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The arrows point in the positive $x$ direction, with largest speed $v_0$ on $y=0$ and rapidly decreasing length away from that line. Since only $v_x$ is nonzero,The vorticity points towards positive $z$ above the centre line and negative $z$ below it. A small fluid element therefore has local angular velocity $(\nabla \times v)/2$: anticlockwise above and clockwise below when viewed from positive $z$. At $y=0$, its speed is maximal but its local vorticity is zero. Translation speed and local rotation are different properties.

Let the angular velocity vector be the spatially constant $\mathbf{\omega }$. Choose an origin on the fixed rotation axis. The velocity field of a rigid body is $v=\mathbf{\omega }\times r$. In components,Taking the curl givesAn added spatially uniform translational velocity has zero curl, so it does not affect this relation. Equivalently, the vorticity of rigid rotation is twice the angular velocity; the local rotational part of a general smooth flow has angular velocity $\frac{1}{2}\nabla \times v$.

In cylindrical coordinates, $F=r^{1-2n}\hat{\theta }$. The tangential coefficient is positive for every real $n$ and $r>0$, so the large-scale circulation is always counterclockwise when viewed from positive $z$.

By solution, the infinitesimal local angular velocity is $(1-n)r^{-2n}\hat{z}$. Its sign gives“Zero local spin” does not mean zero velocity: at $n=1$ a marked small element is still carried around the axis while its infinitesimal rotational component is zero. Orbital motion around the axis and local vorticity are different measurements.

The equation of motion in a rotating frame with constant angular velocity $\mathbf{\omega }$ isWith $\mathbf{F}=-4m{\omega }^2\mathbf{x}$ and $\mathbf{\omega }=(0,0,\omega )$, this becomesThe given initial data make $z=0$. For the complex coordinate $\zeta =x+iy$, the planar equations combine intoIts characteristic roots are $i\omega$ and $-3i\omega$. Applying $\zeta (0)=1$, $\dot{\zeta }(0)=0$ gives $\zeta =\frac{3}{4}{e}^{i\omega t}+\frac{1}{4}{e}^{-3i\omega t}$. The triple-angle identities reduce this astroid motion of a harmonic oscillator in a rotating frame toIt traces an astroid. Differentiating gives the relative speedFor $\omega >0$, its maxima occur at $t=(2n+1)\pi /(4\omega )$, withThis is speed measured in the specified rotating frame. The same physical motion has inertial complex coordinate $e^{i\omega t}\zeta =\cos (2\omega t)+(i/2)\sin (2\omega t)$, an ellipse; its initial velocity is the frame-rotation contribution even though the initial relative velocity is zero.

Work with the complex representation and take its real part to obtain the physical magnetic field. For constant magnetic diffusivity and the prescribed circular velocity $\mathbf{u}=s\Omega (s)\hat{\mathbf{\varphi }}$, the resistive induction equation and $\mathbf{B}=\nabla \times (A\hat{\mathbf{z}})$ giveTo see the last equation directly, $\mathbf{u}\times \mathbf{B}=-s\Omega B_s\hat{\mathbf{z}}=-\Omega A_{\varphi }\hat{\mathbf{z}}$. Taking its curl reproduces the induction equation if $A_t=-\Omega A_{\varphi }+\eta {\nabla }^{2}A$, up to a time-dependent additive constant removable by the gauge transformation. This is the scalar form of advection and magnetic diffusion in cylindrical coordinates.

Put $\psi =\varphi -\Omega (s)t$. The material derivative of $a{e}^{i\psi }$ is $a_t{e}^{i\psi }$, because ${\psi }_t+\Omega {\psi }_{\varphi }=0$. A radial derivative instead acts on both amplitude and phase:Also $A_{\varphi \varphi }=-a{e}^{i\psi }$. Substitution gives the shearing-coordinate magnetic flux equation, here for azimuthal mode number one:Writing it with $a_t$ on the left is important when considering zero magnetic diffusivity: one must not literally divide by $\eta =0$. In the ideal limit, $a_t=0$, so $A(s,\varphi ,t)=A(s,\varphi -\Omega (s)t,0)$. This is magnetic flux freezing: the scalar flux pattern is carried with each rotating fluid ring. It does not make the magnetic field stationary. Indeed,so differential rotation winds the field and can make its azimuthal component grow linearly in time. Rigid rotation, for which ${\Omega }^{\prime }=0$, only rotates the pattern.

For the specified interior angular velocity, ${\Omega }^{\prime }=-{\Omega }_0/d$ and ${\Omega }^{\prime \prime }=0$ away from the circulation edge. Let $T={\Omega }_{0}t$, $x=s/d$ and $\mathrm{Rm}={\Omega }_0{d}^2/\eta$. The amplitude equation becomesFor a smooth initial amplitude on the cell scale, at $x$ bounded away from zero and the edge, the $-T^{2}a$ term dominates when $T\gg 1$. Its accumulated damping is order one at $T=O({\mathrm{Rm}}^{1/3})$, whereas the integrated terms proportional to $T$ are only $O({\mathrm{Rm}}^{-1/3})$ and the terms independent of $T$ are $O({\mathrm{Rm}}^{-2/3})$. The resulting cubic-time resistive damping in a differentially rotating cell isThe magnetic field has this leading exponential decay envelope. Its components also contain the algebraic winding factor visible in $B_{\varphi }$, so the claim concerns the cubic exponent, not an exact amplitude with no prefactor. Equivalently the radial wavenumber grows as $|k_s|\simeq |{\Omega }^{\prime }|t$ and the integrated resistive damping is ${\int }_0^t\eta k_s^{2}dt=\eta ({\Omega }^{\prime }{)}^2{t}^3/3$.

This is phase mixing in magnetic flux expulsion. The circulation creates ever finer gradients, allowing weak magnetic diffusion to act much sooner than its unsheared time $t_{\eta }=d^2/\eta$:Rapid winding followed by resistive smoothing expels the imposed field from the cell interior. The maintained exterior magnetic field is redistributed into a circulation-edge boundary layer. The result is an interior high-magnetic Reynolds number transient, not a uniform assertion at the axis and the nonsmooth edge, nor the exact $t\to \infty$ solution with a continuously imposed far field. The diffusion length at ${\tau }_c$ is $\sqrt{\eta {\tau }_c}/d=3^{1/6}{\mathrm{Rm}}^{-1/3}$, identifying where a local bulk approximation can fail. The linear profile also needs smoothing at a regular rotation axis and at $s=d$ if used as a fully smooth global flow.

In photospheric magnetoconvection, circulating and diverging convection transports weak magnetic flux towards converging downflow regions. Solar granulation concentrates field in intergranular lanes, while larger-scale supergranulation helps form the solar magnetic network. Flux expulsion explains why a highly conducting photosphere can have relatively weak field in cell interiors and intermittent concentrations at their edges: high electrical conductivity delays diffusion on the original scale, but does not prevent diffusion on the fine scales created by advection. A long-lived circulation must last long enough for the expulsion time; rapidly changing cells need not reach this idealized state. Once the field becomes strong, the Lorentz force modifies the convection, so this prescribed-flow calculation alone neither determines the final field strength nor describes strong-field sunspots.

The wind integrals require a steady axisymmetric magnetohydrodynamic wind in ideal magnetohydrodynamics, with no viscous azimuthal torque or imposed toroidal loop voltage. Axisymmetry by itself is insufficient. Assume smooth connected poloidal flux surfaces with nonzero poloidal magnetic field; functions of the flux label below are understood on each such branch. In cylindrical coordinates, the given poloidal magnetic flux function obeysFor a steady ideal wind, the electric field is $\mathbf{E}=-\mathbf{u}\times \mathbf{B}$ and $\nabla \times \mathbf{E}=0$. Its toroidal component satisfies ${\partial }_z{E}_{\varphi }=0$ and ${\partial }_s(s{E}_{\varphi })=0$, so $E_{\varphi }=C/s$. Regularity at the axis, or equivalently the absence of an imposed loop voltage in an annular domain, gives $C=0$. Thus $({\mathbf{u}}_p\times {\mathbf{B}}_p{)}_{\varphi }=0$ and ${\mathbf{u}}_p=f{\mathbf{B}}_p$. The continuity equation and $\nabla \cdot {\mathbf{B}}_p=0$ now yieldConsequently $\kappa$ is the magnetohydrodynamic mass loading. Including the toroidal magnetic field, rewrite the full velocity asSince $\mathbf{u}\times \mathbf{B}=\omega \nabla \chi$, steady ideal induction gives $0=\nabla \omega \times \nabla \chi$. Hence $\omega =\omega (\chi )$, establishingThe invariant $\omega$ is the field-line angular velocity of an axisymmetric wind, not generally the fluid's angular velocity $\Omega$. When an ideal magnetic surface is anchored to a rigidly rotating star, $\omega$ equals the stellar angular velocity. With no poloidal outflow this reduces to Ferraro's law of isorotation.

To calculate the magnetic torque, use the azimuthal momentum equation. Axisymmetric pressure and Newtonian gravity have no azimuthal components; the magnetic pressure has none either. The azimuthal acceleration and magnetic tension contain the curvature terms characteristic of cylindrical coordinates:Multiplication by $s$ gives the Maxwell torque conservation in an axisymmetric wind equationUsing $u_{\varphi }=s\Omega$, $\rho {\mathbf{u}}_p=\kappa {\mathbf{B}}_p$ and ${\mathbf{B}}_p\cdot \nabla \kappa =0$ provesThis is the total angular momentum transport per unit poloidal magnetic flux. The first term is the material contribution and the second is the contribution of the Maxwell stress tensor. The specific magnetohydrodynamic angular-momentum invariant is $L=\ell /\kappa$ when $\kappa \ne 0$; its normalization differs from the paper's $\ell$.

Define the poloidal Alfvén Mach number bySubstituting $\Omega =\omega +\kappa B_{\varphi }/(\rho s)$ into $\ell$ givesAt a smooth crossing of the Alfvén surface, $M_A^2=1$ and the toroidal magnetic field is finite. The Alfvén-surface regularity condition for an axisymmetric wind is thereforeOne can also expose the apparent singularity explicitly:Finite passage through the Alfvén surface requires the numerator to vanish with the denominator. Equality of the numerator zeros is necessary; a globally smooth wind must also satisfy its other dynamical critical conditions. Here $s_A$ is the cylindrical distance from the rotation axis, not automatically the spherical Alfvén radius.

For a tube carrying outward mass flux $d\dot{M}=\kappa {\mathbf{B}}_p\cdot d\mathbf{S}$, the outward angular momentum flux isThus the stellar torque is ${\dot{J}}_{\star }=-\int \omega s_A^{2}d\dot{M}$, with the integral over the escaping wind. Magnetic stresses give each mass element an effective angular-momentum lever arm equal to the cylindrical Alfvén crossing radius. In a strongly sub-Alfvénic region the fluid approximately corotates with the magnetic surface; beyond that region it can carry the accumulated angular momentum away. For an outward field and positive $\omega$, a super-Alfvénic expanding tube with $s>s_A$ has $B_{\varphi }<0$, so the magnetic stress transports positive angular momentum outwards. If $s_A$ greatly exceeds the stellar surface lever arm, a modest mass-loss rate can cause substantial wind-driven magnetic braking of a solar-type star. An unmagnetized outflow would instead carry approximately the material angular momentum fixed at its launch radius.

Let $\mathcal{G}$ be the outward viscous torque in an accretion disk. For a declining angular velocity, the vertically integrated viscous shear stress givesIn a steady accretion disk without radiated angular momentum, the net inward angular-momentum flux $Fh-\mathcal{G}$ is constant and equals the swallowed flux $F{h}_0$. Thus conservation of angular momentum gives $F(h-h_0)=\mathcal{G}$. Substituting $F=2\pi r\Sigma u$ cancels the surface density and provesThis uses the stipulated phenomenological viscous shear law; it does not replace it by a different fully relativistic stress prescription.

For the orbital calculation let a dot denote a derivative with respect to proper time. Substituting the geodesic energy and specific angular momentum first integrals into the timelike normalization in Schwarzschild spacetime givesDifferentiate with $h$ and $ϵ$ constant along the geodesic. Where $\dot{r}\ne 0$, division by $2\dot{r}$ givesThis equation also holds at turning points and on circular orbits. To justify that without dividing by zero, put $A=1-2m/r$ in the equatorial geodesic Lagrangian $L=(A{c}^2{\dot{t}}^2-A^{-1}{\dot{r}}^2-r^2{\dot{\varphi }}^2)/2$. Its radial Euler-Lagrange equation is $\ddot{r}=(A^{\prime }/2A){\dot{r}}^2-A{A}^{\prime }{c}^2{\dot{t}}^2/2+Ar{\dot{\varphi }}^2$. Substitution of the timelike normalization cancels the ${\dot{r}}^2$ terms and yields $\ddot{r}=-A^{\prime }{c}^2/2+(A/r-A^{\prime }/2)h^2/r^2$, exactly the displayed radial equation, including when $\dot{r}=0$.

A circular orbit requires $\ddot{r}=\dot{r}=0$. Thereforechoosing the positive sense of rotation. Differentiating the square is simpler than differentiating $h$:It is negative for $3m<r<6m$ and positive for $r>6m$, so the global minimum over timelike circular orbits isAs a stability check, the second radial derivative of $W=(1-2m/r)(c^2+h^2/r^2)$ at a circular orbit, holding $h$ fixed for the perturbation, is $W^{\prime \prime }=2m{c}^2(r-6m)/[r^3(r-3m)]$. Thus the exterior branch is stable, and $6m$ is the innermost stable circular orbit. It is the natural matching radius for the idealized zero-torque inner boundary condition.

Choose an explicit orientation convention, since the supplied PDF does not display the referenced diagram. Let ${\mathbf{e}}_z$ point upward and ${\mathbf{e}}_2$ complete a right-handed frame with the horizontal flow direction ${\mathbf{e}}_1$. Usewhere $\mathbf{i}$ points from the head towards the tail. A tail point has position $\mathbf{r}=(a+s)\mathbf{i}$. Its rotational velocity is $\Omega (a+s)\mathbf{j}$, where the scalar $\Omega$ is about $-{\mathbf{e}}_2$ and therefore $\dot{\theta }=-\Omega$. Keeping that sign is essential for stability.

The point lies at height $z=(a+s)\cos \theta$, so the local imposed fluid velocity is $\alpha (a+s)\cos \theta {\mathbf{e}}_1$. Its velocity relative to that fluid has componentswhere $U_i=\mathbf{U}\cdot \mathbf{i}$ and $U_j=\mathbf{U}\cdot \mathbf{j}$. These follow by projection; rigid rotation has no tangential component.

Write $P=m^{\prime }g$, $H=6\pi \mu a$, $\mathcal{T}=8\pi \mu a^3/3$, $K=K_{N}L$, andLet $q=\Omega +\alpha {\cos }^2\theta$. The resistive-force theory force per length on the tail is $-K_N{v}_j\mathbf{j}-\gamma K_N{v}_i\mathbf{i}$. Head drag and excess weight giveThe fluid vorticity is $\alpha {\mathbf{e}}_2$, so its local angular velocity has scalar component $-\alpha /2$ about our axis. Sphere rotation resistance therefore contributes $-\mathcal{T}(\Omega +\alpha /2)$. Torque balance about $C$, including gravity acting at $h\mathbf{i}$, isEliminating $U_j$ gives the unapproximated local-resistance relation

Now retain head and tail translation resistance at the same order, as intended by $\sigma =O(L/a)$, but use $a/L\ll 1$ in the lever-arm integrals. Thus $I_1\simeq L^2/2$, $I_2\simeq L^3/3$. Sphere rotation resistance is smaller than tail rotation resistance by order $(a/L{)}^2$, so omit $\mathcal{T}$ at fixed scaled shear. Also neglect $h$ relative to the tail's hydrodynamic force-centre distance $K_N{I}_1/(H+K)=O(L)$. Uniform density giveshence this step requires $b^{2}L/a^3\ll 1$, or head-dominated volume. It is consistent with the distinguished logarithmic ordering, but $b\ll a\ll L$ by itself would not ensure it. The model further neglects hydrodynamic interaction between head and tail, and uses low-Reynolds-number local drag.

Under these approximations,soBecause $\dot{\theta }=-\Omega$, its equilibria satisfy $\alpha (1-{\sin }^2\theta )=2\beta \sin \theta$. Definewith $s_{\ast }=0$ at zero shear. The two orientations modulo $2\pi$ are ${\theta }_s=\arcsin s_{\ast }$ and ${\theta }_u=\pi -{\theta }_s$. The derivative of $\dot{\theta }=\alpha {\cos }^2\theta -2\beta \sin \theta$ at an equilibrium isThe second factor is positive, so the branch $\cos {\theta }_s>0$ is stable and the other is unstable. On the stable branch the tail tip is above $C$: the head is lower than the far tail end.

The corresponding leading translation velocities areThe second formula uses $q=2\beta \sin \theta$ in the normal force balance. In the equilibrium orientation $\alpha {\cos }^2\theta =2\beta \sin \theta$, and $U_z=U_i\cos \theta +U_j\sin \theta$ simplifies toThis is shear-dependent settling of a sphere-and-tail body. Since $0<\gamma <1$ for a slender filament, $s_{\ast }^2$ increases with the magnitude of shear and the signed upward-coordinate velocity $U_z$ increases towards zero. Equivalently, increasing shear decreases the downward sedimentation speed. The formula remains negative throughout this leading model. If $\gamma =1$ there is no shear dependence; reversing the drag anisotropy would reverse the trend. At very large shear, the omitted sphere rotation term proportional to $\mathcal{T}\alpha$ can become important, so the approximation is not uniform in arbitrarily large $\alpha$.

The usual elastic secondary flow around a rotating sphere is inward towards the equator, then poleward beside the sphere and outward near both poles. It is superposed on the primary azimuthal rotating sphere in Stokes flow. Tensile normal stresses along curved azimuthal streamlines act as hoop stresses, pulling liquid towards the rotation axis. Incompressibility supplies the axial return flow. Unlike the primary rotation, this meridional circulation does not reverse when the rotation is reversed.

The direction assumes the usual polymeric sign $c={\psi }_1+2{\psi }_2>0$ and negligible inertia. A general second-order fluid has circulation proportional to $c$: a negative value reverses the arrows and $c=0$ eliminates the leading elastic circulation. Inertial secondary circulation is a different contribution, so slow rotation alone does not determine which contribution dominates. The sketch shows the elastic contribution for $c>0$:

For a quantitative check of the sketch, take sphere radius $a$, angular velocity $\Omega$ and $K=\Omega a^3$. The primary azimuthal velocity is $v_{\varphi }=K\sin \theta /r^2$, with $\theta$ measured from the rotation axis. Its only nonzero strain components in the spherical orthonormal basis are $E_{r\varphi }=E_{\varphi r}=-3K\sin \theta /(2r^3)$. The leading elastic secondary circulation around a rotating sphere has the stream functionwith $v_r=H(3{\cos }^2\theta -1)/r^2$ and $v_{\theta }=-H^{\prime }\sin \theta \cos \theta /r$. The bracket is $(1-a/r{)}^2(1+2a/r)>0$ outside the sphere. Thus $v_r<0$ at the equator and $v_r>0$ near the poles when $c>0$, and both components vanish on the sphere. The diagram uses these velocities, not independently chosen arrows.

Two other manifestations of the same elastic normal-stress differences are rod climbing, in which inward hoop stresses raise a free surface around a rotating rod, and die swell, in which relaxation of the axial stress and stored polymer deformation allows an extrudate to expand sideways after leaving the die. These are not explained by a scalar shear viscosity alone. A Newtonian extrudate can also swell through velocity-profile rearrangement, so the elastic contribution should not be identified with every instance of swelling.

The second-order fluid is a slow, weak-flow expansion of the objective constitutive history of an isotropic simple fluid. Material frame indifference requires objective tensor rates. For an incompressible flow, the leading isotropic stress is arbitrary pressure, the linear deviatoric term is $2\mu E$, and at the next order the independent contributions can be represented by $E^2$ and an objective rate of $E$. Isotropic scalar contributions are absorbed into the pressure. Their coefficients can be expressed through the two low-rate normal-stress differences, givingHere $\stackrel{△}{E}$ is the lower-convected derivative, not a division by $E$. This local expansion is appropriate when deformation and temporal change over the relevant viscoelastic relaxation time are small: both the Weissenberg number and the appropriate Deborah number must be small. It captures leading normal stresses in slow flows, but not arbitrary finite-strain history, rapid transients or strongly nonlinear shear thinning. In particular its restrictions are stronger than small-amplitude linear oscillatory response at arbitrary frequency.

For steady simple shear flow $\mathbf{v}=(\dot{\gamma }y,0,0)$, use component-first $L_{ij}={\partial }_j{v}_i$, so the paper's $\nabla v$ is $L^T$. The relevant tensors areSubstitution givesThus the first and second normal-stress differences are $N_1={\psi }_1{\dot{\gamma }}^2$ and $N_2={\psi }_2{\dot{\gamma }}^2$, and the shear stress is $\mu \dot{\gamma }$.

For the velocity-preservation results, the definitions imply $\stackrel{△}{E}=\stackrel{\circ }{E}+2E^2$. Consequently the non-Newtonian stress can be writtenwhere $\stackrel{\circ }{E}$ is the Jaumann derivative. Let ${\mathbf{v}}_N$ solve the steady incompressible Stokes flow problem with the prescribed boundary velocities. The supplied identity says that $\nabla \cdot {\stackrel{\circ }{E}}_N$ is curl-free. If ${\psi }_1+2{\psi }_2=0$, then $\nabla \cdot S_N=\nabla \Phi$ for a scalar $\Phi$ on the ordinary simply connected flow domain. The momentum equation at ${\mathbf{v}}_N$ isHence the Newtonian velocity itself is a solution of the second-order constitutive model, with altered pressure $p=p_N+\Phi$. For nontrivial topology one also needs the curl-free force to admit a single-valued pressure; this is implicit in the usual boundary-value problem.

On the regular weakly non-Newtonian branch, Uniqueness of Stokes flow fixes the velocity correction. At first order it solves homogeneous Stokes flow, with zero boundary velocity, after the gradient force has been absorbed into pressure. Multiplying its equation by the correction and integrating gives $2\mu \int E_1:E_{1}dV=0$, hence the correction vanishes. Higher perturbative corrections vanish in turn because the previous velocities remain ${\mathbf{v}}_N$. The velocity is $\mathbf{v}={\mathbf{v}}_N$ in this second-order-fluid solution; only the pressure and stresses change. This is Newtonian velocity preservation in a second-order fluid. It does not assert global uniqueness of distant nonlinear solution branches or exact equality for a real fluid whose higher-order constitutive terms have been omitted.

For a planar incompressible velocity $(u(x,y),v(x,y),0)$, the rate-of-strain tensor has the formAll quantities are independent of $z$, so $\nabla \cdot E_N^2=\nabla (a^2+b^2)$. The extra quadratic term is therefore also a pressure gradient. Together with the supplied identity for the Jaumann derivative, the entire non-Newtonian force is a gradient for arbitrary ${\psi }_1,{\psi }_2$. The same boundary-value and weak-branch argument proves $\mathbf{v}={\mathbf{v}}_N$ for every such planar second-order-fluid flow with prescribed boundary velocities, without requiring ${\psi }_1+2{\psi }_2=0$. Specifying stresses or a free surface instead of velocities would not give the same conclusion automatically, because the extra stresses affect those boundary conditions.

A symplectic manifold is a smooth even-dimensional manifold $M$ with a differential two-form $\omega$ that is closed, $d\omega =0$, and nondegenerate at every point. Nondegeneracy identifies vector fields with differential one-forms by the interior product $X\mapsto {\iota }_X\omega$. In this solution use the explicit Hamiltonian vector field conventionThis contraction sign differs from the negative-sign convention sometimes used for Hamiltonian vector fields; all subsequent signs are fixed by the displayed choice. In Darboux coordinates with $\omega ={\sum }_{a}d{x}_a\wedge d{y}_a$ it givesThe Poisson bracket is bilinear and antisymmetric, and the ordinary product rule gives $\{f,gh\}=\{f,g\}h+g\{f,h\}$. Thus it remains to check the Jacobi identity for the Poisson bracket.

The Cartan formula for the Lie derivative and closedness of the symplectic form give ${\mathcal{L}}_{X_f}\omega =d({\iota }_{X_f}\omega )+{\iota }_{X_f}d\omega =d^{2}f=0$. Henceand nondegeneracy givesApplying both sides to an arbitrary smooth function $h$ gives $\{f,\{g,h\}\}-\{g,\{f,h\}\}=\{\{f,g\},h\}$, which is exactly the Jacobi identity for the Poisson bracket. Consequently the smooth functions, with their pointwise product and this Poisson bracket, form a Poisson algebra. Constants have zero Hamiltonian vector field. If $f,g$ are components of moment maps, the same displayed relation relates the brackets of their Hamiltonian vector fields to the Poisson bracket of those components. In the opposite contraction convention the corresponding relation is the Hamiltonian Lie algebra homomorphism with the compatible bracket sign.

For the point vortices, remove the collision diagonals from $({\mathbb{R}}^2{)}^N$, since the logarithmic Hamiltonian function is singular there. Set $s_{ab}=(x_a-x_b{)}^2+(y_a-y_b{)}^2>0$ for $a\ne b$. Differentiating each pair contribution givesThe Hamiltonian vector field therefore gives the equations of motionEach contribution is perpendicular to the displacement from vortex $b$ to vortex $a$ and has magnitude $2/|{\mathbf{r}}_a-{\mathbf{r}}_b|$. This is the velocity induced by a point vortex of positive circulation, with the normalization $\Gamma /(2\pi )=2$. Each vortex is advected by the velocity of the others, without a self-induced singular velocity. Pairwise contributions cancel in the total velocity, so ${\sum }_a{x}_a$ and ${\sum }_a{y}_a$ are conserved. For two vortices the separation is constant and they rotate counterclockwise about their midpoint, as expected for identical positive point vortices.

For $L=\frac{1}{2}{\sum }_a(x_a^2+y_a^2)$, compute its Poisson bracket with $H$ directly:The last sum vanishes by pairing $(a,b)$ with $(b,a)$, since their numerators are opposite and denominators equal. Thus $L$ Poisson commutes with $H$ and is conserved. Equivalently, the Hamiltonian function depends only on pairwise distances and is invariant under simultaneous rotations.

The standard positive generator of the simultaneous circle group $SO(2)$ action isIts interior product with the symplectic form isUse the moment map convention $d⟨\mu ,\xi ⟩=-{\iota }_{Y_{\xi }}\omega$ for the usual positive rotation generator. Identifying $\mathfrak{so}(2{)}^{\ast }$ with $\mathbb{R}$, this proves $L$ is the moment map for the standard rotation action. It is invariant under the action, hence equivariant because circle group $SO(2)$ is abelian. Additive constants are possible; the natural normalization is $L=0$ at the origin of the ambient space. On the collision-free space this polynomial is its restriction. With the solution's positive contraction convention, the rotation generator is $Y=-X_L$. Confusing these two signs would reverse the final augmented Hamiltonian.

Let ${\mathcal{R}}_{\phi }$ denote simultaneous counterclockwise rotation through $\phi$, and introduce the co-rotating coordinates by $\mathbf{r}(t)={\mathcal{R}}_{\Omega t}{\mathbf{r}}^{\prime }(t)$. Both the symplectic form and $H$ are rotation-invariant, so pulling back the inertial Hamiltonian vector field leaves $X_H$ unchanged. Differentiating the rotation gives the co-rotating equationThe co-rotating positions are constant precisely when this Hamiltonian vector field vanishes. By nondegeneracy of the symplectic form, that is equivalent toConversely a collision-free critical point of this augmented Hamiltonian function produces a rigidly rotating solution. This is a rotating point-vortex relative equilibrium.

As a sign and scaling check, scaling every position by $s>0$ changes $H$ by $-N(N-1)\log s$. Therefore ${\sum }_a{\mathbf{r}}_a\cdot {\nabla }_{a}H=-N(N-1)$. At a rotating point-vortex relative equilibrium, ${\nabla }_{a}H+\Omega {\mathbf{r}}_a=0$, so $2\Omega L=N(N-1)$. For $N\ge 2$ the angular velocity is positive in this circulation convention. An equilateral three-vortex configuration of circumradius $\rho$ has $L=3{\rho }^2/2$ and $\Omega =2/{\rho }^2$, agreeing directly with the equations of motion.

Write the gravitational potential with spherical symmetry as $\Phi (r)$ and let $\mathbf{n}$ be the unit vector normal to a circular orbit, oriented in the direction of its angular velocity. Put $\mathbf{\Omega }=\Omega \mathbf{n}$ and denote the specific angular momentum by $\mathbf{h}=h\mathbf{n}$. Circular force balance and the specific orbital energy giveDifferentiating along the family of circular orbits,A change in orbital orientation contributes $hd\mathbf{n}$ to $d\mathbf{h}$, but $\mathbf{n}\cdot d\mathbf{n}=0$. ConsequentlyThis circular-orbit energy gradient accounts for radial and inclination changes together; spherical symmetry makes the orbital energy independent of the plane.

For either variable-mass particle, $\mathbf{H}=M\mathbf{h}$ and $E=M\epsilon$. Thus $d\mathbf{H}=Md\mathbf{h}+\mathbf{h}dM$, soConservation of mass and conservation of angular momentum impose $d{M}_2=-d{M}_1$, $d{\mathbf{H}}_2=-d{\mathbf{H}}_1$. Adding the two particle contributions provesTo interpret the signs, label the inner orbit 1 and the outer orbit 2. For $f(r)=\epsilon -\Omega h$ the first identity gives $f^{\prime }=-h{\Omega }^{\prime }$. The assumed outward decrease of angular velocity therefore makes $f_1<f_2$, so the mass-exchange part lowers the energy when $d{M}_1>0$: mass is transferred inward. In a coplanar, co-rotating pair, ${\Omega }_1>{\Omega }_2$ and an outward angular momentum transfer has $d{\mathbf{H}}_1=-dH\mathbf{n}$ with $dH>0$. Its contribution is $-({\Omega }_1-{\Omega }_2)dH<0$.