4.4 Matching of horizon and outer solutions

Now, we match the two types of solutions R ν0 and R νC. Note that both of them are convergent in a very large region of r, namely for πœ–κ < ˆz < ∞. We see that both solutions behave as ˆzν multiplied by a single-valued function of ˆz for large |ˆz|. Thus, the analytic properties of R ν 0 and R ν C are the same, which implies that these two are identical up to a constant multiple. Therefore, we set
ν ν R 0 = K νRC. (160 )

In the region πœ–κ < ˆz < ∞, we may expand both solutions in powers of ˆz except for analytically non-trivial factors. We have

( ˆz ) −s−iπœ–+ ∑∞ ∞∑ R ν0 = eiπœ–κe−iˆz(πœ–κ)−ν−iπœ–+ ˆzν+iπœ–+ ---− 1 Cn,jˆzn− j πœ–κ n=− ∞ j=0 ( ) −s−iπœ–+ ∑∞ ∞∑ = eiπœ–κe−iˆz(πœ–κ)−ν−iπœ–+ ˆzν+iπœ–+ ˆz--− 1 C zˆk, (161 ) πœ–κ n,n− k k=− ∞ n=k ( )− s− iπœ– ∞ ∞ ν −iˆz ν −s−iπœ–+ ν+iπœ–+ -ˆz- + ∑ ∑ n+j R C = e 2 (πœ–κ) zˆ πœ–κ − 1 Dn,jˆz n=−∞ j=0 ( )− s− iπœ–+ ∑∞ ∑k = e−iˆz2ν(πœ–κ)−s−iπœ–+zˆν+iπœ–+ -ˆz-− 1 Dn,k−nˆzk, (162 ) πœ–κ k=−∞ n=− ∞
where
----Γ-(1 −-s-−-2iπœ–+)-Γ (2n-+-2ν-+-1)-- Cn,j = Γ (n + ν + 1 − iτ )Γ (n + ν + 1 − s − iπœ–) × (−-n-−-ν-−-iτ)j(− n-−-ν-−-s-−-iπœ–)j(πœ–κ)−n+jfn, (163 ) (− 2n − 2ν)j(j!) Γ (n + ν + 1 − s + iπœ–)(ν + 1 + s − iπœ–)n Dn,j = (− 1)n(2i)n+j -------------------------------------- Γ (2n + 2ν + 2 ) (ν + 1 − s + iπœ–)n (n-+-ν-+-1-−-s-+-iπœ–)j- × (2n + 2ν + 2)(j!) fn. (164 ) j
Then, by comparing each integer power of ˆz in the summation, in the region πœ–κ β‰ͺ ˆz < ∞, and using the formula Γ (z)Γ (1 − z) = πβˆ•sin πz, we find
( r ) −1 ( ∞ ) iπœ–κ s−ν −ν ∑ ∑ K ν = e (πœ–κ) 2 Dn,r−n Cn,n−r n= −∞ n=r eiπœ–κ(2 πœ–κ)s− ν−r2−sir Γ (1 − s − 2iπœ–+)Γ (r + 2 ν + 2) = ----------------------------------------------------------- Γ (r + ν + 1 − s + iπœ–) Γ (r + ν + 1 + iτ )Γ (r + ν + 1 + s + iπœ–) ( ∞∑ ) × (− 1)nΓ (n-+-r-+-2ν-+-1)-Γ (n-+-ν-+-1-+-s +-iπœ–)-Γ (n-+-ν-+-1-+-iτ)fν n=r (n − r)! Γ (n + ν + 1 − s − iπœ–) Γ (n + ν + 1 − iτ) n ( ) −1 ∑ r (− 1)n (ν + 1 + s − iπœ–)n ν × --------------------------------------fn , (165 ) n=−∞ (r − n)!(r + 2ν + 2)n(ν + 1 − s + iπœ–)n
where r can be any integer, and the factor K ν should be independent of the choice of r. Although this fact is not manifest from Equation (165View Equation), we can check it numerically, or analytically by expanding it in terms of πœ–.

We thus have two expressions for the ingoing wave function Rin. One is given by Equation (116View Equation), with pν in expressed in terms of a series of hypergeometric functions as given by Equation (120View Equation) (a series which converges everywhere except at r = ∞). The other is expressed in terms of a series of Coulomb wave functions given by

Rin = K νR νC + K −ν−1R −Cν−1, (166 )
which converges at r > r+, including r = ∞. Combining these two, we have a complete analytic solution for the ingoing wave function.

Now we can obtain analytic expressions for the asymptotic amplitudes of Rin, Btrans, Binc, and Bref. By investigating the asymptotic behaviors of the solution at r → ∞ and r → r +, they are found to be1

( πœ–κ )2s 2lnκ ∑∞ Btrans = --- eiκπœ–+(1+ 1+κ ) fνn, (167 ) ω n=− ∞ ( ) Binc = ω− 1 K ν − ie−iπνsinπ-(ν −-s +-iπœ–)K −ν−1 A ν+e− i(πœ–lnπœ–− 1−κ2 πœ–), (168 ) sinπ (ν + s − iπœ–) ref − 1− 2s( iπν ) ν i(πœ–lnπœ–− 1−2κπœ–) B = ω K ν + ie K −ν−1 A − e . (169 )
UpdateJump To The Next Update Information

Incidentally, since we have the upgoing solution in the outer region (159View Equation), it is straightforward to obtain the asymptotic outgoing amplitude at infinity trans C from Equation (153View Equation). We find

trans −1−2s ν i(πœ–lnπœ–− 1−κπœ–) C = ω A − e 2 . (170 )

  Go to previous page Go up Go to next page