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The strain tensor writes $\epsilon_{ij}=\frac{1}{2}\Big(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_i}{\partial x_j}\Big)$ with $u_i$ the displacement in the $i$ direction.

Then $\frac{\partial \epsilon_{ij}}{\partial t}=\frac{1}{2}\Big(\frac{\partial }{\partial x_j}\frac{ \partial u_i}{\partial t}+\frac{\partial }{\partial x_i}\frac{ \partial u_j}{\partial t}\Big)$.

Is it right to identify $\frac{ \partial u_i}{\partial t}$ with the velocity $v_{i}$ of the particles of the material, thus yielding the tensorial strain-velocity relation : $\frac{\partial \mathbf{\epsilon}}{\partial t}=\frac{1}{2}(\nabla \mathbf{v}+(\nabla \mathbf{v})^T)$ ?

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Actually your equation is pretty close. The difference in velocity between two neighboring material points in a deforming fluid is given by $$\Delta \mathbf{v}=(\nabla \mathbf{v})^T\centerdot d\mathbf{s}$$ where $d\mathbf{s}$ is the differential position vector between the two material points and $(\nabla \mathbf{v})^T$ is the transpose of the velocity gradient tensor. The rate of strain tensor (which factors out the effect of rotation of the fluid parcels) is given by: $$\mathbf{E}=\frac{[(\nabla \mathbf{v})+(\nabla \mathbf{v})^T]}{2}$$such that $$\frac{D(ds)^2}{Dt}=d\mathbf{s}\centerdot \mathbf{E} \centerdot d\mathbf{s}$$where D/Dt is the material derivative.

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  • $\begingroup$ Oh actually I saw I made a little mistake, I'm correcting it in the main text. I hope this time it's ok, I actually got that $E=\partial_t \epsilon$ and then the strain-velocity relation is written in many places. So it has to mean that $v_i=\partial_t u_i$ ! $\endgroup$ – J.A Nov 17 '18 at 18:28

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