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} &amp;= \frac{\partial \vec \omega}{\partial t} + (\vec V \cdot \vec \nabla) \vec \omega \\<br /> &amp;= (\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<br /> \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/