LECTURE NOTES 9 TRACELESS SYMMETRIC TENSOR APPROACH TO LEGENDRE POLYNOMIALS AND SPHERICAL HARMONICS

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Eectromagnetism II October 7, 202 Prof. Aan Guth LECTURE NOTES 9 TRACELESS SYMMETRIC TENSOR APPROACH TO LEGENDRE POLYNOMIALS AND SPHERICAL HARMONICS In these notes I wi describe the separation of variabe technique for soving Lapace s equation, using spherica poar coordinates. The soutions wi invove Legendre poynomias for cases with azimutha symmetry, and more generay they wi invove spherica harmonics. I wi construct these soutions using traceess symmetric tensors, but in Lecture Notes 8 I describe how the soutions in this form reate to the more standard expressions in terms of Legendre poynomias and spherica harmonics. (Logicay Lecture Notes 8 shoud come after these notes, athough they were posted first.) If you are starting from scratch, I think that the traceess symmetric tensor method is the simpest way to understand this mathematica formaism. If you aready know spherica harmonics, I think that you wi find the traceess symmetric tensor approach to be a usefu addition to your arsena of mathematica methods. The symmetric traceess tensor approach is particuary usefu if one needs to extend the formaism beyond what we wi be doing for exampe, there are anaogues of spherica harmonics in higher dimensions, and there are aso vector spherica harmonics that are usefu for expanding vector functions of ange. Vector spherica harmonics are used in the most genera treatments of eectromagnetic radiation, athough we wi not be introducing them in this course. I don t know a reference for the traceess symmetric tensor method, which is the main reason I am writing these notes. For the standard method, imited to the case of azimutha symmetry, our textbook by Griffiths shoud be sufficient. If you woud ike to see an additiona reference on spherica harmonics, which are needed when there is no azimutha symmetry, then I woud recommend J.D. Jackson, Cassica Eectrodynamics, rd Edition (John Wiey & Sons, 999), Sections.,.2,.5, and.6.. LAPLACE S EQUATION IN SPHERICAL COORDINATES: In spherica coordinates, Lapace s equation can be written as 2 ϕ ϕ(r, θ, φ) = r 2 + r2 r r r 2 2 θ ϕ =0, (9.) where the anguar part is given by 2 ϕ 2 ϕ θ ϕ sin θ + sin θ θ θ sin 2 θ φ 2. (9.2)

2 8.07 LECTURE NOTES 9, FALL 202 p. 2 It woud be more ogica to write 2 θ as 2 θ,φ, but it woud be tiresome to write it that way. It is sometimes usefu to rewrite the first term in Eq. (9.) using ) 2 ϕ 2 (r = (rϕ). (9.) r 2 r r r r 2 To use the method of separation of variabes, we can seek a soution of the form Then Lapace s equation can be written as ϕ(r, θ, φ) =R(r)F (θ, ϕ). (9.4) r 2 d 0= 2 dr ϕ = r 2 + RF R dr dr F 2 θ F. (9.5) Since the first term on the right-hand side depends ony on r, and the second term depends ony on θ and φ, the ony way that the equation can be satisfied is if each term is a constant. Thus we can write F 2 θ F = C θ, (9.6) d ( R r 2 d ) = C θ. (9.7) R dr dr 2. THE EXPANSION OF F (θ, φ): We now wish to find the most genera soution to the equation 2 θ F = C θ F, (9.8) which is a rewriting of Eq. (9.6). If such a function F can be found, we say that F is an eigenfunction of the operator 2 θ,witheigenvaue C θ. A function of anges (θ, φ) can equivaenty be thought of as a function of the unit vector nˆ that points in the direction of θ and φ, which can be written expicity as nˆ =sinθ cos φ ê +sinθ sin φ ê 2 +cosθ ê, (9.9) where ê,ê 2,andê can aso be written as ê x,ê y,andê z. I am abeing the unit vectors using the numbers, 2, and when I am thinking about summing over the indices, and otherwise I use x, y, andz. I now caim that the most genera function of (θ, φ) can be written as a power series in nˆ, or more precisey as a power series in the components of nˆ. I wi not prove this,

3 8.07 LECTURE NOTES 9, FALL 202 p. but it is true at east for square-integrabe piece-wise continuous functions F (θ, φ). Such a power series can be written as F (ˆn) =C (0) + C () i nˆi + C ( ij nˆinˆ j C ) i i 2...i nˆi nˆi2...nˆi +..., (9.0) where repeated indices are summed from to (as Cartesian coordinates). Note that C i i 2...i nˆ i nˆi 2...nˆi represents the genera term in the series, where the first three terms correspond to =0, =,and = 2. The indices i, i 2,..., i represent different indices, ike i and j; since they are repeated, they are each summed from to. The coefficients C i i 2...i are caed tensors, and the number of indices is caed the () rank of the tensor. Note that C (0), C i,andc ij are specia cases of tensors, athough they can aso be considered a scaar, a vector, and a matrix. It is possibe to impose restrictions on the coefficients of Eq. (9.0) without actuay restricting what can appear on the right-hand side of the equations. In particuar, we wi insist that ) The tensors C i i 2...i are symmetric under any reordering of the indices: C i i where {j,j 2,...,j } is any permutation of {i,i 2,...,i }. C i i 2...i C 2...i = j j 2...j, (9.) 2) The tensors are traceess, in the sense that if any two indices are set equa to each other and summed, the resut is equa to zero. Since the tensors are aready assumed to be symmetric, it does not matter which indices are summed, so we can choose the ast two: C i i 2...i =0. (9.2) To expain why these restrictions on the C s do not impose any restriction on the righthand side of Eq. (9.0), I wi use the exampe of C ij, but I think you wi be abe to see that that the argument appies to a. The insistence that C ij is symmetric can be seen to make no difference to the right-hand side of Eq. (9.0), because C ij mutipies the symmetric tensor nˆinˆj. Thus,ifC ij had an antisymmetric part, it woud not contribute to the right-hand side of Eq. (9.0). The requirement of traceessness is ess obvious, but suppose that were not traceess. Then we coud write C ij 2jj C ii = λ =0. (9.)

4 8.07 LECTURE NOTES 9, FALL 202 p. 4 We coud then define a new quantity, ij = C ij λδ ij. (9.4) C It foows that C ij is traceess: C ii = C ii λδii = λ λδii =0, (9.5) since δ ii =. The origina term C ij nˆinˆj can then be expressed in terms of C ij : [ ˆ ˆ = C ij n i n j C ij + λδ ij ] nˆinˆ j = C ij nˆinˆj + λ, (9.6) whereweusedthefactthatδ ij nˆinˆj =,sincenˆ is a unit vector. The extra term, λ, can then be absorbed into a redefinition of C (0) : Finay, we can write C (0) = C (0) + λ. (9.7) C (0) + () () (2 C nˆi + ) i C ij nˆ i nˆj = C (0) + C i nˆi + C ij nˆinˆj, (9.8) so we can insist that the tensor that mutipies nˆinˆj be traceess with no restriction on what functions can be expressed in this form. I wi ca the th term of this expansion F (ˆn), so F (ˆn) = i i 2...i i i 2 i C nˆ nˆ...nˆ. (9.9). EVALUATION OF 2 θ F(ˆn): To evauate 2 θ F (ˆn), we are going to take advantage of a convenient trick. Instead of deaing directy with F (ˆn), we wi instead introduce a radia variabe r, usingitto define a coordinate vector r = rn ˆ. (9.20) Foowing the notation of Griffiths (see his Eq. (.9)), I wi denote the coordinates of r by x, y, andz, or in index notation I wi ca them x i.so r = x i ê i = x ê + x 2 ê 2 + x ê. (9.2)

5 8.07 LECTURE NOTES 9, FALL 202 p. 5 Then, given any F (ˆn) of the form given in Eq. (9.9), we can define a function F ( r )by F ( r )=C i i 2...i x i x i 2...x i = r F (ˆn). (9.22) Note that C i i...i isthesamerank traceess symmetric tensor used to define F 2 (ˆn), but we are defining F ( r ) by mutipying C i i...i by x i x i2...x i and then summing over 2 indices, instead of mutipying by nˆ i nˆi2...nˆi and then summing. Now we can make use of Eq. (9.), which reates the fu Lapacian 2 to the anguar Lapacian, 2 θ. We wi find that in this case the fu Lapacian and the radia derivative piece of Eq. (9.) wi both be simpe, so we wi be abe to determine the anguar Lapacian by evauating the other terms in Eq. (9.). To evauate 2 F ( r ), we start with = 0. Ceary 2 F 0( r) = 2 C (0) =0, (9.2) since the derivative of a constant vanishes. Simiary for =, 2 F 2 () ( r) = C i x i =0, (9.24) since the first derivative produces a constant, so the second derivative vanishes. The first nontrivia case is =2: 2 F ( r )= 2 2 C ij x i x j = C ij (x i x j ) x m x m (9.25) = C ij [δ im x j + δ jm x i ] x m =2 C ij [ δ im δ jm ]=2C ii =0, where in the ast step we used the a-important fact that C ij is traceess. We wi ook at one more case, and then I hope it wi be cear that it generaizes. For =, 2 F r 2 () = C ijk x i x j x k = () C ijk (x i x j x k ) x m x m ) = C () ijk (δ im x j x k + (terms that symmetrize in ijk) x m ) = C () ijk (δ im δ jm x k + (terms that symmetrize in ijk) ) = C () ijk (δ ij x k + (terms that symmetrize in ijk) = () C iik x k + (terms that symmetrize in ijk) =0, (9.26)

6 8.07 LECTURE NOTES 9, FALL 202 p. 6 where again it is the traceessness of C i i 2...i that caused the term to vanish. Thinking about the genera term, one can see that after the derivatives are cacuated, there are 2 factors of x i that remain, but there are sti indices on C i i 2...i. Since a indices are summed, there are aways two indices on C i i 2...i which are contracted (i.e., set equa to each other) and summed, which causes the resut to vanish by the traceessness condition. The bottom ine, then, is that 2 F ( r) =0fora. (9.27) To see what this says about 2 θ F (ˆn), reca that F ( r) =r F (ˆn). Using Eq. (9.), we can write 2 2 F 0= F ( r) (r) = r + 2 θ r F ( r) 2 r r r 2 ( d ) = r 2 dr F (ˆn)+ r r 2 dr dr r 2 2 θ F (ˆn) = r 2[ ] ( +)F (ˆn)+ 2 θ F (ˆn), and therefore 2 θ F (ˆn) = ( +)F (ˆn). (9.28) ( ) Thus, we have found the eigenfunctions F (ˆ) = n C i i 2...i nˆi nˆi2...nˆi and eigenvaues ( +) of the differentia operator 2 θ. This is a very usefu resut! 4. GENERAL SOLUTION TO LAPLACE S EQUATION IN SPHERICAL COORDINATES: Now that we know the eigenfunctions of 2 θ, we can return to the soution to Lapace s equation by the separation of variabes in spherica coordinates. We now know that in Eq. (9.6), the ony aowed vaues of C θ are ( +), where is an integer. Thus, Eq. (9.7) becomes ( d r 2 dr ) = ( +)R, (9.29) dr dr and we can ook for soutions by trying R(r) =r p. We find consistency provided that p(p +)=( +), (9.0)

7 8.07 LECTURE NOTES 9, FALL 202 p. 7 which is a quadratic equation with two roots, p = and p = ( + ). Since we found two soutions to a second order inear differentia equation, we know that any soution can be written as a inear sum of these two. Thus we can write R (r) =A r B +, (9.) r + aowing for different vaues of A and B for each. The most genera soution to Lapace s equation, in spherica coordinates, can then be written as Φ( r )= B A r + =0 r + C i i 2... i nˆi nˆi 2...nˆi, (9.2) where the A s and B s are arbitrary constants, and each C i i 2...i is an arbitrary traceess symmetric tensor. In Lecture Notes 8 it is shown expicity that the th term here, when compared with the standard expansion in the spherica harmonic functions Y m (θ, φ), corresponds to the sum of a terms with the same, but for a m. TheY m s are defined for integer vaues of m from to, sothereare2 + terms for each vaue of. 5. COUNTING THE NUMBER LINEARLY INDEPENDENT TRACELESS SYMMETRIC TENSORS: Using the fact that C i i 2...i is traceess and symmetric, we can determine how many ineary independent such tenso rs exist, for any given. In other words, how many rea constants are needed to parameterize the most genera traceess symmetric tensor of rank? We begin by cacuating N sym, the number of ineary independent symmetric tensors of rank. Note that we are postponing the consideration of traceessness. For grounding, we can start by saying that it takes one number to specify S (0),arank0 symmetric tensor, because it is just a number. (I am using S for symmetric tensors, () whie reserving C for traceess symmetric tensors.) It takes numbers to specify S i, () () () since the vaues S, S 2,andS can each be specified independenty. For S ij, however, we see the constraints of symmetry: S ij has to equa S ji, so there are fewer (2 (2 than 9 independent vaues; there are 6, which can be taken to be S, 2, ) ) S S, S 22, S 2, S,with S 2, S,and S 2 determined by symmetry. Since the order of the indices does not matter, we can aways ist the indices in ascending order (as I did for

8 8.07 LECTURE NOTES 9, FALL 202 p. 8 S ij ) and then each independent entry wi occur once. When the indices are so written in ascending order, then the index vaues are competey determined if we just specify how many indices are equa to, how many are equa to 2, and how many are equa to. If it heps, we can imagine three hats, abeed, 2, and, and indistinguishabe bas, representing the indices of a rank tensor. There is then a : correspondence between independent tensor eements and the different ways that the bas can be put into the hats. For exampe, if there are 9 bas, with in the first hat, 2 in the second, and 4 in the third, then this arrangement of bas corresponds to S (9) 22.Sonowwejusthave to figure out how many different ways we can put indistinguishabe bas into hats. One way of counting the bas-in-hats probem is to imagine first abeing each ba with a number, so they are no onger indistinguishabe. We aso introduce 2 dividers, where 2 is one ess than the number of hats. Initiay we wi aso assign numbers to the 2 dividers. Thinking of the bas and dividers together, we have +2 distinguishabe objects. We can imagine isting them in a possibe orderings, and with + 2 distinguishabe objects there are ( + 2)! orderings. For each ordering there is an equivaent bas-inhats assignment. The bas to the eft of the eft-most divider are assigned to hat, the bas between the two dividers are assigned to hat 2, and the bas to the right of the right-most divider are assigned to hat. We have of course overcounted, since many different orderings of our + 2 objects wi ead to the same number of bas in each hat. However, we can see exacty by how much we have overcounted. We can re-order the bas without changing the number of bas in each hat, and we can interchange the two dividers. So we have overcounted by a factor of 2!. Thus, ( +2)! N sym = = ( +)( +2). (9.) 2! 2 It is easiy checked that this gives N sym (0) =, N sym () =, and N sym = 6, consistent with the exampes we started with. To impose traceessness, we require that our traceess tensors aso satisfy S i...i 2 jj =0. (9.4) How many conditions is this? One can see that the tensor on the eft is a symmetric tensor of rank 2, with free indices i...i 2. The number of conditions is then Nsym( 2). The number of ineary indendent traceess symmetric tensors is then given by N traceess sym =N sym N sym ( 2) = ( +)( +2) ( ) 2 2 = 2 +. (9.5) So the correspondence with the standard Y m s is consistent, as it woud have to be. The spherica harmonic expansion is just a rewriting of Eq. (9.2), with a particuar choice of basis for the 2 + independent traceess symmetric tensors of rank.

9 8.07 LECTURE NOTES 9, FALL 202 p SPECIAL CASE: AZIMUTHAL SYMMETRY: Azimutha symmetry means symmetry under rotation about an axis, which we wi take to be the z-axis. Equivaenty, we can say that a probem is azimuthay symmetric if nothing depends on the coordinate φ. To speciaize the genera expansion (9.0) for a function of (θ, φ) to the azimuthay symmetric case, we need to construct traceess symmetric tensors which are invariant under rotations about the z-axis. One straightforward way to this is to buid the traceess symmetric tensor from the vector ẑ, the unit vector in the z direction. Note that the x-, y-, and z- componentsofẑ are 0, 0, and, respectivey, so ẑ i = δ i. [You may have noticed an inconsistency in my notation, as earier (e.g., Eq. (9.9)) Iusedê or ê z for the unit vector in the z direction. In this case my inconsistency was intentiona, with two motivations. First, previousy we never needed a notation for the components of a unit basis vector, but here we wi. It woud be a rea pain to write (ê z ) i. Second, in Lecture Notes 8 I describe a convenient way to construct a basis for the traceess symmetric tensors, which invoves the use of a basis for vectors consisting of ẑ and two compex vectors û + and û. So, the use of ẑ rather than ê z wi remind us that we are thinking about the (û +, û, ẑ) basis, rather than the (ê x, ê y, ê z ) basis.] Arank tensor can be constructed from ẑ simpy by taking the product ẑ i ẑ i2...ẑ i. This is symmetric, and can be made traceess by extracting the traceess part. Extracting the traceess part means subtracting terms proportiona to one or more Kronecker δ- functions in such a way that the resut is traceess. It gets rather compicated to describe how this can be done for a genera symmetric tensor of arbitrary rank, so I wi just iustrate it by exampe. Tensors of rank 0and (i.e., scaars and vectors) are by definition traceess. I wi use cury brackets {...} to denote the traceess symmetric part of... Thus, { } =, { ẑ i } =ẑ i. (9.6) But for rank 2, the trace of ẑ i ẑ j is equa to ẑ i ẑ i =ẑ ẑ =. But we can subtract a constant times δ ij so that the resut is traceess: { ẑ i ẑ j } =ẑ i ẑ j δ ij. (9.7) The coefficient is /, because the trace of δ ij is δ ii =. (9.8)

10 8.07 LECTURE NOTES 9, FALL 202 p. 0 For rank, ẑ i ẑ j ẑ k has trace ẑ i ẑ i ẑ k =ẑ k, but we can make it traceess with a subtraction { ẑ i ẑ j ẑ k } =ẑ i ẑ j ẑ k 5( ẑi δ jk +ẑ j δ ik +ẑ k δ ij ). (9.9) The subtraction must of course be symmetrized, as shown, since we are trying to construct a traceess symmetric tensor. To verify that /5 is the right coefficient to make the expression traceess, we can take its trace. Since it is symmetric we can sum over any pair of indices. I wi choose to sum over i and j: δ ij { ẑ i ẑ j ẑ k } =ẑ i ẑ i ẑ k ( 5 z ˆ i δ ) ik +ẑ i δ ik +ẑ k δ ii (9.40) =ẑ k (ẑ k +ẑ k +ẑ k )=0. 5 For rank 4 there is the option of subtracting terms with either one or two Kronecker δ-functions, and both are needed to give a traceess resut. We can start with arbitrary coefficients, and see what they have to be: ( { ẑ i ẑ j ẑ k ẑ m } =ẑ i ẑ j ẑ k ẑ m + c ẑi ẑ j δ km +ẑ i ẑ k δ mj +ẑ i ẑ m δ jk +ẑ j ẑ k δ im ) (9.4) +ẑ j ẑ m δ ik +ẑ k ẑ m δ ij + c2 δij δ km + δ ik δ jm + δ im δ jk. Within each set of parentheses, the terms are chosen to make the expression symmetric in i, j, k, andm. If we cacuate the trace over i and j, we find ( δ ij { ẑ i ẑ j ẑ k ẑ m } =ẑ k ẑ m + c δkm +ẑ k ẑ m +ẑ k ẑ m +ẑ k ẑ m ) +ẑ k ẑ m +ẑ k ẑ m + c2 δkm + δ km + δ km (9.42) = +7c ẑ k ẑ m + c +5c 2 δ km. For the expression to vanish for a k and m, the c = /7 andc 2 =/5. Thus, two terms must vanish separatey, so { ẑiẑ j ẑ k ẑ m } =ẑ i ẑ j ẑ k ẑ m ( 7 ẑ i ẑ j δ km +ẑ i ẑ k δ mj +ẑ i ẑ m δ jk +ẑ j ẑ k δ im ) ( ) +ẑ j ẑ m δ ik +ẑ k ẑ m δ ij + δ ij δ km + δ ik δ jm + δ im δ jk. 5 (9.4) It can be shown that any traceess symmetric tensor of rank that is invariant under rotations about the z-axis is proportiona to { ẑ i...ẑ i }. (To see this, note that in Lecture Notes 8 we construct a compete (2 + )-dimensiona basis for the traceess symmetry tensors of rank. They depend on the azimutha ange φ as z m (φ) e imφ, with m taking integer vaues from to. Since these functions of φ are orthogona in

11 8.07 LECTURE NOTES 9, FALL 202 p. the sense that 2π zm (φ)z 0 m (φ)dφ =2πδ m m, any traceess symmetric tensor of rank that is independent of φ must proportiona to the m = 0basis tensor.) Since Eq. (9.0) tes us how to expand an arbitrary function of θ and φ in terms of traceess symmetric tensors, we can now say that functions of θ aone (i.e., azimuthay symmetric functions of nˆ) can be expanded as F (θ) =c 0 + c { ẑ i }nˆi + c 2 { ẑ i ẑ j } nˆinˆj c { ẑ i...ẑ i } nˆi...nˆi +..., (9.44) where the c s are constants. This corresponds to what is standardy caed an expansion in Legendre poynomias. In Lecture Notes 8 I show exacty how to reate these terms to the standard conventions for normaizing the Legendre poynomias, but we can see here exacty what these functions are. Using ẑ nˆ =cosθ, wehave { } = { ẑ i } nˆi =cosθ 2 { ẑiẑ j } nˆ inˆj =cos θ (9.45) { ẑ i ẑ j ẑ k } nˆinˆjnˆk =cos θ cos θ { ẑ i ẑ j ẑ k ẑ m } nˆ inˆ jnˆknˆm =cos θ cos θ Up to a normaization convention described in Lecture Notes 8, these are the Legendre poynomias P (cos θ). 7. THE MULTIPOLE EXPANSION: The most genera soution to Lapace s equation, in spherica coordinates, was given as Eq. (9.2). We now wish to appy that resut to a common situation: suppose we have a charged object, and we wish to describe the potentia outside of the object. Let s say for definiteness that the charge of the object is entirey contained within a sphere of radius R, centered at the origin. In that case Lapace s equation wi hod for a r>r,sothere shoud be a soution of the form of Eq. (9.2) that is vaid throughout this region. At infinity the potentia of a ocaized charge distribution wi aways approach a constant, which we can take to be zero, we can see that the A coefficients that appear in Eq. (9.2) must a vanish. The B factors can be absorbed into the definition of C i...i,sowecan write the expansion as Φ( r )= r + =0 C i...i n ˆi...nˆi. (9.46)

12 8.07 LECTURE NOTES 9, FALL 202 p. 2 Since each successive term comes with an extra factor of /r, at arge distances the sum is dominated by the first term or maybe the first few terms. A the information about the charge distribution of the object is contained in the C i...i, so knowedge of the first few C i the ob...i is enough to describe the fied at arge distances, no matter how compicated ject. The first few terms of this series have specia names: the = 0term is the monopoe term, the = term is the dipoe, the = 2 term is the quadrupoe, and the =term is the octupoe. If we want to cacuate the C i...i in terms of the charge distribution, we can start with the genera equation for the potentia of an arbitrary charge distribution: ρ( r ) V ( r) = d x. (9.47) 4πɛ0 r r The mutipoe expansion can then be derived by expanding / r r in a power series in r. I begin by doing it as Griffiths does, which gives the simpest but not the most usefu form of the mutipoe expansion. Griffiths rewrote the denominator as = =, (9.48) r r r 2 + r 2 2 r r r 2 + r 2 2rr cos θ where r and r are the engths of the vectors r and r, respectivey, and θ is the ange between these vectors. Next he used the fact that the Legendre poynomias can be defined by the generating function g(x, λ) =, (9.49) +λ 2 2λx which means that the Legendre poynomias P (x) can be obtained by expanding g(x, λ) in a power series in λ: g(x, λ) = = λ P(x). (9.50) + λ 2 2λx =0 Eq. (9.50) is sometimes taken as the definition of the Legendre poynomias, and sometimes it is derived from another definition. In any case, if we accept Eq. (9.50) as vaid, then r = = P (cos θ). (9.5) r r r + r 2 r r 2 r cos θ r =0 r

13 8.07 LECTURE NOTES 9, FALL 202 p. Inserting this reation into Eq. (9.47), we find V ( r) = 4 πɛ 0 r + =0 r ρ( r )P(cos θ )d x. (9.52) This is the easiest way that I know to show that there is an expansion of V ( r) inpowersof /r, but the compication is that cos θ appears inside the integra. If we coud impement Eq. (9.46), we woud be abe to cacuate (or maybe measure) a sma number of the quantities C i...i, and then we woud be abe to evauate V ( r ) at arge distances in any direction. To use Eq. (9.52) directy, however, one woud have to repeat the integration for every direction of r. Griffiths works around this probem by massaging the formua to extract the monopoe and dipoe terms, and in Probem of Probem Set 5 you had the opportunity to carry this out for the quadrupoe and octupoe terms. The standard method of improving Eq. (9.52) is to use spherica harmonics, but here I wi derive the equivaent reations using the traceess symmetric tensor approach. Instead of expanding / r r in powers of r, we wi think of it as a function of three variabes the components r i of r, and we wi expand it as a Tayor series in variabes. To make the formaism cear, I wi define the function f( r ). (9.5) r r The function can then be expanded in a power series using the standard muti-variabe Tayor expansion: f 2 f f( r )=f( 0)+ x i + x i x +..., (9.54) x j i r x x = 0 2! i j where the repeated indices are summed. To separate the anguar behavior, we write r = 0 x i = r nˆ i, (9.55) so Eq. (9.54) becomes f( r )=f( 0)+r f x i r 2 2 f + nˆ n ˆ ! x x i j r =0 i j r = 0 (9.56) The notation can now be simpified by noting that since f is a function of r r,the derivatives with respect to x i can be repaced by derivatives with respect to x i with a change of sign: f = = =. (9.57) r r x i r r r x = 0 i r x i x i r = 0

14 8.07 LECTURE NOTES 9, FALL 202 p. 4 This aows us to write the derivatives in the expansion (9.56) much more simpy. The th derivative is found by repeating the above operation times: f =( ). (9.58) x i... x i r = x i... x i r 0 Combining Eqs. (9.58) with (9.56), we can write ( ) r ( ) = nˆ r r! x... x r i i =0 i...nˆ i. (9.59) Note that the quantity in parentheses in the equation above is traceess, because ( ) = 2 x i x i x i+2... x i r x i+2... x i r (9.60) = 2 =0, xi+2... x i r because 2 (/ r ) = 0except at r = 0. So we can see the traceess symmetric tensor formaism emerging. To evauate this quantity, we wi work out the first severa terms unti we recognize the pattern. We write r rn ˆ, (9.6) and adopt the abbreviation i. (9.62) x i It is usefu to start by evauating the derivatives of the basic quantities r and nˆi: / r = i(x j x j ) 2 xi i = ( xkx k ) /2 i (x j x j )= 2xjδ ij = =ˆn i, 2 2r r ( xj ) δ ij nˆ = = x r = δ nˆ nˆ. r r r 2 r i j i j i ij i j (9.6) It is then straightforward to show that i = nˆi, r r 2 i j = r { r nˆinˆj }, (9.64) 5 i j k = { nˆiˆjnˆk n }, r r4

15 8.07 LECTURE NOTES 9, FALL 202 p. 5 where {}denotes the traceess symmetric part, and the reevant cases are shown expicity in Eqs. (9.6) (9.9). It becomes cear that the genera formua, which can be proven by induction, is ( ) (2 )!! = { nˆii...nˆi x i... x r + }, (9.65) r where i (2)! (2 )!! (2 )(2 )(2 5)...=, with ( )!!. (9.66) 2! Inserting this resut into Eq. (9.59), we find = (2 )!! r { nˆi...nˆ r r i! r } + nˆi...nˆ i. (9.67) =0 One can write this more symmetricay by writing )!! r = (2 nˆ...nˆ nˆ...nˆ r r { i i! r }{ i i }, (9.67) + =0 since { nˆ i...nˆ i } differs from nˆ i...nˆ i by terms proportiona to Kronecker δ-functions, which vanish when summed with the traceess tensor { nˆi...nˆi }. Starting with Eq. (9.67), one can if one wishes drop the cury brackets around either factor (but not both!). Inserting this expression for / r r into Eq. (9.47), we have the fina resut V ( r )= 4πɛ0 C r + i...i n =0 ˆi...nˆi, (9.68) where (2 C i...i = )!! ρ( r ) { r i!... r i } d x. (9.69) Note that the coefficient in the above expression can aso be written as (2 )!! (2)! =. (9.70)! 2 (!) 2

16 8.07 LECTURE NOTES 9, FALL 202 p. 6 For purposes of iustration, I wi write out the first two terms the monopoe and dipoe terms in a bit more detai. The monopoe term can be written as Q V mono ( r) =, (9.7) 4πɛ 0 r where Q = C (0) = ρ( r )d x. (9.72) The dipoe term is where p nˆ V dip ( r) =, (9.7) 4πɛ 0 r 2 p i = C () i = ρ( r ) r i d x. (9.74)

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