Past exam of the short options paper shared by Prelims and Parts A and B of the physics course of the University of Oxford / 2015 / S25

The tropopause is the transition between the lower, often convecting troposphere and the more stably stratified atmosphere above, where the decrease of temperature with altitude weakens or reverses. Its definition concerns the thermal profile, not a fixed universal altitude. An optical depth is a dimensionless path-integrated extinction, ${\tau }_{\nu }=\int {\kappa }_{\nu }\rho ds$ for mass extinction coefficient ${\kappa }_{\nu }$; without emission into the ray its direct intensity is attenuated by $e^{-{\tau }_{\nu }}$.

In hydrostatic equilibrium, the upward pressure force balances weight:The second equality uses the ideal gas law with specific gas constant $R_a$. For constant $T$ and $g$,For variable temperature the corresponding result is $p(z)=p(z_0)\exp [-{\int }_{z_0}^{z}g/(R_{a}T(z^{\prime }))d{z}^{\prime }]$. The fixed-temperature derivative does not require the entire column to be isothermal.

Assume a well-mixed ideal gas with constant $g,c_p$, no latent heat, and sufficiently strong surface heating to maintain dry convection near the surface. An adiabatically displaced unit mass has specific volume $V=1/\rho$, so the stated relation and hydrostatic equilibrium giveThis is the dry neutral lapse rate; infrared absorption alone does not guarantee that a radiative profile is steep enough to trigger convection. Combining the same equations with the ideal gas law yields $T\propto p^{R_a/c_p}$ along that adiabat.

High above the main infrared-emitting region, optical depth to space becomes small. A transparent-to-sunlight atmosphere has no substantial direct solar heating there; a thin grey layer absorbs the upwelling planetary infrared flux $Q=(1-\alpha )S_0/4$ and emits from both faces. In the optically thin radiative-equilibrium limit,The factor four in $Q$ is projected-area versus whole-sphere averaging; the additional factor two counts emission upward and downward by the tenuous layer. For Earth-like $S_0\simeq 1360\mathrm{W}{\mathrm{m}}^{-2}$ and $\alpha \simeq 0.30$, this is about $214\mathrm{K}$. Thus temperature approaches a height-independent skin value as $\tau \to 0$, rather than continuing the dry adiabat to zero. This is an asymptotic grey-atmosphere approximation, not a proof that an entire finite stratosphere is exactly isothermal.

Let $q$ be the mass mixing ratio of the uniformly mixed infrared absorbers, and let $k$ be their constant mass absorption coefficient. Hydrostatic integration gives downward optical depth $\tau =kqp/g$. The useful fixed-tropopause-optical-depth approximation requires a sufficiently optically thick lower column, unchanged absorbed sunlight and upper radiative profile in optical-depth coordinates, and the same convective lapse law after the perturbation. Then the matching to the radiative region occurs at the same ${\tau }_t$ and temperature $T_t$. Doubling all those absorbers givesAssume also that the surface pressure is unchanged. A dry adiabat satisfies $T_t/T_s=(p_t/p_s{)}^{\kappa }$ with $\kappa =R_a/c_p=0.287$. Consequentlyfor an initial Earth-like surface temperature $288\mathrm{K}$. Its first-order logarithmic estimate, $\Delta T_s\simeq \kappa T_s\ln 2$, is about $57\mathrm{K}$; the finite-doubling formula is preferable because the perturbation is large. This estimate is conditional on fixed ${\tau }_t,T_t$ and dry convective adjustment, not on carbon dioxide's measured spectral behavior.

There is also a consistency limitation to the most literal opacity model. A self-contained plane-parallel grey atmosphere with constant mass opacity has $T^4\propto \tau +2/3$ and $\tau \propto p$. Its radiative logarithmic gradient isIt is therefore subadiabatic for these dry heat capacities, so it does not by itself establish the assumed deep dry convective column. The preceding warming is the intended fixed-matching-level, convectively prescribed estimate. A fully self-consistent model needs opacity varying with pressure, other heating structure or another closure to supply that convection; the two statements should not be silently identified.

The actual Earth response to doubling carbon dioxide is expected to be much smaller than this roughly $60\mathrm{K}$ all-grey-absorbers estimate. Carbon dioxide absorbs in selected spectral bands rather than doubling opacity across the entire thermal spectrum, and doubling it does not double the other infrared absorbers. Saturated band centres and overlap with water absorption further distinguish it from uniform opacity multiplication. Changes in water vapour, clouds and the lapse rate introduce important climate feedbacks, but do not justify treating the real atmosphere as the stipulated dry, wavelength-independent absorber. The idealized number is therefore not an estimate of real-world equilibrium climate sensitivity.

Use a positive stabilizing feedback coefficient: the net flux is $N=F_{\mathrm{ext}}-\lambda \Delta T$. The Planck feedback increases outgoing thermal radiation as temperature rises and therefore contributes positively to $\lambda$. Water-vapor feedback raises infrared opacity as a warmer atmosphere holds more water, tending to reduce $\lambda$; ice-albedo feedback also tends to reduce it as bright ice is lost. These signs depend on the explicitly stated convention.

For doubled carbon dioxide forcing $F_{2\times }>0$, equilibrium givesA symmetric positive interval $[{\lambda }_0-{\sigma }_{\lambda },{\lambda }_0+{\sigma }_{\lambda }]$ maps toThe upward excursion is larger than the downward one because inversion is nonlinear. If the interval reaches zero, it supplies no finite upper bound on equilibrium warming within this linear stable model.

Take the logarithmic forcing law with its normalization fixed by the given doubling value. Let $x$ be the positive magnitude of cooling and $f$ the positive magnitude of the earlier negative forcing:At $C=212\mathrm{ppm}$, $f_0=1.5051\mathrm{W}{\mathrm{m}}^{-2}$. Equilibrium then yieldsThe concentration endpoints $185$ and $244\mathrm{ppm}$ imply forcing magnitudes $2.242$ and $0.7445\mathrm{W}{\mathrm{m}}^{-2}$, approximately $f_0\pm 0.75$. Define normalized variables $u=f/f_0$ and $v=x/x_0$. Independent approximately Gaussian uncertainties give the one-standard-deviation covariance ellipseIts centre is $(1,1)$ and its principal directions are the two coordinate axes. Warming is $2.492(v/u)\mathrm{K}$, so rays from the origin have constant inferred warming. The bounding rays tangent to this ellipse have slopes $m$ satisfying $(1-m{)}^2=\frac{1}{4}(1+m^2)$, orThis gives the useful asymmetric one-parameter profile estimateThe concentration data supply only marginal intervals, not a full likelihood. This result explicitly approximates their logarithmic forcing uncertainty as Gaussian. Linear error propagation would instead give fractional uncertainty $\sqrt{0.5^2+0.5^2}=0.707$, or roughly $2.49\pm 1.76\mathrm{K}$, and misses the large reciprocal asymmetry. A unit-radius two-dimensional covariance ellipse itself contains about $39\%$ of a bivariate Gaussian, not $68\%$; the quoted interval uses the one-parameter $\Delta {\chi }^2=1$ profile/ratio construction, rather than claiming that ellipse is a $68\%$ joint region.

The quadratic coefficient must have units $\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-2}$; the printed ${\mathrm{K}}^{-1}$ makes its term dimensionally incompatible with a radiative flux. Interpret its numerical value as ${\lambda }_2=0.05\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-2}$, with ${\lambda }_1=1.5\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-1}$. The total imbalance isA quadrupling gives $F_{\mathrm{ext}}=2(3.75)=7.5\mathrm{W}{\mathrm{m}}^{-2}$. The two equilibrium roots areFor positive heat capacity, stability requires $\partial N/\partial T=-{\lambda }_1+2{\lambda }_2\Delta T<0$. Thus only the lower root is stable. A saddle-node bifurcation occurs atThe logarithmic forcing relation gives $C_{\mathrm{crit}}=2802^{11.25/3.75}=2240\mathrm{ppm}$. Strict linear stability holds below this concentration on the lower branch; the equality is marginal and is the supremum, not another strictly stable equilibrium. Sustained forcing above it causes warming without a balancing equilibrium in this quadratic model. At such large changes its local feedback expansion can fail, and other physics can intervene; it does not prove unlimited real-planet warming or locate a measured Earth runaway threshold.

Put $\vartheta =T-T_0$ and assume $C>0$, $\lambda >0$. The linear differential equation for a constant forcing step isMultiplying by the integrating factor $e^{\lambda t/C}$ and integrating from zero to $t$ givesThe response initially grows at rate $F_0/C$ and tends to the equilibrium warming $F_0/\lambda$. Heat capacity per unit area and radiative-feedback units make $k_T$ an inverse time.

For logarithmic forcing $F=a\ln (C_a/C_{a0})$, where $C_a$ denotes atmospheric carbon-dioxide concentration rather than heat capacity, $\dot{F}=(a/C_a){\dot{C}}_a$. Approximately constant $\alpha$ in $\dot{F}=\alpha E$ therefore requires a roughly constant conversion from emissions to persistent atmospheric burden and a concentration range small enough to hold $a/C_a$ fixed. Carbon removal must be negligible on the modeled horizon, or be replaced by a fixed retained fraction; a realistic multiscale carbon-cycle response generally violates the permanent-step approximation.

A one-tonne pulse at time $t_e$ is $E(t)=\delta (t-t_e)$ in tonne-normalized units. It gives a persistent forcing step $F=\alpha \Theta (t-t_e)$. Convolving that step with the thermal response yieldsThere is no instantaneous temperature jump for finite $C$, although forcing jumps. Both persistence and constant coefficients are deliberate simplifications, rather than an assertion that all emitted carbon dioxide remains airborne forever.

Insert the pulse response into $D=\beta (T-T_0)$ and write $u=t^{\prime }-t_e$. For a positive discount rate $k_D$,ThereforeWith $k_D=0.03\mathrm{yea}{\mathrm{r}}^{-1}$, $k_T=0.05\mathrm{yea}{\mathrm{r}}^{-1}$ and the supplied one-tonne normalization of $A$,This is valued at the time of emission because the discount factor is relative to that date. Valuing an emission at a future date from an earlier reference date would require an additional discount factor. The infinite-horizon integral converges for $k_D>0$ under the assumed persistent forcing.

Let $Q$ be annual emissions abatement in tonnes per year. The given marginal abatement cost isUnder the stipulated global, stationary, unconstrained cost-benefit model, the optimal interior abatement equates that marginal cost to the avoided marginal damage of approximately $166.7\mathrm{USD}$ per tonne. ThusTotal mitigation expenditure is the area under the marginal-cost curve, not its endpoint value times all abated tonnes:The result also presumes baseline emissions permit this much reduction. Rational policy need not reproduce it: different discounting, damage uncertainty and risk aversion, income weights, omitted ecosystem impacts, adaptation and local air-quality benefits alter valued benefits; technology learning and adjustment constraints alter costs. Separate rational governments may abate less because some benefits accrue abroad, while cooperative policy can internalize those benefits. Future climate or technology changes also undermine a fixed present marginal-damage curve. These are changes in the optimization problem, not evidence of irrational decision-makers.