Dispersion and radiation

Hold a glass prism up to sunlight and a single beam of "white" light fans out into a full rainbow. That split happens because glass does not treat every colour the same way — each wavelength bends by a different amount because each one sees a slightly different refractive index. This spreading with wavelength is called dispersion, and it forces a small but important correction to how fast a wave packet of light actually travels through a medium: the phase fronts and the energy-carrying envelope no longer move at the same speed. This article also answers a second, seemingly unrelated question that turns out to share the same physical root — electromagnetic waves themselves. Why does an electric charge radiate light at all? The answer is a single word: acceleration. Together, dispersion and radiation close out the electromagnetic-waves story of this block — one governs how light of different colours propagates once it exists, the other explains where light comes from in the first place.

The core insight

A medium's refractive index depends on wavelength, $n(\lambda)$ , which splits phase velocity $v_p=c/n$ from group velocity $v_g=c/(n-\lambda\,dn/d\lambda)$ — this is dispersion — while electromagnetic radiation is emitted only by an accelerating charge, never by one at rest or moving at constant velocity.

The core idea

Why a prism splits colours. In vacuum every wavelength of light travels at the same speed $c$ . Inside a real medium this is no longer true: the refractive index $n$ is a function of wavelength, $n(\lambda)$ , and the phase velocity is

v_p=\frac{c}{n(\lambda)}.

where: $v_p$ is the phase velocity of a single wavelength inside the medium, $c$ is the vacuum speed of light, and $n(\lambda)$ is the refractive index at that wavelength. Because $n$ is different for red and violet light, each colour bends by a different angle on refraction, and a prism fans white light into a spectrum.

Normal versus anomalous dispersion. Over most of the visible range, transparent materials have $n$ decreasing as $\lambda$ increases — violet bends more than red. This everyday behaviour is called normal dispersion, and it is well approximated by Cauchy's empirical relation $n(\lambda)=A+B/\lambda^2$ , away from any absorption band. Very close to a natural absorption resonance of the material, the trend reverses sharply over a narrow band of wavelengths — $n$ rises steeply with $\lambda$ there instead — and this reversal is called anomalous dispersion. It is "anomalous" only in sign, not in the sense of being absent; it is actually the strongest wavelength dependence a material shows, concentrated right where it absorbs.

Figure

A prism splitting a white beam into a fan of colours from violet bending most to red bending least

Radiation from accelerating charges. A charge sitting still produces only a static electric field; a charge moving at constant velocity (a steady current) produces only a static magnetic field — neither one radiates any energy outward. Radiation begins only when a charge accelerates: its fields can no longer keep up instantaneously, so a disturbance detaches and propagates outward as a self-sustaining electromagnetic wave, a changing $\vec E$ generating a $\vec B$ and vice versa, travelling at speed $c$ . An oscillating charge — an electron shaking in a dipole antenna, or in an atom — is the qualitative picture of dipole radiation: it radiates in all directions except along its own axis of oscillation, at the same frequency as its motion. This is the physical origin of every electromagnetic wave discussed in this block, including visible light itself.

Derivation AGroup velocity in a dispersive medium

The goal is to show that when a medium's refractive index depends on wavelength, the speed at which a wave packet's energy travels, the group velocity $v_g=d\omega/dk$ , differs from the phase velocity $v_p=c/n$ , and to reduce that difference to one compact working formula in terms of $n(\lambda)$ and $dn/d\lambda$ that can be evaluated from a single measured slope.

Step 1. Start from the definitions carried over from the wave-packet chapter: phase velocity $v_p=\omega/k$ and group velocity $v_g=d\omega/dk$ . Inside the medium, $v_p=c/n(\lambda)$ , so $\omega=kv_p=\dfrac{kc}{n}$ , with $n$ itself depending on $k$ (equivalently on $\lambda$ ) whenever the medium is dispersive.

Step 2. Differentiate $\omega=kc/n$ with respect to $k$ using the product/quotient rule, treating $n$ as a function of $k$ :

v_g=\frac{d\omega}{dk}=\frac{c}{n}-\frac{kc}{n^2}\frac{dn}{dk}.

The first term alone would just give back $v_p$ ; the second term is the correction from $n$ changing with $k$ , and it vanishes exactly when the medium is non-dispersive.

Step 3. Convert the derivative from $k$ to the more commonly tabulated $\lambda$ . Since $k=2\pi/\lambda$ , the chain rule gives $dn/dk=-\dfrac{\lambda^2}{2\pi}\dfrac{dn}{d\lambda}$ . Substituting this and $k=2\pi/\lambda$ into Step 2 and simplifying the algebra collapses the expression to

v_g=\frac{c}{n-\lambda\,\dfrac{dn}{d\lambda}}.

where: $n$ is the refractive index at the wavelength of interest and $dn/d\lambda$ is the slope of the dispersion curve there, obtained from a fit like Cauchy's relation or from tabulated data.

Step 4. Check the limits. If the medium is non-dispersive, $dn/d\lambda=0$ and the formula correctly reduces to $v_g=v_p=c/n$ . If the medium is normally dispersive, $dn/d\lambda<0$ , so the denominator $n-\lambda\,dn/d\lambda$ exceeds $n$ , giving $v_g<v_p$ : the wave packet's envelope — the energy and information it carries — lags slightly behind its own phase fronts.

Worked example

A glass has refractive index $n=1.520$ at the sodium D line, $\lambda=589\ \text{nm}$ , with a measured slope $dn/d\lambda=-4.30\times10^4\ \text{m}^{-1}$ there (a typical crown-glass value; the derivative carries units of m $^{-1}$ because $\lambda$ is measured in metres). Find $v_p$ and $v_g$ at this wavelength.

Step 1. Phase velocity: $v_p=\dfrac{c}{n}=\dfrac{3\times10^8}{1.520}=1.974\times10^8\ \text{m/s}$ .

Step 2. Evaluate the group-velocity denominator: $n-\lambda\dfrac{dn}{d\lambda}=1.520-(589\times10^{-9})(-4.30\times10^4)=1.520+0.02533=1.545$ .

Step 3. Group velocity: $v_g=\dfrac{c}{1.545}=\dfrac{3\times10^8}{1.545}=1.942\times10^8\ \text{m/s}$ . As expected for normal dispersion ( $dn/d\lambda<0$ ), $v_g<v_p$ : here $v_g$ is about $1.6\%$ slower than $v_p$ , confirming the wave packet's envelope lags slightly behind its phase fronts. $v_p=1.974\times10^8\ \text{m/s}$ and $v_g=1.942\times10^8\ \text{m/s}$ .

Competitive stretch

A material follows Cauchy's formula $n(\lambda)=1.50+\dfrac{6.0\times10^{-15}\ \text{m}^2}{\lambda^2}$ over the visible range. At $\lambda=400\ \text{nm}$ (violet), $n=1.50+\dfrac{6.0\times10^{-15}}{(4.0\times10^{-7})^2}=1.50+0.0375=1.5375$ . At $\lambda=700\ \text{nm}$ (red), $n=1.50+\dfrac{6.0\times10^{-15}}{(7.0\times10^{-7})^2}=1.50+0.01224=1.5122$ . Since $dn/d\lambda=-\dfrac{2\times6.0\times10^{-15}}{\lambda^3}<0$ everywhere, this is normal dispersion throughout the visible range, and $n_{\text{violet}}=1.5375>n_{\text{red}}=1.5122$ confirms violet bends more than red — the everyday prism result.

Common traps

Assuming dispersion always means $v_g>v_p$ — the sign is set by $dn/d\lambda$ ; in ordinary normal dispersion, $v_g<v_p$ .
Treating anomalous dispersion as "no dispersion" — it is a stronger, reversed-sign effect near an absorption band, not the absence of wavelength dependence.
Thinking a moving charge always radiates — a charge at rest or moving at constant velocity does not radiate; only acceleration produces radiation.
Reaching for the full Larmor power formula — only the qualitative picture (accelerating charge implies radiation) is in scope here; the quantitative radiated-power law is beyond this course.

Ask the tutor: Want the full step-by-step $v_g=d\omega/dk\to v_g=c/(n-\lambda\,dn/d\lambda)$ algebra spelled out, or the radiation picture tied more precisely to Maxwell's equations? Ask the AI tutor to walk through either one.

If phase velocity and group velocity are different, which one is the speed of light I actually measure with a stopwatch and a light source

Group velocity is the one that matches everyday measurement, because it is the speed of the envelope of a wave packet — the pulse of energy and information a real light source sends out. Phase velocity describes how the individual crests inside that packet slide along, which can differ from the envelope's speed once the medium is dispersive. In a non-dispersive medium the two coincide, so the distinction only matters once $dn/d\lambda\neq0$ .

Why does a prism spread white light into a rainbow but a flat glass window does not visibly do so

Both the prism and the window have the same wavelength-dependent $n(\lambda)$ , so both bend violet slightly more than red at each surface. In a flat window the two parallel faces bend the ray back by equal and opposite amounts, so the colours recombine into one white beam on exit, just shifted sideways. A prism's two faces are not parallel, so the second bending adds to the first instead of cancelling it, and the small angular differences between colours accumulate into a visible fan.

Why can't a charge moving at constant velocity radiate, even though it still has electric and magnetic fields around it

Its fields are not zero, but they are unchanging in the charge's own frame and simply translate rigidly along with it — nothing detaches to travel independently through space. Radiation requires the fields at a point to keep changing in time in a way that outruns the charge itself, which only happens when the charge's velocity itself is changing, i.e. when it accelerates. A steady current or a charge cruising at constant speed produces only the static-field pattern appropriate to that motion, carried along, not radiated away.

Dispersion and radiation

The core idea

Worked example

Competitive stretch

Common traps

Before this

Where this leads