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The difference between the angular momentum of circular coplanar orbits at fixed semi-major axes and the system's actual angular momentum along the reference normal. With circular momenta $L_j=m_j\sqrt{G{M}_{\star }{a}_j}$ it is ${\sum }_j{L}_j(1-\sqrt{1-e_j^2}\cos I_j)$. For small planar orbital eccentricities its quadratic part is $\frac{1}{2}{\sum }_j{L}_j|z_j{|}^2$. Its conservation in $\dot{z}=iAz$ requires $L_j{A}_{jk}=L_k{A}_{kj}$.

Let a disk contain fraction $f_d$ of its host mass and let its edge specific angular momentum be $f_j{R}_h{v}_h$. With halo relation $v_h^2=G{M}_h/R_h$ and disk geometry coefficient $\kappa =v_d^2{R}_d/(G{M}_d)$, combining centrifugal balance and the edge angular momentum gives the displayed estimates. The coefficient records the radial mass profile and thickness: enclosed mass does not determine a disc rotation curve. Treating $\kappa =1$ as a monopole estimate must not be mistaken for an exact thin-disk theorem.

The rate at which angular momentum crosses a surface, including advection and stresses. With positive inward mass accretion rate, the net inward flux in a steady accretion disk is $\dot{M}h-\mathcal{G}$.

A Keplerian disk with surface density $\Sigma \sim r^{-p}$ at infinity has finite outer mass for $p>2$ and finite outer angular momentum for $p>5/2$, since the relevant integrals are $\int \Sigma rdr$ and $\int \Sigma r^{3/2}dr$. The borderline $p=5/2$ has logarithmically divergent angular momentum even though its mass converges.

A transformed radial equation containing $L^2/(2q^2)$ becomes a full planar central-force orbit by choosing $d\bar{\theta }/d\tau =L/q^2$. For $d\tau /dt=(q/r)dq/dr$ and old angular velocity $L/r^2$, this gives $d\bar{\theta }/d\theta =(r/q)dq/dr$. Keeping the old angle generally fails to preserve the stated transformed angular momentum.

For initially well-separated Kerr black holes with no orbital angular momentum, exact axisymmetry gives $J_f=J_1+J_2$. Put $X=(A_1+A_2)/(8\pi )$. The black-hole area theorem and the irreducible mass identity implyIndeed, $X\ge |J_1|+|J_2|\ge |J_f|$, and $\frac{1}{2}(Y+J_f^2/Y)$ is nondecreasing for $Y\ge X$. Opposite spin signs lower $J_f^2$ at fixed initial areas and hence raise the allowed radiated-energy bound; this need not be achievable in an actual merger.

An exact axial Killing vector field makes Komar angular momentum independent of the surrounding vacuum surface. Therefore gravitational waves in a configuration with exact axisymmetry carry no net angular momentum about that symmetry axis, although they can carry energy. This is specific to axial angular momentum; it is not an absence of radiation.

During binary mass-ratio reversal, an initially more massive donor star loses enough material that its companion becomes the more massive member. The stars retain their different prior nuclear evolution, resolving the Algol paradox. Conservative binary mass transfer shrinks the orbit before equal masses and widens it after reversal, because $a\propto (M_d{M}_a{)}^{-2}$ at fixed total mass and orbital angular momentum.

For a circular orbit, fixed total mass, transfer from $M_2$ to $M_1$, negligible spin and $R_L\propto a(M_2/M{)}^{1/3}$, the orbital angular momentum $J=M_1{M}_2\sqrt{Ga/M}$ givesIf the donor follows a radius sequence with stellar radius response exponent $\zeta$, has no additional radius change at fixed mass, and maintains contact, equating its radius derivative to this expression proves the displayed contact equation. Negative external $\dot{J}$ drives donor mass loss when $\zeta +5/3-2q>0$. The appropriate equilibrium or adiabatic response and its timescale must be specified separately; the equation alone is not a universal dynamical-stability criterion.

The black-hole no-hair principle says that radiation carries away the higher multipole structure of a collapsing isolated body, leaving a stationary black hole specified by conserved asymptotic charges such as mass, angular momentum, and electric charge.

Coherent prograde gas accretion increases the angular momentum of a black hole. For a zero-torque thin disc, the rest mass $d{m}_0$ entering from the innermost stable circular orbit changes the hole by $dM=E_{\mathrm{ISCO}}d{m}_0$ and $dJ={\ell }_{\mathrm{ISCO}}d{m}_0$. Its dimensionless spin obeys $d{a}_{\ast }/d\log M=c{\ell }_{\mathrm{ISCO}}/(GM{E}_{\mathrm{ISCO}})-2a_{\ast }$. Starting from zero spin and neglecting captured radiation, the formal extremal endpoint has $M_f=\sqrt{6}{M}_i$ and requires rest-mass supply approximately $1.8464M_i$. Photon capture leads instead to the Thorne spin limit.

A central potential depends only on distance from a fixed centre, $V(\mathbf{x})=V(|\mathbf{x}|)$. Its force is radial, and rotational symmetry implies conservation of angular momentum.

For a circular Kepler orbit, neglecting stellar spin, the orbital angular momentum of a binary star is $J=\mu a^2\Omega =\mu \sqrt{GMa}$, where $M$ is total mass, $\mu$ is the reduced mass, and $a$ is separation. Kepler's third law also gives $J=G^{2/3}{P}^{1/3}{M}_1{M}_2/[(2\pi {)}^{1/3}{M}^{1/3}]$. Distances from the center of mass are $a{M}_2/M$ and $a{M}_1/M$; adding their individual orbital angular momenta proves the formula.

A bounded trajectory in the damped Binet equation with nonzero angular momentum asymptotically becomes circular. Energy loss is nonnegative in magnitude and vanishes only for zero radial velocity. The centrifugal effective potential energy excludes collision; bounded radius forces indefinitely increasing forward angular distance, so the reciprocal-radius oscillation decays. Escaping trajectories are excluded by the boundedness hypothesis.

For $V(r)=km{r}^2$, fixed nonzero angular momentum $h$ gives the effective potential $V_{\mathrm{eff}}=h^2/(2m{r}^2)+km{r}^2$. A positive-radius circular orbit exists when $k>0$, at the displayed radius, and $V_{\mathrm{eff}}^{\prime \prime }(r_0)=8km>0$ proves radial stability by the effective potential stability criterion. Its angular speed has magnitude $\sqrt{2k}$ and its small radial oscillation frequency is $\sqrt{8k}$. If $k\le 0$, no such orbit with nonzero angular momentum exists; for $h=0,k>0$, the sole minimum is at the origin, not a positive-radius circular orbit.

For unit mass with specific angular momentum $l$ in $V(r)=-e^{-r}$, the effective potential is $-e^{-r}+l^2/(2r^2)$. A positive-radius circular orbit satisfies the displayed equation. Since $r^3{e}^{-r}$ has its unique maximum $27/e^3$ at $r=3$, exactly two radii occur for $0<l^2<27/e^3$. Their radial stability follows from $V_{\mathrm{eff}}^{\prime \prime }=e^{-r}(3/r-1)$: the inner orbit is stable at fixed angular momentum, the outer unstable. The case $l=0$ has no finite-radius circular orbit despite satisfying the upper bound.

Collisionless matter approaching a black hole generally cannot dissipate enough orbital energy to remain bound, so direct capture requires sufficiently small angular momentum. For nonrelativistic incident speed $v$, a Schwarzschild black hole has direct test-particle capture cross-section approximately $16\pi G^2{M}^2/(c^2{v}^2)$. A Newtonian pericentre estimate using the Schwarzschild radius instead gives $4\pi G^2{M}^2/(c^2{v}^2)$. Both show that capture at a given mass density is suppressed by order $(v/c{)}^2$ relative to supersonic Bondi--Hoyle--Lyttleton accretion. Tidal disruption events provide a distinct way to accrete some stellar debris.

With fixed total mass and orbital angular momentum and $R_L\propto a(M_d/M{)}^{1/3}$, the Roche-lobe radius response exponent is $2q-5/3$. A donor with stellar radius response exponent ${\zeta }_{\ast }$ is stable to infinitesimal mass loss when ${\zeta }_{\ast }>2q-5/3$, giving $q<({\zeta }_{\ast }+5/3)/2$. The threshold depends on the donor response.