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A plane-parallel atmosphere with isothermal sound speed $c_s$ in gravity $-{\Omega }^{2}z$ has mass density ${\rho }_0{e}^{-z^2/(2H^2)}$, $H=c_s/\Omega$. Above a rigid base, vertically directed adiabatic fluid perturbations satisfy $w^{\prime \prime }-z{w}^{\prime }/H^2+({\sigma }^2/{\Omega }^2-1)w/(\gamma H^2)=0$ and $w(0)=0$. Set $x=z/H$ and $\lambda =({\sigma }^2/{\Omega }^2-1)/\gamma$. The substitution $y=e^{-x^2/4}w$ converts finite Gaussian-weighted kinetic energy into square integrability for $-y^{\prime \prime }+x^{2}y/4=(\lambda +1/2)y$. Odd extension across the base and the oscillator lowering operator force $\lambda$ to be an odd positive integer. Thus $w_n$ is proportional to the Probabilists' Hermite polynomial ${\mathrm{He}}_{2n+1}(z/H)$. These are vertical acoustic waves; a general stably stratified atmosphere also admits internal gravity waves with horizontal structure.

For the regular isothermal Lane-Emden equation, the first fixed-mass pressure turning point occurs at ${\xi }_{\mathrm{crit}}\simeq 6.451$. Thus the boundary radius there is $R_{\mathrm{crit}}\simeq 1.820c_s/\sqrt{G{\rho }_c}$, where $c_s^2=K$ is the squared isothermal sound speed. The approximate choice ${\xi }_{\mathrm{crit}}=6.5$ gives coefficient $1.83$. Comparing with the Jeans length requires specifying the reference density and remembering that the latter is a wavelength rather than a boundary radius.

A steady spherical subsonic isothermal breeze with isothermal sound speed $a$ obeys $(u-a^2/u)u^{\prime }=2a^2/r-GM/r^2$. Since $0<u<a$, its finite-radius speed maximum lies at $r_M=GM/(2a^2)$. Spherical mass conservation then gives the displayed density, where ${\mathcal{M}}_M=u(r_M)/a\in (0,1)$ is the peak Mach number. This free parameter is selected by a boundary condition, not by $T$, $M$ and $\dot{M}$ alone. Therefore ${\rho }_M>\dot{M}{a}^3/(\pi G^2{M}^2)$, while a peak Mach number of order unity gives that expression as a characteristic scale.

A direct demonstration of the free parameter comes from the first integral $u^2/2-a^2\ln (u/a)=2a^2\ln (r/r_M)+GM/r+C$. Its additive constant remains free for a breeze; imposing the regular transonic branch would be a different condition.

Assume a passive disk has local isothermal sound speed squared $c_s^2=aT$, disk scale height $H=c_s/{\Omega }_K$, and irradiated height $z_s=fH$ with constant $f$. Its irradiation grazing angle of a disk and Stefan–Boltzmann law give $q^{\prime }=C{q}^8/R^3$ for $q=z_s/R$, where $C=4\pi \sigma (GM/a{)}^4/(L_{\ast }{f}^8)$. Integrating gives $q^{-7}=R_0^2/R^2+B$, with $R_0^2=7C/2$. Where the first term dominates and the thin disk assumptions hold, $q\simeq (R/R_0{)}^{2/7}$, $H\propto R^{9/7}$ and $T\propto R^{-3/7}$. The law cannot be extrapolated into a geometrically thick region using the same grazing-angle approximation.

In an optically thick steady spherical stellar wind of constant mean molecular weight ${\mu }_{\mathrm{mol}}$, let $v$ be the outward speed and $a_g^2=k_{B}T/({\mu }_{\mathrm{mol}}{m}_u)$ the isothermal sound speed squared. Mass conservation gives ${\rho }^{\prime }/\rho =-2/r-v^{\prime }/v$, where a prime denotes differentiation with respect to radius $r$. Use $P_{\mathrm{gas}}=\rho a_g^2$ and radiative diffusion to write $P_{\mathrm{rad}}^{\prime }/\rho =-g\Gamma$, where $g=G{M}_r/r^2$ and $\Gamma =L_r/L_{\mathrm{Edd}}$ uses the local Eddington luminosity $L_{\mathrm{Edd}}=4\pi cG{M}_r/\kappa$ and opacity $\kappa$. Substituting into the momentum equation yields the displayed wind equation. Smooth sonic point crossing requires both its denominator and numerator to vanish. The radiation-pressure gradient already supplies radiative acceleration; adding that force separately would count it twice.

Across a thin steady transition with negligible external impulse, mass conservation and momentum conservation give the displayed relations. If the gas stays isothermal while neutral atomic hydrogen becomes molecular hydrogen, the mean particle mass doubles and $a_2^2=a_1^2/2$ for its isothermal sound speed. Let $J=u_1+a_1^2/u_1$. The downstream subsonic branch has $u_2=[J-\sqrt{J^2-2a_1^2}]/2$, ${\rho }_2/{\rho }_1=u_1/u_2$, and $p_2-p_1={\rho }_1{u}_1(u_1-u_2)>0$.

For $u_1\ll a_1$, the leading behavior is $u_2\simeq u_1/2$, ${\rho }_2\simeq 2{\rho }_1$ and $p_2\simeq p_1$, at unchanged temperature. Radiation of the molecular binding energy prevents that released energy from necessarily heating the gas; it does not, by itself, establish an isothermal thermal closure. If temperature evolution is allowed, an energy balance and a cooling model are required to determine all four changes uniquely.

In aligned steady ideal flow with $p=c_s^2\rho$ and constant isothermal sound speed, the Lorentz force density has no field-parallel component. Projecting momentum along the magnetic field makes $u^2/2+\Phi +c_s^2\ln (\rho /{\rho }_0)$ constant on each connected magnetic surface. The arbitrary reference density changes only the additive constant.

A steady spherical Parker wind with constant isothermal sound speed $c$ and negligible wind self-gravity obeys $4\pi r^2\rho u=\dot{M}$ and $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$. A regular sonic point has $u=c$ at $r_s=GM/(2c^2)$. With $x=r/r_s$ and $w=u^2/c^2$, integration gives the displayed equation for the smooth accelerating transonic branch. Its slope at the sonic point is $u^{\prime }(r_s)=c/r_s$. Maintaining constant gas temperature during expansion generally requires heat input, so this model is distinct from an adiabatic wind with $\gamma >1$.

For constant isothermal sound speed, time-independent shearing-sheet tidal potential ${\Phi }_t$, and inviscid flow, define $E=\Sigma u^2/2+c_s^2\Sigma \ln (\Sigma /{\Sigma }_{\mathrm{ref}})+\Sigma {\Phi }_t$. The barotropic energy density and kinetic energy balances combine to give ${\partial }_{t}E+\nabla \cdot [(E+P)\mathbf{u}]=0$. Coriolis acceleration does no work. Integral conservation requires vanishing net boundary energy flux.

An isothermal shock joins two states at the same temperature and isothermal sound speed. Mass and momentum conservation imply $u_1{u}_2=c_s^2$ for a nontrivial planar jump, so the compression ratio is the upstream isothermal Mach number squared. Heat exchange replaces adiabatic energy conservation across the gas alone.

For a thin disk in the harmonic vertical gravitational potential ${\Omega }_K^2{z}^2/2$, hydrostatic equilibrium and $p=c_s^2\rho$ with height-independent isothermal sound speed give the displayed Gaussian density. For perturbations with adiabatic index $\gamma$, the Brunt–Väisälä frequency is $N^2=(1-1/\gamma ){\Omega }_K^2{z}^2/H^2$. An isothermal background can therefore support buoyancy for adiabatic perturbations; isothermal perturbations instead have $\gamma =1$ and zero buoyancy frequency.

For a strong monatomic shock, the planar shock compression of a finite slab gives bulk speed $3V_s/4$ and kinetic energy $E_{\mathrm{bulk}}=9M_c{V}_s^2/32$. The initial internal energy is $E_{\mathrm{th}}=3M_{c}p/(2\rho )$. A bulk-ballistic escape estimate $E_{\mathrm{bulk}}\gtrsim W$ uses positive binding-energy magnitude $W$, not negative gravitational potential energy. With adiabatic sound speed $c_{\mathrm{ad}}^2=5p/(3\rho )$ it yields the first displayed coefficient. Using the isothermal sound speed $c_T^2=p/\rho$ instead yields $16/3\simeq 5.3$. The definitions obey ${\mathcal{M}}_T^2=(5/3){\mathcal{M}}_{\mathrm{ad}}^2$. Heating, subsequent expansion and energy loss can alter the physical escape threshold, so the coefficient is not independent of the energy budget and Mach convention.

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.

Let $a_T^2=k_{B}T/(\mu m_u)$ be the constant isothermal sound speed squared. The stationary galactic gas obeys hydrostatic equilibrium in the specified gravitational potential:Integrating from the pressure-matching radius $r_0$ gives $\rho (r)=\rho (r_0)\exp [-(\psi (r)-\psi (r_0))/a_T^2]$. Since $\rho (r_0)=p_0/a_T^2$ and $\psi =-A{r}^{-1/2}$,The exponent is dimensionless: the quantity in parentheses times $A$ is a difference of gravitational potential, of units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$. Density increases inward, as expected. The idealized potential is singular at the centre, so this formal profile should not be extrapolated into a region where the prescribed potential or the neglect of gas self-gravity fails.

Assume a steady spherical outflow of fixed composition. Let $\bar{m}$ be its mean particle mass, and define the constant isothermal sound speed by $a^2=k_B{T}_0/\bar{m}$. Subsonic here refers to this isothermal closure, so $0<u<a$. Mass conservation and radial momentum balance giveLogarithmically differentiating the first equation yields $d\ln \rho /dr=-2/r-u^{\prime }/u$. Substitution eliminates the density gradient and gives the isothermal Parker wind equationAt a smooth speed maximum, $u^{\prime }=0$, while the left prefactor is finite and negative. The right side must vanish, henceFor $r<r_M$ both the right side and the prefactor are negative, so $u^{\prime }>0$; for $r>r_M$, $u^{\prime }<0$. This is a true maximum on a subsonic isothermal breeze, not the accelerating transonic branch.

There is an underdetermination in the requested density estimate. Introduce the peak Mach number ${\mathcal{M}}_M=u_M/a$, where $0<{\mathcal{M}}_M<1$. The density at the maximum speed of an isothermal breeze is exactlyThe hypotheses do not fix ${\mathcal{M}}_M$. A boundary density or pressure, or an equivalent wind integration constant, is needed. The definite statement following from subsonicity is the lower boundIf a maximum speed of order $a$ is intended as an extra estimate, then ${\rho }_M\sim {\rho }_{\mathrm{char}}\propto \dot{M}{T}_0^{3/2}{M}^{-2}$. A very slow breeze can have a much larger density, so setting $u_M=a$ as an exact equality would contradict the stated always-subsonic assumption. The mean particle mass is another implicit fixed-composition input needed to express a temperature as a squared sound speed.

One can see the free parameter directly by integrating the wind equation:Different $C$ give different subsonic maxima at the same radius. The original plot illustrates this nonuniqueness. These breeze branches require nonzero ambient pressure at large radius; they are not the usual transonic escape into a low-pressure exterior.

Maintaining an isothermal expanding wind generally requires heat exchange; isothermal motion is not adiabatic motion.^[ref]

Retain the preceding isothermal assumption, so the gas temperature remains $T_0$. When two neutral hydrogen atoms become one molecular hydrogen molecule, the mean particle mass doubles. At fixed mass density and temperature the ideal gas law would halve the pressure, but the flowing gas is free to compress and change speed. It cannot in general keep density, pressure, and velocity all unchanged.

To describe a thin steady transition, label the atomic and molecular sides by $1$ and $2$ and neglect the integrated external force and radiative recoil across its small thickness. Local mass conservation and momentum conservation requireThe isothermal sound speed changes from $a_1^2=k_B{T}_0/m_H$ to $a_2^2=a_1^2/2$. Dividing the momentum flux by $j$ gives $u_2+a_2^2/u_2=J$, with $J=u_1+a_1^2/u_1$. The subsonic continuation is thereforeThus the gas becomes denser and slower. In the well-subsonic limit, where ram stresses are small compared with thermal pressure, the useful leading description of isothermal flow through molecular hydrogen formation isThe exact momentum condition gives a small pressure increase at low Mach number, rather than an imposed discontinuous factor-of-two pressure drop. The full expressions above apply when the subsonic speed is not asymptotically small.

Radiation of the molecular binding energy prevents that released energy from automatically heating the gas. It does not, by itself, determine the temperature: the constant-temperature result uses the inherited isothermal closure. If that assumption is relaxed, an energy equation including cooling and molecular degrees of freedom is needed; the wording about emitted binding energy alone does not uniquely determine all four changes.

Supersonic communication. A point source moving at speed $V>c_s$ outruns its small acoustic waves. After a time $t$, the source has moved $Vt$, while a disturbance emitted at its old position has radius $c_{s}t$. The envelope of these spherical wavefronts is the Mach cone, with half-angle $\mu$ satisfyingInformation is restricted to the downstream cone, so upstream gas cannot smoothly anticipate a supersonic obstacle. Finite compressions can also steepen because their characteristics converge. Such disturbances form a shock wave, across which the fluid variables change rapidly. A perfectly uniform supersonic flow need not contain a shock: the cone expresses the communication constraint, while a disturbance or compression supplies the reason for a discontinuity.

Conserved fluxes. Work in the shock frame, take velocities normal to the front, and integrate the conservation laws through a thin steady control volume of unit area. Mass conservation gives constant flux $j=\rho u$, soThe normal momentum flux is the advected part $\rho u^2$ plus the pressure stress $p$. With negligible body-force impulse across the thin front,For an ideal gas with constant adiabatic index $\gamma$, the specific internal energy is $e=p/[(\gamma -1)\rho ]$. The energy flux iswhere $h_{\mathrm{sp}}$ is the specific enthalpy. No radiation or other heat flux escapes in this adiabatic shock model. Conservation of the energy flux, divided by the nonzero mass flux, givesThese are the Rankine-Hugoniot conditions for a perfect gas. The shock is adiabatic but irreversible: its entropy increases, so an isentropic relation cannot be imposed across the discontinuity.

Strong-shock compression. Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and neglect $p_1$ and the upstream enthalpy compared with ${\rho }_1{u}_1^2$ and $u_1^2$. Momentum conservation gives $p_2={\rho }_1{u}_1^2(1-1/\chi )$. Substitute this into energy conservation:ThusThe root $\chi =1$ describes no shock. The compressive branch of the Strong-shock Rankine-Hugoniot conditions isIt also gives $p_2=2{\rho }_1{u}_1^2/(\gamma +1)$. A monatomic gas has $\gamma =5/3$, hence $\chi =4$.

The finite slab in its initial rest frame. Denote the shock's lab speed by $V_s$. The upstream gas is initially stationary, so $u_1=V_s$ and the downstream shock-frame speed is $u_2=V_s/4$. Transforming to the initial slab frame givesThis is planar shock compression of a finite slab. In the uniform post-shock approximation, breakout occurs at $t_b=L/V_s$. The original front face has then moved $(3V_s/4)t_b=3L/4$, while the shock reaches the originally stationary far face at $L$, leaving thickness $L/4$. Equivalently, the fixed mass per area $\rho L$ becomes $4\rho L_2$. The result assumes a maintained approximately constant shock speed and no important boundary rarefaction during the crossing; an unconfined finite cloud need not remain uniformly compressed.

Escape estimate and the sound-speed convention. Let $M_c$ be the slab mass, $W>0$ its binding-energy magnitude in the star's field, and $E_{\mathrm{th}}$ its initial internal energy. Treat the potential as nearly uniform across the slab. For the bulk-ballistic escape estimate, requireThe Mach-number convention in shock unbinding estimates matters here. With the usual upstream adiabatic sound speed, $c_{\mathrm{ad}}^2=\gamma p/\rho =5p/(3\rho )$, define ${\mathcal{M}}_{\mathrm{ad}}=V_s/c_{\mathrm{ad}}$. ThenIf instead the shock speed is normalized by the isothermal sound speed, $c_T^2=p/\rho$, then ${\mathcal{M}}_T^2=(5/3){\mathcal{M}}_{\mathrm{ad}}^2$ andThe original PDF's “around five” is consistent with the latter convention or an order-of-magnitude sound-speed estimate. It is not the exact coefficient obtained with the conventional adiabatic Mach number and the stated bulk kinetic-energy criterion. Here $W=-E_{\mathrm{grav}}$ if the potential is zero at infinity; using negative gravitational energy directly on the right would be incorrect. Subsequent conversion of shock-generated internal energy to expansion, external pressure, or radiative losses would change the physical escape threshold, so this numerical factor is not universal.

Hydrostatic equilibrium with self-gravity. A resting fluid obeys $\nabla p=-\rho \nabla \Psi$. Spherical symmetry and the Poisson equation for Newtonian gravity giveEliminate the Newtonian gravitational potential derivative to obtainEquivalently, the shell theorem gives ${\Psi }^{\prime }=GM(r)/r^2$ and $M^{\prime }(r)=4\pi r^2\rho$, which leads to the same hydrostatic equilibrium equation.

A uniform-density liquid. With constant $\rho$, $M(r)=4\pi \rho r^3/3$. There is no central point mass, so the interior gravitational acceleration magnitude is $4\pi G\rho r/3$. ThusUse $p(R)=0$ at a vacuum boundary, neglecting surface tension. The uniform-density planetary pressure profile isThe central value can also be writtenIt has units pascals and decreases smoothly from its maximum at the centre to zero at the surface.

An isothermal gas. Here $c$ is the constant isothermal sound speed, with $p=c^2\rho$. Try $p=A/r^2$, so $\rho =A/(c^2{r}^2)$. The two sides of the equilibrium equation becomeThus $A=c^4/(2\pi G)$, giving the singular isothermal sphere,This does not describe a finite gaseous sphere in vacuum. Both pressure and mass density diverge at the centre, and $p(R)>0$ for every finite outer radius $R$, so it cannot meet $p=0$ there. Extending it to infinity removes the finite-radius pressure jump but makes the total mass infinite. It is useful as an idealized profile over an intermediate radial range, or as a truncated model with nonzero external pressure. The vacuum boundary obstruction for an isothermal sphere also applies to the regular, nonsingular isothermal solution; removing the central singularity does not produce a finite vacuum surface.

Steady outflow. In the wind calculation, take the central mass $M$ to be fixed and neglect the wind's own self-gravity. Constant mass flow rate and the isothermal equation of state giveThe steady radial Euler equation isSubstituting the density derivative givesThis is the isothermal Parker wind equation. A smooth sonic point requires both sides to vanish when $u=c$, soMaintaining constant temperature during expansion generally requires heat input; it is an isothermal model, not the adiabatic blast-wave model of Question 5.

The escaping branch at large radius. Integrate the wind equation, using dimensionless logarithm arguments:Set $x=r/r_s$ and $w=u^2/c^2$. ThenFor the accelerating transonic branch, $w(1)=1$, which fixes $C_0=-3$. At $x\gg 1$ on its supersonic branch, $w\to \infty$ and $\ln w$ is smaller than $w$. ConsequentlyA first correction makes the ordering explicit:This is the large-radius asymptotics of an isothermal Parker wind. The leading expression also holds for other supersonic branches with finite $C_0$.

The source implicitly selects an escaping wind. Positive steady outward flow alone does not determine that branch: a subsonic isothermal breeze also satisfies the differential equation. Its $w\ll 1$ limit gives $w\sim e^{-C_0}{x}^{-4}$, hence $u\propto r^{-2}$ and $\rho \to {\rho }_{\infty }>0$. Such a breeze needs nonzero pressure at infinity. Selecting a regular outflow from a subsonic base into vanishing ambient pressure instead gives the accelerating transonic wind used above.

The asymptotic speed has the same units as $c$; the logarithm is dimensionless. Its unlimited growth is a formal feature of indefinite isothermal heating and a prescribed point-mass potential. Real heating, surrounding pressure, and eventually the neglected wind mass limit that extrapolation.

Let $M(r)$ be the enclosed mass. For spherical symmetry, hydrostatic equilibrium and mass conservation giveThe isothermal equation of state $p=K\rho$ therefore impliesMultiply by $r^2$, differentiate and eliminate $M$:or equivalently $r^{-2}d[r^2{\rho }^{-1}{\rho }^{\prime }]/dr=-4\pi G\rho /K$. This is the density equation for an isothermal sphere. Here $K$ has units ${\mathrm{m}}^2{\mathrm{s}}^{-2}$ and equals the squared isothermal sound speed.

A regular isothermal sphere does not reach zero density at a finite radius. The finite body discussed later must therefore be truncated by a nonzero external pressure, as in a pressure-confined isothermal sphere. A finite vacuum boundary would not be compatible with this equation of state.

Using the supplied first critical value ${\xi }_c\simeq 6.5$, the critical radius of a pressure-confined isothermal sphere isNo radius in metres can be specified without $K$ and a central mass density. Here $c_s$ is the isothermal sound speed, consistent with the fixed-temperature variations in part (d). At a given central mass density, equilibria truncated at $\xi <{\xi }_c$ lie on the first stable branch, while increasing the dimensionless outer radius beyond this first threshold gives an unstable configuration. Numerical integration gives ${\xi }_c\simeq 6.451$, consistent with the paper's rounded $6.5$.

For a transparent comparison with the given Jeans length, set the uniform comparison density ${\rho }_0={\rho }_c$ and use the same speed of sound. ThenBoth scales express the competition between pressure support and self-gravity, with characteristic length $c_s/\sqrt{G\rho }$. Their numerical closeness under this density choice does not mean that the models impose the same threshold: $r_c$ is a boundary radius, whereas ${\lambda }_J$ is the wavelength of a perturbation. Comparing $r_c$ with ${\lambda }_J/2$ would instead give a factor about $2.07$. Using the sphere's mean or boundary density instead of its central density changes the coefficient again, so the reference density must always be stated.

The pressure-confined isothermal sphere gives an actual finite, centrally concentrated hydrostatic equilibrium and an explicit confining pressure, making its radial stability threshold more self-consistent for a roughly spherical isothermal cloud. The uniform Jeans instability model gives a simpler local criterion, includes perturbations at different wavelengths, and does not require constructing a spherical equilibrium. Its ideal uniform static infinite background requires the Jeans swindle or an explicitly specified treatment of background gravity. Neither model includes rotation, magnetic fields, turbulent support or a realistic temperature evolution, so neither is a universal collapse criterion.