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1-4forces.tex.\\

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\section{Forces in the Fluid}
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There are two types of forces acting on a fluid element:
\begin{enumerate}
\item Short-range force (surface force): It is molecular in origin, and decreases
rapidly with the distance between interacting elements. This force  is appreciable
only if fluid elements are in contact, therefore  exists only on the boundary, and
is called    a "surface force."  
Surface force per unit area $\vec{\Sigma}$ is called the stress, which  depends on time, on
the location, $\vec{x}$, and on the orientation of the surface element, i.e., or its unit normal
$\vec{n}$.
\item Long-range force (volume force,    body force ):  The origin of the force is far
away from the zone of interest.  The strength of such forces varies very slowly,
and acts uniformly on all parts of a fluid parcel.  Therefore, the total force is
proportional to the volume of fluid. We define $\vec{f}$ to be the "body force"
acting on a unit volume, then the "body force" on a given mass $\rho dV$ is
$\rho\vec{f}dV$. A typical body force is  gravity.
\end{enumerate}

\subsection{Surface force and stresses}

Consider a cubic element, depicted in Figure \ref{fig4.1:stresses}.  On any surface element there are three components of
stresses $\vec{\Sigma}$.  We denote the component in the direction of $j$ acting on
the surface element whose normal is in $i$ direction by $\sigma_{ij}$, thus the stress
components form a square array: 
\be \lcb \sigma_{ij}\rcb= \lcb \begin{array}{ccc} 
  \sigma_{11} & \sigma_{12} & \sigma_{12} \\
   \sigma_{21} & \sigma_{22} & \sigma_{23} \\
   \sigma_{31} & \sigma_{32} & \sigma_{33} \end{array}\rcb  \ee
\begin{figure}\vspace{2in}\hspace{1in}
\includegraphics[scale=0.75]{f1-4-1.eps}
\caption{Stress components on a fluid element}
\label{fig4.1:stresses}
\end{figure}
The first subscript indicates
the direction of the   unit normal to the surface element, and the second subscript
indicates the direction of the stress component. The entire array $\{\sigma_{ij}\}$ is
called   the stress tensor  (see notes on Tensors).

The diagonal terms $\sigma_{ii} \quad i = 1,2,3,$ are the normal stress  components;
the off-diagonal terms $\sigma_{ij} \quad i \neq j \quad 1,2,3,$ are the shear stress
components. 

We shall first show that  the stress tensor is symmetric
\be \sigma_{ij} = \sigma_{ji}, \label{Eq:4.1}\ee
 Let  the length of each side of a cubic element be $\Delta \ell$.   Conservation of angular momentum
requires  :
\[ I\f{d\omega}{dt} =-\sigma_{12} \Delta \ell^2 \cdot \Delta \ell + \sigma_{21} \Delta
\ell^2 \Delta \ell \] 
From left to right, the  terms represent, respectively,   the  angular inertia, the 
torque due to shear stress on the two vertical surfaces and the torque due to shear on
two horizontal  surfaces.  Now the moment of inertia $I \propto \rho (\Delta \ell)^5$.  Hence, as $\Delta
\ell \rightarrow 0, \sigma_{12} = \sigma_{21}$ as long as $d\omega/dt \neq \infty$. 
After similar arguments for all other off-diagonal components, we prove 
(\ref{Eq:4.1}).  Thus among nine components, only six
can be distinct.
 
\subsection{Cauchy's theorem}
   Are $\sigma_{ij}$, defined on three mutually orthogonal surfaces in a chosen coordinate system,  capable of describing stresses on any surface?  In other words,
can the stress components on any surface be expressed in terms of $\sigma_{ij}$ ? To answer this question let us consider a
tetrahedron shown in Figure 1.4.2. Three  sides are formed by  orthogonal surface elements $dS_1 = BCO, dS_2 = ACO, dS_3 = ABO$ whose unit normals are $-\vec e_1, -\vec e_2,-\vec e_3,$ respectively. The fourth side $dS = ABC$ is inclined with the unit normal $\vec n=(n_1, n_2,n_3)$ pointing in an arbitrary direction.
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\begin{figure}[h]
\vspace{2.2in}\hspace{2in}
\includegraphics[scale=1]{f1-4-2.eps}
\caption{Stress components on any   surface element}
\label{fig4.2:tetrahedron}
\end{figure}
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First, let's show that
\begin{eqnarray*}
  dS_1 &=& n_1dS = \cos \lp \vec{n},\vec{e}_1 \rp dS \\ dS_2 &=& n_2dS = \cos \lp
\vec{n},\vec{e}_2 \rp dS \\ dS_3 &=& n_3dS = \cos \lp \vec{n},\vec{e}_3 \rp dS 
 \end{eqnarray*} In other words 
\be   dS_i = n_idS, \quad i = 1,2,3. \ee 
Consider an {\em  arbitrary constant} vector $\vec{A}$.  By Gauss' theorem
\[ \3int_V \nabla \cdot \vec{A}dS = \2int_S \vec{A}\cdot \vec n d  {S} , \] where
$V  
$ is the  volume of the tetrahedron and $S   $ the total surface of tetrahedron.  Since $\vec{A}$ is a
constant vector, $\nabla \cdot \vec{A} = 0.$
\[ 0 = \2int_S \vec{A} \cdot\vec n  d{S}   =
\vec{A}
\cdot \2int_S \vec{n}dS = 0 . \] Since $\vec{A}$ is arbitrary
\[ \2int_S \vec{n} dS = 0 =  \vec{n} dS - \vec{e}_1 dS_1 - \vec{e}_2 dS_2 -
\vec{e}_3 dS_3 . \] Hence
\begin{eqnarray*}
\vec{n} \cdot \vec{e}_1 dS & = & n_1dS = dS_1 \\
\vec{n} \cdot \vec{e}_2 dS & = & n_2dS = dS_2 \\
\vec{n} \cdot \vec{e}_3 dS & = & n_3dS = dS_3 
\end{eqnarray*}
Note that the  volume of the tetrahedron is $ \f{1}{3}h dS$ where $h$ is the
vertical distance from the origin to the surface $dS$.


Next we consider the force balance in the $x_1$ direction, i.e., $F_x = ma_x$: 
\be   -\sigma_{11}n_1dS - \sigma_{21}n_2dS - \sigma_{31}n_3dS   + \Sigma_1 dS + \rho f_x
\f{h}{3} dS =\rho \f{h}{3} dS \f{dq_1}{dt} . 
\ee where $f_x$ is the $x$ component of the body force per unit volume. As the
tetrahedron shrinks to a point $h \to 0$, the last two terms  
diminish much faster than the rest by a factor $h$ for any finite $f_i$ and  $dq_1/dt$, $i=1,2,3$, hence,
\begin{eqnarray*}
\Sigma_1 & = & \sigma_{11}n_1 + \sigma_{21}n_2 + \sigma_{31}n_3 \\ & = & \sigma_{11}n_1 +
\sigma_{12}n_2 + \sigma_{13}n_3 \qquad (\sigma_{21}=\sigma_{12}) \\ & = & \sigma_{1j}n_j .
\end{eqnarray*} Similar results are obtained by  considering the force balance in
two other  directions. In summary, we have
\be \Sigma_i = \sigma_{ij}n_j \label{eq:Cauchy}\ee 
which states that the stress $\vec \Sigma$ on a surface element of any orientation is a linear superposition of the stress components defined on a cube in some Cartesian coordinate system. This result is  called Cauchy's formula.   


Note that  $\Sigma_i$ is  tensor of rank 1  and $\vec n=\lcb n_j\rcb$ is an arbitrary tensor of rank 1 (because the surface
$dS$ is arbitrary). Equation (\ref{eq:Cauchy}) states that   the scalar product of  the array of numbers $\sigma_{ij}$    with an arbitrary  tensor $n_j$  of  rank 1 is another tensor   $\Sigma_i$ of rank 1.  By the quotient law the array $\lcb \sigma_{ij} \rcb$ must be  a   tensor of rank two.


\end{document}