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Work per unit axial length, with $y$ vertically upward and $x$ across the thin gap. In the frame of the downward-translating cylinder, the wall moves upward at speed $U$ and the cylinder surface is stationary. Let $\eta$ be viscosity and use the leading lubrication theoryThe transverse pressure gradient is negligible; solving $\eta u_{xx}=p_y$ with $u(0)=U$, $u(h)=0$ givesThe flux $q$ is constant by incompressibility. Equal ambient pressure at both ends implies $\int p_{y}dy=0$. With $I_j={\int }_{-\infty }^{\infty }{h}^{-j}dy$ and $I_2/I_3=4h_0/3$, this yieldsThe tangential traction on the cylinder isSince $I_2=I_1/(2h_0)$, its integrated force is $-2\eta U{I}_1+4\eta U{h}_0{I}_2=0$. Pressure acts radially on a circular cylinder and contributes no torque; the tangential traction has moment arm $a$ to leading order. HenceFor clarity, the total vertical hydrodynamic force is not zero. Its leading pressure contribution isupward, while the leading shear contribution just calculated vanishes. Thus this model predicts no induced rolling direction: a freely rotating cylinder can remain nonrotating, rather than rolling like a wheel in contact with the wall. This is the torque-free cylinder in a lubrication gap cancellation; finite ends or different boundary conditions would require a different model.

For the initially homogeneous sphere, $V_i=-3G{M}^2/(5R_i)$ and $K_i=0$. Conservation of energy and the virial theorem imply $E_f=V_f/2=V_i$, hence $V_f=2V_i$. With the usual homologous-sphere approximation, $V\propto -1/R$, soThese are the conventional collapse estimates. A final virial equilibrium is not, by itself, guaranteed to remain homogeneous: if $V_f=-{\alpha }_{f}G{M}^2/R_f$, the actual relation is $R_f/R_i={\alpha }_f/(2{\alpha }_i)$ with ${\alpha }_i=3/5$. Thus the factor of two applies exactly to the gravitational radius, and to the outer radius only under the stated structural approximation.

Here “non-interacting” means collisionless apart from gravity. Baryonic gas can collide, convert ordered motion to thermal energy, and radiate it. If radiative cooling is effective, the fixed-energy virial state loses pressure support and contracts further. Baryonic cooling can destabilize the collisionless virial state; baryons are not inevitably unstable if cooling is slow or heating and pressure support maintain equilibrium.

For a reversible change, the first law of thermodynamics is $TdS=dU+pdV$. Using the monatomic internal energy,Along the constant-volume leg, $Q_{AB}=\frac{3}{2}{V}_0(2p_0-p_0)=\frac{3}{2}{p}_0{V}_0>0$, so heat is absorbed. Along the constant-pressure return leg, $Q_{CA}=\frac{5}{2}{p}_0(V_0-2V_0)=-\frac{5}{2}{p}_0{V}_0$, so heat is emitted. The work signs are zero on $AB$ and negative, or work on the gas, on $CA$.

Write $Q=U_j{x}_j$ and use repeated-index summation. Direct differentiation givesTracing yields ${\partial }_i{u}_i=0$. Since ${\nabla }^2(1/r)=0$ for $r>a$, and ${\nabla }^2(Q{x}_i/r^3)=2U_i/r^3-6Q{x}_i/r^5$, we getHence the incompressible Stokes equations hold everywhere outside the bubble.

The Newtonian viscous stress tensor $T_{ij}=-p{\delta }_{ij}+\mu ({\partial }_j{u}_i+{\partial }_i{u}_j)$ simplifies toAt $r=a$ with $n=x/a$, $u=\frac{1}{2}[U+(U\cdot n)n]$, so $u\cdot n=U\cdot n$: the interface is impermeable in the bubble frame. Also $Tn=-(3\mu /a)(U\cdot n)n$ is purely normal, so the tangential traction vanishes, appropriate to a clean bubble with negligible internal viscosity. The tangential fluid velocity need not equal the bubble's velocity. At infinity $u,p\to 0$. The calculation assumes the usual spherical, negligible-deformation limit; exact normal-stress balance includes interior pressure, surface tension and any buoyancy sustaining the motion.

Integrate the outer-fluid traction with the normal pointing from the bubble into the fluid:Thus $\boxed{F_{\mathrm{drag}}=-4\pi \mu aU}$, smaller than the no-slip rigid-sphere Stokes drag law $-6\pi \mu aU$.

For a simple homogeneous thermodynamic system, internal energy is extensive: $E(\lambda S,\lambda V,\lambda N)=\lambda E(S,V,N)$. Euler theorem for homogeneous functions and $dE=TdS-pdV+\mu dN$ give $E=TS-pV+\mu N$, henceAlso $dG=-SdT+Vdp+\mu dN$, so at fixed $T,N$,At coexistence gas and liquid have the same temperature, pressure and chemical potential. Integrating $d\mu =vdp$ along the intervening isotherm gives $\int vdp=0$. Integration by parts, with endpoints at the same coexistence pressure $p_{\ast }$, gives the Maxwell constructionThe horizontal line at $p_{\ast }$ replaces the mechanically unstable loop of the van der Waals isotherm. The extensivity argument presupposes the ordinary thermodynamic limit and neglects macroscopic interface contributions.

The expansion about the liquid-gas critical point isDifferentiating at fixed $t$ gives the requested $-9{\varphi }^2/2+O({\varphi }^3)+t[-6+O(\varphi )]$. At coexistence $\varphi =O(\sqrt{-t})$, so to leading nontrivial order keep the cubic odd part $\bar{p}=1+4t-6t\varphi -3{\varphi }^3/2$.

Let $a={\varphi }_l$, $b={\varphi }_g$ for this leading cubic model. Equality of endpoint pressures gives $a^2+ab+b^2=-4t$. The Maxwell condition in pressure coordinates is ${\int }_a^b(1+\varphi ){\bar{p}}^{\prime }(\varphi )d\varphi =0$. Since the endpoint pressures agree, its first term vanishes, leavingIf $a+b\ne 0$, these two equations would imply $(a-b{)}^2=0$, contrary to distinct phases. Therefore $a=-b$, and the pressure equality gives $a^2=b^2=-4t$. HenceThus the coexistence gap has exponent one half. The printed equality ${\varphi }_l(t)=-{\varphi }_g(t)$ is a leading-order statement, not an exact identity for the full Van der Waals equation. Discarded asymmetric terms shift both phase volumes at higher order. This distinction is needed even though the leading cubic calculation has exact antisymmetry within its truncation.

Assume incompressible inviscid layers, gravity $-g{\mathbf{e}}_y$, no surface tension, and irrotational perturbations of the uniform base flows. The full bulk equations are $\nabla \cdot {\mathbf{u}}_j=0$ and ${\partial }_t{\mathbf{u}}_j+{\mathbf{u}}_j\cdot \nabla {\mathbf{u}}_j=-{\rho }_j^{-1}\nabla p_j-g{\mathbf{e}}_y$. At the rigid walls $u_{1y}(h)=u_{2y}(-h)=0$. At the material interface $y=\eta (x,t)$ each layer obeys $u_{jy}={\eta }_t+u_{jx}{\eta }_x$, and $p_1=p_2$. There is no tangential-velocity matching condition across this vortex sheet.

Write ${\mathbf{u}}_j=(U_j,0)+\nabla {\varphi }_j$, with $U_1=U$, $U_2=-U$. Then ${\nabla }^2{\varphi }_j=0$, and the wall conditions are ${\partial }_y{\varphi }_1(h)={\partial }_y{\varphi }_2(-h)=0$. Linearizing the interface conditions at zero givesThe second follows from linear Bernoulli pressure $p_j^{\prime }=-{\rho }_j({\partial }_t+U_j{\partial }_x){\varphi }_j$ and the hydrostatic pressure shift $-{\rho }_{j}g\eta$.

For $k>0$, use amplitudes ${\varphi }_1=B_1\cosh k(y-h)e^{ikx+\sigma t}$ and ${\varphi }_2=B_2\cosh k(y+h)e^{ikx+\sigma t}$. Kinematics gives $B_1=-(\sigma +ikU)A/[k\sinh kh]$ and $B_2=(\sigma -ikU)A/[k\sinh kh]$. The dynamic condition becomesCompleting the square yieldsA positive real part occurs precisely whenThe function $\tanh (kh)/k$ decreases strictly from $h$ at $k=0$ to zero at infinity. Indeed its derivative has numerator $kh{\mathrm{sech}}^2(kh)-\tanh (kh)<0$, since $\sinh (2kh)>2kh$. Therefore the minimum speed producing instability at every finite nonzero wavenumber isAt equality every $k>0$ still satisfies the strict growth inequality, since $\tanh (kh)<kh$; the limiting infinite-wavelength mode is marginal. A strict uniform long-wave margin requires $U>U_{\mathrm{crit}}$.

For a thin spherical shell, the radial pressure force per unit volume is $-dP/dr$. By the shell theorem, the inward gravitational acceleration is $GM(r)/r^2$, because exterior spherical shells contribute no force. Balancing force densities in hydrostatic equilibrium givesThe second equation is mass conservation; it closes the pressure-support equation once an equation of state is supplied.

Regularity at the centre requires finite ${\rho }_c,P_c$ and $M(0)=0$. Then $M(r)=4\pi {\rho }_c{r}^3/3+o(r^3)$, so $P^{\prime }(0)=0$. At a free stellar surface, $P(R)=0$; more generally it matches the external pressure. For $P=K{\rho }^2$ with $K>0$ and vacuum outside, this also gives $\rho (R)=0$.

Inside the star, cancel positive $\rho$ to obtain $2K{\rho }^{\prime }=-GM/r^2$. Differentiate and use mass conservation:Thus the Lane-Emden equation of polytropic index one isFor example, the regular solution is $\rho ={\rho }_c\sin x/x$, with its central value interpreted by continuity, and the first vacuum surface is at $x=\pi$. This also verifies the stated central and surface boundary conditions.

Assume an isothermal atmosphere in hydrostatic equilibrium, with constant gravity $g$, mean molecular weight $\mu$, and an ideal gas equation of state. Use $h$ for the atmospheric scale height, to distinguish it from the total observable height $H$. Thenso the cloud-top pressure in an isothermal atmosphere isAn opaque, approximately grey exoplanet cloud deck can flatten the exoplanet transmission spectrum if the remaining gas above it has little wavelength-dependent extinction. If its top nearly reaches the observable atmospheric extent, $H_c\simeq H$, this gives $P_c\simeq P_s{e}^{-H/h}$. More generally, a top at $N$ atmospheric scale heights has $P_c/P_s=e^{-N}$; illustrative values $N=5$ and $10$ give $6.7\times 10^{-3}$ and $4.5\times 10^{-5}$.

A flat exoplanet transmission spectrum and $P_s$ alone do not determine a unique cloud-top pressure. The temperature, gravity, mean molecular weight and cloud altitude, or equivalent opacity information, are also needed. A large mean molecular weight can itself suppress spectral features, so the exoplanet cloud deck interpretation is conditional.

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.