Helicity is Conserved

Question

In fluid mechanics, the helicity is defined as $$\int_{R^3} u(x,t)\cdot \omega(x,t),$$ where $u(x,t)$ is a smooth solution of the Euler equations $$\partial_tu + (u \cdot \nabla) u = -\nabla p$$ $$\nabla \cdot u = 0,$$ and $\omega$ is the vorticity $\omega = \nabla \times u$. We need to show that helicity is a conserved quantity.

What I tried to do: We also have the Euler equations in vorticity form $$\partial_t \omega + ((u\cdot \nabla )\omega) = (\omega \cdot \nabla)u.$$

So, if we write in component form, and then multiply the first equation by $\omega_j$, the second equation by $u_j$, and then sum over $j$, the left hand side has a term $\frac{d}{dt}\int_{R^3} (u \cdot \omega)$, which is what we are trying to show is $0$. But then the rest of the terms don't seem to cancel (I would also think some sort of integration by parts should help here).

Is there a different approach that you suggest works?

Answer 1

The rate of change of the helicity is given by

$$\frac{dH}{dt} = \frac{d}{dt}\int_{\mathbb{R}^3} [u\cdot \omega]\,{\rm d}^3x = \int_{\mathbb{R}^3} \left[\frac{\partial u}{\partial t}\cdot \omega + u \cdot \frac{\partial \omega}{\partial t}\right]\,{\rm d}^3x$$

First we can use the Euler equation and the vorticity equation

$$\frac{\partial \omega}{\partial t} = (\omega\cdot\nabla)u - (u\cdot\nabla)\omega$$

to get rid of the time-derivatives. This leaves us with

$$\frac{dH}{dt} = \int_{\mathbb{R}^3} \left[-\nabla p \cdot \omega - \omega\cdot(u\cdot \nabla)u + u \cdot(\omega\cdot\nabla)u - u \cdot(u\cdot\nabla)\omega\right]\,{\rm d}^3x$$

The first term can be rewritten as $-\nabla \cdot [p \omega]$ since $\nabla\cdot \omega = 0$ and for the rest of the terms we find

$$\begin{align}\frac{1}{2}\nabla\cdot(|u|^2\omega) &= u\cdot (\omega\cdot \nabla)u \\ \nabla\cdot[u(u\cdot\omega)] &= \omega\cdot(u\cdot \nabla)u + u\cdot (u\cdot\nabla)\omega\end{align}$$

where we have used $\nabla\cdot\omega = \nabla\cdot u = 0$. Putting it all togeather gives us

$$\frac{dH}{dt} = \int_{\mathbb{R}^3} \nabla\cdot\left[-p \omega - u(u\cdot\omega) + \frac{1}{2}|u|^2\omega\right]\,{\rm d}^3x$$

and the desired result, $\frac{dH}{dt} = 0$, follows by the divergence theorem.

Answer 2

One way to write the Euler equation is via the so called magnetization variables (see Chorin, A. J.: Vorticity and Turbulence, or http://faculty.missouri.edu/~stephen/preprints/mo_smpok.html) $$ \frac{\partial m}{\partial t} = - u \cdot \nabla m - m \cdot (\nabla u)^T = -\mathcal L_u^{(1)} m ,$$ where $u$ can be computed from $m$ via the Hodge decomposition $m = u + \nabla q$, where $\nabla \cdot u = 0$. Here $\mathcal L_u^{(1)}$ represents the Lie derivative along $u$ of a 1-form. Thus, taking curls (equivalently the $d$ operator on exterior forms), we get the equation $$ \frac{\partial w}{\partial t} = -\mathcal L_u^{(2)} w ,$$ where $w = \text{curl} \, m = \text{curl} \, u = dm$. Here $\mathcal L_u^{(2)}$ represents the Lie derivative along $u$ of a 2-form.

Since we are in 3D, and since the flow is incompressible, the Hodge duality naturally identifies 2-forms with vectors: $$ w = \begin{bmatrix} 0 & -w_3 & w_2 \\ w_3& 0 & -w_1 \\ -w_2 & w_1 & 0 \end{bmatrix} \leftrightarrow \begin{bmatrix} w_1\\w_2\\w_3\end{bmatrix} =: \tilde w. $$ Here we denote the vorticity as a 2-form by $w$, and the vorticity as a vector as $\tilde w$. Then $$ \frac{\partial \tilde w}{\partial t} = - \mathcal L_u \tilde w = - u \cdot \nabla \tilde w + \tilde w \cdot \nabla u .$$ Here $\mathcal L_u$ represents the Lie derivative along $u$ of a vector.

Next, we see that $m \cdot \tilde w = m \wedge w$ is a three form, which in 3D naturally identifies with scalar functions. (Actually volume forms, but since the flow is incompressible this is the same as a scalar function.) Thus $m \cdot \tilde w$ obeys the simple transport equation $$ \frac{\partial}{\partial t} (m \cdot \tilde w) = - u \cdot \nabla (m \cdot \tilde w) .$$ The transport equation leaves the distribution invariant, that is, $$ \text{measure}\{x:(m(x,t) \cdot \tilde w(x,t)) \in A\} $$ is constant in time for any Borel set $A$. In particular $$ H = \int_{\mathbb R^3} m \cdot \tilde w $$ is constant in time. But $$ \int_{\mathbb R^3} (\nabla q) \cdot \tilde w = 0$$ because $\nabla \cdot \tilde w = d(dm) = 0$. Hence $$ H = \int_{\mathbb R^3} u \cdot \tilde w $$ is actually the helicity.