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Problem interpreting the divergence result

  1. Feb 21, 2012 #1
    1. The problem statement, all variables and given/known data
    I take the divergence of the function:
    [tex]V=x^2 \boldsymbol{\hat {x}}+3xz^2\boldsymbol{\hat {y}}-2xz\boldsymbol{\hat {z}}[/tex]
    And get zero. the answer doesn't make sense, since i expect to get a zero divergence only for a function that looks like the one in the drawing attached.
    3. The attempt at a solution
    [tex]\nabla \cdot V=2x+0-2x=0[/tex]
    I test to see whether the function V behaves like in the drawing.
    The function V at an arbitrary point, (1,1,1) is:
    [tex]V_(1,1,1)=1\boldsymbol{\hat {x}}+3\boldsymbol{\hat {y}}-2\boldsymbol{\hat {z}}[/tex]
    And, on another arbitrary point, lets say (2,2,2):
    [tex]V_(2,2,2)=4\boldsymbol{\hat {x}}+6\boldsymbol{\hat {y}}-6\boldsymbol{\hat {z}}[/tex]
    The vectors are different, not the same like i expected.
     

    Attached Files:

  2. jcsd
  3. Feb 21, 2012 #2

    tiny-tim

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    Hi Karol! :smile:

    Think of a vector field V as showing the velocity of a fluid at each point.

    If the fluid has constant density ("incompressible"), then divV = 0.

    divV = 0 is another way of saying that, for a fixed region of space, the flow in equals the flow out.

    Alternatively, if we look at a fixed mass of fluid, divV = 0 says that that fixed mass may change shape, but it will always have the same volume. :wink:
     
  4. Feb 21, 2012 #3
    Does Div V=0 also mean that the vector V at point B must have the same length as the vector at point A, but may change it's direction, like in the new drawing attached?
    But the function V doesn't even support this state since the magnitude of the two vectors i have calculated, at points (1,1,1) and (2,2,2) isn't the same!
    I see no regularity in V, but, of course, i cannot know from inspecting only 2 points
     

    Attached Files:

  5. Feb 22, 2012 #4

    tiny-tim

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    Hi Karol! :smile:

    (just got up :zzz:)
    No.

    Imagine V is the velocity of water flowing through a pipe that's wider near the end …

    the water will slow down, but it won't change density …

    |V| is less, but divV = 0 …

    A disc of water will get wider but thinner … same volume, different shape :wink:

    If you drew arrows to show the flow, you would have to draw shorter arrows, but more of them.​
     
  6. Feb 22, 2012 #5
    Thanks, Tim, but the divergence is in a point, and you are talking about volumes of water
     
  7. Feb 22, 2012 #6

    tiny-tim

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    Agreed, but div is a derivative, so …

    i] I'm allowed to talk about an infinitesimal volume

    ii] I'm allowed to integrate over a finite volume :wink:

    (also, you are talking about divV = 0 everywhere: so divV = 0 over a volume)
     
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