Returning to our molecular picture of shear stresses, we can think of the momentum flux (shear stress) as being related to the movement (swapping) of molecules across a plane. We might write this flux (for Z molecules that moved) as:

$\displaystyle{\tau = - \sum_{n=1}^Zm_n\left (v_x|_{y-}-v_x|_{y+}\right)}$

where the |_{y-} refers to the value at the location
*just below* our plane. If we relate this value (and the +
variety) to our slope of the velocity (shear rate or rate of
strain) by:

$\displaystyle{v_x|_{y-} = v_x|_{y_0}-\left (\frac{dv_x}{dy} \right)|_{y_o}\delta}$

$\displaystyle{v_x|_{y+} = v_x|_{y_0}+\left (\frac{dv_x}{dy} \right)|_{y_o}\delta}$

we can then combine these equations to give

$\displaystyle{\tau = 2 \sum_{n=1}^Zm_n\frac{dv_x}{dy}|_{y_o}\delta}$

Here, we can think of $\delta$ as the average distance between
molecular collisions. This can be related to a statistical
mechanics quantity, the *mean free path* as
$\delta=(2/3)\lambda$ so that our sum becomes (for molecules of
the same mass):

$\displaystyle{\tau = \frac{4}{3}\lambda Z m\frac{dv_x}{dy}|_{y_o}}$

which we can show gives us

$\displaystyle{\mu = \frac{4}{3}\lambda Z m}$

Z can be related to the molecular concentration N and the average random molecular velocity C as Z = NC/4. Using this and the statistical mechanics expressions for the hard-sphere mean free path and average velocity:

$\displaystyle{\lambda = \frac{1}{\sqrt{2}\pi N d^2}}$

$\displaystyle{C = \sqrt{\frac{8\kappa T}{\pi m}}}$

where d is the molecular diameter. Combining these expressions gives the hard sphere viscosity:

$\displaystyle{\mu = \frac{2}{3\pi^{3/2}}\frac{\sqrt{m\kappa T}}{d^2}}$

Explain the molecular origins of fluid viscosity and shear stresses