Vortex lines
We have already written down the general equations for the flow of an incompressible fluid when there may be vorticity. They are \begin{align*} \text{I.}&\quad\FLPdiv{\FLPv}=0,\\[1.5ex] \text{II.}&\quad\FLPOmega=\FLPcurl{\FLPv},\\[1ex] \text{III.}&\quad\ddp{\FLPOmega}{t}+ \FLPcurl{(\FLPOmega\times\FLPv)}=\FLPzero. \end{align*} The physical content of these equations has been described in words by Helmholtz in terms of three theorems. First, imagine that in the fluid we were to draw vortex lines rather than streamlines. By vortex lines we mean field lines that have the direction of $\FLPOmega$ and have a density in any region proportional to the magnitude of $\FLPOmega$. From II the divergence of $\FLPOmega$ is always zero (remember—Section 3-7—that the divergence of a curl is always zero). So vortex lines are like lines of $\FLPB$—they never start or stop, and will tend to go in closed loops. Now Helmholtz described III in words by the following statement: the vortex lines move with the fluid. This means that if you were to mark the fluid particles along some vortex lines—by coloring them with ink, for example—then as the fluid moves and carries those particles along, they will always mark the new positions of the vortex lines. In whatever way the atoms of the liquid move, the vortex lines move with them. That is one way to describe the laws.  It also suggests a method for solving any problems. Given the initial flow pattern—say $\FLPv$ everywhere—then you can calculate $\FLPOmega$. From the $\FLPv$ you can also tell where the vortex lines are going to be a little later—they move with the speed $\FLPv$. With the new $\FLPOmega$ you can use I and II to find the new $\FLPv$. (That’s just like the problem of finding $\FLPB$, given the currents.) If we are given the flow pattern at one instant we can in principle calculate it for all subsequent times. We have the general solution for nonviscous flow.  We would like to show how Helmholtz’s statement—and, therefore, III—can be at least partly understood. It is really just the law of conservation of angular momentum applied to the fluid. Suppose we imagine a small cylinder of the liquid whose axis is parallel to the vortex lines, as in Fig. 40-13(a). At some time later, this same piece of fluid will be somewhere else. Generally it will occupy a cylinder with a different diameter and be in a different place. It may also have a different orientation, say as in Fig. 40-13(b). If, however, the diameter has decreased as shown in Fig. 40-13, the length will have increased to keep the volume constant (since we are assuming an incompressible fluid). Also, since the vortex lines are stuck with the material, their density will go up as the cross-sectional area goes down. The product of the vorticity $\FLPOmega$ and area $A$ of the cylinder will remain constant, so according to Helmholtz, we should have \begin{equation} \label{Eq:II:40:21} \Omega_2A_2=\Omega_1A_1. \end{equation}  Now notice that with zero viscosity all the forces on the surface of the cylindrical volume (or any volume, for that matter) are perpendicular to the surface. The pressure forces can cause the volume to be moved from place to place, or can cause it to change shape; but with no tangential forces the magnitude of the angular momentum of the material inside cannot change. The angular momentum of the liquid in the little cylinder is its moment of inertia $I$ times the angular velocity of the liquid, which is proportional to the vorticity $\Omega$. For a cylinder, the moment of inertia is proportional to $mr^2$. So from the conservation of angular momentum, we would conclude that \begin{equation*} (M_1R_1^2)\,\Omega_1=(M_2R_2^2)\,\Omega_2. \end{equation*} But the mass is the same, $M_1=M_2$, and the areas are proportional to $R^2$, so we get again just Eq. (40.21). Helmholtz’s statement—which is equivalent to III—is just a consequence of the fact that in the absence of viscosity the angular momentum of an element of the fluid cannot change.  There is a nice demonstration of a moving vortex which is made with the simple apparatus of Fig. 40-14. It is a “drum” two feet in diameter and two feet long made by stretching a thick rubber sheet over the open end of a cylindrical “box.” The “bottom”—the drum is tipped on its side—is solid except for a $3$-inch diameter hole. If you give a sharp blow on the rubber diaphragm with your hand, a vortex ring is projected out of the hole. Although the vortex is invisible, you can tell it’s there because it will blow out a candle $10$ to $20$ feet away. By the delay in the effect, you can tell that “something” is travelling at a finite speed. You can see better what is going on if you first blow some smoke into the box. Then you see the vortex as a beautiful round “smoke ring.”  The smoke ring is a torus-shaped bundle of vortex lines, as shown in  Fig. 40-15(a).  Since $\FLPOmega=\FLPcurl{\FLPv}$, these vortex lines represent also a circulation of $\FLPv$ as shown in part (b) of the figure. We can understand the forward motion of the ring in the following way: The circulating velocity around the bottom of the ring extends up to the top of the ring, having there a forward motion. Since the lines of $\FLPOmega$ move with the fluid, they also move ahead with the velocity $\FLPv$. (Of course, the circulation of $\FLPv$ around the top part of the ring is responsible for the forward motion of the vortex lines at the bottom.)  We must now mention a serious difficulty. We have already noted that Eq. (40.9) says that, if $\FLPOmega$ is initially zero, it will always be zero. This result is a great failure of the theory of “dry” water, because it means that once $\FLPOmega$ is zero it is always zero—it is impossible to produce any vorticity under any circumstance.  Yet, in our simple demonstration with the drum, we can generate a vortex ring starting with air which was initially at rest. (Certainly, $\FLPv=\FLPzero$, $\FLPOmega=\FLPzero$ everywhere in the box before we hit it.) Also, we all know that we can start some vorticity in a lake with a paddle. Clearly, we must go to a theory of “wet” water to get a complete understanding of the behavior of a fluid.  Another feature of the dry water theory which is incorrect is the supposition we make regarding the flow at the boundary between it and the surface of a solid.  When we discussed the flow past a cylinder—as in Fig. 40-11, for example—we permitted the fluid to slide along the surface of the solid. In our theory, the velocity at a solid surface could have any value depending on how it got started, and we did not consider any “friction” between the fluid and the solid. It is an experimental fact, however, that the velocity of a real fluid always goes to zero at the surface of a solid object. Therefore, our solution for the cylinder, with or without circulation, is wrong—as is our result regarding the generation of vorticity. We will tell you about the more correct theories in the next chapter. 