3.2 Horizon solution; z β‰ͺ 1

In this section, we first consider the solution near the horizon, which we call the horizon solution, based on [85Jump To The Next Citation Point]. To do so, we assume z β‰ͺ 1 and treat πœ– as a small number, but leave the ratio zβˆ• πœ– arbitrary. We change the independent variable to x = 1 − zβˆ•πœ– and the wave function to
( πœ–)3 -X-β„“-- (πœ–-)2------πœ–ξβ„“----- Z = z Atrans= z Atranseiπœ–(lnπœ–−1). (73 ) β„“ β„“
Note that the horizon corresponds to x = 0. We then have
x (x − 1)Z′′ + [2(3 − iπœ–)x − (1 − 2iπœ–)]Z ′ + [6 − β„“(β„“ + 1) − 5iπœ– + πœ–2(2 − 3x + x2)]Z = 0, (74 )
where a prime denotes differentiation with respect to x. We look for a solution which is regular at x = 0.

First, we consider the lowest order solution by setting πœ– = 0 in Equation (74View Equation). The boundary condition (72View Equation) requires that Z = 1 at x = 0. The solution that satisfies the boundary condition is

∑β„“− 2 Z = (2-−-β„“)n(β„“ +-3)nxn, (a) = Γ (a-+-n-). (75 ) n! n Γ (a) n=0
Thus, the lowest order solution is a polynomial of order β„“ − 2 in x = 1 − zβˆ•πœ–.

Next, we consider the solution accurate to π’ͺ(πœ–). We neglect the terms of π’ͺ(πœ–2) in Equation (74View Equation). Then, the wave equation takes the form of a hypergeometric equation,

′′ ′ x(x − 1)Z + [(a + b + 1)x − c]Z + abZ = 0, (76 )
with parameters
a = − (β„“ − 2) − iπœ– + π’ͺ (πœ–2), b = β„“ + 3 − iπœ– + π’ͺ (πœ–2), (77 ) c = 1 − 2iπœ–.
The two linearly independent solutions are F(a,b;c;x ) and x1−cF (a + 1 − c,b + 1 − c;2 − c;x), where F is the hypergeometric function. However, only the first solution is regular at x = 0. Therefore, we obtain
(z )2 ( z) ξβ„“(z) = Atrβ„“ansπœ–−1eiπœ–(lnπœ–− 1) -- F a,b;c;1 − -- . (78 ) πœ– πœ–

The above solution must be matched with the solution obtained from the post-Minkowski expansion of Equation (70View Equation), which we call the outer solution, in a region where both solutions are valid. It is the region where the post-Newtonian expansion is applied, i.e., the region πœ– β‰ͺ z β‰ͺ 1. For this purpose, we rewrite Equation (78View Equation) as (see, e.g., Equation (15.3.8) of [1Jump To The Next Citation Point])

trans −1 iπœ–(lnπœ–−1)[(z-)β„“+iπœ– Γ (c)Γ (b-−-a) ( πœ–) ξβ„“ = A β„“ πœ– e πœ– Γ (b)Γ (c − a)F a,c − b;a − b + 1; z ( ) ( )] + z-− β„“−1+iπœ– Γ-(c)Γ (a −-b)F b,c − a;b − a + 1; πœ– . (79 ) πœ– Γ (a)Γ (c − b) z
This naturally allows the expansion in πœ–βˆ•z. It should be noted that the second term in the square brackets of the above expression is meaningless as it is, since the factor Γ (a − b) diverges for integer β„“. So, when evaluating the second term, we first have to extend β„“ to a non-integer number. Then, only after expanding it in terms of πœ–, we should take the limit of an integer β„“. One then finds that this procedure gives rise to an additional factor of π’ͺ (πœ–). For πœ–βˆ•z β‰ͺ 1, it therefore becomes π’ͺ (πœ–2β„“+2) higher in πœ– than the first term. Then, we obtain
(2β„“)! zβ„“ [ (β„“ − 2)(β„“ + 2) πœ– ] ξβ„“(πœ– β‰ͺ z β‰ͺ 1) = ----------------β„“+1 1 + iπœ–(aβ„“ + ln z) − ----------------+ π’ͺ (πœ–2) , (80 ) (β„“ − 2)!(β„“ + 2)!πœ– 2β„“ z
where
a = 2γ + ψ (β„“ − 1) + ψ (β„“ + 3) − 1, (81 ) β„“
and ψ(n) is the digamma function,
n−1 ∑ −1 ψ (n) = − γ + k , (82 ) k=1
and γ ≃ 0.57721 is the Euler constant.

As we will see below, the above solution is accurate enough to determine the boundary condition of the outer solution up to the 6PN order of expansion.


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