4
1 C o n g r u e n c e s o f W o r l d L i n e s
in (s) is transformed by infinitesimal, linear transformation to (δ a
b + A a b ds)η b in
(s + ds). This linear transformation is generated by A a b . We see from (1.11) that
A ab = u a;b − ˙
u a u b and so we immediately have A ab u a = 0 = A ab u b . We can
decompose A ab into linearly independent parts by first identifying its symmetric
and skew-symmetric parts:
A ab = ϑ ab + ω ab ,
(1.12)
with
ϑ ab = A (ab) =
1
2
(A ab + A ba ) = ϑ ba ,
(1.13)
and
ω ab = A [ab] =
1
2
(A ab − A ba ) = −ω ba .
(1.14)
Then we can subtract the trace from the symmetric part, remembering to preserve
the orthogonality with u a , by defining
σ ab = ϑ ab −
1
3
ϑ h ab ,
(1.15)
with ϑ = ϑ a a . Then σ ab = σ ba and σ a
a = 0. Hence we have
A ab = σ ab +
1
3
ϑ h ab + ω ab = u a;b − ˙
u a u b ,
(1.16)
with
ϑ = u
a ;a ,
(1.17)
σ ab = u (a;b) − ˙
u (a u b) −
1
3
ϑ h ab ,
(1.18)
ω ab = u [a;b] − ˙
u [a u b] .
(1.19)
We emphasise that here, as in (1.13) and (1.14), round brackets denote symmetrisation while square brackets denote skew-symmetrisation. It is also important to point
out that (1.16) can be viewed as a decomposition of u a;b . We have mentioned above
that A ab generates a linear map from (s) to + ds). It is interesting to interpret
geometrically the influence of ϑ, ω ab and σ ab separately on this linear map.
Taking ω ab = 0, ϑ = 0, σ ab = 0 the transport law (1.11) becomes
h
a
b ˙
η
b
= ω
a
b η
b .
(1.20)
1 C o n g r u e n c e s o f W o r l d L i n e s
in (s) is transformed by infinitesimal, linear transformation to (δ a
b + A a b ds)η b in
(s + ds). This linear transformation is generated by A a b . We see from (1.11) that
A ab = u a;b − ˙
u a u b and so we immediately have A ab u a = 0 = A ab u b . We can
decompose A ab into linearly independent parts by first identifying its symmetric
and skew-symmetric parts:
A ab = ϑ ab + ω ab ,
(1.12)
with
ϑ ab = A (ab) =
1
2
(A ab + A ba ) = ϑ ba ,
(1.13)
and
ω ab = A [ab] =
1
2
(A ab − A ba ) = −ω ba .
(1.14)
Then we can subtract the trace from the symmetric part, remembering to preserve
the orthogonality with u a , by defining
σ ab = ϑ ab −
1
3
ϑ h ab ,
(1.15)
with ϑ = ϑ a a . Then σ ab = σ ba and σ a
a = 0. Hence we have
A ab = σ ab +
1
3
ϑ h ab + ω ab = u a;b − ˙
u a u b ,
(1.16)
with
ϑ = u
a ;a ,
(1.17)
σ ab = u (a;b) − ˙
u (a u b) −
1
3
ϑ h ab ,
(1.18)
ω ab = u [a;b] − ˙
u [a u b] .
(1.19)
We emphasise that here, as in (1.13) and (1.14), round brackets denote symmetrisation while square brackets denote skew-symmetrisation. It is also important to point
out that (1.16) can be viewed as a decomposition of u a;b . We have mentioned above
that A ab generates a linear map from (s) to + ds). It is interesting to interpret
geometrically the influence of ϑ, ω ab and σ ab separately on this linear map.
Taking ω ab = 0, ϑ = 0, σ ab = 0 the transport law (1.11) becomes
h
a
b ˙
η
b
= ω
a
b η
b .
(1.20)