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For displacement $u(X)$ in reference coordinates, the displacement gradient is $H_{ij}=\partial u_i/\partial X_j$. It records how nearby points move differently. The deformation gradient is $F=I+H$.

In small-strain kinematics, $ϵ=(H+H^T)/2$ describes strain, while the antisymmetric part gives infinitesimal rotation. For a rigid translation, $H=0$. For $u=(\alpha X_1,0,0)$, $H_{11}=\alpha$ describes uniform axial extension. For finite rotations, the linearized strain is not an exact rotation-invariant measure, so a finite-deformation strain tensor is needed.

The displacement gradient has components $(\nabla \mathbf{u}{)}_{ij}={\partial }_i{u}_j$ and records local changes of the displacement field.

This tensor is a local derivative of the displacement field; it enters the deformation gradient and strain measures.

For a static base state with zero incremental body force, homogeneous incremental tractions or homogeneous Dirichlet boundary conditions eliminate the boundary term in integration by parts. The displayed Rayleigh quotient then gives real squared frequencies because of major symmetry. Positive moduli give strictly positive squared frequencies for modes with nonzero displacement gradient. Pure dead loading leaves constant translations at zero; fixed displacement conditions remove those modes.

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.

The infinitesimal strain tensor is the symmetric part of a displacement gradient tensor. It is the linear approximation to a finite strain measure for small displacement gradients. An infinitesimal rigid rotation contributes no strain, so the elastic energy depends on the symmetric part rather than the entire gradient.

A normalized measure of consistency of complex radar returns over a neighborhood. Low coherence makes interferometric phase unreliable and can arise from vegetation, surface changes, viewing differences or extreme displacement gradients.

Linearize $\det F=h(\theta )$ at $F=I$, $\theta ={\theta }_0$, $h({\theta }_0)=1$, and zero reaction. With displacement gradient $H$ and temperature increment $\vartheta$, one has $\mathrm{tr}H=h^{\prime }({\theta }_0)\vartheta$ andwhere $C_{jilk}={\rho }_0{\psi }_{,F_{ij}{F}_{kl}}$, ${\beta }_{ji}={\rho }_0{\psi }_{,F_{ij}\theta }$ and $C_e=-{\rho }_0{\psi }_{,\theta \theta }$ at the reference state. The temperature factor converts this entropy coefficient to a volumetric heat capacity: ${\theta }_0{C}_e$. Strictly constant volume is the case $h^{\prime }=0$.

For a right stretch tensor $U={\sum }_a{\lambda }_a{n}_a\otimes n_a$, define $E^f=f(U)={\sum }_{a}f({\lambda }_a)n_a\otimes n_a$. A standard normalized choice has $f\in C^1(0,\infty )$, $f(1)=0$, $f^{\prime }(1)=1$ and $f$ strictly increasing; requiring $f^{\prime }>0$ gives a nonsingular stretch-to-strain parameterization. This strain vanishes exactly for a rigid rotation, is independent of superposed spatial rotations, and agrees to first order with the infinitesimal symmetric displacement gradient. Indeed $F=I+H$ gives $U=I+(H+H^T)/2+O(\Vert H{\Vert }^2)$, and Taylor expansion of $f$ proves the last statement.

Strain measures local changes of length and angle relative to a chosen reference configuration. For a small displacement field $\mathbf{u}(\mathbf{X})$, the infinitesimal strain tensor isIts diagonal entries are fractional length changes along the coordinate axes. Off-diagonal entries describe changes of angle, with engineering shear strain equal to twice the corresponding tensor component.

A uniform translation has zero strain because its displacement gradient vanishes. An infinitesimal rigid rotation also has zero strain because its displacement gradient is antisymmetric. Finite rotations require finite strain theory, since the linear formula is then insufficient. In tensile testing, the simple ratio of extension to original gauge length is engineering strain; it coincides with the small axial component in the appropriate limit but is not a complete three-dimensional description.