Acoustic Scattering from a Sphere - Steve Turley

8 Dierential cross section for scattering from an acoustically hard sphere of radius 3 wavelengths. 239 Dierential cross section for scattering from an acoustically hard sphere of radius 10 wavelengths. 2410 Total intensity near a 2.5 wavelength radius hard sphere. . . . . . . . . . . . . . . . . . . . . . 2511 Dierential cross section for scattering from a penetrable sphere of radius 0.1 wavelengthswith k d /k = 1.5 and ρ d /ρ = 1.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2712 Dierential cross section for scattering from a penetrable sphere of radius 1 wavelength withk d /k = 1.5 and ρ d /ρ = 1.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2813 Dierential cross section for scattering from a penetrable sphere of radius 3 wavelengths withk d /k = 1.5 and ρ d /ρ = 1.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 2914 Dierential cross section for scattering from a penetrable sphere of radius 10 wavelengths withk d /k = 1.5 and ρ d /ρ = 1.5. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 3015 Total near eld for scattering from a penetrable sphere of radius 2.5 wavelengths and withk d = 1.2 and ρ d = 1.2. . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 311 IntroductionThis article derives the forumula for scalar 2d scattering from a sphere. It uses my earlier article on scalar2d scattering from a circle[1] as a starting point. That article in turn is based on a derivation on general 2dscattering[2]. Another derivation of many of these formulas can be found in an article by R. Kress availablein print[3] and on the web[4]. Since this article is intended to be more of a summary than derivation, I referthe read to the above articles and references therein for additional details and proofs.The problem of scattering from a homogenous bounded object inside another homogenous region involvessolving the Helmholtz Equation,(∇ 2 + k 2 )u(x) = 0, (1)where the pressure p(x, t) is related to u by the expressionp(x, t) = R { u(x)e iωt} . (2)The quantity k in Equation 1 is called the wave number and is related by the wavelength λ, angular frequencyω and speed of sound in the medium c by the relationsk = 2π λ(3)= ω c . (4)The wave number is actually a vector whose direction is the direction of propagation of the place wavesolutions to Equation 1. It is helpful to divide the total eld u into a sum of an incident plane wave u i anda scattered wave u s withu = u i + u s . (5)Since Equation 1 is linear and u i satises this equation, u s must satisfy it as well. Equation 1 has a uniquesolution in the regions outside the scatterer if u s satises the Sommerfeld radiation condition at innity( )∂uslimr→∞ ∂r − iku s = 0 (6)and appropriate boundary conditions on the surface of the scatterer.1.1 Boundary ConditionsIn this article, I will treat three boundary conditions. The rst is for scattering from a sound-soft obstaclewhere the excess pressure is zero on the boundary (Dirichlet boundary condition). This isn't the problem ofimmediate interest, but produces a good and simple test case. A similarly simple situation arises when theobject is a hard scatterer so that the velocity is zero at the boundary. The nal object I will treat is wherethe scatterer is a homogenous bounded scatterer. In that case, there are two boundary conditions which2

elate the wave inside and outside the scatter. Let u D be u inside the scatter. Continuity of the pressureand the normal velocity across the interface requires thatu = u D (7)1 ∂uρ ∂ˆn = 1 ∂u D(8)ρ D ∂ˆnat the boundary. The quantities ρ and ρ D are the respective densities of the material outside and inside thescatterer. The quantity∂ˆn ∂u is the normal derivative of u,where ˆn is the normal to the surface (pointing into the scatterer),∂u= ˆn · ∇u . (9)∂ˆnIn the case of a balloon, the problem which motivated this article, the boundary condition in Equation 8seems reasonable, but Equation 7 is probably not going to hold. The membrane on the surface of theballoon provides a gradient in the pressure. To a rst approximation, I'll assume the membrane doesn't haveany surface waves (the uids on either side can't exert transverse forces) and that the pressure boundarycondition is just a constant pressure dierence.where p b is a constant on the surface of the scatterer.1.2 Green's FunctionThe Green's function of Equation 1 isu = u D − p b , (10)G(x, x ′ ) = 1 e ik|x − x′ |4π |x − x ′ . (11)|Application of Green's rst and second identities to Equation 1 allow one to show that∫ {u(x) = u(x ′ ) ∂G(x, x′ )∂ˆn(x ′ − ∂u}) ∂ˆn (x′ )G(x, x ′ ) ds(x ′ ) , (12)∂Dwhere the integral is over the surface of D (which I'll represent as ∂D) and x is in the region outside of thescatterer.1.3 Far FieldThe eld u can be found far away from the scatterer by expanding the Green's function in Equation 11 ina Taylor series in |x|/|x ′ |. It is found that the wave is a spherical wave modulated by a factor f(ˆx) whichonly depends on the direction of observation ˆx.u(x) = eik|x|f(ˆx) (13)|x|f(ˆx) = 1 ∫}{u(x ′ ∂e−ikˆx·x′)4π∂ˆn(x ′ − ∂u) ∂ˆn (x′ )e −ikˆx·x′ ds(x ′ ) (14)∂DNote that this expansion need not be applied to compute the eld from the scattering problem. Equation 12can be solved for the eld anywhere in space outside of scatterer once u and its normal derivative on thesurface are known.3

1.4 Solutions in Spherical CoordinatesA separable solutions to Equation 1 can be found in spherical coordinates. There are two classes of solutions.The rst set of solutions are valid everywhere in space, but don't satisfy the radiation condition. They arecalled the complete solutions.V m n (x) = j n (kx)Y m n (ˆx) , (15)where x = |x|, j n are the regular spherical Bessel functions, and Ynm are the spherical harmonics. The secondclass of solutions satisfy the Sommerfeld radiation condition, but are not regular at the origin. They arecalled the radiating solutions.Wn m (x) = h (1)n (kx)Yn m (ˆx) , (16)where h n are the spherical Hankel functions withh (1)n (z) = j n (z) + iy n (z) . (17)The spherical Bessel functions are related to the regular Bessel functions by the relationf n (z) =√ π2z F n+1/2(z) , (18)where f n is one of j n , y n , or h n and F is one of J, Y , or H. The radiating solutions are complete in thesense that any radiating solution can be expanded as a linear combination of the terms in Equation 16.u(x) =∞∑n∑n=0 m=−na m n h (1)n (kx)Y m n (ˆx) . (19)The coecients a m n are determined by the bondary conditions in the problem. Once these coecients havebeen determined, the scattering amplitude in Equation 14 can be expressed asf(ˆx) = 1 k∞∑n=01i n+1n∑m=−na m n Y m n (ˆx) . (20)It will be useful later to have an expansion of the Green's function in Equation 11 in terms of these twosolutions. It is∞∑ n∑G(x, x ′ ) = ik Wn m (x)Vn m (x ′ ) , (21)n=0 m=−nwhere the overbar denotes the complex conjugate of an expression.2 PotentialsIt is convenient to solve scattering problems in terms of acoustic single-layer and acoustic double-layerpotentials. Physically, these correspond to continuous layers of monopoles and dipoles on the surface ofthe scatterer. The potentials are solutions to Equation 1 inside and outside the scatterer and satisfy theSommerfeldt radiation condition. The single-layer potential u and double-layer potential v away from thesurface are∫u(x) = ρ(x ′ )G(x, x ′ ) ds(x ′ ) (22)v(x) =∂D∫ρ(x ′ ) ∂G(x, x′ )∂ˆn(x ′ ds(x ′ ) .)(23)∂DExamination of Equation 12 shows that solutions of Equation 11 can be expressed as a combination ofEquations 22 and 23 with the densities ρ being the values of u or its normal derivative on the boundary ofthe scatterer.4

We are interested in the values of the potentials at the surface of the scatter. Taking the limits approachingthe surface requires some care because of a discontinuity in k. Taking this limit carefully, one nds that onthe surface∫u(x) = ρ(x ′ )G(x, x ′ ) ds(x ′ ) (24)∂D∂u ±∂ˆn∫∂D(x) = ∫v ±, (x) =∂Dρ(x ′ ) ∂G(x, x′ )∂ˆn(x ′ )ρ(x ′ ) ∂G(x, x′ )∂ˆn(x ′ )ds(x ′ ) ∓ 1 ρ(x) (25)2ds(x ′ ) ± 1 ρ(x) (26)2In addition, the normal derivative of v is continuous across the boundary ∂D. The + or - sign in the aboveformulas refers to the direction in which the limit is taken on the normal derivatives. A plus sign is the limitfrom the side in the direction of the normal to the surface. Formally,∂u ±∂ˆn ≡ lim ˆn(x) · ∇u(x ± hˆn(x)) . (27)h→+0With these limiting values of the potentials, it is simple to derive the following jump relations relatingthese potentials and their derivatives to each other on either side of the boundary.u + = u − (28)∂u +∂ˆn − ∂u −∂ˆn= −ρ (29)v + − v − = ρ (30)∂v +∂ˆn = ∂v −∂ˆn(31)We will use these relations to enforce the boundary conditions for our various cases. The integrals appearingin the Equations 22, 23, 24, 25, and 26 can be written as the operators∫(Sρ)(x) = 2 G(x, x ′ )ρ(x ′ ) ds(x ′ ) (32)∫(Kρ)(x) = 2∫(K ′ ρ)(x) = 2∂D∂D∂D∂(T ρ)(x) = 2∂ˆn(x)with these denitions, the jump relations can be written as∂G(x, x ′ )∂ˆn(x ′ ) ρ(x′ ) ds(x ′ ) (33)∂G(x, x ′ )ρ(x ′ ) ds(x ′ ) (34)∂ˆn(x)∫∂G(x, x ′ )∂ˆn(x ′ ) ρ(x′ ) ds(x ′ ) (35)∂Du ± = 1 Sρ (36)2∂u ±∂ˆn = 1 2 K′ ρ ∓ ρ (37)v ± = 1 Kρ ± ρ (38)2∂v ±∂ˆn = 1 2 T ρ (39)Using this short-hand notation, the conditions on the solutions to Equation 1 for the region inside thescatterer can be written as (u∂u/∂ˆnFor the exterior region this relation is(u∂u/∂ˆn)=)=( ) (−K S−T K ′( K −ST′) (u∂u/∂ˆnu∂u/∂ˆn). (40)). (41)5

3 Scattering TheoryOne of the rst things a mathematician worries about in scattering problems is whether the problem asposed as a solution and whether that solution is unique. As physicists, the existence question may seem abit quaint. Physically, it would seem imposible for there to be no solutions. Also, as we anticiated a singlephysical result, it would be unusal to have more than one possible solution.However, there is a special case, that is physically meaningful and can cause numerical problems. Considera system where the frequency of the sound matches an internal resonance in the scatterer. Since thesedisturbances go to zero on the boundary of the scatterer, they don't radiate and therefore don't contributeto the scattered wave. Any amount of these resonances added to the solution is therefore also a solution,and the equations will not have a unique answer.3.1 Integral EquationsThere are a number of dierent ways people use to solve the Equation 1 for scattering problems. One wayto classify two general approaches are dierential equation methods and intergral equation methods. I willfocus on integral equation methods in this article for a number of reasons:1. By applying Green's theorem, integral equations can reduce the problem domain from R 3 (3 dimensionalspace) to R 2 (two dimensional space) on the boundary of the scatter. (This only applies to homogenousmedia.)2. The exact Sommerfeld radiation condition is automatically taken into account. Dierential equationmethods need to do this by approximation radiation boundary conditions applied in the near eld.3. Finite dierence approaches used in dierential equation techniques involve small dierences betweenlarge numbers. This can cause numerical diculties unless the scattering surface is modeled accuratelyand great care is taken in doing the numerical derivatives.The integral equations developed in Section 1 and the potentials dened in Section 2 provide a good startingpoint. Applying Equation 41 to u s and Equation 40 to u i ,u s = Ku s − S ∂u s∂ˆnu i = −Ku i + S ∂u i∂ˆn . (43)∂u s∂ˆn = T u s − K ′ ∂u s∂ˆn(44)∂u i∂ˆn = −T u i + K ′ ∂u i∂ˆn(45)Subtractring Equation 42 from Equation 43 and using Equation 5, one hasu2u i = Ku + S ∂u∂ˆn + u . (46)Subtracting Equation 44 from Equation 45, one has2 ∂u i∂u= −T u + K′∂ˆn ∂ˆn + ∂u∂ˆn . (47)The existance of solutions to these equations depend on the smoothness of the surface, and in particularlythe functions u and∂ˆn ∂u on those surfaces. In the case of a sphere, we expect our surface to be smooth andnot cause diculties. We still face a problem of the uniqueness of the solutions near interior resonances(where Equation 46 has solutions for u i = 0 or Equation 47 has solutions for∂ˆn ∂u = 0). These solutions canbe added to any other solution of the non-homogeneous equations to also give a valid solution. Luckily, thesesolutions don't radiate, and won't change our nal computed scattered eld. We will need to take some carewith considering these limiting cases, however. One way the challenges of these internal resonances can beavoid is by solving a linear combination of Equation 46 and Equation 47.2 ∂u i∂ˆn − 2iu i = −T u + K ′ ∂u∂ˆn + ∂u∂u− iKu − iS − iu . (48)∂ˆn ∂ˆn(42)6

3.2 Applying Boundary ConditionsThere several classes of boundary conditions of interest in this case. I will rst consider the simplest one,called the exterior Dirichlet problem. In this case, the function u has a known value f on the surface of thescattering. (For the simplest possible Dirichelt problem, f = 0.) This would be the acoustically soft spherementioned earlier. It is treated in Section 3.2.1.Another interesting problem is where the normal discplacement is zero at the boundary of the scatterer.This would be the case with a hard scatterer. It is treated in Section 3.2.2.The nal problem (and the one of current interest) would be the general case of a penetrable scatterer.This is the most complicated of the three cases because it involves considering both the interior and exteriorsolutions. It is treated in Section 3.2.3.3.2.1 Dirichlet ProblemIntegral Equation Approach For an acoustically soft sphere with u = 0 on the boundary, Equation 46reduces to2u i = S ∂u∂ˆn . (49)This problem could be solved in several straightforward ways. I will illustrate use a somewhat involved approachthat has value of being generalizable for other boundary conditions and problems requiring numericalsolutions.The goal is to solve Equation 49 for ∂u/∂ˆn on the surface of the sphere. For notational simplicity, letσ = ∂u/∂ˆn on the surface of the sphere ∂D. Substituting this expression into Equation 49 with S as denedin Equation 32,∫u i = G(x, x ′ )σ(x ′ ) ds(x ′ ) . (50)∂DSince the Green's function has an expansion in terms of spherical harmonics, it makes sense to expand bothG and σ in spherical harmonics. Let the sphere have radius a. From Equations 21, 15, and 16 one hasG(aˆx, aˆx ′ ) = ik∞∑n∑n=0 m=−nj n (ka)h (1)n (ka)Y m n (ˆx)Y m n (ˆx ′ ) . (51)I have also used the fact the j n is real. We'll use the following expansion for σ.∞∑ n∑σ(ˆx ′ ) = s mn Yn m (ˆx ′ ) (52)n=0 m=−nSince the Y m n functions are orthogonal to each other with the relation∫Y m n (ˆx)Y m′n ′ (ˆx) dˆx = δ m m ′δ n n ′ (53)the integral in Equation 50 can be done explicitly. The Kronaker δ's appearing in Equation 53 reduce thequadruple sum to a double sum.u i = ik∞∑n∑n=0 m=−ns nm j n (ka)h (1)n (ka)Y m n (ˆx) (54)The constants s nm can be found by expanding the plane wave u i in spherical coordinates as well. Thisexpansion can be written asu i = e ik·x ∞∑n∑= 4π i n j n (kx) Yn m (ˆx)Y n m (ˆk) . (55)n=0m=−nComparing the terms in Equation 54 with those in Equation 55, we see thats nm = 4π k−mYn(ˆk)in−1h (1)n (ka) . (56)7

Substituting the value for s nm from Equation 56 into Equation 52, we have a solution for the normalderivative of u on the boundary.Equation 59 uses the identity∂u∂ˆn (aˆx′ ) = σ(ˆx ′ ) (57)= 4π ∞∑ i n−1 n∑Ykn=0 h (1)n m (ˆx ′ )Yn m (ˆk)n (ka) m=−n(58)= 1 ∞∑i n (2n + 1) P n(ˆx ′ · ˆk)ik(ka)(59)n=02n + 14π P n(ˆx · ŷ) =n∑m=−nh (1)nY m n (ˆx)Y m n (ŷ) . (60)Once σ is known, the far eld can be computed from Equation 14. Using the expansion in Equation 55,f(ˆx) = − 4π ∫ [∑ ∞i n−1 n∑Yk ∂D n=0 h (1)n m (ˆx ′ )Yn (ˆk)] ⎡ ⎤∞∑∑n ′m ⎣ (− n′ j n ′(ka) Y m′ m′n ′ (ˆx)Ynn (ka)) ⎦ dˆx (61)′m=−nn ′ =0m ′ =−n ′= 4πi ∑ j n (ka)km,n h (1)n (ka) Y n m (ˆx)Y n m (ˆk) (62)= i ∞∑(2n + 1) j n(ka)kh (1)n (ka) P n(ˆx · ˆk) . (63)n=0A more general solution good in the near and far eld can be computed from Equation 12 with the expansionin Equation 21 for the Green's function.u(x) =∞∑ i n−1n,n ′−4πin,n h (1)′ n (ka) j n ′(ka)h n ′(kx) ∑∫Yn m (ˆk)Y n m′Y ′ n m (ˆx ′ )Yn m′) dˆx ′m=−n,m ′ =−n ′ ∂D(64)=∞∑−4π i n h (1)n (kx) j n(ka)n∑Yn=0 h (1)n m (ˆk)Y n m (ˆx)n (ka) m=−n(65)=∞∑− (2n + 1)i n h (1)n (kx) j n(ka)h (1)n (ka) P n(ˆx · ˆk) . (66)n=0Series Approach A simpler, but less generalizable way to solve this problem is to apply the boundarycondition to series expansions of the incident plane wave and scattered wave at the surface os the sphere.An the surface,e ik·aˆx′ = 4π ∑ n,mi n j n (ka)Y m n (ˆx ′ )Y m n (ˆk) (67)u s (aˆx ′ ) = ∑ n,ma nm h (1)n (ka)Y m n (ˆx ′ ) . (68)For the soft sphere problem, the sum of these two series should be zero. Since the Ynmindependent, the terms in each series must be equal to minus each other.functions are linearlya nm = −4πi n j n(ka)h (1)n (ka) Y m n (ˆk) (69)u s (ˆx) = −4π ∑ i n j n(ka)h (1)n(ka) h(1)= − ∑ i n (2n + 1) j n(ka)h (1)nn (kx)Y m n (ˆk)Y m n (ˆx) (70)(ka) h(1)n (kx)P n (ˆk · ˆx) (71)8

3.2.2 Neuman ProblemIntegral Equation ApproachFor the acoustically hard sphere with ∂u/∂ˆn = 0, Equation 47 reduces to2 ∂u i= −T u . (72)∂ˆnIt can be solved using a similar approach to the one given in Section 3.2.1. Substituting the expression forthe operator T from Equation 35 into Equation 72,∂u i∂r∫∂D= ∂ 2 G(x, x ′ )∂r∂r ′ u(x ′ ) ds(x ′ ) . (73)Exanding the Green's function in spherical harmonics and taking the derivatives explicitly,∂ 2 G(aˆx, aˆx ′ ) ∑∞ n∑∂r∂r ′ = ik 3 j n(ka)h ′ (1)′n (ka)Yn m (ˆx)Y n m (ˆx ′ ) . (74)Exanding u in spherical harmonics,u(ˆx ′ ) =n=0 m=−nUsing the same orthogonality relations as before,∞∑n∑n=0 m=−nt nm Y m n (ˆx ′ ) . (75)∂u i∂r = ∑ ik3 t nm j n(ka)h ′ (1)′n (ka)Yn m (ˆx ′ ) . (76)n,mExpanding the radial derivative of the plane wave as in Equation 55,∞∂u i∂r = 4πk ∑n∑i n j n(ka)′ Yn m (ˆx ′ )Yn m (ˆk) . (77)n=0m=−nComparing the series in Equation 76 with the expansion in Equation 77, one can solve for t nm .t nm = 4πk−mYnin−12h (1)′nSubstituting Equation 78 back into Equation 75 yields an expression for u on the surface.u(ˆx ′ ) = 4π ∑ ∞i n−1 n∑k 2Yh (1)′n m (ˆk)Y n m (ˆx ′ ) (79)n (ka)n=0m=−n(ˆk)(ka)Plugging Equation 79 into Equation 14 with the plane wave expanded as before,f(ˆx) === i k4πik4πik∞∑n=0∞∑n,n ′ =0∞∑n=0i n (−i) n′ j′ n ′(ka)h ′ n(ka)j ′ n(ka)h ′ n(ka)n∑m=−nn,n ′∑Yn m (ˆk)Y n m′′m,m ′ =−n,−n ′(ˆx) ∫∂D(78)Y m n (ˆx ′ )Y m′n ′ (ˆx′ ) ds(ˆx ′ ) (80)Y m n (ˆk)Y m n (ˆx) (81)(2n + 1) j′ n(ka)h ′ n(ka) P n(ˆx · ˆk) . (82)The general solution which is also valid in the near eld comes from substituting Equation 79 into Equation 12with the expansion in Equation 21 for the Green's function.u(x) = −4π ∑ i n h (1)nn,m∞∑= − (2n + 1)i n h (1)n=0(kx) j′ n(ka)h (1)′n (ka) Y m n (ˆk)Y m n (ˆx) (83)n(kx) j′ n(ka)h (1)′n (ka) P n(ˆx · ˆk) . (84)9

Series Approach As the the soft sphere, the hard sphere problem can be solved by requiring that thesum of the series for the normal derivative of the incident and scattered waves by zero.3.2.3 Penetrable Scatterer∂∂ˆn u i = − ∂ eik·aˆx′ ∂x ′ = −4πk ∑ i n j n(ka)Y ′ n m (ˆx ′ )Yn m (ˆk) (85)n,m− ∂∂ˆn u s = ∂ ∑∂x ′ a nm h (1)n (kx ′ )Yn m (ˆx ′ ) (86)= k ∑ n,mn,ma nm = −4πi n j′ n(ka)a nm h (1)′n (ka)Y m n (ˆx ′ ) (87)h (1)′n(ka) h(1)u s (ˆx) = −4π ∑ i n h (1)nn,m∞∑= − (2n + 1)i n h (1)n=0n Y m n (ˆk) (88)(kx) j′ n(ka)h (1)′n (ka) Y m n (ˆk)Y m n (ˆx) (89)n(kx) j′ n(ka)h (1)′n (ka) P n(ˆx · ˆk) (90)Integral Equation Approach The penetrable sphere can be computed with the series approach just aswas done with the soft and hard spheres. Because of the orthogonality of the spherical harmonics, it leadsto the same solution as the direct series solution which is outlined below. Since the extra algebra does littleto enlighten the reader about the characteristics of the solution, it will be omitted here.Series Approach I will break the eld u at the boundary into three components: u i (the initial planewave), u s (the scattered wave), and u d (the wave inside the scatterer). Expressing each of these as a seriesas before,u i = 4π ∑ n,mi n j n (ka)Y m n (ˆx ′ )Y m n (ˆk) (91)u s = ∑ n,ma nm h (1)n (ka)Y m n (ˆx ′ ) (92)u d = ∑ n,mb nm j n (k d a)Y m n (ˆx ′ ) . (93)Neglecting the pressure dierence enabled by the balloon (imagine scattering from bubbles in water, forinstance),u i + u s = u d . (94)We can recover our acoustically soft sphere solution by setting b nm = 0. Since each term in the sum mulipliesby the same factor of Y m n (ˆx ′ ), each term must be equal to zero.4πi n j n (ka)Y m n (ˆk) + a nm h (1)n (ka) − b nm j n (k d a) = 0 (95)4πi n j n (ka)Yn m (ˆk) + a nm h (1)n (ka)j n (k d a)= b nm (96)With our without the balloon, the radial velocities must match at the boundary as well (Equation 8).1 ∂ρ ∂x ′ (u i + u s ) = 1 ∂ρ d ∂x ′ u d (97)10

Setting the sum of the derivative terms equal to zero yields4πkρ in j ′ n(ka)Y m n (ˆk) + k ρ a nmh (1)′n (ka) = k dρ db nm j ′ n(k d a) (98)4πkρ in j n(ka)Y ′ n m (ˆk) + k ρ a nmh (1)′n (ka)k dρ dj n(k ′ d a)= b nm . (99)Setting the expression for b nm in Equation 96 equal to the expression for b nm in Equation 99,4πi n j n (ka)Yn m (ˆk) + a nm h (1)n (ka)j n (k d a)a nm[h (1)n (ka)j n (k d a) − kρ dh (1)′n (ka)k d ρj n(k ′ d a)Plugging Equation 100 into Equation 92 (with a going to x),]= 4πkρ di n j n(ka)Y ′ n m (ˆk) + kρ d a nm h (1)′n (ka)k d ρj n(k ′ d a)[= 4πi n kρdYn m j ′(ˆk)n(ka)k d ρj n(k ′ d a) − j ]n(ka)j n (k d a)a nm = 4πi n Yn m j n (k d a)kρ d j ′(ˆk)n(ka) − k d ρj n(k ′ d a)j n (ka)k d ρj n(k ′ d a)h (1)n (ka) − j n (k d a)kρ d h (1)′n (ka) . (100)u s (x) = 4π ∑ i n j n(k d a)kρ d j n(ka) ′ − k d ρj n(k ′ d a)j n (ka)n,m k d ρj n(k ′ d a)h (1)n (ka) − j n (k d a)kρ d h (1)′n (ka) h(1) n (kx)Yn m (ˆk)Y n m (ˆx) (101)= ∑ i n kρ d j(2n + 1)n(ka)j ′ n (k d a) − k d ρj n(k ′ d a)j n (ka)nk d ρj n(k ′ d a)h (1)n (ka) − kρ d h (1)′n (ka)j n (k d a) h(1) n (kx)P n (ˆk · ˆx) (102)f(ˆx ′ ) = − i ∑ kρ d j(2n + 1)n(ka)j ′ n (k d a) − k d ρj n(k ′ d a)j n (ka)kk d ρj n(k ′ d a)h (1)n (ka) − kρ d h (1)′n (ka)j n (k d a) P n(ˆk · ˆx) . (103)nNote that if k = k d and ρ = ρ d , the scattered wave has an amplitude of zero as would be expected. In thelimit of k d small, Equation 102 reduces to that for scttering from a hard sphere. In the limit of k d large,Equation 102 reduces to that for scattering from a soft sphere.4 Balloon ScatteringIn this section, I'll generalize the results from Section 3 to that from an idealized balloon. To make generalizationsof this calculations explicit, I'll start by outlining the assumptions of the theory in this section.1. The balloon is a perfect sphere.2. The balloon membrane is neglibily thin and has the sole eect of providing a pressure dierence betweenthe gas on the outside of the balloon and inside the balloon.3. The gasses inside and outside the balloon have uniform density and temperature.5 ResultsIn this section, I have included some numerical results for the formas derived in this paper. They are dividedinto three groups, the acoustically soft sphere, the acoustically hard sphere, and penetrable scatterers.5.1 Soft SphereThe following Matlab functions were used to plot scattering from a soft sphere of radius a. The rst thingI needed were some short functions to computer spherical Bessel functions. The function in the sphj.mcomputes the spherical Bessel function of the rst kindj n (x) =√ π2x J n+1/2(x) . (104)11

function j=sphj(n,x)% compute the spherical bessel function of the first kind% of order n at the point xj=sqrt(pi/(2*x))*besselj(n+0.5,x);endSimilarly, the le function in sphh.m computes the spherical Hankel function of the rst kindh n (x) =√ π2x H n+1/2(x) . (105)function h=sphh(n,x)% compute the spherical hankel function of the first kind% of order n at the point xh=sqrt(pi/(2*x))*besselh(n+0.5,x);end5.1.1 Far FieldThe function soft.m computes the cross section |f| 2 for a soft sphere of radius a (measured in wavelengths).function [theta,intensity]=soft(a)% *** soft(a) ***% compute scattering from a soft sphere of radius a% at the angle theta using nmax as the maximum value% of n in the sum.% The terms in the series oscillate up to about n=ka and then% decrease rapidly. Taking 2ka terms, should be more than sufficient% for large spheres. It always uses a minimum of 5 terms.theta=0:0.01:pi; % scattering angle in radiansk=2*pi; % wave number with wavelength of 1nmax=2*k*a+5;nary=0:nmax;nterm=2*nary+1;jn=sphj(nary,k*a);hn=sphh(nary,k*a);% legendre calculates the associated legendre polynomials.% I'm only interested in the m=0 onefor term=narylpn=legendre(term,cos(theta));pn(term+1,:)=lpn(1,:);endfor term=1:size(theta,2)intensity(term)=abs(i/k*sum(nterm.*jn./hn.*pn(:,term)'))^2;endendIt is somewhat slow to run this program in Matlab for large spheres, but works ne for spheres up to 10wavelengths or so. Figures 1 through 4 are the squares of the scattering amplitude for acoustically softspheres of various wavelengths.These plots generally look okay to me, but I'm worried about a factor of 2 in the normalization. For alarge sphere, I would expect the total cross section∫σ = |f(Ω 2 dΩ (106)12

0.016Soft Sphere, a=0.10.014d sigma/d omega0.0120.010.0080.0060.0040 20 40 60 80 100 120 140 160 180theta, degreesFigure 1: Dierential cross section for scattering from an acoustically soft sphere of radius 0.1 wavelengths.13

20Soft Sphere, a=1.0181614d sigma/d omega1210864200 20 40 60 80 100 120 140 160 180theta, degreesFigure 2: Dierential cross section for scattering from an acoustically soft sphere of radius 1.0 wavelength.14

Soft Sphere, a=3.010 4 theta, degrees10 3d sigma/d omega10 210 110 00 20 40 60 80 100 120 140 160 180Figure 3: Dierential cross section for scattering from an acoustically soft sphere of radius 3.0 wavelengths.15

Soft Sphere, a=10.010 6 theta, degrees10 5d sigma/d omega10 410 310 210 10 20 40 60 80 100 120 140 160 180Figure 4: Dierential cross section for scattering from an acoustically soft sphere of radius 10.0 wavelengths.16

to be equal to the cross sectional area of the sphereA = πa 2 . (107)If I integrate | 2 /π numerically for spheres of radius 10 and 20, I get 10.28 2 and 20.02 2 .> > [theta10,dsigma10]=soft(10);> > [theta20,dsigma20]=soft(20);> > sqrt(sum(dsigma10.*sin(theta10))*(theta(2)-theta(1)))ans =10.2574> > sqrt(sum(dsigma20.*sin(theta20))*(theta(2)-theta(1)))ans =20.0186I can also do these integrals analytically by taking the large argument expansions for the Bessel and Hankelfunctions. Abramowitz and Stegun[5] formula 10.1.16 gives a series expansion for h (1)n (z) ash (1)∑ nn (z) = i −n−1 z −1 e iz (n + 1/2, k)(−2iz) −k . (108)k=0For large x, the dominant term in this series will be the k=0 term. Therefore, the asymptotic values for j nand h (1)n for large x areh n (x) = ei(x−π/2−nπ/2)(109)xj n (x) = cos ( x − π 2 − )nπ2. (110)xNote that the asymptotic expansion for h n in Equation 109 can be subsituted into Equation 63 to giveEquation 66 and into Equation 82 to give Equation 84, providing a good cross check on those sets offormulas.The integrated cross section can be computed using the orthogonality relation for the Legendre polynomials∫ 1The integrated square of Equation 66 is then∫σ =∫ dσdΩ dΩ =P n (x)P m (x) =−1|f| 2 dΩ = 2 k 222n + 1 δ n,m . (111)∞∑n=0(2n + 1) (j n(ka)) 2∣∣h (1)(ka) ∣ 2 . (112)I'd like to make a further reduction of this analytically, but haven't been able to gure it out yet.5.1.2 Near FieldI also implemented a computation of the near eld intensity in the le softnear.m. It takes a very longtime to run in matlab and is much more practical in a compiled language.function [x,y,intensity]=softnear(a)% *** soft(a) ***% compute scattering from a soft sphere of radius a% at the angle theta using nmax as the maximum value% of n in the sum.% The terms in the series oscillate up to about n=ka and then% decrease rapidly. Taking 2ka terms, should be more than sufficient.n17

% It always uses a minimum of 5 terms.x=-10:.1:10;y=x;k=2*pi; % wave number with wavelength of 1nmax=2*k*a+5;nary=0:nmax;nterm=2*nary+1;jn=sphj(nary,k*a);hn=sphh(nary,k*a);% legendre calculates the associated legendre polynomials.% I'm only interested in the m=0 onefor ix=1:size(x,2)for iy=1:size(y,2)r=sqrt(x(ix)^2+y(iy)^2);if(r

Figure 5: Near eld intensity for scattering from a soft sphere of radius 2.5 wavelengths.19

5.2.1 Far FieldThe function hard computes the dierential cross section for scattering from a hard sphere.function [theta,intensity]=soft(a)% *** soft(a) ***% compute scattering from a soft sphere of radius a% at the angle theta using nmax as the maximum value% of n in the sum.% The terms in the series oscillate up to about n=ka and then% decrease rapidly. Taking 2ka terms, should be more than sufficient.% It always uses a minimum of 5 terms.theta=0:0.01:pi; % scattering angle in radiansk=2*pi; % wave number with wavelength of 1nmax=2*k*a+5;nary=0:nmax;nterm=2*nary+1;jnp=sphjp(nary,k*a);hnp=sphhp(nary,k*a);% legendre calculates the associated legendre polynomials.% I'm only interested in the m=0 onefor term=narylpn=legendre(term,cos(theta));pn(term+1,:)=lpn(1,:);endfor term=1:size(theta,2)intensity(term)=abs(i/k*sum(nterm.*jnp./hnp.*pn(:,term)'))^2;endendFor large spheres, this gives an integrated result close to that of the small sphere. Figures 6 through 9 arethe squares of the scattering amplitudes for acoustically hard spheres.5.2.2 Near FieldThe total intensity in the near eld can be computed using the same FORTRAN program which did thenear eld for the soft sphere. The results for a 2.5 radius hard sphere are shown in Figure 10. The incidenteld is coming from the left.5.3 Penetrable Scatters5.3.1 Far FieldI implemented Equation 101 in the Matlab le penetrate.m. This hasn't been as extensively checked as thesoft.m and hard.m. It may not agree with them well in the cases where k d is very large and very small asit should. However, the FORTRAN version of this used for near eld calculations agrees very well with thecases of k d small, unity, and large.function [theta,intensity]=penetrate(a,kf,rhof)% *** penetrate(a) ***% compute scattering from a penetrable sphere of radius a% kf=k_d/k, here k is the wave number (2 pi/lambda) of the gas in the% exterior volume and k_d is the wave number inside the sphere.% rhof=rho_d/rho where rho is the density of the gas in the% exterior volume and rho_d is the density of the gas inside the sphere.% The terms in the series oscillate up to about n=ka and then20

Hard Sphere, a=0.18 x 10−4 theta, degrees76d sigma/d omega5432100 20 40 60 80 100 120 140 160 180Figure 6: Dierential cross section for scattering from an acoustically hard sphere of radius 0.1 wavelengths.21

6Hard Sphere, a=1.05d sigma/d omega432100 20 40 60 80 100 120 140 160 180theta, degreesFigure 7: Dierential cross section for scattering from an acoustically hard sphere of radius 1 wavelength.22

Hard Sphere, a=3.010 3 theta, degrees10 2d sigma/d omega10 110 010 −110 −20 20 40 60 80 100 120 140 160 180Figure 8: Dierential cross section for scattering from an acoustically hard sphere of radius 3 wavelengths.23

Hard Sphere, a=10.010 5 theta, degrees10 4d sigma/d omega10 310 210 110 00 20 40 60 80 100 120 140 160 180Figure 9: Dierential cross section for scattering from an acoustically hard sphere of radius 10 wavelengths.24

Figure 10: Total intensity near a 2.5 wavelength radius hard sphere.25

% decrease rapidly. Taking 2ka terms, should be more than sufficient.% It always uses a minimum of 5 terms.theta=0:0.01:pi; % scattering angle in radiansk=2*pi; % wave number with wavelength of 1ka=k*a;kda=k*kf*a;nmax=2*k*a+5;nary=0:nmax;nterm=2*nary+1;jn=sphj(nary,ka);hn=sphh(nary,ka);jnp=sphjp(nary,ka);hnp=sphhp(nary,ka);jnd=sphj(nary,kda);jnpd=sphjp(nary,kda);% legendre calculates the associated legendre polynomials.% I'm only interested in the m=0 onefor term=narylpn=legendre(term,cos(theta));pn(term+1,:)=lpn(1,:);endfor term=1:size(theta,2)num=rhof*jnp.*jnd-kf*jnpd.*jn;denom=kf*jnpd.*hn-rhof*hnp.*jnd;intensity(term)=abs(-i/k*sum(nterm.*num./denom.*pn(:,term)'))^2;endendFigures 11 through 14 are the squares of the scattering amplitudes for penetrable spheres.5.3.2 Near FieldThe same FORTRAN program used to compute near eld intenisities for the hard and soft spheres was alsoused with a penetrable sphere. Figure 15 shows the results of running this program for a sphere is radius 2.5wavelengths with k_d=1.2 and \rho_d=1.2. An interesting feature is the focussing of the incident beamwith a maximum intensity about 4 wavelengths beyond the center of the sphere.5.4 BalloonReferences[1] S. Turley, Scattering from a Circle, Internal Hughes Research Laboratory Notes, Aug. 1992.[2] S. Turley, 2d Scalar Surface Scattering, Internal Hughes Research Laboratory Notes, May 1992.[3] R. Kress, Acoustic Scattering, in Scattering - Scattering and Inverse Scattering in Pure and AppliedScience, edited by E. Pike and P. Sabatier, chapter 2, Academic Press, London, 2001.[4] R. Kress, Acoustic Scattering, November 2006, a copy of sections 2.1 and 2.2 can be found athttp://diogenes.iwt.uni-bremen.de/vt/laser/papers/Kress-acustic-scatter-ed-book.ps.[5] M. Abramowitz and I. A. Stegun, Handbook of Mathematical Functions, Applied Mathematics Series,National Bureau of Standards, Cambridge, 1972.26

Penetrable Sphere, a=0.1,kf=1.5,rhof=1.51.6 x 10−4 theta, degrees1.41.2d sigma/d omega10.80.60.40.200 20 40 60 80 100 120 140 160 180Figure 11: Dierential cross section for scattering from a penetrable sphere of radius 0.1 wavelengths withk d /k = 1.5 and ρ d /ρ = 1.5.27

20Penetrable Sphere, a=1.0,kf=1.5,rhof=1.5181614d sigma/d omega1210864200 20 40 60 80 100 120 140 160 180theta, degreesFigure 12: Dierential cross section for scattering from a penetrable sphere of radius 1 wavelength withk d /k = 1.5 and ρ d /ρ = 1.5.28

Penetrable Sphere, a=3.0,kf=1.5,rhof=1.510 3 theta, degrees10 210 1d sigma/d omega10 010 −110 −210 −310 −40 20 40 60 80 100 120 140 160 180Figure 13: Dierential cross section for scattering from a penetrable sphere of radius 3 wavelengths withk d /k = 1.5 and ρ d /ρ = 1.5.29

Penetrable Sphere, a=10.0,kf=1.5,rhof=1.510 6 theta, degrees10 510 4d sigma/d omega10 310 210 110 010 −110 −20 20 40 60 80 100 120 140 160 180Figure 14: Dierential cross section for scattering from a penetrable sphere of radius 10 wavelengths withk d /k = 1.5 and ρ d /ρ = 1.5.30

Figure 15: Total near eld for scattering from a penetrable sphere of radius 2.5 wavelengths and with k d = 1.2and ρ d = 1.2.31