\documentclass[12pt,a4paper]{article}
\usepackage{amsbsy}
\textwidth=6.25truein
\textheight=9.65truein
\voffset=-1truein
\hoffset=-0.5truein


\newcommand{\bx}{{\bf x}}
\newcommand{\bu}{{\bf u}}
\newcommand{\bn}{{\bf n}}
\newcommand{\qword}[1]{\quad\hbox{\rm #1}\quad}
\newcommand{\cd}{\!\cdot\!}
\newcommand{\pd}[2]{{\partial #1\over\partial #2}}
\newcommand{\sq}[1]{\;#1\;}
\newcommand{\Frac}[2]{{\textstyle{#1\over#2}}}

\setcounter{section}{2}
\setcounter{subsection}{2}

\begin{document}
\subsection{Stress tensor}
\subsubsection{Tractions}

\large
In continuum mechanics, there are two types of force.
\begin{enumerate}
\item[(i)]{{\sl Volume\/} or body forces, which are long ranged, 
and which are
proportional to the volume of the fluid element, e.g.~gravity.}
\item[(ii)]{{\sl Surface tractions\/}, which are short ranged, and which are
proportional to the surface area of the fluid element, e.g.~pressure and
viscous friction forces.}
The force exerted by the side into which $\bn$ points is
\end{enumerate}
\[
 \boldsymbol{\tau}(\bx,t;\bn)\delta S .
\]

\subsubsection{Tensor}

\parbox{0.65\linewidth}{
Take $\bn \delta S$ to be the inclined surface of a small
tetrahedron, whose other surfaces coincide with the coordinate planes.
Consider
the force balance on this small tetrahedron. 
The volume and acceleration forces
are $O(\rho g L^3)$, where $L$ is the linear size of the tetrahedron.
The surface forces are $O(pL^2)$, with typical pressure $p=O(\rho g H)$,
where $H$ is the height of the atmosphere. 
Hence for small tetrahedra $L \ll H$, the surface forces are $O(H/L)$ larger
than the volume and acceleration forces, and so must balance amongst
themselves, i.e.}
\[
 \boldsymbol\tau(\bn)\delta S \;+\;  \boldsymbol\tau({\bf -1})\delta S_1 \;+\; 
   \boldsymbol\tau({\bf -2})\delta S_2 \;+\; 
    \boldsymbol\tau({\bf -3})\delta S_3 \;=\; 0.
\]
But by the geometry of the tetrahedron
\[
  \delta S_i = n_i \delta S.
\]
Hence the surface traction is linear in the direction $\bn$
\[
 \tau_i(\bx,t;\bn) = \sigma_{ij}(\bx,t) n_j ,
\]
where the second-order {\sl stress tensor\/} is
$\sigma_{ij} = \tau_i({\bf j})$ is the $i$th component of force exerted
across a unit area in the $j$th direction by the $+$ve side.

\subsubsection{Symmetry}

Consider the angular momentum balance on a small volume $V$, taking moments
about a fixed point in $V$.
The moment of the surface forces is
\[
 \int \epsilon_{ijk}x_j\sigma_{kl} n_l \, dS ,
\]
which is $O(pL^3)$.
But the moments of the volume and acceleration forces are $O(\rho g L^4)$.
So again the surface forces must balance amongst themselves, i.e.
\[
0\quad = \qword{above} \stackrel{\hbox{\rm Div Thm}}= \quad
 \int_V {\partial\over\partial x_l}
    \left(\epsilon_{ijk}x_j\sigma_{kl}\right) \,dV
 \quad=\quad \int_V \epsilon_{ijk}
    \left(\sigma_{kj} + x_j {\partial\sigma_{kl}\over\partial x_l}
    \right) \, dV .
\]
The second term in the last integral is $O(\rho g L^4)$ and so can be ignored
for small tetrahedra. Hence
\[
 \epsilon_{ijk}\sigma_{kj} = 0 \qword{or} \sigma_{ij} = \sigma_{ji}
\]

\pagebreak
\subsection{Constitutive equation for a Newtonian fluid}

In an incompressible ($\nabla\cd\bu=0$) viscous fluid, the stress
tensor has two parts.
\begin{enumerate}
\item[(i)]{an isotropic {\sl pressure\/} term, as in IB's inviscid
fluids, which produces a traction normal to the surface with magnitude
independent of the direction of the surface.}
\item[(ii)]{a frictional part due to neighbouring fluid sliding past
 -- called the {\sl deviatoric\/} part.} 
\end{enumerate}
\[
 \sigma_{ij} = -p\delta_{ij} + \sigma^{\rm dev}_{ij}\left(\pd{u_k}{x_l}\right),
\qword{with} \sigma^{\rm dev}_{ii} = 0 .
\]
One can use the first term in the Taylor series, $\nabla\bu$, if 
$\ell_{\rm micro} \ll \ell_{\rm lab}$.

\bigskip
For common viscous liquids, we can make two further assumptions.
\begin{enumerate}
\item[(a)]{$\sigma^{\rm dev}$ is linear and instantaneous in 
$\partial u_k/\partial x_l$, which is good if $t_{\rm micro} \ll t_{\rm lab}$.
Thus
\[
 \sigma^{\rm dev} = A_{ijkl}\pd{u_k}{x_l}
\]
}
\item[(b)]{The fluid is isotropic. The general fourth-order isotropic tensor is
\[ 
A_{ijkl} = \mu'\delta_{ij}\delta_{kl} + \mu''\delta_{ik}\delta_{jl} + 
   \mu'''\delta_{il}\delta_{kj} .
\]
}
\end{enumerate}
Thus
\[
 \sigma^{\rm dev}_{ij} \sq= \mu'\delta_{ij}\pd{u_k}{x_k}
 \sq+ \mu''\pd{u_i}{x_j} \sq+ \mu'''\pd{u_j}{x_i} .
\]
But $\nabla\cd\bu = 0$ and $\sigma_{ij}$ is symmetric $i\leftrightarrow j$, so
$\mu'' = \mu''' = \mu$, say. Hence the expression for the stress tensor in a
Newtonian viscous fluid is
\[
 \sigma_{ij} = - p\delta_{ij} + \mu\left(\pd{u_i}{x_j} + \pd{u_j}{x_i}\right),
\]
or
\[
 \fbox{$ \sigma_{ij} = - p\delta_{ij} + 2\mu e_{ij}$ },
\]
using the strain-rate tensor of \S2.2.

Note that the stress does not dependent of the vorticity, as expected a solid
body rotation does not stress the fluid, stretching is instead required.


\end{document}
\bye