Derivation of vorticity equation (incompressible flow)

Question

In the following derivation of the vorticity equation, I do not understand how $\nabla \cdot v=0$ implies $\frac{1}{\rho^2}\nabla \rho \times \nabla p=0$.
We start with the Euler equation $$\frac{\partial \vec{v}}{\partial t}+(\vec{v}.{\nabla})\vec{v}=-\frac{1}{\rho}{\nabla}p + \vec{F},$$
where $\vec{v}$, $p$, $\rho$ and $\vec{F}$ are velocity, pressure, density, and body force, respectively. Now, we take the curl of this equation and obtain:
$$\frac{\partial \vec{w}}{\partial t}={\nabla}\times(\vec{v}\times\vec{w})+\frac{1}{\rho^2}\nabla \rho \times \nabla p,$$
Where, $\vec{w}=\nabla \times \vec{v}$ is the vorticity field. For incompressible fluids we have that ${\nabla \cdot \vec{v}}=0$ and the vorticity equation should become:
$$\frac{\partial \vec{w}}{\partial t}= {\nabla}\times(\vec{v}\times\vec{w}).$$

I don't understand how $\nabla \cdot v=0$ implies $\frac{1}{\rho^2}\nabla \rho \times \nabla p=0$. Is this generally true or there is some extra hypothesis (other than $\nabla \cdot v=0$) that is needed to obtain the above vorticity equation?

Answer 1

Acheson = Acheson, D.J.: Elementary Fluid Dynamics, Oxford: Clarendon Press, 2005.
Falkovich = Falkovich, G: Fluid mechanics, 2nd ed., Cambridge: Cambridge University Press, 2018.
The vorticity equation
The barotropic version: $\frac{D \pmb{\omega}}{Dt}=(\pmb{\omega}\cdot\nabla)\pmb{u}$ [Acheson, p.17, (1.25)].
Remark. The incompressible, constant density [Acheson, p.6, §1.3, (ii)] version is a special case of the barotropic version.
The generalized version: $\frac{D}{Dt}(\frac{\pmb{\omega}}{\rho})=(\frac{\pmb{\omega}}{\rho}\cdot \nabla)\pmb{u}-\frac{1}{\rho}\nabla (\frac{1}{\rho})\times \nabla p$ [Acheson, p.25, (1.39)].
I. The barotropic version is a special case of the generalized version.
Proof.
$\nabla p=p'(\rho)\nabla \rho$.
$\nabla (\frac{1}{\rho})=-\rho^{-2}\nabla \rho$.
Consequently, $\nabla (\frac{1}{\rho})\times \nabla p=\pmb{0}$.
$\frac{D}{Dt}(\frac{\pmb{\omega}}{\rho})=\frac{1}{\rho}\frac{D\pmb{\omega}}{Dt}$ [we assume that $\rho$ is constant in $t$].
II. Proof of the generalized version.
Proof.
$\frac{\partial \pmb{u}}{\partial t}+\pmb{\omega}\times \pmb{u}+\frac{1}{2}\pmb{u}^2=-\frac{1}{\rho}\nabla p-\nabla \chi$ [Acheson, p.24, l.$-$1].
Taking the curl of the above equality, we have
$\frac{\partial \pmb{\omega}}{\partial t}+\nabla\times (\pmb{\omega}\times \pmb{u})=-\nabla\times (\frac{1}{\rho}\nabla p)$
$=-\nabla (\frac{1}{\rho})\times \nabla p$ [Acheson, p.348, (A.8)]. Thus,
$\frac{\partial \pmb{\omega}}{\partial t}=\nabla\times (\pmb{u}\times \pmb{\omega})-\nabla (\frac{1}{\rho})\times \nabla p$ ($*$).
If we substitute [Falkovich, p.17, (1.20)] into $\frac{\partial \pmb{\omega}}{\partial t}$ given in [Falkovich, p.17, $-$11], we obtain [Falkovich, p.17, (1.22)]. However, if we substitute ($*$) instead into $\frac{\partial \pmb{\omega}}{\partial t}$ given in [Falkovich, p.17, $-$11], we obtain [Acheson, p.25, (1.39)].
Remark. With the generalized version at hand, we may easily recognize what conditions we should impose on its hypothesis in order to get a specialized version.
This answer is excerpted from §1.12.(A), Remark 4 in https://sites.google.com/view/lcwangpress/%E9%A6%96%E9%A0%81/papers/quantum-mechanics.