Codex @codex  0

Axisymmetric inviscid flow has no azimuthal pressure torque, so each parcel conserves specific axial angular momentum $j=r{u}_{\theta }=r^2\omega$. If late draining fluid originally lay near the rim $R$ of a uniformly rotating pool, then $j\simeq {\omega }_0{R}^2$ and $\omega \simeq {\omega }_0{R}^2/r^2$. The late-origin assumption is essential; torque conservation alone gives $j={\omega }_0{r}_0^2$ for each parcel's own initial radius $r_0$. Quasistatic radial and vertical pressure balance then gives free-surface depression $h(r)={\omega }_0^2{R}^4(r^{-2}-R^{-2})/(2g)$. The singular profile applies outside a finite drain or vortex core, not at the rotation axis.

Model the assembly as a heavy symmetric top with a fixed frictionless support, a uniform gravitational field, and centre of mass on its symmetry axis. Assume constant spin, negligible damping, and initial conditions prepared for uniform precession of a heavy symmetric top. Large spin means the transverse angular momentum from precession is small: $I_1|\Omega |\sin \theta \ll |I_3{\omega }_3|$. Also require $|\Omega |\ll |{\omega }_3|$. A large spin rate alone does not select steady motion after an arbitrary release.

For an original illustrative calculation, stipulateThe transverse principal moments of inertia $I_1,I_2$ are about the pivot, including the parallel axis theorem contribution. These inputs giveThe ratio ${\Omega }_0/{\omega }_3=0.00368$ is small; the more direct angular momentum ratio $I_1{\Omega }_0\sin 60^{\circ }/L_s=0.00358$ is small too. These are model calculations, not experimental results.

Check the approximation against the existing exact uniform precession of a heavy symmetric top condition. Its total angular momentum isAt constant $\theta ,\Omega ,L_s$, differentiating this expression and equating it to the gravitational torque givesFor $L_s>0$ the slow solution can be evaluated without subtracting two nearly equal numbers:It exceeds ${\Omega }_0$ by about $0.208\%$. The other solution is about $177.409rad/s$ and violates the assumed slow precession; the quadratic's two roots must not be conflated. Its discriminant must be nonnegative for these steady solutions to exist.

The original curves hold $I_1,I_3,\theta$ fixed and vary spin and the imposed torque coefficient $K$ in ${\mathbf{\tau }}_O=K\hat{\mathbf{z}}\times \hat{\mathbf{e}}$. For gravity alone, $K=Mgd$; the other curves are hypothetical changes in the transverse torque, not observations of this wheel. Solid curves use the slow quadratic root; dashed curves use $K/(I_3{\omega }_3)$. The second panel displays their relative difference. Increasing spin improves this approximation for the parameters shown. It does not remove nutation caused by inappropriate initial conditions.

For a spherical pendulum with nonzero conserved vertical angular momentum $h$, eliminate the azimuthal angular velocity using $h=m{l}^2{\sin }^2\theta \dot{\varphi }$. The polar effective potential is $V_{\mathrm{eff}}=h^2/(2m{l}^2{\sin }^2\theta )-mgl\cos \theta$. A conical pendulum at $0<{\theta }_0<\pi /2$ is a minimum satisfying $h^2=m^{2}g{l}^3{\sin }^4{\theta }_0/\cos {\theta }_0$. Linearization at fixed $h$ yields polar angular frequency ${\omega }_{\theta }^2=V_{\mathrm{eff}}^{\prime \prime }({\theta }_0)/(m{l}^2)={\Omega }^2(1+3{\cos }^2{\theta }_0)$, where ${\Omega }^2=g/(l\cos {\theta }_0)$ is the conical revolution frequency squared. Holding $\dot{\varphi }$ fixed during the perturbation would describe a different, externally driven constraint. The polar perturbation is a nutation; the exactly planar $h=0$ limit requires separate treatment.

For a central potential $V=\beta r^k$ and fixed nonzero angular momentum $L$, the effective potential $L^2/(2m{r}^2)+\beta r^k$ has a nondegenerate stable circular orbit when $\beta k>0$ and $k>-2$. Its radius obeys $r_0^{k+2}=L^2/(m\beta k)$. The radial small oscillation frequency satisfies ${\omega }_r^2=(k+2){\Omega }^2$, where $\Omega =|L|/(m{r}_0^2)$. Rational frequency ratios close the linearized near-circular sketch, but generally not the exact finite-amplitude orbit; Bertrand's theorem distinguishes the exceptional harmonic and Kepler potentials.

A particle constrained to the graph $z=h(x,y)$ has kinetic energy $m[|\dot{\mathbf{r}}{|}^2+(\nabla h\cdot \dot{\mathbf{r}}{)}^2]/2$ and potential energy $mgh$. For a gently varying graph and velocities for which its geometric kinetic correction is higher order, the leading Euler-Lagrange equations are the displayed planar gradient-force equations. Small slope alone at arbitrarily large speed or curvature does not justify discarding the normal inertial acceleration. For a radial height, the leading planar angular momentum is conserved, allowing an effective potential reduction.

In the circular-Earth-orbit, unpowered patched conic approximation, the least Earth-relative excess speed for direct solar-surface contact lowers the heliocentric tangential speed to that of an ellipse with apocentre $a$ and pericentre $R_{\odot }$. The vis-viva equation gives $v_a=v_E\sqrt{2R_{\odot }/(a+R_{\odot })}$, so the retrograde excess magnitude is $v_E-v_a$ and the terrestrial surface launch speed is the title expression. Cancelling all heliocentric angular momentum instead produces radial fall to a point Sun and the larger estimate $\sqrt{v_{e,E}^2+v_E^2}$. Rotation, planetary gravity assists and atmospheric drag are outside this direct-flight model.

The laboratory expectation value of the axial electric quadrupole moment in the stretched nuclear quantum state $|I,M=I⟩$; $M\hslash$ is the laboratory $z$ component of total angular momentum. It differs from the intrinsic nuclear quadrupole moment $Q_0$. In an axial nuclear rotational band with $K=0$, projection onto definite angular momentum givesHere $K\hslash$ denotes the component of total angular momentum along the intrinsic symmetry axis. Consequently a positive $Q_0$ for a prolate spheroid corresponds to a negative $Q_s(2)$ in this model. The sign of the laboratory moment alone must be interpreted together with the band assignment.

The $I=0$ result has a short symmetry explanation. A spin-zero state has no preferred spatial direction. Its second charge moments satisfy $⟨x^2⟩=⟨y^2⟩=⟨z^2⟩$, so $⟨2z^2-x^2-y^2⟩=0$. Equivalently, averaging an intrinsic axis uniformly over directions gives $⟨P_2(\cos \theta )⟩=0$, where $P_2(u)=(3u^2-1)/2$ is a Legendre polynomial. This is a statement about a quantum expectation value, not a classical picture of a spinning object sweeping through orientations.

For an independent check of the $I=2$ sign, the $K=0$, $M=2$ rotor orientation probability is proportional to $|Y_2^2{|}^2\propto {\sin }^4\theta$, using a spherical harmonic. With $u=\cos \theta$,A zero static laboratory moment does not make every quadrupole observable zero. With electric multipole operator ${\mathcal{M}}_{2\mu }=\int {\rho }_{\mathrm{ch}}{r}^2{Y}_2^{\mu }{d}^{3}r$, define the reduced transition strength using a reduced matrix element:In the axial $K=0$ rotor, angular projection gives $B(E2;0^{+}\to 2^{+})=5Q_0^2/(16\pi )$ and $B(E2;2^{+}\to 0^{+})=Q_0^2/(16\pi )$. The factor of five is the ratio of the numbers of magnetic substates. Thus an electric quadrupole transition can reveal collective deformation even when $Q_s(0)=0$. These strengths determine $|Q_0|$ within the model; their squares alone cannot distinguish its sign.

A point mass constrained to a fixed distance $r$ from a pivot and free to move in three dimensions under uniform gravitational acceleration. With the vertical axis downward and polar angle $\theta$ measured from it, the Lagrangian is the expression in the title. The azimuth $\varphi$ is a cyclic coordinate, so $m{r}^2{\sin }^2\theta \dot{\varphi }$ is conserved angular momentum about the vertical axis. A rigid massless rod enforces the constraint without the loss-of-tension limitation of a string.

A rank-$k$ spherical tensor operator has components $T_q^{(k)}$, $-k\le q\le k$, that transform under rotation like an angular momentum $k$. Equivalently, their commutators with the angular momentum operator satisfyFor a vector operator, the rank is one and a conventional choice is $T_0=V_z$, $T_{\pm 1}=\mp (V_x\pm i{V}_y)/\sqrt{2}$. The Wigner-Eckart theorem separates the matrix elements of a spherical tensor operator into Clebsch-Gordan coefficients and a reduced matrix element.

Purdue's Bicycle Wheel Gyroscope demonstration uses a bicycle wheel on a long axle, with a ball at one end seated in a freely rotating socket on a heavy stand. Spin the wheel by hand or with the supplied motor, seat the ball, and release the nearly horizontal axle. The readout is the axle sweeping around the vertical while the wheel continues spinning. Compare with the stopped wheel, which falls when one end is released; reverse the spin with the same support and the precession reverses. These are the page's documented apparatus, presentation, and reversal checks. ^[ref]

An alternative in Physics Demonstrations by Julien Clinton Sprott (section 1.3, Materials and Procedure) suspends one axle end by stainless steel wire and a fishing-line swivel. The swivel allows precession without winding up the suspension. Sprott describes the axle bobbing as nutation and precession becoming faster as spin decreases. A suspended support can move, whereas the calculation here assumes a fixed $O$; use the socket version for the closest realization. ^[ref]

A proposed quantitative extension is to record an overhead view of an axle marker against a fixed angle scale, together with a side view for $\theta (t)$. Time successive overhead passages to estimate $\Omega =2\pi /T$; count passages of a rim mark to estimate $|{\omega }_3|=2\pi f$. Record the spin sense too. Measure $M$ and the centre of mass offset $d$ of the whole moving assembly, and determine its axial moment of inertia $I_3$ independently. Wheel, axle, support, release angle, and observation interval should stay the same between spin trials. Measuring both $\Omega$ and $\theta (t)$ distinguishes azimuthal precession from nutation; an overhead movie alone can miss the latter. These camera readouts and parameter checks are proposed here, not reported measurements from the demonstration pages.

For slow precession, compare $\Omega I_3{\omega }_3$ with $Mgd$, and test whether reversing ${\omega }_3$ reverses $\Omega$ with nearly unchanged magnitude. A brief interval minimizes spin loss from friction; repeat at several starting spin rates rather than interpreting an entire slowing run as one steady state. A suspension restoring torque governed by its torsion constant, socket friction, support motion, a misplaced centre of mass, and a push at release can all supply extra torque or change the initial conditions. The swivel reduces the twist that activates the suspension's torsion constant; it does not demonstrate zero torque.

Steady precession keeps $\theta$ constant; nutation changes $\theta$ while the axle may also precess. Sudden release generally excites both. A small initial drop is consistent with conservation of energy and conserved vertical angular momentum: some gravitational potential energy supplies the motion around the vertical, and the axial contribution to vertical angular momentum adjusts. Bearing friction may damp the bobbing, but an ideal undamped heavy symmetric top need not settle into steady precession. Chapter 20: Rotation in space explicitly distinguishes uniform precession from release transients and nutation (section 20-3, paragraphs following equation 20.15).^[ref]

A spin operator describes intrinsic angular momentum. Its components obey angular-momentum commutators; for spin one half, ${\hat{S}}_j=\hslash {\sigma }_j/2$ in terms of the Pauli matrices.

The spin-rotation coupling contribution to a rotating-frame Hamiltonian couples angular velocity to angular momentum. When discussing spin alone, identify which part of the total angular momentum enters the effective model.

Spin-triplet p-wave pairing has pair spin $S=1$ and orbital angular momentum $L=1$, allowing multiple order parameters beyond conventional BCS theory with zero orbital angular momentum. ^[ref]

Specifying this symmetry does not determine its microscopic attraction mechanism. A short exchange-symmetry check explains why these quantum numbers fit fermions. The spin-one-half triplet states $|\uparrow \uparrow ⟩$, $(|\uparrow \downarrow ⟩+|\downarrow \uparrow ⟩)/\sqrt{2}$ and $|\downarrow \downarrow ⟩$ are symmetric under exchange. A stipulated orbital factor proportional to relative momentum $k_x$ changes sign under $\mathbf{k}\mapsto -\mathbf{k}$. Multiplying any symmetric spin function by that odd orbital factor gives an antisymmetric pair wave function, as required by Fermi-Dirac statistics. By comparison, an even s-wave orbital factor needs the antisymmetric spin singlet state. This checks compatibility with exchange statistics; it neither selects a stable phase nor derives its transition temperature.

For an incompressible vertical-field mode, frozen-in magnetic tension supplies a spring of frequency ${\omega }_A$. Its displacement satisfies ${\ddot{\xi }}_x-2\Omega {\dot{\xi }}_y-4\Omega A{\xi }_x=-{\omega }_A^2{\xi }_x$ and ${\ddot{\xi }}_y+2\Omega {\dot{\xi }}_x=-{\omega }_A^2{\xi }_y$. Their determinant gives the ideal magnetorotational dispersion relation. A weak spring transfers angular momentum from the faster inner element to the slower outer element, causing inward and outward migration that increase their separation when angular velocity decreases outward. A sufficiently strong spring instead supplies a restoring force.

// Target: physics.bigb

In static coordinates on anti-de Sitter spacetime, write $f=1+r^2/L^2$, conserved energy per unit rest mass $E=f\dot{t}$, and conserved specific angular momentum $h=r^2\dot{\varphi }$. Timelike normalization gives ${\dot{r}}^2=E^2-V_{\mathrm{eff}}$ with $V_{\mathrm{eff}}=f(1+h^2/r^2)$. A circular geodesic at $r_0>0$ obeys $V_{\mathrm{eff}}^{\prime }=0$, so $h^2=r_0^4/L^2$ and $E=f(r_0)$. Thus $\dot{t}=1$ and $\dot{\varphi }=\pm 1/L$. Both its static-coordinate period and its own proper time period are $2\pi L$, independent of radius. Stability follows from $V_{\mathrm{eff}}^{\prime \prime }(r_0)=8/L^2>0$. A static clock at the orbital radius measures the different period $2\pi L\sqrt{f(r_0)}$, which does depend on radius.

A Stäckel potential allows additive separation of the Hamilton-Jacobi equation in an orthogonal coordinate system. In axisymmetric prolate spheroidal coordinates, the relative-potential form is $\psi =[G(\lambda )-G(\mu )]/(\lambda -\mu )$. It gives energy, axial angular momentum and an independent third separation integral. Here $G$ is a one-variable generating function, not the gravitational constant. Generic bounded orbits are regular invariant-torus motions; resonance and coordinate degeneracy require separate treatment.

A passing star perturbs a planetary or cometary system through its differential Newtonian gravitational field. A distant encounter is predominantly tidal; a sufficiently close one can substantially change specific orbital energy and angular momentum, eject a comet, or shift its periapsis into or out of the planetary region.

A change in a star's mean orbital radius through exchange of angular momentum, particularly at a corotation resonance with a rotating perturbation. For the $k_R=0$ harmonic in action-angle variables, the leading change of radial action vanishes even while azimuthal angular momentum changes. Migration can therefore rearrange a galactic disk with much less heating than exchange at a Lindblad resonance.

The 1922 interpretation concerned space quantization of atomic magnetic moments in the old quantum theory. The modern electron spin proposal came in 1925; the original observers did not identify intrinsic electron spin as the cause of their splitting. An orbital interpretation then seemed plausible, but the later spin explanation gives a different account of the same two-component evidence.^[ref]

Modern atomic data assign neutral silver the ground state electron configuration $[\mathrm{Kr}]4d^{10}5s$ and electronic term ${}^2{S}_{1/2}$. The filled electron shells contribute zero net electronic angular momentum; the outer unpaired electron has orbital angular momentum $L=0$, spin $S=1/2$, and electronic total angular momentum $J=1/2$. The magnetic quantum number has two values $m_J=\pm 1/2$.^[ref]

In the effective electronic description,The minus sign matters: for positive $G$, the component with $m_J=+1/2$ has negative electronic magnetic moment and deflects toward negative $z$ in our coordinates. Calling the upper spot “spin up” without specifying the magnetic field and channel calibration can therefore reverse the assignment. The total angular momentum operator here sums electronic orbital angular momentum and electronic spin; $J$ is not a claim that the atomic nucleus has no spin. Hyperfine structure and nuclear magnetic moments can require additional internal levels, so a two-channel electronic approximation must be justified for the field and resolution being used.

This is a neutral atomic beam, not a beam of free electrons. The deflected mass in the flight equations is the entire silver chemical atom; its dominant electronic magnetic moment supplies the state-dependent force. Replacing that atom by a free electron would also introduce the charged particle's Lorentz force and a different preparation problem. The ideal two-state model is useful because it isolates the connection between discrete internal components and spatially resolved outputs; it does not retroactively turn the 1922 deposit into a direct contemporary identification of intrinsic electron spin.

To construct an original example rather than imitate GW150914, set $G=6.67430\times 10^{-11}{\mathrm{m}}^3\mathrm{k}{\mathrm{g}}^{-1}{\mathrm{s}}^{-2}$, $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$, and adopt the rounded solar mass $M_{\odot }=1.98847\times 10^{30}\mathrm{kg}$. Compare $10+10$ and $20+20$ solar mass binaries starting at $20\mathrm{Hz}$. Stop each at the stipulated orbital compactness thresholdThis is a conservative pedagogical stopping rule, not a claimed error bound: the relative speed already reaches $\sqrt{0.10}c$. Higher-order corrections accumulate wave phase well before merger, and precision inference requires better models. The two cutoffs are approximately $102.2$ and $51.1\mathrm{Hz}$ respectively; doubling the chirp mass multiplies $\dot{f}$ at fixed $f$ by $2^{5/3}$.

For a nearby stipulated source at distance $D$, neglect redshift and define inclination $\iota$ as the angle between the orbital angular momentum and the line of sight. The leading quadrupole formula yieldsThe numerator of $A$ has units ${\mathrm{m}}^5{\mathrm{s}}^{-4}$, matching $c^{4}D$, so both polarizations are dimensionless. Its $f^{2/3}$ growth and the rising frequency produce the tightening, strengthening oscillations. For face-on viewing the polarization components have equal magnitudes in phase quadrature; for edge-on viewing the cross component vanishes, agreeing with circular-binary quadrupole radiation.

Shifting an arrival time aligns the chirp evolution, but different polarization projections need not overlay after only a sign flip. The last panel divides the ideal projected records by their known envelope factors to expose this phase difference. A physical source model must jointly explain those differences, not erase them by arbitrary rescaling. Independent integration of the separation equation, the radiated power and $d\Phi /dt=2\pi f$ checks the analytic chirp against orbital energy balance over the entire stipulated interval.