The packet that will not keep its shape

A pulse on a stretched string travels down it and arrives looking exactly as it left. A pulse of deep-water waves travels across a pond and arrives longer, lower and stretched out. Both are superpositions of sinusoids; both obey a linear wave equation. The difference is one number — the second derivative of the dispersion relation — and it is the number the group velocity discards.

What the expansion says after the first term #

A packet is a superposition,

$$\psi(x,t) = \int A(k)\, e^{i(kx - \omega(k) t)}\, dk,$$

with $A(k)$ concentrated near a carrier $k_0$. Expanding $\omega$ about $k_0$,

$$\omega(k) = \omega_0 + v_g\,(k - k_0) + \tfrac12\beta\,(k - k_0)^2 + \dots, \qquad v_g = \left.\frac{d\omega}{dk}\right|_{k_0}, \quad \beta = \left.\frac{d^2\omega}{dk^2}\right|_{k_0},$$

and substituting. The zeroth term is a phase common to everything, which moves the crests. The first term collects into $e^{ik(x - v_g t)}$ and translates the envelope bodily — that is the group velocity, and if the expansion stopped there the packet would be rigid.

The second term is the one this essay is about. It multiplies each component by a phase quadratic in $(k - k_0)$, which is not a translation of anything: it puts the components out of step with one another in a way that grows with time, and the envelope broadens.

For a Gaussian packet of initial width $\sigma_0$ the result is exact:

$$\sigma(t) = \sigma_0\sqrt{1 + \left(\frac{t}{\tau}\right)^2}, \qquad \tau = \frac{\sigma_0^2}{|\beta|}.$$

Why the short one goes first #

The spreading time is $\sigma_0^2/|\beta|$, which is quadratic in the initial width. Halving the length of a packet therefore divides its spreading time by four, and the initial length is inversely related to the bandwidth, so this is the same statement twice over:

$$\tau \approx \frac{1}{|\beta|\,(\Delta k)^2}.$$

A packet with no bandwidth would never spread and would also never begin or end. A packet with a wide spectrum is short and comes apart almost at once. There is no way to have a short packet that lasts.

The reciprocity at the root of it is that a spread of wavenumbers and the envelope they build have a fixed product of widths. Narrowing the spectrum widens the packet and vice versa — so a short packet is necessarily built from a wide range of wavenumbers, and a wide range is exactly what a dispersive medium separates fastest. The short one goes first because it contains more of what the medium can pull apart.

That the relation is $\Delta x\,\Delta k \gtrsim 1$ is a property of Fourier transforms and of nothing else. It is the same inequality that appears in quantum mechanics with $p = \hbar k$ substituted, and there it is usually presented as a statement about measurement. It is not one here: nothing is being measured, no observer is involved, and the water does not know about Planck’s constant. What quantum mechanics adds is the identification of $\hbar k$ with momentum, and everything else in the inequality was already true of ripples.

The medium that does not do it #

If $\omega(k)$ is a straight line, $\beta = 0$ and there is no spreading at all — at any bandwidth, for ever.

The condition is exact and it is also fragile. A medium is non-dispersive only where $\omega(k)$ is straight, and no real medium’s is straight everywhere: air’s departs at gigahertz frequencies where relaxation processes lag, glass’s departs at every wavelength, and even a string departs once its stiffness begins to matter alongside its tension. What is meant by a non-dispersive medium is always a non-dispersive band, and the band’s edges are where the packets stop arriving intact.

That is why sound is intelligible. A word is a packet with a bandwidth of several kilohertz on a carrier of a few hundred hertz, which is an enormous fractional bandwidth; if air were as dispersive as deep water, a sentence would arrive as a rising tone with no consonants in it. The absence of dispersion in air over the audible range is a stronger condition than it looks, and it is why the medium’s near-constancy of sound speed is worth checking rather than assuming.

On deep water the phase speed goes as $\sqrt{\lambda}$ and the group speed is half of it; in shallow water both are the same and neither depends on the wavelength. That is the medium that does not spread a packet — and it is the exception rather than the rule. Non-dispersive propagation requires the relation between frequency and wavenumber to be a straight line through the origin, which is a coincidence rather than a default.

Two speeds and a third thing #

Three quantities have now been extracted from one function $\omega(k)$, and it is worth listing them together because they are constantly confused.

$\omega/k$ is the phase velocity — the speed of the crests, which carries nothing and may exceed $c$ without embarrassment. On deep water it is twice the group velocity, which is why crests appear at the back of a group, run forward through it, and vanish at the front. $d\omega/dk$ is the group velocity — the speed of the envelope, and where the energy goes. $d^2\omega/dk^2$ has no name in ordinary use and is the spreading: it is not a velocity, it has units of area per time, and it says how fast the packet loses its identity.

Two components give a carrier inside a slow envelope, and with only two there is no spreading at all — because two points always lie on a straight line, so the relation between frequency and wavenumber is exactly linear over the pair. Spreading needs a third component to be off that line, which is the cleanest statement of what dispersion is: a curvature, not a slope.

That last observation is worth taking seriously. The beat picture — two frequencies, an envelope moving at $\Delta\omega/\Delta k$ — is the standard introduction to group velocity, and it is structurally incapable of showing dispersion. Anyone who learns group velocity from two waves has learned the first derivative and has been given no reason to suspect there is a second.

The three quantities also fail together in a useful way. Where $\omega(k)$ is straight, all three are simple: phase and group velocity are equal and the spreading is zero. Where it curves, the two speeds separate and the packet begins to come apart, and the size of both effects is set by the same curvature. So a medium in which the crests visibly outrun the group is a medium in which packets visibly spread, and the two observations are one observation.

The spectrum never changes #

There is one statement that makes the reversibility above concrete rather than a claim about equations, and it can be read straight off the superposition.

Each component is multiplied by $e^{-i\omega(k)t}$, which has modulus one at every $k$ and every $t$. So $|A(k)|$ is untouched: whatever the packet does, its spectrum at the end is bit for bit the spectrum it started with. Dispersion is a phase operation and nothing else. A packet that has spread by a factor of ten contains exactly the same wavenumbers in exactly the same proportions as the one that left, and every difference between them is in the relative phases.

That is why the envelope broadens and flattens with the area under the intensity preserved, and it is why nothing has been lost. It also gives the spreading a shape more specific than “broader”. The quadratic phase means the instantaneous frequency varies linearly along the packet: the components that travel fastest arrive first, so the leading edge carries one end of the spectrum and the trailing edge the other. The packet is chirped, and the chirp is the record of everything that happened to it.

The vocabulary that goes with this is worth having, because it names the thing rather than the process. A packet whose duration is the shortest its spectrum allows — no chirp, all components in step — is transform-limited. Spreading takes a transform-limited packet and makes it not, without changing the limit itself, so the packet is always at least as long as its spectrum permits and is usually longer. Anything that shortens a packet is therefore doing one of exactly two things: broadening its spectrum, or removing its chirp. Nothing else is available.

Running it backwards on purpose #

Once the spreading is understood as an accumulated quadratic phase, undoing it is an engineering problem rather than a paradox: pass the packet through something that applies the opposite quadratic phase, and the components come back into step.

A length of glass with the opposite sign of $\beta$ will do it. So will a pair of diffraction gratings arranged so that the long wavelengths travel further than the short ones, which is a geometric way of making a delay that depends on wavelength, adjustable by moving one grating.

The most consequential use of it inverts the whole problem. The obstacle to amplifying an extremely short pulse is that a short pulse of useful energy has an enormous peak power and destroys whatever it passes through. So: stretch it first, deliberately, by sending it through a long dispersive path, so that its peak power falls by the same factor its duration rises; amplify the long, mild pulse, which the amplifier survives; then compress it back with the opposite dispersion. Because the stretching changed only the phases, the compressed pulse can be as short as the original — and it now carries the amplified energy.

That is why the spreading being phase-only matters practically rather than philosophically. If dispersion did anything to $|A(k)|$, the trick would be impossible: the compressed pulse’s duration is set by the spectrum, and stretching had better not touch it. The whole technique rests on the modulus-one factor two paragraphs above.

Spreading, and the direction of time #

The equation that produced all of this is time-reversible. Running $t$ backwards in the superposition gives a perfectly good solution, and it describes a packet that narrows — components arriving from all over, converging, forming a short pulse, and dispersing again.

So spreading is not irreversible in the sense the word usually carries. Nothing is lost: the phase relations are all still there, and applying the opposite quadratic phase reassembles the packet exactly. That is not a thought experiment. It is what a pulse compressor does, and it works.

What makes spreading look one-way in practice is that packets are usually made by something local — a stone, a spark, a switch — and a narrowing packet would have to be made by arranging phases across a wide region in advance. It is the same asymmetry that makes an outgoing ripple ordinary and an ingoing one a piece of stagecraft: the equation permits both and the initial conditions available in the world do not.

Where it matters #

Optical fibres. A pulse in glass spreads because the index depends on wavelength, and the amount is quoted directly as $\beta_2$ in picoseconds squared per kilometre. The bandwidth a fibre can carry over a given distance is set by the requirement that adjacent pulses not overlap on arrival — so the data rate falls as the square of the length, and every long link either compensates the dispersion with a length of fibre of the opposite sign or works at the wavelength where $\beta_2$ passes through zero.

That zero is worth a sentence, because it is not a place where the medium stops being dispersive. The phase velocity still depends on wavelength there, and strongly; what vanishes is the second derivative, so the first-order spreading disappears and the third-order term takes over. A pulse at the zero-dispersion wavelength still comes apart, more slowly and asymmetrically, and the design question is which term is cheaper to live with rather than whether any of them can be removed.

The numbers are worth having, because they are quoted in a unit that hides the second derivative. A telecommunications fibre’s dispersion is given as $D$, in picoseconds of spread per nanometre of bandwidth per kilometre of length, and standard fibre at 1550 nm has $D$ around 17. So a pulse a nanometre wide in spectrum picks up 17 picoseconds of spread per kilometre, which over a hundred kilometres is 1.7 nanoseconds — vastly longer than the pulse itself, and enough to smear thousands of bits into one another. $D$ and $\beta$ are the same quantity in different clothes, related by the wavelength squared over $2\pi c$.

The wavelength where $D$ passes through zero in ordinary fibre is around 1310 nm, which is not where the loss is lowest. Between the wavelength with the least attenuation and the wavelength with the least dispersion, the industry chose the first and fixed the second — by building fibre with the opposite sign of $D$ and splicing the right length of it into the line, which is the compression trick above applied to a signal rather than to a pulse.

Radio through the ionosphere. The plasma dispersion relation $\omega^2 = \omega_p^2 + c^2k^2$ has a curvature that falls with frequency, so a pulse from a distant source arrives with its low frequencies delayed. Measuring that delay against frequency gives the total electron content along the path.

A single wavenumber extends for ever, with no beginning and no end, and it cannot spread because nothing about it is localised. That is worth ending the mechanism on: spreading is not something a medium does to a wave, it is something a medium does to the relationship between the components of a superposition — and a wave with one component has no relationships to disturb.

Water waves. A storm at sea emits a broad spectrum at once, and by the time it reaches a distant shore the long components have arrived and the short ones have not, so the swell arrives as a slow downward sweep in frequency over hours. That is not spreading of a packet so much as the packet being dismantled entirely, and it is the same square-root dispersion relation doing it.

A free particle has $\omega = \hbar k^2/2m$, so the dispersion coefficient is $\hbar/m$ exactly and every matter wave spreads. The arithmetic is identical to the classical case and the interpretation is not: the spreading of a probability amplitude is a growing uncertainty about where the particle is, rather than a pulse physically getting wider. Same equation, and a quite different thing being described.

Where the model stops #

The expansion is truncated. Keeping $\beta$ and dropping the cubic term is an approximation whose error grows with bandwidth, and the figures measure it: 0.3% at 10% bandwidth, 3.3% at 18%, 8.5% at 28%. Beyond that the packet does not merely broaden but becomes asymmetric, developing a tail on one side, which no symmetric formula can describe.

The medium is linear. In a nonlinear medium the packet’s own amplitude alters the local wave speed, and if the nonlinearity has the right sign it can exactly balance the spreading. The result is a soliton — a packet that does not spread at all, at one particular amplitude, for as long as the balance holds. That is a different subject and its existence is the strongest possible statement that spreading is not inevitable.

The initial packet is Gaussian. The closed form is exact for a Gaussian envelope and only approximate for anything else, because a Gaussian is the one shape whose Fourier transform is the same shape. A square pulse acquires ringing as it spreads rather than merely broadening, and its second moment does not follow a hyperbola. The figures use Gaussians for that reason, and a real pulse from a real source is not one.

And the medium is assumed not to change along the path. Every expression here takes one $\beta$, evaluated once, and applies it for the whole journey. A packet crossing from deep water to shallow, or a pulse passing through a splice between two kinds of fibre, meets a different curvature partway and accumulates the phase of each in turn. That is not a complication so much as the reason the compensation described above is possible at all: what accumulates is a sum, and a sum can be made to come to zero.

The packet is one-dimensional. In two or three dimensions a packet spreads transversely as well, by diffraction, and for a beam that effect usually dominates. The spreading treated here is along the direction of travel and is a separate mechanism from the spreading of a beam through an aperture.

What the pictures cannot show #

The width plotted is a second moment, which is one number extracted from a whole envelope. Two packets with the same second moment can look quite different — one Gaussian, one with long tails — and the measure cannot tell them apart. It was chosen because it is what the closed form predicts, and because a half-height width would have been ambiguous once the envelope stops being Gaussian.

The frame is moving with the group, so the figures show no motion at all. Everything drawn is what a swimmer keeping pace with the packet would see, and the fact that the whole arrangement is travelling at 1.25 m/s across the pond is invisible.

And nothing here shows the phase. The components are getting out of step, which is the entire mechanism, and a plot of the envelope’s width has integrated the phase away. The two snapshot figures show it indirectly — the crests have moved through the envelope between them — and that is the closest a figure of this kind gets.

Where the ladder goes next #

The three rungs of this ladder have taken one function and differentiated it twice. The first derivative gave a second speed different from the crests’; the water-wave case gave the dispersion relation a concrete form; this one takes the second derivative and finds that the packet has a lifetime.

Above are the rungs where the expansion is abandoned. Solitons, where nonlinearity cancels the spreading exactly. Pulse compression, where a deliberately chirped packet is sent through a dispersive medium chosen so that the spreading runs backwards and the packet arrives shorter than it left — which is how a radar reconciles range resolution with transmitted energy, and how the shortest laser pulses are made.

The habit is to keep expanding. A first-order description that works is an invitation to ask what the second-order term does, and the answer is very often not a small correction to the same phenomenon but a different phenomenon with its own timescale.