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

(1)

\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}},

(2)

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} .

(3)

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 .

(4)

Thus

ϵ = M^{a} L^{b}

has

a = - 1, b = 1 .

(5)

ii

 0  0

Solution

 0  0

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},

(1)

and hence

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

(2)

The radiative equation can be written

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

(3)

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} .

(4)

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} .

(5)

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},

(6)

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

 0  0

Solution

 0  0

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} .

(1)

Therefore

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

(2)

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 )} .

(3)

Thus the model is locally a stellar polytrope with

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

(4)

Since

y = β / (1 - β)

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

(5)

The radiation-pressure and gas-pressure limits are

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

(6)

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Past exam of the mathematics course of the University of Cambridge / 2022 / iii / Paper 317 / 1

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