The plane a deformation leaves alone

The walls a strain permits asks which planes two domains of a ferroelastic crystal may meet on, and answers it with the linearised strain: the difference of the two strain tensors has to vanish on the wall, which is a quadratic form being zero and picks out a cone of directions.

That is the right answer for a small strain and it is not the right answer for a large one. A transition that changes a cell by six per cent is not a small strain, and the linearised condition and the exact one disagree about which planes exist and about whether any exist at all. This essay does the exact one.

The distinction is worth stating once as a general point, because it recurs. A linearised condition is a statement about the derivative of an exact condition at zero, and it is reliable exactly where the quantity being linearised is small. A ferroelastic domain wall separates two variants whose strains differ by a fraction of a per cent, and there the linearisation is excellent. A martensitic interface separates a variant from an undeformed parent, and there the quantity is six per cent — small enough that a linearised formula still returns an answer, and large enough that the answer is wrong in kind rather than in the third decimal place.

What an interface demands #

Take two regions of a crystal, each uniformly deformed — one by F and the other by G, meaning that a material vector x becomes Fx on one side and Gx on the other. If the two regions meet across a plane, every vector lying in that plane has to be sent to the same place by both, or the material tears.

Write n for the plane’s normal. Every x with x · n = 0 must satisfy Fx = Gx, so F − G annihilates a two-dimensional subspace, so it has rank at most one:

F − G = a ⊗ n.

That is the whole of the condition and it is called a rank-one connection. It says nothing about energies, nothing about which materials do it, and nothing about how thick an interface is. It is a statement that two linear maps agree on a plane.

A rotation of a whole region costs nothing and moves no interface, so the useful question is whether F and G can be made rank-one connected after rotating one of them: whether there is a rotation Q with

Q F − G = a ⊗ n.

Two things about that equation are worth fixing before any algebra. The first is that n is a normal in the reference configuration — the undeformed one — so it is a plane of the material rather than a plane of either deformed state. That matters when the answer is quoted as a habit plane, because a habit plane observed in a micrograph is a plane of the deformed crystal and the two differ by whichever deformation carried it there.

The second is that only the symmetric part of a deformation can obstruct anything. Any invertible F splits uniquely as a rotation times a symmetric positive-definite stretch — the polar decomposition — and the rotation is free: it moves the region without straining it, and a region can always be rotated into place. So every question here is a question about stretches, and the rotation is what gets solved for at the end rather than what has to be arranged in advance. That is the same division the strain that arrives with the transition makes when it separates the spontaneous strain from the orientation of the domain that carries it.

The condition is a signature #

The rotation is a nuisance and there is a standard way to remove it. Multiply each side of QF = G + a ⊗ n by its own transpose. On the left Qᵀ Q is the identity and the rotation vanishes; on the right the product expands, and writing c for Gᵀa:

FᵀF − GᵀG = c ⊗ n + n ⊗ c + |a|² n ⊗ n.

Set d = c + (|a|²/2) n and the right-hand side is d ⊗ n + n ⊗ d. A symmetric matrix of that shape is easy to recognise: it is zero on everything perpendicular to both d and n, and on the plane they span it has determinant −(d · n)²… which is negative unless d and n are parallel. So it has one positive eigenvalue, one negative one, and one that is exactly zero.

That is the criterion, and it is worth having in that form rather than in the usual one because it is checkable by computing three numbers.

There is a reading of the signature condition that makes it feel less like algebra. The matrix FᵀF − GᵀG measures, for each direction, how much more the first deformation stretches it than the second does. A plane can be shared only if some direction is stretched more by the first, some direction is stretched less, and one direction is stretched exactly the same — because the plane has to contain two independent directions of equal stretch, and the only way to have two is to have a whole zero eigenvalue and one crossing from the other two. Positive, negative and zero is that sentence written as a signature.

For G the identity — one region undeformed — the matrix is FᵀF − I, whose eigenvalues are the squares of the principal stretches minus one. So the condition “middle eigenvalue zero” reads: the middle principal stretch is exactly one. That is the classical statement of the invariant-plane condition, and here it arrives as a special case of a signature rather than as a rule to be remembered.

Once the signature is right the solution is not a search. The eigenvectors of FᵀF − GᵀG give n and d in closed form up to a sign, |a|² comes out of a quadratic, and there are exactly two solutions — the two signs. Nothing is fitted and nothing is optimised.

One consequence of the closed form deserves to be drawn out, because it is why this is a classification and not a numerical study. The two solutions depend continuously on the two deformations, and they exist or fail to exist according to whether one number is zero. So the set of deformation pairs that can share a wall is a hypersurface — a codimension-one condition — inside the space of all pairs. Two deformations picked at random never share a wall. The ones that do are the ones a symmetry put there, which is why every interface in a crystal has a group-theoretical explanation and why the wall has a group of its own.

It is worth noticing how little freedom the two solutions represent. A rank-one connection between two given deformations is not a family; it is at most two planes, and the two are mirror images in the plane of the extreme eigenvectors. So a pair of deformations does not “have an interface” in the sense of having a range of them to choose from — it has two, or none, and which of the two occurs is settled by something outside this calculation. That rigidity is what makes the theory predictive: a twin is a symmetry the crystal lacks identifies the twin law, and this fixes the plane it can act on without any further input.

The transition that cannot fit #

Now apply it to a real situation. A cubic crystal transforming to a tetragonal one stretches two of its axes by η₁ and the third by η₃, and there are three variants because there are three axes to choose.

The middle stretch of every variant is η₁, and η₁ is one only if the transition does nothing to those two axes. So the answer is not “a plane is hard to find”; it is that there is no plane, for any variant, for any orientation of the interface.

This is the point at which the linearised theory and the exact one part company, and the disagreement is instructive rather than technical. The walls a strain permits finds walls between two variants, and that answer survives here. What the linearised theory cannot see is the interface between a variant and the undeformed parent, because at linear order the condition it imposes is on the difference of two strains, and comparing a variant with the parent compares a finite strain with zero — exactly the case in which the linearisation is not valid.

There is a temptation to look for a way round the prohibition by allowing the interface to be curved, or diffuse, or to carry defects. None of those helps, and it is worth saying why in one paragraph rather than leaving it as an assumption. A curved interface is locally a plane, so the condition applies at every point of it and fails at every point. A diffuse interface interpolates between the two deformations, and the interpolation has to be compatible at every stage, which is a stronger requirement rather than a weaker one. Defects do help — an array of dislocations accommodates a mismatch, which is exactly what a semicoherent interface is — and at a six per cent strain the density required is so high that the “interface” is no longer describable as two crystals meeting. The exact condition is a statement about what a coherent interface can be, and coherence is what the rest of this essay is about.

A last observation about the three variants before leaving them. They are the same stretch with its axes permuted, which means they are related by the parent’s threefold axes as well as by its mirrors, and it means the obstruction is identical for all three. That is not automatic. A transition to a lower symmetry can produce variants whose stretches are not related by a parent operation — that happens when the order parameter has more than one component and the variants are not a single orbit — and then the variants have different obstructions and different walls. Two hundred and forty-seven descents is where this collection counts which descents give which orbits, and the guarantee below holds for the ones that give a single orbit.

Variants against each other #

Between two variants, the same equation is the twinning equation, and it behaves entirely differently.

Every pair solving is not luck and it is not a property of these particular numbers. Two variants of one transition are related by an operation of the parent’s symmetry — that relation is what makes them variants rather than unrelated deformations — and a pair related that way always satisfies the condition. How many domains a transition makes counts the variants by exactly that relation, so the count of variants and the guarantee that they twin are the same fact stated twice.

The guarantee is not about the number of variants either. A cubic-to-orthorhombic transition produces six of them and fifteen pairs, and every one of the fifteen solves for the same reason — which the next rung shows, because there the count matters for a different question.

So the situation after this essay is a genuine impasse, and stating it plainly is the point of stopping here. A variant cannot meet the parent. Two variants can meet each other. A crystal transforming from the parent to a variant therefore has no way to start — and it does start, in every material that does this.

There is one more thing the impasse is not. It is not a statement that the transition cannot happen, and it is not a statement that the parent and the product cannot coexist. Crystals do both. What it says is that a coherent planar interface between the parent and a single homogeneously deformed variant does not exist, so whatever the two phases do when they meet, it is not that. The rest of the crystallographic theory of martensite is the business of finding out what it is instead, and the answer is entirely within the same equation.

It is also worth measuring the failure rather than only noting it, because the size decides whether the impasse is real. A middle stretch of 1.062 means the best available plane changes lengths in it by six per cent, and six per cent is two orders of magnitude beyond the elastic strain a crystal carries before it yields. So the prohibition is not one a material can quietly violate; it is one it has to route around, and that is why the route around it is observable.

The third of those is the one worth pausing on. Two identical deformations are refused a wall, which sounds wrong — surely a homogeneous crystal has planes in it. It does, and none of them is an interface: the equation F − G = a ⊗ n with F = G forces a = 0, and a wall with no jump across it is not a wall. The machinery reporting “the two deformations are the same” rather than “every plane works” is the distinction between a boundary and a plane drawn on paper.

One further reading of the twin planes, since {110} is a family this collection has met before. The twin planes here come out at the components of a face diagonal because the two variants are exchanged by a mirror in that plane, and that mirror is an operation the parent has and the product does not. So the twin plane is a lost mirror, which is the definition a twin is a symmetry the crystal lacks gives — and here it is not assumed but returned by an equation that was told only about two stretches. Two derivations that share no step giving the same plane is the strongest kind of agreement available.

The shears are worth a sentence too, because their size is what makes the walls physical. A twin wall in this transition carries a shear of order a tenth: that is enormous by the standards of elasticity, and it is why a twin boundary is a feature a microscope sees rather than a bookkeeping surface. It is also why the twins in a martensite plate are fine — a thin lamella carries the shear over a short distance, and fineness is how the structure pays for it.

It is worth recording what this construction does not need, because the list is short and surprising. It does not need a crystal structure — only two stretches. It does not need the symmetry of either phase, except to explain why the variants are related. It does not need an energy, a temperature, or a modulus. And it does not need the interface to be flat over any particular distance: the condition is pointwise, so a wall that is planar in patches satisfies it in each patch. Everything that comes out of it therefore holds for any material with these two stretches, which is why the results are quoted as facts about a transition rather than about a substance. Epitaxy is a measurement makes the same point about a different interface: what a geometry forbids, no chemistry supplies.

What is missing #

The impasse is resolved by something the equation above allows and this essay has not used: the deformation on one side does not have to be a single variant. A fine mixture of two variants has an average deformation that is neither of them, and the average has a free parameter in it — the volume fraction. The next rung follows that parameter and finds that it passes through the condition twice, which is why martensite plates are internally twinned and why their volume fractions are predictable before anything is looked at.

That resolution is worth one advance warning, because it changes what “the deformation” means. Up to here F has been an actual deformation gradient at a point. In the laminate it is an average over a length scale much larger than the twin spacing and much smaller than the plate — and the whole construction is exact only in the limit of infinitely fine twins. That limit is where the theory’s predictions live, and it is the one assumption in it that is not arithmetic.