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
EDIT UPDATE:
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.
$\frac{\partial u_i}{\partial u_j}$
) for the strain tensor? $\endgroup$