- THE EQUIVALENCE OF Q AND Q 131
Using (29.18), we compute that the components of the second covariant deriv-
ative of v are\i'~'lj;V = V1f;1f;,
V'^2 sin2'lj;00 v = voo - - 2 - v'!f;,
\i'~ 0 v = vo'!f; +tan 'ljJ v 0.
For the third covariant derivative, we obtain\i'~'lj;'lj; V = V1f;1f;'lj;,
\i'^3 3 sin 2'lj;^2
000 v = vooo - 2 V1f;() - 2 sin 'l/Jvo,V' 'lj;()()v^3 = V()()'lj; - -2-sin^2 'lj; V1f;1f; + 2 ta n 'ljJ VI)() - V1f;)
\i'~'!f;'!f;v = V1f;1f;O + 2 ta n 'ljJ V()1f; + 2 tan^2 'l/Jvo.
Therefore0: = V1f;1f;'lj; - 2V1f; - 3 sec^2 'ljJ ( V()()'lj; - sin2'lj; -2-V1f;1f; + 2 tan 'ljJ VI)() - V1f; )
andcos 'ljJ · f3 = sec^2 'lj;v 000 - 3V1f;1f;O - 9 tan 'l/JV1f;() - 2 ( 4 tan^2 'ljJ + 1) v 0.
We now eva luate 0:00--^1 and (300--^1 using vou-^1 = cpfJ. Since cp is independent
of (), we have that
VI) o o--l = <pfJo,
VI)() o o--l = <pV()(),VI)()() o a--l = <pV()()().
Recall that ~~ = -,. 2 ~ 1. We compute that
_ 1 2 _ 8r _
v'!f;oa- =---v,. + 2 v,
r^2 + 1 (r^2 + 1)and hence
and
1 2 8r
V1f;()() o a- = -,,..2 + 1 Vr()() + (r2 + 1)2 VI)().
We then compute that
(29.211)
_ 1 _ 6r _ 4 ( 3r^2 - 1) _
V1f;1f; o O" = Vrr - ~l Vr + 2 V
r + (r^2 + 1)and hence
1 6r 4 ( 3r^2 - 1)
V1f;1f;Ooo- =v,.,.o- r2+1v,.o+ (r2+1) 2 vo
and
1 r^2 +1 9r^2 -5 4r(3r^2 -5)_
V1f;1f;'lj; o O" = --
2
- Vrrr + 3rv,.,. - 2 l Vr + 2 2 V.
r + (r + 1)