Green–Lagrange strain tensor

Remarks

With the displacement $u(X) = \varphi(X) - X$, so that $F = I + \nabla u$, one has $$E = \tfrac12 \left( \nabla u + (\nabla u)^T + (\nabla u)^T \nabla u \right).$$ Dropping the quadratic term gives the infinitesimal strain tensor $\varepsilon = \tfrac12 (\nabla u + (\nabla u)^T)$. For the simple shear with $F$ having rows $(1, \gamma, 0)$, $(0, 1, 0)$, $(0, 0, 1)$, the tensor $E$ has rows $(0, \gamma/2, 0)$, $(\gamma/2, \gamma^2/2, 0)$, $(0, 0, 0)$. For a rotation by the angle $\theta$ about the third axis, $E = 0$. But $\varepsilon$ has diagonal entries $\cos\theta - 1$, $\cos\theta - 1$, $0$, which is nonzero for $\theta \notin 2\pi\Z$. So the small-strain tensor fails to vanish for large rotations. The tensor $E$ is also called the Green–St. Venant or Lagrangian finite strain tensor.