text: Q:
  
  How normal shear stress vanish at the stagnant surface
  
  While I had always thought that it is so obvious that $\tau_ii$ (shear normal to x, y, or z planes) for a Newtonian, incompressible fluid is zero, now I find it pretty struggling to show it mathematically.
  Imagining a fluid occupying upper half of 3D space. Let's assume Cartesian coordinates for simplicity. Incompressibility says that $\sum(dV_i/di)=0$ , $i=x,y,z$. applying no penetration and no-slip also gives $V_i=0$ at $z=0$.
  I tried to somehow relate the incompressibility to the shear stresses so I looked at it by multiplying by $2\mu$ so I have $\sum\tau_ii=0$
  now we know that the sum of normal shears are zero. I can think of a fully developed flow so I can assume each of $dV_i/di=0$ so that each of $\tau_ii=0$, but I believe there is a way to show it for a far more general case.
  
  A:
  
  At a stationary surface the flow is locally a simple shear flow. Let $\mathbf{u}(\mathbf{x})$ be the velocity at position $\mathbf{x}$. If $\mathbf{x}_0$ is the position of a point on the surface then expanding the velocity about that point we get:
  $$\mathbf{u}(\mathbf{x}_0+\delta\mathbf{x})=\mathbf{u}(\mathbf{x}_0)+\delta\mathbf{x}\cdot\nabla\mathbf{u}|_{\mathbf{x}_0}+O(\delta\mathbf{x}^2)\\
  =\delta\mathbf{x}\cdot\nabla\mathbf{u}|_{\mathbf{x}_0}+O(\delta\mathbf{x}^2)$$
  because the surface is stationary. Neglecting terms of higher order, and since $\mathbf{u}|_{\mathbf{x}_0}$ is a constant tensor, we se that $\delta\mathbf{x}\cdot\nabla\mathbf{u}|_{\mathbf{x}_0}$ is the velocity field of a simple shear flow.
  You can convince yourself that normal stresses are zero in a simple shear flow. 
  
  