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

  1. Aug 25, 2009 #1
    I'm trying do derive the vorticity equation

    [tex]\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

    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:

    [tex]+ \frac{1}{\rho^2}\vec \nabla \rho \times \vec \nabla p[/tex]

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

    [tex]- \frac{1}{\rho} \vec \nabla p[/tex]

    Looking up useful vector identities:

    [tex]\nabla \times (\psi\mathbf{A}) = \psi\nabla \times \mathbf{A} - \mathbf{A} \times \nabla\psi [/tex]

    I don't see how to obtain this term.
  2. jcsd
  3. Aug 25, 2009 #2
    I've figured it out:

    Last edited: Aug 25, 2009
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