 2τ c 4. The retarded Green’s function for a semi-inﬁnite string with a ﬁxed end, σ and τ constant, and V = 0 is c |x − x | |x + x | GR (x, t; x , t ) = θ t−t − −θ t−t − . 2τ c c 5. The retarded Green’s function for a semi-inﬁnite string with a free end, σ and τ constant, and V = 0 is c |x − x | |x + x | GR = θ t−t − +θ t−t − . 2τ c c 6. The retarded Green’s function can be written in terms of the eigen functions as √ ∗ sin λn (t − t ) GR (x, t; x , t ) = un (x)un (x ) √ . n λn6.13 ReferencesA good reference is [Stakgold67b, p246ﬀ]. This material is developed in three dimensions in [Fetter80, p311ﬀ].
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Chapter 7Surface Waves andMembranes 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 IntroductionIn this chapter we formulate physical problems which correspond toequations involving more than one dimension. This serves to moti-vate the mathematical study of N -dimensional equations in the nextchapter. 107
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108 CHAPTER 7. SURFACE WAVES AND MEMBRANES ¨ ¨ ¨ ¨ h(x) ¨ ¨ ¨ ¨ & x& b(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 .pr:surf1 This situation is represented in ﬁgure 7.1. The height of equilibriumﬁg:7.1 is h(x) and the width of the channel is b(x). The height of the wavepr:hww1 z(x, t) can then be written as z(x, t) = h(x) + u(x, t) where u(x, t) is the deviation from equilibrium. We now assume the shallow wave case u(x, t) h(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)pr:lambda2 λ where λ is the wavelength. In this case the motion of the water 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 ∂ ∂ ∂2 − gS(x) u + b(x) 2 u(x, t) = 0, ∂x ∂x ∂t 1 This material corresponds to FW p. 357–363.
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7.3. TWO DIMENSIONAL PROBLEMS 109which is equivalent to the 1-dimensional string, where σ(x) ⇒ b(x) and 17 Feb p4τ (x) ⇒ gS(x). Consider the case in which b(x) is independent of x: ∂ ∂ ∂2 − gh(x) u + 2 u(x, t) = 0. (7.1) ∂x ∂x ∂tThis corresponds to σ = 1 and τ = gh(x). eq7shallow Propagation of shallow water waves looks identical to waves on astring. For example, in problem 3.5, h(x) = x gives the Bessel’s equa-tion, with the identiﬁcation τ (x) = x and V (x) = m2 /x. As another example, take h(x) to be constant. This gives us wavepropagation with c = τ /σ, τ = gh, and σ = 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 thanthe trough.7.3 Two Dimensional Problems 17 Feb p5 2We now look at the 2-dimensional problem, that of an elastic membrane .We denote the region of the membrane by R and the perimeter (1- pr:elmem1dimensional “surface”) by S. The potential energy diﬀerential for anelement of a 1-dimensional string is 2 1 du dU = τ (x) dx. 2 dxIn the case of a 2-dimensional membrane we replace u(x) with u(x, y) =u(x). In this case the potential energy diﬀerence is (see section 2.4.2) 2 2 1 du du dU = τ (x, y) + dxdy (7.2) 2 dx dy 1 = τ (x)( u)2 dx (7.3) 2 2 This is discussed on p. 271–288 of FW.
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110 CHAPTER 7. SURFACE WAVES AND MEMBRANES where τ is the tension of the membrane. Note that there is no mixed d d term dx dy u since the medium is homogeneous. The total potential energy is given by the equation 1 U= dx τ (x)( u)2 , R 2 where τ (x) is the surface tension. In this equation dx = dx dy and ( u)2 = ( u) · ( u). The total kinetic energy is 2 1 1 ∂u T = mv 2 = dx σ(x) . 2 R 2 ∂t 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 V (x)u(x, t)2 . We 2 also add an additional force which will add to the potential energy: force mass f (x, t)σ(x)dxu(x, t) = (length) (displacement) . mass lengthpr:lagr1 The Lagrangian is thus 1 L= dx× 2 R 2 ∂ σx u − τ (x)( u)2 − V (x)u(x)2 − f (x, t)σ(x)dxu(x, t) . ∂t Notice the resemblance of this Lagrangian to the one for a one dimen-19 Feb p1 sional string (see section 2.4.2).19 Feb p2 We apply Hamiltonian Dynamics to get the equation of motion: ∂2 L0 + σ u(x, t) = σ(x)f (x, t) ∂t2 where L0 = − (τ (x) ) − V (x). This is identical to the equation of motion for a string except now in two dimensions. It is valid everywhere for x inside the region R.
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7.3. TWO DIMENSIONAL PROBLEMS 111 (0, b) (a, b) (0, y) r r(a, y) (0, 0) (a, 0) Figure 7.2: The rectangular membrane. 7.3.1 Boundary Conditionspr:bc2 Elastically Bound Surface The most general statement of the boundary condition for an elastically bound surface is [ˆ · n + κ(x)]u(x, t) = h(x, t) for x on S. (7.4) In this equation the “surface” S is the perimeter of the membrane, n is sone ˆ 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 We now consider the case of a rectangular membrane, illustrated in pr:pbc2 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 ∂2 u(x, y) = u(x, y) for 0 ≤ y ≤ b, ∂x2 x=0 ∂x2 x=a
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112 CHAPTER 7. SURFACE WAVES AND MEMBRANES ∂2 ∂2 u(x, y) = u(x, y) for 0 ≤ x ≤ a. ∂x2 y=0 ∂x2 y=b In this case we can consider the region R to be a torus. 7.4 Example: 2D Surface Waves19 Feb p4 We now give one last 2-dimensional example3 . We consider a tank of 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), with λ h(x) for the shallow water case and u h(x). Thus (using 7.1) our equation of motion is ∂2 − · (gh(x) ) + u(x, t) = f (x, t). ∂t2 Note that in this equation we have σ = 1. For this problem we take the Neumann natural boundary condition: n· ˆ u(x, t) = 0 and ∂ u⊥ = −g ⊥ u|S ∂t for x on S. 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 and τ (x) = gh(x).19 Feb p5 The formula for all these problems is just ∂2 L0 + σ(x) 2 u(x, t) = σ(x)f (x, t) for x in R ∂t where L0 = − (τ (x) ) + V (x). (Note that τ (x) is not necessarily tension.) We let x = (x1 , . . . , xn ) and = (∂/∂x1 , . . . , ∂/∂xn ). The boundary conditions can be elastic or periodic. 3 This one comes from FWp. 363–366.
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7.5. SUMMARY 1137.5 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 ∂2 L0 + σ u(x, t) = σ(x)f (x, t) ∂t2 where L0 = − (τ (x) ) − V (x), subject to either an elastic boundary condition, [ˆ · n + κ(x)]u(x, t) = h(x, t) for x on S, or a periodic boundary condition.7.6 ReferencesThe material on surface waves is covered in greater depth in [Fetter80,p357ﬀ], while the material on membranes can be found in [Fetter80,p271].
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Chapter 8Extension 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 22 Feb p1 case. • Write the solution for the N -dimensional problem in terms of the Green’s function. • Describe the method of images.8.1 IntroductionIn the previous chapter we obtained the general equation in two dimen-sions: ∂2 L0 + σ(x) 2 u(x, t) = σ(x)f (x, t) for x in R (8.1) ∂twhere eq8usf L0 = − · τ (x) + V (x).This is immediately generalizable to N -dimensions. We simply letx = (x1 , . . . , xn ) and = (∂/∂x1 , . . . , ∂/∂xn ). and we introduce the 115
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116 CHAPTER 8. EXTENSION TO N -DIMENSIONS notation n · u ≡ ∂u/∂n. R is now a region in N -dimensional space ˆ and S is the (N − 1)-dimensional surface of R.pr:bc3 The boundary conditions can either be elastic or periodic: 1. Elastic: The equation for this boundary condition is [ˆ · n + K(x)]u(x, t) = h(x, t) for x on S. (8.2)eq8.1b The term K(x) is like a spring constant which determines the 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).pr:pbc3 2. Periodic: For the two dimensional case the region looks like 7.2 and the periodic boundary conditions are 7.5 and following. In the N -dimensional case the region R is an N -cube. Connecting matching periodic boundaries of S yields an N -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 x in R (8.3) ∂ (x, t)|t=0 = u1 (x) for x in R. (8.4) ∂t 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.pr:intprob1 1. Interior problem: Here R is enclosed in a ﬁnite region bounded by 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.
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8.3. EXAMPLES OF N -DIMENSIONAL PROBLEMS 117pr:extprob1 2. Exterior problem: Here R extend to inﬁnity in all directions but is 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: Here R extends 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 ; λ = ω 2 ) = G(x, x ; λ) → GR (x, t; x , t ) retarded → u(x, t) General Response. 8.3.2 Normal Mode Problem pr:NormMode3 The normal mode problem is given by the homogeneous diﬀerential equation ∂2 L0 + σ(x) 2 u(x, t) = 0. ∂t Look for solutions of the form u(x, t) = e−iωn t un (x). √ The natural frequencies are ωn = λn , where λn is an eigen value. The normal modes are eigen functions of L0 . Note: we need RBC to ensure that L0 is Hermitian.
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118 CHAPTER 8. EXTENSION TO N -DIMENSIONS 8.3.3 Forced Oscillation Problem24 Feb p2pr:fhop2 The basic problem of steady state oscillation is given by the equation ∂2 L0 + σ u(x, t) = e−iωt σ(x)f (x) for x ∈ R ∂t2 and (for example) the elastic boundary condition (ˆ · n + K)u = h(x)e−iωt for x ∈ S. We look for steady state solutions of the form u(x, t) = e−iωt u(x, ω). The value of ω is chosen, so this is not an eigen value problem. We assert u(x, ω) = dx G(x, x ; λ = ω + iε)σ(x )f (x ) x ∈R + dx τ (x )G(x, x ; λ = ω + iε)σ(x)h(x ). x ∈S 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 ), we have u(x, t) = e−iωt G(x, x ; λ = ω 2 ). 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 S, L0 u = dxS ∗ (x)L0 u(x). (8.5)eq8tst
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8.5. THE RETARDED PROBLEM 119 8.4.1 Green’s First Identitypr:G1Id2 The derivation here generalizes the derivation given in section 2.1. S, L0 u = dxS ∗ (x)[− · (τ (x) ) + V (x)]u(x) (8.6) R integrate 1st term by parts = dx[− · (S ∗ τ (x) u) + ( S ∗ )τ (x) u + S ∗ V u] R integrate 1st term using Gauss’ Theorem = − dS n · (S ∗ τ (x) u) + ˆ [S ∗ V u + ( · S ∗ τ (x) u)]dx S R This is Green’s First Identity generalized to N -dimensions: S, L0 u = − dS n·(S ∗ τ (x) u)+ ˆ [S ∗ V u+( ·S ∗ τ (x) u)]dx. (8.7) S R Compare this with 2.3. 8.4.2 Green’s Second Identity pr:G2Id2 We now interchange S and u. In the quantity S, L0 u − L0 S, u the 22 Feb p3 symmetric terms will drop out, i.e., the second integral in 8.7 is can- celled. We are left with S, L0 u − L0 S, u = dS n · [−S ∗ τ (x) u + uτ (x) S ∗ ]. ˆ S 8.4.3 Criterion for Hermitian L0 pr:HermOp2 If u, S ∗ satisfy the RBC, then the surface integral in Green’s second identity vanishes. This leaves S, L0 u − L0 S, u , 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
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120 CHAPTER 8. EXTENSION TO N -DIMENSIONS 8.3, and 8.4. We look again at GR which is the response of a system to a unit force: ∂2 L0 + σ(x) 2 GR (x, t; x , t ) = δ(x − x )δ(t − t ) for x, x in R. ∂t (8.8)eq8tB We also require the retarded Green’s function to satisfy RBC and the22 Feb p4 initial condition GR = 0 for t < t . We now use the result from problem 4.2: t u(x, t) = dx dt GR (x, t; x , t )σ(x )f (x , t ) x ∈R 0 t + dx τ (x ) dt GR (x, t; x , t )h(x , t ) x ∈S 0 + dx σ(x ) GR (x, t; x , 0)u1 (x ) x ∈R ∂ − GR (x, t; x , t )u0 (x ) (8.9) ∂teq8ugen The ﬁrst line gives the volume sources, the second gives the surfaceR Horn says sources, and the third and fourth gives the contribution from the initial ˙eval GR at x conditions. Since the deﬁning equations for GR are linear in volume,’=0 surface, and initial condition terms, we were able to write down the ask Baker solution u(x, t) as a linear superposition of the GR .about limits Note that we recover the initial value of u(x, t) in the limit t → t . This is true since in the above equation we can substitute lim GR (x, t; x , t ) = 0 t→t and δ(x − x ) lim GR (x, t; x , t ) = . t→t σ(x ) 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 that GR could by determined by using a Fourier Transform. Here we follow the same procedure generalized to N -dimensions. The Fourier transform of GR in N -dimensions is dω −iω(t−t ) ˜ GR (x, t; x , t ) = e G(x, x ; ω) L 2π
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8.5. THE RETARDED PROBLEM 121where L is a line in the upper half plane parallel to the real axis (c.f., 22 Feb p5 ˜section 6.4), since G is analytic in the upper half plane (which is dueto the criterion GR = 0 for t < t ). Now the problem is simply to evaluate the Fourier Transform. Bythe same reasoning in section 6.3 the Fourier transform of GR is iden-tical to the Green’s function for the steady state case: ˜ G(x, x ; ω) = G(x, x; λ = ω 2 ).Recall that the Green’s function for the steady state problem satisﬁes [L0 − λσ]G(x, x ; λ) = δ(x − x ) for x, x ∈ R, RBC (8.10)We have reduced the general problem in N dimensions (equation8.1)to the steady state Green’s function problem in N -dimensions. 22 Feb p68.5.3 Reduction to Eigenvalue Problem pr:efp3The eigenvalue problem (i.e., the homogeneous equation) in N -dimensionsis L0 un (x) = λn σun (x) x ∈ R, RBC (8.11)The λn ’s are the eigen values of L0 . Since L0 is hermitian, the λn ’s Lots of workare real. The un (x)’s are the corresponding eigenfunctions of L0 . Wecan also prove orthonormality (using the same method as in the singledimension case) dx uα∗ (x)uα (x) = 0 if λn = λm . n m R αThis follows from the hermiticity of L0 . Note that because we are nowin N -dimensions, the degeneracy may now be inﬁnite. 22 Feb p7 By using the same procedure as in chapter 4, we can write a solutionof eq. 8.8 expanded in terms of a solution of 8.11: un (x)u∗ (x ) n G(x, x ; λ) = . n λn − λNote 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
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122 CHAPTER 8. EXTENSION TO N -DIMENSIONS representation δ(x − x ) = σ(x ) un (x)u∗ (x ), n n which is also called the completeness relation. All we have left is to24 Feb p1 discuss the physical interpretation of G. 8.6 Region R24 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 un (x)u∗ (x ) n G(x, x ; λ = ω 2 ) = n λn − ω 2 8.6.2 Exteriorpr:extprob2 For the exterior problem the sums become integrals and we have a continuous spectrum: α un (x)u∗ (x ) n G(x, x ; λ) = dλn . λn λn − λ In this case we take λ = ω 2 + iε. G now 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 λ axis from above or below. We choose ε > 0 to correspond to the physical out going wave ask Baker solution.about omittedmaterial24 Feb p4 8.7 The Method of Images24 Feb p5 We now present an alternative method for solving N -dimensional prob-pr:MethIm1 lem which is sometimes useful when the problem exhibits suﬃcient
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8.7. THE METHOD OF IMAGES 123symmetry. It is called the Method of Images. For simplicity we con-sider a one dimensional problem. Consider the GR problem for periodicboundary conditions with constant coeﬃcients. ∂2 L0 − λσ GR = δ(x − x )δ(t − t ) 0 ≤ x ≤ l. ∂t28.7.1 Eigenfunction MethodWe have previously solved this problem by using an eigen functionexpansion solution (equation 6.31) un (x)u∗ (x ) GR = √ n sin λn (t − t ). n λnFor this problem the eigen functions and eigen values are 1/2 1 un (x) = e2πinx/l n = 0, ±1, ±2, . . . l 2 2πin λn = n = 0, ±1, ±2, . . . l8.7.2 Method of ImagesThe method of images solution uses the uniqueness theorem. Put im-ages over −∞ to ∞ in region of length λ. c |x − x | Φ= θ t−t − . 2τ cThis is not periodic over 0 to l. Rather, it is over all space. Our GR is ∞ c |x − x − nl| GR = θ t−t − . 2τ n=−∞ cNotice that this solution satisﬁes ∞ ∂2 L0 + σ GR = δ(x − x − nl) − ∞ < x, x < ∞. ∂t2 n=−∞
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124 CHAPTER 8. EXTENSION TO N -DIMENSIONSHowever, we only care about 0 < x < l ∞ ∂2 L0 + σ GR = δ(x − x − nl) 0 < x, x < l. ∂t2 n=−∞Since the other sources are outside the region of interest they do notaﬀect 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 S, L0 u = − dS n ·(S ∗ τ (x) u)+ ˆ [S ∗ V u+( ·S ∗ τ (x) u)]dx, S R S, L0 u − L0 S, u = dS n · [−S ∗ τ (x) u + uτ (x) S ∗ ]. ˆ s 3. The solution for the N -dimensional problem in terms of the Green’s function is t u(x, t) = dx dt GR (x, t; x , t )σ(x )f (x , t ) x ∈R 0 t + dx τ (x ) dt GR (x, t; x , t )h(x , t ) x ∈S 0 + dx σ(x ) GR (x, t; x , 0)u1 (x ) x ∈R ∂ − GR (x, t; x , t )u0 (x ) . ∂t
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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 ReferencesThe method of images is covered in most electromagnetism books, forexample [Jackson75, p54ﬀ], [Griﬃths81, p106ﬀ]; a Green’s functionapplication is given in [Fetter80, p317]. The other material in thischapter is a generalization of the results from the previous chapters.
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Chapter 9Cylindrical 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. 29 Feb p1 • 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.9.1 IntroductionIn the previous chapter we considered the Green’s function equation (L0 − λσ(x))G(x, x ; λ) = δ(x − x ) for x, x ∈ Rwhere L0 = − · (τ (x) ) + V (x)subject to RBC, which are either for the elastic case (ˆ · n + κ(S))G(x, x ; λ) = 0 for x ∈ S (9.1) 127
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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 =ˆ +ˆ ix jy.pr:CartCoord1 Instead of the coordinate pair (x, y) we may choose polar coordinates (r, φ). The transformation to cartesian coordinates is x = r cos φ y = r sin φ while the transformation to polar coordinates is (for tan φ deﬁned onpr:r1 the interval −π/2 < φ < π/2) r= x2 + y 2 tan−1 (y/x) for x > 0, y > 0 φ = tan−1 (y/x) + π for x < 0 tan−1 (y/x) + 2π for x > 0, y < 0 A diﬀerential of area for polar coordinates is related by that for carte-pr:jak1 sian coordinates by a Jacobian (see Boas, p220): dxdy = dA x, y = J drdφ r, φ ∂(x, y) = drdφ ∂(r, φ) (∂x/∂r) (∂x/∂φ) = drdφ (∂y/∂r) (∂y/∂φ) cos φ −r sin φ = drdφ sin φ r cos φ = rdrdφ
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9.1. INTRODUCTION 129By expanding dx and dy in terms of dr and dφ, we can write thediﬀerential of arc length in polar coordinates (see Boas, p224) ds = dx2 + dy 2 = dr2 + r2 dφ2 .The diﬀerential operator becomes (see Boas, p252,431) pr:grad1 ∂u ˆ 1 ∂u gradient u=r ˆ +φ , ∂r r ∂φ 1 ∂ 1 ∂ divergence ·B= (rBr ) + (Bφ ). r ∂r r ∂φLet B = τ (x) x, then 1 ∂ ∂u 1 ∂ ∂u · (τ (x) u(x)) = rτ (x) + τ (x) . (9.2) r ∂r ∂r r ∂φ ∂φ9.1.2 Delta Function 29 Feb p2The N -dimensional δ-function is deﬁned by the property pr:DeltaFn2 f (x) = d2 xf (x)δ(x − x ).In polar form we have (since dxdy = rdrdφ) f (x) = f (r, φ) = r dr dφ f (r , φ )δ(x − x ).By comparing this with f (r, φ) = dr dφ f (r , φ )δ(r − r )δ(φ − φ )we identify that the delta function can be written in polar coordinatesin the form pr:delta1 δ(r − r ) δ(x − x ) = δ(φ − φ ). r 29 February 1988
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130 CHAPTER 9. CYLINDRICAL PROBLEMS ˆ S φt φQ ˆ # r t n ˆ a R "! Figure 9.1: The region R as 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), and V (x) = V (r). Thus 9.2 becomes 1 ∂ ∂u τ (r) ∂ 2 u (τ (r) u(x)) = rτ (r) + 2 . r ∂r ∂r r ∂φ2 The equation for the Green’s function 1 (L0 − λσ)G(r, φ; r , φ ) = δ(r − r )δ(φ − φ ) r becomes (for r, φ ∈ R) 1 ∂ ∂ τ (r) ∂ 2 1 − rτ (r) − 2 ∂φ2 + V (r) − λσ(r) G = δ(r−r )δ(φ−φ ). r ∂r ∂r r r (9.3) Here R may 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 R to be the interior of a circle of radius a (see ﬁgure 9.1) and apply the elastic boundary condition 9.1.ﬁg10a We now deﬁne the elasticity on the boundary S, κ(S) = κ(φ). We must further specify κ(φ) = κ, a constant, since if κ = κ(φ), then we might not have cylindrical symmetry. Cylindrical symmetry implies n · = ∂/∂r so that the boundary condition ˆ (ˆ · n + κ(S))G(r, φ; r , φ ) = 0 for r = a is now ∂ + κ G(r, φ; r , φ ) = 0 for r = a. ∂r
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9.3. EXPANSION IN TERMS OF EIGENFUNCTIONS 131We also need to have G periodic under φ → φ + 2π. So 29 Feb p3 G(r, 0; r , φ ) = G(r, 2π; r , φ )and ∂G ∂G = . ∂φ φ=0 ∂φ φ=2πWe have now completely respeciﬁed the Green’s function for the caseof cylindrical symmetry.9.3 Expansion in Terms of EigenfunctionsSince the Green’s function is periodic in φ and since φ only appears inthe operator as ∂ 2 /∂φ2 , we use an eigenfunction expansion to separateout the φ-dependence. Thus we look for a complete set of eigenfunctions pr:efexp1which solve ∂2 − 2 um (φ) = µm um (φ) (9.4) ∂φfor um (φ) periodic. The solutions of this equation are 1 um (φ) = √ eimφ for m = 0, ±1, ±2, . . . 2πand the eigenvalues are µm = m 2 for m = 0, ±1 ± 2, . . ..(Other types of regions would give diﬀerent eigenvalues µm ). Since pr:CompRel4this set of eigenfunctions is complete it satisﬁes the expansion δ(φ − φ ) = um (φ)u∗ (φ ). m m9.3.1 Partial ExpansionWe now want to ﬁnd Gm (r, r ; λ) which satisﬁes the partial expansion(using the principle of superposition) pr:partExp1 G(r, φ; r , φ ) = um (φ)Gm (r, r ; λ)u∗ (φ ). m (9.5) m
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132 CHAPTER 9. CYLINDRICAL PROBLEMS We plug this and 9.4 into the partial diﬀerential equation 9.3: ask Baker where thi 1 d d µm τ comes from um (φ) − r2 + 2 + V (r) − λσ(r) Gm u∗ (φ ) m m r dr dr r 29 Feb p4 1 = δ(r − r ) um (φ)u∗ (φ ) m r mpr:rlo1 We now deﬁne the reduced linear operator d d µm τ (r) Lµm ≡ rL0 = − 0 rτ (r) +r + V (r) (9.6) dr dr r2eq9rLo so Gm (r, r ; λ) must satisfy (Lµm − λrσ(r))Gm (r, r ; λ) = δ(r − r ), 0 for 0 < r, r < a and the boundary condition ∂ + κ Gm (r, r ; λ) = 0 for r = a, 0 < r < a. ∂r Comments on the eigenvalues µm : The RBC will always lead to µn > 0. If µm < 0 then the term µm τ (r)/r2 in 9.6 would act like an attractive sink and there would be no stable solution. Since µm > 0,29 Feb p5 this term instead looks like a centrifugal barrier at the origin. Note that the eﬀective “tension” in this case is rτ (r), so r = 0 is a singular point. Thus we must impose regularity at r = 0: |G(r = 0, r ; 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 − λrσ(r))Gm (r, r ; λ) = δ(r − r ), 0 for 0 < r, r < a, ∂ + κ Gm (r, r ; λ) = 0 for r = a, 0 < r < a, ∂r |G(0, r ; λ)| < ∞.
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9.4. EIGEN VALUE PROBLEM FOR L0 1339.4 Eigen Value Problem for L0To solve the reduced Green’s function problem which we have just ob-tained, we must solve the reduced eigen value problem pr:efp4 Lµm u(m) (r) = λ(m) rσ(r)u(m) (r) 0 n n n for 0 < r < a, du(m) n + κu(m) (r) = 0 n for r = a, dr |u(m) (r)| < ∞ n at r = 0.In these equation λ(m) is the nth eigenvalue of the reduced operator n (µ ) (µ )L0 m and u(m) (r) is the nth eigenfunction of L0 m . From the general ntheory of 1-dimensional problems (c.f., chapter 4) we know that u(m) (r)u∗(m) (r ) n n Gm (r, r ; λ) = (m) for m = 0, ±1, ±2, . . . n λn −λIt follows that (using 9.5) 29 Feb p6 u(m) (r)u∗(m) (r ) G(r, φ, r , φ ; λ) = um (φ) n (m) n u∗ (φ ) m m n λn −λ un (r, φ)u∗(m) (r (m) n ,φ ) = (m) n,m λn − λwhere u(m) (r, φ) = um (φ)u(m) (r). Recall that G satisﬁes (L0 − λσ)G = n nδ(x − x ). with RBC. Thus we can conclude L0 u(m) (r, φ) = λ(m) σ(r)u(m) (r, φ) n n n RBC.These u(m) (r, φ) also satisfy a completeness relation n ﬁgure this part out δ(x − x ) 29 Feb p7 u(m) (r, φ)u∗(m) (r , φ ) = n n m,n σ(x) δ(r − r )δ(φ − φ ) = r σ(r ) 2 Mar p1 The radial part of the Green’s function, Gm , may also be constructeddirectly if solutions satisfying the homogeneous equation are known,
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134 CHAPTER 9. CYLINDRICAL PROBLEMS where one of them also satisﬁes the r = 0 boundary condition and the other also satisﬁes the r = a boundary condition. The method from chapter 3 (which is valid for 1-dimensional problems) gives u1 (r< )u2 (r> ) Gm (r, r ; λ) = − rτ (r)W (u1 , u2 )τ (r)? where (Lµm − λσr)u1,2 = 0 0 |u1 | < ∞ at r = 0 ∂u2 + κu2 = 0 for r = a. ∂r The eﬀective mass density is rσ(τ ), the eﬀective tension is rτ (r), and the eﬀective potential is r(µm τ /r2 + V (r)). 9.5 Uses of the GF Gm(r, r ; λ)2 Mar p3 9.5.1 Eigenfunction Problempr:efp5 Once Gm (r, r ; λ) is known, the eigenvalues and normalized eigenfunc- tions can be found using the relation λ→λ (m) u(m) (r)u(m) (r ) n n Gm (r, r ; λ) ∼n (m) . λn −λ The eigen values come from the poles, the eigen functions come from the residues. 9.5.2 Normal Modes/Normal Frequenciespr:NormMode4 In the general problem with no external forces the equation of motion is homogeneous ∂2 L0 + σ u(x, t) = 0 + RBC. ∂t2this section is We look for natural mode solutions:still rough
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9.5. USES OF THE GF GM (R, R ; λ) 135 (m) u(x, t) = e−iωn t (m) un (r, φ).The natural frequencies are given by (m) (m) ωn = λn .The eigen functions (natural modes) are (cf section 9.4) u(m) (r, φ) = u(m) (r)um (φ). n nThe normal modes are (m) u(m) (x, t) = e−iωn n t (m) un (r, φ).The normalization of the factored eigenfunctions u(m) (r) and um (φ) is n a ∗(m) drrσ(r)u(m) (r)un n (r ) = δn,n for n = 1, 2, . . . 0 2π dφum (φ)u∗ (φ) = δm,m . m 0The overall normalization of the (r, φ) eigen functions is pr:normal3 2π a 2 Mar p4 ∗(m ) dφ dr(rσ(r))u(m) (r, φ)un n (r, φ) = δn,n δm,m 0 0or 2π a ∗(m ) rdrdφσ(r)u(m) (r, φ)un n (r, φ) = d3 xσ(x)u∗ (x)u(x) 0 0 R = δn,n δm,m .9.5.3 The Steady State Problem pr:sss2This is the case of a periodic driving force: ∂2 L0 + σ u(r, φ, t) = σf (r, φ)e−iωt ∂t2 ∂ + κ u(r, φ, t) = h(φ)e−iωt ∂r
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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) and h(φ). It is not necessary to have circularly symmetric forces. The solution is (using 9.5) u(r, φ) = um (φ) m a 2π × r σ(r )dr Gm (r, r ; λ = ω 2 + iε) dφ u∗ (φ )f (r , φ ) m 0 0 2π +Gm (r, a; λ = ω 2 + iε) adφ τ (a)u∗ (φ )f (r , φ ) . m 0 In this equation 02π dφ u∗ (φ )f (r , φ ) is the mth Fourier coeﬃcient of m the interior force f (r , φ ) and 02π adφ τ (a)u∗ (φ )f (r , φ ) is the mth m Fourier coeﬃcient of the surface force τ (a)h(φ ). 9.5.4 Full Time Dependence2 Mar p5pr:GR3 For the retarded Green’s function we have GR (r, φ, t; r , φ , t ) = um (φ)GmR (r, t, r , t )u∗ (φ ) m m where dω −iω(t−t ) GmR (r, t; r , t ) = e Gm (r, r ; λ = ω 2 ) L 2π 1 u1 (r< )u2 (r> ) Gm (r, r ; λ = ω 2 ) = − . rτ (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 ( → , δn,n → δ(n−n ), etc.,) hold. 9.6 The Wedge Problempr: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 R is now 0 < r < a, 0 < φ < γ and its boundary is formed by φ = 0, φ − 2π, and r = a.
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9.6. THE WEDGE PROBLEM 137 & & ee & & f & f & &γ f r Figure 9.2: The wedge.9.6.1 General CaseSee the ﬁgure 9.2. The angular eigenfunction equation is again 9.4: ﬁg10.3 ∂2 − um (φ) = µm um (φ) RBC. ∂φ2Note that the operator ∂ 2 /∂φ2 is positive deﬁnite by Green’s 1st iden-tity. The angular eigenvalues are completely determined by the angularboundary conditions. For RBC it is always true the µm > 0. This isphysically important since if it were negative, the solutions to 9.4 wouldbe real exponentials, which would not satisfy the case of periodic bound-ary conditions. The boundary condition is now (ˆ · n + κ)G = 0 x ∈ S.This is satisﬁed if we choose κ1 (r) = κ1 /r, κ2 (r) = κ2 /r, and κ3 (φ) =κ3 , with κ1 , κ2 ≥ 0. The boundary condition (ˆ · + κ)G = 0 becomes n ∂ − + κ1 G = 0 for φ = 0 ∂φ ∂ + κ2 G = 0 for φ = γ ∂φ ∂ + κ3 G = 0 for r = a, 0 < φ < γ. ∂rWe now choose the um (φ) to satisfy the ﬁrst two boundary conditions.The rest of the problem is the same, except that Lµm gives diﬀerent µm 0eigenvalues.
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138 CHAPTER 9. CYLINDRICAL PROBLEMS 9.6.2 Special Case: Fixed Sides2 Mar p6 The case κ1 → ∞ and κ2 → ∞ corresponds to ﬁxed sides. We thus have G = 0 for φ = 0 and φ = γ. So the um eigenvalues must satisfy ∂2 − u m = µm u m ∂φ2 and um = 0 for φ = 0, γ. The solution to this problem is 2 mπφ um (φ) = sin γ γ with 2 mπ µm = m = 1, 2, . . . . γ The case m = 0 is excluded because its eigenfunction is trivial. As ask Baker γ → 2π we recover the full circle case.why not γ →2π?2 Mar p7 9.7 The Homogeneous Membranepr:membrane1 Recall the general Green’s function problem for circular symmetry. By4 Mar p1 substituting the completeness relation for um (φ), our diﬀerential equa- tion becomes δ(r − r ) (L0 − λσ)G(x, x ; λ) = δ(x − x ) = um (φ)u∗ (φ ) m r m where 1 L0 um (φ) = Lµm um (φ), r 0 d d µm τ (r) Lµm = − 0 rτ (r) +r + V (r) . dr dr r24 Mar p2 We now consider the problem of a complete circle and a wedge.
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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 ; λ), we want tosolve the problem d d µm λr 1 − r + − 2 Gm (r, r ; λ) = δ(r − r ) dr dr r c τwith G = 0 and r = a, which corresponds to ﬁxed ends. This problemwas solved in problem set 3: π J√µm (r< λ/c2 ) √ Gm = J µm (r> λ/c2 )N√µm (a λ/c2 ) 2τ J√µm (a λ/c2 ) − J√µm (aλ/c2 )N√µm (r> λ/c2 ) . (9.7)Using 9.5, this provides an explicit solution of the full Green’s functionproblem. Now we consider the angular part, where we have γ = 2π, so 4 Mar p3 √that µm = ±m which means the angular eigenfunctions are the sameas for the circular membrane problem considered before: 1 um = √ eimφ for µm = m2 , m = 0, ±1, ±2, . . .. 2πThe total answer is thus a sum over both positive and negative m ∞ G(r, φ, r , φ ; λ) = um (φ)Gm (r, r ; λ)u∗ (φ ). m m=−∞ We now redo this with κ → ∞ and arbitrary γ. This implies thatthe eigen functions are the same as the wedge problem considered before 2 mπφ um (φ) = sin , γ γ 2 mπ µm = . γ
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140 CHAPTER 9. CYLINDRICAL PROBLEMS We now get Jmπ/γ (r λ/c2 ) and Nmπ/γ (r λ/c2 ). We also get the orig- Show why inal expansion for G: (new Bessel op ∞ G(r, φ; r , φ ; λ) = um (φ)Gm (r, r ; λ)u∗ (φ ) m m=1 9.7.1 The Radial Eigenvaluespr:efp6 The poles of 9.7 occur when J√µm (a λ/c2 ) = 0. We denote the nth zero of J√µm by x√µm n This gives us x √ µm n c 2 λmn = for n = 1, 2, . . . a where J√µm (x√µm ,n ) = 0 is the nth root of the µm Bessel function. To ﬁnd the normalized eigenfunctions, we look at the residues of λ→λ n u(m) (r)u(m) (r ) n n Gm −→ (m) . λn −λ4 Mar p4 We ﬁnd r 2 J√µm (x√µm n a ) um (r) = n . σa2 J√µm (x√µm n ) Thus the normalized eigen functions of the overall operator L0 u(m) (r, φ) = σλ(m) un (r, φ) n n are u(m) (r, φ) = u(m) (r)um (φ) n n where the the form of um (φ) depends on whether we are considering a wedge or circular membrane.
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9.8. SUMMARY 1419.7.2 The PhysicsThe normal mode frequencies are given by the radial eigenvalues (m) c ωm,n = λn = x√µ,n . aThe eigen values increase in two ways: √ n increases and as m increases. asFor small x (i.e., x 1), J√µm ∼ (x) µm which implies that for largerµm the rise is slower. As m increases, µm increases, so the ﬁrst root occurs at larger x. Aswe increase m, we also increase the number of angular nodes in eimφ orsin(mnφ/γ). This also increases the centrifugal potential. Thus ωm,2 is this true?increases with m. The more angular modes that are present, the moreangular kinetic energy contributes to the potential barrier in the radialequation. Now consider behavior with varying γ for a ﬁxed m. µm increases 4 Mar p5as we decrease γ, so that ωm,n increases. 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 throwninto 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. The δ-function for circular coordinates is δ(r − r ) δ(x − x ) = δ(φ − φ ). r 3. The Green’s function equation for circular coordinates is 1 ∂ ∂ τ (r) ∂ 2 u − rτ (r) − 2 + V (r) − λσ(r) G = δ(x − x ). r ∂r ∂r r ∂φ2
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142 CHAPTER 9. CYLINDRICAL PROBLEMS 4. The partial expansion of the Green’s function for the circular problem is G(r, φ; r , φ ) = um (φ)Gm (r, r ; λ)u∗ (φ ). m m 5. The radial Green’s function for circular coordinates satisﬁes (Lµm − λrσ(r))Gm (r, r ; λ) = δ(r − r ), 0 for 0 < r, r < a, where the reduced linear operator is d d µm τ (r) Lµm ≡ rL0 = − 0 rτ (r) +r + V (r) , dr dr r2 and the boundary condition ∂ + κ Gm (r, r ; λ) = 0 for r = a, 0 < r < a. ∂r9.9 ReferenceThe 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].
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Chapter 10Heat Conduction Chapter Goals: • Derive the conservation law and boundary condi- tions appropriate for heat conduction. 7 Mar p1 • Construct the heat equation and the Green’s func- tion equation for heat conduction. • Solve the heat equation and interpret the solution.10.1 Introduction pr:heat1 1We now turn to the problem of heat conduction. The following physi-cal parameters will be used: mass density ρ, speciﬁc heat per unit masscp , temperature T , and energy E. Again we consider a region R withboundary S and outward normal n. ˆ10.1.1 Conservation of EnergyThe speciﬁc heat, cp , gives the additional amount of thermal energywhich is stored in a unit of mass of a particular material when it’stemperature is raised by one unit: ∆E = cp ∆T . Thus the total energycan be expressed as Etotal = E0 + d3 xρcp T. R 1 The corresponding material in FW begins on page 408 143
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144 CHAPTER 10. HEAT CONDUCTION Diﬀerentiating with respect to time gives dE ∂T = d3 xρcp . dt R ∂t There are two types of energy ﬂow: from across the boundary S and from sources/sinks in R. 1. Energy ﬂow into R across S. This gives dE =− n · jn dS = − ˆ dx · jn dt boundary Rpr:heatcur1 where the heat current is deﬁned jn = −kT T. kT is the thermal conductivity. Note that since T points 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 R due to sources or sinks, dE =− d3 xρq ˙ dt sources R where q is the rate of energy production per unit mass by sources ˙ inside S. Thus the total energy is given by ∂T dE d3 xρcp = R ∂t dt dE dE = + dt boundary dt sources = d3 x(ρq − ˙ · jn ). R By taking an arbitrary volume, we get the relation ∂T ρcp = · (kT T ) + ρq. ˙ (10.1) ∂t
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10.1. INTRODUCTION 145 10.1.2 Boundary Conditionspr:bc47 Mar p2 There are three types of boundary conditions which we will encounter: 1. T given on S. This is the case of a region surrounded by a heat bath. 2. n · T for x ∈ S given. This means that the heat current normal ˆ to the boundary, n · jn , is speciﬁed. In particular, if the boundary ˆ is insulated, then n · T = 0. ˆ 3. −kT (x)ˆ · n T = α(T − Texternal ) for x ∈ 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 α give α 1 =⇒ T ≈ Texternal → #1 α 1 =⇒ n · T ≈ 0 → #2 with insulated boundary. ˆ We now rewrite the general boundary condition (3) as [ˆ · n T + θ(S)]T (x, t) = h(x, t) for x ∈ S (10.2) where θ(S) = α/kT (S) and h(S, t) = (α/kT (S))Texternal . In the limit θ 1, T is 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 1, n · T is given. ˆ 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, [ˆ · n + κ(x)]u(x, t) = h(x, t) we identify u(x, t) → T (x, t) and κ(x) → θ(x).
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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-pr:heateq1 miliar by writing it in our standard diﬀerential equation form ∂ L0 + ρcp T = ρq(x, t) ˙ for x ∈ G (10.3) ∂teq10DE where the linear operator is 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-7 Mar p3 tion for all time. 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 pq(x, t) = δ(x − x )δ(t − t ) and h(S, t) = 0): ˙ ∂ L0 + ρcp G(x, t; x , t ) = δ(x − x )δ(t − t ), ∂t [ˆ · n + θ(S)]G(x, t; x , t ) = 0 for x ∈ S, G(x, t; x , t ) = 0 for t < t .
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10.2. THE STANDARD FORM OF THE HEAT EQ. 147We lose symmetry in time since only the ﬁrst time derivative appears.We evaluate the retarded Green’s functions by applying the standardFourier transform technique from chapter 6: dω −iω(t−t ) ˜ G(x, t; x , t ) = e G(x, x ; ω). L 2π ˜We know by G = 0 for t < t that G is analytic in the Im ω > 0 plane.Thus we take L to be a line parallel to the real ω-axis in the upper halfplane. The Fourier Transform of the Green’s function is the solution ofthe problem ˜ (L0 − ρcp iω)G(x, x ; ω) = δ(x − x ), (ˆ · n ˜ + θ(x))G(x, x ; ω) = 0 for x ∈ S,which is obtained by Fourier transforming the above Green’s functionproblem.10.2.3 Laplace Transform pr:LapTrans1We note that this problem is identical to the forced oscillation Green’sfunction problem with the substitutions σ → ρcp and τ → kT . Thuswe identify ˜ G(x, x ; ω) = G(x, x ; λ = iω).The single time derivative causes the eigenvalues to be λ = iω. Toevaluate this problem we thus make the substitution s = −iω. Thissubstitution results in the Laplace Transformation. Under this trans- 7 Mar p4formation the Green’s function in transform space is related by ˜ G(x, x ; ω = is) = G(x, x ; λ = −s). ˜G is now analytic in the right hand side plane: Re (s) > 0. This variablesubstitution is depicted in ﬁgure 10.1. The transformed contour islabeled L . The Laplace transform of the Green’s function satisﬁes ﬁg10athe relation i ˜ G(x, t; x , t ) = dsG(x, x ; λ = −s )es(t−t ) 2π L↓
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148 CHAPTER 10. HEAT CONDUCTION L ω E =⇒ cL s Figure 10.1: Rotation of contour in complex plane. or, by changing the direction of the path, we have 1 ˜ G(x, t; x , t ) = dsG(x, x ; λ = s )es(t−t ) . (10.4) 2πi L↑ In the following we will denote L ↑ as L. The inversion formula ∞ ˜ G(ω) = dτ eiωτ G(x, x , τ = t − t ) 0 is also rotated to become ∞ ˜ G(s) = dτ e−sτ G(x, x ; τ ). 0 ˜ 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 Expansions7 Mar p5 We now solve the Green’s function by writing it as a bilinear sum of eigenfunctions: un (x)u∗ (x ) n G(x, x ; λ) = . (10.5) n λn − λ The eigenfunctions un (x) solve the problem L0 un (x) = λn ρcp un (x) for x ∈ R
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10.2. THE STANDARD FORM OF THE HEAT EQ. 149 s-plane Im s C C T ¢ C2 1 C ¢ ¢ Re s q q q 1 f λ3 λ2 λ f f t t t s Figure 10.2: Contour closed in left half s-plane.where L0 = − · (kT (x) ) with the elastic boundary condition (ˆ · n + θ(s))un = 0 for x ∈ S.Because of the identiﬁcation ˜ G(x, x ; s) = G(x, x ; λ = −s)we can substitute 10.5 into the transform integral 10.4 to get ds un (x)u∗ (x ) s(t−t ) n G(x, t; x , t ) = e L 2πi n λn + s 1 ds s(t−t ) = un (x)u∗ (x) n e . n 2πi L λn + sThis vanishes for t < t . Close the contour in the left half s-planefor t − t > 0, as shown in ﬁgure 10.2. This integral consists of ﬁg10a1contributions from the residues of the poles at −λn , where n = 1, 2, . . ..So 1 ds s(t−t ) e = e−λn (t−t ) . 2πi Cn λn + sThus G(x, t; x , t) = un (x)u∗ (x )e−λn (t−t ) . n (10.6) n
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150 CHAPTER 10. HEAT CONDUCTION We now consider the two limiting cases for t. Suppose that t → t . Then 10.6 becomes t→t δ(x − x ) G−→ un (x)u∗ (x ) = n . n ρcp7 Mar p6 Thus we see that another interpretation of G is as the solution of an initial value problem with the initial temperature δ(x − x ) T (x, 0) = ρcpwhy is this? and no forcing term. Now suppose we have the other case, t − t 1. We know λn > 0 for all n since L0 is 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∗ (x )e−λ1 (t−t ) 1 (t − t ) 1. In particular, this formula is valid when (t − t ) > 1/λ2 . We may thus interpret 1/λn = τn as the lifetime of these states. After (t − t ) τN , all contributions to G from eigen values with n ≥ N are 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 for t − t 1 is because for higher N there 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 Calculation9 Mar p1 We now consider the heat equation in one dimension.
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10.3. EXPLICIT ONE DIMENSIONAL CALCULATION 151 10.3.1 Application of Transform Methodpr: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 ) for −∞ < x < ∞. We found that the solution for this wave equation is √ 1 ei λ|x−x |/c G(x, x ; λ) = √ . 2 λ σc Transferring from the wave equation to the heat equation as discussed √ above, we substitute τ → kT , σ → ρ, c = τ /σ → κ where κ = √ √ KT /ρcp is the thermal diﬀusivity, and λ → i s which means Im λ > pr:kappa1 0 becomes Re s > 0. The substitutions yield kT → KT and √ c → cp ? 1 √ − s |x−x | √ ˜ G(x, x ; s) = √ ρcp κ e κ 2 s or √ 1 ˜ x ; s) = √ e− s/κ|x−x | . ρcp G(x, 2 κs √ We see that κ plays the role of a velocity. Now invert the transform 9 Mar p3 to obtain the free space Green’s function for the heat equation: ds s(t−t ) ˜ ρcp G(x, t; x , t ) = e ρcp G(x, x ; s) L 2πi √ ds es(t−t )− s/κ|x−x | = √ . L 2πi 2 sκ 10.3.2 Solution of the Transform Integral √ Our result has a branch on s. We parameterize the s-plane: s = |s|eiθ for −π < θ < π √ This gives us Re s = |s|1/2 cos(θ/2) > 0. We choose the contour of integration based on t. 9 Mar p5
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152 CHAPTER 10. HEAT CONDUCTION s-plane Im s √ C L1 s = i |s| ( E E T √ & ) ¢ L2 s = −i |s| ¢ ¢ E L3
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Re s f f f L4 t Lt5t s Figure 10.3: A contour with Branch cut. For t < t we have the condition G = 0. Thus we close the contourin the right half plane so that √ √ s→∞ exp s(t − t ) − s|x − x|/ κ −→0since both terms are increasingly negative. Since the contour enclosesno poles, we recover G = 0 as required. For t−t > 0, close contour in the left half plane. See ﬁgure 10.3. Weknow by Cauchy’s theorem that the integral around the closed contourL + L1 + L2 + L3 + L4 + L5 vanishes. We perform the usual Branchcut evaluation, by treating the diﬀerent segments separately. For L3 itis convenient to use the parameterization s = εeiθ for −π < θ < π asshown in ﬁgure 10.3. In this case the integral becomes 1 1 π √ √ ε→0 √ dθ| ε|[1 + O(ε(t − t )) + O(ε1/2 |x − x |/ κ)]−→0. 2π 2 κ −πIn this equation we assert that it is permissible to take the limit ε → 0before the other quantities are taken arbitrarily large. √ For the contour L2 above the branch cut we have s = i |s|, √and for the contour L4 below the branch cut we have s = −i |s|
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10.3. EXPLICIT ONE DIMENSIONAL CALCULATION 153Combining the integrals for these two cases gives −ε ds es(t−t ) 2 cos |s|/κ|x − x | lim √ . ε→0 −∞ 2π 2 κ |s| For L1 and L5 the integral vanishes. By letting s = Reiθ where−π < θ < π we have √ √ exp [− s|x − x |/ κ] exp − |x−x | R2 cos 1 φ R→∞ √ κ 2 √ ≤ √ −→ 0. s R Our ﬁnal result is 1 ∞ ds ρcp G(x, t; x , t ) = √ √ e−s(t−t ) cos s/κ(x − x ). 2π κ 0 sSubstituting s = u2 gives ∞ 2udu u 2 ρcp G(x, t; x , t ) = √ cos √ |x − x |e−u (t−t ) 0 2πu κ κor 1 √ ρcp G(x, t; x , t ) = √ I(t − t , |x − x |/ κ) κwhere (since the integrand is even) 1 ∞ du −u2 t+iuy I(t, y) = e . 2 −∞ πThis can be made into a simple Gaussian by completing the square: pr:gaus1 2 /(4t) ∞ du ±(u− iy )2 I(t, y) = e−y e 2t . −∞ 2πBy shifting u → u + iy/2t, the result is 9 Mar p6 2 e−y /(4t) I(t, y) = √ . 4πtThe free space Green function in 1-dimension is thus 1 −(x−x )2 ρcp G(x, t; x , t ) = e 4κ(t−t ) . (10.7) 4πκ|t − t |
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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 = t = 0, we have 2 e−x /4κt ρcp G(x, t) = √ . 4πκt At the initial time we have t→0 ρcp G−→δ(x − x ) = δ(x). 2 1. For x√ > 4κt, the amplitude is very small. Since G is √ small for x ≥ 4κt, diﬀusion proceeds at rate proportional to t, not t as in wave equation. The average propagation is proportional to9 Mar p7 t1/2 . This is indicative of a statistical process (Random walk). It 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 x until time x/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. κ is a fundamental parameter whose role for the heat equation is analogous to the role of c for the wave equation. It deter- mines the rate of diﬀusion. κ (= kT /ρcp ) has the dimensions of (distance)2 /time, whereas c has 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: ∂2 d −kT 2 + ρcp T (x, t) = 0 for −∞ < x, x , ∞ ∂x dt
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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) = dx T0 (x )ρcp G(x , 0; x, t) −∞ 1 ∞ 2 /(4κt) = √ dx e−(x−x ) T0 (x ) (10.8) 4πκt −∞10.3.5 Special CasesInitial δ-functionSuppose T0 (x) = δ(x − x ). Then we have T (x, t) = ρcp G(x , 0; x, t).Thus we see that G is the solution to the IVP with the δ-function asthe initial condition and no forcing term.Initial Gaussian Function pr:gaus2We now consider the special case of an initial Gaussian temperature 11 Mar p3 2distribution. Let T0 (x)√ (a/π)1/2 e−ax . The width of the initial dis- =tribution is (∆x)0 = 1/ a. Plugging this form of T0 (x) into 10.8 gives 1 ∞ 2 /(4κt)−ax 2 T (x, t) = √ dx e−(x−x ) π 4κta −∞ 1 1 x2 T (x, t) = √ e− (1/a)+4κt π (1/a) + 4κt 1 1 2 = √ e(x/∆x) π ∆xwhere ∆x = (∆x0 )2 + 4κt. The packet is spreading as (∆x)2 =4κt + (∆x0 )2 . Again, ∆x ∼ t1/2 like a random walk, (again non-dynamical). Suppose t τ ≡ (∆x0 )/4κ. This is the simplest quantitywith dimensions of time, so τ is the characteristic time of the system.We rewrite ∆x = (∆x0 ) 1 + t/τ . Thus for t τ, ∆x ∼ t/τ (∆x0 ).
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156 CHAPTER 10. HEAT CONDUCTION τ = (∆x0 )2 /4κ is a fundamental unit of time in the problem. Since ask Baker the region is inﬁnite, there does not exist any characteristic distance about omitted11 Mar p4 for the problem.11 Mar p5 10.4 Summary 1. Conservation of energy for heat conduction is given by the equa- tion ∂T ρcp = · (kT T ) + ρq, ˙ ∂t where ρ is the mass density, cp is the speciﬁc heat, T is the temper- ature, kT is the thermal conductivity, and q is the rate of energy ˙ production per unit mass by sources inside the region. 2. The general boundary condition for heat conduction is [ˆ · n T + θ(S)]T (x, t) = h(x, t) for x ∈ S. 3. The heat equation is ∂ L0 + ρcp T = ρq(x, t) ˙ for x ∈ G, ∂t where the linear operator is L0 = − · (kT (x) ). 4. The Green’s function equation for the heat conduction problem is ∂ L0 + ρcp G(x, t; x , t ) = δ(x − x )δ(t − t ). ∂t 5. The solution of the heat equation for the initial value problem in one dimension is 1 ∞ 2 /(4κt) T (x, t) = √ dx e−(x−x ) T0 (x ), 4πκt −∞ which is a weighted integration over point sources which individ- ually diﬀuse with a gaussian shape.
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10.5. REFERENCES 15710.5 ReferencesA similar treatment (though more thorough) is given in [Stakgold67b,p194ﬀ]. See also [Fetter80, p406ﬀ]. The deﬁnitive reference on heat conduction is [Carslaw86].
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Chapter 11Spherical Symmetry Chapter Goals: • Derive the form of the linear operator in spherical coordinates. • Show that the angular part of the linear operator Lθφ is hermitian. 28 Mar p1 (17) m • Write the eigenvalue equations for Yl . • Write the partial wave expansion for the Green’s function. • Find the Green’s function for the free space prob- lem.Our object of study is the Green’s function for the problem [L0 − λσ(x)]G(x, x ; λ) = δ(x − x ) (11.1)with the regular boundary condition (RBC) eq11.1 [ˆ · n + K(S)]G(x, x ; λ) = 0for x in a region R and x in the regions boundary S. The term x is aﬁeld point, and x is a source point. The unit vector n is the outward ˆnormal of the surface S. The operator L0 is deﬁned by the equation L0 = − · (τ (x) ) + V (x). 159
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160 CHAPTER 11. SPHERICAL SYMMETRY z (r, θ, ϕ) U r θ y ll ϕ l l l x 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 Coordinates28 Mar p2 We now treat the problem in three dimensions. For this we use sphericalpr:spher1 coordinates (since we will later assume angular independence). A point 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 = r cos θ, x = r sin θ cos ϕ, y = r sin θ sin ϕ.ﬁg11.1 This relationship is illustrated in ﬁgure 11.1 For an arbitrary volume
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11.1. SPHERICAL COORDINATES 161element we have d3 x = (dr)(rdθ)(r sin θdϕ) = r2 dΩdrwhere Ω is the solid angle, and an inﬁnitesimal of solid angle is dΩ = pr:Omega1sin θdθdϕ. We further deﬁne the delta function f (r, θ, ϕ) = f (x) = d3 x f (x ) δ(x − x ) 2 = dr r sin θ dθ dϕ f (r , θ , ϕ ) δ(x − x ).From this we can extract the form of the δ-function for spherical coor-dinates: 1 δ(x − x ) = δ(r − r )δ(θ − θ )δ(ϕ − ϕ ) r2 sin θ δ(r − r ) = δ(Ω − Ω ) r2where the solid angle δ-function is 28 Mar p3 pr:delta2 δ(θ − θ )δ(ϕ − ϕ ) δ(Ω − Ω ) = . sin θ We want to rewrite equation 11.1 in spherical coordinates. First wedeﬁne the gradient pr:grad2 ∂ ˆ θ ∂ ϕ ∂ ˆ =r ˆ + + . ∂r r ∂θ r sin θ ∂ϕSee [Boas] for derivations of identities involving . The divergence is 87’ notes have Gauss’ law, 1 ∂ 2 1 ∂ 1 ∂ p18 ·A= 2 ∂r (r Ar ) + (sin θAθ ) + Aϕ . r r sin θ ∂θ r sin θ ∂ϕWhen we apply this to the case A = τ (x) .
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162 CHAPTER 11. SPHERICAL SYMMETRY the result is 1 ∂ ∂ 1 ∂ τ ∂ · (τ ) = 2 ∂r r2 τ + sin θ r ∂r r sin θ ∂θ r ∂θ 1 ∂ τ ∂ + (11.2) r sin θ ∂ϕ r sin θ ∂ϕeq11.2 where τ = τ (x) = τ (r, θ, ϕ).28 Mar p4 Now we can write L0 . We assume that τ , σ, and V are 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 1 ∂ ∂ τ (r) L0 = − r2 τ (r) + Lθϕ + V (r) (11.3) r2 ∂r ∂r r2eq11.2b where 1 ∂ ∂ 1 ∂2 Lθϕ = − sin θ − , sin θ ∂θ ∂θ sin2 θ ∂ 2 ϕpr:Lthph1 which is the centrifugal term from equation 11.2. In the next few sec- tions we will study the properties of L0 . 11.2 Discussion of Lθϕ Note that Lθϕ is 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 d3 xS ∗ (x)L0 u(x) = d3 x (u∗ (x)L0 S(x))∗ (11.4)eq11.3 where u and S satisfy RBC. We use this fact to show the hermiticity28 Mar p5 of Lθϕ . Consider the functions S(x) = S(r)S(θ, ϕ) and u(x) = u(r)u(θ, ϕ) where u and S satisfy RBC. Such functions are a subset of the functions which satisfy equation 11.4. Choose u(θ, ϕ) and S(θ, ϕ) to be periodic in the azimuthal angle ϕ: u(θ, ϕ) = u(θ, ϕ + 2π), S(θ, ϕ) = S(θ, ϕ + 2π).
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11.3. SPHERICAL EIGENFUNCTIONS 163Now substitute d3 x = r2 drdΩ and L0 (as deﬁned in equation 11.3) intoequation 11.4. The term 1 ∂ ∂ − 2 ∂r r2 τ (r) + V (r) r ∂rin L0 is hermitian so it cancels out in 11.4. All that is left is τ (r) r2 drS ∗ (r) u(r) dΩS ∗ (θ, ϕ)Lθϕ u(θ, ϕ) = r2 τ (r) r2 drS ∗ (r) 2 u(r) dΩ (u∗ (θ, ϕ)Lθϕ S(θ, ϕ))∗ rThis can be rewritten as 28 Mar p6 τ (r) r2 drS ∗ (r) u(r)dΩ S ∗ (θ, ϕ)Lθϕ u(θ, ϕ) r2 − dΩ (u∗ (θ, ϕ)Lθϕ S(θ, ϕ))∗ = 0.The bracket must then be zero. So dΩ(S ∗ (θ, ϕ)Lθϕ u(θ, ϕ) = dΩ (u∗ (θ, ϕ)Lθϕ S(θ, ϕ))∗ . (11.5)This is the same as equation 11.4 with d3 x → dΩ and L0 → Lθϕ . Thus eq11.4Lθϕ is hermitian. If the region did not include the whole sphere, wejust integrate the region of physical interest and apply the appropriateboundary conditions. Equation 11.5 can also be obtained directly from 28 Mar p7the form of Lθϕ by applying integration by parts on Lθϕ twice, butusing L0 = L∗ is much more elegant. 0 ∂2 Note that the operators ∂ϕ2 and Lθϕ commute: ∂2 , Lθϕ = 0. ∂ϕ2Thus we can reduce equation 11.1 to a one dimensional case and expandthe Green’s function G in terms of a single set of eigenfunctions whichare valid for both −∂ 2 /∂ϕ2 and Lθϕ . We know that L0 and Lθϕ arehermitian operators, and thus the eigenfunctions form a complete set.For this reason this method is valid.
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164 CHAPTER 11. SPHERICAL SYMMETRY 11.3 Spherical Eigenfunctions 28 Mar p8 We want to ﬁnd a common set of eigenfunctions valid for both Lθϕ and −∂ 2 /∂ϕ2 . Note that ∂ 2 eimϕ eimϕ − √ = m2 √ m = 0, ±1, ±2, . . . . ∂ϕ2 2π 2π So (2π)−1/2 eimϕ are normalized eigen functions of −∂ 2 /∂ϕ2 . We deﬁnepr:sphHarm1 the functions Ylm (θ, ϕ) as the set of solutions to the equation Lθϕ Ylm (θ, ϕ) = l(l + 1)Ylm (θ, ϕ) (11.6)eq11.5 for eigen values l(l + 1) and periodic boundary conditions, and the equation ∂2 m − Y (θ, ϕ) = m2 Ylm (θ, ϕ) m = 0, ±1, ±2, . . . . (11.7) ∂ϕ2 leq11.6 We can immediately write down the orthogonality condition (due topr:orthon3 the hermiticity of the operator Lθϕ ):is this true? dΩYlm ∗ (θ, ϕ)Ylm (θ, ϕ) = δl,l δm,m . (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 Equation28 Mar p9 We now separate the eigenfunction into the product of a ϕ-part and apr:ulm1 θ-part: eimϕ Ylm (θ, ϕ) = √ um (cos θ) l (11.8) 2πeq11.6b (which explicitly solves equation 11.7, the diﬀerential equation involv- ing ϕ) so that we may write equation 11.6 as 1 ∂ ∂ m2 − sin θ + 2 um (cos θ) = l(l + 1)um (cos θ). sin θ ∂θ ∂θ sin θ l l
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11.3. SPHERICAL EIGENFUNCTIONS 165All we have left to do is solve this eigenvalue equation. The originalregion was the surface of the sphere because the solid angle representsarea on the surface. We make a change of variables: x = cos θ.The derivative operator becomes d dx d d = = − sin θ dθ dθ dx dxso 1 d d − = . sin θ dθ dxThe eigen value equation for u becomes d d m2 − (1 − x2 ) + 2 um (x) = l(l + 1)um (x) l l (11.9) dx dx 1−xdeﬁned on the interval −1 < x < 1. Thus x = 1 corresponds to θ = 0, eq11.7and x = −1 corresponds to θ = π. Note that τ , which representsthe eﬀective tension, is proportional to 1 − x2 , so both end points aresingular points. On account of this we get both regular and irregular 28 Mar p10solutions. A solution occurs only if the eigen value l takes on a specialvalue. Requiring regularity at x = ±1 implies l is an integer. Note also Verify thisthat equation 11.9 represents an inﬁnite number of one dimensionaleigenvalue problems (indexed by m), which makes sense because westarted with a partial diﬀerential equation eigenvalue problem. The way to solve the equation near a singular point is to look forsolutions of the form xp ·[power series], as in the solutions to Bessel’s change thisequation. 30 Mar p111.3.2 Determination of um (x) l 30 Mar p2We now determine the function um (x) which is regular at x = ±1. lSuppose that it is of the form um (x) = (1 − x2 )β (power series). l (11.10)We want to determine the power term β. First we compute eq11.8
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166 CHAPTER 11. SPHERICAL SYMMETRY d (1 − x2 ) (1 − x2 )β = β(−2x)(1 − x2 )β , dx and then d d (1 − x2 ) (1 − x2 )β = β 2 (1 − x2 )β−1 (−2x)2 dx dx + β(−2x)(1 − x2 )β + . . . . For the case x → 1 we can drop all but the leading term: d d (1 − x2 ) (1 − x2 )β ≈ 4x2 β 2 (1 − x2 )β−1 x → 1. (11.11) dx dxeq11.9 Plugging equations 11.10 and 11.11 into equation 11.9 gives (1 − x2 )β−1 A[−4β 2 + m2 ] ≈ (l + 1)lA(1 − x2 )β , x → 1. where A is the leading constant from the power series. Note however that (1 − x2 )β approaches zero faster that (1 − x2 )β−1 as x → 1. So we get m2 = 4β 2 , or m β=± . 2 We thus look for a solution of the form um (x) = (1 − x2 )m/2 Cm (x), l (11.12)eq11.9a where Cm (x) is a power series in x with implicit l dependence. We30 Mar p3 expect regular and irregular solutions for Cm (x). We plug this equation into equation 11.9 to get an equation for Cm (x). The result is −(1 − x2 )Cm + 2x(m + 1)Cm − (l − m)(l + m + 1)Cm (x) = 0. (11.13)eq11.9b We still have the boundary condition that um (x) is ﬁnite. In the case l that m = 0 we have −(1 − x2 )C0 + 2xC0 − l(l + 1)C0 = 0. (11.14)eq11.10 This is called Legendre’s equation. We want to ﬁnd the solution ofpr:LegEq1 this equation which is regular at x = 1. We deﬁne Pl (x) to be such
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11.3. SPHERICAL EIGENFUNCTIONS 167a solution. The irregular solution at x = 1, called Q(x), is of interestif the region R in our problem excludes x = 1 (which corresponds tocos θ = 1 or θ = 0). Note that we consider l to be an arbitrary complexnumber. (From the “general theory”, however, we know that the eigenvalues are real.) By convention we normalize: Pl (1) = 1. We knowC0 (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 Pl must be ﬁnite (regular) at x = −1, wethen have B(l) = 0 for l = 0, 1, 2, . . .. 30 Mar p4 We now take Legendre’s equation, 11.14, with C0 (x) replaced byPl (x) (the regular solution), and diﬀerentiate it m times using Leibnitzformula m pr:LeibForm1 dm m di f dm−i g (f (x)g(x)) = . dxm i=0 i dxi dxm−iThis yields d2dm d dm −(1 − x2 ) P (x) + 2(m + 1)x m l Pl (x) dx2 dx dx dxm m d − (l − m)(l + m + 1) m Pl (x) = 0. (11.16) dxThus (dm /dxm )Pl (x) is also a solution of equation 11.13. We thus see eq11.11that dm Cm (x) = α m Pl (x), (11.17) dxwhere α still needs to be determined. Once we ﬁnd out how to chose l eq11.11bso Pl (x) is 0 at l, Pl for l = l will also be zero. So we see that once wedetermine constraints on l such that the Pl (x) which solves the m = 0equation is zero at zero at x = ±1, we can generate a solution for thecase m = 0. We now calculate a recurrence relation for Pl (x). Set x = 1 in pr:recrel1equation 11.16. This gives dm+1 dm 2(m+1) Pl (x) = (l−m)(l+m+1) Pl (x) . (11.18) dxm+1 x=1 dxm x=1
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168 CHAPTER 11. SPHERICAL SYMMETRY This tells us the (m + 1)th derivative of Pl (x) in terms of the mth eq11.12 derivative. We can diﬀerentiate l times if l is an integer. Take l to be an integer. The case m = l yields dl+1 Pl (x) = 0. (11.19) dxl+1 x=1ask Baker isn’t So all derivatives are zero for m > l at x = 1. This means that P (x)this valid only is an lth order polynomial, since all of its Taylor coeﬃcients vanish forat x = 1 m > l. Since Pl is regular at x = 1 and is a polynomial of degree l,30 Mar p5 it must be regular at x = −1 also. If l were not an integer, we would obtain a series which diverges at x = ±1. Thus we conclude that l must be an integer. Note that even for the solution which is not regular at x = ±1, for which l is 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 dm um (x) = α(1 − x2 )m/2 m Pl (x). l (11.20) dx This equation holds for m = 0, 1, 2, . . . . Furthermore this equation dm solves equation 11.13. Note that because dxm Pl (x) is regular at x = 130 Mar p6 and x = −1, um (x) is also. l We now compute the derivative. Using equation 11.18, for m = 0 we get d 2 Pl (x)|x=1 = l(l + 1)Pl (1) = l(l + 1). dx By repeating this process for m = 1, 2, . . . and using induction, we ﬁndyet to be veri- that the following polynomial satisﬁes equation 11.18ﬁed 1 dl 2 Pl (x) = (x − 1)l . (11.21) 2l l! dxleq11.13 This is called Rodrigues formula for the Legrendre function.pr:rodform1 We deﬁne the associated Legendre polynomial dm Plm (x) = (−1)m (1 − x2 )m/2 Pl (x) m≥0 dxm (1 − x2 )m/2 dl+m 2 = l l! l+m (x − 1)l , m ≥ 0. (11.22) 2 dx
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11.3. SPHERICAL EIGENFUNCTIONS 169eq11.13b For m > l, Plm (x) = 0. So the allowed range of m is −l ≤ m ≤ l. Thus the value of m aﬀects what the lowest eigenvalue, l(l + 1), can be. 11.3.3 Orthogonality and Completeness of um (x) l We want to choose um (x) to be normalized. We deﬁne the normalized pr:normal4 l eigen functions as the set of eigen functions which satisﬁes the condition (with σ = 1) 1 dx (um (x))∗ um (x) = 1. l l (11.23) −1 1 We need to evaluate −1 dx|Plm (x)|2 . Using integration by parts and eq11.13c equation 11.22 we get (a short exercise) 1 1 2 (l + m)! dx|Plm (x)|2 = dx (Plm (x))∗ Plm (x) = , (11.24) −1 −1 2l + 1 (l − m)! so the normalized eigenfunctions are 30 Mar p7 2 (l + m)! m um (x) = l P (x). (11.25) 2l + 1 (l − m)! l The condition for orthonormality is eq11.14 1 pr:orthon4 um (x), um (x) l l = dx (um (x))∗ l um (x) l = δll , (11.26) −1 The corresponding completeness relation is (as usual, with σ = 1) ∞ um (x)um (x ) = δ(x − x ). l l (11.27) l=m The problem we wanted to solve was equation 11.6, so we substitute back in x = cos θ into the completeness relation, which gives ∞ um (cos θ)um (cos θ ) = δ(cos θ − cos θ ) l l l=m = δ((θ − θ )(− sin θ)) δ(θ − θ ) = . sin θ
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170 CHAPTER 11. SPHERICAL SYMMETRY In the second equality we used a Taylor expansion for θ near θ , which yields cos θ − cos θ = −(θ − θ ) sin θ. In the third equality we used the δ-function property δ(ax) = |a|−1 δ(x). The completeness condition for um (cos θ) is thus l ∞ δ(θ − θ ) um∗ (cos θ)um (cos θ ) = l l . (11.28) l=m sin θ 130 Mar p8 Similarly, the orthogonality condition becomes (since −1 d(cos θ) = π 0 sin θdθ) π dθ sin θum (cos θ)um (cos θ) = δll . l l (11.29) 0 11.4 Spherical Harmonicspr:spherH11 Apr p1a We want to determine the properties of the functions Ylm , 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) eimϕ Ylm (θ, ϕ) = √ um (cos θ) l 2π for integer l = m, m + 1, m + 2, . . . and m ≥ 0, where we have found (equation 11.25) m (l − m)! 2l + 1 d um (cos θ) = l (sin θ)m Pl (cos θ). (lm )! 2 d cos θ We deﬁne the Yl−m (θ, ϕ), for m > 0, as Yl−m (θ, ϕ) ≡ (−1)m Ylm (θ, ϕ)∗ −imϕ me = (−1) √ um (cos θ). l 2π −imϕ The term (−1)m is a phase convention and e√2π is an eigen function.pr:consho1 This is often called the Condon-Shortley phase convention.
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11.4. SPHERICAL HARMONICS 17111.4.1 Othonormality and Completeness of YlmWe saw that the functions um satisfy the following completeness con- ldition: ∞ δ(θ − θ ) um (cos θ)um (cos θ ) = l l for all m l=m sin θwhere m is ﬁxed and positive. We know that ∞ eimϕ e−imϕ √ √ = δ(ϕ − ϕ ). (11.30) m=−∞ 2π 2π 1Multiply sin θ δ(θ − θ ) into equation 11.30, so that eq11.15 1 GApr 1b ∞ imϕ ∞ −imϕδ(ϕ − ϕ )δ(θ − θ ) e |m| |m| ∗ e = √ ul (cos θ)ul (cos θ) √ sin θ m=−∞ 2π l≥|m| 2π ∞ ∞ = Ylm (θ, ϕ)Ylm ∗ (θ, ϕ ), m=−∞ l≥|m|since eimϕ |m| Ylm = (−1)m √ ul (cos θ) for m < 0, 2πand −imϕ m∗ me |m| Yl = (−1) √ ul (cos θ) for m < 0. 2πThus we have the completeness relation ∞ l δ(Ω − Ω ) = Ylm (θ, ϕ)Ylm ∗ (θ , ϕ ). (11.31) l=0 m=−lWe also note that Lθϕ has (2l + 1)–fold degenerate eigenvalues l(l + 1) eq11.16in Lθϕ Ylm (θ, ϕ) = l(l + 1)Ylm (θ, ϕ).Thus m is like a degeneracy index in this equation. Next we look at the orthogonality of the spherical harmonics. The pr:orthon5
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172 CHAPTER 11. SPHERICAL SYMMETRY orthogonality relation becomes 2π dϕ 1 dΩYlm (θ, ϕ)Ylm (θ, ϕ) = δmm d cos θum (cos θ)um (cos θ) l l 0 2π −1 = δll δmm , where dΩ = dϕdθ sin θ on the right hand side, Because the u’s are orthogonal and the e−imϕ ’s are orthogonal, the right hand side is zero when l = l or m = m . 11.5 GF’s for Spherical Symmetry1 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 ; λ) = δ(x − x ). (11.32)eq11.16a By substituting the L0 for spherically symmetric problems, which weI don’t know found in equation 11.3, we havehow to ﬁx this. 1 d d τ (r) − 2 dr r2 τ (r) + Lθϕ + V (r) − λσ Glm (x, x ; λ ) r dr r2 δ(r − r ) = δ(Ω − Ω ) (11.33) r2 δ(r − r ) ∞ l = Ylm (θ, ϕ)Ylm∗ (θ , ϕ ). r2 l=0 m=leq11.17 where the second equality follows from the completeness relation, equa-pr:ExpThm2 tion 11.31. Thus we try the solution form ∞ l G(x, x ; λ) = Ylm (θ, ϕ)Glm (r, r ; λ )Ylm ∗ (θ, ϕ). (11.34) l=0 m=l
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11.5. GF’S FOR SPHERICAL SYMMETRY 173eq11.17b Note that the symmetry of θ, φ and θ , φ in this solution form means that Green’s reciprocity principle is satisﬁed, as required. Substituting this into equation 11.33 and using equation 11.6 results in Lθφ being replaced by the eigenvalue of Ylm , which is l(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 Ylm ’s from both sides to get the following radial equation 1 d d τ (r) δ(r − r ) − 2 dr r2 τ (r) +2 l(l + 1) + V (r) Glm (r, r ; λ ) = . r dr r r2 (11.35) The linear operator for this equation has no m dependence. That is, m eq11.18 is just a degeneracy index for the 2l + 1 diﬀerent solutions of the Lθφ equation for ﬁxed l. Thus we can rewrite our radial Green’s function Glm as Gl and deﬁne the radial operator as (l) d d τ (r) L0 = − r2 τ (r) + r2 l(l + 1) + V (r) . (11.36) dr dr r2 We have reduced the three dimensional Green’s function to the standard eq11.18b single dimensional case with eﬀective tension r2 τ (r), eﬀective potential energy V (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 are 1 Apr 2b not spherically symmetric, we need to take account of the angles, i.e., [ˆ · n + κ(S)]G(x x , λ) = 0, for x on S and x in R. Consider a spherical region as shown in ﬁgure 11.2. For spherically ﬁg11b symmetric boundary conditions, we can set κint (S) = ka and κext (S) = kb . Thus ∂ − + ka G = 0 for r = a, ∂r ∂ + kb G = 0 for r = b. ∂r
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174 CHAPTER 11. SPHERICAL SYMMETRY 9 6 S2 E $ R S1 E a &% b 8 7 E Figure 11.2: The general boundary for spherical symmetry. If we insert G from equation 11.34 into these conditions, we ﬁnd the following conditions on how Gl behaves: ∂ − + ka Gl = 0 for r = a, ∂r ∂ + kb Gl = 0 for r = b, ∂r for all l. These equations, together with equation 11.35, uniquely de- termine Gl . 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 and b → ∞.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
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11.5. GF’S FOR SPHERICAL SYMMETRY 175symmetric: τ (r), σ(r), V (r), and κ(S) = ka . Our boundary conditionis ∂ − + ka G(x, x ; λ) = 0, ∂rwhere r = a for all θ, ϕ (that is, |x | > |x| = a). This implies 4 Apr p2,3 ∂ − + ka Gl (r, r ; λ) = 0 for r > r = a. ∂rThe other boundary condition is that Gl is bounded as r → ∞. We now want to solve Gl (r, r ; λ). Recall that we have seen twoways of expressing the Green’s function in terms of solutions of thehomogeneous equation. One way is to write the Green’s functions asa product of the solution satisfying the upper boundary condition andthe solution satisfying the lower boundary condition, and then divideby the Wronskian to ensure continuity. Thus we write 1 ul (r< , λ)ul (r> , λ) 1 2 Gl (r, r ; λ) = − , (11.37) r2 τ (r) W (ul , ul ) 1 2where ul and ul satisfy the equations 1 2 eq11.18c [Ll − λσ(r)r2 ]ul (r, λ) = 0, 0 1 [Ll − λσ(r)r2 ]ul (r, λ) = 0, 0 2and the boundary conditions ∂ − + ka ul = 0 for r = a, 1 ∂r ul (r, λ) < ∞ when r → ∞. 2 The other way of expressing the Green’s function is to look at howit behaves near its poles (or branch cut) and consider it as a sum ofresidues. This analysis was performed in chapter 4 where we obtainedthe following bilinear sum of eigenfunctions: 4 April p4 u(l) (r)u(l) (r ) n n Gl (r, r , λ) = (l) . (11.38) n λn − λ
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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 n should be replaced by integrals over λn . The u(l) (r) solve the corresponding eigen value problem n (l) L0 u(l) (r) = λ(l) r2 σ(r)u(l) (r) n n n with the boundary conditions that ∂ − + ka u(l) = 0 for r = a, n ∂r and u(l) is ﬁnite as r → ∞. The interior problem, with ul ﬁnite as n n r → 0, has a discrete spectrum. The normalization of the u(l) (r) is given by the completeness relation n δ(r − r ) u(l) (r)u(l) (r ) = n n . n r2 σ(r)4 Apr p5 We insert equation 11.38 into 11.34 to get u(l,m) (x)u(l,m) (x ) n n G(x, x ; λ) = (l) (11.39) n λn − λeq11.20 where u(l,m) (x) = Ylm (θ, ϕ)u(l) (r). n n and λ(l) is the position of the nth pole of Gl . So the eigenvalues λn are n the λ(l) determined from the r-space eigenvalue problem. The corre- n sponding eigenfunctions u(l,m) (x) satisfy n (l) L0 u(l,m) (x) = λ(l) r2 σ(r)u(l,m) (x). n n n The completeness relation for u(l,m) (x) is found by substituting equation n 11.39 into 11.32 and performing the same analysis as in chapter 4. The result is δ(x − x ) u(l) (r)un (r ) = n (l) . n σ(x)
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11.6. EXAMPLE: CONSTANT PARAMETERS 177 11.6 Example: Constant Parameters4 Apr p6Old HW#4 We now look at a problem from the homework. We apply the above analysis to the case where V = 0 and τ and σ are constant. Our operator for L becomes (c.f., equation 11.36) (l) d d L0 = τ − r2 + l(l + 1) . dr dr The equation for the Green’s function becomes (after dividing by τ ): d d 1 − r2 + l(l + 1) − k 2 r2 Gl (r, r ; λ) = δ(r − r ), dr dr τ where k 2 = λσ/τ = λ/c2 . 11.6.1 Exterior Problem We again have the boundary conditions ∂ − + ka Gl (r, r ; λ) = 0, (11.40) ∂r where r = a, r > a, and Gl bounded as r → ∞. As usual, we assume (eq11.21d the solution form (l) (l) 1 u1 (r< , λ)u2 (r> , λ) Gl (r, r , λ) = − 2 τ (r) (l) (l) . (11.41) r W (u1 , u2 We solve for u1 and u2 : eq11.21a 4 Apr p7 d d (l) − r2 + l(l + 1) − k 2 r2 u1,2 = 0, dr dr where u1 and u2 satisfy the boundary conditions as r = a and r → ∞ respectively and k ≡ ω/c = λ/c2 . This is the spherical Bessel equa- tion from the third assignment of last quarter. We found the solution R(r) u(r) = √ , r
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178 CHAPTER 11. SPHERICAL SYMMETRY where R(r) satisﬁes the regular Bessel equation d d (l + 1 )2 − r + 2 − k 2 r R(r) = 0. dr dr r The solutions for this equation are: R ∼ Jl+ 1 , Nl+ 1 . 2 2pr:sphBes By deﬁnition π jl (x) = J 1 (x), 2x l+ 2 π nl (x) = N 1 (x), 2x l+ 2 (1) π hl (x) = H 1 (x), 2x l+ 2 where jl (kr) is a spherical Bessel function, nl (kr) is a spherical Neu- (1) mann function, and hl (kr) is a spherical Hankel function. So we can write u1 = Ajl (kr< ) + Bnl (kr< ), u2 = h1 (kr> ), lIs this correct? Note that the u2 solution is valid because it is bounded for large r: (1) i lim hl (x) = − eix (−i)l . x→∞ x 11.6.2 Free Space Problem4 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. (1) (2) The solutions ul and ul are then (1) ul = jl (kr), (2) (1) ul = hl (kr). (1)4 Apr p9 Let λ be an arbitrary complex number. Note that since hl (x) =
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11.6. EXAMPLE: CONSTANT PARAMETERS 179jl (x) + inl (x), we have (1) i W (jl (x), hl (x)) = iW (jl (x), nl (x)) = . x2The last equality follows immediately if we evaluate the Wronskian forlarge r using jl (x) ≈ cos(x − (l + 1)π/2) x → ∞, nl (x) ≈ sin(x − (l + 1)π/2) x → ∞,and recall that τ W is a constant (for the general theory, c.f., problem1, set 3) where “τ ” is r2 τ for this problem. In particular we have (1) i W (jl , hl ) = . (kr)2We then get from equation 11.41 (1) 1 jl (kr< )hl (kr> ) Gl = 2 i r τ k (kr)2 ik (1) = jl (kr< )hl (kr> ). (11.42) τ (eq11.21c We have thus found the solution of 2 λ δ(x − x ) − − 2 G= , (11.43) c τwhich is the fundamental three-space Green’s function. We found (eq11.22 4 Apr p10 ik m m∗ G(x, x , λ) = Yl (θ, ϕ)jl (kr< )hl (kr> )Yl , (11.44) τ lmwhich is a simple combination of equation 11.34 and 11.42. In the ﬁrst eq11.22ahomework assignment we solve this by a diﬀerent method to ﬁnd anexplicit form for equation 11.43. Here we show something related. Inequation 11.43 we have G = G(|x − x |) since V = 0 and σ and τ
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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 s as r → 0. Only the l = 0 term survives, since as r → 0 we have 4 Apr p11 l=0 −→ jl (0) = 1, l=0 −→ jl (0) = 0. Thus we have ik 0 2 G(x, x ; λ) = |Y | h0 (kr). τ 0 Since l = 0 implies m = 0, we get Y00 = const. = 1/4π since it satisﬁes the normalization dΩ|Y |2 = 1. We also know that (1) i h0 (x) = − eix . x This gives us 1 ikr G(x x ; λ) = e . τ 4πr We may thus conclude that eik|x−x | G(|x − x |) = . (11.45) 4π|x − x |τeq11.25stuﬀ omitted 11.7 Summary 1. The form of the linear operator in spherical coordinates is 1 ∂ ∂ τ (r) L0 = − 2 ∂r r2 τ (r) + Lθϕ + V (r), r ∂r r2 where 1 ∂ ∂ 1 ∂2 Lθϕ =− sin θ − . sin θ ∂θ ∂θ sin2 θ ∂ 2 ϕ
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11.8. REFERENCES 181 2. Lθφ is hermitian. 3. The eigenvalue equations for Ylm are Lθϕ Ylm (θ, ϕ) = l(l + 1)Ylm (θ, ϕ), ∂2 m − Y (θ, ϕ) = m2 Ylm (θ, ϕ) m = 0, ±1, ±2, . . . . ∂ϕ2 l 4. The partial wave expansion for the Green’s function is ∞ l G(x, x ; λ) = Ylm (θ, ϕ)Glm (r, r ; λ )Ylm ∗ (θ, ϕ). l=0 m=l 5. The Green’s function for the free spce problem is eik|x−x | G(|x − x |) = . 4π|x − x |τ11.8 ReferencesThe preferred special functions reference for physicists seems to be[Jackson75]. Another good source is [Arfken85]. This material is developed by example in [Fetter81].
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Chapter 12Steady State Scattering Chapter Goals: • Find the free space Green’s function outside a circle of radius a due 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 p1We now look at the important problem of steady state scattering.Consider a point source at x with sinusoidal time dependence pr:sssc1 f (x , t) = δ(x − x )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 spacewith a point source at x satisﬁes ∂2 L0 − σ u0 (x, t) = δ(x − x )e−iωt , ∂t2and was solved in terms of the Green’s function, u0 (x, ω) = G0 (x, x , λ = ω 2 + i )e−iωt (12.1) 183
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184 CHAPTER 12. STEADY STATE SCATTERING $ s # x $ ¨ ¨ e ¡ e s ¡ ¡ E ¡ x e ! d d e % d " ! & % Figure 12.1: Waves scattering from an obstacle. eik|x−x |−iwt = (12.2) 4πτ |x − x |eq11.25b where the Green’s function G0 satisﬁes the equation 11.43 and thepr:uoxo1 second equality follows from 11.45. The equation for G0 can be written 2 1 [− − k 2 ]G0 (x, x ; λ) = δ(x − x ) (12.3) τeq11.25c with the deﬁnition k = λ/c2 = ω/c (Remember that λ = ω 2 + iε).6 Apr p3 We combine these observations to get 1 u0 (x, x ; ω) = e−iω(t−(x−x )/c) . 4πτ |x − x | If there is an obstacle (i.e., interaction), then we have a new steady state response u(x, x ; ω) = G(x, x ; λ = ω 2 + iε)e−iωt , where [L0 − λσ]G(x, x ; λ) = δ(x − x ) RBC. (12.4) This is the steady state solution for all time. We used u0 and G0 for the free space problem, and u and G for 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 GS e−iωt = (G − G0 )e−iωt .pr:GS16 Apr p4
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12.2. PLANE WAVES 18512.2 Plane WavesWe now look at the special case of an incident plane wave instead ofan incident spherical wave. This case is more common. Note that oncewe solve the point source problem, we can also solve the plane waveproblem, 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)ˆ . This is a solution of the homogeneous wave equation pr:Phi0 n 2 1 ∂2 − + Φ0 = 0. c2 ∂t2So Φ0 is the solution to the equation without scattering. Let Φ be the wave when an obstacle is present, 2 1 ∂2 − + Φ=0 c2 ∂t2which solves the homogeneous wave equation and the regular boundarycondition at the surface of the obstacle. To obtain this plane wave problem, we let x go to −∞. We nowdescribe this process. To obtain the situation of a plane wave approach-ing the origin from −∞ˆ, we let x = −rˆ, as r → ∞. We want to ﬁnd z zout what eﬀect this limit has on the plane wave solution we obtainedin equation 11.45, eik|x−x | G0 = . 4πτ |x − x |We deﬁne the angles γ and θ as shown in ﬁgure 12.2. From the ﬁgure pr:gamma1we see that ﬁg12a |x − x | = r + r cos θ = r − r cos γ, |x | → ∞.We further recall that the dot product of unit vectors is equal to thecosine of the separation angle, x·x cos γ = = − cos θ. rr
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186 CHAPTER 12. STEADY STATE SCATTERING I x 0 x − x r x γ θ E z ˆ r origin r cos θ Figure 12.2: Deﬁnition of γ and θ.. So the solution is eik|x−x | G0 (x, x ; λ) = 4πτ |x − x | 1 ik(r−r cos γ) = e 4πτ r 1 ikr ikx·(−ˆ ) x = e e 4πτ r eikr ik·x = e 4πτ r where k = w (−ˆ ). Note that we have used the ﬁrst two terms of the c x approximation in the exponent, but only the ﬁrst term in the denomi- nator. Thus we see that in this limit the spherical wave u0 in free space due to a point source is lim u0 = G0 e−iωt |x |→∞ eikr = Φ0 4πr τ where Φ0 = e−i(ωt−k·x) . 12.3 Relation to Potential Theorypr:PotThy18 Apr p1 Consider the problem of ﬁnding the steady state response due to a point source with frequency ω located at x outside a circular region of radius a. The steady state response must satisfy the regular boundary condition ∂ − G + κa G = 0 for r = a. (12.5) ∂r
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12.3. RELATION TO POTENTIAL THEORY 187In particular we want to ﬁnd this free space Green’s function outside a eq11A1circle of radius a, where V = 0 and σ and τ are constant. The Green’s function G satisﬁes the inhomogeneous wave equation 2 δ(x − x ) [− − k 2 ]G(x, x ) = τwhere k 2 = λσ/τ = λ/c2 . The solution was found to be (from problem1 of the ﬁnal exam of last quarter) ∞ 1 G= eim(ϕ−ϕ ) [Jm (kr< ) + Xm Hm (kr< )]Hm (kr> ). (12.6) (1) (1) 4πτ m=−∞with eq11A3 [kJm (ka) − Ka Jm (ka)] Xm = − (1) (1) . (12.7) pr:Xm1 [kHm (ka) − Ka Hm (ka)] (1) √Note that Hm (kr> ) comes from taking Im λ > 0. If we consider (Eq.BE) √ (2)Im λ < 0, we would have Hm instead. All the physics is in thefunctions Xm . Note that from this solution we can obtain the solution 8 Apr p2to the free space problem (having no boundary circle). Our boundarycondition is then then G must be regular at |x| = 0: G regular, |x| = 0. (12.8)(That is, equation 12.5 becomes 12.8.) How do we get this full space eq11A2solution? From our solution, let a go to zero. So Xm in equation 12.6goes the zero as a goes to zero, and by deﬁnition G → G0 . The freespace Green’s function is then ∞ i G0 = eim(ϕ−ϕ ) Jm (kr< )Hm (kr> ). (1) (12.9) 4πτ m=−∞This is the 2-dimensional analog of what we did in three dimensions. eq11A4We use this to derive the plane wave expression in 3-dimensions. We may now obtain an alternative expression for the free spaceGreen’s function in two dimensions by shifting the origin. In particular,we place the origin at x . This gives us r = 0, for which (1) (1) H0 (kr) m = 0 Jm (kr )Hm (kr)|r =0 = 0 else.
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188 CHAPTER 12. STEADY STATE SCATTERING Thus equation 12.9 reduces to i (1) G0 (r) = H (kr), 4πτ 0 which can also be written as i (1) G0 (|x − x |) = H (k|x − x|). (12.10) 4πτ 0eq11A5 This is the expression for the two dimensional free space Green’s func- tion. In the process of obtaining it, we have proven the Hankel function addition formula ∞ (1) H0 (k|x − x|) = eim(ϕ−ϕ ) Jm (kr< )Hm (kr> ). (1) m=−∞ We now have the free space Green’s functions for one, two, and three dimensions: i ik|x−x | G1D (|x − x |) = 0 e , 2kτ i (1) G2D (|x − x |) = 0 H0 (k|x − x|), 4πτ eik|x−x | G3D (|x − x |) = 0 . (12.11) 4πτ |x − x |eq11A6 We can interpret these free space Green’s functions physically as fol-pr:fsp2 lows. The one dimensional Green’s function is the response due to a8 Apr 5 plane 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 | = 1 + ik|x − x |, and thus we recover the correct potential respectively for a sheet of charge, a line charge, and a point charge.
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12.4. SCATTERING FROM A CYLINDER 18912.4 Scattering from a CylinderWe consider again the Green’s function for scattering from a cylinder,equation 12.6 ∞ 1 G= eim(ϕ−ϕ ) [Jm (kr< ) + Xm Hm (kr< )]Hm (kr> ). (12.12) (1) (1) 4πτ m=−∞ (1)What is the physical meaning of [Jm (kr< ) + Xm Hm (kr< )] in this equa- eq11A6btion? This is gives the ﬁeld due to a point source exterior to the cylin-der: u = G(x, x , λ = ω 2 + iε)e−iωt = G0 e−iωt + (G − G0 )e−iωt (12.13) u0 usNote that Xm contains the physics of the boundary condition, eq11A6c ∂G − + κa G = 0 for r = a. ∂rFrom equations 12.12 and 12.13 we identify the scattered part of thesolution, us , as ∞ e−iωt us = eim(φ−φ ) Xm Hm (kr)Hm (kr ). (1) (1) (12.14) 4πτ m=0 (1)So we have expanded the total scattered wave in terms of Hm , where eq11A7Xm gives the mth amplitude. Why are the r> and r< in equation 12.11but not in equation 12.14? Because there is a singular point at x = xin equation 12.11, but us will never have a singularity at x = x . (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,
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190 CHAPTER 12. STEADY STATE SCATTERINGthen at large distances G must have the form of equation 12.11, eik|x−x | G0 (|x − x |) = . 4πτ |x − x |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 a due to point source is ∞ 1 G= eim(ϕ−ϕ ) [Jm (kr< ) + Xm Hm (kr< )]Hm (kr> ). (1) (1) 4πτ m=−∞ with [kJm (ka) − Ka Jm (ka)] Xm = − (1) (1) . [kHm (ka) − Ka Hm (ka)] 2. The free space Green’s function in one, two, and three dimensions is i ik|x−x | G1D (|x − x |) = 0 e , 2kτ i (1) G2D (|x − x |) = 0 H0 (k|x − x|), 4πτ eik|x−x | G3D (|x − x |) = 0 . 4πτ |x − x | 3. For the problem of scattering from a cylinder, the total response u is easily decomposed into an incident part a scattered part us , where us contains the coeﬃcient Xl .12.6 ReferencesSee any old nuclear of high energy physics text, such as [Perkins87].
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Chapter 13Kirchhoﬀ ’s FormulaAs a further application of Green’s functions to steady state problems,let us derive Kirchhoﬀ’s formula for diﬀraction through an aperture. pr:Kirch1Suppose we have a point source of sound waves of frequency ω at somepoint x0 in 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 x in the right-half plane. ﬁg19aIf we look at the screen directly, we see the aperture is the yz-planeat x = 0 with a hole of shape σ as shown in the ﬁgure. The solutionG(x, x0 ; ω) satisﬁes the equations 2 −[ + k 2 ]G(x, x0 ; ω) = δ(x − x0 ), ∂G = 0. ∂x x=0 x∈σ We will reformulate this problem as an integral equation. Thisintegral equation will have as a kernel the solution in the absence ofthe hole due to a point source at x in the R.H.P. This kernal is thefree space Green’s function, G0 (x, x ; ω), which satisﬁes the equations 2 −[ + k 2 ]G0 (x, x ; ω) = δ(x − x ), ∂G0 = 0. ∂x x=0 all y,zThis we solve by the method of images. The boundary condition is pr:MethIm2 191
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192 CHAPTER 13. KIRCHHOFF’S FORMULA S R x x0 v v σ Figure 13.1: A screen with a hole in it.satisﬁed by adding an image source at x ∗ , as shown in ﬁgure 13.2 Thusthe Green’s function for this boundary value problem is 1 eik|x−x | e−ik|x−x | G(x, x ) = + 4π |x − x | |x − x | 2Now by taking L0 = −( + k 2 ) we may apply Green’s second identity (S ∗ L0 u − uL0 S ∗ ) = dS n · [u S ∗ − S ∗ u] ˆ x∈R x∈Swith u = G(x, x0 ) and S ∗ = G0 (x, x ) where R is the region x > 0 andS is the yz-plane. This gives us L0 u = 0 for x > 0, ∂u = 0, ∂x x=0 x∈σand L0 S ∗ = δ(x − x ), ∂S ∗ = 0. ∂x x=0
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193 x v B ¨ ∗ ¨¨¡ x ¨¨ x ¡! x0 v v v x v x=0 Figure 13.2: The source and image source.These identities allow us to rewrite Green’s second identity as ∂ − dxG(x, x0 )δ(x − x ) = dy dzG0 (x, x ) − G(x, x0 ) ∂x x=0and therefore ∂ G(x , x0 ) = − dy dzG0 (x, x ) G(x, x0 ) . (13.1) ∂x x=0Thus the knowledge of the disturbance, i.e., the normal component of eq19athe velocity at the aperture, determines the disturbance at an arbitrary ∂point x in the right half plane. We have then only to know ∂x G at theaperture to know G everywhere. Furthermore, if x0 approaches the aperture, equation 13.1 becomesan integral equation for G for which we can develop approximationmethods. ∂ G(x , x0 ) = − dy dzG0 (x, x ) G(x, x0 ). (13.2) x∈σ ∂xThe conﬁgurations for the diﬀerent G’s are shown in ﬁgure 13.3. Note eq19b ﬁg19b 1 eik|x−x | G0 (x, x )|x =0 = , 2π |x − x |and equation 13.2 becomes 1 eik|x−x | ∂ G(x , x0 ) = − dy dz G(x, x0 ) . 2π x∈σ |x − x | ∂x x=0
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194 CHAPTER 13. KIRCHHOFF’S FORMULA x x x0 x0 v v v v x x v v G0 (x, x ) G(x0 , x ) G(x, x0 ) Figure 13.3: Conﬁgurations for the G’s. Now suppose that the size a of 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 1 we can write 1 eik|x−x | ∂ ∂ eik|x−x | G(x, x0 ) ≈ − dy dz G(x, x0 ) , 2π x∈σ |x − x | ∂x x=0 ∂x 4π|x − x | x=0 where we have used the substitution ∂ ∂ eik|x−x | G(x, x0 ) = . ∂x x=0 ∂x 4π|x − x | x=0 This equation then gives us an explicit expression for G(x , x0 ) in terms of propagation from the source at x to the ﬁeld point x of the velocity ∂ eik|x−x | disturbance at x of the velocity distribution ∂x 4π|x−x | produced by freeﬁg19c propagation to x from the point x0 of the disturbance. This yields Huy-pr:Huyg1 gen’s principle and other results of physical optics (Babenet’s principle, etc.). 13.1 References See [Fetter80, pp327–332] for a discussion of these results.
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Chapter 14Quantum Mechanics Chapter Goals: • State the Green’s function equation for the inho- mogeneous Schr¨dinger equation. o • State the Green’s function for a bound-state spec- 8 Apr p8 tra in terms of eigen wave functions. pr:QM1 • State the correspondence between classical wave theory and quantum particle theory.The Schr¨dinger equation is o ∂ H − i¯ h ψ(x, t) = 0 ∂twhere the Hamiltonian H is given by h2 ¯ 2 H=− + V (x). 2mThis 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 ψ be a regular function.The expression |ψ(x, t)|2 is the probability given by the probabilityamplitude ψ(x, t). For the time dependent Schr¨dinger equation, we ohave the same form as the heat equation, with ρcp → i¯ . In making the h 195
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196 CHAPTER 14. QUANTUM MECHANICS V (x) i E>0 x i ii 2 $ 2 E <0 $2 h $ h h g gg ¡ f ¢ f ¢ e ¡ x Figure 14.1: An attractive potential. transition from classical mechanics to quantum mechanics, we use H = p2 /2m + V (x) with the substitution p → (¯ /i) ; this correspondence h 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.pr:sss3 We now look at the steady state form. Steady state solutions will be of the form ψ(x, t) = e−iωt ψω (x), where ψω (x) satisﬁes the equation [H − hω]ψω (x) = 0. ¯pr:enLev1 The allowed energy levels for hω are the eigen values E of H: ¯8 Apr p9 Hψ = Eψ. (14.1)eq13.1 So the allowed frequencies are ω = E/¯ , for energy eigenvalues E. The h energy spectrum can be either discrete or continuous. Consider theﬁg13.1 potential shown in ﬁgure 14.1. For E = En < V (0), the energy levels 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
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14.1. QUANTUM MECHANICAL SCATTERING 197 E-plane Im E u u u uu Re E Figure 14.2: The complex energy plane.branch cut on the positive real axis. Note that where as in this problemthere are a ﬁnite number of discrete levels, for the coulomb potentialthere are instead an inﬁnite number of discrete levels. If, on the otherhand, we had a repulsive potential, then there would be no discretespectrum. The Green’s function solves the Schr¨dinger equation with an inho- omogeneous δ-function term: 8 Apr p10 (H − E)G(x, x ; E) = δ(x − x ) (14.2)where E is a complex variable. The boundary condition of the Green’s eq13.2function 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) isenforced by an appropriate choice of the potential (e.g., V = ∞ forr > a). Once we have obtained the Green’s function, we can look at itsenergy spectrum to obtain the ψ’s and En ’s, using the formula obtainedin chapter 4: ψn (x)ψn (x ) G(x, x; E) = . n En − EThis formula relates the solution of equation 14.1 to the solution ofequation 14.2. ﬁg13.2 11 Apr p1 11 Apr p214.1 Quantum Mechanical Scattering 11 Apr p3We now look at the continuum case. This corresponds to the prob- pr:QMS1lem of scattering. We use the Green’s function to solve the problem of
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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 solution u for the classical wave problem is interpreted as a velocity potential, whereas the solution ψ for the quantum mechanical problem is interpreted as the probability amplitude. For the classical wave prob- lem, u2 is interpreted as intensity, while for quantum mechanics, |ψ|2 is 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 ), with λ = ω 2 + iε for causality, to obtain the steady state response due to a point source, u = e−iωt G(x, x; λ = ω 2 + iε). This steady state solution solves the time dependent classical wave equation, ∂2 L0 + σ 2 u(x, t) = δ(x − x ). (14.3) ∂teq13.3 We may decompose the solution u into two parts, u = u0 + uscat , where e−i(ωt−kR) u0 = e−iωt G0 = 4πτ R where R = |x − x |, and uscat = e−iωt [G − G0 ].pr:uscat1 Note that u0 is the steady state solution for a point source at x = x , solution for a point source at x, G0 is the solution for the free case, and us is the solution for outgoing scattered waves. Note also that the outgoing scattered waves have no singularity at x = x .
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14.2. PLANE WAVE APPROXIMATION 19914.2 Plane Wave Approximation pr:PlWv1If one solves the problem of the scattering of the spherical wave from 11 Apr p4a point source, that is, for the Green’s function problem, then we alsohave the solution for scattering from a plane wave, merely by letting|x | → ∞. We now deﬁne Φ(x, t) as the solution of equation 14.3 forthe special case in which x → −∞ˆ, z ∂2 L0 + σ Φ(x, t) = 0. (14.4) ∂t2 eq13.4 For the steady state solution 11 Apr p5 Φ(x, t) = e−iωt Φ(x, ω),we get the equation [L0 − σω 2 ]Φ(x, ω) = 0.How is this equation solved for positive ω? There isn’t a unique solutionfor 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 theGreen’s function with the source point going to inﬁnity in the abovemethod, but it does not satisfy the boundary condition appropriate for 11 Apr p6scattering. In order to determine the unique scattering solution, we must in-troduce a new boundary condition appropriate for scattering. This isnecessary because although our previous equation [L0 − σλn ]un = 0 (14.5)with RBC gave the eigenfunctions, it does not give unique physical eq13abcsolutions for the case of scattering. To ﬁnd the un ’s in equation 14.5 we would extract them from theGreen’s function using, for the discrete case, un u ∗ n G(x, x ; λ) = , n λn − λ
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200 CHAPTER 14. QUANTUM MECHANICS or for the continuum case, 1 un = [G(x, x ; λ = λn + iε) − G(x, x ; λ = λn − iε)] 2πi 1 = Im [G(x, x ; λ = λn + iε)]. π But these un ’s are not solutions corresponding to the scattering bound- ary condition, since they contain both incoming and outgoing waves. 14.3 Quantum Mechanics11 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/¯ )t ψ0 . h We want the total wave to be a superposition of an incident plane wave and a scattered wave. ψ = eik·x−iωt + ψs where ω = E/¯ . Note that eik·x corresponds to an incident plane wave, h11 Apr p9 and ψs corresponds to outgoing waves. To get this form, we use the Green’s function for the free space problem (no potential) eikR m eikR G0 = = , 4πRτ 2π¯ 2 R h where λσ E 2mE p2 k2 = = 2 = = 2. τ h /2m ¯ h2 ¯ h ¯ Take the limit |x | → ∞, the free space Green’s function becomes m eikr e−i(E/¯ )t G0 (x, x ; E) = ψ0 h , 2π¯ 2 r h11 Apr p9 where ψ0 = e−ik·x−i(E/¯ )t . h
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14.4. REVIEW 201 14.4 Review13 Apr p1 We have been considering the steady state response problem ∂2 L0 + σ u(x, t) = δ(x − x )e−iωt . ∂t2 The steady state response for outgoing waves (i.e., that which satisﬁes the boundary condition for scattering) is u(x, t) = e−iωt G(x, x , λ = ω 2 + iε) (14.6) where G solves eq14.g [L0 − σλ]G(x, x ; λ) = δ(x − x ). We want to get the response Φ(x, t) for scattering from a plane wave. We only need to let |x | go to inﬁnity: eikr lim u(x , t) = Φ(x, t). |x |→∞ 4πr This gives scattering from a plane wave. Φ is the solution of ∂2 [L0 + σ ]Φ(x, t) = 0 ∂t2 which satisﬁes the boundary condition of scattering. For the case of steady state response we can write Φ(x, t) = e−iωt Φ(x, ω) where Φ(x, ω) solves the equation [L0 − σω 2 ]Φ(x, ω) = 0. This equation satisﬁes the boundary condition of scattering: 13 Apr p2 Φ(x, ω) = eik·x + Φs (x), where Φs (x) has only outgoing waves. Note that ω2σ ω2 k2 = = 2 and k = kˆ , n τ c where σ = limr→∞ σ(r) and τ = limr→∞ τ (r).
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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 − σω 2 ]uα (x, ω 2 ) = 0. In this equation uα is a positive frequency eigen function with degener- acy number α and eigen value ω 2 . The eigen functions can be obtained directly from the Green’s function by using 1 Im G(x, x ; λ = ω 2 + i ) = uα (x, ω 2 )u∗ (x , ω 2 ). α π α For the case of spherical symmetry we have uα (x, ω 2 ) = ulm (x, ω 2 ) = Ylm (θ, ϕ)ul (r, ω 2 )13 Apr p3 where l = 0, 1, . . . and m = −l, . . . , 0, . . . , l. The eigenvalues ω 2 are continuous: 0 < ω 2 < ∞. Because of the degeneracy, a solution of the diﬀerential equation may be any linear combination of the degenerate eigen functions: Φ(x, ω) = cα uα , where the cα ’s are arbitrary coeﬃcients. This relates the eigen function to the plane wave scattering solution. In the next chapter we will see how the cα ’s are to be chosen. 14.6 Comparison of Classical and Quan- tumpr: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
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14.6. COMPARISON OF CLASSICAL AND QUANTUM 203u2 = energy density |ψ|2 = probability densityωn = natural frequencies En = energy eigenvaluesNormal modes: Stationary states un (x)e−iωn t = un (x, t) ψn (x)e−i(En /¯ )t = ψn (x, t) h 2L0 un = σωn un Hψn (x) = En ψn (x)L0 is positive deﬁnite: H + = H: 2 ωn > 0 En ∈ R 13 Apr p4 The scattering problem is like the eigen value problem but we lookat the region of continuous spectrum. For scattering, we require thatEn > 0. In this continuum case, look at the eigen values from: (H − E)ψα (x, E) = 0, (14.7)where α labels wave functions with degenerate eigenvalues. Rather eq13.25than a wave we have a beam of particles characterized by some energyE. The substitution from classical mechanics to quantum mechanics isas follows: h2 ¯ E 2mE p 2 ω 2 σ → E, τ→ , k2 → 2 = 2 = . 2m h /2m ¯ h ¯ h ¯This last equation is the De Broglie relation. pr:DeBr1 Now we want to look at the solution for quantum mechanical scat- 13 Apr p5tering using Green’s functions. Suppose we have a beam of particlescoming in. This incident free wave has the form eik·x−iEt/¯ which solve hthe free space hamiltonian h2 ¯ 2 H0 = . 2mWe want the solution to equation 14.7 which corresponds to scattering.That is, we want the solution for (H − E)Φ(x, E) = 0which is of the form Φ(x, E) = e−k·x + Φswhere Ψs has only outgoing waves.
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204 CHAPTER 14. QUANTUM MECHANICS To solve this we look at the Green’s function. (H − E)G(x, x ; E) = δ(x − x ) for ImE > 0, with the appropriate boundary conditions. We make the substitution ψ(x, t) = e−i(E/¯ )t G(x, x ; E + iε). h This corresponds not to a beam of particles but rather to a source of13 Apr p6 particles.Stuﬀ omitted 14.7 Summary 1. The Green’s function equation for the inhomogeneous Schr¨dinger o equation is (H − E)G(x, x ; E) = δ(x − x ), where h2 ¯ 2 H=− + V (x). 2m 2. The Green’s function for a bound-state spectra in terms of eigen wave functions is ψn (x)ψn (x ) G(x, x; E) = . n 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.
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Chapter 15Scattering 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. 15 Apr p1 • 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.We have seen that the steady state case reduces the Green’s functionproblem to the equation [L0 − λσ]G(x, x ; λ) = δ(x − x ) 205
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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), and τ (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 ; λ) = Ylm (θ, ϕ)Gl (r, r ; λ)Ylm∗ (θ , ϕ ). lm 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 sphericalOnly partly symmetry.true. Now consider the case of spherical symmetry. From chapter 3 we know that the radial Green’s function can be written ul (r< , λ)ul (r> , λ) Gl (r, r ; λ) = − 12 2 . (15.1) r τ (r)W (ul , ul ) 1 2eq14.0 The u’s solve the same radial eigenvalue equation 1 d d τ (r)l(l + 1) − 2 dr r2 τ (r) + + V (r) − λσ(r) ul = 0, (15.2) 1,2 r dr r2eq14.1 but diﬀerent boundary conditions. The eigenfunction u1 satisﬁes the15 Apr p2 boundary condition ∂ l u − κul = 0 for r = a ∂r 1 1 and as a → 0 we replace this with the boundary condition ul (r) ﬁnite 1 at r = 0. The eigenfunction u2 satisﬁes the boundary condition u2 ﬁnite as r → ∞. 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, τ (r)l(l + 1) Veﬀ = + V (r). r2pr:Veﬀ1 In quantum mechanics we have h2 l(l + 1) ¯ Veﬀ = + V (r). 2mr2
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15.1. ANGULAR MOMENTUM 20715.1 Angular MomentumThe above spherical harmonic expansion for the Green’s function wasobtained by solving the corresponding eigenfunction equation for theangular part, h2 Lθϕ Ylm = h2 l(l + 1)Ylm . ¯ ¯The diﬀerential operator Lθϕ can be related to angular momentum byrecalling that the square of the angular momentum operator satisﬁesthe equation L2 Ylm = h2 l(l + 1)Ylm . op ¯Thus we identify h2 Lθϕ as L2 , the square of the angular momentum ¯ opoperator. h2 Lθϕ ≡ L2 . ¯ opIn the central potential problem of classical mechanics it was foundthat L2 Veﬀ = + V (r), 2mr2where pr:bfL1 L=x×pand L2 = L · L.In quantum mechanics the momentum operator is p = (¯ /i) , so that h h ¯ L = x × p → Lop = x× iand thus h ¯ h ¯ L2 = op x× · x× i i = h2 Lθϕ . ¯This gives the relation between angular momentum in classical mechan-ics and quantum mechanics.
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208 CHAPTER 15. SCATTERING IN 3-DIM $ s # x $ ¨ ¨ e ¡ e s ¡ ¡ E ¡ x e ! d d e % d " ! & % 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-pr:ExpScat1 perimental scattering measurements and is shown in ﬁgure 15.1. Weﬁg14a 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 = σλ/τ , where λ = ω 2 + i classically, and λ = E + i for the quantum case. In classical mechanics we then15 Apr p3 have k = ω/c and in quantum mechanics we have k = p/¯ . h Our incident wave is from a point source, but by taking the source- to-target separation r big, we have a plane wave approximation. After making these approximations, equation 15.2 becomes 1 d d l(l + 1) − 2 dr r2 + − k 2 ul = 0. 1,2 (15.3) r dr r2eq14.2 If we neglect the term l(l + 1)/r2 , compared to k 2 we would need kr 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) as r → ∞, and since we also have to consider l large, we don’t want to kill it. In this limit the Green’s function is proportional to a product of the u’s, lim Gl (r, r ; λ) = Aul (r , λ)ul (r, λ). r→∞ 1 2 (15.4)
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15.2. FAR-FIELD LIMIT 209We assume the the point source is not in the region where things are eq14.3really happening, but far away. In this case V (r ) = 0, σ(r ) = σ, andτ (r ) = τ , for large r . Thus we are looking at the far ﬁeld solutionwhere the point source is outside the region of interaction. 15 Apr p4 We already know the explicit asymptotic solution to the radial equa-tion: (1) ul (r> ) ≈ hl (kr> ) 2 for kr> 1, (15.5) (1) (1) ul (r< ) 1 ≈ jl (kr< ) + Xl hl (kr< ) for kr< 1. (15.6)Xl contains all the physics, which arises due to the boundary condition. eq14.5In general, Xl must be evaluated numerically. For speciﬁc cases such asin the problem set, V (r) = 0 so equation 15.3 is valid everywhere, andthus we may obtain Xl explicitly. For our present situation we haveassumed the far ﬁeld approximation and an interaction-free source, forwhich the asymptotic form of the Green’s function may be written interms of equation 15.5 and 15.6 as (1) (1) lim Gl (r, r ; λ) = A[jl (kr ) + Xl hl (kr )]hl (kr). r→∞We have not yet speciﬁed r > r , only that r 1 and r 1. To obtain the scattered wave at large r, we look at G − G0 . Inparticular we will evaluate Gl − Gl0 . So we look at Gl (r, r ; λ) for 0r r . Recall that Gl0 has the form 15 Apr p5 ik (1) Gl0 (r, r ; λ) = jl (kr< )hl (kr> ). τThis is for the free problem; it solves all the way to the origin. Nowtake the diﬀerence. ik (1) (1) (1) Gl − Gl0 = A − jl (kr< )hl (kr> ) + AXl hl (kr< )hl (kr> ). τThis equation assumes only that we are out of the range of of interac- (1)tion. The term hl (kr> ) gives the discontinuity on dG/dr at r = r .We now assert that A must equal ik/τ because the scattering wave What does thisGl − Gl0 has to be nonsingular. Now look at the case r > r : mean? 15 Apr p6 pr:scatWv1
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210 CHAPTER 15. SCATTERING IN 3-DIM ik (1) (1) Gl − Gl0 = Xl hl (kr )hl (kr). τ 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= cm um (r, ϕ). m=−∞15 Apr p7 At large distances we have Φ → eixt + outgoing waves.18 Apr p118 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 and σ and τ are 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 1) point source in a homogeneous medium incident on a target, which scatter and are detected by detectors far away (r 1). 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 explicitlypr:scGF1 calculate the scattered Green’s function GlS (r, r ; λ): ik GlS = Gl (r, r , λ) − Gl0 (r, r , λ) = Xl h1 (kr)h1 (kr ), l l c
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15.5. SCATTERING AMPLITUDE 211 √with k = λσ/τ = λ/c, where c is the speed. The value of Xldepends on k and is obtained from the behavior of u1 at large r = |x|.We already know that ul (r) must be of the form 1 18 Apr p3 lim ul (r) = jl (kr) + Xl (k)h1 (kr), r→∞ 1 lsince this is the asymptotic form of the solution of the diﬀerential equa- What does thistion. Thus scattering reduces to this form. All we need is Xl , which mean Physi- lis obtained from the behavior of ul (r). In particular, we don’t need to 1 cally? Isn’t U1know anything about ul (r) if we are only interested in the scattering for distances 2problem, because at large distances it cancels out. less than The large distance behavior of the function which satisﬁes the bound- rsource ? Don’tary condition at small distances is what determines the scattering so- we need r < r?lution. In this case r and r are both large enough that we are inessentially a homogeneous region.15.5 Scattering Amplitude pr:scAmp1Consider the special problem where V = 0, σ = const., and τ = const.,with the boundary condition ∂u1 + ku1 = 0 for r = a. ∂rIn the problem set we found Xl by satisfying this condition. The resultwas [kj (ka) − κjl (ka)] Xl = − 1l . [khl (ka) − κh1 (ka)] lThis equation is valid for r > a. 18 Apr p4 Given Xl we can calculate the diﬀerence, Gl −Gl0 so we can calculateGlS . Thus we can determine the scattered wave, which we now do. We calculate the scattered piece by recalling the expansion in termsof spherical harmonics, GS = G − G0 = Ylm (θ, ϕ)[Gl (r, r ; λ) − Gl0 (r, r ; λ)]Ylm∗ (θ , ϕ ). l,m (15.7)We substitute into this the radial part of the the scattered Green’s eq14.6
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212 CHAPTER 15. SCATTERING IN 3-DIM I x 0 x − x r x γ θ E z ˆ r origin r cos θ Figure 15.2: The geometry deﬁning γ and θ. function, ∞ ik (1) (1) G(r, r ; λ) − G0 (r, r ; λ) = Xl (k)hl (kr)hl (kr ), (15.8) l=0 ceq14.7 and the spherical harmonics addition formula18 Apr p5 l 2l + 1pr:addForm1 Ylm (θ, ϕ)Ylm∗ (θ , ϕ ) = Pl (cos γ) m=−l 4π (−1)l = (2l + 1)Pl (cos θ) (15.9) 4πeq14.8 where cos γ = x · x . The geometry is shown in ﬁgure 15.2. The result ˆ ˆﬁg14b of plugging equations 15.8 and 15.9 into equation 15.7 is ∞ ik 1 1 (−1)l GS = G − G0 = Xl (k)hl (kr)hl (kr ) Pl (cos θ)(2l + 1). l=0 τ 4π 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 x→∞ i h(1) (x) −→ − (−i)l eix . x Thus in the far ﬁeld limit the scattered Green’s function becomes ∞ ik eikr (−i) G − G0 → Xk h1 (kr )(i)l (2l + 1)Pl (cos θ). l τ kr 4π l=018 Apr p6 This in the case for a detector very far away. We can write also this as
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15.7. PLANE WAVE SCATTERING 213 eikr ˜ G − G0 = f (θ, r , k), rwhere ∞ ˜ 1 f (θ, r , k) = (2l + 1)(i)l Pl (cos θ)Xl (k)h1 (kr ). l (15.10) 4πτ l=0This 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 isgiven 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 ∼ |f |2 . dtdΩ 18 Apr p7 We know the radial diﬀerential ds = r2 dr of the volume dV = dsdΩfor a spherical shell, so that we get dE dE ds = dtdΩ dtds drNote that dimensionally we have dtds = r12 and dr = r2 , so that dtdΩ is dE ds dEdimensionless. Thus it is the 1/r term in the scattered spherical wavewhich assures conservation of energy. pr:ConsE115.7 Plane Wave ScatteringWe now look at scattering from a plane wave. Let r = |x | go toinﬁnity. This gives us |kr | (1) r →∞ l+1 e hl (kr ) −→ (−i) . krIn this limit equation 15.10 becomes |kr | ˜(θ, r , k) → e f f (θ, k) 4πτ r
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214 CHAPTER 15. SCATTERING IN 3-DIM where 18 Apr p8 ∞ ekr i f (θ, k) = − (2l + 1)Pl (cos θ)Xl . (15.11) r k l=0eq14fth f (θ, k) is called the scattering amplitude for a ﬁeld observer from anpr:ftrk1 incident plane wave. We can now compute the total wave for the far ﬁeld limit with incident plane wave. It is u = e−iωt G = e−iωt [G − G0 + G0 ] eikr eikr = e−iωt − eik·x + f . 4πτ r r In this equation the term eik·x corresponds to a plane wave and the ikr term e r f corresponds to an outgoing scattered wave. So dE |f 2 | ∼ dtdΩ This is a problem in the problem set. 15.8 Special Cases20 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 , 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 after20 Apr p3 which we may ignore the potential. 15.8.1 Homogeneous Source; Inhomogeneous Ob- server In this case x is in a region where V (r) ≈ 0, and σ and τ are constant. We deﬁne u0 to be the steady state solution to the point source problem without a scatterer present, i.e., u0 is the free space solution. u0 = e−iωt G0 ,
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15.8. SPECIAL CASES 215where eik|x−x | G0 = . 4πτ |x − x |Further we deﬁne the scattered solution us ≡ u − u0 .To ﬁnd us we use equation 15.1 to get the spherical wave expansion ik (1) G0l = jl (kr< )hl (kr> ). τThus us = e−iωt (G − G0 ).For the case of a homogeneous source and an inhomogeneous observerr> = r , r< = r = 0. We take (l) (1) u2 (r ) = hl (kr ).Remember that r is outside the region of scattering, so ul solves the 20 Apr p4 2free space equation, 15.3, 1 d d l(l + 1) − 2 dr r2 + − k 2 u2 = 0, (15.12) r dr r2where k 2 = λσ/τ with the condition ul (r) ﬁnite as r → ∞. The general eq14.10 2solution to equation 15.12 is (l) (1) u2 (r ) = hl (kr ).We still need to solve the full problem for u1 with the total eﬀectivepotential V (r) = 0.15.8.2 Homogeneous Observer; Inhomogeneous SourceIn this case the source point is in the interior region. We want to ﬁnd 20 Apr p5u for x inside the medium, but we cannot use the us method as we didin case 1. The reason why it is not reasonable to separate u0 and uS inthis case is because the source is still inside the scattering region.
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216 CHAPTER 15. SCATTERING IN 3-DIM We replace r> → r and r< → r so that (l) (1) u2 (r) → hl (kr) (l) and u1 (r) satisﬁes the full potential problem So once again we only (l) need to solve for u1 (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 | we make this a special case of the previous case. Thus we have r> → r and r< → r . Now both u1 and u2 satisfy the reduced ordinary diﬀerential equation 1 d d l(l + 1) − 2 dr r2 + − k 2 u1,2 = 0, (15.13) r dr r2eq14rad where u1 satisﬁes the lower boundary condition and u2 satisﬁes the upper boundary condition. As we have seen, the asymptotic solutions to this equation are (l) (1) u1 (r ) → jl (kr ) + Xl hl (kr ). (15.14) (l) (1) u2 (r) → hl (kr). (15.15)eq14.11,12 To obtain Xl we must solve20 Apr p6 1 d d l(l + 1) − 2 dr r2 τ (r) + + V (r) τ (r) − λσ u1 = 0, r dr r2 and then take r 1. 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 ik −iωt (1) (1) Gl − G0 → l e Xl hl (kr)hl (kr ). τ We see that the ﬁeld at x is due to source waves u0 and scattered waves uS .
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15.8. SPECIAL CASES 217Homogeneous Source and Observer, Far FieldFor this case the source and the ﬁeld point are out of the region ofinteraction. We take r > r and r → ∞. For these values of r and r we have V = 0 and τ and σ constant.In this case eik|x−x |−iωt e−iωt G → e−iωt G0 = (15.16) 4πτ |x − x | (Eq.f ) ∞ (2l + 1) G0 = (−1)l Pl (cos θ)G0 l (15.17) l=0 4πwhere (Eq.g) ik (1) G0 l = jl (kr< )hl (kr> ) (15.18) τ (Eq.h) −iωt −iωt u=e G=e G0 + us (15.19)where (Eq.i) 22 Apr p3 us = e−iωt (G − G0 ) ik −iωt ∞ (2l + 1)(−1)l (1) (1) = e Xl Pl (cos θ)hl (kr)hl (kr ) τ l=0 4πwhere for large r, (1) u1 → jl (kr) + Xl hl (kr) (15.20)This is the large r behavior of the solution satisfying the small r bound- (Eq.k)ary condition.15.8.4 Both Points in Interior RegionWe put x very far away, next to a detector. The assumption that x liesin the vicinity of a detector implies kr 1. This allows us to makethe following simpliﬁcation from case 2: (1) (−i) ikr hl (kr) → (−i)l e . (15.21) krThus we can rewrite u1 . We have (Eq.m) 20 Apr p7
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218 CHAPTER 15. SCATTERING IN 3-DIM us = e−iωt (G − G0 ) (15.22) and the simpliﬁcation (Eq.n) (1) Gl → jl (kr )hl (kr). 0 (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 , u2 everywhere cases 1, 2 need to know u1 everywhere case 3 need to know u1 at large r only22 Apr p4 We now look at two more special cases. 15.8.6 Far Field Observation Make a large r expansion (r → ∞): (1) −i −ikr hl (kr) → (−i)l ( )e (15.24) kr(EQ.l) e−i(ωt−k|x−x |) e−i(ωt−kr) ˜ u= + f (θ, r , k). (15.25) 4πτ |x − x | r e−i(ωt−kr) ˜(Eq.m) The term r f (θ, r , k) is explicitly just the outgoing wave. We found ∞ ˜ 1 (1) f (θ, r , k) = (2l + 1)(i)l Pl (cos θ)Xl hl (kr ) (15.26) 4πτ l=0(Eq.n) ˜ The term f (θ, r , k) is called the scattering amplitude for a point source22 Apr p5 ˜ at r . The ﬂux of energy is proportional to f 2 .
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15.9. THE PHYSICAL SIGNIFICANCE OF XL 219 15.8.7 Distant Source: r → ∞ Let the distance of thee source go to inﬁnity. Deﬁne k = k(−ˆ ) x (15.27)(Eq.o) ˜ and in f , let r → ∞. This gives us eikr e−i(ωt−kr) u→ e−i(ωt−k·x) + f (θ, k) (15.28) 4πr τ r We can then get (Eq.p) ∞ 22 Apr p6 −i f= (2l + 1)Pl (cos θ)Xl (15.29) k l=0 ˜ This equation is seen in quantum mechanics. f is called the scattering (Eq.q) amplitude at angle θ, and does not depend on ϕ due to symmetry. The basic idea is that plane waves come in, and a scattered wave goes out. The wave number k comes from the incident plane wave. 22 Apr p7 15.9 The Physical signiﬁcance of Xl Recall that Xl is determined by the large distance behavior of the solu- pr:Xl1 tion which satisﬁes the short distance boundary condition. Xl is deﬁned by (1) (1) ul (kr) → jl (kr) + Xl (k)hl (kr). (15.30) This equation holds for large r with V = 0 and σ, τ constant. By using eq14.20 the identity 1 (1) (2) jl (kr) = hl (kr) + hl (kr) , 2 we can rewrite equation 15.30 as (1) 1 (2) (1) ul (kr) → hl (kr) + (1 + 2Xl )hl (kr) . (15.31) 2 We now deﬁne δl (k) by eq14.21 1 + 2Xl = e2iδl (k) ,
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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 (1) 1 (2) (1) ul (kr) → hl (kr) + e2iδl (k) hl (kr) , 2 or (1) 1 (2) (1) ul (kr) = eiδl e−iδl hl + eiδl hl . (15.32) 2eq14.22 The solution ul satisﬁes a real diﬀerential equation. The boundary 1 condition at r → 0 gives real coeﬃcients. Thus ul is real up to an 1 overall constant factor. This implies δl real. Another way of seeing this (1) (2) is to note that by the deﬁnition of hl and hl we have (2) (1) ∗ hl (kr) = hl (kr) . Thus the bracketed expression in equation 15.32 is an element plus its complex conjugate, which is therefore real. If ul (kr) ∈ R, then 1Ask Baker δl (kr) ∈ R.22 Apr p8 We now look at the second term in equation 15.32 for far ﬁelds, (1) r→∞ −i eiδl hl (kr) −→eiδl (k) (−i)l eikr . kr Note that (−i)l = e−iπl/2 . This gives (1) i i(kr−πl/2+δl ) eiδl hl (kr) = − e kr So 1 ul (kr) ∼ 1 sin(kr − πl/2 + δl (k)) r → ∞. (15.33) kreq14.23 Thus δl (k) is the phase shift of the lth partial wave at wave number k. In the case that V = 0 we have ul (kr) → ul (kr). 1 1,0 If there is no potential, then we have Xl (k) → 0,
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15.9. THE PHYSICAL SIGNIFICANCE OF XL 221 (l) (l) u1 (r) u1,0 (r) u (0) Rn Rn r Figure 15.3: Phase shift due to potential.22 Apr p9 and by using the asymptotic expansion of j, we see that equation 15.30 becomes 1 πl ul (r) ∼ 1,0 sin kr − . (15.34) kr 2 Thus the phase shift δl (k) is zero if the potential is zero. eq14.24 Consider the values of r for which the waves u1 and u1,0 are zero in 25 Apr p1 the far ﬁeld limit. For equation 15.33 and equation 15.34 respectively, 25 Apr p2 the zeros occur when 25 Apr p3 πl kRn − + δl = nπ, 2 and 0πl kRn − = nπ. 2 By taking the diﬀerence of these equations we have 0 k(Rn − Rn ) = −δl (k). (15.35) Thus δl (k) gives the large distance diﬀerence of phase between solutions (eq14.25 with interaction and without interaction. This situation is shown in ﬁg- 0 ure 15.3. For the case shown in the ﬁgure, we have Rn > Rn , which ﬁg14c means δl > 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.
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222 CHAPTER 15. SCATTERING IN 3-DIM We now verify this behavior by looking at the diﬀerential equation25 Apr p4 for the quantum mechanical case. We now turn to the quantum me- chanical case. In this case we set τ (r) = h2 /2m and k 2 (x) = 2mE in ¯ ¯2 h the equation l 1 d d Veﬀ (r) λσ(r) l − 2 dr r2 + − u1 (r) = 0. r dr τ (r) τ (r) So for the radial equation with no interaction potential we have λσ/τ = 2mE/¯ 2 , while for the radial equation with an interaction potential we h have λσ/τ = 2m(E − V )/¯ 2 . Thus the eﬀect of the interaction is to h change the wave number from 2 2mE k0 = , h2 ¯ to an eﬀective wave number 2m(E − V ) k 2 (r) = . (15.36) h2 ¯eq14.26 Suppose we have an attractive potential, V (r) > 0. Then from equation 15.36 we see k 2 (r) > k0 , which means momentum is increasing. 2 Also, since k 2 (r) > 0, increasing k 2 increases the curvature of u, which means the wavelength λ(r) decreases and the kinetic energy increases. Thus the case k 2 (r) > l0 corresponds to an attractive potential pulling 2 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 region25 Apr p5 of variable index of refraction. Now consider a repulsive potential with l = 0, as shown in ﬁgureﬁg14d 15.4. We have V (r) = E for r = r0 , and V (r) > E for r > r0 . In this latter case equation 15.36 indicates that k 2 (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, l28 April, p2b It is possible to calculate δl (k) directly from u1 without calculating Xl (k) as an intermediate step. To do this, let r → ∞ and then compare
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15.10. SCATTERING FROM A SPHERE 223 V (r) E r0 r Figure 15.4: A repulsive potential.this ul with the general asymptotic form from equation 15.33, u ∼ 1sin(kr − lπ/2 + δl (k))/r. A diﬀerent method for calculating δl (k) ispresented in a later section.15.10 Scattering from a Sphere pr:ScSph1We now look at the example of scattering from a sphere, which wasalready solved in the homework. We have the boundary conditions V = 0 at r = a ∂ l u + κu1 = 0 at r = a. ∂r 1We found in problem set 2, that Xl for this problem is 25 Apr p7 [kjl (ka) − κjl (ka)] Xl = (1) (1) (15.37) [khl (ka) − κhl (ka)] by solving the radial equation. Stuﬀ missing Now look at the long wavelength limit, which is also the low energylimit. In this case ka 1 where k = 2π/λ. We know asymptoticallythat jl (ka) ∼ (ka)l ,
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224 CHAPTER 15. SCATTERING IN 3-DIM and (1) 1 hl (ka) ∼ . (ka)l+1 Thus we have ka 1 l − κa Xl (k) −→ (ka)2l+1 1 l + κa since (ka)l (ka)2l+1 = . (ka)−l−125 Apr p8 Again, k = 2mE/¯ 2 . We now look at the phase shift for low energy h 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 = V0 e−r/a . In this case the shape of Veﬀ is similar, except that is has a potential barrier for low values of r. V and Veﬀ for this example are shown inﬁg14e ﬁgure 15.5. The centrifugal barrier increases as l increases, that is, it gets steeper. Thus, as l increases, the scattering phase shift gets smaller and smaller since the centrifugal barrier gets steeper. Recall that a represents the range of the potential and 2l + 1 rep- resents the eﬀect of a potential barrier. We assert that in the long
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15.11. CALCULATION OF PHASE FOR A HARD SPHERE 225 V (r) Veﬀ (r) r r a r0 Figure 15.5: The potential V and Veﬀ for a particular example.wave length limit, that is, low energy scattering, the phases shift goesgenerally as δl (k) ∼ (ka)2l+1 for ka 1. This is a great simpliﬁcation for low energy scattering. It means thatas long as ka 1, we need only consider the ﬁrst few l in the inﬁniteseries for the scattering amplitude f (θ). In particular, the dominantcontribution 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 orphysical optics limit. The low energy limit is useful, for example, in the study of thenuclear force, where the range of the potential is a ∼ 10−13 cm, whichgives ka 1. Note that in the geometrical optics limit, ka 1, it isalso possible to sum the series accurately. The summation is diﬃcult inthe middle region, ka ∼ 1. In this case many terms of the series mustbe retained. 27 Apr p1 27 Apr p2 27 Apr p315.11 Calculation of Phase for a Hard SphereWe use the “special case” from above. Take κ → ∞ (very high elasticconstant, very rigid media, a hard sphere). In this case u → 0 when
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226 CHAPTER 15. SCATTERING IN 3-DIM r = a. Thus we get from equation 15.37 − sin ka ka X0 (k) = ika − ieka −ieika − e−ika = 2ieika 1 = − [1 − e−2ika ]. 2 So e2iδ0 = 1 − [1 − e−2ika ] = e−2ika , and thus δ0 (k) = −ka. (15.38)eq14do In terms of quantum mechanics, this is like having27 Apr p4 V (r) = ∞ for r < a, V (r) = 0 for r > a. Outside we get the asymptotic solution form given in equation 15.33. For l = 0 and substituting equation 15.38, this becomes (0) 1 u1 = sin(kr − ka). r By substituting this into equation 15.13 it is easy to verify that this is an exact solution for ul=0 . This is exactly what we would expect: 1 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 shownﬁg14f in ﬁgure 15.6. 15.12 Experimental Measurementpr:ExpMeas1 We now look at the experimental consequences. Assume that we have solved for u1 and know Xl (k) and thus know δl (k). By writing the scattering amplitude from equation 15.11 in terms of the phase shift27 Apr p5 δl (k), we have
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15.12. EXPERIMENTAL MEASUREMENT 227 l=0 u1 (r) V (r) a r a r Figure 15.6: An inﬁnite potential wall. ∞ 1 e2iδl − 1 f (θ) = (2l + 1)Pl (cos θ) . k l=0 2iTo get δl for the solution for u1 , we look at large r. Note that e2iδl − 1 eiδl [eiδl − e−iδl ] = 2i 2i iδl = e sin δl .So ∞ 1 f (θ) = (2l + 1)eiδl sin δl Pl (cos θ). (15.39) k l=0 eq14.5515.12.1 Cross Section pr:CrSec1This scattering amplitude is the quantity from which we determine theenergy or probability of the scattered wave. However, the scatteringamplitude is not a directly measurable experimental quantity. Recall our original conﬁguration of a source, an obstacle, and adetector. The detector measures the number of particles interceptedper unit time, dN/dt. (It may also distinguish energy of the intercepted pr:N2particle.) This number will be proportional to the solid angle coveredby the detector and the incident ﬂux of particles. If we denote theproportionality factor as σ(θ, φ), then this relationship says that therate at which particles are scattered into an element of solid angle is isdN/dt = jinc dσ = jinc (dσ/dΩ)dΩ. Note that an element of solid angle pr:jinc1is related to an element of area by r2 dΩ = dA. The scattered currentthrough area dA is then dN/dt = jinc (σ(θ, φ)/dΩ)dA/r2 . From this we
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228 CHAPTER 15. SCATTERING IN 3-DIM identify the scattered current density σ(θ, φ) 1 jscat = jinc r. ˆ (15.40) dΩ r2eq14cs1 Now the quantum mechanical current density j is deﬁned in terms of the wave function: h ¯ j(r) = Re ψ † ψ , im where in the far-ﬁeld limit the boundary condition of scattering tells us that the wave function goes as r→∞ eikr ψscat −→N eikz + f (θ, φ) . r The wave function has an incident plane wave part and a scattered spherical wave part. The current density for the incident wave is then hk ¯ jinc = |N |2 z = jinc z , ˆ ˆ m and the current density for the scattered wave is hk |f (θ, φ)|2 ¯ |f (θ, φ)|2 jscat = |N |2 r + O(r−3 ) ≈ jinc ˆ r. ˆ (15.41) m r2 r2eq14cs2 By comparing equations 15.40 and 15.41, we identify the diﬀerential cross section as dσ ≡ |f (θ, k)|2 . (15.42) dΩeq14.57 This relationship between cross section and scattering amplitude agrees with dimensional analysis. Note that the only dimensionful quantity appear in equation 15.39 for f is k: dim(f (θ)) = dim(k −1 ) = dim(length). On the other hand, the dimension of the diﬀerential cross section is dim[dσ/dΩ] = 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 dσ σ(k) = dΩ = dΩ|f (θ, k)|2 . dΩ
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15.12. EXPERIMENTAL MEASUREMENT 22915.12.2 Notes on Cross SectionBy using equation 15.39 we can calculate the diﬀerential cross section: ∞ dσ 1 = 2 (2l + 1)(2l + 1)eiδl sin δl e−iδl sin δl Pl (cos θ)Pl (cos θ) dΩ k l,l =0 (15.43)In this equation we get interference terms (cross terms). These interfer- eq14.60ence terms prevent us from being able to think of the diﬀerential cross 27 Apr p6section as a sum of contributions from each partial wave individually.If we are measuring just σ, we can integrate equation 15.43 to get dσσ = dΩ (15.44) dΩ 1 ∞ = dΩ 2 (2l + 1)(2l + 1)eiδl sin δl e−iδl sin δl Pl (cos θ)Pl (cos θ) k ll =0We can simplify this by using the orthogonality of the Legendre poly- eq14.61nomials: 4π dΩPl (cos θ)Pl (cosθ) = δll . 2l + 1In equation 15.44 the terms eiδl cancel. So we now have σ(k) ≡ dΩ|f (θ, k)|2 ∞ 4π = (2l + 1) sin2 δl . (15.45) k2 l=0From this we can conclude that eq14.64 ∞ σ= σl , l=0where 4π (2l + 1) sin2 δl . σl = (15.46) k2Note that σl is the contribution of the total cross section of scattering eq14.65from the 2l + 1 partial waves which have angular momentum l. Thereare no interference eﬀects, which is because of spherical symmetry. 27 Apr p7
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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/dΩ. On the other hand, in the whole measurement of σ, there is still spherical symmetry, and thus no interference eﬀects. A measurement of σ(k) is much more crude than a measurement of dσ/dΩ. Because sine is bounded by one, the total cross section of the partial waves are also bounded: 4π σl ≤ σlmax = 2 (2l + 1), k or, by using λ = 2π/k, 2 λ σlmax = 4π (2l + 1). (15.47) 2πeq14.67 Note that the reality of δl (k) puts a maximum value on the contribution σl (k) of the lth partial wave on the total cross section. 15.12.3 Geometrical Limitpr:GeoLim1 In the geometrical limit we have ka 1, which is the long wavelength limit, λ a. Recall that in this limit δl ∼ (ka)2l+1 . Thus the dominant contribution to the cross section will come from 4π 4π σ0 = 2 sin2 δ0 = 2 (ka)2 = 4πa2 . (15.48) k k From equation 15.47 we have 2 max λ σ0 ∼ 4π , 2π27 Apr p8 and from equation 15.48 we have a 2 σ0 = 4πλ2 . λ max Comparing these gives us σ0 /σ0 1 for a/λ 1, which is the fraction of the incident beam seen by an observer.
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15.13. OPTICAL THEOREM 23115.13 Optical Theorem pr:OptThm1We now take the imaginary part of equation 15.46: ∞ 1 Imf (θ) = (2l + 1) sin2 δl Pl (cos θ). k l=0In the case that θ = 0 we get ∞ 1 Imf (0) = (2l + 1) sin2 δl . k l=0By comparing this with equation 15.45 we obtain k Imf (0) = σ . 4πThis is called the optical theorem. The meaning of this is that theimaginary part of the energy taken out of the forward beam goes intoscattering. This principle is called unitarity or conservation of momen-tum. The quantity Im f (θ)|θ=0 represents the radiation of the intensityin the incident beam due to interference with the forward scatteredbeam. This is just conservation of energy: energy removed from theincident beam goes into the scattered wave. 29 Apr p115.14 Conservation of Probability Inter- pretation: 29 Apr p215.14.1 Hard Sphere σk/4π as a proportional-For the case of a hard sphere of radius a we found that ity factor δ0 = −ka. pr:HardSph1In the case of (ka) 1, only the lower terms of equation 15.39 matter.Exact scattering amplitude from a hard sphere k = 0? So for a sphereof radius a, we have δl ∼ (ka)2l+1 .
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