Past exam of the physics of atmospheres and oceans course of the University of Oxford / 2018 / 2

Use downward optical depth $t$ measured from the top, with total $D={\tau }_{\infty }$. This is the effective optical-depth/two-stream convention that gives an attenuation factor $e^{-t}$. Constant absorption per unit mass and constant gravity give $t=Dp/p_s$. Write the locally emitted flux source as $S(t)=\pi B(t)=S_0+bt$.

The atmospheric contribution to upward outgoing longwave flux isBecause the source is linear, the effective emission pressure makes this exactly $(1-e^{-D})S(t_{\mathrm{eff}})$, whereThus for $D\gg 1$, $t_{\mathrm{eff}}\to 1$ and $p_{\mathrm{rad}}\simeq p_s/D$. For $D\ll 1$, expansion gives $t_{\mathrm{eff}}=D/2-D^2/12+O(D^4)$, so $p_{\mathrm{rad}}\simeq p_s/2$. In the thin limit $F_{\mathrm{atm}}\simeq D\pi B(p_s/2)$: it is the small atmospheric contribution, not the whole outgoing flux, which this level characterizes. Surface radiation transmitted through the column must be added separately.

Angular convention: these numerical coefficients use an effective optical depth. If $t$ instead means literal vertical monochromatic optical depth and the angular integral is retained exactly, the flux weight is $2E_2(t)$, not $e^{-t}$. Its thick-limit mean depth is $2/3$, giving $p_{\mathrm{rad}}\simeq 2p_s/(3D)$ for a linear source. The coefficient one in the requested expression therefore specifies a two-stream/diffusivity convention; it is not an angle-independent identity. The thin-limit half-column result survives to leading order.

The incident stellar power is intercepted over a disk and distributed over the spherical surface. With a black shortwave surface, no clouds and no shortwave atmospheric absorption, the global mean absorbed flux is $F_{\ast }=L_{⊛}/4$. Define $T_e=(F_{\ast }/\sigma {)}^{1/4}$.

To relate a radiating-level temperature to the ground, assume the intervening well-mixed gas follows a dry adiabat with constant $R_g,c_{pg}$:Emission from $p_{\mathrm{rad}}$ balances absorbed sunlight when $\sigma T(p_{\mathrm{rad}}{)}^4=F_{\ast }$. HenceThe final form uses the optically thick result in part (a). Since $p_{\mathrm{rad}}<p_s$, the surface is warmer than the effective emission temperature. Efficient horizontal redistribution alone does not mathematically fix the vertical lapse rate: the adiabatic/convective assumption is needed to obtain the requested dependence on $R_g/c_{pg}$. A different imposed vertical profile would give a different relation.

Assume the added gas mixes throughout the convective column and retain constant mass absorption coefficient for the original gas. The original absorber column mass is unchanged, so its total effective optical depth $D={\tau }_{\infty }$ is unchanged. Under transparent-gas addition at fixed absorber mass, the absorber mass fraction becomes $q=p_s/p_a$, and consequently $t=Dp/p_a$. The new radiating pressure is $p_{\mathrm{rad},a}\simeq p_a/D$, even though the ratio of surface to radiating pressure stays $D$.

The mixture's mass-specific constants areThese pressure weights represent column mass fractions, not mole fractions. Repeating the adiabatic argument givesAt fixed $D>1$, this is warmer than the original surface precisely ifThe equivalence follows by cross multiplication of the positive heat capacities; it holds for any positive added mass. If the two ratios agree, adding a transparent gas leaves this model's surface temperature unchanged. Increased total pressure alone is not a new opacity in a model that fixes absorber mass and mass absorption coefficient. Pressure broadening or collision-induced absorption could change the answer in a more realistic atmosphere, but neither is part of the stated constant-opacity model.

Let $D\ll 1$ be the remaining absorber optical depth, and retain the mixture's convective exponent ${\kappa }_a=R_a/c_{pa}$. The outgoing surface contribution is $\sigma T_s^4{e}^{-D}\simeq \sigma T_s^4(1-D)$. To first order, atmospheric emission is the column-mean source times $D$. If $4{\kappa }_a\simeq 1$, the adiabat gives $T(p{)}^4\simeq T_s^{4}p/p_a$, a source linear in mass/optical depth. Therefore its mean is $T_s^4/2$, or equivalently it is evaluated at $p_a/2$:Setting this equal to $L_{⊛}/4$ givesThe zero-opacity limit correctly gives $T_e$, while a small positive opacity raises the surface temperature slightly. The expression is a first-order thin-column result, not a formula to extend to its apparent pole at ${\tau }_{\infty }=2$. More generally, direct thin-column averaging gives $F_{\mathrm{atm}}\simeq D\sigma T_s^4/(1+4{\kappa }_a)$; the half-column value is exact at this order when $4{\kappa }_a=1$. The assumed convective thermal structure remains a modelling condition even when the gas becomes radiatively thin.