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Inside the innermost stable circular orbit, nearly circular motion is unstable and gas enters the plunging region of a black-hole accretion disk. Its inflow time becomes shorter than the time on which internal stress can communicate angular momentum back to the disk, motivating the zero-torque inner boundary condition at $r_{\mathrm{ISCO}}$.

In a steady state, the given diffusion equation impliesThe constant mass supply fixes $C=\dot{M}/(6\pi )$, and a second integration givesZero torque means $\bar{\nu }\Sigma =0$ at $r=r_{\mathrm{ISCO}}$, so $C_0=r_{\mathrm{ISCO}}^{1/2}$. Thus the Keplerian accretion disk relation is

Let $R$ be the center of mass and $g$ the uniform gravitational acceleration. The gravitational torque about $R$ is

Choose body-fixed principal axes of inertia. The inertial derivative of the angular momentum isFor torque-free motion $N=0$ and $L_i=I_i{\omega }_i$, giving the Euler equations for a torque-free rigid body

For a symmetric top, let $I_1=I_2=I$ and let $I_3$ be the axial moment. Then ${\omega }_3$ is constant andThus ${\omega }_1^2+{\omega }_2^2$ is constant and the transverse angular-velocity vector rotates uniformly about the body symmetry axis. Its rate is $|\Omega |$; it precesses in the positive axial sense when $(I_3-I){\omega }_3>0$ and in the opposite sense when this product is negative.

Let a change of orthonormal coordinates be represented by a rotation matrix $R$. Since both angular momentum and angular velocity are vectors,Using $L_i=I_{ij}{\omega }_j$ in both frames givesfor every vector $\mathbf{\omega }$. Multiplication by $R^{-1}=R^T$ yieldswhich is exactly the transformation law for a rank-two tensor. Thus $I_{ij}$ is a rank-two tensor.

Put $\mathbf{r}=\mathbf{x}-\mathbf{a}$. The vector triple product givesThereforeAt the centre of mass,Expanding the first formula, the terms linear in $x_i$ vanish because the centre of mass is the origin, while $\rho {\int }_{B}dV=M$. Hence the parallel axis theorem in tensor form is

The particle on a circle has the complete orthonormal basis of energy eigenstates $\{|n⟩:n\in \mathbb{Z}\}$. Its spectral representation givesThe integer $n$ is the quantized angular momentum in units of $\hslash$.

The factor $e^{i{k}_{z}z}$ is an eigenstate of longitudinal momentum with $p_z=\hslash k_z$. The integer $n$ is the azimuthal angular momentum quantum number, since $L_z=-i\hslash {\partial }_{\theta }$ has eigenvalue $n\hslash$.

Fordirect differentiation givesRegularity and cancellation of the $r^{-2}$ term require $\alpha =n$ for $n\ge 0$. Cancellation of the $r^2$ terms requiresassuming $eB>0$. The remaining radial energy is independent of $n$:This is the lowest Landau level. If $eB<0$, the same family uses nonpositive angular-momentum quantum numbers and $|eB|$ replaces $eB$.

A fixed space frame is an inertial orthonormal frame whose axes remain fixed in space. A principal body frame is attached to the rigid body and aligned with the principal axes of its inertia tensor, so that the tensor is diagonal with entries $I_1,I_2,I_3$.

If $e_i(t)$ are the body axes and $\omega$ is the angular velocity, then the derivative of a body-fixed basis vector isIn the principal frame,For torque-free motion, the angular momentum has zero space derivative. ThereforeTaking body-frame components gives the Euler equations for a torque-free rigid body:

For an axisymmetric body, put $I_1=I_2=I$ and let $e_3$ be its symmetry axis. Decompose $\omega ={\omega }_1{e}_1+{\omega }_2{e}_2+{\omega }_3{e}_3$. ThenThus $L$ is a linear combination of $\omega$ and $e_3$, proving that the angular momentum, angular velocity, and symmetry axis are always coplanar. This is the coplanarity in a torque-free axisymmetric rigid body.

A central force conserves angular momentumPut $u=1/r$. Since $\dot{\theta }=\ell u^2/m$,where primes now denote differentiation with respect to $\theta$. Substitution into the radial equation gives the Binet equation

With total mass $M={\sum }_i{m}_i$, defineThe angular momentum about the centre of mass isExpanding, using ${\sum }_i{m}_i(x_i-R)=0$ and $P=M\dot{R}$, gives

At time $t$, the remaining water has angular momentumDuring $dt$, the ejected mass is $-dm$, and its signed tangential speed in the inertial frame is $R\omega +u$. Conservation of angular momentum gives, to first order,After simplification, the angular rocket equation isHere $u$ is signed relative to the direction of rotation; reversing the spray changes its sign.

Write $\Delta \mathbf{u}=(U,W,0)$. The background shear givesThe constant radial and azimuthal components of the dust equation are thereforeWith the Stokes number $\mathrm{St}=\tau {\Omega }_K$, their solution is

The gas is slower than a Keplerian particle, so the particle feels an aerodynamic headwind. Drag force removes its angular momentum, and it drifts radially inward because $U<0$. Its radial speed isDifferentiation with respect to $\mathrm{St}>0$ shows that the unique maximum occurs atwhere $|U|=|V|$.

Chandrasekhar dynamical friction is the drag exerted by the overdense gravitational wake that a massive body creates in a background of lighter particles. It transfers orbital energy and angular momentum to the host, causing satellites, star clusters, and massive black holes to spiral inward and promoting galaxy mergers. In a homogeneous isotropic Maxwellian background its force iswhere $\log \Lambda$ is the Coulomb logarithm in stellar dynamics. Its scaling can be reconstructed from the strong-deflection impact parameter $b_{\mathrm{cr}}\sim GM/v^2$: the encountered mass rate is $\sim \rho v\pi b_{\mathrm{cr}}^2$, and multiplying by momentum change $\sim v$ gives $F_{\mathrm{df}}\sim G^2{M}^2\rho /v^2$.

For a spherical host with a flat galaxy rotation curve,This is a singular isothermal sphere with one-dimensional dispersion $\sigma =v_c/\sqrt{2}$, so $X=1$. DefineFor a constant-mass satellite on a circular orbit, the drag magnitude and its torque areIt follows thatThe quadratic radius dependence and inverse mass dependence explain why massive nearby satellites merge much faster than light or distant ones.

Let the satellite also have a flat internal rotation curve of speed $v_s$. Equating its edge density to the host density gives the tidal radiusBecause $M_s(<R_t)=v_s^2{R}_t/G$, the bound mass decreases linearly:Under the question's literal closure that the derivative of the remaining satellite's total orbital angular momentum equals the frictional torque,and henceIf stripped material is explicitly assigned the satellite's instantaneous specific orbital angular momentum, the balance for the bound remnant is instead $M_{s}d(v_{c}r)/dt=-r{F}_{\mathrm{df}}$; that convention gives $t_{\mathrm{df},\mathrm{strip}}=2t_{\mathrm{df},0}$. Both treatments show the robust point: tidal stripping weakens the drag as the orbit shrinks and substantially delays coalescence. In less idealized profiles the mass can fall faster than linearly, producing dynamical-friction stalling by tidal stripping.

The Killing vectors ${\partial }_t$ and ${\partial }_{\varphi }$ give conserved specific energy and angular momentum,The normalization $g_{\mu \nu }{\dot{x}}^{\mu }{\dot{x}}^{\nu }=-1$ then rearranges toso the displayed radial energy is also constant. This is the effective potential for timelike Schwarzschild geodesics.

For a circular orbit, the derivative ofvanishes. ThusA real finite angular momentum therefore requires

Let $q=(q_1,q_2,q_3)$ and $n=(1,1,1)$. Sincethe Lagrangian depends on $q$ only through $|q{|}^2$ and $(q\cdot n{)}^2$, and on the velocity through $|\dot{q}{|}^2$. Rotations about the $n$-axis preserve all three quantities, proving the invariance.

The axially symmetric isotropic-mass Lagrangian therefore conserves the axial angular momentumup to an irrelevant normalization of $n$. Since $L$ has no explicit time dependence, conservation of energy from time-translation invariance gives the second conserved quantity

The coupled basis consists of the tripletand the singletThese equations list all nonzero Clebsch--Gordan coefficients, each coefficient being $1$, $1/\sqrt{2}$, or $-1/\sqrt{2}$ as displayed, andBecause each irreducible total-spin representation occurs once, an operator commuting with total angular momentum isIn the product basis, the nonzero matrix elements areand

Because $H$ is independent of $\varphi$, Hamilton's equation gives ${\dot{p}}_{\varphi }=0$. PutUsingone obtains $d{L}^2/dt=0$. Physically, $p_{\varphi }=L_z$ is the component of angular momentum along the polar axis, while $L^2$ is the squared magnitude of the angular momentum.

The orbital plane has normal parallel to the conserved angular momentum. If its inclination to the equatorial plane is $\psi$, then $L_z=L\cos \psi$. ThereforesoIt vanishes when $\sin \theta =|\cos \psi |$, equivalently at the greatest northern and southern latitudes reached by the inclined orbit. These are the turning points of the polar motion.

The viscous time isWith zero stellar torque, total angular momentum is conserved and $J\sim M(G{M}_{\star }R{)}^{1/2}$, so $M\propto R^{-1/2}$. HenceandAccreting material loses specific angular momentum; conservation requires a diminishing fraction of the disk to spread outward and carry that angular momentum.