Irrotational fluid flow; the flow past a sphere
Let’s now consider an example which is not really a very good one, because the equations we will use will not really represent the subject with complete generality but only in an artificial idealized situation. We take up the problem of water flow. In the case of the stretched sheet, our equations were an approximation which was correct only for small deflections. For our consideration of water flow, we will not make that kind of an approximation; we must make restrictions that do not apply at all to real water. We treat only the case of the steady flow of an incompressible, nonviscous, circulation-free liquid. Then we represent the flow by giving the velocity $\FLPv(\FLPr)$ as a function of position $\FLPr$. If the motion is steady (the only case for which there is an electrostatic analog) $\FLPv$ is independent of time. If $\rho$ is the density of the fluid, then $\rho\FLPv$ is the amount of mass which passes per unit time through a unit area. By the conservation of matter, the divergence of $\rho\FLPv$ will be, in general, the time rate of change of the mass of the material per unit volume. We will assume that there are no processes for the continuous creation or destruction of matter. The conservation of matter then requires that $\FLPdiv{\rho\FLPv}=0$. (It should, in general, be equal to $-\ddpl{\rho}{t}$, but since our fluid is incompressible, $\rho$ cannot change.) Since $\rho$ is everywhere the same, we can factor it out, and our equation is simply \begin{equation*} \FLPdiv{\FLPv}=0. \end{equation*}  Good! We have electrostatics again (with no charges); it’s just like $\FLPdiv{\FLPE}=0$. Not so! Electrostatics is not simply $\FLPdiv{\FLPE}=0$. It is a pair of equations. One equation does not tell us enough; we need still an additional equation. To match electrostatics, we should have also that the curl of $\FLPv$ is zero. But that is not generally true for real liquids. Most liquids will ordinarily develop some circulation. So we are restricted to the situation in which there is no circulation of the fluid. Such flow is often called irrotational. Anyway, if we make all our assumptions, we can imagine a case of fluid flow that is analogous to electrostatics. So we take \begin{equation} \label{Eq:II:12:28} \FLPdiv{\FLPv}=0 \end{equation}  and \begin{equation} \label{Eq:II:12:29} \FLPcurl{\FLPv}=0. \end{equation}   We want to emphasize that the number of circumstances in which liquid flow follows these equations is far from the great majority, but there are a few. They must be cases in which we can neglect surface tension, compressibility, and viscosity, and in which we can assume that the flow is irrotational. These assumptions are valid so rarely for real water that the mathematician John von Neumann said that people who analyze Eqs. (12.28) and (12.29) are studying “dry water”! (We take up the problem of fluid flow in more detail in Chapters 40 and 41.) Because $\FLPcurl{\FLPv}=0$, the velocity of “dry water” can be written as the gradient of some potential: \begin{equation} \label{Eq:II:12:30} \FLPv=-\FLPgrad{\psi}. \end{equation}  What is the physical meaning of $\psi$? There isn’t any very useful meaning. The velocity can be written as the gradient of a potential simply because the flow is irrotational. And by analogy with electrostatics, $\psi$ is called the velocity potential, but it is not related to a potential energy in the way that $\phi$ is. Since the divergence of $\FLPv$ is zero, we have \begin{equation} \label{Eq:II:12:31} \FLPdiv{(\FLPgrad{\psi})}=\nabla^2\psi=0. \end{equation}  The velocity potential $\psi$ obeys the same differential equation as the electrostatic potential in free space ($\rho=0$). Let’s pick a problem in irrotational flow and see whether we can solve it by the methods we have learned. Consider the problem of a spherical ball falling through a liquid. If it is going too slowly, the viscous forces, which we are disregarding, will be important. If it is going too fast, little whirlpools (turbulence) will appear in its wake and there will be some circulation of the water. But if the ball is going neither too fast nor too slow, it is more or less true that the water flow will fit our assumptions, and we can describe the motion of the water by our simple equations. It is convenient to describe what happens in a frame of reference fixed in the sphere. In this frame we are asking the question: How does water flow past a sphere at rest when the flow at large distances is uniform? That is, when, far from the sphere, the flow is everywhere the same. The flow near the sphere will be as shown by the streamlines drawn in Fig. 12–8. These lines, always parallel to $\FLPv$, correspond to lines of electric field. We want to get a quantitative description for the velocity field, i.e., an expression for the velocity at any point $P$. We can find the velocity from the gradient of $\psi$, so we first work out the potential. We want a potential that satisfies Eq. (12.31) everywhere, and which also satisfies two restrictions: (1) there is no flow in the spherical region inside the surface of the ball, and (2) the flow is constant at large distances. To satisfy (1), the component of $\FLPv$ normal to the surface of the sphere must be zero. That means that $\ddpl{\psi}{r}$ is zero at $r=a$. To satisfy (2), we must have $\ddpl{\psi}{z}=v_0$ at all points where $r\gg a$. Strictly speaking, there is no electrostatic case which corresponds exactly to our problem. It really corresponds to putting a sphere of dielectric constant zero in a uniform electric field. If we had worked out the solution to the problem of a sphere of a dielectric constant $\kappa$ in a uniform field, then by putting $\kappa=0$ we would immediately have the solution to this problem. We have not actually worked out this particular electrostatic problem in detail, but let’s do it now. (We could work directly on the fluid problem with $\FLPv$ and $\psi$, but we will use $\FLPE$ and $\phi$ because we are so used to them.) The problem is: Find a solution of $\nabla^2\phi=0$ such that $\FLPE=-\FLPgrad{\phi}$ is a constant, say $\FLPE_0$, for large $r$, and such that the radial component of $\FLPE$ is equal to zero at $r=a$. That is, \begin{equation} \label{Eq:II:12:32} \left.\ddp{\phi}{r}\right|_{r=a}=0. \end{equation}  Our problem involves a new kind of boundary condition, not one for which $\phi$ is a constant on a surface, but for which $\ddpl{\phi}{r}$ is a constant. That is a little different. It is not easy to get the answer immediately. First of all, without the sphere, $\phi$ would be $-E_0z$. Then $\FLPE$ would be in the $z$-direction and have the constant magnitude $E_0$, everywhere. Now we have analyzed the case of a dielectric sphere which has a uniform polarization inside it, and we found that the field inside such a polarized sphere is a uniform field, and that outside it is the same as the field of a point dipole located at the center. So let’s guess that the solution we want is a superposition of a uniform field plus the field of a dipole. The potential of a dipole (Chapter 6) is $pz/4\pi\epsO r^3$. Thus we assume that \begin{equation} \label{Eq:II:12:33} \phi=-E_0z+\frac{pz}{4\pi\epsO r^3}. \end{equation}  Since the dipole field falls off as $1/r^3$, at large distances we have just the field $E_0$. Our guess will automatically satisfy condition (2) above. But what do we take for the dipole strength $p$? To find out, we may use the other condition on $\phi$, Eq. (12.32). We must differentiate $\phi$ with respect to $r$, but of course we must do so at a constant angle $\theta$, so it is more convenient if we first express $\phi$ in terms of $r$ and $\theta$, rather than of $z$ and $r$. Since $z=r\cos\theta$, we get \begin{equation} \label{Eq:II:12:34} \phi=-E_0r\cos\theta+\frac{p\cos\theta}{4\pi\epsO r^2}. \end{equation}  The radial component of $\FLPE$ is \begin{equation} \label{Eq:II:12:35} -\ddp{\phi}{r}=+E_0\cos\theta+\frac{p\cos\theta}{2\pi\epsO r^3}. \end{equation}  This must be zero at $r=a$ for all $\theta$. This will be true if \begin{equation} \label{Eq:II:12:36} p=-2\pi\epsO a^3E_0. \end{equation}  Note carefully that if both terms in Eq. (12.35) had not had the same $\theta$-dependence, it would not have been possible to choose $p$ so that (12.35) turned out to be zero at $r=a$ for all angles. The fact that it works out means that we have guessed wisely in writing Eq. (12.33). Of course, when we made the guess we were looking ahead; we knew that we would need another term that (a) satisfied $\nabla^2\phi=0$ (any real field would do that), (b) dependent on $\cos\theta$, and (c) fell to zero at large $r$. The dipole field is the only one that does all three. Using (12.36), our potential is \begin{equation} \label{Eq:II:12:37} \phi=-E_0\cos\theta\biggl(r+\frac{a^3}{2r^2}\biggr). \end{equation} The solution of the fluid flow problem can be written simply as \begin{equation} \label{Eq:II:12:38} \psi=-v_0\cos\theta\biggl(r+\frac{a^3}{2r^2}\biggr). \end{equation}  It is straightforward to find $\FLPv$ from this potential. We will not pursue the matter further.