The second order reads the strain again

The strain wave that tips the satellites put two waves in one chain of atoms. One was an occupation wave, the probability that a site holds the larger of two species, rising and falling as a cosine. The other was the displacement that wave drags along with it, because larger atoms push their neighbours apart. The size effect locks the displacement a quarter of a period behind the occupation, and the lock makes the first-order satellite pairs lopsided. The satellite on one side of each main reflection is stronger than the one on the other, by an amount proportional to the size factor $\varepsilon$ and independent of how strong the occupation wave is. Reading the size factor back from the tilt of the pairs was the point of that essay.

It ended by noticing that the same formula gives the second-order satellites as readily as the first, and that for a cosine occupation profile they come entirely from cross terms. An occupation wave with no second harmonic makes no second-order satellites by itself, and an occupation wave is its own Fourier series showed why: the satellites of a pure occupation wave are the Fourier coefficients of its profile, and a cosine has none beyond the first. Whatever appears at second order is the strain’s doing. The question left open was whether those satellites give a second reading of the size factor, and whether the second reading could tell a size effect apart from another way of coupling the two waves.

What appears at second order #

The chain’s sites sit at $x_n = n + u\sin(2\pi t_n + \theta)$ with $t_n = nq + \varphi$ and $q$ irrational, and each is occupied to $c + a\cos 2\pi t_n$. The satellite at $h + mq$ is then

$$F(h + mq) = \sum_j \hat f(m + j)\, J_j(2\pi k u)\, e^{-ij\theta}, \qquad k = h + mq,$$

a sum over the occupation’s Fourier coefficients $\hat f(0) = c$ and $\hat f(\pm 1) = a/2$, weighted by Bessel functions of the displacement. At second order, $m = \pm 2$, the only terms are those with $m + j$ equal to $-1$, $0$ or $1$, and to lowest order in $z = 2\pi k u$,

$$F(h + 2q) \approx -\tfrac{a}{4}\, z\, e^{i\theta} + \tfrac{c}{8}\, z^2 e^{2i\theta}, \qquad F(h - 2q) \approx \tfrac{a}{4}\, z\, e^{-i\theta} + \tfrac{c}{8}\, z^2 e^{-2i\theta} .$$

The leading term is the occupation’s first harmonic carried up one more order by the displacement’s first Bessel function. It needs both waves, $a$ from the occupation and $z$ from the strain. The formula was checked against direct sums over 6,000 sites at eighteen second-order satellites and three different displacements, to within $1.3 \times 10^{-4}$, which is the floor a finite chain sets.

The picture at the head of this essay shows the second order arriving with the strain. Without strain its intensity is about $2 \times 10^{-8}$, the floor of a chain of 6,000 sites. With a size factor of 0.1 it rises to between $10^{-6}$ and $4 \times 10^{-4}$, against a first order near $4 \times 10^{-2}$. The second-order pair is lopsided too, and in the opposite direction from the first order. For this size factor the first-order pair is stronger on the inner side of each main reflection, which is the size effect’s tilt. The second-order pair is stronger on the outer side, because $z$ is proportional to $k$, and $h + 2q$ is further from the origin than $h - 2q$. At $h = 1$ the outer satellite is almost fifty times the inner one for that reason alone.

Two powers of one displacement #

The formula says what each kind of satellite measures, and the two measure different things.

The second-order intensity is $|a z / 4|^2$ to lowest order, proportional to $u^2$ and to $a^2$. The phase $\theta$ enters only as a factor of modulus one, so the intensity cannot see it. The first-order tilt, computed in the earlier essay, is proportional to $u \cos\theta$. Measured on chains with an explicit displacement, the second-order intensity rises with $u$ with a logarithmic slope of 1.91 over a sixteenfold range of amplitudes, and the tilt with a slope of 0.996. The small shortfall from two in the first is the next term of the Bessel expansion beginning to matter at the largest amplitudes. So one experiment supplies two different functions of the displacement: its square from the second order, and its signed projection onto the size-locked phase from the first.

Reading the size factor twice #

A size effect ties the displacement’s amplitude to the occupation’s: $u = \varepsilon a \cot(\pi q) / 2$, a quarter of a period behind. That gives two routes back to $\varepsilon$ from satellites alone, with the composition $c$ as the only other input. The first is the earlier essay’s: the first-order imbalance is $-2\pi h c\, \varepsilon \cot \pi q$, so dividing by $2\pi h c \cot \pi q$ gives $\varepsilon$ with the occupation amplitude cancelled. The second is new. The first-order pair’s mean amplitude is $a/2$, which gives $a$. The second-order pair’s amplitude is $a z / 4$, which then gives $z$ and so $u$, and the size-effect relation turns $u$ into $\varepsilon$.

On five chains that differ in size factor, occupation amplitude and composition, both readings return the size factor each was built with. They agree at three reflections each to within 7.6 per cent in the worst case, and to one or two per cent in most. The two readings use different satellites, different orders of the same expansion and different pieces of the formula. The second reading has no sign of its own, since an intensity has none, and takes its sign from the first. Its size is independent.

The ratio is the phase #

Two independent readings of one number are worth having only if they can disagree. They disagree whenever the displacement is not where a size effect puts it.

Take chains whose displacement has the amplitude a size factor of 0.08 would give, and set its phase $\theta$ away from the quarter-period lag that a size effect imposes. The second-order reading does not move, because the second-order strength cannot see $\theta$. The first-order reading falls as $\cos\theta$, because the tilt is the displacement’s projection onto the size-locked phase. Their ratio follows $\cos\theta$ across the whole range from nought to half a turn, to within 0.013 at every one of thirteen phases. At a quarter of a turn the first-order pair has no tilt at all, although the chain is strained as much as before, and the second order still shows it.

This is the discrimination the earlier essay asked for. A size effect is one mechanism for coupling a displacement to an occupation, and it fixes the phase. Others fix it differently. An elastic interaction that favours a particular modulation of the spacing, or a displacive instability that happens to share the occupation’s wavevector, need not lag by a quarter of a period. The first-order tilt alone cannot tell a weak size effect from a strong coupling at the wrong phase, since both give a small tilt. The two readings together can, because their ratio measures the cosine of the phase directly. A ratio of one says the strain is where the size effect puts it. A ratio below one says how much of the strain a size effect accounts for.

There is a limit on what a ratio of one can prove, and symmetry sets it. Superspace groups in the plane describes a modulated crystal’s symmetry in the space where it is periodic again, with the phase of the modulation as an extra coordinate. If that superspace group has a centre of symmetry, and the centre sits where the occupation wave is at a maximum, then the occupation must be an even function of the extra coordinate and every displacement an odd one. An odd displacement is a sine, which is the size-locked phase or its opposite. In a centrosymmetric modulated structure, therefore, $\theta$ is forced to nought or half a turn by symmetry, whatever mechanism couples the waves, and the ratio of the two readings is forced to plus or minus one. The discrimination works only where symmetry leaves the phase free, in a structure whose superspace group has no centre. In a centrosymmetric one the ratio still reports the sign of the coupling and confirms the formula, and it cannot say what caused the strain.

What the second reading costs #

A second reading that agrees with the first is only useful if it can be measured, and the second-order satellites are weak. The cost can be computed by counting. Give every satellite a Poisson count proportional to its intensity, with enough counting time that each first-order satellite collects about ten thousand counts, form both readings from the noisy counts, and repeat four hundred times.

At $h = 2$, with the occupation amplitude of 0.4, the second-order satellites then collect between one count and thirty-five, depending on the size factor. The first reading scatters by 0.0029 in $\varepsilon$ and the second by 0.010, at every size factor from 0.02 to 0.12. The second is about three and a half times worse, which means about twelve times the counting time to match the first.

The two spreads behave differently, and the reason is worth stating. The first reading is a difference of two nearly equal intensities, and its absolute error is set by the first-order counts, which do not change much with $\varepsilon$. The second reading comes from an amplitude, the square root of a count, and the square root of a Poisson count has an error of a half whatever its mean. So the second reading’s absolute error is fixed by the counting time alone and does not grow as the second order weakens. The consequence is that the second reading’s relative precision is worst for small size factors, where it measures a small number with a fixed error. It becomes competitive when the size effect is large or the counting is long, which is when the first-order tilt is also largest. Every reflection, several times over makes the general point that redundancy buys a smaller random error and nothing else. The second order does buy something else, the phase, and the counting time is its price.

The second-order pair tilts too #

The second-order pair is lopsided mainly because its scattering vectors differ in length. Dividing each intensity by the square of its $k$ removes that, and what is left is a tilt of its own.

The formula predicts it. The cross term between the two parts of each second-order structure factor has opposite signs on the two sides, and after division by $z^2$ it leaves an imbalance of $-\tfrac{c}{a}\, 2\pi u h \cos\theta$, which for a size effect is $-\pi h c\, \varepsilon \cot\pi q$: exactly half the first-order tilt. The computed chains follow it at $h = 1$, $2$ and $3$, where the ratio of the two tilts is within 0.007 of a half. An earlier version of this page reported a departure at $h = 1$, a ratio between 0.30 and 0.45 on chains of 30,000 and 120,000 sites alike, and left it unexplained. It was the finite chain after all. The inner satellite at $h = 1$ lies at $k = 1 - 2q \approx 0.24$, close to the zero-order reflection, whose tail in a plain sum over sites oscillates with the chain’s length rather than dying away, and the two lengths tried happened to give it nearly the same size. The satellite beside the zero-order reflection measures it. The sums behind the figure now weight each site with a smooth window, which removes the tail.

The half-tilt adds a consistency check rather than a third independent reading, since it depends on $u\cos\theta$ exactly as the first-order tilt does. It does confirm that the second-order pair carries the same sign information as the first, which a measurement of the second order alone could use when the first order is contaminated. That happens in practice when a first-order satellite coincides with a main reflection of a second phase.

Where this sits among the satellites #

The satellites that need a second integer found that a displacement wave alone gives satellites of every order, with Bessel-function strengths, unequal on the two sides only because $k$ is. An occupation wave alone gives only the orders its profile has harmonics for. The combined wave is a convolution of the two, and the second order is the first place where that is visible as a separate effect: satellites that neither wave would make at that strength alone, since the displacement alone gives a second order of $J_2(z)$, proportional to $z^2$, while the combined wave gives $a z / 4$, proportional to $z$. At small displacements the cross term is by far the larger. Their ratio is $2a/(cz)$, and for $u = 0.005$ at $h = 2$ that is about eighteen on the outer side and forty on the inner, in amplitude.

The same chain has a description in which none of this is a cross term. The extra dimension that makes it periodic lifts a modulated chain into a periodic structure one dimension up, where the occupation and the displacement are two functions on the same internal coordinate and the satellites are ordinary reflections of the lifted crystal. In that picture a size effect is a relation between two atomic-surface functions, the occupation and the displacement drawn against the internal coordinate, and the quarter-period lag is a statement that one is the derivative of the other. The second order is then the second Fourier component of a product of two first components. Two lattices, one crystal uses the same lift for a composite crystal, where two sublattices modulate each other, and the size effect is the simplest case of that mutual modulation: one lattice, whose atoms of two sizes modulate their own spacing.

What the chain leaves out #

One dimension, one wavevector. The chain has a single modulation along its own length. In a crystal the composition wave runs along one direction, and the displacement it drags may have components both along and across that direction when the lattice is anisotropic. The argument here concerns the longitudinal component only, and a transverse displacement contributes to satellites through the angle between the scattering vector and the displacement.

Point atoms and a linear size effect. Each site’s spacing grows linearly with the mean occupancy of its two neighbours. A real size effect has curvature, and a quadratic term would give the occupation’s displacement a second harmonic of its own, adding a direct contribution to the second-order satellites that the cross-term argument does not include. The size of that contribution relative to the cross term is the thing a measurement of both orders would have to allow for.

An irrational wavevector. The formula replaces the sum over sites by an integral over the modulation’s phase, which is exact only when $q$ is irrational. A commensurate modulation, locked to a rational wavevector, samples the phase at finitely many values, and its satellites then depend on where those samples fall. An experiment sees a Farey sequence shows how a real crystal’s wavevector can lock to fractions, and at a locked fraction with a small denominator the second-order satellites of one wave coincide with first-order satellites of another. The readings above would then mix, and separating them would need the phase of the lock.

Intensities, not structure factors. Every reading uses measured intensities, with no phases, which is what an experiment supplies. The formula’s structure factors are complex, and the phases of the second-order pair relative to the first carry more information than the intensities. A phased measurement, possible with resonant scattering, is not considered.

What the calculation has to refuse #

The refused claim is the textbook reading of satellites for an occupation wave, and it holds only when nothing else is moving. The test that carries the argument is the phase sweep. Without it, agreement between the two readings on size-coupled chains would show only that both formulas are right, not that their ratio discriminates.

Still open: the curved size effect #

Two questions follow. The first is the curvature of the size effect. A quadratic term in the relation between occupancy and spacing gives the displacement a second harmonic, and that harmonic makes second-order satellites directly, in proportion to the curvature rather than to $a$. Separating the two contributions needs the second order at several reflections, since the cross term grows with $h$ as $z$ does and the direct term as $J_1$ of a different argument. Whether two reflections suffice to separate them is a calculation on the same chain with one more parameter.

The second question this page once asked, why the tilt at $h = 1$ came out a third of the first order’s instead of a half, has an answer, and it is not a higher order of the expansion. It is the edge of the chain, and the satellite beside the zero-order reflection gives the measurement.