Bragg peak — where it appears
Named by 7 essays across 2 fields — each of them below, with the objects they name alongside it.
Which inflation factors exist
A tiling grown by substitution has an inflation factor, and it is an eigenvalue of an integer matrix — so it is an algebraic integer, and sharp diffraction demands that its conjugates be small. That condition is an inequality between two integers, and it explains why the golden ratio turns up in every quasicrystal anybody has drawn.
Neither a peak nor a bump
A chain whose strongest reflection grows as the length to the power one and a half. A Bragg peak grows as the square and a diffuse bump grows as the length itself, so this is neither — and the essay that ruled out the first possibility could only say so by quoting a theorem.
How much of the hat is a crystal
Put a scatterer on every corner of a patch of hats and the diffraction pattern repeats exactly, because every corner sits on a lattice. Inside each repeat the strongest reflections are those of an ordinary crystal with partly filled sites, and by Parseval's identity they carry sixty-three per cent of what the pattern holds. The aperiodicity the hat is famous for lives in the remaining third, in reflections a hundred times weaker.
A warm plane crystal has no Bragg peaks
Thermal motion is disorder that moves, and in space it costs a crystal a little of each Bragg peak and gathers it into diffuse scattering underneath. In a plane it costs more than a little. Computed exactly from the phonons of a harmonic square lattice, an atom's mean-square wandering grows by the same amount every time the crystal doubles in size, so no Debye–Waller factor survives, and every Bragg peak fades as a power of the crystal's size — a power proportional to the temperature times the square of the reflection's distance from the origin.
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.
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.
The count of reflected hats is never one out
In a hat tiling one tile in about eight is reflected, and that ratio was only ever quoted here. Four clusters and a table of small integers give it exactly: seven plus three root five over two, the fourth power of the golden ratio. The same table says something the ratio does not. Grow a cluster to two trillion hats and its reflected count is still within half a tile of the exact share, where the same share chosen at random would be out by nearly half a million.
Named alongside it
The objects these essays reach for when they reach for this one.
Average structureDiffuse scatteringDisorderAperiodicityDebye waller factorLong-range orderMonotilePhononSubstitutionThermal diffuse scatteringAlgebraic integerCoincidence condition