From 8f883023e6354648727479aec029f418b30ef2dc Mon Sep 17 00:00:00 2001
From: Prefetch
Date: Thu, 15 Apr 2021 20:53:17 +0200
Subject: Expand knowledge base

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+---
+title: "Viscosity"
+firstLetter: "V"
+publishDate: 2021-04-12
+categories:
+- Physics
+- Fluid mechanics
+- Fluid dynamics
+
+date: 2021-04-12T13:14:16+02:00
+draft: false
+markup: pandoc
+---
+
+# 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](/know/concept/cauchy-stress-tensor/)).
+Formally, the **dynamic viscosity** $\eta$ is defined as follows:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma_{xz}
+        = \eta \dv{v}{z}
+    }
+\end{aligned}$$
+
+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:
+
+$$\begin{aligned}
+    \sigma_{ij} = - p \delta_{ij}
+\end{aligned}$$
+
+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 $\nabla_i \!=\! \pdv*{x_i}$:
+
+$$\begin{aligned}
+    \boxed{
+        \sigma_{ij}
+        = - p \delta_{ij} + \eta (\nabla_i v_j + \nabla_j v_i)
+    }
+\end{aligned}$$
+
+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 $\rho$:
+
+$$\begin{aligned}
+    \boxed{
+        \nu
+        \equiv \frac{\eta}{\rho}
+    }
+\end{aligned}$$
+
+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
+1.  B. Lautrup,
+    *Physics of continuous matter: exotic and everyday phenomena in the macroscopic world*, 2nd edition,
+    CRC Press.
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