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The wave phase labels a physical stage of an oscillation at a spacetime event. Treating it as a Lorentz scalar makes its gradient $K_{\mu }={\partial }_{\mu }\Phi$ a covector; raising the index gives the four-wavevector $K^{\mu }=(\omega /c,\mathbf{k})$. A boost along $x$ gives ${\omega }^{\prime }=\gamma (\omega -v{k}_x)$ and $k_x^{\prime }=\gamma (k_x-v\omega /c^2)$. Thus angular frequency and wavevector transform together rather than independently.

A series circuit comprising electrical resistance $R>0$ and inductance $L>0$, placed in a parallel circuit with capacitance $C>0$, has electrical admittance $Y=(R+i\omega L{)}^{-1}+i\omega C$. Vanishing imaginary part at nonzero angular frequency gives the displayed condition, requiring $L>R^{2}C$. It defines zero voltage-current phase difference, which need not coincide exactly with the maximum of impedance magnitude when loss is present. At this frequency the electrical impedance is real and equals $L/(RC)$.

Let $m_z={\int }_V{M}_{z}dV$ be the axial magnetic dipole moment, where $M_z$ is the axial magnetization. Write $J_z$ for the internal angular momentum associated with that magnetic dipole moment, and $I\dot{\theta }$ for the body's rotational angular momentum, with moment of inertia $I$. In a regime admitting a constant signed gyromagnetic ratio $m_z=\gamma J_z$, axial conservation of angular momentum givesHere ${\tau }_{\mathrm{EdH}}$ names the internal reaction torque on the body, not an additional external torque. The external term includes the suspension. The model assumes that the changing internal angular momentum is transferred to the measured mechanical motion, with other storage or escape channels negligible on the observation timescale. A changing or non-collinear mixture of orbital angular momentum and spin angular momentum need not admit a single constant $\gamma$.

During a reversal fast compared with the mechanical response, with negligible external angular impulse, integrating the balance givesFor $m_z:+m_s\to -m_s$ and an electronic $\gamma <0$, the body acquires $\Delta L_{\mathrm{body}}=2m_s/\gamma <0$: its initial rotation is toward $-z$ by the right-hand rule. Reversing the reversal reverses the impulse. Over a complete periodic cycle, $\Delta m_z=0$, so the net reaction angular impulse vanishes; a resonant alternating response does not imply steadily accumulating angular momentum.

The orbital angular momentum prediction can be derived without a material model. A charge $q$ traveling around a circle of radius $r$ at angular frequency $\omega$ has loop electric current $q\omega /(2\pi )$ and magnetic dipole moment $m=q\omega r^2/2$. Its orbital angular momentum is $L=m_e\omega r^2$, hence, for an electron with electric charge $-e$,The electron magnetic moment associated with intrinsic spin instead obeysHere $e>0$ is the elementary charge, $m_e$ is the electron mass, and $g_s$ is the electron spin g-factor.^[ref] Thus the same change of magnetic dipole moment corresponds to about half as much transferred angular momentum in the pure-spin model as in the orbital angular momentum model. This factor of two is a relation between magnetic dipole moment and angular momentum, not evidence that an electron is literally a rotating charged sphere. In a material with both contributions, the inferred coefficient is a material-dependent magnetomechanical gyromagnetic ratio.

The slow radial normal mode of a Penning trap. Its angular frequency arises from the combined static electric field and magnetic field; it is distinct from the driven Paul trap micromotion.

For fixed spring constant $k$ and linear damping coefficient $b$, replacing mass $m$ by $\lambda m$ changes the exponential envelope rate from $\gamma =b/(2m)$ to $\gamma /\lambda$. The damped angular frequency becomes ${\omega }_{d,\lambda }^2=({\omega }_d^2+{\gamma }^2)/\lambda -{\gamma }^2/{\lambda }^2$. Thus envelope decay times scale exactly as mass, while the familiar inverse-square-root frequency scaling is a weak-damping approximation. A constant gravitational force shifts equilibrium without changing either rate.

Three consecutive masses $m,M,m$ joined by four identical springs of constant $k$ between fixed walls have an antisymmetric normal mode with zero central displacement. The two adjacent spring forces cancel on the central mass, while each outer mass experiences restoring force $-2kx$. This proves that the displayed angular frequency is independent of $M$ for every positive central mass. The two symmetric modes satisfy $(2k-m{\omega }^2)(2k-M{\omega }^2)-2k^2=0$ and generally do depend on $M$. For $M=m$, all three mode shapes are $(1,\sqrt{2},1)$, $(1,0,-1)$ and $(1,-\sqrt{2},1)$, with squared frequencies $(2-\sqrt{2})k/m$, $2k/m$, $(2+\sqrt{2})k/m$.

For the dispersion relation $m\omega =k^2$, take two plane waves with wavenumbers $\bar{k}\pm \delta k$ and corresponding angular frequencies $\bar{\omega }\pm \delta \omega$. Adding their two relations proves the title identity, while subtracting gives $\delta \omega /\delta k=2\bar{k}/m$. Thus the exact two-tone envelope speed differs from twice the midpoint carrier phase velocity by a second-order bandwidth correction. In the narrow-band limit the derivative group velocity is exactly twice the phase velocity on the central branch.

A monochromatic wave has a single angular frequency. Its sinusoidal components can have different wave amplitudes and wave phases while sharing that angular frequency, as in the polarization of waves examples.

A motional sideband is a transition that changes internal excitation and a motional quantum number together. Define the internal energy gap as $\hslash {\omega }_0>0$. A carrier drives $|g,n⟩\leftrightarrow |e,n⟩$; the first red sideband drives $|g,n⟩\leftrightarrow |e,n-1⟩$; the first blue sideband drives $|g,n⟩\leftrightarrow |e,n+1⟩$. Their resonant drive angular frequencies follow by subtracting these energy levels. For a Raman transition, the drive is a beam-difference frequency with a sign fixed by the selected internal transition.

Choose a real carrier Rabi angular frequency $\Omega$ defined so that resonant carrier excitation is ${\sin }^2(\Omega t/2)$. In the Lamb-Dicke regime, isolated first-sideband Hamiltonians in a rotating frame can be written, after choosing the pulse phase, asThese formulas use a rotating-wave approximation: neglected carrier and other-sideband terms oscillate rapidly compared with the retained coupling. The drive's effective width and sideband coupling must be small compared with $\nu$; sufficient suppression of off-resonant carrier excitation must also be checked. The red interaction is a Jaynes-Cummings model with a motional quantum replacing a cavity photon. The blue interaction creates internal excitation and a motional quantum together; the external laser supplies energy. The driven-system rotating-frame Hamiltonian alone is not the total laboratory energy.

A system's natural frequency describes a free oscillation in the absence of external driving. Distinguish the undamped angular frequency ${\omega }_0$ from a damped oscillation frequency and from the driving frequency that maximizes a particular measured response.

For a closed two-level system with spontaneous emission population decay $P_e(t)=P_e(0)e^{-\Gamma t}$ and no additional dephasing, the natural linewidth has full width at half maximum $\Gamma$ in angular frequency, or $\Gamma /(2\pi )$ in hertz. The excited-state lifetime is $1/\Gamma$.

A Newtonian binary chirp is a leading-order model in which loss of orbital energy to gravitational radiation makes an almost circular binary system shrink and its gravitational-wave frequency rise. Here is a derivation with explicit SI units. Let the constant component masses $m_1,m_2$ be in kilograms, $M=m_1+m_2$, $\mu =m_1{m}_2/M$, separation $a$ in metres, and orbital angular frequency $\Omega$ in radians per second. Newtonian gravity, circular motion and the quadrupole formula giveThe factor $\pi$ rather than $2\pi$ appears because the dominant mass quadrupole moment repeats twice per orbit. Since $dE/da>0$, radiative energy loss implies $\dot{a}<0$, henceThe quantity $G\mathcal{M}/c^3$ is a time in seconds. Thus $K$ has units ${\mathrm{s}}^{5/3}$ and $K{f}^{11/3}$ has units ${\mathrm{s}}^{-2}$, as required for frequency change per second. This model assumes weak gravity, slow relative motion, negligible eccentricity of a conic, no spin corrections and adiabatic evolution over an orbit. It does not describe merger or ringdown.

Separating variables gives a finite formal coalescence time $t_c$ and accumulated wave phase $\Phi$:Indeed $d\Phi /dt=2\pi f$; the formal divergence at $t_c$ signals extrapolation outside the approximation. Over a finite frequency interval, the number of gravitational wave cycles is $N=3(f_0^{-5/3}-f_1^{-5/3})/(5K)$. This accumulated phase is why a chirp mass can be constrained even when distance and inclination remain entangled.

Choose one driving electric field component and one responding electric polarization component. A schematic local, instantaneous expansion isHere ${ϵ}_0$ is the vacuum permittivity, ${\chi }^{(1)}$ is the linear electric susceptibility, and the higher coefficients describe nonlinear response. In SI, $P$ has units $\mathrm{C}{\mathrm{m}}^{-2}$, $E$ has units $\mathrm{V}{\mathrm{m}}^{-1}$, and ${\chi }^{(2)}$ has units $\mathrm{m}{\mathrm{V}}^{-1}$. The selected scalar coefficient depends on polarization of waves and crystal orientation; it is not a universal material number independent of geometry.

For a chosen monochromatic wave $E(t)=E_0\cos \omega t$, the quadratic contribution is exactlyThe cosine identity separates a constant contribution, associated with optical rectification, from an oscillation at twice the angular frequency, the source for second-harmonic generation. A term oscillating at $2\omega$ is a local driving source; efficient outgoing light additionally requires propagation and phase matching.

For a material with optical dispersion, the scalar instantaneous expansion is only a teaching model. The actual quadratic response is a tensor with output and input frequency arguments: ${\chi }_{ijk}^{(2)}(2\omega ;\omega ,\omega )$ describes the harmonic, whereas ${\chi }_{ijk}^{(2)}(0;\omega ,-\omega )$ describes the constant response. Their coefficients need not coincide. The cosine calculation establishes which frequencies are available, without assigning measured coefficients to any crystal.

A normal mode, also called an eigenmode, is a pattern of free oscillation in a linear system whose relative displacements stay fixed while a single temporal factor changes their size and sign. For an undamped stable vibrational mode,Here $\varphi$ is the mode shape, $A$ sets the amplitude, $\omega$ is the eigenfrequency expressed as an angular frequency, and $\phi$ is the oscillation phase. Every snapshot is a scalar multiple of the same mode shape; neighbouring regions may move in opposite directions. Normal modes of a double pendulum show that normal modes also describe coupled discrete systems, beyond strings. Their shapes are eigenvectors, rather than spatial eigenfunctions. ^[ref]

For the ideal wave equation on a string, take a uniform string of length $L$ with constant tension $\tau$ and linear mass density $\mu$. Assume small transverse slope, negligible bending stiffness, no damping, and fixed endpoints. Its wave speed is $c=\sqrt{\tau /\mu }$, and the nonzero normal modes areThe angular frequency ${\omega }_n$ is in radians per second; the frequency $f_n$ is in hertz. The wavelength is $2L/n$. The mode shape has $n$ loops separated by nodes of a standing wave, including both fixed endpoints. ^[ref]

As a worked example, choose $L=1\mathrm{m}$, $\tau =1\mathrm{N}$, and $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$. Then $c=1\mathrm{m}{\mathrm{s}}^{-1}$ and $f_1=0.5\mathrm{Hz}$, $f_2=1\mathrm{Hz}$, $f_3=1.5\mathrm{Hz}$. Their oscillation periods are $T_1=2\mathrm{s}$, $T_2=1\mathrm{s}$, and $T_3=2/3\mathrm{s}$. These are deliberately simple model parameters, rather than measurements of a musical instrument. The plot divides displacement by $A$ so that the three mode shapes can be compared, and uses each mode's own oscillation period for its snapshots.

For $\phi =0$, at $t=T_n/4$ every point has zero displacement, but most points have nonzero velocity. Only zeros of the mode shape are permanent nodes of a standing wave. At $t=T_n/2$ the original mode shape has reversed sign, and at $t=T_n$ it returns. The synchronized conversion between kinetic energy and potential energy is shown in normal-mode energy exchange.

The short derivation uses separation of variables: substituting $u=\varphi q$ into the wave equation gives ${\varphi }^{\prime \prime }+k^2\varphi =0$ and $q^{\prime \prime }+c^2{k}^{2}q=0$. The fixed-end boundary conditions remove the cosine spatial term and require $\sin (kL)=0$, hence $k=n\pi /L$ and ${\omega }_n=ck$. Different geometries and boundary conditions give different eigenfunctions; a sine mode shape is a property of this uniform fixed string, not a universal form of an eigenmode.

The usefulness is that a complicated vibration becomes a collection of harmonic oscillators. For sufficiently regular fixed-end initial data, the Fourier sine series givesThe initial displacement and velocity determine $a_n$ and $b_n$. This is the superposition principle in a Fourier sine basis, and modal energy distribution measures how strongly each normal mode is excited. The striking simplification is that each component keeps its identity while their sum can continually change shape. ^[ref]

These independent normal modes belong to the linear model. At larger slopes, changing tension introduces nonlinear effects and can couple the modes. A linearly damped string still admits a Fourier sine basis, but its temporal factors decay and need not oscillate; its energy dissipation identity for a linearly damped string replaces conservation of energy. More general damping need not leave undamped mode shapes independent.

For the undamped first normal mode of a fixed string, start at maximum displacement with $u(x,t)=A\sin (\pi x/L)\cos ({\omega }_{1}t)$. In the small-slope wave equation on a string, its kinetic energy and excess potential energy areThe potential energy comes from the extra length under tension: $\sqrt{1+u_x^2}-1\simeq u_x^2/2$. It depends on slope, so a node of a standing wave need not have zero local potential energy density. These are physical energies, with the mass and tension factors restored to the normalized wave energy. ^[ref]

Integrating ${\sin }^2(\pi x/L)$ and ${\cos }^2(\pi x/L)$ over the string gives $L/2$ in both cases. Using $\mu {\omega }_1^2=\tau {\pi }^2/L^2$ yieldsFor the model $\mu =1\mathrm{kg}{\mathrm{m}}^{-1}$, $\tau =1\mathrm{N}$, $L=1\mathrm{m}$, and $A=0.010\mathrm{m}$, the angular frequency is ${\omega }_1=\pi \mathrm{rad}{\mathrm{s}}^{-1}$, the frequency is $0.5\mathrm{Hz}$, and $E\simeq 2.4674\times 10^{-4}\mathrm{J}$. The maximum slope $A\pi /L\simeq 0.0314$ is small. At $t=0$ all the energy is potential energy; at $t=T_1/4=0.5\mathrm{s}$ the string is straight and all the energy is kinetic energy. At $t=T_1/2=1\mathrm{s}$ the displacement is reversed and all the energy is again potential energy. The energy curves repeat twice during one displacement cycle, while conservation of energy keeps their sum constant.

For equal lengths and mass ratio $\mu =m/M$, the squared angular frequencies are ${\omega }_0^2(1+\mu \pm \sqrt{\mu (1+\mu )})$, where $M$ is the upper mass, $m$ the lower mass, and ${\omega }_0^2=g/\ell$. The equal-mass case $\mu =1$ admits particularly simple mode shapes.

Use the ideal model of the double pendulum, with both angles measured from the downward vertical. Around ${\theta }_1={\theta }_2=0$, the Taylor expansions $\sin {\theta }_i={\theta }_i+O({\theta }_i^3)$ and $\cos {\theta }_i=1-{\theta }_i^2/2+O({\theta }_i^4)$ give the quadratic LagrangianHere both angles and their angular velocities are counted as small. Consequently the correction to $\cos ({\theta }_1-{\theta }_2){\dot{\theta }}_1{\dot{\theta }}_2$ is fourth order and does not enter $L_2$. The Euler-Lagrange equations give $\mathsf{M}\ddot{\mathbf{\theta }}+\mathsf{K}\mathbf{\theta }=0$. The mass matrix is not diagonal: an angular motion of the upper rod carries the lower mass too.^[ref]

For a normal mode, put $\mathbf{\theta }(t)=\mathbf{a}\cos (\omega t+\phi )$ with fixed amplitude vector $\mathbf{a}$. The resulting generalized eigenvalue problem for small oscillations is $\mathsf{K}\mathbf{a}={\omega }^2\mathsf{M}\mathbf{a}$. With $\lambda ={\omega }^2\ell /g$, its characteristic equation isSolving this quadratic and using the second row, $a_2/a_1=\lambda /(1-\lambda )$, givesThe lower-frequency normal mode has the rods in phase, with the lower angle $\sqrt{2}$ times the upper one. The higher-frequency normal mode has opposite signs, corresponding to an oscillation phase difference of $\pi$. These are ratios of absolute angles. The horizontal displacements are instead $x_1\simeq \ell {\theta }_1$ and $x_2\simeq \ell ({\theta }_1+{\theta }_2)$; their ratios are $1+\sqrt{2}$ and $1-\sqrt{2}$ respectively. Thus the lower mass moves horizontally opposite to the upper mass in the faster normal mode.

Both squared angular frequencies are positive. Their ratio satisfies ${\omega }_{+}/{\omega }_{-}=1+\sqrt{2}$, independently of $m$, $g$ and $\ell$. For a constructed model with $\ell =1\mathrm{m}$ and $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$, the angular frequencies are approximately $2.3972$ and $5.7874{\mathrm{s}}^{-1}$, and the oscillation periods $2\pi /{\omega }_{\pm }$ are $2.6211$ and $1.0857\mathrm{s}$. Doubling both masses leaves these unchanged; multiplying both rod lengths by four doubles both oscillation periods.

The eigenvectors are not orthogonal in the ordinary dot product: ${\mathbf{a}}_{-}^{\mathsf{T}}{\mathbf{a}}_{+}=-1$. They satisfy mass-matrix orthogonality of normal modes, sinceWriting $\mathbf{\theta }={\mathbf{a}}_{-}{q}_{-}+{\mathbf{a}}_{+}{q}_{+}$ therefore removes the cross terms from both quadratic energies. The two normal coordinates satisfy ${\ddot{q}}_{\pm }+{\omega }_{\pm }^2{q}_{\pm }=0$. This is why the normal modes are useful: one coupled system becomes two independently evolving harmonic oscillators, and any small release can be decomposed into them. At larger angles, the discarded terms in the double pendulum equations of motion matter and this exact linear superposition no longer applies.

Connect every pair among $n$ moving scalar coordinates and one fixed coordinate by identical springs of stiffness $k$. The coupling pattern is the complete graph $K_{n+1}$ with one vertex fixed. Its potential energy is $k{\sum }_{i<j}(X_i-X_j{)}^2/2$. The moving-coordinate stiffness matrix is the principal submatrix displayed in the header. For equal masses $m$, the uniform normal mode has squared angular frequency $k/m$, while every sum-zero vector has squared angular frequency $(n+1)k/m$. There are $n-1$ such degenerate modes. This follows directly from the Laplacian spectrum of a complete graph and its rank-one matrix form. Constant gravitational forces shift the equilibrium without changing the stiffness.

For a uniform rigid beam of mass $m$ and two identical vertical end springs of stiffness $k$, small vertical end displacements have kinetic energy $m({\dot{x}}_1^2+{\dot{x}}_1{\dot{x}}_2+{\dot{x}}_2^2)/6$ and displacement-dependent potential energy $k(x_1^2+x_2^2)/2$. The mass and stiffness matrices are $\mathsf{M}=(m/6)\left(\begin{array}{cc}2& 1\\ 1& 2\end{array}\right)$ and $\mathsf{K}=k1$. Their generalized eigenvalue problem gives symmetric translation $(1,1)$ and antisymmetric rocking $(1,-1)$ at the displayed angular frequencies. The centre coordinate contains only the symmetric mode. Neglect spring inertia, damping and finite-angle geometric corrections.

A cart of mass $m$ moves horizontally on a spring of stiffness $k$, carrying a simple pendulum of bob mass $M$ and length $L$. Let $x$ be cart displacement and $\varphi$ the small angle from downward vertical. The quadratic Lagrangian isThe kinetic cross term comes from the bob's horizontal velocity $\dot{x}+L\dot{\varphi }$ to first order. The generalized eigenvalue problem has mass matrix $\left(\begin{array}{cc}m+M& ML\\ ML& M{L}^2\end{array}\right)$ and restoring matrix $\mathrm{diag}(k,MgL)$. Their characteristic determinant gives the displayed polynomial in ${\omega }^2$. For $m=M=L=g=1,k=2$, the squared angular frequencies are $2\mp \sqrt{2}$, with mode shapes $(x,\varphi )=(1,\pm \sqrt{2})$. The bob and cart move in phase in the lower normal mode and in opposite phases in the higher normal mode.

Consider three beads of masses $m,m,M$ on a horizontal hoop, connected in cyclic order by identical springs constrained to follow the arcs. Their common natural length affects only a constant in the potential energy, since the sum of the three arc lengths is fixed. About the equally spaced configuration, angular displacements $q_i$ haveThe generalized eigenvalue problem has the rigid rotation $(1,1,1)$ as a zero mode. Exchanging the equal masses separates an antisymmetric normal mode $(1,-1,0)$, with squared angular frequency $3k/m$, from a symmetric normal mode $(1,1,-2m/M)$, with squared angular frequency $k/m+2k/M$. The last mode shape has zero total angular momentum. The hoop radius cancels from both the kinetic energy and the restoring quadratic form. These formulas use arc-constrained springs; straight chord springs have a different potential energy.