Again, via "()" process, the right hand side vanishes because (X-3/2 sin I) = 0.
Thus,
Ao = 0. (4-186)
4.4.2 Ba-terms
We take the e-1 term from Eq. (4-159) and define
-1 aPIII t 3 [ ( PIII t} (4-187)
Qa[C q1 -2 2f 1 4 5 (4-187)
where for computing 9a (p2), Eq. (4-176) should be referred to, and 'III is the
cubic part taken directly from Eq. (4-144). In a generic form Eq. (4-187) can be
expressed as
2 2n k Ano-k 0 _i O\k-p (0 )P
Q" +C-II-EEY 2 0) (4-188)
nl k0 p0 0
where A r ro, and Pnkp(a) is the coefficient of each individual term that
depends on n, k and p as well as on the component index a, with a dimension R'k-1
for a = t, r and Rk for a = 0, As we saw already from the analysis given in
Subsection 4.4.1, we replace 0 Oo by (0 0') JiA/f(r2 + J2) to eliminate
the coupling term A (0 Oo) in PII as the first of the steps to lead to pi for the
denominator on the right hand side of Eq. (4-188). For consistency, 0 Oo in
the numerator should be also replaced by (0 0') JiA/f (r + J2). Then, as
we expand the quantity [(Q ') JiA/f(r + J2)1 raised to the (k p)-th
power, a number of additional terms apart from (0 0/)k-p will be created, and the
computation will be very complicated.
By analyzing the structure of the right hand side of Eq. (4-188), one can
prove that 0 0o may be just replaced by 0 0' without -JiA/f(r2 + J2)
in the numerator (the same idea is found in Mino, Nakano, and Sasaki [26]).
The verification follows. The behavior of the quantity on the right hand side of