# Given no strain and uniform angle of rotation, can the displacement field be expressed as a uniform rotation?

Consider displacement vector $$\boldsymbol{\xi}$$ where $$\boldsymbol{\mathbf{\xi}}=\boldsymbol{\mathbf{\phi}} \times \mathbf{x}$$.
$$\boldsymbol{\phi}$$ is an angle vector along the axis of rotation and $$\mathbf{x}$$ is the position vector. Let $$W_{ij}=\frac{\partial \xi_i}{\partial x_j}$$. Then we have that symmetrical part of the tensor $$\mathbf{W}$$ i.e. $$W_{ij}+W_{ji}$$ is $$0$$. This signifies that strain is $$0$$. While for antisymmetric part $$R_{ij}=\frac{1}{2}(W_{ij}-W_{ji})$$ we have the following relation:

$$R_{ij}=-\epsilon_{ijk}\phi_k$$ and $$\phi_i=-\frac{1}{2}\epsilon_{ijk}R_{jk}$$

We also have that $$\boldsymbol{\phi}=\frac{1}{2}\boldsymbol{\nabla}\times\boldsymbol{\xi}$$.

Question: Now suppose we are given that the strain is $$0$$, i.e. $$W_{ij}+W_{ji}=0$$ and $$\boldsymbol{\phi}=\frac{1}{2}\nabla \times \boldsymbol{\xi}$$ where $$\boldsymbol{\phi}$$ is constant over all of space. Can we prove that $$\boldsymbol{\xi}=\boldsymbol{\phi} \times \mathbf{x}$$

Note 1 : Here I have used einstein summation convention in above equations. $$\epsilon_{ijk}$$ is the levi-civita tensor.

Note 2 : Here the angle $$\phi$$ represents an infinitesimal angle with the axis of rotation being the direction along the vector $$\phi$$

• I am missing the axis of rotation in your equations?. Your equations are for small angle of rotation – Eli Oct 6 '19 at 14:39
• Yes, $\phi$ represents infinitesimal angle. Thanks, I did not notice that. Though, $\phi$ is a vector and its direction gives the axis of rotation. – Sudipta Nayak Oct 7 '19 at 20:16