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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}