B STF multipole decompositions

All formulas in this appendix are either taken directly from Refs. [27] and [45] or are easily derivable from formulas therein.

Any Cartesian tensor field depending on two angles A 𝜃 spanning a sphere can be expanded in a unique decomposition in symmetric trace-free tensors. Such a decomposition is equivalent to a decomposition in tensorial harmonics, but it is sometimes more convenient. It begins with the fact that the angular dependence of a Cartesian tensor TS(𝜃A ) can be expanded in a series of the form

A ∑ L TS (𝜃 ) = TS ⟨L ⟩ˆω , (B.1 ) ℓ≥0
where S and L denote multi-indices S = i1...is and L = j1 ...jℓ, angular brackets denote an STF combination of indices, a ω is a Cartesian unit vector, L j1 jℓ ω := ω ...ω, and L ⟨L⟩ ˆω := ω. This is entirely equivalent to an expansion in spherical harmonics. Each coefficient TS ⟨L⟩ can be found from the formula
∫ T = (2ℓ-+-1)!! T (𝜃A)ˆω dΩ, (B.2 ) S⟨L⟩ 4πℓ! S L
where the double factorial is defined by x!! = x(x − 2) ⋅⋅⋅1. These coefficients can then be decomposed into irreducible STF tensors. For example, for s = 1, we have
T = Tˆ(+1)+ 𝜖j ˆT(0) + δ Tˆ(− 1), (B.3 ) a⟨L ⟩ aL a⟨iℓ L−1⟩j a⟨iℓ L −1⟩
where the ˆT(n)’s are STF tensors given by
ˆT(+1):= T⟨L+1⟩, (B.4 ) L+1 ˆT(0):= --ℓ--Tpq⟨L−1𝜖i⟩pq, (B.5 ) L ℓ + 1 ℓ (−1) 2 ℓ − 1 j ˆTL−1 := 2-ℓ +-1 T jL−1. (B.6 )
Similarly, for a symmetric tensor TS with s = 2, we have
( ) T = STF STF 𝜖p Tˆ(+1) + δ Tˆ(0) + δ 𝜖p Tˆ(−1)+ δ δ ˆT(−2) ab⟨L⟩ L ab aiℓ bpL− 1 aiℓ bL−1 aiℓ biℓ−1 pL −2 aiℓ biℓ−1 L−2 ˆ(+2) ˆ +T abL + δabKL, (B.7 )
where
ˆ(+2 ) T L+2 := T⟨L+2⟩, (B.8 ) ˆ(+1 ) --2ℓ- pq T L+1 := ℓ + 2 SLT+F1 (T⟨piℓ⟩qL− 1𝜖iℓ+1 ), (B.9 ) Tˆ(0):= --6ℓ(2ℓ-−-1)---STF (T⟨ji⟩jL−1), (B.10 ) L (ℓ + 1)(2ℓ + 3) L ℓ (− 1) 2(ℓ − 1)(2ℓ − 1) TˆL−1 := ----------------STF (T ⟨jp⟩qjL− 2𝜖iℓ−1pq), (B.11 ) (ℓ + 1)(2ℓ + 1) L−1 ˆ(− 2) 2ℓ-−-3 jk T L−2 := 2ℓ + 1T ⟨jk⟩ L− 2 (B.12 ) ˆ 1 j KL := 3T jL. (B.13 )
These decompositions are equivalent to the formulas for addition of angular momenta, J = S + L, which results in terms with angular momentum ℓ − s ≤ j ≤ ℓ + s; the superscript labels (±n ) in these formulas indicate by how much each term’s angular momentum differs from ℓ.

By substituting Eqs. (B.3View Equation) and (B.7View Equation) into Eq. (B.1View Equation), we find that a scalar, a Cartesian 3-vector, and the symmetric part of a rank-2 Cartesian 3-tensor can be decomposed as, respectively,

T (𝜃A) = ∑ ˆA ˆωL, (B.14 ) L ℓ∑≥0 ∑ [ ] T (𝜃A) = ˆB ˆω + ˆC ωˆL −1 + 𝜖i Dˆ ˆωjL−1 , (B.15 ) a L aL aL− 1 aj iL−1 ℓ≥0∑ ℓ≥1∑ ∑ [ ] T(ab)(𝜃A) = δab ˆKL ˆωL + EˆL ˆωabL + ˆFL− 1⟨aˆωb⟩L−1 + 𝜖ij(aˆωb)iL−1GˆjL −1 ℓ≥0 ℓ≥0 ℓ≥1 ∑ [ ] + HˆabL −2ˆωL− 2 + 𝜖ij(aˆIb)jL− 2ωˆiL −2 . (B.16 ) ℓ≥2
Each term in these decompositions is algebraically independent of all the other terms.

We can also reverse a decomposition to “peel off” a fixed index from an STF expression:

(ℓ + 1)STF T = T + ℓ STF T − --2ℓ-- STF T j δ . (B.17 ) iL i⟨L⟩ i⟨L⟩ L iℓ⟨iL− 1⟩ 2ℓ + 1 L ⟨jL− 1⟩ iℓi

In evaluating the action of the wave operator on a decomposed tensor, the following formulas are useful:

c L cL ---ℓ-- c⟨i1 i2...iℓ⟩ ω ˆω = ˆω + 2ℓ + 1δ ˆω , (B.18 ) ωcωˆcL = -ℓ +-1-ˆωL, (B.19 ) 2ℓ + 1 ℓ(ℓ +-1) r∂cˆωL = − ℓˆωcL + 2ℓ + 1 δc⟨i1ˆωi2...iℓ⟩, (B.20 ) ∂c∂cˆωL = − ℓ(ℓ +-1)-ˆωL, (B.21 ) r2 ωc∂cˆωL = 0, (B.22 ) (ℓ + 1)(ℓ + 2) r∂cωˆcL = -------------ωˆL. (B.23 ) (2ℓ + 1 )

In evaluating the t-component of the Lorenz gauge condition, the following formula is useful for finding the most divergent term (in an expansion in r):

∑ (ℓ + 1 )(ℓ + 2) ∑ r∂ch (ntc,m) = -------------Bˆ(Ln,m)ˆωL − (ℓ − 1) ˆC(Ln,m )ˆωL. (B.24 ) ℓ≥0 2 ℓ + 1 ℓ≥2
And in evaluating the a-component, the following formula is useful for the same purpose:
r∂bh (n,m)− 1rηβγ∂ h (n,m) [ab 2 a βγ ] ∑ 1 ˆ(n,m ) ˆ(n,m) (ℓ +-2)(ℓ-+-3)-ˆ(n,m ) 1 ˆ(n,m ) L = 2ℓ(K L − A L ) + 2ℓ + 3 E L − 6ℓF L ˆωa ℓ≥0 [ ∑ ℓ(ℓ + 1) (n,m) (n,m ) (ℓ + 1)2(2ℓ + 3) (n,m) + ---------(AˆaL−1 − KˆaL−1) + -----------------ˆFaL− 1 ℓ≥1 2 (2ℓ + 1) 6(2ℓ + 1)(2ℓ − 1) ] ∑ [ (ℓ + 2)2 ] − (ℓ − 2) ˆHa(nL,−m)1 ˆωL −1 + ---------ˆG(dnL,m− )1 − 12(ℓ − 1)Iˆ(dnL,m−1) 𝜖acdˆωcL−1 (B.25 ) ℓ≥1 2(2ℓ + 1 )
where we have defined ˆ (n,m) H a := 0 and ˆ(n,m ) Ia := 0.

The unit vector ωi satisfies the following integral identities:

∫ ˆωLd Ω = 0 if ℓ > 0, (B.26 ) ∫ ωLd Ω = 0 if ℓ is odd, (B.27 ) ∫ ω dΩ = 4πδ{i1i2 ...δiℓ−-1iℓ}-if ℓ is even, (B.28 ) L (ℓ + 1)!!
where the curly braces indicate the smallest set of permutations of indices that make the result symmetric. For example, δ{abωc} = δabωc + δbcωa + δcaωb.


  Go to previous page Go up