BrainFlux

Dispersion and Kramers–Kronig

Drag the square handle (or anywhere on the plot) to set the light's frequency; drag the ● to move the line, up to strengthen it, and a ◇ to widen it

Draw on this plot or the next with the mouse: the other follows. Drag the sliver's handle along the top to see one sliver's share

Experiments

Start here

Light shakes the charges in matter, and how they swing back decides how fast the light travels and how much of it is absorbed. The top plot is the medium's index n (blue) and absorption κ (orange) against the light's frequency; below it, one charge on its spring, shaken at the frequency you choose, and the light's wave entering the medium.

Drag the square handle along the top of the plot (it sweeps by itself until you do) and watch the charge: in step with the field below the resonance, a quarter period behind on it, against the field above it. Drag the round handle to move the resonance (up to make it stronger) and the diamonds to widen it. The buttons under Medium load real materials: glass, sodium, the ionosphere, rock salt.

Look for this: n rises slowly towards the resonance, falls steeply through it, where the medium absorbs, and is below 1 just above it. The other two views, 2 · Kramers–Kronig (the rule that ties absorption to dispersion: draw one, get the other) and 3 · Pulses (a pulse sent through the medium, late or spread or early), continue from here, in the experiments below.

1. A charge on a spring: why $n$ rises, then falls through a line

One line at $\omega_0=2$, $\gamma=0.2$. The light's frequency sweeps slowly across it (drag the square handle to take over). Watch the charge and the two curves beside it, $E(t)$ and $x(t)$. Far below the line the spring wins: the charge moves in step with $E$ ($\delta=1.5^\circ$ at $\omega=0.5$), the dipoles $qx$ point along $E$, $\chi>0$ and $n>1$. At the line the charge swings ten times as far as the static pull ($A=\omega_0/\gamma=10$) and lags by a quarter period, $\delta=90^\circ$: its velocity is then in step with $E$, so the field does work on it all the time, and the light is absorbed ($\kappa=0.92$ there; it is largest, 0.99, just above, at $\omega=2.04$, where $n$ has fallen). Above the line inertia wins: the charge moves against $E$ ($\delta=167^\circ$ at $\omega=2.5$), $\chi<0$ and $n<1$ (0.762 there). That is the whole shape of $n(\omega)$: rising slowly towards the line from below (normal dispersion, $dn/d\omega>0$), falling steeply through it (anomalous dispersion, only where the medium absorbs), and below 1 above it, rising back towards 1.

2. What $n$ and $\kappa$ do to the wave

The light at $\omega=1.6$, below the line: $n=1.290$ and $\kappa=0.057$. In the picture on the right the wave comes in from vacuum. Inside, its crests are $\lambda_0/n= 0.775\,\lambda_0$ apart (compare the two scale bars) and the amplitude falls by $1/e$ in $c/\kappa\omega=2.79\,\lambda_0$ (the orange dot). The surface reflects $R=|(\tilde n-1)/(\tilde n+1)|^2=1.7\%$; the faint ripple of the dashed envelope on the left is the reflected wave beating with the incoming one. Drag $\omega$ onto the line ($\omega=2$): the wave is gone within $0.17\,\lambda_0$. Just above it ($\omega=2.1$), where $n<1$ and $\kappa$ is still large, $R$ peaks at 21%. Further up ($\omega=2.5$) the medium is transparent again but $n=0.762$: the crests inside are further apart than in vacuum ($1.31\,\lambda_0$) and move faster than $c$. That breaks no law: a crest carries no signal (experiments 14 and 15).

3. A metal is a mirror below $\omega_p$

Free electrons, $\varepsilon=1-\omega_p^2/(\omega^2+i\gamma\omega)$, with sodium's $\hbar\omega_p=5.9$ eV. With no spring, nothing pulls an electron back, and at every frequency well above $\gamma$ it moves against $E$ ($\delta=179.6^\circ$ at 3 eV; $135^\circ$ at $\omega=\gamma$, where friction matters): $\chi'<0$ always. Below $\omega_p$ it is large enough to make $\varepsilon'<0$: then $n\approx0$ (0.008 at 3 eV) and $\kappa$ is large (1.69), so no wave travels in. The field dies within $c/\kappa\omega=39$ nm, the skin depth, and $R=99.2\%$: a mirror (the reflected wave makes a standing wave outside). Above $\omega_p$ the electrons' response is too weak to make $\varepsilon$ negative, and the metal turns transparent: $R=17\%$ at 6.5 eV and $3.8\%$ at 8 eV. Sodium, like the other alkali metals, lets ultraviolet light through below about 210 nm (Wood, 1933). Drag $\omega$ across $\omega_p$, or drag the $\omega_p$ handle: it sets the electron density, $\omega_p^2=Ne^2/\varepsilon_0m$.

4. The ionosphere reflects AM radio

The same physics with $10^{12}$ electrons per cubic metre: $f_p=\omega_p/2\pi\approx9$ MHz (frequencies here are $f$ in MHz). AM broadcasts (0.53 to 1.7 MHz) are far below $f_p$ and are reflected ($R=99.8\%$); at 15 MHz $R$ is down to 1.2%, and FM (around 100 MHz) and satellite signals go straight through. This is an idealised sharp surface at normal incidence. The real ionosphere thickens gradually with height, and reflects each frequency where its own $f_p$ reaches it; at a slant, $\theta$ from the vertical, it reflects frequencies up to $f_p/\cos\theta$, which is how shortwave goes round the world. AM is heard far away mostly at night, because by day the lower D layer absorbs it before it gets that high. Drag the $\omega_p$ handle: the density changes through the day and with the sunspot cycle.

5. Glass: why a prism spreads the colours

Fused silica, as three oscillators (Malitson's fit): two resonances in the ultraviolet (10.7 and 18 eV) and one in the infrared (0.13 eV, at the left edge). The visible band (the rainbow strip) lies between them, where nothing absorbs. There the ultraviolet charges move in step with $E$ ($\delta\approx0$: drag the light through the visible), so $n>1$, and $n$ rises gently towards their lines: $n=1.4564$ in the red (656 nm), $1.4631$ in the blue (486 nm), so blue is bent more. The rise steepens towards the lines ($dn/d\omega$ is 0.0092 per eV in the red, 0.0140 at 400 nm), which is why a prism spreads the violet end of its spectrum more than the red.

6. Reststrahlen: a crystal that mirrors one band of the infrared

Rock salt in the infrared (frequencies $\nu$ in THz). The Na⁺ and Cl⁻ ions vibrate against each other at the transverse optical frequency, $\nu_\mathrm{TO}=4.9$ THz; the electrons, whose lines are far higher, respond at once and add the background $\varepsilon_\infty=2.34$, so $\varepsilon(0)=5.90$. Above $\omega_\mathrm{TO}$ the ions move against $E$, and so strongly that $\varepsilon'<0$, up to where $\varepsilon'$ comes back through zero at $\omega_\mathrm{LO}=\omega_\mathrm{TO}\sqrt{\varepsilon(0)/\varepsilon_\infty}$ (Lyddane, Sachs and Teller), at 7.78 THz. In between, like a metal below $\omega_p$, the crystal reflects almost everything: 94% at 6.2 THz. Repeated reflections off such crystals filter out all but these "residual rays", Reststrahlen (Rubens and Nichols, 1897). Widen $\gamma$ to 0.5 THz: damping rounds the plateau, and $R$ at 6.2 THz falls to 75%.

7. The f-sum rule: one area that no reshaping can change

However a line is shaped, one weighted area is fixed by the number of charges alone: $$\int_0^\infty\omega\,\varepsilon''(\omega)\,d\omega=\frac{\pi}{2}\,\omega_p^2,\qquad\omega_p^2=\frac{Nq^2}{\varepsilon_0m}.$$ The read-out shows it, $(\pi/2)\times1.000$ for this line. Drag its diamonds to widen it: it gets lower as it gets wider, and the number does not move. Move the line: still the same (the plain area $\int\varepsilon''d\omega\approx\pi\omega_p^2/2\omega_0$ does change). Only the strength, the round handle up or down, changes it. This is Kramers–Kronig at high frequency: there every charge responds as if it were free, $\chi\to-\omega_p^2/\omega^2$.

8. Kramers–Kronig, one sliver at a time

Causality: the charges cannot respond before they are pushed. The bottom plot is $\chi(t)$, the polarisation after a short kick of $E$ at $t=0$, and it is zero before (shaded). For one line, $\chi(t)$ rings at $\omega_0$ and dies away in $2/\gamma$. Now take one sliver of the absorption, $\mathrm{Im}\,\chi$ near $\omega'$ (the teal strip; drag its handle). On its own it would ring as $\sin\omega't$ for $t<0$ as well as for $t>0$ (experiment 10). To be zero before the kick it needs a partner in $\mathrm{Re}\,\chi$: the dashed curve, its share, positive below $\omega'$ and negative above, falling off as $1/(\omega'-\omega)$. With it, its ring starts at the kick (dashed, below). Add up the shares of all the slivers and you have $\mathrm{Re}\,\chi$: $$\mathrm{Re}\,\chi(\omega)=\frac1\pi\,P\!\int_{-\infty}^{\infty}\frac{\mathrm{Im}\,\chi(\omega')}{\omega'-\omega}\,d\omega' .$$ Below the line every sliver pushes $\mathrm{Re}\,\chi$ up, the more the nearer it is, so $n$ rises towards the line; above it they all pull it down. Near its own sliver a share is large, but there the shares of its neighbours on either side cancel it.

9. Draw the absorption, get the index

$\mathrm{Im}\,\chi$ is a flat band of absorption, drawn by hand, and $\mathrm{Re}\,\chi$ is what causality makes of it: $(h/\pi)\ln|(b^2-\omega^2)/(a^2-\omega^2)|$ for a band of height $h$ from $a$ to $b$, with a spike at each sharp edge, up at the lower one and down at the upper. Draw on $\mathrm{Im}\,\chi$: absorption anywhere gives $d\,\mathrm{Re}\,\chi/d\omega>0$ (normal dispersion) everywhere else. Or draw on $\mathrm{Re}\,\chi$ and see what absorption it needs (orange below zero would be gain). The transforms are done with FFTs on a grid reaching 8 times past the plotted range; for this band they agree with the formula to within 1% of $h$, except right at its edges.

10. Break causality: a medium that responds before it is kicked

A medium that absorbs exactly as the single line does, but with $\mathrm{Re}\,\chi=0$: no dispersion at all. Its $\chi(t)$ is odd in time: it responds before the kick as much as after (100%, says the read-out), and so does the sliver. Kramers–Kronig is causality written in frequency: a response that vanishes for $t<0$ has real and imaginary parts that are each other's Hilbert transform. With Break causality on, draw on either side and watch $\chi(t)$ leak into the past; turn it off and the other side is recomputed, and the past goes quiet.

11. Type χ(ω): the sign of i

$\chi(\omega)=\omega_p^2/(\omega_0^2-\omega^2-i\gamma\omega)$ typed in (as wp^2/(w0^2 - w^2 - i g w)): causal (before the kick, $1.5\times10^{-4}$ of its size after; the small leak comes from cutting its $1/\omega^2$ tail off at the end of the grid). Its poles are at $\omega=-i\gamma/2\pm\sqrt{\omega_0^2-\gamma^2/4}$, in the lower half-plane, as a causal response's must be when fields go as $e^{-i\omega t}$. Change $-i$ to $+i$: the poles move to the upper half-plane and $\chi(t)$ is the same ring run backwards, entirely before the kick. (With $e^{+i\omega t}$, the engineers' convention, the signs are the other way round.) Its $\mathrm{Im}\,\chi$ is negative too, as for gain; but a real amplifier, a laser's inverted atoms, has $\chi\to-\chi$, still causal. Try the others: a constant is an instantaneous response, $\delta(t)$; absorption with no dispersion is not causal. A conductor, $i\sigma/\omega$ (i s/w), comes out not quite causal either: $\chi(t)=\frac{\sigma}{2}\,\mathrm{sgn}\,t$. The causal one is $i\sigma/(\omega+i0)$, whose real part has a $\pi\sigma\,\delta(\omega)$ that no formula sampled on a grid can show; with it, $\chi(t)=\sigma$ for $t>0$ and 0 before.

12. A pulse is late: phase and group velocity

A pulse below the line ($\omega_c=1$; its spectrum is the grey hump). The middle plot is the pulse inside the medium at one moment, against the same pulse in vacuum (dashed). The dot rides one crest, at the phase velocity $c/n=0.866\,c$; the triangle rides the peak of the envelope, at the group velocity $c/n_g=0.800\,c$, where $$n_g=n+\omega\frac{dn}{d\omega}=c\,\frac{dk}{d\omega}$$ (green on the top plot). Since $n$ rises with $\omega$ here, $n_g>n$: the crests run forward through the pulse, appearing at its back and vanishing at its front. The bottom plot is what a detector at $z=120$ records: the peak comes $29.8$ after light through vacuum would, as $z/v_g-z/c=z(n_g-1)/c=30.0$ says. The tail of the line still absorbs a little ($\kappa=0.005$): the peak is down to 54%. Drag the carrier towards the line: $n_g$ grows, the pulse is later and spreads, its two sides seeing different $n_g$; its front turns red, since here lower frequencies go faster (the colour is the local frequency).

13. A short pulse spreads, red first

A 10 fs pulse at 800 nm through 5 mm of fused silica, watched moving with the pulse. In the visible and near infrared glass has normal dispersion, so red light (lower $\omega$, lower $n_g$) runs ahead of blue: the pulse spreads and is chirped, red at its front and blue at its back. Its r.m.s. length at the detector grows 1.416 times in 1 mm, 2.004 times in 1.73 mm and 5.111 times in 5 mm (set $z$), close to $\sqrt{1+(z/L)^2}$ (1.414, 2.000, 5.099) with $L=T_0^2/\beta_2=1.0$ mm, for $\beta_2=d^2k/d\omega^2=36\ \mathrm{fs^2/mm}$ ($T_0=T/2\sqrt{\ln2}$). This is why femtosecond lasers need prism or grating compressors. Meanwhile the whole pulse falls 7.79 ps behind light in vacuum ($n_g=1.467$). Make it 20 fs long: $L$ grows as $T^2$, and after 5 mm it is only 1.60 times longer.

14. Faster than light? The peak exits early

A long pulse tuned to the middle of a weak absorption line, where $n$ falls so steeply with $\omega$ that $n_g=-0.975$: negative. At the detector, the peak of what comes out arrives before the peak of a pulse through vacuum: 6.7 early at $z=5$ and 12.9 at $z=10$ (the group velocity alone would say 9.9 and 19.7), but much reduced: 42% as high at $z=5$, 18% at $z=10$ (the curve is enlarged to be seen). Nothing goes faster than light. The charges need time (about $2/\gamma=10$) to build up the motion that absorbs the wave, so the front of the pulse gets through better than its back, and what comes out is a small copy built from the front, which was there, in the smooth Gaussian, long before the peak. Cut the front off sharply and see (15). Make the pulse short ($T=10$): its spectrum is now as wide as the line, the pulse is distorted, and its peak is only 1.6 early at $z=5$.

15. ...and a front never arrives early

The same pulse with a sharp front: before $t_f=-24$ (24 before its peak) there is no field at all. What comes out now piles up just behind the front (its peak is at $-22.9$), and the front itself arrives exactly when light through vacuum would: before it the field is below $10^{-9}$ of its peak (the read-out; switch on log |E| to see it). Information travels with fronts, and fronts travel at $c$: a front contains every frequency, and as $\omega\to\infty$ every medium has $n\to1$ (the charges cannot follow). (The cut is smoothed over a couple of time steps of the computation, a small fraction of a period.)

16. Precursors: what arrives first when a wave is switched on

A carrier switched on suddenly at $t=0$ into a dense medium (Brillouin's own numbers, 1914). The first thing to arrive, exactly at $t=z/c$, is the Sommerfeld precursor (a precursor is the faint field that runs ahead of the main signal): the highest frequencies, which see $n\approx1$, wiggling slower and slower (the period grows from 0.26 to 0.64 in the first three time units at $z=10$). Next comes the Brillouin precursor, the lowest frequencies, a slow swing near $t=n(0)z/c$ ($n(0)=1.5$). Only then does the field settle into the carrier, weakened by absorption to $e^{-\kappa\omega z/c}=0.125$: at $z=10$ it has 80% of that by $t-z/c=7$, well before the group delay $z(n_g-1)/c=14.4$ (green), and all of it by about 24. In so dense a medium the switch-on does not travel at the group velocity. Move the carrier above the line, into the band where it is absorbed (at $\omega=5$, $e^{-\kappa\omega z/c}=e^{-53}$ by $z=10$), and only the precursors get through. Frequencies are in units of $10^{16}\ \mathrm{s^{-1}}$, so $z=10$ is 0.3 µm.

How it works

Light shakes the charges in matter, and the shaken charges radiate too. Here each kind of charge is a mass on a spring with friction (the Lorentz model): charge $q$, mass $m$, natural frequency $\omega_0$, damping rate $\gamma$, pushed by the light's field $E_0\cos\omega t$, $$m\ddot x+m\gamma\dot x+m\omega_0^2x=qE_0\cos\omega t .$$ Once it has settled, $x=Ax_s\cos(\omega t-\delta)$: it swings $A$ times as far as a steady field would pull it ($x_s=qE_0/m\omega_0^2$, the static pull) and lags behind $E$ by $\delta$, with $$A=\frac{\omega_0^2}{\sqrt{(\omega_0^2-\omega^2)^2+\gamma^2\omega^2}},\qquad \tan\delta=\frac{\gamma\omega}{\omega_0^2-\omega^2},$$ the resonance curve of the driven oscillator in mechanics. $N$ such charges per volume make the polarisation $P=Nqx=\varepsilon_0\chi E$. In complex notation, with fields going as $e^{-i\omega t}$ (as in Griffiths and Jackson), $$\chi(\omega)=\frac{\omega_p^2}{\omega_0^2-\omega^2-i\gamma\omega}=\frac{\omega_p^2}{\omega_0^2}\,A\,e^{i\delta},\qquad \omega_p^2=\frac{Nq^2}{\varepsilon_0m},$$ so the part of the motion in step with $E$ ($A\cos\delta$) makes $\mathrm{Re}\,\chi$, and the part a quarter period late ($A\sin\delta$) makes $\mathrm{Im}\,\chi>0$, the absorption. The index is $\tilde n=n+i\kappa=\sqrt{\varepsilon}=\sqrt{1+\chi}$, and a wave in the medium goes as $$e^{i(\tilde n\omega z/c-\omega t)}=e^{-\kappa\omega z/c}\,e^{i(n\omega z/c-\omega t)}:$$ its crests are $\lambda_0/n$ apart and move at $c/n$, and its amplitude falls by $1/e$ in $c/\kappa\omega$. (Griffiths writes the wave number as $k+i\kappa$; his $\kappa$ is our $\kappa\omega/c$.) A medium may have several kinds of charge, each with its own line; $\omega_0=0$ is a free electron (no spring: a metal or a plasma).

The page has three views. 1 · Medium: $n$ and $\kappa$ (or $\varepsilon'$ and $\varepsilon''$) against frequency, one charge and the light's wave at the frequency of the square handle, and the reflectance. Drag the square handle (or anywhere on the plot) to change the light's frequency; drag a line's round handle sideways to move it and up or down to make it stronger, and its small diamonds to make it wider. 2 · Kramers–Kronig: draw $\mathrm{Im}\,\chi$ or $\mathrm{Re}\,\chi$ and get the other; one sliver of the absorption at a time; the response in time. 3 · Pulses: a pulse inside the medium at one moment, and what a detector at depth $z$ records, computed exactly, frequency by frequency. With $c=1$, a frequency unit $\omega_u$ makes $1/\omega_u$ the unit of time and $c/\omega_u$ that of distance; the real materials are in eV, MHz or THz, with times and distances to match.

Approximations: the medium is linear and its response local (no spatial dispersion); a pulse is a plane wave already inside the medium (no reflection at its surface); the time grid is periodic and wide enough that nothing wraps round.

New in 2026 for BrainFlux. Source: github.com/gallicchio/brainflux, MIT licence.