Fluid Dynamics - material derivative of vorticity/density

[Deadstar]

Homework Statement

Show that, for an inviscid fluid, we have

\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

Homework Equations

Euler's equation

\frac{\partial \mathbf{u}}{\partial t} + \mathbf{\omega} \times \mathbf{u} + \nabla(\mathbf{u}^2/2) = -\frac{1}{\rho}\nabla p - \nabla \chi

Conservation of mass

\frac{D \rho}{Dt} + \rho \nabla \dot \mathbf{u} = 0

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.

\frac{\partial \mathbf{\omega}}{\partial t} + \nabla \times (\mathbf{\omega} \times \mathbf{u}) + 0 = \nabla \times (-\frac{1}{\rho}\nabla p) - 0

Using vector identities now we have

\frac{\partial \mathbf{\omega}}{\partial t} + (\mathbf{u} \cdot \nabla)\mathbf{\omega} + \mathbf{\omega}\nabla \cdot \mathbf{u} - \mathbf{u}\nabla \cdot \mathbf{\omega} - (\mathbf{\omega} \cdot \nabla)\mathbf{u} = - \nabla \bigg{(}\frac{1}{\rho}\bigg{)} \times \nabla p

Now the 4th term is equal to zero since \nabla \cdot (\nabla \times \mathbf{\omega}) = 0 and I think we need to divide by \rho 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 \nabla \cdot \mathbf{u} with \frac{1}{\rho}\frac{D \rho}{Dt} as then apparently it should go to zero but I fail to see why this is.?.Basically my thoughts are...

The (\mathbf{\omega} \cdot \nabla) \mathbf{u} term stays as it is with the dividing by \rho 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 \frac{\mathbf{\omega}}{\rho} (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.

 

[fzero]

Since you're finding this so confusing, why not start from

\frac{D }{D t} \left(\frac{\mathbf{\omega}}{\rho}\right) = \left( \frac{\partial}{\partial t} + \mathbf{u}\cdot \nabla \right) \left(\frac{\mathbf{\omega}}{\rho}\right) .

Use Euler's equation to rewrite \partial \mathbf{\omega}/\partial t and the mass conservation equation to rewrite D\rho/Dt.

 

[Deadstar]

That's the way that I first tried to do this but all I still do not see anyway to use conservation of mass.

All that happens is I end up with this circular argument where from the above steps I end up with

\mathbf{\omega}\frac{D}{Dt}\bigg{(}\frac{1}{\rho}\bigg{)} + \frac{1}{\rho} \frac{D\mathbf{\omega}}{Dt}

The way I was doing it in the first post is the 'right way' in terms of how we are expected to solve it. I know, there is no 'right way' but I don't think anyone should abandon a method without knowing why it is wrong. So what goes wrong where in my first post? Is there some 'physical argument' I am missing (i.e, rather than manipulating formula or using a mathematical method, we can eliminate/change some term by knowing properties of the system we are dealing with)

EDIT:

Ok I found this online which seems to be the same way I'm doing it.

http://www.princeton.edu/~gkv/geoturb/turbch.pdf

Page 4, equation 1.17 is where I'm at and the very next step is the answer I'm looking for.

It's just this statement

"The divergence term may be eliminated with the aid of the mass-conservation equation"

that I don't follow.

 

[fzero]

Your original approach should work. I think the only thing you didn't do was argue that (\mathbf{\omega} \cdot \nabla)\mathbf{u}=0. Otherwise, the substitutions you describe in the OP are leading to the right result, but you didn't bother to show what you got when you put them all together.

 

[Deadstar]

Your original approach should work. I think the only thing you didn't do was argue that (\mathbf{\omega} \cdot \nabla)\mathbf{u}=0. Otherwise, the substitutions you describe in the OP are leading to the right result, but you didn't bother to show what you got when you put them all together.

Oh right I thought it was the \mathbf{\omega}\nabla \cdot \mathbf{u} term that would go to zero... Hmm. Ok, I'll post what I think is happening and you can tell me what, if anything, is correct.

So going from the last equation I wrote in OP... Divide everything by \rho.

The first two terms become...

\frac{\partial \mathbf{\omega}}{\partial t}\bigg{(}\frac{1}{\rho}\bigg{)} + (\mathbf{u} \cdot \nabla)\mathbf{\omega}\frac{1}{\rho} = \frac{D}{Dt}\bigg{(}\frac{\mathbf{\omega}}{\rho}\bigg{)} (this is what I'm most unsure about.)

\nabla \cdot (\nabla \times \mathbf{u}) is equal to zero (note this was a mistype in OP) so we have...

\frac{D}{Dt}\bigg{(}\frac{\mathbf{\omega}}{\rho}\bigg{)} + \frac{1}{\rho}\mathbf{\omega}\nabla \cdot \mathbf{u} - \frac{1}{\rho}(\mathbf{\omega} \cdot \nabla)\mathbf{u} = -\frac{1}{\rho}\nabla \bigg{(}\frac{1}{\rho}\bigg{)} \times \nabla p

So that's what I think it reduces down to, now the last part is just to use mass conservation but I can't quite see how it helps.

 

[fzero]

Oh right I thought it was the \mathbf{\omega}\nabla \cdot \mathbf{u} term that would go to zero...

[STRIKE]That's zero as well. The easiest way is to write it in components using the \epsilon_{ijk} symbol.[/STRIKE]

Edit: Actually, I'm not able to show that \mathbf{\omega}\nabla \cdot \mathbf{u} =0, and I don't think there's any reason for it to be. Those terms cancel out in the calculation.

Right now I'm left with an extra term proportional to (\mathbf{\omega} \cdot \nabla) \mathbf{u} which I haven't been able to explicitly show vanishes.

Hmm. Ok, I'll post what I think is happening and you can tell me what, if anything, is correct.

So going from the last equation I wrote in OP... Divide everything by \rho.

The first two terms become...

\frac{\partial \mathbf{\omega}}{\partial t}\bigg{(}\frac{1}{\rho}\bigg{)} + (\mathbf{u} \cdot \nabla)\mathbf{\omega}\frac{1}{\rho} = \frac{D}{Dt}\bigg{(}\frac{\mathbf{\omega}}{\rho}\bigg{)} (this is what I'm most unsure about.)

Not quite:

\frac{\partial \mathbf{\omega}}{\partial t}\bigg{(}\frac{1}{\rho}\bigg{)} + (\mathbf{u} \cdot \nabla)\mathbf{\omega}\frac{1}{\rho} = \frac{1}{\rho} \frac{D\mathbf{\omega}}{Dt}+\mathbf{\omega}(\mathbf{u} \cdot \nabla)\frac{1}{\rho}

There are a couple of steps to be added to obtain (D/Dt)(\mathbf{\omega}/\rho).

It probably would have been easier to have made the substitution

\frac{\partial \mathbf{\omega}}{\partial t}\bigg{(}\frac{1}{\rho}\bigg{)} = \frac{\partial }{\partial t} \left( \frac{\mathbf{\omega}}{\rho} \right) + \frac{\mathbf{\omega }}{\rho^2} \frac{\partial \rho}{\partial t}

before anything else.