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A Hopf bifurcation creates or destroys a small periodic orbit near an equilibrium point as a pair of complex conjugate eigenvalues crosses the imaginary axis. It explains how a resting state can give way to sustained oscillations: the linearization of a dynamical system supplies rotation and growth, while a nonlinear term can saturate that growth at a nonzero amplitude.

The simplest supercritical Hopf bifurcation is the exact Hopf normal formHere $z$ is a complex number, $\mu$ is the bifurcation parameter, and $\omega$ is the angular frequency. In polar coordinates $z=r{e}^{i\theta }$, the equations become $\dot{r}=\mu r-r^3$ and $\dot{\theta }=\omega$ for $r>0$. At $\mu <0$, each turn shrinks towards the equilibrium point. At $\mu >0$, small radii grow and large radii shrink towards $r=\sqrt{\mu }$, producing a limit cycle. For example, $\mu =0.04$ and $\omega =2$ give radius $0.2$, period $\pi$, and frequency $1/\pi$ cycles per unit time. The limiting angular frequency stays nonzero while the amplitude tends to zero: a tiny newborn oscillation can still repeat at a finite rate.

For a precise generic statement, consider a smooth function $f$ defining an autonomous differential equation $\dot{x}=f(x,\alpha )$ on ${\mathbb{R}}^n$, with a smooth branch of equilibrium points $x_{\ast }(\alpha )$ and critical parameter $\alpha =0$. Require:Then a locally unique branch of small periodic orbits, up to phase, emerges on the side $\beta a<0$. On the two-dimensional centre manifold, the branch attracts for $a<0$ and repels for $a>0$. Attraction in the full phase space additionally requires all remaining eigenvalues to have negative real part; any transverse unstable direction makes the periodic orbit unstable there.^[ref]

An eigenvalue-crossing bifurcation test alone cannot establish these nonlinear conclusions. For example, the purely linear equation $\dot{z}=(\mu +i\omega )z$ has the same crossing but no isolated periodic orbit: at $\mu =0$ every nonzero radius is constant, and at $\mu \ne 0$ none is. If the cubic coefficient vanishes, higher-order terms must be examined; a generalized Hopf bifurcation is one possible degeneracy. The usefulness of the generic theorem is that a large system can have its small oscillatory onset understood through one Hopf normal form, subject to these hypotheses and local limits.

The Hopf normal form is a local normal form of a dynamical system on the critical two-dimensional centre manifold. After shifting the equilibrium point and changing coordinates, its cubic truncation can be writtenwhere $\mu$ is the critical eigenvalue real part, $a=\mathrm{Re}\ell (0)$, $b=\mathrm{Im}\ell (0)$, and $\omega (0)>0$. Smooth parameter dependence of the cubic coefficients and higher-order terms are omitted in this truncation. The real coefficient $a$ controls radial saturation; the imaginary coefficient $b$ changes the angular frequency with amplitude. This finite-order reduction, rather than literal equality to a cubic equation everywhere, is the meaning of the Hopf normal form for a general system.^[ref]

In the cubic truncation, $r_{\ast }^2=-\mu /a$ whenever $\mu a<0$. Substituting into the angular equation gives $\Omega =\omega (\mu )-b\mu /a$. For the full smooth system, the leading normal form radius satisfies $r_{\ast }^2=-\mu /a+o(|\mu |)$ and the period tends to $2\pi /\omega (0)$. The exact circle and exactly constant angular frequency of the special model $a=-1$, $b=0$, $\omega (\mu )=\omega$ are conveniences of that model. In the original coordinates the small periodic orbit need not be a circle. Radial scaling can normalize a nonzero $a$ to $-1$ or $+1$, so $\sqrt{|\mu |}$ is a normalized leading amplitude, not a universal physical radius.

Let $K={\partial }_v$ generate a Killing horizon with constant surface gravity $\kappa$, so ${\nabla }_{K}K=\kappa K$. In geometric optics, an affinely transported light wavevector on the same generator is $k=q(v)K$. The geodesic equation gives $q^{\prime }+\kappa q=0$. Observers related by the isometry generated by $K$ have the same $-K\cdot W$ at their crossings, where $W$ is their four-velocity, so their measured angular frequencies $\omega =-k\cdot W$ obey the displayed relation. A degenerate Killing horizon gives ratio one for equally prepared observers; changing their crossing velocities can produce a separate Doppler effect.

Insert a transverse plane wave $\mathbf{u}={\mathbf{u}}_0{e}^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$ into the inertial-wave vorticity equation, with $\mathbf{k}\cdot {\mathbf{u}}_0=0$. It gives $\omega \mathbf{k}\times {\mathbf{u}}_0=2i(\mathbf{\Omega }\cdot \mathbf{k}){\mathbf{u}}_0$. Applying $\mathbf{k}\times$ again and using transversality gives the title dispersion relation. If $\theta$ is the angle to the rotation axis, $|\omega |=2|\Omega \cos \theta |$. At fixed direction the angular frequency does not depend on wavelength. Parallel wavevectors give the maximum $2|\Omega |$; perpendicular ones have zero frequency and belong to the nonoscillating balanced limit.

Wave interference is the reinforcement or cancellation that occurs when overlapping wave amplitudes add by the superposition principle. For two monochromatic waves of the same frequency, mutual optical coherence means that their relative wave phase is stable during the observation. Equal wave phases give constructive interference; a half-cycle phase difference gives the destructive interference intensity minimum. Wave amplitudes add before their squares determine wave intensity. ^[ref].

Consider two overlapping beams in a homogeneous medium, with the same polarization of an electromagnetic wave and individual local time-averaged light intensities $I_1,I_2$. Their phase difference $\delta$ is measured in radians. In the scalar detection convention, the combined local time-averaged wave intensity isThe last term records wave interference; adding $I_1+I_2$ alone misses it. The formula describes irradiance when both fields propagate along the same direction, or when the same detector geometry factor is included in all three intensities. For crossing beams it describes the squared electric field response, proportional to scalar wave intensity; a specified component of the Poynting vector also requires the propagation geometry. The ideal crossing plane waves below make that distinction explicit.

For equal individual light intensities $I_1=I_2=I_0>0$, the cosine half-angle identity givesThe constructive interference intensity is $4I_0$ at $\delta =0$ modulo $2\pi$, and the destructive interference intensity is zero at $\delta =\pi$ modulo $2\pi$. At phase quadrature, $\delta =\pi /2$, it is $2I_0$. With $I_1=1$ and $I_2=0.25$ in the same arbitrary wave intensity unit, the maximum is $2.25$, the minimum is $0.25$, and the phase quadrature value is $1.25$: unequal wave amplitudes cannot cancel completely.

Both curves use the same wave intensity unit and a horizontal angle in radians. The dashed levels are the averages over a full phase difference cycle. Equal beams have interferometric visibility one; the unequal example has interferometric visibility $0.8$.

A spatially varying phase difference turns the same law into interference fringes. Take two ideal plane electromagnetic waves with common wavelength $\lambda$, equal individual light intensities $I_0$, and propagation directions at angles $+\theta$ and $-\theta$ from the $z$ axis in the $xz$ plane, with $0<\theta <\pi /2$. Both have linear polarization along $y$, perpendicular to both propagation directions. Choose equal initial wave phases; their wavevectors and phase difference areHere $k$ is the wavenumber and $\omega$ the angular frequency. The common $z$ and time terms cancel, so the interference fringes stay at fixed $x$ for every $z$. Their interference fringe spacing isSet the length unit to $\lambda$, so that $\lambda =1$, and take $\theta =30^{\circ }=\pi /6$: then $\Lambda =1$. The constructive interference intensity bands lie at integer $x/\lambda$; the destructive interference intensity bands lie at half-integer $x/\lambda$.

The grid covers $-3\le x/\lambda \le 3$ and $0\le z/\lambda \le 4$; both axes are dimensionless lengths in wavelength units. The colour scale is $I/I_0$ from zero to four, using the same single-beam normalization as the equal-beam curve above. This is an ideal plane wave superposition, with no slits or point sources and no near-field approximation. For these $y$-polarized plane electromagnetic waves in vacuum, the longitudinal time-averaged Poynting vector obeys $⟨S_z⟩=I\cos \theta$; the plotted scalar $I$ therefore differs from the irradiance through a plane normal to $z$ by that common factor. This relation follows by adding the magnetic fields as well as the electric fields.

The bright peaks do not create energy. Averaging over one interference fringe spacing, or equivalently over one full phase difference cycle, givesbecause the integral of the cosine is zero, with sine as its antiderivative. Equal beams thus average to $2I_0$, despite bright bands at $4I_0$. For the crossing-wave geometry, the average longitudinal energy flux is correspondingly $2I_0\cos \theta$. Bright and dark regions redistribute the incoming energy; conservation of energy concerns the full energy flux, rather than the height of one peak.

Wave interference is useful because phase differences convert small changes in optical path length into measurable changes of wave intensity, as in a Mach-Zehnder interferometer. Its beauty is that the same superposition principle produces both complete cancellation and fourfold brightening; adding more coherent contributions also produces the narrow peaks of a diffraction grating. Feynman's treatment connects this many-source pattern directly to addition of complex amplitudes. ^[ref].

A short proof outline uses local real fields $E_j(t)=A_j\cos (\omega t+{\varphi }_j)$ with parallel linear polarizations. Write $I=C⟨(E_1+E_2{)}^2{⟩}_t$ and $I_j=C{A}_j^2/2$ for the same positive conversion factor $C$. Averaging over many wave periods while the phase difference stays fixed uses $⟨{\cos }^2(\omega t+{\varphi }_j){⟩}_t=1/2$ and $⟨\cos (\omega t+{\varphi }_1)\cos (\omega t+{\varphi }_2){⟩}_t=\cos \delta /2$, giving the stated law. Equivalently, add the complex amplitudes and take the squared complex modulus. ^[ref].

For different fixed linear polarizations in a common transverse plane, the cross term acquires the dot product of their real unit vectors, equal to $\cos \alpha$ for relative angle $\alpha$. Orthogonal linear polarizations remove the cross term in polarization-insensitive scalar detection; different orientations generally only reduce it. A common polarizer can select overlapping components and recover wave interference. Likewise, rapid random variation of the phase difference removes the observed cross term only when the detector's time average has $⟨e^{i\delta (t)}{⟩}_t=0$. Slowly varying wave phases can instead move resolved interference fringes, and biased or incompletely sampled wave phases can leave residual contrast. Finite optical coherence therefore depends on the observation time as well as the fields.

For a fixed amount of substance and composition of classical ideal gas, internal energy depends only on temperature: $U=U(T,n)$, with no independent volume dependence. This does not require constant heat capacity at constant volume. Between equilibrium endpoints,Here $C_V$ is total heat capacity at constant volume, $C_{V,\mathrm{m}}$ is molar heat capacity, and $U/m$ for sample mass $m$ is specific internal energy, whose change is ${\int }_{T_1}^{T_2}{c}_V(T)dT$. These endpoint relations apply even when the intervening process is not at fixed volume; they do not say that transferred heat always equals $\Delta U$. With constant $C_V$, the safe statement is $U(T)=U(T_{\ast })+C_V(T-T_{\ast })$. Writing $U=C_{V}T$ additionally chooses the reference energy; it does not prove that a physical gas remains classical and ideal down to zero temperature. ^[ref]

A thermodynamic proof outline uses the first law of thermodynamics in equilibrium form, $dU=TdS-pdV$, and the Maxwell relation $(\partial S/\partial V{)}_T=(\partial p/\partial T{)}_V$:The last step uses the ideal gas law and the molar gas constant $R$, with fixed composition and amount of substance. Microscopically, separated ideal gas particles store no intermolecular interaction energy; their translational kinetic energy and any internal molecular energy are determined by temperature. Negligible intermolecular interactions still allow a molecule to have internal vibrational potential energy.

An original analytical model makes the distinction between an ideal gas and constant heat capacity concrete. Assume classical translation, two fully active rotational terms, and one independent quantum harmonic oscillator per diatomic molecule, with fixed angular frequency $\omega$. Define the vibrational temperature scale $\Theta =\hslash \omega /k_B$, using the reduced Planck constant $\hslash =h/(2\pi )$ and Boltzmann constant $k_B$. Summing its thermal occupations givesThe constant $U_{\ast }$ includes the chosen reference and zero-point energy, whose temperature derivative vanishes. When $T\ll \Theta$ the vibrational contribution to heat capacity is exponentially small; when $T\gg \Theta$ it tends to $R$ per mole. This model assumes rotations remain active throughout the interval and excludes dissociation, electronic excitation and real gas corrections. Its low-$T/\Theta$ edge is not a model of rotational freezing or of an actual substance at absolute zero.^[ref]

For this stipulated model, heating from $T_1=0.2\Theta$ to $T_2=\Theta$ gives $\Delta U/(nR\Theta )\simeq 2.57519$. Holding the frozen-vibration value $C_{V,\mathrm{m}}=5R/2$ throughout would give $2$, underestimating the required internal energy increase by about $22.3\%$ of the model's full value. The integral measures all thermal storage activated along the interval. At fixed volume with only heat input, this is also the supplied heat; at fixed pressure, expansion work must be added through Mayer's relation.

Joule's free-expansion experiment tests the independence from volume in a different way: an insulated rigid vessel allows gas into an evacuated compartment, so no external work is delivered and ideally no heat is exchanged. The first law of thermodynamics gives $\Delta U=0$; positive $C_V$ then predicts equal initial and final equilibrium temperatures for the ideal gas. The apparatus, thermal-capacity controls and limitations of this inference are described in Joule's free-expansion experiment and temperature sensitivity of a free-expansion experiment. This endpoint test differs from the reversible adiabatic expansion of a loaded piston, which delivers work and lowers internal energy.

At a step in buoyancy frequency without a jump in background mass density, vertical velocity and fluid pressure are continuous. For a fixed horizontal wavenumber and angular frequency, these are continuity of $W$ and $W^{\prime }$. They determine reflection and transmission between propagating internal gravity waves and evanescent waves.

The classical isotropic harmonic oscillator has the same angular frequency $\Omega >0$ in every Cartesian direction. Its equation of motion is $\ddot{\mathbf{r}}=-{\Omega }^2\mathbf{r}$, with solution $\mathbf{r}=\mathbf{A}\cos \Omega t+\mathbf{B}\sin \Omega t$. The trajectory lies in the span of $\mathbf{A},\mathbf{B}$ and is an ellipse centred on the force centre, degenerating to a line for dependent vectors. For nonzero angular momentum, $r^2$ oscillates at $2\Omega$ unless the trajectory is exactly circular. This distinguishes radial and Cartesian frequencies.

Consider a neutral, isotropic, unpolarized three-dimensional spin-one-half Fermi liquid, with only $F_0^s>0$ nonzero. A longitudinal density wave with angular frequency $\omega$ and wavevector magnitude $q$ has $s=\omega /(q{v}_F)=c_0/v_F$. The collisionless angular equation for the energy displacement $\nu (u)$ of the Fermi surface isAveraging givesThe bracket decreases from infinity to zero on $1<s<\infty$: each $F_0^s>0$ gives one undamped root above the small-$q$ particle-hole continuum.^[ref]

For an original example, $F_0^s=2$ gives $s=1.16474145$. Since $F_1^s=0$ here, the Principle of Galilean relativity gives $m^{\ast }=m$ and taking the hydrodynamic limit before $T\to 0$ gives $c_1/v_F=\sqrt{(1+F_0^s)/3}=1$. Stipulating $v_F=50\mathrm{m}{\mathrm{s}}^{-1}$ gives $c_0\simeq 58.24\mathrm{m}{\mathrm{s}}^{-1}$ and $c_1=50\mathrm{m}{\mathrm{s}}^{-1}$. These are model choices, not helium-3 measurements. At large $F_0^s$, expanding the logarithm gives $s\sim \sqrt{F_0^s/3}$; for small positive $F_0^s$, the root approaches $1$ exponentially closely, so numerical resolution matters. A charged electron liquid needs its long-range interaction included; this neutral model is not its long-wavelength density mode.

A useful stipulated dispersive medium obeys the Klein-Gordon equation $u_{tt}-c^2{u}_{xx}+{\omega }_0^{2}u=0$, with $c>0$ and ${\omega }_0>0$. Its positive-angular frequency dispersion relation isHere ${\omega }_0$ is the gap at zero wavenumber, not the central angular frequency ${\omega }_c=\omega (k_0)$ of a wave packet; the parameter in the preceding formula is $A={\omega }_0/c$. For $k>0$, the phase velocity satisfies $v_p>c$, the group velocity satisfies $0<v_g<c$, and $v_p{v}_g=c^2$. In this stipulated medium $c$ is the limiting propagation speed of the model; it need not mean the vacuum speed of light. The same square-root dispersion relation appears in waveguide phase and group velocities, with a waveguide cutoff frequency playing the role of ${\omega }_0$.^[ref]

For the original dimensionless choice $c={\omega }_0=k_0=1$, a carrier wave has phase velocity $v_p=\sqrt{2}$, while a narrow wave packet has leading group velocity $v_g=1/\sqrt{2}$. After time $t=8$, the carrier's zero-wave phase reference has moved to $x=8\sqrt{2}\simeq 11.31$, whereas the leading wave envelope centre is at $x=8/\sqrt{2}\simeq 5.66$. A carrier wave crest therefore moves through the wave envelope; it does not carry an identifiable piece of the envelope with it.

The origin-to-curve secant on the dispersion diagram gives phase velocity, while its local tangent gives group velocity. As $k\to 0^{+}$, $v_p\to \infty$ and $v_g\to 0$; at $k=0$ the phase velocity itself is undefined. As $k\to \infty$, both approach $c$, from opposite sides. On the same positive-angular frequency branch with $k<0$, both signed velocities are negative and their product remains $c^2$. These conventions distinguish direction from phase speed. Reversing both $k$ and $\omega$ represents the same real plane wave; changing just one sign changes its direction.

For an original finite-bandwidth check, take equal-amplitude plane waves at $k_1=0.8$ and $k_2=1.2$, still with $c={\omega }_0=1$. The exact two-tone wave envelope speed iswhile its carrier wave has phase velocity $(\sqrt{1+0.8^2}+\sqrt{1+1.2^2})/2\simeq 1.421337$. Their product is exactly $1$: rationalizing the difference gives $v_{\mathrm{beat}}=2/({\omega }_1+{\omega }_2)$. Nevertheless $v_{\mathrm{beat}}\ne {\omega }^{\prime }(1)$; the derivative describes the limit of vanishing separation, not an arbitrary pair of tones. The two-tone wave envelope repeats indefinitely, so it has no isolated first arrival and cannot measure front velocity.

The Lamb-Mössbauer factor $f$ is the fraction in the coherent elastic line. In the harmonic-lattice approximation it is a Debye–Waller factor. Write the displacement of the emitting or absorbing atomic nucleus relative to its equilibrium site as $\mathbf{u}$, put $u_{\parallel }=\hat{\mathbf{k}}\cdot \mathbf{u}$, and define the angular wavenumber $k=E_{\gamma }/(\hslash c)$. Choose the equilibrium site so that $⟨u_{\parallel }⟩=0$. Its directional displacement variance is ${\sigma }_{\parallel }^2=⟨u_{\parallel }^2⟩$. A harmonic solid in thermal equilibrium has a Gaussian distribution for this displacement, whose characteristic function gives the elastic amplitudeTaking the squared magnitude gives the intensity fractionThe factor $1/2$ belongs to the amplitude exponent; it disappears when the amplitude is squared. For an isotropic solid, $⟨u_{\parallel }^2⟩=⟨|\mathbf{u}{|}^2⟩/3$, so an equivalent expression is $f=\exp [-k^2⟨|\mathbf{u}{|}^2⟩/3]$. Directional and three-dimensional mean squared displacements must not be substituted for one another without this factor.

An Einstein solid approximation replaces the vibrational spectrum by one positive angular frequency $\omega$. For the directional quantum harmonic oscillator, with the same effective mass $M$ as in $E_R$,Here $k_B$ is the Boltzmann constant and $T$ is temperature. This follows by taking the expected value of $u_{\parallel }^2$ in each oscillator energy eigenstate, $\hslash (n+1/2)/(M\omega )$, and averaging with the Boltzmann distribution. Substitution givesThe zero-temperature limit retains zero-point energy and a nonzero displacement variance. Cooling therefore increases the recoil-free fraction towards a limit below one for finite $E_R/(\hslash \omega )>0$. At high temperature, the exponent tends to $-2E_R{k}_{B}T/(\hslash \omega {)}^2$. This is a single-frequency approximation to a harmonic lattice's coherent elastic weight, not a complete description of all zero-net-energy processes of an isolated oscillator. Real vibrational spectra, directional binding and anharmonic motion can change its predictions.

Extend the stipulated teaching system by choosing $\hslash \omega =0.020\mathrm{eV}$, so $\eta =E_R/(\hslash \omega )=0.1$. Then $f(0)=0.90484$, while $f=0.80542$ at $k_{B}T=\hslash \omega$. If source and absorber are independent, the fraction of events eligible for recoil-free emission followed by recoil-free absorption contains the product $f_s{f}_a$; this gives $0.81873$ for two zero-temperature copies. It is not the total probability of absorption, which also depends on the number of resonant atomic nuclei, geometry and spectral line overlap.

The angular frequency of free Larmor precession in a uniform magnetic field of magnitude $B$, with gyromagnetic ratio $\gamma$. Interactions can shift a collective resonance away from this free-spin scale.

Laser detuning is the difference between a laser angular frequency and a specified atomic resonance angular frequency. Red detuning has $\delta <0$; blue detuning has $\delta >0$. The sign matters in laser cooling: a moving atom can see a red-detuned laser closer to resonance because of the Doppler effect.

Zhu, Miao, Xu and Cao's 2007 experiment drove a triangular metal frame electromagnetically with a sinusoidal generator. In a roughly $3\mathrm{cm}$-deep, temperature-regulated pool, it produced nearly plane capillary waves. A $632.8\mathrm{nm}$ laser illuminated the surface obliquely; the moving ripples acted as a reflection diffraction grating. One split beam monitored laser stability, while a CCD recorded diffraction orders on a screen $6.76\mathrm{m}$ away. An air-supported optical table reduced ambient vibration. ^[ref]

The frequency sweep was $100$–$290\mathrm{Hz}$ in $10\mathrm{Hz}$ steps, using distilled water and kerosene at $20{}^{\circ }\mathrm{C}$. For each setting they averaged at least three measurements of first-order spot separation. Known optical geometry and the grating equation yielded $k=2\pi /\lambda$; generator readings yielded $\omega =2\pi f$. Agreement with the predicted log–log slope $3/2$ tests the capillary law, and its intercept gives surface tension if mass density is known. The authors identify surface contamination as a major error source; noncontact optical detection reduces additional disturbance. ^[ref]

The optical measurement can be understood from the grating equation. Let ${\lambda }_L$ be the laser wavelength, $\alpha$ the incident grazing angle above the mean surface, and ${\beta }_m$ the outgoing grazing angle of diffraction order $m$, with the sign of $m$ chosen so thatThus calibrated spot positions determine outgoing angles, which determine the ripple wavelength $\lambda$, independently of the mechanical dispersion relation. This optical geometry treats the surface as a reflection diffraction grating at the laser wavelength. ^[ref]

The inferred surface tension in the full deep-water model isThis formula makes useful experimental controls explicit. Measure mass density at the sample's temperature; check depth with $kh$; keep the driven wave amplitude small enough that $ka\ll 1$; and distinguish $f$ from angular frequency $\omega$. For the stipulated parameters, the $100$–$290\mathrm{Hz}$ window corresponds to calculated wavelengths about $3.62$–$1.76\mathrm{mm}$ and $kh\simeq 52$–$107$ at $h=30\mathrm{mm}$. Depth corrections are negligible here, but the gravity contribution is about $4.5\%$–$1.1\%$ of the capillary contribution. These are ideal predictions, not digitized results from the experiment.

Omitting gravity would infer ${\gamma }_{\mathrm{cap}}=\rho {\omega }^2/k^3=\gamma +\rho g/k^2$, biasing surface tension upward. In the capillary limit, the differential estimate $d\gamma /\gamma \simeq d\rho /\rho +2d\omega /\omega -3dk/k$ also shows why careful spot-spacing calibration matters. A clean surface and weak viscosity are modeling assumptions: appreciable damping, surfactant stresses or large wave amplitudes call for additional physics. The experiment tests the short-wave regime; its drive window does not establish the crossover or the long-wave surface gravity wave limit.

An ideal electrical circuit exchanges energy between an inductor and a capacitor. With no electrical resistance or applied drive, the electric charge obeys $L{q}^{\prime \prime }+q/C=0$, giving angular frequency ${\omega }_0=1/\sqrt{LC}$. A series RLC circuit adds dissipation and can include an external drive.

A nearest-neighbour chain with the Lennard-Jones potential $V(x)=4ϵ[(\sigma /x{)}^{12}-(\sigma /x{)}^6]$ has equilibrium spacing $a=2^{1/6}\sigma$ and spring constant $K=V^{\prime \prime }(a)=72ϵ/a^2$. The monatomic-chain dispersion relation gives speed of sound $v=a\sqrt{K/m}=\sqrt{72ϵ/m}$ and maximum angular frequency $2v/a$. The cancellation of $\sigma$ from the sound speed holds in this chain model; it does not remove the spacing from the cutoff frequency.

Each LIGO instrument is a Michelson interferometer with approximately $4\mathrm{km}$ arms containing Fabry–Pérot interferometers. A stabilized $1064\mathrm{nm}$ laser illuminates suspended $40\mathrm{kg}$ fused-silica mirrors in vacuum. Four-stage suspensions and active isolation suppress ground motion; low-loss materials reduce thermal noise. Power recycling increases circulating power; signal recycling shapes the response to the gravitational wave.^[ref]

Why measure two perpendicular arms? In the long-wavelength limit, take a gravitational wave with plus polarization normally incident on ideal perpendicular arms. Its transverse-traceless tensor has $h_{xx}=h_{+}$ and $h_{yy}=-h_{+}$. Expanding each proper length to first order givesHere $h_{+}$ is dimensionless, whereas $\Delta L$ is a length. A stipulated $h_{+}=10^{-21}$ and $L=4000\mathrm{m}$ imply $\Delta L=4\times 10^{-18}\mathrm{m}$, with individual arm changes of opposite sign and magnitude $2\times 10^{-18}\mathrm{m}$. These are equivalent differential arm-length changes, not a photograph of mirrors moving against a rigid external ruler. The observable compares returning light phases; coordinate systems can distribute the description between mirror coordinates and light propagation.

For unit arm directions $\mathbf{u},\mathbf{v}$, define the detector tensor $D^{ij}=(u^i{u}^j-v^i{v}^j)/2$. If $e_{ij}^{+},e_{ij}^{\times }$ are the transverse polarization tensors, thenThe dimensionless factors $F_A$ encode sky direction and the polarization basis. Thus even identical sources can have different observed amplitudes and wave phases. Motion changing both arms equally cancels in this ideal differential observable, although real asymmetries can couple common disturbances into it.

Suspension supplies an inertial reference above its resonances. For example, a single ideal harmonic oscillator with support motion $y$ obeys $\ddot{x}+{\omega }_0^2(x-y)=0$. Its response at angular frequency $\omega$ is $x/y={\omega }_0^2/({\omega }_0^2-{\omega }^2)$, which decreases in magnitude as $({\omega }_0/\omega {)}^2$ well above resonance. The mirror follows slow support drift but approximately free motion in the high-frequency limit. Real multistage suspensions also require damping and alignment control; resonant frequencies are not regions of good isolation.

Rainer Weiss's 1972 report proposed suspended masses, multiple-pass optical delay lines, and a servo holding an interference fringe. Its noise analysis separately treats photon-counting fluctuations, laser frequency instability, mechanical thermal noise, radiation pressure and ground motion. Those delay lines should be distinguished from the resonant arm cavities of Advanced LIGO.^[ref]

Let the input electric potential difference be $s(t)=A\cos (\omega t+\varphi )$, with peak amplitude $A\ge 0$, angular frequency $\omega >0$, and phase difference $\varphi$ relative to $\cos \omega t$. For an observation interval $[0,T]$ containing an integer number $N\ge 1$ of cycles, each lasting one period of a function, $T=2\pi N/\omega$, define the lock-in quadratures in a peak convention:These integrals average the products with dimensionless references $2\cos \omega t$ and $-2\sin \omega t$; $X$ and $Y$ have the input's units. The minus sign makes positive $\varphi$ give positive $Y$. The trigonometric functions giveThe oscillating terms have zero integral over the stated interval, proving $X=A\cos \varphi$ and $Y=A\sin \varphi$. Their orthogonality turns an oscillating signal into two constant coordinates.

Represent those coordinates as a vector, or as the complex number $Z=X+\mathrm{i}Y$. Its modulus and complex argument recover the peak amplitude and phase difference:Here $atan2(Y,X)$ selects the angle with the correct signs of both coordinates, rather than applying the arctangent to $Y/X$ and losing the quadrant. When $X=Y=0$, the amplitude is zero and the phase difference is undefined. When $\varphi =\pi /2$, $X=0$ despite a nonzero signal; measuring $Y$ prevents this blind spot. Rotating both references by a fixed angle rotates the coordinate vector but preserves $R$.

The factors of two above deliberately recover peak amplitude. The SR830 instead reports the selected sinusoidal component in root mean square units: in a calibrated, zero-offset display, $R_{\mathrm{rms}}=R/\sqrt{2}$. For $R=0.4\mathrm{V}$ this is $0.282843\mathrm{V}$, while the peak-to-peak span of that component is $0.8\mathrm{V}$. Display normalization, reference sign, gain and angle units must be checked for the particular instrument.^[ref]

For alternating masses $m_A,m_B$, nearest-neighbour spacing $a$ and equal spring constants $\alpha$, the zone-centre optical angular frequency and acoustic group velocity obeyTherefore $\alpha =m_A{m}_B{\omega }_{\mathrm{op}}^2(0)/[2(m_A+m_B)]$ and $a=v_s\sqrt{(m_A+m_B)/(2\alpha )}$. These infer effective model parameters from a measured intercept and initial slope of a phonon dispersion relation. Fitting a real polar crystal to a nearest-neighbour scalar model does not guarantee that every measured point or its structural spacing is reproduced exactly.

The Lorentz force is the force exerted by an electric field $\mathbf{E}$ and a magnetic field $\mathbf{B}$ on a body with electric charge $q$ and velocity $\mathbf{v}$. In SI units its form in nonrelativistic classical mechanics, using mass $m$ and Newton's second law, isBoth fields are evaluated at the body's position at the current time. The electric field contributes $q\mathbf{E}$ even at rest, whereas the magnetic force depends on velocity through a cross product. ^[ref]; ^[ref].

For a concrete charged particle in a uniform magnetic field, use nondimensionalization to stipulate $m=q=B=1$, magnetic field $\mathbf{B}=(0,0,1)$, zero electric field, initial position $(0,0,0)$, and initial velocity $(1,0,0)$. These are rescaled quantities, not an assertion that the mass, electric charge, and magnetic field have the same physical units. At the start, $(1,0,0)\times (0,0,1)=(0,-1,0)$, so the magnetic acceleration points toward $-y$. The resulting circular motion isAt $t=\pi /2$ the position is $(1,-1,0)$ and the velocity is $(0,-1,0)$; at $t=\pi$ the position is $(0,-2,0)$ and the velocity is $(-1,0,0)$. Thus the circle has center $(0,-1,0)$ and radius one. Viewed from $+z$ toward the $xy$ plane, the motion is clockwise. Changing only the electric charge to $q=-1$ gives position $(\sin t,1-\cos t,0)$ and reverses the bending to counterclockwise.

Adding initial velocity $v_z=0.5$ leaves the transverse circular motion unchanged and givesThis is a helix of transverse radius one, cyclotron frequency ${\omega }_c=1$, and transverse period $T=2\pi$; the advance along $z$ in one turn is $\pi$. The parallel component of velocity stays constant because the magnetic force has no parallel component: there is no magnetic acceleration along the magnetic field. If the initial velocity were entirely parallel to the magnetic field, the magnetic force would vanish and the path would be straight. ^[ref].

More generally, with zero electric field, no other forces, and a uniform constant magnetic field of magnitude $B>0$, decompose velocity into perpendicular and parallel components. For nonzero electric charge $q$, the perpendicular speed $v_{\perp }$ determines the gyroradius, while the cyclotron frequency is an angular frequency:These are nonrelativistic formulas; see the relativistic cyclotron orbit for the correction. The beauty is that increasing perpendicular speed increases the radius in exactly the proportion needed to leave the period unchanged. Conversely, known magnetic field, perpendicular speed, and measured radius determine the magnitude of the electric charge divided by mass. This makes the same geometry useful for comparing bodies with different electric charge and mass.

Two short checks explain the result. First, the cross product gives $\mathbf{v}\times \mathbf{B}=(-\sin t,-\cos t,0)$ for both stipulated positive-electric charge paths, exactly their acceleration. Second, the transverse magnetic force supplies the centripetal acceleration, so $m{v}_{\perp }^2/r=|q|v_{\perp }B$; for $v_{\perp }>0$ this gives the gyroradius, and ${\omega }_c=v_{\perp }/r$ gives the cyclotron frequency.

Magnetic forces do no work: the instantaneous power of the magnetic part of the Lorentz force vanishes because a cross product is perpendicular to its first argument. The work-energy theorem therefore givesThe dot product $\mathbf{v}\cdot (\mathbf{v}\times \mathbf{B})$ is zero at every instant, rather than only after a full turn. With zero electric field and no other forces, the kinetic energy and speed remain constant; the electric field can increase or decrease kinetic energy according to the sign of $q\mathbf{E}\cdot \mathbf{v}$. In the two stipulated examples the constant kinetic energy is respectively $K=1/2$ and $K=5/8$ in rescaled units. The extra parallel velocity raises kinetic energy without changing the gyroradius or the cyclotron frequency.

For an electric example in rescaled units, stipulate $m=q=1$, zero magnetic field, constant electric field $\mathbf{E}=(2,0,0)$, no other forces, and rest at $t=0$. The Lorentz force gives acceleration $(2,0,0)$ and velocity $(2t,0,0)$, hence kinetic energy $K=2t^2$ and power $dK/dt=4t=q\mathbf{E}\cdot \mathbf{v}$. At $t=1$, the kinetic energy has risen from zero to two: the electric field has done positive work.

The existing electromagnetic field tensor formulation expresses the same interaction covariantly. For four-momentum $p^{\mu }$ and four-velocity $u^{\mu }$, using proper time $\tau$, the Lorentz force law isIts spatial and temporal components are