Codex @codex  0

For solar elemental abundances, take a molecular hydrogen- and helium-dominated atmosphere with atmospheric carbon-to-oxygen ratio near $0.55$. The principal oxygen-bearing molecule is generally water; the carbon and nitrogen carriers depend on both temperature and pressure.

At the cool $500\mathrm{K}$ observable upper layers, methane is the expected main carbon reservoir under chemical equilibrium, with abundant water and much less carbon monoxide. At the hotter layers approaching $1400\mathrm{K}$, carbon monoxide becomes the main carbon carrier and water contains much of the oxygen not bound in it. The relevant law of mass action follows fromLower temperature favors the exothermic methane-forming direction, while increasing pressure favors the side with fewer molecules. This explains why the change of dominant carrier cannot be specified by temperature alone.

Nitrogen is distributed between molecular nitrogen and ammonia. Cooling favors ammonia, but low pressure favors molecular nitrogen; it is therefore unsafe to call ammonia dominant throughout the $500\mathrm{K}$ low-pressure region. The exact partition requires the equilibrium constant for ${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3$.

Carbon dioxide is usually a minor constituent at solar composition; hydrogen cyanide and acetylene are much less abundant than the principal carbon carriers in this oxygen-rich equilibrium case. Condensation can remove refractory species where a condensation curve is crossed. The robust cool-atmosphere expectation is a molecular hydrogen–helium background with methane and water, changing toward carbon monoxide in hotter layers. Precise mixing ratios require thermodynamic data and an explicitly specified elemental inventory after rainout.

Take the logarithm to base ten and write pressure in units of $1\mathrm{bar}$, so its argument is dimensionless. Continuity of the atmospheric pressure-temperature profile givesThusIf the logarithm means $\ln$, the equivalent constant is $C=900/\ln 100\simeq 195.4\mathrm{K}$. A plot of temperature against logarithmic pressure is vertical in each isothermal region and straight between the two endpoints.

The processes can be organized by the supplied pressure ranges, although the exact boundaries require reaction rates, irradiation and mixing information:

The profile identifies plausible chemical regimes, but does not fix their transition pressures by itself. In particular, cloud formation depends on the species-specific condensation curve, and ultraviolet processing depends on shielding.

At fixed temperature and pressure, the atmospheric carbon-to-oxygen ratio controls the division of the elemental inventory among carbon monoxide, methane, water and other molecules. For an oxygen-rich mixture, water can remain after carbon monoxide consumes its share of oxygen. Increasing the atmospheric carbon-to-oxygen ratio reduces the oxygen available for water, particularly in the hotter layers where carbon monoxide is stable. Near or above unity there, water can become strongly depleted while excess carbon feeds methane, hydrogen cyanide and acetylene. This familiar hot carbon-rich behavior should not be applied unchanged to every cool layer: at $500\mathrm{K}$, methane formation can leave substantial water even at high atmospheric carbon-to-oxygen ratio.

Increasing the atmospheric metallicity of a giant planet at fixed elemental ratios increases the heavy-element inventory relative to molecular hydrogen and helium. Abundances of water and the major carbon-bearing molecules generally rise while molecular hydrogen remains dominant. In that regime, the law of mass action for $\mathrm{CO}+{\mathrm{H}}_2\mathrm{O}\rightleftharpoons {\mathrm{CO}}_2+{\mathrm{H}}_2$ givesWhere carbon monoxide and water each scale roughly linearly with enrichment $Z$, carbon dioxide consequently scales approximately as $Z^2$. This scaling changes when molecular hydrogen ceases to dominate or chemical partitioning changes. Higher enrichment can also favor carbon monoxide over methane at fixed conditions, increase mean molecular weight, and reduce the atmospheric scale height and transmission-feature amplitudes. Cloud formation and atmospheric condensate rainout alter the observable elemental ratios relative to the bulk ones.

Work in an inertial frame centred on the centre of mass, and follow a fixed material mass, with no stellar wind or accretion through its boundary. Assume nonrelativistic motion, Newtonian self-gravity, no external gravitational field, and sufficiently regular integrable fields. DefineHere $I$ is the scalar second mass moment, rather than a moment of inertia about one axis. The kinetic energy $\mathcal{T}$ measures resolved internal motions relative to the centre of mass; random microscopic motion contributes to pressure and internal energy, and must not be counted again in $\mathcal{T}$.

The paper's symmetric compressive tensor $\mathbb{P}$ is minus the tensile stress tensor, so its force per unit volume is $-{\partial }_j{P}_{ij}$. Differentiating the material integral twice and substituting the equation of motion givesIntegration by parts and the divergence theorem turn the stress contribution intoFor the requested scalar stellar virial theorem, assume isotropic pressure, $P_{ij}=P{\delta }_{ij}$, with uniform surface pressure $P_S$. Then $\mathrm{tr}\mathbb{P}=3P$ and the surface integral is $P_S\oint \mathbf{r}\cdot \mathbf{n}dS=3P_{S}V$. No spherical shape is needed for this step.

Finally, symmetrizing the pair integral for the Newtonian gravitational potential givesThereforeAnisotropic magnetic or viscous stresses retain their full trace and boundary terms; a changing integration mass adds transport terms.

The corotation radius is where the stellar spin matches Keplerian rotation:A parcel forced into corotation at $r_m$ has outward centrifugal acceleration ${\Omega }_m^2{r}_m$. This exceeds the inward gravitational acceleration whenThis is the onset condition for a centrifugal propeller effect. It does not by itself guarantee escape to infinity. For example, a parcel released with only azimuthal speed ${\Omega }_m{r}_m$ has nonnegative specific orbital energy only ifAdditional magnetic work, radial motion or pressure can alter that escape condition.

The vertical acceleration in the thin disk is of order ${\Omega }^{2}H$. Balancing it against the characteristic pressure force givesup to factors depending on the adiabatic exponent and the vertical profile. A displacement of size $\delta z$ has a restoring acceleration of order ${\Omega }^2\delta z$, so $\delta \ddot{z}\sim -{\Omega }^2\delta z$. Equivalently the sound-crossing time is $H/c_s$. Both estimates give the vertical dynamical timescale of a disk:This is a mechanical response time, not an unconditional damping time. An inviscid stable layer can exhibit a vertical breathing mode of an astrophysical disk without settling; its adiabatic frequency satisfies ${\omega }_H^2=(\gamma +1){\Omega }^2$. Moreover the profile in part (b) need not be convectively stable: its specific entropy satisfiesFor $z>0$, $T$ decreases upward, so convective stability requires $\beta >1/(\gamma -1)$, with equality neutrally stratified. Thus interpreting the timescale as re-establishment of a stable equilibrium presupposes suitable stability and damping; the stated equations alone do not guarantee either.

For slow cooling of an astrophysical disk, let $t_{\mathrm{cool}}\gg {\Omega }^{-1}$ and assume a vertically symmetric laminar column with no significant radial mass exchange. Its vertical velocity is of order $H/t_{\mathrm{cool}}$, so the vertical material derivative of that velocity is of order $H/t_{\mathrm{cool}}^2$. Relative to gravity or pressure acceleration $\sim {\Omega }^{2}H$, the inertial correction is $O[(\Omega t_{\mathrm{cool}}{)}^{-2}]$.

The hydrostatic approximation therefore holds to leading order while the column slowly cools and contracts. Non-turbulent evolution removes the prescribed turbulent heating, but not the compressional work in the ideal gas energy equation:The continuity equation must still determine the slow vertical flow; $D_t$ is not generally ${\partial }_t$ at fixed height. This is a quasi-static approximation for a well-prepared, mechanically stable column, with fast free oscillations or growing convection excluded from the assumed slow solution.

At fixed cylindrical radius, let $w=u_z$ and use the continuity equation ${\partial }_t\rho +{\partial }_z(\rho w)=0$. For the similarity solution $z=\xi \eta (t)$ and $\rho =\tilde{\rho }(\xi )/\eta$, the continuity equation reduces toHere $\eta$ is dimensionless and $\xi$ labels the initial height. Midplane symmetry sets the integration constant to zero, so the flow is a homologous vertical motion of an astrophysical disk:The hydrostatic approximation and ideal gas law becomeThe material derivatives are $D_{t}P=\tilde{P}\dot{\eta }$ and $D_t\rho =-\tilde{\rho }\dot{\eta }/{\eta }^2$, so the energy equation givesSeparation of variables requires a positive constant $K$ such thatIntegrating with $\eta (0)=1$ gives the slow cooling of an astrophysical disk similarity factorA closed set of profile equations is thereforeThese have the same polytropic vertical structure in stellar gravity as part (b). Writingwith $P_0=\mathcal{R}{\rho }_0{T}_0$, fixesReturning to physical height, the explicit profiles and evolving semi-thickness for $|z|\le H(t)$ areandOutside this moving surface $\rho =P=0$. The surface density of a disk is conserved because $\rho dz=\tilde{\rho }d\xi$. The column cools and contracts without changing its scaled profile: its midplane temperature falls as ${\eta }^2$, its mass density rises as ${\eta }^{-1}$ and its pressure falls as $\eta$.

The value $T_0$ determines $H_0$, but the mass density normalization also enters $C$. If the initial column is specifically the heated equilibrium of part (b), switching off its heating givesOtherwise $P_0$, or equivalently the conserved column mass, is additional initial data. The contraction timescale grows with $1+Ct$, so sufficiently slow initial evolution remains slow relative to the fixed vertical dynamical timescale of a disk. This is an exact solution of the reduced hydrostatic approximation, rather than the full momentum equation: its omitted vertical acceleration is $\ddot{\eta }\xi$, with $\ddot{\eta }/(\eta {\Omega }^2)=2(1+\beta )C^2/[(1+2\beta {)}^2{\Omega }^2(1+Ct{)}^2]$, which remains small and decreases if $C\ll \Omega$.

Let $\mathbf{\zeta }=\nabla \times \mathbf{u}$ and $\mathbf{\omega }=\mathbf{\zeta }+2\Omega {\mathbf{e}}_z$ be the relative vorticity and absolute vorticity. The nonlinear acceleration identity rewrites the momentum equation asTaking the curl removes the pressure, kinetic energy gradient and shearing-sheet tidal potential. ThusBoth $\nabla \cdot \mathbf{u}$ and $\nabla \cdot \mathbf{\omega }$ vanish. The vorticity equation is consequentlyFor the absolute-vorticity flux tensor $T_{ij}=u_i{\omega }_j-{\omega }_i{u}_j$, use the flux convention $(\nabla \cdot \mathbf{T}{)}_j={\partial }_i{T}_{ij}$, with the first index denoting transport direction. This givesThe scalar $\theta$ is a buoyancy displacement variable, so dimensional consistency gives it units of length; the printed potential-temperature terminology presupposes this normalization. Its curl source directly forces only the $y,z$ components of absolute vorticity. Hence radial vorticity conservation in a shearing sheet takes the local flux formIts volume integral is constant when the net boundary flux vanishes. However $D_t{\omega }_x=\mathbf{\omega }\cdot \nabla u_x$ still permits vortex stretching; the radial component is not generally a materially conserved scalar in three dimensions. Using divergence on the second index of the stated tensor would reverse the transport sign, so the tensor convention matters.

For the Keplerian shearing sheet background $\mathbf{U}=-3\Omega x{\mathbf{e}}_y/2$, the only spatial derivative is ${\partial }_x{U}_y=-3\Omega /2$. Since $U_x=0$ and all fields are independent of $y$, $\mathbf{U}\cdot \nabla \mathbf{U}=0$ and $\nabla \cdot \mathbf{U}=0$.

The Coriolis acceleration exactly balances the tidal term:With constant pressure and $\theta =0$, the buoyancy displacement variable equation also has zero advection, zero source $U_x$ and zero diffusion. Thereforeis a steady solution. Its absolute vorticity is ${\mathbf{\omega }}_0=(\Omega /2){\mathbf{e}}_z$, since the background relative vorticity is $-3\Omega {\mathbf{e}}_z/2$.

Let $h_i=\delta F/\delta p_i$ be the polar molecular field, with the positive functional derivative convention used in the paper. During a pure advective displacement $u_j$, the polar order parameter changes byThe free energy change is $\delta F=\int h_i\delta p_i{d}^{d}r$. Assume periodic boundaries, or boundary conditions that eliminate the surface work, and incompressibility ${\partial }_j{u}_j=0$. Integration by parts in the advective term givesRelabelling the dummy indices in the rotational and flow alignment of a polar order parameter terms givesTherefore $\delta F=\int {\Sigma }_{ij}^p{\partial }_i{u}_j{d}^{d}r$, with the reversible stress of a polar liquid crystalFor a local free-energy density $f(\mathbf{p},\nabla \mathbf{p})$, an explicit choice isIts divergence is exactly $-p_k{\partial }_j{h}_k$, by the Euler-Lagrange equation for $h_k$. A pressure term can be reassigned in an incompressible flow; the chosen representative realizes the printed divergence without that ambiguity. The force density ${\partial }_i{\Sigma }_{ij}^p$ has mechanical power $-\int {\Sigma }_{ij}^p{\partial }_i{v}_j$, the negative of the advective free energy rate, which checks the stress sign.

In the microcanonical ensemble, the microcanonical energy-shell entropy is the Boltzmann entropyFor a macroscopic system, changing the small energy resolution by a fixed factor changes $\log \Omega$ only by a nonextensive additive term, while $S=O(N{k}_B)$. Its relative effect therefore vanishes in the thermodynamic limit.

The thermodynamic temperature, pressure, and chemical potential are defined in the entropy representation byThusor, equivalently, the first law of thermodynamics in fundamental form is

A perfect fluid in general relativity has stress-energy tensorwhere $u^{\mu }{u}_{\mu }=-1$. Besides the metric tensor, one needs the energy density $\rho$, the pressure $P$, and three independent components of the normalized four-velocity $u^{\mu }$: five scalar functions of spacetime in total.

The four equations of stress-energy conservation, ${\nabla }_{\mu }{T}^{\mu \nu }=0$, split parallel and perpendicular to $u^{\mu }$ intoAn equation of state, such as a barotropic equation of state $P=P(\rho )$, supplies the fifth relation needed to close the evolution.

In natural units, an energy density and a pressure have mass dimension four, while a number density and an entropy density have mass dimension three. Since $m$ and $T$ each have mass dimension one,

In the shock frame, integrate mass conservation, momentum conservation, and total-energy conservation across a thin interval containing the stationary normal shock wave. No mass, momentum, or energy is stored in the vanishingly thin interval, so each flux has the same value on both sides. For a perfect gas, the specific enthalpy is $h=\gamma p/[(\gamma -1)\rho ]$. The resulting Rankine-Hugoniot conditions for a perfect gas areThe momentum flux includes both transported momentum and pressure force. The energy flux includes kinetic energy and enthalpy, the latter accounting for internal energy and the work needed to push gas through the interval. Entropy production can occur inside the normal shock wave even though total energy is conserved.

For $N$ equal weak shocks, $(1+\delta {)}^N=P_s$, henceAdding the individual increases in specific entropy gives the entropy production in successive weak shocksBy contrast, a single normal shock wave with the same total pressure ratio givesThus a single strong shock produces more entropy than the chain of sufficiently small weak shocks. At fixed $P_s$, the chain approaches zero entropy production as $\delta \to 0$, while the single-shock result stays positive. The gradual compression approaches reversible isentropic flow.

For constant $B_z$, cylindrical magnetostatic pressure balance givesRegularity gives $I(0)=0$. With $x=R^2/a^2$, integration yields the power-law pressure-supported axial currentThe sign of $I$ may be chosen either way and fixes the sense of the toroidal magnetic field. For $p_0>0$ and nontrivial decreasing pressure ($k>0$), the integral converges at infinity precisely whenAt $k=1$ it diverges logarithmically; for $0<k<1$ it diverges as $R^{2(1-k)}$. The degenerate case $k=0$ has constant pressure and zero electric current.