Hydrostatics
The subject of the flow of fluids, and particularly of water, fascinates everybody. We can all remember, as children, playing in the bathtub or in mud puddles with the strange stuff. As we get older, we watch streams, waterfalls, and whirlpools, and we are fascinated by this substance which seems almost alive relative to solids. The behavior of fluids is in many ways very unexpected and interesting—it is the subject of this chapter and the next. The efforts of a child trying to dam a small stream flowing in the street and his surprise at the strange way the water works its way out has its analog in our attempts over the years to understand the flow of fluids. We have tried to dam the water up—in our understanding—by getting the laws and the equations that describe the flow. We will describe these attempts in this chapter. In the next chapter, we will describe the unique way in which water has broken through the dam and escaped our attempts to understand it.  We suppose that the elementary properties of water are already known to you. The main property that distinguishes a fluid from a solid is that a fluid cannot maintain a shear stress for any length of time. If a shear is applied to a fluid, it will move under the shear. Thicker liquids like honey move less easily than fluids like air or water. The measure of the ease with which a fluid yields is its viscosity. In this chapter we will consider only situations in which the viscous effects can be ignored. The effects of viscosity will be taken up in the next chapter.  We begin by considering hydrostatics, the theory of liquids at rest. When liquids are at rest, there are no shear forces (even for viscous liquids). The law of hydrostatics, therefore, is that the stresses are always normal to any surface inside the fluid. The normal force per unit area is called the pressure. From the fact that there is no shear in a static fluid it follows that the pressure stress is the same in all directions (Fig. 40-1). We will let you entertain yourself by proving that if there is no shear on any plane in a fluid, the pressure must be the same in any direction.  The pressure in a fluid may vary from place to place. For example, in a static fluid at the earth’s surface the pressure will vary with height because of the weight of the fluid. If the density $\rho$ of the fluid is considered constant, and if the pressure at some arbitrary zero level is called $p_0$ (Fig. 40-2), then the pressure at a height $h$ above this point is $p=p_0-\rho gh$, where $g$ is the gravitational force per unit mass. The combination \begin{equation*} p+\rho gh \end{equation*} is, therefore, a constant in the static fluid. This relation is familiar to you, but we will now derive a more general result of which it is a special case.  If we take a small cube of water, what is the net force on it from the pressure? Since the pressure at any place is the same in all directions, there can be a net force per unit volume only because the pressure varies from one point to another. Suppose that the pressure is varying in the $x$-direction—and we take the coordinate directions parallel to the cube edges. The pressure on the face at $x$ gives the force $p\,\Delta y\,\Delta z$ (Fig. 40-3), and the pressure on the face at $x+\Delta x$ gives the force $-[p+(\ddpl{p}{x})\,\Delta x]\,\Delta y\,\Delta z$, so that the resultant force is $-(\ddpl{p}{x})\,\Delta x\,\Delta y\,\Delta z$. If we take the remaining pairs of faces of the cube, we easily see that the pressure force per unit volume is $-\FLPgrad{p}$. If there are other forces in addition—such as gravity—then the pressure must balance them to give equilibrium.  Let’s take a circumstance in which such an additional force can be described by a potential energy, as would be true in the case of gravitation; we will let $\phi$ stand for the potential energy per unit mass. (For gravity, for instance, $\phi$ is just $gz$.) The force per unit mass is given in terms of the potential by $-\FLPgrad{\phi}$, and if $\rho$ is the density of the fluid, the force per unit volume is $-\rho\,\FLPgrad{\phi}$. For equilibrium this force per unit volume added to the pressure force per unit volume must give zero: \begin{equation} \label{Eq:II:40:1} -\FLPgrad{p}-\rho\,\FLPgrad{\phi}=\FLPzero. \end{equation} Equation (40.1) is the equation of hydrostatics. In general, it has no solution. If the density varies in space in an arbitrary way, there is no way for the forces to be in balance, and the fluid cannot be in static equilibrium. Convection currents will start up. We can see this from the equation since the pressure term is a pure gradient, whereas for variable $\rho$ the other term is not. Only when $\rho$ is a constant is the potential term a pure gradient. Then the equation has a solution \begin{equation*} p+\rho\phi=\text{const}. \end{equation*} Another possibility which allows hydrostatic equilibrium is for $\rho$ to be a function only of $p$. However, we will leave the subject of hydrostatics because it is not nearly so interesting as the situation when fluids are in motion. 