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4.4 Matching of horizon and outer solutions

Now, we match the two types of solutions Rn0 and RnC. Note that both of them are convergent in a very large region of r, namely for ek < ^z < oo. We see that both solutions behave as ^zn multiplied by a single-valued function of ^z for large |^z|. Thus, the analytic properties of Rn 0 and Rn C are the same, which implies that these two are identical up to a constant multiple. Therefore, we set
n n R 0 = KnR C. (160)

In the region ek < ^z < oo, we may expand both solutions in powers of z^ except for analytically non-trivial factors. We have

( ^z ) -s-ie+ sum oo oo sum Rn0 = eieke-i^z(ek)-n-ie+ ^zn+ie+ ---- 1 Cn,j^zn- j ek n=- oo j=0 ( ) -s-ie+ sum oo oo sum = eieke-i^z(ek)-n-ie+ ^zn+ie+ ^z--- 1 C z^k, (161) ek n,n- k k=- oo n=k ( )- s- ie oo oo n -i^z n -s-ie+ n+ie+ -^z- + sum sum n+j R C = e 2 (ek) z^ ek - 1 Dn,j^z n=- oo j=0 ( )- s- ie+ sum oo sum k = e-i^z2n(ek)-s-ie+z^n+ie+ -^z-- 1 Dn,k-n^zk, (162) ek k=- oo n=- oo
where
----G(1---s---2ie+)-G(2n-+-2n-+-1)---- Cn,j = G(n + n + 1- it )G(n + n + 1- s- ie) × (--n---n---it)j(-n---n---s---ie)j(ek)-n+jf , (163) (-2n - 2n)j(j!) n G(n + n + 1 - s + ie)(n + 1 + s- ie) Dn,j = (- 1)n(2i)n+j ------------------------------------n- G(2n + 2n + 2) (n + 1- s + ie)n (n + n + 1 - s + ie)j × --------------------fn. (164) (2n + 2n + 2)j(j!)

Then, by comparing each integer power of z^ in the summation, in the region ek « ^z < oo, and using the formula G(z) G(1 - z) = p/ sin pz, we find

( r ) -1 ( oo ) iek s-n -n sum sum Kn = e (ek) 2 Dn,r-n Cn,n-r n= - oo n=r eiek(2ek)s- n-r2-sir G(1 - s - 2ie+)G(r + 2n + 2) = ----------------------------------------------------------- G(r + n + 1 - s + ie) G(r + n + 1 + it )G(r + n + 1 + s + ie) ( oo sum ) × (- 1)nG(n-+-r-+-2n-+-1)-G(n-+-n-+-1-+-s +-ie)-G(n-+-n-+-1-+-it)fn n=r (n- r)! G(n + n + 1 - s- ie) G(n + n + 1 - it) n ( ) -1 sum r (-1)n (n + 1 + s - ie)n n × --------------------------------------fn , (165) n=- oo (r - n)!(r + 2n + 2)n(n + 1 - s + ie)n
where r can be any integer, and the factor K n 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 e.

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

Rin = KnRnC + K -n-1R -Cn-1, (166)
which converges at r > r+, including r = oo. 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 --> oo and r-- > r+, they are found to be

(ek )2s sum oo Btrans = --- eie+lnk fnn, (167) w n=- oo ( ) Binc = w -1 Kn - ie- ipnsinp(n----s +-ie)K -n- 1 An+e -ielne, (168) sinp(n + s- ie) Bref = w -1-2s(K + ieipnK )An eielne. (169) n -n-1 -

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

trans -1- 2s ielne n C = w e A - . (170)


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