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Del operator and divergence

  1. Jan 11, 2012 #1
    I'm trying to understand why the del operator is working a certain way.

    So in my literature there is a term:

    [itex]\nabla \cdot \rho_a \mathbf{v}[/itex]

    but then after saying that


    the term can somehow become

    [itex]\rho (\mathbf{v}\cdot \nabla w_a)[/itex]

    I do not understand how nabla and the velocity, v, get flipped.. Is there some assumption that needs to be made for this to be true?
  2. jcsd
  3. Jan 11, 2012 #2
    What literature is that?
    You must know something about nabla and on what operates nabla
  4. Jan 11, 2012 #3
    It's literature on fluid dynamics and chemical reactions. The equation I am looking at gives

    [itex]\frac{\partial \rho_a}{\partial t}+(\nabla \cdot \rho_a \mathbf{v})=0[/itex]

    and then after defining that rho_a=rho*w_a they rearrange the equation to give

    [itex]\rho \left ( \frac{\partial w_a}{\partial t} + \mathbf{v}\cdot \nabla w_a\right )=0[/itex]

    so rho is a density, t is time, w is a mass fraction, and v is velocity.

    I'm just not understanding what rule you follow, if any, to flip [itex]\nabla \cdot
    \mathbf{v}[/itex] without it meaning something completely different.

    Does this make sense?
    Last edited: Jan 11, 2012
  5. Jan 11, 2012 #4
    here is something in attachment

    Attached Files:

  6. Jan 11, 2012 #5
    Thanks I think my misunderstanding is resolved. They are the same because in the product rule the divergence term drops when incompressibility is assumed. Thank you.
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