Helicity is Conserved 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?
 A: 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.
A: 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.
