When the particle is the size of the wave

A clear sky is blue, a cloud is white, and both are made of water and air. The cloud is not a different substance and the light is not different light; the droplets in a cloud are simply about a thousand times larger than the density fluctuations that scatter the blue sky, and that single change of scale removes the colour entirely.

The quantity that decides everything is the size parameter

$$x = \frac{2\pi a}{\lambda},$$

the sphere’s circumference measured in wavelengths. It is one of the small number of dimensionless numbers that carry a whole subject — the same kind of ratio that decides whether a slit diffracts — and the sky-versus-cloud question is the question of whether it is small.

What the small limit says, and what it is a limit of #

The inverse fourth power comes out of a two-step argument. A small sphere in an oscillating field acquires a dipole moment proportional to the field. An oscillating dipole radiates power proportional to the square of its acceleration, hence to the fourth power of the frequency. Square the amplitude, count the frequencies, and the scattered power carries $\omega^4$ — or $\lambda^{-4}$.

That assumption is what “small” means, and it is the only thing the fourth power depends on. It has nothing to do with the material, nothing to do with air, and nothing to do with molecules as such. Any small dielectric sphere obeys it. What happens when the sphere is not small has to be computed rather than argued, because the field is then different at different parts of the particle at the same instant, and the scattered wavelets from different parts interfere.

The full computation, and the shape it has #

The exact solution for a sphere is Mie’s, from 1908: expand the incident plane wave and the scattered field in vector spherical harmonics, match them at the surface, and read off two coefficients per order. It is not deep, but it is unavoidable — there is no closed form for the sum, and the boundary conditions being matched are the same two that fix every reflection at a surface.

Three features of that curve are worth naming.

The rise is steep and then stops. Below $x \approx 0.5$ the efficiency grows as $x^4$, which means that in a fog of fixed total water content, doubling the droplet size while halving their number increases the scattering enormously — up to a point. Past the first maximum, adding size buys nothing.

The first maximum is an interference. Light passing through the sphere is retarded relative to light going round it, by a phase $2x(n-1)$. When that phase reaches about four radians the two combine constructively in the forward direction, and the efficiency peaks. It is the same accounting as a thin film’s, with the sphere as the film.

The limit is two, not one. A large sphere removes twice as much light from a beam as it blocks. Half of that is the light it intercepts and half is diffraction around its edge — light that missed it entirely and is nonetheless deflected out of the beam. This is the extinction paradox, and its resolution is that a “beam” is defined by an angular acceptance: the diffracted light is still going almost forward, and an instrument with a wide enough acceptance would collect it and measure an efficiency of one.

What the coefficients are, and why the sum has to be done #

The two coefficients per order, $a_n$ and $b_n$, are ratios of combinations of Riccati–Bessel functions evaluated at $x$ and at $mx$, where $m$ is the index ratio. Each order describes one multipole of the sphere’s response: $a_1$ is the electric dipole that Rayleigh’s argument keeps, $b_1$ the magnetic dipole, $a_2$ the electric quadrupole, and so on. In the small limit $a_1$ dominates everything by many orders and reduces exactly to Rayleigh’s polarisability; the whole of the classic argument is the first term of the first coefficient.

How many terms are needed is set by the size parameter itself. A sphere responds appreciably in every multipole up to about $x + 4x^{1/3} + 2$, which is Wiscombe’s criterion and is not a rule of thumb but a statement about where the Bessel functions turn over. At $x = 0.1$ that is three terms; at $x = 100$ it is a hundred and twenty-two. So the number of independent ways a sphere can respond to a wave is roughly the number of wavelengths around its circumference, which is the same counting that decides how many modes a cavity has and how many terms a diffraction pattern needs.

One numerical detail deserves recording because it is the sort of thing that produces a plausible wrong answer. The logarithmic derivative of the Bessel function inside the sphere has to be computed by downward recurrence, from an order well above the one needed, and not upward from the bottom. Upward, the recurrence is unstable: the errors grow geometrically, and past a size parameter of ten or so the resulting efficiency is smooth, plausible and wrong. Nothing in the output announces the failure, which is why the check that matters is against the two limits the series must reproduce — $x^4$ below, and two above.

Where the light goes #

The strength of the scattering is only half the story. The other half is the direction, and it changes even more dramatically.

This forward tipping is why clouds behave as they do. A cloud lit from behind is brilliant, and the same cloud with the sun behind the observer is a dull grey — the droplets send the light onward rather than back. It is also why fog defeats headlights, and the mechanism is worth being precise about: the light that comes back to the driver is not reflected. It is scattered many times, each deflection small and forward, and the random walk of many small forward steps eventually turns some of it round.

A photon in fog does not travel until it hits something and then bounce. It is deflected slightly, many times over — and the number of scatterings needed to reverse its direction goes as one over one-minus-the-asymmetry-parameter. At $g = 0.85$ it takes about seven forward-biased scatterings to accomplish what one isotropic one would, which is why fog is bright rather than dark and why a transport mean free path is a different length from a scattering mean free path.

The measurement that tells the sizes apart #

The cleanest experimental discriminator is not colour but polarisation, and it costs one filter.

How that is measured is a polariser and a protractor. Against the blue sky ninety degrees from the sun the transmitted intensity swings by a factor of several between two orientations; against a white cloud in the same part of the sky it barely moves at all. Small scatterers polarise strongly at right angles and large ones hardly at all — the instrument is the same in both cases, and the difference is entirely in the size of what is doing the scattering.

The reason the polarisation is so much more diagnostic than the colour is that it depends on the shape of the scattering pattern rather than on the total, and the shape starts changing at a smaller size parameter than the magnitude does. A haze of particles a tenth of a micrometre across is already visibly depolarising while its colour bias is still nearly Rayleigh’s.

Where each regime is found #

Sorting the world by size parameter produces a list that crosses several subjects.

Air’s density fluctuations, effectively a few tenths of a nanometre, sit at $x \sim 10^{-3}$: pure Rayleigh, strong colour bias, complete polarisation at ninety degrees, blue sky. Cigarette smoke at 50 nm sits near $x \sim 0.5$: still bluish, which is why smoke exhaled straight from a cigarette looks blue against a dark background while the same smoke exhaled from a lung — where the particles have grown by taking up water — looks white. Fog and cloud droplets at 5–20 µm sit at $x \sim 50$: white, forward-throwing, unpolarised. Milk’s fat globules, at about a micrometre, sit between, which is why skimmed milk is faintly blue and full milk is not.

The other half of the Rayleigh story is what survives a long path rather than what is removed from a short one. At sunset the light crosses tens of times as much atmosphere, the blue is scattered out along the way, and the remainder is red — the same fourth-power law read from the transmitted side rather than the scattered one. Both readings fail together the moment the scatterers become comparable with the wavelength, which is why a sunset seen through cloud is grey rather than redder.

Extinction along a path is an exponential in the product of number density, cross-section and distance, and nothing in that relation cares which regime the individual scatterer is in. The size parameter decides the cross-section; the cross-section then enters an exponential that is the same for a molecule and for a droplet. That is why a haze and a fog attenuate by the same formula with different numbers in it, and why measuring transmission alone cannot tell them apart.

The white of a cloud, assembled #

Putting the pieces together gives the cloud’s whiteness by three separate routes at once, and it is worth seeing that they are separate.

The cross-section is nearly the same for every visible wavelength, because a droplet’s size parameter is fifty and the efficiency curve is flat and oscillating about two by then. That alone removes the colour bias. The number of scatterings is enormous — a cloud a hundred metres thick has an optical depth of tens, so a photon meets tens of droplets before emerging — and multiple scattering washes out whatever angular structure a single droplet has, which is why a cloud has no rainbow in it despite being made of the same drops that make one. And the absorption is negligible: liquid water absorbs weakly across the visible, so essentially every photon that goes in comes out somewhere, and a body that scatters everything and absorbs nothing is white by definition.

Change any one of the three and the appearance changes accordingly. Add absorption and the cloud is grey — which is what a thick storm cloud is, not because it scatters differently but because so few photons find their way out of the bottom. Make the droplets uniform in size and the angular structure survives multiple scattering just enough to produce a corona around the sun. And take away the multiplicity, leaving a single layer of drops, and the rainbow comes back.

The rings a cloud makes, and the one that needs the whole series #

One sphere of water, illuminated by sunlight, produces three quite different displays at three different angles — and the three require three different levels of theory, which makes them a good check on where each level applies.

At forty-two degrees there is the rainbow, and it is a ray phenomenon. Tracing rays through the drop, finding the stationary deviation, and adding a wave correction at the caustic accounts for it completely. It needs drops large enough for rays to be meaningful, which is why rain produces one and cloud does not.

At small angles there is the corona: coloured rings immediately around the sun or moon seen through thin cloud. That is diffraction. A droplet diffracts light in the same way an opaque disc of the same size does, so the first dark ring sits at about 0.61 wavelengths over the radius — which for ten-micrometre droplets is a couple of degrees. Measuring a corona’s angular radius therefore measures the droplet size directly, to about ten per cent, with nothing but a camera and a protractor.

A corona is only visible when the droplets are nearly all the same size, because otherwise rings of different radii overlap and wash out. A well-developed one is therefore evidence about the distribution as well as the mean, and it is commonest in freshly condensed cloud where the droplets have not had time to grow apart.

At a hundred and eighty degrees there is the glory: coloured rings centred exactly on the antisolar point, most often seen around the shadow of an aircraft on the cloud below, and occasionally around one’s own shadow from a mountain top.

The glory is the interesting one, because no ray explanation of it exists. Trace rays through a sphere and essentially nothing comes back at exactly 180 degrees; the geometrical optics of a droplet predicts that the backward direction is dark. The rings are there anyway, and they are bright.

What produces them is a mechanism with no counterpart in ray optics at all: light striking the drop at grazing incidence excites a wave that travels around the surface of the droplet, shedding radiation tangentially as it goes, and the contributions that leave in the backward direction interfere. In the Mie series that shows up as the very high-order terms — those with $n$ near the size parameter — contributing coherently, which is precisely the part of the sum that a small-particle expansion discards and a ray treatment never had.

So a droplet produces one phenomenon that rays explain, one that diffraction explains, and one that requires the full solution. The glory is the strongest available argument that summing the series is not mere completeness.

What is actually scattering #

The essay says the sky is scattered by density fluctuations rather than by molecules, and the distinction deserves more than the phrase, because Rayleigh’s own calculation used molecules and the correction to it explains three things at once.

Rayleigh’s argument treats the air as $N$ independent molecules, each acquiring a dipole moment and each radiating. Their phases are random, so the intensities add and the total is $N$ times one molecule’s. That gives the fourth power and the right magnitude, and it is the calculation everybody learns.

It cannot be right in general, and the counterexample is easy. Arrange the same molecules on a perfect lattice and the scattered wavelets from different sites interfere destructively in every direction except the Bragg ones — which is exactly why a perfect crystal is transparent, and why a piece of glass is clearer than the same mass of the same substance as a powder. The molecules have not changed; their arrangement has.

Smoluchowski and then Einstein supplied the correct source in 1908 and 1910. What scatters is not the molecules but the fluctuations in their density over regions small compared with a wavelength — the departures from uniformity, which are what an incoming wave sees as a variation in refractive index. The scattered intensity is proportional to the mean square of that fluctuation, and statistical mechanics gives that as the temperature times the isothermal compressibility.

Three consequences follow from one expression. For a dilute gas the compressibility is that of an ideal gas, the formula reduces exactly to Rayleigh’s $N$-molecule answer, and the blue sky is recovered — which is why Rayleigh’s route worked. For a liquid the compressibility is far smaller, so water scatters far less than $N$ independent molecules would, which is why the sea is not as bright as the sky.

And near a critical point the compressibility diverges. The fluctuations grow without bound, the scattering grows with them, and a fluid that was perfectly transparent turns milky within a fraction of a degree of its critical temperature. Critical opalescence was known and unexplained; Einstein’s paper predicted its magnitude, and it is the same formula that makes the sky blue evaluated where one of its factors runs away.

Where the model stops #

The Mie solution is exact, which makes its assumptions unusually easy to state, and there are three.

The particle is a sphere. Real aerosol is not: ice crystals, soot aggregates, mineral dust and pollen are not even approximately spherical, and non-spherical particles depolarise light in ways a sphere never does — which is exactly how a lidar tells ice cloud from water cloud, since the sphere’s exact zero is a signature that only water can produce.

The particles scatter independently. The series describes one sphere. A cloud has $10^8$ of them per cubic metre, and the total is their sum only if they are far enough apart to be illuminated by the incident wave rather than by each other. That holds in fog and fails in a dense colloid, where the scattering is less than the sum because neighbouring particles’ contributions interfere — which is why blood, packed with cells, is not opaque at the thickness it would be if each cell scattered alone.

The index is real. Everything drawn here uses water’s, which absorbs negligibly in the visible. A complex index adds absorption to scattering, changes the shape of the resonances, and for a strongly absorbing particle like soot removes the ripples entirely.

What the pictures cannot show #

The efficiency curve here is smooth on the scale drawn, and the true Mie efficiency for a single sphere of one exact size has fine structure on it — sharp resonances where a wave circulates around the sphere’s inside and interferes with itself. They are averaged away in any real population, because no two droplets are exactly the same size, and drawing them would suggest a feature no cloud has.

The angular figure shows one plane and is normalised per curve, so it says nothing about how much more a large particle scatters in total; that is what the efficiency curve is for, and the two have to be read together.

And nothing here shows a colour. Every statement about blue and white is a ratio of two numbers at two wavelengths, and the step from those numbers to a perceived colour is a fact about an eye rather than about the light.

The instrument the ratio became #

A quantity that varies steeply and monotonically with size, and can be measured without touching anything, is an instrument waiting to be built, and this one has been built several times over.

The simplest version reads the ratio of scattering at two wavelengths, which the first figure here plots directly: on the steep part of that curve, between about 50 and 500 nanometres of radius, the ratio determines the size to within a few per cent. That is the principle of the sun photometer, and the exponent obtained by fitting the measured extinction to a power of wavelength — the Ångström exponent — is a single number summarising the size distribution of everything between the instrument and the sun. Two is a fine haze; near zero is dust or cloud.

A second version reads the angle. Because the phase function tips forward at a rate that depends on size, the ratio of intensity at fifteen degrees to intensity at forty-five is a size estimate, and a laser diode, two photodiodes and no moving parts make a particle counter on that basis. A third reads the polarisation, which is what the figure above is about and what a lidar uses, and its particular strength is that it is a null measurement: the sphere’s zero is exact, so any depolarisation at all is evidence of something non-spherical rather than of a differently sized sphere.

All three measure a distribution rather than a size, and none of them can measure the number of particles and their size separately from a single reading — a few large scatterers and many small ones produce the same extinction. Separating them takes a second observable, which is why every one of these instruments works at two wavelengths or at two angles rather than at one.

Where the ladder goes next #

One dimensionless number crossing one has turned a colour-selective, symmetric, polarising scatterer into a colourless, forward-throwing, depolarising one, with nothing changing about the material. The next rung on this ladder is what happens once the scattering is dense enough that a photon meets many particles: the transport equation, the diffusion approximation, and the fact that a thick cloud’s brightness depends on its total optical depth rather than on the droplets at all.

The habit worth taking away is about limits. Rayleigh’s law is not an approximation that gradually loses accuracy; it is the first term of a series, and the series’ second term arrives at a definite place — $x \sim 0.5$ — where the answer starts being wrong by factors rather than by per cent. A law derived in a limit carries the limit with it, and the useful question about any such law is not how accurate it is but what dimensionless number it assumed to be small.