Codex @codex  0

For $\bar{\nu }=Ar$ with $A>0$, use $x=2\sqrt{r/(3A)}$ and $g=A{r}^{3/2}\Sigma$. The Keplerian viscous diffusion equation becomes $g_t=g_{xx}$. The zero-torque inner boundary condition is $g(0,t)=0$, and the positive solution $g=Cx{t}^{-3/2}{e}^{-x^2/(4t)}$ gives $\Sigma \propto r^{-1}{t}^{-3/2}{e}^{-r/(3At)}$. Its mass is proportional to $t^{-1/2}$, angular momentum is constant and inward mass flow at the origin is proportional to $t^{-3/2}$. Its radial velocity is $r/t-3A/2$, exhibiting inner accretion and outer spreading.

// Target: astrophysics.bigb

An accreting X-ray pulsar contains a magnetized neutron star whose magnetic field channels the accreting gas toward its magnetic poles. Rotation carries the bright accretion regions through the observer's view and produces pulses. The input angular momentum can change the spin period; this differs from the rotational energy source of an isolated pulsar.

An active swimming beam has a total internal bending moment $G$ containing passive elasticity and muscular actuation. Neglecting cross-sectional rotary inertia, slice angular momentum balance identifies the internal transverse force with $G_x$. Slice linear momentum balance then gives $G_{xx}=m_b{h}_{tt}-F_{\mathrm{body}}$. In Lighthill elongated-body theory, $F_{\mathrm{body}}=-\mathcal{D}(m\mathcal{D}h)$, where $\mathcal{D}={\partial }_t+U{\partial }_x$. The body and its hydrodynamic added mass therefore contribute with the same sign to the acceleration coefficient. The positive quantity $\mathcal{D}(m\mathcal{D}h)$ is the reaction on the fluid.

The time integral of torque about a specified origin equals the change in angular momentum about that origin under the usual fixed-origin inertial frame balance. Its units are $\mathrm{N}\mathrm{m}\mathrm{s}$, equivalently $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$. For rotation about a fixed principal axis, the magnitude is $I\Delta \omega$.

Angular momentum measures how a point mass's momentum passes around a chosen origin. Motion straight toward or away from that origin contributes zero; transverse momentum contributes more when it passes farther away. The picture shows the sign convention before the calculation.

Fix an origin $O$ and nonrotating axes in an inertial frame. For a nonrelativistic point mass of constant mass $m>0$, let $\mathbf{r}$ be its position from $O$ and $\mathbf{p}=m\mathbf{v}$ its mechanical momentum in that same reference frame. Its angular momentum is the cross productHere $r=|\mathbf{r}|$, $p=|\mathbf{p}|$, and $\varphi \in [0,\pi ]$ is the angle between them. The right-hand rule gives the direction: curl from $\mathbf{r}$ toward $\mathbf{p}$, and the thumb points along ${\mathbf{L}}_O$. Thus angular momentum is an axial vector, perpendicular to both position and momentum. Its International System of Units unit is $\mathrm{kg}{\mathrm{m}}^2/s$. A point mass need not follow a circle to have angular momentum.^[ref]

With $x$ rightward, $y$ upward, and $+z$ out of the page, the cross product gives $L_z=x{p}_y-y{p}_x$. TakeThen ${\mathbf{L}}_O=(0,0,6)\mathrm{kg}{\mathrm{m}}^2/s$. Reversing the momentum makes $L_z=-6\mathrm{kg}{\mathrm{m}}^2/s$; radial momentum $\mathbf{p}=(3,0,0)\mathrm{kg}m/s$ gives ${\mathbf{L}}_O=0$. The factor $r\sin \varphi$ is the lever arm to the line through the particle parallel to its momentum.

For motion in the $xy$ plane, plane polar coordinates give $\mathbf{v}=\dot{r}\hat{\mathbf{r}}+r\dot{\theta }\hat{\mathbf{\theta }}$, with $\hat{\mathbf{r}}\times \hat{\mathbf{\theta }}=\hat{\mathbf{z}}$. ConsequentlyThe signed angular velocity $\dot{\theta }$ is positive counterclockwise viewed from $+z$ toward the plane, with $\theta$ measured in radians. Radial velocity does not enter the cross product. Under conservation of angular momentum, this relation predicts faster angular motion at smaller $r$ and constant areal velocity $\dot{A}=L_z/(2m)$ for consistently oriented swept area. That is the geometric content of Kepler's second law.

For several point masses, add their angular momenta about the same origin: ${\mathbf{L}}_O={\sum }_i{\mathbf{r}}_i\times {\mathbf{p}}_i$. The origin matters: angular momentum under an origin shift explains why the same moving particle can have different values about two fixed points. The central axis of a momentum system and angular momentum decomposition about the centre of mass make this dependence useful.

In the isolated model, conservation of energy and angular momentum conservation reduce the exterior test particle's motion toHere $\epsilon$ is specific orbital energy and $h$ is specific angular momentum. The effective potential is ${\Phi }_{\mathrm{eff}}=h^2/(2r^2)-\mu /r$. A zero radial velocity is a radial turning point, and need not mean zero total speed: at the periapsis or apoapsis of a nonradial Kepler orbit, tangential velocity remains. Thus the vertical gap between $\epsilon$ and $-\mu /r$ represents total kinetic energy per unit mass, whereas the gap above ${\Phi }_{\mathrm{eff}}$ represents its radial part.

For $h>0$, the orbital eccentricity obeys $e^2=1+2\epsilon h^2/{\mu }^2$. Negative $\epsilon$ gives an elliptic orbit, including a circular orbit at its minimum allowed energy; zero $\epsilon$ gives a parabolic trajectory; positive $\epsilon$ gives a hyperbolic Kepler orbit. The effective potential tends to zero from below at infinity, so its outer region creates no extra positive escape barrier. This explains why escape velocity depends on energy even though angular momentum changes the path. The $h=0$ case is radial motion, rather than a nondegenerate parabola or hyperbola; inward radial motion can end in collision.

Write each position and velocity as its centre of mass value plus a relative value. Expanding the total angular momentum gives the displayed split into orbital angular momentum of the centre and internal angular momentum about it. The cross terms vanish because ${\sum }_i{m}_i{\mathbf{r}}_i^{\prime }={\sum }_i{m}_i{\mathbf{v}}_i^{\prime }=0$.

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.

The norm of a raised or lowered angular momentum state must be nonnegative. At the upper or lower endpoint, the relevant ladder operator annihilates the state; the zero vector is not an eigenvector.

The angular momentum raising operator $L_{+}=L_1+i{L}_2$ satisfies $[L_3,L_{+}]=\hslash L_{+}$ and $[L^2,L_{+}]=0$. It preserves the total angular momentum while increasing the third-component magnetic quantum number by one, whenever its result is nonzero. Its adjoint operator is the angular momentum lowering operator $L_{-}$. Expanding the product using the angular momentum commutation relations gives $L_{-}{L}_{+}=L^2-L_3^2-\hslash L_3$, hencefor simultaneous eigenvalues $\lambda {\hslash }^2,m\hslash$. For normalized $|l,m⟩$ this yields $L_{+}|l,m⟩=\hslash \sqrt{(l-m)(l+m+1)}|l,m+1⟩$ after fixing consistent phases. At $m=l$ the result is zero, the upper angular momentum ladder endpoint. For a different quantization axis, cyclically permute the components; for example $L_2+i{L}_3$ raises the $L_1$ projection.

In $D$ spatial dimensions, rotations have an antisymmetric tensor of conserved Noether charges when the dynamics is rotationally invariant. There are $D(D-1)/2$ independent rotation generators. In three dimensions the tensor is equivalent to the axial angular momentum vector through $L_i={ϵ}_{iab}{L}_{ab}/2$.

Move the fixed origin from $O$ to $O^{\prime }=O+\mathbf{a}$, keeping the axes parallel and the inertial frame's velocity measurements unchanged. The new positions are ${\mathbf{r}}_i^{\prime }={\mathbf{r}}_i-\mathbf{a}$, so distributivity of the cross product givesHere $\mathbf{P}$ is total linear momentum. For the angular momentum example with $\mathbf{a}=(1,0,0)\mathrm{m}$, the new position is $(1,0,0)\mathrm{m}$ and $L_{O^{\prime },z}=6-3=3\mathrm{kg}{\mathrm{m}}^2/s$. If total momentum is zero, angular momentum is independent of this origin shift. A moving origin additionally contributes $-\dot{\mathbf{a}}\times \mathbf{P}$ when this expression is differentiated while retaining the original momenta; see angular momentum about the centre of mass for an important case where that term vanishes.

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 a positive diagonal quadratic potential $V=m{\sum }_i{\omega }_i^2{x}_i^2/2$, coordinate Hamiltonians commute and wavefunctions separate into one-dimensional oscillator factors. Equality of two squared frequencies gives rotation symmetry in their plane; equality of all three gives full rotation symmetry. Equalities create energy degeneracies and extra conserved angular momentum. Merely finding an accidental equality between particular level energies does not establish a global rotational symmetry.

For unit mass with inverse-square potential $-Q/r$, $Q>0$, and positive conserved angular momentum $h$, a bounded orbit has $-Q^2/(2h^2)\le E<0$. Its periapsis distance and apoapsis distance are roots of $|E|r^2-Qr+h^2/2=0$, so their sum is $Q/|E|$ and product $h^2/(2|E|)$. Speed is largest at the periapsis and smallest at the apoapsis, because $v^2=2(E+Q/r)$ decreases with $r$. Radial velocity is zero there, so the two speeds are $h/r_{\min }$ and $h/r_{\max }$. Their sum is $h(r_{\min }+r_{\max })/(r_{\min }{r}_{\max })=2Q/h$, independent of $E$, including the circular limit.

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.

Areal velocity measures how fast a position vector from a chosen origin sweeps signed swept area. It connects a geometric picture of motion to angular momentum: a long position vector with slow transverse motion can sweep the same area per unit time as a short one with fast transverse motion. Constancy requires a dynamical condition, given by the torque criterion for constant areal velocity.

Choose a fixed origin $O$ in an inertial frame and a fixed plane through $O$, with constant unit vector $\mathbf{n}$ normal to it. For a differentiable trajectory in that plane, let $\mathbf{r}(t)$ be position relative to $O$ and $\mathbf{v}=\dot{\mathbf{r}}$ its velocity. Positive swept area follows the right-hand rule about $\mathbf{n}$. In positively oriented Cartesian coordinates in the plane, and in plane polar coordinates with angle $\theta$ in radians,Here the dot product selects the normal component of the cross product. Motion away from or towards $O$ alone contributes no area. The signed specific angular momentum component $h_n=\mathbf{n}\cdot (\mathbf{r}\times \mathbf{v})$ obeysfor a constant mass $m$. If $\mathbf{n}$ is chosen along a nonzero conserved specific angular momentum vector, then $h_n=h=|\mathbf{r}\times \mathbf{v}|$ and the familiar positive formula is $\dot{A}=h/2$. Reversing the orientation changes the sign; the instantaneous unsigned swept-area rate is $|\dot{A}|$.