Page 162 - Intro to Tensor Calculus

157

2
2
2
2
where g 11 = −1, g 22 = −ρ , g 33 = −ρ sin θ, g 44 = c and g ij =0 for i 6= j. The negative signs are
2 2 2
ds
introduced so that = c − v is positive when v< c and the velocity is not greater than c. However,
dt
this metric will not work since the curvature tensor vanishes. The spherical symmetry of the problem suggest
1
3
2
4
that g 11 and g 44 change while g 22 and g 33 remain fixed. Let (x ,x ,x ,x )= (ρ, θ, φ, t) and assume
2
2
2
u
g 11 = −e , g 22 = −ρ , g 33 = −ρ sin θ, g 44 = e v (1.5.159)
where u and v are unknown functions of ρ to be determined. This gives the conjugate metric tensor
−1 −1
11 −u 22 33 44 −v
g = −e , g = , g = 2 , g = e (1.5.160)
ρ 2 ρ sin θ
2
and g ij =0 for i 6= j. This choice of a metric produces
v
2
2
2
2
2
2
u
2
ds = −e (dρ) − ρ (dθ) − ρ sin θ(dφ) + e (dt) 2 (1.5.161)
together with the nonzero Christoffel symbols

1 1 du 3 1
= =
11 2 dρ 2 1 13 ρ
=

1 −u 12 ρ 3 cos θ 4 1 dv
= − ρe = =
22 2 1 23 sin θ 14 2 dρ
= (1.5.162)

1 −u 2 21 ρ 3 1 4 1 dv
= − ρe sin θ = = .
33 2 31 ρ 41 2 dρ
= − sin θ cos θ

1 1 v−u dv 33 3 cos θ
= e =
44 2 dr 32 sin θ
The equation (1.5.158) is used to calculate the nonzero G ij and we find that
2
1 d v 1 dv 2 1 du dv 1 du
G 11 = + − −
2 dρ 2 4 dρ 4 dρ dρ ρ dρ

1 dv 1 du u
−u
G 22 =e 1+ ρ − ρ − e
2 dρ 2 dρ
(1.5.163)
1 dv 1 du u 2
−u
G 33 =e 1+ ρ − ρ − e sin θ
2 dρ 2 dρ
2 !
2
1 d v 1 du dv 1 dv 1 dv
v−u
G 44 = − e − + +
2 dρ 2 4 dρ dρ 4 dρ ρ dρ
and G ij =0 for i 6= j. The assumption that G ij =0 for all i, j leads to the differential equations
2
d v 1 dv 2 1 du dv 2 du
+ − − =0
dρ 2 2 dρ 2 dρ dρ ρ dρ
1 dv 1 du u
1+ ρ − ρ − e =0 (1.5.164)
2 dρ 2 dρ
2
d v 1 dv 2 1 du dv 2 dv
+ − + =0.
dρ 2 2 dρ 2 dρ dρ ρ dρ