Low frequency expansion of the hypergeometric expansion

4.5 Low frequency expansion of the hypergeometric expansion

So far, we have considered exact solutions of the Teukolsky equation. Now, let us consider their low frequency approximations and determine the value of $ν$ . We solve Equation (136

) with $n = 0$ ,

with a requirement that $ν → ℓ$ as $𝜖 → 0$ .

To solve Equation (174 ), we first note the following. Unless the value of $ν$ is such that the denominator in the expression of $ανn$ or $γνn$ happens to vanish, or $βνn$ happens to vanish in the limit $𝜖 → 0$ , we have $αν = 𝒪 (𝜖) n$ , $γ ν= 𝒪 (𝜖) n$ , and $β ν= 𝒪(1) n$ . Also, from the asymptotic behavior of the minimal solution $ν fn$ as $n → ±∞$ given by Equation (134 ), we have $Rn (ν) = 𝒪 (𝜖)$ and $L −n(ν) = 𝒪 (𝜖)$ for sufficiently large $n$ . Thus, except for exceptional cases mentioned above, the order of $aνn$ in $𝜖$ increases as $|n|$ increases. That is, the series solution naturally gives the post-Minkowski expansion.

First, let us consider the case of $R (ν) n$ for $n > 0$ . It is easily seen that $αν = 𝒪(𝜖) n$ , $γν = 𝒪 (𝜖) n$ , and $ν βn = 𝒪 (1)$ for all $n > 0$ . Therefore, we have $Rn(ν) = 𝒪 (𝜖)$ for all $n > 0$ .

On the other hand, for $n < 0$ , the order of $L−n (ν )$ behaves irregularly for certain values of $n$ . For the moment, let us assume that $L −1(ν) = 𝒪 (𝜖)$ . We see from Equations (124 ) that $α ν0 = 𝒪 (𝜖)$ , $γ0ν= 𝒪(𝜖)$ , since $ν = ℓ + 𝒪 (𝜖)$ . Then, Equation (174 ) implies $βν0 = 𝒪 (𝜖2)$ . Using the expansion of $λ$ given by Equation (110 ), we then find $ν = ℓ + 𝒪 (𝜖2)$ (i.e., there is no term of $𝒪 (𝜖)$ in $ν$ ). With this estimate of $ν$ , we see from Equation (128 ) that $L −1(ν) = 𝒪 (𝜖)$ is justified if $L −2(ν)$ is of order unity or smaller.

The general behavior of the order of $L− n(ν )$ in $𝜖$ for general values of $s$ is rather complicated. However, if we assume $s$ to be a non-integer and $ℓ ≥ |s|$ , and $τ = (𝜖 − mq )∕κ = 𝒪 (1 )$ , it is relatively easily studied. With the assumption that $ν = ℓ + 𝒪 (𝜖2)$ , we find there are three exceptional cases:

For $n = − 2ℓ − 1$ , we have $αn = 𝒪(𝜖)$ , $2 β = 𝒪 (𝜖 )$ , and $γn = 𝒪 (𝜖)$ .
For $n = − ℓ − 1$ , we have $αν = 𝒪 (1∕𝜖) n$ , $βν = 𝒪 (1∕𝜖) n$ , and $γ = 𝒪 (𝜖) n$ .
For $n = − ℓ$ , we have $αn = 𝒪 (𝜖)$ , $βn = 𝒪 (1∕𝜖)$ , and $γn = 𝒪 (1∕𝜖)$ .

These imply that $L−2ℓ−1(ν) = 𝒪 (1∕𝜖)$ , $L −ℓ−1(ν) = 𝒪 (1)$ , and $2 L−ℓ(ν) = 𝒪 (𝜖 )$ , respectively. To summarize, we have

ν R (ν) = -fn--= 𝒪 (𝜖) for all n > 0, n fνn−1 ν L (ν) = -f−ℓ--= 𝒪 (𝜖2), −ℓ fν−ℓ+1 ν L (ν) = f−ℓ−1-= 𝒪 (1), (172 ) −ℓ−1 fν−ℓ ν L (ν) = f−2ℓ−1 = 𝒪 (1∕𝜖), −2ℓ−1 fν−2ℓ ν Ln (ν) = -fn--= 𝒪 (𝜖) for all the other n < 0. fνn+1

With these results, we can calculate the value of $ν$ to $𝒪 (𝜖)$ , which is given by

Now one can take the limit of an integer value of $s$ . In particular, the above holds also for $s = 0$ . Interestingly, $ν$ is found to be independent of the azimuthal eigenvalue $m$ to $2 𝒪 (𝜖 )$ .

The post-Minkowski expansion of homogeneous Teukolsky functions can be obtained with arbitrary accuracy by solving Equation (123 ) to a desired order, and by summing up the terms to a sufficiently large $|n |$ . The first few terms of the coefficients $fν n$ are explicitly given in [68 ]. A calculation up to a much higher order in $𝒪(𝜖)$ was performed in [98 ], in which the black hole absorption of gravitational waves was calculated to $8 𝒪 (v )$ beyond the lowest order.