Editing Infinitesimal strain theory (section)

== Relation to infinitesimal rotation tensor {{anchor|Infinitesimal rotation tensor}} ==
{{see also|Spin tensor (mechanics)}}

The infinitesimal strain tensor is defined as
<math display="block"> \boldsymbol{\varepsilon} = \frac{1}{2} [\boldsymbol{\nabla}\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})^T]</math>
Therefore the displacement gradient can be expressed as
<math display="block"> \boldsymbol{\nabla}\mathbf{u} = \boldsymbol{\varepsilon} + \boldsymbol{W}</math>
where
<math display="block"> \boldsymbol{W} := \frac{1}{2} [\boldsymbol{\nabla}\mathbf{u} - (\boldsymbol{\nabla}\mathbf{u})^T]</math>
The quantity <math>\boldsymbol{W}</math> is the '''infinitesimal rotation tensor''' or '''infinitesimal angular displacement tensor''' (related to the ''[[infinitesimal rotation matrix]]''). This tensor is [[skew symmetric]]. For infinitesimal deformations the scalar components of <math>\boldsymbol{W}</math> satisfy the condition <math>|W_{ij}| \ll 1</math>. Note that the displacement gradient is small only if {{em|both}} the strain tensor and the rotation tensor are infinitesimal.

=== The axial vector ===
A skew symmetric second-order tensor has three independent scalar components. These three components are used to define an '''axial vector''', <math>\mathbf{w}</math>, as follows
<math display="block"> W_{ij} = -\epsilon_{ijk}~w_k ~;~~ w_i = -\tfrac{1}{2}~\epsilon_{ijk}~W_{jk} </math>
where <math>\epsilon_{ijk}</math> is the [[permutation symbol]]. In matrix form
<math display="block"> \underline{\underline{\boldsymbol{W}}} = \begin{bmatrix} 0 & -w_3 & w_2 \\ w_3 & 0 & -w_1 \\ -w_2 & w_1 & 0\end{bmatrix} ~;~~ \underline{\mathbf{w}} = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix} </math>
The axial vector is also called the '''infinitesimal rotation vector'''. The rotation vector is related to the displacement gradient by the relation
<math display="block"> \mathbf{w} = \tfrac{1}{2}~ \boldsymbol{\nabla} \times \mathbf{u} </math>
In index notation
<math display="block"> w_i = \tfrac{1}{2}~\epsilon_{ijk}~u_{k,j} </math>
If <math>\lVert\boldsymbol{W}\rVert \ll 1 </math> and <math>\boldsymbol{\varepsilon} = \boldsymbol{0}</math> then the material undergoes an approximate rigid body rotation of magnitude <math>|\mathbf{w}|</math> around the vector <math>\mathbf{w}</math>.

=== Relation between the strain tensor and the rotation vector ===
Given a continuous, single-valued displacement field <math>\mathbf{u}</math> and the corresponding infinitesimal strain tensor <math>\boldsymbol{\varepsilon}</math>, we have (see [[Tensor derivative (continuum mechanics)]])
<math display="block">\boldsymbol{\nabla}\times\boldsymbol{\varepsilon} = e_{ijk}~\varepsilon_{lj,i}~\mathbf{e}_k\otimes\mathbf{e}_l
 = \tfrac{1}{2}~e_{ijk}~[u_{l,ji} + u_{j,li}]~\mathbf{e}_k\otimes\mathbf{e}_l </math>
Since a change in the order of differentiation does not change the result, <math>u_{l,ji} = u_{l,ij}</math>. Therefore
<math display="block"> e_{ijk} u_{l,ji} = (e_{12k}+e_{21k}) u_{l,12} + (e_{13k}+e_{31k}) u_{l,13} + (e_{23k} + e_{32k}) u_{l,32} = 0 </math>
Also
<math display="block"> \tfrac{1}{2}~e_{ijk}~u_{j,li} = \left(\tfrac{1}{2}~e_{ijk}~u_{j,i}\right)_{,l} = \left(\tfrac{1}{2} ~ e_{kij}~u_{j,i}\right)_{,l} = w_{k,l} </math>
Hence
<math display="block"> \boldsymbol{\nabla} \times \boldsymbol{\varepsilon} = w_{k,l}~\mathbf{e}_k\otimes\mathbf{e}_l = \boldsymbol{\nabla}\mathbf{w} </math>

=== Relation between rotation tensor and rotation vector ===
From an important identity regarding the [[Tensor derivative (continuum mechanics)|curl of a tensor]] we know that for a continuous, single-valued displacement field <math>\mathbf{u}</math>,
<math display="block"> \boldsymbol{\nabla}\times(\boldsymbol{\nabla}\mathbf{u}) = \boldsymbol{0}. </math>
Since <math>\boldsymbol{\nabla}\mathbf{u} = \boldsymbol{\varepsilon} + \boldsymbol{W}</math> we have
<math display="block"> \boldsymbol{\nabla}\times\boldsymbol{W} = -\boldsymbol{\nabla}\times\boldsymbol{\varepsilon} = - \boldsymbol{\nabla} \mathbf{w}. </math>