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Helioseismology infers the Sun's structure and dynamics from its stellar oscillations. Spatial projection of resolved surface observations onto spherical harmonics identifies angular degree and azimuthal order; a temporal Fourier transform measures mode frequencies. Stellar acoustic modes constrain the adiabatic sound speed, while rotational splitting constrains stellar rotation. Their finite wavelength, surface reflection phases and finite observational coverage limit inversion resolution.

The Lamb frequency is the horizontal acoustic angular frequency of a spherical-harmonic component of degree $\ell$. With adiabatic sound speed $c_s$, $S_{\ell }^2=\ell (\ell +1)c_s^2/r^2$. It helps determine inner acoustic turning points and vanishes for radial modes.

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.

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 an adiabatic outflow without additional heating, the reservoir's specific enthalpy supplies both escaping kinetic energy and the work against gravity. For a monatomic ideal gas, $\gamma =5/3$ and $h=c_s^2/(\gamma -1)=3c_s^2/2$, where $c_s$ is the reservoir's adiabatic sound speed. The Bernoulli equation, measured from a reservoir initially at rest at radius $r$, givesThe maximum terminal speed uses all available specific enthalpy, so $h_{\infty }\to 0$ andA freely escaping adiabatic flow needs a nonnegative radicand; equality is marginal escape. If the radicand is negative, this reservoir cannot supply an unheated escaping outflow. The initial rest condition is a stagnation-reservoir approximation: an exactly stationary finite-density surface cannot simultaneously carry a nonzero steady mass flow rate. Further heating, radiation forces or an imposed confining pressure would change the terminal-energy budget.

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.

A uniform conducting fluid supports three propagating magnetohydrodynamic wave branches. Ideal magnetohydrodynamics couples fluid velocity to the magnetic field through the induction equation; the Lorentz force then supplies magnetic tension and magnetic pressure restoring forces. Let $\theta$ be the angle between wave vector $\mathbf{k}$ and background field ${\mathbf{B}}_0$, and define Alfvén speed $v_A=B_0/\sqrt{{\mu }_0{\rho }_0}$ and adiabatic sound speed $c_s^2=\gamma p_0/{\rho }_0$.

The shear Alfvén wave hasIts velocity perturbation is perpendicular to both $\mathbf{k}$ and ${\mathbf{B}}_0$. It bends field lines without first-order compression, so magnetic tension is the restoring force. Its group velocity follows the background field; the displayed phase speed along $\mathbf{k}$ is not the full group speed.

The fast magnetosonic wave and slow magnetosonic wave involve coupled compression and magnetic perturbations, with displacement in the plane of $\mathbf{k},{\mathbf{B}}_0$:The plus sign is the fast branch. Parallel propagation gives characteristic speeds $c_s,v_A$; perpendicular propagation gives fast speed $\sqrt{c_s^2+v_A^2}$ while the other two phase frequencies vanish. Degenerate directions require care in assigning polarizations, rather than claiming three distinct nonzero frequencies everywhere.

These waves transport energy and momentum through magnetized stellar atmospheres and astrophysical plasmas. Their restoring mechanisms distinguish them from ordinary acoustic waves. Resistivity, viscosity, ion-neutral friction and kinetic effects introduce damping or change the model; an ideal undamped dispersion relation alone does not calculate coronal heating.

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.

Take incidence from A into B and first specify pressure amplitude. For a plane acoustic wave traveling normally to the interface, the acoustic impedance is $Z=\rho c_s$. A right-moving wave has pressure amplitude $P=ZU$, where $U$ is its fluid-velocity amplitude; a left-moving reflected wave has $P=-ZU$.

Let $P_i,P_r,P_t$ denote incident, reflected, and transmitted pressure amplitudes. Continuity of pressure and normal velocity at the interface givesSolving the two linear equations gives the acoustic reflection and transmission at an interface coefficientsFor the same perfect gas, take a common adiabatic index $\gamma$. The adiabatic sound speed satisfies $c_s^2=\gamma p/\rho$. The equal equilibrium pressure therefore implies $Z_B/Z_A=\sqrt{{\rho }_B/{\rho }_A}=\sqrt{\delta }$, givingThe word amplitude is otherwise ambiguous. For fluid velocity, or displacement at the same frequency, the corresponding answer isThese are consistent because pressure and velocity amplitudes differ by the impedance. As an energy check, the time-averaged acoustic intensity is $|P{|}^2/(2Z)$, so its transmission coefficient is $\mathcal{T}=4\sqrt{\delta }/(1+\sqrt{\delta }{)}^2\le 1$, even when the transmitted pressure amplitude exceeds the incident one.

Equal mean molecular weight alone does not logically guarantee equal $\gamma$. If the gases have different values of the adiabatic index, replace $\sqrt{\delta }$ in these formulas by $\sqrt{{\gamma }_B\delta /{\gamma }_A}$; a density-only answer requires the common-$\gamma$ assumption.

Work in the rest frame of a thin planar shock wave, with normal upstream and downstream velocities $u_1,u_2>0$. Assume a steady flow, negligible gravitational change across its thickness, no radiative energy loss, and negligible viscous stress and heat flux at the two remote faces of a control volume. Dissipation within the shock is permitted and is essential to the physical discontinuity.

Integrating mass conservation through that volume gives equal mass flux per unit area:The dimensions of $j$ are $\mathrm{kg}{\mathrm{m}}^{-2}{\mathrm{s}}^{-1}$. The normal momentum entering minus that leaving must balance the pressure forces. Equivalently, the conserved normal momentum flux is $p+\rho u^2$, givingFor conservation of energy, matter advects kinetic and internal energy while pressure performs work at rate $pu$ per unit area. Writing $ϵ$ for specific internal energy, the energy flux is $\rho u(u^2/2+ϵ)+pu$. With specific enthalpy $h=ϵ+p/\rho$, its conservation givesThe division by $j$ is valid for a genuine through-flow shock. Together these are the three Rankine-Hugoniot conditions. The conserved energy flux has units $\mathrm{W}{\mathrm{m}}^{-2}$, and both momentum fluxes have units of pressure.

For a calorically perfect ideal gas, $ϵ=p/[(\gamma -1)\rho ]$ and $h=\gamma p/[(\gamma -1)\rho ]$. The three jump conditions can be combined explicitly. Put $r={\rho }_2/{\rho }_1$, so $u_2=u_1/r$ andSubstitute these into the enthalpy equation and use ${\mathcal{M}}_1^2={\rho }_1{u}_1^2/(\gamma p_1)$, with the upstream adiabatic sound speed. Besides the trivial no-shock root $r=1$, the physical compression root isFor unchanged composition, $T_2/T_1=(p_2/p_1)/r$. A compressive shock has ${\mathcal{M}}_1>1$ and increases the specific entropy. Thus an adiabatic shock conserves total energy but is not an isentropic process.

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.

Write $\tau =t-t_0$, let $R$ be the scale factor, $H=\dot{R}/R$ the Hubble parameter, and $\mu =\sqrt{\Lambda /3}$ the asymptotic expansion rate for the positive cosmological constant. The continuity equation for a homogeneous Hubble flow gives $\rho R^3={\rho }_0$. Isotropy and the Poisson equation for Newtonian gravity give a relative acceleration proportional to displacement from any chosen origin:This uses the local relative gravitational acceleration, rather than trying to sum an absolutely convergent potential over an infinite universe. A fluid element with constant comoving coordinate $\mathbf{x}$ has $\mathbf{r}=R\mathbf{x}$. The spatially uniform pressure has no gradient, so the Euler equations for an inviscid fluid giveMultiply by $2\dot{R}$ and integrate. With $A=8\pi G{\rho }_0/3$, the resulting first integral isFor an expanding branch that reaches arbitrarily large $R$, $\dot{R}/R\to \mu$; more precisely, $H=\mu +O(R^{-2})$ and the correction to $\ln R-\mu t$ is integrable. Therefore $R\sim C{e}^{\mu t}$. Positive $\Lambda$ alone does not make every initially expanding branch escape: sufficiently large positive $\kappa$ can give a turning point or a limiting static radius. Under the stated vacuum-dominance inequalities at $R=1$, however, $A/R$ and $|\kappa |$ remain small compared with ${\mu }^2{R}^2$ for the future expanding branch, and $R\simeq e^{\mu \tau }$ is the required approximation.

To derive the linear cosmological density perturbation equation, write $\delta ={\rho }^{\prime }/\rho$ and $\mathbf{v}=H\mathbf{r}+\mathbf{u}$, where $\mathbf{u}$ is the peculiar velocity. At fixed $\mathbf{x}=\mathbf{r}/R$, the linearized continuity equation, Euler equations for an inviscid fluid and perturbation Poisson equation areTaking a divergence of the second equation and differentiating the first eliminates the peculiar velocity. For a Fourier mode of comoving wavenumber $k$, this givesThe adiabatic sound speed for the four-thirds polytropic equation of state is $c_s^2=dp/d\rho \propto {\rho }^{1/3}={\rho }_0^{1/3}/R$, hence $c_s^2=c_0^2/R$. Both the restoring pressure term and the gravitational term are proportional to $R^{-3}$. Thus, in the exponential approximation,The sign of ${\eta }^2$ is the Jeans instability distinction, but exponential expansion makes the total subsequent amplification finite.

Remove Hubble friction by setting $\epsilon =e^{-\mu \tau }y$. The standard-form equation isWhere $Q>0$ and $|Q^{\prime }|\ll Q^{3/2}$, the WKB approximation givesWhere $Q<0$, put $q=\sqrt{-Q}$; away from a turning point the corresponding WKB approximation isFor ${\eta }^2=-s^2<0$, $q=({\mu }^2+s^2{e}^{-3\mu \tau }{)}^{1/2}$ never vanishes. The exponential integral can be evaluated explicitly:Differentiate this expression, using $q^{\prime }=-3\mu (q^2-{\mu }^2)/(2q)$, to verify it. Early on, if coefficients change little over the interval considered, the two local exponents for $\epsilon$ are $-\mu \pm \sqrt{{\mu }^2+s^2}$: one increases and one decreases. At late times $q\to \mu$ and the displayed integral is $\mu \tau +O(1)$. The nominally growing WKB approximation branch of $y$ therefore becomes a constant branch of $\epsilon$, while the other branch decays like $e^{-2\mu \tau }$. The WKB approximation is not automatically accurate at all intermediate times; its stated derivative criterion must be checked.

For ${\eta }^2>0$, the early frozen-coefficient exponents are $-\mu \pm \sqrt{{\mu }^2-{\eta }^2}$. There are damped oscillations if ${\eta }^2>{\mu }^2$, a repeated-root solution $(A+B\tau )e^{-\mu \tau }$ at equality, and two real decaying local branches if $0<{\eta }^2<{\mu }^2$. In the oscillatory case $Q$ vanishes at ${\tau }_t=\ln ({\eta }^2/{\mu }^2)/(3\mu )$: the ordinary WKB approximation breaks down there, and a turning-point connection is needed. After that transition the solutions again approach a constant or decay. A positive pressure-minus-gravity restoring term does not imply indefinitely continuing oscillation; its physical frequency is redshifted away.

The limiting conclusion does not depend on the accuracy of WKB approximation. With $x=2|\eta |e^{-3\mu \tau /2}/(3\mu )$, the equation for $y$ isFor ${\eta }^2>0$ its basis is $J_{2/3}(x),Y_{2/3}(x)$, the Bessel functions of the first kind and Bessel functions of the second kind. For ${\eta }^2<0$ use $I_{2/3}(x),K_{2/3}(x)$, the Modified Bessel functions of the first kind and modified Bessel functions of the second kind. Their small-$x$ branches are proportional to $x^{2/3}$ and $x^{-2/3}$. Multiplication by $e^{-\mu \tau }$ gives the decaying and constant branches respectively. At ${\eta }^2=0$ the exact solution is $A+B{e}^{-2\mu \tau }$. Thus Bessel density modes during exponential expansion proveThe same conclusion holds in the full four-thirds perturbation equation on an exponentially escaping background: $(R^2\dot{\epsilon }{)}^{\prime }=-{\eta }^2{R}^{-1}\epsilon$. Its integrated equation has an absolutely integrable Volterra kernel, bounded by a constant times $e^{-3\mu \tau }$. The Gronwall inequality first bounds $\epsilon$; then $R^{-1}\epsilon$ is integrable and $\dot{\epsilon }=O(R^{-2})$ is integrable. A finite limiting amplitude follows without setting $H$ exactly constant. This argument requires positive $\Lambda$ and the escaping branch; it is not a claim about $\Lambda =0$ or recollapse.

The requested strict increase of that limit needs an initial-velocity qualification. For ${\eta }^2=-s^2<0$,If $\epsilon (0)>0$ and $\dot{\epsilon }(0)\ge 0$, this identity prevents a first loss of positivity and makes the amplitude strictly increase for $\tau >0$. Integrating it twice givesThis is monotone Jeans growth with positive cosmological constant. Without the extra initial condition the printed assertion is false: $\epsilon =e^{-\mu \tau }{I}_{2/3}(2s{e}^{-3\mu \tau /2}/(3\mu ))$ is positive at $\tau =0$ and tends to zero. An arbitrarily small multiple remains an infinitesimal linear cosmological density perturbation, so linearity does not remove the counterexample.

Take $u<0$ for inward radial velocity, and assume a constant specific-heat ratio $\gamma >1$. The isentropic flow has $p=K{\rho }^{\gamma }$ and adiabatic sound speed $c^2=\gamma K{\rho }^{\gamma -1}$. Steady spherical mass conservation and the radial Euler equation for fluid motion giveEliminating the mass density derivative gives $(u-c^2/u)u^{\prime }=2c^2/r-GM/r^2$, orDifferentiating $c^2\propto {\rho }^{\gamma -1}$ gives the corresponding sound speed equation:At a smooth sonic point, $u_s^2=c_s^2$. A finite derivative requires the numerator of the flow equation to vanish too. ThusThis regularity requirement concerns a smooth transonic branch, rather than an arbitrary singular solution reaching unit Mach number.

Integrating the momentum equation using the specific enthalpy $c^2/(\gamma -1)$ yields the Bernoulli equationFor a finite positive mass accretion rate, its large-radius solution hasThese expressions satisfy both the flow equations asymptotically and show $u\to 0$ and $\rho \to {\rho }_{\infty }$. In particular $u^{\prime }/u\sim -2/r$ agrees with the first boxed equation. Combining the Bernoulli equation with sonic regularity additionally gives $c_s^2=2c_{\infty }^2/(5-3\gamma )$, so a finite positive sonic point occurs in the range identified below.

For the small-radius inward free fall branch, neglecting enthalpy against $GM/r$ in the Bernoulli equation predicts $u^2\sim 2GM/r$. The continuity equation then fixes $\rho \propto r^{-3/2}$, and consequently $c^2\propto r^{-3(\gamma -1)/2}$. The neglected-to-leading ratio isIt tends to zero precisely whenThis makes the assumed supersonic flow and leading free fall self-consistent. At equality, the ratio is of constant order rather than vanishing; above it, the proposed asymptotic balance fails. The isothermal limit $\gamma =1$ also permits free fall, with logarithmic rather than power-law enthalpy.

For the stellar funnel, write the magnetic scalar potential as $\psi =\mu \cos \theta /r^2$, where $R=r\sin \theta$ and $z=r\cos \theta$. Its magnetic dipole field has $B_r=-2\mu \cos \theta /r^3$ and $B_{\theta }=-\mu \sin \theta /r^3$, so a magnetic-field-line equation isThe edge of the polar flux tube meets the stellar surface at ${\theta }_{\ast }\simeq a/R_{\ast }$. Thus, near the pole, ${\theta }^2\simeq {\theta }_{\ast }^{2}r/R_{\ast }$, and its transverse radius is $R\simeq r\theta$. This proves the dipolar flux-tube area lawEquivalently magnetic flux conservation gives $A|\mathbf{B}|=\mathrm{constant}$ and $|\mathbf{B}|\propto r^{-3}$. The stipulated small cross section keeps the tube close to the polar axis, where its direction is approximately radial; the approximation is not valid all the way to arbitrarily large $r$.

Outside the star the prescribed field is a gradient, hence $\nabla \times \mathbf{B}=0$ and $\mathbf{j}=0$. Its Lorentz force density is therefore $\mathbf{j}\times \mathbf{B}=0$. The magnetic-pressure and magnetic-tension parts cancel in this exact current-free dipole. Field-aligned velocity also has $\mathbf{u}\times \mathbf{B}=0$, consistent with a steady prescribed field. The following calculation uses the narrow polar funnel approximation.

Now steady mass conservation reads $\rho uA=\mathrm{constant}$. With $A\propto r^3$, replace the spherical factor 2 by 3:The Bernoulli equation is unchanged because the magnetic field supplies no work or force. On a gravity-dominated free fall branch, $u^2\sim 2GM/r$, but $\rho \propto r^{-5/2}$ and $c^2\propto r^{-5(\gamma -1)/2}$. HenceThe same bound follows from regular transonic accretion in a power-law tube: $u_s^2=c_s^2=GM/(3r_s)$ and the reservoir-matched Bernoulli equation gives $c_s^2=2c_{\infty }^2/(7-5\gamma )$. The stellar surface cuts off the mathematical small-$r$ limit; the criterion refers to a funnel that becomes gravity dominated and crosses its sonic point before reaching that surface.

Let the equilibrium ideal gas be at rest, with uniform ${\rho }_0,p_0$ and ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$. This local wave calculation neglects gravity, viscosity and magnetic diffusion. Define the adiabatic sound speed $c\equiv v_s=(\gamma p_0/{\rho }_0{)}^{1/2}$, the Alfvén speed $v_a=B_0/({\mu }_0{\rho }_0{)}^{1/2}$, and a wavevector $\mathbf{k}=k(\sin \theta ,0,\cos \theta )$, where $k=|\mathbf{k}|$ and $\theta$ is its angle to the equilibrium magnetic field. Perturbations have dependence $e^{i\mathbf{k}\cdot \mathbf{x}-i\omega t}$ and phase speed $v=\omega /k$.