357. Vortex line representation. Coulomb interaction of vortex lines

Posted by Dmitry

April 16, 2009

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After brief introduction into vortex line representation we are probably ready to discuss the interaction of vortex lines between each other. But before I proceed to the actual derivation, let me focus for a bit on not so terribly popular (but powerful) formulation of ideal hydrodynamics – Hamiltonian formulation.

The Lagrangian of incompressible fluid (I set $\rho=1$ for simplicity) is

${\cal L}=\frac{1}{2}\int\frac{d^{3}\mathbf{k}}{(2\pi)^{3}}|\mathbf{v}_{\mathbf{k}}|^{2}$ , (1)

and, as usual, we define the canonical momentum as

$\mathbf{p}=\frac{\delta{\cal L}}{\delta\mathbf{v}}$ .

Vorticity field $\mathbf{\Omega}(\mathbf{r},t)$ can be written in terms of momentum as

$\mathbf{\Omega}={\rm curl}\mathbf{p}$ ,

and the Hamiltonian is

${\cal H}(\mathbf{\Omega})=\left(\int d\mathbf{r}\left(\frac{\delta{\cal L}}{\delta\mathbf{v}}\cdot\mathbf{v}\right)-{\cal L}\right)|_{\mathbf{v}=\mathbf{v}(\mathbf{\Omega})}$ .

It is easy to show that the e.o.m. for the vorticity field is given by

$\frac{\partial\mathbf{\Omega}}{\partial t}={\rm curl}\left({\rm curl}\left(\frac{\delta{\cal H}}{\delta\mathbf{\Omega}}\times\mathbf{\Omega}\right)\right)$ .

Exercise: check it out explicitly.

The Hamiltonian written in terms of vorticity field has a remarkably simple form:

${\cal H}_{M}(\mathbf{\Omega})=\frac{1}{2}\int\frac{d\mathbf{k}}{(2\pi)^{3}}\frac{|\mathbf{\Omega}_{\mathbf{k}}|^{2}}{k^{2}}$ ,

i.e.,

${\cal H}_{M}(\mathbf{\Omega})=\frac{1}{2}\int\int\frac{(\mathbf{\Omega}(\mathbf{r}_{1})\cdot\mathbf{\Omega}(\mathbf{r}_{2}))}{|\mathbf{r}_{1}-\mathbf{r}_{2}|}d\mathbf{r}_{1}d\mathbf{r}_{2}$ .

Although the Coulomb interaction is present , probably you don’t quite see yet how separate vortex filaments interact with each other. To show this explicitly, let me finally use the vortex line representation. I express vorticity field as

$\mathbf{\Omega}(\mathbf{r},t)=\int d^{2}\nu\oint\delta(\mathbf{r}-\mathbf{R}(\nu,l,t))\frac{\partial\mathbf{R}}{\partial l}dl$ ,

where $\nu$ are 2d Lagrangian coordinates marking vortex lines and is affine parameter along the given line. Substituting (2) into (1), I finally find

${\cal H}=\frac{\Gamma^{2}}{2}\oint\oint\frac{(\mathbf{R}'(l_{1})\cdot\mathbf{R}'(l_{2}))}{|\mathbf{R}(l_{1})-\mathbf{R}(l_{2})|}dl_{1}dl_{2}$ ,

where $\Gamma$ is circulation (it is conserved due to the Kelvin theorem I have discussed in the previous post).

Let us discuss the physics of this Hamiltonian a bit.

1. Suppose that we have just a single vortex filament. In this case, $\mathbf{R}$ does not depend on line marker $\nu$ at all, and the circulation $\Gamma$ is simply given by the integral $\int d^{2}\nu$ , presumed to be finite. The overall flow is potential (that is, $\mathbf{p}=\nabla\Phi$ ) – the fluid is circulating around the center of the filament. Dynamics is still quite non-trivial, though: small pieces of the filament interact with each other by Coulomb interaction.

2. This Hamiltonian describes an infinitely thin vortex filament – string , and its self-energy is clearly infinite due to the Coulomb-like divergence. If we want to deal with it in a practical fashion, we will have to regularize it somehow (starting presumably from the initial Lagrangian (1) or introducing viscosity).

3. If we wait for some time, we will find that the self-induced velocity of the vortex filament becomes infinite as well (unless, as I said above, we don’t regularize the Coulomb Hamiltonian) This shows that viscosity is very important in turbulence – even if we start in the regime where the Reynolds number is extremely large (so that the viscosity is effectively zero), finite time collapse of the vortex lines will lead to the appearance of the localized regions in the flow, where dissipation should be huge. That’s what we were talking about in the post about four puzzles in physics of turbulence.

Hydrodynamics, Master Mason, Turbulence

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