Curl of - \frac{1}{\rho} \vec \nabla p

[John Creighto]

I'm trying do derive the vorticity equation

$$\begin{align}\frac{D\vec\omega}{Dt} &= \frac{\partial \vec \omega}{\partial t} + (\vec V \cdot \vec \nabla) \vec \omega \\ &= (\vec \omega \cdot \vec \nabla) \vec V - \vec \omega (\vec \nabla \cdot \vec V) + \frac{1}{\rho^2}\vec \nabla \rho \times \vec \nabla p + \vec \nabla \times \left( \frac{\vec \nabla \cdot \underline{\underline{\tau}}}{\rho} \right) + \vec \nabla \times \vec B \end{align}$$

based on the notes give here.

I agree with the result obtained for the LHS of the equation but I am having trouble with one term on the right hand side of the equation:

$$+ \frac{1}{\rho^2}\vec \nabla \rho \times \vec \nabla p$$

which as far as I can understand should be the curl of:

$$- \frac{1}{\rho} \vec \nabla p$$

Looking up useful vector identities:

$$\nabla \times (\psi\mathbf{A}) = \psi\nabla \times \mathbf{A} - \mathbf{A} \times \nabla\psi$$

I don't see how to obtain this term.

 

[John Creighto]

I've figured it out:

This is nearly equivalent to the form of the vorticity equation shown in Wikipedia except for this term:

$$- \nabla \times \frac{1}{\rho} \vec \nabla p$$

The following identity is needed:$$\nabla \times (\psi\mathbf{A}) = \psi\nabla \times \mathbf{A} - \mathbf{A} \times \nabla\psi$$

Therefore:$$- \nabla \times \frac{1}{\rho} \vec \nabla p = \frac{1}{\rho} \nabla \times \vec \nabla p - \vec \nabla p \times \nabla \frac{1}{\rho}$$

but since the curl of a gradient is equal to zero:$$- \nabla \times \frac{1}{\rho} \vec \nabla p = - \vec \nabla p \times \nabla \frac{1}{\rho}$$

Now applying the chain rule:$$- \nabla \times \frac{1}{\rho} \vec \nabla p = - \vec \nabla p \times \frac{1}{\rho^2} \nabla \rho$$

Reversing the order of the cross product changes the sign. Consequently:

$$- \nabla \times \frac{1}{\rho} \vec \nabla p = \frac{1}{\rho^2} \nabla \rho \times \vec \nabla p$$

http://earthcubed.wordpress.com/2009/08/25/64/