 Solution

Write $\mathbf{\omega }=\nabla \times \mathbf{u}$ and use the vector identity $(\mathbf{u}\cdot \nabla )\mathbf{u}=\nabla (|\mathbf{u}{|}^2/2)-\mathbf{u}\times \mathbf{\omega }$. Taking the curl of the momentum equation eliminates all pure gradients, including gravity and the $\nabla (\nabla \cdot \mathbf{u})$ viscous contribution. Before invoking barotropy, the pressure term givesThis is baroclinic vorticity generation. For a barotropic fluid, $p=p(\rho )$, so $\nabla p$ is parallel to $\nabla \rho$ and this source vanishes. Constant $\nu$ permits the curl and Laplacian to commute. The result isNo incompressibility assumption was used. Since $\nabla \cdot \mathbf{\omega }=0$, an equivalent form isThe terms represent vortex stretching/tilting, compression/dilatation and viscous diffusion. In particular, removing the baroclinic source does not make the vorticity of a three-dimensional parcel constant. A spatially varying entropy-dependent $K$ in $p=K{\rho }^{\gamma }$ need not define a single barotropic law; the common-pressure-law assumption matters.