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Use the Burgers reduction of weakly nonlinear rightgoing acoustics, with $\theta =t-x$, $X=Mx$, ${\partial }_t={\partial }_{\theta }$ and ${\partial }_x=-{\partial }_{\theta }+M{\partial }_X$. The equation of state expands asAt order one, mass conservation and momentum conservation give $({\rho }_0-u_0{)}_{\theta }=0$ and $(u_0-p_0{)}_{\theta }=0$. For the pure rightgoing acoustic wave with no independent leading mean density or velocity shift, ${\rho }_0=u_0=p_0$. This is the background choice implicit in the requested traveling-wave expansion; additional mean components would modify its phase convention. The source's duplicated phrase about expanding $p$ should refer to expanding $\rho$ as well as $u$.

At order $M$, the mass conservation equation becomeswhereas the momentum conservation equation becomesIn the latter equation, the two nonlinear products ${\rho }_0{u}_{0,\theta }$ and $-u_0{u}_{0,\theta }$ cancel. Substitute the equation of state relation into the former equation and add the two equations to eliminate the first corrections. The result isThe same calculation gives $u_1-p_1=-(\gamma +1)p_0^2/4+C(X)$, so the first corrections may be chosen without an explicit secular factor of $\theta$. This is why the retarded-time description can remain uniform for $|\theta |=O(M^{-1})$, provided the leading profile and its relevant derivatives remain bounded over that range and the slow propagation interval.

This Inviscid Burgers equation can develop a shock, so the source's uniformity assertion needs a pre-shock qualification. If $p_0(\theta ,0)=F(\theta )$ and $a=(\gamma +1)/2$, its method of characteristics givesThe smooth construction ends when this denominator vanishes; if $\max F^{\prime }>0$, the first such distance is $X_s=[a\max F^{\prime }{]}^{-1}$.

For viscosity write $\beta =M\nu$ with bounded positive $\nu$. Since $u_{0,xx}=p_{0,\theta \theta }+O(M)$, the order-$M$ momentum conservation equation instead reads $(u_1-p_1{)}_{\theta }+p_{0,X}=\nu p_{0,\theta \theta }$. Addition now yields the viscous Burgers equationIts positive diffusion resolves the inviscid steepening rather than supporting an unqualified smooth inviscid expansion through shock formation.

Estimate characteristic bulk terms. Put $ϵ=H/r\ll 1$ and assume smooth radial variation on scale $r$, secular evolution, and $\alpha \lesssim 1$. Vertical support gives $c_s^2\sim p/\rho \sim {\Omega }^2{H}^2$, while the alpha disk prescription and slow accretion implyThese velocity estimates follow from the viscous radial evolution scale and mass conservation. They are estimates for a slowly evolving thin disc, not a claim that geometric thinness excludes independently forced rapid vertical motion.

For vertical balance, the exact central force expands asAt $z\sim H$, the neglected correction has magnitude $\rho {\Omega }^{2}H{ϵ}^2$, compared with the retained force $\rho {\Omega }^{2}H$. Radial pressure support similarly changes the rotation from its Keplerian value byVertical meridional inertia scales as $u_r{u}_z/r$ or $u_z^2/H$, both $O({\alpha }^2{ϵ}^4{\Omega }^{2}H)$. Meridional viscous accelerations can scale as $\nu u_z/H^2$ or $\nu u_r/(rH)$, both $O({\alpha }^2{ϵ}^2{\Omega }^{2}H)$. Thus the gravity correction and possible viscous correction giveSome omitted terms are smaller than this leading order.

For the heating equation, the retained dissipation is $q_0^{+}\sim \rho \nu {\Omega }^2\sim \alpha p\Omega$. Radial pressure support and the finite-height correction change the dominant shear by a relative $O({ϵ}^2)$. Smooth vertical variation of the rotation gives ${\partial }_z(r\Omega )=O(ϵ\Omega )$, whose additional squared shear contributes $O({ϵ}^2{q}_0^{+})$. Meridional shear of order $u_r/H$ gives an additional relative $O({\alpha }^2{ϵ}^2)$ contribution.

Thermal advection and compressional work are of order $\rho u_r{c}_s^2/r$ and $p{u}_r/r$, respectively. Relative to $\alpha p\Omega$ they are $O({ϵ}^2)$. Thermal storage on the viscous evolution timescale $t_{\nu }\sim r^2/\nu$ has the same order, since $p/t_{\nu }\sim \alpha p\Omega {ϵ}^2$. For radiative diffusion with the same local diffusivity in different directions, smooth gradients give $F_r/F_z\sim H/r$, soConsequentlyThis is the thin-accretion-disk truncation error for the secular, radially smooth approximation. Neglect of disc self-gravity and radiation pressure, optical thickness, the opacity model and the idealized zero surface boundary have their own validity conditions; their errors do not become small solely because $H/r$ does. Near a sharp radial boundary layer or during thermal evolution on the heating timescale, the above estimates need not apply.

Take positive volumetric flow rate in the direction of increasing $x$. For an arterial pressure wave, the characteristic admittance of an arterial wave givesThe backward wave contributes the opposite flow sign. Write $p_j$ for each junction's complex pressure amplitude, suppressing the common $e^{i\omega t}$ factor, and put $r=RY$, $C_{\beta }=\cos \beta$, $S_{\beta }=\sin \beta$. The topology is the two paths $K\to L\to N$ and $K\to M\to N$; their intermediate and common distal junctions each have a hydraulic resistance. The coordinate convention in the source puts the upstream end at $-\ell$ and the downstream end at zero in each segment. Consequently the wave transfer matrix of an arterial segment isEquivalently $q_u=Y(C_{\beta }{p}_u-p_d)/(i{S}_{\beta })$ and $q_d=Y(p_u-C_{\beta }{p}_d)/(i{S}_{\beta })$, with their continuous limits understood when $S_{\beta }=0$.

Symmetry gives $p_L=p_M=p_s$. Distal mass conservation and the resistive relation $Q_N=p_N/R$ giveAt $L$, the incoming flow supplies both its bed and segment $D$, soChoose the incident amplitude $P_A$ at $K$. If $P_{AR}$ is the reflected amplitude in vessel $A$, pressure continuity equation gives $p_K=P_A+P_{AR}$, while inlet flow is $2Y(P_A-P_{AR})=2Y(2P_A-p_K)$. Equating this to the two outgoing segment flows yieldsEliminating $p_s,p_K$ from these three relations gives the wave transmission through a resistively loaded arterial loop resultFor large $r$ at fixed phase, $p_N=2P_A{e}^{-2i\beta }+O(r^{-1})$ and $p_M=C_{\beta }{p}_N+O(r^{-1})$. HenceThese leading terms must not be used as nonzero denominators near a phase at which the normal intermediate flow vanishes.

Put $U=Q/{\alpha }_0$, $R_0=R({\alpha }_0)$, $c_0^2={\alpha }_0{\tilde{P}}^{\prime }({\alpha }_0)$ and $\nu =\omega -kU$. Let the area and velocity perturbations have amplitudes $a$ and $w$. Linearizing mass conservation and momentum, and using $R_0^{\prime }=-n{R}_0/{\alpha }_0$, givesFor nonzero $k$, eliminate $w=\nu a/(k{\alpha }_0)$ to obtain the friction-induced instability of a collapsible-tube flow relationAt the stated threshold $nU=c_0$, it factors asTherefore one mode has real $\boxed{\omega =k(U+c_0)}$, while the other has $\omega =k(U-c_0)-i{R}_0$ and decays. The PDF's real-frequency statement is valid for the neutral branch, not for both roots. The Fourier convention $e^{i(kx-\omega t)}$ makes positive imaginary frequency mean growth.

Steady mass conservation makes $\alpha u=Q$. For uniform area, $u=U=Q/{\alpha }_0$ is constant and the pressure-area law of a collapsible tube gives $p_x=p_e^{\prime }(x)$. The acceleration vanishes, so momentum balance requiresThe external pressure must therefore decrease linearly along a positive-flow tube. Its additive constant is free; the constant transmural pressure is $\tilde{P}({\alpha }_0)$.

The first equation is mass conservation for an incompressible flow whose area may change: local area storage balances the divergence of the volume flux $\alpha u$. The two left-hand momentum terms are local acceleration and convective acceleration, while $-p_x$ is the driving pressure gradient and $-R(\alpha )u$ models viscous drag per unit mass. The pressure-area law of a collapsible tube expresses transverse wall balance using internal minus prescribed external pressure.

For example, choose a reference area ${\mathcal{A}}_{\ast }$, length $L$ and velocity $U_{\ast }$. If dimensional variables are $\mathcal{A},s,t_d,v,P$, takeThe dimensionless external pressure uses the same scale. A dimensional friction rate $r_f(\mathcal{A})$ becomes $R(\alpha )=L{r}_f({\mathcal{A}}_{\ast }\alpha )/U_{\ast }$, and the dimensional wall pressure law is divided by $\rho U_{\ast }^2$. These choices make both inertia and the pressure-gradient coefficient unity; other equivalent reference scales are possible. On the mechanically stable wall branch, increasing area requires increased transmural pressure, whereas an enlarged tube reduces resistance. Thus $\boxed{{\tilde{P}}^{\prime }(\alpha )>0,R^{\prime }(\alpha )<0}$.

Complete the square in the dispersion relation:If the imaginary part of the square root has magnitude $b$, thenOne root grows exactly when $b>R_0/2$. For $k\ne 0$, squaring the equivalent positive inequality gives $n^2{U}^2>c_0^2$. Thus positive flow is unstable whenever $\boxed{U>c_0/n}$. This proves the entire stated range, not just a small neighborhood of threshold. Zero wavenumber has only a neutral uniform-area change and a damped velocity mode.

Near threshold write $nU=c_0(1+\delta )$ and follow the neutral branch as $\nu =k{c}_0+\delta {\nu }_1+O({\delta }^2)$. Substitution yieldsTaking its imaginary part gives the growth rateArea increases reduce friction, causing a velocity increase that feeds the area disturbance through mass conservation. The pressure-driven base flow supplies the energy; the positive friction coefficient does not by itself prevent instability of that maintained flow.

In the travelling coordinate $X=x-ct$, mass conservation becomes $[\alpha (u-c){]}^{\prime }=0$. Thus the conserved relative flux is $Q$ andFor a periodic peristaltic pumping solution, the neglected-inertia momentum equation isTake ${\alpha }_0$ to be the spatial mean area, so $⟨{\alpha }_1⟩=⟨{\alpha }_2⟩=0$. This normalization, and zero net pressure drop over a period, are implicit in the specified periodic forcing. The constant leading state has no pressure gradient, giving $Q_0=-{\alpha }_{0}c$. Put $B={\tilde{P}}_0^{\prime }$ and $d=R_{0}c/{\alpha }_0$. At first order,Period averaging gives $Q_1=0$, and the zero-mean solution isFor the second-order mean balance, expand the velocity and friction:Averaging the exact pressure derivative gives zero, so $⟨R_0{u}_2+R_1{u}_1⟩=0$. Hence the weak peristaltic pumping of a collapsible tube result isThe mean laboratory volumetric flow rate is $c⟨\alpha ⟩+Q={ϵ}^2{Q}_2+O({ϵ}^3)$, forward with the wave when $c>0$. The negative leading relative flux is therefore consistent with positive mean pumping.

Measure $z$ from the rigid tube wall towards the air core, so $0<z<h$. To leading order in $h_0/a$ and small interface slope, the curvature entering the Young–Laplace equation isThe air pressure is spatially uniform at the order retained. Thus the linear liquid pressure perturbation associated with thickness amplitude $H$ is $p_1=-{\sigma }_0(a^{-2}-k^2)H$. The circumferential curvature favors growth at long wavelength; the curvature opposes it at short wavelength.

The lubrication theory momentum equation is $\mu u_{zz}=p_x$. With the wall no-slip boundary condition $u(0)=0$ and zero surface shear stress, $u_z(h)=0$, its solution and flux areLeading thin-film mass conservation is $h_t+q_x=0$. For a mode $e^{\beta t+ikx}$, this gives the capillary instability of an annular liquid liningTherefore the unstable band is $\boxed{0<k^2<a^{-2}}$. Maximization with respect to $k^2$ gives $k_{\ast }^2=1/(2a^2)$ andThese curvature coefficients use $a$ instead of $a-h_0$ consistently at leading order in the thin-lining approximation.

The Boussinesq approximation assumes that the fractional mass density variations are small. Replace density by a constant reference value in inertia, pressure scaling and mass conservation, but retain its small anomaly in the gravitational force: the large hydrostatic force otherwise cancels, leaving precisely the buoyancy-driving term. For temperature-driven motion, write $\rho ={\rho }_r[1-\alpha (T-T_r)]$ with $|\alpha (T-T_r)|\ll 1$. The resulting velocity is an incompressible flow; compressional heating and acoustic effects are omitted in the shallow, low-compressibility regime.

For this convection problem, use the strong Boussinesq approximation in the constant-property sense: the thermal expansion coefficient, kinematic viscosity, thermal diffusivity and heat capacity are also independent of temperature over the layer. This additional assumption makes the conduction profile linear and the perturbation coefficients constant. This usage of “strong” is the convection convention described in courses.physics.ucsd.edu/2021/Spring/physics218c/Spiegel%20Convection.pdf . Some stratified-flow literature uses “strong” instead to emphasize a globally constant reference density; that condition is also satisfied here. The equations below specify precisely which assumptions are being used.

Introduce the thermal diffusivity $\kappa$ and positive thermal expansion coefficient $\alpha$, which are necessary material data even though the question names only $\nu$ explicitly. Measure $z$ upward from the bottom. The conductive state of Rayleigh-Bénard convection is $T_b=T_0+\Delta T(1-z/d)$ with zero velocity. Let $\theta =T-T_b$ and let $\pi$ be perturbation pressure divided by the reference density, after subtracting the hydrostatic base pressure. Linearizing givesThe sign in the last equation is positive because upward motion carries warmer base-state fluid into cooler surroundings. Taking the divergence of momentum gives ${\nabla }^2\pi =g\alpha {\theta }_z$. Apply a Laplacian to its vertical component and subtract the vertical derivative of that pressure identity. This eliminates pressure:Apply the temperature operator to obtain the vertical-velocity equationThis is the pressure- and temperature-eliminated linear stability analysis equation for Rayleigh-Bénard convection.

At each boundary, impermeability gives $w=0$, the stress-free boundary condition gives $w_{zz}=0$, and perfect thermal conduction gives $\theta =0$. To obtain the second condition, zero tangential traction gives ${\partial }_z{u}_h+{\nabla }_{h}w=0$; since $w$ vanishes identically along the boundary, differentiate horizontal incompressibility to get $w_{zz}=0$. For the fully eliminated velocity equation, $\theta =0$ and the preceding momentum relation also imply $w_{zzzz}=0$ there. Thus suitable formulations areor, for the sixth-order vertical equation alone, $w=w_{zz}=w_{zzzz}=0$ at both boundaries. Fixed boundary temperature means a zero perturbation there, not zero perturbation heat flux.

Use normal modes $w=W\sin (n\pi z/d)e^{\sigma t+i\mathbf{k}\cdot {\mathbf{x}}_h}$, $n\ge 1$, $k=|\mathbf{k}|>0$. Put $K^2=k^2+n^2{\pi }^2/d^2$. Substitution yields the stress-free convection growth-rate polynomialAt stationary marginal stability, $\sigma =0$. In terms of the Rayleigh number $\mathrm{Ra}=g\alpha \Delta T{d}^3/(\nu \kappa )$ and dimensionless horizontal wave number $a=kd$, the free-slip convection neutral curve isWriting $y=a^2$ and differentiating gives $d\log {\mathrm{Ra}}_n/dy=3/(y+n^2{\pi }^2)-1/y$. Its zero is $a^2=n^2{\pi }^2/2$, with minimum $27n^4{\pi }^4/4$. The smallest vertical mode is $n=1$, so the critical temperature difference isThe first unstable horizontal wavelength is $2\sqrt{2}d$. To check that an oscillatory mode cannot destabilize earlier, expand the growth-rate equation: its linear coefficient is $(\nu +\kappa )K^2>0$, and its constant term changes sign exactly at the stationary neutral curve. For positive temperature difference the discriminant is $(\nu -\kappa {)}^2{K}^4+4g\alpha \Delta T{k}^2/(d{K}^2)>0$, so the crossing root is real. Below the global critical value every convection mode decays; at criticality the selected mode is marginal; above it some modes grow and eventually require nonlinear advection, producing convective heat transport. This is exchange of stabilities in stress-free convection.

Strictly, the linearized equations remain the correct small-amplitude equations on either side of onset. What fails above $\Delta T_{\max }$ is stability of the conductive state and long-time validity of a small-perturbation description, not the algebra of linearization. The threshold also presumes that the required small density variations and constant-property approximation hold at that temperature difference.

In the thin-current approximation, vertical momentum balance is hydrostatic: $p=p_{\mathrm{atm}}+\rho g(h-z)$. Neglect inertia and radial viscous derivatives compared with the vertical ones. The radial Stokes equation is $\mu u_{zz}=p_r=\rho g{h}_r$, with no slip at $z=0$ and zero tangential stress $u_z=0$ at $z=h$. Integration givesIts depth-integrated radial flux is $Q_r={\int }_0^{h}udz=-\rho g{h}^3{h}_r/(3\mu )$. Axisymmetric mass conservation $h_t+r^{-1}{\partial }_r(r{Q}_r)=0$ therefore givesThe remaining conditions are conserved volume $2\pi {\int }_0^{R(t)}rhdr=V$, regularity and zero radial flux at the center, $h(R(t),t)=0$, zero thickness outside the front, and the initially concentrated volume at the origin. The advancing edge obeys $R^{\prime }={\lim }_{r\uparrow R}{Q}_r/h$ when this limit exists. One must not impose a finite zero slope at this compact-support edge.

Put $h=t^{a}F(r{t}^{-b})$ with dimensional factors to be restored. Constant volume gives $a+2b=0$, while balancing the equation gives $a-1=4a-2b$. Hence $a=-1/4,b=1/8$. In normalized similarity variables $\eta =r/(\beta V^{3}t{)}^{1/8}$ and $h=V^{1/4}(\beta t{)}^{-1/4}F(\eta )$, the equation becomesMultiplying by $\eta$ and integrating from the regular center yields $\eta F^3{F}^{\prime }=-{\eta }^{2}F/8$. Where $F>0$, this integrates to $F^3=3({\eta }_0^2-{\eta }^2)/16$. Restoring physical variables giveswith zero thickness outside. Write $h=A(t)(1-r^2/R^2{)}^{1/3}$. Its volume is $3\pi A{R}^2/4$, so $A=4V/(3\pi R^2)$. Combining this with $A^3=3R^2/(16\beta t)$ givesThe center height scales as $t^{-1/4}$, while the front scales as $t^{1/8}$. The ideal profile has a singular slope at the edge; microscopic contact-line physics is excluded from the lubrication model, which applies to the sufficiently spread bulk current.

At $\gamma =5/3$, the sound-speed relation becomes $\rho /{\rho }_0=(c/c_0{)}^3$. With $\mathcal{M}=v/c$, mass conservation gives $c^4=\dot{M}{c}_0^3/(4\pi {\rho }_0{r}^2\mathcal{M})$. ThusMultiply $v^2/2+3c^2/2=3c_0^2/2+GM/r$ by $r$ and substitute. This provesFor the curves, define $x=r{c}_0^2/(GM)$, $\lambda =\dot{M}{c}_0^3/(4\pi G^2{M}^2{\rho }_0)$ and $h(\mathcal{M})={\mathcal{M}}^{3/2}/2+3{\mathcal{M}}^{-1/2}/2$. ThenThe function decreases up to Mach one, increases thereafter, and has minimum $h(1)=2$. The original sketch uses rates relative to ${\dot{M}}_{\max }$, that is $\eta =4\lambda$:

For $0<\lambda <1/4$, the minimum of the formal radius curve lies at negative radius. Its physical parts are separate subsonic and supersonic branches extending to $r=0$. Only the subsonic branch satisfies the specified outer state: as $r\to \infty$, it has $c\to c_0$, $\rho \to {\rho }_0$ and $v\sim \dot{M}/(4\pi {\rho }_0{r}^2)\to 0$. The supersonic branch instead has $\rho \to 0$, $c\to 0$ and $v^2\to 3c_0^2$, so it is incompatible with that boundary condition.

At $\lambda =1/4$, the two branches meet at $(r,\mathcal{M})=(0,1)$. The outer-rest branch remains subsonic at every positive radius and approaches Mach one at the centre. For $\lambda >1/4$, they meet at a positive minimum radius $r_{\min }=(2GM/(3c_0^2))(2\sqrt{\lambda }-1)$, and no real solution exists below it. Such a curve cannot describe a global smooth inflow to a point mass.

Thus globally defined positive-rate solutions satisfying the given outer condition are the subsonic branches with $0<\lambda \le 1/4$. These constitute monatomic Bondi accretion. The outer condition alone does not select a unique member; the conventional maximal Bondi solution is the limiting sonic one, while a specific inner accretor condition is additional physical information. Zero rate is the separate hydrostatic limit, not a positive accretion flow.

Let $\xi$ be the fluid Lagrangian displacement, so $\delta u={\partial }_t\xi$ about the static background. A displaced material volume changes by fractional amount $\nabla \cdot \xi$. Mass conservation therefore gives the fluid Lagrangian perturbation of density $\Delta \rho =-\rho \nabla \cdot \xi$. Converting to the fluid Eulerian perturbation by $\Delta =\delta +\xi \cdot \nabla$ givesAdiabatic conservation of $p/{\rho }^{\gamma }$ gives $\Delta p/p=\gamma \Delta \rho /\rho$, and consequently $\delta p=-\xi \cdot \nabla p-\gamma p\nabla \cdot \xi$.

Integrating the linearized ideal induction equation for a perturbation generated by the displacement gives $\delta B=\nabla \times (\xi \times B)$. Since the equilibrium field is solenoidal, expansion yieldsThis is the displacement form of magnetic flux freezing. It also preserves $\nabla \cdot \delta B=0$.

Rewrite the magnetic force as $-\nabla (B^2/(2{\mu }_0))+{\mu }_0^{-1}B\cdot \nabla B$. Linearizing the full momentum equation, subtracting static force balance, and setting $\delta \Pi =\delta p+B\cdot \delta B/{\mu }_0$ givesTogether these are the linear displacement equations for a magnetized atmosphere. The density multiplying acceleration is the equilibrium value, because a perturbation of that factor would multiply a first-order acceleration and contribute only at second order.

Use $\mathcal{R}$ for the stellar gas constant, reserving $R$ for the stellar radius. The specific internal energy of a monatomic gas obeying the ideal gas law is $e=P/[(\gamma -1)\rho ]=3P/(2\rho )$. At fixed enclosed mass, the first law of thermodynamics gives the heat available to the outgoing luminosity as $ϵ=-{\partial }_{t}e-P{\partial }_t(1/\rho )$ when there is no stellar nuclear fusion heating. ThereforeThe second coefficient includes both the mass density dependence of $e$ and pressure work. This is gravitational energy generation in a homologously contracting ideal-gas star.

Write the characteristic scales asThen $\rho ={\rho }_{0}b$, $P=P_{0}p$, $T=T_{0}p/b$, $m=Mq$ and $L_r=Ll$. Since $d/dr=R^{-1}d/dx$, direct substitution in the stellar hydrostatic equation and mass conservation givesFor radiative diffusion in a star, dividing the temperature equation by $T_0/R$ givesIn particular the fourth power here is of $\mathcal{R}$, not of the radius.

Because $q(x)$ is fixed and $M$ is constant, a fixed mass label has fixed $x$. Thus stellar homology gives ${\partial }_t\rho =-3\rho \dot{R}/R$ and ${\partial }_{t}P=-4P\dot{R}/R$. The heating rate reduces to $ϵ=-3P\dot{R}/(2\rho R)$, positive during Kelvin-Helmholtz contraction. Substitution in $d{L}_r/dr=4\pi r^2\rho ϵ$ now yieldsThe dimensionless profiles fix $D$ and $E$. In a common stellar homology family with fixed stellar composition and opacity normalization, the expression for $D$ impliesIt is independent of radius and constant in time for a particular fixed-mass star. The formula for $E$ gives $\dot{R}=-2EL{R}^2/(3G{M}^2)$, which integrates toWhen the initial-radius term is negligible,This is the Kelvin-Helmholtz contraction time law in the prescribed stellar homology model. Infinite initial radius is a limiting idealization; a finite initial radius retains the first term.

At main sequence arrival, CNO cycle heating with fixed stellar composition has luminosityThe integral is dimensionless and fixed within the stellar homology family. The constant-opacity radiative scaling still gives $L\propto M^3$. Equating the stellar nuclear fusion and transported luminosities givesFinally the Kelvin-Helmholtz contraction law evaluated at $R_{\mathrm{MS}}$ gives $t_{\mathrm{MS}}\propto M^2/(L{R}_{\mathrm{MS}})\propto M^{-34/19}$. In a coeval star cluster, the mass just arriving at the main sequence consequently satisfiesThese results comprise constant-opacity homologous contraction to CNO ignition; their mass exponents compare fixed-composition models with common dimensionless profiles.

Consider a small outward displacement $\delta r$ of a fluid element. It adjusts to the ambient pressure while retaining its specific entropy and stellar composition, so an adiabatic process gives $d\log {\rho }_{\mathrm{parcel}}=d\log P/\gamma$. Its excess mass density over the new surroundings is, to first order,Restoring buoyancy requires this bracket to be positive. With $\gamma =5/3$, the Schwarzschild criterion is thereforeEquality is neutral stability. For uniform mean molecular weight, the ideal gas law gives $d\log \rho =d\log P-d\log T$. Since hydrostatic equilibrium pressure decreases outward, the inequality is equivalent to $\nabla =d\log T/d\log P<{\nabla }_{\mathrm{ad}}=2/5$. Dividing the radiative diffusion in a star and hydrostatic equilibrium stellar structure equations givesThis is the stellar radiative temperature gradient test for stellar convective stability.