Question

Derivation of vorticity equation from momentum equation

Answer 1

The vorticity equation of fluid dynamicsdescribes evolution of the vorticity ω of a particle of a fluid as it moves with its flow, that is, the local rotation of the fluid (in terms of vector calculus this is the curl of the flow velocity). The equation is:





{\displaystyle {\begin{aligned}{\frac {D{\boldsymbol {\omega }}}{Dt}}&={\frac {\partial {\boldsymbol {\omega }}}{\partial t}}+(\mathbf {u} \cdot \nabla ){\boldsymbol {\omega }}\\&=({\boldsymbol {\omega }}\cdot \nabla )\mathbf {u} -{\boldsymbol {\omega }}(\nabla \cdot \mathbf {u} )+{\frac {1}{\rho ^{2}}}\nabla \rho \times \nabla p+\nabla \times \left({\frac {\nabla \cdot \tau }{\rho }}\right)+\nabla \times \left({\frac {B}{\rho }}\right)\end{aligned}}}

where D/Dt is the material derivative operator, uis the flow velocity, ρ is the local fluid density,p is the local pressure, τ is the viscous stress tensor and B represents the sum of the external body forces. The first source term on the right hand side represents vortex stretching.

The equation is valid in the absence of any concentrated torques and line forces, for a compressible Newtonian fluid.

In the case of incompressible (i.e. low Mach number) and isotropic fluids, withconservative body forces, the equation simplifies to the vorticity transport equation

{\displaystyle {\frac {D{\boldsymbol {\omega }}}{Dt}}=\left({\boldsymbol {\omega }}\cdot \nabla \right)\mathbf {u} +\nu \nabla ^{2}{\boldsymbol {\omega }}}