This is the Helmholtz theorem. Vortex lines never end on the fluid.
What is the physical reason for this?
1 Answer
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This is simply another way of saying $\nabla\cdot \nabla \times\vec{u} = 0$ for a vector field at any point where $\vec{u}$ is $C^2$. The vorticity $\vec{\omega} = \nabla\times \vec{u}$ is a divergenceless field because it is the curl of the velocity field.
answered Feb 16, 2017 at 20:45
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1$\begingroup$ In other words, there is no physical reason. This is a kinematical theorem and as such belongs in the realm of mathematics. $\endgroup$– PirxCommented Feb 16, 2017 at 21:19
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$\begingroup$ @Pirx If you have the nice habit of smoking, move your fingers through the smoke and you will see that the smoke reattaches again. There must be some physical reason for this. $\endgroup$– user143115Commented Feb 16, 2017 at 21:30
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$\begingroup$ I don't smoke, but I'm a theoretical fluid dynamicist. What do you think those whiffs of smoke have to do with vortex lines? $\endgroup$– PirxCommented Feb 16, 2017 at 21:32
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$\begingroup$ Thank you, I agree with you. But what is the reason for this? $\endgroup$– user143115Commented Feb 16, 2017 at 21:34
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$\begingroup$ @Pirx I smoke I like it or not because I am a welder. You know better than me that this smoke is deep in the regime of turbulence. $\endgroup$– user143115Commented Feb 16, 2017 at 21:40