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I am trying to summarise the concept of divergence. Say I have a

  1. Oct 5, 2011 #1
    I am trying to summarise the concept of divergence.

    Say I have a vector field, that is radially spreading outwards from the (0,0), but all vectors are equal in each point. So there are no deviations in magnitude in vectors(is that even possible?), but the field lines are spreading like in point charge.

    Is this positive divergence? (At any point except the origin)

    This question mainly popped out because, at electric field, you have spreading field lines. But this spreading is being compensated by the inverse-square law. Thus divergence is 0 at any point, except the origin.

    This spreading of field lines is confusing me, how does this affect divergence?

    Can you say, that the, lets say Electric flux is more "dense", if the electric field is stronger at that point? Is that right way of thinking?
     
    Last edited: Oct 5, 2011
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  3. Oct 6, 2011 #2

    HallsofIvy

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    Re: Divergence-spreading

    A vector pointing directly from the origin to point (x,y,z) is a scalar times [itex]x\vec{i}+ y\vec{j}+ z\vec{z}[/itex]. In order to make that a constant, say, 1, we can divide by its length [itex]\sqrt{x^2+ y^2+ z^2}[/itex]. That is, we want
    [tex](x^2+ y^2+ z^2)^{1/2}(x\vec{i}+ y\vec{j}+ z\vec{k})[/tex].
    (If you want a constant other than 1, multiply this by that constant.)

    Take the divergence of that.
     
  4. Oct 6, 2011 #3
    Re: Divergence-spreading

    Ok I will try that. But still this doesn't answer my question, how does spreading out affect divergence.
     
  5. Oct 6, 2011 #4
    Re: Divergence-spreading

    So about that divergence you told me.

    my vector form is:

    [itex] \vec{A}(x,y,z)=\frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}} \vec{i}+\frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}} \vec{j}+\frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}} \vec{k} [/itex]

    [itex] div \vec{A}= \frac{\partial \frac{x}{\sqrt{x^{2}+y^{2}+z^{2}}}}{\partial x}+\frac{\partial \frac{y}{\sqrt{x^{2}+y^{2}+z^{2}}}}{\partial y}+\frac{\partial \frac{z}{\sqrt{x^{2}+y^{2}+z^{2}}}}{\partial z}= \frac{2}{\sqrt{x^{2}+y^{2}+z^{2}}} [/itex]

    Assuming that I did derivatives well. From this I conclude, that divergence is positive, for any x,y,z(zero not defined). Correct?
     
  6. Oct 6, 2011 #5

    HallsofIvy

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    Re: Divergence-spreading

    More, since the divergence is proportional to 1/r, the "thickness" of the spread decreases proportional to 1/r- the further you are from the center, the less you the "rays" you will feel.

    If your vector field consisted of flying bits of strawberry jam, the further out you stand, the less strawberry jam you will get.
     
  7. Oct 6, 2011 #6
    Re: Divergence-spreading

    I am getting mixed up here with flux and all. So my filed does diverge, and even though its constant. But in that volume definition sense, strength of the field is same when it entered and exited that infinitesimally small volume. How come I got positive divergence?
     
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