Why does this vortex bubble ring reflect off the water-air boundary?

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    $\begingroup$ Please put more details in the description. $\endgroup$ – Mauricio Jul 4 '18 at 12:00
  • $\begingroup$ the video is out of topic after that bubble. The question should be reversed as "why do bubbles not reflect from the surface" and the answer is because the surface tension around the bubble is at a narrow equillibrium, and hitting the surface it breaks . This one in the beginning of the video just was lucky ( within the probabilities of water membrane+water membrane collision $\endgroup$ – anna v Jul 4 '18 at 12:16
  • $\begingroup$ Nice demonstration of underwater vortex rings. Hope you don't mind I reworded question, added additional tags to help receive better attention. Although I don't have a definitive answer I can offer that vortex rings are a type of traveling wave called a 'soliton', an isolated nonlinear impulse. Consider other types of waves are able to reflect, refract off boundaries of media density changes. But I wouldnt have expected a vortex ring to do that. For light waves there is a critical angle at which you transition from reflection to refraction, and I wonder if rings ... $\endgroup$ – docscience Jul 4 '18 at 13:50
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    $\begingroup$ ... would behave in the same manner. Unlike most waves vortex rings transport matter as well as energy. In the air, smoke rings transport smoke particles and air molecules, and in water, air molecules. The property of reflection at a very basic level, due to the conservation of energy. But since the ring carries mass, is there also a conservation of momentum? Not sure, but interesting question. Because the rings are nonlinear waves, their analysis a bit more difficult than linear waves. $\endgroup$ – docscience Jul 4 '18 at 13:54

I think this diagram represents what ought to be happening (I invite others to comment / critique / improve):

enter image description here

It depicts a section through the vortex ring, moving from left to right. The upper part is rotating counterclockwise, while the lower part is rotating clockwise. As the ring gets closer to the surface, the water being pulled by the upper vortex will accelerate, and due to Bernoulli the increased flow velocity lowers the pressure. This should result in dissipation of the vortex ring - not reflection.

In the limit where the surface is considered to be rigid, you can apply a boundary condition that can be resolved by stipulating the presence of an "image vortex" that ensures there is no flow normal to the surface. But such a vortex would again tend to dissipate (rather than reflect) the original ring.

A lot of information is given in "Vortex Ring Interaction with a Free Surface" by Song et al, in the proceedings of the 18th symposium on naval hydrodynamics (1991(. In their experiments, the vortex ring dissipates when it reaches the surface.

All this says - the observation in the video clip is anomalous, this is not what you would expect for reflection off a still free surface. But then this is a pool, and there are waves (ripples) on the surface. I suspect that these "hit the vortex ring just right" so it was reflected rather than dissipated. It would be interesting to do this experiment many times, with a still surface. I suspect the result would be different.


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