Homework # 4
ρ
2
P = Kρ 1 +
ρ0
The equation of state
gives the enthalpy h defined by dh = dP/ρ
3 ρ
h = 2Kρ 1 +
4 ρ0
if h vanishes where P and ρ vanish. This is a quadratic equation for ρ(h).
Note that hydrostatic equilibrium with this equation of state has an analytic solution,
which can be seen as follows: We know that the gradient of ρ vanishes near the origin, so
the behavior must be quadratic there. Assume for the moment that
ρ = βρ0 1 − αr 2 + . . .
Then we have
m (r) =
Z
0
r
3 2
4π
3
4πρr dr =
βρ0 r 1 − αr .
3
5
2
Hydrostatic equilibrum gives
1 dP
3
2
βρ0 αr
= − 4K 1 + β 1 − αr
ρ dr
2
4πG
3 2
Gm
βρ0 r 1 − αr .
= − 2 =−
r
3
5
This is therefore a solution, since each size has a term linear in r and one cubic in r.
Equating terms of the same dimension in r, one finds
α=−
2 πG
,
15 K
β = 1.
Therefore ρ0 is also the central density. We want the density to vanish for r = R so
R2 = α−2 . The total mass becomes
3/2
8π
9K
M=
ρ0
.
25
2πG
The values K = 1014 cgs and ρ0 = 4 g cm−3 give 1 M⊙ and 1 R⊙ for the star’s mass and
radius. (R determines α and therefore K, and then the equation for M determines ρ0 .)