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Use circulation to summarize how much a velocity field runs along an oriented closed curve $C$, such as a circle around a fluid vortex. Its definition is the line integralwhere $d\mathbf{l}$ is tangent to $C$ in the chosen direction. Reversing that direction reverses the sign of circulation. Its units are ${\mathrm{m}}^2{\mathrm{s}}^{-1}$, because velocity is multiplied by a length. Counterclockwise circulation in the horizontal plane is positive when the chosen unit normal vector is $+{\mathbf{e}}_z$, according to the right-hand rule.

For solid-body rotation $\mathbf{u}=(-\Omega y,\Omega x,0)$, a counterclockwise circle of radius $R$ has constant tangential velocity $\Omega R$. Multiplying by its circumference givesFor the potential line vortex $\mathbf{u}=K{\mathbf{e}}_{\theta }/r$, the same circle instead gives ${\Gamma }_C=(K/R)(2\pi R)=2\pi K$, independent of $R$. Stipulate $\Omega =1{\mathrm{s}}^{-1}$, $R=1\mathrm{m}$ and $K=1{\mathrm{m}}^2{\mathrm{s}}^{-1}$: both velocity fields have ${\Gamma }_C=2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}\approx 6.283185{\mathrm{m}}^2{\mathrm{s}}^{-1}$, although their vorticities in the fluid are $2{\mathbf{e}}_z{\mathrm{s}}^{-1}$ and zero respectively. On a circle of radius $2\mathrm{m}$, their circulations are $8\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$ and $2\pi {\mathrm{m}}^2{\mathrm{s}}^{-1}$. The comparison plot in vorticity shows this difference in radius dependence.

The bridge between circulation and vorticity is Stokes theorem:Here the velocity field must be continuously differentiable on a neighborhood of the entire spanning surface $S$ inside the fluid, and its boundary orientation must match the unit normal vector $\mathbf{n}$. For solid-body rotation, the disk is admissible and its surface integral is $2\Omega \pi R^2$. For the potential line vortex, the origin is singular and excluded from the fluid: a disk crossing it is not admissible. On an annulus, Stokes theorem instead gives ${\Gamma }_{R_{\mathrm{out}}}-{\Gamma }_{R_{\mathrm{in}}}=0$, with both circulations here measured counterclockwise; the induced inner boundary direction is clockwise. Nonzero circulation therefore does not imply nonzero local curl in the available fluid. ^[ref]

This is useful because Stokes theorem lets a local vorticity measurement determine a loop's circulation when a suitable surface exists, while Kelvin's circulation theorem tracks circulation along a moving material curve. The beauty is the precise link between a local derivative and a global line integral, with the domain deciding which conclusions are valid.

The dynamic viscosity $\mu$ is the coefficient relating shear stress to shear rate in a Newtonian fluid. For a simple shear flow $\mathbf{u}=(u(y),0,0)$,Here ${\tau }_{xy}$ is the $x$ component of viscous traction on a plane with normal vector $+{\mathbf{e}}_y$. The shear rate $\dot{\gamma }=du/dy$ has units ${\mathrm{s}}^{-1}$. Thus dynamic viscosity measures the stress required to sustain a given rate of deformation. It is the shear coefficient; the separate coefficient for volume change is volume viscosity. ^[ref]

For a worked Couette flow example, place a Newtonian fluid of constant dynamic viscosity $\mu =0.10\mathrm{Pa}\mathrm{s}$ between parallel plates separated by $h=0.0010\mathrm{m}$. Hold the lower plate fixed and move the upper plate at $U=0.020\mathrm{m}{\mathrm{s}}^{-1}$. Assume steady flow, no streamwise pressure gradient, and the no-slip boundary condition. ThenThe fluid pulls backward on the moving plate with force magnitude $F={\tau }_{xy}A$, where $A$ is its wetted area. For $A=0.010{\mathrm{m}}^2$ this is $0.020\mathrm{N}$. Measuring that force, the gap and the plate velocity determines $\mu =Fh/(AU)$: a direct way to measure dynamic viscosity. The linear profile and opposing wall tractions are the standard plane Couette flow solution. ^[ref]

The velocity gradient matters, rather than the velocity itself. Adding the same constant velocity to both plates and the fluid leaves $du/dy$, and hence shear stress, unchanged. In uniform flow, every layer translates together and $du/dy=0$. More generally, even a nonzero velocity gradient need not generate viscous stress: solid-body rotation has zero strain-rate tensor, as the viscous stress tensor example shows. Thin gaps amplify shear stress for a fixed plate-speed difference, making dynamic viscosity useful for predicting the force needed to move lubricated surfaces.

During a spherical stellar homology contraction at fixed mass and angular momentum, a constant inertia coefficient gives $I=\alpha M{R}^2$ and solid-body rotation gives $\Omega =J/I\propto R^{-2}$. A fixed fluid label $s=r/R$ retains its enclosed mass and polar angle, so centrifugal acceleration scales as $R^{-3}$ and gravity as $R^{-2}$. Their ratio increases as $R^{-1}$. This assumes spin conservation, not continuous synchronization to an external orbit.

For a homogeneous Newtonian fluid with constant kinematic viscosity $\nu >0$, assume the same steady purely azimuthal flow. In cylindrical coordinates, the azimuthal Navier-Stokes equation has no pressure forcing or convective term in this state, and becomesEquivalently, ${\Omega }^{\prime \prime }+3{\Omega }^{\prime }/r=0$, or $(r^3{\Omega }^{\prime }{)}^{\prime }=0$. Two integrations give the steady circular Couette profile,This is the basic Taylor-Couette flow between rotating cylinders. The constants depend on boundary angular velocities; for radii $R_{<},R_{>}$ and prescribed angular velocities ${\Omega }_{<},{\Omega }_{>}$,If the domain includes $r=0$, regularity requires $B=0$, leaving solid-body rotation. On an annulus a nonzero $B$ is allowed and wall torques maintain the shear.

The restriction concerns a steady profile. An arbitrary initial angular velocity does not cease to exist when viscosity is introduced; within the purely azimuthal ansatz it evolves according to ${\partial }_t\Omega =\nu ({\Omega }^{\prime \prime }+3{\Omega }^{\prime }/r)$ until compatible boundary conditions and angular-momentum constraints determine its long-time state.

Use the thin-annulus approximation stated in the source, so $u_r^{\prime }+ik{u}_z=0$. LetHere ${\kappa }_{\mathrm{ep}}^2$ is the conventional Rayleigh discriminant, also the squared local radial epicyclic frequency when positive. For $k\ne 0$ and $\omega \ne 0$, the azimuthal and axial mode equations giveThe thin-gap incompressible flow condition then gives $\pi =i\omega u_r^{\prime }/k^2$. Substitute these into the radial equation:ThusTo put this into exactly the form printed in the paper, defineThe narrow-gap centrifugal stability equation is consequentlyThe two rigid cylinders are impermeable, so the no-penetration boundary condition givesThere is no inviscid no-slip condition on the tangential or axial components.

The sign deserves emphasis: the source's printed plus sign makes its $\Phi$ the negative of the conventional Rayleigh discriminant. Using the conventional positive discriminant instead requires a minus sign in front of that potential term. These are equivalent conventions; replacing one without adjusting the other would falsely predict instability of solid-body rotation. The $k=0$ and marginal $\omega =0$ cases, excluded by these eliminations, must be treated separately or by limits.

Let $u_r$ be a nonzero admissible mode, multiply the equation of part (d) by its complex conjugate and integrate between the cylinders. The endpoint conditions remove the boundary term in integration by parts, giving the Rayleigh stability integral identityRearrange to obtain a real Rayleigh quotient:The denominator is strictly positive. Consequently, with the paper's convention,The weak inequality $\Phi \le 0$ excludes exponentially growing axisymmetric modes, but permits marginal cases and possible algebraic growth when the restoring effect vanishes. Strict negativity avoids that degeneracy. Equivalently, the conventional Rayleigh discriminant is positive, or the squared specific angular momentum increases outward:This is the Rayleigh centrifugal stability criterion. In the suggested time-independent Schrödinger equation analogy, a growing mode has ${\omega }^2<0$; if $\Phi \le 0$, its effective potential $k^2\Phi /{\omega }^2$ is nonnegative. The nonnegative kinetic term and that potential cannot have the required negative eigenvalue $-k^2$, which gives the same exclusion.

As a sign check, for solid-body rotation ${\kappa }_{\mathrm{ep}}^2=4{\Omega }^2$ and $\Phi =-4{\Omega }^2$. In a gap of width $d=R_{>}-R_{<}$, the radial modes $u_r\propto \sin [n\pi (r-R_{<})/d]$ haveas expected for inertial waves. The argument establishes axisymmetric inviscid stability within the stated thin-gap model, rather than a claim about all perturbations of a viscous flow.

To test radial stability of an axisymmetric circular orbit, let $\Phi (r)$ be the time-independent gravitational potential per unit stellar mass in the midplane. Axial symmetry conserves the specific angular momentum $j=r^2\dot{\phi }$. The radial equation isA circular orbit at $r_0$ obeys $j_0^2=r_0^3{\Phi }^{\prime }(r_0)$. Choose the guiding circular orbit with the same conserved angular momentum as the perturbed star. Setting $r=r_0+x$ and expanding at fixed $j=j_0$ giveswhere the radial epicyclic frequency isHere ${\Omega }^2(r)={\Phi }^{\prime }(r)/r$ and $j_c=r^2\Omega$ refer to the family of circular orbits. They are differentiated as functions of radius only after the perturbation equation has been expanded at conserved stellar $j$.

For a radially stable circular orbit, ${\kappa }^2>0$, so to linear orderThus the orbit performs small simple harmonic radial oscillations about its guiding radius.

Symmetry alone is insufficient to guarantee stability: ${\kappa }^2<0$ gives exponentially growing displacement, and ${\kappa }^2=0$ is marginal. The intended statement assumes a stable unperturbed orbit.

For a Kepler potential $\kappa =\Omega$; for a flat circular-speed curve $\kappa =\sqrt{2}\Omega$; for solid-body rotation $\kappa =2\Omega$. Conservation of $j$ also gives $\dot{\phi }\simeq {\Omega }_0(1-2x/r_0)$, explaining the accompanying azimuthal epicycle rather than a purely radial motion in the inertial plane.

This is the velocity field of solid-body rotation about the axis parallel to $\mathbf{\omega }$ through the origin. A fluid-following body's centre moves around that axis at angular speed $|\mathbf{\omega }|$, keeping its axial coordinate and perpendicular distance fixed. Its orientation rotates at the same angular velocity. The symmetric part of the velocity-gradient tensor is zero, so the prescribed flow produces no local strain or change of shape. Indeed, the separation $\mathbf{d}$ between two fluid points obeys $d|\mathbf{d}{|}^2/dt=2\mathbf{d}\cdot (\mathbf{\omega }\times \mathbf{d})=0$.

This is a kinematic prediction for an entrained body or material object. A finite inertial body initially at rest cannot instantaneously acquire the prescribed velocity and rotation without forces and torques; its transient after placement would also require density, drag and coupling data. Neglecting gravity alone does not specify that transient. On the rotation axis its centre can remain fixed while its orientation still rotates.

To connect the prescribed interfacial drag coefficient to the energy supply, state the lumped interfacial power model for entrainment explicitly:It estimates the power transmitted by lid-driven flow to the mixing interface. Constant geometric and transfer factors are included in $c_D$ or $C_W$. It is the local stress-power interpretation required for the depth laws in this question, not an identity equating arbitrary lid shaft power to interfacial power. With the potential energy of two-layer entrainment from (a),In model S, $u_I=\Omega R$ and $B$ are constant, soThus the depth increases at a constant rate proportional to the inverse bulk Richardson number, with $h=h_0+2C_W{c}_D\Omega Rt/{\mathrm{Ri}}_B$, until the layer reaches $H$. Here the prescribed $u_I$ must be read as a characteristic speed. Literal pure solid-body rotation has zero radial velocity, so the PDF's parenthetical description of $u_I$ as radial cannot simultaneously be a strict component statement and the assumption $u_I=\Omega R$.

If the phrase about lid work is taken literally as shaft power $P_{\mathrm{lid}}=\Omega {\mathcal{T}}_{\mathrm{lid}}$, the interfacial stress alone does not specify the lid torque or its energy transfer to the interface. The preceding closure is therefore a visible model assumption, not a hidden consequence of solid-body rotation.

Under the Boussinesq approximation, total upper-layer kinetic energy is $K={\rho }_{l}Sh{u}_U^2/2$, where $u_U$ is its volume root-mean-square speed. For solid-body rotation of angular speed $\omega (t)$, direct radial averaging gives $u_U^2={\omega }^2{R}^2/2$, consistent with that energy expression. A constant $K$ implies $h{u}_U^2=h_0{u}_U(0{)}^2$. Define $C_K=u_U(0)/(\Omega R)$ to obtain the fixed-energy turbulent mixed layer relationWith $u_I=C_I{u}_U$, the interfacial power model for entrainment used in (b) givesMultiply by $h^{3/2}$ and integrate from $h_0$. This yieldsThe fixed-energy two-layer entrainment law is thereforeHere ${\mathrm{Ri}}_{B,0}=B/({\Omega }^2{R}^2)$ uses the same lid-speed normalization as part (b). All numerical geometry factors depend on the chosen characteristic stress/speed convention. The formula ceases to apply once $h=H$.

The exponent $2/5$ requires the local power to decrease as $u_I^3\propto h^{-3/2}$. It does not follow from constant upper-layer kinetic energy and a fixed fraction of unspecified lid power alone. For example, an imposed constant shaft power together with a constant fraction going to potential energy would give constant $\dot{h}$, not this exponent. The appropriate distinction between local interface power and global forcing appears explicitly in the primary study Entrainment and mixed layer dynamics of a surface-stress-driven stratified fluid, equations (3.6)–(3.7); the calculations above use that local interpretation.

Let $\Psi$ be the gravitational potential, with ${\nabla }^2\Psi =4\pi G\rho$, and let ${\varpi }^2=x^2+y^2$ measure distance from the rotation axis. Uniform solid-body rotation contributes the centrifugal potential $-{\Omega }^2{\varpi }^2/2$. In the corotating frame, static force balance therefore gives the uniformly rotating stellar hydrostatic equilibrium equationsThe final coefficient follows because ${\nabla }^2(x^2+y^2)=4$.

A tangent displacement along a regular connected equipotential surface has zero dot product with $\nabla \varphi$, and hence with $\nabla P$. Thus $P=P(\varphi )$ locally. Taking the curl of the hydrostatic equilibrium equation gives $\nabla \rho \times \nabla \varphi =0$, so $\rho =\rho (\varphi )$ on such surfaces as well. The Poisson equation then shows that ${\nabla }^2\varphi$ is constant on each equipotential surface.

The magnitude $g=|\nabla \varphi |$ need not be constant. For instance $\varphi =A(x^2+y^2)+B{z}^2$, with positive $A\ne B$, has constant Laplacian but different pole and equator gradient magnitudes on an ellipsoidal equipotential. In a rotating star, flattening and centrifugal acceleration likewise make the effective gravity latitude dependent.

With uniform composition, an equation of state determines $T$ from $P$ and $\rho$; for the supplied ideal gas plus radiation pressure law its right side increases monotonically with $T$. Hence $T=T(\varphi )$, and the radiative conductivity $\chi (\rho ,T)$ is also constant on an equipotential surface. Put $f(\varphi )=-\chi dT/d\varphi$. Then $\mathbf{F}=f(\varphi )\nabla \varphi$. On the stellar surface, below mass shedding, the outward normal is $\nabla \varphi /g$, soThis is the radiative gravity-darkening law. The effective temperature measures emergent flux; it is not the same quantity as the interior material temperature that was constant on an equipotential.

A rapidly rotating star is oblate, with reduced effective gravity and cooler effective temperature around the equator. The poles are hottest. The original sketch uses a Roche model of a uniformly rotating star only to illustrate the shape and surface-flux variation:

The obstruction to purely local radiative equilibrium is visible by taking the divergence:On an equipotential surface, $\rho ϵ$, $f$, $f^{\prime }$ and the second term are constant, but $g^2$ generally varies. Unless $f^{\prime }=0$ or another special circumstance removes the mismatch, $\nabla \cdot \mathbf{F}=\rho ϵ$ cannot hold at every latitude. This is the radiative-equilibrium obstruction in a rotating barotropic star; heat advection by stellar meridional circulation resolves the missing energy transport in the leading hydrostatic treatment.

For the steady circulation, mass conservation gives $\nabla \cdot (\rho \mathbf{v})=0$. The specific enthalpy identity and hydrostatic equilibrium giveConsequentlyBoth $h$ and $\varphi$ are constant on the bounding equipotential surface. By the divergence theorem and steady continuity equation,Integrating the given thermal energy equation therefore gives equipotential luminosity conservation with stellar circulation:Individual circulation streams can cross the surface; only their net steady mass flux has to vanish.

Since $f$ is constant on that surface, integrating the Poisson equation givesThus $dT/d\varphi =-L/[\chi (4\pi Gm-2{\Omega }^{2}V)]$, while $dP/d\varphi =-\rho$. Dividing these derivatives and inserting $\chi =4ac{T}^3/(3\kappa \rho )$ yields the radiative gradient in a uniformly rotating star:The bracket must be positive for the regular outward-gravity configuration being used.

For an interior equipotential, both mass factors must be the enclosed mass $m$. The printed final bracket uses $M$, but its derivation gives $m$; at the outer surface $m=M$ and the printed bracket is recovered. For a direct check, a uniform-density model has $m=\rho V$, so the true correction ${\Omega }^{2}V/(2\pi Gm)={\Omega }^2/(2\pi G\rho )$ is independent of the chosen enclosed volume. Replacing $m$ by a fixed total mass would incorrectly make it volume dependent.

Let $M=M_1+M_2$, and let $\mu =M_1{M}_2/M$ be the reduced mass. The stellar distances from the center of mass are $a_1=a{M}_2/M$ and $a_2=a{M}_1/M$. For a circular Kepler orbit, ${\Omega }^2=GM/a^3$, so summing the two orbital contributions gives the circular-binary orbital angular momentumThe separation $a$ is inside the square root. This also has the required dimensions of angular momentum.

For the homologous rotating collapse, label a fluid element by $s=r/R$ and its polar angle $\theta$. Its enclosed mass $m(s)$ is constant during stellar homology, as are the dimensionless density profile and the inertia coefficient $\alpha$. Conservation of angular momentum and solid-body rotation giveThe magnitude of the centrifugal acceleration is ${\Omega }^{2}r\sin \theta$, while the Newtonian gravitational field has magnitude $Gm(s)/r^2$. ThereforeUsing only the radial component of centrifugal force replaces $\sin \theta$ by ${\sin }^2\theta$ and leaves the scaling unchanged. The centrifugal force vanishes on the rotation axis; the central point is understood through the limiting field rather than a ratio of two zero forces.

At the equator of the cloud, critical rotation of a spherical cloud means ${\Omega }_c^2{R}_c=GM/R_c^2$, where $R_c=R_{\mathrm{crit}}$. ConsequentlyImmediately after the first fission, each daughter has mass $M/2$ and radius $a/2$ because the spheres touch. Their orbital moment of inertia is $M{a}^2/4$, while their combined spin moment of inertia is $\alpha M{a}^2/4$. With a common spin and orbital frequency ${\Omega }_f=\sqrt{GM/a^3}$, the synchronous fission model for binary formation givesThe common frequency can change during fission; corotation requires equal instantaneous frequencies, not an unchanged frequency from before the rearrangement. Equating the two values of $J$ yieldsFor a nonzero positive inertia coefficient, the necessary and sufficient condition in this algebraic model for $a<R_c$ isFor example, a moment of inertia of a uniform solid sphere has $\alpha =2/5$ and gives $a/R_c=64/49>1$; it fails the required compact-fission condition. A sufficiently centrally concentrated cloud can have smaller $\alpha$.

For the next collapse, neglect exchange of spin with the unchanged outer orbit, so each daughter's spin angular momentum is separately conserved. Immediately after the first fission that spin isFor a daughter of mass $m=M/2$ to reach critical rotation of a spherical cloud at radius $R_{c,d}$, the same relation gives $S_d=\alpha m\sqrt{Gm{R}_{c,d}}$. HenceEquivalently, its centrifugal-to-gravity ratio starts at $1/4$ when its radius is $a/2$ and reaches one after contraction by a factor four. Applying the same fission rule to this daughter produces an inner pair with separationCorotation at this second fission is local to each newly formed inner pair; it is not a single common frequency for the entire four-star hierarchy.

This spin assumption matters. If tidal locking instead kept each shrinking daughter synchronized to the fixed outer orbital period throughout the intervening collapse, its angular velocity would remain fixed. Its centrifugal-to-gravity ratio would then scale as $R_d^3$ and decrease, so it would never reach the proposed second fission. This is a counterexample to extending the corotation assumption through the whole contraction. The printed ratio describes the separately spin-conserving interpretation of the repeated fission hierarchy.

The model can generate a compact hierarchy algebraically, but is not a general physical account of multiple-star formation. In the allowed range $\alpha <1/3$, one has $a^{\prime }/a<1/8$, suggesting well-separated inner and outer scales. However, rapidly rotating gas deforms, touching daughters are tidally distorted, and pressure, gas flows, dissipation and spin-orbit torques cannot generally be ignored. The assumed identical profiles, equal mass splits, instantaneous corotation and later torque-free contractions are restrictive. Star formation can involve gravitational fragmentation and redistribution of angular momentum; the toy budget neither proves that fission occurs nor predicts the population of real multiple systems.

Use a local, homogeneous razor-thin disk approximation in a frame rotating with constant $\Omega$. The unperturbed planar velocity is zero in this frame, and the large-scale gravitational and centrifugal forces balance. Neglect viscosity, magnetic fields, thickness and background gradients across a wavelength. Take small planar disturbances, with wavelength short compared with the galaxy's background scale but long enough for a fluid description. The barotropic closure of a razor-thin disk is $P=K{\Sigma }^{\gamma }$, with $K$ fixed and a positive derivativeSolid-body rotation has no shear and has radial epicyclic frequency $\kappa =2|\Omega |$. By rotational symmetry of the local model choose a wavevector along $x$, and write perturbations proportional to $e^{i(kx-\omega t)}$.

Let ${\Sigma }_1,u,v,{\Phi }_1$ be the surface density, $x$-velocity, $y$-velocity and Newtonian gravitational potential amplitudes. The linearized continuity equation and Euler equations with the Coriolis force areThe perturbing Newtonian gravitational potential solves the Poisson equation for Newtonian gravityIts decaying solution is proportional to $e^{-|k||z|}$. The jump in its derivative is $-2|k|{\Phi }_1(0)=4\pi G{\Sigma }_1$, giving the razor-thin disk Poisson kernelEliminating $v$ from the two momentum equations, and using ${\Sigma }_1=k{\Sigma }_{0}u/\omega$, gives the density-wave branchThe complete linear system also has a zero-frequency balanced mode; it is not the growing density-wave branch, and division by $\omega$ in this elimination excludes it. The result is the uniformly rotating gas-sheet dispersion relation: pressure opposes compression at large wavenumber, self-gravity promotes compression, and rotation provides epicyclic support.

A mode is exponentially unstable when ${\omega }^2<0$. With $\Omega =0$ and $c>0$, this occurs atThe uniform $k=0$ perturbation is marginal rather than growing in this local calculation. With $c=0$, rotation fails to stabilize sufficiently short waves:These limits show why both pressure and rotation are needed for stability at all wavelengths.

For nonzero pressure and rotation, put $q=|k|$. Complete the square:Its minimum lies at $q_{\ast }=\pi G{\Sigma }_0/c^2$. Thus a growing mode exists precisely whenThe printed inequality has the opposite physical meaning: it is the condition for no exponentially growing density wave. Equality is marginal. In the usual gas Toomre stability criterion, $Q=\kappa c/(\pi G{\Sigma }_0)=2|\Omega |c/(\pi G{\Sigma }_0)$, so instability is $Q<1$ and stability is $Q\ge 1$. The unstable wavenumber band of a rotating gas sheet iswhen $Q<1$.

If ${\Sigma }_0$ and $\Omega \ne 0$ remain fixed while $c^2$ decreases slowly, first reach marginality atThe first wavelength to become unstable just below this threshold is the marginal fragmentation wavelength of a rotating sheet:Density maxima are separated by approximately this wavelength. The expected fragment size is of this order; an overdense half-wave has width about ${\ell }_{\mathrm{crit}}/2$. Linear theory fixes a preferred wavelength, not an exact nonlinear clump radius or shape. Further cooling shifts the fastest-growing wavelength to $2c^2/(G{\Sigma }_0)$. A rough fragment mass is consequently of order ${\Sigma }_0{\ell }_{\mathrm{crit}}^2$, with a geometrical factor depending on the nonlinear fragmentation pattern.

In a stationary axisymmetric gravitational potential, conservation of angular momentum gives $L_z=R^2\dot{\varphi }$. Eliminating $\dot{\varphi }$ introduces the effective potentialFor an equatorial circular orbit, radial balance is ${\partial }_R{\Phi }_{\mathrm{eff}}(R_c,0)=0$, givingVertical balance also requires ${\Phi }_{,z}(R_c,0)=0$.

For epicyclic motion, assume a twice differentiable stationary gravitational potential symmetric under $z\mapsto -z$, small displacements $|x|,|z|\ll R_c$, and a stable circular orbit. Choose the epicyclic guiding center using the conserved $L_z$. Reflection symmetry makes ${\Phi }_{,Rz}(R_c,0)=0$, eliminating radial-vertical coupling at first order. Expanding the equations about $(R_c,0)$ givesThe derivatives are evaluated at the guiding centre, and stability requires ${\kappa }^2,{\nu }^2>0$. Without midplane symmetry, a mixed Hessian term can couple the two oscillations. The radial epicyclic frequency and vertical epicyclic frequency are the frequencies of these independent linear oscillations.

Set $\Omega =L_z/R_c^2$. Along the family of equatorial circular orbits, ${\Phi }_{,R}=R{\Omega }^2$. Differentiating this relation and adding the centrifugal contribution givesTo interpret the common frequency range, let $q=d\log v_c/d\log R$. Since $v_c=R\Omega$, one has ${\kappa }^2=2(1+q){\Omega }^2$. A Keplerian disk has $q=-1/2$ and $\kappa =\Omega$, a flat galaxy rotation curve has $q=0$ and $\kappa =\sqrt{2}\Omega$, and solid-body rotation has $q=1$ and $\kappa =2\Omega$. Typical galactic rotation curves lie between these slopes. Thus $\Omega \lesssim \kappa \lesssim 2\Omega$ is a useful galactic range, not a theorem for every possible axisymmetric potential. As a precise sufficient example, epicyclic frequency bounds for monotone spherical density follow from ${\kappa }^2={\Omega }^2+4\pi G\rho$ and $0\le \rho \le 3M/(4\pi R^3)$.

Use a Cartesian frame rotating with the epicyclic guiding center, with $x$ pointing radially outwards and $y=R_c(\varphi -\Omega t)$ in the direction of rotation. To first order, conservation of angular momentum givesThe radial harmonic oscillator solution and its azimuthal integral areA constant in $y$ merely changes the azimuthal origin of the guiding centre. The epicyclic ellipse obeys $x^2/X^2+y^2/Y^2=1$. In the usual frequency range it is elongated azimuthally, and the star travels clockwise when $x$ points right and $y$ up: its small motion relative to the prograde guiding centre is retrograde.

For the Oort constants, subtract the defining expressions to obtain $\Omega =A-B$ and insert $R{\Omega }^{\prime }=-2A$ into the radial epicyclic frequency formula:The solar-neighbourhood values give $\Omega =26.5\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, $\kappa =\sqrt{1272}\mathrm{km}{\mathrm{s}}^{-1}\mathrm{kp}{\mathrm{c}}^{-1}$, andThe solar epicyclic ellipse is therefore about $1.49$ times longer azimuthally than radially.

A complete radial oscillation takes $T_r=2\pi /\kappa$. The oscillatory part of $\dot{\varphi }$ has zero average over this interval, so the epicyclic azimuthal advance isThis describes the advance between consecutive radial turning points of the same type; it need not be a full $2\pi$ revolution. For the Sun,The azimuthal advance estimate uses the same linear epicyclic motion approximation as the axis ratio.

Equality in the Cauchy-Schwarz inequality requires $\sqrt{r}v(r)=\Omega r^{3/2}$ for one constant $\Omega$. Thus $v(r)=\Omega r$ andThis is solid-body rotation. At fixed $L$, the constant is $\Omega =2L/(V{a}^2)$. The value at the single axis point $r=0$ does not affect the integrals; if $\omega$ is continuous, it has the same constant value there as well.

For rotational motion about a fixed axis, let $r$ be cylindrical radius, $v$ the signed tangential velocity, and $dm$ the mass measure. Define the kinetic energy, axial angular momentum, and moment of inertia byThe Cauchy-Schwarz inequality gives $L^2\le (\int v^{2}dm)(\int r^{2}dm)=2KI$. More precisely, direct expansion gives the nonnegative energy gapThus the minimum at fixed $L$ and fixed mass distribution occurs exactly when $v=(L/I)r$ almost everywhere with respect to $dm$. This means constant angular velocity $\Omega =L/I$, or solid-body rotation. If the profile is continuous, the equality holds at every point in the mass support.

For a uniform unit-density cylinder of radius $a$, height $h$, and volume $V=\pi a^{2}h$, the moment of inertia is $I=V{a}^2/2$. Therefore $K\ge L^2/(V{a}^2)$, with equality for $\omega (r)=2L/(V{a}^2)$. Multiplying by a uniform nonunit density changes the mass measure and $I$ consistently.

The spin of a star about its axis, which can be solid-body rotation or vary through the interior. Angular momentum transport and rotational mixing can change the stellar composition profile and available fuel. An observed surface rotation rate does not determine the entire interior profile.

For a homogeneous inviscid razor-thin disk approximation with solid-body rotation and barotropic sound speed $c$, the compressive wave branch obeys ${\omega }^2=c^2{k}^2-2\pi G{\Sigma }_0|k|+4{\Omega }^2$. The razor-thin disk Poisson kernel supplies the self-gravity term, and the radial epicyclic frequency is $2|\Omega |$. Growing modes have ${\omega }^2<0$; a separate zero-frequency balanced mode can also exist.

Uniform solid-body rotation adds the centrifugal potential $-{\Omega }^2{\varpi }^2/2$ to the gravitational potential. In the corotating frame, hydrostatic equilibrium becomes $\nabla P=-\rho \nabla \varphi$. The Poisson equation gives ${\nabla }^2\varphi =4\pi G\rho -2{\Omega }^2$. On regular connected equipotential surfaces, $P$ is constant; taking the curl shows that $\rho$ is constant too. Uniform composition and an invertible equation of state then make $T$ constant on each such surface. These statements do not require $|\nabla \varphi |$ to be constant there.

The viscous stress tensor is the part of the Cauchy stress tensor associated with deformation rates. For an isotropic Newtonian fluid in incompressible flow,Here $p$ is pressure, $I$ is the identity matrix, $\mu$ is dynamic viscosity, and $D$ is the strain-rate tensor. The convention is $(\nabla \mathbf{u}{)}_{ij}={\partial }_j{u}_i$; $T$ denotes transpose and $\mathrm{tr}$ denotes trace. Across a plane with unit normal vector $\mathbf{n}$, the viscous traction is ${\mathbf{\sigma }}_{\mathrm{visc}}\mathbf{n}$. The tensor makes this rule work for any plane orientation. ^[ref]

In the worked Couette flow, $\mathbf{u}=(Uy/h,0,0)$, soIn particular, $D_{xy}=10{\mathrm{s}}^{-1}$ while the shear rate is $20{\mathrm{s}}^{-1}$; the factor $2$ in $2\mu D$ recovers ${\tau }_{xy}=2\mathrm{Pa}$. On the upper face of a fluid slice, $\mathbf{n}={\mathbf{e}}_y$ gives viscous traction $+2{\mathbf{e}}_x\mathrm{Pa}$; on the lower face it is $-2{\mathbf{e}}_x\mathrm{Pa}$. Constant stress can therefore transmit a force between the plates even though its divergence, and the net viscous force on each slice, vanish.