I'm trying to prove that for $f$ being a scalar function with D$_tf=0$, where D$_t$ is the material derivative $D_t=\partial_t + v\cdot\nabla$, with $v$ solution of the incompressible Euler equations, then $D_t(\omega\cdot\nabla f)=0$, where $\omega=\nabla \times v$. I have that:
\begin{align} D_t(\omega \cdot \nabla f)= D_t\omega\cdot \nabla f+ \omega \cdot D_t(\nabla f) \end{align}
So, if $D_t\nabla f=\nabla D_tf=0$, then \begin{align} D_t(\omega \cdot \nabla f)= D_t\omega\cdot \nabla f. \end{align} Then, as $D_t\omega=(\omega\cdot\nabla)v$, we have: \begin{align} D_t(\omega \cdot \nabla f)= ((\omega\cdot\nabla)v)\cdot \nabla f. \end{align} I don't know how I can continue with this, any suggestion? Thank you so much.
I will use the more traditional notation $\frac{\mathrm D}{\mathrm Dt}$. $$\frac{\mathrm D}{\mathrm Dt}(\boldsymbol \omega\cdot\nabla f)=\boldsymbol \omega\cdot\frac{\mathrm D}{\mathrm Dt}(\nabla f)+(\nabla f)\cdot \frac{\mathrm D\boldsymbol \omega}{\mathrm Dt}$$ Let's break this down: $$\frac{\mathrm D}{\mathrm Dt}(\nabla f)=\frac{\partial}{\partial t}(\nabla f)+(\boldsymbol u\cdot \nabla)(\nabla f) \\ =\nabla(\partial_t f)+(\boldsymbol u\cdot \nabla)(\nabla f)$$ Breaking into components using index notation, $$\left(\frac{\mathrm D(\nabla f)}{\mathrm Dt}\right)_k=\nabla_k (\partial_t f)+u^j\nabla_j\nabla_k f$$ We live in flat space, so we can commute the covariant derivatives, and so $$\left(\frac{\mathrm D(\nabla f)}{\mathrm Dt}\right)_k=\nabla_k (\partial_t f)+u^j\nabla_k\nabla_j f$$ For a moment consider $$\nabla_k(u^j\nabla_j f)=u^j\nabla_k\nabla_j f+(\nabla_j f)(\nabla_k u^j)$$ Therefore, $$\left(\frac{\mathrm D(\nabla f)}{\mathrm Dt}\right)_k=\nabla_k (\partial_t f)+u^j\nabla_k\nabla_j f=\nabla_k(\partial_t f+u^j\nabla_j f)-(\nabla_k u^j)(\nabla_j f)$$ In other words, $$\frac{\mathrm D(\nabla f)}{\mathrm Dt}=\nabla\left(\frac{\mathrm Df}{\mathrm Dt}\right)-(\nabla\boldsymbol u)\cdot(\nabla f)$$ So the relation $$\frac{\mathrm D(\nabla f)}{\mathrm Dt}=\nabla\left(\frac{\mathrm Df}{\mathrm Dt}\right)$$ Holds iff $\nabla f=0~,~\nabla\boldsymbol u=0$, or both.
EDIT:
Wow, this is dumb. Don't know why I said that. It is (obviously) possible for the dot product to be always zero even when neither of the fields are always zero.
Application: If we have an incompressible linear isotropic fluid with no external forces, it obeys the INSE: $$\frac{\mathrm D\boldsymbol u}{\mathrm Dt}-\nu\nabla^2 \boldsymbol u=-\nabla p \tag{1A}$$ $$ \nabla\cdot\boldsymbol u=0\tag{1B}$$ If the fluid happens to be irrotational as well, then there exists a scalar field $\phi$ such that $\boldsymbol u=\nabla\phi$. Plugging this into $(1\mathrm B)$, we get $$\nabla\cdot\nabla\phi=\Delta\phi=0$$ So $\phi$ satisfies Laplace's equation. Additionally, plugging this into $(1\mathrm A)$, $$\frac{\mathrm D(\nabla\phi)}{\mathrm Dt}-\nu \Delta (\nabla\phi)=-\nabla p$$
We already have an expression for $\frac{\mathrm D(\nabla\phi)}{\mathrm Dt}$, and since in flat space, we have $\nabla\Delta\phi=\Delta\nabla\phi$, we get $$\nabla\left(\frac{\mathrm D\phi}{\mathrm Dt}\right)-(\nabla \phi)\cdot(\nabla\boldsymbol u)-\underbrace{\nu\nabla(\Delta \phi)}_{=0}=-\nabla p$$ We switch to index notation now: $$\nabla_i\big(\partial_t \phi+u^j\nabla_j\phi\big)-(\nabla_j\phi)(\nabla^j u_i)=-\nabla_i p$$ Substitute $u_k=\nabla_k\phi$: $$\nabla_i\big(\partial_t\phi +(\nabla^j\phi)(\nabla_j\phi)\big)-(\nabla_j\phi)(\nabla^j\nabla_i\phi)=-\nabla_i p$$ And again, in flat space, we can switch the order of the derivatives, i.e $\nabla^j\nabla_i=\nabla_i\nabla^j$, so the above becomes $$\nabla_i\big(\partial_t\phi +(\nabla^j\phi)(\nabla_j\phi)\big)-(\nabla_j\phi)~\nabla_i\big((\nabla^j\phi)\big)=-\nabla_i p$$ Expanding the left derivative, $$\nabla_i\big(\partial_t\phi+(\nabla^j\phi)(\nabla_j\phi)\big)=\nabla_i(\partial_t\phi)+\nabla_i\big((\nabla^j\phi)(\nabla_j\phi)\big) \\ =\nabla_i(\partial_t\phi)+(\nabla^j\phi)\nabla_i(\nabla_j\phi)+(\nabla_j\phi)\nabla_i(\nabla^j\phi)$$ Hence our momentum equation is now $$\nabla_i(\partial_t\phi) +(\nabla^j\phi)\nabla_i(\nabla_j\phi)=-\nabla_i p$$ And since $$(\nabla^j\phi)\nabla_i(\nabla_j\phi)=\frac{1}{2}\nabla_i\big(|\nabla\phi|^2\big)$$ We get $$\nabla\left(\partial_t\phi+\frac{1}{2}|\nabla\phi|^2\right)=-\nabla p$$ And finally, under the assumption that the fluid does no work, this becomes Bernoulli's equation(s) for incompressible fluids
$$\partial_t\phi+\frac{1}{2}|\nabla\phi|^2+p=0\tag{2A}$$ $$\Delta\phi=0\tag{2B}$$