A looser fibre, more piezoelectric

A fibre of achiral crystals can rotate light worked with a perfect fibre, every grain with one crystal direction exactly along the sample axis and turned at random about it. It then stated, without computing, what happens when the grains are not perfectly aligned. The degree-two part of the gyration tensor is multiplied by the average of the second Legendre polynomial of each grain’s tilt, which is one for perfect alignment and nought for a uniform spread. So the rotation of light falls smoothly from the grain’s value to the powder’s as the fibre loosens.

What a texture permits supplied the other half of the idea. Every property of a crystal is a pile of spherical harmonics: a vector is degree one, a symmetric second-rank tensor is degrees nought and two, the piezoelectric moduli are two vectors and an octupole, and the elastic constants reach degree four. Put the two together and the statement extends to everything. Under a tilt spread, the degree-$\ell$ part of any property is multiplied by $\langle P_\ell(\cos\beta)\rangle$, one number per degree. A property made of a single degree shrinks. A property made of several changes shape, because its parts shrink at different rates.

This essay computes that, checks it against a direct average over orientations, and follows it into the one property where it matters most, the piezoelectric response of a polar fibre. There it produces two effects that no single grain shows and that nothing about perfect alignment suggests. A fibre can become more piezoelectric as its alignment gets worse. And a coefficient can change sign on the way to nothing.

One number per degree #

A fibre texture here means grains whose chosen crystal direction is tilted from the sample axis by an angle $\beta$ drawn from some distribution, with the azimuth of the tilt random and each grain’s rotation about its own direction random too. Averaging a property over such a texture is an integral over three angles. For a single spherical harmonic the two azimuthal integrals can be done at once. Rotating a harmonic of degree $\ell$ and averaging over both azimuths leaves only its part symmetric about the sample axis, and the addition theorem for spherical harmonics says that part is the perfect fibre’s multiplied by $P_\ell(\cos\beta)$. The remaining integral over the tilt is the average of that polynomial:

$$\langle T \rangle = \sum_\ell \langle P_\ell(\cos\beta) \rangle \; T_\ell ,$$

where $T_\ell$ is the degree-$\ell$ part of the perfect fibre’s tensor. Nothing here depends on the shape of the tilt distribution. It is used below as a Gaussian of width $\sigma$ on the sphere, but any distribution only changes the numbers $\langle P_\ell \rangle$.

The four averages fall from one at perfect alignment to nought at a uniform spread, and they fall in order of degree. The degree-one average, $\langle\cos\beta\rangle$, halves at a spread of 52°. The degree-two average halves at 28°, degree three at 20° and degree four at 15°. For narrow spreads each behaves like $\exp(-\ell(\ell+1)\sigma^2/2)$, so the rate of loss grows roughly as the square of the degree. A fibre loses its properties in order of their complexity. A pyroelectric fibre, whose polarisation is a vector of degree one, keeps half its response at spreads where the highest harmonic of its elastic anisotropy has almost gone.

The formula against the average #

The formula can be checked against the definition, because both sides can be computed. The degree parts of a piezoelectric tensor are computed from its components. The degree-three part is the fully symmetric, traceless part of the tensor, and whatever is left is degree one; for a polar fibre the degree-two part is zero. The brute average rotates the whole tensor through a grid of tilts and both azimuths, weights each rotation, and adds.

The two agree to a few parts in a thousand, which is the size of the quadrature grid’s own error, across two tensors and six spreads. This is the least interesting figure in the essay and the one the rest depends on. It confirms that the degree parts were extracted correctly, and that the factor multiplying each is the average Legendre polynomial and not something that only resembles it.

Two vectors and an octupole #

The piezoelectric tensor of a polar fibre, one whose grains share a polar direction along the axis, has the symmetry of a cone, $\infty m$. It keeps three coefficients: $d_{33}$ along the axis, $d_{31}$ across it, and the shear coefficient $d_{15}$. In degrees it is two vectors and an octupole, the decomposition what a texture permits counted: degree one twice, degree three once, and in the polar case no degree two. So every coefficient of a loosened polar fibre is

$$d(\sigma) \;=\; a\,\langle P_1 \rangle + b\,\langle P_3 \rangle ,$$

with $a$ and $b$ read off the perfect fibre. The vectors and the octupole shrink at the two rates of the first figure, 52° and 20° to half.

That is only interesting when the two parts have opposite signs, and for real polar crystals they often do. Two tensors show it. The first has $d_{33} = 90$, $d_{31} = -33$ and $d_{15} = 560$ pC/N, a shear coefficient six times the longitudinal one, the shape of tetragonal barium titanate’s coefficients. The second has $d_{33} = 12$, $d_{31} = -5$ and $d_{15} = -8$, the shape of a wurtzite crystal such as zinc oxide. Both are used as shapes, not as data about those materials. In the first, $d_{33}$ is $264.8$ of degree one and $-174.8$ of degree three. In the second, $d_{15}$ is $+2$ of degree one and $-10$ of degree three.

A fibre that gains longitudinal response #

For the first tensor, the octupole’s negative contribution to $d_{33}$ decays faster than the vectors’ positive one. So as the grains spread, the negative part is lost first and $d_{33}$ rises.

The picture at the head of this essay is that rise. At perfect alignment the fibre’s longitudinal coefficient is the grain’s, 90 pC/N. At a spread of 30° it is 168, nearly double. Past that it falls, as every coefficient must, towards nothing at a uniform spread. Meanwhile $d_{31}$ grows more negative before it too falls, and $d_{15}$ falls steadily from 560. A fibre drawn or poled with its grains well but not perfectly aligned is, along its axis, a considerably better piezoelectric than a perfect one would be.

The rise is not a statement about the texture alone. It follows from the grain’s own anisotropy, and a single grain shows where it comes from. A grain’s longitudinal coefficient measured along a direction at angle $\theta$ to its polar axis is

$$d_{33}^*(\theta) = d_{33}\cos^3\theta + (d_{31} + d_{15})\sin^2\theta\cos\theta ,$$

and with a shear coefficient this large, the second term wins over a wide range of angles.

Along the polar axis the grain gives 90. Measured 51° off the axis it gives 223, two and a half times as much, which is where the experimental literature on barium titanate places the direction of largest response. A fibre’s longitudinal coefficient is exactly this curve averaged over its grains’ tilts: each grain contributes its oblique coefficient along the sample axis. The computation confirms the identity to six figures at a spread of 0.6 radians. So loosening a fibre moves weight from the axis, where the grain is weak, towards the oblique directions where it is strong. The fibre can never exceed the grain’s best oblique value, since an average cannot exceed its maximum, but a moderate spread gets three quarters of the way there.

The two ways of saying it agree because they are the same statement. The oblique curve is a degree-one term plus a degree-three term in $\cos\theta$, and averaging it over a tilt distribution replaces them with $\langle P_1 \rangle$ and $\langle P_3 \rangle$. The harmonic picture explains which grains show the effect: exactly those whose degree-three part opposes their degree-one part in $d_{33}$. For this shape of tensor that means $d_{31} + d_{15}$ large and positive compared with $d_{33}$.

A coefficient that changes sign #

The second tensor has the opposite arrangement in its shear coefficient. Its $d_{15}$ is $-8$, made of $+2$ of degree one and $-10$ of degree three. The negative octupole dominates at perfect alignment and decays first.

The shear coefficient of the fibre passes through zero at a spread of 34° and is positive beyond it. A fibre textured to that degree has no shear piezoelectric response at all, although every grain in it has one. The other two coefficients fall towards zero without crossing. $d_{31}$ is entirely octupole, so it falls fastest, and $d_{33}$ is mostly octupole too and follows it.

A zero of this kind is invisible to any argument from symmetry. The texture has the same symmetry, $\infty m$, at every spread, and $\infty m$ permits a shear coefficient. Neumann’s principle and its extension to textures in what a texture permits say which coefficients may be non-zero. They cannot say that one happens to be zero at a particular spread, any more than they can say what its value is. The zero is an accident of the numbers, but a structured one, located by the ratio of the two degree parts and the ratio of the two Legendre averages. For a Gaussian spread it falls where $\langle P_3\rangle / \langle P_1\rangle = 2/10$.

Which grains do it: two inequalities #

Both effects can be stated as conditions on the grain’s three coefficients, and the conditions are short. The degree-three part of each coefficient comes out of the symmetrised, traceless part of the tensor:

$$b(d_{33}) = \tfrac25\,(d_{33} - d_{31} - d_{15}), \qquad b(d_{15}) = \tfrac25\,(d_{31} + d_{15} - d_{33}),$$

and the degree-one part is whatever remains. For a narrow spread each Legendre average is $1 - \ell(\ell+1)\langle\beta^2\rangle/4$, so the degree-three part falls six times as fast as the degree-one part. A fibre’s longitudinal coefficient therefore rises under a small spread exactly when

$$d_{31} + d_{15} \;>\; \tfrac32\, d_{33} .$$

This is the same inequality that makes a single grain’s oblique coefficient grow as the measuring direction leaves the polar axis, which is the harmonic argument and the geometric one arriving at one line. The large-shear tensor satisfies it by a factor of four: 527 against 135. A tensor of the shape usually quoted for lead zirconate titanate, with $d_{33} = 374$, $d_{31} = -171$ and $d_{15} = 584$, gives 413 against 561 and fails it. The fibre’s $d_{33}$ then falls monotonically, as the sweep that produced the figures confirms.

The sign change has an equally short condition. At a uniform spread only the degree-one part of a coefficient survives to the last. So $d_{15}$ ends with the sign of $\tfrac35 d_{15} - \tfrac25 d_{31} + \tfrac25 d_{33}$, and it changes sign on the way exactly when that differs from the sign of $d_{15}$ itself. The small tensor gives $+2$ against $-8$. Both conditions were tested against the numerical sweep on sixty tensors with random coefficients in the ranges real polar crystals occupy. Thirty-six of them rise, seven flip, and the inequalities predict every one of the sixty with no exceptions.

What a measurement of the fibre reads #

The same decomposition runs backwards. A fibre’s three measured coefficients are three linear combinations of two unknowns, $\langle P_1 \rangle$ and $\langle P_3 \rangle$, with the grain’s degree parts as coefficients. Given the single-crystal tensor, the three numbers determine the two averages by least squares and leave one residual. The earlier essay observed that an achiral fibre’s optical rotation reads $\langle P_2 \rangle$. A polar fibre’s piezoelectric response reads $\langle P_1 \rangle$ and $\langle P_3 \rangle$ together, two moments of the texture, and the residual tests whether the texture really is a fibre. At a spread of half a radian the computation recovers $\langle P_1 \rangle = 0.787$ and $\langle P_3 \rangle = 0.238$ with a residual of $10^{-14}$. A texture that is not symmetric about the axis would leave a residual, because its averaged tensor would no longer have the three-coefficient shape of $\infty m$.

This is what makes the degree picture useful beyond explanation. A pole figure measured by diffraction gives the whole orientation distribution, and a piezoelectric measurement gives only these two numbers. But they are the two numbers that decide the fibre’s piezoelectric response, and they can be measured on a sample too small or too coarse-grained for a pole figure.

What the degree ordering says about other properties #

The same arithmetic applies to every property a fibre carries, and the ordering of the first figure is the whole of what it predicts. Thermal expansion and dielectric anisotropy are degree two apart from their isotropic part, so they decay at the second rate and cannot change sign on their own. The isotropic part does not decay at all. A property whose parts are all of one degree, such as a pyroelectric coefficient, only shrinks.

The elastic constants carry two quadrupoles and a hexadecapole. So the anisotropy of a loosened fibre’s stiffness changes shape as well as size, its degree-four part vanishing at spreads where its degree-two part is still substantial. This is why the elastic anisotropy of a textured material is often well described by its quadrupole alone once the texture is moderate.

The gyration tensor of the earlier essay is the simplest case. For an achiral class it is purely degree two, so the rotation of light falls exactly as $\langle P_2 \rangle$, halving at 28°, with no change of shape. The earlier essay stated that result, and the first figure here supplies its numbers.

What these fibres leave out #

The tilt distribution is a Gaussian on the sphere. Real textures are measured as pole figures, and their distributions have tails and asymmetries that change the numbers $\langle P_\ell \rangle$ and nothing else. The formula takes any distribution, and a measured pole figure supplies the averages directly.

The average itself is the one that assumes every grain feels the same field, the Voigt bound in the language of composites. A real polycrystal’s grains constrain one another, and the effective coefficients of a ceramic lie between that bound and the opposite one where every grain feels the same stress. For piezoelectric coefficients the difference can be considerable. The degree ordering is a property of the averaging and survives any linear homogenisation. The specific values of 168 and 34° belong to the uniform-field average and are not predictions for a ceramic.

Nor do the figures show domains. A ferroelectric grain is not a single domain, and poling a ceramic reorients domains within grains as well as selecting among grains. The tilt spread here is a spread of whole single-domain grains. The rise of $d_{33}$ with spread is related to the enhancement that domain engineering produces in real crystals, and the oblique curve’s maximum is the same fact in both. But nothing on this page models a domain wall.

What the fibre has to refuse #

The first refused claim is the intuition this essay’s title contradicts: that every coefficient of a fibre falls as its alignment gets worse. For the large-shear tensor, $d_{33}$ goes from 90 to 168. The second refused claim is the overcorrection: that a fibre can do better than any single direction of its own grain. It cannot, because its coefficient is an average of the grain’s oblique coefficient, whose maximum, 223, bounds every fibre made of that grain.

Still open: the best texture, not the best spread #

The spread here is one-parameter and symmetric about the axis, and the question it answers is how a given texture degrades. The inverse question asks which orientation distribution maximises a fibre’s $d_{33}$ for a given grain. The answer is a distribution concentrated at the angle where the oblique curve peaks: a cone of grains at 51° rather than a cap around the axis. Such a cone texture is not what drawing or poling produces, but it is a definite target. How close a two-parameter texture, a cone angle with a spread about it, comes to the grain’s 223, and how much of that survives the grains’ mutual constraint, has not been computed here.