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Within a fixed isolated electronic $J$ multiplet, the electronic field at the nucleus is a vector operator. Rotational symmetry and the Wigner-Eckart theorem imply that its projection into this multiplet is proportional to $\mathbf{J}$:Its detailed orbital, spin-dipolar and contact contributions are included in the reduced coefficient $b_J$. Consequently the projected magnetic-dipole hyperfine interaction becomesin the same dimensionless angular-momentum convention as a. The proportionality is a statement about matrix elements within the chosen electronic manifold, not an identity for the electron field operator on the entire atomic Hilbert space.

Define $\mathbf{F}=\mathbf{I}+\mathbf{J}$. In zero external field the quantum states obtained by addition of angular momentum are labelled $|\gamma J,I;F,m_F⟩$, where $\gamma$ specifies the electron configuration and other fixed labels. The good quantum numbers include $I,J,F,m_F$, with $F=|I-J|,\dots ,I+J$ and $m_F=-F,\dots ,F$. In a description using LS coupling the fixed electronic labels can also include $L,S$. The energies depend on $F$ but not $m_F$:This neglects coupling to other electronic manifolds, electric-quadrupole hyperfine interactions and the Zeeman effect from external fields. Those additions can shift the intervals or change which quantum numbers remain useful.

The orbital angular momentum $L_3=X_1{P}_2-X_2{P}_1$ generates rotations in the plane. The canonical commutation relations giveUse the commutator product rule $[A,B^2]=[A,B]B+B[A,B]$. Since different coordinates commute,Thus, for arbitrary positive angular frequencies,At ${\omega }_1={\omega }_2$, this becomes $\boxed{[L_3,H]=0}$. The rotational symmetry of the two-dimensional isotropic harmonic oscillator allows a basis of simultaneous energy eigenstates and orbital angular momentum eigenstates.

The radial and azimuthal Hamilton equations giveAlso $\dot{\theta }=p_{\theta }/(m{r}^2)$ and ${\dot{p}}_{\theta }=p_{\varphi }^2\cos \theta /(m{r}^2{\sin }^3\theta )$. Differentiating the angular-momentum square,Therefore both $p_{\varphi }$ and $\boxed{J^2=p_{\theta }^2+p_{\varphi }^2/{\sin }^2\theta }$ are constants. Rotational symmetry conserves the full angular-momentum vector; these are its axial component and squared magnitude in spherical canonical variables.

For the isotropic zero-field classical Heisenberg model of a ferromagnet, the analytic vector Landau expansion uses the Cartesian components of the vector order parameter near $\mathbf{m}=0$. rotational symmetry makes its scalar terms functions of $\mathbf{m}\cdot \mathbf{m}$. Global spin inversion symmetry also requires $f(\mathbf{m})=f(-\mathbf{m})$. Through fourth order this givesA positive quartic coefficient stabilizes the unconstrained truncated free energy. The disordered phase has $b>0$ and its minimum at $m=0$; the aligned phase has $b<0$ and a minimum at nonzero $m$. At the continuous transition $b=0$. The attractive alignment interaction supplies an energetic tendency toward negative quadratic curvature; orientational entropy favors the disordered phase at high temperature, so the sign of $b$ is a free-energy question rather than a consequence of the pair energy alone.

The analyticity assumption matters: the variable printed as $m$ is a nonnegative magnitude, so reversing $\mathbf{m}$ does not send that scalar to $-m$. Terms proportional to $|\mathbf{m}|$ or $|\mathbf{m}{|}^3$ would be rotationally invariant but nonanalytic at the vector origin. Standard analytic Landau theory excludes them. Without that assumption, the stated pair symmetry alone would not exclude odd powers of the magnitude.^[ref]

Use the angular coordinate $\theta$ for the position on the ring. The kinetic energy is $T=m{a}^2{\dot{\theta }}^2/2$. The gravitational potential energy is constant because the ring is horizontal, and the ideal constraint force has no tangential component. Thus the Lagrangian has no dependence on $\theta$ itself: it is a cyclic coordinate.

The Euler-Lagrange equation givesEquivalently, the conserved conjugate momentum $p_{\theta }=m{a}^2\dot{\theta }$ is the vertical angular momentum, as required by rotational symmetry.

A common rotation changes neither the separation nor the spring energy. Its infinitesimal direction is $(1,1)$, exactly the zero-frequency normal mode. Thus the continuum of rotated equilibrium points requires that zero mode; a nonzero restoring frequency in that direction would contradict rotational symmetry.

The second mode has $I_1{\eta }_1+I_2{\eta }_2=0$, so it changes the relative angle while leaving the inertia-weighted common angle unchanged. It therefore carries zero perturbation in the conserved angular momentum associated with common motion. Indeed the coordinatesseparate the Lagrangian exactly:The cyclic coordinate $q$ is free, while $\delta$ has restoring or inverted curvature. This also explains why all members of either equilibrium family have the same frequencies. Reflection symmetry makes the potential even in $\delta$, and the stable oscillatory mode is consistent with time reversal in classical mechanics through its two opposite frequency signs.

Let $F(x,y)=(y,x-\Omega y-x^3-x^{2}y)$. Every term changes sign when both variables change sign, so $F(-x,-y)=-F(x,y)$. Consequently the derivative of $(-x(t),-y(t))$ equals $F(-x(t),-y(t))$, proving the stated rotational symmetry.

This is equivariance under a half-turn, not time reversibility. Reversibility of an autonomous differential equation requires a reverser $R$ satisfying $F(Rz)=-DR(z)F(z)$, so that $Rz(-t)$ is a solution. The half-turn $R=-I$ would require $F(-z)=F(z)$, the opposite of what holds. The usual mechanical candidate $R(x,y)=(x,-y)$ also fails when damping is present: its second equation would require $(\Omega +x^2)y=-(\Omega +x^2)y$ identically. The symmetry relates two orbits with the same direction of time, and does not reverse their arrows.

If a condensate order parameter stays nonzero along a closed loop, its phase changes by an integer multiple of $2\pi$. Broken rotational symmetry does not destroy this single-valuedness constraint. In an inhomogeneous ring the winding integer need not label an angular-momentum eigenstate.

Consider a parent atomic nucleus at rest. Write its signed nuclear polarization along a fixed axis $\hat{\mathbf{z}}$ as $P=⟨J_z⟩/(I\hslash )$, where $I\hslash$ is its maximal spin component. Let $\theta$ be the angle of the emitted electron momentum to that axis and let $\beta =v/c$. Use $\mathcal{P}$ for the parity operator, so it is not confused with $P$.

For an axially symmetric leading-dipole model, retain a constant and the term linear in the polarization vector $\mathbf{P}=P\hat{\mathbf{z}}$. Rotational symmetry makes the directional term proportional to $\mathbf{P}\cdot \hat{\mathbf{p}}$. Define its dimensionless coefficient as $A\beta$. At fixed electron energy the relative intensity and normalized probability density function per solid angle are thereforeThe factor $\beta$ is the conventional separation of the speed dependence; rotational symmetry alone does not determine that dependence or $A$. The coefficient $A$ contains the decay dynamics. This leading model omits higher angular terms and detailed recoil and radiative effects. A partially oriented higher-spin ensemble can also have alignment that is not specified by its vector nuclear polarization alone.

The unit-integral condition follows without any model of a detector:Thus $f$ has unit angular average, while the probability density function in the polar angle alone contains the Jacobian determinant factor:In particular, $f$ itself is neither a normalized probability density function in $\theta$ nor a detector count. Positivity on $-1\le u\le 1$ gives $|a|\le 1$.

Under full spatial reflection, $\mathbf{p}\mapsto -\mathbf{p}$ but $\mathbf{J}\mapsto \mathbf{J}$. The latter follows already for orbital angular momentum from $\mathbf{r}\times \mathbf{p}\mapsto (-\mathbf{r})\times (-\mathbf{p})=\mathbf{r}\times \mathbf{p}$; spin has the same axial vector transformation. Hence $\mathbf{P}\cdot \hat{\mathbf{p}}$ is a pseudoscalar, and $\cos \theta \mapsto -\cos \theta$. If parity conservation held for this model, equal probabilities for the inverted process would force $a=0$. Conversely, $a=0$ in an unpolarized sample proves nothing about $A$.

Integrating over the two hemispheres gives a useful original check:Choose $A=-0.8$, $P=0.5$, and $\beta =0.75$, so $a=-0.3$. Then the forward and backward hemisphere probabilities are $0.425$ and $0.575$, and their counting asymmetry is $-0.15$. These are chosen teaching values, not measured values from the Wu experiment. At a single energy, the angular shape determines the product $AP\beta$; it cannot separately determine $A$ without a polarization calibration.

For a radial weight $F$ with the required absolutely integrable moments, rotational symmetry makes the second and fourth moments isotropic tensors. The second moment is $A{\delta }_{ij}$; its trace gives $3A=4\pi {\int }_0^{\infty }{r}^{4}F(r)dr$. Complete symmetry of the fourth moment forces equal coefficients of the three Kronecker delta pairings. Its double tensor contraction gives $15B=4\pi {\int }_0^{\infty }{r}^{6}F(r)dr$, proving the displayed expression. Odd moments vanish under $x\mapsto -x$. Choosing $F$ to be the indicator function of a spherical shell gives all the corresponding shell moments without separate angular integrations.

A regular icosahedron is a Platonic solid with twenty equilateral triangular faces, twelve vertices and thirty edges. Five faces meet at each vertex. Its orientation-preserving rotational symmetry has order sixty: the orbit-stabilizer theorem gives twenty faces times the three rotations stabilizing a chosen face.

The orientation-preserving rotational symmetry group of a regular icosahedron has order sixty. Its conjugacy classes have sizes $1,15,20,12,12$: identity; half-turns about opposite-edge axes; rotations through $\pm 2\pi /3$ about opposite-face axes; and rotations through $\pm 2\pi /5$ and $\pm 4\pi /5$ about opposite-vertex axes, respectively. Conjugation carries an axis to its rotated axis and preserves the rotation angle. The transitive group action on edges, faces and vertices gives the indicated classes, with reversal of the axis identifying the two signs. A normal subgroup must contain whole conjugacy classes; none of the proper nonidentity sums of these class sizes is a divisor of sixty. Thus this group is a simple group.

The orientation-preserving symmetries of a regular tetrahedron form the alternating group on its four vertices. There are twelve elements: the identity, eight rotational symmetries of order three, and three half-turns.

Consider a mechanically free isotropic model of a saturated ferromagnet, at fixed temperature and fixed magnitude of magnetization. Let $\mathbf{m}=\mathbf{M}/M_s$ be a unit vector, where $M_s$ is the saturation magnetization. The magnetostrictive strain is an infinitesimal strain tensor, measured relative to a reference with randomly oriented magnetic domains. This reference convention matters: it sets the orientational average of the preferred strain to zero. Define ${\lambda }_s$, the saturation magnetostriction, as the longitudinal relative length change in this model.

Rotational symmetry permits a symmetric second-rank tensor made from $\mathbf{m}$ of the form $a{\delta }_{ij}+b{m}_i{m}_j$, where ${\delta }_{ij}$ is the Kronecker delta. If we restrict attention to a change of shape with zero first-order volume change, its trace gives $3a+b=0$. Requiring the longitudinal component to equal ${\lambda }_s$ gives $a+b={\lambda }_s$. HenceThe tensor product $\mathbf{m}\otimes \mathbf{m}$ is symmetric and unchanged under $\mathbf{m}\mapsto -\mathbf{m}$. It uses two copies of the magnetic direction, so the third-rank isotropic tensor obstruction in isotropic linear magnetostriction vanishes does not apply. More generally, a small-magnetization expansion can contain $a{M}^2{\delta }_{ij}+b{M}_i{M}_j$; isotropy alone does not require its trace to vanish. The displayed saturated formula isolates the volume-preserving orientational contribution, rather than asserting that every material has zero volume magnetostriction.

Along a measuring unit vector $\mathbf{n}$, with $\mathbf{n}\cdot \mathbf{m}=\cos \theta$, the fractional length change isFor a stipulated ${\lambda }_s=20\times 10^{-6}$ and reference length $L=100\mathrm{mm}$, alignment gives $+2\mu \mathrm{m}$ and a perpendicular orientation gives $-1\mu \mathrm{m}$. Rotating from perpendicular to parallel therefore changes the length by $3\mu \mathrm{m}$, even though the saturated longitudinal value relative to the random reference is $2\mu \mathrm{m}$. A negative ${\lambda }_s$ reverses these elongation and contraction signs.

The eigenvalues of this infinitesimal strain tensor are ${\lambda }_s,-{\lambda }_s/2,-{\lambda }_s/2$, so $\Delta V/V=\mathrm{tr}{\epsilon }^{\mathrm{mag}}=0$ to first order. This is not exact finite incompressibility: using the illustrative deformation gradient $F=I+{\epsilon }^{\mathrm{mag}}$ givesA finite-deformation law would need higher-order corrections if exact volume conservation were required.

For the outward unit normal on a three-dimensional sphere, the off-diagonal integrals vanish by reflection symmetry. The diagonal integrals are equal by rotational symmetry, and their sum is ${\int }_{S^2}|\mathbf{n}{|}^{2}d\Omega =4\pi$. This proves the displayed matrix identity, where ${\delta }_{ij}$ is the Kronecker delta. Contracting it with a constant vector gives ${\int }_{S^2}(\mathbf{U}\cdot \mathbf{n})\mathbf{n}d\Omega =(4\pi /3)\mathbf{U}$, useful for integrating far-field stress forces.

Two equal circular rings of radius $R$ in parallel planes at $z=\pm L$ provide a boundary for an open film. In the ideal axisymmetric model, a stationary connected film has the catenoid profileThe catenoid existence threshold for equal rings explains why this equation can have two solutions, one solution, or none. The cancellation of principal curvatures holds for both solutions whenever they exist; the mean curvature equation alone does not select a stable minimal surface.

This ideal geometry uses $R=1$ and $L/R=0.60$. The two profiles have different waist radii $a$, although both have $H=0$. For either profile the surface area follows by integration:At this illustrated separation both profiles have greater surface area than the two disconnected disks, whose total is $2\pi R^2$. This comparison disproves a claim of global area minimization in a class allowing disconnected competitors; it does not determine the sign of the second variation for small variations of a connected film.

A reported experiment used coaxial wire rings of radius $4\mathrm{cm}$, drawn from a surfactant solution and slowly separated until collapse. The solution contained tetradecyltrimethylammonium bromide in deionized water, with glycerol added to vary viscosity. Fluorescein enabled fluorescence imaging under cyan LED arrays; a Phantom V641 camera recorded 4,000 or 5,600 frames per second. Extracted contours gave neck radius, attachment angle, and inferred surface area. They showed neck narrowing and eventual breakup, rather than indefinite persistence of a stationary catenoid. Comparing solutions across viscosities helped distinguish liquid dissipation from the surrounding air's inertia, but the ring displacement beyond the last stable state was not precisely controlled. These are collapse observations, not direct measurements of equilibrium $H$. ^[ref]

A proposed classroom measurement would instead hold two equal rings coaxial on a translation stage, leave both sides open to the same gas, and photograph the settled silhouette beside a scale. Fit $r(z)=a\cosh (z/a)$ and inspect fit residuals before inferring cancellation of the principal curvatures. Repeat at several separations and with the camera rotated about the axis to test rotational symmetry. Keep solution composition, ring wetting, and elapsed drainage time comparable, and shield the film from drafts. Film weight, drainage, evaporation, surfactant gradients, and rupture can invalidate the ideal soap film pressure balance. A silhouette fit alone neither measures the pressure jump nor tests all possible perturbations of a stable minimal surface.

Let the inner-centre displacement from a fixed outer centre have magnitude $r=\Delta e$, with $0\le e<1$. The leading full-film journal bearing resistance to translation in the displacement direction and perpendicular to it isFor $h=\Delta (1-e\cos \theta )$, the Reynolds lubrication equation gives radial squeezing flux $a{v}_r\sin \theta$ and tangential squeezing flux $a{v}_t[-\cos \theta +3e/(2+e^2)]$. The additive constant enforces periodic pressure. Integrating the pressure traction gives these coefficients; at $e=0$ they coincide, as required by rotational symmetry.

For an axisymmetric planar lamina of total mass $M$, let $I_z$ be its moment of inertia about the central axis normal to its plane. The perpendicular axis theorem and rotational symmetry give $I_x=I_y=I_z/2$ about central diameters. An in-plane tangent line a perpendicular distance $R$ from a parallel central diameter therefore has moment of inertia $I_{\tan }=I_z/2+M{R}^2$ by the parallel axis theorem. An axis normal to the lamina through the same rim point instead has $I_z+M{R}^2$; specifying the point without the axis direction is insufficient.

For an isotropic, integrable background phase-space distribution function, rotational symmetry makes the integral radial in velocity space. Its divergence with respect to $\mathbf{V}$ is $4\pi f(V)$ by the Dirac delta identity for the divergence of the inverse-square vector field. Integrating over a velocity-space sphere and using the divergence theorem proves the displayed identity. Exterior velocity shells have zero contribution, just as in the spherical shell theorem. This explains why the leading-log Chandrasekhar dynamical friction integral contains only particles slower than the test body.