How can i proove that the traceless part of linear strain tensor $e$ in the Euler description:

$$e_{i,j}={ 1 \over 2 } \left({ \partial u_i \over \partial x_j}+{ \partial u_j \over \partial x_i} \right)$$

is alway a pure shear deformation i.e. it does conserve volume. In case it is not clear this is the Euler strain tensor with the assumption $ { \partial u_i \over \partial x_j} << 1$, which means the part $\sum \limits_k{ \partial u_k \over \partial x_j} { \partial u_k \over \partial x_i}$ is neglected.

This apperently always has to hold when decomposed into traceless part $ e^t$ and a generally not traceless part $e^l$:

$$u= \underbrace{{ 1 \over 3 } Tr[e] \mathbb{I}}_{e^l} +\left( \underbrace{e -{ 1 \over 3 } Tr[e] \mathbb{I}}_{e^t} \right) $$

Even just the direction of how this can be done would be appreciated.

EDIT NOTE: the notation in my script is confusing so i changed it


One hint might be that if we look at small local deformations the strain tensor can probably be approximated as equal in all directions in first order. So we can write $e= \mathbb{I} \epsilon+...$. The first term must be nonzero for small deformations, and so the trace in that order does not vanish, but ist there a better more general Argument? This seems very hand waving.

  • $\begingroup$ Do you mean $\frac{\partial u_i}{\partial u_j}$ ($\frac{\partial u_i}{\partial u_j}$) for the strain tensor? $\endgroup$ – Emilio Pisanty Feb 23 '15 at 14:20
  • $\begingroup$ After realizing the post was really full of mistakes i hope it is corrected now. $\endgroup$ – pindakaas Feb 23 '15 at 14:46

Consider a volume element $dV$ at a certain point $\vec{x}$. Let the strain tensor at it be given by \begin{equation} e_{ij} = \frac{\partial u_i}{\partial x_j} + \frac{\partial u_j}{\partial x_i} \end{equation} Let us diagonalize the strain tensor at this point and let its diagonal entries be $u^{(i)}$. Since the trace of tensor is invariant, \begin{equation} e_{ii} = u^{(1)} + u^{(2)} + u^{(3)} \end{equation} In terms of the diagonalized strain tensor, the volume element $dV = dx_1 dx_2 dx_3$ is deformed as $dV^\prime = dx_1(1 + u^{(1)})dx_2(1 + u^{(2)})dx_3(1 + u^{(3)})$. Ignoring higher than linear terms, $dV^\prime - dV = u^{(1)} + u^{(2)} + u^{(3)}$ Clearly, $dV = dV^\prime$ means that $u^{(1)} + u^{(2)} + u^{(3)}$ is zero, which is equivalent to vanishing of trace of the strain tensor.

  • $\begingroup$ I don't see why this would not make sense. We already assumed very small deformations i.e. ${ \partial u_i \over \partial x_j} << 1$ so we should ignore the higher order terms.(And the tensor can always be diagonalized since it is symmetric and real). Reading your answer brightened my day a litte ^^ $\endgroup$ – pindakaas May 9 '15 at 13:36

this website may help you I fonund this on Google books


  • 1
    $\begingroup$ As a general rule Stack Exchange sites tend to consider link-only answers as being rather poor answers. They are fragile in the event of link rot and don't offer anything in the way of annotation. Why is that link good? What does it cover and how does it go about it? $\endgroup$ – dmckee --- ex-moderator kitten Oct 29 '15 at 3:52

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