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There is so much to unpack here. I am going to start this post and will keep adding to it as I get more thoughts.

First, let me reiterate the most important thing - I do not understand the physics behind this derivation at all. I keep trying to come up with some explanation that would make any sense (in particular why the chosen Lagrangian is what it is).

Second, in the mean time we have to do the most obvious thing: test the derivation by reducing it to as many standard solutions as possible. Eowyn has already gave a great start - the derivation of the Webster equation. Personally, I had never seen it in my life. But since he managed to arrive at something that had been already established, we can be reasonably confident the derivation is correct and we should keep trying getting other existent solutions.

My bigger goal currently is to do Reynold's decomposition in Eowyn's framework and see how MKE and TKE work out, see how turbulence energy dissipation term works out for standard flows (wake, jet, shear flow), see how Taylor scale works out, see if we can derive convective scales (for vortex dynamics) and see if we can derive Kolmogorov scales.

**The first baby step**, though, is to do the simplest thing possible: let's try to derive an equation for an infinite steady inviscid uniform flow in the absence of the gravitational force in a three dimensional "flat" space.

Assume, we do not experience any coordinate transformation ever. In this case, covariant derivative in flat space (which is our case) is equal to a regular derivative.

Since the flow is inviscid, the vector potential must be zero $\vec V (\vec x, t) = 0$. Note, I do not understand why Eowyn expresses the vector potential in the phase space. Helmholtz decomposition is performed on the given velocity field $\vec v (\vec x, t)$. Therefore, both flow potential and vector potential must be the functions of $(\vec x, t)$. Before Lagrangian is "applied" to the vector potential, it must be written as a function of regular Decart (i.e., Cartesian) coordinates, I think. And that is what I did.

At this stage we have $D^{\mu}(D^{\mu}(\phi)) = \frac{\partial}{\partial \mu}(\frac{\partial}{\partial \mu}(\phi)) =  0$, where $\mu$ can be either $x$, or $y$, or $z$, or $t$.

Since the flow is assumed steady, $\mu$ reduces to $\mu = \vec x = (x,y,z)$ and our expression reduces to

$$D^{\mu}(D^{\mu}(\phi)) = \frac{\partial}{\partial \mu}\left(\frac{\partial}{\partial \mu}(\phi)\right) = \frac{\partial}{\partial \vec x}\left(\frac{\partial}{\partial \vec x}(\phi)\right) = grad (grad (\phi)) = \nabla(\nabla(\phi)) = \nabla (\vec v)$$

The last equality came from Helmholtz decomposition where the irrotational part of the velocity field is given by $v^{\phi} = \nabla \phi$.

And we end up with the following equation for our uniform flow derived in the Eowyn's framework:

$$\nabla (\vec v) = 0$$

Let's see if we get the same from the Navier-Stokes equation:

$$\rho \left(\frac{\partial\vec v}{\partial t} + \vec v \cdot \nabla\vec v\right) = -\nabla P + \rho\vec g + \nabla \cdot \tau$$

Here: $\tau=0$ because the flow is inviscid, $\rho\vec g = 0$ because the absence of gravity is assumed, $\nabla P =0$ because we need curved surface for a flow to have pressure gradient, which is not the case in our problem, $\partial\vec v / \partial t = 0$ because the flow is assumed steady. And we are left with the equation for our uniform flow

$$\nabla \vec v = 0$$

Which is exactly the same as derived in Eowyn's framework.