# Vector calculus

1. Feb 27, 2014

### Matt atkinson

1. The problem statement, all variables and given/known data
A vector field $$\vec{u}=(u_1,u_2,u_3)$$
satisfies the equations;
$$\Omega\hat{z} \times \vec{u}=-\nabla p , \nabla \bullet \vec{u}=0$$
where p is a scalar variable, $\Omega$ is a scalar constant. Show that $\vec{u}$ is independant of z.
Hint ; how can we remove p from the equations

2. Relevant equations
Included above in question.

3. The attempt at a solution
I know that it means that $\vec{u}$ doesnt have a z component and therefore is only described by x,y but I have no idea where to begin.
I tried removing p but I can't.

I have made some progress I took the curl of the longer equation and got rid of $\nabla p$ using the curl of a scalar gradient = 0, but from then I just have ;
$\nabla \times ( \vec{u} \times \Omega \hat{z})=0$

Last edited: Feb 27, 2014
2. Feb 27, 2014

### vela

Staff Emeritus
How did you try removing p?

3. Feb 27, 2014

### Matt atkinson

I just updated it then because I realised I hadn't included what I'd done, kinda jumped the gun a little :) sorry, not sure where to go from there now though.

4. Feb 27, 2014

### vela

Staff Emeritus
Your textbook should have a list of vector identities. (If not, google it.) You can rewrite the curl of a cross product in a different form that'll let you use the fact that the divergence of u is 0.

5. Feb 27, 2014

### Matt atkinson

$\nabla \times ( \vec{u} \times \Omega \hat{z}) = (\Omega \hat{z} \bullet \nabla)\vec{u} - (\vec{u} \bullet \nabla)\Omega \hat{z} +\vec{u}(\nabla \bullet \Omega \hat{z}) - \Omega \hat{z}(\nabla \bullet \vec{u})$
This one? and then the turns with $\nabla \bullet \vec{u}$ cancel?
thanks for your help I think I Know what to do from here. If thats right ;D

6. Feb 27, 2014

### vela

Staff Emeritus
Yup, looks good.

7. Feb 27, 2014

### Matt atkinson

Thanks alot vela