A stack lends each sheet a tension

A flapping sheet keeps its peaks in the plane took a warm sheet one atom thick, gave its atoms freedom to move out of the plane against a bending stiffness, and computed its Bragg peaks exactly. The sheet wandered far. With bending stiffness alone its mean-square height grew as its area, and that wandering destroyed every reflection with a component along the normal faster than any power of the sheet’s size. Reflections lying in the plane were untouched. They only felt the in-plane motion, which a warm plane crystal has no Bragg peaks had already shown costs a power of the size and no more.

That essay ended on a stack of such sheets. Graphite is one, and so is a layered clay, and so is a lipid multilayer. In a stack each sheet is held by its neighbours through the layers between them, and the prediction was that 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.”

This essay computes the stack, and the prediction comes out half right. Inside a thick stack each sheet wanders exactly as a single sheet under a tension $2\sqrt{\kappa B}$ would, where $\kappa$ is its bending stiffness and $B$ the coupling to its neighbours. The stacking reflections that a single sheet could not keep come back, as peaks that fall as powers of the crystal’s size, with the power growing as the square of the order. But a stack of finitely many sheets has one long wave the coupling cannot touch, the whole stack bending together. At the longest wavelengths a finite stack is a free sheet again, a thick one, and its reflections are lost at a width that grows only as the square root of its thickness.

A sheet among neighbours #

The model is the single sheet’s, repeated. Each sheet is the same $N \times N$ harmonic square lattice on a torus, with a height $h_n(R)$ at each site $R$ of sheet $n$, and each sheet’s energy is its bending term, $\kappa$ times the square of the lattice Laplacian of its heights. To that is added a coupling between neighbouring sheets: a spring of stiffness $B$ on the difference $h_{n+1}(R) - h_n(R)$ at every site. The sheets are a distance $d$ apart, taken equal to the lattice spacing, and the heights are measured in lattice spacings.

A mode of the stack has a wavevector $q$ within the sheets and a pattern across them. Its mean-square amplitude is the temperature divided by its stiffness, $t / (\kappa \lambda(q)^2 + B\mu)$, where $\lambda(q)$ is the lattice Laplacian’s eigenvalue and $\mu$ the stacking pattern’s. Two stackings are computed. A periodic stack, closed on itself, has patterns $e^{ikn}$ with $\mu = 2 - 2\cos k$, and stands for the bulk of a thick crystal with no surface. A stack of $M$ sheets with two free faces has standing patterns with $\mu = 2 - 2\cos(\pi j/M)$, including the pattern $j = 0$ in which every sheet moves alike.

The stacking reflections are those with no component in the plane, $(0, 0, l)$, the points of the reciprocal lattice on the line through the origin along the normal, at a momentum transfer $2\pi l/d$. They see nothing but the heights. Their intensity in a harmonic crystal is an exact sum over pairs of sites of the exponential of minus half the momentum transfer squared times the mean-square height difference between the pair. Each sum here is done exactly. The single sheet’s sum is not reused: the stack’s is a different sum, and a stack of one sheet must reproduce the single sheet to the last digit. It does, at every size, which is the first of the checks below.

Throughout, $\kappa = 1$, $B = 0.1$ and the temperature $t = kT/k_1 = 0.02$, the single sheet’s values with a coupling a tenth of the in-plane spring.

The tension the neighbours lend #

The first question is how far one sheet inside a thick stack wanders. The figure gives the mean-square difference in height between two sites of the same sheet, $R$ apart along a row, in a stack of 4,096 layers and 256 sites a side.

The answer is a logarithm. The difference grows by 0.0100 for every factor of $e$ in the distance. A free sheet’s difference grows as the square of the distance, and at four sites apart it is already 0.10, twice the top of the scale. The sheet in the stack is nearly seven times quieter there, and the gap widens as the square of the distance, because beyond it the stack’s sheet grows only logarithmically.

The prediction said the neighbours act as a tension, and the reason is short. A sheet inside a thick stack feels every stacking pattern at once. Adding the mode stiffnesses over all stacking wavevectors gives, for each in-plane wavevector, an effective stiffness $\sqrt{a^2 + 4aB}$ with $a = \kappa\lambda^2$. At long wavelengths $a$ is small and this is $2\sqrt{\kappa B}\,\lambda$, which is exactly the form a tension takes. So the coupling lends each sheet a tension $\gamma = 2\sqrt{\kappa B}$, here 0.632, and a sheet under a tension wanders logarithmically.

The second curve in the figure tests that literally. It is a single sheet under tension $0.632$ and no neighbours at all, and its height difference grows at 0.0100 per factor of $e$, the same rate. The continuum version of the argument predicts $t/(2\pi\sqrt{\kappa B})$, which is 0.0101. The two curves differ by a nearly constant offset, about 0.004, and that offset is the short-wavelength part of the story, where $a$ is not small and the stack’s spectrum is not yet a tension’s.

So the first half of the prediction holds as stated. Within a thick stack, at long wavelengths, a sheet behaves as a tense sheet does. The coupling sets the tension through the geometric mean of the two stiffnesses. Neither the bending stiffness nor the coupling alone sets it, and halving either one lowers the tension by the same factor of $\sqrt{2}$.

Across the layers, at half the rate #

The second question is how far apart two sheets wander at one site. The same exact sum gives it, and it is again a logarithm, growing by 0.0050 for every factor of $e$ in the number of layers between them. That is half the in-plane rate, to three figures.

Half is not a coincidence of these numbers. In the continuum the stack’s energy weighs a bend as $\kappa q^4$ and a stretch between layers as $Bk^2$, so a distance $R$ along the sheets and a distance $n$ across them are worth the same when $R^2$ is comparable with $n\sqrt{\kappa/B}$. Distance across the layers counts as the square of distance along them. A logarithm of $R^2$ is twice a logarithm of $R$, so the rate per factor of $e$ across is half the rate along. The two lines in the figure cross near eight sites and eight layers, and the length $\sqrt{\kappa/B}$, here 3.2 lattice spacings, sets where.

This anisotropy decides the shape of crystal in which the scattering behaves simply. A crystal that grows equally in every direction mixes the two logarithms in proportions that change with its size. One that grows $N$ sites wide and a number of layers proportional to $N^2$ keeps them in fixed proportion. The crystals below are of that shape, $N^2/16$ layers for a width $N$, from 4 layers at a width of 8 to 1,024 layers at 128.

Every order falls as a power, and the power is the order squared #

The picture at the head of this essay is the result the stack was built for. A single sheet loses its stacking reflections faster than any power of its size. A thick stack does not. Each order falls as a power of the crystal’s size, so on logarithmic axes each is a straight line, and the slopes grow steeply with the order: 0.20 for the first, 0.80 for the second, 1.79 for the third and 3.18 for the fourth.

The continuum theory of layered crystals predicts exactly this. A height difference growing as a logarithm turns the exponential in the intensity into a power of the distance, and the power is half the momentum transfer squared times the logarithm’s coefficient. The momentum transfer of the $l$-th order is $2\pi l/d$, so the power is proportional to $l^2$, with exponent

$$\eta_l = \frac{\pi l^2 t}{d^2 \sqrt{\kappa B}}.$$

Here that is 0.199, 0.795, 1.788, 3.179 and 4.967. The intensity per site squared per layer squared of a crystal of this shape falls as $N^{-\eta_l}$, as long as that is slower than the crystal’s volume grows.

The table compares the prediction with the measured exponent of the fall between each pair of successive widths. The first four orders settle on it to three decimals: 0.199 against 0.199, 0.795 against 0.795, 1.789 against 1.788, 3.178 against 3.179. The convergence is from above, as it should be, because the smallest crystals are not yet long enough for the logarithm to have taken over from the short-wavelength offset seen in the first figure.

The fifth order is different, and it is the most informative row. Its predicted exponent, 4.97, is larger than four, and a crystal $N$ wide and $N^2/16$ thick has a volume growing as $N^4$. The intensity per site squared per layer squared cannot fall faster than one over the number of sites times layers. That is the rate at which a sum of uncorrelated contributions falls, the rate of the diffuse part of the scattering that the average scatters sharply and the rest does not separates from the Bragg part. The measured exponents for the fifth order are 4.16, 4.08, 4.04 and 4.02, approaching four from above. The fifth order of this stack is not a peak at all. There is scattering at $(0, 0, 5)$, but no more of it than at a point between reflections.

That is the sense in which a warm layered crystal has a finite number of stacking orders, a limit set by the temperature rather than by the wavelength, which is what limits how many reflections there are in a rigid crystal. Every order loses its peak eventually as the temperature rises or the coupling weakens, and the order squared in the exponent means the higher orders go first and go much faster. Lipid multilayers are the cleanest real case. Their layers are soft and far apart, and the number of stacking orders seen falls as they swell with water. The exponent for their first order is typically of the order of a tenth, close to the value here.

A finite stack is a free sheet again #

The second half of the prediction does not survive, and the reason is a single mode.

A stack of $M$ sheets with free faces has a stacking pattern the bulk calculation never sees: $j = 0$, every sheet displaced alike. Its stiffness has no coupling in it at all, because the springs between layers measure only differences. It is the whole stack bending as one sheet with bending stiffness $M\kappa$, and it carries one $M$-th of a single free sheet’s wandering. A free sheet’s wandering grows as the square of the distance, so however thick the stack, at a large enough width that one mode dominates every other, and the stack’s stacking reflections collapse exactly as a single sheet’s did.

The figure shows that happening for stacks of 1 to 32 sheets. The single sheet is the curve already known: its first stacking reflection at a width of 128 is 0.0019 of its value at zero temperature. Two sheets hold out a little further, and each doubling of the thickness pushes the collapse to a larger width. Every curve ends the same way.

So at the longest wavelengths a finite stack behaves like the free sheet, not like the tense one, which is the reverse of the prediction’s order. The prediction is right in the middle range of wavelengths, short enough that the stack’s own thickness is not felt and long enough for the coupling’s tension to have taken over. It is wrong at both ends. At the very shortest wavelengths a sheet bends against its own stiffness and the coupling barely matters. At the very longest the stack bends as a whole and the coupling does not matter at all.

Width over the square root of the thickness #

How far a stack holds out can be read off the bending mode alone. Its share of the height difference at a distance $R$ is a free sheet’s divided by $M$, roughly $R^2/M$ times a logarithm. The collapse sets in when that reaches a fixed size, so at a width proportional to $\sqrt{M}$ up to the logarithm’s slow drift.

Plotted against the width divided by $\sqrt{M}$, the six curves should gather on the wide side, and they do. One sheet 32 wide, four sheets 64 wide and sixteen sheets 128 wide keep 0.085, 0.071 and 0.061 of the first reflection. Holding the thickness at one sheet and doubling the width from 32 to 64 divides the reflection by eight; doubling the width and quadrupling the sheets together changes it by less than a fifth at each step. Doubling the width costs what four times the sheets gives back, up to the logarithm’s drift.

The narrow side does not gather, and it is not meant to. At the smallest widths the thicker stacks are lower, not higher. A narrow sheet is a small plate, and the springs between plates hold them only as a chain of $M$ masses on springs holds its ends. A chain of that kind has no long-range order of its own along its length: the mean-square displacement between plates grows in proportion to the number of plates between them. A stack of 32 plates eight sites wide keeps 0.48 of the first reflection where a single plate of that width keeps 0.80. That is the one-dimensional version of the same loss, and it fades as the plates widen and each spring becomes $N^2$ springs side by side.

What the harmonic stack leaves out #

Every number here belongs to a harmonic model, and the model leaves out things a real stack has.

The springs between layers are springs on height differences at the same site, with no shear between layers and no in-plane motion coupled to the bending. In a real layered crystal the interlayer forces also resist sliding, which does not change the stacking reflections’ exponents, since those reflections see only heights, but does change the in-plane ones. A free sheet’s bending is also stiffened by its own thermal crumpling at long wavelengths, which this model omits for a single sheet as the earlier essay did. Inside a stack the lent tension removes the long waves that the stiffening acts on, so the omission matters less here than it did there. For the finite stack’s bending mode it matters again.

The figures cannot show a line shape. Each reports the height of a peak, and the continuum theory also predicts the peak’s shape: a power-law cusp, with the intensity falling away from the reflection as a power of the distance from it rather than as a Lorentzian. That cusp is how the effect is recognised in the laboratory, and none of the computations here samples off the reflection. The shape of scattering away from a reflection is the Fourier transform of the correlations, as the order a diffuse pattern measures found for occupational disorder, and here the correlations are the logarithms of the second and third figures.

The convention the numbers depend on is the crystal’s shape. The exponents are exponents of a crystal growing $N$ wide and $N^2/16$ thick. A crystal grown at fixed thickness, or as a cube, falls at different rates, and the fifth order’s verdict depends on the shape too. What does not depend on it is the ratio of the orders’ exponents, which is fixed at $1 : 4 : 9 : 16$ by the $l^2$, and that ratio is what a measurement can test against a single crystal.

Landau, Peierls and Caillé #

That one-dimensional order in a three-dimensional crystal is destroyed logarithmically was argued by Landau and by Peierls in the 1930s, in the same papers that ruled out two-dimensional crystals. A smectic liquid crystal, a stack of liquid layers, is the case they had in mind, and its stacking order is lost exactly as the stack’s is here.

The exponent and its $l^2$ were derived by A. Caillé in 1972 for the smectic phase, and the power-law line shape was first observed by J. Als-Nielsen, R. J. Birgeneau and their colleagues in 1980, in X-ray scattering from a smectic liquid crystal with enough resolution to see the cusp. Since then the same analysis has become the standard way to read the bending and compression stiffnesses of lipid multilayers off their diffraction. The reading is the calculation here run backwards: from the exponent of each order to $t/\sqrt{\kappa B}$.

The stack of solid sheets is the same physics with a lattice in each layer. The in-plane reflections of each sheet keep the in-plane behaviour of a warm plane crystal, and the stacking reflections behave as the smectic’s do. The two kinds of order are lost in different ways in one crystal.

What the calculation has to refuse #

The checks are chosen so that a mistake would show. A stack of one sheet computed by the stack’s sum must reproduce the single sheet’s peak computed by the earlier essay’s sum, and it does to every printed digit at widths 8 to 64. The sum over a periodic stack’s wavevectors is done in closed form, through the ring’s lattice Green’s function, and it must match the direct sum term by term, which it does to one part in $10^{10}$. The first three orders must fall as the powers the continuum predicts. And the claim the essay exists to refuse is refused: a stack held together by its neighbours does not keep sharp stacking reflections. Its first order falls at every doubling of the crystal, with a local exponent of 0.20 each time.

Still open: the reflections off both axes #

The stacking reflections lie on the normal and see only heights. The in-plane reflections lie in the plane and see only in-plane displacements, as the flapping sheet found. A reflection $(h, k, l)$ with both components nonzero sees both. Its exponent should be the sum of an in-plane part from the warm plane crystal and a stacking part from here, if the two kinds of motion stay independent. In the harmonic stack they do. But the sum of two logarithms with different anisotropies is not a logarithm of any single distance. The crystal shape that keeps both in fixed proportion is not the one used here, and the fall of a general reflection with size is a computation that has not been made. A layer is not a wallpaper counts the symmetries a stack’s layers can have, and a layer group with mirrors in the sheet forbids some of these mixed reflections outright. Which survive the warmth, and at what rate, would join the two calculations.