Page 164 - Intro to Tensor Calculus

159

By the chain rule we have

2
2
2
2
2
d ρ d ρ dφ 2 dρ d φ d ρ c 2 4 dρ −2c 2 4
= + = +
2
ds 2 dφ 2 ds dφ ds 2 dφ ρ 4 dφ ρ 5
and so equation (1.5.176) can be written in the form
2
d ρ 2 dρ 2 c 2 ρ 2 c 2 c 2
− + + − 1 − ρ =0. (1.5.177)
dφ 2 ρ dφ 2 c 2 2 ρ
4
The substitution ρ = 1 reduces the equation (1.5.177) to the form
u
2
d u c 2 3 2
+ u − 2 = c 2 u . (1.5.178)
dφ 2 2c 4 2
Multiply the equation (1.5.178) by 2 du and integrate with respect to φ to obtain
dφ
2
du 2 c 2 3
+ u − u = c 2 u + c 6 . (1.5.179)
dφ c 2
4
where c 6 is a constant of integration. To determine the constant c 6 we write the equation (1.5.161) in the
special case θ = π and use the substitutions from the equations (1.5.174) and (1.5.175) to obtain
2
2 2 2 2
dρ u dρ dφ 2 dφ v dt
u
e = e =1 − ρ + e
ds dφ ds ds ds
or
2 2 4
dρ c 2 2 c 2 c 5 ρ
+ 1 − ρ + 1 − − =0. (1.5.180)
dφ ρ ρ c 2 c 2
4
The substitution ρ = 1 reduces the equation (1.5.180) to the form
u
2 2
du 2 3 1 c 2 c 5
+ u − c 2 u + − u − =0. (1.5.181)
2 2
dφ c 2 c 2 c c
4 4 4
Now comparing the equations (1.5.181) and (1.5.179) we select
2
c 5 1
c 6 = − 1 2
c 2 c 4
so that the equation (1.5.179) takes on the form
2 2
du 2 c 2 c 5 1 3
+ u − 2 u + 1 − 2 = c 2 u (1.5.182)
dφ c 4 c 2 c 4
Now we can compare our relativistic equation (1.5.182) with our Newtonian equation (1.5.143). In order
that the two equations almost agree we select the constants c 2 ,c 4 ,c 5 so that
2
c 5
c 2 = 2GM and 1 − c 2 = E . (1.5.183)
c 2 h 2 c 2 h 2
4 4
The equations (1.5.183) are only two equations in three unknowns and so we use the additional equation

dφ dφ ds
2 2
lim ρ = lim ρ = h (1.5.184)
ρ→∞ dt ρ→∞ ds dt