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Proof of identity involving del

  1. Nov 2, 2012 #1
    Prove that ∇.(u×v) = v.(∇×u) - u.(∇×v), where "." means dot product and u,v are vectors.

    So by scalar product rule, A.(B×C) = C.(A×B)
    So applying same logic to above identity, shouldnt the left hand side just be equal to v.(∇×u)?
    Or just to -u.(∇×v), since A.(B×C) = -B.(A×C) ?
    1. The problem statement, all variables and given/known data



    2. Relevant equations



    3. The attempt at a solution
     
  2. jcsd
  3. Nov 2, 2012 #2

    SammyS

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    [itex]\displaystyle \vec{\nabla}[/itex] is not just any vector. It's a vector differential operator .

    So it's no more true that [itex]\displaystyle \vec{\nabla}\cdot(\vec{u}\times\vec{v})=\vec{v} \cdot(\vec{\nabla}\times\vec{u})[/itex] than it is that [itex]\displaystyle \vec{\nabla}\cdot\vec{u}=\vec{u}\cdot\vec{\nabla}[/itex] or that [itex]\displaystyle \vec{\nabla}\times\vec{u}=-\vec{u}\times\vec{\nabla}\ .[/itex]
     
  4. Nov 3, 2012 #3
    Oh ok...but then when I broke everything into components it comes out to just be zero..?
     
  5. Nov 3, 2012 #4

    SammyS

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    It is not zero.
     
  6. Nov 3, 2012 #5

    dextercioby

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    The only proof I can think of is the one using cartesian tensors.

    [tex] \partial_{i}\left(\epsilon_{ijk}u_{j}v_{k}\right) =... [/tex]
     
  7. Nov 3, 2012 #6

    Thats definately not how we're supposed to prove that lol, we did not learn that yet
     
  8. Nov 3, 2012 #7
    I have attached my proof that it is...what am I doing wrong?


    edit: oops nvm nvm
     

    Attached Files:

  9. Nov 3, 2012 #8
    ok starting with the right hand side I arrived to left side but with additional factor of 2...
     

    Attached Files:

  10. Nov 3, 2012 #9
    Could someone please give hint as to how to prove this (not using tensors)? Breaking into components does not work for some reason
     
  11. Nov 3, 2012 #10
    if no one knows the answer, perhaps someone knows a book with proofs of these things? I cant find anything whatsoever. I found one proof of this involving tensors, but again, this is not how we're supposed to prove this. Ive also attached another picture showing that the right hand side is actually equal to twice the left hand side...perhaps the last picture was not clear enough..I started with the right side and arrived at twice the left..is there perhaps some way to cancel this factor of 2????
     

    Attached Files:

  12. Nov 3, 2012 #11

    SammyS

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    How do you figure that
    [itex]\displaystyle \left\langle \frac{\partial}{\partial x},\,\frac{\partial}{\partial y},\,\frac{\partial}{\partial z}\right\rangle\cdot\left\langle u_yv_z-u_zv_y,\,u_zv_x-u_xv_z,\,u_xv_y-u_yv_x \right\rangle[/itex]​
    is zero?

    Expand that out. Don't forget to use the product rule.

    Added in Edit:

    Remember that [itex]u_x[/itex] is the x component of vector u, and not [itex]\displaystyle \frac{\partial u}{\partial x}\,,\ [/itex] etc.
     
    Last edited: Nov 3, 2012
  13. Nov 3, 2012 #12
    yea I know in my previous edit I said nvm about it being zero. I did in fact mix up my own notation. But if you read my posts after that, I show picture of how I got twice the identity I was supposed to get...there is factor of two for some reason and I dont know how to get rid of it. I am pretty sure I am doing everything right tho...
     
  14. Nov 3, 2012 #13

    SammyS

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    You can't move things from the left of the partial derivative operator to the right of it.

    So the 4th line of the following (the third line which starts with an '=' sign)

    attachment.php?attachmentid=52591&d=1351955400.jpg

    should start out as
    [itex]\displaystyle =
    v_x \frac{\partial}{\partial y}u_z-v_x \frac{\partial}{\partial z}u_y
    +v_y \frac{\partial}{\partial z}u_x-v_y \frac{\partial}{\partial x}u_z \ \
    \dots[/itex]​

    Then use the product rule in reverse to collect terms.
     
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