[Physics] shear and normal stresses

fluid dynamics

When we talk about shear stresses in a fluid, we find that the shear stress is given by $$\tau_{xy} = \mu(\partial_y u + \partial_x v) = \tau_{yx}$$ This relation we get when only looking at one side of our fluid-"cube". Now, in order to take into account the opposite side we assume that the fluid element is so small that the shear stress is constant, leading to the average $$\tau_{xy} = \frac{1}{2}2\mu(\partial_y u + \partial_x v) = \mu(\partial_y u + \partial_x v) = \tau_{yx}$$ Applying the same logic to the normal stresses gives me $$\tau_{xx} = \frac{1}{2}\mu(\partial_x u + \partial_x u) = \mu(\partial_x u)$$ However, in my textbook (White) it is given as
$$\tau_{xx} = 2\mu(\partial_x u)$$ Where does this extra factor of 2 come from in the normal stress?

Best Answer

The definition of the stress tensor is (in Einstein summation notation):

$$ \tau_{ij} = \mu \left(\frac{\partial u_i}{\partial x_j}+\frac{\partial u_j}{\partial x_i} \right)$$

So, if you look at $\tau_{ii}$, you get $2\frac{\partial u_i}{\partial x_i}$. That's really where the factors come from, not from "averaging" over any fluid element or anything like that. It's just due to the symmetry of the tensor.

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