Solution to Problem Set 3
1. The mean molecular weights per particle, per electron and per ion are given by
µ≡
ρ
1
ρ
1
ρ
2
, µi ≡
, µe ≡
≈
≈
≈
nmp
2X + 0.75Y + 0.56Z
ni mp
X + Y /4 + Z/16
n e mp
1+X
(1)
• For a fully pure hydrogen gas, µ = 0.5, µe = 1, µi = 1
• For the solar composition, µ = 0.62, µi = 1.30, µe = 1.18
• For primeval chemical composition, we have µ = 0.59, µi = 1.21, µe = 1.14.
Since the total number density is equal to the sum of the electron and ion number densities
(n = ni + ne ), clearly the three mean molecular weights have to satisfy 1/µe + 1/µi = 1/µ
2. At the center of the Sun, the temperature is Tc = 1.6×107 K and density is ρc = 1.5×105 kg m−3 .
For the given chemical composition, X = 0.3, Y = 0.68, Z = 0.02, we have µi = 2.12, µe = 1.54. So
the ion and electron pressures are:
Pi =
ρ
ρ
kTc = 9.35 × 1015 Pa, Pe =
kTc = 1.29 × 1016 Pa
µi mp
µe mp
(2)
The radiation pressure is
1
Pr = aT 4 = 1.3 × 1013 Pa
3
Notice that the gas pressure dominates over the radiation pressure
(3)
3. For a non-relativistic, degenerate electron gas,
P = 0.99 × 107 (ρ/µe )5/3 = 4.6 × 1023 µe −5/3 Pa
(4)
The electron number density is related to the density by ne = ρ/(µe mp ) = 5.99 × 1036 /µe m−3 .
From the lecture notes, for the electrons to be degenerate, its temperature must be lower than
ne
T 10 K
32
10 m−3
7
2/3
= 1.53 × 1010 Kµe −2/3 .
(5)
Throughout this exercise, we have retained the mean molecular weight per electron. Notice that,
from problems 1-3, how different molecular weights enter the equation of state.
4. For the linear stellar model, the density at radius r is given by
ρ(r) = ρc (1 − r/R)
1) The mass enclosed within radius r is
Z
Mr =
0
r
4π
u
r
4πr ρ(r)dr =
ρc r 3 1 − 3
, u≡ .
3
4
R
2
(6)
The total mass M is just the mass contained within radius R, which, from the previous equation,
is given by M = πρc R3 /3. Thus the central density is given by
ρc =
3M
.
πR3
(7)
The central density is therefore a factor of 4 higher than the average density, ρ̄ = M/(4πR3 /3).
2) The hydrostatic equilibrium equation is given by
dP
GMr
π
= − 2 ρ(r) = − GRρ2c u(4 − 7u + 3u2 )
dr
r
3
(8)
Since the density at the surface is zero, the pressure must be zero as well. So the pressure at
radius r can be obtained by integrating the previous equation from the stellar surface to radius r,
and hence
Z
r
P (r) = P (R) +
R
dP
π
r
dr = GR2 ρ2c (1 − u)2 (5 + 10u − 9u2 ), u ≡
dr
36
R
(9)
The central pressure is
P (r = 0) =
5π 2 2
5 GM 2
Gρc R =
36
4π R4
(10)
which is a factor of 5/(4π) of our rough estimate −GM 2 /R4 .
3) The total binding energy is given by
E=−
Z
R
GMr
0
dMr
26π 2
26 GM 2
=
GR5 ρ2c = −
,
dr
315
35 R
(11)
very close to our rough estimate GM 2 /R.
4) For an ideal gas, we have
P (r) =
ρ(r)
kT.
µmp
(12)
Combining eqs. (12) and (9), one obtains the run of temperature with radius
T (r) =
µmp P (r)
π Gµmp ρc R2
=
(1 − u)(5 + 10u − 9u2 )
kρ(r)
36
k
5. For an ideal gas, [2 marks]
P1 =
(13)
ρ
kT,
µmp
For a non-relativistic and degenerate gas, the pressure is dominated by electron degenerate
pressure [3 marks]
P2 = K1 ρ5/3 ,
K1 = 0.99 × 107 µ−5/3
,
e
The transition occurs when P1 = P2 , which gives [2 marks]
ρ
kT ≈ .99 × 107 ρ5/3 µ−5/3
−→ T ≈ 103 ρ2/3 µµ−5/3
K
e
e
µmp
At the late evolutionary stage of the Sun, helium nuclear burning occurs in a degenerate core
[1 mark]. The energy generation rate is extremely sensitive to the temperature ( ∝ T 40 ) [1 mark].
When the helium burning reaction starts, the energy liberated increases the temperature locally,
but because of the degeneracy, the rising temperature does not increase the pressure, the latter
means the core cannot expand and cool itself. The increasing temperature means the reaction
3α →12 C runs faster, and so on, so we have a thermal runaway, called helium flash. [3 marks]