Equation of Motion

In one dimensional problems, if the applied force is $F(x,t)$, we solve

$$m{dv\over dt} = F(x,t), \qquad (v = {dx\over dt}).$$

In three dimensional situations we have, in cartesians coordinates, we have

$$\begin{eqnarray*} m{dv_{x}\over dt} & = & F_{x}(x,y,z,t), \\ m{dv_{y}\over dt... ...= & F_{y}(x,y,z,t), \\ m{dv_{z}\over dt} & = & F_{z}(x,y,z,t). \end{eqnarray*}$$

This is written in concise form by using vector notation so that

$$m{d{\bf v}\over dt} = {\bf F}.$$ (1.11)

In a continuuous medium, a variation of pressure, $p(x,y,z,t)$, from one location to another produces a pressure force. $p$ is the force per unit area exerted on a surface by a fluid (liquid or gas) or plasma. Thus, if $p=p(x)$ the force on a face is $p(x)dy dz$. This is illustrated in Figure 1.5.

Figure 1.5: The pressure force acting on a face of area $dydz$.

Consider a fluid element, lying between the points $x$ and $x + \delta x$. On the left hand side the pressure force is

$$p(x) dydz,$$

and on the right hand side the force is

$$-p(x+\delta x) dydz.$$

The minus sign arises because the force on the right hand side acts towards the left. Therefore, the net force in the ${\bf i}$ direction PER UNIT VOLUME is

$${p(x) - p(x+\delta x)\over \delta x} = -{\left (p(x + \delt... ... p(x)\right )\over \delta x} \rightarrow -{dp\over dx}{\bf i}.$$

Thus, the equation of motion (1.11) per unit volume for a liquid, gas or plasma of density $\rho$ (mass per unit volume) becomes

$$\rho {dv_{x}\over dt} = - {dp\over dx}.$$ (1.12)

Pressure gradients produce a force in the opposite direction to the actual gradient. Thus, the pressure gradient force acts from areas of high pressure to areas of low pressure and is equal in magnitude to the gradient of the pressure. This is illustrated in Figure 1.6.

Figure 1.6: A sketch of pressure as a function of $x$ and the direction of the force in relationship to the isobars.

In three dimensional problems, where $p = p(x,y,z,t)$, (1.12) generalises to give three components as

$$\begin{eqnarray*} \rho {dv_{x}\over dt} & = & -{\partial p \over \partial x}, \... ...\\ \rho {dv_{z}\over dt} & = & -{\partial p \over \partial z}, \end{eqnarray*}$$

or in vector form

$$\rho {d{\bf v}\over dt} = - \nabla p.$$ (1.13)