LECTURE NOTES 8 THE TRACELESS SYMMETRIC TENSOR EXPANSION AND STANDARD SPHERICAL HARMONICS

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1 MASSACHUSETTS INSTITUTE OF TECHNOLOGY Physics Department Physics 8.07: Eectromagnetism II October, 202 Prof. Aan Guth LECTURE NOTES 8 THE TRACELESS SYMMETRIC TENSOR EXPANSION AND STANDARD SPHERICAL HARMONICS These notes are an addendum to Lecture 4, Wednesday October 0, 202. The notes wi describe a topic that I did not have time to incude in the ecture: the reation between the traceess symmetric tensor expansion and the standard spherica harmonics. Using traceess symmetric tensors, we can expand any function of ange as () C i i 2...i F (ˆn) = () C i n i 2...i nˆi ˆi2...nˆi F (ˆn) = =0 (8.) = C (0) + () C i nˆi + (2) C ij nˆinˆj + (3) C ijk nˆinˆjnˆk +..., where the are traceess symmetric tensors, the indices i, i 2,...i are summed from to 3 as Cartesian indices, and nˆ(θ, φ) =sinθ cos φ ê +sinθ sin φ ê 2 +cosθ ê 3, (8.2) where ê,ê 2,andê 3 can aso be written as ê x,ê y,andê z. In the more standard approach, an arbitrary function of (θ, φ) is expanded in spherica harmonics: F (ˆn) = = m am Y m (θ, φ). (8.3) = We have shown that [ ] 2 () θ C i i 2...i nˆi nˆi2...nˆ i = ( +) ( ) C i i 2...i nˆi nˆi2...nˆi, (8.4) where ) 2 2 θ (sin = θ + sin θ θ θ sin 2 θ φ 2. (8.5) In the standard approach one woud show that 2 θ Y m(θ, φ) = ( +)Y m (θ, φ), (8.6) so apparenty has the same meaning in both formaisms. (I am not trying here to derive the standard formaism, but instead I wi simpy adopt the equations from standard textbooks, and show that we can express these functions in terms of traceess symmetric

2 8.07 LECTURE NOTES 8, FALL 202 p. 2 tensors. A good exampe of such a standard textbook is J.D. Jackson, Cassica Eectrodynamics, 3rd Edition (John Wiey & Sons, 999), Sections 3., 3.2, 3.5, and 3.6. That means that there must be some particuar traceess symmetric tensor, which we wi (,m) ca which is equivaent to Y m ( θ, φ). That is, C i...i (,m) C i...i (,m) nˆi...nˆi = Y m (θ, φ). (8.7) Our goa is to construct C i...i expicity. We have aready shown that the number of ineary independent traceess symmetric tensors of rank (i.e., with indices) is given by 2 +, which is not surprisingy equa to the number of Y m functions for a given. The quantity m is an integer from to, sothereare2 + possibe vaues. We consider first the case of azimutha symmetry, where F (ˆn) is invariant under rotations about the z-axis, and hence independent of φ. In that case, within the standard treatment, the most genera function can be expanded in Legendre poynomias, F (ˆn) = a P (cos θ). (8.8) =0 The P functions are the same as the Y 0 functions, except that they are normaized differenty: 2 + Y 0 (θ, φ) = P(cos θ). (8.9) 4π The Legendre poynomias can be written expicity using Rodrigues formua: P (x) = ( ) d [(x 2 ) ]. (8.0) 2! dx In the traceess symmetric tensor formaism, the azimutha symmetry case must be described by traceess symmetric tensors that are invariant under rotations about the z-axis. () It is easiest to begin by thinking about =, where we are seeking a tensor C i.since () C i has one index, it is a vector, which is the same as a tensor of rank. It is obvious that the ony vector that is invariant under rotations about the z-axisisavectorthat points aong the z axis. I wi et ẑ be a unit vector in the z-direction (which I have aso caed ê z and ê 3 ), and then for azimutha symmetry we have () C i =constẑ i, (8.) where ẑ i = δ i3 is the i th component of ẑ. The resuting function of nˆ is then F (ˆn) = () C i nˆi =constẑ i nˆi =constẑ nˆ =constcosθ, (8.2)

3 8.07 LECTURE NOTES 8, FALL 202 p. 3 which certainy agrees with P (cos θ) =cosθ. To generaize to arbitrary, we can construct a tensor of rank that is invariant under rotations about the z-axis by considering the product ẑ i ẑ i2...ẑ i. This is ceary symmetric, but it is not traceess. However, we can make it traceess by taking its traceess part, which I denote by cury brackets. { ẑ i ẑ i2...ẑ i } traceess symmetric part of ẑ i ẑ i2...ẑ i. (8.3) The traceess symmetric part is constructed by starting with the origina expression and then subtracting terms proportiona to one or more Kronecker δ-functions, where the subtractions are uniquey determined by the requirement that the expression be traceess. For exampe, { } = { ẑ i } =ẑ i { ẑ i ẑ j } =ẑ i ẑ j δ 3 ij ( { ẑiẑjẑ k } =ẑiẑjẑk z 5 ˆiδ ) jk +ẑ j δ ik +ẑ k δ ij ( { ẑiˆjẑ z k ẑ m } =ẑ i ẑ j ẑ k ẑ m 7 ẑi 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 +ẑkẑ m δ ij + 35 δ ij δ km + δ ik δ jm + δ im δ jk, (8.4) where the coefficients are a determined by the requirement of traceessness. We wi argue ater that Eq. (8.4) gives the ony traceess symmetric tensors that are invariant under rotations about the z-axis, and therefore the function F (ˆn) ={ ẑ i...ẑ i } nˆi...nˆi (8.5) is the ony function, up to a mutipicative constant, that is azimuthay symmetric and satisfies 2 θf (ˆn) = (+)F (ˆn). Since P (cos θ) is azimuthay symmetric and satisfies 2 θ P (cos θ) = ( +)P (cos θ), we must have where the constant is yet to be determined. P (cos θ) =const{ ẑ i...ẑ i } nˆi...nˆi, (8.6) Both sides of Eq. (8.6) are poynomias in cos θ, where the highest power is cos θ. If we can find the coefficients of this highest power on each side of the equation, we can determine the constant. On the right-hand side, the highest power comes entirey from the ẑ i...ẑ i term in { ẑ i...ẑ i }, since a the other terms contain Kronecker δ- functions which resut in factors of the form nˆ nˆ =, reducing the number of nˆ factors avaiabe to give powers of cos θ. So, the eading term on the right-hand side is simpy

4 8.07 LECTURE NOTES 8, FALL 202 p. 4 const(ẑ nˆ) =constcos θ. For the eft-hand side, we can use Rodrigues formua to extract the highest power: ( ) d [ P (x) = (x 2 ) ] 2! dx ( ) d [ ] = x 2 +(owerpowers) 2! dx ( ) d [ ] = (2)x 2 +(owerpowers) 2! dx ( ) 2 d [ = (2)(2 )x 2 2 ] +(owerpowers) 2! dx [ = (2)(2 )...( +)x ] +(owerpowers) 2! (2)! = x +(owerpowers). 2 (!) 2 (8.7) Matching these coefficients, we see that (2)! P (cos θ) = { ẑ i...ẑ i } nˆ... 2 i (!) 2 nˆi. (8.8) Now we can return to the genera case, in which there is no azimutha symmetry, and the expansion requires the spherica harmonics, Y m.they m arechosentohavea very simpe dependence on φ, namey Y m (θ, φ) e imφ. (8.9) This property can be described in terms of how the functions transform under a rotation of the coordinate system about the z-axis. Under a rotation by an ange ψ about the z-axis, the ange φ changes by ψ, andy m changes by a factor e iψ. I have not been carefu here about specifying the sign of this rotation, because it wi be easy to fix the sign conventions at the end. The important point here is that if we want to match the conventions of the spherica harmonics, we need to construct traceess symmetric tensors that are modified by a rotation ony by a mutipicative phase factor. That is, we are ooking for tensors that are compex, and that are eigenvectors of the rotation operator. Naturay we begin by considering a vector (a tensor with one index, or a rank tensor), which under a rotation about the z-axis transforms as v x v y = v x cos ψ v y sin ψ = v x sin ψ + v y cos ψ. (8.20)

5 8.07 LECTURE NOTES 8, FALL 202 p. 5 We thus seek an eigenvector of the matrix ( ) cos ψ sin ψ R = sin ψ cos ψ. (8.2) The eigenvaues λ of the matrix are determined by the characteristic equation det(r λi) =0, (8.22) where I is the identity matrix, which can be expanded as ( ) cos ψ λ sin ψ det =0= λ 2 2λ cos ψ +=0, (8.23) sin ψ cos ψ λ for which the soutions are λ =cosψ ± cos 2 ψ =cosψ ± i sin ψ = e ±iψ. (8.24) The eigenvectors then satisfy ( cos ψ e ±iψ sin ψ sin ψ cos ψ e ±iψ )( vx v y ) =0, (8.25) which simpifies to ( )( ) i sin ψ sin ψ vx =0, (8.26) sin ψ i sin ψ from which we see that v y = iv x. Constructing normaized eigenvectors, we can define v y û () û + = (ê x + iêy) 2 û (2) û = (ê 2 x iê y ), (8.27) which are orthonorma in the sense that û (i) û (j) = δ ij. (8.28) We can compete a basis for three-dimensiona vectors by adding û (3) ẑ =ê z. (8.29) You might ask how one shoud visuaize a vector with imaginary components. What direction does it point? It certainy points in a definite direction in compex threedimensiona space, which is equivaent to a six-dimensiona rea-vaued space, but for our

6 8.07 LECTURE NOTES 8, FALL 202 p. 6 purposes we do not need to have any geometric picture of these vectors. We are simpy going to use them to form dot products to construct (compex-vaued) functions of θ and φ. Note that compex conjugation, as used in Eq. (8.28), is essentia for defining a positive definite norm for compex vectors. The quantities û (+) û (+) and û ( ) û ( ),by contrast, are in fact equa to zero. This eads to the convenient fact that is both traceess and symmetric. Since û + i...û + i (8.30) û + nˆ = sin θe iφ, (8.3) 2 we can construct functions proportiona to e imφ,form>0, by incuding m factors of û +, and arranging for each of them to be dotted into nˆ. This can be done by considering the function defined by F m (θ, φ) {û + i...û + i m ẑ im+...ẑ i } nˆi...nˆi. (8.32) To see that in this expression every û + is dotted into an nˆ, reca that û + ẑ =û + û + =0. So, when the right-hand side is expanded and a the indices are summed to give dot products, the ony terms that survive are those for which every û + is dotted into ẑ. Thus, the right-hand side of Eq. (8.32) is proportiona to e imφ. From Eq. (8.4), we know that the right-hand side of Eq. (8.32) is an eigenfunction of 2 θ with eigenvaue ( +). I wi argue beow that any such eigenvector that is proportiona to e imφ is necessariy proportiona to Y m. We wi return to the question of uniqueness, but et us first assume that uniqueness hods, so that F m (θ, φ) Y m (θ, φ). (8.33) As in the previous derivation for Legendre poynomias, we can determine the constant of proportionaity by matching the eading term in the expansions of both sides of the equation. F m (θ, φ) can be written as (sin θ) m e imφ times a poynomia in cos θ, sowe can use the highest power of cos θ to determine the matching. It is easy to extract the eading term from Eq. (8.32), because it comes from the first term in the expansion of { û + i...û + i ẑ im+ } =û + ˆ+ i...u i z i m m ˆ m+...ẑ i + terms δ ip i q. (8.34) The first term gives the highest power of cos θ, because the Kronecker δ-functions that appear in a ater terms cause one or more nˆ s to dot with other nˆ s, reducing the number of nˆ s avaiabe to appear in the form nˆ ẑ =cosθ. Thus, F m (θ, φ) =2 m/2 (sin θ) m e imφ [ (cos θ) m +(owerpowersofcosθ) ]. (8.35)

7 8.07 LECTURE NOTES 8, FALL 202 p. 7 To compare Eq. (8.35) with the eading term in the expansion for the standard function Y m, we need a formua for Y m (θ, φ). It is given in Jackson as Eq. (3.53), p. 08, as Y m (θ, φ) = 2 +( m)! P m (cos θ)e imφ, (8.36) 4π ( + m)! where P m (cos θ) is the associated Legendre function, which can be defined by Jackson s Eq. (3.50), P m ( ) m 2 m/2 d+m 2 (x) = ( x ) (x ). (8.37) 2! dx +m Using the same technique as in Eq. (8.7), we find d +m (x 2 ) =(2)...( +)( )...( m +)x m +(owerpowers) dx +m (2)! = x m +(owerpowers). ( m)! (8.38) Matching the coefficients of these eading terms, we find that we can write (for m 0) where with Y m (θ, φ) = C (,m) i...i nˆi...n ˆi, (8.39) (,m) C i i d m u +...u + z i...z i, 2...i = { ˆi ˆim ˆ m+ ˆ } (8.40) ( ) m (2)! 2 m (2 +) d m =. (8.4) 2! 4π ( + m)! ( m)! For negative vaues of m, the cacuation is identica, except that we use û instead of û +. The resut is (,m) (, m ) i ˆ ˆ C i i...i = d m { û i...û z i...z 2 m + i } = C m i, i 2...i (8.42) where to aow for negative m we need to write d m as ( ) m (2)! 2 m (2 +) d m = 2! 4π ( + m)! ( m)!. (8.43) It is worth mentioning that the cury brackets indicating traceess symmetric part can be put on either factor or both in expressions such as Eq. (8.32). That is, { û + i...û + i ẑ im+...ẑ i } nˆi...nˆi = { û + i...û + i ẑ i... m m m + ẑ i }{nˆi...nˆi } (8.44a) =û + i...ui z ˆi m m+...ẑ i { nˆi...nˆi }, ˆ+ (8.44b)

8 8.07 LECTURE NOTES 8, FALL 202 p. 8 where the top ine is justified because { nˆi...nˆi } differs from nˆi...nˆi ony by terms proportiona to Kronecker δ-functions, which give no contribution when summed with the traceess symmetric tensor { û + i...û + i ẑ i...ẑ m+ i }. Simiary, once the second factor is written in traceess symmetric fo m rm, there is no onger a need to take the traceess symmetric part of the first term, since { û + i...u + zˆ i...z ˆim m+ ˆ i } differs from û + + i...u ˆi z ˆi m m+...ẑ i ony by terms proportiona to Kronecker δ-functions, which vanish when summed with the traceess symmetric tensor { nˆi...nˆi }. Finay, we can return to the question of uniqueness. In asserting that F m (θ, φ) Y m (θ, φ), we knew that both functions are proportiona to e imφ, and that both are eigenfunctions of 2 θ with eigenvaue ( + ). We caimed that, up to a mutipicative constant, there is ony one function that has these properties. Assuming that the power series representation of Eq. (8.) aways exists, the uniqueness that we need is easy to see. We showed in ecture that the number of ineary independent traceess symmetric (,m) tensors of rank is 2+, and now we have constructed 2+ such tensors: the C i...i,for m =,...,. These are ceary ineary independent, since they are each eigenfunc tions of rotations about the z-axis with different eigenvaues. Thus, any traceess symmetric tensor of rank must be a inear sum of the tensors in our basis. When we aso specify that the tensor being sought is an eigenvector of rotations about the z-axis, with a specific eigenvaue, then ony one of the tensors in our basis can contribute. The above argument is soid, but one might sti wonder what happens if we try, for exampe, to construct a different tensor by using both û + s and û s in the same expression. For exampe, we might consider { û + i û j }, which is invariant under rotations about the z-axis. With a itte work, however, one can show that { û + i û j } = 2 { ẑ iẑ j }. (8.45)

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