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Vector calculus

  1. Feb 27, 2014 #1
    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, [itex] \Omega [/itex] is a scalar constant. Show that [itex] \vec{u} [/itex] 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 [itex] \vec{u} [/itex] 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.

    [edit]
    I have made some progress I took the curl of the longer equation and got rid of [itex] \nabla p [/itex] using the curl of a scalar gradient = 0, but from then I just have ;
    [itex] \nabla \times ( \vec{u} \times \Omega \hat{z})=0 [/itex]
     
    Last edited: Feb 27, 2014
  2. jcsd
  3. Feb 27, 2014 #2

    vela

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    How did you try removing p?
     
  4. Feb 27, 2014 #3
    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.
     
  5. Feb 27, 2014 #4

    vela

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    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.
     
  6. Feb 27, 2014 #5
    [itex] \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}) [/itex]
    This one? and then the turns with [itex]\nabla \bullet \vec{u} [/itex] cancel?
    thanks for your help I think I Know what to do from here. If thats right ;D
     
  7. Feb 27, 2014 #6

    vela

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    Yup, looks good.
     
  8. Feb 27, 2014 #7
    Thanks alot vela
     
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