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When we talk about shear stresses in a fluid, we find that the shear stress is given by

[tex]

\tau_{xy} = \mu(\partial_y u + \partial_x v) = \tau_{yx}

[/tex]

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

[tex]

\tau_{xy} = \frac{1}{2}2\mu(\partial_y u + \partial_x v) = \mu(\partial_y u + \partial_x v) = \tau_{yx}

[/tex]

Applying the same logic to the normal stresses gives me

[tex]

\tau_{xx} = \frac{1}{2}\mu(\partial_x u + \partial_x u) = \mu(\partial_x u)

[/tex]

However, in my textbook (White) it is given as

[tex]

\tau_{xx} = 2\mu(\partial_x u)

[/tex]

Where does this extra factor of 2 come from in the normal stress?

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# Deriving normal and shear stresses

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