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 $\phi$

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

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