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In a steady axisymmetric magnetohydrodynamic wind, axisymmetric pressure and Newtonian gravity exert no azimuthal torque. The azimuthal momentum equation is $\rho {\mathbf{u}}_p\cdot \nabla (s{u}_{\varphi })={\mu }_0^{-1}{\mathbf{B}}_p\cdot \nabla (s{B}_{\varphi })$. Mass conservation and the solenoidal vector field condition imply $\nabla \cdot {\mathbf{F}}_J=0$. With magnetohydrodynamic mass loading $\rho {\mathbf{u}}_p=\kappa {\mathbf{B}}_p$, the transported angular momentum per unit poloidal magnetic flux is $\ell =\kappa s{u}_{\varphi }-s{B}_{\varphi }/{\mu }_0$. It is constant on connected regular poloidal flux surfaces. The usual specific magnetohydrodynamic angular-momentum invariant is $L=\ell /\kappa$, provided $\kappa \ne 0$; confusing these two normalizations loses a factor of mass loading in the wind torque.

The momentum thickness weights the streamwise velocity deficit by the local velocity carrying it:Thus $\rho U^2\theta$ measures a momentum flux deficit per unit span relative to transporting the same local mass flow rate at $U$. It is not $\rho \int (U^2-u^2)dy$, which compares a different reference flux. A Blasius boundary layer has $\theta \simeq 0.664115\sqrt{\nu x/U}$; the ratio of displacement thickness to momentum thickness is about $2.591$.^[ref]

For the steady flow over an impermeable flat plate at constant $U$, integrating the boundary-layer equation and using mass conservation givesThis connects growth of the momentum thickness to local wall shear stress. Consequently the drag per unit span on one face up to $x=L$ is $D^{\prime }=\rho U^2\theta (L)$ in the ideal Blasius boundary layer with $\theta (0)=0$. Measured velocity profiles can therefore estimate integrated drag without differentiating noisy near-wall readings. The identity assumes no wall suction and zero outer pressure gradient.^[ref]

A localized temperature adjustment separates regions with different fluid density, viscosity and Darcy speed. Mass conservation in the moving frame gives the displayed plateau-depth relation. A speed stated as $V_T=\Gamma u_c$ with $u_c$ a Darcy velocity has effective retardation parameter $\varphi \Gamma$. Setting both laboratory-frame mass fluxes equal neglects storage associated with passage of the temperature front.

For a triangular-profile line plume with centreline density $\rho$, speed $W$, half-width $b$, ambient density ${\rho }_0$, and inward relative edge speed $e$, volume, mass and momentum conservation giveSubtracting mass conservation from ${\rho }_0$ times volume conservation also gives buoyancy conservation. Batchelor entrainment supplies a closure for $e$.

A localized release of actual fluid volume $V$ per unit width on a reflecting half-line $x\ge 0$ has the Barenblatt solution $h=[L^2-x^2{]}_{+}/(6\lambda t)$, where $L^3=9\lambda Vt/\varphi$ and $\lambda =k\Delta \rho g/(\varphi \mu )$. The mass conservation normalization is $\varphi {\int }_0^{L}hdx=V$. Zero slope at the reflecting boundary gives zero Darcy flux. For symmetric spreading on the full line, the same formula has $L^3=9\lambda Vt/(2\varphi )$ when $V$ denotes the total released volume. A finite initial patch is not determined by volume alone; this exact source profile has a concentrated initial release.

Assume nonzero upstream speed, so $F>0$. For steady constant-density flow, mass conservation gives the constant volume fluxThe Bernoulli equation states that $u^2/2+p/\rho +gz$ is constant along each streamline of a steady inviscid constant-density flow under gravity. It need not have the same value across unrelated streamlines in a general rotational flow. Here uniform upstream velocity supplies the same head, and the slowly varying, nearly horizontal channel permits hydrostatic pressure $p=p_{\mathrm{atm}}+\rho g(h-z)$. Neglecting losses, the Bernoulli equation therefore givesSubstitute $u=UWH/(wh)$ and set $s=h/H$, $F=U^2/(gH)$, the square of the upstream Froude number. Rearrangement givesFor positive depth, this cubic increases to its maximum at $s_{\ast }=(F+2)/3$ and then decreases until it reaches zero at $(F+2)/2$. Its maximum is $G(s_{\ast })=(F+2{)}^3/(27F)$. Consequently the critical width of a rectangular open-channel contraction isAt equality the two positive-depth branches coalesce and the local Froude number equals one.

When $w>w_c$ everywhere, a smooth solution connected to the upstream state cannot change branch without passing through that coalescence. At $w=W$ the upstream depth is $s=1$, with $G^{\prime }(1)=2(F-1)/F$. If $F<1$, this point is on the descending, subcritical branch. At a wider location $w>W$, the left-hand side falls below $1=G(1)$, so the depth on that branch must increase: $h>H$ for subcritical upstream flow $F<1$. If $F>1$, the upstream state is on the ascending, supercritical branch and decreasing the left-hand side decreases $s$: $h<H$ for supercritical upstream flow $F>1$.

For $F=1$, $w_c=W$ and the upstream state is critical. A literally constant-width upstream region then conflicts with the strict assumption $w>w_c$ everywhere. If the upstream state is interpreted only as an asymptotic critical limit, a widened channel has one solution with $h<H$ and one with $h>H$; a further branch condition is required. Merely solving the cubic at one location, without retaining the upstream-connected branch, would not determine the physically continued depth.

Assume axisymmetry in a thin disk orbiting a fixed central mass $M$. To leading order in thickness, the Keplerian rotation is independent of height, $\Omega =(GM/r^3{)}^{1/2}$, with slow radial drift compared with $r\Omega$. Neglect radial pressure support in the orbital balance, changes in the central potential, mass loss through the disk surfaces and applied surface torques. The Cauchy stress tensor can represent viscous or effective turbulent/magnetic transport; only its vertically integrated azimuthal component enters the conservation of angular momentum equation.

Define the surface density and its mass-weighted radial velocity byand set $W_{r\varphi }=\int T_{r\varphi }dz$. Integrating mass conservation vertically givesThe azimuthal equation in cylindrical coordinates, multiplied by $r$ and combined with mass conservation, is the conservative equation for specific angular momentum $l=r^2\Omega$. The cylindrical stress tensor divergence contributes $r^{-1}{\partial }_r(r^2{T}_{r\varphi })+{\partial }_z(r{T}_{\varphi z})$. With vanishing surface torque, vertical integration therefore givesBecause $l(r)$ is fixed in time, subtracting $l$ times the mass equation yieldsDefine the effective viscosity defined by disk stress byThis sign convention treats $T_{r\varphi }$ as the shear stress appearing with $+\nabla \cdot \mathbf{T}$. For the Newtonian fluid stress tensor it reduces to the density-weighted viscosity of a disk, $\bar{\nu }={\Sigma }^{-1}\int \rho \nu dz$, rather than an unweighted height average.

Using $l=\sqrt{GMr}$ and $W_{r\varphi }=-(3/2)\bar{\nu }\Sigma \Omega$ in the conservation of angular momentum equation givesSubstitution into mass conservation proves the Keplerian viscous diffusion equation:The outward viscous torque in an accretion disk is $\mathcal{G}=-2\pi r^2{W}_{r\varphi }=3\pi \sqrt{GM}\sqrt{r}\bar{\nu }\Sigma$.

For $\bar{\nu }=Ar$ with $A>0$, make the heat-equation transform for linear-radius disk viscosityWriting $y=\sqrt{r}$, the diffusion equation becomes $g_t=(3A/4)g_{yy}$; the displayed rescaling of $y$ gives exactly $\boxed{g_t=g_{xx}}$, with the original time unchanged. The torque is $3\pi \sqrt{GM}g$, so the zero-torque inner boundary condition at the origin becomes $g(0,t)=0$ on the half-line $x>0$.

A positive similarity solution obeying this boundary condition isFor example, substitution of $g=t^{-1}F(x/\sqrt{t})$ reduces the heat equation to $F^{\prime \prime }+(\eta /2)F^{\prime }+F=0$, and $F=C\eta e^{-{\eta }^2/4}$ satisfies it and vanishes at zero. A replacement of $t$ by $t+t_0$ gives a finite-mass initial profile at time zero if desired.

To determine the global integrals, transform the mass and angular momentum:The Gaussian integrals giveThus the accreting similarity disk with viscosity proportional to radius loses mass as $t^{-1/2}$ while conserving angular momentum; its characteristic radius grows proportional to $At$. The accretion rate at the origin is $-{\dot{M}}_d=2\pi C\sqrt{3/A}{t}^{-3/2}$. Its radial velocity ${\bar{u}}_r=r/t-3A/2$ shows inward flow at small radius and outward spreading at large radius. The integrable inner density singularity is part of the idealized extension to $r=0$.

For the other profile, direct differentiation givesIt also satisfies the heat equation, but has $g(0,t)=t^{-1/2}\ne 0$ and $g_x(0,t)=0$. It therefore has a nonzero inner torque and zero mass flow at the origin, not the previous zero-torque inner boundary condition. HereAn arbitrary positive amplitude multiplies these quantities. This constant-mass disk with viscosity proportional to radius spreads outward under a central torque that supplies angular momentum, with no ongoing central mass supply. It has the reflecting Neumann boundary condition for the heat equation and can be regarded as the spreading of an initially central mass distribution driven by an applied torque. It is distinct from the accreting zero-torque solution; the ideal origin does not model a finite stellar surface or a relativistic inner disk.

Let $H=h_{00}+\zeta$ and write the horizontal shallow water equations on the constant f-plane asKeeping only terms linear in the disturbance about rest gives the linearized shallow water equationsDefine the horizontal divergence $\delta =u_x+v_y$ and relative vorticity $q=v_x-u_y$. Taking respectively the divergence and vertical curl of the momentum equation yieldsTogether with ${\zeta }_t=-h_{00}\delta$, this provesThe nonlinear shallow-water potential vorticity is $Q=(f+q)/H$. The nonlinear vertical vorticity equation gives $D(f+q)/Dt=-(f+q)\delta$, while mass conservation gives $DH/Dt=-H\delta$; their ratio is therefore materially conserved. Expanding about rest,Advection of its constant background contributes nothing at first order, explaining why the displayed linear anomaly is conserved at each fixed position.

A further time derivative eliminates $q$ and $\zeta$ from the divergence equation:For a plane wave with horizontal wavevector $(k,l)$, write $K=\sqrt{k^2+l^2}$. More directly, substituting the plane wave into the three linear equations gives a coefficient matrix with determinant proportional to $\omega [{\omega }^2-f^2-c_0^2{K}^2]$. Thus the full linear rotating shallow-water dispersion relation has one balanced branch and two wave branches:The latter are inertia-gravity waves. The positive branch starts at $f$ with zero slope and tends to the straight line $c_{0}K$; the negative branch is its reflection. The requested sketch, including the stationary branch, is

For the positive-frequency branch, the magnitudes of the phase velocity and group velocity areThe group velocity vector is $c_0^2(k,l)/\omega$; the negative branch reverses its direction. Near $\omega =f$, namely $K{R}_D\ll 1$ with the Rossby deformation radius $R_D=c_0/f$, $c_p\sim f/K$ is large while $c_g\sim c_0^{2}K/f$ is small. The $K=0$ limit is an inertial oscillation, for which a spatial phase speed is not defined. Far above $f$, both speeds approach $c_0$, the shallow water gravity-wave speed, with $c_p>c_0$ and $c_g<c_0$ at finite $K$.

On the zero-frequency branch, geostrophic balance givesFor a nonconstant plane wave, $\hat{q}=-(g{K}^2/f)\hat{\zeta }$: relative vorticity and elevation have opposite signs. They can both be nonzero and spatially structured, whereas the horizontal divergence vanishes identically. A constant elevation is also a zero-frequency solution, with zero velocity and zero relative vorticity.

For the initially resting ridge, set ${\zeta }_0(x)=ϵh_{00}{e}^{-x^2/L^2}$ and assume an unbounded domain, uniform in $y$. Initial potential vorticity fixesIn the balanced remainder, insert $q_s=(g/f){\zeta }_s^{\prime \prime }$ to obtainThe decaying Green's function of this modified Helmholtz equation gives the Geostrophic adjustment of a Gaussian height ridge explicitly:The initial outward pressure force creates a cross-ridge current; the Coriolis acceleration turns that current and builds the along-ridge geostrophic flow. The remainder is radiated inertia-gravity waves, not a dissipative disappearance of energy.

For precision, use the Fourier transform $\hat{\zeta }(k)=\int \zeta (x)e^{-ikx}dx$. Initial rest implies ${\zeta }_t(x,0)=0$, and the conserved anomaly givesConsequentlyThese formulas satisfy both initial conditions. They also show that the waves carry no linear potential vorticity anomaly: $q=(f/h_{00})(\zeta -{\zeta }_0)$ and $\delta =-{\zeta }_t/h_{00}$ reconstruct the remaining fields, with $u$ and $v$ obtained from their derivatives. On the infinite line, dispersive wave packets propagate away and decay locally, leaving the balanced ridge and its oppositely directed flanking currents. Conservation of energy still holds globally. In a reflecting finite basin, standing waves may persist, so a permanent pointwise relaxation is not guaranteed without dissipation.

If $L\gg R_D$, most of the initial elevation remains balanced: ${\zeta }_s\simeq {\zeta }_0$ with small smoothing corrections. If $L\ll R_D$, the balanced ridge spreads to width $R_D$, and away from the original narrow coreThus the final central height is much smaller than the initial one and the wave component is substantial. The integral of the balanced elevation equals the initial integral, since the Green's function has unit integral; the oscillatory component has zero net added volume.

Integrating the given expansion equation gives $R^{3/2}=t/t_0$, with the integration constant fixed by $R(t_0)=1$. Thus the scale factor and Hubble parameter are $R=(t/t_0{)}^{2/3}$ and $H=2/(3t)$. The Newtonian derivation of the flat Friedmann equation gives $H^2=8\pi G\rho /3$, soUse the comoving coordinate $\mathbf{q}=\mathbf{x}/R$ and the density contrast $\delta ={\rho }_1/\rho$. Linearized mass conservation, the peculiar-velocity Euler equation for fluid motion and the Poisson equation for the peculiar gravitational potential areDifferentiate the first equation, substitute the second and use the third. A Fourier component of comoving wavenumber $k$ obeysSince $c^2{k}^2/R^2=k^2{c}^2(t_0)t_0^{2\gamma -2/3}{t}^{-2\gamma +2/3}$, the normalization is $\Lambda =kc(t_0)t_0^{\gamma -1/3}$, as printed in the PDF. The resulting linear cosmological density perturbation equation isTo reduce it to a Bessel equation, put $\nu =\gamma -4/3$, $z=\Lambda /(\nu t^{\nu })$ and $\epsilon =t^{a}y(z)$. As $\dot{z}=-\nu z/t$, multiplying the substituted equation by $t^{2-a}$ givesTaking $a=-1/6$ makes the first two coefficients proportional to those in the Bessel equation, and the remaining constant is $-25/36$. HenceFor noninteger $\lambda$, the Bessel density modes of a polytropic expanding universe areThere is an exceptional case within the allowed interval: if $\lambda =n$ is an integer, $J_{-n}=(-1{)}^n{J}_n$, so these are not two independent solutions. The full solution then uses $t^{-1/6}[A{Y}_n(z)+B{J}_n(z)]$, with the Bessel function of the second kind providing the missing branch. For example $\gamma =29/18$ gives $\lambda =3$. A basis using $J_{\lambda },Y_{\lambda }$ is valid for every allowed exponent.

For early times, $z\to \infty$. The large-argument asymptotic expansion of the Bessel function of the first kind has oscillatory phase $z-\pi a/2-\pi /4$ and envelope $(2/(\pi z){)}^{1/2}$ for order $a$; the Bessel function of the second kind has the same envelope. ThusThis decreases because $3\gamma <5$. At late times $z\to 0$, the leading powers are $J_{\pm \lambda }(z)\propto z^{\pm \lambda }$ for noninteger order, giving $t^{-1}$ and $t^{2/3}$. The $Y_n$ branch gives the same growing power in the integer case. Therefore a generic perturbation grows as $t^{2/3}$; an initial condition with zero growing-mode coefficient instead has the decaying $t^{-1}$ mode.

To find the growth scale including its dependence on $\nu$, set $\epsilon =t^{2/3}h(t)$. The equation becomesSubstitute $h=1+a{\Lambda }^2{t}^{-2\nu }+\cdots$. The leading coefficient equation is $-2\nu (5/3-2\nu )a+1=0$. The pressure correction to a growing polytropic cosmological mode is consequentlySince $ck/R=\Lambda t^{-\nu -1}$ and $t^{-2}=6\pi G\rho$, the correction becomes small when $(ck/R{)}^2{t}^2$ is small compared with $2\nu (5/3-2\nu )$. The factor $2(5/3-2\nu )$ is between two and $10/3$, giving the stated order-of-magnitude onset conditionFor a well-developed asymptotic growing mode, replace $\lesssim$ by $\ll$. This is a Jeans instability interpretation: the physical acoustic frequency must be sufficiently low compared with the gravitational expansion scale, or equivalently the physical wavelength must be sufficiently long for self-gravity to overcome pressure support. It is an approximate condition for the dust-like growing branch, not an exact initial-condition-independent instant at which any perturbation begins increasing. The instantaneous sign change of the coefficient in the original equation instead occurs at $c^2{k}^2/R^2=4\pi G\rho$.

In the dominant central Newtonian gravitational potential, conservation of energy for gas falling from rest at infinity gives $u^2/2=G{M}_c/R$, hence $u=a{R}^{-1/2}$ with $a=(2G{M}_c{)}^{1/2}$. The characteristic stellar speed has the same radial scaling. For random stellar directions, the cross term in $|\mathbf{u}-\mathbf{v}{|}^2$ averages to zero, so the typical relative speed is also $V=b{R}^{-1/2}$. With the stated kinetic-energy estimate, an rms convention gives $V^2\simeq 4G{M}_c/R$; another convention changes only the constant $b$.

In the steady spherical inflow approximation, mass conservation gives $A=4\pi R^2\rho u$. ThusThis leading calculation treats the inward gas flux as essentially undepleted over the region used. Substantial distributed removal by stars would make the local flux a function of $R$, an important limitation discussed below.

Insert these scalings in the Bondi--Hoyle--Lyttleton accretion rate. The factors $R^{3/2}$ from $V^{-3}$ and $R^{-3/2}$ from $\rho$ cancel:This is quadratic accretion in a central Keplerian potential. Write its fixed coefficient as $\lambda =\Lambda$. Integrating $d(1/M)/dt=-\lambda$ givesThe cancellation of radius means that orbital variation of $R$ alone does not change this coefficient, provided the assumed steady gas profile, central mass and relative-velocity scaling remain fixed.

Conserve the number of stars along these deterministic mass trajectories. Inverting the map and taking its Jacobian determinant givesThe mass-spectrum transport under quadratic accretion is thereforeFor a narrow but nonzero initial mass interval, the mapping stretches the high end especially strongly. If its endpoints are $m_1<m_2$, their evolved ratio is $(m_2/m_1)(1-\lambda m_{1}t)/(1-\lambda m_{2}t)$, which can become large as $t$ approaches $1/(\lambda m_2)$ from below. In an intermediate high-mass range with $\lambda Mt\gg 1$, mapped back into a portion of the initial interval where $F_0$ changes little,Here $M_{\ast }$ is near $1/(\lambda t)$ on the populated side of the initial interval. Finite initial support gives a finite upper mass cutoff at every time before the earliest blow-up; this is an approximate broad-range power law, not a literal infinite tail of a finite pre-blow-up population. A strictly monochromatic initial population remains monochromatic under identical accretion coefficients and does not generate a spectrum.

The limitations of this accretion model are substantial. The star cluster must remain centrally dominated, and both the central mass and gas supply must remain effectively fixed. Stellar capture depletes the gas flux and eventually invalidates the $R^{-3/2}$ profile. Gas pressure, angular momentum, shocks, radiative feedback and winds can reduce capture relative to the cold Bondi–Hoyle accretion estimate; a broad relative-speed distribution is not faithfully represented by one typical speed, especially because capture scales as $V^{-3}$. A growing star can perturb the surrounding potential, overlap capture regions, change its orbit through momentum loading, or compete with its neighbors for the same gas. Finally, number conservation excludes mergers, new star formation and stellar losses. The formal finite-time mass divergence cannot persist under a finite available gas mass, so the inferred mass-spectrum slope applies only while these approximations remain adequate.

Fix the sign convention first: depth $z$ increases downward, while the vertical displacement amplitude $\xi$ is positive upward. Write $\Pi =\delta p$ for the Lagrangian pressure perturbation, and $p_1,{\rho }_1$ for Eulerian fluid perturbations. Hydrostatic equilibrium gives $dp/dz=\rho g$, and the adiabatic relation gives $\delta \rho =\Pi /c^2$. Since the vertical displacement vector has component $-\xi$ along increasing $z$,In the Cowling approximation, horizontal momentum gives ${\mathbf{\xi }}_h=i\mathbf{k}{p}_1/(\rho {\omega }^2)$. The mass conservation equation $\delta \rho =-\rho (i\mathbf{k}\cdot {\mathbf{\xi }}_h-{\xi }^{\prime })$ and vertical momentum ${\omega }^2\rho \xi =-p_1^{\prime }+g{\rho }_1$ then eliminate the horizontal displacement. The plane-parallel Cowling displacement-pressure equations areFor ${\omega }^2=gk$, the second equation admits $\Pi =0$ and the first reduces to ${\xi }^{\prime }=-k\xi$. Thus the plane-parallel stellar f-mode hasIts Eulerian eigenfunctions are $p_1=\rho g\xi$ and ${\rho }_1={\rho }^{\prime }\xi$, while $\delta \rho =0$. Hence it has no Lagrangian compression even in a stratified star. It decays inward but grows upward, explaining the failure of a pure small-amplitude continuation to $z\to -\infty$.

For the lower part of the matched chromospheric interfacial mode, put ${\omega }^2=gk(1-\epsilon )$, $\xi =e^{-k(z-z_0+\epsilon \varphi )}$ and $\Pi =\epsilon f\xi$. The pressure amplitude is of order $\epsilon$ because the limiting f-mode has identically zero Lagrangian pressure amplitude; the frequency correction perturbs that zero solution. Expand $g{k}^2/{\omega }^2=k(1+\epsilon )+O({\epsilon }^2)$ and ${\omega }^2-g^2{k}^2/{\omega }^2=-2gk\epsilon +O({\epsilon }^2)$. The leading equations areExclude the lower homogeneous component $f\propto e^{2kz}$, which would make $\Pi$ grow into the star. Integrating the remaining equation with its integrating factor givesThe accompanying correction is $\varphi (z)={\int }_{z_0}^z[1+(f/\rho )(1/g-1/(k{c}^2))]ds$, with the integrand evaluated at $s$. This fixes $\varphi (z_0)=0$ and determines the lower eigenfunctions to the stated order.

For the upper part, distinguish the rescaled sound speed by writing $\rho =\epsilon \hat{\rho }$ and $c^2={\epsilon }^{-1}{\hat{c}}^2$. Set $\Pi =\epsilon \psi$, $\psi =A{e}^{k(z-z_0+\epsilon \varphi )}$ and $\xi =\mu \psi$. To leading order the first displacement equation becomes ${\xi }^{\prime }=-k\xi -k\psi /(g\hat{\rho })$, while the pressure equation gives ${\psi }^{\prime }=k\psi +\epsilon [k\psi +2gk\hat{\rho }\xi ]+O({\epsilon }^2)$. ConsequentlyThe component $\mu \propto e^{-2kz}$ would give $\xi \propto e^{-kz}$, which diverges upward. Remove it by the upper boundary condition. The leading upper solution isIntegrating ${\varphi }^{\prime }$ with $\varphi (z_0)=0$ gives the first correction to the upper exponential. The displayed $\xi =\mu \psi$ satisfies the required decay only for compatible upper density profiles; integral convergence alone is not the full boundary condition.

To match, define the positive weighted integralsThe lower normalization is $\xi (z_0)=1$ and $f(z_0)=-2gk{I}_b$. Continuity of Lagrangian pressure perturbation makes the upper $A=f(z_0)$; continuity of fluid displacement then requires $\mu (z_0)A=1$. Since $1/\hat{\rho }=\epsilon /\rho$, the upper integral gives $\mu (z_0)=-(k/g)\epsilon I_a$. The two matching conditions therefore giveAs a sign and normalization check, two incompressible constant-density layers give $I_a=(2k{\rho }_a{)}^{-1}$ and $I_b={\rho }_b/(2k)$, so $\epsilon \simeq 2{\rho }_a/{\rho }_b$. This agrees with expansion of the exact interfacial gravity-wave dispersion relation ${\omega }^2=gk({\rho }_b-{\rho }_a)/({\rho }_b+{\rho }_a)$.

For validity, require $\epsilon \ll 1$, convergent weighted integrals, a matching level separating the low-density high-sound-speed atmosphere from the denser envelope, and small accumulated corrections $\epsilon k|\varphi |$ over the regions carrying the matching integrals. The upper expansion also needs $g/(k{c}^2)\ll 1$. The resulting upper displacement must tend to zero: for example if $\hat{\rho }\propto e^{a(z-z_0)}$ at great height, its leading amplitude is proportional to $e^{(k-a)(z-z_0)}$, requiring $a<k$; the weaker condition $a<2k$ merely makes the upper integral converge. Plane geometry, constant $g$ and the Cowling approximation must be appropriate on the wavelength scale. The physical mode amplitude must remain small compared with both its wavelength and local background scale heights, so the adiabatic fluid perturbation remains linear. If these requirements fail, neither the matching formula nor continuation of the leading expansions is controlled.

A stellar polytrope has a spatially constant coefficient $K$ in the relation $P=K{\rho }^{1+1/n}$; $n$ is its polytropic index. This relates pressure to mass density and need not describe an adiabatic gas. Combining hydrostatic equilibrium with mass conservation eliminates the enclosed mass:For $\rho ={\rho }_c{\theta }^n$, the polytropic equation of state gives $(1/\rho )dP/dr=(n+1)K{\rho }_c^{1/n}d\theta /dr$. Choose the Lane-Emden variables for a stellar polytrope withSubstitution then gives the dimensionless Lane-Emden equation:A regular stellar centre has $\theta (0)=1$, ${\theta }^{\prime }(0)=0$ and $m(0)=0$. In particular the regular central expansion is $\theta =1-{\xi }^2/6+O({\xi }^4)$. For an isolated finite-radius stellar polytrope with negligible external pressure, the surface is the first zero ${\xi }_1$ of $\theta$, and $R=\alpha {\xi }_1$. A core embedded in an envelope can instead end at a positive matching pressure, before this zero.

The integrated Lane-Emden equation is ${\xi }^2{\theta }^{\prime }=-{\int }_0^{\xi }{s}^2\theta (s{)}^{n}ds$. Thus mass conservation gives the Lane-Emden mass formula directly:For the proposed functional form, differentiation givesChoosing $\gamma =-1/2$ removes the extra polynomial factor; equality to $-{\theta }^5$ then requires $3\beta =1$. The regular polytrope of index five is thereforeIts enclosed mass and enclosed mean mass density areThere is no finite zero: its isolated radius is infinite. Nevertheless its total mass converges, since the outer mass density falls as $r^{-5}$:The zero mean mass density is a limiting volume average, not a statement that the material has zero mass density at finite radius.