Solution to Problem Set 3

1. The mean molecular weights per particle, per electron and per ion are given by

µ ≡ ρ

nmp≈ 1

2X + 0.75Y + 0.56Z, µi ≡

ρ

nimp≈ 1

X + Y/4 + Z/16, µe ≡

ρ

nemp≈ 2

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 = 1.5×105kg m−3.For the given chemical composition, X = 0.3, Y = 0.68, Z = 0.02, we have µi = 2.12, µe = 1.54. Sothe ion and electron pressures are:

Pi =ρ

µimpkTc = 9.35× 1015Pa, Pe =

ρ

µempkTc = 1.29× 1016Pa (2)

The radiation pressure is

Pr =13aT 4 = 1.3× 1013Pa (3)

Notice that the gas pressure dominates over the radiation pressure

3. For a non-relativistic, degenerate electron gas,

P = 0.99× 107(ρ/µe)5/3 = 4.6× 1023µe−5/3 Pa (4)

The electron number density is related to the density by ne = ρ/(µemp) = 5.99 × 1036/µe m−3.From the lecture notes, for the electrons to be degenerate, its temperature must be lower than

T � 107 K(

ne

1032m−3

)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

Mr =∫ r

04πr2ρ(r)dr =

4π

3ρcr

3(

1− 3u

4

), u ≡ r

R. (6)

The total mass M is just the mass contained within radius R, which, from the previous equation,is given by M = πρcR

3/3. 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

dr= −GMr

r2ρ(r) = −π

3GRρ2

cu(4− 7u + 3u2) (8)

Since the density at the surface is zero, the pressure must be zero as well. So the pressure atradius r can be obtained by integrating the previous equation from the stellar surface to radius r,and hence

P (r) = P (R) +∫ r

R

dP

drdr =

π

36GR2ρ2

c(1− u)2(5 + 10u− 9u2), u ≡ r

R(9)

The central pressure is

P (r = 0) =5π

36Gρ2

cR2 =

54π

GM2

R4(10)

which is a factor of 5/(4π) of our rough estimate −GM2/R4.

3) The total binding energy is given by

E = −∫ R

0GMr

dMr

dr=

26π2

315GR5ρ2

c = −2635

GM2

R, (11)

very close to our rough estimate GM2/R.

4) For an ideal gas, we have

P (r) =ρ(r)µmp

kT. (12)

Combining eqs. (12) and (9), one obtains the run of temperature with radius

T (r) =µmpP (r)

kρ(r)=

π

36GµmpρcR

2

k(1− u)(5 + 10u− 9u2) (13)

5. For an ideal gas, [2 marks]P1 =

ρ

µmpkT,

For a non-relativistic and degenerate gas, the pressure is dominated by electron degeneratepressure [3 marks]

P2 = K1ρ5/3, K1 = 0.99× 107µ−5/3

e ,

The transition occurs when P1 = P2, which gives [2 marks]

ρ

µmpkT ≈ .99× 107ρ5/3µ−5/3

e −→ T ≈ 103ρ2/3µµ−5/3e K

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 (ε ∝ 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 lattermeans the core cannot expand and cool itself. The increasing temperature means the reaction3α→12 C runs faster, and so on, so we have a thermal runaway, called helium flash. [3 marks]