The strain wave that tips the satellites

A modulated crystal has two simple kinds of wave. In a displacive modulation every site holds a whole atom, and the atoms are pushed back and forth by a wave whose wavelength is not a whole number of cells. The satellites that need a second integer worked out what that does to diffraction: a satellite beside every main reflection, at $h + mq$ for every whole $m$, with strength given by a Bessel function of the displacement amplitude. In an occupational modulation the sites stay put and a wave decides how much of an atom sits on each. An occupation wave is its own Fourier series worked out that case: the satellite of order $m$ is the $m$-th Fourier coefficient of the occupation profile, the same on both sides of every main reflection.

Real modulations are rarely either one alone. The earlier essay ended on why. A composition wave strains the lattice, because atoms of different sizes want different spacings, so an occupation wave drags a displacement wave with it. The measured signature is a satellite pair that is not a pair: the one on one side of a main reflection is stronger than the one on the other. Neither pure calculation can produce it. The occupation alone makes the two sides equal, and the displacement alone makes them unequal only because the two sides sit at different distances from the origin of reciprocal space.

This essay puts both waves into one chain and computes the result exactly. The answer is a formula, a convolution of the two earlier results. It gives an imbalance with three properties worth knowing. It follows the phase between the two waves, and the size effect happens to fix that phase at the value that maximises it. It grows in proportion to the index of the main reflection. And, to first order, it does not depend on how strong the occupation wave is at all. A strain wave locked to a composition wave by atomic size tips every satellite pair by an amount set by the size mismatch alone. That makes the tilt a ruler for how much larger one species is than the other, and its sign says which.

Two waves in one chain #

The chain is the simplest one that holds both waves. Sites sit near the integers. Site $n$ has a phase $t_n = nq + \varphi$ along the wave, where $q$ is irrational, here $1/\varphi^2 \approx 0.382$ with $\varphi$ the golden ratio. It is occupied to $f(t_n) = c + a\cos 2\pi t_n$, with mean $c = 0.5$ and amplitude $a = 0.4$, and it is displaced from its lattice point by $u \sin(2\pi t_n + \theta)$. The amplitude $u$ and the phase $\theta$ are free. The size effect will fix them.

The lower half of the picture is the useful view, and it is the superspace view of the extra dimension that makes it periodic. Plotted against the phase of the wave at each site, sixty sites of an incommensurate chain fill out two smooth curves, one for how full the site is and one for how far it has moved. Everything about the diffraction is decided by those two curves, because an irrational $q$ makes the phases $t_n$ fill the interval evenly, and the sum over sites becomes an integral over the phase.

A convolution of the two earlier answers #

Doing that integral with both waves present is short. The structure factor at $k = h + mq$ is the sum over sites of the occupancy times a phase factor, and the displacement contributes a factor $e^{-2\pi i k u \sin(2\pi t + \theta)}$. That factor expands as a sum of Bessel functions, $\sum_j J_j(2\pi k u)\, e^{-ij(2\pi t + \theta)}$. Multiplied by the occupancy and integrated over $t$, each term picks out one Fourier coefficient of the profile:

$$\frac{F(h + mq)}{N} \;\to\; e^{2\pi i m\varphi} \sum_j \hat f(m + j)\; J_j(2\pi k u)\; e^{-ij\theta}.$$

The satellite is the convolution of the profile’s Fourier coefficients with the displacement’s Bessel functions. With no displacement only $j = 0$ survives, and the satellite is $\hat f(m)$, the earlier occupation result. With a uniform occupancy only $\hat f(0)$ survives, so $j = -m$, and the satellite is $c\,J_{-m}(2\pi k u)$, the earlier displacement result. With both, the terms mix.

The formula is checked against the direct sum over a chain of six thousand sites, for thirty-six satellites of first and second order at three settings of the displacement. The largest disagreement is $1.6 \times 10^{-4}$, which is what remains when an infinite average is replaced by six thousand terms. Nothing in the formula is fitted.

For a cosine profile only three coefficients are non-zero: $\hat f(0) = c$ and $\hat f(\pm 1) = a/2$. The first-order satellites are therefore easy to read. To first order in the displacement, where $J_0 \approx 1$ and $J_{\pm 1}(z) \approx \pm z/2$, the pair on either side of $h$ is

$$|F(h \pm q)| \;\approx\; \frac{a}{2} \mp c\,\pi k u \cos\theta .$$

The occupation contributes the same $a/2$ to both sides. The displacement contributes a term that adds on one side and subtracts on the other, in proportion to $k$ and to the cosine of the phase between the waves. That cross term is the interference neither wave has alone, and it is what tips the pair.

Two symmetric waves, one lopsided pair #

The picture at the head of this essay sets the three cases side by side at the first four main reflections.

The occupation wave alone gives $0.2000$ on both sides of every reflection, which is $a/2$ exactly, as the earlier essay predicted. The strain wave alone, with every site half full, gives satellites ten times weaker. Those grow from one reflection to the next, and the high side is always a little stronger because it is further out in $k$. Together, with a displacement of $0.0093$ of a lattice spacing, the pair around $h = 1$ is $0.209$ against $0.180$, and around $h = 4$ it is $0.252$ against $0.135$. The low side has gained what the high side has lost, and the gain grows steadily with $h$.

This imbalance does not belong to either wave. The strain wave’s own imbalance points the other way, towards the high side. It is a property of the two waves together, carried by the cross term in the formula. That is the first thing a crystallographer reading a measured imbalance has to know. A lopsided pair is not evidence of a large displacement, since the displacement here is under a hundredth of a spacing. It is evidence of a displacement locked to the occupation.

The phase decides everything #

The cross term carries $\cos\theta$, the cosine of the phase between the two waves, so the imbalance should swing with that phase. It does.

With the displacement in step with the occupation, $\theta = \pi/2$, so that atoms move towards $+x$ where the sites are fullest, the cross term vanishes. The two sides are then nearly equal again, closer than with the displacement alone, because the occupation’s equal contribution dilutes the displacement’s small imbalance. With the displacement a quarter of a period away, $\theta = 0$ or $\pi$, the cross term is as large as it can be. The imbalance reaches $-0.95$ on one side of that and $+0.85$ on the other. The phase between the waves therefore decides whether a strain of a given size tips the pair at all.

That phase is not a free parameter in a real crystal. The size effect decides it.

Size fixes the phase #

Suppose the spacing between two neighbouring sites grows with how full they are on average, by a size factor $\varepsilon$: positive when the atom that fills the sites is larger than what it replaces, negative when it is smaller. Then each spacing is $1 + \varepsilon\big((f_n + f_{n+1})/2 - c\big)$, and each position is the sum of the spacings before it. The sum of a cosine is a sine. Summing the occupation wave over the sites gives a displacement

$$u \sin 2\pi t_n , \qquad u = \frac{\varepsilon\, a \cos \pi q}{2 \sin \pi q},$$

with $\theta = 0$: exactly a quarter of a period behind the occupation. The sites are displaced most where the occupation crosses its mean, and not at all where it is fullest or emptiest. The reason is that the displacement accumulates. By the time a run of full sites is over, its extra spacing has all been added. Fitting a sine and a cosine to the displacement of a six-thousand-site chain built spacing by spacing confirms it. The sine amplitude matches the formula to five significant figures and the cosine amplitude is $6 \times 10^{-11}$.

So the physical mechanism that couples the two waves also puts them in quadrature, at the phase where the imbalance is largest. A composition wave that strains its lattice tips its satellites as far as its strain allows, for a reason as simple as the fact that positions are sums of spacings.

Driving the size factor through zero shows the imbalance as an odd function of it. For every $h$ its slope is steady out to a strain of twenty per cent, and its sign changes with the sign of $\varepsilon$. When the atom that fills the sites is the larger one, the low side of each main reflection gains. When it is the smaller one, the high side gains. This is the same lean that Warren, Averbach and Roberts found in 1951 in the diffuse scattering of alloys, towards the low-angle side of each Bragg peak when the larger atom is the stronger scatterer. The satellites of a modulated crystal are the sharp version of the same thing. In the diffuse pattern of a disordered alloy the lean is spread over reciprocal space. Here it is concentrated into one pair of peaks.

The tilt measures the size, not the wave #

The first-order pair formula, with the size effect’s $u$ substituted, shows the most useful property. The imbalance is the difference of the two intensities over their sum. The difference is $2 \cdot (a/2) \cdot 2c\pi k u$, summed over the two values of $k$, and the sum is $2(a/2)^2$. Since $u$ is proportional to $a$, the amplitude cancels:

$$\frac{I_+ - I_-}{I_+ + I_-} \;\approx\; -\,2\pi h\, c\, \varepsilon \cot \pi q .$$

The tilt of a satellite pair depends on the size mismatch, the mean occupancy, the wavevector and the reflection, and not on how strong the composition wave is. A weak wave and a strong one with the same size factor tip their satellites by the same fraction. The weak wave’s satellites are weaker, but they lean the same way by the same amount.

The measurement agrees. At $h = 2$ and $\varepsilon = 0.1$ the chain with occupation amplitude $0.1$ gives an imbalance of $-0.246$, and the chains with amplitudes $0.2$, $0.3$ and $0.4$ give $-0.247$ each, against $-0.244$ from the first-order formula. What is left over is second order in the displacement. The practical consequence is that the ratio of a satellite pair measures $\varepsilon$ with no need for the modulation amplitude, which a structure refinement would otherwise have to supply. The same ratio at two main reflections tests the formula’s proportionality to $h$, and so tests the assumption that the size effect is what couples the waves.

Reading the size back #

The formula is only worth having if it can be run backwards, from a measured pair to a size factor. For the reading to be fair, the chain that makes the data must be built independently of the formula: spacing by spacing, from a size factor the reading does not know. The reading then uses only the imbalance at each reflection, the wavevector and the mean occupancy, and divides: $\varepsilon = -\,\text{imbalance} / (2\pi h c \cot \pi q)$.

Three chains test it. The first has a size factor of $0.08$ and an occupation amplitude of $0.3$, and its four main reflections read back $0.0814$, $0.0808$, $0.0799$ and $0.0785$. The second has a negative size factor, $-0.06$, where the atom that fills the sites is the smaller one, and a weaker wave. It reads back $-0.0591$ to $-0.0579$, with the sign right at every reflection. The third returns to $0.08$ with a mean occupancy of $0.3$ instead of a half and reads back $0.0810$ to $0.0800$. Through the third reflection every reading is within three per cent of the truth. The drift at higher reflections is the second-order term the first-order formula leaves out, and its size is itself a check: it grows with $h$ as $(2\pi k u)^2$ does.

Two things are needed that the satellites do not supply. The mean occupancy $c$ enters the formula directly, because the displacement’s contribution scales with how much scattering matter is on the sites on average. It comes from the main reflections, whose strength is the mean occupancy. The wavevector comes from where the satellites sit. The modulation amplitude, which a refinement would otherwise have to fit, is not needed at all. That makes the tilt of a satellite pair one of the few quantities in a modulated structure that can be read without first solving the structure.

When the satellites stop being a portrait #

The earlier essay’s central result was that satellites are a portrait of the occupation profile: the strength of a pair is the profile’s Fourier coefficient, read directly. The question it left was how large a strain wave must be before that stops being true. The formula’s answer is surprisingly small, because the displacement’s contribution carries a factor $c\pi k$ that the occupation’s lacks. The mean occupancy $c$ is usually larger than the wave’s amplitude, and $k$ grows with every reflection.

With a size factor of five per cent the displacement is $0.0039$ of a lattice spacing, and by the third main reflection one of the two satellites is more than ten per cent away from the Fourier coefficient. With a size factor of fifteen per cent, a displacement of $0.012$ of a spacing, the portrait fails at the first reflection. Reading an occupation profile from satellite intensities is therefore safe only where the strain is below a few thousandths of a spacing, or at the lowest reflections. Averaging each pair helps, since the cross term cancels from the mean to first order. The departures in the figure are for the worse of the two sides, which is what a crystallographer reading one satellite would see.

What the picture cannot show #

The chain is one-dimensional with a cosine profile, and both simplifications are chosen for clarity. A crenel profile, the step that the earlier essay showed is the strongest occupation wave there is, has Fourier coefficients at every order. The convolution then mixes every occupation coefficient with every Bessel function, and satellites of higher order tip in patterns that alternate with $m$. The formula covers that case and nothing here draws it. In three dimensions the displacement has a direction, and only its component along the scattering vector enters the Bessel functions. So the tilt depends on which way the strain points relative to each reflection. That is a real complication, and it is how the direction of a size strain is measured in practice.

The size coupling is also the simplest possible: a spacing linear in the mean occupancy of neighbouring sites. Real crystals relax further than nearest neighbours, and the displacement a composition wave produces is then a sine filtered by the elastic response at wavevector $q$. It stays in quadrature for any coupling that is symmetric in space, since a symmetric response cannot shift a sine’s phase, but its amplitude changes. The amplitude-free formula for the tilt then carries that response in place of $\cot \pi q$.

What the combined wave has to refuse #

The first refused claim is the natural reading of a lopsided pair: that the imbalance is simply the displacement’s own. The displacement alone tips the pair around $h = 2$ by $+0.37$, towards the high side. Combined with the occupation it tips it by $-0.30$, the other way. The second refused claim is the one the earlier essay’s portrait invited: that first-order satellites read the occupation profile whatever the strain. At a size factor of ten per cent they fail by the second reflection.

Still open: the second order, and three dimensions #

The formula gives the second-order satellites as readily as the first. With a cosine profile they come entirely from cross terms, $\hat f(\pm1)$ times $J_{\mp1}$ plus $\hat f(0)$ times $J_{\mp2}$, so an occupation wave with no second harmonic acquires second-order satellites from its strain alone, with an imbalance of their own. Whether the ratio of the second-order imbalance to the first gives a second, independent measure of the size factor, one that would tell a size effect apart from any other mechanism coupling the two waves, is the obvious next computation, and it has not been made here.

The other direction is the displacement’s orientation. A composition wave running along one axis of a real crystal pushes its atoms along that axis, and across it, when the lattice is anisotropic. The tilt of each satellite pair then depends on the angle between the scattering vector and the strain. Mapping that angle dependence would separate a longitudinal size effect from a transverse one, and nothing here goes beyond one dimension.