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For an unstable Taylor–Goldstein equation mode in a finite channel, take $a=1$ in the power-transformed Taylor–Goldstein energy identity. Its imaginary part givesThus the real phase velocity lies strictly between the extremes of a nonconstant smooth shear profile. Equality would force the regular eigenfunction to vanish on an interval, hence everywhere by uniqueness for its ordinary differential equation.

The phase velocity is the motion of a constant-wave phase surface. For a one-dimensional plane wave $u=a\cos (kx-\omega t+{\varphi }_0)$ with real $k\ne 0$ and real $\omega$, fixing $kx-\omega t+{\varphi }_0$ givesEach wave crest of this monochromatic travelling wave advances at that signed phase velocity. Its magnitude is the phase speed $|c_p|=\lambda /T$, where $\lambda =2\pi /|k|$ is the wavelength and $T=2\pi /|\omega |$ is the temporal period when $\omega \ne 0$. Choosing $\omega >0$ makes positive $k$ mean rightward phase velocity and negative $k$ mean leftward phase velocity. Reversing both signs leaves the real plane wave and its phase velocity unchanged. At $k=0$ there are no spatial crests to track, so $\omega /k$ is undefined; a finite limit as $k\to 0$ is a limiting phase speed, not the speed of a spatially uniform oscillation.

In several dimensions, a plane wave has wave phase $\mathbf{k}\cdot \mathbf{x}-\omega t+{\varphi }_0$. Constant-wave phase planes obey $\mathbf{k}\cdot \mathbf{v}=\omega$. Their normal phase velocity isTangential motion within a plane does not move that wavefront, which is why the normal direction fixes the vector convention. In an anisotropic medium the group velocity need not point along this wavevector. Likewise the horizontal phase velocity of a tilted wavefront is its crossing speed along a horizontal line, rather than a component of the normal phase velocity.

For the nondispersive wave branch $\omega =ck$, a wave crest and a wave envelope both move at $c$. For deep-water surface gravity waves, $\omega =\sqrt{gk}$ gives $c_p=\sqrt{g/k}$ and group velocity $c_g=c_p/2$. With the illustrative choices $g=9.81\mathrm{m}{\mathrm{s}}^{-2}$ and $\lambda =1\mathrm{m}$, the model predicts $c_p\simeq 1.25\mathrm{m}{\mathrm{s}}^{-1}$ and $c_g\simeq 0.625\mathrm{m}{\mathrm{s}}^{-1}$. These are calculated values for negligible surface tension and sufficiently large $kh$, not measurements. A wave crest moves forward through the wave packet, so timing a crest does not time the packet's journey.

There is a direct laboratory meaning to this distinction. MIT's 2005 surface-wave lab uses a long water tank, a controllable hydraulic paddle, and an absorbing beach. Resistance probes convert submerged length into voltage; calibration converts their readouts into free surface elevation versus time. Two probes a known distance apart, backed by a ruler and stopwatch, let students time a wave crest and separately time a feature of the wave packet. The lab records depth and varies frequency and wave amplitude.^[ref]

For a clean measurement of phase velocity, fit the periodic portion of both records or follow one identifiable wave crest. A phase difference is only known modulo $2\pi$: probe spacing below half the shortest relevant wavelength, or continuity while sweeping frequency, avoids assigning the wrong whole-cycle delay. Then compare the timed phase velocity with $\lambda /T$. Tracking pepper grains instead measures fluid-particle motion, which has a different meaning from either phase velocity or group velocity.

The Klein-Gordon packet velocities example makes the distinction especially sharp: its phase velocity exceeds $c$, yet its narrow wave envelope moves below $c$. A carrier wave labels successive oscillations; its speed alone does not establish information velocity. The front velocity depends on the full causal model, including angular frequencies far outside the carrier's narrow range.

For a mode with nonreal phase velocity $c$, let $V=U-c$, choose a continuous branch of $V^a$, and put $\hat{w}=V^{a}q$. Under impermeable boundary conditions, multiplying the transformed Taylor–Goldstein equation by $V^{2a}\overline{q}$ and applying integration by parts givesChoosing $a=1/2$ gives the Miles–Howard theorem; choosing $a=1$ makes the real phase velocity of an unstable mode a weighted mean of $U$.

Refraction changes the direction of a transmitted wave when it crosses between media with different phase velocities. For light at a planar interface, Snell's law relates the angles measured from the normal vector through the refractive indices. At normal incidence the direction remains unchanged even though the wavelength and phase velocity can change. The worked example under geometrical optics distinguishes transmission from total internal reflection.

For light, refraction explains why a submerged object looks nearer the surface and why an optical lens can form an image. A wavefront crossing a boundary must retain a consistent wave phase along that boundary: its normal spacing changes with phase velocity, and oblique light rays change direction. Normal incidence separates these two effects especially clearly.

Take stationary, transparent, homogeneous media with isotropy, linearity of their optical response, and real positive refractive indices, at one chosen frequency $f$. With $c$ the vacuum speed of light and ${\lambda }_0=c/f$ the vacuum wavelength, medium $j$ hasThe frequency is unchanged at a stationary boundary; the phase velocity, wavelength and wavenumber can change. Here $n_j$ is a phase refractive index, not the group refractive index. If there is optical dispersion, evaluate $n_j$ at this same frequency on both sides.^[ref]

For an original idealized air–water example, stipulate $n_{\mathrm{a}}=1$ and $n_{\mathrm{w}}=4/3$, with vacuum wavelength $600\mathrm{nm}$. Then $f=c/(600\mathrm{nm})\approx 4.99654\times 10^{14}\mathrm{Hz}$, the water phase velocity is $3c/4$, and its wavelength is $450\mathrm{nm}$. At an air angle of incidence $30^{\circ }$, Snell's law gives a water angle of refraction $\arcsin (3/8)\approx 22.0243^{\circ }$. These stipulated values illustrate the relations; actual refractive indices depend on material conditions and frequency. By reversibility of an optical ray, the same path emerging from water has outgoing air angle $30^{\circ }$.

Choose $k,m>0$, $K=(k^2+m^2{)}^{1/2}$ and the positive angular frequency $\omega =Nk/K$. Then the phase velocity is $Nk(k,m)/K^3$ and the group velocity is $N(m^2,-km)/K^3$. Their dot product vanishes and their sum is horizontal with magnitude $N/K$. Draw the group velocity from the tip of the phase velocity: the two form the perpendicular sides of a right triangle. The horizontal hypotenuse is their sum, not the horizontal phase velocity $\omega /k$.