Separate small strain from small rotation

Write the motion as $\mathbf x=\mathbf X+\mathbf u(\mathbf X)$. A reference line $d\mathbf X$ becomes $(\mathbf I+\mathbf H)d\mathbf X$, where $H_{ij}=\partial u_i/\partial X_j$. Comparing its squared length before and after deformation produces a term linear in the symmetric part of $\mathbf H$ and a term quadratic in the full gradient. Dropping the quadratic term is an approximation about the gradient, not simply about how many millimetres the specimen moved.

A translated block may move a kilometre without straining, because every point moves together. A tiny object can instead acquire a large strain from a very small absolute displacement if the displacement changes sharply across its size. In the ordinary geometrically linear theory, all components of the displacement gradient must be sufficiently small. This includes small rotations. A separate finite-rotation, small-elastic-strain formulation is possible, but it requires geometric machinery that the linearized tensor alone does not supply.

Take the infinitesimal displacement $u_x=-\theta Y$ and $u_y=\theta X$. The off-diagonal derivatives cancel in the symmetric part, giving zero infinitesimal strain. The antisymmetric part retains the local rotation. This is the expected first-order result: turning an undeformed object should not store ordinary elastic strain energy. A general displacement combines this local rotation with stretch and shear, which is why a displacement gradient is not itself a strain tensor.

For an actual finite rigid rotation $\mathbf F=\mathbf R$, the exact strain vanishes because $\mathbf R^T\mathbf R=\mathbf I$. Inserting $\mathbf H=\mathbf R-\mathbf I$ into the linearized formula instead produces a nonzero term of second order in a small rotation angle, or a large spurious strain for a large angle. This is a useful diagnostic: a strain measure must be paired with the geometric regime for which it was derived, rather than treated as interchangeable with every finite-deformation measure.

In simple shear let $u_x=\gamma Y$ and $u_y=0$. One originally vertical material line tilts toward the horizontal, while an initially horizontal line stays horizontal. To first order their right angle changes by $\gamma$ radians. The tensor component is $\varepsilon_{xy}=\gamma/2$, because the symmetric tensor assigns half the engineering shear to each off-diagonal entry. Mixing these two conventions is a common source of a factor-of-two error in shear modulus or energy.

The trace $\operatorname{tr}\boldsymbol\varepsilon$ gives the first-order fractional volume change. Pure simple shear has zero trace even though it changes shape. More generally, the normal strain along a unit direction $\mathbf n$ is $\mathbf n^T\boldsymbol\varepsilon\mathbf n$. Rotating coordinates changes the components but not this physical directional prediction. Principal strain axes diagonalize the symmetric tensor, exposing three orthogonal directions whose first-order extensions require no shear components in that basis. A proposed strain field must also be compatible with a continuous displacement field if the body is modeled as unbroken and without defects. Assigning six arbitrary functions does not automatically satisfy that geometric requirement. One practical check in a simple example is to differentiate the proposed displacement first, rather than inventing strain components independently.