The second order cannot see a curved size effect alone

The strain wave that tips the satellites built a chain whose occupation varies as a wave and whose spacing follows the occupation, so that a wave of composition drags a wave of displacement a quarter of a period behind it. The first-order satellites then lean, and the lean reads the size factor with the wave’s amplitude cancelled. The second order reads the strain again found that the second-order satellites read it a second time, and the satellite beside the zero-order reflection removed an anomaly in that second reading by windowing the sums. It ended with the question the second order was meant to answer all along.

In every one of those chains the spacing depends on composition along a straight line. Real alloys are not so obliging. Vegard’s law, that the lattice parameter of a solid solution changes linearly with composition, is an approximation, and the departure from it, the curvature of spacing against composition, is a property of the alloy worth measuring. A quadratic term in that relation gives the displacement wave a second harmonic, and a second harmonic of the displacement makes second-order satellites directly. The earlier essay asked whether the second order at two or three reflections could separate that direct contribution from the cross term the straight part makes.

It cannot, at any number of reflections. The two contributions grow with the scattering vector in the same way, so each second-order satellite sees the straight and the curved parts of the size effect in almost the same proportion, and every reflection repeats one equation. From the second-order satellites of three reflections the curvature is known to no better than a hundred times its own size, and its error is perfectly correlated with the straight part’s. The first-order satellites see the straight part with nothing of the curvature at lowest order. Added to the second order, they separate the two, and three reflections then give the curvature to nine per cent.

A chain whose spacing curves #

The chain has a site at every position along a line, occupied with probability $c + a\cos 2\pi t_n$, with $t_n = nq + \varphi$ and $q = 1/\varphi^2 \approx 0.382$ the irrational wavevector the earlier essays used, the one the Fibonacci chain is built on. It is an occupation wave that is its own Fourier series, with the mean occupation $c = 0.5$ and the amplitude $a = 0.4$. The spacing after each site depends on the local occupation, the mean of the two neighbours’ occupations, through its departure $d_n$ from the mean:

$$x_{n+1} - x_n = 1 + \varepsilon_1 d_n + \varepsilon_2 \bigl(d_n^2 - \langle d^2 \rangle\bigr).$$

$\varepsilon_1$ is the size factor of the earlier essays, the slope of spacing against composition. $\varepsilon_2$ is its curvature. The mean of the quadratic term is subtracted so that the average spacing stays one and the main reflections stay at the integers. The values used are $\varepsilon_1 = 0.03$ and $\varepsilon_2 = 0.15$, chosen so that the two terms move atoms by comparable amounts: $d$ ranges over about $\pm 0.145$, so the straight term changes a spacing by up to 0.0043 either way and the curved term, once its mean is removed, by up to 0.0016 either way.

The displacement follows from adding up the spacings. The straight term, summed, gives the first harmonic the earlier essays used, of amplitude $u_1 = \varepsilon_1 a \cos\pi q / (2\sin \pi q)$. The curved term is quadratic in a cosine, and a squared cosine is a constant plus a cosine of twice the angle, so summing it gives a second harmonic:

$$u_2 = \frac{\varepsilon_2\, a^2 \cos^2 \pi q}{4 \sin 2\pi q}.$$

It is locked to the occupation wave as the first harmonic is, a quarter of its own period behind it. The figure fits both harmonics to the positions of a chain of 30,000 sites by least squares. They come out at $2.333 \times 10^{-3}$ and $1.166 \times 10^{-3}$, agreeing with the formulas to four figures. At these coefficients the second harmonic happens to be half the first, which is large enough that its effect on the pattern is not a small correction.

Three terms in a second-order satellite #

To lowest order in the displacement, the structure factor at a second-order satellite, at $k = h \pm 2q$, collects three contributions. The occupation wave multiplied by the first harmonic of the displacement makes one, in proportion to $a\,k\,u_1$. That is the cross term the earlier essays found. The mean occupation multiplied by the second harmonic makes the direct term, in proportion to $c\,k\,u_2$. And the mean occupation multiplied by the square of the first harmonic makes a third, in proportion to $c\,k^2 u_1^2$.

The first two terms are what the measurement is supposed to tell apart, and they grow with the scattering vector $k$ in exactly the same way, linearly. Both are a displacement wave seen through one factor of $k$. So whatever reflection the satellite belongs to, the cross term and the direct term arrive in a fixed proportion, set by $a u_1$ against $c u_2$, and the satellite measures one combination of $\varepsilon_1$ and $\varepsilon_2$. Only the third term, with its extra factor of $k$, depends on the reflection differently, and at a displacement of a few thousandths of a spacing it is the smallest of the three.

The figure measures that proportion directly, satellite by satellite. For each of the six second-order satellites of the first three reflections it divides how much the intensity changes when $\varepsilon_2$ is nudged by how much it changes when $\varepsilon_1$ is nudged. The six ratios lie between 0.241 and 0.269. Every second-order satellite weighs the curvature against the slope in nearly the same ratio. Six satellites are six measurements of nearly one number. The ratio drifts upward on the outer satellites, from 0.259 to 0.269, and downward on the inner, from 0.249 to 0.241, and that drift is the third term. It is all the second order has to separate the two coefficients with.

The size of the drift can be checked against the size of the term that makes it. The third term’s ratio to the cross term is about $(c/a)\,2\pi k u_1$, the phase a site’s displacement adds at the scattering vector. With $u_1 = 0.0023$ that is 0.03 at the first reflection’s outer satellite, $k \approx 1.76$, and 0.07 at the third’s, $k \approx 3.76$. A correction of three to seven per cent to one term, entering the intensity as a cross product, moves the mixing ratio by a few per cent, which is what the six satellites show. The same estimate says what would help: a larger displacement, or a reflection far enough out that $2\pi k u_1$ approaches one. At ten times this displacement the third term would be comparable to the others and the second order could separate the coefficients by itself, but the expansion behind every formula here would then be failing too.

How well each set of satellites does #

The intensities are computed from sums over the 30,000 sites with a Hann window, the window the previous essay found necessary to keep the zero-order reflection’s tail out of the weak satellites. Each intensity is given a two per cent error. The unknowns are the slope, the curvature and the wave’s amplitude $a$, since an experiment knows none of them in advance. Counted as unknowns against observations, even one second-order pair has the two intensities needed for two unknowns once the amplitude is fixed. The count is satisfied and the measurement still fails, because the observations are not independent. The Cramér–Rao bound, the smallest error any unbiased estimate can have, follows from the derivatives of the intensities with respect to the three unknowns. It says what each set of satellites could deliver at best.

The picture at the head of this essay is the result, on a logarithmic scale. From the second-order pair at the first reflection alone the curvature is known to within 14,060 times itself, which is to say not at all, and its error is correlated with the slope’s at 1.0000. Adding the second reflection brings that down to 304 times itself, and the third to 113. Each reflection helps, because each brings a slightly different contribution from the third term, but the correlation stays at 1.0000 to four figures, and the second order alone never comes within two orders of magnitude of a measurement.

The first-order satellites change everything. With the first- and second-order pairs of the first reflection together, the curvature is known to 32 per cent of itself, with a correlation of $-0.996$. With two reflections it is 15 per cent, and with three it is 9 per cent, correlation $-0.983$. The slope, meanwhile, is pinned to a tenth of itself or better from the first reflection on.

Why the first order breaks the tie #

The first-order satellites see the slope and not the curvature, at lowest order. A first-order satellite at $k = h \pm q$ collects the mean occupation times the first harmonic of the displacement and the occupation wave with no displacement at all. The second harmonic of the displacement can reach a first-order satellite only by combining with the occupation wave, a term of the same size as the second-order cross term but now competing against the much larger first-order terms. So the first order fixes $u_1$, and with it the slope. Given the slope, the one combination the second order measures becomes a measurement of the curvature.

That is why the two coefficients come out correlated at $-0.996$ rather than at zero once the first order is added. The slope is fixed well and the curvature is read off the second order relative to it, so any error in the slope passes into the curvature with the opposite sign, scaled by the second order’s mixing ratio.

The ellipses show it in the plane of the two coefficients. The second order alone gives a line. Every point on it fits the second-order intensities equally well, because moving along it trades slope for curvature at the one fixed rate every satellite sees. The first order cuts across that line, and the result is a long, thin ellipse leaning the other way, along the direction in which an error in the slope is absorbed by the curvature. More reflections shorten it. They do not change its direction, because what limits it is still the first order’s precision on the slope.

The smallest curvature the first order lets through #

With both orders in the measurement the curvature is determined, and the next question is how small a curvature can be told from none. The bound answers it by repeating the calculation at other curvatures and other intensity errors, with both orders at the first three reflections.

The error in the curvature hardly depends on the curvature itself. At two per cent intensity error it is $\pm 0.011$ when the true curvature is 0.05, $\pm 0.013$ at 0.15 and $\pm 0.015$ at 0.3. That is what a measurement linear in the unknown looks like, and it means the error is better quoted as an absolute number than as a fraction. At one per cent intensity error every one of those halves, to between $\pm 0.006$ and $\pm 0.007$. So at two per cent a curvature of about 0.04 is three standard errors from zero, and at one per cent about 0.02.

In the chain’s own terms, a curvature of 0.04 bends the spacing away from the straight line by $0.04\,d^2$, about $8 \times 10^{-4}$ at the extremes of the occupation wave, against a straight-line change of $4 \times 10^{-3}$ there. A departure from Vegard’s law of a fifth of the size effect is detectable at two per cent and one of a tenth at one per cent. Throughout, the correlation between the slope’s error and the curvature’s stays between $-0.97$ and $-0.99$, so the statement is always conditional on the first order: a curvature claimed from these satellites is a curvature relative to the slope they measured.

What this means for measuring a departure from Vegard’s law #

The earlier essays found the second-order satellites useful because they gave a second reading of the slope, independent of the first-order tilt. That independence is exactly what is lost once a curvature is allowed. The second order then has two coefficients to explain and one combination to explain them with, and the second reading of the slope becomes a reading of the curvature instead. An experiment that wants both cannot use the second order to check the first. It has to use the first order to make the second order mean anything.

The practical demand is precision on the first order, not more reflections at the second. That runs against the instinct that measuring every reflection several times over is how precision is bought: here repetition at the second order buys almost nothing, and the photons are better spent on the first-order pair. Going from one reflection to three, with both orders, improves the curvature by less than a factor of four. A better measurement of the first-order tilt improves it in proportion, because the curvature is read relative to the slope. And the windowing is not optional. At the first reflection the inner second-order satellite is the weakest in the calculation, and as sharp as the sphere is wide made the general point that a truncated sum adds ripples no improvement in the data removes. A two per cent error assumed for that satellite is only meaningful if the sum it comes from has had the zero-order tail removed first.

What the calculation leaves out #

One kind of curvature. The spacing here is a quadratic function of the local occupation, which is the simplest curved size effect. A real alloy’s spacing also depends on the arrangement of neighbours beyond the nearest, which makes the displacement depend on the occupation wave non-locally. That would change the phases of the harmonics, and with them the mixing ratio. It would not change the argument, which depends only on both contributions growing linearly with $k$.

A single chain. Everything is one dimension, one wavevector and one occupation profile. The satellites that need a second integer generalise to more dimensions, and in a crystal modulated in two directions the second-order satellites include mixed ones, at $h + q_1 + q_2$, whose cross terms involve the two displacement waves differently. Whether mixed satellites can separate a curvature that the pure second order cannot has not been computed.

Bounds, not fits. No refinement is run. The Cramér–Rao bound says how well an unbiased estimator could do, and at a correlation of 1.0000 it says that no estimator, biased or not, can find the curvature from the second order alone without some outside information about the slope.

Who worked out the pieces #

Vegard stated his law of linear interpolation between the lattice parameters of the end members in 1921, and departures from it have been tabulated for alloys ever since. The diffraction theory of displacement waves and the satellites they make goes back to Dehlinger in the 1920s, and its systematic form, the expansion in Bessel functions used here and in the earlier essays, is the standard treatment of incommensurately modulated structures from the 1970s on. The Cramér–Rao bound is from Rao in 1945 and Cramér in 1946.

What the calculation must refuse #

The first test is what makes the rest worth trusting. The displacement harmonics are fitted to the chain the intensities are computed from, and they agree with the formulas to four figures, so the chain carries the second harmonic the argument says it carries, and at the size it says. The two refusals are the two hopes the question started with. More second-order reflections do not separate the curvature, and the reflections do not weight the two coefficients differently enough to matter, with every mixing ratio within twelve per cent of every other.

Still open: mixed satellites in two directions #

The argument turns on a degeneracy of the second-order satellites of a single wave, and a structure modulated by two waves breaks it in principle. The mixed satellites at $h + q_1 + q_2$ come from a product of the two displacement waves, and a curvature in the size effect makes a direct term there too, from the cross product of the two occupation waves inside the squared composition. Whether that direct term grows with the scattering vector differently from the mixed cross terms, so that a two-dimensional modulation can measure the curvature from its second-order satellites alone, is the next calculation, and it needs a two-dimensional chain with two irrational wavevectors.