# Infinitesimal strain theory: Wikis

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# Encyclopedia

Continuum mechanics

In continuum mechanics, the infinitesimal strain theory, sometimes called small deformation theory, small displacement theory, or small displacement-gradient theory, deals with infinitesimal deformations of a continuum body. For an infinitesimal deformation the displacements and the displacement gradients are small compared to unity, i.e., $\|\mathbf u\| \ll 1 \,\!$ and $\|\nabla \mathbf u\| \ll 1 \,\!$, allowing for the geometric linearization of the Lagrangian finite strain tensor $\mathbf E\,\!$, and the Eulerian finite strain tensor $\mathbf e\,\!$, i.e. the non-linear or second-order terms of the finite strain tensor can be neglected. The linearized Lagrangian and Eulerian strain tensors are approximately the same and can be approximated by the infinitesimal strain tensor or Cauchy's strain tensor, $\boldsymbol\varepsilon\,\!$. Thus,

$\mathbf E \approx \mathbf e \approx \boldsymbol \varepsilon = \frac{1}{2}\left((\nabla\mathbf u)^T + \nabla\mathbf u\right)\,\!$

or

$E_{KL}\approx e_{rs}\approx\varepsilon_{ij}=\frac{1}{2}\left(u_{i,j}+u_{j,i}\right)\,\!$

The infinitesimal strain theory is used in the analysis of deformations of materials exhibiting elastic behaviour, such as materials found in mechanical and civil engineering applications, e.g. concrete and steel.

## Infinitesimal strain tensor

For infinitesimal deformations of a continuum body, in which the displacements and the displacement gradients are small compared to unity, i.e., $\|\mathbf u\| \ll 1 \,\!$ and $\|\nabla \mathbf u\| \ll 1 \,\!$, it is possible for the geometric linearization of the Lagrangian finite strain tensor $\mathbf E\,\!$, and the Eulerian finite strain tensor $\mathbf e\,\!$, i.e. the non-linear or second-order terms of the finite strain tensor can be neglected. Thus we have

$\mathbf E =\frac{1}{2}\left(\nabla_{\mathbf X}\mathbf u + (\nabla_{\mathbf X}\mathbf u)^T + \nabla_{\mathbf X}\mathbf u(\nabla_{\mathbf X}\mathbf u)^T\right)\approx \frac{1}{2}\left(\nabla_{\mathbf X}\mathbf u + (\nabla_{\mathbf X}\mathbf u)^T\right)\,\!$

or

$E_{KL}=\frac{1}{2}\left(\frac{\partial U_K}{\partial X_L}+\frac{\partial U_L}{\partial X_K}+\frac{\partial U_M}{\partial X_K}\frac{\partial U_M}{\partial X_L}\right)\approx \frac{1}{2}\left(\frac{\partial U_K}{\partial X_L}+\frac{\partial U_L}{\partial X_K}\right)\,\!$

and

$\mathbf e =\frac{1}{2}\left(\nabla_{\mathbf x}\mathbf u + (\nabla_{\mathbf x}\mathbf u)^T + \nabla_{\mathbf x}\mathbf u(\nabla_{\mathbf x}\mathbf u)^T\right)\approx \frac{1}{2}\left(\nabla_{\mathbf x}\mathbf u + (\nabla_{\mathbf x}\mathbf u)^T\right)\,\!$

or

$e_{rs}=\frac{1}{2}\left(\frac{\partial u_r}{\partial x_s} +\frac{\partial u_s}{\partial x_r}-\frac{\partial u_k}{\partial x_r}\frac{\partial u_k}{\partial x_s}\right)\approx \frac{1}{2}\left(\frac{\partial u_r}{\partial x_s} +\frac{\partial u_s}{\partial x_r}\right)\,\!$

This linearization implies that the Lagrangian description and the Eulerian description are approximately the same as there is little difference in the material and spatial coordinates of a given material point in the continuum. Therefore, the material displacement gradient components and the spatial displacement gradient components are approximately equal. Thus we have

$\mathbf E \approx \mathbf e \approx \boldsymbol \varepsilon = \frac{1}{2}\left((\nabla\mathbf u)^T + \nabla\mathbf u\right) \qquad \text{or} \qquad E_{KL}\approx e_{rs}\approx\varepsilon_{ij}=\frac{1}{2}\left(u_{i,j}+u_{j,i}\right)\,\!$

where $\varepsilon_{ij}\,\!$ are the components of the infinitesimal strain tensor $\boldsymbol \varepsilon\,\!$, also called Cauchy's strain tensor, linear strain tensor, or small strain tensor.

$\varepsilon_{ij}=\frac{1}{2}\left(u_{i,j}+u_{j,i}\right) = \left[\begin{matrix} \varepsilon_{11} & \varepsilon_{12} & \varepsilon_{13} \ \varepsilon_{21} & \varepsilon_{22} & \varepsilon_{23} \ \varepsilon_{31} & \varepsilon_{32} & \varepsilon_{33} \ \end{matrix}\right] = \left[\begin{matrix} \frac{\partial u_1}{\partial x_1} & \frac{1}{2} \left(\frac{\partial u_1}{\partial x_2}+\frac{\partial u_2}{\partial x_1}\right) & \frac{1}{2} \left(\frac{\partial u_1}{\partial x_3}+\frac{\partial u_3}{\partial x_1}\right) \ \frac{1}{2} \left(\frac{\partial u_2}{\partial x_1}+\frac{\partial u_1}{\partial x_2}\right) & \frac{\partial u_2}{\partial x_2} & \frac{1}{2} \left(\frac{\partial u_2}{\partial x_3}+\frac{\partial u_3}{\partial x_2}\right) \ \frac{1}{2} \left(\frac{\partial u_3}{\partial x_1}+\frac{\partial u_1}{\partial x_3}\right) & \frac{1}{2} \left(\frac{\partial u_3}{\partial x_2}+\frac{\partial u_2}{\partial x_3}\right) & \frac{\partial u_3}{\partial x_3} \ \end{matrix}\right]$

or using different notation:

$\left[\begin{matrix} \varepsilon_{xx} & \varepsilon_{xy} & \varepsilon_{xz} \ \varepsilon_{yx} & \varepsilon_{yy} & \varepsilon_{yz} \ \varepsilon_{zx} & \varepsilon_{zy} & \varepsilon_{zz} \ \end{matrix}\right] = \left[\begin{matrix} \frac{\partial u_x}{\partial x} & \frac{1}{2} \left(\frac{\partial u_x}{\partial y}+\frac{\partial u_y}{\partial x}\right) & \frac{1}{2} \left(\frac{\partial u_x}{\partial z}+\frac{\partial u_z}{\partial x}\right) \ \frac{1}{2} \left(\frac{\partial u_y}{\partial x}+\frac{\partial u_x}{\partial y}\right) & \frac{\partial u_y}{\partial y} & \frac{1}{2} \left(\frac{\partial u_y}{\partial z}+\frac{\partial u_z}{\partial y}\right) \ \frac{1}{2} \left(\frac{\partial u_z}{\partial x}+\frac{\partial u_x}{\partial z}\right) & \frac{1}{2} \left(\frac{\partial u_z}{\partial y}+\frac{\partial u_y}{\partial z}\right) & \frac{\partial u_z}{\partial z} \ \end{matrix}\right] \,\!$

Furthermore, since the deformation gradient can be expressed as $\boldsymbol{F} = \boldsymbol{\nabla}\mathbf{u} + \boldsymbol{I}$ where $\boldsymbol{I}$ is the second-order identity tensor, we have

$\boldsymbol\varepsilon=\frac{1}{2}\left(\boldsymbol{F}^T+\boldsymbol{F}\right)-\boldsymbol{I}\,\!$

Also, from the general expression for the Lagrangian and Eulerian finite strain tensors we have

\begin{align} \mathbf E_{(m)}& =\frac{1}{2m}(\mathbf U^{2m}-\boldsymbol{I}) = \frac{1}{2m}[(\boldsymbol{F}^T\boldsymbol{F})^m - \boldsymbol{I}] \approx \frac{1}{2m}[\{\boldsymbol{\nabla}\mathbf{u}+(\boldsymbol{\nabla}\mathbf{u})^T + \boldsymbol{I}\}^m - \boldsymbol{I}]\approx \boldsymbol{\varepsilon}\ \mathbf e_{(m)}& =\frac{1}{2m}(\mathbf V^{2m}-\boldsymbol{I})= \frac{1}{2m}[(\boldsymbol{F}\boldsymbol{F}^T)^m - \boldsymbol{I}]\approx \boldsymbol{\varepsilon} \end{align}

## Geometric derivation of the infinitesimal strain tensor

Figure 1. Two-dimensional geometric deformation of an infinitesimal material element.

Considering a two-dimensional deformation of an infinitesimal rectangular material element with dimensions $dx\,\!$ by $dy\,\!$ (Figure 1), which after deformation, takes the form of a rhombus. From the geometry of Figure 1 we have

\begin{align} \overline {ab} &= \sqrt{\left(dx+\frac{\partial u_x}{\partial x}dx \right)^2 + \left( \frac{\partial u_y}{\partial x}dx \right)^2} \ &= \sqrt{1+2\frac{\partial u_x}{\partial x}+\left(\frac{\partial u_x}{\partial x}\right)^2 + \left(\frac{\partial u_y}{\partial x}\right)^2}dx \ \end{align}\,\!

For very small displacement gradients, i.e., $\|\nabla \mathbf u\| \ll 1 \,\!$, we have

$\overline {ab} \approx dx +\frac{\partial u_x}{\partial x}dx\,\!$

The normal strain in the $x\,\!$-direction of the rectangular element is defined by

$\varepsilon_x = \frac{\overline {ab}-\overline {AB}}{\overline {AB}}\,\!$

and knowing that $\overline {AB}= dx\,\!$, we have

$\varepsilon_x = \frac{\partial u_x}{\partial x}\,\!$

Similarly, the normal strain in the $y\,\!$-direction, and $z\,\!$-direction, becomes

$\varepsilon_y = \frac{\partial u_y}{\partial y} \quad , \qquad \varepsilon_z = \frac{\partial u_z}{\partial z}\,\!$

The engineering shear strain, or the change in angle between two originally orthogonal material lines, in this case line $\overline {AC}\,\!$ and $\overline {AB}\,\!$, is defined as

$\gamma_{xy}= \alpha + \beta\,\!$

From the geometry of Figure 1 we have

$\tan \alpha=\frac{\dfrac{\partial u_y}{\partial x}dx}{dx+\dfrac{\partial u_x}{\partial x}dx}=\frac{\dfrac{\partial u_y}{\partial x}}{1+\dfrac{\partial u_x}{\partial x}} \quad , \qquad \tan \beta=\frac{\dfrac{\partial u_x}{\partial y}dy}{dy+\dfrac{\partial u_y}{\partial y}dy}=\frac{\dfrac{\partial u_x}{\partial y}}{1+\dfrac{\partial u_y}{\partial y}}\,\!$

For small rotations, i.e. $\alpha\,\!$ and $\beta\,\!$ are $\ll 1\,\!$ we have

$\tan \alpha \approx \alpha \quad , \qquad \tan \beta \approx \beta\,\!$

and, again, for small displacement gradients, we have

$\alpha=\frac{\partial u_y}{\partial x} \quad , \qquad \beta=\frac{\partial u_x}{\partial y}\,\!$

thus

$\gamma_{xy}= \alpha + \beta = \frac{\partial u_y}{\partial x} + \frac{\partial u_x}{\partial y}\,\!$

By interchanging $x\,\!$ and $y\,\!$ and $u_x\,\!$ and $u_y\,\!$, it can be shown that $\gamma_{xy} = \gamma_{yx}\,\!$

Similarly, for the $y\,\!$-$z\,\!$ and $x\,\!$-$z\,\!$ planes, we have

$\gamma_{yz}=\gamma_{zy} = \frac{\partial u_y}{\partial z} + \frac{\partial u_z}{\partial y} \quad , \qquad \gamma_{zx}=\gamma_{xz}= \frac{\partial u_z}{\partial x} + \frac{\partial u_x}{\partial z}\,\!$

It can be seen that the tensorial shear strain components of the infinitesimal strain tensor can then be expressed using the engineering strain definition, $\gamma\,\!$, as

$\left[\begin{matrix} \varepsilon_{xx} & \varepsilon_{xy} & \varepsilon_{xz} \ \varepsilon_{yx} & \varepsilon_{yy} & \varepsilon_{yz} \ \varepsilon_{zx} & \varepsilon_{zy} & \varepsilon_{zz} \ \end{matrix}\right] = \left[\begin{matrix} \varepsilon_{xx} & \gamma_{xy}/2 & \gamma_{xz}/2 \ \gamma_{yx}/2 & \varepsilon_{yy} & \gamma_{yz}/2 \ \gamma_{zx}/2 & \gamma_{zy}/2 & \varepsilon_{zz} \ \end{matrix}\right]\,\!$

### Physical interpretation of the infinitesimal strain tensor

From finite strain theory we have

$d\mathbf{x}^2 - d\mathbf{X}^2 = d\mathbf X \cdot 2\mathbf E \cdot d\mathbf X \quad\text{or}\quad (dx)^2 - (dX)^2 = 2E_{KL}\,dX_K\,dX_L\,\!$

For infinitesimal strains then we have

$d\mathbf{x}^2 - d\mathbf{X}^2 = d\mathbf X \cdot 2\mathbf \boldsymbol \varepsilon \cdot d\mathbf X \quad\text{or}\quad (dx)^2 - (dX)^2 = 2\varepsilon_{KL}\,dX_K\,dX_L\,\!$

Dividing by $(dX)^2\,\!$ we have

$\frac{dx-dX}{dX}\frac{dx+dX}{dX}=2\varepsilon_{ij}\frac{dX_i}{dX}\frac{dX_j}{dX}\,\!$

For small deformations we assume that $dx \approx dX\,\!$, thus the second term of the left hand side becomes: $\frac{dx+dX}{dX} \approx 2\,\!$.

Then we have

$\frac{dx-dX}{dX}=\varepsilon_{ij}N_iN_j = \mathbf N \cdot \boldsymbol \varepsilon \cdot \mathbf N\,\!$

where $N_i=\frac{dX_i}{dX}\,\!$, is the unit vector in the direction of $d\mathbf X\,\!$, and the left-hand-side expression is the normal strain $e_{(\mathbf N)}\,\!$ in the direction of $\mathbf N\,\!$. For the particular case of $\mathbf N\,\!$ in the $X_1\,\!$ direction, i.e. $\mathbf N=\mathbf I_1\,\!$, we have

$e_{(\mathbf I_1)}=\mathbf I_1 \cdot \boldsymbol \varepsilon \cdot \mathbf I_1=\varepsilon_{11}\,\!$

Similarly, for $\mathbf N=\mathbf I_2\,\!$ and $\mathbf N=\mathbf I_3\,\!$ we can find the normal strains $\varepsilon_{22}\,\!$ and $\varepsilon_{33}\,\!$, respectively. Therefore, the diagonal elements of the infinitesimal strain tensor are the normal strains in the coordinate directions.

### Volumetric strain

The dilatation (the relative variation of the volume) is the trace of the tensor:

$\delta=\frac{\Delta V}{V_0} = \varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33}\,\!$

Actually, if we consider a cube with an edge length a, it is a quasi-cube after the deformation (the variations of the angles do not change the volume) with the dimensions $a \cdot (1 + \varepsilon_{11}) \times a \cdot (1 + \varepsilon_{22}) \times a \cdot (1 + \varepsilon_{33})\,\!$ and V0 = a3, thus

$\frac{\Delta V}{V_0} = \frac{\left ( 1 + \varepsilon_{11} + \varepsilon_{22} + \varepsilon_{33} + \varepsilon_{11} \cdot \varepsilon_{22} + \varepsilon_{11} \cdot \varepsilon_{33}+ \varepsilon_{22} \cdot \varepsilon_{33} + \varepsilon_{11} \cdot \varepsilon_{22} \cdot \varepsilon_{33} \right ) \cdot a^3 - a^3}{a^3}\,\!$

as we consider small deformations,

$1 \gg \varepsilon_{ii} \gg \varepsilon_{ii} \cdot \varepsilon_{jj} \gg \varepsilon_{11} \cdot \varepsilon_{22} \cdot \varepsilon_{33} \,\!$

therefore the formula.

Real variation of volume (top) and the approximated one (bottom): the green drawing shows the estimated volume and the orange drawing the neglected volume

In case of pure shear, we can see that there is no change of the volume.

### Strain deviator tensor

The infinitesimal strain tensor $\varepsilon_{ij}\,\!$, similarly to the stress tensor, can be expressed as the sum of two other tensors:

1. a mean strain tensor or volumetric strain tensor or spherical strain tensor, $\varepsilon_M\delta_{ij}\,\!$, related to dilation or volume change; and
2. a deviatoric component called the strain deviator tensor, $\varepsilon'_{ij}\,\!$, related to distortion.
$\varepsilon_{ij}= \varepsilon'_{ij} + \varepsilon_M\delta_{ij}\,\!$

where $\varepsilon_M\,\!$ is the mean stress given by

$\varepsilon_M=\frac{\varepsilon_{kk}}{3}=\frac{\varepsilon_{11}+\varepsilon_{22}+\varepsilon_{33}}{3}=\tfrac{1}{3}I^e_1\,\!$

The deviatoric strain tensor can be obtained by subtracting the mean strain tensor from the infinitesimal strain tensor:

\begin{align} \ \varepsilon'_{ij} &= \varepsilon_{ij} - \frac{\varepsilon_{kk}}{3}\delta_{ij} \ \left[{\begin{matrix} \varepsilon'_{11} & \varepsilon'_{12} & \varepsilon'_{13} \ \varepsilon'_{21} & \varepsilon'_{22} & \varepsilon'_{23} \ \varepsilon'_{31} & \varepsilon'_{32} & \varepsilon'_{33} \ \end{matrix}}\right] &=\left[{\begin{matrix} \varepsilon_{11} & \varepsilon_{12} & \varepsilon_{13} \ \varepsilon_{21} & \varepsilon_{22} & \varepsilon_{23} \ \varepsilon_{31} & \varepsilon_{32} & \varepsilon_{33} \ \end{matrix}}\right]-\left[{\begin{matrix} \varepsilon_M & 0 & 0 \ 0 & \varepsilon_M & 0 \ 0 & 0 & \varepsilon_M \ \end{matrix}}\right] \ &=\left[{\begin{matrix} \varepsilon_{11}-\varepsilon_M & \varepsilon_{12} & \varepsilon_{13} \ \varepsilon_{21} & \varepsilon_{22}-\varepsilon_M & \varepsilon_{23} \ \varepsilon_{31} & \varepsilon_{32} & \varepsilon_{33}-\varepsilon_M \ \end{matrix}}\right] \ \end{align}\,\!

## Compatibility equations

For prescribed strain components $\varepsilon_{ij}\,\!$ the strain tensor equation $u_{i,j}+u_{j,i}= 2 \varepsilon_{ij}\,\!$ represents a system of six differential equations for the determination of three displacements components $u_i\,\!$, giving an over-determined system. Thus, a solution does not generally exist for an arbitrary choice of strain components. Therefore, some restrictions, named compatibility equations, are imposed upon the strain components. With the addition of the three compatibility equations the number of independent equations is reduced to three, matching the number of unknown displacement components. These constraints on the strain tensor were discovered by Saint-Venant, and are called the "Saint Venant compatibility equations".

The compatibility functions serve to assure a single-valued continuous displacement function $u_i\,\!$. If the elastic medium is visualized as a set of infinitesimal cubes in the unstrained state, after the medium is strained, an arbitrary strain tensor may not yield a situation in which the distorted cubes still fit together without overlapping.

In index notation, the compatibility equations are expressed as

$\varepsilon_{ij,km}+\varepsilon_{km,ij}-\varepsilon_{ik,jm}-\varepsilon_{jm,ik}=0\,\!$

## Special cases

### Plane strain

Plane strain state in a continuum.

In real engineering components, stress (and strain) are 3-D tensors but in prismatic structures such as a long metal billet, the length of the structure is much greater than the other two dimensions. The strains associated with length, i.e. the normal strain $\varepsilon_{33}\,\!$ and the shear strains $\varepsilon_{13}\,\!$ and $\varepsilon_{23}\,\!$ (if the length is the 3-direction) are constrained by nearby material and are small compared to the cross-sectional strains. The strain tensor can then be approximated by:

$\underline{\underline{\boldsymbol{\varepsilon}}} = \begin{bmatrix} \varepsilon_{11} & \varepsilon_{12} & 0 \ \varepsilon_{21} & \varepsilon_{22} & 0 \ 0 & 0 & 0\end{bmatrix}\,\!$

in which the double underline indicates a second order tensor. This strain state is called plane strain. The corresponding stress tensor is:

$\underline{\underline{\boldsymbol{\sigma}}} = \begin{bmatrix} \sigma_{11} & \sigma_{12} & 0 \ \sigma_{21} & \sigma_{22} & 0 \ 0 & 0 & \sigma_{33}\end{bmatrix}\,\!$

in which the non-zero $\sigma_{33}\,\!$ is needed to maintain the constraint $\epsilon_{33} = 0\,\!$. This stress term can be temporarily removed from the analysis to leave only the in-plane terms, effectively reducing the 3-D problem to a much simpler 2-D problem.

### Antiplane strain

Antiplane strain is another special state of strain that can occur in a body, for instance in a region close to s screw dislocation. The strain tensor for antiplane strain is given by

$\underline{\underline{\boldsymbol{\varepsilon}}} = \begin{bmatrix} 0 & 0 & \varepsilon_{13} \ 0 & 0 & \varepsilon_{23}\ \varepsilon_{13} & \varepsilon_{23} & 0\end{bmatrix}\,\!$

## Infinitesimal rotation tensor

The infinitesimal strain tensor is defined as

$\boldsymbol{\varepsilon} = \frac{1}{2} [\boldsymbol{\nabla}\mathbf{u} + (\boldsymbol{\nabla}\mathbf{u})^T]$

Therefore the displacement gradient can be expressed as

$\boldsymbol{\nabla}\mathbf{u} = \boldsymbol{\varepsilon} + \boldsymbol{\omega}$

where

$\boldsymbol{\omega} := \frac{1}{2} [\boldsymbol{\nabla}\mathbf{u} - (\boldsymbol{\nabla}\mathbf{u})^T]$

The quantity $\boldsymbol{\omega}$ is the infinitesimal rotation tensor. This tensor is skew symmetric. For infinitesimal deformations the scalar components of $\boldsymbol{\omega}$ satisfy the condition $|\omega_{ij}| \ll 1$. Note that the displacement gradient is small only if 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, $\mathbf{w}$, as follows

$\omega_{ij} = -e_{ijk}~w_k ~;~~ w_i = \tfrac{1}{2}~e_{ijk}~\omega_{jk}$

where eijk is the permutation symbol. In matrix form

$\underline{\underline{\boldsymbol{\omega}}} = \begin{bmatrix} 0 & -w_3 & w_2 \\ w_3 & 0 & -w_1 \\ -w_2 & w_1 & 0\end{bmatrix} ~;~~ \underline{\underline{\mathbf{w}}} = \begin{bmatrix} w_1 \\ w_2 \\ w_3 \end{bmatrix}$

The axial vector is also called the infinitesimal rotation vector. The rotation vector is related to the displacement gradient by the relation

$\mathbf{w} = \tfrac{1}{2}~\boldsymbol{\nabla}\times\mathbf{u}$

In index notation

$w_i = \tfrac{1}{2}~e_{ijk}~u_{k,j}$

If $\lVert\boldsymbol{\omega}\rVert \ll 1$ and $\boldsymbol{\varepsilon} = \boldsymbol{0}$ then the material undergoes an approximate rigid body rotation of magnitude $|\mathbf{w}|$ around the vector $\mathbf{w}$.

### Relation between the strain tensor and the rotation vector

Given a continuous, single-valued displacement field $\mathbf{u}$ and the corresponding infinitesimal strain tensor $\boldsymbol{\varepsilon}$, we have (see Tensor derivative (continuum mechanics))

$\boldsymbol{\nabla}\times\boldsymbol{\varepsilon} = e_{ijk}~\varepsilon_{lj,i}~\mathbf{e}_k\otimes\mathbf{e}_l = \tfrac{1}{2}~e_{ijk}~[u_{l,ji} + u_{j,li}]~\mathbf{e}_k\otimes\mathbf{e}_l$

Since a change in the order of differentiation does not change the result, $u_{l,ji} = u_{l,ij}\,$. Therefore

$\,e_{ijk} u_{l,ji} = (e_{12k}+e_{21k}) u_{l,12} + (e_{13k}+e_{31k}) u_{l,13} + (e_{23k} + e_{32k}) u_{l,32} = 0$

Also

$\tfrac{1}{2}~e_{ijk}~u_{j,li} = \left(\tfrac{1}{2}~e_{ijk}~u_{j,i}\right)_{,l} = \left(\tfrac{1}{2}~e_{kij}~u_{j,i}\right)_{,l} = w_{k,l}$

Hence

$\boldsymbol{\nabla}\times\boldsymbol{\varepsilon} = w_{k,l}~\mathbf{e}_k\otimes\mathbf{e}_l = \boldsymbol{\nabla}\mathbf{w}$

### Relation between rotation tensor and rotation vector

From an important identity regarding the curl of a tensor we know that for a continuous, single-valued displacement field $\mathbf{u}$,

$\boldsymbol{\nabla}\times(\boldsymbol{\nabla}\mathbf{u}) = \boldsymbol{0}$

Since $\boldsymbol{\nabla}\mathbf{u} = \boldsymbol{\varepsilon} + \boldsymbol{\omega}$ we have $\boldsymbol{\nabla}\times\boldsymbol{\omega} = -\boldsymbol{\nabla}\times\boldsymbol{\varepsilon} = - \boldsymbol{\nabla}\mathbf{w}$