The path signature collects the ordered iterated Riemann-Stieltjes integrals of a continuous path of bounded variation. Its truncation belongs to the free step-N nilpotent Lie group. It multiplies under concatenation by the Chen identity, is independent of orientation-preserving reparametrisation of the whole path, and records more than its endpoint displacement.

The Stratonovich enhancement of Brownian motion is a geometric $p$-rough path for every $p>2$. Its second level consists of Lévy area and the prescribed symmetric tensor. Its increments satisfy the Chen identity and have finite Hölder rough-path norms at every exponent below $1/2$.

A rough differential equation interprets a differential equation using the higher-order data of its driving rough path. For $2<p<3$, its local expansion includes $V_i(Y_s)X_{s,t}^i+D{V}_j(Y_s)V_i(Y_s){\mathbb{X}}_{s,t}^{ij}$, with a remainder controlled at order greater than one. For a smooth driver it is the usual controlled ordinary differential equation; for enhanced Brownian motion it is the corresponding Stratonovich equation.

A control is a continuous nonnegative function on ordered pairs of times, zero on the diagonal and superadditive as displayed. It measures the size of increments of a rough path, typically through $|X_{s,t}^{(k)}|\le \omega (s,t{)}^{k/p}$. Continuous finite variation gives suitable controls, which let local estimates be stated without tying them to a particular time parametrization.

Fourier transform microwave spectroscopy excites gaseous molecules with a pulse of electromagnetic radiation, then records the emitted time-dependent electric field. A Fourier transform extracts transition frequencies from the decaying signal. In a cavity implementation, two facing mirrors form a resonator; tuned excitation, receiver electronics and digitization replace the molecule-counting readout of molecular beam electric resonance. The pulsed radiation prepares quantum coherence between rotational quantum states; it does not set the whole gas spinning at one classical angular velocity.

For documented laboratory context, Evolution of Microwave Spectroscopy at the National Bureau of Standards (NBS) and the National Institute of Standards and Technology (NIST) by F. J. Lovas and others (§5, printed pp. 280–282) describes the NIST cavity instrument, phase-locked sources, mirror tuning and detector-mixer modifications. That account explains one instrument family; it does not establish the electronics of every later commercial spectrometer.

In a rotating reference frame, this apparent force acts away from the rotation axis. The corresponding centrifugal acceleration is the force per unit mass.

A mass $m$ in a rotating reference frame with varying angular velocity experiences this apparent force. Dividing it by $m$ gives the Euler acceleration.

For a uniformly rotating incompressible fluid of constant mass density, pressure and conservative body forces combine as the gradient of this scalar. The centrifugal term has a negative sign. Pressure enforces incompressible flow, whereas the velocity-dependent Coriolis force remains explicit and cannot be included in a scalar position potential. Adding a spatially constant function of time to $\psi$ does not change the motion.

These roots are taken in an algebraic closure, not merely in ${\mathbb{F}}_q$. If the characteristic of a field is $p$ and $n=p^{a}m$ with $p\nmid m$, then $X^n-1=(X^m-1{)}^{p^a}$; hence there are $m$ distinct roots. They form a cyclic group by the finite multiplicative subgroup of a field is cyclic theorem. In particular they have order $n$ when $\gcd (n,q)=1$.

Let $S$ be a finite multiset of nonzero complex numbers such that taking the $m^2$th powers permutes it for every positive integer $m$. Taking $m=2$ and iterating the resulting permutation shows that each $z\in S$ obeys $z^{4^r}=z$ for some $r\ge 1$, hence is a root of unity. Let $d$ be a common multiple of their orders. The $d^2$th powers are all one, and the assumed permutation then forces every original element to be one. For a unitary matrix, the spectral theorem for normal operators further forces the matrix itself to be the identity.

A square root of unity is an element $x$ of a ring with identity satisfying $x^2=1$. In a field of characteristic different from two, the only possibilities are $1$ and $-1$: the difference of squares identity $(x-1)(x+1)=0$ has no other roots because a field has no zero divisors. In a general ring more solutions may occur; modulo eight the four residue classes $1,3,5,7$ all have square one.

For a qubit, use the dimensionless Pauli matrices as observables $A={\sigma }_x$, $B={\sigma }_z$. Their Pauli matrix multiplication law givesThese observables have eigenvalues $\pm 1$; the physical spin components are $S_j=(\hslash /2){\sigma }_j$, where $\hslash$ is the reduced Planck constant, so their standard deviations acquire a factor $\hslash /2$.

In the computational basis, prepare $|+y⟩=(|0⟩+i|1⟩)/\sqrt{2}$, an eigenstate of ${\sigma }_y$ with eigenvalue $+1$. Direct multiplication givesBoth Pauli measurements return $+1$ and $-1$ with equal probabilities, and the Robertson uncertainty principle is saturated: $1\cdot 1=1$.

Instead prepare $|+z⟩=|0⟩$. Now the expectation values are $⟨{\sigma }_x⟩=⟨{\sigma }_y⟩=0$, $⟨{\sigma }_z⟩=1$, so $\Delta {\sigma }_x=1$, $\Delta {\sigma }_z=0$. The Pauli measurement of ${\sigma }_z$ is certain, and both sides of the bound vanish. The commutator is still nonzero as an operator: noncommuting observables do not guarantee a positive lower bound independent of the quantum state. This example is useful for distinguishing the algebra of the observables from the expectation values in a particular preparation.

Real chemical bonds can stretch as molecular rotation increases. The resulting larger moment of inertia lowers the energy eigenvalues relative to a rigid rotor. Within one vibrational quantum state, a leading effective correction isBoth $B_v$ and $D_v$ are in frequency units here; $v$ labels the vibrational quantum state. For positive $D_v$, the adjacent-line spacing becomesso the departures grow with $J$. Molecular vibration also changes the effective rotational constant: even the lowest vibrational quantum state does not have motionless atomic nuclei.^[ref]

The quartic correction is a limited expansion, not a model to extrapolate to arbitrary $J$. It is small relative to the rigid contribution only while $(D_v/B_v)J(J+1)\ll 1$; eventually higher terms, changed internal motion, or chemical bond breaking matter. In the synthetic plot, the largest upper-state label is $J=5$, giving a stipulated correction ratio of $0.03$. No universal maximum valid $J$ follows without a particular molecule and precision target.

The quantum numbers also have limits of meaning. Here $J$ describes spin-free rotation. In a molecule with electronic spin, rotation alone is commonly labelled $N$, while $J$ includes electronic angular momentum; including nuclear spin introduces additional total-angular momentum labels and resolved splittings. An applied electric field can mix different field-free $J$ states through the Stark effect. For identical atomic nuclei, the Pauli exclusion principle can restrict combinations of rotational and nuclear-spin states. The simple $2J+1$ degeneracy and every-integer-$J$ ladder therefore describe the stated idealization, not all observed sublevels.^[ref]

For a diatomic molecule, let $m_1,m_2$ be the two masses and $r$ their fixed separation. The distances from the centre of mass are $r_1=m_{2}r/(m_1+m_2)$ and $r_2=m_{1}r/(m_1+m_2)$. A perpendicular axis through the centre of mass therefore has moment of inertiaThe reduced mass $\mu$ here is not the electric dipole moment, for which we use $\mathbf{d}$. For a longer linear molecule, replace $\mu r^2$ by ${\sum }_i{m}_i{\rho }_i^2$, where ${\rho }_i$ is each distance along the axis from the centre of mass. In the point-mass idealization, the axial moment of inertia is zero; turning about that axis changes no nuclear positions, so no independent axial rotational degree of freedom is included.

For the field-free linear rigid rotor, assume a fixed internal quantum state with no electronic angular momentum or electronic spin, and omit couplings to nuclear spin. The rotational Hamiltonian is ${\hat{H}}_{\mathrm{rot}}={\hat{\mathbf{L}}}^2/(2I)$. Its orientation wavefunctions are spherical harmonics $Y_{JM}$, and its angular momentum eigenstates satisfywhere $J=0,1,2,\dots$ and $M=-J,-J+1,\dots ,J$. Each $E_J$ belongs to $2J+1$ angular momentum eigenstates distinguished by the magnetic quantum number $M$, an instance of degenerate energy levels. The Planck constant is $h=2\pi \hslash$; $B$ is the frequency-valued rotational constant. Notice that the energy eigenvalues scale as $J(J+1)$, not $J^2$.

Subtracting consecutive energy eigenvalues givesThus increasing level gaps produce a regularly spaced series of lines at $2B,4B,6B,\dots$, with spacing $2B$. Allowed positions do not guarantee that every line is visible: electric-dipole selection rules, occupation of the initial quantum state, and instrument response also matter.

The Lagrange, Euler, and Kovalevskaya tops are three integrable cases of fixed-point rigid body dynamics. The Euler top has no external torque; the Lagrange top has axial symmetry and its center of mass on the symmetry axis; the Kovalevskaya top has two equal principal moments twice the third and its center of mass in the corresponding principal plane.

In principal body coordinates, Euler's equations express angular momentum balance as $I\dot{\omega }+\omega \times (I\omega )=\tau$, with applied torque $\tau$. Setting the torque to zero gives the following torque-free specialization.

Each massless rod has endpoint masses $m$ at distance $\ell$ from its centre, hence total mass $2m$ and moment of inertia $2m{\ell }^2$. In a symmetric opposite-end collision with centres approaching at speed $V$, a horizontal impulse $J$ changes each centre speed by $J/(2m)$ and each contact-point speed by $J/m$. The coefficient of restitution gives $2J/m-2V=2eV$. Both rods rotate in the same sense for the staggered vertical configuration; their centre speeds after impact are $V(3-e)/4$.

// Target: electromagnetism.bigb

Two identical uniform rigid bodies of rod shape, each of length $2l$ and mass $M$, form a symmetric arch with a free common hinge and both outer ends sliding without friction on a horizontal floor. At angle $\theta$, each center has speed $l|\dot{\theta }|$, giving total kinetic energy $4M{l}^2{\dot{\theta }}^2/3$. The total gravitational potential energy is $2Mgl\sin \theta$. Conservation of energy from rest at $\alpha$ gives the displayed hinge impact speed. Unlike fixing one outer endpoint, allowing both endpoints to slide keeps the arch center horizontally fixed. The floor reaction on each rod is $Mg[1+9{\sin }^2\theta -6\sin \alpha \sin \theta ]/4$, nonnegative throughout this fall.

For a torsion-free linear connection on a manifold, the commutator of two covariant derivatives obeys the tensor-product Leibniz rule: mixed first-derivative terms cancel after antisymmetrizing. It vanishes on scalar fields, so ${\Delta }_{ab}(fQ)=f{\Delta }_{ab}Q$; its action is therefore pointwise tensorial. Its vector action defines the Riemann curvature tensor. Applying the commutator to a scalar contraction proves that every covariant tensor index contributes minus the corresponding curvature action, while every contravariant index contributes plus. Torsion-free does not by itself imply metric compatibility.

The first Bianchi identity follows by applying cyclic curvature commutators to the gradient of an arbitrary scalar field: its symmetric Hessian makes the third-derivative terms cancel, leaving $R_{abc}{}^d+R_{bca}{}^d+R_{cab}{}^d=0$. For the second Bianchi identity, write the vector-bundle derivative as $D_a={\partial }_a+{\Gamma }_a$, so $[D_b,D_c]={\mathcal{R}}_{bc}$. The operator Jacobi identity gives the cyclic sum of ${\partial }_a{\mathcal{R}}_{bc}+[{\Gamma }_a,{\mathcal{R}}_{bc}]$ equal to zero. Restoring the connection terms on the two curvature-form indices gives precisely ${\nabla }_a{R}_{bcd}{}^e+{\nabla }_b{R}_{cad}{}^e+{\nabla }_c{R}_{abd}{}^e=0$: the additional terms cancel because the connection is torsion-free and curvature is antisymmetric in its first pair.