Past exam of the mathematics course of the University of Cambridge / 2024 / iii / Paper 317 / 2 / ii

For

I = A + C μ

, the angular moments are

J = \frac{1}{4 π} \int I d Ω = A,

F = \int μ I d Ω = \frac{4 π}{3} C,

and

P_{r} = \frac{1}{c} \int μ^{2} I d Ω = \frac{4 π}{3 c} A = \frac{4 π}{3 c} J .

Thus this angular form satisfies the Eddington closure approximation,

c P_{r} = 4 π J /3

, and

C = \frac{3 F}{4 π} .

Part (i) gives

S = J = A

. Substitution in

μ d I / d τ = I - S

yields

μ (A^{'} + C^{'} μ) = C μ .

Matching powers of

μ

gives

A^{'} = C

and

C^{'} = 0

, the latter being constant flux. Hence

A = C τ + A_{0}

At the surface there is no incoming intensity. Applying this condition to the outgoing hemisphere in the moment approximation,

F = 2 π \int_{0}^{1} (A_{0} + C μ) μ d μ = π A_{0} + \frac{2 π C}{3},

A_{0} = F / (2 π) = 2 C /3

. Consequently

A = \frac{3 F}{4 π} (τ + \frac{2}{3}) .

Now

A = J = j / κ = σ T^{4} / π

and the effective temperature is defined by

F = σ T_{e}^{4}

T^{4} = \frac{3}{4} T_{e}^{4} (τ + \frac{2}{3}) .

At the top of the grey atmosphere,

T_{0}^{4} = \frac{1}{2} T_{e}^{4}, T_{0} = 2^{- 1/4} T_{e} .

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