Categories: Fluid dynamics, Fluid mechanics, Physics.

# Viscosity

The **viscosity** of a fluid describes how
“sticky” its constituent molecules are;
when one part of the fluid moves, it “drags”
neighbouring parts by an amount proportional to the viscosity.

Imagine a liquid in a canal, flowing in the $x$-direction at a velocity $v(z)$ as a function of depth $z$. Due to the liquid’s viscosity, its molecules are “stuck” to the bottom of the canal $z = 0$, such that it is stationary there $v(0) = 0$. However, at the surface $z = z_s$, there is a flow at $v(z_s) = v_s$.

This difference in $v$ means that there is a velocity gradient across $z$.
Each infinitesimal layer of the liquid
is dragging on the layers above and below it,
meaning there is a nonzero shear stress $\sigma_{xz}$
(see Cauchy stress tensor).
Formally, the **dynamic viscosity** $\eta$ is defined as follows:

This is **Newton’s law of viscosity**,
and fluids obeying it are known as **Newtonian**.
In a Newtonian fluid *at rest*, there are no such shear stresses,
and the Cauchy stress tensor $\hat{\sigma}$ is diagonal:

Where $p$ is the pressure, and $\delta_{ij}$ is the Kronecker delta. If the fluid flows according to a velocity field $\va{v}$, then a more general definition of $\eta$ is as follows, in index notation with $\nabla_i \!=\! \ipdv{}{x_i}$:

$\begin{aligned} \boxed{ \sigma_{ij} = - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i) } \end{aligned}$The double term $\nabla_i v_j + \nabla_j v_i$ comes from the fact that the stress tensor of a Newtonian fluid is always symmetric; this definition of $\sigma_{ij}$ enforces that.

Another quantity is the **kinematic viscosity** $\nu$,
which is simply $\eta$ divided by the density $\rho$:

With this, Newton’s law of viscosity is written using the momentum density $P = \rho v$:

$\begin{aligned} \sigma_{xz} = \nu \dv{P}{z} \end{aligned}$Because momentum is “more fundamental” than velocity, is $\nu$ often more useful than $\eta$. However, this comes at the cost of our intuition: for example, as you would expect, $\eta_\mathrm{water} > \eta_\mathrm{air}$, but you may be surprised that $\nu_\mathrm{water} < \nu_\mathrm{air}$. Since air is less dense, it is easier to set in motion, hence we expect it to be less viscous than water, but in fact air’s molecules are stickier than water’s.

## References

- B. Lautrup,
*Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition, CRC Press.