Conservation of Vorticity in 2D

Starting from the rotating momentum equations in 2D,

$$ \frac{D\vec{u}}{Dt} + f \hat{k}\times \vec{u} = -\frac{1}{\rho}\nabla p $$

prove that the relative vorticity is materially conserved in barotropic, incompressible flow. That is,

$$ \frac{D\zeta}{Dt} = 0 $$

where \(\zeta = \frac{\partial v}{\partial x} – \frac{\partial u}{\partial y}\).

Answer Next
Permanent Link

We need to start from the momentum equations for the zonal and meridional velocities. We'll expand the total derivative:

$$ \frac{\partial\vec{u}}{\partial t} + \vec{u} \cdot \nabla \vec{u} + f \hat{k}\times \vec{u} = -\frac{1}{\rho}\nabla p $$

Barotropic flow means that

$$ \nabla \rho \times \nabla p = 0$$

so isobars and isopycnals are parallel.

We can take the curl of these equations to get an evolution equation for the vertical vorticity: but first, we can rewrite \(\vec{u} \cdot \nabla \vec{u}\) in a more convenient way:

$$ \vec{u} \cdot \nabla \vec{u} = \vec{\omega} \times \vec{u} - \frac{1}{2}\nabla \| \vec{u}\|^2 $$

where \( \vec{\omega} = \zeta \hat{k} \) and \(\hat{k} \) is the unit vector in the vertical direction.

Taking the curl, we get

\begin{align*}&\frac{\partial \vec{\omega}}{\partial t} + f (\nabla \cdot \vec{u})\hat{k} + \nabla \times \left(\vec{\omega} \times \vec{u}\right) \\ &= \frac{1}{\rho^2}\nabla \rho \times \nabla p \end{align*}

In this case, the vorticity vector is simply \(\zeta\hat{k}\), the z-component of vorticity. In 2D flow, the rotation vector cannot be in any direction other than the vertical.

By barotropicity, the RHS of the above equation is equal to zero, and by incompressibility, the Coriolis term with the divergence of velocity disappears as well.

Lastly, we must use one more identity:

\begin{align*} \nabla \times \left(\vec{\omega} \times \vec{u}\right)  &= (\nabla \cdot \vec{u})\vec{\omega}  + \vec{u} \cdot \nabla \vec{\omega} \\ & - (\nabla \cdot \vec{\omega})\vec{u} - \vec{\omega} \cdot \nabla \vec{u} \end{align*}

Three of these terms disappear:

  • \((\nabla \cdot \vec{u})\vec{\omega}\), because the flow is incompressible.
  • \((\nabla \cdot \vec{\omega})\vec{u}\), because \(\vec{\omega}\) is the curl of velocity and we're taking the divergence of it.
  • \(\vec{\omega} \cdot \nabla \vec{u}\), because the (vertical) vorticity is perpendicular to the (horizontal) plane of motion and its gradient.

Thus, the equation reduces to

\begin{align*}\frac{\partial \vec{\omega}}{\partial t} +\vec{u} \cdot \nabla \vec{\omega} &= \vec{0} \\ \implies \frac{D\vec{\omega}}{Dt} &= \frac{D\zeta}{Dt} \hat{k} =\vec{0} \\\implies\frac{D\zeta}{Dt}  &= 0\end{align*}