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Finite strain theory describes changes of shape and orientation without assuming that the displacement gradient is small. Its central variable is the deformation gradient $F=\partial \mathbf{x}/\partial \mathbf{X}$, which maps an infinitesimal material line from the reference configuration to the current configuration. The determinant $J=\det F$ gives its local volume ratio.

Large rotations expose why the infinitesimal approximation fails. For a pure rotation $F=R$, all lengths are unchanged, but $\mathrm{sym}(R-I)$ generally does not vanish. The Green-Lagrange strain tensor $E=(F^{\mathsf{T}}F-I)/2$ instead vanishes because $R^{\mathsf{T}}R=I$. A stretched rubber strip is a practical example in which both large length changes and rotations may matter. The polar decomposition in continuum mechanics separates rotation from stretch tensors, while distinct stress measures distinguish force per original area from force per deformed area.

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.

For two distinct symmetric fibre families use reference directions $l_{\pm }=(\cos \theta ,\pm \sin \theta ,0{)}^T$, with $0<\theta <\pi /2$. The second sign is needed for the orthogonality conclusion: the PDF prints only one direction. If both families literally had that same direction, their images under an invertible deformation would remain parallel, so part (ii) would be false. The following derivation states the intended two-family geometry explicitly.

Write $c=\cos \theta$, $s=\sin \theta$ and take positive $\lambda ,\alpha$, as required for a positive stretch tensor. Applying the inextensible fibre constraint to either family givesWith $b={\alpha }^2$, multiplication by $b$ yields $c^2{b}^2-\lambda b+s^2=0$. Its discriminant is ${\lambda }^2-4c^2{s}^2={\lambda }^2-{\sin }^{2}2\theta$, so the two-family inextensible-fibre pure stretch hasThis also shows why the angle endpoints need separate treatment: at $\theta =0$ the admissible root is ${\alpha }^2=\lambda$, and at $\theta =\pi /2$ it is ${\alpha }^2=1/\lambda$. The nominal two families then represent the same unoriented fibre direction, and the two-branch/orthogonality assertions do not apply.

The principal stretches are the eigenvalues of a stretch tensor, or equivalently the singular values of the deformation gradient. They measure local length ratios along the principal reference directions.

The ratio of current to reference lengths of an infinitesimal material line element. Along an elastic rod, it is $\alpha =ds/dS=|d\mathbf{r}/dS|>0$. A principal stretch is a stretch ratio along a principal direction of the stretch tensor.

A stretch tensor isolates local changes of material length from the rotation contained in a deformation gradient. For an invertible orientation-preserving $F$, the polar decomposition in continuum mechanics iswhere $R$ is a rotation matrix and $U,V$ are positive-definite symmetric matrices. The right stretch tensor $U$ acts on directions in the reference configuration; the left stretch tensor $V$ acts on directions in the current configuration.

The eigenvalues of either stretch tensor are the principal stretches. A value two means a material line in the corresponding principal direction doubles its length, while a value one means unchanged length. For a pure rotation, both stretch tensors are $I$, even though $F$ need not equal $I$. This makes them useful when an experiment includes substantial specimen rotation that should not be mistaken for strain.