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For a spherical planet, combine the hydrostatic pressure support equation with $dm/dr=4\pi r^2\rho$:Neglect surface pressure. Since $r(m)\le R$, the hydrostatic lower bound on planetary central pressure isFor a physically usual mass density decreasing outward, the mean interior density exceeds the global mean. Hence $r(m)\le R(m/M{)}^{1/3}$, giving the sharper minimum within this class,Uniform mass density attains the sharper bound; direct integration with $m(r)=M{r}^3/R^3$ gives $P_c=3G{M}^2/(8\pi R^4)$. Real planets are centrally concentrated through compression and dense cores, so their central pressure is higher.

Using $M_{\oplus }=5.97\times 10^{24}\mathrm{kg}$, $R_{\oplus }=6.37\times 10^6\mathrm{m}$, $M_J=1.90\times 10^{27}\mathrm{kg}$, and $R_J=7.15\times 10^7\mathrm{m}$, the universal bounds are approximately $5.75\times 10^{10}\mathrm{Pa}$ for Earth and $3.66\times 10^{11}\mathrm{Pa}$ for Jupiter. The uniform-density estimates are respectively $1.72\times 10^{11}\mathrm{Pa}$ and $1.10\times 10^{12}\mathrm{Pa}$, or $1.72$ and $11.0$ megabars. Either comparison gives a Jupiter minimum about 6.4 times the Earth minimum.

A close-in giant is often in synchronous rotation after tidal locking, giving persistent dayside heating and nightside cooling. Its contrast is controlled by the competition between radiative relaxation time in a planetary atmosphere, wind transport characterized by the atmospheric advection time, wave adjustment, and drag. When heat transport is fast compared with radiation, day-night heat redistribution lowers the contrast; when radiation is fast, each hemisphere stays closer to its local radiative balance.

For a rough atmospheric column estimate,At comparable pressure, higher planetary equilibrium temperature sharply shortens radiative relaxation, tending to increase the day-night contrast. Wind speeds, rotation, and magnetic drag can modify this trend; dissociation and recombination can carry additional heat in very hot atmospheres.

At higher altitude, lower pressure generally means shorter radiative relaxation and a larger contrast. Infrared bands with larger opacity probe these higher layers, while lower-opacity windows sample deeper layers with longer cooling times and more effective redistribution. A wavelength-dependent exoplanet thermal phase curve can therefore reveal how the contrast and hot-region displacement change with pressure.

Assume a hydrogen-helium perfect gas with mean particle mass $2.3m_H$, Jupiter gravity $g\simeq 24.8\mathrm{m}{\mathrm{s}}^{-2}$, and mixing over one atmospheric scale height. At $T=1000\mathrm{K}$,The eddy mixing time is ${\tau }_{\mathrm{mix}}\simeq H^2/K_{zz}$. A chemical quench level at one bar requires ${\tau }_{\mathrm{mix}}\lesssim {\tau }_{\mathrm{chem}}=10^5\mathrm{s}$, hence the vertical eddy diffusion coefficient must be of orderThe number depends quadratically on the assumed mixing length; using a fraction of $H$ reduces it accordingly.

Above the chemical quench level, neglect photochemistry, condensation, and molecular diffusion. The quenched atmospheric mixing ratio $f_A=n_A/n$ is approximately constant, while the number density isThus the abundance fraction is frozen, but the absolute number density falls with pressure and altitude. Examples are carbon monoxide–methane quenching, through $\mathrm{CO}+3{\mathrm{H}}_2\rightleftharpoons {\mathrm{CH}}_4+{\mathrm{H}}_2\mathrm{O}$, and nitrogen–ammonia quenching, through ${\mathrm{N}}_2+3{\mathrm{H}}_2\rightleftharpoons 2{\mathrm{NH}}_3$.

Consider an element displaced upward by $\xi$ in hydrostatic equilibrium. Assume it quickly reaches the ambient pressure, but moves adiabatically and retains its composition. Put $\nabla =d\log T/d\log P$, ${\nabla }_{\mu }=d\log \mu /d\log P$, and let ${\nabla }_{\mathrm{ad}}$ be the adiabatic temperature gradient. The equation of state has differentialwhere $\delta =-(\partial \log \rho /\partial \log T{)}_{P,\mu }$ and $\varphi =(\partial \log \rho /\partial \log \mu {)}_{P,T}$.

At the new pressure, the parcel-to-environment density difference isWith positive pressure scale height $H_P=-(d\log P/dr{)}^{-1}$, $\Delta \log P=-\xi /H_P$. The buoyancy acceleration is therefore $\ddot{\xi }=-N^2\xi$, with stellar buoyancy frequencyA negative $N^2$ amplifies the displacement. For a radiative background, $\nabla ={\nabla }_{\mathrm{rad}}$, so the Ledoux criterion for instability isFor uniform composition, ${\nabla }_{\mu }=0$, giving the Schwarzschild criterionEquality is marginal stability. A mean molecular weight increasing inward has ${\nabla }_{\mu }>0$ and supplies a stabilizing composition term when $\varphi ,\delta >0$.

Use the stable parameters $b,\kappa >0$. The bulk stationary points satisfy $f^{\prime }(\varphi )=a\varphi +b{\varphi }^3=0$. For $a<0$, $\varphi =0$ has negative curvature, while the two minima areBoth have $f(\pm {\varphi }_B)=-a^2/(4b)$ and $f^{\prime \prime }(\pm {\varphi }_B)=-2a>0$. To establish minimization at fixed total composition, writeFor mean composition $\bar{\varphi }$ with $|\bar{\varphi }|<{\varphi }_B$, a mixture of the two minima with volume fraction $\lambda =(\bar{\varphi }+{\varphi }_B)/(2{\varphi }_B)$ in the positive phase attains this lower bound in the bulk. Macroscopic slabs have only a subextensive interfacial excess free energy, so the bound is attained per unit volume in the thermodynamic limit. Thus constrained minimization gives phase coexistence in the mean-field approximation; at the endpoints only one phase is needed.

The two equilibrium conditions also follow directly by constrained optimization. Introduce a Lagrange multiplier $\mu$ for $\lambda {\varphi }_{+}+(1-\lambda ){\varphi }_{-}=\bar{\varphi }$ in the mixture's bulk free-energy density. Varying ${\varphi }_{+}$ and ${\varphi }_{-}$ gives $f^{\prime }({\varphi }_{+})=f^{\prime }({\varphi }_{-})=\mu$; varying $\lambda$ gives $f({\varphi }_{+})-\mu {\varphi }_{+}=f({\varphi }_{-})-\mu {\varphi }_{-}$. These equate the chemical potentials and pressures, respectively.

Equivalently, their bulk chemical potentials are both zero and their pressures are equal:These are precisely the conditions of the common-tangent construction for phase coexistence: the horizontal supporting tangent touches both minima. Equal chemical potentials prevent net composition exchange, and equal pressure gives mechanical equilibrium at a flat interface.

Put $D={\partial }_t+U{\partial }_x$, write $p$ for perturbation pressure divided by ${\rho }_0$, and set $b=-g{\rho }^{\prime }/{\rho }_0$. The linearized Boussinesq approximation, incompressibility and material derivative giveTaking the curl eliminates pressure; differentiating the resulting vorticity equation in $x$ and using incompressibility gives $D{\nabla }^{2}w-U^{\prime \prime }{w}_x=b_{xx}$. A further application of $D$ therefore yieldsFor a stationary nonzero horizontal Fourier mode, $D=U{\partial }_x$, so cancellation of ${\partial }_x^2$ gives the stationary Taylor–Goldstein equationThis division requires $U\ne 0$ on the interval considered: a zero of $U$ is a critical level of an internal gravity wave. The mean profiles must be sufficiently smooth for the displayed derivatives, with background hydrostatic pressure and stable density stratification, $N^2>0$, for propagating internal gravity waves. The horizontally uniform component is excluded from the cancellation. For a horizontal wavenumber $k$, local vertical propagation additionally requires $N^2/U^2-U^{\prime \prime }/U>k^2$; this is distinct from the smoothness restrictions. If stability of the background against other disturbances is needed, the Miles–Howard theorem supplies the sufficient condition $N^2\ge (U^{\prime }{)}^2/4$, rather than a necessary condition for deriving the equation.

Use a common factor $e^{i(kx-\omega t)}$, with $k=k_0>0$ and $\omega ={\omega }_0$. For the upward propagating branches with $0<{\hat{\omega }}_2=\omega -k{U}_2<N_2$, defineThese are the inclinations of upward-sloping constant-phase lines of an internal gravity wave to the horizontal. Write the incident, reflected and transmitted complex velocity amplitudes as $A_0,A_r,A_t$. Incompressibility gives $(u_i,w_i)=A_0(\cos {\theta }_1,\sin {\theta }_1)$, $(u_r,w_r)=A_r(\cos {\theta }_1,-\sin {\theta }_1)$, and $(u_t,w_t)=A_t(\cos {\theta }_2,\sin {\theta }_2)$.

The jump conditions for stratified inviscid shear flow require a common interface displacement, not a common vertical velocity: $w=-i\hat{\omega }\eta$ on each side. Since the background mass density is continuous, pressure is continuous without a hydrostatic jump. The horizontal momentum equation gives $p/{\rho }_0=(\hat{\omega }/k)u$. HenceThese two matching equations determine internal-wave transmission across a velocity jump:The amplitude formula printed in the PDF is inconsistent with these material-interface matching conditions. In particular, identical layers have $A_t/A_0=1$ and $A_r/A_0=0$, whereas its expression labelled “reflected” equals one. The displayed results above distinguish reflection from transmission and retain the intrinsic-frequency factors required by the kinematic boundary condition and pressure balance.

For equal buoyancy frequencies and the specified opposing current, ${\theta }_1=\pi /6$, ${\theta }_2=\arcsin (3/4)$, andThe incident phase lines have slope $1/\sqrt{3}$, the reflected lines slope $-1/\sqrt{3}$, and the transmitted lines slope $3/\sqrt{7}$: they steepen above the interface. For the other specified current, ${\hat{\omega }}_2=0$. The transmitted vertical wavenumber then diverges and there is no regular propagating upper-layer wave of that frequency: this is the critical level of an internal gravity wave limit. Approaching it from ${\hat{\omega }}_2>0$ gives $A_r/A_0\to -1$ and vanishing transmitted vertical energy flux; setting the intrinsic frequency to zero directly is outside the regular matching calculation.

Let $u=ps{i}_z$ and $w=-ps{i}_x$, so mass conservation for the two-dimensional incompressible flow is automatic. Write $b$ for buoyancy and introduce the diffusive operatorsThe Linearized Boussinesq equations areTaking the curl of the momentum conservation equations eliminates the pressure and givesApply $D_{\kappa }$ and use $D_{\kappa }b=N^2{\psi }_x$. Since the constant-coefficient linear partial differential operators commute,This is the viscous-diffusive internal gravity wave equation.

The pressure force in the $x$-direction on a rectangle of dimensions $\delta x\times \delta y$ isThe shear stress is ${\tau }_{xy}=\mu u_y$, so the net viscous force isThere is no convective acceleration. Dividing the momentum balance by the area and taking the limit givesThe $y$-momentum balance gives $p_y=0$, so the pressure depends only on $x$.

At the upper wall,This is the shear exerted by the plate on the fluid in the positive $x$-direction. By action and reaction, the flow exerts on the top plate the stress

The same result follows from a force balance on a rectangle of length $\delta x$ and full height $h$. The pressure force is $-G_{0}h\delta x{e}_x$, the bottom shear vanishes, and equilibrium requires the top plate to exert $G_{0}h\delta x{e}_x$ on the fluid. Thus the fluid exerts $-G_{0}h{e}_x=-2\mu U{e}_x/h$ on the plate. These results form the Couette-Poiseuille flow with a stress-free stationary wall.

A velocity potential satisfying uniform flow at infinity and no radial flow through $r=a$ isThereforeOn the cylinder, $u_r=0$ and $u_{\theta }=-2U\sin \theta$. The steady Bernoulli equation, compared with the uniform flow at infinity, givesHencethe pressure in potential flow around a circular cylinder.

The steady Euler equations for an inviscid fluid areTaking the scalar product with $u$ and usinggivesThusThe Bernoulli function $H$ is constant along each streamline: fluid particles in steady flow exchange pressure, kinetic, and potential energy without changing their total mechanical energy density.

Let $y(t)$ be the downward displacement of the free surface from its initial level, and let $U$ be the speed in the tube. Conservation of volume givesThe free surface and the outlet are both at atmospheric pressure, and their vertical separation is $H+h_0-y$. Applying the Bernoulli equation between them, while retaining the small free-surface speed, givesThereforeThe surface reaches the upper tube end when $y=h_0$. Hence the draining time of a uniform tank through a siphon is

When $k_y=0$, $k_x$ is constant and the system becomesThusFor time dependence $e^{-i\omega t}$, the dispersion relation isExplicitly,and $\tilde{\delta }=2i{k}_x{\tilde{u}}_y/\Omega$. This is an axisymmetric inertial-acoustic wave: pressure supplies the $c_s^2{k}_x^2$ restoring term and epicyclic motion supplies the ${\Omega }^2$ term.