6
1 C o n g r u e n c e s o f W o r l d L i n e s
Putting η a = l n a again yields
˙
l
l
=
1
3
ϑ and h
a
b ˙
n
b
= 0 .
(1.28)
Hence ϑ measures the rate of change of length per unit length along ξ α of the
connecting vector η a and the second equation in (1.28) represents propagation along
ξ α without rotation and thus n a is Fermi transported [1, 2] along ξ α . If a set of
neighbouring particles to ξ α initially form a 3-sphere of radius l centred on ξ α then
under (1.27) the sphere remains a sphere and its volume V satisfies
˙
V
V
= ϑ .
(1.29)
We call the scalar ϑ the expansion (if ϑ > 0) or contraction (if ϑ < 0) scalar of the
congruence.
Finally consider the case σ ab = 0, ϑ = 0, ω ab = 0 and (1.11) is now
h
a
b ˙
η
b
= σ
a
b η
b .
(1.30)
Since σ ab = σ ba and σ ab u b = 0 we see that u a is the unit time-like eigenvector
of σ ab with zero eigenvalue. Let n
a
(α) be the three unit space-like eigenvectors with
corresponding eigenvalues μ (α) for α = 1, 2, 3. These latter are mutually orthogonal
and each are orthogonal to u a . Thus we have
σ
a
b n
b
(α) = μ (α) n
a
(α) (α = 1, 2, 3) ,
(1.31)
and, since σ a a = 0,
3
α=1
μ (α) = 0 .
(1.32)
In addition the orthonormality conditions satisfied by n a
(α) are
u a n
a
(α) = 0 and g ab n
a
(α) n
b
(β) = −δ αβ .
(1.33)
Now choose three orthogonal connecting vectors in the directions n a
(α) given by
η
a
(α) = l (α) n
a
(α) (α = 1, 2, 3) .
(1.34)
Each of these must satisfy (1.30) and so we find that
˙
l (α)
l (α)
= μ (α) and h
a
b ˙
n
b
(α) = 0 ,
(1.35)
1 C o n g r u e n c e s o f W o r l d L i n e s
Putting η a = l n a again yields
˙
l
l
=
1
3
ϑ and h
a
b ˙
n
b
= 0 .
(1.28)
Hence ϑ measures the rate of change of length per unit length along ξ α of the
connecting vector η a and the second equation in (1.28) represents propagation along
ξ α without rotation and thus n a is Fermi transported [1, 2] along ξ α . If a set of
neighbouring particles to ξ α initially form a 3-sphere of radius l centred on ξ α then
under (1.27) the sphere remains a sphere and its volume V satisfies
˙
V
V
= ϑ .
(1.29)
We call the scalar ϑ the expansion (if ϑ > 0) or contraction (if ϑ < 0) scalar of the
congruence.
Finally consider the case σ ab = 0, ϑ = 0, ω ab = 0 and (1.11) is now
h
a
b ˙
η
b
= σ
a
b η
b .
(1.30)
Since σ ab = σ ba and σ ab u b = 0 we see that u a is the unit time-like eigenvector
of σ ab with zero eigenvalue. Let n
a
(α) be the three unit space-like eigenvectors with
corresponding eigenvalues μ (α) for α = 1, 2, 3. These latter are mutually orthogonal
and each are orthogonal to u a . Thus we have
σ
a
b n
b
(α) = μ (α) n
a
(α) (α = 1, 2, 3) ,
(1.31)
and, since σ a a = 0,
3
α=1
μ (α) = 0 .
(1.32)
In addition the orthonormality conditions satisfied by n a
(α) are
u a n
a
(α) = 0 and g ab n
a
(α) n
b
(β) = −δ αβ .
(1.33)
Now choose three orthogonal connecting vectors in the directions n a
(α) given by
η
a
(α) = l (α) n
a
(α) (α = 1, 2, 3) .
(1.34)
Each of these must satisfy (1.30) and so we find that
˙
l (α)
l (α)
= μ (α) and h
a
b ˙
n
b
(α) = 0 ,
(1.35)