6.2 The complex center-of-mass world line

Before trying to determine ψ ∗0i 1, several comments are in order:

  1. As mentioned earlier, the right-hand side of Equation (287View Equation) is a function of both τ, via the the GC variables, ξi, ξij and the uret, via the ψ01i,ψ01ij and Ψ. The extraction of the l = 1 part of ψ ∗0 1 must be taken on the constant ‘s’ cuts. In other words, both τ and uret must be eliminated by using Equations (268View Equation) – (271View Equation) and (267View Equation).
  2. This elimination must be done in the linear terms, e.g., from Equation (271View Equation),
    ( √ -- ) (R ) --2-i 0 √ -- ij 0 η (u ret ) = η s − 2 ξR(s)Y1i + 2 ξR (s)Y2ij √ -- 2 ′[ i 0 ij 0 ] ≈ η(s) − ---η (s ) ξR (s)Y 1i − 2ξR (s)Y 2ij 2
    or, from Equation (270View Equation),
    ( ( ) ) √2-- √ -- χ(τ) = χ s + i ---ξiI(s)Y01i − 2ξiIj(s )Y 02ij 2 √ -- [ ] ≈ χ(s) + i--2χ (s )′ ξi(s)Y 0− 2ξij(s)Y 0 . 2 I 1i I 2ij

    In the nonlinear terms we can simply use

    uret = τ = s.

  3. In the Clebsch–Gordon expansions of the harmonic products, though we need both the l = 1 and l = 2 terms in the calculation, we keep at the end only the l = 1 terms for the 0i ψ 1. (Note that there are no l = 0 terms since ψ∗10 is spin weight s = 1.)
  4. The calculation to determine the ∗0i ψ1 for the constant ‘s’ slices was probably the most tedious and lengthy in this work. The importance of the results necessitated the calculation be repeated several times.

Expanding and organizing Equation (287View Equation) with the linear terms given explicitly and the quadratic terms collected in the expression for A, we obtain the long expression with all terms functions of either τ or uret:

0∗ 0i 1 0ij 1 0 i 1 0 ij 1 ψ 1 = ψ1 Y1i + ψ 1 Y2ij − 3Ψ ξ Y1i + 18Ψ ξ Y2ij + A. (288 ) i j 1 0 ij k 1 0 i[ kj -kj] 1 0 A = − 3 ξΨ Y1iY1j + 18 ξ Ψ Y2ijY 1k − 3ξ Ψ − 24ξ Y 1iY2kj (289 ) ij [ kl -kl ] 1 0 +18 ξ Ψ − 24ξ (τ ) Y2ijY 2kl i 1 ij 1 = A Y1i + A Y2ij.

Using Equations (269View Equation) and (271View Equation), the uret and τ in Equation (288View Equation) are replaced by s. On the right-hand side all the variables, e.g., 0k i k ψ 1 ,Ψ ,ξ ,etc., are functions of ‘s’; their functional forms are the same as when they were functions of uret and τ ; the linear terms are again explicitly given and the quadratic terms are collected in the BU + BT + A

0∗ -- 0i 1 0ij 1 0 i 1 0 ij 1 ψ1 (s,ζ,ζ) = ψ1√Y1i + ψ1 Y 2ij − 3Ψ ξ Y1i + 18Ψ ξ Y2√ij + BU + BT + A, (290 ) 2 √ -- 2 BU = − ---ψ0i1′ξkRY10kY11i + 2ψ01i′ξkRlY02klY 11i − ---ψ0i1j′ξkRY 10kY21ij √2--0ij′ 2 + 2ψ1 ξkRlY02klY21ij, √ -- { 1 BT = i3 2Ψ0 − -ξi′ξkI(s)Y10kY11i + ξi′ξkIl (s)Y 02klY11i + 3ξij′ξkIY 01kY 12ij 2 } ij′ kl 0 1 − 6ξ ξI Y 2klY2ij .

To proceed, we use the complex center-of-mass condition, namely, ψ ∗0k(s) = 0 1, and solve for 0k ψ 1 (uret). This is accomplished by first reversing the calculation via

0 0∗ 0∗ 0∗ 0 ψ1 = ψ 1 + 3L ψ2 = ψ 1 + 3L ψ2

and then, before extracting the l = 1 harmonic component, replacing the s by uret, via the inverse of Equation (271View Equation),

√2-- √ -- s = u(rRe)t + ---ξiR (u (Rret) )Y10i − 2ξijR(u(rRe)t )Y20ij, 2

using Equation 280View Equation

-- -- XR (s,ζ,ζ) = GR (s,ζ,ζ ). (291 )
In this process several of the quadratic terms cancel out and new ones arise.

The final expression for ψ0j 1, given in terms of the complex world line ξj expressed as a function of u ret, then becomes our basic equation:

√ -- 0j 0 j 18 i 0ij 108 i ij 3 2 0ij′ i ψ1 = 3Ψ ξ − --ξ ψ2 − ---Ψ ξ − -----ψ1 ξR (292 ) √ -- 5 5 ( √ -5 18--2- 0 ji ji i′ 3---2 i k + 5 Ψ (ξR + iξI )ξ + i 2 ξ Ψ √ -- ) 24 216 2 3 + ---ξmRkψ0m1i ′− -------ξmkψ0m2i + -Ψ0 (ξkR + iξkI)ξi′ 𝜖ikj. 5 5 2
This, which becomes the analogue of the Newtonian dipole expression ⃗D = M ⃗R, is our central relationship. Almost all of our results in the following sections follow directly from it.

We emphasize that prior to this discussion/derivation, the 0j ψ 1 and the ξj were independent quantities but in the final expression the ψ01j is now a function of the ξj.

Note that the linear term

0i 0 i ψ1 = 3Ψ ξ

coincides with the earlier results in the stationary case, Equation (259View Equation). From

2√2G- Ψ0 = − ---2--MB c

we have

√ -- ψ0i = − 6--2G-MB ξi, (293 ) 1 √c2- 6 2G i = − ---2--D ℂ(grav), i ci i i −1 i Dℂ (grav) = MB (ξR + iξI) = (D (mass) + ic J ).
We will see shortly that there is a great deal of physical content to be found in the nonlinear terms in Equation (292View Equation).
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