# Relation between strain and velocity

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 : $$\frac{\partial \mathbf{\epsilon}}{\partial t}=\frac{1}{2}(\nabla \mathbf{v}+(\nabla \mathbf{v})^T)$$ ?

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.
• 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 $v_i=\partial_t u_i$ ! – J.A Nov 17 '18 at 18:28