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In an isothermal atmosphere with constant density scale height $H$, the equation $\delta {\rho }_{tt}-c^2\delta {\rho }_{zz}-(c^2/H)\delta {\rho }_z=0$ loses its first spatial derivative under $\delta \rho =e^{-z/(2H)}\Psi$. Direct differentiation gives ${\Psi }_{tt}-c^2{\Psi }_{zz}+c^2\Psi /(4H^2)=0$. Thus a plane wave has dispersion relation ${\omega }^2=c^2{k}^2+{\omega }_c^2$, with real acoustic cutoff frequency ${\omega }_c=c/(2H)$. The cutoff term is positive; a minus sign would instead predict growing sufficiently long spatial modes. Vertical propagation requires $|\omega |\ge {\omega }_c$ within this ideal constant-coefficient model.

This is the atmospheric instance of a scale height.

For an isothermal ideal-gas atmosphere, $H=k_{B}T/(\mu m_{H}g)$. A strong transmission feature spanning $N_H$ scale heights has approximate transit-depth amplitude $2R_p{N}_{H}H/R_{\ast }^2$.

For the density jump of a magnetic pressure discontinuity in an isothermal atmosphere, with $0<|B_{+}|<|B_{-}|$, the fastest Rayleigh-Taylor instability has $m=0$ and $|k|=K_{\ast }=(B_{-}^2-B_{+}^2)/(4B_{+}^{2}H)$. Its vertical decay length is $K_{\ast }^{-1}$. Hence localization relative to both common scale heights requires $4B_{+}^2\ll B_{-}^2-B_{+}^2$. This also makes its growth speed small compared with the isothermal sound speed. If $B_{+}=0$, the ideal sharp-interface dispersion relation instead has unbounded growth as $|k|\to \infty$: there is no finite maximizing mode.

A plane-parallel magnetostatic equilibrium with constant downward gravitational acceleration $g$, horizontal magnetic field, constant isothermal sound speed $c_s$, and constant Alfvén speed $v_a$ hasIndeed the magnetohydrodynamic total pressure is $(c_s^2+v_a^2/2)\rho$, and its vertical derivative is $-g\rho$. The magnetic pressure therefore increases the scale height above that of an unmagnetized atmosphere.

The exponential factor removes the first spatial derivative. Apply the chain rule explicitly:while ${\partial }_t^2\Delta \rho =e^{-z/(2H)}{\Psi }_{tt}$. Substituting into the supplied density equation, the terms proportional to ${\Psi }_z$ cancel, leavingThis is the atmospheric acoustic cutoff transformation. There is a sign error in the target equation in the original PDF, as well as in the transcription: the last term must be $+{\omega }_0^2\Psi$ for real positive ${\omega }_0$. If the displayed minus sign is retained literally, coefficient matching would require ${\omega }_0^2=-c^2/(4H^2)$, so it would not define a real characteristic oscillation period.

For the corrected equation, a plane wave $\Psi \propto e^{i(kz-\omega t)}$ has dispersion relationThus ${\omega }_0$ is the acoustic cutoff frequency: vertically propagating acoustic waves have $|\omega |\ge {\omega }_0$. The scale height and sound speed have units metres and metres per second, respectively, so ${\omega }_0$ has units ${\mathrm{s}}^{-1}$.

For an order-of-magnitude estimate in the lower terrestrial atmosphere, take $H\simeq 8.0\mathrm{km}$ and $c\simeq 340\mathrm{m}{\mathrm{s}}^{-1}$. These are representative atmospheric values, not additional data specified by the paper. They giveThe isothermal atmosphere approximation is local: a real atmospheric temperature profile and the thermodynamic response of the disturbance modify the precise cutoff and period.

The Gaussian integral over the full vertical velocity line is $\sqrt{2\pi }{\sigma }_z$. Thus the mass density of the self-gravitating isothermal slab isNeglecting radial derivatives makes the Poisson equation for Newtonian gravity ${\Phi }_{zz}=4\pi G\rho (z)$. Substituting $\Phi ={\sigma }_z^2\varphi$ and $z=z_0\zeta$, with $z_0^2={\sigma }_z^2/(8\pi G{\rho }_0)$, givesMultiply by ${\varphi }^{\prime }$ and integrate. The central value and symmetry under $z\mapsto -z$ fix the constant of integration:On the positive-$\zeta$ side, Newtonian gravity makes ${\varphi }^{\prime }\ge 0$. Put $y=e^{\varphi /2}$; then $4(y^{\prime }{)}^2=y^2-1$ with $y(0)=1$. Integration gives $2\mathrm{arcosh}y=\zeta$, and even reflection supplies the other side. ThereforeDirect differentiation verifies both the equation and central conditions. Uniqueness of the smooth initial value problem rules out other branches meeting those conditions. The surface density is $4{\rho }_0{z}_0$, and the conventional slab scale height is $2z_0$; the factor of two is only a definition of the length scale.

Write $c_T^2=\mathcal{R}{T}_0$ for the squared isothermal sound speed, distinguishing the gas constant $\mathcal{R}$ from a Rayleigh number, and put $H=c_T^2/g$. A uniform magnetic field has no bulk Lorentz force, so hydrostatic equilibrium and the ideal gas equation giveAt the interface, it is magnetohydrodynamic total pressure, not gas pressure alone, that is continuous:Define ${\Delta }_B=B_1^2-B_2^2>0$. The magnetic pressure discontinuity in an isothermal atmosphere therefore producesIn particular, the upper gas is denser at the interface. Both halves have the same scale height $H$, despite their different pressure normalizations.

For the disturbance calculation, write ${\rho }_{-}=\rho$, ${\rho }_{+}=\rho +\delta \rho$, where $\delta \rho ={\Delta }_B/(2{\mu }_0{c}_T^2)$, and ${\Sigma }_{\rho }={\rho }_{-}+{\rho }_{+}$. Work to leading order in the prescribed short-vertical-scale approximation: the bulk motion is locally incompressible potential flow, while the isothermal equation determines the equilibrium density jump. Its scale separation and small compressive correction will be checked at the maximizing mode.

Let the interface displacement be $h{e}^{st+ikx+imy}$ and $K=\sqrt{k^2+m^2}>0$. Using displacement rather than velocity potentials, write ${\mathbf{\xi }}_i=\nabla {\chi }_i$ and ${\mathbf{u}}_i=s{\mathbf{\xi }}_i$. Incompressibility and irrotational flow give $\Delta {\chi }_i=0$. Decay away from the interface and continuity of its normal displacement fixwith the common horizontal and temporal exponential understood. Both satisfy ${\partial }_z{\chi }_i(0)=h$. Thus the disturbance has vertical extent of order $K^{-1}$.

In ideal magnetohydrodynamics, magnetic flux freezing gives ${\mathbf{b}}_i=({\mathbf{B}}_i\cdot \nabla ){\mathbf{\xi }}_i$ for these locally incompressible displacements. If ${\pi }_i=p_i^{\prime }+{\mathbf{B}}_i\cdot {\mathbf{b}}_i/{\mu }_0$ is the total-pressure perturbation, linear momentum isSince $({\mathbf{B}}_i\cdot \nabla {)}^2=-({\mathbf{k}}_h\cdot {\mathbf{B}}_i{)}^2$ on the mode, integration in each half-space yieldsAt the displaced interface, continuity of Lagrangian magnetohydrodynamic total pressure requires ${\pi }_1-{\rho }_{-}gh={\pi }_2-{\rho }_{+}gh$. Inserting the two potentials gives the orthogonal-field magnetic Rayleigh-Taylor dispersion relation:The first term is the destabilizing Rayleigh-Taylor instability of the heavier upper gas; the second is stabilizing magnetic tension. A negative right-hand side gives an oscillatory mode rather than growth.

For fixed $K$, the magnetic penalty isIt is minimized by $m=0$: the wavevector is along the weaker upper magnetic field and perpendicular to the stronger lower field. If $B_2\ne 0$, define $A=g\delta \rho /{\Sigma }_{\rho }$ and $C=B_2^2/({\mu }_0{\Sigma }_{\rho })$. The growth law for this orientation is $s^2=AK-C{K}^2$, a concave quadratic with maximum at $K_{\ast }=A/(2C)$. HenceFor this orientation the unstable interval is $0<K<2K_{\ast }$; the maximum lies halfway along it.

The localization condition for the fastest orthogonal-field interface mode follows directly from the vertical decay length:This is the condition for ${\ell }_{\ast }\ll H$ on both sides. It also controls the locally incompressible approximation, sinceAcoustic adjustment is then fast compared with growth, so bulk density perturbations are higher-order corrections. If this localization condition fails, the constant-background approximation cannot be used to identify the physical fastest mode.

Finally, the printed assumptions allow $B_2=0$. In that case $m=0$ eliminates magnetic tension on both sides and gives $s^2=g\delta \rho |k|/{\Sigma }_{\rho }$. There is no finite maximum growth rate or maximizing wavenumber in the ideal sharp-interface model: growth is unbounded as $|k|\to \infty$. The finite maximum and localization formula above therefore require $0<|B_2|<|B_1|$.

In the growth curves, $\theta$ is the angle of the horizontal wavevector from the $x$-axis. The illustrated ratio $B_1^2/B_2^2=101$ gives $K_{\ast }H=25$, and shows how even a small rotation toward the stronger magnetic field lowers the maximum growth.

Let ${\rho }_0$ denote the midplane mass density and choose the symmetry plane at $z=0$. Isothermal hydrostatic equilibrium and the Poisson equation for Newtonian gravity giveso $c_s^2(d^2/d{z}^2)\ln \rho =-4\pi G\rho$. For $\rho ={\rho }_0{\mathrm{sech}}^2(z/H)$, direct differentiation yieldsThus the profile is a self-gravitating isothermal slab when $H^2=c_s^2/(2\pi G{\rho }_0)$. Integrating over both sides gives $\Sigma =2{\rho }_{0}H$, henceThe corresponding potential is ${\Phi }_0(z)-{\Phi }_0(0)=2c_s^2\ln \cosh (z/H)$, whose gradient vanishes at the midplane and tends to $\pm 2\pi G\Sigma$ far from the slab. The profile parameter $H$ is not its asymptotic exponential mass density scale height: that is $H/2$.

In the thin-sheet regime, the gas scale height is $h_g(R)={\sigma }_g^2(R)/(2\pi G{\Sigma }_{\ast }(R))$. Hence radius-independent gas velocity dispersion givesFor an exponential galactic disk, ${\Sigma }_{\ast }(R)={\Sigma }_{\ast ,0}{e}^{-R/R_d}$, soThe gas therefore exhibits galactic disk flaring as stellar surface density falls outward. Stellar galactic disks, by contrast, commonly have roughly constant thickness over their main optical region; applying the same scale height relation to the stars requires ${\sigma }_{z,\ast }^2(R)\propto {\Sigma }_{\ast }(R)$, or ${\sigma }_{z,\ast }\propto e^{-R/(2R_d)}$ for an exponential galactic disk. This comparison does not follow by imposing a radially constant stellar velocity dispersion: that extra assumption would predict stellar galactic disk flaring too. The constancy used in the vertical calculation is constancy with height locally, not a mandatory constancy with radius.

Assume a steady, locally plane-symmetric galactic disk, with negligible radial tilt term in the vertical Jeans equation. Write $\rho (z)$ for the stellar mass density, and $K_z={\partial }_z\Phi$ for the upward gradient of the Newtonian gravitational potential; the actual vertical gravitational acceleration is $-K_z$. Constant vertical velocity dispersion givesIntegrating the plane-parallel Poisson equation for Newtonian gravity from the symmetry plane gives $K_z(z)=4\pi G{\int }_0^z\rho (z^{\prime })d{z}^{\prime }$. Outside most of the mass, this tends to $2\pi G\Sigma$ for $z>0$, where the surface density includes both sides of the galactic disk. Treating this field as constant and integrating the vertical Jeans equation yieldsThis exponential is the thin-sheet approximation or the outer tail, not the exact profile at every height. To see the distinction without an additional assumption, set $\rho ={\rho }_0{e}^{-u}$ with $u=(\Phi -\Phi (0))/{\sigma }^2$. The Poisson equation for Newtonian gravity becomes $u^{\prime \prime }=(4\pi G{\rho }_0/{\sigma }^2)e^{-u}$, with $u(0)=u^{\prime }(0)=0$, and is solved bySince $\Sigma =2{\rho }_0{z}_0$, the exact self-gravitating isothermal slab has $z_0={\sigma }^2/(\pi G\Sigma )$. At large $|z|/z_0$, ${\mathrm{sech}}^2(z/z_0)\sim 4e^{-2|z|/z_0}$, recovering precisely the boxed scale height; near the plane its slope vanishes, whereas an exponential has a cusp.

Low-mass main sequence stars live for much of the Milky Way's age, retaining representatives of many generations, while phase mixing disperses their birth associations around their orbits. In an approximately axisymmetric galactic disk, a sufficiently complete local census can therefore sample the mixed annular stellar population at the Sun's radius despite its small volume. This is a representative local-radius sample, not automatically an unbiased whole-disc census: different vertical scale heights, selection effects and radial migration require corrections when inferring integrated population fractions.

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.

Adding the two flux solutions givesThe supplied pressure–optical-depth law gives $D-\tau =Dp/p_0$. With the exponential pressure model $p=p_0{e}^{-z/H}$,It approaches the finite atmospheric top temperature $T_{\infty }$ and has the lowermost-air value $T(0)=T_{\infty }(1+D{)}^{1/4}$. Its lapse rate isThe stated approximate sign is appropriate physically: a nonisothermal ideal gas does not have an exactly constant hydrostatic scale height $RT/g$. Treating $H$ as a representative constant is an additional vertical-coordinate approximation. Algebraically the displayed profile is exact for the prescribed exponential pressure and two-stream equations.

The global mean absorbed solar flux is $L_{\infty }=0.8\times 584/4=116.8\mathrm{W}{\mathrm{m}}^{-2}$. Using $\sigma =5.6704\times 10^{-8}\mathrm{W}{\mathrm{m}}^{-2}{\mathrm{K}}^{-4}$ givesFor the requested graphical parcel estimate, start a dry adiabat at the warm ground, $T_{\mathrm{ad}}(z)=240.5-4.4z$ with $z$ in km, and intersect it with the radiative atmospheric profile $179.1[1+1.25e^{-z/10}{]}^{1/4}$. Solving givesThe original diagram below shows the intersection and the initial ground–air temperature discontinuity. A parcel heated towards the ground temperature remains buoyant until nearly one scale height in this model.

The atmospheric radiative lapse rate itself is only about $3.05\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$ at the bottom and decreases with height, smaller than the stated adiabatic $4.4\mathrm{K}\mathrm{k}{\mathrm{m}}^{-1}$. Thus the interior radiative atmosphere is statically stable; the driver of this particular estimate is the hotter ground and discontinuity at the lower boundary, not a radiative lapse rate everywhere steeper than adiabatic. Convection would smooth that discontinuity and redistribute energy.

This is a diagnostic depth using the unadjusted radiative ground and atmospheric temperatures, not a fully energy-conserving radiative–convective equilibrium after adjustment. A self-consistent adjustment changes the surface temperature, infrared fluxes and boundary height. Mars also has spectral CO${}_2$ absorption, dust, day–night variation, seasonal changes and geographically uneven heating; a gray, horizontally averaged, constant-scale-height column with fixed lapse rate captures none of those details. The calculation indicates the possible importance of a deep surface-driven mixed layer under the stated assumptions, not a universal Martian boundary-layer measurement.

Downward-from-space optical depth at wavelength $\lambda$ iswhere ${\kappa }_{\lambda }$ is the mass absorption coefficient. A nearly isothermal, hydrostatic equilibrium, compositionally well-mixed atmosphere with height-independent ${\kappa }_{\lambda }$ has $\rho \propto e^{-z/H}$ and $\tau \propto e^{-z/H_{\tau }}$ with $H_{\tau }=H=R_{a}T/g$. Thus temperature, mean molecular mass and gravity set the basic scale height. Changes in the absorber mixing ratio, pressure broadening, temperature dependence of line strengths, collision-induced absorption and departures from hydrostatic equilibrium modify it. For example if local absorption scales as ${\rho }^2$ in an isothermal atmosphere, $H_{\tau }=H/2$.

Neglect disk self-gravity and take $|z|\ll r$ around a central point mass $M_{\star }$. Its vertical gravitational acceleration isVertical hydrostatic equilibrium is $dP/dz=\rho g_z$. For a vertically isothermal ideal gas, $P=\rho c_s^2$ and $c_s$ is independent of $z$, givingHere $v_K=r\Omega =\sqrt{G{M}_{\star }/r}$, $c_s^2=k_{B}T/(\mu m_p)$ is the isothermal sound speed squared, and $h$ is the Gaussian scale height. A different convention writing $e^{-z^2/h^2}$ changes the quoted height by $\sqrt{2}$ but not the physics.

Hydrostatic balance and the ideal gas law give $dp/dz=-pg/(R_{a}T)$, hence approximately $p=p_0{e}^{-z/H}$ with $H=R_{a}T/g$ for weak temperature variation. For well-mixed absorber abundance, ${\rho }_{\mathrm{abs}}\propto p/T$, while pressure broadening gives $k\propto p$. Thus $k{\rho }_{\mathrm{abs}}\propto e^{-2z/H}$ and its integral above height $z$ has the same scale. ThereforeThe factor two depends on both pressure dependence of absorption and constant mixing ratio; it is not the ordinary density scale height.

This is the constant scale height of an isothermal atmosphere. The pressure and density decay length in an isothermal atmosphere is $H=k_{B}T/(mg)$.

A plane-symmetric, self-gravitating stellar population with constant vertical velocity dispersion satisfies the vertical Jeans equation and the Poisson equation for Newtonian gravity. With ${\Phi }^{\prime }(0)=0$, their solution isIts surface density is $\Sigma =2{\rho }_0{z}_0$, so $z_0={\sigma }_z^2/(\pi G\Sigma )$. Its outer density tail is exponential with scale height $z_0/2$; its central profile is not exponential.