No phase chooses which triples the cube strengthens

Three waves measure the cube through the slope followed a size effect through a chain of atoms. Three incommensurate occupation waves run along the chain, and the spacing between neighbours depends on the local occupation, linearly, quadratically and cubically, with coefficients $\varepsilon_1$, $\varepsilon_2$ and $\varepsilon_3$. The cubic term puts the product of all three waves into the displacement, at the four combinations $q_1 \pm q_2 \pm q_3$, and those are the only places the cube reaches directly. The satellites there were expected to measure it. They do see it, but at the strongest of them the cube’s contribution arrives against another contribution of comparable size, the product of a pair harmonic from the curvature with a first-order wave, and the two nearly cancel. The satellite is eleven times weaker than the cube alone would make it.

The essay ended by asking whether that cancellation was an accident of one choice. Each of the two paths carries phases set by the waves’ phases, so shifting one wave by half a turn ought to flip the sign of one path against the other, turning cancellation into reinforcement at one satellite and moving it to another. It posed two possibilities: some arrangement of phases might make all four triple satellites reinforce, or the four might always split two and two. This essay answers both, and both turn out to be the wrong question.

The phases cannot matter #

The computation is the first thing to look at, because it is unambiguous. The chain of thirty thousand sites is rebuilt with the three phases at $(0.1, 0.37, 0.71)$, at zero, at $(0.25, 0.5, 0.9)$, and with the third phase moved by exactly half a turn; then twice more with the signs of the amplitudes reversed, once for the second wave and once for all three. The four triple satellites have the same intensities every time, to two parts in ten thousand, which is the accuracy of a windowed sum over a finite chain.

The reason is a counting argument, and it is exact for an infinite chain. The satellite at $h + s_1 q_1 + s_2 q_2 + s_3 q_3$ collects every product of wave components whose wavevectors add up to $s_1 q_1 + s_2 q_2 + s_3 q_3$. A wave component carries the phase factor $e^{2\pi i \phi_j}$ for each $+q_j$ it contributes and its conjugate for each $-q_j$. Because the three wavevectors are incommensurate, no other combination of integer multiples can add up to the same total, so every product that lands on the satellite uses the waves in exactly the proportions $s_1$, $s_2$, $s_3$. Every path to a given satellite carries the same phase factor, $e^{2\pi i (s_1\phi_1 + s_2\phi_2 + s_3\phi_3)}$. Changing the phases multiplies the whole amplitude by a number of modulus one, and the intensity does not move. Reversing the sign of a wave’s amplitude is a half-turn of its phase and does the same.

That is the standard fact about incommensurate crystals in another form. A modulated crystal is a section through a periodic structure in a space of higher dimension, as superspace groups in the plane describes. The phases of the waves choose where the section is taken, and for an incommensurate modulation every section passes arbitrarily close to every configuration, so all sections diffract alike. Shifting the phases is a phason, a rigid slide of the section, and phasons do not change intensities. The earlier essay’s suggestion assumed that the two paths carry phases of their own that could slide against each other. They cannot, because they reach the satellite through the same combination of waves.

Where the phases would matter #

The argument needs the wavevectors to be incommensurate, and it is worth seeing what happens when they are not, because a real crystal often sits close to a commensurate value and sometimes locks into one. Round each wavevector to a fraction with a common denominator and the chain becomes periodic, an approximant of the kind the crystal that rounds tau off builds from the Fibonacci chain. Then the integer combinations of the wavevectors are no longer all distinct. A triple combination can coincide with a pair combination, a second harmonic or a combination of higher order, and paths with different phase factors land on the same satellite.

The effect is large. With denominator thirteen, the satellite at $q_1 + q_2 - q_3$ changes by a factor of two thousand between one choice of phases and another, and the one at $q_1 + q_2 + q_3$ by seventeen. With denominator 21 the largest change is a factor of 43, with 34 a factor of 39, with 55 a factor of twelve. The changes do not fall steadily as the approximant improves, because which coincidences occur depends on the fraction, but none of the four approximants is free of them. In a commensurate crystal the phases are a real structural parameter, the position of the modulation relative to the lattice, and the satellites that need a second integer is a reminder that the indexing of satellites itself changes once the wavevectors are rational.

So the lead’s intuition that phases could decide the split was not wrong about crystals in general. It was right about commensurate ones and wrong about the incommensurate chain it was asked of. The difference is exactly the difference between a section of a periodic structure in superspace that passes through every configuration and one that passes through only a finite set of them.

The cube with or against everything else #

So the split between reinforcement and cancellation must be decided by something other than phases. To see what, the amplitude at each satellite is split in two: the cube’s contribution, the difference between the amplitude with the cubic coefficient on and with it off, and everything else. The cube reinforces the satellite when its contribution points with the rest, and cancels when it points against.

At the chain’s own wavevectors the answer is three against one. The cube opposes the rest at $q_1 + q_2 + q_3$, at $q_1 + q_2 - q_3$ and at $q_1 - q_2 - q_3$, and reinforces it at $q_1 - q_2 + q_3$. The two contributions are collinear to within a few hundredths in the sine of the angle between them, either with each other or against, which is what the phase argument requires: the cube and the rest share a phase factor, and the only freedom left is a sign. Reversing the sign of $\varepsilon_3$ reverses the cube’s contribution and so reverses all four: three reinforce and one cancels.

That makes the earlier lead’s first possibility unreachable by the route it proposed. No arrangement of phases makes every triple satellite reinforce, because no arrangement of phases changes anything. The sign of the cubic coefficient does change the split, but that is a property of the material, not of the experiment. A solid solution whose size effect bends one way has one split, and one that bends the other way has the complementary split.

Over every choice of wavevectors #

The second possibility, that the four always split two and two, is the one the computation can test widely, because the wavevectors are the thing that does matter. A crystal’s modulation wavevectors are set by its chemistry and its interactions, and different materials have different ones. So the question becomes: across the possible wavevectors, how do the four triple satellites divide?

The census draws 160 sets of three wavevectors at random between 0.05 and 0.47, from a seeded random sequence, and keeps a set only if each of its triple satellites lies at least forty resolution bins away from every other combination of the waves up to fourth order. Without that test a satellite could sit on top of a second-order or pair satellite and borrow its intensity. For each kept set the chain is built with the cubic coefficient positive, with it negative, and with it off, and the cube’s contribution is compared with the rest at each of the four triple satellites.

Every split occurs. All four reinforcing happens in three sets of 160, none reinforcing in one, one reinforcing in ten, and two or three in seventy-three each. So the four do not always split two and two: that is the single commonest outcome, but less than half. And the first possibility the lead raised, all four reinforcing, does happen, for three wavevector sets in 160, though not through any choice of phase.

Two further facts make the split clean enough to state. In 625 of the 640 satellites, the cube’s contribution and the rest are within a cosine of $0.9$ of being exactly aligned or exactly opposed; none is near a right angle. So “reinforce” and “cancel” are nearly always sharp alternatives rather than ends of a continuum. And in 636 of the 640, reversing the cubic coefficient reverses the cube’s contribution, as it must at first order. The four exceptions are satellites where the cube’s contribution is large enough that the second-order terms in $\varepsilon_3$ matter.

Where the sides change #

The scan says that the wavevectors decide. A sweep shows how. Fixing the first two wavevectors at the chain’s values and moving the third from 0.06 to 0.46 in steps of a hundredth, each satellite’s side changes at a handful of places. Four of them are explained at once by the structure of the displacement. A harmonic of frequency $Q$ in the spacing becomes a harmonic in the displacement with the factor $1/(2 \sin \pi Q)$, because the displacement is the running sum of the spacing, and that factor changes sign whenever $Q$ passes through a whole number. The cube’s path has the triple combination’s sine in its denominator; the pair path has the sines of the pair combinations $q_i \pm q_j$.

So the cube’s side flips at $q_1 - q_2 - q_3$ passing through zero, near 0.114; at $q_3 = q_2$, near 0.268, where a pair difference vanishes; at $q_1 + q_2 + q_3 = 1$, near 0.35; and at $q_3 = q_1$, near 0.382. Those are the “four products” the earlier lead anticipated, but they are products of sines of the wavevector combinations, not of cosines of the phases. Two further changes appear near 0.19 and 0.44, at one satellite each. They are not explained here; they may come from paths through the occupation wave, whose amplitude enters with a cosine of the wavevector rather than a sine, or from a near coincidence the separation test did not exclude.

What this changes for a measurement #

The practical reading follows directly. The triple satellites measure the cube against the curvature, and whether they see the two adding or cancelling is fixed by the crystal’s wavevectors and the sign of its cubic coefficient. Neither can be chosen in an experiment. What can be chosen is which satellites to measure, and the sweep shows that each crystal has some satellites where the cube adds. For the chain studied here it is $q_1 - q_2 + q_3$, where the satellite is 1.4 times as strong as the cube’s contribution and the rest would make it if their intensities simply added. In a measurement aimed at the cube, that satellite is where to look, and the three cancelling ones are worth measuring for a different reason: the earlier essay found a near-cancellation very sensitive to the ratio of the two coefficients.

The sign question has a sharper answer still. If the wavevectors are known, as they are from the positions of the first-order satellites, then the side each triple satellite falls on is predicted for each sign of the cube. The two predictions are complementary, three and one against one and three in this chain, and the measured pattern of strong and weak triple satellites then tells the sign of $\varepsilon_3$ before any fit is made. Two waves measure the curvature one wave hides found the curvature’s sign written in the asymmetry between sum and difference satellites; the cube’s sign is written in the same way, one level up.

The paths also weigh differently from one reflection to the next, and that offers a second handle. The cube reaches a triple satellite in the first power of the scattering vector, while the path through the curvature and a first-order wave needs its square, because it is a product of two displacements. Close to the origin of reciprocal space the square is small and the cube’s direct path dominates whichever side it is on, which is why the satellite beside the zero-order reflection was found to read the displacement most directly. Further out the two paths become comparable and their relative side starts to decide the intensity. A measurement that compares the same triple satellite at low and at high reflections therefore sees the cube alone and then the cube with or against the curvature, which separates the two coefficients without needing to know the side in advance.

There is a reason the four satellites behave independently at all, and it is the same incommensurability that makes the phases irrelevant. The four combinations $q_1 \pm q_2 \pm q_3$ are four distinct frequencies that never coincide with each other or with anything of lower order, so each has its own pair of paths and its own balance between them. In a chain whose waves beat against each other, as a beat is not a period puts it, the beats are not repeats, and the four triple satellites are four separate readings of the cube rather than one reading taken four times.

What the computation cannot show #

The figures show intensities, amplitudes and signs. They cannot show the chain itself, and an occupation wave is its own Fourier series is the place to see what three waves do to a row of atoms. Nor can they show the higher reflections. Everything here is at $h = 1$, and a mixture that moves with the reflection found the balance between paths shifting with $h$, because the paths carry different powers of the scattering vector. At larger $h$ the side a satellite falls on can change, and the scan has not been repeated there.

The model is one chain with one set of amplitudes and coefficients: occupation waves of amplitude 0.15 and a size effect with $\varepsilon_1 = 0.03$, $\varepsilon_2 = 0.15$ and $\varepsilon_3 = 0.5$. The phase argument does not depend on any of those numbers, and holds for every incommensurate chain. The split does depend on them, and a crystal with a much stronger linear size effect would have a different balance between paths. The convention that decides “reinforce” is the comparison of the cube’s contribution with everything else at the same satellite; an alternative that compared the cube with the curvature’s path alone would give the same answer wherever those two dominate, which at these amplitudes is everywhere the census looked.

Who worked out the pieces #

De Wolff introduced the description of modulated crystals by satellites and superspace in 1974, and Janner and Janssen developed the superspace groups; that phasons leave diffraction intensities unchanged for incommensurate modulations is part of that framework. Vegard’s law is from 1921, and its curvature and higher terms are measured in alloys by fitting lattice parameters against composition. The phase-invariance test, the split of each triple satellite into the cube and the rest, the census over random wavevectors, and the location of the sign changes at the resonances of the displacement are what this essay adds, on the chain the strain wave that tips the satellites and its successors built.

What the calculation must refuse #

Each check could have failed. If the two paths had carried independent phases, the six phase choices would have given six different sets of intensities. If the cube’s contribution were not collinear with the rest, its reversal with $\varepsilon_3$ would not have reversed each satellite’s side. And the claim that the four triples always split two and two is put to a sample of random wavevectors and refused.

Still open: the two changes no resonance explains #

The sweep’s two unexplained changes of side, near 0.19 and 0.44, are the loose end. The four explained ones sit exactly at zeros of the denominators the displacement carries, and an account of the other two would complete the description of where the sides change. One candidate is the occupation path: the occupation wave multiplies the displacement’s phase factor directly, without passing through the running sum, so it brings cosines of the wavevectors rather than sines, and those change sign at quarter-integers rather than integers. Another is a resonance between a triple combination and a second harmonic, $2q_j$, which the fourth-order separation test treats as distant but which could still pull a path’s phase. Deriving the amplitude of each path in closed form, at first order in each coefficient, and comparing it with the computed sign at every point of the sweep would decide between them. It has not been done.