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For $|z|\ll r$, the gravitational potential has the Taylor expansionHence the vertical gravity is $-{\partial }_z\Phi \simeq -{\Omega }_K^{2}z$, where ${\Omega }_K^2=G{M}_{\star }/r^3$. Vertical hydrostatic equilibrium balances a pressure gradient of order $P/H$ against $\rho {\Omega }_K^{2}H$. Since $c_s^2\sim P/\rho$,The thin disk condition $H/r\ll 1$ is therefore equivalent to a highly supersonic orbital speed.

Hydrostatic equilibrium requiresFor the polytrope $p=K{\rho }^{1+1/n}$, the specific enthalpy isThus $h+\Phi$ is constant. Since $\rho =p=0$ at $r=R$, this constant is ${\Omega }^2{R}^2/2$, andfor $0\le r\le R$, with both fields zero outside the model star.

For a homologous star, hydrostatic equilibrium, mass conservation, and the ideal gas equation give the central scalingswhere $\mu$ is the mean molecular weight. Integrating the nuclear energy-generation law over a fixed homologous profile givesRadiative stellar structure gives independentlyEquating the two luminosities yieldsAt fixed zero-age composition, the stars are therefore homologous with

The effective temperature satisfies $L=4\pi R^2\sigma T_e^4$, so $T_e\propto M^{3/8}$. The zero-age main sequence consequently hasIt is a steep line rising toward high luminosity and high temperature on a Hertzsprung-Russell diagram.

For fully ionized hydrogen and helium with $Y=1-X$,while $\kappa \propto 1+X$. At fixed mass,and the corresponding radius relation is

At the pure-hydrogen zero-age point $X=1$,whereasHenceandAs hydrogen is consumed, $X$ falls, so both $L$ and $T_e$ rise. In the usual diagram with temperature increasing leftward, the evolutionary track initially moves upward and leftward from the zero-age main sequence.

Let $c_s^2\sim P_g/\rho$. The assumed magnetic pressure givesso the plasma beta obeys $\boxed{\beta =P_g/P_m\sim {\mathrm{Ma}}^{-1}}$. In the magnetically dominated regime, vertical hydrostatic equilibrium gives $P_m/\rho \sim {\Omega }^2{H}^2$. Combining this with $P_m/\rho \sim r\Omega c_s$ yields

Let $p=K{\rho }^{1+1/m}$. Hydrostatic equilibrium in the uniform gravitational field givesSubstitution and integration with $\rho =0$ at $z=0$ giveswhere $H$ fixes the normalization. Integrating the hydrostatic equation from $z$ to the free surface gives the pressure directly:This indeed has $p\propto {\rho }^{1+1/m}$ and $p(0)=0$, as required for a polytropic atmosphere.

A stellar polytrope of index $n$ obeysCombining hydrostatic equilibrium, $dP/dr=-Gm\rho /r^2$, with mass conservation, $dm/dr=4\pi r^2\rho$, gives the Lane-Emden equationRegularity and normalization at the stellar centre requireIntegrating the Lane-Emden equation from the centre then gives the enclosed mass

For $n=5$, direct substitution into the equation findsso $\beta =1/3$ and $\gamma =-1/2$. This profile has no finite first zero and hence has infinite radius, but its total mass converges:Its mean density over the full, infinite configuration is consequently zero.

For a perfect gas, $T\propto P/\rho \propto {\rho }^{1/5}$, so $T/T_c=\theta$. The luminosity from CNO cycle burning with $ϵ={ϵ}_0\rho T^{11}$ is thereforewhereThus $A$ is of order unity. The model is physically poor because the $n=5$ polytrope has infinite radius and zero mean density, while strongly temperature-sensitive CNO burning changes the thermal gradient and commonly creates convection. Real cores also have evolving composition, non-polytropic opacity and energy transport, and boundaries supplied by the surrounding star.

In stellar homology, two stars have identical dimensionless density, pressure, temperature, and luminosity profiles. Mass conservation, hydrostatic equilibrium, and the perfect-gas equation of state then giveThe chemical composition is initially fixed here, so $\mu$ is constant.

For opacity $\kappa ={\kappa }_0{\rho }^a{T}^b$, radiative diffusion gives the homology scalingNuclear burning with $ϵ={ϵ}_0\rho T^n$ instead givesEquating the two luminosities yieldsand the mass-luminosity relationFor usual main-sequence opacity and burning laws, $\eta >1$, so massive stars are much more luminous and exhaust a fuel supply proportional to $M$ in a time $t_{\ast }\propto M/L\propto M^{1-\eta }$.

At sufficiently high mass, radiation pressure and radiative acceleration become important. Hydrostatic balance requires the luminosity to remain below the Eddington luminosityThe upper envelope is linear in $M$, so the relation must flatten toward slope one; the corresponding nuclear lifetime approaches a weakly mass-dependent or roughly constant value rather than continuing the steep decline predicted by gas-pressure homology.

During core hydrogen burning, conversion of hydrogen into helium reduces the number of free particles per unit mass and raises the mean molecular weight. The core contracts and heats to retain pressure support, while the envelope expands and the luminosity generally rises. Homology ultimately fails because composition becomes strongly nonuniform, an inert helium core and hydrogen-burning shell appear, and the core and envelope acquire qualitatively different equations of state, transport regimes, and radial scales as the star leaves the main sequence.

The first equation is vertical hydrostatic equilibrium. In a thin disk, stellar gravity has vertical component $-{\Omega }^{2}z$, giving $dP/dz=-\rho {\Omega }^{2}z$. The second is local energy conservation: Keplerian shear has $rd\Omega /dr=-3\Omega /2$, so viscous dissipation per unit volume isThe third equation is radiative diffusion. With radiation energy density $a_{\mathrm{R}}{T}^4=4\sigma T^4/c$, an optically thick medium carrieswhich rearranges to the stated gradient. Thomson scattering makes $\kappa$ approximately independent of density and temperature. The final equation adds perfect-gas pressure $k\rho T/({\mu }_m{m}_p)$ and radiation pressure $a_{\mathrm{R}}{T}^4/3=4\sigma T^4/(3c)$.

Here an elastic fluid means a gas or vapour. Such a fluid is mobile, occupies the accessible space of its container, and is appreciably compressible. Its pressure acts normally on a confining surface. In static equilibrium the stress at a point is isotropic: a fluid has no equilibrium resistance to a sustained shear deformation. Pressure can still vary with height through hydrostatic equilibrium; isotropy does not mean the same pressure at every position.

Reducing a confined gas's volume at fixed temperature normally raises its pressure. Heating at fixed volume normally raises the pressure, and gases exchange material through diffusion. Their particles are in continual thermal motion, whose momentum transfer gives a microscopic account of gas pressure in the kinetic theory of gases.

For a dilute gas away from condensation, the approximate ideal gas law isIt contains Boyle's law at fixed temperature and the corresponding temperature-expansion laws. These simple proportionalities are approximations, not exact properties of every elastic fluid at every density. Intermolecular interactions matter in a dense gas, and a vapour can condense; the classification does not promise that every sample remains gaseous under arbitrary compression or cooling.

For an isolated approximately spherical mass distribution outside its luminous extent, retain the gravitational force $GMm/r^2$ and use the specified low-acceleration inertial law. circular motion then givesThe radius cancels when enclosed luminous mass is effectively constant. This is the asymptotic modified Newtonian dynamics result associated with the baryonic Tully-Fisher relation; an arbitrary extended disk cannot everywhere be replaced exactly by a point mass.

At the disk edge,Using $G=6.67\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$ givesThis is an order-of-magnitude isolated, low-acceleration estimate. The edge acceleration is below $a_0$, but not extremely small compared with it; a full interpolation law, disk geometry and an external gravitational field can affect a more precise mass fit.

For the spherical mass profile for a flat rotation curve, assume spherical symmetry where the halo dominates. A flat rotation curve gives $M(<r)=v^{2}r/G$, and differentiating $M(<r)=4\pi {\int }_0^r\rho (s)s^{2}ds$ yieldsHere $\rho$ is the density of the spherical gravitating model; it is not an exact deprojection of a thin luminous disk or a proof that the central density must diverge. At the outermost stated measured radius,This gives the requested galaxy-plus-halo mass estimate if the model is truncated at the outermost observed radius $2R$. The observation itself constrains enclosed mass within $2R$; any additional outer halo remains unconstrained. Extending $\rho \propto r^{-2}$ to infinity would give divergent mass. Subtracting the MOND-inferred luminous mass to obtain a Newtonian halo mass would mix independent model-dependent estimates without an independently measured luminous mass.

In the Bullet Cluster collision, most observed baryonic mass is hot gas. ram pressure slows and shocks the gas, while the galaxies pass through much more freely. weak gravitational lensing maps the projected gravitational mass without requiring that the gas be in hydrostatic equilibrium. The dominant lensing peaks lie closer to the galaxies than to the displaced X-ray gas. This supports a dominant weakly collisional dark matter component under the usual gravity model. It is stronger evidence than a mass-to-light discrepancy alone because distinct components separate during the same encounter. The inference needs background-source shape/redshift calibration, projection control and uncertainty in the lensing reconstruction; it does not identify a WIMP species or by itself exclude every modified-gravity theory.^[ref]

For dark-matter direct detection, consider XENON100 at Gran Sasso. A halo WIMP can undergo elastic scattering from a xenon atomic nucleus, leaving a recoil of only a few to tens of keV. The expected rate is very low, so a large target, low threshold and long exposure must be combined with background rejection. The nuclear-recoil kinematics of WIMP scattering gives $E_{R,\max }=2{\mu }_{\chi N}^2{v}^2/m_N$ in the nonrelativistic limit, where ${\mu }_{\chi N}$ is the reduced mass and masses and velocities use consistent units. This explains why low-energy response calibration matters.

The dual-phase xenon detector measures prompt particle scintillation S1 and delayed electroluminescence S2 from ionization electrons extracted into gas. Their time separation gives interaction depth; the S2 light pattern supplies transverse position. An inner fiducial volume uses xenon's self-shielding to suppress external backgrounds. The S2/S1 distribution statistically separates nuclear recoils from electron recoils; it is not perfect event-by-event identification. Underground siting reduces cosmic-ray backgrounds, while passive shielding, low-radioactivity components, xenon purification and a surrounding active detector veto reduce other events.^[ref]

Neutrons can also make nuclear recoils, so charge-to-light discrimination alone cannot eliminate them. Multiple-scatter rejection, shielding and background modelling remain necessary. Gamma-source and neutron-source calibrations map electron- and nuclear-recoil distributions, energy scale and selection acceptance. Position-dependent light collection and charge losses need correction; photon-counting fluctuations and uncertain low-energy yields limit threshold sensitivity. Blind event selection and independent sidebands guard against tuning cuts to apparent candidates. The 224.6-live-day, 34 kg search found two benchmark candidates against about one expected background event, consistent with background. Its interaction limit therefore depends on the assumed halo distribution, WIMP mass and nuclear response; it is not a detection or an exclusion of all dark matter.^[ref]

Flow visualization and the idealization. A streamline is everywhere tangent to the instantaneous velocity field at one fixed time. A streakline is the locus, at one observation time, of particles that previously passed through a specified injection point. A marked particle's own trajectory is a pathline. These coincide geometrically in steady flow, but need not do so in an unsteady draining flow.

The intended ideal liquid is approximately incompressible and inviscid: viscosity and associated shear stresses are negligible on the scales and times being considered, the Mach number is small, and the density is effectively constant. This does not require zero vorticity; the initially rotating state is already a counterexample. It also does not justify ignoring viscosity in an arbitrarily thin boundary layer or at a real drain.

Spiralling flow and angular momentum. In an axisymmetric inviscid flow, pressure and gravity exert no torque about the vertical axis. A parcel therefore conserves axial specific angular momentum,The late, shallow remnant in the intended draining picture is supplied by parcels initially near the outer rim, so $r_0\simeq R$. The slow-drain angular-momentum amplification then givesThe assumption about the parcels' origin is necessary: torque conservation by itself does not assign $r_0=R$ to every parcel at every time.

A continuous dye feed near the outer free surface produces inward spirals that wind increasingly rapidly near the drain. The figure uses a late quasisteady model, $u_r=-D/r$ and $u_{\varphi }=C/r$ with positive constants $D,C$, giving $\varphi ={\varphi }_0-(C/D)\ln (r/r_0)$. Its streaklines and streamlines approximately coincide over an observation interval short compared with the overall draining time. Strictly, individually marked particles at the initial surface trace pathlines, not a continuously injected streakline.

The free-surface depression. Let $z=H(r)$ be the surface. Slow radial draining makes radial inertial corrections small compared with the azimuthal centripetal acceleration. The radial Euler equation and vertical hydrostatic equilibrium giveThus $g{H}^{\prime }(r)=u_{\varphi }^2/r=C^2/r^3$, with $C={\omega }_0{R}^2$. Integrating from $r$ to the rim givesFor $r\ll R$ this becomesThe surface rises outward because the inward pressure gradient force must sustain the circular motion. The dimensions are metres, and the $1/r^2$ profile is not valid inside the finite drain or vortex core.

The terrestrial estimate. Use the vertical component of Earth's rotation as the ideal initial angular speed, ${\omega }_0\simeq {\Omega }_E\sin \lambda$. At Cambridge, taking $\lambda \simeq 52.2^{\circ }$ and ${\Omega }_E=7.292\times 10^{-5}{\mathrm{s}}^{-1}$, this gives ${\omega }_0\simeq 5.76\times 10^{-5}{\mathrm{s}}^{-1}$. Hence, for $R=1\mathrm{m}$, $r=0.01\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$,Taking the whole terrestrial angular speed rather than its local vertical projection would give about $2.7\mu \mathrm{m}$. The Cambridge estimate uses the projection appropriate to the vertical vortex. This very small head assumes terrestrial rotation supplies the only initial swirl; imposed residual rotation would change it quadratically.

Curl of the azimuthal velocity. For axisymmetric $u_{\varphi }=r\omega (r)$ with no vertical dependence, the rotational part of the vorticity is vertical:The ideal quasisteady radial sink has no curl outside its core. The source's symbol $\Omega$ denotes this vertical vorticity component, not the fluid's angular velocity. In general its magnitude is the absolute value of the bracket.

The two limiting states. For initial uniform rotation, $\omega ={\omega }_0$, whereas the late approximation has $\omega =C/r^2$. ThereforeThe late flow can have large angular velocity and nonzero circulation while being locally irrotational. The drain core is excluded from the domain of the $1/r$ velocity field.

What is actually conserved. Inviscid does not mean that each parcel's vorticity magnitude is fixed in three dimensions. For an incompressible barotropic liquid with conservative body forces, the vorticity equation isVertical compression of a material column changes its vorticity through vortex stretching, even while its volume is conserved by horizontal expansion. In the columnar shallow-layer limit, vorticity dilution in a shallow draining layer givesThus a remnant much shallower than its original column can have much smaller vertical vorticity. At finite depth there is generally a residual value; the exactly irrotational late profile is a leading limiting approximation, not an exact parcelwise conversion of nonzero vorticity to zero in finite time.

Kelvin's circulation theorem conserves circulation around a material loop, not around a fixed spatial circle. A loop whose original radius is near $R$ has circulation about $2\pi {\omega }_0{R}^2$, consistent with the late $u_{\varphi }=C/r$ field. Fluid also leaves through the drain, so the remaining tank contents are not a fixed material system. If the calculation is expressed explicitly in the Earth-fixed rotating frame, the corresponding shallow-water potential vorticity is $(\zeta +f)/H$, with $f=2{\Omega }_E\sin \lambda$; the conserved quantity involves absolute vorticity.

Newton's second law takes its familiar form $m\mathbf{a}=\mathbf{F}$ in an inertial frame. Axes fixed to a turning apparatus or to the Earth form a rotating reference frame. Extra inertial forces account for the acceleration of these coordinates. Their values depend on the chosen reference frame; physical interactions such as gravity, tension and contact forces must still be included separately.

Let the rotating origin have inertial acceleration $\mathbf{A}$, and let its axes have angular velocity $\mathbf{\Omega }(t)$. For any vector $\mathbf{q}$, the rotating-frame derivative formula isEach rotating basis vector changes by $\mathbf{\Omega }\times {\mathbf{e}}_i$, which proves this rule by expanding $\mathbf{q}$ in that basis. Apply it twice to the position relative to the moving origin. With ${\mathbf{v}}^{\prime }$ and ${\mathbf{a}}^{\prime }$ measured in the rotating axes,Consequently the equation in the rotating reference frame isThe additional terms are the translational inertial force, Coriolis force, Euler force and centrifugal force, respectively. The last two simplify when the angular velocity is constant: the Euler force vanishes, while the other rotation terms remain.

The centrifugal force points away from the rotation axis. If ${\mathbf{r}}_{\perp }$ is the perpendicular displacement from that axis, thenFor uniform rotation it has potential energyA mass at rest relative to a turntable can therefore be described as having an inward tension or friction balancing its outward centrifugal force. In an inertial frame, the inward physical force instead produces the required centripetal acceleration. These are two descriptions of the same motion; one must not insert the rotating-frame outward term into the inertial equation as an additional physical interaction.

A rotating liquid gives a useful quantitative example. For a liquid in hydrostatic equilibrium relative to a vessel rotating uniformly about the vertical axis,Along the free surface the pressure is constant, so ${\Omega }^{2}sds-gdz=0$ andThe surface is a elliptic paraboloid. This is the balance of gravity and the centrifugal force in the rotating description. On the Earth, the same outward acceleration reduces effective gravity, most strongly at the equator, and contributes to the planet's equatorial bulge.

The Coriolis force depends on relative motion:It is perpendicular to both $\mathbf{\Omega }$ and ${\mathbf{v}}^{\prime }$, vanishes for motion along the rotation axis, and reverses when the relative velocity reverses. The dot product givesThus the Coriolis force does no work in the rotating reference frame: it changes the direction of motion rather than directly changing the rotating-frame kinetic energy. For example, on a horizontal turntable rotating anticlockwise as viewed from above, an outward-moving slider has a clockwise Coriolis acceleration, whereas an inward-moving slider has an anticlockwise one. The deflection can be observed by comparing a laboratory camera with a camera rotating with the table.

For slow horizontal motion on the Earth at latitude $\lambda$, keep the vertical component ${\Omega }_E\sin \lambda$ of its angular velocity. In east-north coordinates $(u,v)$ the horizontal Coriolis acceleration isIt deflects motion to the right in the northern hemisphere and to the left in the southern hemisphere. In large-scale atmospheric and oceanic motion it can balance a horizontal pressure gradient, giving geostrophic balance. With $\hat{\mathbf{z}}$ upward, the steady horizontal equation without friction isThe resulting velocity runs along lines of equal pressure. This approximation requires rotation to dominate local inertial changes; it is inappropriate close to the equator where $f$ tends to zero.

The Foucault pendulum gives another terrestrial example. In the small-angle approximation, its horizontal coordinates satisfyTo first order in the small ratio ${\Omega }_E/{\omega }_0$, the oscillation plane precesses relative to the ground at angular rate $-{\Omega }_E\sin \lambda$: clockwise in the northern hemisphere, zero at the equator, and one revolution per sidereal day at a pole. The Coriolis force explains this changing orientation without requiring energy input. Together these examples show why inertial forces are useful whenever the apparatus or environment supplies naturally rotating coordinates.

Put the rigid lid at $z=0$, the interface at $z=-h(x,y,t)$ and ${\rho }_1={\rho }_0-\Delta \rho$. Assume hydrostatic vertical balance and a resting deep layer. Its pressure is $p_2=C-{\rho }_{0}gz$, with horizontally uniform $C$. Pressure continuity at the interface then fixes the upper-layer pressure:Hence its horizontal pressure gradient at fixed $z$ is ${\nabla }_h{p}_1=\Delta \rho g{\nabla }_{h}h$, independent of depth. Taking the horizontal momentum equation and a vertically uniform upper-layer velocity givesThe exact density denominator for upper-layer acceleration is ${\rho }_1$; the familiar ${\rho }_0$ denominator is the small-density-contrast Boussinesq approximation. The restoring pressure is weaker than for a free surface because only the density difference contributes.

The horizontal shallow-layer equations use the vertical component of planetary rotation: $f=2\Omega \sin \phi$. Thus the source's vector notation is interpreted as the projected vector $f\mathbf{k}$ after the traditional approximation. Taking the full magnitude $2\Omega$ as a midlatitude Coriolis parameter would discard the latitude dependence. Together with layer-volume conservation, the model has ${\partial }_{t}h+{\nabla }_h\cdot (h\mathbf{u})=0$.

Both descriptions are depth-averaged hydrostatic shallow-layer models. In a one-layer free-surface shallow water equations, a homogeneous layer overlies a rigid bottom and its upper surface moves. Surface elevation changes give a restoring acceleration based on full gravity $g$, with external-wave speed $\sqrt{g{H}_0}$. Barotropic coastal storm surges, long surface gravity waves or a basin seiche are examples.

In the rigid-lid reduced gravity model, the movable boundary is a density interface beneath the active upper layer, while the much deeper lower layer remains at rest. The internal interface mode instead has speed $\sqrt{g^{\prime }{h}_0}$ and represents, for example, thermocline displacements and an upper-ocean baroclinic eddy. The mathematical shallow-water form is retained with $g$ replaced by $g^{\prime }$ and with the appropriate interface-depth convention. A freely moving two-layer surface would support both external and internal modes; the rigid lid filters out the fast external mode, and the resting abyss removes its independent dynamics.

For a unit mass of ideal gas, the first law of thermodynamics in enthalpy form is $\delta Q=c_{p}dT-vdp$. A reversible adiabatic displacement has $\delta Q=0$, and the ideal gas law gives $v=RT/p$. ThusThe atmospheric profile approaches this dry adiabat when convection efficiently mixes potential temperature and there is no condensation or other strong diabatic source. The first law by itself determines a parcel path, not every possible environmental temperature profile; the convective mixing assumption is needed. Hydrostatic balance gives the corresponding height lapse rate $dT/dz=-g/c_p$.

For rigid linear molecules with translation and rotation active but vibration frozen, the molecular heat capacities are $C_V=(5/2)R_u$ and $C_P=(7/2)R_u$, hence $R/c_p=2/7$. Earth is dominated by the diatomic gases nitrogen and oxygen. Mars is dominated by carbon dioxide, which is triatomic but linear and has the same two rotational degrees of freedom. In the rigid-molecule approximation it therefore has the same ratio. The value is not exact for carbon dioxide when its bending vibrations contribute appreciably to heat capacity, and variable heat capacities give a nonconstant dry-adiabatic exponent. Here $\gamma$ denotes $R/c_p$, not the distinct adiabatic index $c_p/c_v$.

A hydrostatic pressure interval contains $dp/(gM)$ moles of background gas per area. Its absorbing column is $x(p)dp/(gM)$, so for a nadir rayThe assumed Planck-temperature power and dry adiabat give $B(p)/B_0=[T(p)/T_0{]}^{3.5}=p/p_0$. To first order in optical depth, $-d\mathfrak{T}\simeq d\tau$, and the radiance solution becomesFor a uniform mixing ratio this also evaluates to $L\simeq B_0[1-k_{0}x{p}_0/(6gM)]$. Its total optical depth is ${\tau }_0=k_{0}x{p}_0/(2gM)$, so the deficit is $B_0{\tau }_0/3$.

A pressure-resolved perturbation defines the infrared absorber sensitivity kernel by $\delta L=\int K_p(p)\delta x(p)dp$, withIt is zero at the top, negative in the interior, and zero again at the surface. The magnitude is largest where $d{K}_p/dp=0$, namelySign and coordinate qualification. Literally the maximum of the signed kernel is zero at the endpoints; $p_0/2$ is its minimum and the maximum of its absolute value. The derivative with respect to a single vertically uniform $x$ is the integrated constant $-k_0{B}_0{p}_0/(6gM)$ and has no pressure location. For equal-width logarithmic pressure perturbations the kernel is instead $K_{\log p}=p{K}_p$ and its magnitude peaks at $2p_0/3$. The stated pressure-interval convention is therefore essential.

The low top-of-atmosphere sensitivity reflects the small pressure-broadened absorption coefficient and small molecular column per $dp$ there; the low surface sensitivity reflects small thermal contrast. The printed phrase that temperature decreases with pressure reverses the actual stipulated relation: with positive exponent, temperature decreases as pressure decreases upward. The derivation uses the explicit adiabatic formula.

Take an isothermal hydrostatic reference atmosphere, ${\rho }_0(z)={\rho }_s{e}^{-z/H}$, and linearize about rest. The horizontal acceleration and pressure-gradient force retain the equatorial Coriolis parameter $f\simeq \beta y$, with $\beta =2\Omega /a$. The traditional approximation retains only its vertical component. Quadratic perturbation advection is neglected. The vertical acceleration is balanced by perturbation buoyancy and pressure gradient after removing the reference hydrostatic balance.

The continuity constraint is the anelastic approximation, $\nabla \cdot ({\rho }_0\mathbf{u})=0$, giving $u_x+v_y+w_z-w/H=0$. It retains stratification-driven expansion while filtering acoustic density fluctuations. It is therefore not the full compressible continuity equation. Adiabatic displacement in a stably stratified background gives $b_t+N^{2}w=0$, with $N$ the Brunt–Väisälä frequency; upward displacement develops restoring negative buoyancy.

Pressure convention matters. With the entropy-buoyancy equation just described, a consistent reduced-pressure form writes the pressure acceleration as $-\nabla \Phi$, where $\Phi =p^{\prime }/{\rho }_0$. Its vertical term is $-{\partial }_z(p^{\prime }/{\rho }_0)$, not simply $-{\rho }_0^{-1}{p}_z^{\prime }$. The original source explicitly specifies variable ${\rho }_0(z)$ but prints the latter. Those differ by $p^{\prime }/({\rho }_{0}H)$, so the two cannot silently be equated. Parts b and c display the consequence and then solve the explicitly corrected reduced-pressure model that yields the requested wave equation. This is a mathematical inconsistency in the printed set, not missing source material.

Work with the explicitly corrected reduced-pressure system in b. Write $d=1/(2H)$ and use the complex wave amplitude, taking real parts at the end. The vertical factor satisfies $({\partial }_z^2-H^{-1}{\partial }_z)w=-(m^2+d^2)w$, so the wave equation givesFor a Kelvin wave, continuity, zonal momentum and the buoyancy equation then determine all polarizations:In particular $u,w,b,\Phi$ have amplitude proportional to $e^{z/(2H)}$, whereas pressure $p^{\prime }$ has amplitude proportional to $e^{-z/(2H)}$. Directly substituting these expressions into vertical momentum gives ${\Phi }_z=-i\omega (m^2+d^2)w/k^2=i(\omega -N^2/\omega )w$, verifying both the sign and the dispersion relation.

Meridional momentum is $\beta yu+{\Phi }_y=0$. Since $\Phi =cu$, it integrates to a Gaussian $\exp [-\beta y^2/(2c)]$. ThusThe eastward intrinsic branch $c>0$ is selected by decay away from the equator; choosing $c<0$ would make the meridional amplitude grow without bound. Although the squared dispersion relation has both frequency signs, a meridionally trapped Kelvin wave has this one phase-propagation direction.

For short vertical wavelengths, $m^2\gg k^2+d^2$, so ${\omega }^2\simeq N^2{k}^2/m^2$, $c\simeq N/|m|$ and $L_D^2\simeq N/(\beta |m|)$. The wave is a slowly oscillating, hydrostatic internal gravity wave with strong vertical variation and relatively narrow trapping. Here $u\simeq -(m/k)w$ and $p^{\prime }/{\rho }_0\simeq -(\omega m/k^2)w$; the buoyancy expression above remains valid.

For long vertical wavelengths, $|m|\ll \sqrt{k^2+d^2}$, the exact limiting speed is $c=N/\sqrt{k^2+d^2}$. To obtain the stated height-controlled limit also requires long horizontal wavelength, $|k|H\ll 1$. With both $|m|H\ll 1$ and $|k|H\ll 1$,The finite density scale height prevents the hydrostatic $1/|m|$ phase speed from diverging as $m\to 0$. The polarizations tend to $u\simeq -iw/(2Hk)$, $\Phi \simeq -iNw/k$ and $b\simeq -iNw/(2Hk)$. Long vertical wavelength alone at fixed large $kH$ would not justify $c=2NH$. In the opposite horizontally short limit the frequency approaches $N$ instead. The growth $e^{z/(2H)}$ compensates decreasing reference density in wave-energy density; a radiating wave train is a local modal description, not a finite-total-energy eigenfunction over an infinite vertical domain.

Take the convection to maintain a dry, nearly reversible adiabat, with specific gas constant $R$. Hydrostatic equilibrium and an ideal-gas equation of state will be used separately to convert pressure to height. For an adiabatic displacement, the first law of thermodynamics in enthalpy form givesThe adiabat with linear heat capacity follows by integrating from the surface: this gives the monotone implicit temperature–pressure relationAn explicit form, using the positive real Lambert W branch, isAs $B\to 0$, $T=T_s(p/p_s{)}^{R/A}$. At fixed $p<p_s$,Larger heat capacity therefore keeps the upper atmosphere warmer at the same pressure. Equivalently $d\ln T/d\ln p=R/(A+BT)$ becomes smaller. The original normalized sketch below shows the resulting family of profiles; no particular gas constants or surface temperature are assumed there.

The height equations areIntegrating the latter givesThe physically selected root starts at $T_s$ and decreases. Its formal upper boundary isAt that height $T\to 0$ and $p=p_s(T/T_s{)}^{A/R}{e}^{B(T-T_s)/R}\to 0$. The $B=0$ limit is $z_{\ast }=A{T}_s/g$. This finite-height zero is a property of the assumed polytropic-like model, not a claim that a real atmosphere reaches absolute zero: molecular heat capacities, composition, transport and the continuum ideal-gas description cease to support that extrapolation.

Interpret the requested mass path as the vapour column; no condensed-water loading is prescribed. Hydrostatic equilibrium gives $dp=-\rho gdz$, so if $q_B={\rho }_B/\rho$ is vapour mass fraction,For the mixture $q_B=ϵe_s/[p+(ϵ-1)e_s]$. Under the dilute conditions of part (a), $q_B\simeq ϵe_s/p$. Introduce $\zeta =-\ln (p/p_s)$, so $T=T_s{e}^{-x\zeta }$ and $-dp=pd\zeta$. Integrating the constant-$L$ Clausius-Clapeyron equation relative to the surface givesThereforeThe explicit $p_s$ cancels: decreasing vapour mass fraction in a denser dry column compensates its greater total mass. This cancellation belongs to the dilute, dry-adiabatic approximation, and is not a general property of moist atmospheres. With suspended condensate, an additional and unspecified condensate mass path would have to be added.