AA Review on post-Newtonian theory
Gihyuk ChoNovember 18, 2019
Contents a r X i v : . [ g r- q c ] N ov .1 Exchanges of the operators . . . . . . . . . . . . . . . . . . . . . . . . . . . 144.1.1 Exchange of Multipolar expansion and Instanteneous integral . . . . . 144.1.2 Exchange of PN Expansion and the Retarded Integration . . . . . . . 174.2 The Matching . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 174.3 Completion of R L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 18 → Linearized metric . . . . . . . . . . . . . . . . . . . . . . . . . 205.1.1 Derivation of G L . . . . . . . . . . . . . . . . . . . . . . . . . . . . . 215.2 Linearized metric → Non-Linear Contribution . . . . . . . . . . . . . . . . . 245.3 The Source Moments revisited . . . . . . . . . . . . . . . . . . . . . . . . . . 245.3.1 Generation of Gravitational waves from PN source . . . . . . . . . . 25
A Multipolar expansion of retarded Green function 26
A.1 General expression of multipole expanded solution . . . . . . . . . . . . . . . 312
Preliminary
Before performing post Newtonian formalism, we need to recast exact Einstein equationinto a form of the wave equation. So, Introduce following new gravitational field variables h µν as h µν := √− gg µν − η µν , (1.1)and note that I DO NOT impose the numerical conditions such as | h | (cid:28)
1. If we take thenumerical condition then the exact theory will be reduced the linearized theory.With ‘harmonic’ gauge condition ∂ ν h µν = 0 , (1.2)the exact Einstein equation G µν = 8 πGc T µν (1.3)becomes (cid:3) h µν = 16 πGc τ µν , (1.4)where τ µν := ( − g ) T µν + c πG Λ µν [ h ] . (1.5)In this harmonic gauge, and if considering energy-momentum conservation, we can find thatthe following is satisfied, ∂ ν τ µν = 0 . (1.6) In the previous section, I mention the relaxed EE. If we can solve the equation exactlyand analytically, everything is fine. But the world is not naive. Only a little special caseswhich have many geometric symmetries allow us to get exact solution. But the physicsistsare not naive neither. Usually, when the exact calculation is stucked then we bypass itvia perturbation. I will introduce two kinds of perturbation : post-Newtonian expansion &multiplor expansion. 3n this section, I introduce post-Newtonian expansion, which expand the motion of matteror metric field configuration in powers of 1 /c . What is the constraints of this perturbation?Because all perturbation theory is valid only in an appropriate domain, we have to narrowdown considering physical situation. First, if matter source is expanded as T µν = + ∞ (cid:88) m c n T µν ( n ) (2.1)it means that since each term of T are physical quantities such as energy and momentum,the matter source is non-relativistic, i.e. v/c (cid:28)
1. Additionally, if assume self gravity whichimplies R s d ∼ v/c, (2.2)where R s is size of matter component, and d is a size of matter distribution, thus mattersource undergoes weak self-gravitity. When these conditions(slow and weak self gravitation)are satisfied we call the matter source as post-Newtonian source .Then what about metric field? In this case, since metric field is generated from very slowmatter, the change of metric field in time is smaller than spatial gradient of metric field.Thus the d’Alembert operator becomes a Laplacian operator ∇ − c ∂ ∂t = ∇ (1 + O ( v/c ) ) . (2.3)It is natural result because our matter source is non-relativistic, the metric field is given byinstant potential field as Newton gravity. It also means that the retarded effect is very small,so post-Newtonain expansion include F ( t − rc ) = F ( t ) − rc F (1) ( t ) + r c F (2) ( t ) + · · · (2.4)as you see, for this perturbation not to break down, the radius r should be small. So thefinal physical constraint is small radius. Let λ be the typical gravitational wave, for smallretardation effect, the post-Newtonian expansion is valid when r < λ . Now, in the region that we set in the above subsection, we expand h in the powers of c ,¯ h µν = ∞ (cid:88) n =2 c n h µν ( n ) , (2.5)and ¯ τ µν = ∞ (cid:88) n = − c n τ µν ( n ) . (2.6)4ut these expansions in the relaxed EE (1.4), and equating terms with the same powersof c , then we get a recursive series of Poisson-type equations, ∇ h µν ( n ) = 16 πGτ µν ( n − + ∂ t h µν ( n − . (2.7)For ∇ U = ρ , the most well-known inverse of Laplacian, i.e. [ ∇ ] − is Poisson Integration, U = [ ∇ ] − ( ρ )( x ) = − π (cid:90) d x (cid:48) | x − x (cid:48) | ρ ( x (cid:48) ) (2.8)but the Poisson integration diverges when ρ does not vanish at spatial infinity r → ∞ . This divergence is supposed to occur when we repeat the recursive integration. Andthe Post-Newtonian expansion is valid only in small retardation region as defined earlier.Thus we can conclude that Poisson Integration seems not proper for our purpose in thatintegrands and integral domain are not ours.To resolve this problem, we introduce r a where a is negative and sufficiently large so that r a ρ vanishes at spatial infinity, then the following Poisson integration can be finite[ ∇ ] − ( r a ρ )( x ) = − π (cid:90) d x (cid:48) | x − x (cid:48) | | x (cid:48) r | a ρ ( x (cid:48) ) . (2.9)where r is a regularization parameter. Laurent-ly expanding the integral as [ ∇ ] − ( r a ρ )( x ) = (cid:80) ∞ n = −∞ a n I ( n ) . Because we should again put a = 0, ignoring negative order of a , we can getthe finite and intstaneous inversion of Laplacian which denoted as ∆ − [ ρ ] := FP [ ∇ ] − ( ρ ).Now, we can get a finite and particular solution of Poisson integral. But for gettinggeneral solution, we need homogeneous solution, that is, we need general solution of ∆ φ = 0, φ = B L ( t )ˆ x L + C L ( t ) ˆ ∂ L ( 1 r ) , (2.10)it is obvious that when r >
0, then ∆ r = 0. The first term is ∂ i ∂ i ( x i ··· i l ) = ∂ i ( δ i i x i ··· i l + · · · ) = δ i i δ ii x ··· · · · . (2.11)so every term includes trace of any two indexes. Because every trace should be vanishing,the first term is a solution of Laplace equation. Thus, the solution of (2.7), with generalhomogeneous part requring only non-singular term at r = 0 , can be expressed as h µν ( n ) = 16 πG ∆ − τ µν ( n − + ∂ t ∆ − h µν ( n − + ∞ (cid:88) l =0 B µν ( n ) ,L ( t )ˆ x L . (2.12)But only ˆ x L terms are needed because we need a solution singular at r = 0, i.e. C ( n ) ,L = 0.Now proceed iteration, until we get the following expression, h µν ( n ) = 16 πG [ n/ − (cid:88) k =0 ∂ kt ∆ − k − τ µν ( n − − k ) + ∞ (cid:88) l =0 [ n/ − (cid:88) k =0 ∂ kt B µν ( n − k ) ,L ∆ − k (ˆ x L ) . (2.13) Because we are dealing with near zone field. ¯ h µν = 16 πGc (cid:3) − ¯ τ µν + ∞ (cid:88) l =0 ∞ (cid:88) k =0 c k ∂ kt ( + ∞ (cid:88) n =0 B µν ( n +2) ,L c n +2 )∆ − k (ˆ x L ) (2.14)where (cid:3) − := + ∞ (cid:88) k =0 ( ∂c∂t ) k ∆ − k − . (2.16)Changing variable from B to A , + ∞ (cid:88) n =0 B µν ( n +2) ,L c n +2 = − ∂ l +1 t A µνL ( t ) c l +1 (2 l + 1)!! , (2.17)and using the following identity∆ − k (ˆ x L ) = (2 l + 1)!!(2 l )!!(2 l + 2 k + 1)!! r k + l ˆ n L , (2.18)and ˆ ∂ L ( r λ ) = ˆ n L r λ − l ( λ )!!(2 l )!! , (if λ ≥ l, otherwise 0) . (2.19) ¯ h µν = + ∞ (cid:88) n =2 c n h µν ( n ) (2.15)= 16 πG + ∞ (cid:88) n =2 [ n/ − (cid:88) k =0 c n ∂ kt ∆ − k − τ µν ( n − − k ) + (cid:88) l + ∞ (cid:88) n =2 [ n/ − (cid:88) k =0 c n ∂ kt B ( n − k,L ) ∆ − k (ˆ x L )= 16 πG ∞ (cid:88) k =0 + ∞ (cid:88) n =2+2 k c n ∂ kt ∆ − k − τ µν ( n − − k ) + ∞ (cid:88) k =0 + ∞ (cid:88) n =2+2 k (cid:88) l c n ∂ kt B ( n − k,L ) ∆ − k (ˆ x L )= 16 πG ∞ (cid:88) k =0 + ∞ (cid:88) n =0 c n +2 k +2 ∂ kt ∆ − k − τ µν ( n − + ∞ (cid:88) k =0 + ∞ (cid:88) n =0 (cid:88) l c n +2 k +2 ∂ kt B ( n +2 ,L ) ∆ − k (ˆ x L )= 16 πGc ∞ (cid:88) k =0 ( ∂ t /c ) k ∆ − k − ( + ∞ (cid:88) n =0 c n − τ µν ( n − ) + (cid:88) l ∞ (cid:88) k =0 c k ∂ kt ( + ∞ (cid:88) n =0 B ( n +2 ,L ) c n +2 )∆ − k (ˆ x L )= 16 πG [ n/ − (cid:88) k =0 ∂ kt ∆ − k − τ µν ( n − − k ) + (cid:88) l [ n/ − (cid:88) k =0 ∂ kt B ( n − k ) ∆ − k (ˆ x L ) . ∞ (cid:88) l =0 ∞ (cid:88) k =0 c k ∂ kt ( + ∞ (cid:88) n =0 B µν ( n +2) ,L c n +2 )∆ − k (ˆ x L ) = − ∞ (cid:88) l =0 ∞ (cid:88) k =0 ∂ l +2 k +1 t A µνL ( t ) c l +2 k +1 ˆ ∂ L r l +2 k l !(2 l + 2 k + 1)! k !2 k − l (2.20)= − ∞ (cid:88) l =0 ∞ (cid:88) k = l ∂ k +1 t A µνL ( t ) c k +1 ˆ ∂ L r k (2 k + 1)! (2.21)then the above solution has the following form:¯ h µν = 16 πGc (cid:3) − ¯ τ µν − Gc ∞ (cid:88) l =0 ( − ) l l ! ˆ ∂ L (cid:2) A µνL ( t − rc ) − A µνL ( t + rc )2 r (cid:3) . (2.22)Note that the homogeneuos part have an essential role in radiation reaction and tail effect. In this section, I consider the multipole expansion of exact solution of the relaxed EE.If we avoid r = 0 point, multipole expansion can be acheived easily without any physicalconstraints. This is different from post-Newtonian expansion at this point because pN ex-pansion gives both physical condition and constructing way. Although the post-Newtonianconstruction is good tool in the domain of its validity, it does not work at far region ( r > λ )as mentioned in the previous argument. So one need to develop new construction of approx-imate expression of exterior field, say post-Minkoswkian construction. Usually this methodequip multipole expansion, so also called MPM expansion. Because the pM construction only concern vaccum solution, There is no need to considerany physical constraints of matter. Let us start from expanding exact solution h µν ext = + ∞ (cid:88) m =1 G m h µν ( m ) . (3.1)So that, (cid:3) h µν ( m ) = Λ µν ( m ) [ h (1) , h (2) , h (3) · · · h ( m − ] , (3.2) ∂ µ h αµ ( m ) = 0 . (3.3)7ote that it is not perturbation because G is not small paramter, it is just bookkeepingparameter. Rather, this is a kind of iteration method. As PN case, the source term Λ doesnot have a compact support, so we need to introduce a renormalization process.Let us define the integration operator (cid:3) − (cid:3) − ρ ( t, x ) = FP B =0 (cid:16) − π (cid:90) d x (cid:48) | x − x (cid:48) | | x (cid:48) r | B ρ ( t − | x (cid:48) − x | /c, x (cid:48) ) (cid:17) . (3.4)It is obvious that (cid:3) (cid:16) − π (cid:82) d x (cid:48) | x − x (cid:48) | | x (cid:48) r | B ρ ( t − | x (cid:48) − x | /c, x (cid:48) ) (cid:17) = ( rr ) B ρ ( t, x ). Since the finite partof the right hand side of it is ρ ( t, x ), thus (cid:3) · (cid:3) − ρ ( t, x ) = ρ ( t, x ) . We conclude that (cid:3) − isreally an inverse operator of (cid:3) . From this particular solution, we can find a general solutionwith a homogeneous solution h µν ( m ) ( t, x ) = (cid:3) − Λ µν ( m ) ( t, x ) + + ∞ (cid:88) l =0 ˆ ∂ L (cid:0) X µν ( m ) L ( t − r/c ) r (cid:1) . (3.5)Before moving forward, I need to claim that the above general solution has a following nearzone structrure, i.e. when r → h µν ( m ) ( t, x ) = (cid:88) ˆ n L r q (ln r ) p F L,q,p,m ( t ) + O ( r N ) , (3.6)or, h ( m ) ∈ L m − where q ≤ q ≤ N , with q is a negative intger, and p ≤ m − h (1) ∈ L , and second, we aregoing to check that if f ∈ L n , then (cid:3) − f ∈ L n +1 . Let us prove the latter.(Proof) Since f ∈ L n we can write that f = (cid:88) l, p, q ˆ n L r q (ln r ) p F ( t ) + O ( r N ) , (3.7)with p ≤ n . Thus (cid:3) − ( r B f ) = (cid:88) l, p, q ( ∂∂B ) p (cid:3) − (cid:0) ˆ n L r B + q F ( t ) (cid:1) + (cid:3) − r B O ( r N ) (3.8)where (cid:3) − is the usual retarded integral. Because (cid:3) = ∆ − ( ∂c∂t ) , we can write (cid:3) − (cid:0) ˆ n L r B + q F ( t ) (cid:1) = F ( t )∆ − (cid:0) ˆ n L r B + q (cid:1) + 1 c (cid:3) − ( F (2) ( t ) ∆ − (ˆ n L r B + q )) . (3.9)this identitiy is easily checked by taking (cid:3) operator on both sides. By s -th iterations of thisprocess, we can write that (cid:3) − (cid:0) ˆ n L r B + q F ( t ) (cid:1) = s (cid:88) k =0 F (2 k ) ( t )∆ − k − (cid:0) ˆ n L r B + q (cid:1) + 1 c s +2 (cid:3) − ( F (2 s +2) ( t ) ∆ − s − (ˆ n L r B + q )) . (3.10)8rom this, it becomes with taking the finite part operator, FP B =0 (cid:3) − ( r B f ) (3.11)= FP B =0 (cid:88) l, p, q ( ∂∂B ) p s (cid:88) k =0 F (2 k ) ( t )∆ − k − (cid:0) ˆ n L r B + q (cid:1) + FP B =0 (cid:88) l, p, q ( ∂∂B ) p c s +2 (cid:3) − ( F (2 s +2) ( t ) ∆ − s − (ˆ n L r B + q ))+ FP B =0 (cid:3) − r B O ( r N ) . Because ∆ − (cid:0) ˆ n L r B + q (cid:1) = ˆ n L r B + q +2 ( B + q + 2 − l )( B + q + 3 + l ) (3.12)we can deduce that every ∆ − m (cid:0) ˆ n L r B + q (cid:1) has only simple poles at each positions. Thus, at B = 0, because it has simple pole structure, (If it has a pole at B = 0, and we are consideringthe case.) we can write it as FP B =0 ( ∂∂B ) p ∆ − m (cid:0) ˆ n L r B + q (cid:1) = FP B =0 ( ∂∂B ) p ( D ( B ) B e B ln r ) r q +2 m (3.13)where D ( B ) is a rational function of B and analytic at B = 0. Because of the presence of thedifferential operators ( ∂∂B ) p , we need to find the coefficient of B p +1 of a Laurent expansionof D ( B ) e B ln r which gives a regularized expression. Therefore, the first term can be writtenas, FP B =0 (cid:88) l, p, q ( ∂∂B ) p s (cid:88) k =0 F (2 k ) ( t )∆ − k − (cid:0) ˆ n L r B + q (cid:1) = (cid:88) l, p ≤ n, q 3. The k = 2 case requires a particular concern. (cid:3) − [ r − ˆ n L H , L ( u )] = − c ˆ n L r (cid:90) ∞ dy Q l ( r + cyr ) H , L ( t − r + cyc ) (3.29)where Q l is the Legendre function of the second kind. Using the asymptotic behaviour of Q l ( x ) as x → (cid:3) − [ r − ˆ n L H , L ( u )] = c ˆ n L r (cid:90) ∞ dy H , L ( t − rc − y )[log( cy r ) + 2 a l ] + O ( log rr ) . (3.30)Before getting rid of this log r , direct computations yields that H ( u, n ) := ˆ n L H ,L ( u ) has afollowing structure that H µν ( u, n ) = k µ k ν σ (2) ( u, n ) , (3.31)where k µ = (1 , n i ). Hence by choosing that ξ µ (2) := (cid:3) − [2 r − k µ (cid:90) u −∞ dv σ (2) ( v, n )] . (3.32)Let us claim that the following expression does not have any logarithms : (cid:3) − [ r − H µν ( u, n )] + ∂ µ ξ ν (2) + ∂ ν ξ µ (2) − η µν ∂ ρ ξ ρ (2) . (3.33)First, if derivatives of ∂ µ ξ ν (2) + ∂ ν ξ µ (2) acts on u in integration, it yields ∂ µ ξ ν (2) + ∂ ν ξ µ (2) ∼ − (cid:3) − [ r − H µν ( u, n )] (3.34)hence we can cancle (cid:3) − [ r − H µν ( u, n )]. Additionally, the part of ∂ ρ ξ ρ (2) which the derivationacts on u likewise, is zero ( ∵ k µ k µ = 0). And the remaining derivatives give rise to anotherfactor of r , and thus no logarthm.Therefore choosing H µ (2) ≡ (cid:3) ξ µ (2) completes the second iteration of the MPM constructionin the radiative coordinate ( i.e. logarithm-free ) as˜ h µν rad(2) ( t, x ) = ¯ h µν + ∂ µ ξ ν (2) + ∂ ν ξ µ (2) − η µν ∂ ρ ξ ρ (2) . (3.35)In principle, we can stop here and do same things at the higher orders iteratively. But forfurther convenience, we define new set of multipole moments M of which the monopole iszero, such that h µν rad(1) [ M ] is a general outgoing homogeneous solution of dAlembert equationand its leading coefficient of 1 /r expansion cancles that of ˜ h µν rad(2) ( t, x ), h µν rad(1) [ M ] = − Z µν ( u, n ) r + O ( 1 r ) , (3.36)12here ˜ h µν rad(2) ( t, x ) = Z µν ( u, n ) r + O ( 1 r ) , (3.37)so that new solution h µν rad(2) , which is different from ˜ h µν rad(2) only by a homogeneous solution, h µν rad(2) ( t, x ) = ˜ h µν rad(2) + h µν rad(1) [ M ] = O ( 1 r ) (3.38)starts from the order of r . Hence now for the cases of n ≥ 3, there are no logarithms. Bythis algorithm, it is possible to get the general expression of metric in radiative coordinatewithout any logarithms. Now we come to the conclusion that Now we attempt to find out coordinate transformation rule as x (cid:48) µ = x µ + G ¯ ξ µ (1) ( x ) + G ¯ ξ µ (2) ( x ) + O ( G ) , (3.39)where x is a harmonic coordinate while x (cid:48) stands for a radiative coordinate. For exampleup to quadractic order, the results are displayed. After the change of coordinates from x to x (cid:48) , g h ( x ) = √− g g ( x ) = η + (cid:80) n =1 G n h ( n ) ( x ) becomes g rad ( x (cid:48) ) = √− g g ( x (cid:48) ) = η + (cid:80) n =1 G n h rad( n ) ( x (cid:48) ) as follows : h µν rad(1) ( x (cid:48) ) = (cid:104) h µν (1) + ∂ µ ¯ ξ ν (1) + ∂ ν ¯ ξ µ (1) − η µν ∂ ρ ¯ ξ ρ (1) (cid:105) x = x (cid:48) , (3.40) h µν rad(2) ( x (cid:48) ) = (cid:104) h µν (2) + ∂ µ ¯ ξ ν (2) + ∂ ν ¯ ξ µ (2) − η µν ∂ ρ ¯ ξ ρ (2) − ¯ ξ ρ (1) ∂ ρ h µν (1) (cid:105) x = x (cid:48) , (3.41) · · · . One can easily get that ¯ ξ (1) = ξ (1) , which has been already given. And I conjecture that wedo not need to find explicit expressions of ¯ ξ ( n ) ( n ≥ 2) for getting gravitational waveforms(the coefficient of r ) because they vanish at null future infinity , i.e. ¯ ξ ( n ) ∼ O ( log rr ) while¯ ξ (1) ∼ log r . At the radiative region ( r >> i.e. (cid:80) ∞ n =1 h rad( n ) = H + O ( H ). As seen before, in harmonic gauge, thelinearized metric can be expressed generally as in Eq.() with two multipole moments U, V (which are STF tensors), and hence gravitational waveforms are described as follows H T Tij = 4 Gc r P ijab ( n ) ∞ (cid:88) l =2 c l l ! (cid:26) ˆ n L − U abL − ( u ) − lc ( l + 1) ˆ n cL − (cid:15) cd ( a V b ) dL − ( u ) (cid:27) + O ( 1 r ) . (3.42)13his expression is derived in harmonic gauge but in the linearized theory, the TT part is agauge-invariant, so we can regard H T Tij as bieng expressed in radiative coordinate. Now, by the post-Newtonian and post-Minkowskian Construction, we can construct thesolution of relaxed EE in the expression of Near-zone and Far-zone expansion. It means thateach solution have different domain of validity, but unfortunately two ways of constructioninclude other global operators, (cid:3) − and (cid:3) − . It is not trivially true that, for example,the far-zone expansion operator and (cid:3) − inst commute. Because by (cid:3) − we integrateentire space domain, so the post-Newtonian constructed solution ¯ h which is same with theexact solution h (which we do not know how to construct) only in r < R . So, (cid:3) − ¯ h and (cid:3) − h results in different solution. Even though after it we expand (cid:3) − h in far zone style,i.e. (cid:3) − h , it does not have coincidence with (cid:3) − ¯ h .This section includes the algebraic results of exchanges of expansion operator and inte-gration operators which involves in constructing solutions. In this subsection, I am going to prove that M ( (cid:3) − (¯ τ )) = (cid:3) − ( M (¯ τ )) − π ∞ (cid:88) l =0 ( − ) l l ! ˆ ∂ L (cid:8) F ( t − rc ) + F ( t + rc )2 r (cid:9) , (4.1)Note that, M ( (cid:3) − (¯ τ )) is singular at | x | = 0 while (cid:3) − ( M (¯ τ )) is regular at any points.Before proceeding retarded integral, we consider the Poisson integral, I ( t, x ) := ( − π )[∆ − ¯ τ ( t, x )] = FP B=0 (cid:90) d y | y | B | x − y | ¯ τ ( t, y ) . (4.2)We rewrite the integration I ( t, x ) = (cid:90) | y | < | x | d y | y | B | x − y | ¯ τ ( t, y ) + (cid:90) | y | > | x | d y | y | B | x − y | ¯ τ ( t, y ) (4.3)= (cid:88) l ˆ ∂ L ( 1 | x | ) (cid:90) | y | < | x | d y | y | B + l ˆ y L ¯ τ ( t, y ) + (cid:90) | y | > | x | d y | y | B | x − y | M (¯ τ ( t, y )) . The second equality holds only unless | x | (cid:54) = 0. The changing of the first term is because thatif | y | has a finite value then integration and multipole expansion can be converted. See that14 ( | x − y | ) = (cid:80) l ( − ) l l ! ˆ ∂ L ( | x | ) y L . So the first one is a multipole expanded homogeneous solutionof the Poisson equation.Furthermore, the integrand of the first integration becomes diverging when | y | → ∞ since thepN source ¯ τ diverges, while the integrand of the second integration becomes diverging when | y | = 0 since the multipole expansion of the source ¯ τ diverges. Thus by B -regularzation,those divergences are removed. I becomes in terms of entire integals M · I ( t, x ) = (cid:88) l ˆ ∂ L ( 1 | x | ) (cid:90) d y | y | B + l ˆ y L ¯ τ ( t, y ) + (cid:90) d y | y | B | x − y | M (¯ τ ( t, y )) . (4.4)Note that, the fianl form is the multipole exapnded. Because, the first term is the mulitpoleexpanded homogenerous solution. And the second term, since (cid:90) d Ω y ˆ y L | x − y | ∼ ˆ x L (4.5)also has multipole exanded form. Thus I can regard the above expression as the mulitpoleexpansion of M · I ( t, x ). That is, we come to the conclusion that M (∆ − (¯ τ )) = ∆ − ( M (¯ τ )) + ˆ ∂ L ( 1 | x | ) (cid:90) d y | y | B + l ˆ y L ¯ τ ( t, y ) (4.6)Getting rid of M in front of the second term is from : See footnote 3.Now go back to our goal M ( (cid:3) − (¯ τ )), let us change the global operators interating theabove identities. Because the local operator can be exchanged trivially , M ( (cid:3) − (¯ τ )) = + ∞ (cid:88) k =0 ( ∂c∂t ) k M (∆ − k − ¯ τ ) (4.10)= (cid:3) − M (¯ τ ) − π + ∞ (cid:88) k =0 (cid:88) l k (cid:88) i =0 ∆ − i [ ˆ ∂ L ( 1 | x | )]FP B=0 (cid:90) d y ∆ i − k (ˆ y L ) | y | B ( ∂c∂t ) k ¯ τ ( t, y ) . I claim that the mulitpolar expansion of source gives far zone expansion of integral. Because If one splitthe second integral again, as (cid:90) d y | y | B | x − y | M (¯ τ ) = (cid:88) (cid:90) | x | d y | y | B | x − y | M (¯ τ ) + (cid:90) + ∞| x | d y | y | B | x − y | M (¯ τ ) (4.7)with the far-zone structure of M (¯ τ ) = (cid:80) ˆ y L y a (ln y ) p τ ( t ), (cid:90) d y | y | B | x − y | M (¯ τ ) ∼ (cid:88) ˆ x L | x | l +1 (cid:90) | x | dy y B + a + l +2 (ln y ) p + (cid:88) ˆ x L (cid:90) ∞| x | dy y B + a − l +1 (ln y ) p (4.8)= (cid:88) ˆ x L | x | l +1 ( ddB ) p [ | x | B + a + l +3 B + a + l + 2 ] + (cid:88) ˆ x L ( ddB ) p [ −| x | B + a + l +3 B + a + l + 2 ]thus taking the finite part operator on both sides yields the far-zone structutre : FP B =0 (cid:90) d y | y | B | x − y | M (¯ τ ) ∼ (cid:88) ˆ x L | x | a (ln x ) p +1 . (4.9) 15r by manipulating indices, it can be written as M ( (cid:3) − (¯ τ )) (4.12)= (cid:3) − M (¯ τ ) − π + ∞ (cid:88) i =0 (cid:88) l ∞ (cid:88) k = i ∆ − i [ ˆ ∂ L ( 1 | x | )]FP B=0 (cid:90) d y ∆ i − k (ˆ y L ) | y | B ( ∂c∂t ) k ¯ τ ( t, y )= (cid:3) − M (¯ τ ) − π + ∞ (cid:88) i =0 ( ∂c∂t ) i (cid:88) l ∆ − i [ ˆ ∂ L ( 1 | x | )] ∞ (cid:88) k =0 FP B=0 (cid:90) d y ∆ − k (ˆ y L ) | y | B ( ∂c∂t ) k ¯ τ ( t, y )Using ∆ − i [ ˆ ∂ L ( 1 | x | )] = ˆ ∂ L ( r i − (2 i )! ) (4.13)and defining new function as, F L ( t ) := + ∞ (cid:88) j =0 c j FP B=0 (cid:90) d y ∆ − j (ˆ y L ) | y | B ( ∂∂t ) j ¯ τ ( t, y ) , (4.14)it can be rewritten as M ( (cid:3) − (¯ τ )) (4.15)= (cid:3) − M (¯ τ ) − π + ∞ (cid:88) l =0 ˆ ∂ L + ∞ (cid:88) i =0 r i − (2 i )! ( ∂c∂t ) i F L ( t )finally, we arrive at the desirable result, M ( (cid:3) − (¯ τ )) = (cid:3) − ( M (¯ τ )) − π ∞ (cid:88) l =0 ( − ) l l ! ˆ ∂ L (cid:8) F ( t − rc ) + F ( t + rc )2 r (cid:9) . (4.16)Or, use ∆ − j (ˆ y L ) = | y | j ˆ y L (cid:90) +1 − dz z j (2 j )! δ l ( z ) , (4.17) I use an identity that (cid:90) d z | z | B ˆ z L ∆ − ¯ τ = (cid:90) d z | z | B ˆ z L (cid:90) d y | y | B ¯ τ ( t, y ) | z − y | (4.11)= (cid:90) d y | y | B ¯ τ ( t, y ) (cid:90) d z ˆ z L | z | B | z − y | = (cid:90) d y | y | B ¯ τ ( t, y )∆ − ˆ y L . δ l ( z ) = (2 l + 1)!!2 l +1 l ! (1 − z ) l . (4.18)so that F L can be written in more compact form, F L ( t ) = FP B=0 (cid:90) d y ˆ y L | y | B (cid:90) +1 − dz δ l ( z ) ¯ τ ( t ± z | y | /c, y ) . (4.19) In this subsection , the exchange of PN expansion and the integrals is introduced. But Ido not take a rigorous derivation but rather an intuitive way. If you want to see the rigorousway, see [2], or my another note. Now I claim that (cid:3) − [ M ( τ )] = (cid:3) − M ( τ ) − π ∞ (cid:88) l =0 ( − ) l l ! ˆ ∂ L (cid:40) R L ( t − r/c ) − R L ( t + r/c )2 r (cid:41) (4.20)Here is a rough proof. Fistly, both (cid:3) − [ M ( τ )] and (cid:3) − M ( τ ) are the particular solutionsof (cid:3) A = M ( τ ). Thus, they are different up to some homogeneous solutions. And secondly,since (cid:3) − [ M ( τ )] diverges as | x | → ∞ while (cid:3) − M ( τ ) does not diverge, the homogeneouspart must be divergent at infinity. The existence of the buffer region d < r < λ where both PN and MPM approximationswork. Let h be an exact solution. Then outside of sources region, its multipole expandedone M ( h ) which is constructed by MPM iteration, is numerically equal i.e. h = M ( h ) ( r > d ) (4.21)Note that the numerical equality does not hold in r < d and also M ( h ) diverges at r = 0.Mathematically, multipole expansioned functions converge to the original one only except r = 0. But here our multipole expansion is a vacuum solution. That is why there is arestriction of r > d . And the pN expansion h is numerically same with h in r < λ , h = h ( r < λ ) . (4.22)Thus in d < r < λ , the following numerically holds. h = M ( h ) . (4.23)17owerver, by M ( h ) = M ( h ) (4.24)I mean that two expansions are formally equivalent term by term. Note that M ( h ), the pNexpansion c → ∞ of M ( h ), is a Talyor expansion as r → M ( h ) is made of blockslike I L ( t − r/c ). And M ( h ) is a kind of far zone expansion r → ∞ . This might be justifiedintuitively, because in the Electrostatics or Newtonian gravity, the mulitpole expansion of | x − x | is nothign but far-zone expansion. Thus both are a Laurent series of r , i.e. M ( h ) = M ( h ) = (cid:88) m,l,p ˆ n L r m (ln r ) p F L,m,p ( t ) . (4.25)Before we are going to get more concrete expression, recall the result of previous sections,¯ h µν = 16 πGc (cid:3) − ¯ τ µν − Gc ∞ (cid:88) l =0 ( − l l ! ∂ L (cid:2) A L ( t − rc ) − A L ( t + rc )2 r (cid:3) (4.26) M ( h µν ) = (cid:3) − [ M (Λ µν )] − Gc ∞ (cid:88) l ( − l l ! ˆ ∂ L (cid:8) X µνL ( t − rc ) r (cid:9) . Applying M ( h ) = M ( h ) and the exchanges of the operators derived before, gives rise to16 πGc (cid:3) − ( M (¯ τ )) − Gc ∞ (cid:88) l =0 ( − ) l l ! ˆ ∂ L (cid:8) F L ( t − rc ) + F L ( t + rc )2 r (cid:9) (4.27) − Gc ∞ (cid:88) l =0 ( − ) l l ! ˆ ∂ L (cid:2) A L ( t − rc ) − A L ( t + rc )2 r (cid:3) = (cid:3) − M ( ¯Λ) − Gc ∞ (cid:88) l =0 ( − l ! ˆ ∂ L (cid:8) R L ( t − rc ) − R L ( t + rc )2 r (cid:9) − Gc ∞ (cid:88) l ˆ ∂ L (cid:8) X µνL ( t − rc ) r (cid:9) , because πGc M ( τ ) = M (Λ), since they are made by the MPM construction, we come to theconclusion that X L ( t ) = F L ( t ) , (4.28) A L ( t ) = R L ( t ) + F L ( t ) . R L To complete the expression R L , I am going to make a rather sloppy argument, since eventhough there is an explict way to R L it is quite messy and complicated to get. The physical18eaning of F L is the vacuum part of graviational field moving across the surface of lightcone. I write it again, F L ( t ) = (cid:90) d y | y | B ˆ y L (cid:90) +1 − dz δ l ( z ) ¯ τ µν ( y, t − z | y | ) . (4.29)Sicne the non-linearity of graviation makes transferring of physical information within light-cones, we need more contribution other than F L . Let us denote it ˜ R L tentatively,˜ R L ( t ) = (cid:90) d y | y | B ˆ y L (cid:18)(cid:90) + ∞ +1 dzδ l ( z )¯ τ µν ( y, t − z | y | ) + (cid:90) − −∞ dzδ l ( z )¯ τ µν ( y, t − z | y | (cid:19) . (4.30)Since the latter terms is from future, we let it removed. And the weight function δ l ( z )which is normalized as (cid:82) +1 − dzδ l ( z ) = 1, must be replaced by suitable weight fucntion. Let γ l ( z ) = − δ l ( z ) which satisfies, (cid:90) ∞ +1 dz γ l ( z ) = 1 . (4.31)Finally, I come to an induction that˜ R L ( t ) = (cid:90) d y | y | B ˆ y L (cid:90) + ∞ +1 dzγ l ( z )¯ τ µν ( y, t − z | y | ) , (4.32)but R L arose when we compute (cid:3) − M ( τ ), it is desirable to change PN source to MPMsource. Finally we have R L ( t ) = (cid:90) d y | y | B ˆ y L (cid:90) + ∞ +1 dz γ l ( z ) M ( τ ) µν ( y, t − z | y | ) . (4.33) In this section, let us turn our head to thinking M ( h ) = (cid:80) ∞ n =0 G n h ( n ) ( t, x ) as construc-tion method. Thus it is followed that ∞ (cid:88) n =1 G n h ( n ) ( t, x i ) = (cid:3) − [ M (Λ µν )] − Gc ∞ (cid:88) l ( − l l ! ˆ ∂ L (cid:8) F µνL ( t − rc ) r (cid:9) , (5.1)where M (Λ)[ h ] = Λ[ M ( h )]. It is our aim to match h ( n ) ( t, x ) which arises in practical com-putations, with a part of the general structure.19 .1 PN source → Linearized metric Let u µν ≡ (cid:3) − [ M (Λ µν )], then because ∂ µ M ( h ) µν = 0, we get ∂ ν u µν = 4 Gc ∞ (cid:88) l ( − l l ! ∂ ν ˆ ∂ L (cid:8) F µνL ( t − rc ) r (cid:9) . (5.2)Thus we can find that ∂ ν u µν is a homogeneous solution i.e. (cid:3) ∂ ν u µν and also timedepndence should be given in terms of t − r/c . Thus schematically this has the followingstrucute, ∂ ν u µν = (cid:88) l ( − ) l l ! ˆ ∂ L (cid:8) G µL ( t − r/c ) r (cid:9) . (5.3)By direct integration we get v µν ( t, x ) which satisfies ∂ ν ( u µν + v µν ) = 0 and (cid:3) v = 0. v = 4 Gc (cid:26) − cr (cid:90) G + ∂ a ( 1 r (cid:20) c (cid:90) G a + c (cid:90) (cid:90) G a − G bab (cid:21) ) (cid:27) , (5.4) v i = 4 Gc (cid:110) − r (cid:20) c (cid:90) G i − c ˙ G aai (cid:21) + c ∂ a (cid:18) r [ (cid:90) G ia − (cid:90) G ai ] (cid:19) (5.5) − (cid:88) l ≥ ( − ) l l ! ˆ ∂ L − (cid:18) r G iL − (cid:19) (cid:111) ,v ij = 4 Gc (cid:40) r G ( ij ) + 2 (cid:88) l ≥ ( − ) l l ! ˆ ∂ L − (cid:18) rc ¨ G aijaL − (cid:19) (5.6)+ (cid:88) l ≥ ( − ) l l ! (cid:104) ˆ ∂ L − ( 1 rc ˙ G aijL − ) + ∂ aL − ( 1 r G aijL − ) + 2 δ ij ˆ ∂ L − ( 1 r G aaL − ) − ∂ L − i (cid:18) r G aj ) aL − (cid:19) − ∂ L − (cid:18) r G ( ij ) L − (cid:19) (cid:105)(cid:41) . Thus if the STF tensor G is determined then v is determined. Now let us rewrite the generalMPM solution as M ( h ) = u + v − v − Gc ∞ (cid:88) l ( − l l ! ˆ ∂ L (cid:8) F µνL ( t − rc ) r (cid:9) . (5.7)As the construction both u + v and − v − Gc (cid:80) ∞ l ( − l l ! ˆ ∂ L (cid:8) F µνL ( t − rc ) r (cid:9) satisfy the harmonicitycondition. Thus, the linearized potential can be assigned as G h µν (1) ( t, x ) = − Gc ∞ (cid:88) l ( − l l ! ˆ ∂ L (cid:8) F µνL ( t − rc ) r (cid:9) − v µν . (5.8)In order to get the explicit expression of h µν (1) ( t, x ), let us find the explicit expression for v .20 .1.1 Derivation of G L First let us compute that ∂ F µ ( t − r/c ) = dc du F µ ( t − r/c ) . (5.9)21rom the definition of F we have gotten earlier, we get ∂ F µ (5.10)= (cid:90) d y | y | B ˆ y L (cid:90) +1 − dz δ l ( z ) ∂ ¯ τ µ ( t − rc + z | y | c , y i ) ( ∵ ∂ ν ¯ τ νµ = 0)= − (cid:90) d y | y | B ˆ y L (cid:90) +1 − dz δ l ( z ) (cid:18) ddy j ¯ τ jµ ( T ( y ) , y i ) − ∂T ( y ) ∂y j ∂ T ¯ τ jµ ( T, y i ) (cid:19) ( T := t − rc + z | y | c )= (cid:90) d y ∂∂y j ( | y | B ˆ y L ) (cid:90) +1 − dz δ l ( z ) ¯ τ jµ ( T, y i ) (integration by parts) − (cid:90) d y | y | B ˆ y L (cid:90) +1 − dz ddz δ l +1 ( z )(2 l + 3) z ∂T ( y ) ∂y j ∂ T ¯ τ jµ ( T, y i ) ( ∵ dδ l +1 dz = − (2 l + 3) zδ ( z ) l )=: G µL ( t − r/c ) + (cid:90) d y | y | B l δ j The linear metric, which is now exaactly determined by the PN source, G h (1) µν ( t, x ) = − Gc ∞ (cid:88) l ( − l l ! ˆ ∂ L (cid:8) F µνL ( t − rc ) r (cid:9) − v µν , (5.19)determines the second order h (2) µν ( t, x ) as the following algorithm:1. Put G h (1) µν ( t, x ) into u = (cid:3) − M (Λ)[ h ].2. Get G order of u , u = G u (2) + O ( G ).3. From u (2) calculate G (2) L not refering to eq.(5.11), or from the definition eq.(5.11) whichrequires the knowledge about h (1) at some order, we can get G contribution of G L .4. Get v (2) via eq.(5.4).5. Determine h (2) := u (2) + v (2) .The higher orders are determined similarly. Mathematically, as generic symmetric 2-rank tensors h µν which satisfies ∂ µ h µν , linearizedmetric h (1) is also decomposed into h (1) µν = k µν + ∂ µ ϕ ν + ∂ ν ϕ µ − η µν ∂ λ ϕ λ , (5.20)24ith k = − c (cid:88) l =0 ( − ) l l ! ˆ ∂ L (cid:18) I L ( t − rc ) r (cid:19) , (5.21) k i = 4 c (cid:88) l =1 ( − ) l l ! ˆ ∂ L − (cid:40) I (1) iL − ( t − rc ) r + ll + 1 (cid:15) iab ∂ a (cid:18) J bL − ( t − rc ) r (cid:19)(cid:41) ,k ij = − c (cid:88) l =2 ( − ) l l ! ˆ ∂ L − (cid:40) I (2) ijL − ( t − rc ) r + 2 ll + 1 ∂ a (cid:32) (cid:15) ab ( i J (1) j ) L − ( t − rc ) r (cid:33)(cid:41) ,ϕ = 4 c (cid:88) l =0 ( − ) l l ! ˆ ∂ L (cid:18) W L ( t − rc ) r (cid:19) , (5.22) ϕ i = − c (cid:88) l =0 ( − ) l l ! ˆ ∂ iL (cid:18) X L ( t − rc ) r (cid:19) , − c (cid:88) l =1 ( − ) l l ! ˆ ∂ L − (cid:40) Y (1) iL − ( t − rc ) r + ll + 1 (cid:15) iab ∂ a (cid:18) Z bL − ( t − rc ) r (cid:19)(cid:41) . By matching two expressions of h (1) , − Gc ∞ (cid:88) l ( − l l ! ˆ ∂ L (cid:8) F µνL ( t − rc ) r (cid:9) − v µν = G k µν + G ( ∂ µ ϕ ν + ∂ ν ϕ µ − η µν ∂ λ ϕ λ ) , (5.23)we can get the explict expressions of { I, J, W, X, Y, Z } . This set of 6 mulitpole momentsis called the source mulitpole moments , which parametrizes 6 degrees of freedom in h (1) . Ilist their explicit expressions in terms of PN sourc below. (5.24) We have already known that the relation between { I, J, W, X, Y, Z } and { M, S } , andthe other relation between { M, S } and { U, V } perturbatively. Thus we know how to connectPN source ¯ τ and the gravitational waveform h tt , i.e. h tt [¯ τ ], only perturbatively not in a closedform. 25 Multipolar expansion of retarded Green function The goal of this section is to find a solution of the following equation: (cid:3) φ ( t, x ) = − πρ ( t, x ) , (A.1)when the source is expanded in multipolarity ρ = (cid:80) ˆ n L ρ L , by means of, so-called, retradedintegration, i.e. φ ( t, x ) = − π (cid:3) − [ ρ ( t, x )] . (A.2)Additionally, we compare the above solution and the well-known and general solution outsidethe source ( r > d ) which is extended in multipolarity φ ( t, x ) = ∞ (cid:88) l =0 ( − l l ! ∂ L (cid:8) F L ( t − r/c ) r (cid:9) , (A.3)because (cid:3) ( F L ( t − r ) r ) = 0 is always satisfied whatever F L s are.Now let us consider a general solution of retarded integration for both the inside and theoutside of the source. And this result is well-known as retarded green function method: φ ( t, (cid:126)x ) = (cid:90) d x (cid:48) G R ( x − x (cid:48) ) ρ ( x (cid:48) ) , (A.4)where G R ( x − x (cid:48) ) = δ ( t − t (cid:48) − | (cid:126)x − (cid:126)x (cid:48) | ) | (cid:126)x − (cid:126)x (cid:48) | (A.5)= δ ( t − t (cid:48) − √ r + r (cid:48) − rr (cid:48) (cid:126)n · (cid:126)n (cid:48) ) √ r + r (cid:48) − rr (cid:48) (cid:126)n · (cid:126)n (cid:48) Before we expand the retarded Green function as the expasion of Legendre function, F ( (cid:126)n · (cid:126)n (cid:48) ) = 12 + ∞ (cid:88) l =0 (2 l + 1)!! l ! ˆ n L ˆ n (cid:48) L (cid:90) +1 − dzF ( z ) P l ( z ) , (A.6)This formula comes from the fact thatˆ n L ˆ n (cid:48) L = l !(2 l − P l ( (cid:126)n · (cid:126)n (cid:48) ) . (A.7)26ow, expand retarded Green function, G R ( x − x (cid:48) ) = 12 + ∞ (cid:88) l =0 (2 l + 1)!! l ! ˆ n L ˆ n (cid:48) L (cid:90) +1 − dz δ ( t − t (cid:48) − √ r + r (cid:48) − rr (cid:48) z ) √ r + r (cid:48) − rr (cid:48) z P l ( z ) (A.8)= Θ( t − t (cid:48) )Θ(1 − ν )2 + ∞ (cid:88) l =0 (2 l + 1)!! l ! ˆ n L ˆ n (cid:48) L (cid:90) +1 − dz δ ( z − ν ) rr (cid:48) P l ( z )= Θ( t − t (cid:48) )Θ(1 − ν )2 rr (cid:48) + ∞ (cid:88) l =0 (2 l + 1)!! l ! ˆ n L ˆ n (cid:48) L P l ( ν ) , where ν = r (cid:48) + r − ( t − t (cid:48) ) rr (cid:48) . (A.9)Furthermore, if we assume that the source has the single multipolarity l , that is ρ ( x ) = (cid:88) l ˆ n L ( θ, ϕ ) ρ L ( r, t ) , (A.10)because the, the angular integration becomes easy in that: (cid:90) d Ω( (cid:126)n )ˆ n L ˆ n K = 4 πl !(2 l + 1)!! δ lk , (A.11)more generally, for multipole expansion, f ( (cid:126)n ) = (cid:80) ∞ l =0 f L ˆ n L , f L = (2 l + 1)!!4 πl ! (cid:90) d Ω ˆ n L f ( (cid:126)n ) (A.12)then the above equation becomes φ ( x ) = (cid:90) d x (cid:48) Θ( t (cid:48) − t )Θ(1 − ν )2 rr (cid:48) + ∞ (cid:88) k =0 (2 k + 1)!! k ! ˆ n K ˆ n (cid:48) K P k ( ν )ˆ n (cid:48) L ( θ (cid:48) , ϕ (cid:48) ) ρ ( r (cid:48) , t (cid:48) ) (A.13)= 2 π ˆ n L r (cid:90) dt (cid:48) dr (cid:48) Θ( t (cid:48) − t )Θ(1 − ν ) r (cid:48) P l ( ν ) ρ L ( r (cid:48) , t (cid:48) )= 2 π ˆ n L r (cid:90) D du (cid:48) dv (cid:48) ( v (cid:48) − u (cid:48) ) P l ( ν ) ρ L ( 12 ( v (cid:48) − u (cid:48) ) , 12 ( u (cid:48) + v (cid:48) )) , in last line, I change variables as u = t − r , v = t + r . And the domain is D = { u (cid:48) , v (cid:48) | u (cid:48)
28n the last lines, I used the following identity holds, l (cid:88) k =0 k !( v − u ) l ( v (cid:48) − u (cid:48) ) k l − k (cid:88) t =0 l − k − t (cid:88) s =0 ( − ) k + s + t ( l + k + s )!( l − s − k − t )!( s + k )! t ! s ! ( u (cid:48) − u ) l − t ( u (cid:48) − v ) k + t (A.16)= l (cid:88) k =0 k !( v − u ) l ( v (cid:48) − u (cid:48) ) k l − k (cid:88) t =0 ( − ) k ( l + k )!( l − k )! l !( l − k )!( k + t )!( l − t )! t !( l − k − t )! ( u (cid:48) − u ) l − t ( u (cid:48) − v ) k + t Here is a proof. This identity holds if the following holds, l − k − t (cid:88) s =0 ( − ) k + s + t ( l − k − t )!( l + k + s )!( l − s − k − t )!( s + k )! s ! = ( − ) l ( l + k )! l !( k + t )!( l − t )! . (A.17)in the notation of ( l ) k = l ( l + 1)( l + 2) · · · ( l + k − l − k − t (cid:88) s =0 ( − ) k + t ( − l + k + t ) s ( l + k + 1) s ( l + k )!( k + 1) s k ! s ! (A.18)= ( − ) k + t ( l + k )! k ! ∞ (cid:88) s =0 ( − l + k + t ) s ( l + k + 1) s ( k + 1) s s != ( − ) k + t ( l + k )! k ! lim z → − F ( − l + k + t ; l + k + 1; k + 1; z )= ( − ) k + t ( l + k )! k ! Γ( k + 1)Γ( − t − k )Γ( − l )Γ( l − t + 1) = ( − ) l ( l + k )! k ! Γ( k + 1)Γ( l + 1)Γ(1 + t + k )Γ( l − t + 1)= ( − ) l ( l + k )! l !( k + t )!( l − t )!in the second line, the upper bound of summation vanishes because ( − l + k + t ) s = 0 when s > l − k − t , and F is hypergeometry series . In the third line, I use the limit of hypergeometyfunction given as lim z → − F ( α ; β ; γ ; z ) = Γ( γ )Γ( γ − α − β )Γ( γ − α )Γ( γ − β ) , (A.19)and also using Γ( (cid:15) − n ) = ( − n − Γ( − (cid:15) )Γ( (cid:15) + 1)Γ( n + 1 − (cid:15) ) . (A.20)29Ith this result, the Legendre function can be written as,ˆ n L P l ( ν ) = ˆ n L ( − l l ! ( v − u ) l +1 ( v (cid:48) − u (cid:48) ) l ( ∂∂u ) l (cid:110) ( u − u (cid:48) ) l ( u − v (cid:48) ) l ( v − u ) l +1 (cid:111) (A.21)= ˆ n L ( − l l ! ( v − u ) l +1 ( v (cid:48) − u (cid:48) ) l ( ∂∂u ) l ( ∂∂v ) l (cid:110) ( u − u (cid:48) ) l ( u − v (cid:48) ) l ( v − u ) (cid:111) = 12 l ! ( v − u )( v (cid:48) − u (cid:48) ) l ˆ ∂ L ( ( u − u (cid:48) ) l ( u − v (cid:48) ) l r ) . Alternatively, the above relation also can be proved by invoking the Riemann function[4] of Euler-Poisson-Darboux equation, ∂ uv f + l ( l + 1)( v − u ) f = 0 . (A.22)if P l ( ν ) is the Riemann function of the EPD equation, there are requirements that, first P l ( ν )itself is the solution(because the EPD equation is self-adjoint), and ∂ u P l ( ν ) = 0 when v = v (cid:48) and also ∂ v P l ( ν ) = 0 when u = u (cid:48) . And this is also the case of( − l l ! ( v (cid:48) − u (cid:48) ) l +1 ( v − u ) l ( ∂∂u (cid:48) ) l (cid:110) ( u (cid:48) − u ) l ( u (cid:48) − v ) l ( v (cid:48) − u (cid:48) ) l +1 (cid:111) . (A.23)That is, both expressions are the Reimann function of the EPD equation and because theReimann function is unique thus they are same.Anyway, the expression of solution becomes φ ( x ) = − π (cid:90) u du (cid:48) (cid:90) uv dv (cid:48) l !( v (cid:48) − u (cid:48) ) l − ˆ ∂ L ( ( u − u (cid:48) ) l ( u − v (cid:48) ) l r ) ρ L ( 12 ( v (cid:48) − u (cid:48) ) , 12 ( u (cid:48) + v (cid:48) ))(A.24)= − π ˆ ∂ L (cid:104) r − (cid:90) u du (cid:48) (cid:90) uv dv (cid:48) ( u − u (cid:48) ) l ( u − v (cid:48) ) l l !( v (cid:48) − u (cid:48) ) l − ρ L ( 12 ( v (cid:48) − u (cid:48) ) , 12 ( u (cid:48) + v (cid:48) )) (cid:105) . (note that I was able to place ˆ ∂ L before intergals. See Eq.(A.34)) By changing the variable v (cid:48) = u (cid:48) + 2 r (cid:48) , (A.25) φ ( x ) = − π l − l ! ˆ ∂ L (cid:104) r − (cid:90) u −∞ du (cid:48) (cid:90) u − u (cid:48) v − u (cid:48) dr (cid:48) ( u − u (cid:48) ) l ( u − u (cid:48) − r (cid:48) ) l l ! r (cid:48) l − ρ L ( r (cid:48) , u (cid:48) + r (cid:48) ) (cid:105) (A.26)additional change of variable as u (cid:48) = u + ( z − r (cid:48) , (A.27)yields 30 ( x ) = ( − ) l π l − l ! ˆ ∂ L (cid:104) r − (cid:90) +1 − dz (cid:90) r dr (cid:48) (1 − z ) l r (cid:48) l +2 ρ L ( r (cid:48) , u + zr (cid:48) ) (cid:105) (A.28)( ρ = 0 at r (cid:48) > r ) re-input that ρ L ( r, t ) = (2 l + 1)!!4 πl ! (cid:90) d Ωˆ n L ρ ( (cid:126)x, t ) , (A.29) φ ( x ) = (2 l + 1)!!( − ) l π l − l ! ˆ ∂ L (cid:104) r − (cid:90) +1 − dz (cid:90) d x (cid:48) (1 − z ) l r (cid:48) l ˆ n (cid:48) L ρ ( x (cid:48) , u + zr (cid:48) ) (cid:105) , (A.30)finally we get a result F L ( u ) = (2 l + 1)!!2 l +1 l ! (cid:90) d x (cid:48) ˆ x (cid:48) L (cid:90) +1 − dz (1 − z ) l ρ ( x (cid:48) , u + zr (cid:48) ) . (A.31) A.1 General expression of multipole expanded solution