Past exam of the mathematics course of the University of Cambridge / 2017 / iii / Paper 315 / 1 / a

Take $P_0>0$, real $\alpha$, and the real branch $T\ge T_0$, with physical temperature $T>0$. In hydrostatic equilibrium the pressure decreases with height $z$:Thus an atmospheric thermal inversion means $dT/dz>0$, equivalently $dT/dP<0$. For $\alpha \ne 0$, the logarithm of the given relation yieldsThe sign restriction must be retained after squaring. In particular, the inverted branch has $0<P\le P_0$, whereas the non-inverted branch has $P\ge P_0$. By differentiation, on the interior of either branch,Consequently,At $T=T_0$, the inverse temperature gradient vanishes at the branch endpoint, rather than producing a finite isothermal interval. If $\alpha =0$, the relation becomes $P=P_0$ independently of $T$; it is not a nontrivial vertical atmospheric pressure-temperature profile under hydrostatic equilibrium. These distinctions prevent the two sides of the squared parabola from being interpreted as one permissible atmosphere.

Assume a homogeneous ideal gas, constant specific-heat ratio $\gamma >1$, and an upward displacement that is an adiabatic process and maintains pressure balance with its surroundings. If the ambient temperature falls faster with decreasing pressure than the parcel's adiabatic temperature gradient, the parcel becomes hotter and less dense than its surroundings; buoyancy amplifies the displacement. This is the Schwarzschild criterion:The square-root exponential atmospheric profile hasAn atmospheric thermal inversion with $\alpha <0$ has $\nabla <0$ and is stable to these ordinary adiabatic displacements. For $\alpha >0$, put $s=\sqrt{T-T_0}$. The condition becomes the quadratic inequalityintersected with $s\ge 0$ and $T_0+s^2>0$. In the usual case $T_0>0$, the maximum gradient occurs at $s=\sqrt{T_0}$, namely $T=2T_0$, and is $1/(\alpha \sqrt{T_0})$. Hence a convectively unstable interval exists precisely whenEquality in the existence condition gives a single neutrally stable point. A molecular-hydrogen-dominated hot Jupiter with rotational modes active but vibrational excitation and dissociation negligible has approximately $\gamma =7/5$, so $A=2/7$. The existence condition is then $\alpha \sqrt{T_0}<7/2$. For $T_0\le 0$ the general inequality above remains valid, but its roots must be intersected with the positive-temperature domain; the positive-$T_0$ maximum formula must not be reused. Composition gradients require the Ledoux criterion, and dissociation or variable heat capacities change $A$. Sustained convection normally adjusts a superadiabatic profile toward an adiabatic temperature gradient.

Use the usual atmospheric plotting convention: temperature on the horizontal axis, logarithmic pressure increasing downward. To illustrate a physically plausible local range for a hot Jupiter, set $T_0=1000\mathrm{K}$ and $|\alpha |=0.05{\mathrm{K}}^{-1/2}$. The sign is not fixed by the fact that the planet is a hot Jupiter; irradiation and opacity determine whether an atmospheric thermal inversion is present.

The original sketch shows both admissible branches of the square-root exponential atmospheric profile, with separate assumed normalizations $P_0=0.01\mathrm{bar}$ for the outward-cooling example and $P_0=1\mathrm{bar}$ for the inverted example. Each extends from $1000$ to $2600\mathrm{K}$ and has $|\log (P/P_0)|\le 2$. For $\alpha >0$, increasing depth increases both pressure and temperature, so the curve bends rightward downward. For $\alpha <0$, moving upward lowers pressure and raises temperature, so the curve bends rightward upward. Each branch meets $T_0$ with $dT/d\log P=0$; it must not be continued through that endpoint onto the other branch.

For the outward-cooling example, $\alpha \sqrt{T_0}=1.58<3.5$, so a section of the imposed profile is unstable for a homogeneous diatomic ideal gas. It should be interpreted as an illustrative retrieved radiative profile that would be modified by convection if realised, rather than a self-consistent deep convective atmosphere. The inverted branch is stable under the same assumptions.

The functional form is not specific to exoplanets. With positive pressure normalization and physical temperature, a square-root exponential atmospheric profile can approximate a monotone interval of the terrestrial atmosphere. An outward-cooling interval, such as part of the troposphere or mesosphere, requires $\alpha >0$. An upward-warming interval, such as part of the stratosphere heated by ultraviolet absorption or the thermosphere heated by high-energy radiation, requires $\alpha <0$.

The local atmospheric lapse rate follows from the ideal gas relation $P=\rho k_{B}T/(\mu m_H)$ and hydrostatic equilibrium:For dry terrestrial air, approximately $\gamma =1.4$ and $\mu \simeq 29$ give the dry-adiabatic lapse rate $g/c_p\simeq 9.8\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$. The familiar mean tropospheric value near $6.5\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ is less steep; a local fit must satisfy $\nabla \lesssim 2/7$ to be dry-convectively stable. Moist convection needs the moist parcel thermodynamics instead, and the terrestrial atmosphere is not uniformly dry or chemically homogeneous at all heights.

The squared-logarithm shape cannot reproduce an exactly constant nonzero lapse rate over an arbitrary thick region, all the alternating atmospheric layers, or a finite exactly isothermal region. It is a local parametrization with a fixed sign of the temperature gradient; it has no terrestrial universality.