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Take a static, nonrotating, nonmagnetic spherical equilibrium with $\nabla P=-\rho \nabla \Phi$ and ${\nabla }^2\Phi =4\pi G\rho$. Use a fluid displacement $\mathbf{\xi }(\mathbf{r})e^{-i\omega t}$. Write $p_1,{\rho }_1,{\varphi }_1$ for Eulerian fluid perturbations. The Lagrangian pressure perturbation and corresponding mass density change obey $\Delta P=p_1+\mathbf{\xi }\cdot \nabla P$, $\Delta \rho ={\rho }_1+\mathbf{\xi }\cdot \nabla \rho$. Linearizing mass conservation, the Euler equations for an inviscid fluid and Poisson equation for Newtonian gravity givesThere is no equilibrium acceleration to multiply a perturbed mass density. The adiabatic process condition, with composition carried by the parcel, closes the system:These are the complete linear adiabatic stellar oscillation equations, including the perturbation of self-gravity. Neglect of heat exchange is appropriate when oscillation periods are short compared with relevant thermal relaxation times; it does not determine nonadiabatic excitation or damping.

For the pressure and buoyancy modes, separate angular dependence using spherical harmonics:The horizontal amplitude ${\xi }_h$ has dimensions of length. Angular differentiation gives $\nabla \cdot \mathbf{\xi }=[r^{-2}(r^2{\xi }_r{)}^{\prime }-\ell (\ell +1){\xi }_h/r]Y_{\ell }^m$, and tangential momentum gives ${\omega }^{2}r{\xi }_h=p_{\ell }/\rho +{\varphi }_{\ell }$. Define the adiabatic sound speed, stellar buoyancy frequency and Lamb frequency byThe adiabatic mass density relation becomes ${\rho }_{\ell }/\rho =p_{\ell }/(\rho c_s^2)+(N^2/g){\xi }_r$. Substitution produces a radial form of the full oscillation equations for nonzero $\omega$:The apparent $N^2/g$ factor is evaluated through its defining gradient at the centre rather than by dividing two zeros.

Regularity at the centre excludes singular solutions. At a free surface the Lagrangian pressure perturbation vanishes, $\Delta P=0$; outside the star the gravitational perturbation decays as $r^{-\ell -1}$. For a model with mass density tending to zero at its surface, continuity of ${\varphi }_{\ell }$ and its radial derivative gives ${\varphi }_{\ell }^{\prime }(R)=-(\ell +1){\varphi }_{\ell }(R)/R$. If the equilibrium mass density jumps to vacuum, include the displaced-surface mass sheet: the outward-minus-inward derivative jump is $4\pi G\rho (R){\xi }_r(R)$, so the interior condition is ${\varphi }_{\ell }^{\prime }(R)=-(\ell +1){\varphi }_{\ell }(R)/R-4\pi G\rho (R){\xi }_r(R)$. An atmospheric boundary condition can replace the ideal free surface.

These conditions make ${\omega }^2$ an eigenvalue, not an arbitrary local sound frequency. With conservative boundary conditions the adiabatic operator is self-adjoint, giving real ${\omega }^2$; negative values describe instability. The frequencies depend on $P(r),\rho (r),{\Gamma }_1(r)$, self-gravity, stratification and boundaries, as well as angular degree $\ell$ and radial order. In a spherical nonrotating star they are degenerate in $m$. For homologous equilibrium structures, their scale isthus the typical oscillation time measures inverse square root of mean mass density, while individual frequencies probe the interior adiabatic sound speed and stellar buoyancy frequency profiles.

For radial modes, write ${\xi }_r=r\eta$. Eliminating the pressure and gravitational perturbations gives the radial stellar pulsation equationThis is a Sturm-Liouville problem. Multiplication by $\eta$ and integration, with vanishing boundary terms, gives its Rayleigh quotientFor constant ${\Gamma }_1>4/3$, $P^{\prime }<0$ makes both numerator contributions nonnegative. At ${\Gamma }_1=4/3$, a homologous displacement is neutral; for constant ${\Gamma }_1<4/3$ the same trial displacement makes the quotient negative. For the uniform-density stellar model, $\eta =$ constant is an exact mode and ${\omega }^2=(3{\Gamma }_1-4)GM/R^3$. With varying ${\Gamma }_1$, the integral criterion, rather than a universal pointwise threshold, controls radial stability.

The Cowling approximation neglects ${\varphi }_{\ell }$ while retaining the equilibrium gravitational field. It is useful for short-wavelength modes, but is not needed for the full derivation above. In a locally slowly varying region, take both remaining amplitudes proportional to $e^{i\int k_{r}dr}$ and retain the leading derivative terms. ThenEliminating either amplitude gives the acoustic-gravity propagation relationPositive $k_r^2$ is oscillatory propagation; negative $k_r^2$ means an evanescent wave. The high-frequency branch, ${\omega }^2>S_{\ell }^2,N^2$, describes stellar acoustic modes, restored chiefly by compressibility and pressure. At frequencies well above $N$, this gives ${\omega }^2\simeq c_s^2[k_r^2+\ell (\ell +1)/r^2]$. The low-frequency propagating branch in stable stratification, ${\omega }^2<N^2,S_{\ell }^2$, instead describes stellar gravity modes, restored by buoyancy. At $\ell =0$ there is no such nonradial gravity-wave cavity.

A stellar acoustic mode is trapped between an inner turning point near $\omega =S_{\ell }$ and an outer reflecting region. Low-$\ell$ modes penetrate deeply; radial modes reach the centre. Higher-degree modes turn farther out. Standing waves require the WKB quantization conditionwhere the phase $\alpha$ depends on the central or turning-point behaviour and surface reflection. For high radial order and small degree, the leading acoustic travel-time result is the large frequency separationThe $\ell /2$ term is the leading central angular phase shift; smaller frequency separations depend on detailed interior gradients. The near-surface mass density stratification sets the acoustic cutoff frequency. In a plane-parallel isothermal atmosphere with mass density scale height $H_{\rho }$, ${\omega }_{\mathrm{ac}}\simeq c_s/(2H_{\rho })$. Modes below this cutoff can reflect and form a cavity; waves sufficiently above it escape and need an outgoing-wave boundary condition. Nonadiabatic damping, driving and rotation alter real-star mode properties, but the adiabatic frequency problem isolates their dependence on the equilibrium structure.

A toroidal magnetic field introduces an inward hoop force. In cylindrical coordinates, the radial magnetic tension is $-B_{\varphi }^2/({\mu }_{0}s)$, so cylindrical magnetostatic pressure balance in SI units becomesAt the field-free boundary $P(s_0)=p_e$. HenceRegularity on the axis gives $B_{\varphi }=O(s)$, which makes the integral finite and gives $B_{\varphi }(0)=0$. Evaluating at the axis proves the central-field identity:The positive hoop-tension contribution allows the central axial field to exceed $B_p$ without negative pressure. For an explicit regular example, put $x=s^2/s_0^2$, choose a constant $a>0$ and $0<\epsilon <1$, and take inside the flux tubeSubstitution verifies the radial balance, with continuous zero field at $s=s_0$ and positive gas pressure. For $a=3$, $\epsilon =1/4$, the central value is $B_z(0{)}^2=3B_p^2/2$.

To obtain the cross-sectional average, multiply the radial balance by $s^2$ and integrate by parts:The toroidal contributions cancel. The resulting flux-tube axial field virial identity isThis is at most $B_p^2$, and is strictly smaller whenever the weighted pressure integral is positive. In particular, continuous matching with $p_e>0$ guarantees strictness. If only $p\ge 0$ is assumed without a nondegenerate gas or continuous positive boundary pressure, the justified conclusion is the non-strict bound; the identity specifies exactly when equality occurs.

In SI units, magnetostatic equilibrium without gravity is $\nabla p={\mu }_0^{-1}(\nabla \times \mathbf{B})\times \mathbf{B}$. For a purely axial magnetic field, the magnetic tension term $(\mathbf{B}\cdot \nabla )\mathbf{B}$ vanishes. Thus the radial balance isMatching the magnetohydrodynamic total pressure to the field-free exterior gives the gas pressure and axial field bound:The inequality follows directly from nonnegative gas pressure. It is the magnetic pressure limit for an untwisted flux tube, without inward hoop tension.

Write $D_t={\partial }_t+\mathbf{u}\cdot \nabla$ for the material derivative. For any specific quantity $f$, conservation of mass givesDotting the momentum equation with $\Sigma \mathbf{u}$ produces the kinetic energy balanceThe Coriolis acceleration does no work because $\mathbf{u}\cdot ({\mathbf{e}}_z\times \mathbf{u})=0$. Since the shearing-sheet tidal potential is time-independent, its advected potential-energy density obeysFor the isothermal equation of state, define the barotropic energy density $U=c_s^2\Sigma \ln (\Sigma /{\Sigma }_{\mathrm{ref}})$, with fixed positive reference surface density ${\Sigma }_{\mathrm{ref}}$. The continuity equation impliesAdding these balances combines the two pressure terms into $-\nabla \cdot (P\mathbf{u})$. This establishes isothermal shearing-sheet energy conservation. Thus the conserved energy density and its flux areChoosing the density unit so that ${\Sigma }_{\mathrm{ref}}=1$ reproduces the printed logarithm. Changing the reference adds a multiple of the conserved mass to $E$ and its advective energy flux. The barotropic energy density is the mathematical energy of this fixed-isothermal sound speed closure; it is not the microscopic thermal energy of a thermally isolated gas. Maintaining an isothermal process can require heat exchange.

For a local axisymmetric Fourier mode, the Toomre stability criterion balances three contributions to the squared oscillation frequency:The radial epicyclic frequency supplies rotational restoration at long wavelengths; isothermal sound speed and pressure stabilize short wavelengths; disk self-gravity destabilizes intermediate wavelengths. Minimizing over $|k|$ gives $|k|=\pi G{\Sigma }_0/c_s^2$ and ${\omega }_{\min }^2={\kappa }^2(1-Q^{-2})$. Axisymmetric gravitational instability occurs when $Q<1$.

In a centrally dominated Keplerian disk, $\kappa =\Omega$ and vertical hydrostatic equilibrium gives $c_s\simeq H\Omega$. Using ${\Omega }^2=G{M}_{\ast }/r^3$ and a local disk-mass estimate $M_D\sim \pi r^2{\Sigma }_0$,This is the disk mass form of the Toomre criterion. An actual enclosed mass depends on the radial surface density profile and changes an order-one coefficient. In particular, a profile proportional to $r^{-3/2}$ has $M_D(<r)\simeq 4\pi r^2{\Sigma }_0$ when its inner cutoff is negligible.

For the protosolar estimate, take $M_{\ast }\simeq M_{\odot }\simeq 2\times 10^{33}\mathrm{g}$. This solar mass is implicit in identifying the central star with the young Sun. The supplied constant disk aspect ratio givesUsing the printed approximate astronomical unit and gravitational constant, one can also obtain $\Omega (1\mathrm{AU})\simeq 3.7\times 10^{-7}{\mathrm{s}}^{-1}$ and $c_s\simeq 3.7\times 10^5{\mathrm{cm}\mathrm{s}}^{-1}$; they give the same $Q$. The gravitational constant cancels from the mass form.

At one astronomical unit the disk is very stable, with $Q\simeq 640$. Formal extrapolation givesThis is far beyond the planetary region and any plausible extent of the minimum-mass solar nebula. Moreover, the extrapolated enclosed disk mass is already a substantial fraction of a solar mass, so the centrally dominated approximation becomes questionable. The formal radius is not a prediction of a real unstable outer nebula. Direct gas fragmentation by gravitational instability is unlikely to have formed Solar System planets in this model. Core accretion is the more natural route; an earlier substantially more massive or colder disk would be a different model. Even $Q<1$ alone does not guarantee fragmentation, because sufficiently rapid cooling is also needed.

First eliminate the density-dependent definition of the auxiliary $\mathbf{g}$. The Newtonian gravitational stress tensor can be written without derivatives of its normalization:Here ${\delta }_{ij}$ is the Kronecker delta, and repeated indices are summed. Differentiating givesThe first and third terms cancel because mixed partial derivatives commute; the final step uses the Poisson equation for Newtonian gravity. Therefore the gravitational force density is a stress divergence:The identity holds for spatially varying mass density: substituting the definition of $\mathbf{g}$ before differentiating prevents erroneous extra density-gradient terms.

For the gravitational stress contribution to angular-momentum transport, the anisotropic term $\rho g_i{g}_j$ supplies angular momentum transport. In cylindrical components its radial–azimuthal entry isIn the momentum conservation equation, $r{T}_{r\varphi }$ is the radial stress contribution to the angular-momentum flux, whose sign depends on the correlated radial and azimuthal field components. It vanishes for a perfectly axisymmetric potential but can be nonzero for spiral disturbances.

The other term is isotropic and acts as an effective negative pressure, $P_g=-\rho g^2/2=-|\nabla \Phi {|}^2/(8\pi G)$. Its force contribution is $-\nabla P_g=+\nabla (\rho g^2/2)$, modifying normal compression and force balance. It has no off-diagonal shear component and therefore no direct radial angular momentum transport. In an axisymmetric averaged disk it supplies no azimuthal torque. The complete symmetric Newtonian gravitational stress tensor also expresses conservation of angular momentum without an internal couple.

When the imposed magnetic field vanishes, the dynamical dispersion relation becomesThe nonstationary hydrodynamic branch therefore has ${\sigma }^2=-(N^2+{\Omega }^2)$. The radial Solberg–Høiland instability criterion in this local model isThe radial buoyancy frequency supplies either restoration or a driving force, while the radial epicyclic frequency $\kappa =\Omega$ supplies rotational restoration. Zero sum is marginal, and positive sum gives stable oscillations. The neutral roots do not themselves signify exponential growth.

Because both equilibrium pressure and dimensionless specific entropy decrease outward, their radial gradients have the same sign. The leading minus sign in the expression for $N^2$ therefore makes $N^2<0$: the radial stratification is adverse. For smooth profiles varying over radius $r$, $|{\partial }_{r}P|\sim P/r$ and $|{\partial }_{r}S|\sim 1/r$, soFor a thin disk, the magnitude of this negative radial buoyancy frequency squared is much less than ${\Omega }^2$. Rotation stabilizes the hydrodynamic mode despite the adverse entropy gradient. This estimate assumes gradients on the global radial scale; a sharp thermal feature may be different. It addresses the ideal nondiffusive equations here. Processes such as convective overstability require additional thermal relaxation and are not ruled out by this particular criterion.

Take $z$ positive upwards and let $f$ denote axial body force per unit volume. With inertia and surface tension omitted, the extensional equations for a slender Newtonian column areMass conservation gives the first equation. The second uses the Trouton ratio three: the leading radial velocity is $u_r=-r{w}_z/2$, the radial normal traction gives $p=p_{\mathrm{ext}}-\mu w_z$, and the axial stress is $-p_{\mathrm{ext}}+3\mu w_z$. An axial control volume balance then gives the displayed equation. The approximation requires a slender column and slowly varying radius.

Let $h=R-a$. Since the container base is closed, the total volume flux at every section is zero. The column carries flux $\pi a^{2}w$, so the annulus carries $-\pi a^{2}w$ and has mean speed of order $a|w|/h$. Its interfacial shear stress is therefore of order $\lambda \mu a|w|/h^2$. Transmitting this stress through the core produces an axial velocity variation of order $\lambda a^2|w|/h^2$. The plug-flow criterion for a column in a low-viscosity annulus is consequently

Use modified pressure $P=p_{\mathrm{ext}}+\rho gz$, removing the annular fluid's hydrostatic pressure. In a local planar gap, with $y=0$ at the column and $y=h$ at the container wall, lubrication theory givesThe required return flux per unit circumference is $q_g=-aw/2$ to leading order in $h/a$. Its pressure-driven component is larger than the Couette component by $a/h$, so the annular return flow around a slumping column givesThe annular shear contribution to the integrated axial force is smaller than this pressure-gradient term by $h/a$. Retaining the leading terms and substituting $f=-(\rho +\Delta \rho )g$ gives the dimensional equationsThe boundary conditions are $w(0,t)=0$ at the base and $w_z(L(t),t)=0$ at the free upper end; the latter expresses zero excess axial extensional stress.

For $\lambda =0$ and initially constant $a=a_0$, put $G=\Delta \rho g/(3\mu )$. Then $w_{zz}=G$, so

For $\lambda >0$, defineThis function records the leading thin-annulus resistance; differences from curvature and the Couette term enter at the already neglected relative order $h/a$. Choose the extensional screening length of a slumping column and the associated scales asHere $h_0=R-a_0$ denotes the initial annular gap in this question. With $Z=z/\hat{z}$, $T=t/\hat{t}$ and $W=w/\hat{w}$, the dimensional equations reduce to

Initially $\varphi ={\varphi }_0$ and ${\Lambda }_0=L_0/\hat{z}$, so $W_{ZZ}=1+W$, with $W(0)=0$ and $W_Z({\Lambda }_0)=0$. The initial velocity profile of an annularly confined column isThe column thickens everywhere below its stress-free top, while that initial thickening rate is zero at the top.

For a tall column, ${\Lambda }_0\gg 1$, the limiting forms away from exponentially small end corrections areThe bulk falls at approximately $-\hat{w}$, where excess weight balances annular hydraulic resistance. Extensional stresses adjust this velocity to zero in a basal boundary layer of thickness $\hat{z}$; almost all thickening occurs there.

For a short column on this scale, ${\Lambda }_0\ll 1$, while still slender, the limits areThe annular-resistance term $W$ is small, so excess weight is balanced predominantly by extensional viscous stress. The top speed is $-\hat{w}{\Lambda }_0^2/2$, recovering the zero-annular-viscosity result. Thus $\hat{z}$ is the vertical distance over which extensional stress communicates the basal constraint before annular resistance screens it. The sketches below use separate natural normalizations for the two limits.

Use pressure with the ambient hydrostatic pressure removed, and let $\Delta P$ be the pressure below the falling sphere minus that above it. Away from the sphere the tube flow is Hagen-Poiseuille flow. The upper and lower clear lengths add to $L$ to leading order in $a/L$, so their pressure drops add:The large exterior reservoir returns this flux with negligible pressure loss compared with the long narrow tube. The sphere is assumed far enough from the ends that entrance effects and its occupied length give only lower-order corrections.

Now use a frame translating downwards with the sphere, and take $z$ positive upwards from its center. The sphere is stationary and the tube wall moves upwards with speed $U$. The narrow gap isIt has axial length scale $a\sqrt{ϵ}$, making lubrication theory appropriate. The sphere-frame flux in a tube is constant: the upward relative flux through the whole section is $\pi a^{2}U-Q$, so its flux per unit circumference isWith $y=0$ on the sphere and $y=h$ on the wall, Couette-Poiseuille flow in a thin gap givesSince $\Delta P=-\int p_{z}dz$, the pressure-jump relation isThe parabolic inner gap can be extended to infinite $z$ because these integrals are dominated by $|z|=O(a\sqrt{ϵ})$. The lubrication resistance integrals for a nearly occluding sphere areSetting $\xi =\tan \theta$ evaluates $J_1=\pi$, $J_2=\pi /2$, $J_3=3\pi /8$, hence

For the force, enclose the fluid between sections just above and below the sphere, including the annular layer. The net upward pressure force is $\pi a^2\Delta P$. The tube wall exerts upward shear traction $\mu u_y(h)$ on this fluid; the sphere exerts a downward force equal to the upward hydrodynamic force $F$ on itself. From the gap solution,The axial control volume force balance therefore gives the control-volume drag formula for a sphere in a tube:Using a control volume avoids having to integrate the curved-sphere normal pressure and tangential shear separately.

With $Q^{\ast }=Q/(\pi a^{2}U)$, $\Delta P^{\ast }=a\Delta P/(6\mu U)$, $q^{\ast }=2q/(aU)$, $F^{\ast }=F/(6\pi \mu aU)$, $I_n^{\ast }=a^{n-1}{I}_n$ and $L^{\ast }=4L/(3a)$, the three flux/pressure equations becomeSolving these linear relations givesThe force relation is $F^{\ast }=\Delta P^{\ast }+\frac{4}{3}{I}_1^{\ast }-q^{\ast }{I}_2^{\ast }$. Substitution first yieldsBecause $I_2^{\ast }/I_3^{\ast }=2ϵ/3$ and $I_1^{\ast }/I_3^{\ast }=2{ϵ}^2/3$, the terms $-2L^{\ast }{I}_2^{\ast }$ and $\frac{4}{3}{L}^{\ast }{I}_1^{\ast }$ are lower order than $L^{\ast }{I}_3^{\ast }$. Keeping the uniformly relevant leading terms givesThe squared $I_2^{\ast }$ is essential; replacing it by $I_2^{\ast }{I}_3^{\ast }$ would give the wrong sign and scaling. This establishes the three drag regimes for a nearly occluding sphere.

For $L^{\ast }\ll {ϵ}^{-1/2}$,The sphere nearly carries the tube's fluid down with it: the far-field flux is close to the piston displacement rate. The relative annular leakage is small in total volume, and the dominant force comes from the local annular shear term. The pressure force associated with the overall tube resistance is smaller.

For ${ϵ}^{-1/2}\ll L^{\ast }\ll {ϵ}^{-5/2}$,The flow remains approximately piston-like, but the dominant force is now the pressure needed to drive Hagen-Poiseuille flow through the long clear tube. Both displayed terms in $q^{\ast }$ are retained because their relative size changes inside this regime without producing an additional leading drag regime.

For $L^{\ast }\gg {ϵ}^{-5/2}$,The large tube resistance makes its net through-flow negligible. The sphere's displaced fluid instead passes upwards relative to it through the thin annular gap, with gap velocity of order $U/ϵ$. Its pressure drop and resulting pressure force dominate.

Finally consider two identical co-moving spheres, with nonoverlapping lubrication regions and the same total clear-tube length to leading order. The common through-flux is not doubled. In the middle regime it is still approximately $\pi a^{2}U$, so the same total tube pressure drop is shared equally between the two identical spheres. The load sharing between co-moving spheres in a tube gives half the single-sphere force on each sphere in regime (ii). In regime (i), drag is set locally by the nearly pressure-balanced annular shear and is unchanged on each sphere. In regime (iii), each sphere must pass essentially its own displacement flux through its annular gap, giving the same local pressure drag as before; adding another pressure jump changes the already small tube flux but not either leading drag. The force is unchanged in regimes (i) and (iii).