Why a litre of water is not blue for the reason the sky is

A cubic wavelength of air at sea level contains about four million molecules. Every one of them scatters, and the standard account of why the sky is blue adds up their contributions by multiplying one molecule’s cross-section by the number of molecules. That step is not obviously allowed. Amplitudes add, not intensities, which is the first thing a wave does when it meets another, and a hundred amplitudes that happen to agree in phase give ten thousand times one amplitude’s intensity rather than a hundred times.

The reason the multiplication is allowed sideways is that the molecules are in random positions, so the path differences to a distant observer are random compared with a wavelength and the phases average away. But the same molecules are not randomly phased in the forward direction, and what they do there has a name that does not sound like scattering at all.

Forward, the phases always agree #

Consider light going straight through. Each molecule re-radiates, and the phase of its contribution at a point far downstream depends on the total path from source to molecule to observer. For a molecule displaced sideways, that path is longer; for one displaced along the beam, the extra distance to the molecule is exactly cancelled by the reduced distance from it. In the exactly forward direction the total path is the same for every molecule in a thin slab, whatever its position.

So the forward sum is coherent, and it is the upper curve of the figure above.

That agreement is the essay’s first point. The refractive index of air and the blueness of the sky are not two facts about air. They are one polarisability, summed twice: once with the phases lined up, which produces a delay and therefore an index, and once with the phases scrambled, which produces a haze. A measurement of how slowly light crosses a room predicts the colour of the sky over an ocean.

There is a way to see that the forward direction has to be special without any algebra. Scattering removes energy from a beam, and energy is conserved, so whatever is scattered sideways must be missing from what goes forward. A sum over random phases cannot do that bookkeeping — it would have to know about the other directions — so the removal has to come from a term that survives the averaging, and the only such term is the one where every phase agrees.

And the two are not independent in a subtler way either. The forward-scattered wave is what removes energy from the beam: it interferes destructively with the incident wave, and the amount removed is exactly what went sideways. That is the optical theorem, and it is why an index and an absorption are one analytic function, and it is the reason an index and a cross-section have to be two readings of one quantity rather than two properties that happen to correlate.

The sky’s brightness counts the molecules #

There is a use for the pairing that is worth more than the consistency check, and Rayleigh made it before anybody had weighed an atom.

The two sums contain the same two unknowns in different combinations. The forward sum — the refractive index — is proportional to the number density times the polarisability, $N\alpha$, because each molecule contributes a delay in proportion to how polarisable it is and the contributions add. The sideways sum — the scattering — is proportional to $N\alpha^2$, because each molecule contributes an intensity in proportion to the square of its polarisability and those add instead. Two measurements, two equations, two unknowns.

Divide one by the other and $\alpha$ drops out, leaving $N$. So a measurement of how blue the sky is, together with a measurement of the refractive index of air, returns the number of molecules in a cubic metre — and with the density of air and the mass of a mole, Avogadro’s number. Rayleigh did exactly that in 1899, from the observed brightness of the sky, and got within a factor of two of the modern value with an instrument that was mostly a judgement about how bright a patch of sky looked against a standard lamp.

That is the deepest content of the two-sums idea. Molecules are invisible individually, and their existence shows up in two macroscopic quantities that depend on the count differently — one linearly and one through the square of a per-molecule property. Whenever that happens the count is recoverable, and it is the same logic Einstein used on Brownian motion a few years later and Perrin used on sedimentation a few years after that.

Why the sky is blue rather than white #

The optical depth computed above is not merely a number to check against a table. It decides what the sky looks like.

An optical depth of 0.0877 at 550 nm means that about eight per cent of the light on a vertical path is scattered out of it, and that almost all of what is scattered has been scattered exactly once before it reaches an eye. A single scattering carries the $\lambda^{-4}$ intact, so the sky’s colour is the spectrum of that one event, undiluted.

Make the medium thicker and that stops being true. Light scattered twice has had the $\lambda^{-4}$ applied twice to a beam already depleted of blue, and light scattered fifty times has forgotten the wavelength dependence entirely, because every colour has been scattered so many times that all of them are certain to have been scattered. A cloud is that medium — an optical depth of tens or hundreds — and it is white for exactly that reason, not because a droplet scatters all colours equally, though it does that too.

So the thinness of the atmosphere is a condition of its being coloured. A planet with ten times Earth’s surface pressure would have a vertical optical depth near one in the blue and its sky would be washed toward white overhead; and the reddening of a sunset is the same effect along the one path where the atmosphere does become thick, a path of thirty-eight air masses at the horizon rather than one.

What happens when the molecules stop being independent #

Now condense the gas. The molecules are a thousand times closer, so a naive count says the scattering per unit volume should rise by the same factor and a metre of water should be very hazy indeed. It is not, and the reason is that the phases have stopped being random.

A liquid is nearly incompressible — far stiffer than any gas at the same density —, which is a statement about positions: put a small volume anywhere in it and the number of molecules inside is very nearly the same every time. Pack the scatterers on a perfect lattice and the sideways sum would cancel exactly, as it does in a good crystal, which is why a flawless piece of glass is clear and why a regular array diffracts instead of scattering. A liquid is not a lattice, but it is a great deal more ordered than a gas, and what is left over to scatter is the residual wobble in the local density.

That is Einstein’s and Smoluchowski’s answer, and it changes the subject of the calculation. A liquid does not scatter because it is made of molecules; it scatters because its density is not perfectly uniform, and the size of the departure is fixed by the compressibility. Everything about the molecules that is not in the polarisability has dropped out.

The check that it is right. Put water’s own compressibility into the expression and it returns a scattering coefficient at 550 nm of 1.86 × 10⁻³ per metre, against a tabulated 1.9 × 10⁻³. Two per cent, from a calculation that treats water as a structureless jelly and never mentions a hydrogen bond.

Two ways of saying the same thing. One is the phase argument above: a liquid’s molecules are correlated, so the cross terms in the sideways sum no longer average away and the near-cancellation between neighbours removes most of the scattering. The other is thermodynamic: the scattering is proportional to the mean square fluctuation of the dielectric constant in a volume small compared with a wavelength, and that fluctuation is proportional to the compressibility because it costs work to squeeze a liquid. They are the same statement because a compressibility is a measure of how far the positions are free to wander.

The thermodynamic form has one great advantage: it contains nothing about molecules at all. It applies to a solution, to a polymer melt, to a mixture near its consolute point — anything with a free energy and a fluctuating density — and the intensity it predicts is a way of measuring a susceptibility with light.

And even that is not what makes water blue #

Having computed the scattering of water carefully, the honest thing to do next is to compare it with the other way light is removed from a beam.

The numbers are worth holding side by side. A metre of water removes six per cent of green light by absorption and two parts in a thousand by scattering. Ten metres removes forty-eight per cent and two per cent. A hundred metres removes everything at 650 nm and sixty per cent at 450 — and the surviving beam is a deep blue that has never been scattered at all.

So the colour of a swimming pool, a glacier crevasse or an ocean away from the coast is red being subtracted, not blue being scattered. Both mechanisms produce a blue, and they are opposites: the sky is what has been deflected out of the beam, and deep water is what is left in it.

The molecular origin is different too. The sky’s blue comes from the λ⁻⁴ of an electron cloud driven far below its own resonance. Water’s blue comes from an overtone of the O–H stretch — a vibration whose fundamental is in the infrared, with enough anharmonicity that its fourth and fifth overtones reach into the red end of the visible. Water’s vibrations sit where they do for the reason every molecular level sits where it does, and water is essentially the only common substance whose molecular vibrations are visible at all, and that is why it is essentially the only liquid with an intrinsic colour that is not due to a dissolved solute.

Where the model stops #

The fluctuation picture fails where the fluctuations stop being small. ρkTκ_T is the mean square number fluctuation relative to Poisson, and it is derived assuming the fluctuations are small enough that the scattering from each is independent of the rest. Approach the critical point and the compressibility diverges, the factor runs past one, the correlated regions grow to the size of a wavelength, and the fluid goes milk-white — which is where the two phases become one and is the one place where a liquid scatters more than a gas of the same molecules would.

Everything in the liquid calculation assumes the correlation length is far below a wavelength, which for water at room temperature it is by four orders of magnitude. It fails completely, and visibly, within a few millikelvin of the critical temperature: the correlation length grows without limit there, and once it reaches $\lambda/2\pi$ the fluid scatters strongly at every angle and turns milky. That is the one condition under which a fluid is opaque from its own fluctuations, and it is as far from ordinary water as a state of water can be.

“Random phases” is a statement about the observer, not the sample. The sideways sum is incoherent because the path differences are large compared with a wavelength and uncorrelated. Reduce the scattering volume to much less than a wavelength across and every phase agrees again, and the scattering goes as N² — which is what makes a small dense particle a far stronger scatterer than its molecules separately, and is where the size of the thing starts to matter.

And nothing here is about sea water. The blue of the open ocean has this absorption in it, and it also has the scattering from everything suspended — plankton, mineral particles, bubbles — which is larger than the water’s own scattering by orders of magnitude in most of the sea. The calculation above is for water that has been distilled and filtered and left to settle, and the reason it is worth doing is that it establishes the floor: a scattering that no purification can remove, because it comes from the thermal motion of the liquid itself.

The isotropic polarisability is an approximation with a name. Nitrogen and oxygen are not spheres, so a molecule’s induced dipole is not parallel to the driving field, and the scattering is partly depolarised. The correction — the King factor — is about 1.05 for air and 1.16 for water, and it is the honest explanation of the ten per cent left over in the second figure rather than a fudge.

The fluctuations that were frozen in #

The clearest use of the liquid calculation is in a solid, and it decides the reach of every optical fibre in the world.

Silica glass is a liquid that stopped moving. Its density fluctuations are the ones a liquid has, captured at the temperature where the melt became too stiff to rearrange, and they are still there — frozen, permanent, and impossible to anneal away, because annealing only lets the glass find the fluctuations appropriate to a slightly lower temperature. So a fibre scatters, by the mechanism this essay computed for water, with the compressibility and temperature being those of the glass at its fictive temperature rather than at room temperature.

That scattering is the floor. Impurities can be removed — the transition metals and the hydroxyl groups that dominated early fibres are now below a part in a billion — and the frozen fluctuations cannot. What is left falls as $\lambda^{-4}$, from about 1.2 decibels per kilometre in the near infrared at 850 nm down to a tenth of that at 1550 nm, where the factor of eleven is exactly the fourth power of the ratio of wavelengths.

Which is why long-haul fibre runs at 1550 nm and not somewhere longer still. The Rayleigh floor is still falling there, and something else is rising: the infrared absorption of the silica network itself, the tail of a vibration in the same way water’s red absorption is the tail of one. The minimum loss of a fibre, around 0.2 decibels per kilometre, is where those two curves cross — a falling scattering and a rising absorption, meeting at a wavelength.

That is the same figure as the one drawn for water in this essay, with the axes stretched by a factor of three in wavelength and by five decades in path length. Both media are transparent in a window bounded below by scattering and above by a molecular vibration, both windows are narrow, and in both cases the position of the window has nothing to do with anybody’s choice.

The generalisation #

The pattern that has come out of this is not about air or water. It is that a collection of scatterers has two answers, and which one applies depends only on whether the path differences are ordered or not.

Two scatterers are the whole argument in miniature. Where the path difference between them is a whole number of wavelengths the amplitudes add and the intensity is four times one source’s; where it is a half, they cancel. A gas is that figure with $10^{25}$ sources whose path differences are random, so the cross terms average to nothing and the intensities add. A crystal is the same figure with the path differences arranged, so the cross terms survive and produce diffraction instead of haze. A liquid is neither, which is why it needs the fluctuation argument at all.

The cross terms are the whole of it. Writing the total intensity as the square of a sum of N amplitudes gives N diagonal terms and N(N−1) cross terms. The diagonal terms are what “add the intensities” means and they are always there. The cross terms carry the phase differences, and everything in this essay is a statement about what happens to them: random positions make them average to zero, ordered positions make them survive, and a liquid’s partial order makes most of them cancel and leaves the remainder proportional to a compressibility.

Every distinct case in the subject is this question answered differently. Diffraction from a grating is the ordered sum. The blue of the sky is the disordered one. The transparency of glass is the ordered sum arranged to cancel everywhere except forwards, which is where a refractive index comes from. The whiteness of milk is the disordered sum with strong scatterers. And the refractive index of any of them is the forward sum, which is ordered no matter what the sample is.

And the crossing between the two regimes can be drawn as a colour. A droplet far smaller than the wavelength scatters blue several times more strongly than red; by the time it is comparable with the wavelength that ratio has gone to one and the scattering is grey. None of it applies to the second half of this essay, where there is no droplet and no size parameter: the scatterer is a fluctuation, and its size is set by how far the correlations reach rather than by anything with a boundary.

The rung after this one #

Two rungs of this ladder have now taken the scattering strength as given and asked what happens to a beam. What neither has asked is what the light does to its polarisation on the way through — Rayleigh scattering at ninety degrees is almost completely polarised, the sky carries a polarisation pattern with a definite geometry, and the degree of that polarisation is reduced by exactly the anisotropy correction that spoiled the ten per cent above. The next rung is that pattern: how to compute it, how much the depolarisation costs, and what it is used for by anything that can see it.