A flapping sheet keeps its peaks in the plane

A warm plane crystal has no Bragg peaks computed a square lattice of atoms joined by springs, held at a temperature and allowed to move within its plane. It found the mean-square displacement of each atom growing as the logarithm of the crystal’s size. That is enough to make every Bragg peak, divided by the square of the number of atoms, fall as a power of the size, with an exponent proportional to the temperature and to the square of the reflection’s scattering vector. A two-dimensional crystal has quasi-long-range order and no more. Its closing section noted that a real layer, a sheet of graphene or a membrane of lipids, can also move out of its plane. There the modes are softer still, and the mean-square height diverges much faster than the logarithm. It asked how much of each reflection such a sheet keeps.

The answer divides the reflections cleanly in two, and the division is geometric rather than thermal. A reflection whose scattering vector lies in the sheet’s plane cannot see the flapping at all, and keeps exactly the power law of the flat crystal. A reflection with any component along the normal loses its peak faster than any power, unless something stiffens the sheet against flapping. Both halves are exact statements about the harmonic sheet, computed here on tori up to 128 atoms on a side.

A sheet with a height #

The model is the flat crystal’s with one more coordinate. Each atom of the square lattice has a height $h$ above the plane as well as its displacement within it. In the harmonic approximation the two kinds of motion do not mix. A small height changes the lengths of the springs only to second order, so the in-plane springs exert no force on it. The height’s energy comes from bending, the curvature of the sheet, and optionally from a tension that resists any tilt. On the lattice the curvature is the discrete Laplacian, whose eigenvalue for a wave of wavevector $q$ is $\lambda(q) = 2(2 - \cos q_x - \cos q_y)$. Each flexural mode then carries

$$\langle |h_q|^2 \rangle = \frac{kT}{\kappa\,\lambda(q)^2 + \gamma\,\lambda(q)} ,$$

with $\kappa$ the bending stiffness and $\gamma$ the tension. For small $q$, $\lambda \approx q^2$. A free sheet, $\gamma = 0$, has mode energies rising as $q^4$ rather than the $q^2$ of the in-plane modes, and its long waves are very soft. The temperature is the flat crystal’s, $kT/k_1 = 0.02$, and $\kappa = k_1$.

A reflection at scattering vector $Q = (G, Q_z)$, with $G$ in the plane and $Q_z$ along the normal, has the exact harmonic intensity

$$\frac{I(Q)}{N^2} = \frac{1}{N^2}\sum_R \exp\!\Big(-\tfrac12 W_G(R) - \tfrac12 Q_z^2\, H(R)\Big),$$

where $W_G$ is the in-plane correlation the flat crystal already had and $H(R)$ is the mean-square difference in height between two atoms $R$ apart. Every term is a sum over the $N^2$ modes of the torus, done exactly, as in the flat case.

How far a sheet wanders #

The free sheet’s mean-square height grows as the square of its linear size. The logarithmic slope measured from $N = 32$ to $N = 128$ is 1.99, and the height goes from 0.0059 square lattice spacings at $N = 8$ to 1.27 at $N = 128$. At that size the typical height is more than a lattice spacing, in a sheet whose atoms move within the plane by a few hundredths of one. The reason is the $q^4$: summed over a two-dimensional zone, $1/q^4$ diverges as the square of the largest wavelength, where the flat crystal’s $1/q^2$ diverged only as its logarithm.

Under a tension of 0.5 the long waves are stiffened back to $q^2$, and the mean-square height grows as the logarithm, by 0.0063 per doubling of the size once the sheet is larger than a few lattice spacings. The tension sets a length, $\sqrt{\kappa/\gamma}$, about one and a half lattice spacings here. Waves longer than that are held by the tension and shorter ones by bending. At a tension of 0.05 the same crossover sits at four or five spacings, and the logarithm has not yet taken over by $N = 64$. A tense sheet wanders out of its plane as slowly as the flat crystal wanders within it, and for the same reason.

What the plane cannot see #

The first consequence is exact and needs no computation beyond checking it. A reflection with $Q_z = 0$ has no term in $H$ at all. The height enters the intensity only through $Q_z^2 H(R)$, because a displacement along the normal changes the phase of the scattered wave only by $Q_z h$. So every in-plane reflection of the flapping sheet is identical to the flat crystal’s, with or without tension, at every size. On the tori computed here the two agree to the last digit, a difference of exactly nought, because the flexural term is multiplied by zero before it is added.

That is the refused claim below, and it runs against the natural picture. A sheet whose atoms rise and fall by more than a lattice spacing looks too disordered to diffract sharply, and it would be easy to suppose that its in-plane spots blur as it flaps. They do not. An in-plane scattering vector measures positions within the plane, which the flapping leaves alone to first order. The in-plane spots fade only as the flat crystal’s do, by the power law set by the in-plane wandering, falling as the size to the power $-0.17$ for the (1, 0) reflection at this temperature.

Along the rod above a reflection #

A sheet one atom thick has no periodicity along its normal, so its reciprocal lattice is not a lattice of points but of rods, one through each in-plane reciprocal vector and running parallel to the normal. The reciprocal lattice of a periodic crystal has points because the crystal repeats in every direction. A sheet repeats in two, and in the third its transform is a line. A flat sheet scatters equally all along each rod. Tilting the sheet in a beam moves the scattering vector up the rod, which is how the height along a rod is measured.

The picture at the head of this essay shows the rods. At height zero both panels show the same line, the flat crystal’s power law. Above the plane they part. Under tension every height falls as a power too, a little faster than in the plane. For the free sheet the higher points on the rod fall faster with each doubling of the size, and at height 0.3 the peak per site squared of a sheet 128 atoms across is $2.3 \times 10^{-2}$, against 0.38 under tension.

A fall no power describes #

A power law has a constant exponent, and the measured local exponents show which falls are power laws. Under tension the exponent at height 0.2 on the rod is −0.177, −0.177, −0.178 and −0.178 across four doublings, a power law. For the free sheet the same exponents are −0.20, −0.30, −0.67 and −1.88. Each doubling of the size costs more than the last, and at height 0.3 the last doubling costs a factor of seven. The fall is Gaussian in the size rather than a power of it. The intensity is dominated by $\exp(-\tfrac12 Q_z^2 \langle h^2 \rangle)$, and $\langle h^2\rangle$ grows as $N^2$. A large free sheet has, in the strict sense, no peak anywhere off its plane.

Under tension the exponent is the in-plane one raised a little, from 0.17 in the plane to 0.178 at height 0.2 and 0.19 at height 0.3. The rise grows roughly as $Q_z^2$, because the tense sheet’s height wanders logarithmically exactly as its in-plane displacement does, and adds its own logarithm, weighted by $Q_z^2$, to the in-plane one. A tense sheet is a flat crystal with a third direction of logarithmic wandering, and every reflection on every rod keeps a power law.

What a tilt series sees #

Read along a rod at a fixed size, the same numbers describe an experiment. A flat sheet’s rod would be uniform, equally bright at every tilt. A sheet under tension dims slowly as the tilt grows. A free sheet dims as a Gaussian in the height on the rod, with a width that shrinks as $1/\sqrt{\langle h^2 \rangle}$, inversely with the sheet’s size. The rod contracts towards its in-plane point, and spots that are sharp at normal incidence broaden and fade as the sheet is tilted. That is the signature observed in suspended graphene by transmission electron diffraction, where spots broaden with tilt, and it was read there as evidence that a free-standing sheet is rippled. The calculation here says what a harmonic sheet would show. A real sheet also has anharmonic coupling between bending and stretching, which stiffens the long waves as tension does, and so real graphene sits between the two panels.

The loss depends on one number #

The free sheet’s loss off the plane has a simple structure that the rods’ numbers show once they are divided by the in-plane peak. At height 0.1 on the rod, a free sheet keeps 0.998 of its in-plane intensity at 8 atoms across, 0.992 at 16, 0.969 at 32, 0.885 at 64 and 0.623 at 128. At height 0.2 it keeps 0.968 at 16, 0.884 at 32 and 0.622 at 64. The two rows are the same row shifted by one doubling: halving the height on the rod doubles the size at which the same fraction is lost. Checked at three pairs of height and size, the fractions agree to within a hundredth. They are 0.885 and 0.884, 0.886 and 0.885, and 0.623 and 0.622.

The reason is in the formula. The loss is set by $Q_z^2 \langle h^2 angle$, and $\langle h^2 angle$ grows as $N^2$, so the loss depends on $Q_z N$ alone. A sheet twice as large looks, along its rods, like a sheet of the original size seen at twice the tilt. That is a practical rule for reading a tilt series. The tilt at which a free sheet’s spots fade tells its size, or rather the size over which it is free to flap, which in a real membrane is set by whatever holds its edges. Under tension there is no such rule. The fraction kept stays above 0.97 everywhere computed, and the loss depends on the height and the size separately and weakly.

At the sizes real sheets have #

The sizes here are small, a hundred and twenty-eight lattice spacings at most, and real suspended sheets are thousands or millions. The scaling makes the extrapolation simple for the harmonic free sheet. A sheet ten thousand spacings across, at height 0.1 on the rod, is equivalent to a sheet of 128 at height 7.8, far beyond anything computed here and far into the collapse: its peak there is gone. It keeps what the 128-atom sheet keeps at height 0.1 only at height 0.00128, a tilt eighty times smaller. So an enormous free sheet keeps its spots only in a narrow band of tilts about the plane, a band whose width is inversely proportional to its size. For the flat crystal’s in-plane peaks, the earlier essay found that the power law with exponent 0.17 leaves a peak visible at any practical size. The free sheet’s rods are the opposite case: at practical sizes they are visible only very close to the plane.

The same sum, in two directions #

The contrast between the two halves of the result is one sum evaluated twice. The average scatters sharply and the rest does not split any disordered crystal’s scattering into the average structure’s peaks and the variance’s diffuse background, and thermal motion is the case where the average is over time. The height of every peak is set by how far atoms wander from their average positions along the scattering vector. In the plane that distance grows as $\ln N$ for a flat sheet or a flapping one alike. Along the normal it grows as $\ln N$ for a tense sheet and as $N^2$ for a free one. What the size does to a peak depends on the direction of its scattering vector, and a sheet is anisotropic in exactly the way the reflections are.

The intensity lost from the peaks is not lost. As in the flat crystal, the total scattering over a zone is fixed at one unit per atom, and what leaves a peak moves into the diffuse scattering around it. For the free sheet the diffuse scattering along each rod gathers into a broad ridge rather than a sharp spot, spread along the normal by the flapping. It is the thermal counterpart of the short-range-order ridges that the order a diffuse pattern measures reads for occupational disorder, with the correlation measured along the normal instead of across the lattice.

A powder of sheets #

Much of what is known about layered materials comes from powders, not single sheets, and a powder of flat flakes has a characteristic line shape. Averaging a rod over every orientation smears it into a line with a sharp edge on its low-angle side and a long tail towards high angle, the sawtooth that Warren described in 1941 for the reflections of turbostratic carbon. What a powder pattern loses counts the information a powder discards, and the rods are an extreme case, a whole line of reciprocal space folded into one profile. The tail comes from the rod’s length, the intensity far up the rod. A free sheet’s rod contracts as the sheet grows, so the flapping should shorten the tail and sharpen the line, a change in shape a powder measurement could read. That consequence is argued here, not computed, since it needs the rod’s full profile averaged over orientations.

What the harmonic sheet leaves out #

Anharmonic coupling. A real membrane that bends must also stretch slightly, and the coupling between the two stiffens long bending waves. The effective bending stiffness grows with the wavelength. That is the mechanism by which a free-standing crystalline membrane stays flat on average at large sizes, and it softens the $N^2$ growth of the height to a slower power. The harmonic free sheet here is the limiting case with no such coupling, and its runaway height is an upper bound. Curvature does the same job more directly. A sheet rolled into a tube, as in the tube has a screw no lattice allows, cannot bend without stretching, and its flapping is stiffened by its own shape.

Temperature is small and fixed. At $kT/k_1 = 0.02$ every in-plane peak keeps most of its intensity at the sizes computed. At higher temperatures the in-plane exponents grow in proportion, and the free sheet’s heights grow in proportion too. The division between the two kinds of reflection does not depend on the temperature.

One reflection. The rods are computed above (1, 0). Every in-plane reflection has its own rod, and the in-plane part of its fading grows as $|G|^2$, as in the flat crystal. The out-of-plane part depends only on the height on the rod and is the same for every rod.

What the calculation has to refuse #

The test the rest depends on is the growth of the height, since both the runaway and the power law come from it. The mean-square height has to grow as the area for the free sheet and as the logarithm under tension, and the computed sums have to show it: slope 1.99, and 0.0063 per doubling. The exactness of the in-plane result is a check on the bookkeeping. If the flexural modes leaked into an in-plane peak by even a rounding error, the two sheets would differ at $Q_z = 0$.

Still open: the layer that has a thickness #

A single sheet has rods. A stack of sheets, graphite or a layered clay or a lipid multilayer, has rods broken into points by the stacking, and the flapping of each sheet is coupled to its neighbours’ through the layers between them. The coupling acts like a tension on long waves, so a stack should behave like the tense sheet at long wavelengths and like the free one at short. A layer is not a wallpaper counts the symmetries such a stack’s layers can have. How the stack’s reflections fade with the number of layers and with their lateral size is the same exact sum with the layers’ coupling added to the flexural energy. The crossover from free to tense behaviour would then be set by the stiffness between layers.