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$$\epsilon=\frac{\nabla u+ \nabla u^T}{2},$$

$u$ is vector displacement, and $\nabla u$ is the gradient matrix of $u$.

Now for a Newtonian, incompressible fluid, this describes the shear stress forces on a surface element. the problem is, I can see why $\nabla u$ is needed, the gradient in velocity of the direction parallel to the force creates a drag force on surface element due to viscosity.

But why is ${\nabla u^T}$ needed? As far as i can tell there is no physical intuition for this term?

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  • $\begingroup$ the tensor $\epsilon$ is symmetrical $\endgroup$
    – hyportnex
    Commented Jul 24, 2022 at 0:21
  • $\begingroup$ I am aware. but this is ad hoc from angular momentum. and does not give any physical intuition to the second term regarding forces and shear $\endgroup$
    – mr chap
    Commented Jul 24, 2022 at 0:22
  • $\begingroup$ one man's ad hoc is another man's propter hoc... $\endgroup$
    – hyportnex
    Commented Jul 24, 2022 at 0:29
  • $\begingroup$ how does a sideways velocity gradient for upwards velocity, through a surface element with normal vector upwards, pull the element sideways? $\endgroup$
    – mr chap
    Commented Jul 24, 2022 at 0:39
  • $\begingroup$ Does this answer your question? Viscous stress tensor for an incompressible Newtonian fluid $\endgroup$
    – Chemomechanics
    Commented Jul 24, 2022 at 21:55

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