Horizon solution; z ≪ 1

3.2 Horizon solution; $z ≪ 1$

In this section, we first consider the solution near the horizon, which we call the horizon solution, based on [85

]. To do so, we assume $z ≪ 1$ and treat $𝜖$ as a small number, but leave the ratio $z∕ 𝜖$ arbitrary. We change the independent variable to $x = 1 − z∕𝜖$ and the wave function to

Note that the horizon corresponds to $x = 0$ . We then have

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 $𝜖 = 0$ in Equation (74 ). The boundary condition (72 ) requires that $Z = 1$ at $x = 0$ . The solution that satisfies the boundary condition is

Thus, the lowest order solution is a polynomial of order $ℓ − 2$ in $x = 1 − z∕𝜖$ .

Next, we consider the solution accurate to $𝒪(𝜖)$ . We neglect the terms of $𝒪(𝜖2)$ in Equation (74 ). Then, the wave equation takes the form of a hypergeometric equation,

with parameters

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

The above solution must be matched with the solution obtained from the post-Minkowski expansion of Equation (70 ), 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 $𝜖 ≪ z ≪ 1$ . For this purpose, we rewrite Equation (78 ) as (see, e.g., Equation (15.3.8) of [1 ])

trans −1 i𝜖(ln𝜖−1)[(z-)ℓ+i𝜖 Γ (c)Γ (b-−-a) ( 𝜖) ξℓ = A ℓ 𝜖 e 𝜖 Γ (b)Γ (c − a)F a,c − b;a − b + 1; z ( ) ( )] + z-− ℓ−1+i𝜖 Γ-(c)Γ (a −-b)F b,c − a;b − a + 1; 𝜖 . (79 ) 𝜖 Γ (a)Γ (c − b) z

This naturally allows the expansion in $𝜖∕z$ . 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 $ℓ$ . So, when evaluating the second term, we first have to extend $ℓ$ to a non-integer number. Then, only after expanding it in terms of $𝜖$ , we should take the limit of an integer $ℓ$ . One then finds that this procedure gives rise to an additional factor of $𝒪 (𝜖)$ . For $𝜖∕z ≪ 1$ , it therefore becomes $𝒪 (𝜖2ℓ+2)$ higher in $𝜖$ than the first term. Then, we obtain

where

and $ψ(n)$ is the digamma function,

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.