# Earth Cubed

## Deriving The Vorticy Equation

As I mentioned here wikipedia has two good links on the vorticity equation:

At the moment the derivation seems to be missing but there is enough information provided to do it:

From wikipedia’s link Derivation of Navier Stokes Equations the following equation is given:

$\rho\frac{D\mathbf{v}}{D t} = -\nabla p + \nabla \cdot\mathbb{T} + \mathbf{f}$

Dividing each side by the density you get roughly form of stokes equations given here (Note, we have $\vec B$ instead of $\rho \Vec B$ . We get this by letting B be the acceleration per unit mass,  so the vorce times unit mass is the acceleration multiplied by the density. Then when you divide both thids of the above by the density we are left with just $\vec B$ this will work out to the correct result in the end.

$\frac{D \vec V}{D t} = \frac{\partial \vec V}{\partial t} + (\vec V \cdot \vec \nabla) \vec V = - \frac{1}{\rho} \vec \nabla p + \vec B + \frac{\vec \nabla \cdot \underline{\underline{\tau}}}{\rho}$

Now the vorticy equation is derived by taking the cross product of both sides. This will give on the Left hand side of the equation:

$LHS=\nabla \times \frac{D \vec V}{D t} = \nabla \times \left( \frac{\partial \vec V}{\partial t} + (\vec V \cdot \vec \nabla) \vec V \right)$

$LHS= \frac{\partial \vec \omega}{\partial t} + \nabla \times ( \vec V \cdot \vec \nabla ) \vec V$

using the identity (see the thread: Vector Identities and Derivations)

$\vec V \cdot \vec \nabla \vec V = \vec \nabla (\tfrac{1}{2} \vec V \cdot \vec V) - \vec V \times \vec \omega$

We get:

$LHS= \frac{\partial \vec \omega}{\partial t} + \nabla \times \left( \vec \nabla (\tfrac{1}{2} \vec V \cdot \vec V) - \vec V \times \vec \omega \right)$

since the curl of the divergence is equal to zero this reduces to:

$LHS= \frac{\partial \vec \omega}{\partial t} - \nabla \times \left( \vec V \times \vec \omega \right)$

using the identity:

$\vec \nabla \times (\vec V \times \vec \omega ) = -\vec \omega (\vec \nabla \cdot \vec V) + (\vec \omega \cdot \vec \nabla ) \vec V - (\vec V \cdot \vec \nabla) \vec \omega$

We arrive at:

$LHS= \frac{\partial \vec \omega}{\partial t} + \vec \omega (\vec \nabla \cdot \vec V) - (\vec \omega \cdot \vec \nabla ) \vec V + (\vec V \cdot \vec \nabla) \vec \omega$

Now combining this with the cross product of the right hand side one obtains:

$\frac{\partial \vec \omega}{\partial t} + \vec \omega (\vec \nabla \cdot \vec V) - (\vec \omega \cdot \vec \nabla ) \vec V + (\vec V \cdot \vec \nabla) \vec \omega = - \nabla \times \left( \frac{1}{\rho} \vec \nabla p \right) + \nabla \times \vec B + \nabla \times \frac{\vec \nabla \cdot \underline{\underline{\tau}}}{\rho}$

rearanging:

$\frac{\partial \vec \omega}{\partial t} + (\vec V \cdot \vec \nabla) \vec \omega = (\vec \omega \cdot \vec \nabla ) \vec V - \vec \omega (\vec \nabla \cdot \vec V) - \nabla \times \left( \frac{1}{\rho} \vec \nabla p \right) + \nabla \times \vec B + \nabla \times \frac{\vec \nabla \cdot \underline{\underline{\tau}}}{\rho}$

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:

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

substituting this result back into the vorticity equation gives:

$\frac{\partial \vec \omega}{\partial t} + (\vec V \cdot \vec \nabla) \vec \omega = (\vec \omega \cdot \vec \nabla ) \vec V - \vec \omega (\vec \nabla \cdot \vec V) + \frac{1}{\rho^2} \nabla \rho \times \vec \nabla p + \nabla \times \vec B + \nabla \times \frac{\vec \nabla \cdot \underline{\underline{\tau}}}{\rho}$

The following simplifications will be useful depending on the application:

# In case of conservative force, $\vec \nabla \times \vec B = 0$
# For barotropic fluid, $\vec \nabla \rho \times \vec \nabla p = 0$. This is also true for a constant density fluid where $\vec \nabla \rho = 0$

# For inviscid fluids, $\underline{\underline{\tau}} = 0$.

## Vector Identities and Derivations

August 25, 2009 - Posted by | Navier Stokes

1. As a side note, I have vectors symbols above some of the Del operators but not over them all. I wonder if it is better style with or without these symbols above the dell operator.

Comment by s243a | August 26, 2009 | Reply

2. I thought I goggled this before but I tried goggling it again and found someone Else’s derivation:

http://ocw.mit.edu/NR/rdonlyres/Earth–Atmospheric–and-Planetary-Sciences/12-800Fall-2004/66DEA521-B845-4F55-8977-CF85E00C1426/0/lecture13.pdf

Hopefully mine is better. If you go though the directors at that web site there is probably also a lot of other good PDFs of lectures to be found.

Comment by s243a | August 26, 2009 | Reply