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Use the usual homogeneous, local, analytic Landau-Ginzburg theory; rotational symmetry and Z2 symmetry alone do not specify locality, analyticity, or the signs of the coefficients. The local derivative expansion contains every rotational scalar with an even total number of fields:The omitted terms include all allowed higher field powers and contracted spatial derivatives. Terms differing by a total derivative are equivalent for bulk behavior with suitable boundary conditions. The displayed gradient energy assumes spatial homogeneity. If only rotations about an origin are imposed, rotationally invariant position-dependent coefficients are also possible.

In the mean-field approximation, take a spatially uniform order parameter and minimize its Landau free energy. For $\gamma >0$, $u>0$ and $r=a(T-T_c)$ with $a>0$, keeping the quadratic and quartic terms givesThe single minimum splits into two minima as the temperature passes below the critical temperature. Choosing one minimum gives spontaneous symmetry breaking of the Z2 symmetry and a continuous phase transition. The order parameter is the expectation in a selected ordered state; averaging equally over both states gives zero. A Z2 symmetry does not guarantee a continuous phase transition: a negative quartic coefficient stabilized by a positive sextic coefficient can instead give a first-order phase transition.

Relaxing the Z2 symmetry permits odd terms. A nonzero linear source generally rounds this ordinary continuous phase transition. A cubic term can instead give a first-order phase transition: for $V(\varphi )=r{\varphi }^2/2+w{\varphi }^3/3+u{\varphi }^4/4$, $u>0$ and $w\ne 0$, equal minima occur atThe order parameter jumps from zero to ${\varphi }_{\ast }$. A continuous phase transition without an exact Z2 symmetry is still possible with additional tuning; odd terms do not logically exclude it.

For the multicritical even Landau potential, set $r={\mu }^2=a(T-T_c)$ and assume an integer $n\ge 2$, ${\lambda }_n>0$ and $\gamma >0$. For $n>2$, all lower even interactions must be absent or tuned to zero to obtain these mean-field critical exponents. The stationary equation isIts nonzero solutions are minima because $V^{\prime \prime }({\varphi }_0)=-(2n-2)r>0$. Thus the order-parameter critical exponent isSubstituting into the minimum free energy density givesThe singular heat capacity per unit volume is $c_s=-T{\partial }_T^2{f}_{\min }$. Smooth variation of the other coefficients does not change the leading power. ConsequentlyFor $n=2$, this means a finite heat capacity jump, with ${\alpha }_{\mathrm{MF}}=0$; for $n=3$, the tricritical point has ${\beta }_{\mathrm{MF}}=1/4$ and ${\alpha }_{\mathrm{MF}}=1/2$. The case $n=1$ would contain only a quadratic potential and would not stabilize an ordered phase.

The Gaussian fluctuation correction near a critical point is more singular than the mean-field approximation contribution whenThis is the upper critical dimension of an even scalar interaction. Equivalently, its coupling has engineering dimension $2n-(n-1)d$, positive below $d_c$. Dominant fluctuations signal breakdown of the mean-field approximation; the Gaussian expression is not then a calculation of the exact interacting heat-capacity critical exponent. At $d=d_c$, equality of the powers allows logarithmic corrections.

The nematic order parameter is a symmetric second-rank tensor and a traceless second-rank tensor describing orientational anisotropy. For molecular unit axes $\mathbf{s}$, $Q_{ij}$ is proportional to $⟨s_i{s}_j-{\delta }_{ij}/3⟩$. It vanishes in an isotropic phase and is unchanged by head-tail reversal $\mathbf{s}\mapsto -\mathbf{s}$, unlike a polar order parameter. For uniaxial nematic order, the nematic director is thus an unoriented axis.

One convention for the Landau-de Gennes free energy through fourth order, respecting rotational symmetry, isFor a traceless second-rank tensor in three dimensions, the Cayley-Hamilton theorem gives $\mathrm{Tr}{Q}^4=(\mathrm{Tr}{Q}^2{)}^2/2$, so there is only one independent quartic rotational invariant of a symmetric traceless tensor. The only possible linear scalar is $\mathrm{Tr}Q=0$; without an external anisotropy no linear term survives.

The cubic rotational invariant of a symmetric traceless tensor distinguishes prolate and oblate forms of uniaxial nematic order. Head-tail reversal leaves $Q$ unchanged and does not impose $Q\mapsto -Q$. Thus a cubic term is allowed in three dimensions and generically produces a first-order phase transition, rather than a symmetry-enforced continuous onset. In two dimensions a symmetric traceless second-rank tensor has eigenvalues $s,-s$, hence $\mathrm{Tr}{Q}^3=0$: the cubic invariant vanishes identically in two dimensions. This is a statement about the Landau free energy; fluctuations in two dimensions require a separate treatment.

The spin raising operator and spin lowering operator areThe angular momentum commutation relations implyso $S_{\pm }|s,\sigma ⟩$ is proportional to $|s,\sigma \pm 1⟩$. Sinceits squared norm isChoosing the conventional positive phase gives

Using the spin ladder operators,and thereforeIn the ordered product basis of the tensor product of quantum systemsthe matrix representation isThe two parallel-spin states are already eigenvectors. Diagonalizing the central block gives the four energy eigenvaluesThis is the spectrum of the Two-spin Heisenberg Hamiltonian in an opposing longitudinal field.

The asserted ground-state ordering requires the antiferromagnetic case $\alpha >0$. As $B\to 0$,The corresponding state at $E_{-}$ is the unique spin-one-half singlet statewhereas the three states at the first excited energy form the spin-one-half triplet state.

Indeed, for the total angular momentum operatorone has at $B=0$The Hamiltonian then has rotational symmetry, so energy depends only on the total-spin sector. The singlet has total spin $s=0$ and multiplicity one; the triplet has $s=1$ and multiplicity $2s+1=3$. This rotational symmetry explains why the triplet energy eigenspace has dimension three.

For a uniform field the gradient term vanishes and minimization over $p=|\mathbf{p}|$ givesWhen $a<0$, the nonzero minimum haswhile its direction is arbitrary by rotational symmetry. A sudden quench creates many independently oriented domains. Their mismatches contain gradients and topological defects, whose motion and pair annihilation are slow, so the uniform state is not reached promptly.

Parametrize the loop by $r^{\prime }=R{\hat{e}}_r(\varphi )$ and $d{\ell }^{\prime }=R{\hat{e}}_{\varphi }(\varphi )d\varphi$. At the observation point $r=z{\hat{e}}_z$, the Biot-Savart law for a line current givesThe components perpendicular to the axis cancel by rotational symmetry, while the axial component of the cross product is $R^{2}d\varphi$. ThereforeThe direction is fixed by the right-hand rule and reverses when the electric current reverses. At the centre,This is the on-axis magnetic field of a circular current loop.

Landau-Ginzburg theory assumes a local gradient expansion for a slowly varying order parameter. At zero field the microscopic spin-flip symmetry sends $m\mapsto -m$, so only even powers occur. Rotational symmetry makes the leading derivative term $(\nabla m{)}^2$, while locality and analyticity near the transition organize higher powers and derivatives. Stability requires a positive gradient coefficient and, at an ordinary continuous transition, ${\alpha }_4(T_c)>0$; the coefficient ${\alpha }_2(T)$ changes sign at $T_c$.

For a separated axisymmetric harmonic function $\Psi =R(r)\Theta (\theta )$, multiplying the Laplace equation by $r^2/(R\Theta )$ givesThe two sides depend on different variables, hence both equal a constant. Put $u=\cos \theta$ in the angular equation. Since $\sin \theta {\Theta }^{\prime }=-{\sin }^2\theta d\Theta /du$, it becomes the Legendre differential equationRegularity on both parts of the polar axis restricts $\lambda =\ell (\ell +1)$, with $\ell =0,1,\dots$, and selects the Legendre polynomial $P_{\ell }(u)$. The other angular solution is singular at an endpoint.

The radial Euler differential equation isTrying $R=r^s$ gives $s(s+1)=\ell (\ell +1)$, so $s=\ell$ or $s=-\ell -1$. Superposition of these separated harmonic functions givesThe Legendre polynomials form the angular expansion for regular data with rotational symmetry about the polar axis; the radial domain and the boundary conditions determine which coefficients survive and where the expansion converges.

The disk still has only three degrees of freedom and hence three normal modes: two centre coordinates and one in-plane rotation. Five springs alter the restoring quadratic form but add no disk coordinates. The fivefold rotational symmetry makes the translational stiffness proportional to the two-dimensional identity matrix, and makes translation-rotation coupling vanish. Therefore the two translation modes are always degenerate; the rotational mode is separate, with possible accidental equality of its frequency for particular parameters. The general formulas in small oscillations of a disk attached to prestretched radial springs apply with $N=5$.

Count the pairs $(g,\omega )$ for which $g$ fixes $\omega$. Counting first by group elements gives ${\sum }_g\mathrm{Fix}(g)$, while counting first by points gives ${\sum }_{\omega }|\mathrm{Stab}(\omega )|$. On a group orbit of size $r$, orbit-stabilizer theorem makes each stabilizer have size $|G|/r$, so that group orbit contributes $|G|$ to the second sum. Dividing by $|G|$ proves Burnside's lemma:A tetrahedron has four vertices. A rotational symmetry is determined by the image of one vertex, with four possibilities, and the cyclic orientation of the three vertices around it, with three possibilities. All these choices are realizable, so its rotational symmetry group of a tetrahedron has order $12$. More explicitly, its elements are the identity, the two nontrivial rotational symmetries around each of four vertex-to-opposite-face axes, and the half-turn about each of three axes through opposite edge midpoints. Thus there is one element of order 1, eight of order 3, and three of order 2. The induced permutations of vertices are the even permutations, identifying the group with $A_4$.

Apply the group action to the set of edge colourings. A colouring is fixed by a rotational symmetry precisely when its colour is constant on each cycle of the induced edge permutation. The identity has six edge cycles and fixes $4^6$ colourings. A vertex-axis rotational symmetry has two cycles of length three, fixing $4^2$. An opposite-edge half-turn fixes the two edges on its axis as sets and exchanges the other four in two pairs, giving four cycles and $4^4$ fixed colourings. ThereforeReflections are excluded because the equivalence in the source uses rotational symmetries.

Choose the $z$-axis along the cylinder and the origin at its midpoint, so $x^2+y^2\le b^2$ and $-a\le z\le a$. Reflection symmetry puts the centroid at the origin and makes all products of inertia zero. The mass isCylindrical integration gives ${\int }_A\rho x^{2}dV={\int }_A\rho y^{2}dV=m{b}^2/4$, while the uniform height coordinate has ${\int }_A\rho z^{2}dV=m{a}^2/3$. ConsequentlyThe two equal transverse entries reflect the rotational symmetry about the cylinder axis.

The disc carries mass $M_d=2M/3$ and the thin rim carries $M_r=M/3$. About the wheel's central normal axis, add the moment of inertia of a uniform disc and the thin rim's moment of inertia:Treat the wheel as a thin planar lamina, as implied by the disc-and-thin-rim model. Interpret the requested perpendicular axis as perpendicular to the wheel's original axis and tangent to the rim: if the wheel lies in the $xy$ plane, choose the tangent line $x=R,z=0$, parallel to the $y$ diameter. By rotational symmetry, the two diametral moments of inertia are equal. The perpendicular axis theorem therefore givesThe parallel axis theorem then gives the tangent-axis inertia of an axisymmetric lamina:The axis direction matters. If “perpendicular” were intended to mean perpendicular to the wheel's plane, an axis through a rim point would instead be parallel to the original wheel axis, giving $I_{\mathrm{rim},\mathrm{normal}}=I_z+M{R}^2=5M{R}^2/3$. The $4M{R}^2/3$ answer uses a tangent axis perpendicular to the original axis; the $5M{R}^2/3$ value belongs to the different, explicitly identified normal axis. Finite wheel thickness would require additional information for the in-plane moment of inertia.

Use the usual dumbbell geometry: the uniform thin rod joins the disc centres and is perpendicular to both disc faces. The required axis through the rod midpoint lies in a direction parallel to a diameter of each disc. For either disc, rotational symmetry and the perpendicular axis theorem giveEach disc centre lies a distance $l/2$ from the requested axis. The parallel axis theorem therefore gives a disc contribution $M{a}^2/4+M(l/2{)}^2$. The rod contributesAdd the two disc contributions and the rod's moment of inertia:Using $M{a}^2/2$ for each disc's central contribution would instead use an axis perpendicular to its face, which is parallel to the rod and is not the requested axis.

Take the rotation axis as $z$. A slice of the uniform sphere at coordinate $z$ is a disc of radius $R(z)=\sqrt{a^2-z^2}$ and mass $dm=\rho \pi (a^2-z^2)dz$, where $\rho =3M/(4\pi a^3)$ is its mass density. Applying a to the slice givesThe slices' axes coincide, so no displaced-axis correction is needed. Integrating,Thus the moment of inertia of a uniform solid sphere isThe sphere's rotational symmetry makes this valid about every central axis. A thin spherical shell would have a different mass distribution and a different moment of inertia.

The moment of inertia of a uniform thin ring about its perpendicular central axis is $M_a{a}^2$. Its two in-plane diameters are equivalent by rotational symmetry, so the perpendicular axis theorem gives $M_a{a}^2/2$ about either diameter. The central rod contribution about an axis perpendicular to its length is $M_l{l}^2/12$.

The requested axis is parallel to these central axes and passes through the whole racquet's centre of mass. Applying the parallel axis theorem separately givesAn equivalent reduced expression isBoth refer to the in-plane axis perpendicular to the handle; the ring contribution would be different for an axis normal to the racquet face.

Use the geometric cross-section shown in the source: the outer cylinder has radius $2a$ and centre $O$, and the circular cavity has radius $a$ and centre $C=a\hat{\mathbf{x}}$. Let the electric current be in the $+z$ direction. The conducting cross-sectional area is $3\pi a^2$, so the uniform electric current density is $\mathbf{J}=I\hat{\mathbf{z}}/(3\pi a^2)$.

To obtain the exterior magnetic field of a wire with an eccentric cylindrical cavity, apply the superposition principle to a full cylinder of electric current density $\mathbf{J}$ centred at $O$ and a second cylinder of electric current density $-\mathbf{J}$ centred at $C$. Their electric currents are $4I/3$ and $-I/3$. This reproduces the actual electric current density everywhere, including zero in the cavity. Outside a long circular cylinder, Ampère's law gives $\mathbf{B}={\mu }_0{I}_{\mathrm{cyl}}\hat{\mathbf{z}}\times \mathbf{r}/(2\pi r^2)$. Every point outside the outer boundary is outside both component cylinders. Consequently, writing $\mathbf{r}=x\hat{\mathbf{x}}+y\hat{\mathbf{y}}$,In Cartesian coordinates, the same magnetic flux density isThe cavity breaks rotational symmetry. Thus the magnetic flux density is generally neither constant in magnitude nor tangential on a circle centred at $O$; replacing its line integral by $2\pi rB$ would be unjustified. Its line integral is nevertheless ${\mu }_{0}I$, as Ampère's law requires. At distances $r\gg a$, the leading term becomes ${\mu }_{0}I\hat{\mathbf{\varphi }}/(2\pi r)$, with corrections of order ${\mu }_{0}Ia/r^2$.

Assume the electrical conductor is nonmagnetic, with magnetic permeability ${\mu }_0$, and neglect end effects. Rotational symmetry and translational symmetry make the magnetic flux density azimuthal and constant along a coaxial circle. The magnetostatic Ampère law therefore givesUniform current density is $j=I/(\pi a^2)$, so $I_{\mathrm{enc}}(r)=j\pi r^2=I{r}^2/a^2$. ThusThe direction follows the right-hand rule around the electric current. For a uniform linear magnetic material, replace ${\mu }_0$ by its magnetic permeability.

orbital angular momentum generates spatial rotations and is $\mathbf{L}=\mathbf{r}\times \mathbf{p}$. With $\mathbf{p}=-i\hslash \nabla$, its components satisfyThus $L^2$ and one component, conventionally $L_z$, have simultaneous eigenstates, while all three components cannot generally have sharp values. On ordinary single-valued spatial wavefunctions,The spherical harmonics form an orthonormal angular basis. The integer restriction is specific to orbital wavefunctions; half-integer spin is an intrinsic degree of freedom, not the angular dependence of a scalar spatial wavefunction.

The angular momentum ladder operators obeyThe ladder ends at $m=\pm \ell$, producing $2\ell +1$ states. A state with specified $m$ has a sharp projection $\hslash m$, but the angular momentum magnitude is $\hslash \sqrt{\ell (\ell +1)}$, not $\hslash \ell$. No classical picture of a vector with simultaneously fixed Cartesian components reproduces all these measurement properties.

In a central potential, rotational symmetry gives $[H,L_i]=0$. separation of variables yields $\psi (\mathbf{r})=R_{\ell }(r)Y_{\ell m}(\theta ,\varphi )$. The radial Schrödinger equation includes a centrifugal potential ${\hslash }^2\ell (\ell +1)/(2m{r}^2)$, which suppresses small-radius amplitude for nonzero $\ell$. The energy is independent of $m$ when rotational symmetry is intact, but need not be independent of $\ell$. A magnetic field can break that degeneracy. Spatial inversion gives $Y_{\ell m}(-\hat{\mathbf{r}})=(-1{)}^{\ell }{Y}_{\ell m}(\hat{\mathbf{r}})$, connecting orbital angular momentum to parity and to the even/odd rotational restriction for the hydrogen spin isomers.

Set $\hat{B}={\hat{a}}_x^{†}{\hat{a}}_y-{\hat{a}}_y^{†}{\hat{a}}_x$. Taking the Hermitian adjoint reverses the order of products, so ${\hat{B}}^{†}=-\hat{B}$. Complex conjugation also reverses the sign of $-i$, and thereforeThus $\hat{Q}$ is a self-adjoint operator on the usual oscillator states.

Using the coefficients in part (a), products containing two position operators or two momentum operators cancel in the difference. Cross-coordinate operators commute, leavingHenceIt is the component of orbital angular momentum perpendicular to the plane, with units $\mathrm{kg}{\mathrm{m}}^2{\mathrm{s}}^{-1}$ (equivalently $\mathrm{J}\mathrm{s}$). The rotational symmetry of the two-dimensional isotropic harmonic oscillator makes this a conserved observable; directly, each term in $\hat{B}$ preserves $N_x+N_y$, so $[\hat{H},\hat{Q}]=0$.

An inverse-square point-source field has constant flux through enclosing spheres because their area grows as $r^2$. In electrostatics, Gauss's law is ${\oint }_S\mathbf{E}\cdot d\mathbf{S}=Q_{\mathrm{enc}}/{\epsilon }_0$, or $\nabla \cdot \mathbf{E}={\rho }_e/{\epsilon }_0$. For a spherical charge distribution, rotational symmetry makes the exterior field radial and constant in magnitude on a concentric sphere, giving $\mathbf{E}=Q\hat{r}/(4\pi {\epsilon }_0{r}^2)$. Symmetry is needed to take the field outside the surface integral; Gauss's law alone does not make an arbitrary charge distribution spherical.

For gravity the corresponding relations are ${\oint }_S\mathbf{g}\cdot d\mathbf{S}=-4\pi G{M}_{\mathrm{enc}}$ and $\nabla \cdot \mathbf{g}=-4\pi G{\rho }_m$. A spherical mass has exterior field $\mathbf{g}=-GM\hat{r}/r^2$. Inside a uniform sphere, $M_{\mathrm{enc}}(r)=4\pi {\rho }_m{r}^3/3$, so $\boxed{\mathbf{g}=-(4\pi G{\rho }_m/3)r\hat{r}}$. Inside an empty spherical shell the enclosed mass is zero and symmetry gives zero field. These examples distinguish an inverse-square exterior field from a field inside distributed matter.

The law also treats other symmetries, but their field-distance dependence differs: an infinitely long line charge has cylindrical rather than spherical spreading and a $1/r$ field. Flux conservation and geometry must both be used, rather than assuming every Gaussian surface leads to an inverse-square answer.

Let $I_{\perp }$ denote the moment of inertia normal to the disc and $I_d$ that about any central diameter. Rotational symmetry makes the two perpendicular diameter values equal. By the perpendicular axis theorem, $I_{\perp }=2I_d$. The intended angular impulse has magnitude $J=I_{\perp }{\omega }_0$; delivering that magnitude about a diameter instead gives $J=I_d\omega$. ThusThese are principal axes, so angular momentum is parallel to angular velocity in both cases. Equal delivered angular impulse means equal magnitude of the change in angular momentum, not equal rotational kinetic energy; the actual rotational kinetic energy is twice the intended value.