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Homework Statement
Show that, for an inviscid fluid, we have
[tex]\frac{D}{Dt}\bigg{(}\frac{\mathbf{\omega}}{\rho}\bigg{)} = \bigg{(}\frac{\mathbf{\omega}}{\rho}\cdot \nabla \bigg{)}\mathbf{u} - \frac{1}{\rho}\nabla\bigg{(}\frac{1}{\rho}\bigg{)} \times \nabla p[/tex]
Homework Equations
Euler's equation
[tex]\frac{\partial \mathbf{u}}{\partial t} + \mathbf{\omega} \times \mathbf{u} + \nabla(\mathbf{u}^2/2) = -\frac{1}{\rho}\nabla p - \nabla \chi[/tex]
Conservation of mass
[tex]\frac{D \rho}{Dt} + \rho \nabla \dot \mathbf{u} = 0[/tex]
EDIT: The above equation isn't showing for me but perhaps it's just me. I'm sure you know it but it's missing a dot u = 0 for what I can see.
The Attempt at a Solution
From Euler's equation, we take the curl of both sides giving.
[tex]\frac{\partial \mathbf{\omega}}{\partial t} + \nabla \times (\mathbf{\omega} \times \mathbf{u}) + 0 = \nabla \times (-\frac{1}{\rho}\nabla p) - 0[/tex]
Using vector identities now we have
[tex]\frac{\partial \mathbf{\omega}}{\partial t} + (\mathbf{u} \cdot \nabla)\mathbf{\omega} + \mathbf{\omega}\nabla \cdot \mathbf{u} - \mathbf{u}\nabla \cdot \mathbf{\omega} [/tex] [tex]- (\mathbf{\omega} \cdot \nabla)\mathbf{u} = - \nabla \bigg{(}\frac{1}{\rho}\bigg{)} \times \nabla p[/tex]
Now the 4th term is equal to zero since [tex]\nabla \cdot (\nabla \times \mathbf{\omega}) = 0[/tex] and I think we need to divide by [tex]\rho[/tex] as well.
That's sort of where I am right now although I have ideas as to where to go. A hint I'm given is to replace [tex]\nabla \cdot \mathbf{u}[/tex] with [tex]\frac{1}{\rho}\frac{D \rho}{Dt}[/tex] as then apparently it should go to zero but I fail to see why this is.?.Basically my thoughts are...
The [tex](\mathbf{\omega} \cdot \nabla) \mathbf{u}[/tex] term stays as it is with the dividing by [tex]\rho[/tex] creating the 1st term on the right hand side of the equation I am trying to derive. The first two terms are just an expression for the material derivative of [tex]\frac{\mathbf{\omega}}{\rho}[/tex] (hugely unsure of this though) so the third term should be zero...
It's the use of the conservation of mass that is confusing me as I don't quite know how it's supposed to help me here.