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A superconductor with an s-wave orbital pairing component: isotropic zero orbital angular momentum in the rotationally invariant model, or the corresponding fully symmetric orbital representation in a crystal. Crystal anisotropy can still make the gap magnitude angle dependent. In the usual centrosymmetric spin-singlet model its orbital parity is even.

Use the material-first nominal stress $P_{Ii}$ and $F=RU$. The symmetric Biot stress conjugate to the Biot strain tensor is $T^{(1)}=\mathrm{sym}(PR)$. To prove it, set $M=P^T$ and differentiate $F=RU$. Conservation of angular momentum makes $M{F}^T$ symmetric, so the contraction of $R^{T}M$ with $R^T\dot{R}U$ is zero: it is a contraction of a symmetric tensor with a skew tensor after cyclically moving $U$. Thus $M:\dot{F}=\mathrm{sym}(R^{T}M):\dot{U}=T^{(1)}:{\dot{E}}^{(1)}$. No assumption that $PR$ itself is symmetric is needed.

Consider a symmetric top with fixed vertex $Q$, axial unit vector $\mathbf{e}$, principal moments of inertia $A,A,C$, and centre of mass at $h\mathbf{e}$. Let a force $\kappa \ell (\mathbf{k}-\mathbf{e})$ act at $\ell \mathbf{e}$, where $\mathbf{k}$ is the upward vertical unit vector. Together with gravity, it produces torque $(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ and potential energy $mgh\mathbf{k}\cdot \mathbf{e}+\kappa {\ell }^2(1-\mathbf{k}\cdot \mathbf{e})$.

The axial angular velocity of a symmetric top $n$, vertical angular momentum $\mathbf{k}\cdot \mathbf{L}$ and total mechanical energy are conserved. For steady precession, $\dot{\mathbf{e}}=\Omega \mathbf{k}\times \mathbf{e}$ and $\mathbf{k}\cdot \mathbf{e}=\cos \alpha$. Substituting $\mathbf{e}\times \ddot{\mathbf{e}}={\Omega }^2\cos \alpha \mathbf{e}\times \mathbf{k}$ into $A\mathbf{e}\times \ddot{\mathbf{e}}+Cn\dot{\mathbf{e}}=(\kappa {\ell }^2-mgh)\mathbf{e}\times \mathbf{k}$ gives the condition in the title for nonvertical inclination. If the total angular velocity itself is $\Omega \mathbf{k}$, then $n=\Omega \cos \alpha$, and the condition becomes $\kappa {\ell }^2=mgh+(A-C){\Omega }^2\cos \alpha$.

Writing the unit symmetry axis as $n$ and axial spin momentum as $s=I_3{\omega }_3$, angular momentum is $L=I_{1}n\times \dot{n}+sn$. Taking its derivative and applying gravitational torque gives $I_{1}n\times \ddot{n}+s\dot{n}=-mg\ell n\times \hat{z}$.

A tabletop mass $m$ joined through a hole by a massless inextensible string of length $l$ to a hanging mass $M$ has radial inertia $M+m$ and conserved angular momentum $J=m{r}^2\dot{\theta }$. The radial energy is $\frac{1}{2}(M+m){\dot{r}}^2+U_J(r)$, with $0<r<l$. Its effective potential diverges at the hole for $J\ne 0$ and has a unique minimum at $r_0^3=J^2/(mMg)$. Since $U_J^{\prime \prime }(r_0)=3Mg/r_0$, small radial oscillations at fixed $J$ have squared angular frequency $3Mg/[(M+m)r_0]$.

Let fuel of remaining mass $M_f$ be carried at radius $a$ on a rotor whose dry moment of inertia is $I_d$. Then $I=I_d+a^2{M}_f$. Fuel leaves tangentially backwards with speed $v_e$ relative to the rim, so its inertial tangential velocity is $a\omega -v_e$. Conserving angular momentum of rotor plus newly emitted fuel during $d{M}_f<0$ givesSince $dI=a^{2}d{M}_f$, integration yields ${\omega }_f-{\omega }_i=(v_e/a)\log (I_i/I_f)$. Including the fuel's co-rotation is essential: the rotor alone is an open system and its angular momentum is not conserved. For three radial rods of mass $m$ and length $a$, each carrying a point rocket of dry mass $m$ and fuel mass $m_f$, $I_i=(4m+3m_f)a^2$ and $I_f=4m{a}^2$.

In LS coupling, an atomic term symbol specifies total electronic spin $S$, total orbital angular momentum $L$, and total angular momentum $J$. The superscript $2S+1$ is the spin multiplicity; the letters $S,P,D,F,\dots$ denote $L=0,1,2,3,\dots$. Coupling gives $J=|L-S|,\dots ,L+S$, and a fixed-$J$ level has $2J+1$ magnetic substates before an external field resolves them. For example, ${}^{3}P$ has $S=L=1$ and levels $J=0,1,2$, containing $1+3+5=9$ states altogether. A term symbol does not itself specify the radial electron configuration.

A quantum vortex becomes energetically favourable in a rotating reference frame when $\Delta E-\Omega \Delta L_z<0$. For a singly quantized vortex in a long radially Thomas–Fermi condensate, the cutoff-shell estimate isIntegrating the trapped number density against $u_{\theta }^2$ produces the logarithm and constant; integrating it against angular momentum per particle gives $\Delta L_z\simeq \pi H{n}_0\hslash R_r^2/2$. Resolved-core corrections can alter the nonlogarithmic constant. This criterion compares equilibrium rotating-frame energies and does not remove a dynamical vortex-entry barrier.

Consider an original hypothetical atom with a lower angular momentum multiplet $F=1$, $m=-1,0,+1$, and one opposite-parity excited level $F^{\prime }=0,m^{\prime }=0$. Stipulate that all spontaneous emission returns to these three lower quantum states, with no other decay channel. This is an isolated four-state model, not the actual sodium or rubidium D-line structure. In particular, the physical rubidium cell above contains multiple hyperfine structure levels.

The electric-dipole selection rule permits $\Delta F=-1$ here; ${\sigma }^{+}$ absorption with $\Delta m=+1$ excites only $m=-1$. Decay from $F^{\prime }=0$ has three equal integrated branching probabilities $b_{-}=b_0=b_{+}=1/3$: rotational symmetry gives equal squared Clebsch-Gordan coefficients. The emitted photons carry the three permitted angular momentum projections; equal integrated branching does not imply equal intensity in every observation direction. Thus $m=0$ and $m=+1$ are both dark. The chosen beam cannot prepare a unique pure $m=+1$ state.

Let $R\ge 0$ be the absorption rate of an atom in $m=-1$, and let $\Gamma$ be the excited-state spontaneous emission rate, both in inverse seconds. Assume $R,\gamma \ll \Gamma$, negligible stimulated emission, and no maintained quantum coherence between sublevels. Eliminating the short-lived excited population gives normalized lower populations $p_{-}+p_0+p_{+}=1$ to this approximation. At each scattering event the atom leaves $m=-1$ with probability $2/3$; the remaining $1/3$ simply returns it there.

Model depolarizing collisions and other population relaxation by resets to the unpolarized distribution at rate $\gamma \ge 0$. The rate equations areThis relaxation prescription assumes that Zeeman splitting is small compared with thermal energy, so the unpolarized distribution approximates thermal equilibrium. It is not an energy-independent relaxation law at arbitrarily low temperature.

Every transfer has a nonnegative rate: optical transfers $-\to 0$ and $-\to +$ each have rate $R/3$, and relaxation transfers between each distinct pair have rate $\gamma /3$. These define a continuous-time Markov chain. The three derivatives sum to zero, and at $p_i=0$ the corresponding derivative cannot be negative. Hence the model conserves population and preserves nonnegative populations. Optical cycles that return to $m=-1$ contribute to scattering even though they cause no net population transfer.

Tidal torque theory explains the early angular momentum of a protohalo through the misalignment of its mass second-moment tensor $I_{ij}$ and the Hessian matrix $T_{ij}={\partial }_i{\partial }_j\Phi$ of the surrounding peculiar gravitational potential. At leading order,

Torque measures a force's turning effect about a specified origin. The geometry in the angular momentum diagram applies with force replacing momentum: transverse forces give positive or negative torque, while a radial force gives zero. A longer perpendicular lever arm makes the same force more effective, explaining why pushing near the end of a handle is useful.

For a force $\mathbf{F}$ applied at position $\mathbf{r}$ from a fixed origin $O$ in an inertial frame, its torque, or moment of force, isHere $F=|\mathbf{F}|$, $\varphi$ is the angle between position and force, and $\ell =r\sin \varphi$ is the perpendicular lever arm. Direction follows the right-hand rule; in the $xy$ plane ${\tau }_z=x{F}_y-y{F}_x$, positive counterclockwise as viewed from $+z$. The International System of Units unit is $\mathrm{N}\mathrm{m}=\mathrm{kg}{\mathrm{m}}^2/{\mathrm{s}}^2$. Although torque and energy have the same physical dimensions, torque is an axial vector and energy is a scalar; their roles differ.

For $\mathbf{r}=(2,0,0)\mathrm{m}$ and $\mathbf{F}=(0,3,0)\mathrm{N}$, ${\tau }_z=+6\mathrm{N}\mathrm{m}$. Applying the same force at $(1,0,0)\mathrm{m}$ gives $+3\mathrm{N}\mathrm{m}$; reversing it gives a negative torque; directing it along $\mathbf{r}$ gives zero. More generally, sum ${\mathbf{r}}_i\times {\mathbf{F}}_i$ to find the total torque about one common origin.

For a constant-mass point mass, the product rule and Newton's second law show why net torque changes angular momentum:The first cross product vanishes because its factors are parallel. For a collection with canceling internal torques, the remaining sum is external torque, giving conservation of angular momentum when it vanishes.^[ref]

A nonzero torque can turn the direction of angular momentum as well as change its magnitude. Indeed, the dot product gives $d|{\mathbf{L}}_O{|}^2/dt=2{\mathbf{L}}_O\cdot {\mathbf{\tau }}_O$: a perpendicular torque changes direction without instantaneously changing the magnitude. This is the geometry behind gyroscope precession. For a rapidly spinning symmetric top with a fixed tip, centre of mass distance $d$ from the tip, mass $M$, and spin-dominated angular momentum magnitude $L\simeq I\omega$, the slow steady precession approximation gives $\Omega \simeq Mgd/L$, where $g$ is gravitational acceleration. ^[ref] This is an approximation with $\Omega \ll \omega$; the uniform precession of a heavy symmetric top retains contributions omitted by the spin-dominated estimate.

Assume a constant positive mass $m$, a twice differentiable trajectory in the fixed plane, a fixed origin in an inertial frame, and Newton's second law for the net force. The product rule gives the short proofThe term $\mathbf{v}\times \mathbf{v}$ vanishes. This is the planar angular momentum balance written in area language.^[ref]

On a connected time interval, areal velocity is constant if and only if ${\tau }_n=0$ throughout. Since both position and acceleration lie in the fixed plane, the net torque has only that normal component. For $r\ne 0$, zero net torque is therefore equivalent to net force parallel to $\mathbf{r}$: its instantaneous direction is central, possibly inward, outward or zero. This diagnoses the net force along this motion, not its dependence on position or time elsewhere. Opposing transverse forces may cancel.

A central force about a fixed origin gives conservation of angular momentum, irrespective of its radial dependence: an inverse-square law is sufficient but unnecessary. With nonzero conserved angular momentum, $\mathbf{r}\cdot \mathbf{L}=0$ fixes the orbital plane. Zero angular momentum gives radial motion and zero areal velocity; it does not select a unique orbital plane. The converse needs the fixed-plane assumption: a constant projected areal velocity in three dimensions only constrains one torque component. Also, $r=0$ makes the instantaneous torque vanish regardless of force and makes plane polar coordinates singular. A moving origin or changing plane requires additional terms.

For an axially symmetric rigid body with $I_1=I_2=I_{\perp }$ and $I_3=I_{\parallel }$, no external torque means that angular momentum $\mathbf{L}$ is fixed in an inertial frame. The axial angular velocity ${\omega }_3$ is constant by Euler equations for a torque-free rigid body, and the symmetry axis obeysIts angle to $\mathbf{L}$ is constant and it precesses in space at $L/I_{\perp }$. The transverse angular-velocity components instead rotate in body coordinates at signed rate $(I_{\parallel }-I_{\perp }){\omega }_3/I_{\perp }$. These are different frame descriptions and must not be substituted for one another.

A useful readout of a tumble is the alignment of the intermediate principal axis ${\hat{e}}_2$ with the conserved angular momentum direction $\hat{L}$. In the principal body frame, write $L_i=I_i{\omega }_i$ and $L=|\mathbf{L}|$. ThenA change from $c_2\simeq +1$ to $c_2\simeq -1$ means that the directed body axis has turned nearly $180^{\circ }$ relative to $\mathbf{L}$. It does not mean that $\mathbf{L}$ reversed in the fixed space frame. The unsigned line through that principal axis hides this distinction, so a marked end matters in an actual demonstration.

For the normalized intermediate-axis model with $\mathbf{\omega }(0)=(0.01,1,0)$, the rotational kinetic energy and squared angular momentum are $E=1.00005$ and $L^2=4.0001$. Hence $L^2-2E{I}_2=-0.0001$: this initial condition lies close to, but off, the intermediate-axis separatrix. An independent calculation evolves ${\dot{\mathbf{L}}}_{\mathrm{body}}={\mathbf{L}}_{\mathrm{body}}\times I^{-1}{\mathbf{L}}_{\mathrm{body}}$ and a rotation matrix taking body components to space components. The intermediate principal axis crosses $c_2=0$ at normalized times approximately $10.3773$ and $31.1319$, alternating long nearly aligned intervals with quicker reversals.

Both calculations use step size $0.005$ without rescaling either invariant. Over $0\le t\le 50$, the tumbling calculation changes rotational kinetic energy and squared angular momentum by less than $4\times 10^{-13}$ relative to their initial values. Reconstructing $\mathbf{L}$ in the fixed space frame gives relative error below $3\times 10^{-12}$. Halving the step changes the body components and rotation matrix entries by less than $3\times 10^{-10}$. These checks distinguish physical tumbling from numerical drift. On the exact intermediate-axis separatrix, approach to an intermediate-axis rotation takes infinite time; the recurrent off-separatrix example here is different.

For free fermions on a ring, each one-particle orbital has integer angular momentum label $l$ and two spin states. Two particles occupy the $l=0$ orbital in a unique singlet state at zero kinetic energy. At the next distinct energy ${\hslash }^2/(2m{R}^2)$, orbitals $0,+1$ or $0,-1$ can be combined with one singlet state or three triplet state spin states, giving total quantum degeneracy eight. The lowest triplet state has six states and its antisymmetric spatial factor vanishes when the particles coincide.

Orbital migration of a low-mass embedded planet driven by its interaction with disk gas, without a deep gap. Spiral wave torques and corotation-region torques exchange orbital angular momentum. The sign depends on the disk gradients and thermodynamics; this regime is not restricted to inward migration.

A one-dimensional condensate on a ring of circumference $2\pi R$ has a uniform winding state $\psi =\sqrt{N/(2\pi R)}{e}^{in\theta }$ when the order parameter is normalized to atom number. Its angular momentum per atom is $n\hslash$, while the flow contribution to its total chemical potential is ${\hslash }^2{n}^2/(2m{R}^2)$.

On the canonical cotangent bundle $T^{\ast }{\mathbb{R}}^n$, $\omega =i\sum d{z}_j\wedge d{\bar{z}}_j$ and $H=z^{†}z$. The unitary group acts by $z\mapsto Uz$, preserving both. For a Hermitian $A$, the generator $-iA$ has Hamiltonian ${\mu }_A=z^{†}Az$. The Poisson bracket satisfies $\{{\mu }_A,{\mu }_B\}={\mu }_{-i[A,B]}$, with no central term. The central generator has moment map $H$; imaginary antisymmetric matrices give ordinary angular momentum, while real symmetric matrices give quadratic observables mixing position and momentum.

The University of Florida's Rotation Vibration Spectrum of the HCl Molecule, University of Florida laboratory protocol, printed pp. 2–3 and 10–11, uses a scanning Michelson interferometer for Fourier transform spectroscopy. An air-filled silicon-window cell supplies the background. Supervised preparation adds concentrated hydrogen chloride liquid; the cleaned cell subsequently receives DCl liquid. Acquisition settings are identical.

The readout is transmission or absorbance against spectroscopic wavenumber, resolving molecular rotation alongside stretching. Background addresses instrument and cell contributions; cleaning addresses carryover. Excessive absorption impairs analysis. Assigning isotopes and rotational branches precedes fitting band parameters to test isotope scaling of diatomic vibration. These are prescribed measurements, not numerical observations supplied here.^[ref]

For a transparent ideal model, add the linear rigid rotor to the harmonic vibrational energy levels and assume the same frequency-valued rotational constant $B$ in both vibrational states:For a spin-free diatomic molecule with no electronic angular momentum, electric-dipole selection rules give $\Delta J=\pm 1$ for this vibrational absorption. Subtracting initial from final energy eigenvalues yieldsThese are two rotational branches about the vibrational band origin, not a single stretch line. For stipulated dimensionless values ${\tilde{\nu }}_{\mathrm{harm}}=100$ and $\tilde{B}=1$, the first three R-branch positions are $102,104,106$, and the first three P-branch positions are $98,96,94$. There is no $J=0\to -1$ transition. The numbers illustrate level subtraction; they are not HCl measurements. Real upper and lower rotational constants differ, and centrifugal distortion shifts lines further.

A spin-one parent cannot decay into two identical spin-zero particles in the same internal state while preserving rotational symmetry and Bose exchange symmetry. Their final total angular momentum is orbital $L=1$, which is antisymmetric under interchange. Bose symmetry requires even $L$ instead. An energetically allowed decay with matching parity can therefore remain forbidden independently of the interaction's coupling strength.