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Taking the scalar product of the Lorentz force equation with the velocity givesThe triple product vanishes because the cross product is perpendicular to its first factor. Thus the kinetic energy is constant, even if the magnetic field varies with position: magnetic forces do no work.

In cylindrical coordinates, the radial, angular and vertical equations areFor a nonstationary circular orbit with constant $r$ and $z$, the first equation gives $-m{\dot{\theta }}^2=qB(r)\dot{\theta }$. Hence its signed angular velocity is $\boxed{\dot{\theta }=-qB(r)/m}$. The sign gives the sense of circulation. The zero root describes a stationary transverse position, not a moving circular orbit. A constant vertical velocity can be superposed to produce a helix.

Define the mechanical angular momentum about the axis by $L=m{r}^2\dot{\theta }$. Multiplying the angular equation by $r$ givesFor the inverse-radius magnetic field this becomes $\dot{L}=-q{B}_0\dot{r}$. Integrating from an initial radius $r_0$ provesLet $c=q{B}_0/m$ and $\ell =(L_0+q{B}_0{r}_0)/m$, so that $L/m=\ell -cr$. Since $\dot{z}$ is constant and the total kinetic energy is constant, the transverse speed squaredis also constant. Substituting the angular momentum relation gives the required radial curve:Only the region $W(r)\ge 0$ is accessible. For $\ell \ne 0$, $W(r)\to -\infty$ as $r\to 0$, and its horizontal asymptote is $v_{\perp }^2-c^2$. Its derivative is $W^{\prime }(r)=2\ell (\ell -cr)/r^3$. Thus, when $\ell c>0$, it has a single maximum $W=v_{\perp }^2$ at $r=\ell /c$. The positive roots of $W$ are the turning points obtained from $r=\ell /(c\pm v_{\perp })$.

For nonzero $c$ and $\ell c>0$, write $r_c=|\ell /c|$. When $0<v_{\perp }<|c|$, two turning points occur atand the radius is confined between them. At $v_{\perp }=|c|$, there is only the finite lower turning point $r_c/2$; $W$ tends to zero from above and an outgoing particle reaches arbitrarily large radius. For $v_{\perp }>|c|$, only the lower turning point remains and the outgoing radial velocity approaches $\sqrt{v_{\perp }^2-c^2}$. The sketch shows these three cases and omits inaccessible negative values of ${\dot{r}}^2$ from the physical interpretation.

For completeness, if $\ell c<0$, $W$ rises monotonically from $-\infty$ to its asymptote; admissible motion then requires $v_{\perp }>|c|$ and is radially unbound after any inward branch turns. If $\ell =0$, $W=v_{\perp }^2-c^2$ is constant; equality gives a circular orbit rather than escape. Above threshold the outgoing branch escapes, whereas an incoming branch reaches the singular axis in finite time and cannot be continued through it without specifying a nonsingular core field. These cases explain why a borderline escape assertion needs an initial-direction qualification.

The transverse escape velocity threshold for radial escape in an inverse-radius axial magnetic field isFor a radial launch at $r_0$, $\ell =c{r}_0$, and equality permits marginal escape; for $\ell \ne 0$, larger transverse speed guarantees radial escape in forward time after any initial inward leg turns. The singular incoming $\ell =0$ branch is the exception described above. More general launches at equality can instead give the circular exception $\ell =0$. A vertical velocity adds to the total speed but cannot pay the transverse escape-energy requirement; if escape merely means unbounded $z$, any nonzero $\dot{z}$ already achieves that. If $q{B}_0=0$, transverse motion is force-free and there is no positive escape-speed threshold.

Put $\mathbf{R}=\mathbf{r}-{\mathbf{r}}^{\prime }$. The magnetic field is the curl of the magnetic vector potential. Since $\mathbf{J}({\mathbf{r}}^{\prime })$ does not depend on the differentiated variable $\mathbf{r}$,Taking the curl under the source integral consequently gives the Biot-Savart law,Outside the compact source this differentiation is ordinary differentiation of a smooth kernel. For regular volume currents, the locally integrable $R^{-2}$ derivative kernel gives the same formula at interior points.

For the loop, choose positive current circulating counterclockwise as viewed from the positive $z$-axis. Parametrize its source position and directed line element byAt $(0,0,z)$ the denominator is the constant $(a^2+z^2{)}^{3/2}$, whileThe transverse terms integrate to zero around the loop. The Biot-Savart law therefore yieldsThe magnetic field is axial, with the stated magnitude for $I>0$; reversing the electric current reverses its direction.

In vacuum magnetostatics, the two relevant Maxwell equations areOrient the unit normal $\mathbf{n}$ from the minus side to the plus side. Integrating Gauss's law for magnetism over a pillbox straddling the current sheet and shrinking its thickness gives $\mathbf{n}\cdot ({\mathbf{B}}_{+}-{\mathbf{B}}_{-})=0$.

To obtain the tangential condition, choose an arbitrary tangent unit vector $\mathbf{t}$ and an infinitesimal rectangle with its long sides of length $\ell$ parallel to $\mathbf{t}$ on the two sides of the sheet. Orient its surface normal as $\mathbf{n}\times \mathbf{t}$, so that its plus-side long edge is traversed along $\mathbf{t}$. The line integral in Ampère's law tends to $\ell \mathbf{t}\cdot ({\mathbf{B}}_{+}-{\mathbf{B}}_{-})$. The enclosed electric current tends to $\ell \mathbf{s}\cdot (\mathbf{n}\times \mathbf{t})$. HenceBecause this holds for every tangent direction and the normal jump is zero,Here $\mathbf{s}\cdot \mathbf{n}=0$, as required for a surface current density.

The force on a current sheet from the mean magnetic field isThe mean removes the sheet's equal and opposite local self-fields, leaving the field that acts on its current through the Lorentz force. Using a one-sided limiting magnetic field would incorrectly include the sheet's own force.

Choose the planar magnetic field convention $\mathbf{B}=\nabla \times (A{\mathbf{e}}_z)$, so $B_s=s^{-1}{A}_{\varphi }$ and $B_{\varphi }=-A_s$. The resistive induction equation, with constant magnetic diffusivity and a suitable time-dependent gauge constant removed, reduces to advection and diffusion of this Cartesian magnetic flux function:For the specified differential rotation, its mth azimuthal Fourier mode obeysThe physical field is the real part of the complex mode. Substitute $a=s^{m}g(t)\exp [-im{\Omega }_{0}t-i{s}^{2}f(t)]$. A direct radial differentiation givesOn the left, rigid rotation cancels, leaving $\dot{g}/g-i{s}^2\dot{f}+im\omega s^2$. Equating the constant and quadratic coefficients yieldsThe initial condition fixes $f(0)=0$ and $g(0)=1$, since the constant C was already factored out. Set ${\mu }^2=4i\eta m\omega$. Then the exact quadratic-shear magnetic flux solution follows by differentiation:Indeed $\dot{f}=m\omega {\mathrm{sech}}^2(\mu t)$ and $4i\eta f^2=m\omega {\tanh }^2(\mu t)$. Also $\dot{g}/g=-(m+1)\mu \tanh (\mu t)=-4i\eta (m+1)f$. For an unspecified initial normalization the second formula has a multiplicative constant $C^{\prime }$, but here $C^{\prime }=1$.

For $\eta >0$ and $\omega >0$, choose $\mu =(1+i)\sqrt{2\eta m\omega }$. The asymptotic condition is $\mathrm{Re}(\mu )t\gg 1$: then $f$ tends to the finite constant $m\omega /\mu$ and $g\sim 2^{m+1}\exp [-(m+1)\mu t]$. At any fixed s, differentiating A to obtain the magnetic field multiplies it by factors involving $m/s$ and $sf$, which have no additional exponential time dependence. Therefore its envelope decays asUsing the paper's magnetic Reynolds number definition gives the precise scalingThus the decay time scales as ${\mathrm{Rm}}^{1/2}/{\Omega }_0$ for positive ${\Omega }_0$ and fixed m. The mechanism is magnetic phase mixing under differential rotation: winding produces progressively fine radial structure on which magnetic diffusion acts efficiently. For negative $\omega$, choose the square root with positive real part, replacing $\omega$ by $|\omega |$ in the decay time. If $\omega =0$, the initial harmonic mode simply rotates, with $f=0$, $g=1$; there is no shear-enhanced decay. The printed positive-root and Reynolds-number expressions implicitly use $\omega >0$.

Write the poloidal magnetic flux function and toroidal amplitude explicitly in spherical components:Dotting the magnetostatic equilibrium equation with $\mathbf{B}$ gives $\mathbf{B}\cdot \nabla p=0$. Since all fields are axisymmetric, the poloidal magnetic field is tangent to the contours of $\chi$ in a meridional section. Thus, locally on regular connected flux surfaces, $p=p(\chi )$. This is a local flux-surface statement; disconnected surfaces with the same numerical value of $\chi$ need not have the same pressure without further global assumptions.

Taking the azimuthal component of the Lorentz force givesIts numerator is the meridional Jacobian determinant of $\psi$ and $\chi$. Vanishing therefore implies locally $\psi =\psi (\chi )$ on the same regular surfaces. DefineA direct curl calculation now givesSince ${\mathbf{e}}_{\varphi }\times {\mathbf{B}}_{\mathrm{pol}}=\nabla \chi /(r\sin \theta )$, the remaining force isConsequently the spherical magnetostatic pressure balance equation following from the stated physical force balance isThe PDF's displayed scalar equation has the wrong sign and omits ${\mu }_0$ for its stated pressure p. This is not corrected by setting ${\mu }_0=1$, which leaves the sign discrepancy. For example, $\chi =r^4{\sin }^2\theta$, $\psi =0$, $p=p_0-10\chi /{\mu }_0$ satisfies the stated vector force balance and the corrected scalar equation, because ${\Delta }_{\ast }\chi =10r^2{\sin }^2\theta$; it fails the printed scalar equation. One could obtain the printed form by replacing its pressure variable by $-{\mu }_0{p}_{\mathrm{physical}}$, but that is not the variable in the given force balance.

Now set $\psi =\alpha \chi$, $p=p_0+P\chi$, and $\chi =F(r){\sin }^2\theta$. The angular operator gives $-2F{\sin }^2\theta /r^2$, soFor the trigonometric form in the question, direct substitution yieldsFor $\alpha \ne 0$, this is a solution providedThe same coefficient relation makes the trigonometric pair regular at the origin: $\cos \alpha r-\sin \alpha r/(\alpha r)=-{\alpha }^2{r}^2/3+O(r^4)$. Thus all components of the magnetic field are finite there. With $p_0=0$, this explicitly constructs the requested proportional-pressure family; adding $p_0$ does not change force balance. The coefficient c is the one called C in the final sentence of the PDF.

For a confined linear-pressure spherical magnetic equilibrium, the condition that the entire magnetic field vanish on $r=a$ is precisely $F(a)=F^{\prime }(a)=0$. In components these conditions readThey eliminate both the radial and meridional components; the toroidal component vanishes too because $\psi =\alpha \chi$. For a nontrivial solution with $A\ne 0$, substituting $B=-A/\alpha$ givesThe last equation selects the allowed nonzero values of $\alpha a$. Equivalently, when division is legitimate, $\tan x=3x/(3-x^2)$; the undivided equation also handles vanishing denominators correctly. Since $\chi (a,\theta )=0$, $\boxed{p(a,\theta )=p_0}$ is independent of latitude. In the strictly proportional convention $p=P\chi$, this is zero. The zero-field solution is a trivial possibility; the displayed conditions describe the nontrivial confined family.

A solar-type main sequence star has a radiative interior and an outer convection zone. Rotating convection and differential rotation continually regenerate its magnetic field by dynamo action. The field is therefore not simply an initially stored field undergoing passive magnetic diffusion. In an alpha-Omega dynamo, the Omega effect winds a poloidal magnetic field into a toroidal magnetic field, while the alpha effect associated with helical motions can regenerate poloidal flux. In a solar dynamo, the Babcock-Leighton mechanism provides another poloidal source through the emergence and dispersal of tilted active regions. Shear near the tachocline and transport through the convection zone help organize the cycle; no single dynamo model is needed to claim that rotation and convection are essential ingredients.

Rotation acts on convective motions through the Coriolis force, altering their correlations and helicity. A useful activity measure is the stellar activity Rossby number, ${\mathrm{Ro}}_{\ast }=P_{\mathrm{rot}}/{\tau }_c$, where ${\tau }_c$ is a convective turnover time. At a given stellar structure, faster rotation ordinarily increases rotational influence and produces greater magnetic activity in the unsaturated regime. Young solar-type main sequence stars commonly rotate rapidly and show strong sunspots, flares, and chromospheric and coronal emission, although their initial rotation rates have substantial scatter. Very rapid rotators enter a saturated regime, so activity does not increase without limit as rotation increases. The saturation mechanism is not uniquely fixed by the correlation itself. The observed unsaturated correlation and saturated regime are documented in primary rotation–activity measurements.

The reverse interaction is wind-driven magnetic braking of a solar-type star. An ionized stellar wind is coupled to the star's open magnetic field. Matter carries angular momentum, and the Maxwell stress tensor also transmits a torque. Magnetic coupling provides an effective lever arm comparable to the Alfvén radius, rather than only the stellar radius. Writing ${\dot{M}}_w>0$ for the outward mass-loss rate, a schematic angular-momentum equation iswhere a geometry factor can be absorbed into an effective $R_A$. This lever-arm interpretation follows from the angular-momentum invariant of a magnetized wind; it is the central result of the original magnetized solar-wind calculation. Magnetic braking can be effective despite a small mass-loss rate, because $R_A$ can greatly exceed the stellar radius.

One can derive the familiar rotation-age scaling in a simple unsaturated model. For a radial open field, take $B_r(R_A)\simeq B_{\ast }(R_{\ast }/R_A{)}^2$, ${\dot{M}}_w\simeq 4\pi R_A^2{\rho }_A{v}_w$, and $v_w^2\simeq B_r(R_A{)}^2/({\mu }_0{\rho }_A)$. Eliminating ${\rho }_A$ givesIf the unsaturated large-scale field satisfies $B_{\ast }\propto \Omega$ and the other factors vary slowly, this gives $\dot{J}=-K{\Omega }^3$ with positive K. Approximating I as constant on this part of the main sequence, integration yieldsThus $\boxed{\Omega \propto t^{-1/2},P_{\mathrm{rot}}\propto t^{1/2}}$ in the late-time part of this approximation. This Skumanich rotation law captures the classic observed secular decline of rotation and magnetic activity; its observational basis is the original rotation and chromospheric-emission analysis. It is a scaling model, not an exact trajectory for every star. With a saturated field, the same simplified wind model instead gives a torque approximately proportional to $-\Omega$, so initially rapid rotation can decay more nearly exponentially until the unsaturated regime is reached.

Over the main sequence lifetime, wind loss therefore generally makes the star rotate more slowly, increases its stellar activity Rossby number, and reduces its unsaturated mean activity. Later changes of moment of inertia, wind properties and open-field topology alter the simple power law. Internal transport of angular momentum can also feed the braked envelope from the radiative interior, so surface rotation is not automatically a measure of the entire star's angular momentum. In a star with weak braking, this secular history differs from the ideal cubic-torque model; an age law alone is not a fundamental dynamo equation.

The stellar rotation-activity feedback also acts on shorter time scales. A growing magnetic field exerts a Lorentz force that changes differential rotation and convective motions, extracting kinetic energy and limiting further growth through dynamo quenching. Conversely, varying shear and poloidal regeneration change the magnetic cycle. The present solar-like regime can sustain repeated solar cycles, with polarity reversals and activity modulation superposed on the much slower rotational evolution. Rotation promotes dynamo activity; dynamo-generated fields brake rotation through the wind and react back on the motions that generate them. This coupled feedback explains the broad progression from young, rapidly rotating, strongly active stars toward older, more slowly rotating, less active solar-type stars, while allowing cycles, irregular modulation and different initial histories.

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.

Put $\kappa =qB/m$, retaining its sign. The Lorentz force equations areThus $z=0$ remains invariant when $z(0)=\dot{z}(0)=0$. Also $d|\dot{\mathbf{r}}{|}^2/dt=2\dot{\mathbf{r}}\cdot \ddot{\mathbf{r}}=0$, since a vector is perpendicular to its cross product with the magnetic field. Choose the initial planar velocity along the positive $x$ axis and the initial location at $(0,0)$. Integration givesThis is a circle centred at $(0,-u_0/\kappa )$ with radius $u_0/|\kappa |$. The cyclotron frequency is $|qB|/m$. Equivalently, choosing this centre as the polar origin givesThe rotation is clockwise viewed from positive $z$ when $qB>0$, and reverses when $qB<0$. In the PDF hint, $\dot{r}=\ddot{r}=0$ are radial conditions with this centred origin; they do not require zero particle velocity. If $qB=0$, the limiting motion is a straight line instead.

With linear drag, put $a=\mu /m>0$. The governing equations for damped cyclotron motion becomeTaking the scalar product with the velocity gives $\frac{1}{2}d|\dot{\mathbf{r}}{|}^2/dt=-a|\dot{\mathbf{r}}{|}^2$, henceFor the planar path, let $w=\dot{x}+i\dot{y}$ and $\zeta =x+iy$. Then $\dot{w}=-(a+i\kappa )w$, so for arbitrary initial planar dataLet ${\zeta }_{\infty }={\zeta }_0+w_0/(a+i\kappa )$. The position relative to this limiting point isIts radius shrinks exponentially while its angle rotates uniformly: the path is a logarithmic spiral into ${\zeta }_{\infty }$, not a succession of circles about the original vacuum guiding centre. The weak-drag assumption is $\mu \ll |qB|$, so many revolutions occur before appreciable contraction.

In Cartesian coordinates the magnetic field is $(bx,by,-2bz)$. Its divergence is $b+b-2b=0$, and all components of its curl vanish. Thus it satisfies the static vacuum magnetic equations $\nabla \cdot \mathbf{B}=0$ and $\nabla \times \mathbf{B}=0$, with no electric current in the region.

Fix clockwise orientation as viewed from above, looking down the positive $z$-axis. Its positive area normal is $-{\mathbf{e}}_z$. The loop's external magnetic flux is thereforeWith the opposite normal its flux has the opposite sign. On the circle $\mathbf{B}=ba{\mathbf{e}}_r-2bz{\mathbf{e}}_z$ and clockwise line element is $d\mathbf{\ell }=-a{\mathbf{e}}_{\phi }d\phi$. The Lorentz force element isThe radial terms cancel around the circle, soThis fixes the current/flux/force signs together.

Neglect self-inductance initially, so Faraday's law and Ohm's law give $RI=-K\dot{z}$. Including gravity then givesFor $R>0$ and $K\ne 0$, the terminal solution is $z=z_0-vt$ withThe general velocity relaxes exponentially to $-v$ on time scale $mR/K^2$. In the terminal state $I=Kv/R$, and the Joule-heating rate isThus all the lost gravitational potential energy becomes resistive heat when kinetic energy is constant.

Within the zero-inductance approximation $v\to 0$ as $R\to 0$; finite-speed motion would require an unbounded current. This limit is singular, so it is important to describe the physical ideal-conductor limit separately. With finite self-inductance $\mathcal{L}$, the circuit equation is $\mathcal{L}\dot{I}+RI=-K\dot{z}$. At $R=0$, total linked flux $\mathcal{L}I+Kz$ is conserved and there is no Joule heating. If $I=0$ initially at $z_0$, thenThis lossless limit of magnetic loop braking has undamped oscillations about $z_0-mg\mathcal{L}/K^2$, with magnetic energy storing and returning mechanical energy. It has no generic dissipative terminal descent. Thus the vanishing terminal speed is the formal resistive-model limit, not a justification for dropping inductance in a perfectly conducting moving loop.

Linearizing the ideal magnetohydrodynamic induction equation about a static magnetic field and using $u^{\prime }={\partial }_t\xi$ gives the fluid Lagrangian displacement-generated magnetic field perturbation $b=\nabla \times (\xi \times B)$. The two solenoidal vector field conditions reduce the curl identity toIts divergence vanishes because it is a curl. Directly, the two products of first derivatives in ${\partial }_i{b}_i$ cancel after exchanging their indices, while the remaining terms differentiate $\nabla \cdot \xi$ or $\nabla \cdot B$, both zero. Similarly the integrated linear continuity equation gives ${\rho }^{\prime }=-\nabla \cdot (\rho \xi )$, so $\boxed{{\rho }^{\prime }=-\xi \cdot \nabla \rho }$.

Let $\Pi =p+B^2/(2{\mu }_0)$ be the magnetohydrodynamic total pressure. Magnetostatic equilibrium and the momentum equation in the linearized ideal magnetohydrodynamic equations areThere is no perturbed gravitational potential here because the field is fixed. Multiply by ${\xi }_i^{\ast }$ and integrate. The pressure term vanishes by the assumed boundary cancellation and $\nabla \cdot {\xi }^{\ast }=0$. Thuswhere $\mathcal{M}=-\int {\xi }_i^{\ast }(B_j{\partial }_j{b}_i+b_j{\partial }_j{B}_i)dV$.

Integrate the first magnetic term along $B$, using $\nabla \cdot B=0$, then insert the expression for $b$. This gives the positive term $\int |(B\cdot \nabla )\xi {|}^{2}dV$ and two mixed derivative terms. For clarity, the troublesome term transforms asThe first term on the right cancels the other mixed term. Combining the remaining derivative with the term containing ${\partial }_j{B}_k{\partial }_k{B}_i$ givesDifferentiating the equilibrium relation now combines the gradient of mass density and magnetostatic equilibrium terms:We obtain the incompressible magnetic displacement energy identityThe two Hessian matrices are real symmetric, making their Hermitian form evaluations real. The last term is a squared norm, and $I>0$ for a nonzero displacement in fluid with positive mass density. Thus $\boxed{{\sigma }^2\text{ is real}}$. It may be negative, in which case one temporal branch has exponential growth; real squared frequency is not itself a proof of stability.

For the atmosphere with a horizontal magnetic field, $\Phi =gz$ has zero Hessian matrix and the equilibrium equation isConsequently the interchange stability of an incompressible magnetized atmosphere is governed by the numeratorFor ${\xi }_z=\sin ky$, the magnetic tension term is zero: these fluid Lagrangian displacements interchange neighboring magnetic field lines without varying along them. A mass density increasing with height gives negative energy and instability; decreasing mass density gives positive energy for these vertical interchanges. If ${\rho }_z\le 0$ everywhere, the displayed form is nonnegative for every admitted incompressible displacement. If the gradient changes sign, suitably localized divergence-free interchange trials detect an unstable inverted layer; For example, take ${\xi }_z=f(z)e^{iky}$ and ${\xi }_y=i{f}^{\prime }(z)e^{iky}/k$, with $f$ supported in that layer. Their divergence is zero, and neither component varies along the magnetic field.

For ${\xi }_z=\sin kx$, magnetic field lines are bent. Averaging over a full horizontal wavelength gives the trial Rayleigh quotientunder the stipulated admissible boundary/integrability conditions. It is a trial quotient, not a claim that this displacement is an exact eigenfunction for every profile. mass density decreasing upward is stable here too. For inverted mass density, magnetic tension can stabilize sufficiently short waves along the magnetic field, but does not remove the transverse interchange instability. Locally the two contributions have the familiar scale $-g{\rho }_z/\rho +k^2{v}_A^2$, with the Alfvén speed $v_A=B/\sqrt{{\mu }_0\rho }$.

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 $i{\omega }_A\mathbf{b}$.