Blundell Thermal Physics · Problem 37.8

Earth's atmosphere

Physics — Statistical Mechanics · radiative-equilibrium-greenhouse

Problem

This question explores a purely radiative equilibrium model, which grossly simplifies the radiative transfer in the atmosphere. Assume that the radiative transfer equations for upwards and downwards radiation can be written (following eqns 37.23 and 37.24 and dropping the frequency subscript) −dF↑dχ∗+F↑=πBdF↓dχ∗+F↓=πB.(37.38)\begin{aligned} -\frac{\mathrm{d}F^\uparrow}{\mathrm{d}\chi^*} + F^\uparrow &= \pi B \tag{37.38}\\ \frac{\mathrm{d}F^\downarrow}{\mathrm{d}\chi^*} + F^\downarrow &= \pi B. \end{aligned} In radiative equilibrium, F↑−F↓F^\uparrow - F^\downarrow is a constant, which we will call ϕ\phi. Writing ψ=F↑+F↓\psi = F^\uparrow + F^\downarrow, show that dψdχ∗=ϕ(37.40)\frac{\mathrm{d}\psi}{\mathrm{d}\chi^*} = \phi \tag{37.40} and dϕdχ∗=ψ−2πB(37.41)\frac{\mathrm{d}\phi}{\mathrm{d}\chi^*} = \psi - 2\pi B \tag{37.41} and hence that ψ=2πB\psi = 2\pi B. At the top of the atmosphere, χ∗=0\chi^* = 0 and F↓=0F^\downarrow = 0. Hence show that B=ϕ2π(χ∗+1),(37.42)B = \frac{\phi}{2\pi}(\chi^* + 1), \tag{37.42} and sketch F↑F^\uparrow, F↓F^\downarrow and πB\pi B as a function of optical depth χ∗\chi^* on the same graph. If the optical depth at the bottom of the atmosphere increases because of increasing CO2\mathrm{CO}_2, what happens to the surface temperature in this model? (Remember that BB is a function of TT.)

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