Past exam of the mathematics course of the University of Cambridge / 2012 / iii / Paper 61 / 1 / ii

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- Solution ii

Solution

Take the positive constant ratio

λ = p_{r} / p_{g}

. The equation of state implies

ρ = \frac{4 σ}{3 c λ R} T^{3}, p = (1 + λ) R ρT \propto T^{4} .

This is a polytropic equation of state of index three. Differentiating the pressure and using vertical hydrostatic equilibrium gives

4 (1 + λ) R ρ T_{z} = - ρ Ω^{2} z

. Hence

T = T_{0} - \frac{Ω ^{2} z ^{2}}{8 ( 1 + λ ) R} .

Define the semi-thickness

H

by

T (H) = 0

and put

s = 1 - z^{2} / H^{2}

. Then the constant gas-radiation ratio vertical disc structure is

T = T_{0} s, ρ = ρ_{0} s^{3}, p = p_{0} s^{4}, ∣ z ∣ < H,

with

T_{0} = \frac{Ω ^{2} H ^{2}}{8 ( 1 + λ ) R}, ρ_{0} = \frac{σ Ω ^{6} H ^{6}}{384 c λ ( 1 + λ ) ^{3} R ^{4}}, p_{0} = \frac{ρ _{0} Ω ^{2} H ^{2}}{8} = \frac{σ Ω ^{8} H ^{8}}{3072 c λ ( 1 + λ ) ^{3} R ^{4}} .

The surface density is the integral across both faces. The supplied polynomial integral gives

\int_{0}^{1} (1 - x^{2})^{3} d x = 16/35

, so

Σ = (32/35) ρ_{0} H

. Substitution yields

420 λ (1 + λ)^{3} Σ = (\frac{μ _{m} m _{p}}{k})^{4} \frac{σ}{c} H^{7} Ω^{6} .

All coefficients here are local in radial position; the derivation fixes the vertical profile rather than a global accretion rate. The zero-temperature surface is a formal boundary of this bulk model; optically thick diffusion ultimately fails in a thin photospheric layer.

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