7.2. STARS 417
3.Derivation of formula (7.2.14).To deduce (7.2.14) we have to derive the gravitational
potentialα. The first equation of (7.1.79) can be rewritten as
dM
dr= 4 πr^2 ρ 0 −
c^2
4 Gr^2 ψ′φ′
α ψ, M=
c^2 r
2 G( 1 −
1
α),
It gives the solution as
M=
4
3
πr^3 ρ 0 −c^2
4 G∫r0r^2 ψ′φ′
α ψ
dr, α=(
1 −
2 MG
c^2 r)− 1
.
Byρ 0 =m/^43 πr^30 , and in the nondimensional form(r→r 0 r), we get
(7.2.16) α= ( 1 −δr^2 −η)−^1 for 0≤r≤ 1 ,
whereη,δare as in (7.2.12) and (7.2.13). By (7.1.81) we have
(7.2.17) η( 1 ) = 0 , (i.e.η(r 0 ) = 0 ).
Then, the formula (7.2.14) follows from (7.2.16).- Derivation ofσ-factor (7.2.11). By (7.2.10) we need to calculateT′andψ′. By
(7.1.80),T′can be expressed in the form
dT
dr=−
1
κr^2∫r0r^2 αQdr+a
r^2,
whereais a determined constant. By (7.1.84) we obtain
a=−Ar^20 +1
κ∫r 00r^2 αQdr.In the nondimensional form, we have
(7.2.18)
dT
dr=−
A
r^2+
1
κr^2∫ 1rr^2 αQdr for 0≤r≤ 1 , A> 0.To considerψ′, by the second equation of (7.1.79) we obtain(7.2.19) ψ=
k
reζ(r),whereζ(r)is as in (7.2.12),kis a to-be-determined constant. In view of (7.1.81), i.e.ψ( 1 ) =
1 −δ, we have
k= ( 1 −δ)e−ζ(^1 ).
Then, it follows from (7.2.19) that
(7.2.20)
dψ
dr=
( 1 −δ)eζ(r)
eζ(^1 )r^0(
α− 1
r^2+rξ)
for 0≤r≤ 1 ,