Crystallography & Symmetry
patterns derived from their groups
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The theme: From the diffraction back — page 5

Nobody has seen a space group. They are inferred from where a crystal scatters and, just as informatively, from where it does not.
Three difference maps of one substructure, and the map they approximate. Patterson syntheses over 1,681 reflections of a structure with 120 light atoms and two anomalous scatterers, from squared Bijvoet differences at one wavelength, squared dispersive differences between two wavelengths, and the two normalised and summed; the fourth panel is the Patterson of the anomalous atoms' own intensities. Circles mark the two vectors between the anomalous atoms. Each map's correlation with the exact one is printed beneath it: Bijvoet 0.775, dispersive 0.838, combined 0.999. How it is known

A second wavelength completes the map

The Bijvoet difference carries the anomalous substructure's intensity times the sine squared of a phase; the dispersive difference between two wavelengths carries it times the cosine squared. Normalised and added, the two make the substructure's own Patterson, correlation 0.999. It needs no fewer reflections than either half, because the threshold was never the average. What the dispersive half brings is a way of failing the Bijvoet half cannot have: two datasets, and a scale between them.

 Along the rod above a reflection, a free sheet loses its peak. The Bragg intensity per site squared of a warm square-lattice sheet at the in-plane reflection (1, 0) and at heights 0.1, 0.2 and 0.3 along the rod above it, against the sheet's size, on logarithmic axes. In the plane the peak falls as a power of the size whatever the sheet does out of it. Above the plane a free sheet's peak collapses faster and faster, to 2.3e-2 at size 128 and height 0.3; under tension every height falls as a power, to 0.38. How it is known

A flapping sheet keeps its peaks in the plane

A warm sheet one atom thick wanders out of its plane far more than within it: a free sheet with only bending stiffness has a mean-square height that grows with its area. That wandering cannot touch a reflection whose scattering vector lies in the plane, which fades exactly as it would if the sheet were held flat. Every reflection with a component along the normal loses its peak faster than any power of the size. Tension puts it back on a power law.

 Almost no homometric pair on the torus is a reversed factor. For ten plane groups at several numbers of atoms on a 6 × 6 torus, the structures, the pairs of structures sharing one Patterson, how many of those pairs reversing one factor of a sumset explains, and how many are a patchwork of Galois classes. Factor reversal explains 2 of 3522; the patchwork explains all 3522. How it is known

A homometric twin is the structure moved class by class

On a ring, every homometric pair a small search found came from one construction: factor the structure as a sumset and reverse one factor. Imposing a plane group on a torus made homometry more common, and the obvious guess was that symmetry makes structures easier to factor. Tested pair by pair, factor reversal explains two of 3,522 homometric pairs on a 6 × 6 torus, and none at all at six atoms. What explains all 3,522 is a different mechanism: the partner is the structure's own Fourier transform moved independently on each Galois class of characters, by a shift, a sign or an inversion.

 Where the third wavelength should go. The median Cramér–Rao bound on the phase difference for two hundred reflections, measured at the inflection and the high remote and at a third wavelength whose energy runs across the edge, twice: with f′ known everywhere, and with f′ at the edge energies refined from the data. Known, the best third sits 1 eV above the edge at 6.26°, and a repeat of either wavelength is nearly as good. Refined, both repeats determine nothing, and the best third moves to +2 eV at 6.30°. How it is known

The third wavelength is there to measure the edge

A three-wavelength MAD experiment is usually justified as more anomalous signal: a third look at the edge, placed where the first two left the most out. Counted with the scattering factors known, a third wavelength buys nothing a longer look at the best pair would not buy better. Its real job appears only when f′ at the edge has to be found from the data, which it always does. Two wavelengths then carry one equation too few, and any third energy that is not a repeat supplies it, for two hundredths of a degree.

 Success of the flipping solver against how far off the subtracted lead is. The share of forty starts on the cell of nineteen carbons and one lead that the flipping solver solves, with the lead's density subtracted at a position displaced from the true one by up to sixteen grid steps, half the cell. The rate does not fall with the offset: the points scatter without a trend about the pooled rate of 40%, the band marking two binomial standard deviations, and the original solver, with nothing subtracted, solves none. Refining the position from the map does no better than holding it. How it is known

The lead is wherever it is taken out

Charge flipping solves a cell of nineteen carbons and one lead once the lead's own density is taken out before each flip and put back after. The obvious worry was precision: a lead subtracted a fraction of a grid step from where it really is leaves a residual larger than any carbon. The residual is there, larger than a carbon at a tenth of a step. It does not matter. Subtracted two steps away, or half the cell away, the lead still lets the solver succeed at the same rate, because the solver builds the whole structure round wherever it is put.

 A scale error between two wavelengths has a sign. The contrast of the substructure's vectors in a dispersive difference Patterson, the weaker of its two peaks over the strongest other peak, as the second dataset is scaled wrongly, with no metal atom and with two metal atoms that do not scatter anomalously. An error that weakens the second dataset loses the map at 3.7% with no metal; one that strengthens it at 50%. The dashed line is where the substructure's peak is predicted to reach zero, a quarter of the fractional change in f′ between the wavelengths. The metals shorten the strong side and hardly move the weak one. How it is known

A scale error between two wavelengths has a sign

A dispersive difference compares two datasets, so a wrong scale factor between them adds the whole structure to it. On a structure whose own Patterson is a haze, the substructure's vectors survived a scale error of fifty per cent, and heavy atoms that do not scatter anomalously were expected to take over much sooner. They do, but only in one direction. Scale the second dataset the other way and the substructure is erased at four per cent whatever else is in the crystal, because the error's linear term cancels the signal. Scaling the datasets onto each other is itself an error of that sign.

 Every stacking reflection of a warm stack of sheets falls as a power of its size. The Bragg peak per site squared per layer squared of the stacking reflections (0, 0, l), l = 1 to 5, of a stack of flapping square-lattice sheets with bending stiffness 1 and coupling 0.1 between neighbouring layers, at kT/k₁ = 0.02, against the stack's width, with the number of layers growing as the width squared. Each order falls as a fixed power of the size, and the powers grow as the square of the order: 0.20, 0.80, 1.79 and 3.18 for the first four. The fifth falls as the fourth power, as fast as the crystal's volume grows, which is what a reflection with no peak at all does. How it is known

A stack lends each sheet a tension

A single warm sheet wanders so far out of its plane that every reflection with a component along its normal is lost. Stack the sheets and couple each to its neighbours, and inside the stack each sheet wanders only as a sheet under tension does, logarithmically. The stacking reflections survive as peaks that fall as powers of the crystal's size, and the power grows as the square of the order, so the fifth order of the stack computed here has no peak at all. A stack of finitely many sheets keeps even the first only out to a width growing as the square root of its thickness.

 No triple of wavelengths is best for both the phase and the edge. Every triple of energies on a grid across a model selenium edge, placed by two Cramér–Rao bounds with the edge's position and white line refined from the data: the median phase error across, and the error in the edge's position up, on a logarithmic scale. The front of triples that no other beats on both runs from +1, +2, +150 eV (6.15°, ±0.066 eV) to -120, -1, +3 eV (7.54°, ±0.022 eV). The classic peak, inflection and high remote sits on it, at 6.42° and ±0.026 eV. How it is known

The wavelengths that phase best do not find the edge

A MAD experiment refines the anomalous atom's scattering factors so that it can solve for the phases, and those same factors report where the absorption edge sits and how tall its white line stands, which is chemistry. Scored for the edge instead of the phase, the best triple of wavelengths moves: two straddling the step and one far below it, where the phase-best triple put two on the step and one far above. No triple is best at both, and the textbook peak, inflection and remote turns out to be the compromise between them.

 Four flipping rules on four cells. The best success rate of the flipping solver, pooled over five arrangements and eight starts, on four cells: twenty carbons, a quarter sulphur, one sulphur among twenty and one lead among twenty. Four rules for what is flipped: below a number of standard deviations of the map, below a height set by the lightest atom's weight, the same height above the map's mean, and the original rule with the known heavy atom's density removed first. On the lead cell only the last solves anything, 55% against nothing; on both sulphur cells it raises the rate to near certainty; on carbon alone it is the original rule. How it is known

The lead taken out solves what no threshold could

Charge flipping never solved a cell of nineteen carbons and one lead, and two repairs were proposed: a threshold set by the composition rather than by the map's spread, and the lead's own density removed before each flip. The threshold does nothing for the lead cell at any setting, measured from zero or from the map's mean. Removing the lead takes it from nought to fifty-five per cent, and lifts both cells with sulphur in them to near certainty. The trouble was never where the threshold sat.

 Three-species arrangements whose sums are parallel at every spot, with and without a centre. Every arrangement of three species, all present, on a cyclic cell of four to nineteen sites. The third column counts those whose two phase sums are parallel at every superlattice spot, the fourth those among them with a centre of inversion, the fifth those without one, and the last how many kinds the exceptions make up to translation, inversion, relabelling the spots and renaming the species. Primes are shaded. At 5, 7 and 11 sites every parallel arrangement has a centre; at 13 sites 936 do not, and they are all one arrangement; at 17 there are 17 kinds and at 19 there are 48. Operations

Thirteen sites obey Friedel's law without a centre

At five and seven sites the earlier census found that a three-species structure has a contrast that darkens a spot exactly when it has a centre, and asked whether that is a theorem. It is not. The two phase sums are parallel exactly when the A–B pairs are as common at every separation forwards as backwards, and exactly when no anomalous scatterer can produce a Bijvoet difference anywhere in the pattern. At 5, 7 and 11 sites only a centre achieves that. At 13 sites one arrangement does it without one.

 Success of the flipping solver against the error in the vector between two subtracted leads. The share of 80 starts on a cell of eighteen carbons and two leads that the flipping solver solves, with both leads' density subtracted, the first at its true place and the second displaced from its true place by an error in a random direction. The rate holds to a quarter of a grid step (21, 19, 18, 11, 0, 0, 0 successes at errors of 0, 0.1, 0.25, 0.5, 1, 2, 4 steps), falls by half at half a step, and is zero from one step on. With one lead the place did not matter at all; with two, the vector between them does. How it is known

Two leads need their vector to half a step

With one heavy atom in a cell with no symmetry, charge flipping did not care where the atom was subtracted: it rebuilt the structure round whatever place it was given. With two heavy atoms that freedom runs out, as predicted, because no translation can put both under wrong references at once. Subtract one of them a whole grid step from where it belongs and the solver never succeeds. But the tolerance is five times looser than the estimate it was predicted from: up to a quarter of a step the rate does not move, at half a step it halves, and the residual the solver absorbs is three carbons' worth, not one.

 Fitting the outer shells makes it worse, and correcting the fit recovers it. The substructure's weaker peak as a share of its height with no scaling, on data with no error, with the scale error each scaling builds in, for five falloffs and three scalings. Fitting everything loses 42% with no falloff and less as the falloff grows. Fitting only the outer shells loses more, and at the steepest falloffs builds in a large error of the opposite sign. The fit corrected by the ratio the composition expects recovers the whole peak at every falloff, within a few per cent. How it is known

Fit the scale, then put the lean back

Scaling two wavelengths to each other builds in an error of the cancelling sign, because their means differ for a physical reason. The proposal was to fit against the part the edge does not touch, at high resolution. It does the opposite: f₀ falls with angle and f′ does not, so the outer shells lean furthest and fitting them costs more. Taking the scale from the composition alone reproduces the lean as well. What works is fitting as usual and then putting back the difference the composition expects, and that recovers the whole peak.

 Completeness against sweep, for four spindles. Completeness against rotation range for one tetragonal crystal, class 4/mmm, on four different spindles. The four-fold spindle rises fastest at first, then stops at 88.9 per cent at forty degrees and never improves. The two-fold spindles reach completion near 87 degrees. The general spindle is highest at forty-five degrees and reaches completion at the same 87. How it is known

The same measurements, and a hole that stays

How much of the sphere must be swept for a given redundancy? The question has two answers, because it has two parts. The mean number of observations depends only on the sweep, to within two per cent whichever axis the crystal turns on. Completeness depends entirely on the axis, and on one kind of axis no sweep ever finishes: when the Laue group keeps the spindle's line, every copy of a blind reflection is blind too, and fifteen of a tetragonal crystal's 135 reflections are never measured.

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