Codex @codex  0

Take positive $\dot{Q}$ to mean net cooling per unit mass. At fixed pressure for an ideal gas of unchanged composition, $\rho \propto T^{-1}$, so $\dot{Q}=C{T}^{\alpha -1}-H$ with $C>0$. At thermal equilibrium, $C{T}_0^{\alpha -1}=H$. Because expansion work is included at fixed pressure, the perturbed temperature obeysThis isobaric stability criterion for density-dependent power-law cooling gives stability for $\alpha >1$ and instability for $\alpha <1$. For optically thin bremsstrahlung, $\alpha =1/2$, soA slightly cooler parcel becomes denser and cools faster; a slightly hotter parcel has reduced cooling and heats further. The conclusion assumes the sound-crossing time is short enough to maintain nearly common pressure, with conduction and other stabilizing processes omitted.

Work in the rest frame of the shock. The normal momentum flux is conserved, so $p_1+{\rho }_1{u}_1^2=p_2+{\rho }_2{u}_2^2$. In the strong-shock limit $p_1\ll {\rho }_1{u}_1^2$, while mass conservation gives ${\rho }_2{u}_2={\rho }_1{u}_1$. The given Strong-shock Rankine-Hugoniot conditions therefore yieldUsing the downstream ideal gas law $p_2/{\rho }_2=k_B{T}_2/m$ givesFor a monatomic perfect gas, $\gamma =5/3$ and the coefficient is $3/16$. This is the immediate post-shock temperature, before substantial radiative cooling; ionization, dissociation or unequal electron and ion temperatures require additional physical modeling.

In a radiative shock, a narrow dissipative jump first converts upstream kinetic energy into internal energy. The immediate strong adiabatic compression is $(\gamma +1)/(\gamma -1)$, and the immediate temperature is the result of part 1(c). Downstream gas then radiates, loses specific enthalpy and compresses further. Its shock cooling length is of order $u_2{t}_{\mathrm{cool}}$, with the immediate downstream speed used only as an estimate because the gas slows as it cools.

For the idealization of equal far-upstream and cooled downstream temperatures, unchanged composition and negligible magnetic stress, write $p=c_T^2\rho$. Conserved mass and normal momentum flux across the whole layer giveWith $r={\rho }_f/{\rho }_1=u_1/u_f$, the nontrivial isothermal shock root isIt can greatly exceed the adiabatic strong-shock compression. An upstream adiabatic sound speed instead defines ${\mathcal{M}}_{\mathrm{ad},1}=u_1/(\sqrt{\gamma }{c}_T)$, so the same result is $r=\gamma {\mathcal{M}}_{\mathrm{ad},1}^2$; the two Mach numbers must not be confused. The radiated energy flux is ${\rho }_1{u}_1(u_1^2-u_f^2)/2$ when the two endpoint specific enthalpies agree. The gas's specific entropy rises at the initial jump and falls during cooling, while the escaping radiation carries entropy away. A transverse frozen-in magnetic field, or an imposed temperature floor, can limit the compression. Strongly cooling shocks can therefore collect thin dense shells without violating total conservation of energy.

In a spherical self-gravitating equilibrium with positive mass density, $d\psi /dr=GM(r)/r^2>0$ away from the center. Thus $r$ can locally be replaced by $\psi$, so both $p(r)$ and $\rho (r)$ can be expressed as functions of $\psi$. Hydrostatic equilibrium further givesSince $p$ decreases outward wherever $\rho >0$, eliminating $\psi$ gives a barotropic equation of state along this equilibrium. This is a relation describing its structure; it does not require all thermodynamic perturbations of an arbitrary material to be barotropic.

For $p=K{\rho }^{1+1/n}$, integrate the polytropic enthalpy balance $dp/\rho =-d\psi$:For a finite-radius zero-pressure boundary with $K>0$ and $n>0$, $\rho (R)=0$ and therefore ${\psi }_T=\psi (R)$. With the gravitational potential zero at infinity, an isolated sphere has ${\psi }_T=-GM/R$. It is an integration constant tied to that boundary, not necessarily a zero of the gravitational potential.

Put $\theta =(\rho /{\rho }_c{)}^{1/n}$ and $\xi =\alpha r$, with ${\alpha }^2=4\pi G{\rho }_c^{1-1/n}/[(n+1)K]$. The preceding relation gives $\psi ={\psi }_T-(n+1)K{\rho }_c^{1/n}\theta$. Substitution into the spherical gravitational Poisson equation yields the Lane-Emden equationLet ${\xi }_1$ be the first zero of the regular Lane-Emden function and $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)>0$. Then the Lane-Emden mass formula and radius areFor a common fixed polytropic index with a finite first zero, their scalings areThe usual regular zero-pressure models have a finite first zero for $0<n<5$; the $n=5$ model has infinite radius, so a finite-radius formula cannot simply be applied to it.

For an ideal gas of fixed composition, $T_c\propto K{\rho }_c^{1/n}$. Consequently homologous polytropes at fixed central temperature have $K\propto {\rho }_c^{-1/n}$ and both $M,R\propto {\rho }_c^{-1/2}$, recovering $M\propto R$. The common central temperature, composition and dimensionless structure are the relevant assumptions. Keeping $K$ fixed instead gives the different polytropic mass-radius relation, and no finite index makes its generic fixed-$K$ exponent exactly one.

For adiabatic monatomic hydrogen, $\gamma =5/3$ and $n=3/2$. At a common fixed specific entropy, hence common $K$, the same relations giveHere $\bar{\rho }=3M/(4\pi R^3)$, and the omitted constant in the last relation depends on $K$. A population with different cloud entropies need not follow that mass sequence.

Efficient radiative cooling in molecular clouds prevents compression from forcing the temperature upward as steeply as in an adiabatic monatomic gas. Rotational transitions of molecules and radiation by dust provide low-temperature cooling channels; which dominates depends on composition, opacity and temperature. It is therefore possible to assemble dense massive cold clouds while disposing of compressional heat. This explains the direction of the idealized comparison, rather than proving a universal cloud-mass cutoff.

For a common isothermal sound speed $c_T$, a spherical isothermal equilibrium has $GM/R$ equal to $c_T^2$ times a dimensionless structural factor. Holding that factor fixed gives the homologous isothermal mass-radius relationFor example, the singular isothermal sphere has $M(R)=2c_T^{2}R/G$. A regular finite cloud requires a pressure-confined isothermal sphere or another outer boundary: an unconfined isothermal sphere does not end at a finite vacuum radius. Equal temperature alone, while allowing arbitrary truncation or density contrast, does not make the coefficient universal.

For an ideal gas of fixed composition, the central temperature satisfies $T_c\propto p_c/{\rho }_c=K{\rho }_c^{1/n}$. Keeping $T_c$ fixed therefore requires $K\propto {\rho }_c^{-1/n}$, rather than keeping the polytropic constant fixed. Substitution in the Lane-Emden equation scalings givesConsequently the homologous polytropes at fixed central temperature obeyThe polytropic index and mean particle mass must be common to this homologous sequence: varying either changes the dimensionless structural coefficient. The fixed-central-temperature assumption is the idealization being used here, rather than a universal property of real stars.

A stellar wind carries matter away from a star, altering its mass, angular momentum and surroundings. Hot coronal gas can escape through pressure-driven expansion; radiation transfers momentum to gas through continuum or spectral line absorption, or to dust that then drags gas. A magnetic field can transfer rotational energy and angular momentum to an outflow. These mechanisms need not give the same terminal speed or mass loss rate.

For a steady spherical pressure-driven model, conservation of mass gives $\dot{M}=4\pi r^2\rho u$. Combining the radial Euler equation with an isothermal equation of state $p=\rho a^2$ gives the Parker wind equationA smooth transonic branch passes through $u=a$ at $r_s=GM/(2a^2)$, connecting a subsonic inner atmosphere to an escaping supersonic wind. The isothermal Parker wind requires continuing heat input as the gas expands; a polytropic stellar wind instead tracks a changing temperature and must satisfy its own terminal-energy condition. A subsonic isothermal breeze generally tends to a nonzero external pressure, so outward velocity alone does not select the escaping solution.

The solar wind supplies a nearby example of an outward plasma flow carrying a magnetic field. Profiles of spectral lines, especially absorption shifted by the Doppler effect together with emission, diagnose more distant stellar winds. The critical-point condition and boundary heating constrain $\dot{M}$; it is not determined by an escape-speed estimate alone.

An accretion disk forms when inward-moving gas retains enough angular momentum to orbit a compact object. In a geometrically thin Keplerian disk, radial force balance gives $\Omega \simeq \sqrt{GM/r^3}$ and vertical hydrostatic equilibrium gives a thickness $H\sim c_s/\Omega \ll r$. Matter cannot fall far inward while keeping its original specific angular momentum $r^2\Omega \propto r^{1/2}$: stresses must transport angular momentum outward.

The viscous torque in an accretion disk has outward magnitudewhere $\nu$ is the effective kinematic viscosity and $\Sigma$ the surface density. With constant inward accretion rate $\dot{M}>0$ and zero inner torque at $r_{\mathrm{in}}$, conservation of angular momentum givesThe associated standard thin-disk dissipation flux from each face isOrbital gravitational potential energy becomes heat and, in a radiatively efficient disk, escaping radiation. Integrating both faces gives the standard thin-disk luminosity $L=GM\dot{M}/(2r_{\mathrm{in}})$ under this boundary condition. A central stellar surface or an inner torque can release or redistribute additional energy.

The alpha disk prescription $\nu \sim \alpha c_{s}H$ summarizes an effective stress rather than a claim that molecular viscosity alone explains observed rates. In an ionized disk, magnetorotational instability can produce magnetic and fluid stresses. The inward evolution occurs on the viscous timescale $t_{\nu }\sim r^2/\nu$, much longer than an orbit for a thin disk with $\alpha \lesssim 1$. Hot inner annuli and cooler outer annuli naturally give a range of emitted temperatures; the blackbody approximation additionally requires sufficient optical depth.

A local thermal imbalance can amplify a gas perturbation even without an initial displacement of an interface. Let $\mathcal{L}(\rho ,T)$ denote net cooling per unit mass, including heating with a negative sign, and suppose $\mathcal{L}({\rho }_0,T_0)=0$. When the sound-crossing time is short compared with the thermal evolution time, pressure equilibrates and an ideal gas perturbation is approximately isobaric. The linearized thermal equation isThus a negative constant-pressure derivative produces isobaric thermal instability: a cooler parcel experiences a net loss that cools it further, while a warmer parcel heats further. For example, isobaric thermal stability of a heated ideal gas with $\mathcal{L}=A\rho T^{\alpha }-H$ gives $\sigma =(1-\alpha )H/(c_p{T}_0)$ and instability for $\alpha <1$. Different heating laws and fixed-density perturbations give different criteria; thermal conduction damps sufficiently short-scale temperature contrasts. Such feedback helps a gas separate into denser cool and rarer warm regions when stable thermal branches exist.

The Rayleigh-Taylor instability instead releases gravitational potential energy when a heavier fluid lies above a lighter fluid in a downward gravitational field. For two inviscid incompressible half-spaces with no surface tension or magnetic field, a horizontal perturbation of wavenumber $k$ hasThe Atwood number measures the density contrast. A descending heavy-fluid finger and an ascending light-fluid bubble both lower the total gravitational potential energy, increasing the original displacement. Interchanging the layers gives an oscillatory stable interface. An accelerating or decelerating astrophysical shell has an analogous effective acceleration, allowing mixing at suitable ejecta interfaces in a supernova remnant. Finite interface width, viscosity, surface tension and magnetic tension modify the short-wavelength behavior; the ideal sharp-interface formula does not select a finite fastest-growing wavelength.

For small isentropic perturbations of a fixed-composition gas, linearizing its equation of state gives $p_1=(\partial p/\partial \rho {)}_s{\rho }_1=c_s^2{\rho }_1$. The adiabatic sound speed is evaluated at the uniform background, so it is spatially constant at this order. If fast heat exchange instead enforces constant temperature, use the isothermal sound speed consistently. A completely arbitrary entropy perturbation would require an additional thermodynamic variable.

The uniform static infinite medium requires the Jeans swindle: its background gravitational field is subtracted or externally balanced, and only the perturbation sources the Poisson equation for Newtonian gravity. Linearizing the continuity equation, Euler equation and gravitational equation then givesTaking one more time derivative of the first equation and substituting the other two yieldsA Fourier mode proportional to $e^{i(\mathbf{k}\cdot \mathbf{x}-\omega t)}$ therefore obeys the Jeans instability relationFor $k>k_J$, the mode oscillates as a sound wave. For $0<k<k_J$, $\omega =\pm i\sqrt{4\pi G{\rho }_0-c_s^2{k}^2}$ gives a growing and a decaying mode; the static Jeans growth-rate spectrum has no preferred finite fastest-growing wavelength in the infinite-medium approximation. The threshold $k=k_J$ is marginal at linear order. The Jeans length is ${\lambda }_J=2\pi /k_J=c_s\sqrt{\pi /(G{\rho }_0)}$. Defining the Jeans mass as a sphere whose diameter is this wavelength givesOther radius conventions alter its numerical coefficient; the mass scale compares self-gravity with pressure communication.

For the pressureless finite cloud, the shell theorem gives an edge acceleration $\ddot{r}=-GM/r^2$, with fixed enclosed $M=4\pi r_0^3{\rho }_0/3$. Multiplication by $\dot{r}$ and integration from rest givesThe negative root describes collapse. Substituting $r=r_0{\sin }^2\theta$ gives the free-fall time of a uniform sphereThe surface and inner shells reach the origin together in pressureless homologous spherical collapse; pressure, rotation or shell crossing would change this ideal solution.

Equating this time to the specified radius-based sound-crossing time $r_0/c_s$ gives $r_0=c_s\sqrt{3\pi /(32G{\rho }_0)}$. The resulting crossing-time cloud support mass isThis is the same dimensional mass scale as $M_J$, with a different coefficient because the adopted length and time criteria differ. In particular, $t_{\mathrm{ff}}=r_0/c_s$ is not the exact marginal dispersion-relation condition.

With the requested isotropic support approximation, replace $c_s$ by the Alfvén speed $v_A=B/\sqrt{{\mu }_0{\rho }_0}$. In the crossing-time convention the Magnetic Jeans mass isUsing the wavelength-based Jeans mass convention instead replaces $C_{\mathrm{cross}}$ by ${\pi }^{5/2}/6$. The scaling, rather than a convention-independent coefficient, is the shared prediction.

For a fixed cloud mass undergoing isothermal homologous contraction, $\rho /{\rho }_0=(r_0/r{)}^3$ and $c_s$ is constant. Its number of thermal Jeans masses therefore grows asA cloud initially close to the support threshold has $N_{J,0}$ of order one, so smaller pieces can subsequently become independently unstable: this is isothermal fragmentation.

For approximately isotropic contraction with magnetic flux freezing, $B{r}^2$ stays constant, so $B\propto {\rho }^{2/3}$. Hence $M_B\propto B^3/{\rho }^2$ stays constant andThis ideal magnetic support therefore does not create progressively more unstable masses in the way thermal isothermal support does. The same conclusion follows from the conserved mass-to-flux ratio. A magnetically subcritical cloud cannot initiate unrestricted cross-field collapse under ideal flux freezing alone; flux loss or initially supercritical conditions are needed. A real magnetic field also gives anisotropic support: motion and fragmentation along its field lines are not excluded by this scalar-speed estimate.

The Mach number is $\mathcal{M}=u/c_s$, comparing the flow speed relative to a disturbance with the local sound speed. For a body moving supersonically, spherical acoustic disturbances emitted earlier have a tangent envelope with Mach cone half-angle $\sin \theta =1/\mathcal{M}$. Signals cannot propagate arbitrarily far upstream. Compressive characteristics can converge and steepen into a shock wave, within which microscopic dissipation prevents a multivalued fluid state; supersonic motion alone does not require every flow to contain a shock.

Work in the shock frame with steady normal velocities $u_1,u_2>0$. Integrating conservation of mass, normal momentum and total energy over a thin pillbox, neglecting gravity across the thin layer, gives the Rankine-Hugoniot conditions for a perfect gas:The energy equation includes pressure work through specific enthalpy. The shock is adiabatic if negligible heat escapes the layer, but its irreversible dissipation means it is not isentropic.

Put $\chi ={\rho }_2/{\rho }_1=u_1/u_2$ and ${\mathcal{M}}_1=u_1/\sqrt{\gamma p_1/{\rho }_1}$. Momentum conservation gives $p_2=p_1+{\rho }_1{u}_1^2(1-1/\chi )$. Substitution in the energy equation produces the unshocked root $\chi =1$ and the compressive rootSince ${\rho }_1{u}_1^2/p_1=\gamma {\mathcal{M}}_1^2$, the strong-shock limit has negligible upstream thermal pressure compared with ram pressure. The Strong-shock Rankine-Hugoniot conditions giveFor unchanged composition, the ideal gas law implies $T_2/T_1=(p_2/p_1)/\chi$, soFor $\gamma =5/3$, the compression tends to four and $T_2\simeq 3\bar{m}{u}_1^2/(16k_B)$, where $\bar{m}$ is mean particle mass.

With constant specific heat $c_v$, entropy production in a perfect-gas shock is measured by the change in specific entropyThe strong-shock expression is thereforeThe exact jump is positive for every compressive shock with ${\mathcal{M}}_1>1$, in accordance with the Second law of thermodynamics; the asymptotic expression is only for ${\mathcal{M}}_1\gg 1$. There is a normalization issue in the printed description: this logarithmic formula with constant $c_v$ describes entropy per unit mass. The true entropy density is $\rho s$, so its difference is ${\rho }_2{s}_2-{\rho }_1{s}_1$, rather than the same logarithmic jump with a constant heat capacity per unit volume. A material parcel's irreversible entropy increase should be assessed using $\Delta s$.

Downstream gas is denser and hotter. For optically thin collisional emission the loss per unit volume is typically $n_e{n}_i\Lambda (T)$, whereas stored thermal energy scales approximately as $nT$. Compression can therefore reduce the radiative gas cooling time; excitation and ionization at the higher temperature can open additional cooling channels. The actual efficiency still depends on the astrophysical cooling function, composition and optical depth. In a radiative shock with separated scales, the temperature first rises across a narrow dissipative layer, then decreases through a much broader cooling region of characteristic length ${\ell }_{\mathrm{cool}}\sim u_2{t}_{\mathrm{cool}}$. Subsonic cooling usually brings further compression with approximately constant pressure. The immediate microscopic shock thickness and the macroscopic cooling length are distinct.

Let state $f$ be the fully cooled gas, with $T_f=T_1$ and the same composition. Set $a^2=p_1/{\rho }_1$, the isothermal sound speed squared. Across the entire shock plus cooling layer, mass and momentum are still conserved, but gas energy alone is not. With ${\chi }_f={\rho }_f/{\rho }_1$,The nontrivial isothermal shock therefore hasIf the upstream Mach number were defined with $a$ instead, this would be ${\rho }_f/{\rho }_1={\mathcal{M}}_{\mathrm{iso}}^2$. Retaining the distinction avoids losing the factor $\gamma$.

The gas's specific entropy rises irreversibly at the hot shock and then decreases as escaping radiation removes heat. Since the final temperature equals the initial temperature,This decrease for the gas is consistent with the Second law of thermodynamics: the emitted radiation and its eventual surroundings carry entropy away. It does not make the microscopic compressive shock reversible. The figure illustrates both the hot entropy jump and the subsequent cooling; its spatial cooling profile is schematic because no cooling law has been specified.

Immediately upstream, free fall from rest at infinity gives $u_1^2=2G{M}_0/r_0$. Treat the surface shock as locally planar and stationary, with negligible upstream pressure. The Strong-shock Rankine-Hugoniot conditions for a monatomic gas, $\gamma =5/3$, give ${\rho }_2/{\rho }_1=4$ and $u_2=u_1/4$. momentum conservation then givesUse the source's gas-constant convention $p=\rho (R^{\ast }/\mu )T$. The immediate postshock temperature isHere adiabatic means that the jump does not lose appreciable heat to the surroundings; the shock is nevertheless irreversible and raises entropy. The gas still has velocity $u_1/4$ immediately downstream, so setting all upstream kinetic energy equal to downstream internal energy would give the wrong coefficient. Later settling, radiative cooling or changing ionization can alter the temperature; they are outside this immediate, fixed-composition perfect-gas jump.

A finite-amplitude compression can steepen because different parts of the wave travel at different characteristic speeds. For a right-moving simple wave in a perfect gas, one may write $dx/dt=c_0+(\gamma +1)u/2$ in terms of the local velocity perturbation $u$. Faster compressive characteristics catch slower ones. When the smooth inviscid description would become multivalued, a physical compression shock replaces it. A rarefaction normally spreads rather than forming an entropy-violating expansion shock.^[ref]

Across a locally planar shock, use velocities relative to its moving front. The Rankine-Hugoniot conditions for a perfect gas conserve mass, momentum and total energy flux:where $h=\gamma p/[(\gamma -1)\rho ]$ is the specific enthalpy. For upstream normal Mach number $M_1>1$,Thus a normal compression shock increases pressure, density and temperature, reduces the normal velocity, and changes a supersonic upstream flow to subsonic downstream flow in the shock frame. In the strong-shock limit, compression approaches $(\gamma +1)/(\gamma -1)$, which is four for a monatomic gas. A shock conserves total energy but generates entropy; an adiabatic shock is not an isentropic compression. For a calorically perfect gas, stagnation temperature is conserved while stagnation pressure falls.^[ref]

Real dissipative processes resolve the narrow front; the jump conditions do not require its detailed internal profile. In a continuum-gas experiment a shock tube can produce a moving shock by releasing a pressure discontinuity. pressure probes separated along the tube measure arrival speed and the pressure jump; distinguishing the shock from the following contact surface is essential. A contact surface can have a density/entropy jump with continuous pressure and normal velocity, whereas a shock carries nonzero mass flux through its front.

Write the polytropic law as $p=K{\rho }^{1+1/n}$, with constant $K$ and $n>0$. Its pressure potential, or polytropic enthalpy relative to zero density, isHydrostatic balance is $dH/dr=-d\psi /dr$. If the surface has zero density and potential ${\psi }_0$, integration gives $H={\psi }_0-\psi$. At the centre, $\Delta \psi \equiv {\psi }_0-{\psi }_c=(n+1)K{\rho }_c^{1/n}>0$. ThereforeInsert $\psi ={\psi }_0-\Delta \psi \theta$ and $\rho ={\rho }_c{\theta }^n$ into the spherical Poisson equation for Newtonian gravity:The central conditions are $\theta (0)=1$ and ${\theta }^{\prime }(0)=0$; the latter follows from a regular spherically symmetric centre. This reproduces the supplied coordinate and potential scaling rather than introducing an unrelated normalization.

For the sound-speed task, the adiabatic sound speed of a perfect polytropic gas isA diatomic gas with active rotational but frozen vibrational degrees of freedom has $\gamma =7/5$ and $n=5/2$. With the requested surface-potential gauge ${\psi }_0=0$,The interior potential is then negative. Choosing zero potential at the surface is legitimate; for a finite isolated sphere it is a different additive gauge from setting potential to zero at infinity. Both choices cannot generally be imposed simultaneously.

For the mass-radius task, let ${\xi }_1$ be the first zero of the regular Lane-Emden function and $Y_n=-{\xi }_1^2{\theta }^{\prime }({\xi }_1)$. A finite-radius polytrope hasThusAt fixed central temperature, fixed composition and fixed polytropic index, $M\propto R$. These are homologous polytropes at fixed central temperature. They do not in general have constant $K$; imposing constant entropy instead gives a different polytropic mass-radius sequence. If $n$ or $\mu$ varies between stars, the displayed dimensionless structure factor and composition factor must be retained: equal $T_c$ alone is not a universal exact proportionality for arbitrary stellar models.

Treat the gas as a smooth quasi-one-dimensional flow through a rigid passage, neglecting axial body forces and heat input. It is the confining medium that is incompressible; the flowing gas must remain compressible. With area $\mathcal{A}(x)$ and axial speed $u$, continuity and inviscid momentum balance giveCombining their logarithmic differentials yields the nozzle area-velocity relationA smooth crossing of $\mathcal{M}=1$ requires an area extremum. The given radius is positive and has derivative $r^{\prime }(x)=16r_{0}x\exp [-16x^2/L^2]/L^2$, which vanishes only at $x=0$ within the slab and changes from negative to positive there. Thus this station is the unique throat, with $r_s=r_0/2$ and ${\mathcal{A}}_s=\pi r_0^2/4$. The specified subsonic inlet and supersonic outlet select the smooth accelerating branch, soA throat by itself does not guarantee transonic flow for arbitrary reservoir and back pressures; that branch is specified here. Before the throat a converging section accelerates subsonic flow; beyond it a diverging section accelerates supersonic flow.^[ref]

For the temperature task, evaluate the mass flux and sound speed at the throat using $p=K{\rho }^{\gamma }$:The ideal-gas relation then givesThis expresses the sonic temperature in terms of the entropy constant, geometry, mass flux and composition; no numerical value of $K$ or $\dot{M}$ is provided. Equivalently $c_s^{\gamma +1}=\gamma K(\dot{M}/{\mathcal{A}}_s{)}^{\gamma -1}$ and $T_s=\mu c_s^2/(\gamma R^{\ast })$. The slab length drops out because the sonic thermodynamic state depends on the minimum area, not on the axial distance over which a lossless smooth nozzle contracts. As a check, conserved stagnation enthalpy gives $T_s=2T_{\mathrm{stag}}/(\gamma +1)$.^[ref]

The result assumes cross-sectional thermodynamic uniformity and negligible transverse kinetic energy. Exact multidimensional flow through a sharply varying passage can have a curved sonic surface; the radius profile alone does not solve that boundary-value problem. The original numerical plot below computes the standard quasi-one-dimensional branch, rather than reproducing the supplied passage sketch.

Assume a constant downward gravitational acceleration $g$, constant composition, and an ideal gas law $p=\rho \mathcal{R}T$, where $\mathcal{R}=k_B/(\mu m_u)$ is the gas constant per unit mass. Here $m_u$ is the unified atomic mass unit; this convention makes the specified mean molecular weight dimensionless. Hydrostatic equilibrium givesIntegrating from sea level yieldsDividing by the temperature ratio gives the hydrostatic atmosphere with a constant temperature lapse rate:The model requires $T_0-\alpha z>0$ and is used here only within the troposphere.

Convert the atmospheric lapse rate to $\alpha =6.5\times 10^{-3}\mathrm{K}{\mathrm{m}}^{-1}$. At $11\mathrm{km}$, $T=228.5\mathrm{K}$. With $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, $k_B=1.380649\times 10^{-23}\mathrm{J}{\mathrm{K}}^{-1}$, $m_u\simeq 1.66054\times 10^{-27}\mathrm{kg}$ and $\mu =28$, the density exponent is $4.0825$. HenceThe mass density is therefore approximately one third of its sea-level value in this model.

The original PDF has absolute-value bars around both gradients; they are missing from the transcription. The stability argument must respect the sign of the upward coordinate.

Displace a small parcel upward by $\xi$. Assume it remains in pressure equilibrium with its surroundings but exchanges negligible heat, so it follows an adiabatic process. For a perfect gas with adiabatic index $\gamma$, $T\propto p^{(\gamma -1)/\gamma }$ along that parcel's adiabat. Its temperature gradient is thereforeThe last step uses hydrostatic equilibrium, and $c_p$ is the specific heat capacity at constant pressure. At its new height, the temperature difference between the parcel and the environment is $(T_{\mathrm{ad}}^{\prime }-T^{\prime })\xi$. At equal pressure, its relative density excess is the negative of its relative temperature excess. The upward buoyancy acceleration is thusThis is the squared Brunt–Väisälä frequency. Convective stability requires $N^2>0$, or, in signed-gradient form,For a temperature decreasing upward, this is exactly the PDF's Schwarzschild criterionThe absolute-value condition is also sufficient more generally, although a sufficiently steep temperature inversion remains stable even when that stronger condition fails. Reversing the signed inequality, as the bar-free transcription does, would instead describe an unstable temperature decrease.

A plasma contains mobile charged species whose collective electromagnetic response matters. On scales large compared with the Debye length, positive and negative charge densities nearly cancel: this is quasineutrality, not the absence of electric currents or electric fields. For electrons of temperature $T_e$ and number density $n_e$, the characteristic plasma screening and oscillation scales areA collective plasma description requires many particles within a screening volume; long-wavelength, sufficiently slow behavior can often be described by magnetohydrodynamics with an appropriate fluid closure.

An individual particle feels the Lorentz force $q(\mathbf{E}+\mathbf{v}\times \mathbf{B})$. In a uniform magnetic field, its transverse motion gyrates while its parallel motion remains free, giving a helical path when both components are present. The nonrelativistic cyclotron frequency and gyroradius areWhen the system scale is much larger than $r_L$ and the relevant time scale longer than ${\Omega }_c^{-1}$, the field strongly organizes transport across it, provided collisions do not interrupt the gyromotion. A perpendicular electric field also gives the common guiding-centre drift ${\mathbf{v}}_E=\mathbf{E}\times \mathbf{B}/B^2$.

For a conducting fluid, the macroscopic Lorentz force density has the useful decompositionThese terms are magnetic pressure and magnetic tension. Magnetic tension resists bending, while magnetic-pressure gradients resist nonuniform compression. The plasma beta $\beta =2{\mu }_{0}p/B^2$ compares thermal and magnetic stresses.

With uniform magnetic diffusivity ${\eta }_m$, the resistive induction equation isIf the magnetic Reynolds number $R_m=UL/{\eta }_m\gg 1$, ideal magnetohydrodynamics applies approximately and magnetic flux freezing links the field geometry to fluid motion. Finite diffusivity permits field slippage and magnetic reconnection. Field tension supports Alfvén waves with $v_A=B/\sqrt{{\mu }_0\rho }$; coupled magnetic and gas compression also supports magnetosonic waves. These effects allow a field to transmit forces and angular momentum between distant parts of an astrophysical plasma.

An accretion disk forms when infalling material has enough angular momentum to orbit before it can reach the central object. In a thin disc dominated by a central mass $M$, the orbital motion is approximately Keplerian rotation:Vertical hydrostatic equilibrium gives the disk scale height $H\sim c_s/{\Omega }_K$; a geometrically thin disc has $H/R\ll 1$. The surface density is $\Sigma =\int \rho dz$, and a steady inward mass flow rate is $\dot{M}=-2\pi R\Sigma v_R>0$.

Moving inward requires losing specific angular momentum. Internal stresses transport angular momentum outward; much of the mass can then accrete while an outer fraction spreads to larger radius. Molecular viscosity alone is generally inadequate for rapid astrophysical accretion. Turbulent and magnetic stresses, including those generated by the magnetorotational instability, provide effective transport. A common parametrization is $\nu =\alpha c_{s}H$ for an effective kinematic viscosity, with $\alpha$ a dimensionless stress parameter.