Page 163 - Intro to Tensor Calculus

158

Subtracting the first equation from the third equation gives

du dv
+ =0 or u + v = c 1 = constant. (1.5.165)
dρ dρ
The second equation in (1.5.164) then becomes

du u
ρ =1 − e (1.5.166)
dρ

Separate the variables in equation (1.5.166) and integrate to obtain the result
1
u
e = c 2 (1.5.167)
1 −
ρ
where c 2 is a constant of integration and consequently

v
e = e c 1 −u = e c 1 1 − c 2 . (1.5.168)
ρ
2
The constant c 1 is selected such that g 44 approaches c as ρ increases without bound. This produces the
metrices
−1 2 2 2 2 c 2
g 11 = , g 22 = −ρ , g 33 = −ρ sin θ, g 44 = c (1 − ) (1.5.169)
1 − c 2 ρ
ρ
where c 2 is a constant still to be determined. The metrices given by equation (1.5.169) are now used to
expand the equations (1.5.157) representing the geodesics in this four dimensional space. The differential
equations representing the geodesics are found to be

2
d ρ 1 du dρ 2 −u dθ 2 −u 2 dφ 2 1 v−u dv dt 2
+ − ρe − ρe sin θ + e =0 (1.5.170)
ds 2 2 dρ ds ds ds 2 dρ ds
2
d θ 2 dθ dρ dφ 2
+ − sin θ cos θ =0 (1.5.171)
ds 2 ρ ds ds ds
2
d φ 2 dφ dρ cos θ dφ dθ
+ +2 =0 (1.5.172)
ds 2 ρ ds ds sin θ ds ds
2
d t dv dt dρ
+ =0. (1.5.173)
ds 2 dρ ds ds
The equation (1.5.171) is identically satisfied if we examine planar orbits where θ = π is a constant. This
2
value of θ also simplifies the equations (1.5.170) and (1.5.172). The equation (1.5.172) becomes an exact
differential equation

d 2 dφ 2 dφ
ρ =0 or ρ = c 4 , (1.5.174)
ds ds ds
and the equation (1.5.173) also becomes an exact differential

d dt v dt v
e =0 or e = c 5 , (1.5.175)
ds ds ds
where c 4 and c 5 are constants of integration. This leaves the equation (1.5.170) which determines ρ. Substi-
tuting the results from equations (1.5.174) and (1.5.175), together with the relation (1.5.161), the equation
(1.5.170) reduces to
2
d ρ c 2 c 2 c 2 4 c 2 c 2 4
+ + − (1 − ) =0. (1.5.176)
ds 2 2ρ 2 2ρ 4 ρ ρ 3