Taking the curl of the incompressible Navier–Stokes equation results in the elimination of pressure. This is especially easy to see if 2D Cartesian flow is assumed (like in the degenerate 3D case with u z = 0 and no dependence of anything on z ), where the equations reduce to: ρ ρ ( ∂ ∂ u x ∂ ∂ t + u x ∂ ∂ u x ∂ ∂ x + u y ∂ ∂ u x ∂ ∂ y ) = − − ∂ ∂ p ∂ ∂ x + μ μ ( ∂ ∂ 2 u x ∂ ∂ x 2 + ∂ ∂ 2 u x ∂ ∂ y 2 ) + ρ ρ g x ρ ρ ( ∂ ∂ u y ∂ ∂ t + u x ∂ ∂ u y ∂ ∂ x + u y ∂ ∂ u y ∂ ∂ y ) = − − ∂ ∂ p ∂ ∂ y + μ μ ( ∂ ∂ 2 u y ∂ ∂ x 2 + ∂ ∂ 2 u y ∂ ∂ y 2 ) + ρ ρ g y .