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=
D 1 =
( κ g + κ n ) D 1 +( κ g
κ n τ g ) D 2 +( κ g τ g + κ n ) D 3 ,
D 1 = G K U K ,
D 1 = G K,L U K U L + G K U K ,
(E.9)
D 1 = G K,LM U K U L U M + G K,L ( U K U L + U K U L )+ G K,L U K U L + G K U K ,
G K,LM =
= N
G N + N
G N,M + H KL,M G 3 + H KL G 3 ,M ,
KL
KL
,M
G K,LM =
(E.10)
H KL H MM 2 G M 2 K 2 G K 2 +
=
+ N 1
M 2
N 1 M
G M 2 + N 1
H N 1 ,M G 3 + N
,M G N + H KL,M G 3 ,
KL
KL
KL
D 1 =
= U M +(2 U K U L + U K U L ) M
+ U K U L U M 2
KL
M
KL
,M 2 + M 1
M
M 1 M 2
(E.11)
KL
H KL H M 1 M 2 G M 1 M )] G M +[(2 U K U L + U K U L ) H KL +
+ U K U L U M 2 M 1
H M 1 M 2 + H KL,M 2
G 3 .
KL
M
KL
= M
, “symmetry” of Christoffel symbols leads to
LK
D 1 =
= U M +3 U K U L M
+ U K U L U N
KL
M
KL
,N + Q
M
NQ
H KL H NQ G QM G M +
(E.12)
KL
+ 3 U K U L H KL + U K U L U N Q
H NQ + H KL,N D 3 .
KL
τ g then is obtained as
κ g τ g + κ n =3 U K U L H KL + U K U L U N ( Q
H NQ + H KL,N ) ,
KL
κ n = H KL,N U K U L U N +2 H KL U K U L
κ g τ g = H KL U K U L + Q
H NQ U K U L U N
(E.13)
KL
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