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Take two reservoirs separated by a semipermeable membrane that passes water and retains one neutral solute. Stipulate $T=300\mathrm{K}$, $c_A=10\mathrm{mol}{\mathrm{m}}^{-3}$ and $c_B=30\mathrm{mol}{\mathrm{m}}^{-3}$, equivalent to $10$ and $30\mathrm{mmol}{\mathrm{L}}^{-1}$. Treat both chemical solutions as dilute and ideal. These are invented model conditions, not measurements or recommended biological conditions.

For pressures measured at the same height beside the membrane, define $\Delta P=P_B-P_A$. The model predictsAt $\Delta P=0$, net water transfer favors A to B. At $\Delta P\simeq 49.89\mathrm{kPa}$, the opposing pressure balances the osmotic pressure difference, giving zero net transfer with unequal concentrations. At $\Delta P=75\mathrm{kPa}$, the sign reverses and net water transfer favors B to A, toward the more dilute reservoir. Raising pressure changes the total water potential, rather than changing the membrane into a one-way gate.

Fixed concentrations require large maintained reservoirs or an instantaneous comparison. In a closed membrane osmometer, entering water causes dilution of the receiving chemical solution and changes its volume; the final balance must use final concentrations. A supported cell wall provides another way to develop the opposing pressure, as explained under turgor pressure.

Osmotic pressure is the excess pressure needed on a chemical solution to maintain osmotic thermodynamic equilibrium with pure solvent across a semipermeable membrane. For a nearly incompressible solvent, equality of chemical potential gives $\Pi \simeq -(RT/{\bar{V}}_w)\ln a_w$. It is an equilibrium property; an observed initial swelling rate is a different measurement.^[ref]

For a dilute ideal chemical solution of retained particles,Here $n$ is the amount of substance of independently dissolved solute particles, measured in moles; $V$ is the final volume of chemical solution, not the starting volume of pure solvent. Count the resulting ions separately if a solute dissociates; do not multiply twice if $n$ already counts them. With $c_{\mathrm{osm}}$ in $\mathrm{mol}{\mathrm{m}}^{-3}$ and $R$ in $\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, $RT{c}_{\mathrm{osm}}$ has units $\mathrm{J}{\mathrm{m}}^{-3}=\mathrm{Pa}$. This resembles the ideal gas law, although the particles here are dissolved in a liquid.

The assumptions are uniform temperature, a dilute ideal chemical solution, a nearly incompressible solvent, and a semipermeable membrane that retains the counted solutes while passing solvent. Association, incomplete dissociation and interactions change the relation between added material and independent particles; nonideal thermodynamic activities invalidate simple particle counting. The formula does not describe a concentrated brine merely because the apparatus contains a membrane.

A short derivation explains why particles are counted. In an ideal mixture, $a_w=x_w=n_w/(n_w+n)$, with $n_w$ the amount of substance of water. If $n\ll n_w$, then $-\ln x_w=\ln (1+n/n_w)\simeq n/n_w$ and $V\simeq n_w{\bar{V}}_w$. Substitution gives $\Pi \simeq RTn/V$. This is a dilute-limit approximation, not a claim that water molecules stop moving or that dissolved particles are a gas.

The partial pressure of a component of a gas mixture is the pressure it would exert alone at the same temperature and total volume. For an ideal mixture, $p_i=n_{i}RT/V$ and $p={\sum }_i{p}_i$. The chemical potential of an ideal gas uses $p_i$ for that component rather than the total pressure $p$.

The pascal is the SI unit of pressure:For a uniform normal force of $10\mathrm{N}$ over area $0.02{\mathrm{m}}^2$, the pressure is $500\mathrm{Pa}$. The unit expresses a force per area, rather than the total force alone. ^[ref].

With pressure amplitude $P=ZV$ and velocity $V$ directed into a boundary, the mean absorbed acoustic energy flux is $\frac{1}{2}\mathrm{Re}(P{V}^{\ast })=\frac{1}{2}\mathrm{Re}Z|V{|}^2$. A passive boundary cannot supply net energy, so $\mathrm{Re}Z\ge 0$ at a real frequency. Its reflection from a propagating acoustic plane wave has magnitude at most one. A pole continued to a complex incidence angle or frequency must be interpreted separately from real-frequency passive scattering.

For constant density $\rho$, the Euler equations for an inviscid fluid with a conservative force per unit mass areSet $\mathbf{\omega }=\nabla \times \mathbf{u}$. The vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$ and vanishing curl of a gradient givewhere both divergence terms vanish. Thus the vorticity equation isThe partial time derivative measures Eulerian change; $(\mathbf{u}\cdot \nabla )\mathbf{\omega }$ is advection; the right-hand side describes stretching and tilting of vortex lines by velocity gradients. A conservative body force supplies no curl source, and constant density removes any baroclinic pressure source.

For planar velocity independent of $z$, vorticity is perpendicular to the plane, $\mathbf{\omega }=(0,0,\omega )$, and $(\mathbf{\omega }\cdot \nabla )\mathbf{u}=\omega {\partial }_z\mathbf{u}=0$. Hence each material particle conserves its vorticity, $D\omega /Dt=0$. Choose the stream function convention $u={\psi }_y$, $v=-{\psi }_x$. ThenIn a steady flow, ${\psi }_y{\omega }_x-{\psi }_x{\omega }_y=0$, so vorticity is constant along streamlines. Where $\nabla \psi \ne 0$, use $\psi$ as a local transverse coordinate: this equality makes $\omega$ independent of the coordinate along a streamline. ConsequentlyThis is steady planar vorticity as a local function of stream function.

A global single-valued $F$ additionally requires compatible vorticity values on all components of each stream-function level set. That condition is not supplied by steady Euler flow alone. For example, $\psi =x^3-x$ gives $(u,v)=(0,1-3x^2)$, a smooth steady incompressible shear with zero advective acceleration and constant pressure. Yet $\psi (0,y)=\psi (1,y)=0$ while $\Delta \psi (0,y)=0$ and $\Delta \psi (1,y)=6$. Thus the unrestricted global deduction is false; the local result, or a suitable connected-level-set hypothesis, is the valid conclusion.

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.

The energy-conserving self-similar blast wave requires a stage in which the original stellar radius, ejecta mass, ambient pressure and cooling have ceased to supply important scales. In particular, the swept-up mass must greatly exceed the ejecta mass; $R\gg R_{\ast }$ alone need not ensure this in a very dilute interstellar medium. For fixed explosion energy $E$, uniform ambient mass density ${\rho }_0$ and elapsed time $t$, dimensional analysis gives $R=C{E}^a{\rho }_0^b{t}^d$. Matching mass, length and time gives $a=1/5$, $b=-1/5$, $d=2/5$. Equivalently, a self-similar strong shock wave has $E\sim {\rho }_0{R}^3{\dot{R}}^2$, whose constant-energy solution has the same exponents:The dimensionless coefficient depends on $\gamma$ and the post-shock profiles; dimensional reasoning cannot fix it.

For completeness, obtain the Strong-shock Rankine-Hugoniot conditions in the shock frame. Let $v_1=V-u$ denote downstream speed relative to the outward-moving front. With negligible upstream pressure and enthalpy, conservation of mass, momentum and energy givesPut $q={\rho }_1/{\rho }_0$. Substituting $v_1=V/q$ into the last two equations and selecting the compressive, nontrivial branch yields $q=(\gamma +1)/(\gamma -1)$. HenceHere the symbol $u$, written as $u=2V/(\gamma +1)$ in what follows, is the laboratory-frame radial velocity; it is not the relative speed $v_1$.

Use the thin-shell approximation for a spherical blast wave. Set $W=4\pi R^3/3$ and $M_s={\rho }_{0}W$. Equating the swept-up mass to $4\pi R^{2}d{\rho }_1$ gives $d/R=(\gamma -1)/[3(\gamma +1)]$. The growing shell's momentum is $M_{s}u$. Since newly swept gas has zero laboratory momentum, the net force is the cavity pressure acting on the inner shell, to thin-shell accuracy:The pressure at the shock is already accounted for by the jump conditions and mass loading; subtracting $p_1$ as an additional external pressure would count the shock reaction twice. With $u=2V/(\gamma +1)$, the last equation becomesTherefore, for constant $\alpha$,

The shell volume is $M_s/{\rho }_1=W/q$. Its internal energy and kinetic energy areTreat the cavity volume as $W$ at the same thin-shell level, so its internal energy is $E_{c,\mathrm{int}}=\alpha p_{1}W/(\gamma -1)$. Their sum is proportional to $R^3{V}^2$ for constant $\alpha$ and $\gamma$. Constant total energy thus requires $3+6(\alpha -1)=0$, giving $\alpha =1/2$. ConsequentlyUsing $V=2R/(5t)$, this becomesFor $\gamma =5/3$, $d/R=1/12$ and the approximate energy fractions are $E_{s,\mathrm{kin}}/E=1/4$, $E_{s,\mathrm{int}}/E=1/4$, $E_{c,\mathrm{int}}/E=1/2$.

This is an approximate normalization, not an exact treatment of finite shell thickness. If instead one subtracts the shell volume and uses $W_c=W(1-1/q)$ for the cavity, while retaining the other shell assumptions, the energy coefficient becomes $2(2\gamma -1)/[(\gamma -1)(\gamma +1{)}^2]$ and ${\xi }_0^5=75(\gamma -1)(\gamma +1{)}^2/[32\pi (2\gamma -1)]$. The difference is an order-$d/R$ geometric correction of this crude shell model; a consistent exact coefficient requires the full self-similar profiles.

In the idealized mass budget all swept gas is assigned to the shell, so the cavity inertia is neglected. The hot cavity's residual mass density is not specified, so its kinetic energy cannot be assigned a unique numerical fraction from the stated data. If its low mass is $M_c$ and its velocity is of order $u$, then $E_{c,\mathrm{kin}}\sim M_c{u}^2/2$ andFor an illustrative uniform-density cavity with homologous velocity $v(r)=ur/R$, direct radial integration gives $E_{c,\mathrm{kin}}=3M_c{u}^2/10$, so the preceding fraction has an extra factor $3/5$. The cavity can carry substantial internal energy despite negligible kinetic energy because it is hot. Neglect of its inertia is justified in the swept-mass-dominated stage, not merely by its small initial radius.

The calculation describes the adiabatic expansion stage of a supernova remnant, giving an estimate of its age or explosion energy from its radius, expansion speed and ambient mass density. For example $t\simeq 2R/(5V)$, and $E\propto {\rho }_0{R}^3{V}^2$. Such blasts inject energy and momentum into the interstellar medium and redistribute stellar ejecta. The early ejecta-dominated stage, radiatively cooled late shell, nonuniform ambient medium, finite ambient sound speed, magnetic fields, nonspherical geometry and energy transfer to cosmic rays violate different assumptions. Even in the adiabatic spherical stage, the real profiles are not a uniform thin shell and a uniform-pressure cavity, so ${\xi }_0$ here is an estimate.

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.

Use the real magnetic field $\mathcal{B}$ in the coupled equations; its leading periodic approximation will be $\mathrm{Re}(\mathbf{B}{e}^{i\omega t})$. In the conducting incompressible flow, the magnetohydrodynamic momentum equation, resistive induction equation and solenoidal magnetic-field constraint areHere $p$ is pressure, $\mathbf{j}$ is electric current density, and $\eta$ and $\nu$ are magnetic diffusivity and kinematic viscosity. The moving-conductor Ohm law gives $\mathbf{E}=-\mathbf{u}\times \mathcal{B}+\eta \nabla \times \mathcal{B}$.

In the insulating exterior, neglecting displacement current, the magnetic field is curl-free away from the wire and solenoidal. The imposed electric current density suppliesThe exterior electric field also satisfies Faraday's law. At the rigid wall, impose the no-slip boundary condition $\mathbf{u}=0$. At finite conductivity, the interface conditions for electromagnetic fields give continuous normal magnetic field, continuous tangential magnetic field for equal permeability and no prescribed singular surface current, and continuous tangential electric field. The normal electric current density is zero at an insulating wall; it is automatic for the present two-dimensional fields with current only along $z$. The induced fields decay far from the wire and wall, and initial conditions, or selection of the periodic state after transients, complete the specification. In the limiting perfect conductor approximation a surface current can emerge, permitting a tangential-field jump.

The fully coupled solution need not remain monochromatic: a harmonic magnetic field produces both a mean and a twice-frequency Lorentz force density, and the resulting velocity can generate further harmonics. The following parts consistently use the leading magnetic response with fluid motion neglected.

Write $P(x)$ for the outer pressure extrapolated to the wall. The normal balance in the thin boundary layer is $p_y={\overline{F}}_y$, givingThe spatially varying magnetic pressure deficit drives a tangential pressure gradient. For fixed forcing as $\omega \to \infty$, layer inertia is small compared with viscous dissipation; explicitly the ratio is $U{\delta }^2/(\nu b)$. The leading tangential balance isThe outer gradient $P^{\prime }$ contributes only at higher inner order. Matching the leading outer shear gives $u_y\to 0$, and the physical no-slip boundary condition gives $u(x,0)=0$. Two integrations yield the magnetic skin-layer streaming slip:For the present field amplitude,Using incompressibility and $v(x,0)=0$ also givesHence $v/u=O(\delta /b)$ in the layer, except at symmetry zeros where a componentwise ratio is inappropriate. In the matching region $\delta \ll y\ll b$, $u\sim U$ and $v\sim -y{U}^{\prime }$, up to the smaller displacement term $\delta U^{\prime }/2$.

The effective outer boundary conditions are tangential slip $u_{\mathrm{bulk}}(x,0)=U(x)$ and zero leading normal velocity $v_{\mathrm{bulk}}(x,0)=0$. These describe the bulk flow extrapolated through the unresolved magnetic skin layer; the actual wall remains at rest. For $x>0$ the slip is negative, and for $x<0$ it is positive. Fluid converges along the wall toward the wire, turns upward near $x=0$, and returns outward farther above the wall.

The following original streamline sketch uses the small-Reynolds number bulk Stokes flow for this slip. It illustrates that circulation without claiming that the unspecified bulk Reynolds number fixes a unique complete flow. With $X=x/b$, $Y=y/b$, $Z=X+iY$, a dimensionless stream function isIts factor in brackets is holomorphic in $Y>0$, so $\Psi$ solves the biharmonic equation; its wall derivative is ${\Psi }_Y(X,0)=-X/(1+X^2{)}^3$. Thus it satisfies both effective wall conditions and gives an exact creeping-flow illustration of the derived slip.

Let $K$ be the kinetic energy, $M$ the magnetic energy and $E=K+M$:Dot the magnetohydrodynamic momentum equation with $\mathbf{v}$ and integrate. Incompressibility and the zero boundary velocity remove the advective and pressure work. The viscous term integrates to $-\rho \nu {\int }_D|\nabla \mathbf{v}{|}^2$. ThusThe ideal magnetohydrodynamic induction equation gives, with no boundary contribution,The Lorentz force density therefore transfers energy between the field and the fluid without creating total energy. Since $\nabla \cdot \mathbf{v}=0$ and $\mathbf{v}=0$ on the boundary, another integration by parts gives $\int |\nabla \mathbf{v}{|}^2=\int |{\mathbf{\omega }}_v{|}^2$, where ${\mathbf{\omega }}_v=\nabla \times \mathbf{v}$ is vorticity. The viscous magnetic-relaxation energy identity isThe conserved nonzero magnetic helicity prevents all the field energy from disappearing. Fix a bounded inverse-curl convention for the magnetic vector potential, so that $\Vert \mathbf{A}{\Vert }_2\le C_D\Vert \mathbf{B}{\Vert }_2$. The Cauchy-Schwarz inequality gives the magnetic-energy lower bound from helicity,Consequently $E$ decreases to a finite positive limit andUnder the intended regular long-time interpretation of smooth vorticity, the dissipation cannot keep producing peaks of fixed height and arbitrarily short duration. More precisely, if $d(t)=\Vert {\mathbf{\omega }}_v(t){\Vert }_2^2$ is uniformly continuous, uniformly continuous integrable dissipation tends to zero proves $d(t)\to 0$. The Poincaré inequality and the identity above then imply $\Vert \mathbf{v}(t){\Vert }_2\to 0$.

To identify the asymptotic state, suppose the regular relaxation has a limit, or work on a compact invariant limiting set on which the energy is continuous. On such a set the energy is constant, so the displayed identity forces ${\mathbf{\omega }}_v=0$, and the zero boundary velocity forces $\mathbf{v}=0$. The induction equation is then stationary, and the momentum equation becomesThis is magnetostatic equilibrium, with nonzero field allowed and required here by conserved magnetic helicity. It is not necessarily a force-free magnetic field: a pressure gradient can balance the magnetic force. If the field develops a current sheet, equilibrium is understood with the corresponding interface force balance.

There is a mathematical qualification to the smoothness premise. Smoothness at each finite time alone does not prove uniform long-time bounds, uniform continuity of dissipation or convergence to one limiting field. The energy calculation proves finite total dissipation unconditionally for a smooth solution; the settling conclusion is the regular-relaxation argument just given. A rigorous global convergence theorem requires that additional long-time control. In particular, smooth vorticity does not require smooth electric current density in the limiting state.