Phase velocity, group velocity, and dispersion

Start with two moving marks

At t = 0 s, two chosen marks in our supplied wave model are at x = 0 m. After 0.2 s, one is at x = 0.406 m and the other at x = 0.802 m. Both move to the right, but they track different features. Their different speeds are the question this lesson will explain.

The circle follows one fixed phase of a rapidly oscillating factor, and sits on the actual displacement curve. The square follows one maximum of a broader modulation, and sits on its positive bounding curve. The square’s height is not generally the displacement at that location. Neither mark identifies a moving piece of material.

Algebra, functions, graphs, and logarithms prepares substitution and ratios. Geometry, trigonometry, and coordinate systems prepares cosine and phase; the needed identity is introduced below. Units, standards, and physical constants is a published return path for quantity notation.

Begin with the two marks and their table. Derivatives, Taylor series, and local approximation is an optional published return path for the later derivative and narrow-band discussion. No unpublished wave lesson is required.

Declare the waves and their dispersion relation

Let x be position in metres, positive to the right; t is time in seconds, with t ≥ 0. The signed displacement y is in metres. For a sinusoid, k is angular wave number in rad/m and ω is angular frequency in rad/s. Its phase is kx − ωt. A dispersion relation specifies which ω goes with each k.

Use the stipulated positive-k relation ω(k) = ck + βk³, where c = 1 m/s and β = 0.01 m³/s. Radians have dimension one in SI, so both terms have the units of angular frequency. This exact mathematical law is not a measured material specification.

Choose two contributions of equal amplitude A = 1 mm, wave numbers k₁ = 9 rad/m and k₂ = 11 rad/m, and zero initial phases. Substitution gives ω₁ = 16.29 rad/s and ω₂ = 24.31 rad/s. Define y_j(x,t) = A cos(k_j x − ω_j t) and assume linear superposition y = y₁ + y₂.

The waves are defined for all real x and all t ≥ 0. These two indefinitely extended sinusoids are not a localized pulse or a finite-duration transmission. We use them to compare precisely defined features.

Compare the two moving features

The identity cos a + cos b = 2 cos((a−b)/2) cos((a+b)/2) separates the sum into two factors. Let k̄ = (k₁+k₂)/2 = 10 rad/m, ω̄ = (ω₁+ω₂)/2 = 20.3 rad/s, Δk = k₂−k₁ = 2 rad/m and Δω = ω₂−ω₁ = 8.02 rad/s.

Define the carrier phase Φ = k̄x − ω̄t and modulation phase Ψ = (Δk x − Δω t)/2. Then y = B cos Φ, with B = 2A cos Ψ. B is signed. Its nonnegative magnitude is M = |B|, so the displacement is always between −M and +M. The signed curves +B and −B exchange which one is upper when B changes sign.

For compact formulas, let ξ = x/(1 m) and τ = t/(1 s). The exact sum is y = (2 mm) cos(ξ − 4.01τ) cos(10ξ − 20.3τ). Choose the circle by Φ = 0 and the square by Ψ = 0. Solving these level equations gives their positions, rather than assuming that a visible crest must move at one universal wave speed.

The selected circle moves at v_C = ω̄/k̄ = 2.03 m/s. The selected square moves at v_M = Δω/Δk = 4.01 m/s. At t = 0.2 s, the circle’s displacement is about 1.845 mm, while the square is at the positive bound of 2 mm. Other periodic copies of these features also exist.

Follow one component’s phase

For a single component, hold its phase fixed: kx − ωt = a constant. Since k is nonzero, x changes at v_phase(k) = ω(k)/k. This is phase velocity: the motion of a fixed phase of that sinusoid, rather than its local transverse displacement rate.

For the supplied law, v_phase(k) = c + βk². The 9 rad/m component has phase velocity 1.81 m/s, while the 11 rad/m component has 2.21 m/s. Their different phase velocities are dispersion in this example.

The carrier factor combines both components. Its speed 2.03 m/s is not either component’s speed. Evaluating the law at k = 10 rad/m instead gives ω(10) = 20 rad/s and phase velocity 2 m/s. This hypothetical single component is different from using the actual mean frequency ω̄ = 20.3 rad/s.

Separate finite ratios from derivatives

The finite modulation quotient uses two distinct wave numbers. A derivative uses the limiting slope at one wave number. Define v_g(k₀) = dω/dk at k₀, called group velocity. Differentiating the supplied polynomial gives v_g(k₀) = c + 3βk₀². At k₀ = 10 rad/m, this is 4 m/s, not the finite value 4.01 m/s.

To see the limit without hiding the difference, choose k₁ = k₀−h and k₂ = k₀+h, with h > 0 in rad/m and k₀ > h. Expanding the cubic gives v_C = c + β(k₀² + 3h²) and v_M = c + β(3k₀² + h²). The extra terms are 3βh² and βh² respectively.

As h approaches zero, the finite carrier speed approaches ω(k₀)/k₀ and the finite modulation speed approaches dω/dk at k₀. At a finite spacing, retain the finite formulas. The table below describes the local single-wave ratio and derivative at each k, not the two selected feature speeds.

Control a local packet approximation

A localized wave packet would require more than our two extended sinusoids. For a later superposition concentrated near k₀, a first-order Taylor approximation to ω makes a slowly varying factor translate at v_g(k₀). Keeping its shape this way requires the neglected higher-order phase to stay small over the stated time interval.

For the cubic law, write k = k₀+δ. The exact expansion is ω(k₀+δ) = ω(k₀) + (c+3βk₀²)δ + 3βk₀δ² + βδ³. The final two terms are the omitted remainder in a first-order approximation.

Take k₀ = 10 rad/m and suppose a later packet’s spectral support is within |δ| ≤ Δ, with 0 < Δ ≤ 2 rad/m. Over 0 ≤ t ≤ T, the omitted phase is at most Tβ(3k₀Δ² + Δ³). A chosen dimensionless phase tolerance ε therefore requires this bound to be no larger than ε. A broader band or longer time can invalidate the approximation.

This is a conditional phase-error check, not a constructed packet or a guarantee for spectral tails outside the declared band. No packet bandwidth, tolerance or observation horizon has been supplied for a numerical packet example here. It does not establish an energy, information or front velocity.

Compare a nondispersive law

In a separate comparison, set c = 2 m/s and β = 0, keeping k₁ = 9 rad/m and k₂ = 11 rad/m. The law is now ω = ck. The two frequencies are 18 rad/s and 22 rad/s, and each component has phase velocity 2 m/s.

The mean ratio is 20/10 = 2 m/s; the difference ratio is 4/2 = 2 m/s; the derivative is also 2 m/s. Every contribution depends on x − ct, so their sum has the same translated spatial shape in this exact comparison.

The coincidence follows from this linear relation, not from the word “wave.” Returning to the original cubic law restores the distinctions among individual phase velocities, finite feature speeds and the local derivative.

State which feature your speed describes

Before reporting a speed, name the feature and its definition: a phase of one component, the carrier level Φ = 0, the modulation level Ψ = 0, or a derivative at a stated k₀. A mark on a bound answers a different question from a mark on the displacement curve.

The diagram contains two spatial snapshots of the mathematical model. The exact table supplies intermediate feature coordinates; no connecting line turns either mark into a material trajectory. The plotted window also does not make these whole-line sinusoids into a localized packet.

There is no universal ordering of phase, group, energy and front speeds supplied by this lesson. Our chosen law has a positive derivative larger than its single-wave phase ratio, but that is a result of these coefficients. Questions about energy transport or a newly sent signal need an appropriate physical model and additional conditions.

Worked examples

Try it yourself

1. For ω(k) = ck + βk³, with k in rad/m, ω in rad/s and dimension-one radians, what units must c and β have?

   Show the answer and reasoning

   c has units m/s; β has units m³/s.

   Both ck and βk³ must have the units of angular frequency. Multiplying their stated units by m⁻¹ and m⁻³ respectively gives s⁻¹.

2. For the main law at k = 9 rad/m, find ω and the speed of a fixed phase of that single component.

   Show the answer and reasoning

   ω = 16.29 rad/s and v_phase = 1.81 m/s.

   Substitution gives 9 + 0.01(9³) = 16.29 rad/s. Divide by 9 rad/m, rather than by a wave-number difference.

3. For the main law, evaluate v_g at k = 10 rad/m. Is it exactly the main pair’s finite modulation speed?

   Show the answer and reasoning

   v_g = 4 m/s. It differs from the main pair’s 4.01 m/s.

   Use c + 3βk² for the derivative. The finite pair instead uses its nonzero frequency and wave-number differences.

4. Keep the main c and β, but use k₁ = 8 rad/m and k₂ = 12 rad/m. Calculate ω₁, ω₂, v_C and v_M; compare with the local phase velocity and v_g at 10 rad/m.

   Show the answer and reasoning

   ω₁ = 13.12 rad/s; ω₂ = 29.28 rad/s; v_C = 2.12 m/s; v_M = 4.04 m/s. The local values at 10 rad/m remain 2 and 4 m/s respectively.

   The mean frequency is 21.2 rad/s and mean wave number 10 rad/m. The differences are 16.16 rad/s and 4 rad/m. A wider finite spacing changes the ratios without changing the derivative at the same reference point.

5. Under the separate law c = 2 m/s and β = 0, why do component, carrier, modulation and derivative speeds all equal 2 m/s for the 9 and 11 rad/m pair?

   Show the answer and reasoning

   Because ω is exactly proportional to k with the same coefficient 2 m/s.

   Individual ratios, the mean ratio, the difference ratio and the derivative all return that coefficient. Every declared component translates as a function of x−ct.

6. At t = 0.2 s, a square is at x = 0.802 m on the positive bound. Does this identify the actual displacement there, a material trajectory, or an exact localized-packet or energy speed?

   Show the answer and reasoning

   No. It identifies the selected Ψ = 0 modulation feature on a bound, moving at the finite value 4.01 m/s.

   The actual displacement at that location is different. The model contains two extended sinusoids, and no material trajectory, localized packet or energy-transport model has been supplied.

What to carry forward

• A velocity needs a named feature and a declared phase or modulation condition.
• For a single nonzero-k component, phase velocity is ω/k; group velocity at a point is dω/dk.
• Finite two-component ratios agree with local ratios and derivatives only in the appropriate limit or special relation.
• A narrow-band packet interpretation needs controlled higher-order phase error over its stated horizon.