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A hyperelastic energy well is a set of deformation gradients minimizing the stored-energy function at a fixed temperature. Material frame indifference makes each well invariant under left multiplication by proper rotations. A phase-changing crystal can have a single austenitic rotational well at high temperature and several martensitic wells below a transition. Continuous deformations cannot freely join arbitrary wells: planar interfaces require a rank-one connection between the selected gradients.

A martensitic variant is one of the lattice deformations corresponding to the low-temperature crystal phase. The parent crystal's discrete symmetries generate different variants. In a nonlinear elasticity model each representative $U_i$ supplies a hyperelastic energy well $SO(3)U_i$. A representative need not itself be a symmetric stretch tensor; multiplication by a proper rotation leaves the same well.

Represent a point by the column of homogeneous coordinates $(X,Y,Z,1{)}^T$. The actions of the four matrices are:

The last operation produces $(X,Y,Z,Z/f{)}^T$. It requires a perspective divide: divide the first three components by the last. For $Z\ne 0$,This is why the last component of homogeneous coordinates cannot simply be discarded. Points with $Z=0$ have no finite projected position, and a visible scene normally uses $Z>0$ under this forward-axis convention. The first three transformations leave the last component equal to one and need no nontrivial divide. The displayed $K$ retains no depth variation after division; it describes projection onto the image plane, rather than a complete depth-buffer mapping.

The call constructs dimensions $(d,w,h)$ along the old $(x,y,z)$ axes. To obtain target dimensions $(w,h,d)$ and make an old layer $z=k\text{step}$ become a new layer $y^{\prime }=k\text{step}$, use the cyclic permutationThe new coordinate ranges are immediately $[0,w]$, $[0,h]$, $[0,d]$. Its spatial block is orthogonal with determinant $+1$, since a three-cycle is an even permutation. Hence this is a proper rotation, through $-2\pi /3$ about the axis in direction $(1,1,1)$ under the right-hand convention. No translation is required.

In the solid-modelling convention used here, a rigid-body transformation is an orientation-preserving rigid transformation, represented by a proper rotation and a translation. The proposed upper-left block $S$ is orthogonal but hasIt is a reflection in the plane $x=y$. It preserves distances, but reverses handedness, so it cannot describe continuous motion of a rigid solid through proper rotations and translations. Some mathematical uses of rigid transformation include this orientation-reversing case; the determinant is the distinction relevant to the warning.

In a solid modeller, triangle winding and outward normal vectors must agree. A reflection reverses the winding: for edge vectors $a,b$,Simply transforming stored normal vectors by $S$ while retaining the original vertex order therefore makes the geometric and stored normals disagree. Back-face culling, inside/outside tests and Phong reflection model calculations can then use the wrong side. Code assuming a positive oriented volume can likewise interpret the reflected solid incorrectly. A modeller can support reflections if it deliberately corrects these orientation conventions.

A suitable alternative rotates by $-\pi /2$ about $z$ and then translates upward by height. Writing $h=\text{height}$,Its rotation matrix has determinant $+1$. The source bounds $0\le x\le h$, $0\le y\le w$, $z=0$ become $0\le x^{\prime }\le w$, $0\le y^{\prime }\le h$, $z^{\prime }=0$, where $w=\text{width}$. A horizontal stripe $y=k\text{step}$ becomes the vertical stripe $x^{\prime }=k\text{step}$, so even the stripe positions from the left edge are preserved. The face's $+z$ normal vector is also preserved.

Directly multiplying the columns' scalar products gives $C^{T}C=I$, and direct evaluation gives $\det C=1$. Thus the operation is a proper rotation. Its trace is zero. A three-dimensional proper rotation has eigenvalues $1,e^{i\phi },e^{-i\phi }$, hence $\mathrm{tr}C=1+2\cos \phi$. ThereforeThis determines the usual unsigned angle in $[0,\pi ]$ without changing coordinates. The axis can be checked, if desired, from $C(0,1,-1{)}^T=(0,1,-1{)}^T$. The determinant test must precede the trace formula: an improper rotation instead has trace $-1+2\cos \phi$.

Multiplication by $A$ rotates the $(x,y)$ components by $\theta$ counterclockwise when viewed from positive $z$ toward the origin, leaving $z$ unchanged. It is a proper rotation about the $z$ axis, with determinant $+1$. Multiplication by $B$ sends $(x,y,z)$ to $(x,y,-z)$: it is a reflection in the $xy$ plane, with determinant $-1$. Neither operation changes a vector's Euclidean norm.

Use a temperature-dependent strain energy density $W(F,\vartheta )$ with different minimizing deformation gradients in the two crystal phases. Above the transition temperature its wells describe austenite; below it the wells are unions $SO(3)U_i$ of martensitic variants. Left rotational orbits reflect material frame indifference, while the crystal's discrete material symmetries relate the variants. Minimizing the integral of this elastic energy, together with loading and boundary constraints, models the change of lattice shape. Compatibility restricts which low-energy gradients can meet continuously; fine compatible mixtures can accommodate macroscopic boundary shapes.

For the jump condition, write the two affine pieces as $Ax+c_{+}$ and $Bx+c_{-}$. Continuity on the interface meansTaking differences along the interface gives $(A-B)t=0$ for every tangent vector $t$ with $t\cdot N=0$. Decompose any vector $v$ as $t+(v\cdot N)N$. ThenWith $a=(A-B)N$ this provesConversely this rank-one connection permits continuity by choosing $c_{+}-c_{-}=-ka$. A coherent twin interface is such a compatible planar boundary between different martensitic variants, possibly after a rigid rotation of one variant. The rank-one condition is therefore the geometric compatibility condition for a simple twin.

For the unrotated pair, direct subtraction givesso $R=I$ is one solution, with reference interface normal $e_1$.

Put $d=1+{\delta }^2$, $c=(1-{\delta }^2)/d$ and $s=2\delta /d$. Since $c^2+s^2=1$, the stated $\overline{R}$ is a proper rotation. Multiplication givesHere $c+\delta s=1$ and $s-\delta c=\delta$. ConsequentlyThis is the second rank-one connection, now with normal $e_3$. Because $\delta >0$, the two rotations are distinct. The permitted assertion that there are at most two proves completeness.

Finally suppose $Q{U}_1$ is rank-one connected to $\overline{R}{U}_2$, with $Q\in SO(3)$. Left multiplication of their difference by $Q^T$ preserves rank and gives $U_1-Q^T\overline{R}{U}_2$. The classification just proved forces $Q^T\overline{R}$ to be either $I$ or $\overline{R}$, hence $Q=\overline{R}$ or $Q=I$. The two requested matrices are thereforeTheir connections are explicit: the first was calculated above, and $\overline{R}{U}_1-\overline{R}{U}_2=2\delta (\overline{R}{e}_3)\otimes e_1$. If desired the second matrix is

An orientation-preserving orthogonal transformation. In three dimensions it has an invariant axis and acts as a planar rotation through angle $\theta$ perpendicular to that axis. Its eigenvalues are $1,e^{i\theta },e^{-i\theta }$, so $\mathrm{tr}R=1+2\cos \theta$. In higher dimensions a proper rotation can involve several mutually orthogonal planes and need not fix an axis.

An orientation-preserving rigid transformation, composed of a proper rotation and a translation. This solid-modelling convention distinguishes continuous rigid motion from orientation-reversing distance-preserving maps such as a reflection.