Helicity and Clebsch decomposition

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The discussion centers on the relationship between helicity and Clebsch decomposition in fluid dynamics. It is established that a velocity field with a Clebsch decomposition has zero helicity in any closed vortex tube. The main question posed is whether the converse is true: if a velocity field has zero helicity in every closed vortex tube, does it necessarily have a Clebsch decomposition? The author expresses skepticism about this converse, suggesting that zero helicity in closed vortex tubes may not provide enough information to determine the overall form of the velocity field. A request for proof regarding this relationship is made, highlighting the need for further exploration of the topic.
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It is well-known that using the Clebsch decomposition of a velocity field, the helicity contained in an any closed vortex tube is zero.

Does the converse hold?
That is, given zero helicity in any closed vortex tube, does this imply that the velocity field has a Clebsch decomposition?
 
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Hmm..no one has replied as yet :cry:

Perhaps I was a bit kryptic, so I'll give a few definitions and results.
(I'll assume constant density \rho)
1. Helicity
Given a velocity field \vec{v} we define the helicity pr. unit volume as:
h=\vec{v}\cdot\vec{c},\vec{c}=\nabla\times\vec{v}

That is, the helicity (pr. unit volume) is a rough measure of the extent to which the fluid particles traverse in a helical path around the local rotation vector, \frac{1}{2}\vec{c}

For general, inviscid flow, the helicity H is conserved for any given, closed vortex tube \mathcal{V}_{c}
That is:
\frac{dH}{dt}=0, H=\int_{\mathcal{V}_{c}}hdV

2. Clebsch decomposition:
A Clebsch decomposition of a velocity field takes the form:
\vec{v}=\nabla\Phi+\alpha\nabla\beta
with \Phi,\alpha,\beta scalar functions.
The "neat" thing about the Clebsch decomposition is that for an inviscid fluid, we may replace the equations of motion with the following 3 equations:
\frac{D\alpha}{dt}=\frac{D\beta}{dt}=0
p=-\rho(\Phi_{t}+\alpha\beta_{t}+\frac{1}{2}\vec{v}^{2}+gz)

That is, we gain a generalized Euler equation for the pressure p in the fluid.
In addition, the intersection of surfaces where \alpha,\beta are constant are vortex lines, since we have:
\vec{c}=\nabla\alpha\times\nabla\beta

3. Helicity in a Clebsch fluid:
We have, for a Clebsch fluid:
h=\nabla\cdot(\Phi\vec{c})
Hence, the helicity H is not only constant in a closed vortex tube in a Clebsch
fluid, but in fact equal to 0.
This shows that Clebsch fluids are only special cases of inviscid flow.

4. The converse:
What I asked, was if the helicity was zero for every closed vortex tube, does it then follow that there exists a Clebsch decomposition of the velocity field?

Personally, I doubt the existence of theconverse, since a bounded region does not only consist of closed vortex tubes, but may also have vortex tubes connecting different parts of the boundary.
Hence, it would be rather strange if information from only part of the region (zero helicity on the closed vortex tubes) should be sufficient in specifying the form of the velocity field in the whole region (that is, as a Clebsch decomposition).

However, I would really like a proof of this matter, the one way or the other..
 
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