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In a Keplerian shearing sheet, an epicyclic guiding center $x_0$ carries rotating-frame energy $-3{\Omega }^2{x}_0^2/8$, plus nonnegative radial and vertical oscillation energies. Inelastic collisions dissipate the total Jacobi energy in a shearing sheet, while momentum conservation preserves $\sum x_0$. Thus $\sum x_0^2$ must grow: the ring spreads about its fixed mean radius. The effect is an angular momentum transport process, with some particles moving inward and others outward.

In a steady axisymmetric star, a torque due to Reynolds stress, Maxwell stress tensor or viscosity must balance advection of specific angular momentum by stellar meridional circulation. Writing the nonadvective angular momentum transport flux as ${\mathbf{F}}_J$ gives the displayed balance. Thus stresses that maintain differential rotation also drive a meridional response. The stellar thermal-wind balance supplies a separate leading force constraint, not a replacement for this torque equation.

Stellar rotation is a problem of coupled force balance, heat transport and angular momentum transport. A star need not rotate as a rigid body: in the Sun, the convective envelope rotates faster near the equator than near the poles, while much of the radiative interior rotates more nearly uniformly. The tachocline is the thin transition near the base of the convection zone. Surface tracking and Doppler effect measurements constrain the exterior flow; solar rotational mode splitting in helioseismology constrains interior stellar rotation through integrals of the form $\delta {\omega }_{n\ell m}\simeq m\int K_{n\ell m}\Omega dV$. Multiple kernels are needed, so surface rotation is not a measurement of every depth.

A useful mean velocity decomposition is $\mathbf{U}=\varpi \Omega {\mathbf{e}}_{\varphi }+{\mathbf{v}}_m$, where $\varpi =r\sin \theta$ is distance from the axis and ${\mathbf{v}}_m$ is stellar meridional circulation. In a frame of constant angular velocity ${\mathbf{\Omega }}_0$, use the relative velocity $\mathbf{V}=\mathbf{U}-{\mathbf{\Omega }}_0\times \mathbf{r}$. Its momentum equation includesThe Coriolis force changes the direction of a parcel's motion but does no work. Thermal driving, release of differential rotational kinetic energy, contraction, or external torques supply the energy of the flows. Solid-body stellar rotation in a suitable hydrostatic equilibrium does not by itself require a meridional flow.

For slowly circulating, approximately axisymmetric material, the leading mechanical balance in an inertial frame is $\nabla p/\rho =-\nabla \Phi +\varpi {\Omega }^2{\mathbf{e}}_{\varpi }$. Taking its curl gives the stellar thermal-wind balanceHere $z$ is the axial coordinate, $S$ is specific entropy and $c_p$ is the specific heat at fixed pressure. The last form uses approximately radial hydrostatic equilibrium and uniform composition. If the material is barotropic, the right-hand side vanishes and angular velocity is constant on cylinders. Small latitudinal entropy differences can instead support the observed noncylindrical differential rotation. Stresses and inertial terms modify this leading balance. The equation relates shear to baroclinicity; it does not give the circulation speed or explain how the shear is maintained.

That second question requires an angular-momentum budget. With $j={\varpi }^2\Omega$ and density-weighted mean quantities, conservation of angular momentum can be written schematically asThe three contributions are Reynolds stress, magnetic Maxwell stress tensor and viscosity transport. Appropriate density weighting replaces the simple correlations in a strongly compressible average. Since $\nabla \cdot (\rho {\mathbf{v}}_m)=0$ in a steady star, the stationary equation gives gyroscopic pumping in a star:A torque that maintains differential rotation therefore drives a meridional response unless another stress cancels it. Conversely the circulation transports angular momentum and feeds back on the rotation. An assumed meridional pattern without its driving torque is not a closed dynamical model.

In the convection zone, outward heat transport drives convection. Rotation deflects the rising and falling parcels and makes their correlations anisotropic. Nondiffusive convective angular momentum transport can drive shear even from an initially uniform rotation rate; treating every Reynolds stress as a positive eddy viscosity would only smooth shear and miss its source. A limiting parcel that conserves $j$ when it moves outwards slows relative to uniform rotation, whereas rapid rotational constraint and correlated horizontal motions change the sign and direction of the net transport. Thus the solar equator-fast pattern requires the actual convective correlations, not a universal parcel argument. The Rossby number $u_c/(2\Omega {\ell }_c)$ compares convective inertia with rotational deflection: deep, slower, larger-scale convection is more rotationally constrained than the fastest photospheric motions. Turnover times ${\ell }_c/u_c$ range from minutes near the visible surface to weeks or longer at greater depth. On these times the Reynolds stresses can redistribute angular momentum and drive circulation, while the star's total spin changes much more slowly.

In a radiative region, rotation distorts equipotential surfaces. Under uniform rotation and uniform composition, leading hydrostatic equilibrium makes pressure, density and temperature constant on each such surface, but effective gravity varies along it. Radiative diffusion then produces a flux $\mathbf{F}=f({\Phi }_{\mathrm{eff}})\nabla {\Phi }_{\mathrm{eff}}$. Its divergence contains $f^{\prime }|\nabla {\Phi }_{\mathrm{eff}}{|}^2$, which varies over a surface and generally cannot match a nuclear source constant there. This is the radiative-equilibrium obstruction in a rotating barotropic star. Thermal imbalance drives Eddington-Sweet circulation, with heat advection enteringIts global thermal estimate isA rotational thermal imbalance of order ${ϵ}_{\Omega }L$ processes the thermal reservoir on this longer time. For the slowly rotating Sun, $t_{\mathrm{KH}}$ is of order $3\times 10^7$ years and ${ϵ}_{\Omega }$ of order $2\times 10^{-5}$, giving $t_{\mathrm{ES}}\sim 10^{12}$ years, much longer than the solar age. Stable composition gradients impede the flow further. Thus classical radiative Eddington-Sweet circulation cannot by itself establish the Sun's interior rotation rapidly. In a rapidly rotating star the same estimate is shorter and rotational mixing can compete with its evolutionary lifetime. Torque-driven convection-zone circulation and radiative thermal circulation have different controlling balances.

Magnetic fields introduce a further coupling. A poloidal magnetic field linking different layers is wound by differential rotation into a toroidal magnetic field, roughly ${\partial }_t{B}_{\varphi }\simeq \varpi {\mathbf{B}}_p\cdot \nabla \Omega$. The resulting Maxwell stress tensor transport angular momentum. Communication along a field occurs on an Alfvén speed crossing time $t_A\sim \ell \sqrt{{\mu }_0\rho }/B$, which can be far shorter than molecular diffusion if a coherent connecting field exists. In steady ideal axisymmetric induction, absence of continuing winding requires ${\mathbf{B}}_p\cdot \nabla \Omega =0$; this is constancy along the field, not automatically solid-body rotation throughout the star. The topology and strength of an interior field matter. Solar dynamo action, especially the interaction of shear with poloidal-field regeneration, couples stellar rotation to the magnetic cycle. Magnetic feedback can change shear and circulation on dynamical or cycle times; an approximately eleven-year activity cycle and twenty-two-year magnetic polarity cycle are distinct from secular spin-down.

A magnetized stellar wind also supplies an external torque. It enforces approximate corotation out to an effective Alfvén radius $R_A$, giving wind-driven magnetic braking of a solar-type star $\dot{J}\sim -{\dot{M}}_w\Omega R_A^2$. Its spin-down time is $t_J\sim I/({\dot{M}}_w{R}_A^2)$, much longer than a convective turnover time. For slowly changing moment of inertia and an unsaturated field scaling that gives $\dot{J}=-K{\Omega }^3$, integration gives ${\Omega }^{-2}={\Omega }_0^{-2}+2Kt/I$, the Skumanich rotation law. A growing or contracting star also changes $I$ and can spin up or down even without torque. Differential spin-down can create shear between layers unless internal transport couples them.

Finally, shear can excite Kelvin-Helmholtz instability, and convection can launch internal gravity waves into the stable radiative interior. Wave damping deposits angular momentum; the net transport depends on excitation and selective absorption and is not fixed by a wave-crossing time alone. Such stresses, magnetic coupling and circulation are candidate agents for the nearly uniform radiative rotation and the narrow tachocline. Molecular kinematic viscosity has a diffusion time ${\ell }^2/\nu$ generally too long to couple the entire solar interior over its age, whereas a turbulent effective kinematic viscosity can act much faster but is not appropriate to every stable layer. A circulation has an advection time $\ell /U_m$, thermal diffusion has time ${\ell }^2/\chi$, and magnetic diffusion has time ${\ell }^2/{\eta }_m$; these must be compared with the local stellar rotation and evolutionary times rather than conflated with them. The solar dynamical time $\sqrt{R^3/(GM)}$ is about half an hour, its rotation period is of order a month, and secular magnetic braking takes an evolutionary time. Rapid force adjustment, convective maintenance of shear, cyclic magnetic feedback and slow global spin evolution are separate processes. Their coupled conservation equations and timescale hierarchy are the basis of a consistent account of solar rotation.

Set $S=T=0$ and choose $\boxed{\bar{\nu }(r)={\nu }_{0}r}$, with ${\nu }_0>0$ constant. Define $y=\sqrt{r}$ and $g=r^{3/2}\Sigma$. Then $\sqrt{r}\bar{\nu }\Sigma ={\nu }_{0}g$, and ${\partial }_r=(2y{)}^{-1}{\partial }_y$. Directly transforming the Keplerian viscous diffusion equation givesThe heat-equation transform for linear-radius disk viscosity is completed by $x=2y/\sqrt{3{\nu }_0}$:Equivalently one can use $g=\sqrt{r}\bar{\nu }\Sigma$, which differs only by a constant. This is the viscosity law making the transformed diffusivity constant under this square-root-radius reduction. Initial and boundary conditions transform too; the change of variables does not supply unspecified disc-edge data.

The physical diffusion arises from angular momentum transport. Differential rotation creates a viscous shear stress, which transfers angular momentum between neighboring annuli. Gas that loses angular momentum moves inward, while gas that gains it can move outward. Combining conservation of angular momentum with mass conservation turns the stress gradient into a radial mass flux and its divergence into a second radial derivative. Positive kinematic viscosity smooths the appropriately weighted profile; matter need not all move in one direction, and an accreting disc can simultaneously spread outward. Viscosity dissipates orbital energy but redistributes angular momentum internally except for boundary losses.

The additional conditions are $\boxed{\nabla \cdot \mathbf{u}=0,\nabla \cdot \mathbf{B}=0}$: incompressible flow and the solenoidal magnetic field constraint. The scalar $\Psi$ combines pressure per unit density with magnetic pressure per unit density and any conservative gravitational and centrifugal potentials. In particular, the Lorentz force identity separates magnetic tension from $-\nabla (B^2/2{\mu }_0\rho )$. Adding this latter term to the rotating-flow pressure potential gives the pressure appearing in the stated equation; the Coriolis force remains explicit.

Work in magnetic velocity units $\mathbf{B}/\sqrt{{\mu }_0\rho }$, and write $\mathbf{U}=-2Ax{\mathbf{e}}_y$ for the background shearing sheet flow. Its advective acceleration is zero, and $2\Omega {\mathbf{e}}_z\times \mathbf{U}=4\Omega Ax{\mathbf{e}}_x=-\nabla (-2\Omega A{x}^2)$, so the stated quadratic background pressure potential gives an exact balance. A spatially uniform mean magnetic field must evolve asThis is stretching of radial field into azimuthal field by differential rotation.

For the shearing wave phase $\Theta =\mathbf{k}(t)\cdot \mathbf{x}$, advection by the background givesIt vanishes identically ifThe changing wavevector describes the shear of initially planar wavefronts. The mean-field stretching and wavevector evolution cancel in the Alfvén frequency:Hence $\boxed{{\omega }_A=\mathbf{k}\cdot \bar{\mathbf{b}}\text{is constant}}$.

These solutions are exact finite-amplitude disturbances, not merely a linearization. To see this, let ${\mathbf{u}}^{\prime }=\mathrm{Re}(\mathbf{v}{e}^{i\Theta })$ and ${\mathbf{B}}^{\prime }=\mathrm{Re}(\mathbf{b}{e}^{i\Theta })$ in magnetic velocity units. The divergence constraints give $\mathbf{k}\cdot \mathbf{v}=\mathbf{k}\cdot \mathbf{b}=0$. Thus both real disturbances are pointwise perpendicular to k, whereas all their spatial derivatives are along k. Every quadratic term ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }$, ${\mathbf{B}}^{\prime }\cdot \nabla {\mathbf{B}}^{\prime }$, ${\mathbf{u}}^{\prime }\cdot \nabla {\mathbf{B}}^{\prime }$ and ${\mathbf{B}}^{\prime }\cdot \nabla {\mathbf{u}}^{\prime }$ vanishes, including terms involving conjugate amplitudes. This extends the exact incompressible shearing wave to an exact magnetic shearing wave. The gas pressure can compensate quadratic magnetic-pressure terms even though the combined $\Psi$ contains only its stated background and wave harmonic.

The remaining wave terms are therefore preciselyThe shear terms arise from perturbations acting on $\nabla \mathbf{U}$, the magnetic terms from the uniform field acting on the wave, and the last terms from diffusion. Pressure enforces the moving-wavevector velocity constraint: differentiating it givesUsing this pressure leaves $\mathbf{k}\cdot \mathbf{v}=0$ invariant. Differentiating $\mathbf{k}\cdot \mathbf{b}$ in the induction equation similarly leaves it zero. Thus the evolution is consistent with both constraints for initially transverse amplitudes.

Now take positive $\Omega$, so the stipulated condition implies $0<A<2\Omega$, and take magnetic Prandtl number $\nu /\eta =1$, with $\nu =\eta >0$. For a vertical wavevector $\mathbf{k}=k{\mathbf{e}}_z$, use the equal-diffusivity magnetorotational channel formThe two horizontal momentum equations, after adding them, give $\dot{V}=(A-\nu k^2)V$. Either component then gives $i{\omega }_{A}B=(A-2\Omega )V$. The first induction equation gives $({\partial }_t+\nu k^2)B=i{\omega }_{A}V$. These equations are simultaneously solved byThe second induction component gives the same relation, since $B_y=-B_x$. Exponential growth occurs when $\boxed{\nu k^2<A}$. The quarter-cycle phase between magnetic and velocity perturbations is required by the factor i in the Fourier equations.

For Keplerian rotation, expanding the local shear gives $2A=-(r{\Omega }^{\prime })=3\Omega /2$, hence $A=3\Omega /4$. Therefore ${\omega }_A^2=15{\Omega }^2/16$. With $v_{Az}={\bar{B}}_z/\sqrt{{\mu }_0\rho }$ and ${\omega }_A=k{v}_{Az}$, the growth condition for this prescribed polarization becomesA weaker field would require a shorter wavelength to retain this mode's selected Alfvén frequency, making diffusive damping too strong. The required wavelength must also fit any actual vertical domain; no vertical boundary conditions are specified here.

The field bound is for the requested equal-component mode, rather than a universal necessary condition for all magnetorotational instability. This distinction matters if all vertical wavelengths are admitted. Indeed, let $s=\gamma +\nu k^2$ for a general horizontal polarization. Eliminating its two magnetic amplitudes gives $(s+{\omega }_A^2/s)v_x-2\Omega v_y=0$ and $(s+{\omega }_A^2/s)v_y+[2(\Omega -A)-2A{\omega }_A^2/s^2]v_x=0$. Their determinant givesFor a Keplerian disk, ${\kappa }^2={\Omega }^2$. As a concrete counterexample to a universal reading of the bound, in consistent dimensionless units take $\Omega =1$, $A=3/4$, $\nu =\eta =1$, $v_{Az}=0.1$, and $k=0.01$. The positive root gives $s\simeq 0.00173205$ and $\gamma =s-0.0001>0$, even though $v_{Az}^2=0.01<1.25$. This mode has a different polarization and a much longer wavelength. Real finite-height discs restrict such wavelengths; the PDF supplies no such restriction.

Finally, the horizontally averaged radial flux of azimuthal momentum is the sum of Reynolds stress and magnetic tension stress:Multiplication by the local radius gives an outward angular-momentum flux. Equal radial and azimuthal velocities give the largest positive product at fixed horizontal kinetic energy, while opposite magnetic components give the largest positive $-B_x{B}_y$ at fixed horizontal magnetic energy. Both follow from $2|ab|\le |a{|}^2+|b{|}^2$. This stress-maximizing magnetorotational channel polarization therefore makes both contributions reinforce outward angular momentum transport. It extracts shear energy at rate $-T_{xy}{\partial }_x{U}_y=2A{T}_{xy}>0$, explaining why these correlations are especially effective in driving accretion.

Consider a local shearing sheet centered at a circular orbit. Take uniform mass density, background velocity ${\mathbf{u}}_0=-2Ax{\mathbf{e}}_y$, rotation $\mathbf{\Omega }=\Omega {\mathbf{e}}_z$ and a weak uniform vertical magnetic field ${\mathbf{B}}_0=B_0{\mathbf{e}}_z$. HereUse ideal magnetohydrodynamics, neglect viscosity, magnetic diffusion and vertical stratification, and take incompressible axisymmetric disturbances proportional to $e^{ikz-i\omega t}$. They have horizontal velocity and field perturbations. Let $\mathbf{b}=\delta \mathbf{B}/\sqrt{{\mu }_0\rho }$ be the magnetic perturbation in velocity units, $v_A=B_0/\sqrt{{\mu }_0\rho }$ the Alfvén speed and ${\omega }_A=k{v}_A$ the Alfvén frequency. Magnetic pressure can be included in the perturbation's total pressure; the horizontal Lorentz force for this mode is the magnetic tension term $i{\omega }_A\mathbf{b}$.

Linearizing momentum and induction givesThe term $-2A{b}_x$ winds a radial magnetic perturbation into an azimuthal one. Omitting it would remove the shear's magnetic energy source and give the wrong stability criterion.

A useful elimination that also covers neutral modes introduces the fluid displacement $\mathbf{\xi }$. Frozen-in flux gives $\mathbf{b}=i{\omega }_A\mathbf{\xi }$, while the Eulerian perturbation velocity is $u_x={\dot{\xi }}_x$, $u_y={\dot{\xi }}_y+2A{\xi }_x$, because the displacement samples the gradient of the basic flow. The equations become the spring model of magnetorotational instability:For the stated time dependence, their determinant isThus the ideal magnetorotational dispersion relation isIt follows without dividing by $\omega$, so the neutral stability boundary is included.

To analyze it, put $a={\omega }_A^2$ and $Y={\omega }^2$. ThenFor a hydrodynamically stable rotation law, ${\kappa }_r^2>0$, the sum of the two roots is positive. Their product is negative precisely when $0<a<4\Omega A$. One squared frequency is then negative, giving an exponentially growing mode $\omega =i\gamma$ and its decaying partner. Therefore the magnetorotational instability, or MRI, has criterionAngular velocity must decrease outwards, although specific angular momentum may increase outwards and satisfy Rayleigh's circulation criterion. This is why a Keplerian disc can be stable to the axisymmetric hydrodynamic disturbances of Question 2 yet unstable once magnetic coupling is introduced. At zero magnetic frequency the unstable root becomes neutral, while sufficiently large magnetic frequency restores stability. For increasing angular velocity, $A<0$ with $\Omega >0$, this ideal vertical-field instability is absent.

The growth rate isFor $0<A<\Omega$, differentiation with respect to $a$ sets $\sqrt{{\kappa }_r^4+16{\Omega }^{2}a}=4{\Omega }^2$, yielding the maximum growth rate of ideal magnetorotational instability:For Keplerian rotation, $A=3\Omega /4$, soThus the fastest growth is on an orbital timescale. The field strength selects its wavelength, $k_{\max }\propto 1/v_A$, rather than the maximum ideal growth rate. At fixed $k$, growth does approach zero as $B_0\to 0$; retaining the finite maximum in that limit requires shorter wavelengths, which magnetic diffusion can eventually exclude. Unstable wavelengths satisfy $\lambda >2\pi v_A/\sqrt{4\Omega A}$ in this local model. A finite disc must accommodate such a wavelength, so a sufficiently strong field can suppress the available vertical modes through excessive tension.

Physically, a bent field line acts like a spring joining neighboring fluid elements. The inner element rotates faster and pulls the outer element forward, transferring angular momentum outward. The inner element loses angular momentum and moves inward; the outer element gains it and moves outward. In a rotation law with outward-decreasing angular velocity, these motions increase their angular-velocity difference and the magnetic coupling produces further separation. A weak restoring spring therefore creates a feedback instability by enabling angular-momentum exchange. Without that exchange, each displaced element keeps its angular momentum and a positive epicyclic frequency restores its orbit. Strong magnetic tension instead prevents the separation. The free energy is the differential rotation, not the initial magnetic energy.

The instability provides an important route to turbulence and angular momentum transport in sufficiently conducting accretion discs. The radial flux of azimuthal momentum includes the stressThe first term is a Reynolds stress and the second is the magnetic contribution from the Maxwell stress tensor, with the sign appropriate to momentum flux. For the growing branch, take $u_x$ real and positive. The linear equations give $u_y/u_x=({\gamma }^2+a)/(2\Omega \gamma )>0$ and $b_y/b_x=({\gamma }^2+a-4\Omega A)/(2\Omega \gamma )<0$. The last sign follows from $({\gamma }^2+a)({\gamma }^2+a-4\Omega A)+4{\Omega }^2{\gamma }^2=0$. Thus both the velocity and magnetic contributions carry angular momentum outward. The associated energy extraction is $-T_{xy}rd\Omega /dr=2A{T}_{xy}>0$. Inward accretion can then accompany outward angular-momentum transport, and dissipation of the resulting fluctuations heats the disc. This gives a physical mechanism behind an effective turbulent-viscosity description such as an alpha disk, whereas microscopic viscosity alone is generally too weak for rapid accretion.

MRI is an exponentially growing magnetic instability of outward-decreasing angular velocity; hydrodynamic shearing-wave amplification is transient and does not by itself establish sustained turbulence. Nonlinear saturation, field regeneration and transport efficiency depend on field geometry, boundaries and thermodynamics. Poor conductivity or important nonideal magnetic effects can suppress or modify the ideal instability, so the local dispersion relation is a mechanism and growth criterion rather than a universal prescription for the saturated stress.

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.

Initially every particle has $A=B=0$, so its Jacobi energy in a shearing sheet is $-3{\Omega }_0^2{x}_0^2/8$. During an inelastic collision, positions are fixed at the instant of impact, while momentum conservation preserves the sum of the tangential velocities. Hence the sum of $p_y=\dot{y}+2{\Omega }_{0}x$, and therefore the sum of the epicyclic guiding center positions $x_0=2p_y/{\Omega }_0$, is unchanged. Between collisions, these quantities are individually conserved.

The collision dissipates kinetic energy without changing the instantaneous tidal potential. Consequently the total Jacobi energy in a shearing sheet decreases. At any later time, writing $x_{0j}$ for the particles' current epicyclic guiding center positions givesIf $D\ge 0$ is the accumulated energy dissipated in the inelastic collisions, comparison with the initial circular orbits givesThus the mean guiding-center position stays fixed, while its variance grows. The initial ensemble has no preference for positive or negative $x$, and the local equations and collision law preserve the symmetry $(x,y,z)\mapsto (-x,-y,z)$. The spreading is therefore symmetric in the ensemble average: angular momentum transport moves some particles inward and others outward. A particular finite random realization need not be exactly symmetric. This is the microscopic energy argument for dissipative spreading of a planetary ring.

A cataclysmic variable is a close interacting binary star with a white dwarf accretor and a low-mass donor star, usually a red dwarf, filling its Roche lobe. Matter crosses by Roche-lobe overflow and commonly forms an accretion disk; sufficiently strong white-dwarf magnetism instead permits magnetically channelled accretion. Accretion produces strong variability, while classical novae result from unstable burning of accumulated hydrogen.

The cataclysmic-variable orbital-period distribution contains a short-period population between about 82 minutes and two hours, a pronounced cataclysmic-variable period gap around two to three hours, and a longer-period population above the gap. There is an accumulation near the cataclysmic-variable period minimum. These describe ordinary hydrogen-rich systems; evolved donors and helium-rich systems can have shorter periods. Selection by luminosity and outburst activity affects the observed counts, so $N(P)$ is not simply the intrinsic distribution.

For the contact red dwarf, $R_2=R_L\propto a(M_2/M{)}^{1/3}$. Using its supplied thermal-equilibrium mass-radius relation, $R_2\propto M_2$, gives $a\propto M_2^{2/3}{M}^{1/3}$. Kepler's third law then givesindependent of $M_1$. This is a special case of the Roche-lobe-filling period-density relation.

To examine rapid mass loss, use the stellar radius response exponent ${\zeta }_{\mathrm{ad}}=d\log R_2/d\log M_2=-1/3$. During conservative binary mass transfer, $M$ and $J=M_1{M}_2\sqrt{Ga/M}$ are fixed, and $d{M}_1=-d{M}_2$. ThereforeThe change in overfill is $d\log (R_2/R_L)=({\zeta }_{\mathrm{ad}}-{\zeta }_L)d\log M_2$. Since $d\log M_2<0$, overfill grows, and mass loss runs away, when ${\zeta }_{\mathrm{ad}}<{\zeta }_L$. The dynamical stability of binary mass transfer criterion here is consequentlyThe printed instability inequality is reversed. With the defined $q=M_2/M_1$, the specified radius laws imply the inequality above; $q<q_{\mathrm{crit}}$ is the stable side. Equality is marginal, and the supplied Roche-radius approximation restricts this calculation to its stated mass-ratio range.

Long-term contact is maintained by angular momentum transport out of the orbit. Gravitational-wave emission from a binary system provides one loss mechanism. Magnetic braking of a binary star uses the donor's stellar wind: its magnetic field enforces approximate corotation out to the Alfvén radius, so the wind carries much more specific angular momentum than a nonmagnetic surface outflow. At that radius, roughlyThe geometry contributes order-one factors to the torque. Even when ${\dot{M}}_w\ll |{\dot{M}}_2|$, the large lever arm can remove substantial spin angular momentum. Tidal locking couples donor spin to orbital motion, so the wind torque ultimately brakes the orbit. In practice classical novae can also eject matter; the conservative calculations here isolate the stated idealization.

For the prescribed $\dot{J}/J=-\alpha$, now use the supplied equilibrium law $R_2\propto M_2$ for the slowly evolving donor. Contact implies $\dot{a}/a=(2/3){\dot{M}}_2/M_2$, assuming negligible net wind mass loss. Taking the logarithmic derivative of $J$ givesSince $P\propto M_2$ along this equilibrium sequence,For the dynamically stable mass ratios, this is negative. The result uses thermal equilibrium as well as slow hydrostatic evolution; evolution slow compared only with the dynamical time does not by itself guarantee the radius law $R_2\propto M_2$.

In the disrupted magnetic braking model, the donor above the gap loses mass fast enough to be inflated relative to thermal equilibrium. When the donor becomes a fully convective star, the model assumes a substantial reduction in magnetic braking. The red dwarf then contracts toward equilibrium on its Kelvin-Helmholtz cooling time, becoming smaller than its Roche lobe; accretion largely stops. Gravitational-wave emission from a binary system continues to reduce the separation, carrying the detached binary through the gap until contact is restored near its lower edge.

This explanation requires strong braking immediately above the gap: the donor must already be out of equilibrium and inflated, so there is room to contract and detach when braking drops. With nearly fixed component masses during detachment, $R_L\propto a\propto P^{2/3}$, giving the illustrative inflation factorA weak braking torque that kept the donor in thermal equilibrium just above the gap would not produce this detachment. The model requires a decrease in the torque, not disappearance of the donor's magnetic field.

Define the surface density of a disk, outward mass flux, internal torque, and surface magnetic torque byThe sign convention makes $\mathcal{G}$ positive for outward angular momentum transport in an ordinary Keplerian accretion disk. Vertical integration of mass conservation gives

The specific angular momentum is $h=r{u}_{\varphi }=r^2\Omega$. Multiply the azimuthal equation by $r$, use the continuity equation to put its left-hand side in conservative form, and integrate over $z$. The assumed decay removes the vertical mass and viscous fluxes, whereas the magnetic surface stress remains:Subtracting $h$ times the integrated mass equation yieldsSince $F_M=-(dh/dr{)}^{-1}({\partial }_r\mathcal{G}-r\mathcal{T})$, substitution in mass conservation gives the required one-dimensional advection-diffusion equation

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.