Three waves measure the cube through the slope

A modulated crystal is a row of sites whose occupation rises and falls in waves, with the spacing between sites following the occupation: where a site holds more of the larger atom the next spacing is longer. That dependence is the size effect, and Vegard’s law makes it a straight line. The second order cannot see a curved size effect alone bent the line, and found that a single wave’s satellites could not tell the bend from the slope, because both grew with the scattering vector in the same way. Two waves measure the curvature one wave hides added a second wave. The bend then put the product of the two waves into the displacement, at their sum and difference, and the satellites there saw it with mixtures of opposite sign, which separated it.

That essay ended by asking the next question. A size effect that bends twice, with a cubic term in the spacing, has a product of three waves in it, so a chain modulated by three waves should carry displacement at the sums and differences of all three wavevectors together, $q_1 \pm q_2 \pm q_3$. Neither the slope nor the curvature puts any displacement there directly. If the triply mixed satellites saw the cubic coefficient in a mixture distinct from everything else, a crystal with three modulation waves could measure the third derivative of its spacing against composition.

The calculation has been made, and the answer has three parts, of which only the first was expected. The cubic term does reach the triple satellites, alone, in the displacement. In the intensity it does not arrive alone, and at the strongest triple satellite it is very nearly cancelled. And the best measurement of it comes from somewhere nobody pointed at: the first-order satellites, the strongest in the pattern, where the cube changes each wave’s apparent slope by an amount set by the other two waves.

A chain with three waves #

The chain is the two-wave chain with a third wave added. Each of its 30,000 sites has an occupation of one half plus three cosine waves of amplitude 0.15, smaller than before so that the occupation stays between nought and one. The wavevectors are the golden $q_1 = 1/\varphi^2 \approx 0.382$ and $q_2 = 2 - \sqrt3 \approx 0.268$ of the two-wave chain, and $q_3 = 1/\sqrt5 \approx 0.447$.

The third wavevector was chosen, not taken. Three waves make many combinations, and a satellite that falls close to another combination’s position collects that combination’s intensity too. The first choice tried, $(\sqrt7 - 2)/2$, put the satellite at $q_1 + q_2 + q_3$ within fifteen resolution widths of $-q_1 - 2q_3$, near enough for the window’s leakage to matter. Of a dozen quadratic irrationals tested, $1/\sqrt5$ keeps every satellite used in the calculation, sixteen kinds, at least 196 resolution widths from any combination of the three wavevectors to fourth order, where the resolution width is one over the length of the chain.

The spacing after each site is

$$1 + \varepsilon_1 d + \varepsilon_2 \left(d^2 - \langle d^2 \rangle\right) + \varepsilon_3 \left(d^3 - \langle d^3 \rangle\right),$$

with $d$ the mean of the two neighbouring occupations minus one half, $\varepsilon_1 = 0.03$ and $\varepsilon_2 = 0.15$ as in both earlier chains, and $\varepsilon_3 = 0.5$. The cubic coefficient looks large, but $d$ rarely exceeds a quarter, so its cube is small, and the three terms contribute spacings of comparable order.

The cube in the displacement #

The cube of $d$ is a cube of a sum of three cosines, and among its terms is six times the product of all three, with no other coefficient in front. A product of three cosines is a sum of four cosines, one at each of the combinations $q_1 \pm q_2 \pm q_3$, each with a quarter of the product’s weight. Summing the spacings along the chain turns a cosine of wavevector $Q$ into a displacement wave of amplitude one over twice the sine of $\pi Q$, as in the earlier chains, so the displacement at each triple wavevector is

$$u(Q) = \frac{\varepsilon_3 \cdot \tfrac32\, b_1 b_2 b_3}{2\,|\sin \pi Q|}, \qquad b_i = a_i \cos \pi q_i,$$

where $b_i$ is the amplitude of wave $i$ after averaging two neighbours. Fitted to the chain by least squares, the displacement at the four triple wavevectors matches the formula to within one per cent at every one. The largest of the four is at $q_1 + q_2 + q_3$, a wavevector of about 1.097, which is $0.097$ beyond a whole number, so the sine in the denominator is small and the displacement large.

So the first half of the expectation holds exactly. The cubic coefficient, and only the cubic coefficient, puts displacement at the triple wavevectors. If the satellites there read the displacement and nothing else, they would measure the cube directly.

Two paths to one satellite #

A satellite does not read the displacement alone. The diffracted amplitude at a scattering vector is a sum over sites of the occupation times a phase set by each site’s position, and when the phase is expanded in the displacement, products of several displacement waves and of the occupation with them all appear. The satellite at $h + q_1 + q_2 + q_3$ collects every product whose wavevectors add up to $q_1 + q_2 + q_3$. The direct cubic displacement is one of them. A pair harmonic from the curvature, at $q_1 + q_2$, multiplied by the third wave’s first-order displacement at $q_3$, is another, and so is a product of all three first-order displacement waves, which comes from the slope alone.

The figure at the head of this essay takes them apart by switching coefficients off. With only the slope, the satellite’s intensity at the first reflection is below $10^{-10}$: the triple product of first-order waves is negligible at this size of modulation. With only the curvature it is $1.9 \times 10^{-7}$, and with only the cube $3.1 \times 10^{-7}$, so both of the bending terms reach the satellite with comparable strength. With all three on, the intensity is $2.8 \times 10^{-8}$, eleven times weaker than with the cube alone. The two paths arrive with opposite signs and nearly cancel.

That changes what the satellite can measure. A reflection whose intensity is the near-cancellation of two terms is extremely sensitive to the ratio of the two coefficients, and almost blind to anything that changes both together. The measured sensitivities say exactly that: at a single reflection the errors of the curvature and the cube, estimated from the triple satellites, are correlated to $1.000$ in the third decimal place. The triple satellites measure a combination of $\varepsilon_2$ and $\varepsilon_3$, very well, and cannot say which is which.

A mixture that moves with the reflection #

The two-wave essay’s test of whether satellites can separate two coefficients was the ratio of their sensitivities. One ratio across many satellites means one equation repeated; different ratios mean different equations. Applied to the triple satellites, the test gives two answers.

Against the slope, the ratio moves. At $q_1 + q_2 + q_3$ it is 0.96 at the first reflection, 0.67 at the second, 0.51 at the third and 0.42 at the fourth, and the other three triple satellites fall in the same way. That is the opposite of what defeated the single wave’s second order, where the ratio was fixed because the two terms grew alike with the scattering vector. Here they do not: the direct cubic term is linear in the displacement, and the slope reaches these satellites only through products of several displacement waves, which grow faster with the scattering vector. Several reflections therefore separate the cube from the slope.

Against the curvature, the ratio hardly moves: $-0.44$, $-0.48$, $-0.52$, $-0.57$ at $q_1 + q_2 + q_3$, and $0.06$ to $0.07$ at $q_1 - q_2 + q_3$ at all four reflections. Both paths are linear in a displacement harmonic of the bending kind, so they grow alike, exactly as the slope and the curvature grew alike at a single wave’s second order. The triple satellites alone cannot separate the cube from the curvature, however many reflections are measured. They need the curvature fixed from elsewhere, and the pair satellites at $q_i \pm q_j$, the two-wave essay’s mixed satellites, are where it is fixed.

What each set of satellites says #

The Cramér–Rao bound turns the sensitivities into the best precision any refinement could reach, for a stated error on each intensity. The chain has six parameters, the three size coefficients and the three wave amplitudes, all refined together. With a two per cent error on every intensity, the error the two earlier essays assumed, the triple satellites of one reflection fix the cube to twelve per cent of its value, and of four reflections to 0.44 per cent. The pair satellites of four reflections give 3.1 per cent, the first and second orders together 2.2 per cent, and all of them with the triples 0.02 per cent, a gain of twenty over the best single kind, because the triples’ direction of ignorance and the lower orders’ are nearly perpendicular.

On that count the triple satellites do what the earlier essay hoped. The trouble is the count. Two per cent of every intensity, however weak, means a satellite of $10^{-11}$ is measured as well as one of $10^{-3}$, and real measurements do not behave like that. The four triple satellites at the first reflection have intensities of $2.8 \times 10^{-8}$, $4.0 \times 10^{-9}$, $5.7 \times 10^{-7}$ and $9.9 \times 10^{-12}$, against $5 \times 10^{-3}$ for the first-order satellites and a main reflection of a quarter. The two-wave essay already called its error model generous for satellites four or five orders below the main reflection. The triples are six to ten orders below it, and there the model stops being generous and becomes wrong.

So the bound is computed a second way, with a background floor added to every intensity’s error: an absolute error of $10^{-8}$ of the scattering per site, which still credits the measurement with a dynamic range of more than seven orders of magnitude. With the floor, the triple satellites of four reflections give the cube to 127 per cent of its value, which is to say not at all. The first and second orders give 2.3 per cent, barely changed, and every kind together 0.82 per cent.

Where the triple satellites disappear #

Sweeping the floor shows where the change happens. At a floor of $10^{-12}$ the triples of four reflections still give the cube to 1.4 per cent; at $10^{-11}$, to 6 per cent; at $10^{-10}$, to 16; at $10^{-9}$, to 35; at $10^{-8}$, to 127. The first, second and pair satellites together go from 1.5 per cent with no floor to 2.0 at $10^{-8}$ and 2.5 at $10^{-7}$: they hardly notice, because their intensities sit far above any of these floors.

Every kind together falls between: 0.04 per cent at $10^{-12}$, 0.8 at $10^{-8}$, 2.2 at $10^{-7}$. Below about $10^{-9}$ the triples still add information the lower orders lack, because they pin the combination of curvature and cube that the lower orders leave loosest. Above that they add almost nothing. The crossover is the size of the triple satellites themselves, which is what a floor ought to do and is no surprise. What is a surprise is that the lower orders carry the cube at two per cent on their own.

The cube in the slope #

The first-order satellites at $h \pm q_i$ are the strongest satellites in the pattern, and they were not expected to see the cube. They see the displacement wave at $q_i$, which the slope puts there in proportion to $\varepsilon_1 b_i$. But the cube of $d$ has a term at $q_i$ too. Expanding the cube of a sum of three cosines, the terms at a single wave’s own wavevector come from that wave cubed, $\tfrac34 b_i^3$, and from that wave times the square of each other wave, $\tfrac32 b_i b_j^2$. So the displacement at $q_i$ is proportional to

$$b_i \left[\varepsilon_1 + \varepsilon_3\left(\tfrac34 b_i^2 + \tfrac32 \textstyle\sum_{j \ne i} b_j^2\right)\right].$$

Each wave sees an effective slope, the true slope plus the cube times a weight set by its own amplitude and the other two waves’ amplitudes. With one wave the weight is the same function of the one amplitude that also sets the satellite’s size, and the cube folds into the slope where no measurement can find it. With three waves of different neighbour-averaged amplitudes the weights differ from wave to wave. Here the three $b_i$ are 0.054, 0.100 and 0.025, because averaging two neighbours shrinks a wave by the cosine of $\pi q_i$, and that cosine is small for the third wave’s wavevector near a half. The weights come to 0.018, 0.013 and 0.020, and the measured ratios of the first-order satellites’ sensitivities to the cube and to the slope are 0.018, 0.013 and 0.021, which is the formula.

Three satellites that see three different mixtures of the same two coefficients separate them, the same principle that made the two-wave essay’s sum and difference satellites work, here with the mixtures differing in size rather than in sign. And because the first-order satellites are strong, the separation survives any plausible background. The second-order satellites help in the same way, with mixtures of their own, and together the two kinds give the cube to about two per cent with every amplitude refined.

What this means for a measurement #

A crystal with three incommensurate modulation waves is unusual but not exotic, and the size effect’s curvature and its third derivative are the departures from Vegard’s law that a careful study of a solid solution would want. The calculation says how to measure them, and the answer inverts the plan the two-wave essay proposed. The weak, high-order satellites at the triple combinations are where the cube is isolated in the displacement, and they are where it is hardest to read, because two bending terms cancel there and the satellites sit near the noise. The strong first-order satellites are where the cube is mixed into the slope, and they are where it is most easily read, because three waves give three mixtures.

The design consequence is that the amplitudes of the three waves matter more than which satellites are measured. The cube’s lever on the first-order satellites is the spread of the weights, and the spread comes from waves of unequal neighbour-averaged amplitude. Three waves of equal $b_i$ would give the first-order satellites one mixture, and the cube would again hide in the slope.

What the calculation leaves out #

One chain, one choice of amplitudes and wavevectors. The cancellation at $q_1 + q_2 + q_3$ depends on the relative sizes of the curvature and the cube and on the phases of the waves; other choices would cancel less, or at a different triple satellite. The calculation shows that the two bending paths reach the same satellite with comparable strength, and that they can cancel. It does not show that they always do.

A bound, not a refinement. The Cramér–Rao numbers are the best any unbiased estimate can do with the stated errors and a correct model. A real refinement adds systematic errors, of which a smooth error across reciprocal space is one kind, and the weak satellites are the most exposed to them.

One dimension. The chain is a line. In a real crystal three modulation waves need not run along one direction, and the triple satellites then lie off the line of the main reflections, in the geometry the satellites that need a second integer set out with one index per wave. The effective-slope argument does not depend on direction, since it is about the displacement at each wave’s own wavevector, but the sizes of the satellites do.

The window. Intensities are Hann-windowed sums over the chain, the window the satellite beside the zero-order reflection found necessary, and the choice of the third wavevector keeps every satellite at least 196 window widths from anything else.

What the pictures cannot show is a measurement. Every number in the figures is a sensitivity or a bound computed from a chain whose coefficients are known. The figures show what an ideal refinement could reach. They cannot show what a diffractometer would record, which for the triple satellites is mostly background.

What the calculation must refuse #

The refused claim is the reading the displacement invites: that because only the cube puts displacement at $q_1 + q_2 + q_3$, the cube alone sets the intensity there. It does not. The curvature reaches the same satellite through a pair harmonic times a first-order wave, with comparable size and the opposite sign, and the satellite is eleven times weaker than the cube alone would make it.

Who worked out the pieces #

De Wolff described modulated crystals by their satellites, and Janner and Janssen’s superspace formalism gives each independent wavevector its own index, so that the triple satellites here are reflections with three non-zero satellite indices, a case superspace groups in the plane does not reach. Vegard’s law is from 1921 and its higher derivatives are measured in alloys by fitting lattice parameters against composition. The expansion of a phase factor into products of displacement waves is the standard treatment of displacive modulation in the strain wave that tips the satellites and the second order reads the strain again, and the occupation’s own share of each satellite is the series an occupation wave is its own Fourier series wrote out. The Cramér–Rao bound is from the 1940s. What the calculation adds is the cancellation at the strongest triple satellite, the fixed mixture of curvature and cube there, and the effective slope through which three first-order satellites see the cube.

Still open: a cube the triple satellites can see #

The cancellation at $q_1 + q_2 + q_3$ is a statement about one choice of phases and coefficients. The pair-times-first-order path and the direct cubic path both carry phases set by the waves’ phases, and changing the third wave’s phase by half a turn changes the sign of one path relative to the other, which would turn the cancellation into reinforcement at that satellite and move it to another. Whether some arrangement of phases makes every triple satellite reinforce, so that the triples become as strong as the cube alone would make them, or whether the four combinations always split into two that reinforce and two that cancel, is a question about the signs of four products of cosines. It decides whether a triple satellite could ever rise above a realistic background, and it has not been worked out.