\input{aefm-macro}

1-5momem.tex,  
\setcounter{chapter}{1}
  
\setcounter{section}{4}
\section{Law of Momentum Conservation}
\setcounter{equation}{0}
 Let us apply Newton's second law   to a volume containing the same fluid :
\be  
\f{D}{Dt} \3int_{V} \rho \vec{q}dV =  \2int_{S} \vec{\Sigma} \, dS \; +
  \3int_{V} \rho \vec{f}dV   \label{Eq:5.1} \ee
 where the successive terms represent, from left to right, 
the rate of change of fluid momentum in the volume, the total surface force acting on the fluid,  and the total body force on the fluid.

Using the kinematic transport theorem, the left-hand side becomes
\[ \f{D}{Dt} \3int_V \rho \vec{q}dV = \3int_V \rho \f{D\vec{q}}{Dt} dV ,
\] 
In  index notation, (\ref{Eq:5.1}) can be written
\begin{eqnarray} \3int_V \rho \f {Dq_i}{Dt} dV & = & \2int \Sigma_i dS +
\3int \rho f_i dV \nonumber\\ & = & \2int_S \sigma_{ij}n_j dS + \3int_V
\rho f_i dV . \label{Eq:5.2}
\end{eqnarray} after using Cauchy's formula. 


For a fixed $i$, $\sigma_{ij}$ are three components of a vector.  By Gauss'
theorem of divergence, the surface integral can be turned to a volume
integral:
\[ \2int_S \sigma_{ij}n_j dS = \3int_V \f{\p \sigma_{ij}}{\p x_j} dV . \]
Now  (\ref{Eq:5.2}) can be written
\[ \3int_V dV \lp \rho \f{Dq_i}{Dt} - \f{\p \sigma_{ij}}{\p x_j} - \rho f_i
\rp = 0 . \] Because $V$ is an arbitrary material volume, the integrand
must be zero everywhere
\be 
\rho \f{Dq_i}{Dt} = \f{\p \sigma_{ij}}{\p x_j} + \rho f_i \qquad  i = 1,2,3
. \label{Eq:5.3} \ee 
We now have 10 unknowns : one of $\rho$,  three of $q_i $ and six of
   $ \sigma_{ij} $. Equations {(1.3.4)} and    (\ref{Eq:5.3})
constitute only four scalar equations; six more equations are needed. 

In the first part of this course we shall restrict to  isothermal incompressible
fluids only. 
We then have the {\em equation of state}\footnote{If thermal effects are important, the equation of state is relation among $\rho$, the fluid pressure, and   the fluid temperature which would be  an additional dynamical quantity. This will discussed later. }
\be \f{D\rho}{Dt} =  \f{\p\rho}{\p t} +\bfq\cdot\nabla \rho= 0   \label{Eq:5.4} \ee 
It follows from mass
conservation  that
\be \nabla \cdot \vec{q} = 0 . \label{Eq:5.5} \ee 
Four more conditions are still needed.





\end{document}