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1-3trans-thm.tex, 
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\section{Kinematic transport theorem  }
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In particle mechanics, Newton's second law  requires the balance between the rate of change of particle momentum and the applied forces. In continuum mechanics, one needs  to calculate the rate of change in a volume of fluid.  Now different choices can be made for the control volume, such as a   geometric volume fixed in space, or moving in a prescribed manner, or a material volume consisting of the same body of fluid. Calculation of the rate can be facilitated by the so-called kinematic transport theorem. We  first derive a theorem for any moving volume $V(t)$ bounded by $S(t)$, see Figure {\ref{fig3.1:kin-trans}. 
\newtheorem{theorem}{Theorem}
\begin{theorem}
  Let $G({\bfx}, t)$  be some fluid property per unit volume, then
\be \f{d}{dt}\3int_V G\, dV = \3int_V \f{\p G}{\p t} \, dV + \2int_S GU_n \, dS
\label{eq:ktt}\ee where
$U_n$ is the normal component of the velocity of a point on $S$.
\end{theorem}
Note first that  
\begin{eqnarray*}\lefteqn{
 \lb\3int_V G(\bfx,t)\, dV\rb_{t+dt} = \3int_{V(t+dt)} G(\bfx, t+dt)\, dV }\\ &&  =
\3int_{V(t+dt)} \lb G(\bfx,t) + \f{\p G}{\p t} dt + O(dt)^2 \rb dV \end{eqnarray*} Due to the
movement of $S(t)$ the volume change for a    surface element  $dS$ is $U_n\,dt\,dS$, thus
\[ \3int_{V(t+dt)} = \3int_{V(t)} +\3int_{\Delta V} = \3int_{V(t)} +\2int_{S(t)} U_n\, dS \,dt 
\] We then have
\[ \lb\3int_V G(\bfx,t)\, dV\rb_{t+dt} = \3int_{V(t)}  \lb G(x,t) + \f{\p G}{\p t} dt +
O(dt)^2\rb dV \]
\[+\2int_{S(t)} U_n\, dS\, dt \lb G(x,t) + \f{\p G}{\p t} dt + O(dt)^2 \rb dV \]
\[=\3int_{V(t)}  G(x,t)dV + \lb \3int_{V(t)}\f{\p G}{\p t} dV + \2int_{S(t)} U_n\, dS \,G(x,t)    \rb dt+O(dt)^2
\]
\begin{figure}\vspace{2in}\hspace{2in}
\includegraphics[scale=0.75]{f1-3-1.eps}
\caption{A moving volume in the fluid}
\label{fig3.1:kin-trans}
\end{figure}
Since 
\[\f{d}{dt}\3int_V G\, dV = \lim_{dt\to 0}\f{1}{dt}\lcb\lb \3int_V G(\bfx,t)\, dV\rb_{t+dt} - \3int_{V(t)}    G(x,t) dt\rcb
 \]
(\ref{eq:ktt}) is proven. 

If  $V(t)$ is a material volume containing the  same moving fluid particles, then, $U_n = \bfq\cdot \bfn$ and $d/dt$ is the material derivative. We then have as a corollary,   the following
\begin{theorem}
If   $V(t)$ is  a material volume,  \be \f{D}{Dt}\3int_V G\, dV = \3int_V \f{\p G}{\p t} \, dV + \2int_S G\bfq \cdot \bfn \, dS
\label{kintrans}\ee
\end{theorem}
 This is the {\em kinematic transport theorem}.

Let us apply   this theorem to derive certain differential conservation laws.
 
\subsection{ Transport of Mass }

 Let $G=\rho$ = density and $V(t)$ be a material volume within which there are no mass sources or sinks, then by mass conservation, 
\[ \f{D}{D t} \3int_{V(t)} \rho dV = 0\] by mass conservation. Using (\ref{kintrans}) we get
\be  \3int_V \f{\p \rho}{\p t} + \2int_S \rho \bfq \cdot \bfn dS = \3int_V \lp
\f{\p \rho }{\p t} + \nabla \cdot (\rho \bfq)
\rp dV = 0 \ee

 Because $V$ is arbitrary the integrand must vanish identically : 
\be  \f{\p \rho }{\p t} + \nabla \cdot (\rho \bfq) 
 = 0 \ee
 This is the differential form of mass conservation law, valid at any  point inside the fluid. As an alternate form we may write
\be  \f{\p \rho }{\p t} + \bfq \cdot \nabla\rho + \rho \nabla \cdot  \bfq= \f{D \rho}{D t} +
\rho \nabla \cdot  \bfq= 0 \label{diffmass}\ee
For the special case of an incompressible but non homogeneous fluid: 
\be \f{D\rho }{D t} = \f{\p \rho }{\p t } + \bfq \cdot \nabla \rho = 0 \label{incompressible}\ee It follows that \be
\nabla \cdot\bfq = 0\label{zerodiv} \ee 
which is also known as the continuity equation.   If 
the incompressible fluid is also  homogeneous, then   (\ref{zerodiv}) holds and
\be \rho = \mbox{constant} \ee
replaces (\ref{incompressible}).
Do avoid the common error of assuming  incompressible fluid as being homogeneous. 

\subsection{ Transport of any dynamical property}
 Let $G= \rho F$ where $F$ is some fluid-dynamic
property per unit mass, and
$V$ is a material volume, then,
\be \f{D}{D t} \3int_V \rho F dV = \3int_V \rho \f{D F}{D t}dV \ee

  Thus, the total derivative can pass through the integral sign and  $\rho$. From
(\ref{kintrans})
\begin{eqnarray*}
\f{D}{D t} \3int_V\rho F dV &=& 
\3int_V \f{\p \rho F}{\p t} dV + \2int_S \rho F \bfq \cdot \bfn dS \\ &=& \3int_V \lp
\f{\p \rho}{\p t} F + \rho \f{\p F}{\p t} + \nabla \cdot (\rho \bfq F) \rp dV \\ &=&
\3int_V  \lcb F\lb\f{\p \rho }{\p t} + \nabla \cdot (\rho \bfq)\rb + \rho \lb \f{\p F}{\p t}
+ \bfq \cdot  \nabla F \rb\rcb dV \\ &=& \3int_V \rho \f{D F}{D t}dV \end{eqnarray*} after
using the law of mass conservation (\ref{diffmass}). 



As a special case   the rate of  momentum transport   in  a fluid volume $V$ is
found by taking
$G =\rho
\bfq$,  then, 
\be \f{D }{D t} \3int_V \rho \bfq dV = \3int_V \rho \f{D \bfq}{D t} dV \ee which will be
used later. 



 

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