Apply the formal solution of the radiative transfer equation successively to the two isothermal layers at normal incidence. In local thermodynamic equilibrium,

I_{ν, 2} = I_{ν, 0} e^{- (τ_{1} + τ_{2})} + B_{ν} (T_{1}) (1 - e^{- τ_{1}}) e^{- τ_{2}} + B_{ν} (T_{2}) (1 - e^{- τ_{2}}) .

For optically thin layers this becomes

I_{ν, 2} = I_{ν, 0} + τ_{1} [B_{ν} (T_{1}) - I_{ν, 0}] + τ_{2} [B_{ν} (T_{2}) - I_{ν, 0}] + O (τ^{2}) .

(2)

This assumes no scattering and negligible reflection between layers.

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 0  0

Solution

 0  0

With

I_{ν, 0} = B_{ν} (T_{0})

, the sign of each first-order term is set by whether that layer is hotter or colder than the incident blackbody.

If $T_{2} < T_{1} < T_{0}$ , temperature decreases with altitude. Both layers remove intensity, so frequencies of larger optical depth appear in absorption; the colder upper layer gives the strongest absorption when it controls the opacity.
If $T_{2} > T_{1}$ while $T_{1} < T_{0}$ , the atmosphere has a temperature inversion. The lower layer absorbs, whereas the upper layer partly fills the absorption. If $T_{2} > T_{0}$ , upper-atmosphere bands appear in emission; if $T_{2} < T_{0}$ , they remain in absorption but are shallower.
If $T_{2} > T_{1} > T_{0}$ , both layers add intensity and spectral bands appear in emission, with the hotter upper layer producing the largest contrast.

The corresponding temperature profiles, from bottom to top, are respectively monotone decreasing, decreasing then increasing, and increasing throughout.

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Past exam of the mathematics course of the University of Cambridge / 2023 / iii / Paper 315 / 2 / b

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