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Homework Help: Definition of Divergence

  1. Jul 25, 2009 #1
    The definition of the Divergence is given by the following,
    [tex]\nabla \cdot \vec{V} \equiv lim_{\Delta v \rightarrow 0}(\frac{\int \int _{surface}\vec{V} \cdot \vec{ds}}{\Delta v}),[/tex]
    where [tex]v[/tex] is the unit volume.

    Relevant questions:
    The expression [tex]\vec{V} \cdot \vec{ds}[/tex] on the right side corresponds to the amount of the vector field [tex]V[/tex] diverging in the normal direction of the surface ([tex]\vec{ds}[/tex]). If [tex]V[/tex] is perpendicular in the direction of [tex]\vec{ds}[/tex], then for the particular surface element of the entire surface, the divergence will have a value zero associated with it. Could someone tell me if my understanding is correct? Also, what if [tex]V[/tex], and [tex]\vec{ds}[/tex] are not perpendicular, nor parallel with one another (but in-between). How would the definition define that (if my question, makes any sense)?


    Last edited: Jul 25, 2009
  2. jcsd
  3. Jul 25, 2009 #2
    Divergence is outward flux from a point.
  4. Jul 25, 2009 #3
    [tex]\vec{V} \cdot d\vec{s}[/tex] does not "correspond to the amount of the vector field diverging in the normal direction." The integral of that simply is the flux out of the entire surface S. When you take the limit as [tex]\delta V[/tex] approaches 0, you find the amount of net flux for a given point, noting that a point has no volume. Hence, divergence is often called the flux density.

    Now to your question. Divergence is evaluated at a point. The moment you enclose your surface and before you limit it's volume to 0, it is possible to have the expression [tex]\vec{V} \cdot d\vec{s}[/tex] be zero somewhere on the surface. However, once you take the limit as the volume of the surface approaches 0, that specific place on the surface goes away as what's left is just a point. Divergence equals 0 when you have no net flux in or out of a point. When the the divergence of an entire vector field equals 0, the flux out of any enclosed surface of any size will be 0 for that vector field.
  5. Jul 25, 2009 #4
    Out of curiosity, are you in Calc 3 right now? If so, what textbook does your class use? Is it Stewart? I hear a lot of schools use it nowadays.
  6. Jul 25, 2009 #5
    [tex]\vec{V} \cdot \vec{ds} = (\vec{V} \cdot \hat{n})ds.[/tex] So we project the vector [tex]\vec{V}[/tex] in the normal direction [tex]\hat{n},[/tex] which measures the magnitude of the decompression or divergence for a given surface element (per infinitisimal volume) in the [tex]\hat{n}[/tex] direction. The keyword phrase that answered my question was the projection of the vector in the normal direction.


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