Do vortex rings have an intrinsic forward momentum? (Feynman lectures II.41) I was going through Feynman's lecture vol.II, #41 about non-viscous fluid mechanics, and he makes a point that I do not understand.
Towards the end, he discusses vortex rings (he takes the youtube-favorite setup of a drum with a circular hole as a vortex ring generator).
However, he makes a remark to explain that the ring is moving forward that puzzles me for several reasons:

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 Ω move with the fluid, they also move ahead with the velocity v. (Of course, the circulation of v around the top part of the ring is responsible for the forward motion of the vortex lines at the bottom.)


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*First, it just seems obvious to me that the ring is going forward (you slapped the back of the drum; it's a simple matter of conservation of momentum that you're pushing the fluid out, whether that fluid is shaped as a rotating donut or not). I don't get why we need to look for further explanation

*Second, I simply don't understand the wording. What does he mean by "The circulating velocity around the bottom of the ring extends up to the top of the ring"? Circulation is defined on a closed loop, what does it mean to extend that loop? I just genuinely don't get the image he is trying to convey.

*Third, he seems to imply that the forward motion of the ring comes from the circulation of the vortices themselves; that the vortex ring is intrinsically moving forward, because of its inner rotation (i.e. there couldn't be a vortex ring staying still). Well, I don't get that either. If you take a cross-section of the ring like on Fig41.b., it seems clear that this cross-section has a circular symmetry, and if you were to vector-sum all the velocities along the ring, you would find that they cancel out (the velocity at any point on the circle cancels out with its diametrical opposite). And you can repeat that all along the donut to show that everything should cancel out. Therefore, the rotating motion in itself carries no total momentum forward or backwards. What he explains seems to contradict that.

Can anyone help me understand this mysterious passage?
 A: It's all about velocity induction in ideally irrotational flows (irrotational except for an infinitely small region of the domain, like vortex tubes).

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*I haven't read the page you're referring to, so I'll leave the first point for a further edit of the answer

*The circulation of a vortex ring is defined as the line integral of the velocity field on a path that form a loop around a infinitely small section of the vortex ring (it's "bound" to the vortex ring, passing one time through the hole of the vortex ring).  Helmholtz's theorems state properties of the circulation, roughly speaking constant in time for every path winded once on the vortex ring). This circulation can be interpreted as a result of the viscous, rotational effects inside the core of the vortex ring.

*being the circulation constant along the vortex ring, every section of a circular vortex ring induces velocity in the same direction on all the other regions of the vortex ring. Let's use cylindrical coordinates to descibe the vortex ring and the velocity field, with circulation $\mathbf{\Gamma}(\mathbf{r}_1) = \Gamma \mathbf{\hat{\theta}}(\mathbf{r}_1)$. The velocity induced by the circulation at the point $\mathbf{r}_1$ on the point $\mathbf{r}_2$ has direction determined by the cross product $ \mathbf{\Gamma}(\mathbf{r}_1) \times (\mathbf{r}_2 - \mathbf{r}_1) ≈ d\mathbf{u}_1(\mathbf{r}_2)$. The overall induced velocity on the point $\mathbf{r}_2$ is the integral of all the contributions from the points of the vortex ring (intgeral over coordinate $\mathbf{r}_1$). It should not be hard to prove that all these contributions are positive in the $\mathbf{\hat{z}}$-direction (try to drraw it and use tje right-hand rule for the cross product).

I'll let you think about it. I'll edit the answer further if need it. To get a in-depth knowledge of fluid dynamics is not so easy
A: Imagine a smoke ring as a series of spinning solid disks arranged in a circle. Such a ring could float in space and not move forward.
Now put it in a snug fitting cylinder. The rings roll along the inside of the cylinder. The ring moves forward.
Make two concentric cylinders so the inside cylinder fits snugly against the inside of the rings. Now there must be some slip. But the region of contact along the inside is smaller than along the outside. You would expect the ring to still roll forward. Especially if the inside diameter of the ring is small.
It works in a similar way with a smoke ring in air. The outside of a stationary smoke ring blows against nearby still air. This pushes the ring forward. In the center, the ring blows backward against still air, holding the ring back. The ring velocity that minimizes and balances these forces is when the ring moves forward.
A: You are assuming that the vortex ring is separate from the fluid surrounding it.
If you look at a video of a colored smoke ring (where the density of the ring is about the same as the fluid around it), you'll see it leave a colored trail behind it. The center of the ring is moving backward, and is left behind. More fluid is incorporated from the front. You can't just vector-sum the velocities, because they in fact don't add up, because the content of the vortex is changing. It continually incorporates new material and leaves old material behind.
Air vortex rings in water don't do that. I believe that the major part of air vortex rings are actually water vortex rings, and the air just gets carried along in the smallest-velocity part of the rings. As the remaining ink eventually does in colored vortex rings.
video with colored vortex rings
Looking at Feynman.... his vortex lines are not streamlines, showing where fluid goes. They're more like magnetic lines of force. They are perpendicular to the rotation.
When he says "top" he's talking about the top of the lower ring in Figure 2, and when he says "bottom" he means the bottom of the lower ring. It might make more sense to say "inside" and "outside".
The vortex lines on the inside are compressed compared to those on the outside. So the inside moves forward faster than the outside moves back.
I think that's what he's saying. But he also says

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.

A: 
First, it just seems obvious to me that the ring is going forward

True. Momentum is simply the integral $\rho v$ over the ring and it's immediate surroundings. A ring moving forward has forward momentum.
In fully enclosed spaces the total momentum of the air must time-average to zero, so shooting the ring in a closet will push other air backwards. In tightly confined spaces, such as a tube closed at both ends, the "momentum of the ring" is less meaningful. A similar effect, ground effect, hides the fact that airfoils push air downwards (one reason why airfoils are misunderstood so much).

What does he mean by ...

The curl of the velocity is the vorticity. Since the divergence of the curl is zero, you can trace lines though this field much like magnetic field lines. Each line represents a certain fixed amount of circulation around itself, just as each magnetic field line is a fixed amount of flux. If divergence was non-zero the lines would have "ends" to them (even though the divergence is zero, the lines can still "diverge" away from each-other).
In a vortex ring these lines resemble a bundle of bracelets. Vortex lines move with the fluid flow, and in our case they are moving like a water snake toy slipping out of your hand: a combined forward and "threading the ring" motion. The net forward motion of the vortex lines indicates that the vortex itself is moving forward.

there couldn't be a vortex ring staying still

Not for long! A vortex ring is slightly less dense than air, yet it can push though air with ease. Try throwing a balloon that far! The low friction is because it's outer edges are "rolling" past the air much like the water toy slips out of your hand. A stationary ring would face friction at it's edges which would propel itself forward (weakening the vortex in the process) until the speeds were matched.
