Helicity integral in differential forms

Question

Let $V^{3}(t)$ be a compact region moving with the fluid.

Assume that at $t=0$ the vorticity $2$-form $\omega^{2}$ vanishes when restricted to the boundary $\partial V^{3}(0)$; that is, $i^{*}\omega^{2}=0$, where $i$ is the inclusion of $\partial V$ in $\mathbb{R}^{3}$.

(This does $\textit{not}$ say that $\omega^{2}$ itself vanishes, rather only that $\omega({\vec{u}},{\vec{w}})=0$ for $\vec{u}$,$\vec{w}$ tangent to $\partial V^{3}(0)$.)

Then the $\textbf{helicity}$ integral $\displaystyle{\int_{V(t)}{\vec{v}}\cdot{\vec{\omega}}\ dx\wedge dy\wedge dz}$ can be constant in time.

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How you actually prove that the helicity integral is constant in time?

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Here's my attempt:

$$\frac{d}{dt}\int_{V(t)}{\vec{v}}\cdot{\vec{\omega}}\ dx\wedge dy\wedge dz$$

$$= \frac{d}{dt}\int_{V(t)}{\vec{v}}\cdot{\vec{\omega}}\ dx^{1}\wedge dx^{2}\wedge dx^{3}$$

$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} {\vec{v}}\cdot{\vec{\omega}}\ \text{vol}^{3}$$

$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} \nu^{1}\wedge \omega^{2}$$

$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} \nu^{1}\wedge {\rm d}\nu^{1}$$

$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{V(t)} \nu\wedge {\rm d}\nu$$

$$= \frac{1}{o(x)\sqrt{g}(x)}\frac{1}{2} \frac{d}{dt}\int_{W(t)} (\nu\wedge {\rm d}\nu)+(\nu\wedge {\rm d}\nu)$$

$$= \frac{1}{o(x)\sqrt{g}(x)}\frac{1}{2} \frac{d}{dt}\int_{W(t)} (\nu\wedge {\rm d}\nu)-({\rm d}\nu\wedge \nu)$$

$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\int_{W(t)} {\rm d}(\nu\wedge \nu)$$

$$= \frac{1}{o(x)\sqrt{g}(x)} \frac{d}{dt}\oint_{\partial W(t)} \nu\wedge \nu$$

$$= 0.$$

I think this is wrong because I haven't used the assumption that at $t=0$ the vorticity $2$-form $\omega^{2}$ vanishes when restricted to the boundary $\partial V^{3}(0)$; that is, $i^{*}\omega^{2}=0$, where $i$ is the inclusion of $\partial V$ in $\mathbb{R}^{3}$.

Answer 1

I assume that you are trying to solve problem 4.3(5)(v) in Theodore Frankels' Geometry of Physics (adding this for metadata).

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First of all:

$\nu \wedge d \nu = d \nu \wedge \nu$, not $-d\nu \wedge \nu$, because $d \nu$ is a 2-form.

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Here is my attempt at solution (unfinished). Some facts first:

1. $d \omega^2 = d \; d \nu = 0$
2. $\frac{d}{dt} \omega^2 = 0$, i.e. $\omega^2$ is invariant under the flow. (By Helmholtz's theorem)
3. $\frac{d}{dt} \nu = d(\frac{1}{2}|v|^2 + \phi - \int \frac{dp}{\rho}) \equiv d \psi$, which is just Euler equation ($\psi$ is introduced for later convenience).

The unfinished solution:

By (2): $$ \frac{d}{dt} \int_{V(t)} \nu \wedge \omega^2 = \int_{V(t)} \frac{d}{dt} \nu \wedge \omega^2 + \nu \wedge \frac{d}{dt} \omega^2 = \int_{V(t)} \frac{d}{dt} \nu \wedge \omega^2 = \cdots $$

By (3) and (1): $$ \cdots = \int_{V(t)} d \psi \wedge \omega^2 = \int_{V(t)} d(\psi \wedge \omega^2) - \psi \wedge d \omega^2 = \int_{V(t)} d(\psi \wedge \omega^2) = \cdots $$

Now using Stokes' Theorem: $$ \cdots = \int_{\partial{V}(t)} \psi \wedge \omega^2 $$

Given that we have to pullback $\omega^2$ to the boundary, at $t = 0$ this integral is $0$. I have yet to prove that it stays $0$ at all times, this is why this solution is incomplete.