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For two inertial frames with equal speed in a laboratory, a photon is measured at equal angular frequency by their comoving observers exactly when its laboratory propagation unit vector $\mathbf{n}$ obeys the displayed condition. Indeed, measured angular frequency is $-k\cdot U=\gamma {\omega }_{\mathrm{lab}}(1-\mathbf{n}\cdot \mathbf{v}/c)$, and equal laboratory Lorentz factors cancel. Thus the permitted directions form a great circle on the unit sphere. In the plane of the two laboratory velocities, only the internal angle bisector line remains. The velocity information alone fixes neither direction along that line nor coplanarity of the ray.

A bridge driven between top and bottom nodes has equal resistors on its left arms, upper-right electrical impedance $Z=R_x/(1+i\omega R_x{C}_x)$ and lower-right electrical impedance $W=R_v+1/(i\omega C)$. A zero detector electric current requires $Z=W$. Equating real parts and imaginary parts gives $R_v=R_x(1-C_x/C)$ and ${\omega }_0^{-2}=R_x^2{C}_x(C-C_x)$, requiring $0<C_x<C$. Its open-circuit right-minus-left junction voltage is $V_0(W-Z)/[2(W+Z)]$, whereas a connected ideal ammeter has zero terminal voltage at every angular frequency. This distinction separates a null-current measurement from an unloaded voltage measurement.

The equipartition theorem says that each independent quadratic term in a Hamiltonian described by classical mechanics has mean energy $k_{B}T/2$ in a canonical ensemble. Here $k_B$ is the Boltzmann constant and $T>0$ the absolute temperature. Count terms in the energy, rather than vaguely counting everything that can move: one harmonic oscillator has one position degree of freedom but TWO quadratic terms, one in momentum and one in displacement.

A precise sufficient setting is $H(y,\xi )=a{y}^2/2+H_0(\xi )$ with $a>0$, where $y$ is one of the canonical variables, $\xi$ denotes all remaining variables, and the phase space integration measure factors as $dyd\xi$. The range of $y$ is the whole real line, independently of $\xi$; there is no constraint coupling its allowed values to other variables. The partition function must be finite, and the relevant expected values must exist. The Boltzmann factor then givesThe probability density function ${\rho }_y$ is a normal distribution with variance $k_{B}T/a$. A larger coefficient narrows the fluctuations enough to cancel its greater energy cost. This is why the mean contribution is independent of mass or spring constant.^[ref]

For a proof outline, the one-variable Gaussian integral is $Z_y={\int }_{-\infty }^{\infty }{e}^{-\beta a{y}^2/2}dy=\sqrt{2\pi /(\beta a)}$, so $-{\partial }_{\beta }\ln Z_y=1/(2\beta )$. More generally, integration by parts in a canonical variable gives $⟨y{\partial }_{y}H⟩=k_{B}T$ when its boundary term $[y{e}^{-\beta H}]$ vanishes and the integrals converge. For $a{y}^2/2$ this is twice its mean energy. At a finite hard boundary the boundary term need not vanish; hard-wall correction to harmonic equipartition illustrates the resulting departure. A constraint, a nonquadratic potential energy, or a noncanonical integration measure therefore requires its own calculation. The theorem is an ensemble average, not equality of instantaneous energies, and applying it to a recorded trajectory requires equilibration and sufficient sampling.

For a classical one-dimensional harmonic oscillator, with positive mass $m$ and spring constant $k$,The origin is the equilibrium position. Its heat capacity is $k_B$ at fixed $m,k$, not $k_B/2$. Although the two mean energies agree, their measured amplitudes do not: $⟨p^2⟩=m{k}_{B}T$ and $⟨x^2⟩=k_{B}T/k$. In a harmonic optical trap, the position measurement observes the potential energy contribution; it does not measure both contributions by observing $x$ alone.^[ref]

For $N$ particles in a three-dimensional classical monatomic gas, neglect interactions and thermally accessible internal excitations. Its three translational momentum terms give internal energy $U=3N{k}_{B}T/2$, heat capacity at constant volume $C_V=3N{k}_B/2$, and heat capacity at constant pressure $C_p=C_V+N{k}_B=5N{k}_B/2$. Per mole, replace $N{k}_B$ by the molar gas constant $R$. For example, using stipulated $R=8.314\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$, the molar heat capacity at fixed volume is $12.471\mathrm{J}\mathrm{mo}{\mathrm{l}}^{-1}{\mathrm{K}}^{-1}$. These results assume a dilute, nonrelativistic ideal gas with classical translational statistics, rather than a quantum degeneracy regime.

For an ideal gas of diatomic molecules, modeled as two chemical atoms with a bond, there are three translational quadratic terms, two accessible rotational kinetic energy terms for changes of bond orientation, and one bond-stretching harmonic oscillator with two quadratic terms. Rotation about the bond axis is not an additional active mode in this linear-molecule model. The three familiar limits of heat capacity at constant volume areCorresponding molar heat capacities are $3R/2$, $5R/2$ and $7R/2$. Counting the stretching harmonic oscillator only once as $k_{B}T/2$ would give the wrong last value.^[ref]

The frozen regimes require quantum mechanics. In the simple molecular rotation model, $E_j=k_B{\theta }_{r}j(j+1)$ with state degeneracy $2j+1$, ${\theta }_r={\hslash }^2/(2I{k}_B)$, and moment of inertia $I$. For the bond-stretching quantum harmonic oscillator, ${\theta }_v=\hslash \omega /k_B$ with angular frequency $\omega$ and $\hslash =h/(2\pi )$ from the Planck constant. When ${\theta }_r\ll {\theta }_v$, the three limits above correspond to $T\ll {\theta }_r$, ${\theta }_r\ll T\ll {\theta }_v$, and $T\gg {\theta }_v$, provided translational motion remains classical. They are smooth crossovers, not abrupt switches. Real molecules can have different allowed rotational populations; at high temperature, thermal dissociation, internal excitations and nonharmonic bond motion can invalidate the fixed-molecule model before its last limit is reached.^[ref]

For a quantum harmonic oscillator with fixed $\omega$, put $z={\theta }_v/T$. Its exact mean energy and heat capacity areAt $T\gg {\theta }_v$, the mean energy approaches $k_{B}T$ and $C_{\mathrm{osc}}$ approaches $k_B$. At $T\ll {\theta }_v$, thermal excitation is exponentially suppressed and $C_{\mathrm{osc}}$ tends to zero, although the zero-point energy $\hslash \omega /2$ survives. Frozen thus means little thermal response, not zero motion or zero energy.^[ref]

A motivating measurement is thermal calibration of an optical trap. A focused laser through a microscope objective confines a bead in water; imaging or a quadrant photodetector records a position time series. Gieseler and colleagues describe a polystyrene bead of radius $1.03\mu \mathrm{m}$ at $293.15\mathrm{K}$, a $1064\mathrm{nm}$ trapping laser, and photodetector acquisition at $10^5\mathrm{Hz}$. Their recorded bead continues undergoing Brownian motion; higher trapping power produces a narrower equilibrium position distribution. Infer the centered variance, not a squared distance from an arbitrary camera origin:The normal distribution position histogram and agreement with dynamical calibration methods support the local harmonic optical trap interpretation. For a stipulated example, $T=300\mathrm{K}$ and ${\sigma }_x=20\mathrm{nm}$ imply $k\simeq 1.04\times 10^{-5}N/m=0.0104pN/nm$. This calculated scale is not an experimental reading.^[ref]

Position calibration is essential: if detector output $v$ has conversion $x=s(v-v_0)$, its physical variance is $s^2\mathrm{Var}(v)$. An unknown scale $s$ cannot be determined together with $k$ from that one variance. A complementary calibration demonstrated by Tolić-Nørrelykke and colleagues combines the power spectrum of thermal motion with the bead response to known sinusoidal motion of a translation stage. The driven response supplies an additional length scale, and their method does not require bead size, liquid dynamic viscosity or wall distance as inputs. It still uses a thermal dynamical model; agreement between methods is a consistency check, not a theory-free proof of equipartition theorem.^[ref]

The readout itself can hide fluctuations. Wong and Halvorsen used a $1064\mathrm{nm}$ optical trap and adjustable-exposure CCD imaging of a bead in water. They varied exposure and trapping power, held the bead away from the nearest wall, adjusted illumination to maintain comparable images, and measured tracking noise using a stuck bead. The measured variance fell as exposure increased; corrected results agreed with their time-averaging model. For a stationary overdamped harmonic optical trap, with optical trap relaxation time $\tau$, an exposure of length $W$ measures ${\bar{x}}_W=W^{-1}{\int }_0^{W}x(t)dt$. Its ideal noise-free variance isFor $\alpha >0$, $S(\alpha )<1$: using the blurred variance in the uncorrected formula overestimates $k$ by $1/S(\alpha )$. At $W=\tau$, the calculated factor is $e/2\simeq 1.36$. Additive tracking noise instead raises the measured variance and biases the inferred $k$ downward if left uncorrected.^[ref]

Useful controls are to record over many optical trap relaxation times, compare blocks for drift, calibrate the position scale, measure fixed-bead noise, vary exposure, and check the position histogram for departures from a normal distribution. A sparse sequence of accurate instantaneous positions can estimate an equilibrium variance; a long exposure averages away motion even if many frames are collected. Flow, appreciable laser heating, escape from the harmonic region or nonconservative forces can break the assumed thermal equilibrium. Inferring $k$ by imposing equipartition theorem is a calibration, whereas testing the relation requires an independently established stiffness and position scale. The beautiful practical connection is that microscopic randomness can set a force scale: once calibrated, a small constant applied force displaces the mean by $f/k$. The same quadratic-term counting predicts macroscopic heat capacity and makes microscopic fluctuations useful measurements.

For dimensionless Hermite index zero, the equatorial shallow-water dispersion relation factorizes as $(\omega +k)({\omega }^2-k\omega -1)=0$. The root $\omega =-k$ is generally spurious: for $v=C{e}^{-y^2/2}$, the original equations require ${\delta }^{\prime }-y\delta =i(2k-1/k)v$, where $\delta =u-\eta$. Multiplication by $e^{-y^2/2}$ and integration over the real line shows that decay at both ends requires $C=0$ unless $k^2=1/2$. At that exception the root already belongs to the physical quadratic; $u=\eta =i\omega yv$ is regular and no extra wave branch is created. The assertion concerns nonzero $k$ and angular frequency.

The extraordinary wave's electric displacement field lies in that plane and is perpendicular to the wavevector; its electric field need not be. For angle $\theta$ between the wavevector and optic axis,Thus $n_{\mathrm{ex}}(0)=n_o$ and $n_{\mathrm{ex}}(\pi /2)=n_e$.^[ref]

A short derivation starts from the existing plane wave in an anisotropic dielectric equation. Put $\mathbf{k}=(k_x,0,k_z)$ and $k_0=\omega /c$, with angular frequency $\omega$ and vacuum speed of light $c$. The $y$ component gives the ordinary wave relation $k_x^2+k_z^2=n_o^2{k}_0^2$. The determinant of the $x,z$ system givesSubstituting $k_x=n_{\mathrm{ex}}{k}_0\sin \theta$ and $k_z=n_{\mathrm{ex}}{k}_0\cos \theta$ yields the displayed directional refractive index.

Use the Driven two-level Rabi model in the ordered basis $(|e⟩,|g⟩)$, with Pauli matrices satisfying ${\sigma }_z=\mathrm{diag}(1,-1)$. Let the Rabi detuning be $\delta ={\omega }_0-\omega =2\pi ({\nu }_0-\nu )$, in angular frequency, while ${\nu }_0$ and $\nu$ are ordinary frequencies in hertz. Assume two identical square pulses, equal drive oscillation phases, constant positive Rabi angular frequency $\Omega$, constant $\delta$ during pulses and flight, and no quantum decoherence or loss. The rotating-wave approximation and isolation of the chosen two-level system are required, just as for the Driven two-level Rabi model. Free flight means no transverse drive; the reference oscillator continues running.

The pulse and free-flight Hamiltonian operators in the rotating frame areThe Pauli matrix multiplication law gives $(\delta {\sigma }_z+\Omega {\sigma }_x{)}^2=q^{2}I$. Thus the exact special unitary group $\mathrm{SU}(2)$ pulse matrix for this effective Hamiltonian operator, and the flight matrix, areThe sequence is $U_p{U}_f{U}_p$. Starting in $|g⟩$, matrix multiplication and the Born rule giveThis retains the tilted rotation axis and quantum phase accumulated during each finite pulse. At resonance it gives $P_{g\to e}(0)={\sin }^2(\Omega \tau )$; with no flight it reduces to the single square-pulse response for duration $2\tau$. Both checks would fail if arbitrary ideal $\pi /2$ rotations were silently substituted at nonzero Rabi detuning.

For resonant pulse area $\Omega \tau =\pi /2$, $|\delta |/\Omega \ll 1$ and $\tau \ll T$, the central fringe approachesIts half-height points are $|\nu -{\nu }_0|=1/(4T)$, so the central peak's full width at half maximum is $1/(2T)$ in hertz, or $\pi /T$ in angular frequency. This width assumes an effectively flat pulse envelope across that central peak, one flight time and unit contrast; it is neither a universal finite-pulse width nor a frequency uncertainty. Longer flight lets a smaller frequency difference accumulate the same observable quantum phase.

The two pulses here have the same sign and the plotted outcome is transfer to $|e⟩$. The earlier Ramsey interferometry example uses an inverse analysis pulse and reports return to its initial label. Specifying the measured outcome and pulse oscillation phases reconciles the two conventions. In a beam the lengths and speed determine $\tau$ and $T$; in the Quantronium Ramsey interference experiment electronics set the pulse times for a stationary circuit.

An observation interval of finite duration is not an infinitely sharp frequency selector. Combine the lock-in quadratures intoFor unwanted electric potential difference $s(t)=B\cos ((\omega +\delta )t+\psi )$, where $\delta$ is an angular frequency offset and $\omega +\delta >0$, expansion into complex exponentials gives the exact responseThe sinc function formula follows by integrating the exponential function and factoring $e^{\mathrm{i}\nu T}-1=2\mathrm{i}{e}^{\mathrm{i}\nu T/2}\sin (\nu T/2)$; its value at zero follows from a limit. The first term changes slowly after multiplication; the second oscillates at the sum frequency. It vanishes at exact matching when $T$ is an integer number of reference cycles, each lasting one period of a function, but cannot generally be discarded in a short-window calculation.

For $|\delta |\ll \omega$ and many reference cycles, the first term usually dominates: its modulus is $B|\mathrm{sinc}(\delta T/2)|$ and its complex argument includes $\delta T/2$. Thus $|\delta |T\ll 1$ gives almost no rejection. A longer observation separates closer frequencies, with the first zeros of this term at $|\delta |=2\pi /T$. Those zeros are not a brick-wall cutoff: the sinc function has nonzero lobes beyond them. At $\delta =0$ a synchronized contribution remains however large $T$ becomes.

For a numerical check, use $f_0=1\mathrm{Hz}$, $T=4\mathrm{s}$, $\psi =0$ and an unwanted frequency $f=1.125\mathrm{Hz}$. Then $(f-f_0)T=1/2$ and the exact expression giveswhereas the difference-frequency term alone gives $2/\pi \approx 0.636620$. An interference amplitude $B=0.9\mathrm{V}$ therefore contributes $0.539254\mathrm{V}$ to $|Z_T|$ in this constructed example. A nearby frequency can survive strongly even though the three-times-frequency component in lock-in harmonic rejection cancels exactly.

A practical lock-in amplifier's low-pass filter weights past products differently from this rectangular observation interval. Its time constant and roll-off determine near-reference rejection and settling time; narrowing detection slows the response.^[ref]

Angular-relaxation collision time $\tau$ distinguishes first sound ($\omega \tau \ll 1$) from zero sound ($\omega \tau \gg 1$). Collisions establish local thermodynamic equilibrium in the former.^[ref]

For the figure's stipulated law $\omega \tau =(T/T_{\times }{)}^{-2}$, cooling from $T=5T_{\times }$ to $T=T_{\times }/5$ changes $\omega \tau$ from $0.04$ to $25$, without changing the driving angular frequency. The crossover temperature is defined by $\omega \tau (T_{\times })=1$. Calling zero sound a high-frequency mode therefore means high relative to the collision rate $1/\tau$, while its excitation energy must still remain within the low-energy domain of the Fermi liquid.

Choose $\ell =1.50\mathrm{m}$, $n=4/3$, ${\lambda }_0=550\mathrm{nm}$ and $u=6.00\mathrm{m}{\mathrm{s}}^{-1}$, with $c=299792458\mathrm{m}{\mathrm{s}}^{-1}$. These are illustrative choices, not measured values or a reconstruction of Hippolyte Fizeau's data. Each beam traverses $L=3.00\mathrm{m}$ of water. The nondispersive prediction isThe first number is the offset from rest to one flow setting; the second compares equal opposite settings. The very short equivalent delay explains why interference is useful: a measurable fraction of a wavelength replaces an unattainable direct timing measurement.

The Galilean transformation would give phase velocity magnitudes $c/n\pm u$, hence $N_{\mathrm{full}}\simeq 4\ell u{n}^2/(c{\lambda }_0)$. At the chosen values this is $0.38815$ fringe from rest, larger than the special relativity prediction by $n^2/(n^2-1)=16/7$. A model with no flow-induced change of phase velocity instead gives $N=0$. The three alternatives are experimentally distinguishable even when $u/c$ is tiny.

Hippolyte Fizeau's positive displacement favoured Augustin-Jean Fresnel's partial-drag law over zero or full drag. He explicitly distinguished support for that law from establishing its underlying luminiferous aether hypothesis.^[ref] The Fizeau experiment therefore supplies a nonzero relative-motion measurement, not a null test proving an luminiferous aether's absence. The derivation above exhibits the same first-order law within special relativity, showing why agreement with the coefficient alone cannot select an physical interpretation.

Two limitations matter for precision. The velocity profile sampled by the beam need not equal the cross-sectional average inferred from volumetric flow rate. Also, the Doppler effect gives opposite rest-frame frequencies for the counterpropagating beams. With optical dispersion, the coefficient $n^2-1$ in the interference fringe formula becomeswhere the refractive index and its derivative are evaluated for stationary water at the laboratory angular frequency $\omega =2\pi c/{\lambda }_0$. A wavelength-independent $n$ therefore defines an approximation, not a precision fit to real water.^[ref] Flow reversal removes a fixed phase difference, but a mechanical or temperature change that reverses with the flow can still imitate the signal and must be controlled.

The Hessian of potential energy at a mechanical equilibrium, expressed in specified displacement coordinates. It determines restoring forces to first order. Combining it with the mass matrix gives the dynamical matrix whose eigenvalues are squared mode angular frequencies.

With uniform thermal wind $U=\Lambda z$ above a plane boundary, write the zero-interior-potential vorticity response as ${\psi }^{\prime }=\mathrm{Re}\{A{e}^{-\mu z+i(kx+ly)}\}$, where $\mu =N\sqrt{k^2+l^2}/f$ and $f>0$. Boundary vertical velocity $ϵ\cos (kx+ly-{\omega }_{0}t)$ gives $C=N^2ϵ/(f\mu )$ and natural angular frequency $\omega =\Lambda k/\mu$. An initially zero response has $A=C(e^{-i{\omega }_{0}t}-e^{-i\omega t})/[i(\omega -{\omega }_0)]$. Its continuous resonant limit is $Ct{e}^{-i\omega t}$: forcing in phase with the boundary mode supplies energy every cycle, and linear undamped theory predicts secular growth.

For a supported mass $m$, total spring constant $k$, and viscous coefficient $b$, a harmonic applied force has response $|x|=F_0/\sqrt{(k-m{\omega }^2{)}^2+b^2{\omega }^2}$. Far above the natural angular frequency, light damping gives $|x|\simeq F_0/(m{\omega }^2)$. For a prescribed displacement bound $x_{\max }$, this estimates $m\gtrsim F_0/({\omega }^2{x}_{\max })$, but the finite spring term can make the estimated equality slightly violate the bound. If $F_0/x_{\max }>b\omega$, the high-mass exact threshold isThis model describes a specified external force. Motion transmitted through a moving support instead has a base-displacement forcing term and a different dependence on damping.

Fourier analysis describes a function by the oscillations that compose it. A graph records values at successive positions or times; a frequency spectrum records how much of each frequency is present, together with its wave phase. The same complicated shape can become a short list of simple components. For periodic functions, the components form a Fourier series; on the whole real line, the Fourier transform uses a continuous range of frequencies, and the Fourier inversion theorem puts them back together.

For a concrete example, consider the trigonometric polynomialWith $t$ measured in seconds, its three Fourier harmonics have frequencies $1$, $3$, and $5$ cycles per second and amplitudes $1$, $1/2$, and $1/4$. Each cosine has zero wave phase. Adding them creates the sharper peaks in the figure. The frequency spectrum makes the recipe immediately visible; changing an amplitude changes the shape without changing the component's frequency. Fourier orthogonality lets us recover each Fourier coefficient by averaging against the corresponding sine or cosine, because the other Fourier harmonics cancel over a full period of a function.

The deeper reason these components are so useful is that they simplify operations. By Euler's formula, the exponential function $e^{i\omega x}$ packages a cosine and a sine into one Fourier mode. A translation or a derivative changes only its multiplier:Thus Fourier modes are simultaneous eigenfunctions of these operations. In Fourier series, this gives a literal diagonalization in the Fourier basis; on the real line, the Fourier transform gives the continuous counterpart, with each frequency handled separately. The Translation property of the Fourier transform and the Fourier transform of a derivative turn shifting and differentiating a whole function into multiplication of its Fourier transform. A convolution likewise becomes multiplication by the convolution theorem. This is beautiful because one description simplifies several apparently different operations at once.

For example, the heat equation $u_t=\kappa u_{xx}$, with thermal diffusivity $\kappa >0$, evolves a spatial Fourier mode asFine ripples have larger angular frequencies and decay faster: the formula explains why heating smooths a jagged temperature profile. For a general initial profile with sufficient regularity, the Fourier transform gives $\hat{u}(\omega ,t)=e^{-\kappa {\omega }^{2}t}\hat{u}(\omega ,0)$, and Fourier inversion recovers the temperature. Similarly, multiplying selected parts of a frequency spectrum suppresses unwanted components. Claude E. Shannon's Communication in the Presence of Noise (1949), Section III, describes ideal filtering as a projection that retains some frequency coordinates and removes others.^[ref]

The physical motivation is already visible in Joseph Fourier's Théorie analytique de la chaleur (1822). Chapter III constructs heat profiles using sums of sines and cosines; Chapter IX, articles 355–359, passes from Fourier series to integrals as the spatial interval becomes unbounded. These heat problems explain the motivation for the discrete and continuous descriptions. Their modern formulations specify which functions admit reconstruction and in what sense the expansions converge.^[ref][ref]

A Fourier mode is a single complex oscillatory component $e^{i\omega x}$. Its real and imaginary parts are a cosine and a sine; $\omega$ is the spatial angular frequency. Every constant-coefficient differential operator acts on it by multiplication: $P({\partial }_x)e^{i\omega x}=P(i\omega )e^{i\omega x}$. In particular, ${\partial }_x^2{e}^{i\omega x}=-{\omega }^2{e}^{i\omega x}$. These eigenfunctions reduce a partial differential equation such as the heat equation to separate equations for each frequency. On the real line a pure Fourier mode is not in the L2 space; the Fourier transform uses such modes in a continuous superposition.

A Fourier series describes a periodic function through sines and cosines whose frequencies are integer multiples of one base frequency. These Fourier harmonics give a useful alternative to describing the value at every input: each Fourier coefficient measures how much of one oscillating pattern is present. For a real, integrable function of period $2\pi$, the real form isThe cosine formula includes $n=0$; the sine formula uses $n\ge 1$. Thus $a_0/2$ is the mean value over a period, not $a_0$ itself. The symbol $\sim$ associates a function with its Fourier coefficients; an equality requires a specified kind of convergence, such as pointwise convergence. For general period $T>0$, replace $nx$ by $2\pi nt/T$ and the coefficient prefactor by $2/T$. Fourier orthogonality explains why integrating against one mode removes the others. ^[ref].

For a finite example, takeIts mean is $1/2$, its real Fourier coefficients are $a_0=1$, $a_1=1$, $b_2=1/2$, and all other $a_n,b_n$ vanish. At $x=\pi /4$, the three contributions add to $1/2+1/\sqrt{2}+1/2=1+1/\sqrt{2}\approx 1.70710678$. The function is already a trigonometric polynomial, so retaining these modes reconstructs it exactly. Here $x$ is an angular coordinate in radians; if it instead represents time, a base angular frequency must supply the corresponding units.

The change of viewpoint is useful because many linear problems act separately on these modes. The second derivative maps $\sin (nx)$ to $-n^2\sin (nx)$ and maps $\cos (nx)$ in the same way. Consequently a differential equation that is complicated for a whole profile can become a simpler calculation for each Fourier harmonic. A linear time-invariant system similarly has a separate response to each complex exponential. Superposition combines the responses when the relevant sums and operations are justified. ^[ref].

The infinite case contains a subtlety absent from the finite example. Under pointwise convergence of a piecewise smooth Fourier series, the value at a jump discontinuity is the average of the two one-sided limits. For a square-integrable function, the complex Fourier series instead converges in the $L^2$ norm: the integrated squared error tends to zero. Neither statement says that the error becomes small everywhere at once. A square wave shows the difference vividly through the Gibbs phenomenon.

A historical source for this connection between profiles and modes is Joseph Fourier's 1822 book Théorie analytique de la chaleur. Chapter III develops expansions in sines and cosines while studying the propagation of heat. The heat equation gives a continuing application of this approach. ^[ref].

The four-wavevector combines the angular frequency and spatial wavevector of a plane wave. Phase invariance gives $K^{\prime \mu }={\Lambda }^{\mu }{}_{\nu }{K}^{\nu }$. Under a boost $\mathbf{v}$, ${\omega }^{\prime }=\gamma (\omega -\mathbf{v}\cdot \mathbf{k})$, ${\mathbf{k}}_{\parallel }^{\prime }=\gamma ({\mathbf{k}}_{\parallel }-\omega \mathbf{v}/c^2)$ and ${\mathbf{k}}_{\perp }^{\prime }={\mathbf{k}}_{\perp }$. For metric signature $(+,-,-,-)$, $R_{\mu }{K}^{\mu }=\omega t-\mathbf{k}\cdot \mathbf{r}$ is a Lorentz scalar. A vacuum light wave has $K_{\mu }{K}^{\mu }=0$.

For a free particle of mass $m>0$ in nonrelativistic quantum mechanics, choose the potential energy zero and writeApplying the momentum operator $-i\hslash \nabla$ returns $\mathbf{p}\psi$; this fixes the sign convention. The angular frequency is measured relative to this chosen energy zero. Adding a constant energy changes only a time-dependent global phase, leaving the spatial de Broglie wavelength unchanged. No rest-energy term is included in this nonrelativistic quantum mechanics convention.

The plane wave has $|\psi {|}^2=|A{|}^2$ and probability current $\mathbf{j}=(\mathbf{p}/m)|A{|}^2$. Its real part can be drawn as a sine or cosine, but that graph is a component of a probability amplitude, not a transverse displacement. On infinite space this is a generalized momentum eigenstate, not a state satisfying wavefunction normalization. A narrow wave packet instead moves with group velocity ${\nabla }_{\mathbf{k}}\omega =\mathbf{p}/m$; the phase velocity $\omega /|\mathbf{k}|=|\mathbf{p}|/(2m)$ is not its transport speed.

The equality of canonical momentum and kinetic momentum here assumes a field-free region with the magnetic vector potential chosen zero. For a charged particle in a magnetic field, the kinetic momentum is $\mathbf{\Pi }=\mathbf{p}-q\mathbf{A}$. An arbitrary canonical phase gradient must not then be substituted into $m\mathbf{v}$; the free particle formula needs its stated assumptions.

A free-running circadian rhythm persists without an entraining cycle. Its period $\tau$ separates corresponding circadian phases, such as successive activity onsets, under specified constant conditions.^[ref] A circadian phase tells where the system is within its cycle; a period tells how long one cycle takes. Choosing onset rather than peak as the reference changes the phase label, not the period of an ideal repeating signal.

Consider an original, deliberately idealized model with $\tau =23.5\mathrm{h}$. For seven days prescribe a 12:12 LD cycle with activity starting at hour 12 of each 24-hour day. Then omit the next scheduled dawn and keep the environment dark. Stipulate that the first DD onset retains hour 12, and that every later onset is exactly $23.5\mathrm{h}$ after the previous one. Activity lasts eight hours after each onset. These are chosen model values, not recorded mouse or human data, and the fixed activity window represents an output of the model rather than the molecular mechanism of a circadian clock.

An actogram displays activity against time in successive rows. Here each row spans $T=24\mathrm{h}$, regardless of the circadian rhythm's period. If $t_0$ is the first DD onset and $n$ counts completed free-running cycles, thenThe unwrapped onset displacement $x_n$ is therefore $-0.5n\mathrm{h}$. After eight cycles the onset is four hours earlier than a 24-hour continuation would predict; after sixteen cycles it is eight hours earlier. Because these onsets remain within successive calendar rows in the displayed interval, their row positions are $12,11.5,11,\dots ,4$ hours. The visual slant comes from comparing two different periods, not from cessation of the oscillation.

An alternative phase oscillator description makes the same distinction explicit. Set the circadian phase to zero at $t_0$ and defineThe angular frequency $\omega$ is constant in this model. Onset occurs whenever the unwrapped circadian phase completes another turn. A period longer than $T$ would instead produce later onsets; equality would produce aligned rows even without entrainment of circadian rhythm. Thus aligned rows by themselves do not identify the cause of a daily pattern. Real activity can be noisy, and a free-running circadian rhythm's period can vary with conditions or with time after release; the stipulated straight line assumes a stationary period.

Take two point masses joined by a massless spring of spring constant $k>0$ and equilibrium length $\ell >0$. Assume one-dimensional motion along the spring, no external force, no damping, and an extension $s=x_2-x_1-\ell$ satisfying Hooke's law; keep $x_2>x_1$. For a real spring this is a model within its elastic range. Its potential energy is $U=k{s}^2/2$, soHere $\omega$ is angular frequency and $\nu$ is frequency. Choose $|A|<\ell$ so the model bodies do not cross. The internal normal mode has displacements $u_1=-m_{2}s/M$ and $u_2=m_{1}s/M$ about equilibrium offsets from the centre of mass; its amplitude ratio is $|u_1|/|u_2|=m_2/m_1$. The other normal mode is collective translation with zero frequency and no spring extension. This free pair has different boundary conditions from normal modes of two equal masses between three springs.

For $k=10N/m$ and two $1\mathrm{kg}$ masses, $\omega =\sqrt{20}{\mathrm{s}}^{-1}\approx 4.47214{\mathrm{s}}^{-1}$ and $\nu \approx 0.711763\mathrm{Hz}$. Replacing the second mass by $100\mathrm{kg}$ gives $\omega =\sqrt{10.1}{\mathrm{s}}^{-1}\approx 3.17805{\mathrm{s}}^{-1}$ and $\nu \approx 0.505802\mathrm{Hz}$. Fixing the second endpoint instead would give $\sqrt{10}{\mathrm{s}}^{-1}$: the free heavy partner still moves. The free pair's angular frequency exceeds that fixed-end limit by the factor $\sqrt{1+1/q}$, about $1.004988$ at $q=100$.

Frequency is the number of repetitions of a periodic phenomenon per unit time. For a wave of period $T$, its frequency is $\nu =1/T$, measured in hertz. The angular frequency is $\omega =2\pi \nu$.