Paper 317 (Structure and Evolution of Stars)

i

Solution

The stellar structure equations are

\frac{d M _{r}}{d r} = 4 π r^{2} ρ, \frac{d P}{d r} = - \frac{G M _{r} ρ}{r ^{2}},

\frac{d L _{r}}{d r} = 4 π r^{2} ρ ϵ, \frac{d T}{d r} = - \frac{3 κ ρ L _{r}}{16 π a _{rad} c r ^{2} T ^{3}},

with

P = P_{g} + P_{rad}, P_{g} = \frac{ρ k _{B} T}{μ H}, P_{rad} = \frac{a _{rad} T ^{4}}{3}, κ = κ_{0} ρ T^{- 7/2} .

The last relation is the Kramers opacity law. Uniform energy release per unit mass means

ϵ

is constant, so integration of the mass and luminosity equations gives

L_{r} = ϵ M_{r}

. At the surface,

ϵ = \frac{L}{M} = M^{- 1} L .

Thus

ϵ = M^{a} L^{b}

has

a = - 1, b = 1 .

ii

Solution

Let

y = β / (1 - β) = P_{g} / P_{rad}

. The two pressure laws give

P = \frac{a _{rad}}{3} T^{4} (1 + y), ρ = \frac{a _{rad} μ H}{3 k _{B}} y T^{3},

and hence

κ = \frac{κ _{0} a _{rad} μ H}{3 k _{B}} y T^{- 1/2} .

The radiative equation can be written

\frac{d P _{rad}}{d r} = - \frac{κ ρ L _{r}}{4 π c r ^{2}} .

Dividing it by hydrostatic equilibrium and using

L_{r} / M_{r} = L / M

gives

\frac{d P _{rad}}{d P} = \frac{κ L}{4 π c GM} .

Since

P = P_{rad} (1 + y)

and

P_{rad} \propto T^{4}

\frac{d P}{d P _{rad}} = 1 + y + \frac{T}{4} \frac{d y}{d T} .

With

x = \frac{4 π c GM}{κ _{0} L} \frac{k _{B}}{μ H} \frac{3}{a _{rad}} T^{1/2}, \frac{d x}{d T} = \frac{x}{2 T},

the reciprocal of the preceding moment equation becomes

1 + y + \frac{x}{8} \frac{d y}{d x} = \frac{x}{y} .

(7)

Multiplication by

y

proves

\frac{1}{8} x y \frac{d y}{d x} = x - y (y + 1) .

(8)

iii

Solution

Write

q = 32/31

. The supplied approximation

x = q y (q y + 1)

and

x \propto T^{1/2}

imply

ρ \propto y x^{6}, P \propto (1 + y) x^{8} .

Therefore

\frac{d l o g x}{d l o g y} = \frac{1 + 2 q y}{1 + q y},

and direct logarithmic differentiation gives

1 + \frac{1}{n} = \frac{d l o g P}{d l o g ρ} = \frac{544 y ^{2} + 791 y + 248}{( 1 + y ) ( 416 y + 217 )} .

Thus the model is locally a stellar polytrope with

n (y) = \frac{( 1 + y ) ( 416 y + 217 )}{128 y ^{2} + 158 y + 31} .

Since

y = β / (1 - β)

n (β) = \frac{199 β + 217}{β ^{2} + 96 β + 31} .

The radiation-pressure and gas-pressure limits are

n (0) = 7, n (1) = \frac{13}{4} .

2

i

Solution

For the centrally condensed point-source model,

M_{r} = M

L_{r} = L

, and gas pressure dominates. The equations are

\frac{d P}{d r} = - \frac{GMρ}{r ^{2}}, \frac{d T}{d r} = - \frac{3 κ _{0} ρ ^{2} L}{16 π a _{rad} c r ^{2} T ^{13/2}}, P = R ρT,

where

R = k_{B} / (μ H)

. Dividing the first equation by the second and eliminating

ρ = P / (R T)

gives

P \frac{d P}{d T} = \frac{16 π a _{rad} c GM R}{3 κ _{0} L} T^{15/2} .

The surface conditions

P = T = 0

then yield

P = (\frac{64 π a _{rad} c GM R}{51 κ _{0} L})^{1/2} T^{17/4},

ρ = \frac{1}{R} (\frac{64 π a _{rad} c GM R}{51 κ _{0} L})^{1/2} T^{13/4} .

Substitution into hydrostatic equilibrium cancels the common power of

T

and gives

\frac{d T}{d r} = - \frac{4 GM}{17 R r ^{2}} .

Using

T (R) = 0

T (r) = \frac{4 GM}{17 R} (\frac{1}{r} - \frac{1}{R}) .

ii

Solution

At the core-envelope interface, continuity at temperature

T_{c}

requires the envelope pressure

P = C_{0} (\frac{M}{L})^{1/2} T_{c}^{17/4}, C_{0}^{2} = \frac{64 π a _{rad} c G R}{51 κ _{0}},

to equal the nonrelativistic electron degeneracy pressure

P = K (\frac{ρ}{μ _{e}})^{5/3}, ρ = \frac{P}{R T _{c}} .

Eliminating

ρ

gives

P = (R μ_{e} T_{c})^{5/2} K^{- 3/2} .

Equating the two pressures and solving for luminosity gives Mestel's cooling law

\frac{L}{M} = \frac{64 π a _{rad} c G}{51 κ _{0}} \frac{K ^{3}}{R ^{4} μ _{e}^{5}} T_{c}^{7/2} .

In particular,

L / M \propto T_{c}^{7/2}

3

i

Solution

Assume the material gas is monatomic. Its specific enthalpy together with the radiation enthalpy is

h = \frac{5}{2} R T + \frac{4 a _{rad} T ^{4}}{3 ρ} = R T (\frac{4}{β} - \frac{3}{2}) .

At constant total pressure, logarithmic differentiation of

P = R ρT + a_{rad} T^{4} /3

gives

\frac{d β}{d T}_{P} = - \frac{4 ( 1 - β )}{T} .

Consequently the specific heat capacity at constant pressure is

c_{P} = (\frac{\partial h}{\partial T})_{P} = \frac{R}{2 β ^{2}} (32 - 24 β - 3 β^{2}) .

ii

Solution

The three stellar adiabatic exponents are defined by

Γ_{1} = (\frac{\partial l o g P}{\partial l o g ρ})_{s}, Γ_{3} - 1 = (\frac{\partial l o g T}{\partial l o g ρ})_{s},

\frac{Γ _{2}}{Γ _{2} - 1} = (\frac{\partial l o g P}{\partial l o g T})_{s} .

For the gas-radiation mixture,

χ_{ρ} = (\frac{\partial l o g P}{\partial l o g ρ})_{T} = β, χ_{T} = (\frac{\partial l o g P}{\partial l o g T})_{ρ} = 4 - 3 β,

and

c_{V} = \frac{R}{2 β} (24 - 21 β) .

It follows that

Γ_{3} - 1 = \frac{8 - 6 β}{24 - 21 β},

Γ_{1} = \frac{32 - 24 β - 3 β ^{2}}{24 - 21 β}, Γ_{2} = \frac{32 - 24 β - 3 β ^{2}}{24 - 18 β - 3 β ^{2}} .

Their defining derivatives imply the identity

\frac{Γ _{1}}{Γ _{3} - 1} = \frac{Γ _{2}}{Γ _{2} - 1} .

(7)

iii

Solution

For any equation of state,

c_{P} - c_{V} = \frac{P}{ρT} \frac{χ _{T}^{2}}{χ _{ρ}}, Γ_{3} - 1 = \frac{P χ _{T}}{ρT c _{V}} .

Eliminating

P / (ρT c_{V})

gives the requested ratio of specific heats:

γ = \frac{c _{P}}{c _{V}} = 1 + (Γ_{3} - 1) \frac{( \partial lo g P / \partial lo g T ) _{ρ}}{( \partial lo g P / \partial lo g ρ ) _{T}} .

iv

Solution

In the gas-pressure limit

β \to 1

, the mixture becomes a monatomic ideal gas, so

γ = Γ_{1} = Γ_{2} = Γ_{3} = \frac{5}{3} .

In the radiation-pressure limit

β \to 0

Γ_{1} = Γ_{2} = Γ_{3} = \frac{4}{3},

whereas

χ_{ρ} = β \to 0

and therefore

γ = \frac{c _{P}}{c _{V}} \to \infty .

The divergence reflects the singular constant-pressure heat capacity of pure radiation: its pressure fixes its temperature independently of density.

4

i

Solution

The gas-to-radiation pressure ratio gives

\frac{β}{1 - β} = \frac{P _{g}}{P _{rad}} = \frac{3 ρ k _{B}}{a _{rad} μ H T ^{3}} .

Hence

T (β, ρ) = [\frac{3 k _{B}}{a _{rad} μ H} \frac{1 - β}{β} ρ]^{1/3},

and either pressure component gives

P (β, ρ) = \frac{k _{B}}{μ H β} [\frac{3 k _{B}}{a _{rad} μ H} \frac{1 - β}{β}]^{1/3} ρ^{4/3} .

ii

Solution

The radiative-pressure equation is

\frac{d P _{rad}}{d r} = - \frac{κ ρ L _{r}}{4 π c r ^{2}} .

Dividing by hydrostatic equilibrium and substituting

κ = κ_{0} (M_{r} / L_{r}) (L / M)

gives

\frac{d P _{rad}}{d P} = \frac{κ _{0} L}{4 π c GM},

a constant. Since both pressures vanish at the surface,

P_{rad} = (1 - β) P

and

β

is constant throughout the star. Part i therefore has

P = K ρ^{4/3}, n = 3,

with spatially constant

K

Writing

ρ = ρ_{c} θ^{3}

and

α^{2} = \frac{K}{π G} ρ_{c}^{- 2/3}, r = α ξ,

gives the

n = 3

Lane-Emden equation

\frac{1}{ξ ^{2}} \frac{d}{d ξ} (ξ^{2} \frac{d θ}{d ξ}) = - θ^{3}, θ (0) = 1, θ^{'} (0) = 0.

The surface is the first zero

θ (ξ_{1}) = 0

. The

n = 3

equation has no elementary closed-form solution; among nonnegative indices, the standard analytic cases are

n = 0, 1, 5

. This Eddington standard model approximates chemically homogeneous massive main-sequence stars with substantial radiation pressure.

iii

Solution

At any radius,

\frac{d P _{rad}}{d P} = \frac{κ L _{r}}{4 π c G M _{r}} = \frac{L _{r}}{L _{crit}}, L_{crit} = \frac{4 π c G M _{r}}{κ} .

Both gas and radiation pressures decrease outwards, so

0 < d P_{rad} / d P < 1

. Therefore

L_{r} < L_{crit} = \frac{4 π c G M _{r}}{κ} .

Equality is the Eddington luminosity, where radiation supplies the entire outward force and hydrostatic gas support is lost.

iv

Solution

Stellar homology gives

P_{c} \sim \frac{G M ^{2}}{R ^{4}}, ρ_{c} \sim \frac{M}{R ^{3}} .

For an

n = 3

polytrope,

P = K ρ^{4/3}

, so these scalings imply

K \sim G M^{2/3}, M \propto (K / G)^{3/2} .

The equation of state in part i gives

K^{3} \propto \frac{1 - β}{β ^{4}}

at fixed chemical composition. Substitution into the homologous mass scale proves the Eddington quartic relation

M \propto \frac{( 1 - β ) ^{1/2}}{β ^{2}} .