$$\ \\1/2\left( \nabla\mathbf{u} + (\nabla\mathbf{u}) ^\mathrm{T} \right) \ $$ Why is the strain rate tensor the equation above?


Consider a general displacement of a continuum body. Let's have coordinates of points as $\mathbf x(\mathbf X, t)$, expressed in terms of initial position $\mathbf X$ and time $t$.

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You can represent a general displacement as a motion of a rigid body (translation + rotation) + deformation (strain).

The displacement vector can be written as $\mathbf u(\mathbf X, t) = \mathbf x(\mathbf X,t) - \mathbf X$. The displacement tensor (how displacement changes in space) as $D=\nabla_{\mathbf X}\mathbf u=\nabla_{\mathbf X}\mathbf x-I$. Note, that displacement tensor consist only of rotation tensor and strain tensor, since the translation vector is constant for all points of the body and thus has zero gradient.

The displacement rate will become $\partial_tD=\partial_t\nabla_{\mathbf X}\mathbf x=\nabla_{\mathbf X} \mathbf v$, where $\mathbf v=\partial_t \mathbf x$ is a velocity field. You can represent it as: $$ \nabla_{\mathbf X}\mathbf v=\frac{\nabla_{\mathbf X}\mathbf v-(\nabla_{\mathbf X}\mathbf v)^T}2 + \frac{\nabla_{\mathbf X}\mathbf v+(\nabla_{\mathbf X}\mathbf v)^T}2. $$

The first term here is an antisymmetric tensor which is known to be the time derivative of a rotation matrix. The second term is a symmetric strain tensor. Finally, since we are interested in strain tensor in global coordinate frame (not some a frame associated with body), we select $\mathbf X=\mathbf x$ at the current time and can drop index in nabla: $$ S=\frac12\Big(\nabla \mathbf v+(\nabla\mathbf v)^T\Big) $$

  • $\begingroup$ It might be worth mentioning that the vorticity tensor (rotation tensor) does not contribute to the strain of the fluid. $\endgroup$ – Chet Miller Oct 11 '19 at 11:42

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