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The axial angular momentum per unit volume is $mn(r)r{u}_{\theta }(r)=\hslash n(r)$ for a singly quantized positive quantum vortex. Thus every particle in the prescribed shell contributes $\hslash$:This yieldsThis is $\hslash$ times the shell particle number. For a wall-truncated cloud with $R=R_0<R_r$, the corresponding result is $L_z=\pi H_0\hslash n_0[R^2-a_0^2-(R^4-a_0^4)/(2R_r^2)]$. Reversing the quantized circulation reverses $L_z$ without changing the flow kinetic energy. For winding number $q$, $L_z$ is multiplied by $q$.

In a rotating reference frame at angular frequency $\Omega$, the relevant energy difference is approximately $E_{\mathrm{kin}}-\Omega L_z$. The vortex-free state has zero circulation and zero corresponding angular momentum. For $a_0/R_r\ll 1$, the preceding shell estimates reduce toTherefore the thermodynamic vortex-nucleation frequency for the first positive quantum vortex isThe common factors $H_0,n_0$ cancel explicitly, although the trap radius and core scale still depend on number density. In the strict leading-logarithm limit, ${\Omega }_c\sim 2\hslash \log (R_r/a_0)/(m{R}_r^2)$. The finite constant $-1/2$ is the one obtained from the stated Thomas–Fermi shell integral; a resolved vortex core and readjustment of the number density at fixed particle number can change the nonlogarithmic constant. This is an energetic threshold, not a dynamical guarantee of nucleation: a surface barrier can delay entry. Higher positive winding numbers have flow energy proportional to $q^2$ and angular momentum proportional to $q$, so their analogous threshold is larger in this approximation.

Put $N_a=a^{†}a$ and $N_b=b^{†}b$. The canonical commutation relations give $[N_a,a^{†}]=a^{†}$, $[N_a,a]=-a$, and the analogous relations for $b$, with all cross-mode operators commuting. Consequently $[K_3,K_{+}]=K_{+}$ and $[K_3,K_{-}]=-K_{-}$. Alsoso $[K_{+},K_{-}]=N_a-N_b=2K_3$. This is the Schwinger boson representation of the angular momentum algebra, with $\hslash =1$ in the stated commutators.

The vacuum has $N_a=N_b=0$, and $(a^{†}{)}^2|0⟩$ has occupation numbers $(2,0)$, hence $K_3$ eigenvalue $1$. With $N=N_a+N_b$,Thus $\boxed{K^2(a^{†}{)}^2|0⟩=2(a^{†}{)}^2|0⟩}$. It is the highest-weight state of the spin-one irreducible representation. There are two other states in this multiplet, $a^{†}{b}^{†}|0⟩$ and $(b^{†}{)}^2|0⟩$, with $K_3$ eigenvalues $0,-1$. A normalized basis is $(a^{†}{)}^2|0⟩/\sqrt{2}$, $a^{†}{b}^{†}|0⟩$, $(b^{†}{)}^2|0⟩/\sqrt{2}$. This count assumes the usual two-mode bosonic Fock space; an extra unmentioned vacuum multiplicity could replicate the multiplet.

A central force has the form $\mathbf{F}=f(r)\hat{\mathbf{r}}$, so it always points along the line from a fixed centre to the particle. Its torque about the centre vanishes:Thus angular momentum is conserved. Since $\mathbf{r}\cdot \mathbf{L}=0$, the trajectory lies in the fixed plane through the origin perpendicular to $\mathbf{L}$.

The area swept in time $dt$ is $dA=\frac{1}{2}|\mathbf{r}\times d\mathbf{r}|$, sowhich is constant. This proves Kepler's second law.

Let $q=\sigma /L$, with $L>0$, and let $Z=X^1+i{X}^2$. Direct differentiation gives $\ddot{Z}=Z^{\prime \prime }=-Z/L^2$, while the time coordinate also satisfies the wave equation. Using the Minkowski metric,Thus both Virasoro constraints hold, and the induced worldsheet metric is $g_{\mu \nu }={\cos }^{2}q\mathrm{diag}(-1,1)$. The configuration solves the Nambu–Goto action equations in its interior.

Place the fixed end at $\sigma =0$. The first free endpoint has $Z^{\prime }=e^{it/L}\cos q=0$ at $q=\pi /2$. For the usual single unfolded rotating Nambu–Goto string,This parameter interval need not equal the earlier interval $[0,\pi ]$, since that interval was a coordinate convention. The proper length is measured on a constant-$X^0$ slice; the local motion is perpendicular to the string, so there is no longitudinal Lorentz contraction.

All points have the same phase $t/L$, hence the segment rotates rigidly with angular velocity $\Omega =1/L$. At radius $r=L\sin q$, its speed is $v=r/L$. Therefore the free tip moves at the speed of light. This is consistent with its Neumann boundary condition: $X^{\prime }=0$ and the Virasoro constraints imply ${\dot{X}}^2=0$ at the free tip. The null endpoint is a boundary limit, not a nondegenerate interior point.

In this conformal gauge, $P_m=T{\dot{X}}_m$, so the energy density per $\sigma$ is $-P_0=T$. Equivalently, the energy density per proper length is $T/\sqrt{1-r^2/L^2}$. Thusand the rotational kinetic energy isThe angular momentum about the fixed point isTherefore the Regge trajectory isThe endpoint conditions alone also admit folded extensions with parameter length $(n+\frac{1}{2})\pi L$, $n\ge 0$. Their proper length, energy and angular momentum are $(2n+1)$ times the values above, and $\beta =1/[(2n+1)\pi T]$. The displayed answer uses the implicit unfolded-segment convention.

The magnetic torque transfers angular momentum from the faster rotator to the slower one. When ${\Omega }_m>{\Omega }_K(r_m)$, the neutron star supplies angular momentum to the inner accretion disk and possible propeller flow, so it tends to spin down. When the inner accretion disk rotates faster than the star, the magnetic connection and accreted angular momentum tend to spin it up.

For slowly varying mass accretion rate and magnetic dipole moment, these tendencies favour a magnetic spin equilibrium with the corotation radius near the magnetospheric truncation radius:The precise balance depends on accreted and ejected angular momentum and on the extent of magnetic coupling across the accretion disk; corotation is a scaling estimate, not an exact zero-torque theorem.

Use $\dot{M}>0$ for inward mass accretion rate, and $T_m>0$ for magnetic torque adding angular momentum to the disk. Combining the continuity equation with conservation of angular momentum identifies the inward mass flux as $(dh/dr{)}^{-1}d\mathcal{G}/dr$. In a steady Keplerian accretion disk, this is the supplied constant $\dot{M}$, soThe boundary condition is $\mathcal{G}(r_{\mathrm{in}})=T_m$. Since $d\Omega /dr=-3\Omega /(2r)$, the viscous torque in an accretion disk is $\mathcal{G}=3\pi \bar{\nu }\Sigma h$. Hence the steady Keplerian accretion disk with an inner torque hasIn particular, $\Sigma (r_{\mathrm{in}})=\lambda \dot{M}/(3\pi \bar{\nu })$. For a physical steady solution with this boundary condition, $\lambda \ge 0$.

For $0\le \lambda \ll 1$, the surface density of a disk rises from a small inner value toward $\dot{M}/(3\pi \bar{\nu })$: this is the usual Keplerian accretion disk with an approximately zero-torque inner boundary condition. For $\lambda \gg 1$, the inner surface density of a disk is large and decreases approximately as $r^{-1/2}$, giving a torque-dominated accretion disk. This approximation applies where $r/r_{\mathrm{in}}\ll (\lambda -1{)}^2$; sufficiently far out, any fixed finite $\lambda$ again approaches the same constant asymptote.

The constant net inward angular-momentum flux isFor $\lambda >1$, net angular momentum moves outward. The strong-torque profile is therefore decretion disk-like in its stress-dominated structure, but the specified mass flux remains inward: a genuine net decretion disk requires an outward mass flux, not merely large $\lambda$.

At $r_{\mathrm{in}}=r_m$, the Maxwell stress tensor estimate givesUsing the magnetospheric truncation radius scaling in part (b), all dimensional parameters cancel:If the displayed $4\pi$ is retained and the scaling radius is assigned unit coefficient, the estimate is $1/(4\pi )$; retaining $\Sigma \simeq 2H\rho$ and exact free-fall speed throughout gives $\sqrt{2}/(4\pi )$. Those prefactors are not controlled by the scaling argument. In particular, it does not generate a parametrically large $\lambda \gg 1$.

A principal body frame is an orthonormal basis ${\mathbf{e}}_a(t)$ fixed in the rotating body and aligned with its principal axes. If $R(t)$ is the rotation matrix whose columns are these vectors, differentiating $R^{T}R=I$ shows that $\dot{R}{R}^T$ is skew-symmetric. Every skew-symmetric linear map in three dimensions is cross product with a unique vector $\mathbf{\omega }$, so

For any vector $\mathbf{A}=A_a{\mathbf{e}}_a$,Applying this to the conserved angular momentumgives the Euler equations for a torque-free rigid body:

Multiplying these equations respectively by ${\omega }_i$ and summing proves conservation of kinetic energy; multiplying by $I_i{\omega }_i$ and summing proves conservation of squared angular momentum. Thus the torque-free rigid-body invariants areSolving these two linear equations for the first two squared components givesUsing $I_3{\dot{\omega }}_3=(I_1-I_2){\omega }_1{\omega }_2$ yields the quartic equation for one component of torque-free angular velocity:

For an affine parameter $\lambda$, the geodesic Lagrangian isVarying $S=\int \mathcal{L}d\lambda$ gives the geodesic equations

The rotational Killing vector fields of the spherical symmetry conserve the angular-momentum vector. Its direction is fixed, so every nonradial orbit lies in the plane through the symmetry centre orthogonal to that vector. A spatial rotation can make this the equatorial plane. Equivalently, the initial conditions $\theta =\pi /2$ and $\dot{\theta }=0$ solve the $\theta$ equation uniquely. This is the planarity of geodesics in spherical symmetry; a radial geodesic has zero angular momentum and may be assigned any plane.

On $\theta =\pi /2$, the cyclic coordinates $t$ and $\varphi$ produce geodesic conserved quantities from Killing vectors,For a timelike geodesic parametrized by proper time, the proper-time normalization is $g_{ab}{\dot{x}}^a{\dot{x}}^b=-1$. Substitution of $E$ and $L$ gives the first-order radial equationThis is the timelike geodesic effective potential for static Anti-de Sitter spacetime with curvature radius one.

A geodesic passing through $r=0$ must have $L=0$. If it moves away from the origin, $E>1$, andTaking proper time $s=0$ at departure givesIt turns around at $s=\pi /2$ and first returns to the origin atThis energy-independent refocusing is the characteristic radial timelike geodesic in anti-de Sitter spacetime result; for curvature radius $a$, the answer is $\pi a$.

An isolated collapse emits transient matter and gravitational waves. The exterior is expected to settle to an asymptotically flat stationary spacetime. Because the star is uncharged, the black-hole uniqueness theorem identifies the regular final state with a Kerr black hole. Its higher multipole moments are fixed by its conserved charges: the black-hole no-hair theorem leaves only the total mass $M$ and angular momentum $J$. Thus the late-time spacetime is characterized by $M$ and $J$.

Keep the inner-pair Laplace coefficient unexpanded because $\alpha =a_1/a_2$ is close to one. Put $\ell =L_1/L_2$ andUsing the small-argument Laplace coefficient for the distant planet and $n_2/n_1={\alpha }^{3/2}$ givesThe reverse-coupling expressions follow from the angular momentum identity $L_j{B}_{jk}=L_k{B}_{kj}$. If desired, $L_1/L_2=(M_1/M_2)\sqrt{\alpha }$, $L_1/L_3=(M_1/M_3)\sqrt{\beta }$, and $L_2/L_3=(M_2/M_3)\sqrt{\beta /\alpha }$ express all ratios in the supplied quantities. The diagonal entries are $B_{11}=-(B_{12}+B_{13})$, $B_{22}=-(B_{21}+B_{23})$, and $B_{33}=-(B_{31}+B_{32})$.

For comparable masses and closely spaced inner orbits, $\alpha \simeq 1$, $\ell \sim 1$, and $L_1/L_3,L_2/L_3=O(\sqrt{\beta })$. ThusThe distant couplings $B_{13},B_{23}$ scale as ${\beta }^3$ relative to a fixed inner architecture, and the reverse couplings have an additional $\sqrt{\beta }$ suppression.

Initially the two inner orbital normals coincide, so their combined angular momentum is $L_{12}=L_1+L_2$ along one direction. Use the outer orbital normal as the vertical axis. The inner pair's vector then has perpendicular component $L_{12}\sin \Delta I$ and parallel component $L_{12}\cos \Delta I$. Adding the outer angular momentum givesThe invariable plane is perpendicular to this total vector. Thus the outer orbit's orbital inclination to that plane equals the angle between its own normal and the total normal. The inclination to the invariable plane of two orbital planes isThis uses the small, prograde inclinations for which the denominator is positive; in general use the corresponding two-argument arctangent to choose the quadrant. If $A\gg 1$, $I_3\simeq \Delta I/(A+1)$ for small $\Delta I$: the distant planet carries most of the angular momentum and lies close to the invariable plane.