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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 $\omega =2\pi \nu$.

Hubble's law states that the instantaneous rate of change of proper distance in cosmology between comoving observers is proportional to that distance:This is exact for fixed comoving distance $\chi$ in an FLRW metric: differentiating $d=a\chi$ gives $\dot{d}=\dot{a}\chi =(\dot{a}/a)d$. The dot uses cosmic time, and both distances are compared on spatial slices at the same cosmic time. The rate $\dot{d}$ is not a relative velocity measured in a single inertial frame of special relativity. Thus the instantaneous relation is not an exact observed redshift-distance formula. The associated Hubble flow describes the ideal expansion; peculiar velocities describe departures from it.

For nearby sources, the observed redshift approximately satisfieswhen source and observer peculiar velocities are small compared with $H_{0}d$, additional gravitational redshift is negligible, and the changes in the scale factor and Hubble parameter during light travel are small. Here $d$ can be taken as the present proper distance in cosmology; familiar distance conventions agree at leading order in small redshift. Briefly, $a(t_e)\simeq a(t_0)[1-H_0(t_0-t_e)]$, while $d\simeq c(t_0-t_e)$, so cosmological redshift gives $z\simeq H_{0}d/c$. At greater redshift, use an expansion history and an explicit distance convention instead.

Stipulate the illustrative Hubble constant $H_0=70\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ and speed of light $c=299792.458\mathrm{km}{\mathrm{s}}^{-1}$. A $\mathrm{Mpc}$ is a million parsecs. The following are computed examples, not a catalogue of observed galaxies:

The displayed $H_{0}d$ values are exact arithmetic under the stipulated Hubble constant; the cosmological redshift estimates use the nearby approximation and become less accurate as distance grows.

Hubble's law turns a nearby observed redshift into an approximate distance indicator and makes the uniformity of cosmic expansion tangible: twice the proper distance in cosmology means twice the instantaneous separation rate. Recentring the ideal Hubble flow on another comoving observer preserves the same proportionality, as illustrated under scale factor, so recession in every direction does not select a central observer.

In his 1929 PNAS paper, Edwin Hubble presented a relation between the distances and radial velocities of extragalactic objects. His fits gave coefficients of $465$ and $513\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ for 24 individual objects and nine groups respectively, and he used about $500\mathrm{km}{\mathrm{s}}^{-1}\mathrm{Mp}{\mathrm{c}}^{-1}$ as a rounded representative value. Those historical fits used the paper's distance estimates; they are distinct from the stipulated modern-scale illustration above. ^[ref].

An interferometer combines waves from two or more paths to measure their interference.

A physicist whose research joined waves, optics, acoustics and precise measurement. He maintained a laboratory at Terling and served as Cavendish professor of experimental physics from 1879 to 1884. At Cambridge he developed laboratory courses; later he donated his Nobel award proceeds toward a Cavendish laboratory extension.^[ref]

The Rayleigh-Ramsay argon density experiment, pursued with William Ramsay, connects a small discrepancy in gas weighing to chemical separation. Its lesson is practical: agreement within one preparation method does not ensure that a second method produces the same chemical substance. The gas density model explains how composition and filling conditions both affect the readout.

Linear mass density is mass per unit length, measured in $\mathrm{kg}{\mathrm{m}}^{-1}$. A uniform string of mass $m$ and length $L$ has $\mu =m/L$; with constant tension $T$, its transverse wave speed is $\sqrt{T/\mu }$. Mass density instead measures mass per unit volume.

For a locally oscillatory wave written $u(\mathbf{x},t)=A(\mathbf{x},t)\cos \varphi (\mathbf{x},t)$, the local wavevector is ${\mathbf{k}}_{\mathrm{loc}}=\nabla \varphi$, where $\nabla$ is the spatial gradient. The local wavelength $2\pi /|{\mathbf{k}}_{\mathrm{loc}}|$ measures local wave phase spacing normal to the wavefront. Interpreting it as a crest separation requires the amplitude and local wavevector to vary little over one wavelength, as in the WKB approximation. Along a unit direction $\mathbf{e}$ the corresponding local repeat distance is $2\pi /|{\mathbf{k}}_{\mathrm{loc}}\cdot \mathbf{e}|$, if that dot product is nonzero; oblique sampling measures this projected spacing.

A real snapshot alone does not uniquely specify $A$ and $\varphi$: their meaning requires a chosen oscillatory component or physical approximation. Strong interference, zeros of amplitude, and rapidly changing media can defeat a single-carrier description. If $\nabla \varphi =0$, the local reciprocal formula gives no finite wavelength; it does not establish an exact global spatial period.

A long-wavelength transverse disturbance of an initially isotropic magnetic second-moment tensor behaves as an elastic shear wave when small-scale rearrangements of the tangle are omitted. For $v_A^2=⟨B^2⟩/3$, its fluid displacement obeys ${\partial }_t^2\mathbf{\xi }=v_A^2{\nabla }^2\mathbf{\xi }$ and $\nabla \cdot \mathbf{\xi }=0$ on the oscillatory branch. The tensor perturbation is $\delta M_{ij}=v_A^2({\partial }_i{\xi }_j+{\partial }_j{\xi }_i)$. A zero mean magnetic field does not imply zero restoring magnetic tension.

For propagation perpendicular to an equilibrium magnetic field, an ordinary plasma wave has its electric field parallel to that magnetic field, so the particle response involves no magnetic deflection. With immobile ions, the dispersion relation is ${\omega }^2={\omega }_{pe}^2+c^2{k}^2$. It is a transverse electromagnetic wave, despite its parallel particle motion.

Average the equations in part b and make exactly the prescribed scale-separation approximation: the retained perturbations vary on lengths much greater than $\ell$, and perturbations at the tangle scale are omitted. Products of two retained perturbations can then be neglected. Writing $\mathbf{v}=\delta \mathbf{u}$, $m=\delta M$ and $\pi =\delta p$, the linearized equations areNo unmodelled small-scale response has been inserted into this closure.

Introduce the fluid displacement $\mathbf{\xi }$ by $\mathbf{v}={\partial }_t\mathbf{\xi }$ and choose $\nabla \cdot \mathbf{\xi }=0$. Integrating the magnetic second-moment tensor equation giveswhere $s$ is a time-independent integration tensor set by the initial stress. For a pure oscillatory normal mode, $s=0$. Its restoring force is therefore $v_A^2{\nabla }^2\mathbf{\xi }$: the isotropic magnetic tangle has the same transverse response as an incompressible elastic medium with shear modulus ${\rho }_0{v}_A^2$.

For a plane wave proportional to $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, with $\omega \ne 0$, incompressibility gives $\mathbf{k}\cdot \hat{\mathbf{v}}=0$. The tensor equation and magnetohydrodynamic momentum equation becomeDotting the second equation with $\mathbf{k}$ yields $\hat{\pi }=0$ for $k\ne 0$, since $k_i{k}_j{\hat{m}}_{ij}=0$. Substitution then givesThere are two independent transverse polarizations. The phase speed is $v_A$ in every direction, and the two signs of $\omega$ give group velocities $\pm v_A\hat{\mathbf{k}}$. These magnetoelastic waves in an isotropic magnetic tangle exist because a shear deformation distorts the second moment of the tangled magnetic field, producing a restoring magnetic tension even though its mean is zero.

For comparison, a uniform magnetic field with Alfvén velocity ${\mathbf{B}}_0$ gives the incompressible Alfvén wave relation ${\omega }^2=(\mathbf{k}\cdot {\mathbf{B}}_0{)}^2$. Its frequency depends on the direction of $\mathbf{k}$, and energy propagates along ${\mathbf{B}}_0$. Here the averaged restoring response is isotropic: there is no preferred mean-field direction. Both systems resist transverse deformation through magnetic tension, rather than through compressive sound.

The propagating branch is not the entire perturbation space. For example, $\mathbf{v}=0$, $m_{ij}=f(\mathbf{r}){\delta }_{ij}$ and $\pi =f(\mathbf{r})$ solve the linearized equations without propagation, for any small, slowly varying $f$. Such a static stress perturbation in magnetoelastic wave equations balances a pressure gradient. Thus the requested wave conclusion applies to the oscillatory branch; it must not exclude stationary initial stress perturbations.

A uniform conducting fluid supports three propagating magnetohydrodynamic wave branches. Ideal magnetohydrodynamics couples fluid velocity to the magnetic field through the induction equation; the Lorentz force then supplies magnetic tension and magnetic pressure restoring forces. Let $\theta$ be the angle between wave vector $\mathbf{k}$ and background field ${\mathbf{B}}_0$, and define Alfvén speed $v_A=B_0/\sqrt{{\mu }_0{\rho }_0}$ and adiabatic sound speed $c_s^2=\gamma p_0/{\rho }_0$.

The shear Alfvén wave hasIts velocity perturbation is perpendicular to both $\mathbf{k}$ and ${\mathbf{B}}_0$. It bends field lines without first-order compression, so magnetic tension is the restoring force. Its group velocity follows the background field; the displayed phase speed along $\mathbf{k}$ is not the full group speed.

The fast magnetosonic wave and slow magnetosonic wave involve coupled compression and magnetic perturbations, with displacement in the plane of $\mathbf{k},{\mathbf{B}}_0$:The plus sign is the fast branch. Parallel propagation gives characteristic speeds $c_s,v_A$; perpendicular propagation gives fast speed $\sqrt{c_s^2+v_A^2}$ while the other two phase frequencies vanish. Degenerate directions require care in assigning polarizations, rather than claiming three distinct nonzero frequencies everywhere.

These waves transport energy and momentum through magnetized stellar atmospheres and astrophysical plasmas. Their restoring mechanisms distinguish them from ordinary acoustic waves. Resistivity, viscosity, ion-neutral friction and kinetic effects introduce damping or change the model; an ideal undamped dispersion relation alone does not calculate coronal heating.

A fluid particle follows a Lagrangian trajectory, so its position satisfies $\dot{x}=u(x,t)$ rather than the velocity evaluated permanently at its initial position. Assume $\omega \ne 0$, and writeSet $\alpha =k{X}_0$ and $\theta =\alpha -\omega t$. A Taylor expansion of the velocity about $X_0$ givesThus $\dot{a}=\sin \theta$, andAt the next order,so the required second-order trajectory in a sinusoidal velocity field isBoth corrections vanish at $t=0$. All retained terms have units of length. The secular contribution gives the Stokes driftalthough the Eulerian time average of the given velocity is zero. The particle samples the wave at its displaced position, producing this second-order drift. The expansion is in the dimensionless parameter $|k{u}_0/\omega |\ll 1$, for fixed times; it ceases to be uniform when the accumulated drift changes the wave phase by order unity.

The real Airy function asymptotics used for the Airy turning-point connection formula are, for $z,s\to +\infty$,These real-axis asymptotics fix both normalization and phase.^[ref] Near $a$, $I\sim 2z^{3/2}/3$, $\Phi \sim 2(-z{)}^{3/2}/3$, and ${\kappa }^{-1/2}\sim {\ell }^{1/2}{z}^{-1/4}$. Matching to the constants of c gives $C_A=2\sqrt{\pi }{\ell }^{1/2}D$ and $C_B=\sqrt{\pi }{\ell }^{1/2}G$. ThereforeIn particular, the decaying component matches the cosine with the requested $-\pi /4$ shift, with a factor two in its oscillatory coefficient. The growing component matches a sine with the indicated minus sign. Expanding into travelling waves givesThe factor two is essential when converting the real oscillatory components to incident and reflected travelling waves.

Let $J(x)={\int }_x^b\kappa (s)ds=\eta -I(x)$. The WKB approximation in the barrier can be rewritten asThe right Airy turning-point connection formula givesso $A_{\mathrm{t}}=D_R+i{G}_R/2$ and $A_{\mathrm{back}}=D_R-i{G}_R/2$. Impose no incident wave from the right, $A_{\mathrm{back}}=0$. ThenConsequently, within the leading Airy turning-point connection formula calculation,For an opaque barrier, $\eta \gg 1$,The transmission coefficient is a ratio of Poynting fluxes; the $k^{-1/2}$ normalization already includes their local wavenumber factors. The displayed denominator is useful for consistent matching and Wronskian flux conservation, but does not claim accuracy for subleading terms beyond the WKB approximation, or for merging wave turning points.

For real $k^2(x)$, the wave equation has the conserved Wronskian fluxFor this ordinary plasma wave, the time-averaged Poynting flux is $⟨S_x⟩=c^2\mathcal{J}/(8\pi \omega )$. With no incident wave from the right, evaluating the same Wronskian flux on both sides provesThe coefficients in f also satisfy this identity algebraically, before the opaque-barrier expansion.

A plasma wave resonance makes the ideal cold description singular and requires an appropriate causal regularization. A passive resonant layer can absorb wave energy into particle motion, or convert it into another branch. For the two measured outgoing channels one then haswith $A>0$ when resonant absorption or uncounted mode conversion occurs. A numerical value of $A$ requires the layer physics; a resonance by itself does not specify a universal loss fraction. Counting all converted and absorbed energy preserves conservation of energy.

Take $d$ as the centre-to-centre separation and $ϵ={ϵ}_0{ϵ}_r$. Use the ideal lossless two-conductor transmission line model, with nonmagnetic surroundings and negligible internal-wire inductance. Opposite line charges $+{\lambda }_q$ and $-{\lambda }_q$ approximate the cylindrical conductors when $d\gg a$. Integrating their electric fields, or adding their logarithmic electric potentials, gives the potential differenceThis is the two-wire capacitance at large separation. For opposite electric currents $+I$ and $-I$, the magnetic fields between the wires add. The linked magnetic flux per unit length is approximatelyThus the external inductance of a thin two-wire line is $L^{\prime }\simeq {\mu }_0\ln (d/a)/\pi$. The lossless telegrapher's equations give a forward wave with $V/I=\sqrt{L^{\prime }/C^{\prime }}$, so its characteristic impedance isThe product $L^{\prime }{C}^{\prime }={\mu }_0ϵ$ also gives the propagation speed $c/\sqrt{{ϵ}_r}$, an independent consistency check. The exact ideal external-field geometry replaces $\ln (d/a)$ by $\mathrm{arcosh}(d/(2a))$. If the wire separation were specified between surfaces, its centre separation would be $d+2a$, with the same leading large-separation result. Finite electrical conductivity and internal-wire magnetic energy require a more detailed line model.

Write $\mu ={\mu }_0{\mu }_r$ and $ϵ={ϵ}_0{ϵ}_r$ for homogeneous scalar material constants, and use $\mathbf{J}=\sigma \mathbf{E}$. Taking the curl of Faraday's law of induction and substituting the Ampere-Maxwell law givesThe vector calculus identity for a double curl therefore yieldsThe second term represents electric current response, rather than displacement current. A dispersive $\sigma (\omega )$ is used in the frequency domain; its general time-domain counterpart is a temporal response convolution.

For the convention $e^{i(\mathbf{k}\cdot \mathbf{r}-\omega t)}$, the plane wave equation isDecompose the electric field into components parallel and perpendicular to $\mathbf{k}$. With $S=\mu ϵ{\omega }^2+i\mu \sigma \omega$, the longitudinal eigenvalue is $S$ and each transverse eigenvalue is $S-k^2$; thus the determinant is $S(S-k^2{)}^2$. For a transverse electromagnetic wave, $\mathbf{k}\cdot {\mathbf{E}}_0=0$ andwhen ${\mu }_r=1$ and the background ${ϵ}_r\simeq 1$. A real dissipative electrical conductivity generally makes $k$ complex, with the decaying branch chosen for a passive medium.

For a cold, collisionless, unmagnetized plasma, treat ions as stationary and write $e>0$ for the magnitude of the elementary charge. The electron equation of motion is $-i\omega m_e\mathbf{v}=-e\mathbf{E}$, soThis reactive response includes no collision-induced Joule heating. Substitution into the transverse dispersion relation givesThe quantity ${\omega }_p$ is the electron plasma frequency. The refractive index is real above this cutoff; below it, the wave is evanescent. The electron response is represented here by $\sigma (\omega )$, so it must not also be counted a second time in the background permittivity.

For a nonzero longitudinal component, taking the dot product of the matrix equation with $\mathbf{k}$ instead gives $S(\mathbf{k}\cdot {\mathbf{E}}_0)=0$. The same cold plasma response makes $S=({\omega }^2-{\omega }_p^2)/c^2$, henceAt nonzero $k$, the remaining equation forces ${\mathbf{E}}_0$ to be parallel to $\mathbf{k}$, and Faraday's law of induction gives ${\mathbf{B}}_0=0$. These are electrostatic Langmuir waves, with no $k$ dependence of angular frequency in the zero-temperature model. Equilibrium neutrality does not forbid their perturbed charge density; Gauss's law requires such a density when $\mathbf{k}\cdot {\mathbf{E}}_0\ne 0$.

Write the velocity field as $u=U+u\prime$, $v=v\prime$. The uniform background has zero relative vorticity, and the perturbation has $\xi \prime =v{\prime }_x-u{\prime }_y$. Keeping only terms linear in the perturbations givesThe incompressibility condition permits a streamfunction $\psi$ with $u\prime =-{\psi }_y$, $v\prime ={\psi }_x$, hence $\xi \prime ={\nabla }_h^2\psi$. Thus the Linear Rossby-wave equation here isChoose a plane wave $\psi =\mathrm{Re}\{\hat{\psi }{e}^{i(kx+ly-\omega t)}\}$, with $K^2=k^2+l^2>0$. Substitution givesand therefore the local Rossby-wave dispersion relation in a zonal jetThe $Uk$ term is the background Doppler shift; the intrinsic phase velocity is westward. A real physical wave can always be represented using the conjugate wavevector if necessary to choose $\omega \ge 0$.

The relevant ground-frame zonal phase velocity and group velocity areTime increases downward in the source diagram. The individual phase contours lean towards increasing longitude as time increases, so their phase velocity is eastward. The envelope moves eastward faster: successive crests grow at the downstream edge and decay upstream. This illustrates Rossby-wave downstream development.

For the intended purely zonal wave packet, $l=0$, andConsequently, with $U,\beta >0$ and the positive-frequency choice $k>0$,There is a small interpretation qualification: if “zonally propagating” specified only the plotted direction while allowing an unknown nonzero meridional wavenumber, eastward phase velocity alone would imply $k^2+l^2>\beta /U$, not the stronger bound on $k$ alone. In this uniform model a strictly zonal group velocity with $\beta k\ne 0$ does imply $l=0$, which supplies the intended interpretation.

Choose both waves as sinusoids with a common zero phase and amplitude $A$. The trigonometric sum identity givesThe slowly varying wave envelope has constant phase when $\delta x-ϵt$ is constant. Its speed is $ϵ/\delta$, and in the close-frequency limit the group velocity isIt describes the motion of a narrow-band envelope in a medium where the constituent waves retain a coherent packet interpretation; it is not an unconditional signal-speed theorem in strongly absorbing or anomalously dispersive media. The carrier phase velocity is $v_p=\omega /k$. In a nondispersive medium, $\omega =vk$ with constant $v$, so $v_g=v=v_p$.

Convert the fixed position to $x=0.20\mathrm{m}$. The element's displacement is then $y(t)=0.27\sin (2.4-500t)\mathrm{mm}$. Taking time derivatives at fixed position gives its transverse velocity and acceleration:Equivalently the prefactors in SI units are $-0.135\mathrm{m}{\mathrm{s}}^{-1}$ and $-67.5\mathrm{m}{\mathrm{s}}^{-2}$. The phase is measured in radians. At $t=4\mathrm{s}$,These are the transverse motions of the material element, not the propagation speed of the wave along the string.

For the reflection of a string wave from a freely moving endpoint mass, let $c=\sqrt{T/\mu }$, $k=\omega /c>0$ and use the complex time-harmonic wave conventionThe string pulls its endpoint toward negative $x$, so its transverse force on the attached mass is $-T{y}_x(0,t)$. Newton's second law gives the boundary condition $m{y}_{tt}(0,t)=-T{y}_x(0,t)$. Substitution yieldsHere $Z$ is the characteristic impedance of a stretched string. Since the incident and reflected waves share the same string, their mean power ratio is the squared modulus of this displacement reflection coefficient:There is no remaining fraction of the mean incident energy: the freely moving mass provides no dissipation and no onward transmission channel. It exchanges kinetic energy with the string during each cycle but has zero mean absorbed power. Indeed the energy flux of a transverse string wave satisfieswhose cycle average is zero. As checks, $m\to 0$ gives $R=+1$, the free-end reflection of a transverse string wave, while $m\to \infty$ at fixed $\omega >0$ gives $R=-1$, the fixed-end limit.