Is the Linear Eulerian Deformation Tensor an Exact Measure of Deformation?

[facenian]

Hello, I have difficulty interpreting the following fact (I'm reading Cotinuum Mechanics by Spencer). The relative velocity between two nearby points P and Q in the current configuarion is given by: $dv_i=D_{ik}dx_k + W_{ik}dx_k$
where $D_{ik}=\frac{d}{dt}e_{ik}$ is the rate of deformation tensor being equal to the material derivative of the linear eulerian deformation tensor and the part involving the vorticity tensor W_ik can be interpreted as a pure rotation about an axes through the point P. This expression is "exact" despite the appearance of the linear eulerian tensor. Now comes the paradox: if D_ik=0 then all points near P simply "rotate rigidly" and this strongly suggest that there is no change of the (non linear) eulerian tensor E_ik which is and exact measure of deformation around P however only the linear part of the tensor vanishes so it seems that despite the pure rotation about P there is a change in the eulerian deformation tensor E_ik,ie, this tensor changes but the deformation does not change.
Can someone point out my mistake?

 

[Valerie Park]

Your mistake is in interpreting the linear Eulerian deformation tensor as an exact measure of deformation. The linear Eulerian deformation tensor only measures the linear part of the deformation, i.e., the strain caused by a pure rotation. The non-linear Eulerian deformation tensor, on the other hand, is an exact measure of deformation, and it includes the strain caused by both rotations and translations. So even if the linear Eulerian deformation tensor is zero, the non-linear Eulerian deformation tensor still changes due to the rotational component of the deformation. Therefore, the deformation is still changing despite the fact that the linear Eulerian deformation tensor is zero.