Equation of Motion

Figure A.1: Movement of fluid element
\begin{figure}\centering\leavevmode
\epsfxsize =.45\columnwidth \epsfbox{eps/fluid_element.ps}\end{figure}

Consider a fluid element with mass $m$ as in A.1. From the Newton's second law of mechanics, the acceleration of an element is written as

\begin{displaymath}
\frac{d \mbox{\boldmath${v}$}}{d t}=\mbox{\boldmath${f_m}$}\equiv \frac{\mbox{\boldmath${F}$}}{m},
\end{displaymath} (A.1)

where $m$, $\mbox{\boldmath${v}$}$, $\mbox{\boldmath${F}$}$, and $\mbox{\boldmath${f}$}_m$ represent the mass, the velocity of the element, and force working on the element and the force per mass working on the element, respectively.

In fluid dynamics, using the mass density $\rho$ and the force working on the unit volume $\mbox{\boldmath${f}$}_v$ equation (A.1) is rewritten as

\begin{displaymath}
{\rho}\frac{d \mbox{\boldmath${v}$}}{d t}={\mbox{\boldmath${f}$}_v}.
\end{displaymath} (A.2)

Which kind of force works in a fluid? Gas pressure force does work in any fluids. Beside this, if there is the gravity, $\rho \mbox{\boldmath${g}$}$ should be included in $\mbox{\boldmath${f}$}_v$. If the electric currents is running in the fluid and the magnetic fields exist, the Lorentz force $\mbox{\boldmath${j\times B}$}$ should be added.

To write down the expression of the gas pressure force, consider a fluid element between $x$ and $x+\Delta x$ as shown in Figure A.2. Pressure force exerting on the surface $\S$ at $x$ is $p(x)S$, while that on the opposite side is $-p(x+\Delta x)S$. The net pressure force working on the volume of $S\Delta x$ is equal to $(p(x)-p(x+\Delta x))S$, which is approximated as $-\partial p/\partial x\Delta x S +O(\Delta x^2$).

Figure A.2: Pressure force exerted on an element with volume $S\Delta x$
\begin{figure}\centering\leavevmode
\epsfxsize =.45\columnwidth \epsfbox{eps/p.ps}\end{figure}
Thus, the pressure force working on the unit volume is written $-\partial p/\partial x$. Equation (A.2) can be rewritten as
\begin{displaymath}
\rho\frac{d \mbox{\boldmath${v}$}}{d t}=-\mbox{\boldmath${\nabla}$}p+\rho\mbox{\boldmath${g}$},
\end{displaymath} (A.3)

when gravity $\mbox{\boldmath${g}$}$ is working.

Kohji Tomisaka 2012-10-03