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Vector Calculus proof

  1. Mar 31, 2014 #1
    1. The problem statement, all variables and given/known data
    Use your knowledge of vector algebra to verify the following identity:

    [tex]
    \vec{\Omega} \cdot \nabla n = \nabla \cdot \vec{\Omega} n

    [/tex]
    2. Relevant equations

    Divergence product rule
    [tex]
    \nabla \cdot (\vec{F} \phi) = \nabla (\phi) \cdot \vec{F} + \phi (\nabla \cdot \vec{F})
    [/tex]

    3. The attempt at a solution

    By the product rule,

    [tex]
    \nabla \cdot (\vec{\Omega} n) = \nabla n \cdot \vec{\Omega} + n (\nabla \cdot \vec{\Omega})
    [/tex]

    Therefore,

    [tex]
    \vec{\Omega} \cdot \nabla n = \nabla \cdot \vec{\Omega} n = \nabla \cdot (\vec{\Omega} n) = \nabla n \cdot \vec{\Omega} + n (\nabla \cdot \vec{\Omega})
    [/tex]

    and

    [tex]
    0 = n (\nabla \cdot \vec{\Omega})
    [/tex]

    I'm not quite sure what I'm doing wrong. Maybe it's a grouping thing. Any help would be appreciated.
     
  2. jcsd
  3. Mar 31, 2014 #2

    HallsofIvy

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    Staff Emeritus
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    It would help us if you would explain your notation. What is [itex]\vec{\Omega}[/itex]? Is it a general vector function? And what is n?
     
  4. Mar 31, 2014 #3
    edit: Confirmed that omega is an arbitrary vector and n is a scalar.

    Sorry about that. I believe omega is a general vector function and n is simply a scalar -- it is not specified but imo the problem statement phrases it as like it is a general identity. This is for a reactor physics course and which features
    [tex]
    \vec{\Omega}
    [/tex]
    and
    [tex]
    n
    [/tex]
    but in that case n is no trivial function
    [tex]
    n=f(r,E,\vec{\Omega},t)
    [/tex]
    and
    [tex]
    \vec{\Omega}
    [/tex] is the neutron direction unit vector.
     
    Last edited: Mar 31, 2014
  5. Mar 31, 2014 #4

    LCKurtz

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    That would give your result if ##\nabla\cdot\Omega = 0##. Is there some reason from the physical situation for why that would be true?
     
  6. Mar 31, 2014 #5

    HallsofIvy

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    Staff Emeritus
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    [itex]\vec{\Omega}\cdot\nabla n= \nabla\cdot \left(\vec{\Omega}n\right)[/itex]
    is certainly NOT true in general.
    For example, if [itex]\vec{\Omega}= x\vec{i}+ y\vec{j}+ z\vec{k}[/itex] and [itex]n= x^2[/itex] then [itex]\vec{\Omega}n= x^3\vec{i}+ x^2y\vec{j}+ x^2z\vec{k}[/itex] and [itex]\nabla\cdot \left(\vec{\Omega}n\right)= 3x^2+ x^2+ x^2= 4x^2[/itex] while [itex]\vec{\Omega}\cdot\nabla n= (x\vec{i}+ y\vec{j}+ z\vec{k})\cdot(2x\vec{i})= 2x^2[/itex]
     
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