Taking the same steps as used for A,-term above via Eqs. (4-176), (4-150) and
(4-151) in order, we obtain
a0 (2) t=t 2 (r2 + J2) sin O cos O + 0[(x xo)3]. (4-180)
Then, in a similar manner to that employ, -l in the previous cases, in the limit
A -- 0 Eq. (4-179) becomes
liq2 -3/2 Cos 4
limQ 22 32 JC21/ a P,(cos 0). (4-181)
a (r7 +o 1)2 0 A
The right hand side vanishes through "()" process because (-3/2 cos) 0.
Hence,
A =- 0. (4-182)
A0-term:
It is evident from the particle's motion, which is confined to the equatorial
plane 0o = that no self-force is acting on the particle in the direction perpendic-
ular to this plane. This is due to the fact that both the derivatives of retarded field
and the singular source field with respect to 0 tend to zero in the coincidence limit.
Our calculation of A0 should support this. Through the same process as employ, ,
before, we have
Q[C-2] q -o3 a0 (p2) o (4-183)
with
S(p2) to =-2r sin sin) + O[(x- x)3]. (4-184)
Then, similarly as in the case of A6-term above
q22 -3/2 Sin
lim Q [c-2 -2 sin COS ). (4-185)
a (r 72 ) J3/ 0 A